diff --git a/CHANGELOG.md b/CHANGELOG.md
index 8492bc4485..1245455772 100644
--- a/CHANGELOG.md
+++ b/CHANGELOG.md
@@ -161,6 +161,7 @@ Breaking changes in this release:
- Added pull-based capabilities system for dynamically discovering adapter capabilities at runtime, in PR [#5679](https://github.com/microsoft/BotFramework-WebChat/pull/5679), by [@pranavjoshi001](https://github.com/pranavjoshi001)
- Added Speech-to-Speech (S2S) support for real-time voice conversations, in PR [#5654](https://github.com/microsoft/BotFramework-WebChat/pull/5654), by [@pranavjoshi](https://github.com/pranavjoshi001)
- Added core mute/unmute functionality for speech-to-speech via `useRecorder` hook (silent chunks keep server connection alive), in PR [#5688](https://github.com/microsoft/BotFramework-WebChat/pull/5688), by [@pranavjoshi](https://github.com/pranavjoshi001)
+- 🧪 Added incremental streaming Markdown renderer for livestreaming, in PR [#5799](https://github.com/microsoft/BotFramework-WebChat/pull/5799), by [@OEvgeny](https://github.com/OEvgeny)
### Changed
diff --git a/__tests__/assets/custom-element/event-stream-adapter.ce.html b/__tests__/assets/custom-element/event-stream-adapter.ce.html
new file mode 100644
index 0000000000..7de72a98a9
--- /dev/null
+++ b/__tests__/assets/custom-element/event-stream-adapter.ce.html
@@ -0,0 +1,236 @@
+
+
+
+
+ Event Stream Adapter Custom Element
+
+
+
+
+
diff --git a/__tests__/assets/esm/adapter/LivestreamSession.js b/__tests__/assets/esm/adapter/LivestreamSession.js
new file mode 100644
index 0000000000..ed6a4afd63
--- /dev/null
+++ b/__tests__/assets/esm/adapter/LivestreamSession.js
@@ -0,0 +1,54 @@
+/* eslint-disable security/detect-object-injection */
+/*!
+ * Copyright (C) Microsoft Corporation. All rights reserved.
+ */
+
+const NextLivestreamSequence = Symbol();
+const PreviousActivitySymbol = Symbol();
+const SessionIdSymbol = Symbol();
+
+export default class LivestreamSession {
+ constructor(sessionId) {
+ this[NextLivestreamSequence] = 1;
+ this[PreviousActivitySymbol] = undefined;
+ this[SessionIdSymbol] = sessionId;
+ }
+
+ /**
+ * Last string, useful for decompressing delta-compressed chunks.
+ */
+ get previousActivity() {
+ return this[PreviousActivitySymbol];
+ }
+
+ set previousActivity(value) {
+ this[PreviousActivitySymbol] = value;
+ }
+
+ /**
+ * Activity ID of the session (and the first activity.)
+ *
+ * @type {string}
+ */
+ get sessionId() {
+ return this[SessionIdSymbol];
+ }
+
+ get isConcluded() {
+ return this[NextLivestreamSequence] === Infinity;
+ }
+
+ /** @return {number} */
+ getNextLivestreamSequence(
+ /** @type {boolean | undefined} */
+ isFinal = false
+ ) {
+ if (isFinal) {
+ this.previousActivity = undefined;
+
+ return (this[NextLivestreamSequence] = Infinity);
+ }
+
+ return this[NextLivestreamSequence]++;
+ }
+}
diff --git a/__tests__/assets/esm/adapter/LivestreamSessionManager.js b/__tests__/assets/esm/adapter/LivestreamSessionManager.js
new file mode 100644
index 0000000000..4c7ce02285
--- /dev/null
+++ b/__tests__/assets/esm/adapter/LivestreamSessionManager.js
@@ -0,0 +1,80 @@
+/* eslint-disable security/detect-object-injection */
+
+/*!
+ * Copyright (C) Microsoft Corporation. All rights reserved.
+ */
+
+import LivestreamSession from './LivestreamSession.js';
+
+const ActiveLivestreamSymbol = Symbol();
+
+export default class LivestreamSessionManager {
+ constructor() {
+ this[ActiveLivestreamSymbol] = new Map();
+ }
+
+ *concludeAll() {
+ for (const [sessionId, session] of this[ActiveLivestreamSymbol]) {
+ if (!session.isConcluded) {
+ const { previousActivity } = session;
+ const entitiesWithoutStreamInfo = (previousActivity?.entities ?? []).filter(
+ ({ type }) => type !== 'streaminfo'
+ );
+
+ yield Object.freeze({
+ ...previousActivity,
+ channelData: Object.freeze({
+ ...previousActivity?.channelData,
+ chunkType: undefined,
+ streamId: sessionId,
+ streamSequence: undefined,
+ streamType: 'final'
+ }),
+ entities: Object.freeze([...entitiesWithoutStreamInfo]),
+ id: `${sessionId}/final`,
+ text: previousActivity?.text,
+ type: 'message'
+ });
+ }
+ }
+ }
+
+ has(livestreamSessionId) {
+ return this[ActiveLivestreamSymbol].has(livestreamSessionId);
+ }
+
+ *sequence(livestreamSessionId, activity, isFinal = false) {
+ let livestreamSession = this[ActiveLivestreamSymbol].get(livestreamSessionId);
+
+ if (!livestreamSession) {
+ livestreamSession = new LivestreamSession(livestreamSessionId);
+
+ this[ActiveLivestreamSymbol].set(livestreamSessionId, livestreamSession);
+ }
+
+ if (livestreamSession.isConcluded) {
+ return;
+ }
+
+ const streamSequence = livestreamSession.getNextLivestreamSequence(isFinal);
+ const entitiesWithoutStreamInfo = (activity.entities ?? []).filter(({ type }) => type !== 'streaminfo');
+
+ // We assume the chat adapter will do delta decompression.
+ livestreamSession.previousActivity = activity;
+
+ yield Object.freeze({
+ ...activity,
+ channelData: Object.freeze({
+ ...activity.channelData,
+ chunkType: undefined,
+ streamId: streamSequence === 1 ? undefined : livestreamSessionId,
+ streamSequence: streamSequence === Infinity ? undefined : streamSequence,
+ streamType: streamSequence === Infinity ? 'final' : 'streaming'
+ }),
+ entities: Object.freeze([...entitiesWithoutStreamInfo]),
+ id: streamSequence === 1 ? livestreamSessionId : activity.id,
+ text: livestreamSession.previousActivity.text,
+ type: streamSequence === Infinity ? 'message' : 'typing'
+ });
+ }
+}
diff --git a/__tests__/assets/esm/adapter/createStreamCoalescer.js b/__tests__/assets/esm/adapter/createStreamCoalescer.js
new file mode 100644
index 0000000000..18364c1993
--- /dev/null
+++ b/__tests__/assets/esm/adapter/createStreamCoalescer.js
@@ -0,0 +1,104 @@
+/* eslint-env browser */
+/*!
+ * Copyright (C) Microsoft Corporation. All rights reserved.
+ */
+
+import { getActivityLivestreamingMetadata } from 'botframework-webchat-core';
+
+/**
+ * Creates a `TransformStream` that coalesces interim livestream activities.
+ *
+ * All activities from one network chunk arrive as a burst of microtasks.
+ * Buffering interims until the next macrotask (`setTimeout(32)`) collapses
+ * the burst into a single latest-per-session activity.
+ *
+ * Non-interim activities (first-in-session, informative, contentless, final)
+ * pass through immediately.
+ */
+export default function createStreamCoalescer() {
+ const knownSessions = new Set();
+ const pendingInterim = new Map();
+
+ let controller;
+ let flushScheduled = false;
+ let flushTimer;
+
+ function flushPending() {
+ if (!flushScheduled) {
+ return;
+ }
+
+ flushScheduled = false;
+ clearTimeout(flushTimer);
+
+ for (const activity of pendingInterim.values()) {
+ controller.enqueue(activity);
+ }
+
+ pendingInterim.clear();
+ }
+
+ function scheduleFlush() {
+ if (flushScheduled) {
+ return;
+ }
+
+ flushScheduled = true;
+
+ flushTimer = setTimeout(flushPending, 32);
+ }
+
+ return new TransformStream({
+ start(ctrl) {
+ controller = ctrl;
+ },
+
+ transform(activity) {
+ const meta = getActivityLivestreamingMetadata(activity);
+
+ // Non-livestream activity passes through.
+ if (!meta) {
+ controller.enqueue(activity);
+ return;
+ }
+
+ const { sessionId, type } = meta;
+
+ // Final activity passes through and clears session state.
+ if (type === 'final activity') {
+ pendingInterim.delete(sessionId);
+ knownSessions.delete(sessionId);
+ controller.enqueue(activity);
+ return;
+ }
+
+ // First activity in a session passes through to establish the session downstream.
+ if (!knownSessions.has(sessionId)) {
+ knownSessions.add(sessionId);
+ controller.enqueue(activity);
+ return;
+ }
+
+ // Informative / contentless pass through (low volume, distinct UI slot).
+ if (type !== 'interim activity') {
+ controller.enqueue(activity);
+ return;
+ }
+
+ // Interim activity: buffer the latest per session, flush on next macrotask.
+ pendingInterim.set(sessionId, activity);
+ scheduleFlush();
+ },
+
+ flush() {
+ clearTimeout(flushTimer);
+ flushScheduled = false;
+
+ for (const activity of pendingInterim.values()) {
+ controller.enqueue(activity);
+ }
+
+ pendingInterim.clear();
+ }
+ });
+}
diff --git a/__tests__/assets/esm/adapter/demuxChainOfThought.js b/__tests__/assets/esm/adapter/demuxChainOfThought.js
new file mode 100644
index 0000000000..0a7632c127
--- /dev/null
+++ b/__tests__/assets/esm/adapter/demuxChainOfThought.js
@@ -0,0 +1,121 @@
+/*!
+ * Copyright (C) Microsoft Corporation. All rights reserved.
+ */
+
+import { getActivityLivestreamingMetadata } from 'botframework-webchat-core';
+import { asyncGeneratorWithLastValue } from 'iter-fest';
+
+import findAllCopilotStudioThoughtEntity from './findAllCopilotStudioThoughtEntity.js';
+import LivestreamSessionManager from './LivestreamSessionManager.js';
+
+function* transform(options, livestreamSessionManager, activity) {
+ // eslint-disable-next-line @typescript-eslint/no-explicit-any
+ const livestreamingMetadata = getActivityLivestreamingMetadata(activity);
+ const thoughts = findAllCopilotStudioThoughtEntity(activity);
+
+ // If the activity has no "thought" entity and its livestream session ID is not known as part of a chain of thought,
+ // then, the activity is not a chain of thought, return it as-is.
+ // Notes: the conclusion does not have "thought" entity, thus, we are checking against livestream session ID instead of thought.chainOfThoughtId.
+ if (!thoughts.length && !(livestreamingMetadata && livestreamSessionManager.has(livestreamingMetadata.sessionId))) {
+ yield activity;
+
+ return;
+ }
+
+ for (const thought of thoughts) {
+ const thoughtActivity = {
+ entities: Object.freeze([
+ Object.freeze({
+ '@context': 'https://schema.org',
+ '@id': '',
+ '@type': 'Message',
+ type: 'https://schema.org/Message',
+
+ abstract: thought.title,
+ isPartOf: Object.freeze({
+ '@id': `_:${thought.chainOfThoughtId}`,
+ '@type': 'HowTo'
+ }),
+ // If "Collapsible" keyword is present, the UI should collapse the thought by default. Otherwise, the thought will be expanded by default.
+ keywords: options.collapseByDefault ? ['Collapsible'] : [],
+ position: thought.sequenceNumber
+ })
+ ]),
+ from: activity.from,
+ id: `${thought.chainOfThoughtId}/${thought.sequenceNumber}/${activity.id}`,
+ text: thought.text,
+ timestamp: activity.timestamp ?? new Date().toISOString(),
+ type: 'message'
+ };
+
+ if (livestreamingMetadata) {
+ // Resequence the single chain-of-thought livestream, so every thought is of their own livestream.
+ yield* livestreamSessionManager.sequence(
+ `${thought.chainOfThoughtId}/${thought.sequenceNumber}`,
+ thoughtActivity,
+ livestreamingMetadata.type === 'final activity'
+ );
+ } else {
+ yield thoughtActivity;
+ }
+ }
+
+ // Is this an empty thought carrier activity?
+ if (!activity.text && thoughts.length) {
+ // Special case: if this is a thought carrier activity and the activity is empty, don't pass it.
+ // If we pass this empty carrier, it will reserve a spot in the chat history that is probably not the correct position in the history.
+ return;
+ }
+
+ const soleActivity = Object.freeze({
+ ...activity,
+ entities: Object.freeze((activity.entities ?? []).filter(({ type }) => type !== 'thought')),
+ timestamp: activity.timestamp ?? new Date().toISOString()
+ });
+
+ if (livestreamingMetadata) {
+ yield* livestreamSessionManager.sequence(
+ livestreamingMetadata.sessionId,
+ soleActivity,
+ livestreamingMetadata.type === 'final activity'
+ );
+ } else {
+ yield soleActivity;
+ }
+}
+
+/**
+ * @returns {AsyncIterableIterator}
+ */
+export async function* forIterator(options, activities) {
+ const livestreamSessionManager = new LivestreamSessionManager();
+
+ for await (const activity of activities) {
+ yield* transform(options, livestreamSessionManager, activity);
+ }
+
+ yield* livestreamSessionManager.concludeAll();
+}
+
+/**
+ * This patch will demux thoughts from a single activity into multiple activities.
+ *
+ * Web Chat use the multiple activities format.
+ *
+ * @param turnGenerator
+ * @returns A patched `TurnGenerator` which all chain-of-thought activities are demuxed.
+ */
+function demuxChainOfThought(options, turnGenerator) {
+ const patchTurnGenerator = currentTurnGenerator =>
+ (async function* () {
+ const turnGeneratorWithLastValue = asyncGeneratorWithLastValue(currentTurnGenerator);
+
+ yield* forIterator(options, turnGeneratorWithLastValue);
+
+ return (...args) => patchTurnGenerator(turnGeneratorWithLastValue.lastValue()(...args));
+ })();
+
+ return patchTurnGenerator(turnGenerator);
+}
+
+export default demuxChainOfThought;
diff --git a/__tests__/assets/esm/adapter/findAllCopilotStudioThoughtEntity.js b/__tests__/assets/esm/adapter/findAllCopilotStudioThoughtEntity.js
new file mode 100644
index 0000000000..65b2119e14
--- /dev/null
+++ b/__tests__/assets/esm/adapter/findAllCopilotStudioThoughtEntity.js
@@ -0,0 +1,7 @@
+/*!
+ * Copyright (C) Microsoft Corporation. All rights reserved.
+ */
+
+export default function findAllCopilotStudioThoughtEntity(activity) {
+ return activity.entities?.filter(entity => entity.type === 'thought') ?? [];
+}
diff --git a/__tests__/html2/livestream/html-chunk.html b/__tests__/html2/livestream/html-chunk.html
new file mode 100644
index 0000000000..e4a1abddd0
--- /dev/null
+++ b/__tests__/html2/livestream/html-chunk.html
@@ -0,0 +1,218 @@
+
+
+
+ HTML Streaming: Streaming Support
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/__tests__/html2/livestream/html-chunk.html.snap-1.png b/__tests__/html2/livestream/html-chunk.html.snap-1.png
new file mode 100644
index 0000000000..bd12486652
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.html.snap-1.png differ
diff --git a/__tests__/html2/livestream/html-chunk.html.snap-2.png b/__tests__/html2/livestream/html-chunk.html.snap-2.png
new file mode 100644
index 0000000000..441632b2b2
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.html.snap-2.png differ
diff --git a/__tests__/html2/livestream/html-chunk.html.snap-3.png b/__tests__/html2/livestream/html-chunk.html.snap-3.png
new file mode 100644
index 0000000000..e52b47ada0
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.html.snap-3.png differ
diff --git a/__tests__/html2/livestream/html-chunk.html.snap-4.png b/__tests__/html2/livestream/html-chunk.html.snap-4.png
new file mode 100644
index 0000000000..d8e818def0
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.html.snap-4.png differ
diff --git a/__tests__/html2/livestream/html-chunk.no-streaming.html b/__tests__/html2/livestream/html-chunk.no-streaming.html
new file mode 100644
index 0000000000..096b61b60e
--- /dev/null
+++ b/__tests__/html2/livestream/html-chunk.no-streaming.html
@@ -0,0 +1,10 @@
+
+
+
+ HTML Streaming: No Streaming Support
+
+
+
+
diff --git a/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-1.png b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-1.png
new file mode 100644
index 0000000000..bd12486652
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-1.png differ
diff --git a/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-2.png b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-2.png
new file mode 100644
index 0000000000..441632b2b2
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-2.png differ
diff --git a/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-3.png b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-3.png
new file mode 100644
index 0000000000..e52b47ada0
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-3.png differ
diff --git a/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-4.png b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-4.png
new file mode 100644
index 0000000000..d8e818def0
Binary files /dev/null and b/__tests__/html2/livestream/html-chunk.no-streaming.html.snap-4.png differ
diff --git a/__tests__/html2/markdown/vnext/markdownStreaming.html b/__tests__/html2/markdown/vnext/markdownStreaming.html
new file mode 100644
index 0000000000..7ecfda336e
--- /dev/null
+++ b/__tests__/html2/markdown/vnext/markdownStreaming.html
@@ -0,0 +1,85 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/__tests__/html2/markdown/vnext/markdownStreaming.html.snap-1.png b/__tests__/html2/markdown/vnext/markdownStreaming.html.snap-1.png
new file mode 100644
index 0000000000..69c0256fb4
Binary files /dev/null and b/__tests__/html2/markdown/vnext/markdownStreaming.html.snap-1.png differ
diff --git a/__tests__/html2/markdown/vnext/stream-12k.txt b/__tests__/html2/markdown/vnext/stream-12k.txt
new file mode 100644
index 0000000000..0d429b2b91
--- /dev/null
+++ b/__tests__/html2/markdown/vnext/stream-12k.txt
@@ -0,0 +1,36571 @@
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8","type":"typing","text":"Below","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamSequence":1},"entities":[{"streamType":"streaming","streamSequence":1,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1","type":"typing","text":" is","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":2},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2","type":"typing","text":" a","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":3},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":3,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3","type":"typing","text":" **","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":4},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4","type":"typing","text":"copy","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":5},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":5,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-5","type":"typing","text":"‑","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":6},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":6,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-6","type":"typing","text":"paste","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":7},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":7,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-7","type":"typing","text":"‑","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":8},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8","type":"typing","text":"ready","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":9},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-610","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":611},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":611,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-655","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":656},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":656,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-661","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":662},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":662,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-662","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":663},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":663,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-737","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":738},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":738,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-745","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":746},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":746,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-827","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":828},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":828,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-893","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":894},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":894,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1092","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1093},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1093,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1093","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1094},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1094,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1201","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1202},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1202,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1208","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1209},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1209,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1229","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1230},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1230,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1230","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1231},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1231,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1283","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1284},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1284,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1284","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1285},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1285,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1301","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1302},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1302,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1305","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1306},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1306,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1345","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1346},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1346,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1350","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1351},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1351,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1374","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1375},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1375,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1375","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1376},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1376,"type":"streaminfo","properties":{}}]}
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[1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1517},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1517,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1751","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1752},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1752,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1752","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1753},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1753,"type":"streaminfo","properties":{}}]}
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[1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1794},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1794,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1818","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1819},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1819,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1859","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1860},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1860,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1866","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1867},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1867,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1876","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1877},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1877,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-1877","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":1878},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":1878,"type":"streaminfo","properties":{}}]}
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[1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2175},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2175,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2195","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2196},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2196,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2200","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2201},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2201,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2201","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2202},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2202,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2206","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2207},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2207,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2219","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2220},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2220,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2241","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2242},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2242,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2249","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2250},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2250,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2250","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2251},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2251,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-2772","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":2773},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":2773,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3160","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":3161},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":3161,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3161","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":3162},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":3162,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3566","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":3567},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":3567,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3567","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3936","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":3937},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":3937,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-3937","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4021","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4022},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4022,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4026","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4038","type":"typing","text":" i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":4039},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4039,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4039","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4040},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4040,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4040","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4388","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4389},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4389,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4389","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4390},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4390,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4390","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4391},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4391,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4391","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4392},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4392,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4392","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4393},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4393,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4393","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4394},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4394,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4394","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4395},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4395,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4395","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4396},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4396,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4396","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4397},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4397,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4397","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4398},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4398,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4398","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4399},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4399,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4399","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4400},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4400,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4400","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4401},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4401,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4401","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4402},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4402,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4402","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4403},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4403,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4403","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4404},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4404,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4467","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":4468},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":4468,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-4468","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8125","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8137","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8535","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":8536},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8536,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8541","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":8542},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8542,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8556","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8627","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":8628},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8628,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8628","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8649","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8703","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":8704},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8704,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-8721","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":8722},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":8722,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9019","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9020},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9020,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9020","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9049","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9060","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9084","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9085},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9085,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9090","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9091},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9091,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9091","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9092},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9092,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9096","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9097},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9097,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9097","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9106","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9118","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9119},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9119,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9122","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9123},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9123,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9123","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9124},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9124,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9126","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9141","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9210","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9211},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9211,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9211","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9212},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9212,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9219","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9220},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9220,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9244","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":9245},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":9245,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9251","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9263","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9691","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-9742","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-10340","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":10341},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":10341,"type":"streaminfo","properties":{}}]}
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+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-10341","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":10342},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":10342,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11012","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11206","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11207},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11207,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11220","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11324","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11325},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11325,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11567","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11568},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11568,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11579","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11751","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). 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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11782","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). Exact \\(\\int_0^1x^3dx=1/4\\).\n\n---\n\n## 26) Romberg integration\n\n### Mathematical foundation\nRichardson extrapolation on trapezoids:\n$$\nR_{0,0}=T(h),\\quad R_{k,0}=T(h/2^k)\n$$\n$$\nR_{k,j}=R_{k,j-1}\u002B\\frac{R_{k,j-1}-R_{k-1,j-1}}{4^j-1}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef romberg(f, a, b, levels=5):\n a, b = float(a), float(b)\n R = np.zeros((levels, levels))\n R[0, 0] = trapezoidal(f, a, b, n=1)\n for k in range(1, levels):\n R[k, 0","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11783},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11783,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11816","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). Exact \\(\\int_0^1x^3dx=1/4\\).\n\n---\n\n## 26) Romberg integration\n\n### Mathematical foundation\nRichardson extrapolation on trapezoids:\n$$\nR_{0,0}=T(h),\\quad R_{k,0}=T(h/2^k)\n$$\n$$\nR_{k,j}=R_{k,j-1}\u002B\\frac{R_{k,j-1}-R_{k-1,j-1}}{4^j-1}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef romberg(f, a, b, levels=5):\n a, b = float(a), float(b)\n R = np.zeros((levels, levels))\n R[0, 0] = trapezoidal(f, a, b, n=1)\n for k in range(1, levels):\n R[k, 0] = trapezoidal(f, a, b, n=2**k)\n for j in range(1, k\u002B1):\n R[k, j","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11817},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11817,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11824","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). Exact \\(\\int_0^1x^3dx=1/4\\).\n\n---\n\n## 26) Romberg integration\n\n### Mathematical foundation\nRichardson extrapolation on trapezoids:\n$$\nR_{0,0}=T(h),\\quad R_{k,0}=T(h/2^k)\n$$\n$$\nR_{k,j}=R_{k,j-1}\u002B\\frac{R_{k,j-1}-R_{k-1,j-1}}{4^j-1}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef romberg(f, a, b, levels=5):\n a, b = float(a), float(b)\n R = np.zeros((levels, levels))\n R[0, 0] = trapezoidal(f, a, b, n=1)\n for k in range(1, levels):\n R[k, 0] = trapezoidal(f, a, b, n=2**k)\n for j in range(1, k\u002B1):\n R[k, j] = R[k, j-1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamSequence":11825},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":11825,"type":"streaminfo","properties":{}}]}
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+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-11833","type":"typing","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1](https://www.press.jhu.edu/books/title/10678/matrix-computations)[2](https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ)[3](https://link.springer.com/book/10.1007/978-0-387-40065-5)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[5](https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf) \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6](https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method)[4](https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf)[7](https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). 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+data: {"type":"message","id":"7866415a-21dd-4247-a13c-0b77905dc6be","timestamp":"2026-03-31T20:37:32.46713\u002B00:00","channelId":"pva-studio","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b","name":"cr36e_streamingTestAgent","role":"bot"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"recipient":{"id":"00d048c7-d24a-4e5f-9701-2241ab75f3e2","aadObjectId":"00d048c7-d24a-4e5f-9701-2241ab75f3e2","role":"user"},"membersAdded":[],"membersRemoved":[],"reactionsAdded":[],"reactionsRemoved":[],"locale":"en-US","text":"Below is a **copy‑paste‑ready numerical methods \u0026 algorithms reference** that I built after surveying standard, widely cited references in numerical linear algebra, numerical analysis, and optimization (e.g., **Golub \u0026 Van Loan**, **Burden \u0026 Faires**, **Nocedal \u0026 Wright**, plus well-known CG/HMM tutorials). [1][2][3][4][5] \n\n\u003E **Transparency (important):** Your spec (“comprehensive… do not abbreviate code or math… worked examples with intermediate values… cover all topics”) is **far larger than a single chat response can physically contain** without truncation. So I’m providing a **maximally dense but complete** treatment: every topic appears **in order**, each with **(1) heading, (2) LaTeX foundation, (3) complete Python function**, and **(4) worked example** (small, with explicit intermediate values). \n\u003E If you want *full-length* (textbook-style) expansions per topic, the only practical way is to split into multiple documents/messages—but I will not ask you to wait or promise later; this response is fully usable as-is.\n\n---\n\n## Shared Python Setup (used by all sections)\n\n\u0060\u0060\u0060python\nimport numpy as np\n\ndef _as_float_array(A):\n return np.array(A, dtype=float)\n\ndef _norm_inf(x):\n return float(np.max(np.abs(x)))\n\ndef _safe_div(a, b, eps=1e-15):\n return a / (b if abs(b) \u003E eps else (eps if b \u003E= 0 else -eps))\n\u0060\u0060\u0060\n\n---\n\n# Linear Algebra\n\n## 1) Gaussian elimination\n\n### Mathematical foundation\nSolve \\(A x=b\\) by transforming \\([A|b]\\) to upper triangular via elimination:\n$$\nm_{ik}=\\frac{a_{ik}}{a_{kk}},\\quad\na_{ij}\\leftarrow a_{ij}-m_{ik}a_{kj},\\quad\nb_i\\leftarrow b_i-m_{ik}b_k\n$$\nBack substitution:\n$$\nx_n=\\frac{b_n}{a_{nn}},\\quad\nx_i=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j=i\u002B1}^n a_{ij}x_j\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gaussian_elimination(A, b, pivot=True):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n Ab = np.hstack([A, b.reshape(-1, 1)])\n\n for k in range(n - 1):\n if pivot:\n p = k \u002B np.argmax(np.abs(Ab[k:, k]))\n if p != k:\n Ab[[k, p]] = Ab[[p, k]]\n if abs(Ab[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot encountered.\u0022)\n for i in range(k \u002B 1, n):\n m = Ab[i, k] / Ab[k, k]\n Ab[i, k:] = Ab[i, k:] - m * Ab[k, k:]\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n rhs = Ab[i, -1] - np.dot(Ab[i, i \u002B 1:n], x[i \u002B 1:n])\n x[i] = rhs / Ab[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example (with intermediate values)\nSolve:\n\\[\n\\begin{aligned}\n2x\u002By\u0026=5\\\\\n4x-6y\u0026=-2\n\\end{aligned}\n\\Rightarrow\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\4\u0026-6\u0026-2\\end{array}\\right]\n\\]\nEliminate row2 using row1: \\(m_{21}=4/2=2\\). \nRow2 ← Row2 − 2·Row1:\n\\[\n\\left[\\begin{array}{cc|c}2\u00261\u00265\\\\0\u0026-8\u0026-12\\end{array}\\right]\n\\]\nBack substitute: \\(y=(-12)/(-8)=1.5\\). \nThen \\(2x\u002B1.5=5\\Rightarrow x=1.75\\).\n\n---\n\n## 2) LU decomposition\n\n### Mathematical foundation\nFactor \\(A=LU\\) (Doolittle: diag(L)=1). For \\(k\\le j\\):\n$$\nu_{kj}=a_{kj}-\\sum_{s=1}^{k-1} l_{ks}u_{sj}\n$$\nFor \\(i\u003Ek\\):\n$$\nl_{ik}=\\frac{1}{u_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}u_{sk}\\right)\n$$\nSolve \\(Ly=b\\), then \\(Ux=y\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lu_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.eye(n)\n U = np.zeros((n, n))\n\n for k in range(n):\n for j in range(k, n):\n U[k, j] = A[k, j] - np.dot(L[k, :k], U[:k, j])\n if abs(U[k, k]) \u003C 1e-15:\n raise ValueError(\u0022Zero pivot (use pivoting LU in practice).\u0022)\n for i in range(k \u002B 1, n):\n L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]\n return L, U\n\ndef lu_solve(L, U, b):\n L = _as_float_array(L); U = _as_float_array(U)\n b = _as_float_array(b).reshape(-1)\n n = L.shape[0]\n\n y = np.zeros(n)\n for i in range(n):\n y[i] = b[i] - np.dot(L[i, :i], y[:i])\n\n x = np.zeros(n)\n for i in range(n - 1, -1, -1):\n x[i] = (y[i] - np.dot(U[i, i \u002B 1:], x[i \u002B 1:])) / U[i, i]\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00263\\\\6\u00263\\end{bmatrix}\\). \nCompute \\(U_{11}=4\\). \\(L_{21}=6/4=1.5\\). \nThen \\(U_{12}=3\\). \\(U_{22}=3-1.5\\cdot3= -1.5\\). \nSo:\n\\[\nL=\\begin{bmatrix}1\u00260\\\\1.5\u00261\\end{bmatrix},\\quad\nU=\\begin{bmatrix}4\u00263\\\\0\u0026-1.5\\end{bmatrix}\n\\]\nFor \\(b=\\begin{bmatrix}10\\\\12\\end{bmatrix}\\): \nForward: \\(y_1=10\\), \\(y_2=12-1.5\\cdot 10=-3\\). \nBack: \\(x_2=(-3)/(-1.5)=2\\); \\(x_1=(10-3\\cdot2)/4=1\\).\n\n---\n\n## 3) Cholesky decomposition\n\n### Mathematical foundation\nFor SPD \\(A\\), \\(A=LL^T\\) with \\(L\\) lower triangular:\n$$\nl_{kk}=\\sqrt{a_{kk}-\\sum_{s=1}^{k-1} l_{ks}^2}\n$$\n$$\nl_{ik}=\\frac{1}{l_{kk}}\\left(a_{ik}-\\sum_{s=1}^{k-1}l_{is}l_{ks}\\right),\\quad i\u003Ek\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef cholesky_decomposition(A):\n A = _as_float_array(A)\n n = A.shape[0]\n L = np.zeros((n, n))\n for k in range(n):\n s = np.dot(L[k, :k], L[k, :k])\n val = A[k, k] - s\n if val \u003C= 0:\n raise ValueError(\u0022Matrix not SPD.\u0022)\n L[k, k] = np.sqrt(val)\n for i in range(k \u002B 1, n):\n s2 = np.dot(L[i, :k], L[k, :k])\n L[i, k] = (A[i, k] - s2) / L[k, k]\n return L\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00262\\\\2\u00263\\end{bmatrix}\\). \n\\(l_{11}=\\sqrt{4}=2\\). \n\\(l_{21}=(2)/2=1\\). \n\\(l_{22}=\\sqrt{3-1^2}=\\sqrt2\\approx1.4142\\). \nSo \\(L=\\begin{bmatrix}2\u00260\\\\1\u00261.4142\\end{bmatrix}\\).\n\n---\n\n## 4) QR decomposition\n\n### Mathematical foundation\nOrthogonal \\(Q\\), upper triangular \\(R\\): \\(A=QR\\), \\(Q^TQ=I\\). \nClassical Gram–Schmidt:\n$$\nv_k=a_k-\\sum_{j=1}^{k-1}(q_j^Ta_k)q_j,\\quad q_k=\\frac{v_k}{\\|v_k\\|}\n$$\nThen \\(R_{jk}=q_j^Ta_k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef qr_gram_schmidt(A):\n A = _as_float_array(A)\n m, n = A.shape\n Q = np.zeros((m, n))\n R = np.zeros((n, n))\n for k in range(n):\n v = A[:, k].copy()\n for j in range(k):\n R[j, k] = np.dot(Q[:, j], A[:, k])\n v = v - R[j, k] * Q[:, j]\n R[k, k] = np.linalg.norm(v)\n if R[k, k] \u003C 1e-15:\n raise ValueError(\u0022Linearly dependent columns.\u0022)\n Q[:, k] = v / R[k, k]\n return Q, R\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}1\u00261\\\\1\u00260\\\\0\u00261\\end{bmatrix}\\). \n\\(a_1=(1,1,0)^T\\), \\(\\|a_1\\|=\\sqrt2\\), \\(q_1=(1/\\sqrt2,1/\\sqrt2,0)\\). \n\\(R_{12}=q_1^Ta_2=(1/\\sqrt2)\\cdot1\u002B(1/\\sqrt2)\\cdot0=1/\\sqrt2\\). \n\\(v_2=a_2-R_{12}q_1=(1,0,1)-(1/\\sqrt2)(1/\\sqrt2,1/\\sqrt2,0)=(1-1/2,-1/2,1)=(0.5,-0.5,1)\\). \n\\(\\|v_2\\|=\\sqrt{0.25\u002B0.25\u002B1}= \\sqrt1.5\\). \nSo \\(q_2=v_2/\\sqrt1.5\\).\n\n---\n\n## 5) SVD\n\n### Mathematical foundation\nSingular value decomposition:\n$$\nA=U\\Sigma V^T\n$$\nwhere \\(U,V\\) orthogonal, \\(\\Sigma=\\mathrm{diag}(\\sigma_1\\ge\\cdots\\ge 0)\\). \nConnection:\n$$\nA^TA = V\\Sigma^2 V^T,\\quad AA^T = U\\Sigma^2 U^T\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef svd_decomposition(A):\n A = _as_float_array(A)\n U, s, Vt = np.linalg.svd(A, full_matrices=False)\n Sigma = np.diag(s)\n return U, Sigma, Vt\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}3\u00260\\\\4\u00260\\end{bmatrix}\\). \n\\(A^TA=\\begin{bmatrix}25\u00260\\\\0\u00260\\end{bmatrix}\\Rightarrow \\sigma_1=\\sqrt{25}=5,\\sigma_2=0\\). \n\\(v_1=(1,0)^T\\). \nThen \\(u_1=\\frac{Av_1}{\\sigma_1}=\\frac{(3,4)^T}{5}=(0.6,0.8)^T\\). \nSo \\(A=u_1[5]v_1^T\\).\n\n---\n\n## 6) Eigenvalue decomposition\n\n### Mathematical foundation\nFor diagonalizable \\(A\\):\n$$\nA = V\\Lambda V^{-1},\\quad \\Lambda=\\mathrm{diag}(\\lambda_i)\n$$\nEigenpairs satisfy:\n$$\nA v_i = \\lambda_i v_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef eigen_decomposition(A):\n A = _as_float_array(A)\n w, V = np.linalg.eig(A)\n Lam = np.diag(w)\n Vinv = np.linalg.inv(V)\n return V, Lam, Vinv\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\). \nCharacteristic: \\(\\det(A-\\lambda I)=(2-\\lambda)^2-1=\\lambda^2-4\\lambda\u002B3=0\\). \n\\(\\lambda=3,1\\). \nFor \\(\\lambda=3\\): \\((A-3I)v=0\\Rightarrow \\begin{bmatrix}-1\u00261\\\\1\u0026-1\\end{bmatrix}v=0\\Rightarrow v=(1,1)\\). \nFor \\(\\lambda=1\\): \\(v=(1,-1)\\).\n\n---\n\n## 7) Power iteration\n\n### Mathematical foundation\nFor dominant eigenpair \\((\\lambda_{\\max},v)\\):\n$$\nx_{k\u002B1}=\\frac{Ax_k}{\\|Ax_k\\|},\\quad\n\\lambda_k=\\frac{x_k^TAx_k}{x_k^Tx_k}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef power_iteration(A, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n n = A.shape[0]\n x = np.ones(n) if x0 is None else _as_float_array(x0).reshape(-1)\n x = x / np.linalg.norm(x)\n\n lam_old = 0.0\n for _ in range(max_iter):\n y = A @ x\n x = y / np.linalg.norm(y)\n lam = float(x.T @ (A @ x))\n if abs(lam - lam_old) \u003C tol:\n break\n lam_old = lam\n return lam, x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}2\u00261\\\\1\u00262\\end{bmatrix}\\), \\(x_0=(1,0)\\). \n\\(y_1=Ax_0=(2,1)\\), \\(\\|y_1\\|=\\sqrt5\\), \\(x_1=(2/\\sqrt5,1/\\sqrt5)\\). \nRayleigh: \\(\\lambda_1=x_1^TAx_1=3 - \\frac{2}{5}=2.6\\) (intermediate value). \nNext iteration increases toward \\(3\\).\n\n---\n\n## 8) Jacobi iteration\n\n### Mathematical foundation\nSplit \\(A=D\u002BL\u002BU\\) (D diagonal). Jacobi:\n$$\nx^{(k\u002B1)} = D^{-1}\\bigl(b-(L\u002BU)x^{(k)}\\bigr)\n$$\nComponent form:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\\ne i}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef jacobi_iteration(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n D = np.diag(A)\n R = A - np.diagflat(D)\n\n for _ in range(max_iter):\n x_new = (b - R @ x) / D\n if _norm_inf(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve:\n\\(\\;10x\u002By=11,\\; x\u002B10y=11\\). Start \\(x^{(0)}=(0,0)\\). \n\\(x_1^{(1)}=(11-0)/10=1.1\\), \\(y^{(1)}=(11-0)/10=1.1\\). \nNext: \\(x^{(2)}=(11-1.1)/10=0.99\\), \\(y^{(2)}=(11-1.1)/10=0.99\\). Converges to \\((1,1)\\).\n\n---\n\n## 9) Gauss–Seidel method\n\n### Mathematical foundation\nUse newest values immediately:\n$$\nx_i^{(k\u002B1)}=\\frac{1}{a_{ii}}\\left(b_i-\\sum_{j\u003Ci}a_{ij}x_j^{(k\u002B1)}-\\sum_{j\u003Ei}a_{ij}x_j^{(k)}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_seidel(A, b, x0=None, max_iter=500, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n for _ in range(max_iter):\n x_old = x.copy()\n for i in range(n):\n s1 = np.dot(A[i, :i], x[:i])\n s2 = np.dot(A[i, i\u002B1:], x_old[i\u002B1:])\n x[i] = (b[i] - s1 - s2) / A[i, i]\n if _norm_inf(x - x_old) \u003C tol:\n return x\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame system \\(10x\u002By=11,\\;x\u002B10y=11\\), \\(x^{(0)}=(0,0)\\). \nIter1: \\(x^{(1)}=(11-0)/10=1.1\\). \nThen \\(y^{(1)}=(11-1.1)/10=0.99\\). \nIter2: \\(x^{(2)}=(11-0.99)/10=1.001\\); \\(y^{(2)}=(11-1.001)/10=0.9999\\). Faster than Jacobi.\n\n---\n\n## 10) Conjugate gradient method\n\n### Mathematical foundation\nSolve SPD \\(Ax=b\\) minimizing:\n$$\nf(x)=\\frac12 x^TAx-b^Tx,\\quad \\nabla f = Ax-b=-r\n$$\nIteration (common form):\n$$\nr_0=b-Ax_0,\\; p_0=r_0\n$$\n$$\n\\alpha_k=\\frac{r_k^Tr_k}{p_k^TAp_k},\\quad x_{k\u002B1}=x_k\u002B\\alpha_k p_k\n$$\n$$\nr_{k\u002B1}=r_k-\\alpha_kAp_k,\\quad\n\\beta_k=\\frac{r_{k\u002B1}^Tr_{k\u002B1}}{r_k^Tr_k},\\quad\np_{k\u002B1}=r_{k\u002B1}\u002B\\beta_k p_k\n$$\n(See standard derivations and discussions in CG references.) [6][4][7]\n\n### Python implementation\n\u0060\u0060\u0060python\ndef conjugate_gradient(A, b, x0=None, max_iter=1000, tol=1e-10):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n n = A.shape[0]\n x = np.zeros(n) if x0 is None else _as_float_array(x0).reshape(-1)\n\n r = b - A @ x\n p = r.copy()\n rs_old = float(r @ r)\n\n for _ in range(max_iter):\n Ap = A @ p\n alpha = rs_old / float(p @ Ap)\n x = x \u002B alpha * p\n r = r - alpha * Ap\n rs_new = float(r @ r)\n if np.sqrt(rs_new) \u003C tol:\n return x\n beta = rs_new / rs_old\n p = r \u002B beta * p\n rs_old = rs_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(A=\\begin{bmatrix}4\u00261\\\\1\u00263\\end{bmatrix}\\), \\(b=(1,2)\\), \\(x_0=(0,0)\\). \n\\(r_0=b=(1,2)\\), \\(p_0=r_0\\). \n\\(Ap_0=(4*1\u002B1*2,\\;1*1\u002B3*2)=(6,7)\\). \n\\(\\alpha_0=(r_0^Tr_0)/(p_0^TAp_0)=(1^2\u002B2^2)/(1*6\u002B2*7)=5/20=0.25\\). \n\\(x_1=(0,0)\u002B0.25(1,2)=(0.25,0.5)\\). \n\\(r_1=r_0-\\alpha_0Ap_0=(1,2)-0.25(6,7)=(1-1.5,\\;2-1.75)=(-0.5,0.25)\\). \nContinue similarly; in exact arithmetic CG solves 2×2 in ≤2 steps.\n\n---\n\n## 11) Least squares fitting\n\n### Mathematical foundation\nGiven \\(A\\in\\mathbb{R}^{m\\times n}\\), minimize \\(\\|Ax-b\\|_2^2\\). \nNormal equations:\n$$\nA^TAx=A^Tb\n$$\nQR approach (preferred numerically):\n$$\nA=QR,\\quad \\min \\|QRx-b\\| \\Rightarrow Rx=Q^Tb\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef least_squares_qr(A, b):\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n Q, R = np.linalg.qr(A, mode=\u0022reduced\u0022)\n y = Q.T @ b\n x = np.linalg.solve(R, y)\n return x\n\u0060\u0060\u0060\n\n### Worked example\nFit line \\(y\\approx c_0\u002Bc_1x\\) to points \\((0,1),(1,2),(2,2)\\). \n\\(A=\\begin{bmatrix}1\u00260\\\\1\u00261\\\\1\u00262\\end{bmatrix}\\), \\(b=(1,2,2)\\). \nCompute \\(A^TA=\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\), \\(A^Tb=(5,6)\\). \nSolve:\n\\(\\begin{bmatrix}3\u00263\\\\3\u00265\\end{bmatrix}\\begin{bmatrix}c_0\\\\c_1\\end{bmatrix}=\\begin{bmatrix}5\\\\6\\end{bmatrix}\\). \nEliminate: row2-row1 ⇒ \\(2c_1=1\\Rightarrow c_1=0.5\\). \nThen \\(3c_0\u002B3*0.5=5\\Rightarrow 3c_0=3.5\\Rightarrow c_0=1.1667\\).\n\n---\n\n# Root Finding\n\n## 12) Bisection method\n\n### Mathematical foundation\nIf \\(f(a)f(b)\u003C0\\), root in \\([a,b]\\). Midpoint \\(c=(a\u002Bb)/2\\). \nShrink interval based on sign:\n$$\n(a,b)\\leftarrow\n\\begin{cases}\n(a,c),\u0026 f(a)f(c)\u003C0\\\\\n(c,b),\u0026 f(c)f(b)\u003C0\n\\end{cases}\n$$\nError bound after \\(n\\) steps:\n$$\n|x^*-c_n|\\le \\frac{b-a}{2^{n\u002B1}}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bisection(f, a, b, max_iter=200, tol=1e-12):\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022f(a) and f(b) must have opposite signs.\u0022)\n for _ in range(max_iter):\n c = 0.5 * (a \u002B b)\n fc = f(c)\n if abs(fc) \u003C tol or 0.5 * (b - a) \u003C tol:\n return c\n if fa * fc \u003C 0:\n b, fb = c, fc\n else:\n a, fa = c, fc\n return 0.5 * (a \u002B b)\n\u0060\u0060\u0060\n\n### Worked example\nFind root of \\(f(x)=x^2-2\\) on \\([1,2]\\). \n\\(f(1)=-1\\), \\(f(2)=2\\). \n1) \\(c=1.5\\), \\(f(c)=0.25\u003E0\\Rightarrow [1,1.5]\\) \n2) \\(c=1.25\\), \\(f=-0.4375\u003C0\\Rightarrow [1.25,1.5]\\) \n3) \\(c=1.375\\), \\(f=-0.109375\u003C0\\Rightarrow [1.375,1.5]\\) \n4) \\(c=1.4375\\), \\(f=0.06640625\u003E0\\Rightarrow [1.375,1.4375]\\) … converges to \\(\\sqrt2\\).\n\n---\n\n## 13) Newton–Raphson\n\n### Mathematical foundation\nLinearize:\n$$\nx_{k\u002B1}=x_k-\\frac{f(x_k)}{f\u0027(x_k)}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_raphson(f, df, x0, max_iter=100, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n fx = f(x)\n dfx = df(x)\n if abs(dfx) \u003C 1e-15:\n raise ValueError(\u0022Derivative too small.\u0022)\n x_new = x - fx / dfx\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(f\u0027(x)=2x\\), start \\(x_0=1.5\\). \n\\(x_1=1.5-\\frac{0.25}{3}=1.4166667\\). \n\\(f(x_1)=1.4166667^2-2=0.0069444\\). \n\\(x_2=1.4166667-\\frac{0.0069444}{2.8333334}=1.4142157\\).\n\n---\n\n## 14) Secant method\n\n### Mathematical foundation\nApproximate derivative by secant slope:\n$$\nx_{k\u002B1}=x_k-f(x_k)\\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef secant(f, x0, x1, max_iter=200, tol=1e-12):\n x_prev, x = float(x0), float(x1)\n f_prev, fx = f(x_prev), f(x)\n for _ in range(max_iter):\n denom = fx - f_prev\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Zero secant slope.\u0022)\n x_new = x - fx * (x - x_prev) / denom\n if abs(x_new - x) \u003C tol:\n return x_new\n x_prev, f_prev = x, fx\n x, fx = x_new, f(x_new)\n return x\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2-2\\), \\(x_0=1\\), \\(x_1=2\\). \n\\(x_2=2-2\\cdot(2-1)/(2-(-1))=2-2/3=1.3333333\\). \n\\(f(x_2)=-0.2222222\\). \n\\(x_3=1.3333333-(-0.2222)(1.3333-2)/(-0.2222-2)=1.4\\).\n\n---\n\n## 15) Fixed-point iteration\n\n### Mathematical foundation\nRewrite \\(x=g(x)\\). Iterate:\n$$\nx_{k\u002B1}=g(x_k)\n$$\nConverges if \\(|g\u0027(x^*)|\u003C1\\) near fixed point.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef fixed_point(g, x0, max_iter=500, tol=1e-12):\n x = float(x0)\n for _ in range(max_iter):\n x_new = float(g(x))\n if abs(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSolve \\(x=\\cos x\\). Take \\(g(x)=\\cos x\\), \\(x_0=1\\). \n\\(x_1=\\cos1=0.5403023\\) \n\\(x_2=\\cos0.5403023=0.8575532\\) \n\\(x_3=\\cos0.8575532=0.6542898\\) … converges to ~0.739085.\n\n---\n\n## 16) Müller\u0027s method\n\n### Mathematical foundation\nQuadratic through \\((x_{k-2},f_{k-2}), (x_{k-1},f_{k-1}), (x_k,f_k)\\). \nLet \\(h_1=x_{k-1}-x_{k-2}\\), \\(h_2=x_k-x_{k-1}\\),\n$$\n\\delta_1=\\frac{f_{k-1}-f_{k-2}}{h_1},\\quad\n\\delta_2=\\frac{f_k-f_{k-1}}{h_2}\n$$\n$$\nd=\\frac{\\delta_2-\\delta_1}{h_2\u002Bh_1},\\quad\nb=\\delta_2\u002Bh_2 d,\\quad\nD=\\sqrt{b^2-4 f_k d}\n$$\nChoose denominator with larger magnitude:\n$$\nx_{k\u002B1}=x_k-\\frac{2f_k}{b\\pm D}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef muller(f, x0, x1, x2, max_iter=100, tol=1e-12):\n x0, x1, x2 = float(x0), float(x1), float(x2)\n f0, f1, f2 = f(x0), f(x1), f(x2)\n for _ in range(max_iter):\n h1, h2 = x1 - x0, x2 - x1\n d1 = (f1 - f0) / h1\n d2 = (f2 - f1) / h2\n d = (d2 - d1) / (h1 \u002B h2)\n b = d2 \u002B h2 * d\n D = np.sqrt(b*b - 4.0*f2*d \u002B 0j) # allow complex\n denom = b \u002B D if abs(b \u002B D) \u003E abs(b - D) else b - D\n if abs(denom) \u003C 1e-15:\n raise ValueError(\u0022Denominator too small.\u0022)\n dx = -2.0 * f2 / denom\n x3 = x2 \u002B dx\n if abs(dx) \u003C tol:\n return x3\n x0, x1, x2 = x1, x2, x3\n f0, f1, f2 = f1, f2, f(x2)\n return x2\n\u0060\u0060\u0060\n\n### Worked example (first step)\n\\(f(x)=x^3-x-1\\), guesses \\(x_0=1\\), \\(x_1=1.5\\), \\(x_2=1.2\\). \n\\(f_0=-1\\), \\(f_1=0.875\\), \\(f_2=-0.472\\). \n\\(h_1=0.5\\), \\(h_2=-0.3\\). \n\\(\\delta_1=(0.875-(-1))/0.5=3.75\\). \n\\(\\delta_2=(-0.472-0.875)/(-0.3)=4.49\\). \n\\(d=(4.49-3.75)/(0.2)=3.7\\). \n\\(b=\\delta_2\u002Bh_2 d=4.49\u002B(-0.3)(3.7)=3.38\\). \nThen compute \\(D\\) and update \\(x_3\\).\n\n---\n\n## 17) Brent\u0027s method\n\n### Mathematical foundation\nBrent combines **bisection** (guaranteed bracketing) with **secant/inverse quadratic interpolation** (fast). Maintain bracket \\([a,b]\\) with \\(f(a)f(b)\u003C0\\); propose IQI/secant step if “safe,” otherwise bisection.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef brent(f, a, b, max_iter=200, tol=1e-12):\n a, b = float(a), float(b)\n fa, fb = f(a), f(b)\n if fa * fb \u003E 0:\n raise ValueError(\u0022Root must be bracketed.\u0022)\n c, fc = a, fa\n d = e = b - a\n\n for _ in range(max_iter):\n if abs(fc) \u003C abs(fb):\n a, b, c = b, c, b\n fa, fb, fc = fb, fc, fb\n\n tol1 = 2*np.finfo(float).eps*abs(b) \u002B 0.5*tol\n m = 0.5*(c - b)\n if abs(m) \u003C= tol1 or fb == 0:\n return b\n\n if abs(e) \u003E= tol1 and abs(fa) \u003E abs(fb):\n s = fb / fa\n if a == c:\n p = 2*m*s\n q = 1 - s\n else:\n q = fa / fc\n r = fb / fc\n p = s*(2*m*q*(q - r) - (b - a)*(r - 1))\n q = (q - 1)*(r - 1)*(s - 1)\n if p \u003E 0:\n q = -q\n p = abs(p)\n if 2*p \u003C min(3*m*q - abs(tol1*q), abs(e*q)):\n e, d = d, p / q\n else:\n e, d = d, m\n else:\n e, d = d, m\n\n a, fa = b, fb\n b = b \u002B (d if abs(d) \u003E tol1 else (tol1 if m \u003E 0 else -tol1))\n fb = f(b)\n if fb * fc \u003E 0:\n c, fc = a, fa\n d = e = b - a\n return b\n\u0060\u0060\u0060\n\n### Worked example (key intermediates)\nSame \\(f(x)=x^2-2\\), bracket \\([1,2]\\). Brent will:\n- keep bracket sign change,\n- attempt secant \\(x\\approx 1.3333\\),\n- fall back to bisection when interpolation step would leave bracket,\n- converge quickly to \\(1.41421356\\).\n\n---\n\n# Interpolation\n\n## 18) Lagrange interpolation\n\n### Mathematical foundation\nGiven nodes \\((x_i,y_i)\\):\n$$\np(x)=\\sum_{i=0}^n y_i \\ell_i(x),\\quad\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-x_j}{x_i-x_j}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lagrange_interpolate(x_nodes, y_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n p = 0.0\n for i in range(n):\n li = 1.0\n for j in range(n):\n if j != i:\n li *= (x - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n p \u002B= y_nodes[i] * li\n return p\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((0,1),(1,3),(2,2)\\). Evaluate at \\(x=1.5\\). \n\\(\\ell_0=(x-1)(x-2)/((0-1)(0-2))=(0.5)(-0.5)/(2)= -0.125\\). \n\\(\\ell_1=(x-0)(x-2)/((1-0)(1-2))=(1.5)(-0.5)/(-1)=0.75\\). \n\\(\\ell_2=(x-0)(x-1)/((2-0)(2-1))=(1.5)(0.5)/2=0.375\\). \n\\(p(1.5)=1(-0.125)\u002B3(0.75)\u002B2(0.375)= -0.125\u002B2.25\u002B0.75=2.875\\).\n\n---\n\n## 19) Newton’s divided differences\n\n### Mathematical foundation\nPolynomial:\n$$\np(x)=c_0\u002Bc_1(x-x_0)\u002Bc_2(x-x_0)(x-x_1)\u002B\\cdots\n$$\nDivided differences:\n$$\nf[x_i]=y_i,\\quad\nf[x_i,\\dots,x_{i\u002Bk}]=\\frac{f[x_{i\u002B1},\\dots,x_{i\u002Bk}]-f[x_i,\\dots,x_{i\u002Bk-1}]}{x_{i\u002Bk}-x_i}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_divided_differences(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(y_nodes).reshape(-1).copy()\n n = len(x)\n for j in range(1, n):\n coef[j:n] = (coef[j:n] - coef[j-1:n-1]) / (x[j:n] - x[0:n-j])\n return coef\n\ndef newton_eval(x_nodes, coef, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n coef = _as_float_array(coef).reshape(-1)\n x = float(x)\n n = len(coef)\n p = coef[-1]\n for k in range(n - 2, -1, -1):\n p = coef[k] \u002B (x - x_nodes[k]) * p\n return p\n\u0060\u0060\u0060\n\n### Worked example\nSame nodes \\(x=[0,1,2]\\), \\(y=[1,3,2]\\). \nLevel1: \\(f[x0,x1]=(3-1)/(1-0)=2\\), \\(f[x1,x2]=(2-3)/(2-1)=-1\\). \nLevel2: \\(f[x0,x1,x2]=(-1-2)/(2-0)=-1.5\\). \nSo \\(p(x)=1\u002B2(x-0)-1.5(x-0)(x-1)\\).\n\n---\n\n## 20) Cubic spline interpolation (natural)\n\n### Mathematical foundation\nPiecewise cubic \\(S_i(x)\\) with continuity of \\(S,S\u0027,S\u0027\u0027\\). Natural boundary: \\(S\u0027\u0027(x_0)=S\u0027\u0027(x_n)=0\\). Solve tridiagonal for second derivatives \\(M_i=S\u0027\u0027(x_i)\\):\nFor uniform notation \\(h_i=x_{i\u002B1}-x_i\\),\n$$\nh_{i-1}M_{i-1}\u002B2(h_{i-1}\u002Bh_i)M_i\u002Bh_iM_{i\u002B1}\n=6\\left(\\frac{y_{i\u002B1}-y_i}{h_i}-\\frac{y_i-y_{i-1}}{h_{i-1}}\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef natural_cubic_spline(x_nodes, y_nodes):\n x = _as_float_array(x_nodes).reshape(-1)\n y = _as_float_array(y_nodes).reshape(-1)\n n = len(x) - 1\n h = x[1:] - x[:-1]\n\n A = np.zeros((n\u002B1, n\u002B1))\n rhs = np.zeros(n\u002B1)\n A[0,0] = 1.0\n A[n,n] = 1.0\n\n for i in range(1, n):\n A[i, i-1] = h[i-1]\n A[i, i] = 2*(h[i-1] \u002B h[i])\n A[i, i\u002B1] = h[i]\n rhs[i] = 6*((y[i\u002B1]-y[i])/h[i] - (y[i]-y[i-1])/h[i-1])\n\n M = np.linalg.solve(A, rhs)\n\n def S(xq):\n xq = float(xq)\n i = np.clip(np.searchsorted(x, xq) - 1, 0, n-1)\n hi = h[i]\n a = (x[i\u002B1] - xq)/hi\n b = (xq - x[i])/hi\n return (a*y[i] \u002B b*y[i\u002B1] \u002B\n ((a**3 - a)*M[i] \u002B (b**3 - b)*M[i\u002B1])*(hi**2)/6.0)\n return S, M\n\u0060\u0060\u0060\n\n### Worked example (intermediate system)\nNodes \\((0,0),(1,1),(2,0)\\). \\(h_0=h_1=1\\). Natural: \\(M_0=M_2=0\\). \nEquation for \\(i=1\\):\n\\(1\\cdot M_0 \u002B 2(1\u002B1)M_1 \u002B 1\\cdot M_2 = 6((0-1)/1 - (1-0)/1)=6(-1-1)=-12\\). \nSo \\(4M_1=-12\\Rightarrow M_1=-3\\). \nThen evaluate \\(S(0.5)\\) using the formula in code.\n\n---\n\n## 21) Hermite interpolation\n\n### Mathematical foundation\nGiven \\(y_i=f(x_i)\\) and \\(y_i\u0027=f\u0027(x_i)\\). Hermite basis (for each node):\n$$\nH_i(x)=\\left(1-2L_i\u0027(x_i)(x-x_i)\\right)L_i(x)^2,\\quad\nK_i(x)=(x-x_i)L_i(x)^2\n$$\nHermite polynomial:\n$$\np(x)=\\sum_i \\bigl(y_i H_i(x) \u002B y_i\u0027 K_i(x)\\bigr)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef hermite_interpolate(x_nodes, y_nodes, dy_nodes, x):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n dy_nodes = _as_float_array(dy_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n\n def L(i, xq):\n val = 1.0\n for j in range(n):\n if j != i:\n val *= (xq - x_nodes[j]) / (x_nodes[i] - x_nodes[j])\n return val\n\n def Lprime_at_i(i):\n s = 0.0\n for j in range(n):\n if j != i:\n s \u002B= 1.0 / (x_nodes[i] - x_nodes[j])\n return s\n\n p = 0.0\n for i in range(n):\n Li = L(i, x)\n Lpi = Lprime_at_i(i)\n Hi = (1 - 2*Lpi*(x - x_nodes[i])) * (Li**2)\n Ki = (x - x_nodes[i]) * (Li**2)\n p \u002B= y_nodes[i]*Hi \u002B dy_nodes[i]*Ki\n return p\n\u0060\u0060\u0060\n\n### Worked example\nInterpolate \\(f\\) with \\(f(0)=1,f\u0027(0)=0\\); \\(f(1)=2,f\u0027(1)=1\\). \nCompute at \\(x=0.5\\). \nFor 2 nodes, \\(L_0(x)=1-x\\), \\(L_1(x)=x\\). \n\\(L_0\u0027(0)=-1\\), \\(L_1\u0027(1)=1\\). \nThen \\(H_0=(1-2(-1)(x-0))(1-x)^2=(1\u002B2x)(1-x)^2\\), \\(K_0=x(1-x)^2\\). \n\\(H_1=(1-2(1)(x-1))x^2=(3-2x)x^2\\), \\(K_1=(x-1)x^2\\). \nPlug \\(x=0.5\\) to get numeric intermediate values.\n\n---\n\n## 22) Chebyshev interpolation\n\n### Mathematical foundation\nChebyshev nodes on \\([-1,1]\\):\n$$\nx_k=\\cos\\left(\\frac{(2k\u002B1)\\pi}{2(n\u002B1)}\\right),\\;k=0,\\dots,n\n$$\nInterpolate using barycentric formula:\n$$\np(x)=\\frac{\\sum_{k=0}^n \\frac{w_k}{x-x_k} y_k}{\\sum_{k=0}^n \\frac{w_k}{x-x_k}}\n$$\nChebyshev barycentric weights (first kind): \\(w_k=(-1)^k\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef chebyshev_nodes(n):\n k = np.arange(n\u002B1)\n return np.cos((2*k \u002B 1) * np.pi / (2*(n\u002B1)))\n\ndef barycentric_interpolate(x_nodes, y_nodes, x, weights=None):\n x_nodes = _as_float_array(x_nodes).reshape(-1)\n y_nodes = _as_float_array(y_nodes).reshape(-1)\n x = float(x)\n n = len(x_nodes)\n if weights is None:\n weights = np.ones(n)\n weights = _as_float_array(weights).reshape(-1)\n\n dif = x - x_nodes\n j = np.where(np.abs(dif) \u003C 1e-15)[0]\n if len(j) \u003E 0:\n return float(y_nodes[j[0]])\n\n num = np.sum(weights * y_nodes / dif)\n den = np.sum(weights / dif)\n return float(num / den)\n\ndef chebyshev_interpolate(f, n, x):\n xs = chebyshev_nodes(n)\n ys = np.array([f(xi) for xi in xs], dtype=float)\n ws = np.array([(-1.0)**k for k in range(n\u002B1)], dtype=float)\n return barycentric_interpolate(xs, ys, x, ws)\n\u0060\u0060\u0060\n\n### Worked example\nApproximate \\(f(x)=1/(1\u002B25x^2)\\) with \\(n=4\\). \nCompute nodes:\n\\(x_0=\\cos(\\pi/10)\\approx0.9511\\), \\(x_1=\\cos(3\\pi/10)\\approx0.5878\\), … \nCompute \\(y_k=f(x_k)\\). \nEvaluate at \\(x=0\\): barycentric simplifies since \\(x-x_k=-x_k\\), show numerator/denominator sums explicitly.\n\n---\n\n# Numerical Integration\n\n## 23) Trapezoidal rule\n\n### Mathematical foundation\nSingle interval:\n$$\n\\int_a^b f(x)\\,dx \\approx \\frac{b-a}{2}\\left(f(a)\u002Bf(b)\\right)\n$$\nComposite with step \\(h=(b-a)/n\\):\n$$\n\\int_a^b f(x)\\,dx\\approx h\\left(\\frac12 f(a)\u002B\\sum_{i=1}^{n-1}f(a\u002Bih)\u002B\\frac12 f(b)\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef trapezoidal(f, a, b, n=1):\n a, b = float(a), float(b)\n n = int(n)\n h = (b - a) / n\n s = 0.5 * (f(a) \u002B f(b))\n for i in range(1, n):\n s \u002B= f(a \u002B i*h)\n return h * s\n\u0060\u0060\u0060\n\n### Worked example\nIntegrate \\(f(x)=x^2\\) on \\([0,1]\\) with \\(n=2\\). \\(h=0.5\\). \nValues: \\(f(0)=0\\), \\(f(0.5)=0.25\\), \\(f(1)=1\\). \nComposite trap:\n\\(I\\approx 0.5[0.5*0 \u002B 0.25 \u002B 0.5*1]=0.5[0.25\u002B0.5]=0.375\\). Exact \\(=1/3\\approx0.3333\\).\n\n---\n\n## 24) Simpson’s 1/3 rule\n\n### Mathematical foundation\nFor even \\(n\\), step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{h}{3}\\left[f(x_0)\u002B4\\sum_{i\\ \\text{odd}}f(x_i)\u002B2\\sum_{i\\ \\text{even},\\,i\\ne 0,n}f(x_i)\u002Bf(x_n)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_1_3(f, a, b, n=2):\n a, b = float(a), float(b)\n n = int(n)\n if n % 2 != 0:\n raise ValueError(\u0022n must be even.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n s \u002B= (4 if i % 2 == 1 else 2) * f(a \u002B i*h)\n return (h/3.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^2\\), \\([0,1]\\), \\(n=2\\), \\(h=0.5\\). \n\\(I\\approx (0.5/3)[f(0)\u002B4f(0.5)\u002Bf(1)]=(1/6)[0\u002B4*0.25\u002B1]=(1/6)*2=0.333333\\) exact.\n\n---\n\n## 25) Simpson’s 3/8 rule\n\n### Mathematical foundation\nFor \\(n\\) multiple of 3, step \\(h\\):\n$$\n\\int_a^b f(x)\\,dx\\approx \\frac{3h}{8}\\left[f_0\u002B3(f_1\u002Bf_2)\u002B2(f_3\u002Bf_6\u002B\\cdots)\u002B3(\\cdots)\u002Bf_n\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simpson_3_8(f, a, b, n=3):\n a, b = float(a), float(b)\n n = int(n)\n if n % 3 != 0:\n raise ValueError(\u0022n must be multiple of 3.\u0022)\n h = (b - a) / n\n s = f(a) \u002B f(b)\n for i in range(1, n):\n coef = 2 if i % 3 == 0 else 3\n s \u002B= coef * f(a \u002B i*h)\n return (3*h/8.0) * s\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=x^3\\) on \\([0,1]\\), \\(n=3\\), \\(h=1/3\\). \nValues: \\(f(0)=0\\), \\(f(1/3)=1/27\\), \\(f(2/3)=8/27\\), \\(f(1)=1\\). \n\\(I\\approx (3h/8)[0\u002B3(1/27\u002B8/27)\u002B1]=(1/8)[3*(9/27)\u002B1]=(1/8)[1\u002B1]=0.25\\). Exact \\(\\int_0^1x^3dx=1/4\\).\n\n---\n\n## 26) Romberg integration\n\n### Mathematical foundation\nRichardson extrapolation on trapezoids:\n$$\nR_{0,0}=T(h),\\quad R_{k,0}=T(h/2^k)\n$$\n$$\nR_{k,j}=R_{k,j-1}\u002B\\frac{R_{k,j-1}-R_{k-1,j-1}}{4^j-1}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef romberg(f, a, b, levels=5):\n a, b = float(a), float(b)\n R = np.zeros((levels, levels))\n R[0, 0] = trapezoidal(f, a, b, n=1)\n for k in range(1, levels):\n R[k, 0] = trapezoidal(f, a, b, n=2**k)\n for j in range(1, k\u002B1):\n R[k, j] = R[k, j-1] \u002B (R[k, j-1] - R[k-1, j-1]) / (4**j - 1)\n return R\n\u0060\u0060\u0060\n\n### Worked example\n\\(f(x)=\\sin x\\) on \\([0,\\pi]\\). \n\\(R_{0,0}=T(n=1)=\\pi/2(\\sin0\u002B\\sin\\pi)=0\\). \n\\(R_{1,0}=T(n=2)=\\frac{\\pi}{2}(\\tfrac12*0\u002B1\u002B\\tfrac12*0)=\\pi/2\\). \n\\(R_{1,1}=R_{1,0}\u002B\\frac{R_{1,0}-R_{0,0}}{3}=\\pi/2\u002B\\pi/6=2\\pi/3\\approx2.094\\). \nExact is \\(2\\).\n\n---\n\n## 27) Gaussian quadrature (Gauss–Legendre)\n\n### Mathematical foundation\nFor \\([-1,1]\\):\n$$\n\\int_{-1}^1 f(x)\\,dx \\approx \\sum_{i=1}^n w_i f(x_i)\n$$\nNodes \\(x_i\\) are roots of \\(P_n(x)\\), weights:\n$$\nw_i=\\frac{2}{(1-x_i^2)[P_n\u0027(x_i)]^2}\n$$\nMap \\([a,b]\\) via \\(x=\\frac{b-a}{2}t\u002B\\frac{a\u002Bb}{2}\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gauss_legendre(f, a, b, n=2):\n a, b = float(a), float(b)\n # Use numpy\u0027s built-in nodes/weights for Legendre\n t, w = np.polynomial.legendre.leggauss(int(n))\n xm = 0.5*(b \u002B a)\n xr = 0.5*(b - a)\n return xr * np.sum(w * np.array([f(xm \u002B xr*ti) for ti in t]))\n\u0060\u0060\u0060\n\n### Worked example (n=2)\nNodes \\(t=\\pm 1/\\sqrt3\\), weights \\(w_1=w_2=1\\). \nIntegrate \\(f(x)=x^2\\) on \\([-1,1]\\): \nApprox \\(=1\\cdot f(-1/\\sqrt3)\u002B1\\cdot f(1/\\sqrt3)=2\\cdot(1/3)=2/3\\). \nExact \\(\\int_{-1}^1x^2dx=2/3\\).\n\n---\n\n## 28) Monte Carlo integration\n\n### Mathematical foundation\nFor uniform \\(X_i\\sim U(a,b)\\):\n$$\n\\int_a^b f(x)\\,dx \\approx (b-a)\\frac{1}{N}\\sum_{i=1}^N f(X_i)\n$$\nVariance \\(\\propto 1/N\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef monte_carlo_integral(f, a, b, N=10000, seed=0):\n rng = np.random.default_rng(seed)\n a, b = float(a), float(b)\n xs = rng.uniform(a, b, int(N))\n return (b - a) * float(np.mean([f(x) for x in xs]))\n\u0060\u0060\u0060\n\n### Worked example\nEstimate \\(\\int_0^1 x^2 dx\\). Use samples \\(x=[0.2,0.5,0.9]\\) (toy \\(N=3\\)). \nMean \\(f=\\frac{0.04\u002B0.25\u002B0.81}{3}=0.3666667\\). \nEstimate \\(=1*0.3666667\\) (rough, improves as \\(N\\) increases).\n\n---\n\n## 29) Adaptive quadrature (adaptive Simpson)\n\n### Mathematical foundation\nSimpson on \\([a,b]\\):\n$$\nS(a,b)=\\frac{b-a}{6}\\left(f(a)\u002B4f\\!\\left(\\frac{a\u002Bb}{2}\\right)\u002Bf(b)\\right)\n$$\nRecursively split until:\n$$\n|S(a,m)\u002BS(m,b)-S(a,b)| \u003C 15\\varepsilon\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef adaptive_simpson(f, a, b, eps=1e-10, max_depth=20):\n a, b = float(a), float(b)\n\n def S(a, b):\n m = 0.5*(a\u002Bb)\n return (b-a)/6.0 * (f(a) \u002B 4*f(m) \u002B f(b))\n\n def rec(a, b, eps, whole, depth):\n m = 0.5*(a\u002Bb)\n left = S(a, m)\n right = S(m, b)\n if depth \u003C= 0 or abs(left \u002B right - whole) \u003C 15*eps:\n return left \u002B right \u002B (left \u002B right - whole)/15.0\n return rec(a, m, eps/2, left, depth-1) \u002B rec(m, b, eps/2, right, depth-1)\n\n whole = S(a, b)\n return rec(a, b, eps, whole, max_depth)\n\u0060\u0060\u0060\n\n### Worked example (first split)\nIntegrate \\(f(x)=\\sin x\\) on \\([0,\\pi]\\). \nWhole Simpson: \\(S(0,\\pi)=\\pi/6(0\u002B4\\cdot1\u002B0)=2\\pi/3\\approx2.094\\). \nSplit at \\(m=\\pi/2\\): compute \\(S(0,\\pi/2)\\) and \\(S(\\pi/2,\\pi)\\), compare and refine until tolerance.\n\n---\n\n# ODEs\n\n## 30) Euler’s method\n\n### Mathematical foundation\nFor \\(y\u0027=f(t,y)\\), step \\(h\\):\n$$\ny_{n\u002B1}=y_n\u002Bh f(t_n,y_n)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef euler(f, t0, y0, h, n_steps):\n t = float(t0)\n y = float(y0)\n out = [(t, y)]\n for _ in range(int(n_steps)):\n y = y \u002B h * f(t, y)\n t = t \u002B h\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example\n\\(y\u0027=y\\), \\(y(0)=1\\), \\(h=0.5\\). \nStep1: \\(y_1=1\u002B0.5*1=1.5\\). \nStep2: \\(y_2=1.5\u002B0.5*1.5=2.25\\). Exact at \\(t=1\\) is \\(e\\approx2.718\\).\n\n---\n\n## 31) Heun’s method (improved Euler)\n\n### Mathematical foundation\nPredictor \\(y^*=y_n\u002Bh f(t_n,y_n)\\). Corrector:\n$$\ny_{n\u002B1}=y_n\u002B\\frac{h}{2}\\left[f(t_n,y_n)\u002Bf(t_{n\u002B1},y^*)\\right]\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef heun(f, t0, y0, h, n_steps):\n t = float(t0)\n y = float(y0)\n out = [(t, y)]\n for _ in range(int(n_steps)):\n k1 = f(t, y)\n y_pred = y \u002B h * k1\n k2 = f(t \u002B h, y_pred)\n y = y \u002B (h/2.0) * (k1 \u002B k2)\n t = t \u002B h\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example\n\\(y\u0027=y\\), \\(y(0)=1\\), \\(h=0.5\\). \n\\(k1=1\\), \\(y^*=1.5\\), \\(k2=1.5\\). \n\\(y_1=1\u002B0.25(1\u002B1.5)=1.625\\) (closer to \\(e^{0.5}=1.6487\\)).\n\n---\n\n## 32) Classical Runge–Kutta (RK4)\n\n### Mathematical foundation\n$$\nk_1=f(t_n,y_n)\n$$\n$$\nk_2=f\\left(t_n\u002B\\frac{h}{2},y_n\u002B\\frac{h}{2}k_1\\right),\\;\nk_3=f\\left(t_n\u002B\\frac{h}{2},y_n\u002B\\frac{h}{2}k_2\\right)\n$$\n$$\nk_4=f(t_n\u002Bh,y_n\u002Bhk_3)\n$$\n$$\ny_{n\u002B1}=y_n\u002B\\frac{h}{6}(k_1\u002B2k_2\u002B2k_3\u002Bk_4)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef rk4(f, t0, y0, h, n_steps):\n t = float(t0)\n y = float(y0)\n out = [(t, y)]\n for _ in range(int(n_steps)):\n k1 = f(t, y)\n k2 = f(t \u002B h/2, y \u002B h*k1/2)\n k3 = f(t \u002B h/2, y \u002B h*k2/2)\n k4 = f(t \u002B h, y \u002B h*k3)\n y = y \u002B (h/6.0)*(k1 \u002B 2*k2 \u002B 2*k3 \u002B k4)\n t = t \u002B h\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example (one step)\n\\(y\u0027=y\\), \\(y(0)=1\\), \\(h=0.5\\). \n\\(k1=1\\). \n\\(k2=f(0.25,1.25)=1.25\\). \n\\(k3=f(0.25,1.3125)=1.3125\\). \n\\(k4=f(0.5,1.65625)=1.65625\\). \n\\(y_1=1\u002B0.5/6(1\u002B2*1.25\u002B2*1.3125\u002B1.65625)=1.6484375\\) (very close to 1.64872).\n\n---\n\n## 33) Adaptive RK45 (Dormand–Prince)\n\n### Mathematical foundation\nCompute 4th and 5th order estimates \\(y^{(4)},y^{(5)}\\); error \\(e=\\|y^{(5)}-y^{(4)}\\|\\). Update step:\n$$\nh_{\\text{new}} = h\\cdot \\left(\\frac{\\text{tol}}{e}\\right)^{1/5}\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef rk45_step(f, t, y, h):\n # Dormand–Prince coefficients\n k1 = f(t, y)\n k2 = f(t \u002B h*(1/5), y \u002B h*(1/5)*k1)\n k3 = f(t \u002B h*(3/10), y \u002B h*(3/40*k1 \u002B 9/40*k2))\n k4 = f(t \u002B h*(4/5), y \u002B h*(44/45*k1 - 56/15*k2 \u002B 32/9*k3))\n k5 = f(t \u002B h*(8/9), y \u002B h*(19372/6561*k1 - 25360/2187*k2 \u002B 64448/6561*k3 - 212/729*k4))\n k6 = f(t \u002B h, y \u002B h*(9017/3168*k1 - 355/33*k2 \u002B 46732/5247*k3 \u002B 49/176*k4 - 5103/18656*k5))\n # 5th order\n y5 = y \u002B h*(35/384*k1 \u002B 500/1113*k3 \u002B 125/192*k4 - 2187/6784*k5 \u002B 11/84*k6)\n # 4th order (embedded)\n k7 = f(t \u002B h, y5)\n y4 = y \u002B h*(5179/57600*k1 \u002B 7571/16695*k3 \u002B 393/640*k4 - 92097/339200*k5 \u002B 187/2100*k6 \u002B 1/40*k7)\n return y4, y5\n\ndef adaptive_rk45(f, t0, y0, t1, h0=0.1, tol=1e-8):\n t = float(t0)\n y = float(y0)\n h = float(h0)\n out = [(t, y)]\n while t \u003C t1 - 1e-15:\n if t \u002B h \u003E t1:\n h = t1 - t\n y4, y5 = rk45_step(f, t, y, h)\n err = abs(y5 - y4)\n if err \u003C tol or h \u003C 1e-15:\n t = t \u002B h\n y = float(y5)\n out.append((t, y))\n scale = (tol / max(err, 1e-15))**0.2\n h = h * min(5.0, max(0.2, 0.9*scale))\n return out\n\u0060\u0060\u0060\n\n### Worked example (conceptual intermediates)\nFor \\(y\u0027=y\\), \\(y(0)=1\\), start \\(h=0.5\\). \nCompute \\(y^{(4)}, y^{(5)}\\); suppose \\(err=|y5-y4|=2\\times10^{-7}\\) and tol \\(=10^{-8}\\). \nScale factor \\((10^{-8}/2\\cdot10^{-7})^{1/5}=(0.05)^{0.2}\\approx0.549\\). \nNew \\(h\\approx 0.9*0.549*0.5\\approx0.247\\).\n\n---\n\n## 34) Adams–Bashforth methods\n\n### Mathematical foundation\nExplicit multistep. AB2:\n$$\ny_{n\u002B1}=y_n\u002B\\frac{h}{2}\\left(3f_n-f_{n-1}\\right)\n$$\nAB4:\n$$\ny_{n\u002B1}=y_n\u002B\\frac{h}{24}\\left(55f_n-59f_{n-1}\u002B37f_{n-2}-9f_{n-3}\\right)\n$$\n\n### Python implementation (AB2)\n\u0060\u0060\u0060python\ndef adams_bashforth_2(f, t0, y0, h, n_steps):\n # bootstrap one step with RK4 for y1\n sol = rk4(f, t0, y0, h, 1)\n (t0, y0), (t1, y1) = sol[0], sol[1]\n out = [(t0, y0), (t1, y1)]\n f0 = f(t0, y0)\n f1 = f(t1, y1)\n t, y = t1, y1\n for _ in range(1, int(n_steps)):\n y_next = y \u002B (h/2.0)*(3*f1 - f0)\n t_next = t \u002B h\n f0, f1 = f1, f(t_next, y_next)\n t, y = t_next, y_next\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example\n\\(y\u0027=y\\), \\(h=0.5\\). \nBootstrap \\(y_1\\approx 1.6484375\\) (RK4). \nThen \\(f_0=1\\), \\(f_1=1.6484375\\). \nAB2: \\(y_2=y_1\u002B0.25(3*1.6484375-1)=1.6484375\u002B0.25(3.9453125)=2.6347656\\) (exact at \\(t=1\\) is 2.71828).\n\n---\n\n## 35) Backward Euler\n\n### Mathematical foundation\nImplicit:\n$$\ny_{n\u002B1}=y_n\u002Bh f(t_{n\u002B1},y_{n\u002B1})\n$$\nSolve nonlinear equation each step (e.g., Newton).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef backward_euler(f, dfdy, t0, y0, h, n_steps, newton_iters=20):\n t = float(t0)\n y = float(y0)\n out = [(t, y)]\n for _ in range(int(n_steps)):\n t1 = t \u002B h\n y_guess = y\n for _it in range(newton_iters):\n g = y_guess - y - h*f(t1, y_guess)\n dg = 1.0 - h*dfdy(t1, y_guess)\n y_new = y_guess - g/dg\n if abs(y_new - y_guess) \u003C 1e-12:\n y_guess = y_new\n break\n y_guess = y_new\n y = float(y_guess)\n t = t1\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example\n\\(y\u0027=y\\). Backward Euler: \\(y_{n\u002B1}=y_n\u002Bh y_{n\u002B1}\\Rightarrow y_{n\u002B1}=y_n/(1-h)\\). \nWith \\(h=0.5\\), \\(y_1=1/(0.5)=2\\), \\(y_2=2/0.5=4\\) (stable for stiff decay problems, but grows here).\n\n---\n\n## 36) Crank–Nicolson\n\n### Mathematical foundation\nTrapezoid in time:\n$$\ny_{n\u002B1}=y_n\u002B\\frac{h}{2}\\left(f(t_n,y_n)\u002Bf(t_{n\u002B1},y_{n\u002B1})\\right)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef crank_nicolson(f, dfdy, t0, y0, h, n_steps, newton_iters=20):\n t = float(t0)\n y = float(y0)\n out = [(t, y)]\n for _ in range(int(n_steps)):\n t1 = t \u002B h\n y_guess = y\n fn = f(t, y)\n for _it in range(newton_iters):\n g = y_guess - y - (h/2.0)*(fn \u002B f(t1, y_guess))\n dg = 1.0 - (h/2.0)*dfdy(t1, y_guess)\n y_new = y_guess - g/dg\n if abs(y_new - y_guess) \u003C 1e-12:\n y_guess = y_new\n break\n y_guess = y_new\n y = float(y_guess)\n t = t1\n out.append((t, y))\n return out\n\u0060\u0060\u0060\n\n### Worked example\nFor \\(y\u0027=y\\): \\(y_{n\u002B1}=y_n\u002B\\frac{h}{2}(y_n\u002By_{n\u002B1})\\Rightarrow y_{n\u002B1}=\\frac{1\u002Bh/2}{1-h/2}y_n\\). \nWith \\(h=0.5\\): factor \\(=(1.25)/(0.75)=1.6666667\\). So \\(y_1=1.6667\\).\n\n---\n\n# Optimization\n\n## 37) Gradient descent\n\n### Mathematical foundation\nMinimize \\(f(x)\\):\n$$\nx_{k\u002B1}=x_k-\\alpha_k\\nabla f(x_k)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef gradient_descent(grad, x0, alpha=0.1, max_iter=1000, tol=1e-10):\n x = _as_float_array(x0).reshape(-1)\n for _ in range(max_iter):\n g = _as_float_array(grad(x)).reshape(-1)\n x_new = x - alpha * g\n if np.linalg.norm(x_new - x) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nMinimize \\(f(x)=(x-3)^2\\), \\(\\nabla f=2(x-3)\\), \\(x_0=0\\), \\(\\alpha=0.1\\). \n\\(x_1=0-0.1*2(-3)=0.6\\). \n\\(x_2=0.6-0.1*2(-2.4)=1.08\\). Converges to 3.\n\n---\n\n## 38) Newton’s method for optimization\n\n### Mathematical foundation\nUsing Hessian \\(H\\):\n$$\nx_{k\u002B1}=x_k-H(x_k)^{-1}\\nabla f(x_k)\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef newton_opt(grad, hess, x0, max_iter=50, tol=1e-12):\n x = _as_float_array(x0).reshape(-1)\n for _ in range(max_iter):\n g = _as_float_array(grad(x)).reshape(-1)\n H = _as_float_array(hess(x))\n dx = np.linalg.solve(H, g)\n x_new = x - dx\n if np.linalg.norm(dx) \u003C tol:\n return x_new\n x = x_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame \\(f(x)=(x-3)^2\\). \\(g=2(x-3)\\), \\(H=2\\). \nFrom \\(x_0=0\\): \\(dx=g/H= -3\\). \\(x_1=0-(-3)=3\\) (one step).\n\n---\n\n## 39) BFGS\n\n### Mathematical foundation\nUpdate inverse Hessian approximation \\(B_k\\):\n$$\ns_k=x_{k\u002B1}-x_k,\\quad y_k=\\nabla f_{k\u002B1}-\\nabla f_k\n$$\n$$\nB_{k\u002B1}=(I-\\rho s y^T)B_k(I-\\rho y s^T)\u002B\\rho s s^T,\\quad \\rho=\\frac{1}{y^Ts}\n$$\n(Standard in optimization texts.) [3][8]\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bfgs(grad, x0, alpha=0.1, max_iter=200, tol=1e-10):\n x = _as_float_array(x0).reshape(-1)\n n = len(x)\n B = np.eye(n)\n g = _as_float_array(grad(x)).reshape(-1)\n\n for _ in range(max_iter):\n p = -B @ g\n x_new = x \u002B alpha * p\n g_new = _as_float_array(grad(x_new)).reshape(-1)\n\n s = (x_new - x).reshape(-1, 1)\n y = (g_new - g).reshape(-1, 1)\n ys = float(y.T @ s)\n if abs(ys) \u003C 1e-15:\n return x_new\n rho = 1.0 / ys\n I = np.eye(n)\n B = (I - rho*s@y.T) @ B @ (I - rho*y@s.T) \u002B rho*(s@s.T)\n\n if np.linalg.norm(x_new - x) \u003C tol:\n return x_new\n x, g = x_new, g_new\n return x\n\u0060\u0060\u0060\n\n### Worked example (first update)\nMinimize \\(f(x)= (x-3)^2\\) in 1D. \\(B_0=1\\). \n\\(g_0=2(x_0-3)\\). With \\(x_0=0\\), \\(g_0=-6\\), \\(p=6\\). \nTake \\(\\alpha=0.1\\): \\(x_1=0.6\\). \\(g_1=2(-2.4)=-4.8\\). \n\\(s=0.6\\), \\(y=1.2\\), \\(\\rho=1/(0.72)=1.3889\\). Update \\(B\\) accordingly.\n\n---\n\n## 40) L-BFGS\n\n### Mathematical foundation\nStore last \\(m\\) pairs \\((s_i,y_i)\\). Two-loop recursion gives \\(H_k g_k\\) without storing full matrices.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef lbfgs(grad, x0, m=5, alpha=0.5, max_iter=200, tol=1e-10):\n x = _as_float_array(x0).reshape(-1)\n s_list, y_list, rho_list = [], [], []\n g = _as_float_array(grad(x)).reshape(-1)\n\n for _ in range(max_iter):\n q = g.copy()\n a = []\n for s, y, rho in reversed(list(zip(s_list, y_list, rho_list))):\n ai = rho * np.dot(s, q)\n a.append(ai)\n q = q - ai * y\n # initial H0 scaling\n if len(s_list) \u003E 0:\n s, y = s_list[-1], y_list[-1]\n gamma = np.dot(s, y) / max(np.dot(y, y), 1e-15)\n else:\n gamma = 1.0\n r = gamma * q\n for (s, y, rho), ai in zip(zip(s_list, y_list, rho_list), reversed(a)):\n beta = rho * np.dot(y, r)\n r = r \u002B s * (ai - beta)\n\n p = -r\n x_new = x \u002B alpha * p\n g_new = _as_float_array(grad(x_new)).reshape(-1)\n\n s = x_new - x\n y = g_new - g\n ys = np.dot(y, s)\n if abs(ys) \u003E 1e-15:\n rho = 1.0 / ys\n s_list.append(s); y_list.append(y); rho_list.append(rho)\n if len(s_list) \u003E m:\n s_list.pop(0); y_list.pop(0); rho_list.pop(0)\n\n if np.linalg.norm(x_new - x) \u003C tol:\n return x_new\n x, g = x_new, g_new\n return x\n\u0060\u0060\u0060\n\n### Worked example\nSame 1D example: stores scalars \\(s,y\\) and produces quasi-Newton direction close to Newton step as iterations proceed.\n\n---\n\n## 41) Simplex method (LP)\n\n### Mathematical foundation\nStandard form:\n$$\n\\max c^Tx\\quad\\text{s.t.}\\quad Ax\\le b,\\ x\\ge 0\n$$\nIntroduce slack \\(s\\ge0\\): \\(Ax\u002Bs=b\\). Pivot to improve objective until no negative reduced costs.\n\n### Python implementation (small, pedagogical)\n\u0060\u0060\u0060python\ndef simplex_max(c, A, b, max_iter=100):\n c = _as_float_array(c).reshape(-1)\n A = _as_float_array(A)\n b = _as_float_array(b).reshape(-1)\n\n m, n = A.shape\n T = np.zeros((m\u002B1, n\u002Bm\u002B1))\n T[:m, :n] = A\n T[:m, n:n\u002Bm] = np.eye(m)\n T[:m, -1] = b\n T[m, :n] = -c # maximize\n\n basis = list(range(n, n\u002Bm))\n\n for _ in range(max_iter):\n # entering: most negative in objective row\n j = int(np.argmin(T[m, :-1]))\n if T[m, j] \u003E= -1e-12:\n break\n col = T[:m, j]\n if np.all(col \u003C= 1e-15):\n raise ValueError(\u0022Unbounded.\u0022)\n ratios = np.array([T[i, -1]/col[i] if col[i] \u003E 1e-15 else np.inf for i in range(m)])\n i = int(np.argmin(ratios))\n piv = T[i, j]\n T[i, :] = T[i, :] / piv\n for r in range(m\u002B1):\n if r != i:\n T[r, :] = T[r, :] - T[r, j] * T[i, :]\n basis[i] = j\n\n x = np.zeros(n\u002Bm)\n for i, bi in enumerate(basis):\n x[bi] = T[i, -1]\n return x[:n], T[m, -1]\n\u0060\u0060\u0060\n\n### Worked example\nMaximize \\(z=3x\u002B2y\\) s.t. \\(x\u002By\\le4\\), \\(x\\le2\\), \\(y\\le3\\), \\(x,y\\ge0\\). \nAt optimum: \\(x=2,y=2\\Rightarrow z=10\\). Tableau pivots will reach that BFS; intermediate ratio tests identify leaving variable each pivot.\n\n---\n\n## 42) Lagrange multipliers\n\n### Mathematical foundation\nMinimize \\(f(x)\\) subject to equality \\(g(x)=0\\). Lagrangian:\n$$\n\\mathcal{L}(x,\\lambda)=f(x)\u002B\\lambda^T g(x)\n$$\nStationarity:\n$$\n\\nabla_x \\mathcal{L}=0,\\quad g(x)=0\n$$\n\n### Python implementation (solve KKT for equality constraints)\n\u0060\u0060\u0060python\ndef lagrange_equalities(grad_f, jac_g, x0, lam0, max_iter=50, tol=1e-12):\n x = _as_float_array(x0).reshape(-1)\n lam = _as_float_array(lam0).reshape(-1)\n for _ in range(max_iter):\n g = _as_float_array(jac_g(x)[\u0022g\u0022]).reshape(-1) # constraint values\n J = _as_float_array(jac_g(x)[\u0022J\u0022]) # Jacobian\n gf = _as_float_array(grad_f(x)).reshape(-1)\n\n # Solve linearized KKT:\n # [ I J^T ][dx] = -[gf \u002B J^T lam]\n # [ J 0 ][dl] -[g]\n K = np.block([[np.eye(len(x)), J.T],\n [J, np.zeros((len(g), len(g)))]])\n rhs = -np.concatenate([gf \u002B J.T @ lam, g])\n sol = np.linalg.solve(K, rhs)\n dx = sol[:len(x)]\n dlam = sol[len(x):]\n x_new = x \u002B dx\n lam_new = lam \u002B dlam\n if np.linalg.norm(dx) \u003C tol and np.linalg.norm(g) \u003C tol:\n return x_new, lam_new\n x, lam = x_new, lam_new\n return x, lam\n\u0060\u0060\u0060\n\n### Worked example\nMinimize \\(f(x,y)=x^2\u002By^2\\) s.t. \\(x\u002By-1=0\\). \n\\(\\mathcal{L}=x^2\u002By^2\u002B\\lambda(x\u002By-1)\\). \nStationarity: \\(2x\u002B\\lambda=0\\), \\(2y\u002B\\lambda=0\\Rightarrow x=y\\). \nConstraint: \\(2x=1\\Rightarrow x=y=0.5\\). \n\\(\\lambda=-1\\).\n\n---\n\n## 43) KKT conditions\n\n### Mathematical foundation\nFor inequality constraints \\(g_i(x)\\le0\\), equality \\(h(x)=0\\): \nLagrangian:\n$$\n\\mathcal{L}=f(x)\u002B\\lambda^T h(x)\u002B\\mu^T g(x)\n$$\nKKT:\n- Stationarity: \\(\\nabla f\u002B\\nabla h^T\\lambda\u002B\\nabla g^T\\mu=0\\)\n- Primal feasibility: \\(h(x)=0,\\ g(x)\\le0\\)\n- Dual feasibility: \\(\\mu\\ge0\\)\n- Complementary slackness: \\(\\mu_i g_i(x)=0\\)\n\n### Python implementation (check KKT residuals)\n\u0060\u0060\u0060python\ndef kkt_check(x, grad_f, h, J_h, g, J_g, lam, mu, tol=1e-8):\n x = _as_float_array(x).reshape(-1)\n lam = _as_float_array(lam).reshape(-1)\n mu = _as_float_array(mu).reshape(-1)\n r_station = grad_f(x) \u002B J_h(x).T @ lam \u002B J_g(x).T @ mu\n r_primal_h = h(x)\n r_primal_g = g(x)\n r_dual = np.minimum(mu, 0.0)\n r_comp = mu * r_primal_g\n return {\n \u0022stationarity_inf\u0022: _norm_inf(r_station),\n \u0022eq_feas_inf\u0022: _norm_inf(r_primal_h),\n \u0022ineq_max\u0022: float(np.max(r_primal_g)),\n \u0022dual_neg_min\u0022: float(np.min(r_dual)),\n \u0022comp_inf\u0022: _norm_inf(r_comp)\n }\n\u0060\u0060\u0060\n\n### Worked example\nFor \\(f=x^2\u002By^2\\), constraint \\(x\u002By\\ge1\\) rewrite \\(g(x,y)=1-x-y\\le0\\). \nAt \\((0.5,0.5)\\): \\(g=0\\) active; choose \\(\\mu=1\\) gives stationarity \\(2x- \\mu=0\\Rightarrow 1-1=0\\).\n\n---\n\n## 44) Simulated annealing\n\n### Mathematical foundation\nRandomized search with temperature \\(T\\). Accept worse move \\(\\Delta E\u003E0\\) with probability:\n$$\np=\\exp(-\\Delta E/T)\n$$\nReduce \\(T\\) by schedule.\n\n### Python implementation\n\u0060\u0060\u0060python\ndef simulated_annealing(f, x0, step, T0=1.0, cooling=0.99, iters=5000, seed=0):\n rng = np.random.default_rng(seed)\n x = float(x0)\n fx = float(f(x))\n best_x, best_fx = x, fx\n T = float(T0)\n\n for _ in range(int(iters)):\n x_new = x \u002B step * rng.normal()\n f_new = float(f(x_new))\n dE = f_new - fx\n if dE \u003C= 0 or rng.uniform() \u003C np.exp(-dE / max(T, 1e-15)):\n x, fx = x_new, f_new\n if fx \u003C best_fx:\n best_x, best_fx = x, fx\n T *= cooling\n return best_x, best_fx\n\u0060\u0060\u0060\n\n### Worked example\nMinimize \\(f(x)=x^4-x^2\\). From \\(x_0=0.1\\), propose \\(x_1=0.3\\). \nCompute \\(\\Delta E=f(0.3)-f(0.1)\\). If \\(\\Delta E\u003E0\\), accept with \\(\\exp(-\\Delta E/T)\\).\n\n---\n\n# Statistics \u0026 Probability\n\n## 45) Bayes’ theorem derivation\n\n### Mathematical foundation\nFrom joint probability:\n$$\nP(A\\cap B)=P(A|B)P(B)=P(B|A)P(A)\n$$\nThus:\n$$\nP(A|B)=\\frac{P(B|A)P(A)}{P(B)}\n$$\nAnd:\n$$\nP(B)=\\sum_i P(B|A_i)P(A_i)\n$$\n(HMM/ML references often use this repeatedly.) [5][9]\n\n### Python implementation\n\u0060\u0060\u0060python\ndef bayes_posterior(prior_A, likelihood_B_given_A, evidence_B):\n prior_A = float(prior_A)\n likelihood_B_given_A = float(likelihood_B_given_A)\n evidence_B = float(evidence_B)\n return (likelihood_B_given_A * prior_A) / evidence_B\n\u0060\u0060\u0060\n\n### Worked example\nDisease test: \\(P(D)=0.01\\), \\(P(\u002B|D)=0.99\\), \\(P(\u002B|\\neg D)=0.05\\). \nEvidence:\n\\(P(\u002B)=0.99*0.01\u002B0.05*0.99=0.0099\u002B0.0495=0.0594\\). \nPosterior:\n\\(P(D|\u002B)=0.99*0.01/0.0594=0.0099/0.0594=0.1667\\).\n\n---\n\n## 46) Maximum likelihood estimation (MLE)\n\n### Mathematical foundation\nGiven data \\(x_1,\\dots,x_n\\) with density \\(p(x|\\theta)\\):\n$$\n\\hat{\\theta}=\\arg\\max_\\theta \\prod_{i=1}^n p(x_i|\\theta)\n$$\nUse log-likelihood:\n$$\n\\ell(\\theta)=\\sum_{i=1}^n \\log p(x_i|\\theta)\n$$\n\n### Python implementation (Gaussian mean/variance MLE)\n\u0060\u0060\u0060python\ndef mle_gaussian_params(data):\n x = _as_float_array(data).reshape(-1)\n mu_hat = float(np.mean(x))\n # MLE variance uses 1/n (not unbiased 1/(n-1))\n sigma2_hat = float(np.mean((x - mu_hat)**2))\n return mu_hat, sigma2_hat\n\u0060\u0060\u0060\n\n### Worked example\nData \\([1,2,4]\\). \n\\(\\hat\\mu=(1\u002B2\u002B4)/3=7/3=2.3333\\). \nSquared deviations: \\(( -1.3333)^2=1.7778\\), \\((-0.3333)^2=0.1111\\), \\((1.6667)^2=2.7778\\). \n\\(\\hat\\sigma^2=(1.7778\u002B0.1111\u002B2.7778)/3=1.5556\\).\n\n---\n\n## 47) Linear regression derivation\n\n### Mathematical foundation\nModel \\(y=X\\beta\u002B\\epsilon\\). Least squares:\n$$\n\\hat\\beta=\\arg\\min_\\beta \\|X\\beta-y\\|_2^2\n$$\nNormal equations:\n$$\nX^TX\\hat\\beta=X^Ty\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef linear_regression_fit(X, y):\n X = _as_float_array(X)\n y = _as_float_array(y).reshape(-1)\n beta = np.linalg.solve(X.T @ X, X.T @ y)\n return beta\n\u0060\u0060\u0060\n\n### Worked example\nFit \\(y\\approx c_0\u002Bc_1x\\) with earlier points: gives \\(c_0=1.1667,c_1=0.5\\) (see LS section intermediates).\n\n---\n\n## 48) Logistic regression\n\n### Mathematical foundation\nBinary \\(y\\in\\{0,1\\}\\):\n$$\nP(y=1|x)=\\sigma(w^Tx)=\\frac{1}{1\u002Be^{-w^Tx}}\n$$\nLog-likelihood:\n$$\n\\ell(w)=\\sum_i \\left[y_i\\log\\sigma(z_i)\u002B(1-y_i)\\log(1-\\sigma(z_i))\\right],\\ z_i=w^Tx_i\n$$\nGradient:\n$$\n\\nabla \\ell(w)=\\sum_i (y_i-\\sigma(z_i))x_i\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef logistic_regression_fit(X, y, lr=0.1, max_iter=5000, tol=1e-8):\n X = _as_float_array(X)\n y = _as_float_array(y).reshape(-1)\n w = np.zeros(X.shape[1])\n\n def sigmoid(z):\n return 1.0 / (1.0 \u002B np.exp(-z))\n\n for _ in range(max_iter):\n z = X @ w\n p = sigmoid(z)\n g = X.T @ (y - p)\n w_new = w \u002B lr * g\n if np.linalg.norm(w_new - w) \u003C tol:\n return w_new\n w = w_new\n return w\n\u0060\u0060\u0060\n\n### Worked example (one gradient step)\nOne feature with intercept: \\(X=[[1,0],[1,1]]\\), \\(y=[0,1]\\), \\(w=(0,0)\\). \n\\(z=(0,0)\\), \\(p=(0.5,0.5)\\). \n\\(y-p=(-0.5,0.5)\\). \nGradient \\(g=X^T(y-p)=(0,0.5)\\). \nWith lr=1: \\(w=(0,0.5)\\).\n\n---\n\n## 49) PCA derivation\n\n### Mathematical foundation\nCentered data matrix \\(X\\). Covariance:\n$$\nC=\\frac{1}{n-1}X^TX\n$$\nPCA directions are eigenvectors of \\(C\\):\n$$\nC v_i=\\lambda_i v_i\n$$\nEquivalently via SVD \\(X=U\\Sigma V^T\\), principal axes \\(V\\), variances \\(\\Sigma^2/(n-1)\\).\n\n### Python implementation\n\u0060\u0060\u0060python\ndef pca(X, k=None):\n X = _as_float_array(X)\n Xc = X - np.mean(X, axis=0, keepdims=True)\n U, s, Vt = np.linalg.svd(Xc, full_matrices=False)\n if k is None:\n k = Vt.shape[0]\n components = Vt[:k].T\n explained_var = (s[:k]**2) / (X.shape[0]-1)\n return components, explained_var\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\((1,0),(2,0),(3,0)\\). Centered: \\((-1,0),(0,0),(1,0)\\). \nCovariance \\(C=\\frac{1}{2}\\begin{bmatrix}2\u00260\\\\0\u00260\\end{bmatrix}=\\begin{bmatrix}1\u00260\\\\0\u00260\\end{bmatrix}\\). \nEigenvectors: first axis \\((1,0)\\) with eigenvalue 1.\n\n---\n\n## 50) k-means derivation\n\n### Mathematical foundation\nMinimize within-cluster SSE:\n$$\nJ=\\sum_{i=1}^n \\|x_i-\\mu_{c_i}\\|^2\n$$\nAlternating minimization:\n1) Assign \\(c_i=\\arg\\min_k \\|x_i-\\mu_k\\|^2\\) \n2) Update \\(\\mu_k=\\frac{1}{|C_k|}\\sum_{i\\in C_k} x_i\\)\n\n### Python implementation\n\u0060\u0060\u0060python\ndef kmeans(X, k, max_iter=100, seed=0):\n rng = np.random.default_rng(seed)\n X = _as_float_array(X)\n n = X.shape[0]\n centers = X[rng.choice(n, k, replace=False)]\n\n for _ in range(max_iter):\n d2 = ((X[:, None, :] - centers[None, :, :])**2).sum(axis=2)\n labels = np.argmin(d2, axis=1)\n new_centers = np.array([X[labels == j].mean(axis=0) for j in range(k)])\n if np.linalg.norm(new_centers - centers) \u003C 1e-10:\n break\n centers = new_centers\n return centers, labels\n\u0060\u0060\u0060\n\n### Worked example\nPoints \\(\\{0,2,8,10\\}\\) in 1D, \\(k=2\\). Start centers \\(0,10\\). \nAssign: \\(\\{0,2\\}\\to0\\), \\(\\{8,10\\}\\to10\\). \nUpdate: \\(\\mu_1=1\\), \\(\\mu_2=9\\). Next assignment unchanged → converged.\n\n---\n\n## 51) Gaussian mixture models (GMM)\n\n### Mathematical foundation\nMixture:\n$$\np(x)=\\sum_{k=1}^K \\pi_k \\mathcal{N}(x|\\mu_k,\\Sigma_k)\n$$\nEM: responsibilities\n$$\n\\gamma_{ik}=\\frac{\\pi_k \\mathcal{N}(x_i|\\mu_k,\\Sigma_k)}{\\sum_j \\pi_j \\mathcal{N}(x_i|\\mu_j,\\Sigma_j)}\n$$\nM-step:\n$$\nN_k=\\sum_i\\gamma_{ik},\\quad \\pi_k=N_k/n,\\quad\n\\mu_k=\\frac{1}{N_k}\\sum_i \\gamma_{ik}x_i\n$$\n$$\n\\Sigma_k=\\frac{1}{N_k}\\sum_i \\gamma_{ik}(x_i-\\mu_k)(x_i-\\mu_k)^T\n$$\n\n### Python implementation (full-cov, small K)\n\u0060\u0060\u0060python\ndef gmm_em(X, K, max_iter=100, seed=0, eps=1e-6):\n rng = np.random.default_rng(seed)\n X = _as_float_array(X)\n n, d = X.shape\n mu = X[rng.choice(n, K, replace=False)]\n Sigma = np.array([np.eye(d) for _ in range(K)])\n pi = np.ones(K) / K\n\n def gauss_pdf(x, m, S):\n S = S \u002B eps*np.eye(d)\n det = np.linalg.det(S)\n inv = np.linalg.inv(S)\n diff = x - m\n expo = -0.5 * diff.T @ inv @ diff\n norm = 1.0 / np.sqrt((2*np.pi)**d * det)\n return float(norm * np.exp(expo))\n\n for _ in range(max_iter):\n gamma = np.zeros((n, K))\n for i in range(n):\n denom = 0.0\n for k in range(K):\n gamma[i, k] = pi[k] * gauss_pdf(X[i], mu[k], Sigma[k])\n denom \u002B= gamma[i, k]\n gamma[i, :] /= max(denom, 1e-15)\n\n Nk = np.sum(gamma, axis=0)\n pi = Nk / n\n for k in range(K):\n mu[k] = (gamma[:, k][:, None] * X).sum(axis=0) / max(Nk[k], 1e-15)\n diff = X - mu[k]\n Sigma[k] = (gamma[:, k][:, None, None] * (diff[:, :, None] @ diff[:, None, :])).sum(axis=0) / max(Nk[k], 1e-15)\n return pi, mu, Sigma\n\u0060\u0060\u0060\n\n### Worked example (one E-step)\n1D points \\(x=[0,1,9,10]\\), \\(K=2\\), init \\(\\mu=[0,10]\\), \\(\\sigma^2=1\\), \\(\\pi=0.5\\). \nCompute \\(\\gamma_{i1}\\) using Gaussian pdf at each point; e.g. for \\(x=1\\): pdf near \\(\\mu=0\\) is high, near 10 is tiny → \\(\\gamma\\approx1\\) for cluster1.\n\n---\n\n## 52) Markov chains\n\n### Mathematical foundation\nTransition matrix \\(P\\), \\(P_{ij}=P(X_{t\u002B1}=j|X_t=i)\\). \nState distribution evolves:\n$$\n\\pi_{t\u002B1}=\\pi_t P\n$$\nStationary distribution:\n$$\n\\pi=\\pi P,\\quad \\sum_i \\pi_i=1\n$$\n\n### Python implementation\n\u0060\u0060\u0060python\ndef markov_step(pi, P):\n pi = _as_float_array(pi).reshape(-1)\n P = _as_float_array(P)\n return pi @ P\n\ndef stationary_distribution(P, iters=10000, tol=1e-12):\n P = _as_float_array(P)\n n = P.shape[0]\n pi = np.ones(n) / n\n for _ in range(iters):\n pi_new = pi @ P\n if _norm_inf(pi_new - pi) \u003C tol:\n return pi_new\n pi = pi_new\n return pi\n\u0060\u0060\u0060\n\n### Worked example\n\\(P=\\begin{bmatrix}0.9\u00260.1\\\\0.2\u00260.8\\end{bmatrix}\\), start \\(\\pi_0=(1,0)\\). \n\\(\\pi_1=(0.9,0.1)\\). \n\\(\\pi_2=(0.9,0.1)P=(0.9*0.9\u002B0.1*0.2,\\;0.9*0.1\u002B0.1*0.8)=(0.83,0.17)\\). \nStationary solves \\(\\pi=\\pi P\\Rightarrow \\pi=(2/3,1/3)\\).\n\n---\n\n## 53) Hidden Markov Models (HMMs)\n\n### Mathematical foundation\nHMM parameters: initial \\(\\pi\\), transition \\(A\\), emissions \\(B\\). [9][5][10] \nForward probabilities:\n$$\n\\alpha_1(i)=\\pi_i b_i(o_1)\n$$\n$$\n\\alpha_{t\u002B1}(j)=\\left(\\sum_i \\alpha_t(i)a_{ij}\\right)b_j(o_{t\u002B1})\n$$\nViterbi (most likely path):\n$$\n\\delta_{t\u002B1}(j)=\\max_i \\delta_t(i)a_{ij}\\, b_j(o_{t\u002B1})\n$$\n\n### Python implementation (discrete emissions)\n\u0060\u0060\u0060python\ndef hmm_forward(pi, A, B, obs):\n pi = _as_float_array(pi); A = _as_float_array(A); B = _as_float_array(B)\n obs = list(obs)\n N = len(pi); T = len(obs)\n alpha = np.zeros((T, N))\n alpha[0] = pi * B[:, obs[0]]\n for t in range(T-1):\n alpha[t\u002B1] = (alpha[t] @ A) * B[:, obs[t\u002B1]]\n return alpha, float(np.sum(alpha[-1]))\n\ndef hmm_viterbi(pi, A, B, obs):\n pi = _as_float_array(pi); A = _as_float_array(A); B = _as_float_array(B)\n obs = list(obs)\n N = len(pi); T = len(obs)\n delta = np.zeros((T, N))\n psi = np.zeros((T, N), dtype=int)\n\n delta[0] = pi * B[:, obs[0]]\n for t in range(1, T):\n for j in range(N):\n vals = delta[t-1] * A[:, j]\n psi[t, j] = int(np.argmax(vals))\n delta[t, j] = np.max(vals) * B[j, obs[t]]\n\n path = np.zeros(T, dtype=int)\n path[-1] = int(np.argmax(delta[-1]))\n for t in range(T-2, -1, -1):\n path[t] = psi[t\u002B1, path[t\u002B1]]\n return path, float(np.max(delta[-1]))\n\u0060\u0060\u0060\n\n### Worked example (forward intermediates)\nTwo states \\(S=\\{0,1\\}\\), observations \\(\\{0,1\\}\\). \n\\(\\pi=(0.6,0.4)\\). \n\\(A=\\begin{bmatrix}0.7\u00260.3\\\\0.4\u00260.6\\end{bmatrix}\\). \n\\(B=\\begin{bmatrix}0.5\u00260.5\\\\0.1\u00260.9\\end{bmatrix}\\). \nObs sequence \\(o=[1,0]\\). \n\\(\\alpha_1=(0.6*0.5,\\;0.4*0.9)=(0.3,0.36)\\). \nNext:\n\\(\\alpha_2(0)=((0.3*0.7\u002B0.36*0.4))*0.5=(0.21\u002B0.144)*0.5=0.177\\). \n\\(\\alpha_2(1)=((0.3*0.3\u002B0.36*0.6))*0.1=(0.09\u002B0.216)*0.1=0.0306\\). \nLikelihood \\(P(o)=0.2076\\).\n\n---\n\n# Quick “What you can do next”\nIf you tell me **your preferred formatting target** (Markdown doc, Word, or internal wiki style), I can restructure this into a ready-to-paste **knowledge base layout** (navigation, cross-links, consistent notation), while keeping the same math/code and the same topic order.\n\n[1]: https://www.press.jhu.edu/books/title/10678/matrix-computations \u0022Matrix Computations | Hopkins Press\u0022\n[2]: https://books.google.com/books/about/Numerical_Analysis.html?id=75PBQgAACAAJ \u0022Numerical Analysis - Richard L. Burden, J. Douglas Faires - Google Books\u0022\n[3]: https://link.springer.com/book/10.1007/978-0-387-40065-5 \u0022Numerical Optimization | Springer Nature Link\u0022\n[4]: https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf \u0022An Introduction to the Conjugate Gradient Method Without the Agonizing Pain\u0022\n[5]: https://courses.physics.illinois.edu/ece417/fa2023/slides/lec13.pdf \u0022Baum-Welch and Viterbi - University of Illinois Urbana-Champaign\u0022\n[6]: https://en.wikipedia.org/wiki/Derivation_of_the_conjugate_gradient_method \u0022Derivation of the conjugate gradient method - Wikipedia\u0022\n[7]: https://web.stanford.edu/class/ee364b/lectures/conj_grad_slides.pdf \u0022Conjugate Gradient Method - Stanford University\u0022\n[8]: http://users.iems.northwestern.edu/~nocedal/book/num-opt.html \u0022Numerical Optimization, by Nocedal and Wright\u0022\n[9]: https://hmmlearn.readthedocs.io/en/latest/tutorial.html \u0022Tutorial — hmmlearn 0.3.3.post1\u002Bge01a10e documentation\u0022\n[10]: https://en.wikipedia.org/wiki/Baum–Welch_algorithm \u0022Baum–Welch algorithm - Wikipedia\u0022\n[11]: https://www.geeksforgeeks.org/artificial-intelligence/viterbi-algorithm-for-hidden-markov-models-hmms/ \u0022Viterbi Algorithm for Hidden Markov Models (HMMs)\u0022\n[12]: https://www.mi.fu-berlin.de/wiki/pub/ABI/HiddenMarkovModelsWS13/script.pdf \u00229 Markov chains and Hidden Markov Models - fu-berlin.de\u0022\n[13]: https://www.digitalocean.com/community/tutorials/hidden-markov-models \u0022Hidden Markov Models Explained - DigitalOcean\u0022\n[14]: https://books.google.com/books/about/Matrix_Computations.html?id=X5YfsuCWpxMC \u0022Matrix Computations - Gene Howard Golub, Charles F. Van Loan - Google Books\u0022\n[15]: https://archive.org/details/matrixcomputatio0004golu \u0022Matrix computations : Golub, Gene H. (Gene Howard), 1932-2007 : Free ...\u0022\n[16]: https://www.dsprelated.com/books/1643.php \u0022Matrix Computations (Johns Hopkins Studies in the Mathematical Sciences)\u0022\n[17]: https://www.perlego.com/book/5060144/matrix-computations-pdf \u0022[PDF] Matrix Computations by Gene H. Golub, 4th edition - Perlego\u0022\n[18]: https://github.com/johanastborg/mc-python \u0022GitHub - johanastborg/mc-python: Matrix Computations, Gene Howard Golub ...\u0022\n[19]: https://www.cse.iitm.ac.in/~vplab/downloads/opt/Jorge%20Nocedal,%20Stephen%20Wright-Numerical%20Optimization-Springer%20%282000%29.pdf \u0022Numerical Optimization - Indian Institute of Technology Madras\u0022\n[20]: https://books.google.com/books/about/Numerical_Optimization.html?id=VbHYoSyelFcC \u0022Numerical Optimization - Jorge Nocedal, Stephen Wright - Google Books\u0022\n[21]: https://www.academia.edu/49277110/Numerical_Optimization_Jorge_Nocedal_Stephen_Wright \u0022Numerical Optimization Jorge Nocedal Stephen Wright\u0022\n[22]: https://www.math.kent.edu/~reichel/courses/optimization/Numerical_Optimization.pdf \u0022Second Edition - Kent\u0022\n[23]: https://valenpe7.github.io/numerical_methods/tutorials/04-linear_algebra_contd.html \u0022Tutorial 04 - Linear algebra (cont’d) — 12NME1 - Numerical methods\u0022\n[24]: https://optimization.cbe.cornell.edu/index.php?title=Conjugate_gradient_methods \u0022Conjugate gradient methods - Cornell University\u0022\n[25]: https://rupakmukherjee.github.io/Numerical-Analysis/ \u0022Numerical-Analysis | Numerical Analysis by R L Burden \u0026 J D Faires\u0022\n[26]: https://www.cengageasia.com/TitleDetails/isbn/9789814834285 \u0022AE Numerical Analysis - Cengage Asia\u0022\n[27]: https://studylib.net/doc/27886414/numerical-analysis-burden \u0022Numerical Analysis 10th Edition: Mathematics Textbook by Burden, Faires\u0022\n[28]: https://sites.millersville.edu/rbuchanan/math375/Jacobi.pdf \u0022Jacobi and Gauss-Seidel Iterative Techniques\u0022\n[29]: https://archive.org/details/numericalanalys00burd \u0022Numerical analysis : Burden, Richard L : Free Download, Borrow, and ...\u0022\n[30]: https://openlibrary.org/books/OL28549995M/Numerical_Analysis \u0022Numerical Analysis by Richard L. 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+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-555","type":"typing","text":".reshape","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":556},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":556,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-556","type":"typing","text":"(-","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":557},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":557,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-557","type":"typing","text":"1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":558},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":558,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-558","type":"typing","text":",","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":559},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":559,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-559","type":"typing","text":" ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":560},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":560,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-560","type":"typing","text":"1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":561},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":561,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-561","type":"typing","text":")])\n\n","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":562},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":562,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-562","type":"typing","text":" ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":563},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":563,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-563","type":"typing","text":" for","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":564},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":564,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-564","type":"typing","text":" k","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":565},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":565,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-565","type":"typing","text":" in","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":566},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":566,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-566","type":"typing","text":" range","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":567},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":567,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-567","type":"typing","text":"(n","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":568},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":568,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-568","type":"typing","text":" -","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":569},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":569,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-569","type":"typing","text":" ","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":570},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":570,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-570","type":"typing","text":"1","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":571},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":571,"type":"streaminfo","properties":{}}]}
+
+event: activity
+data: {"id":"6a0ece68-c14b-45ba-a6ab-007698198dc8-571","type":"typing","text":"):\n","from":{"id":"Default-e996fb66-b977-4a40-ae92-c5096c0964e4/660ef082-6c28-f111-88b4-000d3a36b21b"},"conversation":{"id":"3ce13336-2c31-442d-8328-d4c202e77478"},"channelData":{"streamType":"streaming","streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","chunkType":"delta","streamSequence":572},"entities":[{"streamId":"6a0ece68-c14b-45ba-a6ab-007698198dc8","streamType":"streaming","streamSequence":572,"type":"streaminfo","properties":{}}]}
diff --git a/__tests__/html2/markdown/vnext/streamingRenderer.html b/__tests__/html2/markdown/vnext/streamingRenderer.html
new file mode 100644
index 0000000000..4a03d820bc
--- /dev/null
+++ b/__tests__/html2/markdown/vnext/streamingRenderer.html
@@ -0,0 +1,96 @@
+
+
+
+
+
+ Streaming markdown rendering via createStreamingRenderer
+
+
+
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/__tests__/html2/markdown/vnext/streamingRenderer.html.snap-1.png b/__tests__/html2/markdown/vnext/streamingRenderer.html.snap-1.png
new file mode 100644
index 0000000000..e7cc7c64d1
Binary files /dev/null and b/__tests__/html2/markdown/vnext/streamingRenderer.html.snap-1.png differ
diff --git a/__tests__/html2/markdownRenderHTML/default.messageActivity.html.snap-1.png b/__tests__/html2/markdownRenderHTML/default.messageActivity.html.snap-1.png
index b0b6aae76a..9098627fff 100644
Binary files a/__tests__/html2/markdownRenderHTML/default.messageActivity.html.snap-1.png and b/__tests__/html2/markdownRenderHTML/default.messageActivity.html.snap-1.png differ
diff --git a/packages/bundle/src/markdown/createStreamingRenderer.ts b/packages/bundle/src/markdown/createStreamingRenderer.ts
new file mode 100644
index 0000000000..74b1be7479
--- /dev/null
+++ b/packages/bundle/src/markdown/createStreamingRenderer.ts
@@ -0,0 +1,350 @@
+/* eslint-disable no-magic-numbers */
+import katex from 'katex';
+import { compile, parse, postprocess, preprocess } from 'micromark';
+import { gfm, gfmHtml } from 'micromark-extension-gfm';
+import type { Event, Options } from 'micromark-util-types';
+
+import { math, mathHtml } from './mathExtension';
+import betterLinkDocumentMod from './private/betterLinkDocumentMod';
+import extractDefinitionsFromEvents, { type MarkdownLinkDefinition } from './private/extractDefinitionsFromEvents';
+import { pre as respectCRLFPre } from './private/respectCRLF';
+import { createDecorate } from './private/createDecorate';
+
+type StreamingRenderInit = Readonly<{
+ externalLinkAlt: string;
+}>;
+
+type StreamingRenderOptions = Readonly<{
+ markdownRenderHTML?: boolean | undefined;
+ markdownRespectCRLF: boolean;
+}>;
+
+type StreamingNextOptions = Readonly<{
+ container: HTMLElement;
+ containerClassName?: string | undefined;
+ transformFragment?: ((fragment: DocumentFragment) => DocumentFragment) | undefined;
+}>;
+
+type StreamingNextResult = Readonly<{
+ definitions: readonly MarkdownLinkDefinition[];
+}>;
+
+type StreamingRenderer = Readonly<{
+ finalize: (options: StreamingNextOptions) => StreamingNextResult;
+ next: (chunk: string, options: StreamingNextOptions) => void;
+ reset: () => void;
+}>;
+
+// Top-level block token types emitted by micromark.
+// An exit event at depth 0 for one of these types marks a committed block boundary.
+const TOP_LEVEL_BLOCK_TYPES: ReadonlySet = new Set([
+ 'atxHeading',
+ 'blockQuote',
+ 'codeFenced',
+ 'codeIndented',
+ 'content',
+ 'htmlFlow',
+ 'listOrdered',
+ 'listUnordered',
+ 'setextHeading',
+ 'table',
+ 'thematicBreak',
+ 'math'
+]);
+
+type BlockBoundary = {
+ readonly endOffset: number;
+ readonly startOffset: number;
+ readonly type: string;
+};
+
+function findTopLevelBlocks(events: ReadonlyArray): readonly BlockBoundary[] {
+ const blocks: Array<{ endOffset: number; startOffset: number; type: string }> = [];
+ let depth = 0;
+
+ for (const [action, token] of events) {
+ if (!TOP_LEVEL_BLOCK_TYPES.has(token.type)) {
+ continue;
+ }
+
+ if (action === 'enter') {
+ if (!depth) {
+ blocks.push({ endOffset: -1, startOffset: token.start.offset, type: token.type });
+ }
+
+ depth++;
+ } else {
+ depth--;
+
+ if (!depth && blocks.length) {
+ blocks[blocks.length - 1].endOffset = token.end.offset;
+ }
+ }
+ }
+
+ return blocks;
+}
+
+export default function createStreamingRenderer(
+ { markdownRenderHTML, markdownRespectCRLF }: StreamingRenderOptions,
+ { externalLinkAlt }: StreamingRenderInit
+): StreamingRenderer {
+ const micromarkOptions: Options = {
+ allowDangerousHtml: markdownRenderHTML ?? true,
+ allowDangerousProtocol: true,
+ extensions: [gfm(), math()],
+ htmlExtensions: [
+ gfmHtml(),
+ mathHtml({
+ renderMath: (content, isDisplay) =>
+ katex.renderToString(content, {
+ displayMode: isDisplay,
+ output: 'mathml'
+ })
+ })
+ ]
+ };
+
+ const domParser = new DOMParser();
+
+ // Parser state.
+ let activeBlockStartOffset = 0;
+ let previousMarkdown = '';
+ const emptyDefinitions: readonly MarkdownLinkDefinition[] = Object.freeze([]);
+
+ // DOM reconciliation state.
+ let wrapperDiv: HTMLDivElement | null = null;
+ let activeSentinel: Comment | null = null;
+
+ function parseEvents(source: string): Event[] {
+ return postprocess(
+ parse(micromarkOptions)
+ .document()
+ .write(preprocess()(source, undefined, true))
+ );
+ }
+
+ function applyTransform(
+ fragment: DocumentFragment,
+ transformFragment: ((fragment: DocumentFragment) => DocumentFragment) | undefined
+ ): DocumentFragment {
+ return transformFragment ? transformFragment(fragment) : fragment;
+ }
+
+ function ensureWrapper(container: HTMLElement, containerClassName: string | undefined): HTMLDivElement {
+ if (wrapperDiv && container.contains(wrapperDiv)) {
+ wrapperDiv.className = containerClassName || '';
+
+ return wrapperDiv;
+ }
+
+ const wrapper = document.createElement('div');
+
+ wrapper.className = containerClassName || '';
+ container.textContent = '';
+ container.appendChild(wrapper);
+ wrapperDiv = wrapper;
+ activeSentinel = null;
+
+ return wrapper;
+ }
+
+ function renderNext(chunk: string, options: StreamingNextOptions): void {
+ previousMarkdown += chunk;
+
+ if (!previousMarkdown) {
+ activeBlockStartOffset = 0;
+ activeSentinel = null;
+
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+
+ wrapper.replaceChildren();
+
+ return;
+ }
+
+ let processedMarkdown = previousMarkdown;
+
+ if (markdownRespectCRLF) {
+ processedMarkdown = respectCRLFPre(processedMarkdown);
+ }
+
+ // Incremental path: re-parse only from the last committed block boundary.
+ // Guard: if sentinel was lost (e.g. container changed), fall back to full reparse.
+ if (activeBlockStartOffset > 0) {
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+
+ if (activeSentinel && wrapper.contains(activeSentinel)) {
+ const tailEvents = parseEvents(processedMarkdown.slice(activeBlockStartOffset));
+ const tailBlocks = findTopLevelBlocks(tailEvents);
+ const tailHTML = compile(micromarkOptions)(tailEvents);
+
+ if (tailBlocks.length <= 1) {
+ // Fast path: active block grew, no new committed blocks.
+ // Replace only the active zone (after sentinel).
+ const activeDoc = domParser.parseFromString(tailHTML, 'text/html');
+ const activeFragment = activeDoc.createDocumentFragment();
+
+ activeFragment.append(...Array.from(activeDoc.body.childNodes));
+ betterLinkDocumentMod(activeFragment, createDecorate(emptyDefinitions, externalLinkAlt));
+
+ const activeRange = document.createRange();
+
+ activeRange.setStartAfter(activeSentinel);
+ activeRange.setEndAfter(wrapper.lastChild!);
+ activeRange.deleteContents();
+
+ wrapper.append(applyTransform(activeFragment, options.transformFragment));
+ } else {
+ // New block boundary in tail: commit newly-finished blocks, replace active.
+ const newActiveOffsetInTail = tailBlocks[tailBlocks.length - 1].startOffset;
+ const committedTailEvents = tailEvents.filter(([, token]) => token.start.offset < newActiveOffsetInTail);
+ const committedTailHTML = compile(micromarkOptions)(committedTailEvents);
+
+ activeBlockStartOffset += newActiveOffsetInTail;
+
+ const committedDoc = domParser.parseFromString(committedTailHTML, 'text/html');
+ const committedFragment = committedDoc.createDocumentFragment();
+
+ committedFragment.append(...Array.from(committedDoc.body.childNodes));
+ betterLinkDocumentMod(committedFragment, createDecorate(emptyDefinitions, externalLinkAlt));
+
+ const remainingHTML = tailHTML.slice(committedTailHTML.length);
+ const activeDoc = domParser.parseFromString(remainingHTML, 'text/html');
+ const activeFragment = activeDoc.createDocumentFragment();
+
+ activeFragment.append(...Array.from(activeDoc.body.childNodes));
+ betterLinkDocumentMod(activeFragment, createDecorate(emptyDefinitions, externalLinkAlt));
+
+ // Remove old sentinel and active zone.
+ const tailRange = document.createRange();
+
+ tailRange.setStartBefore(activeSentinel);
+ tailRange.setEndAfter(wrapper.lastChild!);
+ tailRange.deleteContents();
+
+ // Append newly committed, new sentinel, active.
+ activeSentinel = document.createComment('');
+
+ wrapper.append(
+ applyTransform(committedFragment, options.transformFragment),
+ activeSentinel,
+ applyTransform(activeFragment, options.transformFragment)
+ );
+ }
+
+ return;
+ }
+
+ // Sentinel lost — reset and fall through to full reparse.
+ activeBlockStartOffset = 0;
+ activeSentinel = null;
+ }
+
+ // Full reparse path.
+ const fullEvents = parseEvents(processedMarkdown);
+ const rawHTML = compile(micromarkOptions)(fullEvents);
+ const parsedDocument = domParser.parseFromString(rawHTML, 'text/html');
+ const fragment = parsedDocument.createDocumentFragment();
+
+ fragment.append(...Array.from(parsedDocument.body.childNodes));
+
+ const blocks = findTopLevelBlocks(fullEvents);
+
+ if (blocks.length >= 2) {
+ activeBlockStartOffset = blocks[blocks.length - 1].startOffset;
+
+ const range = document.createRange();
+
+ range.setStartBefore(fragment.firstChild!);
+ range.setEndBefore(fragment.lastElementChild!);
+
+ const committedFragment = range.extractContents();
+ const decorate = createDecorate(emptyDefinitions, externalLinkAlt);
+
+ betterLinkDocumentMod(committedFragment, decorate);
+ betterLinkDocumentMod(fragment, decorate);
+
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+
+ activeSentinel = document.createComment('');
+
+ wrapper.replaceChildren(
+ applyTransform(committedFragment, options.transformFragment),
+ activeSentinel,
+ applyTransform(fragment, options.transformFragment)
+ );
+
+ return;
+ }
+
+ // Single block — full replace, no sentinel.
+ activeBlockStartOffset = 0;
+ activeSentinel = null;
+
+ betterLinkDocumentMod(fragment, createDecorate(emptyDefinitions, externalLinkAlt));
+
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+
+ wrapper.replaceChildren(applyTransform(fragment, options.transformFragment));
+ }
+
+ return Object.freeze({
+ finalize(options: StreamingNextOptions): StreamingNextResult {
+ if (!previousMarkdown) {
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+
+ wrapper.replaceChildren();
+
+ return Object.freeze({ definitions: Object.freeze([]) });
+ }
+
+ let processedMarkdown = previousMarkdown;
+
+ if (markdownRespectCRLF) {
+ processedMarkdown = respectCRLFPre(processedMarkdown);
+ }
+
+ const fullEvents = parseEvents(processedMarkdown);
+ const rawHTML = compile(micromarkOptions)(fullEvents);
+ const parsedDocument = domParser.parseFromString(rawHTML, 'text/html');
+ const fragment = parsedDocument.createDocumentFragment();
+
+ fragment.append(...Array.from(parsedDocument.body.childNodes));
+
+ const definitions = extractDefinitionsFromEvents(fullEvents);
+
+ betterLinkDocumentMod(fragment, createDecorate(definitions, externalLinkAlt));
+
+ activeBlockStartOffset = 0;
+ activeSentinel = null;
+
+ // Full replace on finalize — no incremental path needed.
+ const wrapper = ensureWrapper(options.container, options.containerClassName);
+ const transformedFragment = applyTransform(fragment, options.transformFragment);
+
+ wrapper.replaceChildren(transformedFragment);
+
+ return Object.freeze({ definitions });
+ },
+
+ next(chunk: string, options: StreamingNextOptions): void {
+ renderNext(chunk, options);
+ },
+
+ reset(): void {
+ previousMarkdown = '';
+ activeBlockStartOffset = 0;
+ activeSentinel = null;
+ wrapperDiv = null;
+ }
+ });
+}
+
+export {
+ type MarkdownLinkDefinition,
+ type StreamingNextOptions,
+ type StreamingNextResult,
+ type StreamingRenderer,
+ type StreamingRenderOptions
+};
diff --git a/packages/bundle/src/markdown/private/createDecorate.ts b/packages/bundle/src/markdown/private/createDecorate.ts
new file mode 100644
index 0000000000..481565cf45
--- /dev/null
+++ b/packages/bundle/src/markdown/private/createDecorate.ts
@@ -0,0 +1,82 @@
+import { onErrorResumeNext } from 'botframework-webchat-core';
+import type { BetterLinkDocumentModDecoration } from './betterLinkDocumentMod';
+import type { MarkdownLinkDefinition } from './extractDefinitionsFromEvents';
+import { sanitizeUri } from 'micromark-util-sanitize-uri';
+
+export const ALLOWED_SCHEMES = ['data', 'http', 'https', 'ftp', 'mailto', 'sip', 'tel'];
+
+export function createDecorate(
+ definitions: readonly MarkdownLinkDefinition[],
+ externalLinkAlt: string
+): (href: string, textContent: string) => BetterLinkDocumentModDecoration {
+ const linkDefinitions = definitions.map(definition =>
+ Object.freeze({
+ ...definition,
+ markupUrl: sanitizeUri(definition.url),
+ parsedUrl: onErrorResumeNext(() => new URL(definition.url))
+ })
+ );
+
+ const allowedSchemas = ALLOWED_SCHEMES.map(scheme => `${scheme}:`);
+
+ return (href: string, textContent: string): BetterLinkDocumentModDecoration => {
+ const decoration: BetterLinkDocumentModDecoration = {
+ rel: 'noopener noreferrer',
+ target: '_blank',
+ wrapZeroWidthSpace: true
+ };
+
+ const ariaLabelSegments: string[] = [textContent];
+ const classes: Set = new Set();
+ const linkDefinition = linkDefinitions.find(({ url, markupUrl }) => url === href || (href && markupUrl === href));
+ const protocol = onErrorResumeNext(() => new URL(href).protocol);
+
+ if (linkDefinition) {
+ ariaLabelSegments.push(linkDefinition.title || linkDefinition?.parsedUrl?.host || linkDefinition.url);
+
+ // linkDefinition.identifier is uppercase, while linkDefinition.label is as-is.
+ linkDefinition.label === textContent && classes.add('render-markdown__pure-identifier');
+ }
+
+ // Let javascript: fell through. Our sanitizer will catch and remove it from .
+ // Otherwise, it will be turn into