MAT9004

0）你先要搞清楚的“硬规则”（不清楚就会白丢分）

• Final 占 60%，而且是 hurdle：期末卷面 <45% 就直接挂科，不管你平时作业多高分[2]Source: asksia-bible-mat9004-bilingual.pdfMAT9004 . MONASH UNIVERSITY 中英双语版 · BILINGUAL EDITION 英文主讲,中文随行 一 考试要点与术语保留英文原词 The final is 60% of your mark and a hurdle - score under 45% on it and you fail the unit, whatever your assignment marks. It is closed-book with no calculator: a formula sheet is supplied, so the exam tests one thing - can you do the method by hand on fresh numbers. This book drills exactly that, across all six examined topics. Independent study companion. Not affiliated with or endorsed by Monash University. Corrections: takedowns@asksia. ai PREFACE - HOW TO USE THIS BOOK By hand, on fresh numbers 手算,用全新的数字 Closed-book, no calculator - the exam tests method, not memory 闭卷、不许用计算器 -- 考试考的是方法,不是记忆 This is not a transcript of the lecture slides. It is a self-contained course in every technique MAT9004 examines - each definition stated plainly, each method shown on a worked example with real arithmetic, each common slip flagged. The final is closed-book with no calculator and a formula sheet is provided, so the examiner cannot test what you remember - only whether you can execute the method by hand under time. That is exactly what these pages drill. 这不是讲义幻灯片的逐字转录。它是一门自成体系的课程,覆盖 MAT9004 所考查的每一种技巧 -- 每个定义都直白陈述, 每种方法都配以带真实运算的例题(worked example)演示,每个常见失误都被标出。期末是闭卷、不可使用计算器且提供公 式表(formula sheet),所以考官无法考你记住了什么 -- 只能考你能否在限时内手动执行方法。而这正是这些篇幅所要反复 操练的。 A 1 . LEARN 1·学习 You haven't seen the lecture yet. Read a chapter top to bottom. Every topic opens with a one-line TL;DR, then define - picture - formula - worked example - trap. The diagrams are original schematics of the standard maths - learn the idea cold. 你还没上过讲课。从头到尾读一 章。每个主题以一行 TL;DR 开 篇,然后是定义→图示→公式 →例题→陷阱。图示是标准数 学的原创示意图 -- 把思想学 到冷启动也能跑。 B 2 . DRILL 2 · 刷题 You've done the lecture and applied class. Cover the worked steps and re-derive each answer by hand - no calculator, just like the exam. Speed and accuracy on fresh numbers is the whole game. 你已上过讲课和应用课。盖住做 过的步骤,亲手重新推导每个答 案 -- 不用计算器,正如考 试。在全新数字上的速度与准确 度才是全部所在。 C 3 · EXAM 3 · 考试 It's SWOTVAC. The TL;DR strips, the recap boxes and the practice bank are your map. The blueprint overleaf shows the 60% hurdle, what the formula sheet gives you, and the recurring question template. 正值 SWOTVAC (考前复习 周)。TL;DR 摘要条、回顾框与 练习题库是你的地图。背面的蓝 图展示了 60% 的及格门槛、公 式表会给你什么,以及反复出现 的题型模板。 AskSia Library · MAT9004 · 双语 Bilingual ! The single most important thing to understand about MAT9004[4]Source: asksia-bible-mat9004-bilingual.pdf关于 MAT9004 最需要理解的一件事 This is six disjoint maths worlds in one paper - single-variable calculus, linear algebra, multivariable optimisation, counting/combinatorics, probability & Bayes, and graphs/trees. They don't build on each other the way a normal maths unit does, so you can't coast on one strength. The exam samples all six, and there is a 45% hurdle on the exam itself: you must clear it to pass, regardless of assignment marks. Spread your revision; don't leave a whole world blank. 这是一张试卷里的六个互不相交的数学世界 -- 单变量微积分、线性代数、多变量优化、计数/组合学、概率与贝叶 斯,以及图/树。它们不像普通数学单元那样层层递进,所以你无法靠单项强项蒙混过关。考试对全部六个取样,且考试 本身设有 45% 及格门槛:无论作业分数如何,你都必须越过它才能通过。摊开你的复习;别把整个世界留作空白。 i How this book was built - and the two-layer rule 这本书是怎样建成的 -- 以及两层规则 Standard mathematical definitions, theorems and formulas are stated plainly (they are universal). The unit's own framing and the lecturer's specific example numbers are paraphrased and re-numbered, never copied from slides or past papers. The course runs on the official MAT9004 lecture notes; no textbook is prescribed. Every arithmetic step here has been checked - but verify dates and weights against your own Moodle (learning. monash. edu), as details can shift between cohorts. 标准的数学定义、定理与公式都直白陈述(它们是普适的)。本单元自身的表述框架与讲师特定的示例数字均经改写并重 新编号,绝不照抄幻灯片或往届试卷。本课程依据官方 MAT9004 讲义运行;不指定教科书。这里的每一步算术都已核 对 -- 但请对照你自己的 Moodle (learning. monash. edu)核实日期与权重,因为细节会随届别变动。 - AskSia Library · MAT9004 · 双语 Bilingual THE BLUEPRINT THE EXAM BLUEPRINT FINAL 60% . HURDLE 60% final, and it's a hurdle 期末占 60%,而且是及格门槛(hurdle) Assignments 20% . Quizzes 20% . Final 60% (must score ≥ 45%) 作业 20% · 测验 20% · 期末 60%(须≥45%) Your mark is built from three pieces, but one dominates and gates the others. The final exam is 60% and a hurdle - under 45% on it and the unit is failed, whatever else you scored. 你的成绩由三部分构成,但其中一部分占主导并卡住其余部分。期末考试(exam)占 60% 且设有及格门槛(hurdle) -- 若 该项低于 45%,无论其余分数如何,本单元即不及格。 60% FINAL EXAM (HURDLE) 期末考试(及格门槛 hurdle) 45% MIN TO CLEAR HURDLE 通过门槛的最低分 20% 5 QUIZZES (4% EACH) 5 次测验(每次4%) 20% 2 ASSIGNMENTS 2 次作业 The three assessment pieces 三项评估构成 Component Weight When / detail Final examination - all six 60% Exam period · ≥45% to pass topics, HURDLE Applied-class quizzes (5 x 30 min) 20%[11]Source: asksia-cheatsheet-mat9004.pdfDETAIL Weight 60% · hurdle ≥45% Coursework 2 assign×10% + 5 quiz×4% Duration ~3 h 10 min . e-exam Materials closed-book . NO calc Formula sheet Template provided in paper -36 questions, fixed slots ~36-slot pattern: ~ 31 short-answer (1. 5-3 marks, answer is an integer or lowest-terms a/b, no spaces, no decimals) + 5 long-answer (6 marks: global extrema · Hessian classify · eigen/diagonalise · free-variable system · two-stage Bayes). This is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const)。

• 闭卷、无计算器、无笔记、无书、不可用 AI；试卷内自带公式表（page 2）[1]Source: asksia-bible-mat9004-bilingual.pdfWeeks 4, 6, 8, 10, 12 Assignment 1 / Assignment 2 20% Wk 7 & Wk 11 . 10% each ★ The exam format - closed-book, by hand 考试形式 -- 闭卷、手算 An e-exam, ~3 hours 10 min: no calculator, no notes, no books, no AI. A formula sheet is provided inside the paper. Two answer styles - short-answer (type the exact value; fractions in lowest terms, no decimals unless asked) and hand-written response (work shown, scanned and uploaded). Past papers follow a stable ~36-slot template: the same skills reappear with fresh numbers each year. 一场电子考试,约3小时10分:不可用计算器、不 可带笔记、不可带书、不可用 AI。试卷内提供一张公 式表。两种作答方式 -- 简答(键入精确值;分数取 最简,除非要求否则不用小数)与手写作答(写出过 程,扫描上传)。往届试卷遵循一个稳定的~36 槽位模 板:相同的技能每年以全新数字重现。 What the formula sheet means for you 公式表(formula sheet)对你意味着什么 On the sheet (don't cram) You must execute cold Standard derivative / integral table All differentiation rules, by hand Binomial / counting formulas Gaussian elimination & back- substitution Probability identities, Bayes 2×2 inverse, determinant, eigen-solve Distribution means & variances Gradient + Hessian classification AskSia Library . MAT9004 . XXia Bilingual ✓ The strategy this dictates 由此决定的策略 Every question is procedural: take a function, matrix, count or probability, apply the right technique, give the exact value. The recurring chains are - differentiate - set f'=0 - classify; row-reduce - back-substitute; Vf=O - Hessian test; condition - Bayes; degree sequence - edges. Show every line - method marks are real. Drill the chains and fresh numbers can't surprise you. 每道题都是程序性的:取一个函数、矩阵、计数或概 率,应用正确技巧,给出精确值。反复出现的链条是 -- 求导→令f'=0→分类;行化简→回代;▽f=0 → 海森检验;条件化→贝叶斯;度序列→边数。写 出每一行 -- 步骤分是实打实的。把这些链条练熟, 全新数字就无法让你意外。 AskSia Library · MAT9004 · 双语 Bilingual CONTENTS - CONTENTS Six worlds, one exam-ready book 六大板块,一本应考即用的书 Calculus & algebra first (the largest block), discrete maths & probability second 微积分与代数先行(最大板块),离散数学与概率其次 Ch Topic Core methods Part 1 . Calculus & optimisation (the exam's largest block)[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or。
• 考试大约 3h10m，并且有 两种答题形态：
  • 短答：输入一个精确值（整数或最简分数 $a/b$，无空格、除非要求否则不用小数）[1]Source: asksia-bible-mat9004-bilingual.pdfWeeks 4, 6, 8, 10, 12 Assignment 1 / Assignment 2 20% Wk 7 & Wk 11 . 10% each ★ The exam format - closed-book, by hand 考试形式 -- 闭卷、手算 An e-exam, ~3 hours 10 min: no calculator, no notes, no books, no AI. A formula sheet is provided inside the paper. Two answer styles - short-answer (type the exact value; fractions in lowest terms, no decimals unless asked) and hand-written response (work shown, scanned and uploaded). Past papers follow a stable ~36-slot template: the same skills reappear with fresh numbers each year. 一场电子考试,约3小时10分:不可用计算器、不 可带笔记、不可带书、不可用 AI。试卷内提供一张公 式表。两种作答方式 -- 简答(键入精确值;分数取 最简,除非要求否则不用小数)与手写作答(写出过 程,扫描上传)。往届试卷遵循一个稳定的~36 槽位模 板:相同的技能每年以全新数字重现。 What the formula sheet means for you 公式表(formula sheet)对你意味着什么 On the sheet (don't cram) You must execute cold Standard derivative / integral table All differentiation rules, by hand Binomial / counting formulas Gaussian elimination & back- substitution Probability identities, Bayes 2×2 inverse, determinant, eigen-solve Distribution means & variances Gradient + Hessian classification AskSia Library . MAT9004 . XXia Bilingual ✓ The strategy this dictates 由此决定的策略 Every question is procedural: take a function, matrix, count or probability, apply the right technique, give the exact value. The recurring chains are - differentiate - set f'=0 - classify; row-reduce - back-substitute; Vf=O - Hessian test; condition - Bayes; degree sequence - edges. Show every line - method marks are real. Drill the chains and fresh numbers can't surprise you. 每道题都是程序性的:取一个函数、矩阵、计数或概 率,应用正确技巧,给出精确值。反复出现的链条是 -- 求导→令f'=0→分类;行化简→回代;▽f=0 → 海森检验;条件化→贝叶斯;度序列→边数。写 出每一行 -- 步骤分是实打实的。把这些链条练熟, 全新数字就无法让你意外。 AskSia Library · MAT9004 · 双语 Bilingual CONTENTS - CONTENTS Six worlds, one exam-ready book 六大板块,一本应考即用的书 Calculus & algebra first (the largest block), discrete maths & probability second 微积分与代数先行(最大板块),离散数学与概率其次 Ch Topic Core methods Part 1 . Calculus & optimisation (the exam's largest block)[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE
  • 手写题：写过程，考后 30 分钟内手机扫描上传；没上传=0 分[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE

• 往年卷子是稳定的“约 36 个槽位”：约 31 个短技能题 + 5 个 6 分大题套路（每年换数字但技能重复）[1]Source: asksia-bible-mat9004-bilingual.pdfWeeks 4, 6, 8, 10, 12 Assignment 1 / Assignment 2 20% Wk 7 & Wk 11 . 10% each ★ The exam format - closed-book, by hand 考试形式 -- 闭卷、手算 An e-exam, ~3 hours 10 min: no calculator, no notes, no books, no AI. A formula sheet is provided inside the paper. Two answer styles - short-answer (type the exact value; fractions in lowest terms, no decimals unless asked) and hand-written response (work shown, scanned and uploaded). Past papers follow a stable ~36-slot template: the same skills reappear with fresh numbers each year. 一场电子考试,约3小时10分:不可用计算器、不 可带笔记、不可带书、不可用 AI。试卷内提供一张公 式表。两种作答方式 -- 简答(键入精确值;分数取 最简,除非要求否则不用小数)与手写作答(写出过 程,扫描上传)。往届试卷遵循一个稳定的~36 槽位模 板:相同的技能每年以全新数字重现。 What the formula sheet means for you 公式表(formula sheet)对你意味着什么 On the sheet (don't cram) You must execute cold Standard derivative / integral table All differentiation rules, by hand Binomial / counting formulas Gaussian elimination & back- substitution Probability identities, Bayes 2×2 inverse, determinant, eigen-solve Distribution means & variances Gradient + Hessian classification AskSia Library . MAT9004 . XXia Bilingual ✓ The strategy this dictates 由此决定的策略 Every question is procedural: take a function, matrix, count or probability, apply the right technique, give the exact value. The recurring chains are - differentiate - set f'=0 - classify; row-reduce - back-substitute; Vf=O - Hessian test; condition - Bayes; degree sequence - edges. Show every line - method marks are real. Drill the chains and fresh numbers can't surprise you. 每道题都是程序性的:取一个函数、矩阵、计数或概 率,应用正确技巧,给出精确值。反复出现的链条是 -- 求导→令f'=0→分类;行化简→回代;▽f=0 → 海森检验;条件化→贝叶斯;度序列→边数。写 出每一行 -- 步骤分是实打实的。把这些链条练熟, 全新数字就无法让你意外。 AskSia Library · MAT9004 · 双语 Bilingual CONTENTS - CONTENTS Six worlds, one exam-ready book 六大板块,一本应考即用的书 Calculus & algebra first (the largest block), discrete maths & probability second 微积分与代数先行(最大板块),离散数学与概率其次 Ch Topic Core methods Part 1 . Calculus & optimisation (the exam's largest block)[11]Source: asksia-cheatsheet-mat9004.pdfDETAIL Weight 60% · hurdle ≥45% Coursework 2 assign×10% + 5 quiz×4% Duration ~3 h 10 min . e-exam Materials closed-book . NO calc Formula sheet Template provided in paper -36 questions, fixed slots ~36-slot pattern: ~ 31 short-answer (1. 5-3 marks, answer is an integer or lowest-terms a/b, no spaces, no decimals) + 5 long-answer (6 marks: global extrema · Hessian classify · eigen/diagonalise · free-variable system · two-stage Bayes). This is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const)。

1）整门课的结构：六个“互不相交的世界”（你要平均铺开）

• (1) 单变量微积分（极限/求导/最值/积分）
• (2) 线性代数（矩阵、消元、行列式、逆、特征值/对角化）
• (3) 多变量微积分与优化（偏导、梯度、Hessian、分类）
• (4) 计数与组合
• (5) 概率与贝叶斯（条件概率、全概率、Bayes、期望方差、分布、连续型 pdf）
• (6) 图与树（度数、握手引理、邻接矩阵、Euler/Hamilton）[4]Source: asksia-bible-mat9004-bilingual.pdf关于 MAT9004 最需要理解的一件事 This is six disjoint maths worlds in one paper - single-variable calculus, linear algebra, multivariable optimisation, counting/combinatorics, probability & Bayes, and graphs/trees. They don't build on each other the way a normal maths unit does, so you can't coast on one strength. The exam samples all six, and there is a 45% hurdle on the exam itself: you must clear it to pass, regardless of assignment marks. Spread your revision; don't leave a whole world blank. 这是一张试卷里的六个互不相交的数学世界 -- 单变量微积分、线性代数、多变量优化、计数/组合学、概率与贝叶 斯,以及图/树。它们不像普通数学单元那样层层递进,所以你无法靠单项强项蒙混过关。考试对全部六个取样,且考试 本身设有 45% 及格门槛:无论作业分数如何,你都必须越过它才能通过。摊开你的复习;别把整个世界留作空白。 i How this book was built - and the two-layer rule 这本书是怎样建成的 -- 以及两层规则 Standard mathematical definitions, theorems and formulas are stated plainly (they are universal). The unit's own framing and the lecturer's specific example numbers are paraphrased and re-numbered, never copied from slides or past papers. The course runs on the official MAT9004 lecture notes; no textbook is prescribed. Every arithmetic step here has been checked - but verify dates and weights against your own Moodle (learning. monash. edu), as details can shift between cohorts. 标准的数学定义、定理与公式都直白陈述(它们是普适的)。本单元自身的表述框架与讲师特定的示例数字均经改写并重 新编号,绝不照抄幻灯片或往届试卷。本课程依据官方 MAT9004 讲义运行;不指定教科书。这里的每一步算术都已核 对 -- 但请对照你自己的 Moodle (learning. monash. edu)核实日期与权重,因为细节会随届别变动。 - AskSia Library · MAT9004 · 双语 Bilingual THE BLUEPRINT THE EXAM BLUEPRINT FINAL 60% . HURDLE 60% final, and it's a hurdle 期末占 60%,而且是及格门槛(hurdle) Assignments 20% . Quizzes 20% . Final 60% (must score ≥ 45%) 作业 20% · 测验 20% · 期末 60%(须≥45%) Your mark is built from three pieces, but one dominates and gates the others. The final exam is 60% and a hurdle - under 45% on it and the unit is failed, whatever else you scored. 你的成绩由三部分构成,但其中一部分占主导并卡住其余部分。期末考试(exam)占 60% 且设有及格门槛(hurdle) -- 若 该项低于 45%,无论其余分数如何,本单元即不及格。 60% FINAL EXAM (HURDLE) 期末考试(及格门槛 hurdle) 45% MIN TO CLEAR HURDLE 通过门槛的最低分 20% 5 QUIZZES (4% EACH) 5 次测验(每次4%) 20% 2 ASSIGNMENTS 2 次作业 The three assessment pieces 三项评估构成 Component Weight When / detail Final examination - all six 60% Exam period · ≥45% to pass topics, HURDLE Applied-class quizzes (5 x 30 min) 20%[8]Source: asksia-bible-mat9004-bilingual.pdf1 Single-variable calculus limits . derivative rules . optimisation . integration → 2 Linear algebra vectors . matrices . Gaussian elim . determinants · eigen → 3 Multivariable calculus partials · gradient . Hessian . 2-var optimisation → Part 2 . Discrete maths & probability 4 Counting & combinatorics product/sum rules . permutations . binomial . incl-excl → 5 Probability & Bayes events . conditional . Bayes . random variables . distributions → 6 Graphs & trees degrees . trees . adjacency matrices . Euler/Hamilton → Part 3 . Walk in ready 7 Glossary & formula map every term EN++ . what's on the sheet → 8 Practice bank & worked solutions the 36-slot exam template, re-numbered → 9 Exam morning the by-hand checklist . timing . traps → i Why this order 为什么是这个顺序 MAT9004 teaches calculus and linear algebra first (Weeks 2-8), then counting, probability and graphs (Weeks 9-12). We keep that order because the calculus/algebra block is the largest share of exam marks and the techniques (derivatives, row-reduction) recur inside the later topics. Part 3 turns the lot into marks: glossary, the re-numbered question template, and a one-page exam-morning drill. MAT9004 先教微积分与线性代数(第2-8周),再教计数、概率与图(第9-12周)。我们保留这个顺序,因为微积 分/代数模块占考试分数的最大份额,且其技巧(导数、行化简)在后面的主题中反复出现。第3部分把这一切转化为分 数:词汇表、重新编号的题型模板,以及一页考试当天的速练。 AskSia Library · MAT9004 · 双语 Bilingual CH 1 . CALCULUS - CHAPTER 1 . SINGLE - VARIABLE CALCULUS

2）最高频“得分链条”（你要练到闭眼都能写）

3）Area 1：单变量微积分（导数/最优化/积分）——最大块之一

3.1 必背定义 & 公式（但重点是会用）

导数定义（切线斜率）

• $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$
• 典型提醒：$|x|$ 在 $0$ 处不可导（会作为“奇异点/不可导点”出现）[18]Source: asksia-cheatsheet-mat9004.pdfThis is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const) xᵇ b. x^(b-1) aª In(a)·a* ex ex ln x 1/x loga X 1/(ln(a)·x) COMBINING RULES (c. f) '=c. f' . (fig) ' =f'±g' product: (fg) ' = f'g + fg' chain: (f(g(x) ) ) ' = g' (x) . f' (g(x)) COMMON CHAINS (e^(cx+d))' = c. e^(cx+d) (Ln(cx+d))' = c/(cx+d) (a^(cx+d))' = c. Ln(a) . a^(cx+d) nth derivative f(n) = differentiate n times (f'=f(2). Worked tangent slope: for h(x)=x. e^(2x), h'=e^(2x) (1+2x) => h'(0)=1 (product rule, then evaluate). 1b . Function Types L1-3 Convex: chord lies on/above plot e f"≥0. Concave: chord below + f"≤0. Lines are both. Transform-plot trick: log-log linear => power law (slope -a); log-lin linear = exponential (slope In a); lin- log = logarithmic.

常见导数表（你要会手算）（试卷给表，但你要会用规则）[1]Source: asksia-bible-mat9004-bilingual.pdfWeeks 4, 6, 8, 10, 12 Assignment 1 / Assignment 2 20% Wk 7 & Wk 11 . 10% each ★ The exam format - closed-book, by hand 考试形式 -- 闭卷、手算 An e-exam, ~3 hours 10 min: no calculator, no notes, no books, no AI. A formula sheet is provided inside the paper. Two answer styles - short-answer (type the exact value; fractions in lowest terms, no decimals unless asked) and hand-written response (work shown, scanned and uploaded). Past papers follow a stable ~36-slot template: the same skills reappear with fresh numbers each year. 一场电子考试,约3小时10分:不可用计算器、不 可带笔记、不可带书、不可用 AI。试卷内提供一张公 式表。两种作答方式 -- 简答(键入精确值;分数取 最简,除非要求否则不用小数)与手写作答(写出过 程,扫描上传)。往届试卷遵循一个稳定的~36 槽位模 板:相同的技能每年以全新数字重现。 What the formula sheet means for you 公式表(formula sheet)对你意味着什么 On the sheet (don't cram) You must execute cold Standard derivative / integral table All differentiation rules, by hand Binomial / counting formulas Gaussian elimination & back- substitution Probability identities, Bayes 2×2 inverse, determinant, eigen-solve Distribution means & variances Gradient + Hessian classification AskSia Library . MAT9004 . XXia Bilingual ✓ The strategy this dictates 由此决定的策略 Every question is procedural: take a function, matrix, count or probability, apply the right technique, give the exact value. The recurring chains are - differentiate - set f'=0 - classify; row-reduce - back-substitute; Vf=O - Hessian test; condition - Bayes; degree sequence - edges. Show every line - method marks are real. Drill the chains and fresh numbers can't surprise you. 每道题都是程序性的:取一个函数、矩阵、计数或概 率,应用正确技巧,给出精确值。反复出现的链条是 -- 求导→令f'=0→分类;行化简→回代;▽f=0 → 海森检验;条件化→贝叶斯;度序列→边数。写 出每一行 -- 步骤分是实打实的。把这些链条练熟, 全新数字就无法让你意外。 AskSia Library · MAT9004 · 双语 Bilingual CONTENTS - CONTENTS Six worlds, one exam-ready book 六大板块,一本应考即用的书 Calculus & algebra first (the largest block), discrete maths & probability second 微积分与代数先行(最大板块),离散数学与概率其次 Ch Topic Core methods Part 1 . Calculus & optimisation (the exam's largest block)[18]Source: asksia-cheatsheet-mat9004.pdfThis is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const) xᵇ b. x^(b-1) aª In(a)·a* ex ex ln x 1/x loga X 1/(ln(a)·x) COMBINING RULES (c. f) '=c. f' . (fig) ' =f'±g' product: (fg) ' = f'g + fg' chain: (f(g(x) ) ) ' = g' (x) . f' (g(x)) COMMON CHAINS (e^(cx+d))' = c. e^(cx+d) (Ln(cx+d))' = c/(cx+d) (a^(cx+d))' = c. Ln(a) . a^(cx+d) nth derivative f(n) = differentiate n times (f'=f(2). Worked tangent slope: for h(x)=x. e^(2x), h'=e^(2x) (1+2x) => h'(0)=1 (product rule, then evaluate). 1b . Function Types L1-3 Convex: chord lies on/above plot e f"≥0. Concave: chord below + f"≤0. Lines are both. Transform-plot trick: log-log linear => power law (slope -a); log-lin linear = exponential (slope In a); lin- log = logarithmic.

• $(x^b)'=b x^{b-1}$，$(e^x)'=e^x$，$(\ln x)'=1/x$
• $\big(\log_a x\big)'=\dfrac{1}{\ln(a),x}$

四大求导规则（手算必熟）[18]Source: asksia-cheatsheet-mat9004.pdfThis is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const) xᵇ b. x^(b-1) aª In(a)·a* ex ex ln x 1/x loga X 1/(ln(a)·x) COMBINING RULES (c. f) '=c. f' . (fig) ' =f'±g' product: (fg) ' = f'g + fg' chain: (f(g(x) ) ) ' = g' (x) . f' (g(x)) COMMON CHAINS (e^(cx+d))' = c. e^(cx+d) (Ln(cx+d))' = c/(cx+d) (a^(cx+d))' = c. Ln(a) . a^(cx+d) nth derivative f(n) = differentiate n times (f'=f(2). Worked tangent slope: for h(x)=x. e^(2x), h'=e^(2x) (1+2x) => h'(0)=1 (product rule, then evaluate). 1b . Function Types L1-3 Convex: chord lies on/above plot e f"≥0. Concave: chord below + f"≤0. Lines are both. Transform-plot trick: log-log linear => power law (slope -a); log-lin linear = exponential (slope In a); lin- log = logarithmic.

• 和差：$(f\pm g)'=f'\pm g'$
• 常数倍：$(cf)'=cf'$
• 乘积：$(fg)'=f'g+fg'$
• 链式：$(f(g(x)))'=g'(x),f'(g(x))$

积分与微积分基本定理（FTC）

• 不定积分：$\int f(x),dx=F(x)+C$（$F'=f$）[26]Source: asksia-cheatsheet-mat9004.pdf- RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x- ln|x|+c e^(ax) (1/a)e^(ax)+c FUNDAMENTAL THEOREM OF CALCULUS [ab f(x)dx = F(b) - F(a) G(x)=fax f(z)dz => G'=f Linearity / (f+g)=/f+[g, fcf=c/f. Additivity Sac=fab+]_bº (piecewise). Rate => total: if E'(x)=rate, total change = [ rate dx. SIA - A definite integral is signed area - area below the x-axis counts negative. Sketch first if signs are in doubt. 3b . Worked . Definite + RENUMBERED FTC Jo2(x3-6x2)dx = [x4/4 - 2x3],2 = 4 - 16 = - 12. Rate->total (battery): cost falls at rate (x+1)-2, start $5. D(4)=5+5. 4 -(x+1)-2dx=5+[(x+1)-1]04 =5 +(1/5- 1) = $4. 20 = 21/5. Antiderivative chosen so D(0)=5. Look-up integrals (used in probability): [xe-xdx = -e -* (x+1)+c; Je^(-x2/2)dx is not elementary (the normal's normaliser uses erf - hence z-tables). J 2(6x2+6x-4)dx-type slots are routine power-rule. 3c · Σ/I Notation SLOTS 1-2 A invertible - Ax=b has unique x = A-1b __ {x=a}^{b}f(x)=f(a)+ . . . +f(b); Il is the product. Empty sum = 0, empty product = 1. e. g. E_{k=1}^{4} k2 = 1+4+9+16 = 30; [1_{k=1}^{4} k = 4! = 24. N={0,1,2, . . . } here (0 is natural). "iff" = if and only if. 4 . Vectors AREA 2 . L8 Rd = column d-tuples; add & scale component-wise. DOT PRODUCT & NORM V. W = V1W1+ . . . +V_dw_d Ilvll = v(v. v) = V(v12+ . . . +v_d2) orthogonal . v. w = 0 Linear comb. w=a,V1+ . . . +anVn. Linearly dependent = one vị is a combo of the rest. Pairwise-orthogonal nonzero vectors are independent.
• 定积分：$$\int_a^b f(x),dx=F(b)-F(a)$$[26]Source: asksia-cheatsheet-mat9004.pdf- RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x- ln|x|+c e^(ax) (1/a)e^(ax)+c FUNDAMENTAL THEOREM OF CALCULUS [ab f(x)dx = F(b) - F(a) G(x)=fax f(z)dz => G'=f Linearity / (f+g)=/f+[g, fcf=c/f. Additivity Sac=fab+]_bº (piecewise). Rate => total: if E'(x)=rate, total change = [ rate dx. SIA - A definite integral is signed area - area below the x-axis counts negative. Sketch first if signs are in doubt. 3b . Worked . Definite + RENUMBERED FTC Jo2(x3-6x2)dx = [x4/4 - 2x3],2 = 4 - 16 = - 12. Rate->total (battery): cost falls at rate (x+1)-2, start $5. D(4)=5+5. 4 -(x+1)-2dx=5+[(x+1)-1]04 =5 +(1/5- 1) = $4. 20 = 21/5. Antiderivative chosen so D(0)=5. Look-up integrals (used in probability): [xe-xdx = -e -* (x+1)+c; Je^(-x2/2)dx is not elementary (the normal's normaliser uses erf - hence z-tables). J 2(6x2+6x-4)dx-type slots are routine power-rule. 3c · Σ/I Notation SLOTS 1-2 A invertible - Ax=b has unique x = A-1b __ {x=a}^{b}f(x)=f(a)+ . . . +f(b); Il is the product. Empty sum = 0, empty product = 1. e. g. E_{k=1}^{4} k2 = 1+4+9+16 = 30; [1_{k=1}^{4} k = 4! = 24. N={0,1,2, . . . } here (0 is natural). "iff" = if and only if. 4 . Vectors AREA 2 . L8 Rd = column d-tuples; add & scale component-wise. DOT PRODUCT & NORM V. W = V1W1+ . . . +V_dw_d Ilvll = v(v. v) = V(v12+ . . . +v_d2) orthogonal . v. w = 0 Linear comb. w=a,V1+ . . . +anVn. Linearly dependent = one vị is a combo of the rest. Pairwise-orthogonal nonzero vectors are independent.
• 定积分是“带符号面积”：$x$ 轴下方算负，必要时先画草图避免符号错[26]Source: asksia-cheatsheet-mat9004.pdf- RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x- ln|x|+c e^(ax) (1/a)e^(ax)+c FUNDAMENTAL THEOREM OF CALCULUS [ab f(x)dx = F(b) - F(a) G(x)=fax f(z)dz => G'=f Linearity / (f+g)=/f+[g, fcf=c/f. Additivity Sac=fab+]_bº (piecewise). Rate => total: if E'(x)=rate, total change = [ rate dx. SIA - A definite integral is signed area - area below the x-axis counts negative. Sketch first if signs are in doubt. 3b . Worked . Definite + RENUMBERED FTC Jo2(x3-6x2)dx = [x4/4 - 2x3],2 = 4 - 16 = - 12. Rate->total (battery): cost falls at rate (x+1)-2, start $5. D(4)=5+5. 4 -(x+1)-2dx=5+[(x+1)-1]04 =5 +(1/5- 1) = $4. 20 = 21/5. Antiderivative chosen so D(0)=5. Look-up integrals (used in probability): [xe-xdx = -e -* (x+1)+c; Je^(-x2/2)dx is not elementary (the normal's normaliser uses erf - hence z-tables). J 2(6x2+6x-4)dx-type slots are routine power-rule. 3c · Σ/I Notation SLOTS 1-2 A invertible - Ax=b has unique x = A-1b __ {x=a}^{b}f(x)=f(a)+ . . . +f(b); Il is the product. Empty sum = 0, empty product = 1. e. g. E_{k=1}^{4} k2 = 1+4+9+16 = 30; [1_{k=1}^{4} k = 4! = 24. N={0,1,2, . . . } here (0 is natural). "iff" = if and only if. 4 . Vectors AREA 2 . L8 Rd = column d-tuples; add & scale component-wise. DOT PRODUCT & NORM V. W = V1W1+ . . . +V_dw_d Ilvll = v(v. v) = V(v12+ . . . +v_d2) orthogonal . v. w = 0 Linear comb. w=a,V1+ . . . +anVn. Linearly dependent = one vị is a combo of the rest. Pairwise-orthogonal nonzero vectors are independent.[27]Source: asksia-cheatsheet-mat9004.pdfln|x|+c e^(ax) (1/a)e^(ax)+c FUNDAMENTAL THEOREM OF CALCULUS [ab f(x)dx = F(b) - F(a) G(x)=fax f(z)dz => G'=f Linearity / (f+g)=/f+[g, fcf=c/f. Additivity Sac=fab+]_bº (piecewise). Rate => total: if E'(x)=rate, total change = [ rate dx. SIA - A definite integral is signed area - area below the x-axis counts negative. Sketch first if signs are in doubt. 3b . Worked . Definite + RENUMBERED FTC Jo2(x3-6x2)dx = [x4/4 - 2x3],2 = 4 - 16 = - 12. Rate->total (battery): cost falls at rate (x+1)-2, start $5. D(4)=5+5. 4 -(x+1)-2dx=5+[(x+1)-1]04 =5 +(1/5- 1) = $4. 20 = 21/5. Antiderivative chosen so D(0)=5. Look-up integrals (used in probability): [xe-xdx = -e -* (x+1)+c; Je^(-x2/2)dx is not elementary (the normal's normaliser uses erf - hence z-tables). J 2(6x2+6x-4)dx-type slots are routine power-rule. 3c · Σ/I Notation SLOTS 1-2 A invertible - Ax=b has unique x = A-1b __ {x=a}^{b}f(x)=f(a)+ . . . +f(b); Il is the product. Empty sum = 0, empty product = 1. e. g. E_{k=1}^{4} k2 = 1+4+9+16 = 30; [1_{k=1}^{4} k = 4! = 24. N={0,1,2, . . . } here (0 is natural). "iff" = if and only if. 4 . Vectors AREA 2 . L8 Rd = column d-tuples; add & scale component-wise. DOT PRODUCT & NORM V. W = V1W1+ . . . +V_dw_d Ilvll = v(v. v) = V(v12+ . . . +v_d2) orthogonal . v. w = 0 Linear comb. w=a,V1+ . . . +anVn. Linearly dependent = one vị is a combo of the rest. Pairwise-orthogonal nonzero vectors are independent. Worked norm: |(3,12,-4)|=(9+144+16)=/169=13; Il(-8,9,-12)|=/289=17. Orthogonal solve: (2,-1,z)-(3,4,1)=0 = 6-4+z=0=>z =- 2. Line interval joining u,v = {au+(1-a)v : a€[0,1]}. Geometrically a vector = displacement (direction + length, no fixed position). 4b . Inverse Functions L2 g=f-1g(f(x))=x and f(g(y))=y. f-1(x)# 1/f(x). To find f-1: solve y=f(x) for x. f has an inverse iff bijective (injective + surjective). Worked: f(x)=x2+4x on [0,00), find f-1(32): x2+4x=32 => x2+4x-32=0 => (x+8)(x-4)=0 => x=4 (take the non- negative root). 4c . Constrained Product SUBSTITUTION Max xy s. t. 2x+5y=100: y=(100-2x)/5 => maximise (1/5) (100x-2x2); deriv 100-4x=0 => x=25, y=10 = xy=250. Substitute the constraint, reduce to one variable, then optimise - also a Lagrange-style multivariable framing. 5 . Matrices & AREA 2 . L8-10 ★ Systems

3.2 期末最爱考的“最优化工作流”（global extrema on $[c,d]$）

题型触发词：给 $f(x)$，在闭区间 $[c,d]$ 求 global max/min（经常是 6 分大题之一）[23]Source: asksia-cheatsheet-mat9004.pdfstat pts + endpoints "max/min/saddle f(x,y)" Vf=0 then det H "solve the system" Gaussian elim "not invertible" set det = 0 "A=PDP-1 / A"" det(A-MI)=0 "how many ways" order? repeat? = cell "committee / choose" C(n,r) "given . . . find Pr(-|-)" total prob + Bayes "E[-] / Var(-) from pdf" normalise, [ xf, fx2f "# edges from degrees" handshaking /2 "walks of length k" (A); "can graph exist?"[29]Source: asksia-cheatsheet-mat9004.pdfGLOBAL EXTREMA ON [C, D] candidates = stationary pts + singular pts (f' undef) + endpoints c, d. Evaluate f at ALL - Largest = max, smallest = min. Shortcut: local min of a convex f is the global min ; local max of concave = global max. Quadratic roots: x2+ax+b=0 => x =- a/2±/(a2/4-b); real iff a2≥4b. (x-u)(x-v)=x2-(u+v)x+uv. SIA - The long-answer "global extrema on [a,b]" question forgets the endpoints at your peril - they are candidates too. Always tabulate f at every candidate. 2b . Worked . Extrema RENUMBERED f'(t)=t3-5t2+6t=t(t-2)(t-3) on [-1,3]. Stationary t=0,2,3. f''=3t2-10t+6: f''(0)=6>0 (min), f''(2) =- 2<0 (max), f''(3)=3>0 (min). Compare fat {-1,0,2,3} = pick global max/min by value. Cubic variant: f(x)=2x3-3x2-12x+4 on [-3,3] => stationary x =- 1,2; global max (-1,11), global min at the endpoint (-3,-41). The endpoint wins - count it. 2c . Worked . Can min surface Volume Ttr2h = 128Tt = h=128/r2. Surface f(r)=2Tt(r2+128/r); f'=2Tt(2r-128/r2)=0 = r3=64 = r=4. f"'>0 (convex) = global min; h=128/16=8. Page-layout variant: printable A=(x-4)(y-6) with xy=294=A=318-6x-1176/x; A' =- 6+1176/x2=0=>x=14, A"<0 (concave, so a max), y=21. Convexity-on-interval trap: for f"=6ax-12 to be neither convex nor concave on (2,3), require f" to change sign there => solve for the parameter range, don't just plug one point. Constrained product, variants: max xy s. t. 2x+3y=60 =150; s. t. x+3y=60 = 300. Same substitute-then- optimise recipe; check it's a max (f''<0), not a min. Singular points (f' undefined, e. g. a corner like |x| at 0) are candidates too - don't only solve f'=0. The Extreme Value Theorem guarantees a continuous f on [c,d] attains both extrema. 2d . RSS / least squares APPLICATION RESIDUAL SUM OF SQUARES - RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x-

配方（你必须完整写出来）：

• 候选点 =
  • 驻点：$f'(x)=0$
  • 奇异点：$f'$ 不存在
  • 端点：$c,d$（超级大陷阱：很多人忘端点）[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。[29]Source: asksia-cheatsheet-mat9004.pdfGLOBAL EXTREMA ON [C, D] candidates = stationary pts + singular pts (f' undef) + endpoints c, d. Evaluate f at ALL - Largest = max, smallest = min. Shortcut: local min of a convex f is the global min ; local max of concave = global max. Quadratic roots: x2+ax+b=0 => x =- a/2±/(a2/4-b); real iff a2≥4b. (x-u)(x-v)=x2-(u+v)x+uv. SIA - The long-answer "global extrema on [a,b]" question forgets the endpoints at your peril - they are candidates too. Always tabulate f at every candidate. 2b . Worked . Extrema RENUMBERED f'(t)=t3-5t2+6t=t(t-2)(t-3) on [-1,3]. Stationary t=0,2,3. f''=3t2-10t+6: f''(0)=6>0 (min), f''(2) =- 2<0 (max), f''(3)=3>0 (min). Compare fat {-1,0,2,3} = pick global max/min by value. Cubic variant: f(x)=2x3-3x2-12x+4 on [-3,3] => stationary x =- 1,2; global max (-1,11), global min at the endpoint (-3,-41). The endpoint wins - count it. 2c . Worked . Can min surface Volume Ttr2h = 128Tt = h=128/r2. Surface f(r)=2Tt(r2+128/r); f'=2Tt(2r-128/r2)=0 = r3=64 = r=4. f"'>0 (convex) = global min; h=128/16=8. Page-layout variant: printable A=(x-4)(y-6) with xy=294=A=318-6x-1176/x; A' =- 6+1176/x2=0=>x=14, A"<0 (concave, so a max), y=21. Convexity-on-interval trap: for f"=6ax-12 to be neither convex nor concave on (2,3), require f" to change sign there => solve for the parameter range, don't just plug one point. Constrained product, variants: max xy s. t. 2x+3y=60 =150; s. t. x+3y=60 = 300. Same substitute-then- optimise recipe; check it's a max (f''<0), not a min. Singular points (f' undefined, e. g. a corner like |x| at 0) are candidates too - don't only solve f'=0. The Extreme Value Theorem guarantees a continuous f on [c,d] attains both extrema. 2d . RSS / least squares APPLICATION RESIDUAL SUM OF SQUARES - RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x-
• 把 $f$ 在所有候选点的函数值都算出来，比较大小，最大就是 global max，最小就是 global min[29]Source: asksia-cheatsheet-mat9004.pdfGLOBAL EXTREMA ON [C, D] candidates = stationary pts + singular pts (f' undef) + endpoints c, d. Evaluate f at ALL - Largest = max, smallest = min. Shortcut: local min of a convex f is the global min ; local max of concave = global max. Quadratic roots: x2+ax+b=0 => x =- a/2±/(a2/4-b); real iff a2≥4b. (x-u)(x-v)=x2-(u+v)x+uv. SIA - The long-answer "global extrema on [a,b]" question forgets the endpoints at your peril - they are candidates too. Always tabulate f at every candidate. 2b . Worked . Extrema RENUMBERED f'(t)=t3-5t2+6t=t(t-2)(t-3) on [-1,3]. Stationary t=0,2,3. f''=3t2-10t+6: f''(0)=6>0 (min), f''(2) =- 2<0 (max), f''(3)=3>0 (min). Compare fat {-1,0,2,3} = pick global max/min by value. Cubic variant: f(x)=2x3-3x2-12x+4 on [-3,3] => stationary x =- 1,2; global max (-1,11), global min at the endpoint (-3,-41). The endpoint wins - count it. 2c . Worked . Can min surface Volume Ttr2h = 128Tt = h=128/r2. Surface f(r)=2Tt(r2+128/r); f'=2Tt(2r-128/r2)=0 = r3=64 = r=4. f"'>0 (convex) = global min; h=128/16=8. Page-layout variant: printable A=(x-4)(y-6) with xy=294=A=318-6x-1176/x; A' =- 6+1176/x2=0=>x=14, A"<0 (concave, so a max), y=21. Convexity-on-interval trap: for f"=6ax-12 to be neither convex nor concave on (2,3), require f" to change sign there => solve for the parameter range, don't just plug one point. Constrained product, variants: max xy s. t. 2x+3y=60 =150; s. t. x+3y=60 = 300. Same substitute-then- optimise recipe; check it's a max (f''<0), not a min. Singular points (f' undefined, e. g. a corner like |x| at 0) are candidates too - don't only solve f'=0. The Extreme Value Theorem guarantees a continuous f on [c,d] attains both extrema. 2d . RSS / least squares APPLICATION RESIDUAL SUM OF SQUARES - RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x-

二阶导判别（局部）：

• $f''>0$ 局部最小；$f''<0$ 局部最大；$f''=0$ 不充分，需要别的方法（符号表等）[1]Source: asksia-bible-mat9004-bilingual.pdfWeeks 4, 6, 8, 10, 12 Assignment 1 / Assignment 2 20% Wk 7 & Wk 11 . 10% each ★ The exam format - closed-book, by hand 考试形式 -- 闭卷、手算 An e-exam, ~3 hours 10 min: no calculator, no notes, no books, no AI. A formula sheet is provided inside the paper. Two answer styles - short-answer (type the exact value; fractions in lowest terms, no decimals unless asked) and hand-written response (work shown, scanned and uploaded). Past papers follow a stable ~36-slot template: the same skills reappear with fresh numbers each year. 一场电子考试,约3小时10分:不可用计算器、不 可带笔记、不可带书、不可用 AI。试卷内提供一张公 式表。两种作答方式 -- 简答(键入精确值;分数取 最简,除非要求否则不用小数)与手写作答(写出过 程,扫描上传)。往届试卷遵循一个稳定的~36 槽位模 板:相同的技能每年以全新数字重现。 What the formula sheet means for you 公式表(formula sheet)对你意味着什么 On the sheet (don't cram) You must execute cold Standard derivative / integral table All differentiation rules, by hand Binomial / counting formulas Gaussian elimination & back- substitution Probability identities, Bayes 2×2 inverse, determinant, eigen-solve Distribution means & variances Gradient + Hessian classification AskSia Library . MAT9004 . XXia Bilingual ✓ The strategy this dictates 由此决定的策略 Every question is procedural: take a function, matrix, count or probability, apply the right technique, give the exact value. The recurring chains are - differentiate - set f'=0 - classify; row-reduce - back-substitute; Vf=O - Hessian test; condition - Bayes; degree sequence - edges. Show every line - method marks are real. Drill the chains and fresh numbers can't surprise you. 每道题都是程序性的:取一个函数、矩阵、计数或概 率,应用正确技巧,给出精确值。反复出现的链条是 -- 求导→令f'=0→分类;行化简→回代;▽f=0 → 海森检验;条件化→贝叶斯;度序列→边数。写 出每一行 -- 步骤分是实打实的。把这些链条练熟, 全新数字就无法让你意外。 AskSia Library · MAT9004 · 双语 Bilingual CONTENTS - CONTENTS Six worlds, one exam-ready book 六大板块,一本应考即用的书 Calculus & algebra first (the largest block), discrete maths & probability second 微积分与代数先行(最大板块),离散数学与概率其次 Ch Topic Core methods Part 1 . Calculus & optimisation (the exam's largest block)[28]Source: asksia-cheatsheet-mat9004.pdfRECURRING THEMES · Saddle points have no 1-var analogue - they're why the 2-var test needs det H, not just fxx · Convexity = global extremum in BOTH 1-var (f"≥0) and 2-var (det H≥0 + fxx≥0) Formula Belt SIDE 1 (xb)'=bx^(b-1) . (ex)'=ex . chain g' . f' (g) 2nd-deriv: f' '>0 min, <0 max, =0 ? [x == x^(a+1)/(a+1) . FTC F(b) -F(a) det2=ad-bc . Ax=b = x=A-1b det(A-AI)=0 . A"=PDOP-1 det H = fxxf_yy-fxy2 (<0 saddle) asksia. ai/cheatsheet/ mat9004 · side 1/2 AskSia CHEATSHEET SERIES AREA 2 . 7 . Eigenvalues & Eigenvectors L11-12 * For square A: Ax = Xx, x # 0 . x = eigenvector, A = eigenvalue. Zero vector is never an eigenvector; }=0 can be an eigenvalue. RECIPE 1 eigenvalues: det(A - AI) = 0 (char. poly) 2 eigenvectors: solve (A - AI)x = 0 (always wo many - scalar multiples) If v is an eigenvector so is cv (c#0). Char-poly degree = n; roots = the eigenvalues. DIAGONALISATION A = P D P-1 . P = (V1 | . . . | Vn), D = diag (Ai) An = P Dn P-1 (D" = diag raised to n) Construct when n distinct eigenvalues. Use: linear recurrences/Markov processes an=V"ao; long run ruled by the largest eigenvalue (=1 for stochastic); PageRank = eigenvector for A=1. SIA > P1 is usually not needed for the long- answer - examiners want eigenvalues, eigenvectors and the assembled P, D. Show det(A-MI)=O explicitly. 7b · Worked . Diagonalise 2×2 RENUMBERED A=[0,2 ;- 1,3]: det(A-λΙ)=λ2-3λ+2=(λ-1) (λ-2) =λ=1,2. A=2: (A-21)x=0 => v=[1,1]T. N=1: v=[2,1]]. So P=[1,2;1,1], D=[2,0;0,1], A=PDP-1. Sanity: trace 0+3=3=1+2 V; det 0. 3-2. (-1)=2=1. 2 V. 6b . Worked . Ax=b RENUMBERED 3x+y=2, 5x+2y=3=> det=1, A-1=[2,-1 ;- 5,3]; x=A-1b= (2-2-1-3, -5. 2+3. 3) = (1,-1) . Reuse A-1 for many b. 6c . Worked . Gaussian RENUMBERED elim Solve x+2y-z=6 ;- x+y+2z=3;x+y-z=8. R2+R1, R3-R1: gives 3y+z=9 and -y=2 = y =- 2; back- sub z=9-3y=15? recheck signs in your own working - the discipline is clear one column at a time, top-to- bottom, then back-substitute and verify in the original. 6d . Free-Variable Systems

加速判断（常用于全局）：

• 若 $f$ 凸（convex，$f''\ge 0$），则局部最小=全局最小；若 $f$ 凹（concave，$f''\le 0$），局部最大=全局最大[18]Source: asksia-cheatsheet-mat9004.pdfThis is a REVISION sheet - you cannot bring it in. The exam gives its own formula sheet, so memorise the METHODS, not the formulas : the recipe wins marks, the formula is handed to you. SIA > Everything must be hand-computable. Practise Gaussian elimination, char-poly eigenvalues and Bayes by hand until fluent - no calculator on the day. 1 . Derivatives . Rules AREA 1 . L4-5 on interval f'(a) = slope of tangent = limx->a (f(x)-f(a))/(x-a). |x| is not differentiable at 0. F(X) F'(X) c (const) xᵇ b. x^(b-1) aª In(a)·a* ex ex ln x 1/x loga X 1/(ln(a)·x) COMBINING RULES (c. f) '=c. f' . (fig) ' =f'±g' product: (fg) ' = f'g + fg' chain: (f(g(x) ) ) ' = g' (x) . f' (g(x)) COMMON CHAINS (e^(cx+d))' = c. e^(cx+d) (Ln(cx+d))' = c/(cx+d) (a^(cx+d))' = c. Ln(a) . a^(cx+d) nth derivative f(n) = differentiate n times (f'=f(2). Worked tangent slope: for h(x)=x. e^(2x), h'=e^(2x) (1+2x) => h'(0)=1 (product rule, then evaluate). 1b . Function Types L1-3 Convex: chord lies on/above plot e f"≥0. Concave: chord below + f"≤0. Lines are both. Transform-plot trick: log-log linear => power law (slope -a); log-lin linear = exponential (slope In a); lin- log = logarithmic.[29]Source: asksia-cheatsheet-mat9004.pdfGLOBAL EXTREMA ON [C, D] candidates = stationary pts + singular pts (f' undef) + endpoints c, d. Evaluate f at ALL - Largest = max, smallest = min. Shortcut: local min of a convex f is the global min ; local max of concave = global max. Quadratic roots: x2+ax+b=0 => x =- a/2±/(a2/4-b); real iff a2≥4b. (x-u)(x-v)=x2-(u+v)x+uv. SIA - The long-answer "global extrema on [a,b]" question forgets the endpoints at your peril - they are candidates too. Always tabulate f at every candidate. 2b . Worked . Extrema RENUMBERED f'(t)=t3-5t2+6t=t(t-2)(t-3) on [-1,3]. Stationary t=0,2,3. f''=3t2-10t+6: f''(0)=6>0 (min), f''(2) =- 2<0 (max), f''(3)=3>0 (min). Compare fat {-1,0,2,3} = pick global max/min by value. Cubic variant: f(x)=2x3-3x2-12x+4 on [-3,3] => stationary x =- 1,2; global max (-1,11), global min at the endpoint (-3,-41). The endpoint wins - count it. 2c . Worked . Can min surface Volume Ttr2h = 128Tt = h=128/r2. Surface f(r)=2Tt(r2+128/r); f'=2Tt(2r-128/r2)=0 = r3=64 = r=4. f"'>0 (convex) = global min; h=128/16=8. Page-layout variant: printable A=(x-4)(y-6) with xy=294=A=318-6x-1176/x; A' =- 6+1176/x2=0=>x=14, A"<0 (concave, so a max), y=21. Convexity-on-interval trap: for f"=6ax-12 to be neither convex nor concave on (2,3), require f" to change sign there => solve for the parameter range, don't just plug one point. Constrained product, variants: max xy s. t. 2x+3y=60 =150; s. t. x+3y=60 = 300. Same substitute-then- optimise recipe; check it's a max (f''<0), not a min. Singular points (f' undefined, e. g. a corner like |x| at 0) are candidates too - don't only solve f'=0. The Extreme Value Theorem guarantees a continuous f on [c,d] attains both extrema. 2d . RSS / least squares APPLICATION RESIDUAL SUM OF SQUARES - RSS = Σι (yi - f(X1))2 Fit f(x)=2x-1 to (2,1),(3,4),(5,2): preds 3,5,9; residuals -2,-1,-7 => RSS = 4+1+49 = 54. Squares are differentiable & punish big errors. Workflow: 1 f', solve f'=0 + find singular pts; @ classify with f" (or sign change); @ compare f at stationary/singular/boundary; use convexity if available. 3 . Integration · FTC AREA 1 . L7 Antiderivative F: F'=f, unique up to +c. [f dx = F(x)+c. F(X) SF DX xª (a¢-1) x^(a+1)/(a+1)+c x-

考试陷阱 1：忘记端点（宝典把它列为四大丢分陷阱之一）[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。。

4）Area 2：线性代数（高斯消元、行列式/逆、特征值与对角化）

4.1 最核心技能：高斯消元解 $Ax=b$（最常考）

标准流程（按步骤分写）：

• 写增广矩阵 $[A|b]$
• 行化简到阶梯形（echelon）
• 自底向上回代（back-substitute）[13]Source: asksia-bible-mat9004-bilingual.pdfAB 的末行。(3,4)·(5,7)= 15+28 = 43; (3,4)·(6,8) = 18+32 = 50. 4 So AB = 19 43 50 , but BA = 23 34 46 different. 19 22 23 34 于是 AB = 50 但 BA = ， 同。 ! Check the inner dimensions before you write anything 动笔之前先核对内层维度 A(2×3)B(3x2) is defined (inner 3's match) and gives 2×2; B(3x2)A(2x3) gives 3x3 - a different size entirely. One product can be defined while the other is not. Match the inner numbers first, and remember each entry is a dot product, not an element-wise multiply. A(2×3)B(3×2) 有定义(内层两个 3匹配)且结果为 2×2;B(3×2)A(2×3)则给出 3x3 -- 尺寸完全不同。一个乘积有定义 而另一个未必。先匹配内层数字,并记住每个元素都是一个点积(dot product),而非逐元素相乘。 AskSia Library · MAT9004 · 双语 Bilingual — 一不 i Why this rule? Matrices are linear maps 为什么有这条法则?矩阵(matrix)是线性映射 AB is the matrix that does "apply B, then apply A. " That is exactly why the inner dimensions must match (B's output feeds A's input) and why order matters (rotate-then-stretch # stretch-then-rotate). A whole dataset X times a weight matrix W - the core operation of a linear model and of one neural-network layer - is just this multiplication. AB 这个矩阵执行的是“先作用 B,再作用A”。这正是为什么内层维度必须匹配(B的输出喂给A的输入),也是为什么 次序重要(先旋转再拉伸≠先拉伸再旋转)。整个数据集 X乘以权重矩阵 W -- 线性模型与神经网络一层的核心运算 一就是这个乘法。 Every big idea downstream is matrix multiplication wearing a costume: solving Ax = b, raising A to a power for a Markov chain, counting walks with Ak in a graph. Get the row-into-column reflex automatic now. 下游每一个大概念都是穿着外衣的矩阵乘法:求解 Ax= b、为马尔可夫链对 A 求幂、用AK在图中计数途径。现在就把 “行乘列”的反射练成本能。 EXAM-MOVE . MAT9004 LECTURES 8-9 AskSia Library · MAT9004 · 双语 Bilingual CH 2 . ELIMINATION - CHAPTER 2 (CONT. ) THE HEADLINE SKILL Systems Ax = b & Gaussian elimination 线性方程组 Ax = b 与高斯消元(Gaussian elimination) Row-reduce the augmented matrix to a staircase, then back-substitute 把增广矩阵行化简成阶梯形,再回代 This is the skill the exam tests most. Write the system as an augmented matrix [A | b], row-reduce to upper-triangular (echelon) form, then back- substitute bottom-up.

三种合法行变换（只做一种，别混）：交换两行；某行乘非零常数；某行加上另一行的倍数[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。

考试陷阱 2：行化简负号错（四大陷阱之一）

• 建议：每一步都带着 $b$ 列；解出来以后把解回代到原方程组检查，能抓到符号翻车[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。。

4.2 维度检查：矩阵乘法先看“内层维度”

规则：$A(m\times n)B(n\times p)$ 才定义，结果是 $m\times p$；顺序不同尺寸可能直接变掉（甚至不定义）[13]Source: asksia-bible-mat9004-bilingual.pdfAB 的末行。(3,4)·(5,7)= 15+28 = 43; (3,4)·(6,8) = 18+32 = 50. 4 So AB = 19 43 50 , but BA = 23 34 46 different. 19 22 23 34 于是 AB = 50 但 BA = ， 同。 ! Check the inner dimensions before you write anything 动笔之前先核对内层维度 A(2×3)B(3x2) is defined (inner 3's match) and gives 2×2; B(3x2)A(2x3) gives 3x3 - a different size entirely. One product can be defined while the other is not. Match the inner numbers first, and remember each entry is a dot product, not an element-wise multiply. A(2×3)B(3×2) 有定义(内层两个 3匹配)且结果为 2×2;B(3×2)A(2×3)则给出 3x3 -- 尺寸完全不同。一个乘积有定义 而另一个未必。先匹配内层数字,并记住每个元素都是一个点积(dot product),而非逐元素相乘。 AskSia Library · MAT9004 · 双语 Bilingual — 一不 i Why this rule? Matrices are linear maps 为什么有这条法则?矩阵(matrix)是线性映射 AB is the matrix that does "apply B, then apply A. " That is exactly why the inner dimensions must match (B's output feeds A's input) and why order matters (rotate-then-stretch # stretch-then-rotate). A whole dataset X times a weight matrix W - the core operation of a linear model and of one neural-network layer - is just this multiplication. AB 这个矩阵执行的是“先作用 B,再作用A”。这正是为什么内层维度必须匹配(B的输出喂给A的输入),也是为什么 次序重要(先旋转再拉伸≠先拉伸再旋转)。整个数据集 X乘以权重矩阵 W -- 线性模型与神经网络一层的核心运算 一就是这个乘法。 Every big idea downstream is matrix multiplication wearing a costume: solving Ax = b, raising A to a power for a Markov chain, counting walks with Ak in a graph. Get the row-into-column reflex automatic now. 下游每一个大概念都是穿着外衣的矩阵乘法:求解 Ax= b、为马尔可夫链对 A 求幂、用AK在图中计数途径。现在就把 “行乘列”的反射练成本能。 EXAM-MOVE . MAT9004 LECTURES 8-9 AskSia Library · MAT9004 · 双语 Bilingual CH 2 . ELIMINATION - CHAPTER 2 (CONT. ) THE HEADLINE SKILL Systems Ax = b & Gaussian elimination 线性方程组 Ax = b 与高斯消元(Gaussian elimination) Row-reduce the augmented matrix to a staircase, then back-substitute 把增广矩阵行化简成阶梯形,再回代 This is the skill the exam tests most. Write the system as an augmented matrix [A | b], row-reduce to upper-triangular (echelon) form, then back- substitute bottom-up.

观念：矩阵乘法是“先做 $B$ 再做 $A$ 的线性映射组合”，所以顺序很关键[13]Source: asksia-bible-mat9004-bilingual.pdfAB 的末行。(3,4)·(5,7)= 15+28 = 43; (3,4)·(6,8) = 18+32 = 50. 4 So AB = 19 43 50 , but BA = 23 34 46 different. 19 22 23 34 于是 AB = 50 但 BA = ， 同。 ! Check the inner dimensions before you write anything 动笔之前先核对内层维度 A(2×3)B(3x2) is defined (inner 3's match) and gives 2×2; B(3x2)A(2x3) gives 3x3 - a different size entirely. One product can be defined while the other is not. Match the inner numbers first, and remember each entry is a dot product, not an element-wise multiply. A(2×3)B(3×2) 有定义(内层两个 3匹配)且结果为 2×2;B(3×2)A(2×3)则给出 3x3 -- 尺寸完全不同。一个乘积有定义 而另一个未必。先匹配内层数字,并记住每个元素都是一个点积(dot product),而非逐元素相乘。 AskSia Library · MAT9004 · 双语 Bilingual — 一不 i Why this rule? Matrices are linear maps 为什么有这条法则?矩阵(matrix)是线性映射 AB is the matrix that does "apply B, then apply A. " That is exactly why the inner dimensions must match (B's output feeds A's input) and why order matters (rotate-then-stretch # stretch-then-rotate). A whole dataset X times a weight matrix W - the core operation of a linear model and of one neural-network layer - is just this multiplication. AB 这个矩阵执行的是“先作用 B,再作用A”。这正是为什么内层维度必须匹配(B的输出喂给A的输入),也是为什么 次序重要(先旋转再拉伸≠先拉伸再旋转)。整个数据集 X乘以权重矩阵 W -- 线性模型与神经网络一层的核心运算 一就是这个乘法。 Every big idea downstream is matrix multiplication wearing a costume: solving Ax = b, raising A to a power for a Markov chain, counting walks with Ak in a graph. Get the row-into-column reflex automatic now. 下游每一个大概念都是穿着外衣的矩阵乘法:求解 Ax= b、为马尔可夫链对 A 求幂、用AK在图中计数途径。现在就把 “行乘列”的反射练成本能。 EXAM-MOVE . MAT9004 LECTURES 8-9 AskSia Library · MAT9004 · 双语 Bilingual CH 2 . ELIMINATION - CHAPTER 2 (CONT. ) THE HEADLINE SKILL Systems Ax = b & Gaussian elimination 线性方程组 Ax = b 与高斯消元(Gaussian elimination) Row-reduce the augmented matrix to a staircase, then back-substitute 把增广矩阵行化简成阶梯形,再回代 This is the skill the exam tests most. Write the system as an augmented matrix [A | b], row-reduce to upper-triangular (echelon) form, then back- substitute bottom-up.。

4.3 行列式与逆（尤其 $2\times 2$）

$2\times2$ 行列式

• 若 $A=\begin{pmatrix}a&b\c&d\end{pmatrix}$，则 $\det(A)=ad-bc$[24]Source: asksia-cheatsheet-mat9004.pdfFree variable = write solution in vector form (point + t. direction). Fewer equations than variables = usually oo many. 6 . Determinant & Inverse AREA 2 . . 10 2×2 DET & INVERSE det[a b; c d] = ad - bc A-1 = (1/det A) . [d -b; - c a] A invertible + det(A) # 0. det(AB)=det(A)det(B); det(I)=1. Identity In: 1s on diagonal, AI=IA=A. SOLVE VIA INVERSE - Worked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1

可逆判定：$A$ 可逆 $\Leftrightarrow \det(A)\ne 0$；若不可逆，$Ax=b$ 不会有唯一解（要么无解要么无穷多解）[24]Source: asksia-cheatsheet-mat9004.pdfFree variable = write solution in vector form (point + t. direction). Fewer equations than variables = usually oo many. 6 . Determinant & Inverse AREA 2 . . 10 2×2 DET & INVERSE det[a b; c d] = ad - bc A-1 = (1/det A) . [d -b; - c a] A invertible + det(A) # 0. det(AB)=det(A)det(B); det(I)=1. Identity In: 1s on diagonal, AI=IA=A. SOLVE VIA INVERSE - Worked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1[25]Source: asksia-cheatsheet-mat9004.pdfWorked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1 5. For powers/recurrences quote A"=PDnp-1 8 · Multivariable Calculus AREA 3 . L13- 16 * Partial derivative fx: differentiate, treat y as constant; f_y treats x as constant. GRADIENT Vf = [fx ; f_y] points in direction of steepest increase Vf 1 the level set through the point Level set of value c = {(x,y): f=c}; level curves never cross. Linear approx: f(x+Ax,y+Ay) = f + fx4x + f_yAy. - STATIONARY POINT

$2\times2$ 逆矩阵公式

• $$A^{-1}=\frac{1}{\det(A)}\begin{pmatrix}d&-b\-c&a\end{pmatrix}$$[24]Source: asksia-cheatsheet-mat9004.pdfFree variable = write solution in vector form (point + t. direction). Fewer equations than variables = usually oo many. 6 . Determinant & Inverse AREA 2 . . 10 2×2 DET & INVERSE det[a b; c d] = ad - bc A-1 = (1/det A) . [d -b; - c a] A invertible + det(A) # 0. det(AB)=det(A)det(B); det(I)=1. Identity In: 1s on diagonal, AI=IA=A. SOLVE VIA INVERSE - Worked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1

4.4 特征值、特征向量、对角化（大题槽位之一）

定义：$Ax=\lambda x$，$x\ne 0$；$\lambda$ 是特征值，$x$ 是特征向量[28]Source: asksia-cheatsheet-mat9004.pdfRECURRING THEMES · Saddle points have no 1-var analogue - they're why the 2-var test needs det H, not just fxx · Convexity = global extremum in BOTH 1-var (f"≥0) and 2-var (det H≥0 + fxx≥0) Formula Belt SIDE 1 (xb)'=bx^(b-1) . (ex)'=ex . chain g' . f' (g) 2nd-deriv: f' '>0 min, <0 max, =0 ? [x == x^(a+1)/(a+1) . FTC F(b) -F(a) det2=ad-bc . Ax=b = x=A-1b det(A-AI)=0 . A"=PDOP-1 det H = fxxf_yy-fxy2 (<0 saddle) asksia. ai/cheatsheet/ mat9004 · side 1/2 AskSia CHEATSHEET SERIES AREA 2 . 7 . Eigenvalues & Eigenvectors L11-12 * For square A: Ax = Xx, x # 0 . x = eigenvector, A = eigenvalue. Zero vector is never an eigenvector; }=0 can be an eigenvalue. RECIPE 1 eigenvalues: det(A - AI) = 0 (char. poly) 2 eigenvectors: solve (A - AI)x = 0 (always wo many - scalar multiples) If v is an eigenvector so is cv (c#0). Char-poly degree = n; roots = the eigenvalues. DIAGONALISATION A = P D P-1 . P = (V1 | . . . | Vn), D = diag (Ai) An = P Dn P-1 (D" = diag raised to n) Construct when n distinct eigenvalues. Use: linear recurrences/Markov processes an=V"ao; long run ruled by the largest eigenvalue (=1 for stochastic); PageRank = eigenvector for A=1. SIA > P1 is usually not needed for the long- answer - examiners want eigenvalues, eigenvectors and the assembled P, D. Show det(A-MI)=O explicitly. 7b · Worked . Diagonalise 2×2 RENUMBERED A=[0,2 ;- 1,3]: det(A-λΙ)=λ2-3λ+2=(λ-1) (λ-2) =λ=1,2. A=2: (A-21)x=0 => v=[1,1]T. N=1: v=[2,1]]. So P=[1,2;1,1], D=[2,0;0,1], A=PDP-1. Sanity: trace 0+3=3=1+2 V; det 0. 3-2. (-1)=2=1. 2 V. 6b . Worked . Ax=b RENUMBERED 3x+y=2, 5x+2y=3=> det=1, A-1=[2,-1 ;- 5,3]; x=A-1b= (2-2-1-3, -5. 2+3. 3) = (1,-1) . Reuse A-1 for many b. 6c . Worked . Gaussian RENUMBERED elim Solve x+2y-z=6 ;- x+y+2z=3;x+y-z=8. R2+R1, R3-R1: gives 3y+z=9 and -y=2 = y =- 2; back- sub z=9-3y=15? recheck signs in your own working - the discipline is clear one column at a time, top-to- bottom, then back-substitute and verify in the original. 6d . Free-Variable Systems

标准配方（必须背成肌肉记忆）：[24]Source: asksia-cheatsheet-mat9004.pdfFree variable = write solution in vector form (point + t. direction). Fewer equations than variables = usually oo many. 6 . Determinant & Inverse AREA 2 . . 10 2×2 DET & INVERSE det[a b; c d] = ad - bc A-1 = (1/det A) . [d -b; - c a] A invertible + det(A) # 0. det(AB)=det(A)det(B); det(I)=1. Identity In: 1s on diagonal, AI=IA=A. SOLVE VIA INVERSE - Worked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1[28]Source: asksia-cheatsheet-mat9004.pdfRECURRING THEMES · Saddle points have no 1-var analogue - they're why the 2-var test needs det H, not just fxx · Convexity = global extremum in BOTH 1-var (f"≥0) and 2-var (det H≥0 + fxx≥0) Formula Belt SIDE 1 (xb)'=bx^(b-1) . (ex)'=ex . chain g' . f' (g) 2nd-deriv: f' '>0 min, <0 max, =0 ? [x == x^(a+1)/(a+1) . FTC F(b) -F(a) det2=ad-bc . Ax=b = x=A-1b det(A-AI)=0 . A"=PDOP-1 det H = fxxf_yy-fxy2 (<0 saddle) asksia. ai/cheatsheet/ mat9004 · side 1/2 AskSia CHEATSHEET SERIES AREA 2 . 7 . Eigenvalues & Eigenvectors L11-12 * For square A: Ax = Xx, x # 0 . x = eigenvector, A = eigenvalue. Zero vector is never an eigenvector; }=0 can be an eigenvalue. RECIPE 1 eigenvalues: det(A - AI) = 0 (char. poly) 2 eigenvectors: solve (A - AI)x = 0 (always wo many - scalar multiples) If v is an eigenvector so is cv (c#0). Char-poly degree = n; roots = the eigenvalues. DIAGONALISATION A = P D P-1 . P = (V1 | . . . | Vn), D = diag (Ai) An = P Dn P-1 (D" = diag raised to n) Construct when n distinct eigenvalues. Use: linear recurrences/Markov processes an=V"ao; long run ruled by the largest eigenvalue (=1 for stochastic); PageRank = eigenvector for A=1. SIA > P1 is usually not needed for the long- answer - examiners want eigenvalues, eigenvectors and the assembled P, D. Show det(A-MI)=O explicitly. 7b · Worked . Diagonalise 2×2 RENUMBERED A=[0,2 ;- 1,3]: det(A-λΙ)=λ2-3λ+2=(λ-1) (λ-2) =λ=1,2. A=2: (A-21)x=0 => v=[1,1]T. N=1: v=[2,1]]. So P=[1,2;1,1], D=[2,0;0,1], A=PDP-1. Sanity: trace 0+3=3=1+2 V; det 0. 3-2. (-1)=2=1. 2 V. 6b . Worked . Ax=b RENUMBERED 3x+y=2, 5x+2y=3=> det=1, A-1=[2,-1 ;- 5,3]; x=A-1b= (2-2-1-3, -5. 2+3. 3) = (1,-1) . Reuse A-1 for many b. 6c . Worked . Gaussian RENUMBERED elim Solve x+2y-z=6 ;- x+y+2z=3;x+y-z=8. R2+R1, R3-R1: gives 3y+z=9 and -y=2 = y =- 2; back- sub z=9-3y=15? recheck signs in your own working - the discipline is clear one column at a time, top-to- bottom, then back-substitute and verify in the original. 6d . Free-Variable Systems

• 1. 写 $\det(A-\lambda I)=0$ 得特征多项式（char poly）
• 1. 解根得到特征值 $\lambda$
• 1. 对每个 $\lambda$ 解 $(A-\lambda I)x=0$ 得特征向量（会出现自由变量，取一个好看的倍数即可）
• 1. 若有 $n$ 个不同特征值：组 $P=(v_1|\cdots|v_n)$，$D=\mathrm{diag}(\lambda_i)$，则 $A=PDP^{-1}$

快速自检（sanity check）：

• 特征值之和 = trace；特征值之积 = det（2×2 特别好用）[25]Source: asksia-cheatsheet-mat9004.pdfWorked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1 5. For powers/recurrences quote A"=PDnp-1 8 · Multivariable Calculus AREA 3 . L13- 16 * Partial derivative fx: differentiate, treat y as constant; f_y treats x as constant. GRADIENT Vf = [fx ; f_y] points in direction of steepest increase Vf 1 the level set through the point Level set of value c = {(x,y): f=c}; level curves never cross. Linear approx: f(x+Ax,y+Ay) = f + fx4x + f_yAy. - STATIONARY POINT

4.5 自由变量系统（free-variable）

触发词：方程少于未知数 / 行化简出现 $0=0$

写法：设自由变量 $t\in\mathbb{R}$，把主元变量写成 $t$ 的函数，然后用向量形式

• $$(x,y,z,w)=(a,b,c,0)+t(p,q,r,1)$$
• 并写清楚 “for all $t\in\mathbb{R}$”[19]Source: asksia-cheatsheet-mat9004.pdf7b · Worked . Diagonalise 2×2 RENUMBERED A=[0,2 ;- 1,3]: det(A-λΙ)=λ2-3λ+2=(λ-1) (λ-2) =λ=1,2. A=2: (A-21)x=0 => v=[1,1]T. N=1: v=[2,1]]. So P=[1,2;1,1], D=[2,0;0,1], A=PDP-1. Sanity: trace 0+3=3=1+2 V; det 0. 3-2. (-1)=2=1. 2 V. 6b . Worked . Ax=b RENUMBERED 3x+y=2, 5x+2y=3=> det=1, A-1=[2,-1 ;- 5,3]; x=A-1b= (2-2-1-3, -5. 2+3. 3) = (1,-1) . Reuse A-1 for many b. 6c . Worked . Gaussian RENUMBERED elim Solve x+2y-z=6 ;- x+y+2z=3;x+y-z=8. R2+R1, R3-R1: gives 3y+z=9 and -y=2 = y =- 2; back- sub z=9-3y=15? recheck signs in your own working - the discipline is clear one column at a time, top-to- bottom, then back-substitute and verify in the original. 6d . Free-Variable Systems PARAMETRISE Underdetermined (fewer equations than unknowns) = a row collapses to 0=0 => one free variable tER. Express each pivot variable in terms of t, then write the solution set as point + t·direction (a line) - e. g. (x,y,z,w) = (a,b,c,0) +t(p,q,r,1). State "for all tER "; two free variables = a plane. Two-variable products (Hadamard) exist but are NOT the matrix product used here - always use the row-column rule. (kA)B = k(AB). Worked 2x2 mult: [1,2;0,1] . [3;4] = [1. 3+2-4; 0-3+1-4] = [11;4]. Mult is defined only when columns of A = rows of B A vector is an m×1 matrix. Identity In acts as 1: Aln=A. If BA=I then AB=I, so each is the other's inverse. A(B+D)=AB+AD distributes. Compiled by AskSia . mapped to the MAT9004 syllabus . asksia. ai/cheatsheet/mat9004 CONTINUOUS . Single-variable calculus . Optimisation . Integration . Linear algebra . Eigenvalues . A=PDP-1 . Multivariable calculus REVISION SHEET . CLOSED-BOOK EXAM Independent revision aid . check learning. monash. edu for exam rules . @ 2026 flip + for side 2 . counting, probability & graphs APPLIED MAT9004 Mathematical Foundations for Data Science & AI MONASH UNIVERSITY . FACULTY OF SCIENCE EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Counting . probability . graphs SIDE 2/2 EULer/Hamilton[20]Source: asksia-cheatsheet-mat9004.pdfPARAMETRISE Underdetermined (fewer equations than unknowns) = a row collapses to 0=0 => one free variable tER. Express each pivot variable in terms of t, then write the solution set as point + t·direction (a line) - e. g. (x,y,z,w) = (a,b,c,0) +t(p,q,r,1). State "for all tER "; two free variables = a plane. Two-variable products (Hadamard) exist but are NOT the matrix product used here - always use the row-column rule. (kA)B = k(AB). Worked 2x2 mult: [1,2;0,1] . [3;4] = [1. 3+2-4; 0-3+1-4] = [11;4]. Mult is defined only when columns of A = rows of B A vector is an m×1 matrix. Identity In acts as 1: Aln=A. If BA=I then AB=I, so each is the other's inverse. A(B+D)=AB+AD distributes. Compiled by AskSia . mapped to the MAT9004 syllabus . asksia. ai/cheatsheet/mat9004 CONTINUOUS . Single-variable calculus . Optimisation . Integration . Linear algebra . Eigenvalues . A=PDP-1 . Multivariable calculus REVISION SHEET . CLOSED-BOOK EXAM Independent revision aid . check learning. monash. edu for exam rules . @ 2026 flip + for side 2 . counting, probability & graphs APPLIED MAT9004 Mathematical Foundations for Data Science & AI MONASH UNIVERSITY . FACULTY OF SCIENCE EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Counting . probability . graphs SIDE 2/2 EULer/Hamilton 10 . Counting . Basic Rules L17 AREA 4 . PRODUCT & SUM RULES AND (sequence) = multiply: |S| = n kj OR (disjoint cases) - add: |S| = 2 complement: | good| = |S| - | bad| Keyword test: "and"/"then" =>x; "or"/disjoint cases => +. n! = 1. 2 . . . . . n; 0! = 1 . Sequences with kj choices each = |S|= nlkj. 11 . Four Selection Types AREA 4 . L17- 18 * Draw r from n - ask: order matter? repeats allowed?[24]Source: asksia-cheatsheet-mat9004.pdfFree variable = write solution in vector form (point + t. direction). Fewer equations than variables = usually oo many. 6 . Determinant & Inverse AREA 2 . . 10 2×2 DET & INVERSE det[a b; c d] = ad - bc A-1 = (1/det A) . [d -b; - c a] A invertible + det(A) # 0. det(AB)=det(A)det(B); det(I)=1. Identity In: 1s on diagonal, AI=IA=A. SOLVE VIA INVERSE - Worked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1

5）Area 3：多变量微积分与优化（梯度 + Hessian 分类）——大题高频

5.1 关键定义

偏导：$f_x$ 把 $y$ 当常数；$f_y$ 把 $x$ 当常数[25]Source: asksia-cheatsheet-mat9004.pdfWorked: det[1,1 ;- 1,1]=1-(-1)=2 = invertible. det[1,1;1,1]=0 = not invertible (repeated row). Singular-parameter Q: set ad-bc=0, solve. If A is square but not invertible, Ax=b has either no solution or infinitely many (never a unique one). 7c · Worked . An recurrence MATRIX POWERS Cn+1=4Cm+4Un, Un+1=Cn+4Un => A=[4,4;1,4]. det(Α-λΙ)=λ2-8+12=(λ-2)(λ-6) =λ=2,6; eigenvectors [2,-1]],[2,1]]. Long-run ratio > dominant eigenvalue 6. 7d . Diagonal Matrices WHY DIAG IS EASY D (D¡ ¡= 0 off-diagonal): D" raises each diagonal entry to n . That is the whole point of A=PDP-1 - push the power onto D where it's trivial, then conjugate back. Char-poly check: for a 2×2, det(A-NI)=)2 - (trace)} + det. So )2-3)+2 came from trace 3, det 2. Sum of eigenvalues = trace; product = det - a fast sanity check. 7e . Worked . Eigenvector SHORT unknown SLOT Given N is an eigenvalue and an eigenvector of the form (x,1,z)T, solve (A-XI)v=0 row-by-row for x and z (a small linear system). Likewise "find the missing entry of A"1": use A-A-1=I and read off one equation. 7f . Eigen Recipe Recap STEP LIST 1. Form A-NI; expand det(A-AI)=0 to the characteristic polynomial 2. Solve for the roots M1, . . . , An (the eigenvalues) 3. For each A, row-reduce (A-AI)x=0; the free variable gives the eigenvector (pick the neatest scalar multiple) 4. Distinct X => assemble P=(v1| . . . |vn), D=diag(2) => A=PDP-1 5. For powers/recurrences quote A"=PDnp-1 8 · Multivariable Calculus AREA 3 . L13- 16 * Partial derivative fx: differentiate, treat y as constant; f_y treats x as constant. GRADIENT Vf = [fx ; f_y] points in direction of steepest increase Vf 1 the level set through the point Level set of value c = {(x,y): f=c}; level curves never cross. Linear approx: f(x+Ax,y+Ay) = f + fx4x + f_yAy. - STATIONARY POINT

梯度：$$\nabla f=\begin{pmatrix}f_x\f_y\end{pmatrix}$$

• 指向“最陡上升方向”；$-\nabla f$ 是最陡下降[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or

驻点（stationary point）：$\nabla f=0$，即 $f_x=0$ 且 $f_y=0$[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or

5.2 Hessian 判别（你要会一套表秒做）

Hessian：$H=\begin{pmatrix}f_{xx}&f_{xy}\f_{xy}&f_{yy}\end{pmatrix}$

计算判别量：$$D=\det(H)=f_{xx}f_{yy}-f_{xy}^2$$

分类规则（每个驻点都要单独算一次）：[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or[10]Source: asksia-bible-mat9004-bilingual.pdf- PRACTICE BANK (CONT. ) Long-answer - gradient, Hessian, classification 大题––梯度(gradient)、海森矩阵(Hessian)、分类判定 (2: find every stationary point of a two-variable function and classify it L2:求二元函数的每一个驻点(stationary point)并加以分类 L2 LONG ANSWER 6 marks . multivariable calculus Let f(x, y) = x3 + y3 - 3xy. Find the gradient Vf and the Hessian H, locate all stationary points, and classify each using the second-derivative (Hessian) test. [6] 设 f(x, y)= x3 + y3 -3xy。求梯度 Vf 与海森矩阵 H,定位所有驻点,并用二阶导(海森)检验对每个分类。[6] L2 Worked solution - gradient + Hessian classification 1 Gradient. fx = 3x2 - 3y, fy = 3y2 - 3x. So Vf = (3x2 - 3y, 3y2 - 3x). 梯度。fx= 3x2-3y, fy = 3y2-3x。所以 Vf =(3x2 -3y,3y2-3x)。 2 Stationary points. Set both to zero: 3x2 = 3y -> y = x2; and 3y2 = 3x -> x = y2. Substitute: x = (x2)2 = x4 -+ x4 - x = x(x3 - 1) = 0 -+ x = 0 or x = 1. This gives (0, 0) and (1,1). 驻点。令两者均为零:3×2=3y→y=x2;且 3y2=3×→×=y2。代入:x=(x2)2= x4 →×4 - x= x(×3 -1)= 0 →x=0 或x=1。由此得(0,0)和(1,1)。 3 Hessian. fxx = 6x, fyy = 6y, fxy = - 3. H = 6x -3 6y -3 , det H = 36xy - 9. 海森矩阵(Hessian)。fxx = 6x, fyy = 6y, fxy =- 3。H= 6x -3 6y -3 det H = 36xy - 9. ， 4 Classify (0, 0). det H = 36(0)(0) - 9 = - 9 < 0 - saddle point. 分类(0,0)。det H = 36(0)(0)-9 =- 9<0→ 鞍点(saddle point)。 Classify (1, 1). det H = 36(1)(1) - 9 = 27 > 0 and fxx = 6 > 0-local minimum. (f(1,1) =1+1-3 =- 1. ) 分类(1,1)。det H = 36(1)(1)-9= 27>0 且 fxx= 6>0→局部最小值。(f(1,1)=1+1-3 =- 1。) - i The Hessian decision rule 海森矩阵(Hessian)判别法则 Compute det H = fxxfyy - fxy2. det H < 0 - saddle. det H > O -> extremum: fxx > O gives a local min, fxx < O a local max. det H = 0 - test is inconclusive. This table is on the exam formula sheet - but apply it cleanly per point. 计算 det H = fxxfyy -fxy2。det H <0→ 鞍点。det H > 0→ 极值: fxx> 0 给局部极小,fxx <0 给局部极大。det H =0→判别法无法判定。此表在考试公式表上 -- 但要逐点干净地套用。 - AskSia Library · MAT9004 · 双语 Bilingual Show every step. The 6-mark long-answers are graded on method: a stated gradient, the simultaneous solve, the Hessian, and a per-point verdict each earns marks - even a small arithmetic slip at the end keeps most of the credit. A bare final answer throws that away. 展示每一步。6分的大题是按方法评分的:写出梯度、联立求解、海森矩阵、对每个点的判定,各自都能得分 -- 即便 末尾有点小算术失误,也能保住大部分分数。光给最终答案则会把这些全丢掉。

• 若 $D<0$：鞍点（saddle）
• 若 $D>0$ 且 $f_{xx}>0$：局部最小
• 若 $D>0$ 且 $f_{xx}<0$：局部最大
• 若 $D=0$：无法判定（inconclusive）

阅卷逻辑：这类 6 分大题按步骤给分：写梯度、联立解驻点、写 Hessian、逐点分类；最后算错一点也能保住大部分分数——所以一定写过程[10]Source: asksia-bible-mat9004-bilingual.pdf- PRACTICE BANK (CONT. ) Long-answer - gradient, Hessian, classification 大题––梯度(gradient)、海森矩阵(Hessian)、分类判定 (2: find every stationary point of a two-variable function and classify it L2:求二元函数的每一个驻点(stationary point)并加以分类 L2 LONG ANSWER 6 marks . multivariable calculus Let f(x, y) = x3 + y3 - 3xy. Find the gradient Vf and the Hessian H, locate all stationary points, and classify each using the second-derivative (Hessian) test. [6] 设 f(x, y)= x3 + y3 -3xy。求梯度 Vf 与海森矩阵 H,定位所有驻点,并用二阶导(海森)检验对每个分类。[6] L2 Worked solution - gradient + Hessian classification 1 Gradient. fx = 3x2 - 3y, fy = 3y2 - 3x. So Vf = (3x2 - 3y, 3y2 - 3x). 梯度。fx= 3x2-3y, fy = 3y2-3x。所以 Vf =(3x2 -3y,3y2-3x)。 2 Stationary points. Set both to zero: 3x2 = 3y -> y = x2; and 3y2 = 3x -> x = y2. Substitute: x = (x2)2 = x4 -+ x4 - x = x(x3 - 1) = 0 -+ x = 0 or x = 1. This gives (0, 0) and (1,1). 驻点。令两者均为零:3×2=3y→y=x2;且 3y2=3×→×=y2。代入:x=(x2)2= x4 →×4 - x= x(×3 -1)= 0 →x=0 或x=1。由此得(0,0)和(1,1)。 3 Hessian. fxx = 6x, fyy = 6y, fxy = - 3. H = 6x -3 6y -3 , det H = 36xy - 9. 海森矩阵(Hessian)。fxx = 6x, fyy = 6y, fxy =- 3。H= 6x -3 6y -3 det H = 36xy - 9. ， 4 Classify (0, 0). det H = 36(0)(0) - 9 = - 9 < 0 - saddle point. 分类(0,0)。det H = 36(0)(0)-9 =- 9<0→ 鞍点(saddle point)。 Classify (1, 1). det H = 36(1)(1) - 9 = 27 > 0 and fxx = 6 > 0-local minimum. (f(1,1) =1+1-3 =- 1. ) 分类(1,1)。det H = 36(1)(1)-9= 27>0 且 fxx= 6>0→局部最小值。(f(1,1)=1+1-3 =- 1。) - i The Hessian decision rule 海森矩阵(Hessian)判别法则 Compute det H = fxxfyy - fxy2. det H < 0 - saddle. det H > O -> extremum: fxx > O gives a local min, fxx < O a local max. det H = 0 - test is inconclusive. This table is on the exam formula sheet - but apply it cleanly per point. 计算 det H = fxxfyy -fxy2。det H <0→ 鞍点。det H > 0→ 极值: fxx> 0 给局部极小,fxx <0 给局部极大。det H =0→判别法无法判定。此表在考试公式表上 -- 但要逐点干净地套用。 - AskSia Library · MAT9004 · 双语 Bilingual Show every step. The 6-mark long-answers are graded on method: a stated gradient, the simultaneous solve, the Hessian, and a per-point verdict each earns marks - even a small arithmetic slip at the end keeps most of the credit. A bare final answer throws that away. 展示每一步。6分的大题是按方法评分的:写出梯度、联立求解、海森矩阵、对每个点的判定,各自都能得分 -- 即便 末尾有点小算术失误,也能保住大部分分数。光给最终答案则会把这些全丢掉。[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE。

6）Area 4：计数与组合（四选型 + 容斥 + 鸽巢）——最容易“选错类型”直接全错

6.1 三条基本法则（读动词）

乘法法则（AND）：分阶段选择，乘起来

加法法则（OR 且互斥）：分情况相加

补集法则（至少一个）：$|S_{\text{good}}|=|S|-|S_{\text{bad}}|$[15]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING - CHAPTER 4 . COUNTING & COMBINATORICS DISCRETE MATHS Counting without listing: rules, selections & coefficients 不靠列举来计数:法则、选取与系数 Product / sum / complement . the four selection types . binomial & inclusion-exclusion 乘法 / 加法/ 补集法则 · 四种选取类型 · 二项式(binomial) 与容斥原理(inclusion-exclusion) Combinatorics answers one question - "how many?" - without ever writing the list out. You break a count into stages and multiply (the product rule), split it into disjoint cases and add (the sum rule), or count the bad ones and subtract (the complement rule). The heart of the topic is deciding, for a selection of r things from n, whether order matters and whether repetition is allowed - four cases, four formulas. Get that decision right and the arithmetic is easy; get it wrong and every later mark is gone. 组合学回答一个问题 -- “有多少?” -- 而无须把列表写出。你把一个计数拆成若干阶段并相乘(乘法法则),拆成不相交 的情形并相加(加法法则),或数出坏的并相减(补集法则)。这个主题的核心是判定:在从 n个中选 r个时,顺序是否重要 以及是否允许重复 -- 四种情形,四个公式。这个判定做对,算术就容易;做错,后面每一分都没了。 ★ What the exam asks here 这里考试考什么 Counting is one fixed exam block. You will be asked to count arrangements and selections - k-digit numbers with digit rules, a committee of k from n, a password of length r from an n-symbol alphabet, buying k items from t types; to decide ordered-vs-unordered and repetition-or-not; to evaluate binomial coefficients C(n,r) exactly; and to apply inclusion-exclusion (|AuBI) and the pigeonhole principle. All by hand, all exact integer answers - leave them as products or factorials if the number is huge (e. g. 3110), do not reach for a decimal. 计数是固定的一个考试板块。会要求你数排列与选取- 一带数字规则的 k位数、从n 人中选 k人的委员会、从 n 符号 字母表取长度 r的密码、从t类中买k件;判断有序还是无序、可否重复;精确求二项式系数 C(n,r);并应用容斥原理 (|AUB)与鸽巢原理。全部手算,全是精确整数答案 -- 若数字巨大(如 3110)就留成乘积或阶乘形式,别去算小数。 4. 1 The three fundamental counting rules 4. 1 三条基本计数法则 Almost every counting problem is built from three primitives. The whole skill is reading the problem and matching its wording to one of them - the keywords "and", "or" and "not" are your signposts. 几乎每个计数问题都由三个基元构建而成。整套技能就是读懂问题并把它的措辞匹配到其中之一 -- 关键词“且(and)”、 “或(or)”与“非(not)”是你的路标。 - AskSia Library · MAT9004 · 双语 Bilingual THE THREE RULES Product rule (AND). A task done in stages with k1, k2, . . . , kn independent choices per stage: |s| = k1 x k2 x . . . x kn Sum rule (OR). Split into cases S1 , . . . , Sm : |s| = | S1 | + . . . + | Sml Complement rule. Easier to count the bad ones : | Sgood| = |s| - | Spadl Read the verb. "Pick a shirt and a tie" => multiply. "The number is even or a multiple of 5" => add (if the cases don't overlap). "At least one X" => complement: total minus none. 读动词。“挑一件衬衫和一条领带”⇒相乘。“这个数是偶数 或5 的倍数”⇒相加(若各情形不重叠)。“至少一个X”⇒ 补集:总数减去零个。 Order matter? YES NO Ordered Unordered no rep[17]Source: asksia-bible-mat9004-bilingual.pdfOR -+ + SUM RULE (DISJOINT) 加法原理(不相交) 2×2 SELECTION GRID CELLS 选取网格的单元格 [n/m] PIGEONHOLE BOUND 鸽巢原理下界 AskSia Library · MAT9004 · 双语 Bilingual i Recap - the counting playbook in seven lines 回顾 -- 计数攻略七行讲清 (1) Read the verb: "and" => multiply, disjoint "or" => add, "at least one" => complement. (2) For a selection, ask order? and repetition? - read the cell off the 2×2 grid: "Pr, n', C(n,r), C(n+r-1,r). (3) Combinations divide permutations by r !. (4) C(n,r) = n!/(r!(n-r)!); compute by cancelling and using symmetry C(n,r)=C(n,n-r). (5) Binomial theorem (x+y)" uses row n of Pascal; raise the whole bracket term. (6) Overlapping sets = inclusion-exclusion (+ singles - pairs + triple). (7) n items, m boxes => some box has ≥ [n/m]. Counting feeds Chapter 5: a uniform sample space has Pr(A) = |A|/ISI, and feature/hyperparameter grids are just product-rule counts. (1) 读动词:“且”⇒相乘,互斥的“或”⇒相加,“至少一个”⇒取补。(2)选取时问是否有序?与是否重复 ?-- 从 2x2 网格读出那一格:"Pr、n'、C(n,r)、C(n+r-1,r)。(3)组合是排列除以 r !。 (4) C(n,r)= n!/(r!(n-r)!);用约分和对称性 C(n,r)=C(n,n-r)计算。(5)二项式定理(x+y)”用帕斯卡三角第 n 行;整个括号项一起乘方。(6)集合重叠⇒ 容斥原理 (+单个 一成对+三重)。(7) n 个物品、m 个盒子⇒ 某盒至少[n/m]。计数喂给第5章:均匀样本空间满足 Pr(A)= IAI/IS],特征/超参数网格不过是乘法法则的计数。 AskSia Library · MAT9004 · 双语 Bilingual CH 5 . PROBABILITY - CHAPTER 5 . PROBABILITY & BAYES DISCRETE MATHS Probability: the language of uncertainty & of machine learning 概率:不确定性与机器学习的语言 Sample spaces . conditional probability . Bayes . expectation & variance . distributions 样本空间(sample space)· 条件概率(conditional probability)· 贝叶斯(Bayes)· 期望(expectation) & 方 差(variance)· 分布 Probability turns "I'm not sure" into a number. A probability space lists every outcome and how likely each is; an event is a set of outcomes, and its probability is just the weight of that set. From three short axioms the whole machinery follows - complements, unions, conditioning, and the crown jewel, Bayes' theorem, which flips a known cause-effect probability into the effect-cause one you actually want. This is exactly how a classifier reasons: prior belief, see evidence, update. Master the two-stage Bayes tree and the expectation/variance arithmetic and you own this topic. 概率把“我不确定”变成一个数。一个概率空间列出每个结果及其各自的可能性;一个事件是一组结果,其概率就是该集合的权 重。从三条简短公理出发,整套机制随之而来 -- 补集、并集、条件化,以及皇冠上的明珠 贝叶斯定理,它把已知的因→果 概率翻转为你真正想要的果→因概率。这正是分类器的推理方式:先验信念,看到证据,更新。掌握两阶段的贝叶斯树以及期 望/方差的运算,你便拥有了这个主题。 ★ What the exam asks here 这里考试考什么 (1) Compute a probability from a sample space (often equally-likely -+ pure counting, linking to Ch 4). (2) Use the complement / union rules and decide if events are mutually exclusive or independent. (3) A conditional probability Pr(A | B). (4) The guaranteed long-answer: a two-stage Bayes problem - law of total probability to get Pr(B), then reverse to Pr(A | B). (5) Expectation E[X] and variance Var(X), and E[aX + bY] by linearity. (6) Identify and use a named distribution (Bernoulli / binomial / geometric / uniform / exponential / normal). (7) Find a pdf's normalising constant so [pdf = 1. Everything by hand, in exact fractions. (1) 从样本空间算概率(常为等可能→纯计数,衔接第4章)。(2)用补/并法则,判断事件是互斥还是独立。(3)条件 概率 Pr(A | B)。(4)必考长答题:一个两阶段贝叶斯(Bayes)问题 -- 用全概率公式求 Pr(B),再反推 Pr(A| B)。(5) 期望 E[X] 与方差 Var(X),以及由线性性求 E[aX+bY]。(6)辨认并使用一个具名分布(伯努利/二项/几何/均匀/ 指数/正态)。(7)求概率密度函数的归一化常数使 Spdf =1。全部手算,用精确分数。 5. 1 Probability spaces, events & the axioms 5. 1 概率空间、事件 & 公理 A probability space is a sample space S (the set of all possible outcomes of a trial) together with a probability function Pr that assigns each outcome a weight in [0, 1], with all weights summing to 1. Each trial produces exactly one outcome. An event A & S is any subset of outcomes, and Pr(A) = ExEA Pr(x). 一个概率空间是一个样本空间(sample space)S (一次试验所有可能结果的集合)连同一个概率函数 Pr,后者为每个结果指 派一个 [0,1] 中的权重,且所有权重之和为1。每次试验产生恰好一个结果。一个事件 A s S 是结果的任意子集,且 Pr(A)= ΣΧΕΑ Pr(x). - AskSia Library · MAT9004 · 双语 Bilingual THE AXIOMS[20]Source: asksia-cheatsheet-mat9004.pdfPARAMETRISE Underdetermined (fewer equations than unknowns) = a row collapses to 0=0 => one free variable tER. Express each pivot variable in terms of t, then write the solution set as point + t·direction (a line) - e. g. (x,y,z,w) = (a,b,c,0) +t(p,q,r,1). State "for all tER "; two free variables = a plane. Two-variable products (Hadamard) exist but are NOT the matrix product used here - always use the row-column rule. (kA)B = k(AB). Worked 2x2 mult: [1,2;0,1] . [3;4] = [1. 3+2-4; 0-3+1-4] = [11;4]. Mult is defined only when columns of A = rows of B A vector is an m×1 matrix. Identity In acts as 1: Aln=A. If BA=I then AB=I, so each is the other's inverse. A(B+D)=AB+AD distributes. Compiled by AskSia . mapped to the MAT9004 syllabus . asksia. ai/cheatsheet/mat9004 CONTINUOUS . Single-variable calculus . Optimisation . Integration . Linear algebra . Eigenvalues . A=PDP-1 . Multivariable calculus REVISION SHEET . CLOSED-BOOK EXAM Independent revision aid . check learning. monash. edu for exam rules . @ 2026 flip + for side 2 . counting, probability & graphs APPLIED MAT9004 Mathematical Foundations for Data Science & AI MONASH UNIVERSITY . FACULTY OF SCIENCE EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Counting . probability . graphs SIDE 2/2 EULer/Hamilton 10 . Counting . Basic Rules L17 AREA 4 . PRODUCT & SUM RULES AND (sequence) = multiply: |S| = n kj OR (disjoint cases) - add: |S| = 2 complement: | good| = |S| - | bad| Keyword test: "and"/"then" =>x; "or"/disjoint cases => +. n! = 1. 2 . . . . . n; 0! = 1 . Sequences with kj choices each = |S|= nlkj. 11 . Four Selection Types AREA 4 . L17- 18 * Draw r from n - ask: order matter? repeats allowed?

关键词：and/then → 乘；disjoint or → 加；at least one → 补集[15]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING - CHAPTER 4 . COUNTING & COMBINATORICS DISCRETE MATHS Counting without listing: rules, selections & coefficients 不靠列举来计数:法则、选取与系数 Product / sum / complement . the four selection types . binomial & inclusion-exclusion 乘法 / 加法/ 补集法则 · 四种选取类型 · 二项式(binomial) 与容斥原理(inclusion-exclusion) Combinatorics answers one question - "how many?" - without ever writing the list out. You break a count into stages and multiply (the product rule), split it into disjoint cases and add (the sum rule), or count the bad ones and subtract (the complement rule). The heart of the topic is deciding, for a selection of r things from n, whether order matters and whether repetition is allowed - four cases, four formulas. Get that decision right and the arithmetic is easy; get it wrong and every later mark is gone. 组合学回答一个问题 -- “有多少?” -- 而无须把列表写出。你把一个计数拆成若干阶段并相乘(乘法法则),拆成不相交 的情形并相加(加法法则),或数出坏的并相减(补集法则)。这个主题的核心是判定:在从 n个中选 r个时,顺序是否重要 以及是否允许重复 -- 四种情形,四个公式。这个判定做对,算术就容易;做错,后面每一分都没了。 ★ What the exam asks here 这里考试考什么 Counting is one fixed exam block. You will be asked to count arrangements and selections - k-digit numbers with digit rules, a committee of k from n, a password of length r from an n-symbol alphabet, buying k items from t types; to decide ordered-vs-unordered and repetition-or-not; to evaluate binomial coefficients C(n,r) exactly; and to apply inclusion-exclusion (|AuBI) and the pigeonhole principle. All by hand, all exact integer answers - leave them as products or factorials if the number is huge (e. g. 3110), do not reach for a decimal. 计数是固定的一个考试板块。会要求你数排列与选取- 一带数字规则的 k位数、从n 人中选 k人的委员会、从 n 符号 字母表取长度 r的密码、从t类中买k件;判断有序还是无序、可否重复;精确求二项式系数 C(n,r);并应用容斥原理 (|AUB)与鸽巢原理。全部手算,全是精确整数答案 -- 若数字巨大(如 3110)就留成乘积或阶乘形式,别去算小数。 4. 1 The three fundamental counting rules 4. 1 三条基本计数法则 Almost every counting problem is built from three primitives. The whole skill is reading the problem and matching its wording to one of them - the keywords "and", "or" and "not" are your signposts. 几乎每个计数问题都由三个基元构建而成。整套技能就是读懂问题并把它的措辞匹配到其中之一 -- 关键词“且(and)”、 “或(or)”与“非(not)”是你的路标。 - AskSia Library · MAT9004 · 双语 Bilingual THE THREE RULES Product rule (AND). A task done in stages with k1, k2, . . . , kn independent choices per stage: |s| = k1 x k2 x . . . x kn Sum rule (OR). Split into cases S1 , . . . , Sm : |s| = | S1 | + . . . + | Sml Complement rule. Easier to count the bad ones : | Sgood| = |s| - | Spadl Read the verb. "Pick a shirt and a tie" => multiply. "The number is even or a multiple of 5" => add (if the cases don't overlap). "At least one X" => complement: total minus none. 读动词。“挑一件衬衫和一条领带”⇒相乘。“这个数是偶数 或5 的倍数”⇒相加(若各情形不重叠)。“至少一个X”⇒ 补集:总数减去零个。 Order matter? YES NO Ordered Unordered no rep[17]Source: asksia-bible-mat9004-bilingual.pdfOR -+ + SUM RULE (DISJOINT) 加法原理(不相交) 2×2 SELECTION GRID CELLS 选取网格的单元格 [n/m] PIGEONHOLE BOUND 鸽巢原理下界 AskSia Library · MAT9004 · 双语 Bilingual i Recap - the counting playbook in seven lines 回顾 -- 计数攻略七行讲清 (1) Read the verb: "and" => multiply, disjoint "or" => add, "at least one" => complement. (2) For a selection, ask order? and repetition? - read the cell off the 2×2 grid: "Pr, n', C(n,r), C(n+r-1,r). (3) Combinations divide permutations by r !. (4) C(n,r) = n!/(r!(n-r)!); compute by cancelling and using symmetry C(n,r)=C(n,n-r). (5) Binomial theorem (x+y)" uses row n of Pascal; raise the whole bracket term. (6) Overlapping sets = inclusion-exclusion (+ singles - pairs + triple). (7) n items, m boxes => some box has ≥ [n/m]. Counting feeds Chapter 5: a uniform sample space has Pr(A) = |A|/ISI, and feature/hyperparameter grids are just product-rule counts. (1) 读动词:“且”⇒相乘,互斥的“或”⇒相加,“至少一个”⇒取补。(2)选取时问是否有序?与是否重复 ?-- 从 2x2 网格读出那一格:"Pr、n'、C(n,r)、C(n+r-1,r)。(3)组合是排列除以 r !。 (4) C(n,r)= n!/(r!(n-r)!);用约分和对称性 C(n,r)=C(n,n-r)计算。(5)二项式定理(x+y)”用帕斯卡三角第 n 行;整个括号项一起乘方。(6)集合重叠⇒ 容斥原理 (+单个 一成对+三重)。(7) n 个物品、m 个盒子⇒ 某盒至少[n/m]。计数喂给第5章:均匀样本空间满足 Pr(A)= IAI/IS],特征/超参数网格不过是乘法法则的计数。 AskSia Library · MAT9004 · 双语 Bilingual CH 5 . PROBABILITY - CHAPTER 5 . PROBABILITY & BAYES DISCRETE MATHS Probability: the language of uncertainty & of machine learning 概率:不确定性与机器学习的语言 Sample spaces . conditional probability . Bayes . expectation & variance . distributions 样本空间(sample space)· 条件概率(conditional probability)· 贝叶斯(Bayes)· 期望(expectation) & 方 差(variance)· 分布 Probability turns "I'm not sure" into a number. A probability space lists every outcome and how likely each is; an event is a set of outcomes, and its probability is just the weight of that set. From three short axioms the whole machinery follows - complements, unions, conditioning, and the crown jewel, Bayes' theorem, which flips a known cause-effect probability into the effect-cause one you actually want. This is exactly how a classifier reasons: prior belief, see evidence, update. Master the two-stage Bayes tree and the expectation/variance arithmetic and you own this topic. 概率把“我不确定”变成一个数。一个概率空间列出每个结果及其各自的可能性;一个事件是一组结果,其概率就是该集合的权 重。从三条简短公理出发,整套机制随之而来 -- 补集、并集、条件化,以及皇冠上的明珠 贝叶斯定理,它把已知的因→果 概率翻转为你真正想要的果→因概率。这正是分类器的推理方式:先验信念,看到证据,更新。掌握两阶段的贝叶斯树以及期 望/方差的运算,你便拥有了这个主题。 ★ What the exam asks here 这里考试考什么 (1) Compute a probability from a sample space (often equally-likely -+ pure counting, linking to Ch 4). (2) Use the complement / union rules and decide if events are mutually exclusive or independent. (3) A conditional probability Pr(A | B). (4) The guaranteed long-answer: a two-stage Bayes problem - law of total probability to get Pr(B), then reverse to Pr(A | B). (5) Expectation E[X] and variance Var(X), and E[aX + bY] by linearity. (6) Identify and use a named distribution (Bernoulli / binomial / geometric / uniform / exponential / normal). (7) Find a pdf's normalising constant so [pdf = 1. Everything by hand, in exact fractions. (1) 从样本空间算概率(常为等可能→纯计数,衔接第4章)。(2)用补/并法则,判断事件是互斥还是独立。(3)条件 概率 Pr(A | B)。(4)必考长答题:一个两阶段贝叶斯(Bayes)问题 -- 用全概率公式求 Pr(B),再反推 Pr(A| B)。(5) 期望 E[X] 与方差 Var(X),以及由线性性求 E[aX+bY]。(6)辨认并使用一个具名分布(伯努利/二项/几何/均匀/ 指数/正态)。(7)求概率密度函数的归一化常数使 Spdf =1。全部手算,用精确分数。 5. 1 Probability spaces, events & the axioms 5. 1 概率空间、事件 & 公理 A probability space is a sample space S (the set of all possible outcomes of a trial) together with a probability function Pr that assigns each outcome a weight in [0, 1], with all weights summing to 1. Each trial produces exactly one outcome. An event A & S is any subset of outcomes, and Pr(A) = ExEA Pr(x). 一个概率空间是一个样本空间(sample space)S (一次试验所有可能结果的集合)连同一个概率函数 Pr,后者为每个结果指 派一个 [0,1] 中的权重,且所有权重之和为1。每次试验产生恰好一个结果。一个事件 A s S 是结果的任意子集,且 Pr(A)= ΣΧΕΑ Pr(x). - AskSia Library · MAT9004 · 双语 Bilingual THE AXIOMS

6.2 四种选取类型（考试头号坑）

下笔前先问两句：顺序要紧吗？允许重复吗？

• “排列/序列/排名”通常是有序；“委员会/子集/手牌”通常无序[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。

宝典明确：这是计数最大丢分点——把 $C(n,r)$、$nP r$、以及允许重复的情况混了，方法就全崩[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。[15]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING - CHAPTER 4 . COUNTING & COMBINATORICS DISCRETE MATHS Counting without listing: rules, selections & coefficients 不靠列举来计数:法则、选取与系数 Product / sum / complement . the four selection types . binomial & inclusion-exclusion 乘法 / 加法/ 补集法则 · 四种选取类型 · 二项式(binomial) 与容斥原理(inclusion-exclusion) Combinatorics answers one question - "how many?" - without ever writing the list out. You break a count into stages and multiply (the product rule), split it into disjoint cases and add (the sum rule), or count the bad ones and subtract (the complement rule). The heart of the topic is deciding, for a selection of r things from n, whether order matters and whether repetition is allowed - four cases, four formulas. Get that decision right and the arithmetic is easy; get it wrong and every later mark is gone. 组合学回答一个问题 -- “有多少?” -- 而无须把列表写出。你把一个计数拆成若干阶段并相乘(乘法法则),拆成不相交 的情形并相加(加法法则),或数出坏的并相减(补集法则)。这个主题的核心是判定:在从 n个中选 r个时,顺序是否重要 以及是否允许重复 -- 四种情形,四个公式。这个判定做对,算术就容易;做错,后面每一分都没了。 ★ What the exam asks here 这里考试考什么 Counting is one fixed exam block. You will be asked to count arrangements and selections - k-digit numbers with digit rules, a committee of k from n, a password of length r from an n-symbol alphabet, buying k items from t types; to decide ordered-vs-unordered and repetition-or-not; to evaluate binomial coefficients C(n,r) exactly; and to apply inclusion-exclusion (|AuBI) and the pigeonhole principle. All by hand, all exact integer answers - leave them as products or factorials if the number is huge (e. g. 3110), do not reach for a decimal. 计数是固定的一个考试板块。会要求你数排列与选取- 一带数字规则的 k位数、从n 人中选 k人的委员会、从 n 符号 字母表取长度 r的密码、从t类中买k件;判断有序还是无序、可否重复;精确求二项式系数 C(n,r);并应用容斥原理 (|AUB)与鸽巢原理。全部手算,全是精确整数答案 -- 若数字巨大(如 3110)就留成乘积或阶乘形式,别去算小数。 4. 1 The three fundamental counting rules 4. 1 三条基本计数法则 Almost every counting problem is built from three primitives. The whole skill is reading the problem and matching its wording to one of them - the keywords "and", "or" and "not" are your signposts. 几乎每个计数问题都由三个基元构建而成。整套技能就是读懂问题并把它的措辞匹配到其中之一 -- 关键词“且(and)”、 “或(or)”与“非(not)”是你的路标。 - AskSia Library · MAT9004 · 双语 Bilingual THE THREE RULES Product rule (AND). A task done in stages with k1, k2, . . . , kn independent choices per stage: |s| = k1 x k2 x . . . x kn Sum rule (OR). Split into cases S1 , . . . , Sm : |s| = | S1 | + . . . + | Sml Complement rule. Easier to count the bad ones : | Sgood| = |s| - | Spadl Read the verb. "Pick a shirt and a tie" => multiply. "The number is even or a multiple of 5" => add (if the cases don't overlap). "At least one X" => complement: total minus none. 读动词。“挑一件衬衫和一条领带”⇒相乘。“这个数是偶数 或5 的倍数”⇒相加(若各情形不重叠)。“至少一个X”⇒ 补集:总数减去零个。 Order matter? YES NO Ordered Unordered no rep。

6.3 容斥原理（inclusion-exclusion）

两集合：$$|A\cup B|=|A|+|B|-|A\cap B|$$

三集合：$$|A\cup B\cup C|=\sum |单个|-\sum |两两交|+|三者交|$$[22]Source: asksia-cheatsheet-mat9004.pdf13 . Inclusion-Exclusion AREA 4 . & Pigeonhole L18 TWO / THREE SETS |AUB | = |A |+|B| - | AnB| | AuBuC] = Σ singles - Σ pairs + | AnBnC | None-of-N (general I-E): | U| - Σ singles + E pairs - Σ triples + . . . + (-1)"(all-N). Same structure as Pr(AuB). PIGEONHOLE n items, m boxes, n > m = some box ≥ 2 generalised - some box ≥ [n/m] Use: "must two share . . . ? " or "can all be distinct?" - e. g. 10 people, degrees in {0, . . . ,9}: 0 and 9 can't coexist > two share a friend-count. Worked I-E: |U|=100, |A|=30, |B|=40, |AnB|=10 = satisfying neither = 100-30-40+10 = 40. (Same add- subtract pattern as Pr(AuB) - the two topics share one identity. ) Three-set "none": | U|-ΣΙΑ,|+ΣΙΑ,ΠΑ; |-/Α, ΛΑΖΠΑ3]. Venn diagrams make the signs obvious. The general rule alternates sign: +singles? no - subtract singles, add pairs, subtract triples . . . 14 . Probability Foundations L19 AREA 5 . Sample space S; Pr:S->[0,1], E Pr(s)=1. Event ACS, Pr(Α)=Σ_{χεΑ}Pr(x). CORE RULES uniform: Pr(A) = |A|/|S| (pure counting) complement: Pr (Ā) = 1 - Pr(A) union: Pr(AuB)=Pr(A)+Pr (B) -Pr (AnB) independent: Pr (AnB)=Pr (A) . Pr (B) Mutually exclusive = Pr(AnB)=0 = Pr(AUB)=Pr(A)+Pr(B). Independent repeated trials: Pr((S1,S2))=Pr(S1)Pr(S2). Worked: Pr(A)=0. 3, Pr(B)=0. 4, Pr(AuB)=0. 5 = Pr(AnB)=0. 3+0. 4-0. 5=0. 2; since 0. 20. 3. 0. 4=0. 12, A and B are not independent. Each trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes

记忆：符号交替（+单个 −成对 +三重 −…）[22]Source: asksia-cheatsheet-mat9004.pdf13 . Inclusion-Exclusion AREA 4 . & Pigeonhole L18 TWO / THREE SETS |AUB | = |A |+|B| - | AnB| | AuBuC] = Σ singles - Σ pairs + | AnBnC | None-of-N (general I-E): | U| - Σ singles + E pairs - Σ triples + . . . + (-1)"(all-N). Same structure as Pr(AuB). PIGEONHOLE n items, m boxes, n > m = some box ≥ 2 generalised - some box ≥ [n/m] Use: "must two share . . . ? " or "can all be distinct?" - e. g. 10 people, degrees in {0, . . . ,9}: 0 and 9 can't coexist > two share a friend-count. Worked I-E: |U|=100, |A|=30, |B|=40, |AnB|=10 = satisfying neither = 100-30-40+10 = 40. (Same add- subtract pattern as Pr(AuB) - the two topics share one identity. ) Three-set "none": | U|-ΣΙΑ,|+ΣΙΑ,ΠΑ; |-/Α, ΛΑΖΠΑ3]. Venn diagrams make the signs obvious. The general rule alternates sign: +singles? no - subtract singles, add pairs, subtract triples . . . 14 . Probability Foundations L19 AREA 5 . Sample space S; Pr:S->[0,1], E Pr(s)=1. Event ACS, Pr(Α)=Σ_{χεΑ}Pr(x). CORE RULES uniform: Pr(A) = |A|/|S| (pure counting) complement: Pr (Ā) = 1 - Pr(A) union: Pr(AuB)=Pr(A)+Pr (B) -Pr (AnB) independent: Pr (AnB)=Pr (A) . Pr (B) Mutually exclusive = Pr(AnB)=0 = Pr(AUB)=Pr(A)+Pr(B). Independent repeated trials: Pr((S1,S2))=Pr(S1)Pr(S2). Worked: Pr(A)=0. 3, Pr(B)=0. 4, Pr(AuB)=0. 5 = Pr(AnB)=0. 3+0. 4-0. 5=0. 2; since 0. 20. 3. 0. 4=0. 12, A and B are not independent. Each trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes。

6.4 鸽巢原理（pigeonhole）

$n$ 个物品放进 $m$ 个盒子（$n>m$）⇒ 至少有一个盒子 $\ge 2$

一般化下界：至少有一个盒子 $\ge \left\lceil \frac{n}{m}\right\rceil$[22]Source: asksia-cheatsheet-mat9004.pdf13 . Inclusion-Exclusion AREA 4 . & Pigeonhole L18 TWO / THREE SETS |AUB | = |A |+|B| - | AnB| | AuBuC] = Σ singles - Σ pairs + | AnBnC | None-of-N (general I-E): | U| - Σ singles + E pairs - Σ triples + . . . + (-1)"(all-N). Same structure as Pr(AuB). PIGEONHOLE n items, m boxes, n > m = some box ≥ 2 generalised - some box ≥ [n/m] Use: "must two share . . . ? " or "can all be distinct?" - e. g. 10 people, degrees in {0, . . . ,9}: 0 and 9 can't coexist > two share a friend-count. Worked I-E: |U|=100, |A|=30, |B|=40, |AnB|=10 = satisfying neither = 100-30-40+10 = 40. (Same add- subtract pattern as Pr(AuB) - the two topics share one identity. ) Three-set "none": | U|-ΣΙΑ,|+ΣΙΑ,ΠΑ; |-/Α, ΛΑΖΠΑ3]. Venn diagrams make the signs obvious. The general rule alternates sign: +singles? no - subtract singles, add pairs, subtract triples . . . 14 . Probability Foundations L19 AREA 5 . Sample space S; Pr:S->[0,1], E Pr(s)=1. Event ACS, Pr(Α)=Σ_{χεΑ}Pr(x). CORE RULES uniform: Pr(A) = |A|/|S| (pure counting) complement: Pr (Ā) = 1 - Pr(A) union: Pr(AuB)=Pr(A)+Pr (B) -Pr (AnB) independent: Pr (AnB)=Pr (A) . Pr (B) Mutually exclusive = Pr(AnB)=0 = Pr(AUB)=Pr(A)+Pr(B). Independent repeated trials: Pr((S1,S2))=Pr(S1)Pr(S2). Worked: Pr(A)=0. 3, Pr(B)=0. 4, Pr(AuB)=0. 5 = Pr(AnB)=0. 3+0. 4-0. 5=0. 2; since 0. 20. 3. 0. 4=0. 12, A and B are not independent. Each trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE

考点：它不考算术，考你“谁是物品、谁是盒子”[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE。

7）Area 5：概率与贝叶斯（两阶段 Bayes 大题是“招牌套路”）

7.1 基础定义（你要能写出来）

样本空间 $S$，事件 $A\subseteq S$；概率函数把每个结果赋权重并总和为 1；

• $$\Pr(A)=\sum_{x\in A}\Pr(x)$$[17]Source: asksia-bible-mat9004-bilingual.pdfOR -+ + SUM RULE (DISJOINT) 加法原理(不相交) 2×2 SELECTION GRID CELLS 选取网格的单元格 [n/m] PIGEONHOLE BOUND 鸽巢原理下界 AskSia Library · MAT9004 · 双语 Bilingual i Recap - the counting playbook in seven lines 回顾 -- 计数攻略七行讲清 (1) Read the verb: "and" => multiply, disjoint "or" => add, "at least one" => complement. (2) For a selection, ask order? and repetition? - read the cell off the 2×2 grid: "Pr, n', C(n,r), C(n+r-1,r). (3) Combinations divide permutations by r !. (4) C(n,r) = n!/(r!(n-r)!); compute by cancelling and using symmetry C(n,r)=C(n,n-r). (5) Binomial theorem (x+y)" uses row n of Pascal; raise the whole bracket term. (6) Overlapping sets = inclusion-exclusion (+ singles - pairs + triple). (7) n items, m boxes => some box has ≥ [n/m]. Counting feeds Chapter 5: a uniform sample space has Pr(A) = |A|/ISI, and feature/hyperparameter grids are just product-rule counts. (1) 读动词:“且”⇒相乘,互斥的“或”⇒相加,“至少一个”⇒取补。(2)选取时问是否有序?与是否重复 ?-- 从 2x2 网格读出那一格:"Pr、n'、C(n,r)、C(n+r-1,r)。(3)组合是排列除以 r !。 (4) C(n,r)= n!/(r!(n-r)!);用约分和对称性 C(n,r)=C(n,n-r)计算。(5)二项式定理(x+y)”用帕斯卡三角第 n 行;整个括号项一起乘方。(6)集合重叠⇒ 容斥原理 (+单个 一成对+三重)。(7) n 个物品、m 个盒子⇒ 某盒至少[n/m]。计数喂给第5章:均匀样本空间满足 Pr(A)= IAI/IS],特征/超参数网格不过是乘法法则的计数。 AskSia Library · MAT9004 · 双语 Bilingual CH 5 . PROBABILITY - CHAPTER 5 . PROBABILITY & BAYES DISCRETE MATHS Probability: the language of uncertainty & of machine learning 概率:不确定性与机器学习的语言 Sample spaces . conditional probability . Bayes . expectation & variance . distributions 样本空间(sample space)· 条件概率(conditional probability)· 贝叶斯(Bayes)· 期望(expectation) & 方 差(variance)· 分布 Probability turns "I'm not sure" into a number. A probability space lists every outcome and how likely each is; an event is a set of outcomes, and its probability is just the weight of that set. From three short axioms the whole machinery follows - complements, unions, conditioning, and the crown jewel, Bayes' theorem, which flips a known cause-effect probability into the effect-cause one you actually want. This is exactly how a classifier reasons: prior belief, see evidence, update. Master the two-stage Bayes tree and the expectation/variance arithmetic and you own this topic. 概率把“我不确定”变成一个数。一个概率空间列出每个结果及其各自的可能性;一个事件是一组结果,其概率就是该集合的权 重。从三条简短公理出发,整套机制随之而来 -- 补集、并集、条件化,以及皇冠上的明珠 贝叶斯定理,它把已知的因→果 概率翻转为你真正想要的果→因概率。这正是分类器的推理方式:先验信念,看到证据,更新。掌握两阶段的贝叶斯树以及期 望/方差的运算,你便拥有了这个主题。 ★ What the exam asks here 这里考试考什么 (1) Compute a probability from a sample space (often equally-likely -+ pure counting, linking to Ch 4). (2) Use the complement / union rules and decide if events are mutually exclusive or independent. (3) A conditional probability Pr(A | B). (4) The guaranteed long-answer: a two-stage Bayes problem - law of total probability to get Pr(B), then reverse to Pr(A | B). (5) Expectation E[X] and variance Var(X), and E[aX + bY] by linearity. (6) Identify and use a named distribution (Bernoulli / binomial / geometric / uniform / exponential / normal). (7) Find a pdf's normalising constant so [pdf = 1. Everything by hand, in exact fractions. (1) 从样本空间算概率(常为等可能→纯计数,衔接第4章)。(2)用补/并法则,判断事件是互斥还是独立。(3)条件 概率 Pr(A | B)。(4)必考长答题:一个两阶段贝叶斯(Bayes)问题 -- 用全概率公式求 Pr(B),再反推 Pr(A| B)。(5) 期望 E[X] 与方差 Var(X),以及由线性性求 E[aX+bY]。(6)辨认并使用一个具名分布(伯努利/二项/几何/均匀/ 指数/正态)。(7)求概率密度函数的归一化常数使 Spdf =1。全部手算,用精确分数。 5. 1 Probability spaces, events & the axioms 5. 1 概率空间、事件 & 公理 A probability space is a sample space S (the set of all possible outcomes of a trial) together with a probability function Pr that assigns each outcome a weight in [0, 1], with all weights summing to 1. Each trial produces exactly one outcome. An event A & S is any subset of outcomes, and Pr(A) = ExEA Pr(x). 一个概率空间是一个样本空间(sample space)S (一次试验所有可能结果的集合)连同一个概率函数 Pr,后者为每个结果指 派一个 [0,1] 中的权重,且所有权重之和为1。每次试验产生恰好一个结果。一个事件 A s S 是结果的任意子集,且 Pr(A)= ΣΧΕΑ Pr(x). - AskSia Library · MAT9004 · 双语 Bilingual THE AXIOMS[22]Source: asksia-cheatsheet-mat9004.pdf13 . Inclusion-Exclusion AREA 4 . & Pigeonhole L18 TWO / THREE SETS |AUB | = |A |+|B| - | AnB| | AuBuC] = Σ singles - Σ pairs + | AnBnC | None-of-N (general I-E): | U| - Σ singles + E pairs - Σ triples + . . . + (-1)"(all-N). Same structure as Pr(AuB). PIGEONHOLE n items, m boxes, n > m = some box ≥ 2 generalised - some box ≥ [n/m] Use: "must two share . . . ? " or "can all be distinct?" - e. g. 10 people, degrees in {0, . . . ,9}: 0 and 9 can't coexist > two share a friend-count. Worked I-E: |U|=100, |A|=30, |B|=40, |AnB|=10 = satisfying neither = 100-30-40+10 = 40. (Same add- subtract pattern as Pr(AuB) - the two topics share one identity. ) Three-set "none": | U|-ΣΙΑ,|+ΣΙΑ,ΠΑ; |-/Α, ΛΑΖΠΑ3]. Venn diagrams make the signs obvious. The general rule alternates sign: +singles? no - subtract singles, add pairs, subtract triples . . . 14 . Probability Foundations L19 AREA 5 . Sample space S; Pr:S->[0,1], E Pr(s)=1. Event ACS, Pr(Α)=Σ_{χεΑ}Pr(x). CORE RULES uniform: Pr(A) = |A|/|S| (pure counting) complement: Pr (Ā) = 1 - Pr(A) union: Pr(AuB)=Pr(A)+Pr (B) -Pr (AnB) independent: Pr (AnB)=Pr (A) . Pr (B) Mutually exclusive = Pr(AnB)=0 = Pr(AUB)=Pr(A)+Pr(B). Independent repeated trials: Pr((S1,S2))=Pr(S1)Pr(S2). Worked: Pr(A)=0. 3, Pr(B)=0. 4, Pr(AuB)=0. 5 = Pr(AnB)=0. 3+0. 4-0. 5=0. 2; since 0. 20. 3. 0. 4=0. 12, A and B are not independent. Each trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes

7.2 条件概率、全概率、贝叶斯（必会）

条件概率：$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$$[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions[22]Source: asksia-cheatsheet-mat9004.pdf13 . Inclusion-Exclusion AREA 4 . & Pigeonhole L18 TWO / THREE SETS |AUB | = |A |+|B| - | AnB| | AuBuC] = Σ singles - Σ pairs + | AnBnC | None-of-N (general I-E): | U| - Σ singles + E pairs - Σ triples + . . . + (-1)"(all-N). Same structure as Pr(AuB). PIGEONHOLE n items, m boxes, n > m = some box ≥ 2 generalised - some box ≥ [n/m] Use: "must two share . . . ? " or "can all be distinct?" - e. g. 10 people, degrees in {0, . . . ,9}: 0 and 9 can't coexist > two share a friend-count. Worked I-E: |U|=100, |A|=30, |B|=40, |AnB|=10 = satisfying neither = 100-30-40+10 = 40. (Same add- subtract pattern as Pr(AuB) - the two topics share one identity. ) Three-set "none": | U|-ΣΙΑ,|+ΣΙΑ,ΠΑ; |-/Α, ΛΑΖΠΑ3]. Venn diagrams make the signs obvious. The general rule alternates sign: +singles? no - subtract singles, add pairs, subtract triples . . . 14 . Probability Foundations L19 AREA 5 . Sample space S; Pr:S->[0,1], E Pr(s)=1. Event ACS, Pr(Α)=Σ_{χεΑ}Pr(x). CORE RULES uniform: Pr(A) = |A|/|S| (pure counting) complement: Pr (Ā) = 1 - Pr(A) union: Pr(AuB)=Pr(A)+Pr (B) -Pr (AnB) independent: Pr (AnB)=Pr (A) . Pr (B) Mutually exclusive = Pr(AnB)=0 = Pr(AUB)=Pr(A)+Pr(B). Independent repeated trials: Pr((S1,S2))=Pr(S1)Pr(S2). Worked: Pr(A)=0. 3, Pr(B)=0. 4, Pr(AuB)=0. 5 = Pr(AnB)=0. 3+0. 4-0. 5=0. 2; since 0. 20. 3. 0. 4=0. 12, A and B are not independent. Each trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes

乘法公式：$\Pr(A\cap B)=\Pr(A|B)\Pr(B)$[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

全概率（典型二分支）：

• $$\Pr(B)=\Pr(B|A)\Pr(A)+\Pr(B|\bar A)\Pr(\bar A)$$[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

贝叶斯：

• $$\Pr(A|B)=\frac{\Pr(B|A)\Pr(A)}{\Pr(B)}$$[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

7.3 两阶段 Bayes（6 分大题套路）

标准配方（强烈建议画树）：

• 第 1 步：用全概率把分母 $\Pr(\text{evidence})$ 算出来
• 第 2 步：用 Bayes 反推后验（某一分支贡献 ÷ 总证据）[7]Source: asksia-bible-mat9004-bilingual.pdfCH 8 . PRACTICE L5 - PRACTICE BANK (CONT. ) Long-answer - two-stage Bayes 大题 -- 两阶段贝叶斯(Bayes) (5: total probability, then the reverse conditional - the signature probability set-piece L5:全概率公式,再求逆向条件概率––招牌式的概率大题套路 L5 LONG ANSWER 6 marks . probability A factory uses two machines. Machine A makes 70% of the parts and 4% of A's parts are defective. Machine B makes the remaining 30% and 10% of B's parts are defective. 一家工厂使用两台机器。机器 A生产70% 的零件,其中A的零件有4% 有缺陷。机器 B 生产其余的30%,其中 B 的零件有10% 有缺陷。 (a) Find the probability a randomly chosen part is defective. [2] (b) Given a part is defective, find the probability it came from machine B. [2] (c) Given a part is good, find the probability it came from machine A. [2] (a)求随机选取的一个零件有缺陷的概率。[2](b)已知一个零件有缺陷,求它来自机器B 的概率。[2] (c)已知一个零件 合格,求它来自机器A的概率。[2] L5 Worked solution - two-stage Bayes 1 Set up. P(A) = 7/10, P(B) = 3/10. Defect rates P(D|A) = 4/100 = 1/25, P(D|B) = 10/100 = 1/10. 设置。P(A)= 7/10,P(B)= 3/10。次品率 P(DIA)= 4/100 = 1/25, P(DB) = 10/100 = 1/10。 (a) Total probability. P(D) = P(A)P(D|A) + P(B)P(D|B) = (7/10)(1/25) + (3/10)(1/10) = 7/250 + 3/100. Common denominator 500: 14/500 + 15/500 = 29/500 (= 0. 058). (a) 全概率。P(D) = P(A)P(DIA) + P(B)P(DB) = (7/10)(1/25) + (3/10)(1/10) = 7/250 + 3/100。通分到 500: 14/500 + 15/500 = 29/500 (= 0. 058) . 3 (b) Reverse conditional, Bayes. P(B|D) = P(B)P(D|B) / P(D) = (3/100) / (29/500) = (15/500) / (29/500) = 15/29. (b)反向条件,贝叶斯。P(BID)= P(B)P(DB) / P(D) =(3/100) /(29/500)=(15/500) /(29/500)= 15/29。 4 (c) Good part from A. P(good) = 1 - 29/500 = 471/500. P(A n good) = P(A)P(good|A) = (7/10)(24/25) = 168/250 = 336/500. P(A|good) = (336/500) / (471/500) = 336/471 = 112/157 (divide top and bottom by 3). (c) 来自 A的合格件。P(合格)=1-29/500= 471/500。P(A ∩ 合格)= P(A)P(合格|A) = (7/10)(24/25) = 168/250 = 336/500。P(AI合格)=(336/500)/(471/500)= 336/471=112/157(分子分母同除以 3)。 AskSia Library · MAT9004 · 双语 Bilingual ✓ The two-step Bayes recipe 两步贝叶斯配方 Step 1 - total probability: sum (branch prob x conditional) over all branches to get P(evidence). Step 2 - reverse: the wanted posterior is (one branch's contribution) + (that total). A tree diagram with the branch probabilities written on each edge makes both steps almost automatic. 第1步––全概率:对所有分支求和(分支概率 ×条件概率)得 P(证据)。第2步 -- 反推:所求后验为(某一分支的 贡献)÷(该总和)。在每条边上写好分支概率的树状图能让两步几乎自动完成。 Same habit across every topic: substitute your answer back. Re-multiply P. D. P-1, plug t into all three equations, confirm the posteriors over each evidence sum to 1. The closed-book exam gives no calculator - that single back-check is where you catch the arithmetic slip before the marker does. 每个主题都要养成同一个习惯:把你的答案代回去验证。重新算 P· D· P-1,把t 代入全部三个方程,确认各证据下的后 验之和为1。闭卷考试不给计算器 -- 这一次回代检查正是你在评卷人之前抓住算术失误的地方。 MARKER'S NOTE . MAT9004 FINAL AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

考试陷阱 3：基率谬误（base-rate fallacy）

• 稀有事件时 $\Pr(\text{cause}|\text{effect})$ 往往远小于 $\Pr(\text{effect}|\text{cause})$，千万别当成一样；分母一定要完整展开[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。。

7.4 期望与方差（含连续型）

线性性：$E[aX+bY]=aE[X]+bE[Y]$（常考）[14]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q26-Q31 - PRACTICE BANK (CONT. ) Probability - expectation, coin game, pdf constant, continuous E/Var 概率––期望(expectation)、拋硬币游戏、概率密度函数常数、连续型 E/Var Q26-Q28 and Q31: the discrete and continuous expectation slots Q26-Q28 与 Q31:离散与连续型期望(expectation)的题位 Q26 E [AX + BY] 3 marks . Probability X and Y are independent fair-die rolls (each uniform on 1-6). Find E[2X + Y]. 027 COIN GAME 3 marks . Probability A fair coin is tossed twice. You win $18 for two heads, $4 for exactly one head, and lose $12 for no heads. Find the expected winnings. Q28 PDF CONSTANT 2 marks . Probability For what constant k is f(x) = k. x(2 - x) on (0, 2) a valid probability density? Q31 CONTINUOUS E & VAR 3 marks . Probability A continuous RV has density f(x) = (3/8)x2 on (0, 2). Find E[X] and Var(X). Q26-Q28, Q31 Worked solutions - probability 1 Q26. A fair die has E[X] = (1+2+3+4+5+6)/6 = 7/2. By linearity E[2X + Y] = 2. (7/2) + 7/2 = 7 + 7/2 = 21/2. Q26。公平骰子有 E[X]=(1+2+3+4+5+6)/6= 7/2。由线性性 E[2X+Y]= 2·(7/2)+7/2=7+7/2=21/2。 Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4. (1/2) + (-12)-(1/4) = 18/4 + 2 - 12/4 = 9/2 + 2 - 3 = 9/2 - 1 = 7/2 (i. e. $3. 50). Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4·(1/2) + (-12) · (1/4) = 18/4 + 2 -12/4 =9/2 +2 -3=9/2-1=7/2(即 $3. 50)。 3 Q28. RequireJo2 k·x(2-x) dx = 1. Jo2(2x - x2) dx = [x2 - x3/3]2 = 4 - 8/3 = 4/3. So k·(4/3) = 1 + k = 3/4. Q28。要求So2k·×(2-x)dx=1. So2(2x-x2) dx=[x2-x3/3]02=4-8/3=4/3。所以 k·(4/3)=1→k= 3/4. 4 Q31. E[X] =[2 x. (3/8)x2 dx = (3/8)[x^/4][2=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5] 2=(3/8)(32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. Q31. E[X] = S02 x·(3/8)x2 dx = (3/8)[x4/4]02=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5]02=(3/8) (32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. - AskSia Library . MAT9004 . XXia Bilingual ! Variance subtracts the SQUARE of the mean, not the mean 方差(variance)减去的是均值的平方,而非均值本身 Var(X) = E[X2] - (E[X])2. In Q31 that is 12/5 - (3/2)2 = 12/5 - 9/4 = 3/20 - subtracting E[X] (not its square) gives the wrong, and often negative, number. Var(X) = E[×2] - (E[X])2。Q31 中即 12/5 -(3/2)2=12/5-9/4= 3/20 -- 减去 E[X](而非其平方)会得到错误、 且常为负的数。 AskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q29-L1 - PRACTICE BANK (CONT. ) Graphs - and the first long-answer[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

方差核心公式：

• $$\mathrm{Var}(X)=E[X^2]-(E[X])^2$$
• 重点陷阱：是“减均值的平方”，不是减均值本身[14]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q26-Q31 - PRACTICE BANK (CONT. ) Probability - expectation, coin game, pdf constant, continuous E/Var 概率––期望(expectation)、拋硬币游戏、概率密度函数常数、连续型 E/Var Q26-Q28 and Q31: the discrete and continuous expectation slots Q26-Q28 与 Q31:离散与连续型期望(expectation)的题位 Q26 E [AX + BY] 3 marks . Probability X and Y are independent fair-die rolls (each uniform on 1-6). Find E[2X + Y]. 027 COIN GAME 3 marks . Probability A fair coin is tossed twice. You win $18 for two heads, $4 for exactly one head, and lose $12 for no heads. Find the expected winnings. Q28 PDF CONSTANT 2 marks . Probability For what constant k is f(x) = k. x(2 - x) on (0, 2) a valid probability density? Q31 CONTINUOUS E & VAR 3 marks . Probability A continuous RV has density f(x) = (3/8)x2 on (0, 2). Find E[X] and Var(X). Q26-Q28, Q31 Worked solutions - probability 1 Q26. A fair die has E[X] = (1+2+3+4+5+6)/6 = 7/2. By linearity E[2X + Y] = 2. (7/2) + 7/2 = 7 + 7/2 = 21/2. Q26。公平骰子有 E[X]=(1+2+3+4+5+6)/6= 7/2。由线性性 E[2X+Y]= 2·(7/2)+7/2=7+7/2=21/2。 Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4. (1/2) + (-12)-(1/4) = 18/4 + 2 - 12/4 = 9/2 + 2 - 3 = 9/2 - 1 = 7/2 (i. e. $3. 50). Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4·(1/2) + (-12) · (1/4) = 18/4 + 2 -12/4 =9/2 +2 -3=9/2-1=7/2(即 $3. 50)。 3 Q28. RequireJo2 k·x(2-x) dx = 1. Jo2(2x - x2) dx = [x2 - x3/3]2 = 4 - 8/3 = 4/3. So k·(4/3) = 1 + k = 3/4. Q28。要求So2k·×(2-x)dx=1. So2(2x-x2) dx=[x2-x3/3]02=4-8/3=4/3。所以 k·(4/3)=1→k= 3/4. 4 Q31. E[X] =[2 x. (3/8)x2 dx = (3/8)[x^/4][2=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5] 2=(3/8)(32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. Q31. E[X] = S02 x·(3/8)x2 dx = (3/8)[x4/4]02=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5]02=(3/8) (32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. - AskSia Library . MAT9004 . XXia Bilingual ! Variance subtracts the SQUARE of the mean, not the mean 方差(variance)减去的是均值的平方,而非均值本身 Var(X) = E[X2] - (E[X])2. In Q31 that is 12/5 - (3/2)2 = 12/5 - 9/4 = 3/20 - subtracting E[X] (not its square) gives the wrong, and often negative, number. Var(X) = E[×2] - (E[X])2。Q31 中即 12/5 -(3/2)2=12/5-9/4= 3/20 -- 减去 E[X](而非其平方)会得到错误、 且常为负的数。 AskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q29-L1 - PRACTICE BANK (CONT. ) Graphs - and the first long-answer[21]Source: asksia-cheatsheet-mat9004.pdfEach trial yields exactly one outcome; the unit's probability is discrete apart from the continuous distributions in L22. 15 . Conditional & Bayes ★ AREA 5 . L20 CONDITIONAL & MULTIPLICATION Pr(A|B) = Pr(AnB)/Pr(B) - Pr (AnB) = Pr(A|B) . Pr (B) TOTAL PROBABILITY & BAYES Pr(B) = Pr(B|A)Pr(A) + Pr(B|Ã)Pr(Ā) Pr(A|B) = Pr(B|A)Pr(A) / Pr(B) Independence + Pr(A)=Pr(A|B). Use a tree diagram; watch the base-rate fallacy (rare-disease tests). -- SIA > The long-answer Bayes is two steps: total probability first (the denominator), then divide. Write both fractions - method marks survive an arithmetic slip. 15b . Worked . Two- stage Bayes RENUMBERED Supplier A: 20% of parts, 20% late. B: 80%, 5% late. Pr(late)=0. 2-0. 2+0. 8-0. 05 = 0. 04+0. 04 = 0. 08. Pr(A|late)=0. 04/0. 08 = 1/2 . Commuter variant: walks 10% (late 60%), drives 90% (late 20%). Pr(walk|late)=0. 06/(0. 06+0. 18)=1/4; Pr(drove|on-time)=0. 72/0. 76=18/19. 16 . Random AREA 5 . L21 ★ Variables RV X: S->R. Distribution = list Pr(X=x). Independent · Pr(X=x,Y=y)=Pr(X=x)Pr(Y=y) for all x,y. EXPECTATION & VARIANCE E[X] = E pixi (weighted average) Var [X] = E[ (X-p)2] = E[X2] - p2 0 = VVar [X] ALGEBRA E[X+Y]=E[X]+E[Y] (even if dependent!) E[kX]=KE[X] . Var[kX]=k2Var[X] indep: E[XY]=E[X]E[Y], Var [X+Y]=Var [X]+Var[Y] Linearity-of-expectation (indicator trick): [[Σ X ] =Σ E[X] works even when the X, are dependent - e. g. expected #adjacent-odd-pairs in a 4-digit PIN = 3-(1/4) = 3/4. 16b . Worked . E & Var RENUMBERED Die X (1-6): E=3. 5. Coin game: 2H->+24, 1H->+2, OH->-8 => E =24. 1/4+2. 1/2-8-1/4 = 6+1-2 = 5. Two indep uniforms: Var(2X+3Y)=4Var(X)+9Var(Y). XE{3,7,11} probs 1/4,1/4,1/2 = E[X]=8. Var uses E[X2]-u2, not E[X]2. 17 . Distributions

连续型 pdf 必考点：

• 合法 pdf：$f(x)\ge 0$ 且 $\int f(x),dx=1$（先用它求归一化常数 $k$）[30]Source: asksia-cheatsheet-mat9004.pdfBinomial(n,p) np np(1-p) Geometric(p) (1-p)/p (1-p)/p Unif {a . . b} (a+b)/2 ((b-a+1)2-1)/12 Unif [a,b] (a+b)/2 (b-a)2/12 Expon(A) 1/2 1/72 Normal(μ,σ) μ Binomial Pr(X=k)=C(n,k)pk(1-p)^(n-k); geometric Pr(X=k)=p(1-p)k (k failures first). Z=(X-u)/o = standard normal score. Recognise the model: single success/fail = Bernoulli; #successes in n trials = Binomial; #failures before first success => Geometric; sums/large counts => tend to Normal (CLT intuition). 17b . Continuous . PDF NORMALISE! PDF REQUIREMENTS f ≥ 0 everywhere AND [ f over domain = 1 Pr(a≤Xsb)=[ab f dx . Pr(X=a)=0 E[X]={ x f dx; Var={ x2f dx - (E[X])2. Find k: h=kx(1-x) on (0,1) = [=k/6=1 = k=6 . Variant kx2(1-x) = k=12. Always set /f=1 first, then compute moments. Worked E,Var: h(x)=(x-1)2 on (0,2): [ h=1 V, E[X]=[x h=1, Var=[x2h-1=1/5. (Integration is the engine for continuous distributions - Area 1 feeding Area 5. ) Continuous uniform on [a,b]: f=1/(b-a). Exponential(A) f=Ne^(-Xx), x≥0-memoryless, the Poisson waiting time. ≤ and < are interchangeable since Pr(X=a)=0.[14]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q26-Q31 - PRACTICE BANK (CONT. ) Probability - expectation, coin game, pdf constant, continuous E/Var 概率––期望(expectation)、拋硬币游戏、概率密度函数常数、连续型 E/Var Q26-Q28 and Q31: the discrete and continuous expectation slots Q26-Q28 与 Q31:离散与连续型期望(expectation)的题位 Q26 E [AX + BY] 3 marks . Probability X and Y are independent fair-die rolls (each uniform on 1-6). Find E[2X + Y]. 027 COIN GAME 3 marks . Probability A fair coin is tossed twice. You win $18 for two heads, $4 for exactly one head, and lose $12 for no heads. Find the expected winnings. Q28 PDF CONSTANT 2 marks . Probability For what constant k is f(x) = k. x(2 - x) on (0, 2) a valid probability density? Q31 CONTINUOUS E & VAR 3 marks . Probability A continuous RV has density f(x) = (3/8)x2 on (0, 2). Find E[X] and Var(X). Q26-Q28, Q31 Worked solutions - probability 1 Q26. A fair die has E[X] = (1+2+3+4+5+6)/6 = 7/2. By linearity E[2X + Y] = 2. (7/2) + 7/2 = 7 + 7/2 = 21/2. Q26。公平骰子有 E[X]=(1+2+3+4+5+6)/6= 7/2。由线性性 E[2X+Y]= 2·(7/2)+7/2=7+7/2=21/2。 Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4. (1/2) + (-12)-(1/4) = 18/4 + 2 - 12/4 = 9/2 + 2 - 3 = 9/2 - 1 = 7/2 (i. e. $3. 50). Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4·(1/2) + (-12) · (1/4) = 18/4 + 2 -12/4 =9/2 +2 -3=9/2-1=7/2(即 $3. 50)。 3 Q28. RequireJo2 k·x(2-x) dx = 1. Jo2(2x - x2) dx = [x2 - x3/3]2 = 4 - 8/3 = 4/3. So k·(4/3) = 1 + k = 3/4. Q28。要求So2k·×(2-x)dx=1. So2(2x-x2) dx=[x2-x3/3]02=4-8/3=4/3。所以 k·(4/3)=1→k= 3/4. 4 Q31. E[X] =[2 x. (3/8)x2 dx = (3/8)[x^/4][2=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5] 2=(3/8)(32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. Q31. E[X] = S02 x·(3/8)x2 dx = (3/8)[x4/4]02=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5]02=(3/8) (32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. - AskSia Library . MAT9004 . XXia Bilingual ! Variance subtracts the SQUARE of the mean, not the mean 方差(variance)减去的是均值的平方,而非均值本身 Var(X) = E[X2] - (E[X])2. In Q31 that is 12/5 - (3/2)2 = 12/5 - 9/4 = 3/20 - subtracting E[X] (not its square) gives the wrong, and often negative, number. Var(X) = E[×2] - (E[X])2。Q31 中即 12/5 -(3/2)2=12/5-9/4= 3/20 -- 减去 E[X](而非其平方)会得到错误、 且常为负的数。 AskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q29-L1 - PRACTICE BANK (CONT. ) Graphs - and the first long-answer
• 然后
  • $$E[X]=\int x f(x),dx,\quad E[X^2]=\int x^2 f(x),dx,\quad \mathrm{Var}=E[X^2]-(E[X])^2$$[14]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q26-Q31 - PRACTICE BANK (CONT. ) Probability - expectation, coin game, pdf constant, continuous E/Var 概率––期望(expectation)、拋硬币游戏、概率密度函数常数、连续型 E/Var Q26-Q28 and Q31: the discrete and continuous expectation slots Q26-Q28 与 Q31:离散与连续型期望(expectation)的题位 Q26 E [AX + BY] 3 marks . Probability X and Y are independent fair-die rolls (each uniform on 1-6). Find E[2X + Y]. 027 COIN GAME 3 marks . Probability A fair coin is tossed twice. You win $18 for two heads, $4 for exactly one head, and lose $12 for no heads. Find the expected winnings. Q28 PDF CONSTANT 2 marks . Probability For what constant k is f(x) = k. x(2 - x) on (0, 2) a valid probability density? Q31 CONTINUOUS E & VAR 3 marks . Probability A continuous RV has density f(x) = (3/8)x2 on (0, 2). Find E[X] and Var(X). Q26-Q28, Q31 Worked solutions - probability 1 Q26. A fair die has E[X] = (1+2+3+4+5+6)/6 = 7/2. By linearity E[2X + Y] = 2. (7/2) + 7/2 = 7 + 7/2 = 21/2. Q26。公平骰子有 E[X]=(1+2+3+4+5+6)/6= 7/2。由线性性 E[2X+Y]= 2·(7/2)+7/2=7+7/2=21/2。 Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4. (1/2) + (-12)-(1/4) = 18/4 + 2 - 12/4 = 9/2 + 2 - 3 = 9/2 - 1 = 7/2 (i. e. $3. 50). Q27. P(2H) = 1/4, P(1H) = 1/2, P(OH) = 1/4. E = 18·(1/4) + 4·(1/2) + (-12) · (1/4) = 18/4 + 2 -12/4 =9/2 +2 -3=9/2-1=7/2(即 $3. 50)。 3 Q28. RequireJo2 k·x(2-x) dx = 1. Jo2(2x - x2) dx = [x2 - x3/3]2 = 4 - 8/3 = 4/3. So k·(4/3) = 1 + k = 3/4. Q28。要求So2k·×(2-x)dx=1. So2(2x-x2) dx=[x2-x3/3]02=4-8/3=4/3。所以 k·(4/3)=1→k= 3/4. 4 Q31. E[X] =[2 x. (3/8)x2 dx = (3/8)[x^/4][2=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5] 2=(3/8)(32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. Q31. E[X] = S02 x·(3/8)x2 dx = (3/8)[x4/4]02=(3/8)(16/4) =(3/8)(4) = 3/2. E[X2] = (3/8)[x5/5]02=(3/8) (32/5) = 12/5. Var = 12/5 - (3/2)2 = 48/20 - 45/20 = 3/20. - AskSia Library . MAT9004 . XXia Bilingual ! Variance subtracts the SQUARE of the mean, not the mean 方差(variance)减去的是均值的平方,而非均值本身 Var(X) = E[X2] - (E[X])2. In Q31 that is 12/5 - (3/2)2 = 12/5 - 9/4 = 3/20 - subtracting E[X] (not its square) gives the wrong, and often negative, number. Var(X) = E[×2] - (E[X])2。Q31 中即 12/5 -(3/2)2=12/5-9/4= 3/20 -- 减去 E[X](而非其平方)会得到错误、 且常为负的数。 AskSia Library · MAT9004 · 双语 Bilingual CH 8 . PRACTICE Q29-L1 - PRACTICE BANK (CONT. ) Graphs - and the first long-answer[30]Source: asksia-cheatsheet-mat9004.pdfBinomial(n,p) np np(1-p) Geometric(p) (1-p)/p (1-p)/p Unif {a . . b} (a+b)/2 ((b-a+1)2-1)/12 Unif [a,b] (a+b)/2 (b-a)2/12 Expon(A) 1/2 1/72 Normal(μ,σ) μ Binomial Pr(X=k)=C(n,k)pk(1-p)^(n-k); geometric Pr(X=k)=p(1-p)k (k failures first). Z=(X-u)/o = standard normal score. Recognise the model: single success/fail = Bernoulli; #successes in n trials = Binomial; #failures before first success => Geometric; sums/large counts => tend to Normal (CLT intuition). 17b . Continuous . PDF NORMALISE! PDF REQUIREMENTS f ≥ 0 everywhere AND [ f over domain = 1 Pr(a≤Xsb)=[ab f dx . Pr(X=a)=0 E[X]={ x f dx; Var={ x2f dx - (E[X])2. Find k: h=kx(1-x) on (0,1) = [=k/6=1 = k=6 . Variant kx2(1-x) = k=12. Always set /f=1 first, then compute moments. Worked E,Var: h(x)=(x-1)2 on (0,2): [ h=1 V, E[X]=[x h=1, Var=[x2h-1=1/5. (Integration is the engine for continuous distributions - Area 1 feeding Area 5. ) Continuous uniform on [a,b]: f=1/(b-a). Exponential(A) f=Ne^(-Xx), x≥0-memoryless, the Poisson waiting time. ≤ and < are interchangeable since Pr(X=a)=0.

8）Area 6：图与树（Euler/Hamilton、握手引理、邻接矩阵）

8.1 必会：握手引理（handshaking）

度数和 = $2|E|$，所以

• $$|E|=\frac{\sum \deg(v)}{2}$$

推论：奇度顶点个数一定是偶数（常用于“图是否可能存在”的判断）[9]Source: asksia-bible-mat9004-bilingual.pdf1 Count odd-degree vertices. Degrees 3 and 3 are odd - that is two odd vertices; the rest (2,2,4) are even. - 数奇数度顶点。度数 3 和 3 为奇数 -- 即两个奇数 度顶点;其余(2,2,4)为偶数。 2 Euler trail? Exactly 2 odd-degree vertices and connected = yes, an Euler trail exists (it must start at one odd vertex and end at the other). 欧拉迹(Euler trail)?恰有 2 个奇数度顶点且连通 ⇒是,存在欧拉迹(它必从一个奇数度顶点出发,终 于另一个)。 3 Euler circuit? A circuit needs all degrees even - here two are odd, so no Euler circuit. 欧拉回路?回路要求所有度数为偶 -- 此处有两个为 奇,故不存在欧拉回路。 ! Euler counts edges, Hamilton counts vertices 欧拉数边,哈密顿数顶点 The classic mix-up. Euler = every edge once, decided instantly by counting odd-degree vertices (0 - circuit, 2 - trail, >2 - neither). Hamilton = every vertex once, with no simple degree test - the degree-2-n/2 rule is only sufficient, never a way to rule a graph out. If asked to disprove Hamiltonicity you must argue by hand, not by a formula. 经典混淆。欧拉= 每条边一次,靠数奇度顶点立即判 定(0→回路,2→迹,>2→都不存在)。哈密顿= 每个顶点一次,没有简单的度判据 -- 度≥n/2 的规 则仅是充分条件,绝不能用来排除一个图。若要求你 证伪哈密顿性,必须手工论证,而非套公式。 AskSia Library · MAT9004 · 双语 Bilingual A E C D numbered edges 1 . . m red = odd degree (exactly 2) B i Recap - Graphs & trees in six lines 回顾 -- 六行讲清图与树 (1) Graph = vertices + edges; simple = no loops/parallel; degree = incident edges. (2) Handshaking: Edeg = 2|El, so |E| = Edeg/2; odd sum = impossible; odd-degree vertices come in pairs. (3) Walk - path - cycle; connected = 1 component. (4) Tree = connected + acyclic => exactly n-1 edges; every connected graph has a spanning tree. (5) Adjacency A symmetric (undirected); |E| = (#1's)/2; (Ak) ¡¡ = #walks of length k. (6) Euler (edges) needs 0/2 odd- degree vertices; Hamilton (vertices) has no easy test. Kn has n(n-1)/2 edges, Km,n has mn. (1) 图 = 顶点+边;简单图=无自环/无重边;度= 关联的边数。(2)握手引理:zdeg = 2|티,故 |티| = zdeg/2; 和为 奇⇒ 不可能;奇度顶点成对出现。(3)途径→路径→圈;连通=1个分量。(4)树= 连通+无圈⇒恰 n-1 条边; 每个连通图都有生成树(spanning tree)。(5)邻接矩阵 A 对称(无向);||=(1的个数)/2;(Ak)j = 长度为 k的途径 数。(6)欧拉(边)需0/2个奇度顶点;哈密顿(顶点)无简单判据。Kn 有 n(n-1)/2条边,Km,n 有 mn 条边。 AskSia Library · MAT9004 · 双语 Bilingual AskSia Library EXAM BIBLE . ASKSIA FACULTY OF IT / SCIENCE SEMESTER 1 . 2026 L 0 3 THE COMPLETE EXAM BIBLE Mathematical Foundations forDataScience&AI 数据科学与人工智能数学基础 FIVE MATHS WORLDS - CALCULUS, ALGEBRA, OPTIMISATION, COUNTING, PROBABILITY, GRAPHS- IN ONE CLOSED-BOOK EXAM. 五大数学世界 · 一场闭卷大考 · 60%且设及格门槛[23]Source: asksia-cheatsheet-mat9004.pdfstat pts + endpoints "max/min/saddle f(x,y)" Vf=0 then det H "solve the system" Gaussian elim "not invertible" set det = 0 "A=PDP-1 / A"" det(A-MI)=0 "how many ways" order? repeat? = cell "committee / choose" C(n,r) "given . . . find Pr(-|-)" total prob + Bayes "E[-] / Var(-) from pdf" normalise, [ xf, fx2f "# edges from degrees" handshaking /2 "walks of length k" (A); "can graph exist?"

8.2 Euler vs Hamilton（最经典混淆）

Euler：每条边恰好一次；用“奇度顶点个数”秒判：

• 0 个奇度 ⇒ Euler circuit
• 2 个奇度 ⇒ Euler trail

• 2 个奇度 ⇒ 都不存在[9]Source: asksia-bible-mat9004-bilingual.pdf1 Count odd-degree vertices. Degrees 3 and 3 are odd - that is two odd vertices; the rest (2,2,4) are even. - 数奇数度顶点。度数 3 和 3 为奇数 -- 即两个奇数 度顶点;其余(2,2,4)为偶数。 2 Euler trail? Exactly 2 odd-degree vertices and connected = yes, an Euler trail exists (it must start at one odd vertex and end at the other). 欧拉迹(Euler trail)?恰有 2 个奇数度顶点且连通 ⇒是,存在欧拉迹(它必从一个奇数度顶点出发,终 于另一个)。 3 Euler circuit? A circuit needs all degrees even - here two are odd, so no Euler circuit. 欧拉回路?回路要求所有度数为偶 -- 此处有两个为 奇,故不存在欧拉回路。 ! Euler counts edges, Hamilton counts vertices 欧拉数边,哈密顿数顶点 The classic mix-up. Euler = every edge once, decided instantly by counting odd-degree vertices (0 - circuit, 2 - trail, >2 - neither). Hamilton = every vertex once, with no simple degree test - the degree-2-n/2 rule is only sufficient, never a way to rule a graph out. If asked to disprove Hamiltonicity you must argue by hand, not by a formula. 经典混淆。欧拉= 每条边一次,靠数奇度顶点立即判 定(0→回路,2→迹,>2→都不存在)。哈密顿= 每个顶点一次,没有简单的度判据 -- 度≥n/2 的规 则仅是充分条件,绝不能用来排除一个图。若要求你 证伪哈密顿性,必须手工论证,而非套公式。 AskSia Library · MAT9004 · 双语 Bilingual A E C D numbered edges 1 . . m red = odd degree (exactly 2) B i Recap - Graphs & trees in six lines 回顾 -- 六行讲清图与树 (1) Graph = vertices + edges; simple = no loops/parallel; degree = incident edges. (2) Handshaking: Edeg = 2|El, so |E| = Edeg/2; odd sum = impossible; odd-degree vertices come in pairs. (3) Walk - path - cycle; connected = 1 component. (4) Tree = connected + acyclic => exactly n-1 edges; every connected graph has a spanning tree. (5) Adjacency A symmetric (undirected); |E| = (#1's)/2; (Ak) ¡¡ = #walks of length k. (6) Euler (edges) needs 0/2 odd- degree vertices; Hamilton (vertices) has no easy test. Kn has n(n-1)/2 edges, Km,n has mn. (1) 图 = 顶点+边;简单图=无自环/无重边;度= 关联的边数。(2)握手引理:zdeg = 2|티,故 |티| = zdeg/2; 和为 奇⇒ 不可能;奇度顶点成对出现。(3)途径→路径→圈;连通=1个分量。(4)树= 连通+无圈⇒恰 n-1 条边; 每个连通图都有生成树(spanning tree)。(5)邻接矩阵 A 对称(无向);||=(1的个数)/2;(Ak)j = 长度为 k的途径 数。(6)欧拉(边)需0/2个奇度顶点;哈密顿(顶点)无简单判据。Kn 有 n(n-1)/2条边,Km,n 有 mn 条边。 AskSia Library · MAT9004 · 双语 Bilingual AskSia Library EXAM BIBLE . ASKSIA FACULTY OF IT / SCIENCE SEMESTER 1 . 2026 L 0 3 THE COMPLETE EXAM BIBLE Mathematical Foundations forDataScience&AI 数据科学与人工智能数学基础 FIVE MATHS WORLDS - CALCULUS, ALGEBRA, OPTIMISATION, COUNTING, PROBABILITY, GRAPHS- IN ONE CLOSED-BOOK EXAM. 五大数学世界 · 一场闭卷大考 · 60%且设及格门槛

Hamilton：每个顶点恰好一次；没有简单的“充分必要”度判据（度 $\ge n/2$ 只是充分条件，不可用来排除）[9]Source: asksia-bible-mat9004-bilingual.pdf1 Count odd-degree vertices. Degrees 3 and 3 are odd - that is two odd vertices; the rest (2,2,4) are even. - 数奇数度顶点。度数 3 和 3 为奇数 -- 即两个奇数 度顶点;其余(2,2,4)为偶数。 2 Euler trail? Exactly 2 odd-degree vertices and connected = yes, an Euler trail exists (it must start at one odd vertex and end at the other). 欧拉迹(Euler trail)?恰有 2 个奇数度顶点且连通 ⇒是,存在欧拉迹(它必从一个奇数度顶点出发,终 于另一个)。 3 Euler circuit? A circuit needs all degrees even - here two are odd, so no Euler circuit. 欧拉回路?回路要求所有度数为偶 -- 此处有两个为 奇,故不存在欧拉回路。 ! Euler counts edges, Hamilton counts vertices 欧拉数边,哈密顿数顶点 The classic mix-up. Euler = every edge once, decided instantly by counting odd-degree vertices (0 - circuit, 2 - trail, >2 - neither). Hamilton = every vertex once, with no simple degree test - the degree-2-n/2 rule is only sufficient, never a way to rule a graph out. If asked to disprove Hamiltonicity you must argue by hand, not by a formula. 经典混淆。欧拉= 每条边一次,靠数奇度顶点立即判 定(0→回路,2→迹,>2→都不存在)。哈密顿= 每个顶点一次,没有简单的度判据 -- 度≥n/2 的规 则仅是充分条件,绝不能用来排除一个图。若要求你 证伪哈密顿性,必须手工论证,而非套公式。 AskSia Library · MAT9004 · 双语 Bilingual A E C D numbered edges 1 . . m red = odd degree (exactly 2) B i Recap - Graphs & trees in six lines 回顾 -- 六行讲清图与树 (1) Graph = vertices + edges; simple = no loops/parallel; degree = incident edges. (2) Handshaking: Edeg = 2|El, so |E| = Edeg/2; odd sum = impossible; odd-degree vertices come in pairs. (3) Walk - path - cycle; connected = 1 component. (4) Tree = connected + acyclic => exactly n-1 edges; every connected graph has a spanning tree. (5) Adjacency A symmetric (undirected); |E| = (#1's)/2; (Ak) ¡¡ = #walks of length k. (6) Euler (edges) needs 0/2 odd- degree vertices; Hamilton (vertices) has no easy test. Kn has n(n-1)/2 edges, Km,n has mn. (1) 图 = 顶点+边;简单图=无自环/无重边;度= 关联的边数。(2)握手引理:zdeg = 2|티,故 |티| = zdeg/2; 和为 奇⇒ 不可能;奇度顶点成对出现。(3)途径→路径→圈;连通=1个分量。(4)树= 连通+无圈⇒恰 n-1 条边; 每个连通图都有生成树(spanning tree)。(5)邻接矩阵 A 对称(无向);||=(1的个数)/2;(Ak)j = 长度为 k的途径 数。(6)欧拉(边)需0/2个奇度顶点;哈密顿(顶点)无简单判据。Kn 有 n(n-1)/2条边,Km,n 有 mn 条边。 AskSia Library · MAT9004 · 双语 Bilingual AskSia Library EXAM BIBLE . ASKSIA FACULTY OF IT / SCIENCE SEMESTER 1 . 2026 L 0 3 THE COMPLETE EXAM BIBLE Mathematical Foundations forDataScience&AI 数据科学与人工智能数学基础 FIVE MATHS WORLDS - CALCULUS, ALGEBRA, OPTIMISATION, COUNTING, PROBABILITY, GRAPHS- IN ONE CLOSED-BOOK EXAM. 五大数学世界 · 一场闭卷大考 · 60%且设及格门槛

9）考场策略：怎样把 45% 门槛稳稳跨过去（同时冲高分）

• 大题按步骤给分；即便最终数字小错，方法清晰仍能拿分；只写最终答案会把步骤分全丢[10]Source: asksia-bible-mat9004-bilingual.pdf- PRACTICE BANK (CONT. ) Long-answer - gradient, Hessian, classification 大题––梯度(gradient)、海森矩阵(Hessian)、分类判定 (2: find every stationary point of a two-variable function and classify it L2:求二元函数的每一个驻点(stationary point)并加以分类 L2 LONG ANSWER 6 marks . multivariable calculus Let f(x, y) = x3 + y3 - 3xy. Find the gradient Vf and the Hessian H, locate all stationary points, and classify each using the second-derivative (Hessian) test. [6] 设 f(x, y)= x3 + y3 -3xy。求梯度 Vf 与海森矩阵 H,定位所有驻点,并用二阶导(海森)检验对每个分类。[6] L2 Worked solution - gradient + Hessian classification 1 Gradient. fx = 3x2 - 3y, fy = 3y2 - 3x. So Vf = (3x2 - 3y, 3y2 - 3x). 梯度。fx= 3x2-3y, fy = 3y2-3x。所以 Vf =(3x2 -3y,3y2-3x)。 2 Stationary points. Set both to zero: 3x2 = 3y -> y = x2; and 3y2 = 3x -> x = y2. Substitute: x = (x2)2 = x4 -+ x4 - x = x(x3 - 1) = 0 -+ x = 0 or x = 1. This gives (0, 0) and (1,1). 驻点。令两者均为零:3×2=3y→y=x2;且 3y2=3×→×=y2。代入:x=(x2)2= x4 →×4 - x= x(×3 -1)= 0 →x=0 或x=1。由此得(0,0)和(1,1)。 3 Hessian. fxx = 6x, fyy = 6y, fxy = - 3. H = 6x -3 6y -3 , det H = 36xy - 9. 海森矩阵(Hessian)。fxx = 6x, fyy = 6y, fxy =- 3。H= 6x -3 6y -3 det H = 36xy - 9. ， 4 Classify (0, 0). det H = 36(0)(0) - 9 = - 9 < 0 - saddle point. 分类(0,0)。det H = 36(0)(0)-9 =- 9<0→ 鞍点(saddle point)。 Classify (1, 1). det H = 36(1)(1) - 9 = 27 > 0 and fxx = 6 > 0-local minimum. (f(1,1) =1+1-3 =- 1. ) 分类(1,1)。det H = 36(1)(1)-9= 27>0 且 fxx= 6>0→局部最小值。(f(1,1)=1+1-3 =- 1。) - i The Hessian decision rule 海森矩阵(Hessian)判别法则 Compute det H = fxxfyy - fxy2. det H < 0 - saddle. det H > O -> extremum: fxx > O gives a local min, fxx < O a local max. det H = 0 - test is inconclusive. This table is on the exam formula sheet - but apply it cleanly per point. 计算 det H = fxxfyy -fxy2。det H <0→ 鞍点。det H > 0→ 极值: fxx> 0 给局部极小,fxx <0 给局部极大。det H =0→判别法无法判定。此表在考试公式表上 -- 但要逐点干净地套用。 - AskSia Library · MAT9004 · 双语 Bilingual Show every step. The 6-mark long-answers are graded on method: a stated gradient, the simultaneous solve, the Hessian, and a per-point verdict each earns marks - even a small arithmetic slip at the end keeps most of the credit. A bare final answer throws that away. 展示每一步。6分的大题是按方法评分的:写出梯度、联立求解、海森矩阵、对每个点的判定,各自都能得分 -- 即便 末尾有点小算术失误,也能保住大部分分数。光给最终答案则会把这些全丢掉。[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE
• 每个 $Ax=b$、每个“根/驻点”尽量回代验证，抓符号/算术错误[5]Source: asksia-bible-mat9004-bilingual.pdfEuler (edges): O odd-deg - circuit, 2 - trail. Hamilton (vertices): no easy 6 test - deg≥n/2 is only sufficient AskSia Library · MAT9004 · 双语 Bilingual CH 9 . EXAM MORNING - CHAPTER 9 . EXAM MORNING (CONT. ) FINAL 60% . HURDLE Traps, timing, and the last word 陷阱、时间分配与最后的叮嘱 Where marks leak, how to spend 3h10m, and why every line earns 分数在哪里流失、如何分配 3小时10分钟,以及为何每一行都有分 9. 2 The four traps that cost the most 9. 2 代价最大的四个陷阱 ! 1 . Selection-type confusion 1 · 选择类型混淆 The single biggest counting error. Before you write a number, answer two questions: does order matter? and is repetition allowed? "Arrangements / sequences / ranked" => ordered; "committee / subset / hand" => unordered. Mixing up C(n,r) with n!/(n-r)! or n' turns a right method into a wrong answer. 头号计数错误。下笔写数字前,先回答两个问题:次 序要紧吗?以及允许重复吗?“排列/序列/排名”⇒ 有序;“委员会/子集/手牌”⇒无序。把C(n,r)与 n!/(n-r)!或 n'弄混,会把对的方法变成错的答案。 ! 3 . Forgetting the endpoints 3 · 遗漏端点 On a closed interval [c,d] the global max/min can sit at an endpoint, not at a stationary point. Candidates are stationary pts, singular pts and c, d - evaluate f at all of them and compare. Same discipline in 2 variables: check the boundary of the region, not just where Vf = O. 在闭区间 [c,d] 上,全局极大/极小可能位于端点而非 驻点。候选点为驻点、奇异点以及c、d -- 在它们全 部处求f 值并比较。二维同样讲究:检查区域的边 界,而不仅是 Vf=0 处。 ! 2 . Sign errors in row reduction 2 · 行化简中的符号错误 Gaussian elimination dies by minus signs. Only the three legal row ops (swap, scale by a nonzero, add a multiple of one row to another) - never two at once. Carry the b column through every step and back- substitute your solution into the original system to catch a flipped sign before the marker does. 高斯消元死于负号。只用三种合法行操作(交换、乘 非零数、把某行的倍数加到另一行) -- 绝不一次做 两个。每步都带着 b 列,并把解回代进原方程组,以 在批改老师之前抓到翻错的符号。 ! 4 . Base-rate fallacy in Bayes 4 · 贝叶斯中的基率谬误 When the effect is rare, P(cause | effect) is usually far smaller than P(effect | cause) - the rare-disease / zombie-test trap. Do not equate the two. Expand the denominator in full: Pr(B) = Pr(B|A)Pr(A) + Pr(B| Å)Pr(Å), and the prior Pr(A) is doing the heavy lifting. 当效应稀少时,P(原因| 效应)通常远小于 P(效应| 原 因) -- 罕见病/僵尸检测陷阱。不要把两者等同。把 分母完整展开:Pr(B)= Pr(B|A)Pr(A) + Pr(B) - A)Pr(A),而先验 Pr(A)起着决定性作用。 9. 3 A timing plan for 3h10m + the upload window 9. 3 3 小时 10分钟+上传窗口的时间规划 1 First 10 min - triage, no calculator panic. Read the whole paper. The formula sheet is on page 2 - locate it now. Tag each question by topic island and difficulty; you do not have to answer in order. 前 10 分钟 -- 分诊,别因没有计算器而慌乱。通读整张卷子。公式表(formula sheet)在第 2页 -- 现在就找到它。按 主题板块和难度给每道题打标签;你不必按顺序作答。[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE

• 约 120 分钟：先扫短答（能立刻匹配触发表就先拿下），答案必须是最简分数/整数、无空格、不写小数[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE
• 约 50 分钟：做手写大题；每一步一行，扫描时方法一眼可见；即便做不完也先写出“触发表第一步”去拿方法分[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE
• 最后 10 分钟：检查约分、检查空白框、回代验证；“猜一个约分分数也比空白强”[12]Source: asksia-bible-mat9004-bilingual.pdfAskSia Library · MAT9004 · 双语 Bilingual 2 ~120 min - short-answer sweep. These return one rational number. Bank the ones you recognise from the trigger table first. Write the answer as an integer or lowest-terms fraction a/b, no spaces, no decimals - reduce before you submit. 约 120 分钟 -- 简答题扫荡。这些题返回一个有理数。先把你从触发表中认得的题攒入囊中。把答案写成整数或最简分数 a/b,不留空格、不用小数 -- 提交前先约分。 - 3 ~50 min - hand-written responses. The longer, working-shown questions. Lay out one line per logical step on your A4 so the method is unmistakable when scanned. Start each at its trigger-row method even if you cannot finish it. 约50分钟 -- 手写解答题。这些是更长、需展示过程的题。在你的A4 纸上每个逻辑步骤占一行,使方法在扫描时清晰无 误。即便做不完,也要从其触发行的方法开始着手每道题。 4 Last ~10 min - sweep for cheap marks. Back-substitute every Ax=b and root; confirm fractions are reduced and spaceless; make sure no short-answer box is empty - a guessed reduced fraction beats a blank. 最后约 10 分钟 -- 扫荡捡分。对每个 Ax=b 与每个根回代验证;确认分数已约分且无空格;确保没有简答框是空的 -- 猜 一个约分分数也胜过留白。 5 +30 min after the exam - the upload. The handwritten work is scanned and uploaded by phone within 30 minutes of the exam ending. Photograph every page, right-side-up and legible, and confirm the upload before you leave. Unsubmitted working scores nothing. 考后+30 分钟 -- 上传。手写部分需在考试结束后 30 分钟内用手机扫描并上传。把每一页拍正、拍清晰,离场前确认上传 成功。未提交的解答得零分。 ✓ Method marks are real - show every line 步骤分是实打实的 -- 写出每一行 A wrong final number with the right method visible still earns. So on every hand-written question, commit the first move from the trigger table to paper: write Vf = 0, set up det(A-AI)=O, state the handshaking sum, name the selection type. Never leave a box blank, never erase a part-correct line, and reduce every fraction. You only need 45 to clear the hurdle - bank the method, breathe, and work one clean line at a time. 最终数字错了但方法清晰可见仍能得分。所以每道手写题,把触发表里的第一步落到纸上:写 Vf=0、列 det(A-入)=0、写出握手之和、点明选取类型。永远别留空白框,别擦掉部分正确的行,并把每个分数化简。你只需 45 分就过门槛 -- 存好方法分,深呼吸,一次写一行干净的算式。 AskSia Library · MAT9004 · 双语 Bilingual CH 4 . COUNTING 4. 9 The pigeonhole principle 4. 9 鸽巢原理 (pigeonhole principle) A deceptively simple idea with surprisingly sharp consequences: if you put more items than boxes, some box must hold more than one item. No arithmetic skill is tested - the marks are for identifying what the items and boxes are. 一个看似简单却有出人意料锋利后果的思想:若你放进的物品多于盒子,则某个盒子必含不止一个物品。它不考算术技能 - 分数在于识别物品与盒子分别是什么。 7 items into 6 holes => at least one hole holds >= ceil(7/6) = 2 overfull hole 1 hole 2 hole 3 hole 4 hole 5 hole 6 Fig 4. 5 - Seven items dropped into six holes force at least one hole to hold ≥ [7/6] = 2. The principle guarantees a collision; it does not say which hole. 图 4. 5 - 七个物品放入六个洞,迫使至少一个洞装≥[7/6]=2 个。该原理保证发生碰撞;但不说是哪个洞。 PIGEONHOLE PRINCIPLE

• “max/min on $[c,d]$”→ 候选点（驻点+奇异点+端点）比较
• “max/min/saddle $f(x,y)$”→ $\nabla f=0$ 然后 Hessian
• “solve the system”→ Gaussian elim
• “not invertible”→ 设 $\det=0$
• “how many ways”→ 先判 order/rep 再选公式
• “given…find $\Pr(\cdot|\cdot)$”→ 全概率 + Bayes[6]Source: asksia-bible-mat9004-bilingual.pdfCHAPTER 9 . EXAM MORNING FINAL 60% . HURDLE The carpark sheet 停车场速查表 Read this once before you walk in - logistics, triggers, traps, timing 走进考场前读一遍 -- 后勤、触发条件、陷阱、时间 60% OF FINAL GRADE 占最终成绩 ≥45% HURDLE TO PASS 通过的及格门槛 3h10m E-EXAM DURATION 电子考试时长 TOPICS, ONE PAPER 众多主题,一张卷子 ★ The four facts that decide everything 决定一切的四个事实 (1) It is a hurdle. The exam is worth 60%, but you must score at least 45/100 on the exam itself to pass MAT9004 - a good in-semester mark cannot rescue a failed paper. (2) Closed book, NO calculator. Authorised materials = none: no calculator, no notes, no dictionary, no books, no online sources, no generative AI. A formula sheet is printed inside the paper (page 2) - you are not asked to memorise formulas, you are asked to use them by hand. Unlimited blank A4 is allowed for working. (3) Two answer types. Short-answer outputs are a rational number - an integer or a lowest-terms fraction a/b, no spaces, no decimals; hand-written-response questions are scanned and uploaded by phone in the 30 minutes after the exam ends. (4) Whole syllabus. The exam is the only assessment hitting all six ULOs - every topic island can appear. (1)它是及格门槛。考试占60%,但你必须在考试本身上至少得 45/100才能通过 MAT9004 -- 平时分再好也救不了 挂掉的卷面。(2)闭卷,无计算器。许可材料=无:无计算器、无笔记、无词典、无书、无在线资源、无生成式 AI。试 卷内印有公式表(第2页) -- 不要求你背公式,而要你手工使用它们。可用无限张空白A4纸演算。(3)两种答题类 型。简答输出为有理数 -- 整数或最简分数 a/b,无空格、无小数;手写作答题在考试结束后 30 分钟内用手机扫描上 传。(4)全考纲。考试是唯一覆盖全部六个ULO的评估 -- 每座主题孤岛都可能出现。 9. 1 If you see X, do Y - the trigger table 9. 1 若你看到 X,就做 Y -- 触发对照表 Read the verb and the object of each question, match the row, run the recipe. This is the fastest route from "stuck" to a method mark. 读每道题的动词与宾语,匹配对应的行,运行配方。这是从“卡住”到拿下步骤分的最快路径。 If the question gives / asks . . . . . . reach for this method Area Calculus - single & multivariable (Areas 1, 3) f(x) on a closed interval [c,d], asked max/min Candidates = stationary pts (f'=0), singular pts, both endpoints c,d; 1 compare f-values f(x,y) given, asked local max/min/saddle Vf = 0 (fx=O AND fy=0), then Hessian D = fxxfyy - fxy2: D>O & fxx>O min, fxx<O 3 max, D<0 saddle "direction of steepest increase" / perpendicular to a level set gradient Vf = [fx; fy]; - Vf is steepest decrease 3 "total cost / total change from a rate", or[23]Source: asksia-cheatsheet-mat9004.pdfstat pts + endpoints "max/min/saddle f(x,y)" Vf=0 then det H "solve the system" Gaussian elim "not invertible" set det = 0 "A=PDP-1 / A"" det(A-MI)=0 "how many ways" order? repeat? = cell "committee / choose" C(n,r) "given . . . find Pr(-|-)" total prob + Bayes "E[-] / Var(-) from pdf" normalise, [ xf, fx2f "# edges from degrees" handshaking /2 "walks of length k" (A); "can graph exist?"

10）你现在最需要做的：把复习切成“可执行清单”

• 1. 你的 Final 是哪天？离现在还有几天？
• 1. 六个板块里你最不稳的 两块 是哪两块：
  • A 单变量微积分（最值/积分）
  • B 线代（消元/特征值对角化）
  • C 多变量（梯度/Hessian）
  • D 计数（四选型/容斥/鸽巢）
  • E 概率（两阶段 Bayes/期望方差/pdf）
  • F 图（Euler/Hamilton/邻接矩阵）
• 1. 你更缺：公式定义记不住，还是“知道公式但不会跑步骤/老算错”？