MATH1961

• Final exam：60%（两条线都考）[1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.
• Mid-sem quiz：13% [1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.
• 10 online quizzes：10% [1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.
• 2 assignments：15%（而且“proof-heavy”）[1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[6]Source: asksia-bible-math1961-bilingual.pdfFormal exam period streams Mid-semester quiz 13% ~ Week 8 (Apr) Two written assignments (proof-heavy) 15% Asgn 1 Wk 3 · Asgn 2 Wk 11 State the MVT Prove Rolle - MVT - Cauchy MVT 10 online quizzes (~1% each) 10% Weekly, on Canvas Compute a Taylor polynomial Bound the Lagrange remainder Tutorial attendance 2% Both tutorial streams Where the marks are won - the proof premium 分数在哪里赢得 -- 证明溢价 A 1061 student can . . . Apply the limit laws Prove a limit from £-o Quote 'continuous => IVT' Prove IVT / EVT via sup Row-reduce a matrix Prove rank-nullity MATH1961 . Mathematics 1A (Advanced) i What the assignments already demand 作业本身已经要求什么 Assignment 1 alone asks for binomial / de Moivre work, an &-style limit proof + an IVT surjectivity
• Tutorials：2% [1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.

• $\varepsilon$–$\delta$ 极限证明（量词顺序是核心）[2]Source: asksia-bible-math1961-bilingual.pdfargument, a cross-product vector sum, and a strong-induction proof plus a vector-space isomorphism (verify all the axioms). That is the level: the continuous assessment is a rehearsal for the proof questions on the final. 仅 Assignment 1 就要求做二项式 / 棣莫弗的运算、 一个 & 式极限证明+一个 IVT 满射性论证、一个叉积 向量求和,以及一个强归纳法证明外加一个向量空间 同构(验证全部公理)。这就是难度水平:平时考核就 是对期末证明题的一次彩排。 ★ The strategy this dictates - the AHA-unit drill 由此决定的策略 -- AHA 单元训练 Every chapter here is built as an AHA-unit: the definition or theorem (typeset exactly), the proof / method as numbered steps, a worked example, and the rigour trap. On the exam: state the definition with the right quantifier order (partial credit lives there), then build the argument in full sentences. The recurring 'prove that . . . ' items are - £-o limit; sup = . . . (both clauses); limit DNE by two sequences; IVT/EVT via completeness; the MVT-+Taylor++FTC chain. 这里的每一章都构建为一个 AHA 单元:定义或定理 (精确排版)、分编号步骤的证明/方法、一个例题, 以及严谨性陷阱。在考试中:以正确的量词顺序陈述 定义(部分分就在这里),然后用完整句子搭建论证。 反复出现的“证明. ”题目有 -- c-ǒ 极限;sup = . . . (两条子句);用两个数列证明极限 DNE;借助完备 性的 IVT/EVT;MVT→泰勒→FTC 链条。 MATH1961 . Mathematics 1A (Advanced) CONTENTS CONTENTS The rigour layer first, then the calculus 先打严格性这一层,再做微积分 Logic, proof and completeness up front - they are the spine of every later theorem 把逻辑、证明与完备性放在最前 -- 它们是后续每个定理的脊梁 Ch Topic Core ideas Part 1 . The rigour layer (what makes 1961 Advanced) 1 Foundations of rigour logic & quantifiers · proof techniques . induction . completeness of R → 2 Limits, continuity & the reals 8-8 · sequences & &-N . DNE by two sequences · IVT / EVT → Part 2 . Calculus, proved (Cirstea stream) 3 Differentiation & the MVT rigorous derivative . Rolle . MVT . Cauchy MVT · L'Hôpital → 4 Taylor & integration Taylor + Lagrange remainder . the Riemann integral . FTC I & II → Part 3 . Algebra & complex numbers (Brownlowe stream) 5 Vectors, spaces, systems Rn · dot / cross · Gaussian elim · span / independence · rank-nullity → 6 Eigenvalues & complex numbers determinants · eigen · diagonalisation · Argand · de Moivre → Part 4 . Walk in ready 7 Glossary & named-proof map every definition . the proofs you must reproduce → 8 Practice bank & worked proofs exam-style 'prove that . . . ' questions, re-numbered → i Why this order[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .
• 完备性：$\sup/\inf$、Least Upper Bound Axiom（后面 IVT/EVT 都靠它）[1]Source: asksia-bible-math1961-bilingual.pdf关于 MATH1961 最需要理解的一件事 You must PROVE, not just compute. MATH1961 is the Advanced twin of mainstream MATH1061 - same broad topics, but with a rigour layer bolted on: the precise 8-o definition of a limit, the completeness of R (sup / inf, the Least Upper Bound Axiom), and formal proofs of every big theorem (Cauchy-Schwarz, IVT, EVT, Rolle, MVT, Taylor's remainder, the FTC, rank-nullity). It runs 4 lecture hours/week vs ~3 for mainstream and is "a deeper look - more proof based. " Mainstream is now only ~60-70% of the Advanced content. The marks that separate you from a 1061 student are the proofs - and a correct-looking answer with no justification scores almost nothing. 你必须去证明,而不只是计算。MATH1961 是主流 MATH1061 的高阶孪生版 -- 主题范围大致相同,但外加了一层严谨 性:极限的精确 8-8 定义、R的完备性(sup/ inf、最小上界公理),以及对每个重大定理的形式化证明(Cauchy- Schwarz、IVT、EVT、Rolle、MVT、泰勒余项、FTC、秩-零化度)。它每周开设4 学时讲课,而主流约为 3学时,是 “更深入的视角 -- 更偏向证明”。主流现在仅覆盖高阶内容的约 60-70%。把你和 1061 的学生区分开来的分数正是这 些证明 -- 而一个看似正确却毫无论证的答案几乎得不到分。 i How this book was built - and the two-layer rule 本书是如何构建的 -- 以及双层规则 Standard mathematical definitions, theorems and proofs are stated plainly (they are universal: an 8-o definition is a fact, the IVT is a fact). The unit's own framing and the lecturers' specific example numbers are paraphrased and re- numbered, never copied from the Daners / Brownlowe notes or past papers. Every figure is our own clean drawing of standard first-year mathematics. The course runs on the bespoke USyd notes (Calculus by D. Daners, Linear Algebra by N. Brownlowe); no commercial textbook is mandated. Verify dates and weights against your own Canvas, as details shift between cohorts. 标准的数学定义、定理和证明均平实陈述(它们是普适的:c-ō 定义是一个事实,IVT 是一个事实)。本单元自身的框架 表述以及讲师们具体的例题数字都经过改述与重新编号,绝不照搬自 Daners / Brownlowe 讲义或往年试卷。每一幅插 图都是我们自己对标准大一数学的清晰绘制。本课程依托 USyd 定制讲义(Calculus, D. Daners 著;Linear Algebra, N. Brownlowe 著);不指定任何商业教材。请对照你自己的 Canvas 核实日期与权重,因为各届细节会有变动。 MATH1961 . Mathematics 1A (Advanced) THE BLUEPRINT - THE EXAM BLUEPRINT FINAL 60% 60% final - and it is a proof exam 期末占 60% -- 而且是一场证明考试 Tutorial 2% . online quizzes 10% . mid-sem quiz 13% . assignments 15% . final 60% 辅导课 2% · 在线测验 10% · 期中测验 13% · 作业 15% · 期末 60% Your mark is built from five pieces, but one dominates: the final exam is 60% and covers both streams (Calculus + Linear Algebra). The continuous-assessment pieces keep you honest week to week; the final is where the Advanced rigour is cashed in. Book status is not stated in the available materials - check your own Canvas. 你的分数由五部分构成,但有一部分占主导:期末考试占 60%,覆盖两条线(微积分+线性代数)。平时持续评估的各部分 让你周复一周保持扎实;而期末才是进阶严格性兑现之处。现有材料中未说明教材状态 -- 请自行查阅你的 Canvas。 60% FINAL EXAM final exam 13% MID -SEM QUIZ mid-sem quiz 15% TWO ASSIGNMENTS 两次 assignments 10% 10 ONLINE QUIZZES 10 个在线测验 The five assessment pieces 五项评估构成 Component Weight When / detail Final examination - both 60%[11]Source: asksia-bible-math1961-bilingual.pdf强(完全)归纳法 -- 当某情形需要多个更早的情形时 Strong induction assumes P(i) for all 0 _ i _ k (not just P(k), then proves P(k + 1): [P(0), . . . , P(k) all true]=>P(k+1) Vn : P(n). Use it for recurrences of order ≥2. A Fibonacci-type relation f(m+2) = f(m+1)+f(m) needs the two previous values, so weak induction is not enough - the function is determined by f(0), f(1) (this exact pattern appears in Assignment 1's vector-space problem). Rigour trap: reaching back to more than the immediately-previous case and only assuming P(k) is a genuine logical gap - switch to strong induction. 强归纳法假设 对所有 成立(而不仅是),然后证明: [P(0), . . . , P(k)all true ]=> P(k+1) => Vn : P(n). 用它处理阶数≥2 的递推关系。一个 Fibonacci 型关系需要前两个值,故弱归纳不够用 -- 该函数由 决定(这一确切 模式出现在 Assignment 1 的向量空间问题中)。严谨性陷阱:回溯到不止前一种情形、却只假设,是一个真正的逻辑漏 洞 -- 要改用强归纳法。 MATH1961 . Mathematics 1A (Advanced) COMPLETENESS 1. 5 . THE REAL NUMBERS & COMPLETENESS AHA - UNIT Supremum, infimum & the Least Upper Bound Axiom 上确界、下确界与最小上界公理 The defining property of R - and the engine behind IVT, EVT, and convergence R 的本质属性 -- 也是 IVT、EVT 与收敛背后的引擎 What makes IR different from Q? Completeness. The rationals have 'holes' (the set {x E Q : x2 < 2} has no rational least upper bound); R has none. This single axiom is what later proves the IVT, the EVT and monotone convergence - it is the most important idea in the unit. 是什么让有别于?完备性。有理数有“洞”(集合没有有理数的最小上界);则一个洞也没有。正是这一条公理在后面证明了 IVT、EVT 与单调收敛 -- 它是本单元中最重要的思想。 Bounded sets & bounds 有界集与界 A nonempty A C R is bounded above if some real beats every element; any such real is an upper bound: 一个非空集合若存在某实数压过它的每个元素,就称其有上 界;任何这样的实数都是一个上界: UPPER BOUND A bdd above > 3bER : a<b VaEA (and then every M > b is also an upper bound) The supremum is the least upper bound - the tightest possible ceiling: 上确界是最小的上界––可能最紧的天花板: SUPREMUM = LEAST UPPER BOUND c = sup A> a ≤ c Va E A ∧ Vbub: c <b (1) c is an upper bound (2) nothing smaller works Infimum inf A is dual: the greatest lower bound.
• 归纳法（弱归纳 + 强归纳）[13]Source: asksia-bible-math1961-bilingual.pdf- 1. 4 . MATHEMATICAL INDUCTION AHA - UNIT The domino principle - and when one domino is not enough 多米诺骨牌原理 -- 以及当一块骨牌还不够时 Examined every semester; strong induction is the upgrade for recurrences 每学期必考;强归纳法是处理递推关系的升级版 Induction proves a statement P(n) for all n E N with two finite checks. The picture: knock the first domino, and prove each domino topples the next; then all of them fall. 归纳法用两次有限的检验证明一个命题对所有成立。图景如下:推倒第一块骨牌,并证明每一块骨牌都推倒下一块;于是它们 全部倒下。 PRINCIPLE OF (WEAK) INDUCTION P(0) true 1 [ Vk: P(k) => P(k+1) =>Vn EN : P(n) The template - two named steps 模板 -- 两个命名的步骤 1 Basis step. Verify P(0) (or P(1)) directly. 基础步骤。直接验证 (或)。 2 Inductive step. Fix an arbitrary k and state the inductive hypothesis explicitly: assume P(k). Then prove P(k + 1), using the hypothesis somewhere. 归纳步骤。固定一个任意的,并明确陈述归纳假 设:假设。然后证明,并在某处用到该假设。 Conclude. By induction, P(n) holds for all n. ■ 作结。由数学归纳法,对一切 成立。 W1 Worked: prove 3 | 47 - 1 for all n ≥ 0 weak induction Basis. n = 0 : 40 - 1 = 0 = 3 . 0, divisible by 3. V 基础步。,可被3整除。 Inductive step. Suppose 3 | 4k - 1, i. e. 4k - 1 = 3m for some m E Z. Then 归纳步。设,即对某个有。那么 4k+1-1=4. 4k -1=4(4k -1) +3=4(3m) +3 = 3(4m +1). So 3 | 4k+1 - 1. Conclusion. By induction the claim holds for all n ≥0. 于是。结论。由归纳法,该断言对所有成立。 MATH1961 . Mathematics 1A (Advanced) ! The hypothesis must actually be USED 假设必须真正被用到 If your inductive step never invokes P(k), you have not done induction - you have re- proved each case from scratch (and probably made an error). Point to exactly where the hypothesis enters. And always state it: 'Suppose P(k) holds for some k ≥ 1 ’ 如果你的归纳步骤从未用到,那你就没有做 归纳 -- 你只是从头重新证明了每一种情形 (而且多半出了错)。要指出假设究竟在何 处进入。并且务必明确陈述:‘设 对某个 成 立。’ i Strong (complete) induction - when P(k +1) needs several earlier cases[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .
• 证明链条：Rolle $\to$ Cauchy MVT $\to$ MVT $\to$（L’Hôpital / Taylor remainder / 单调性）$\to$ FTC [9]Source: asksia-bible-math1961-bilingual.pdfEvaluate limx-+0 e™ - 1 - x 22 . 求。 At x = 0 top and bottom -> 0 (form 0/0): differentiate, 2x ea - 1 - still 0/0. 在分子分母(型):求导, -- 仍是。 2 Apply again: → e 1 2 2 再次应用 :。 3 So the limit is }. . 所以极限为。 ! Check the hypotheses BEFORE you differentiate 在求导之前先检查假设 L'Hôpital applies only to genuine 0/0 (or co/00) forms with g' # 0. On lima-+0 COS z - which is not 0/0 - differentiating gives the wrong answer. Verify the form first, every time. - L'Hopital 法则仅适用于真正的 (或)型且。在上 -- 它不是 -- 求导会给出错误答案。每一次都先核 实其形式。 - THEOREM (ROLLE) SEEDS IT ALL 定理(Rolle)是一切的种子 NAMED MVTS IN THE CHAIN 链条中各个命名的 MVT 0/0 THE FORM L'HÔPITAL NEEDS L'Hopital 所需的形式 ⚠ ALWAYS STATE HYPOTHESES 始终陈述假设 Rolle is the seed; Cauchy MVT is the trunk; the MVT, monotonicity and L'Hôpital are the branches. Prove the seed once and you can grow the whole tree on demand. Rolle 是种子;柯西中值定理是树干;MVT、单调性与洛必达法则是枝条。把种子证明一次,你就能随时长出整棵树。 MATH1961 . THE PROOF YOU REUSE ALL SEMESTER MATH1961 . Mathematics 1A (Advanced) x THE RIEMANN INTEGRAL . THE FTC - THE RIEMANN INTEGRAL C6 . PROOF -HEAVY Trap the area between two sums 把面积夹在两个和之间 Darboux upper & lower sums squeeze the integral from above and below Darboux 上和与下和从上下两侧夹逼积分 The definite integral is not "the antiderivative evaluated at the ends" - that is a theorem (the FTC, p. 3). The definition is about area trapped between rectangles: chop [a, b] into strips, over-estimate with the sup on each strip, under-estimate with the inf, and force the gap to zero. 定积分不是“在两端点处取值的原函数” -- 那是一个定理(FTC,见第3页)。其定义关乎被夹在矩形之间的面积:把分成小 条,用每条上的上确界高估,用下确界低估,再迫使缺口趋于零。 UPPER & LOWER (DARBOUX) SUMS Partition P= {a = x0 <. . . < xn = b}, strip width Axk :[19]Source: asksia-cheatsheet-math1961.pdfWhy unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +co, so U(f,P)=+co for every P and U-L can't be made < = not integrable (e. g. 1/x on (0,1]). 10 . FTC . both parts PROVED A F(x)=[ a" f is continuous when f integrable (|f|≤C + triangle inequality + squeeze). FTC I. f cts = F(x)=[ a" f is differentiable, F'(x)=f(x). Proof: [F(x+h)-F(x)]/h = (1/h)[x^{x+h} f; by EVT/IVT = f(¿), E=[x,x+h], and {->x as h->0 gives f(x). FTC II. G any primitive (G'=f) = [ ab f=G(b)-G(a). Proof: F-G has zero derivative = constant. LEIBNIZ (VARIABLE LIMITS) d/dx [_{g1 (x)}^{g2 (x)} f = f(g2)g2' - f(g1)g1' From FTC: parts /fg' = [fg] - [f'g; substitution [f(g)g' = J_{g(a)}^{g(b)} f(u)du; partial fractions for rationals ( dt/(1+t2)=arctan t). Worked Leibniz: H(x)=[_{x}^{x2} e^{t2} dt => H'(x) = e^{x4} ·2x- e^{x2]. 1. Improper: [. 1 dx/x diverges (unbounded at 0, handled as lim_{&->0+}[_€1). Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors -> A/(x-r) + (Bx+C)/(x2+px+q) + . . . . Recognise / dt/(1+t2)=arctan t and Jdt/t=In|t|. Integration by parts and substitution both descend from the product and chain rules via FTC II - so every technique is, at heart, a theorem you can prove. The integral function F(x)=[ a" f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II). Proof Recap SIDE 1 ε-δ: Ψε 30, 01x-al<δ = |f-1|<ε sup: (1) upper bound (2) least - BOTH Rolle-Cauchy MVT-MVT-L'Hôpital Rn = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 FTC: F'=f . Jabf = G(b) -G(a) Write proofs in full sentences ; state the definition, name the technique, justify each step. Quick proof-chain map: Completeness (LUB) = IVT & EVT => Rolle => Cauchy MVT => MVT = {monotonicity, L'Hôpital, Lagrange remainder}; EVT+IVT => FTC I = FTC II. Knowing the arrows lets you rebuild any proof from the one before it. Revision aid . check the current unit outline for assessment . @ 2026 flip - for side 2 . complex numbers, linear algebra & eigenvalues asksia. ai/cheatsheet/ usyd-math1961 . side 1/2 AskSia CHEATSHEET SERIES REVISION SHEET . ALL TOPICS Compiled by AskSia . mapped to the MATH1961 syllabus . asksia. ai/cheatsheet/usyd- math1961 -
• 线代：rank-nullity、eigenvalue reasoning、对角化“重根陷阱”。[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.

2.1 量词与否定（proof toolkit 的地基）

• 否定规则（非常常考、也非常常丢分）：[16]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone. Assessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3.[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .
  • $\neg(\forall x,P(x)) \equiv \exists x,\neg P(x)$
  • $\neg(\exists x,P(x)) \equiv \forall x,\neg P(x)$
• 关键提醒：量词顺序改变命题真假（老师就扣这个）。[5]Source: asksia-bible-math1961-bilingual.pdf为什么是这个顺序 The Advanced unit hangs everything on a rigour toolkit - logic, the proof methods, induction, and the completeness of R (sup / inf). We put that first (Chs 1-2) because the same machinery proves the IVT, the EVT, the MVT and the FTC later on. Master the spine and the rest of the course is 'apply the toolkit'. The three chapters in this build are Ch 1 Foundations and Ch 2 Limits & continuity, sitting under the front matter you are reading now. 高阶单元把一切都挂在一套严谨性工具箱上 -- 逻辑、各种证明方法、数学归纳法,以及R的完备性(sup/ inf)。我们 把它放在最前(第1-2章),因为同一套机制随后会用来证明 IVT、EVT、MVT 和 FTC。掌握这条主干,课程其余部分 便只是“套用工具箱”。本版本中的三章是 第1章 基础 与 第2 章 极限与连续性,置于你正在阅读的前言之下。 MATH1961 . Mathematics 1A (Advanced) LOGIC 1 . THE RIGOUR TOOLKIT LO1 . RIGOUR Logic, quantifiers & reading a statement exactly 逻辑、量词,以及精确读懂一个命题 Marks are lost for sloppy logic - the quantifier order is part of the answer 逻辑松散会丢分 -- 量词的顺序也是答案的一部分 Advanced mathematics is written in a precise language. Before any proof, you must read a statement exactly. what is being asserted, in what order, and over which set. The lecturer grades this - a misplaced quantifier or a confused implication arrow loses marks even when the underlying idea is right. 进阶数学以一种精确的语言书写。在任何证明之前,你必须精确地读懂一条命题:它断言了什么,以什么顺序,在哪个集合 上。讲师会对此评分 -- 即使底层想法正确,一个错位的量词或一个混淆的蕴含箭头都会丢分。 - 1. 1 Notation you must use correctly 1. 1 你必须正确使用的记号 Symbol Reads as €,¢ is / is not an element of = defined to equal (vs =, a deduced equality) >, <, >> implies . is implied by . iff (all different!) E'A for all . there exists NCZCQCRC C the number-system chain ! Quantifier ORDER changes the meaning 量词顺序会改变含义 "Valy : y > x" (true in R) is not "By Vx : y > x" (false - no single y beats every x). In the &-o definition the order Ve Ed is the whole point: o is allowed to depend on &. Swap them and you have written something false. (在 中为真)不同于“”(为假 -- 不存在单一的 y 胜过每一个 x)。在c-ǒ 定义中,顺序 VE38 才是关键 所在:ō 是允许依赖于c的。把它们对调,你写出的 便是错的。 1. 2 Negating quantified statements

2.2 $\varepsilon$–$\delta$ 极限定义（你必须会“完整写出来”）

• 定义（核心版本）：若 $f$ 在 $a$ 的某个去心邻域上有定义，则
  $$\lim_{x\to a} f(x)=L \iff \forall \varepsilon>0,\exists \delta=\delta(\varepsilon)>0:\ 0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon.$$
  （注意：$\forall \varepsilon$ 在前，$\exists \delta$ 在后，且 $\delta$ 允许依赖 $\varepsilon$。）[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .
• 你还要会它的“变体”（也可考）：$x\to a^\pm$、$x\to\infty$、极限为 $\pm\infty$ 等。[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .

2.3 连续（Continuity）

• 连续定义：$f$ 在 $a$ 连续 $\iff \lim_{x\to a}f(x)$ 存在且等于 $f(a)$。[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF
• 常见陷阱：极限存在不代表连续（如果 $f(a)\neq \lim$）。[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF

2.4 上确界 $\sup$（完备性的核心，必须写“两条子句”）

• 若 $A\subset\mathbb R$ 非空且有上界，则 $c=\sup A$ 的定义要写齐两条：[11]Source: asksia-bible-math1961-bilingual.pdf强(完全)归纳法 -- 当某情形需要多个更早的情形时 Strong induction assumes P(i) for all 0 _ i _ k (not just P(k), then proves P(k + 1): [P(0), . . . , P(k) all true]=>P(k+1) Vn : P(n). Use it for recurrences of order ≥2. A Fibonacci-type relation f(m+2) = f(m+1)+f(m) needs the two previous values, so weak induction is not enough - the function is determined by f(0), f(1) (this exact pattern appears in Assignment 1's vector-space problem). Rigour trap: reaching back to more than the immediately-previous case and only assuming P(k) is a genuine logical gap - switch to strong induction. 强归纳法假设 对所有 成立(而不仅是),然后证明: [P(0), . . . , P(k)all true ]=> P(k+1) => Vn : P(n). 用它处理阶数≥2 的递推关系。一个 Fibonacci 型关系需要前两个值,故弱归纳不够用 -- 该函数由 决定(这一确切 模式出现在 Assignment 1 的向量空间问题中)。严谨性陷阱:回溯到不止前一种情形、却只假设,是一个真正的逻辑漏 洞 -- 要改用强归纳法。 MATH1961 . Mathematics 1A (Advanced) COMPLETENESS 1. 5 . THE REAL NUMBERS & COMPLETENESS AHA - UNIT Supremum, infimum & the Least Upper Bound Axiom 上确界、下确界与最小上界公理 The defining property of R - and the engine behind IVT, EVT, and convergence R 的本质属性 -- 也是 IVT、EVT 与收敛背后的引擎 What makes IR different from Q? Completeness. The rationals have 'holes' (the set {x E Q : x2 < 2} has no rational least upper bound); R has none. This single axiom is what later proves the IVT, the EVT and monotone convergence - it is the most important idea in the unit. 是什么让有别于?完备性。有理数有“洞”(集合没有有理数的最小上界);则一个洞也没有。正是这一条公理在后面证明了 IVT、EVT 与单调收敛 -- 它是本单元中最重要的思想。 Bounded sets & bounds 有界集与界 A nonempty A C R is bounded above if some real beats every element; any such real is an upper bound: 一个非空集合若存在某实数压过它的每个元素,就称其有上 界;任何这样的实数都是一个上界: UPPER BOUND A bdd above > 3bER : a<b VaEA (and then every M > b is also an upper bound) The supremum is the least upper bound - the tightest possible ceiling: 上确界是最小的上界––可能最紧的天花板: SUPREMUM = LEAST UPPER BOUND c = sup A> a ≤ c Va E A ∧ Vbub: c <b (1) c is an upper bound (2) nothing smaller works Infimum inf A is dual: the greatest lower bound.[15]Source: asksia-bible-math1961-bilingual.pdf下确界是对偶概念:最大的下界。 i sup A need not lie in A 未必落在其中 For A = {1 - 1 : n ≥ 1} = {0, 2, 2, . . . } we have sup A = 1 ¢ A but inf A = 0 € A. When the sup is attained, sup A = max A. 对于 我们有 但。当 sup 确实被取到时,。 ★ The Least Upper Bound Axiom - completeness of R 最小上界公理––R的完备性 Every nonempty A C R that is bounded above has a supremum in R. (Dual: the Greatest Lower Bound Axiom gives inf A. ) This is the axiom Q fails - and it is the foundation every later existence proof stands on. 每一个非空且有上界的ACR在R 中都有上确界。 (对偶:最大下界公理给出。)这正是 所不具备的公 理 -- 也是后续每一个存在性证明所立足的根基。 How to PROVE sup A = '- both c clauses ‘ 如何证明“” -- 两个子句 1 Clause (1): c is an upper bound. Take an arbitrary a E A and show a ≤ c. 子句(1): c 是一个上界。取任意 并证明。 2 Clause (2): nothing smaller works. The standard move - fix any & > 0 and exhibit an a € A with a > c - 8. That proves c - & is not an upper bound, so no number below c can be one. 子句(2):没有更小的数行得通。标准做法 -- 固定任意 并 给出一个满足 的。这就证明了 不是上界,故任何小于 的 数都不可能是上界。 MATH1961 . Mathematics 1A (Advanced) ! BOTH clauses, every time 每一次都要写齐两个子句 A 'find / prove the sup' answer that shows only that c is an upper bound is half a proof. You must also kill everything below c - the s-argument in step 2. Forgetting clause (2) is the most common mark-loser here. 一个“求/证 sup”的答案如果只表明 是一个上界,那 只是半个证明。你还必须排除其下方的一切 -- 即步 骤2 中的 论证。漏掉子句(2)是这里最常见的失分原 因。 ✓ The bridge to limits: an approximating sequence 通往极限的桥梁:一个逼近数列 From clause (2) with a = - you get points an € A with c - - < an ≤ c, hence an -> sup A. This is precisely how completeness powers monotone convergence and the IVT in Ch 2. 由子句(2),取 可得到满足 的点 ,因此。这恰恰就 是完备性如何驱动第 2 章中的单调收敛与 IVT。 MATH1961 . Mathematics 1A (Advanced) E-A LIMIT 2 . LIMITS, CONTINUITY & THE REALS AHA - UNIT The precise &-o definition of a limit 极限的精确 8-ǒ 定义 The centrepiece of the Advanced unit - you must PROVE limits, not just apply laws 进阶单元的核心 -- 你必须证明极限,而不只是套用法则 Mainstream calculus says "f (x) gets close to L as x gets close to a. " That is a feeling, not mathematics. MATH1961 replaces it with a challenge-response game with a precise winning condition - the £-o definition, the single most important statement in the calculus stream.
  • (1) $c$ 是上界：$\forall a\in A,\ a\le c$
  • (2) “没有更小的上界”：$\forall b$（若 $b$ 是上界）则 $c\le b$；等价常用写法是
    $$\forall \varepsilon>0\ \exists a\in A:\ a>c-\varepsilon.$$
• 重要提醒：只证明“$c$ 是上界”是半个证明，会丢大分。[15]Source: asksia-bible-math1961-bilingual.pdf下确界是对偶概念:最大的下界。 i sup A need not lie in A 未必落在其中 For A = {1 - 1 : n ≥ 1} = {0, 2, 2, . . . } we have sup A = 1 ¢ A but inf A = 0 € A. When the sup is attained, sup A = max A. 对于 我们有 但。当 sup 确实被取到时,。 ★ The Least Upper Bound Axiom - completeness of R 最小上界公理––R的完备性 Every nonempty A C R that is bounded above has a supremum in R. (Dual: the Greatest Lower Bound Axiom gives inf A. ) This is the axiom Q fails - and it is the foundation every later existence proof stands on. 每一个非空且有上界的ACR在R 中都有上确界。 (对偶:最大下界公理给出。)这正是 所不具备的公 理 -- 也是后续每一个存在性证明所立足的根基。 How to PROVE sup A = '- both c clauses ‘ 如何证明“” -- 两个子句 1 Clause (1): c is an upper bound. Take an arbitrary a E A and show a ≤ c. 子句(1): c 是一个上界。取任意 并证明。 2 Clause (2): nothing smaller works. The standard move - fix any & > 0 and exhibit an a € A with a > c - 8. That proves c - & is not an upper bound, so no number below c can be one. 子句(2):没有更小的数行得通。标准做法 -- 固定任意 并 给出一个满足 的。这就证明了 不是上界,故任何小于 的 数都不可能是上界。 MATH1961 . Mathematics 1A (Advanced) ! BOTH clauses, every time 每一次都要写齐两个子句 A 'find / prove the sup' answer that shows only that c is an upper bound is half a proof. You must also kill everything below c - the s-argument in step 2. Forgetting clause (2) is the most common mark-loser here. 一个“求/证 sup”的答案如果只表明 是一个上界,那 只是半个证明。你还必须排除其下方的一切 -- 即步 骤2 中的 论证。漏掉子句(2)是这里最常见的失分原 因。 ✓ The bridge to limits: an approximating sequence 通往极限的桥梁:一个逼近数列 From clause (2) with a = - you get points an € A with c - - < an ≤ c, hence an -> sup A. This is precisely how completeness powers monotone convergence and the IVT in Ch 2. 由子句(2),取 可得到满足 的点 ,因此。这恰恰就 是完备性如何驱动第 2 章中的单调收敛与 IVT。 MATH1961 . Mathematics 1A (Advanced) E-A LIMIT 2 . LIMITS, CONTINUITY & THE REALS AHA - UNIT The precise &-o definition of a limit 极限的精确 8-ǒ 定义 The centrepiece of the Advanced unit - you must PROVE limits, not just apply laws 进阶单元的核心 -- 你必须证明极限,而不只是套用法则 Mainstream calculus says "f (x) gets close to L as x gets close to a. " That is a feeling, not mathematics. MATH1961 replaces it with a challenge-response game with a precise winning condition - the £-o definition, the single most important statement in the calculus stream.

2.5（线代）线性无关/子空间/对角化判定这些“定义型得分点”

• 线性无关最常见扣分点是必须写“只有平凡解”（trivial-only solution）。[4]Source: asksia-bible-math1961-bilingual.pdfPartial credit on 'state the definition' and 'prove that' questions hinges on the exact wording - the quantifier order in €-o, both clauses of sup, the trivial-only solution for independence, XA before eigenvectors. Drill these until you can write each one cold. “陈述定义”和“证明 "题目的部分分取决于精确的措辞- -8-ō 中量词的顺序、sup的两条子句、线性无关的‘仅有平 凡解’、特征向量之前的XA。 把这些反复操练,直到你能脱稿写出每一条。 MATH1961 . Mathematics 1A (Advanced) PRACTICE . Q1-Q2 - PRACTICE - THE ADVANCED FINAL, END TO END FOREGROUND RIGOUR Prove it, then compute it 先证明它,再计算它 Fourteen fresh problems in the 1961 style: rigorous proofs first, then the computation slots 十四道 1961 风格的全新题目:先做严格证明,再填入计算环节 The one-line takeaway. The MATH1961 final is half "prove that . . . " (induction, 8-o, sup, MVT/Taylor, Cauchy- Schwarz, independence) and half computation (eigenvalues, diagonalisation, de Moivre, integrals, Gaussian elimination). Hand-waving is penalised - every proof below is written in full sentences mixing English and symbols, the style the markers reward. 一句话要点。MATH1961 的期末考试一半是“证明 ……. . (prove that . . . )”(数学归纳法、c-ō、sup、MVT/泰勒、Cauchy- Schwarz、线性无关),另一半是计算(特征值、对角化、棣莫弗、积分、高斯消元法)。含糊其辞会被扣分 -- 下面每一个证 明都用夹杂英文与符号的完整句子写成,正是阅卷人所奖励的风格。 ★ These are FRESH problems in the exam style 这些是考试风格的全新题目 AskSia-authored stems with invented numbers, not real exam questions. Standard definitions and theorems are canonical. Cover each solution, attempt it on paper closed-book, then check the working line by line. 由 AskSia 撰写、用虚构数字的题干,并非真实考题。标准定义与定理是规范的。遮住每个解答,闭卷在纸上尝试,然后 逐行核对过程。 5 marks Q1 PROOF . INDUCTION Prove by induction that for all integers n ≥ 1, n k (k + 1) = n(n+1)(n+2) . k=1 Q1 Worked proof - induction Basis (n = 1). LHS = 1. 2 = 2; RHS = 1. 2. 3/3 = 2. Equal, so P(1) holds. Inductive step. Assume P(k): E,k j(j+1) = k(k+1)(k+2)/3. Then 归纳步骤(Inductive step)。假设 P(k): >1kj(j+1)= k(k+1)(k+2)/3。则 _j(j+1) = k+1 ん13 j=1 k(k+1)(k+2) 3 +(k+1)(k+2)=(k+1)(k+2) +1 = (k +1)(k +2) . k+3 _(k+1)(k+ 2)(k +3) 3 3 , which is exactly P(k+1) (replace n by k+1 in the formula). By induction the identity holds for all n ≥ 1. 这恰好就是 P(k+1)(将公式中的 n 换成k+1)。由数学归纳法,该恒等式对一切n≥1成立。■ MATH1961 . Mathematics 1A (Advanced)[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors
• “是不是子空间/向量空间”：要检查 $0$ 和封闭性/或 8 条公理（按课的要求）。[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors
• “可对角化？”：重复特征值一定要检查几何重数=代数重数（否则是假对角化）。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors

3.1 $\varepsilon$–$\delta$ 证明极限：四步骨架（写作模板）

• 模板：[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF
  • 1）Fix $\varepsilon>0$（写“任取 $\varepsilon>0$”）
  • 2）把 $|f(x)-L|$ 估到某个“含 $|x-a|$ 的上界”
  • 3）选择 $\delta$ 让这个上界 $<\varepsilon$
  • 4）最后验证：若 $0<|x-a|<\delta$ 则 $|f(x)-L|<\varepsilon$
• 例子提示（你可以照抄思路）：$\lim_{x\to a}\sqrt{x}=\sqrt{a}$ 的估计用
  $$|\sqrt{x}-\sqrt{a}|=\frac{|x-a|}{\sqrt{x}+\sqrt{a}} \le \frac{|x-a|}{\sqrt a}$$
  然后取 $\delta=\varepsilon\sqrt a$。[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF

3.2 “极限不存在”标准打法：两条数列法（DNE by two sequences）

• 若能找 $x_n\to a$ 与 $x_n'\to a$，但 $f(x_n)$ 与 $f(x_n')$ 极限不同，则 $\lim_{x\to a}f(x)$ 不存在。[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF
• 关键提醒：考试要求你把两条数列写出来，不能只说“它振荡所以不存在”。[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF

3.3 归纳法：弱归纳 + 强归纳（写作结构必须标准）

• 弱归纳骨架：写清楚 Basis 和 Inductive step，且在 step 里必须真的用到归纳假设。[13]Source: asksia-bible-math1961-bilingual.pdf- 1. 4 . MATHEMATICAL INDUCTION AHA - UNIT The domino principle - and when one domino is not enough 多米诺骨牌原理 -- 以及当一块骨牌还不够时 Examined every semester; strong induction is the upgrade for recurrences 每学期必考;强归纳法是处理递推关系的升级版 Induction proves a statement P(n) for all n E N with two finite checks. The picture: knock the first domino, and prove each domino topples the next; then all of them fall. 归纳法用两次有限的检验证明一个命题对所有成立。图景如下:推倒第一块骨牌,并证明每一块骨牌都推倒下一块;于是它们 全部倒下。 PRINCIPLE OF (WEAK) INDUCTION P(0) true 1 [ Vk: P(k) => P(k+1) =>Vn EN : P(n) The template - two named steps 模板 -- 两个命名的步骤 1 Basis step. Verify P(0) (or P(1)) directly. 基础步骤。直接验证 (或)。 2 Inductive step. Fix an arbitrary k and state the inductive hypothesis explicitly: assume P(k). Then prove P(k + 1), using the hypothesis somewhere. 归纳步骤。固定一个任意的,并明确陈述归纳假 设:假设。然后证明,并在某处用到该假设。 Conclude. By induction, P(n) holds for all n. ■ 作结。由数学归纳法,对一切 成立。 W1 Worked: prove 3 | 47 - 1 for all n ≥ 0 weak induction Basis. n = 0 : 40 - 1 = 0 = 3 . 0, divisible by 3. V 基础步。,可被3整除。 Inductive step. Suppose 3 | 4k - 1, i. e. 4k - 1 = 3m for some m E Z. Then 归纳步。设,即对某个有。那么 4k+1-1=4. 4k -1=4(4k -1) +3=4(3m) +3 = 3(4m +1). So 3 | 4k+1 - 1. Conclusion. By induction the claim holds for all n ≥0. 于是。结论。由归纳法,该断言对所有成立。 MATH1961 . Mathematics 1A (Advanced) ! The hypothesis must actually be USED 假设必须真正被用到 If your inductive step never invokes P(k), you have not done induction - you have re- proved each case from scratch (and probably made an error). Point to exactly where the hypothesis enters. And always state it: 'Suppose P(k) holds for some k ≥ 1 ’ 如果你的归纳步骤从未用到,那你就没有做 归纳 -- 你只是从头重新证明了每一种情形 (而且多半出了错)。要指出假设究竟在何 处进入。并且务必明确陈述:‘设 对某个 成 立。’ i Strong (complete) induction - when P(k +1) needs several earlier cases
• 强归纳：当 $P(k+1)$ 需要用到多个更早情形（例如二阶递推/Fibonacci 型）时用：假设 $P(0),\dots,P(k)$ 全都成立，再证 $P(k+1)$。[11]Source: asksia-bible-math1961-bilingual.pdf强(完全)归纳法 -- 当某情形需要多个更早的情形时 Strong induction assumes P(i) for all 0 _ i _ k (not just P(k), then proves P(k + 1): [P(0), . . . , P(k) all true]=>P(k+1) Vn : P(n). Use it for recurrences of order ≥2. A Fibonacci-type relation f(m+2) = f(m+1)+f(m) needs the two previous values, so weak induction is not enough - the function is determined by f(0), f(1) (this exact pattern appears in Assignment 1's vector-space problem). Rigour trap: reaching back to more than the immediately-previous case and only assuming P(k) is a genuine logical gap - switch to strong induction. 强归纳法假设 对所有 成立(而不仅是),然后证明: [P(0), . . . , P(k)all true ]=> P(k+1) => Vn : P(n). 用它处理阶数≥2 的递推关系。一个 Fibonacci 型关系需要前两个值,故弱归纳不够用 -- 该函数由 决定(这一确切 模式出现在 Assignment 1 的向量空间问题中)。严谨性陷阱:回溯到不止前一种情形、却只假设,是一个真正的逻辑漏 洞 -- 要改用强归纳法。 MATH1961 . Mathematics 1A (Advanced) COMPLETENESS 1. 5 . THE REAL NUMBERS & COMPLETENESS AHA - UNIT Supremum, infimum & the Least Upper Bound Axiom 上确界、下确界与最小上界公理 The defining property of R - and the engine behind IVT, EVT, and convergence R 的本质属性 -- 也是 IVT、EVT 与收敛背后的引擎 What makes IR different from Q? Completeness. The rationals have 'holes' (the set {x E Q : x2 < 2} has no rational least upper bound); R has none. This single axiom is what later proves the IVT, the EVT and monotone convergence - it is the most important idea in the unit. 是什么让有别于?完备性。有理数有“洞”(集合没有有理数的最小上界);则一个洞也没有。正是这一条公理在后面证明了 IVT、EVT 与单调收敛 -- 它是本单元中最重要的思想。 Bounded sets & bounds 有界集与界 A nonempty A C R is bounded above if some real beats every element; any such real is an upper bound: 一个非空集合若存在某实数压过它的每个元素,就称其有上 界;任何这样的实数都是一个上界: UPPER BOUND A bdd above > 3bER : a<b VaEA (and then every M > b is also an upper bound) The supremum is the least upper bound - the tightest possible ceiling: 上确界是最小的上界––可能最紧的天花板: SUPREMUM = LEAST UPPER BOUND c = sup A> a ≤ c Va E A ∧ Vbub: c <b (1) c is an upper bound (2) nothing smaller works Infimum inf A is dual: the greatest lower bound.[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .

3.4 Rolle / Cauchy MVT / MVT：最常被复现的“命名证明链条”

• 课程强调：Rolle 是种子，Cauchy MVT 是树干，MVT/L’Hôpital/Taylor remainder 是枝条；一定要写假设（连续/可微/区间）并点名你用的定理。[9]Source: asksia-bible-math1961-bilingual.pdfEvaluate limx-+0 e™ - 1 - x 22 . 求。 At x = 0 top and bottom -> 0 (form 0/0): differentiate, 2x ea - 1 - still 0/0. 在分子分母(型):求导, -- 仍是。 2 Apply again: → e 1 2 2 再次应用 :。 3 So the limit is }. . 所以极限为。 ! Check the hypotheses BEFORE you differentiate 在求导之前先检查假设 L'Hôpital applies only to genuine 0/0 (or co/00) forms with g' # 0. On lima-+0 COS z - which is not 0/0 - differentiating gives the wrong answer. Verify the form first, every time. - L'Hopital 法则仅适用于真正的 (或)型且。在上 -- 它不是 -- 求导会给出错误答案。每一次都先核 实其形式。 - THEOREM (ROLLE) SEEDS IT ALL 定理(Rolle)是一切的种子 NAMED MVTS IN THE CHAIN 链条中各个命名的 MVT 0/0 THE FORM L'HÔPITAL NEEDS L'Hopital 所需的形式 ⚠ ALWAYS STATE HYPOTHESES 始终陈述假设 Rolle is the seed; Cauchy MVT is the trunk; the MVT, monotonicity and L'Hôpital are the branches. Prove the seed once and you can grow the whole tree on demand. Rolle 是种子;柯西中值定理是树干;MVT、单调性与洛必达法则是枝条。把种子证明一次,你就能随时长出整棵树。 MATH1961 . THE PROOF YOU REUSE ALL SEMESTER MATH1961 . Mathematics 1A (Advanced) x THE RIEMANN INTEGRAL . THE FTC - THE RIEMANN INTEGRAL C6 . PROOF -HEAVY Trap the area between two sums 把面积夹在两个和之间 Darboux upper & lower sums squeeze the integral from above and below Darboux 上和与下和从上下两侧夹逼积分 The definite integral is not "the antiderivative evaluated at the ends" - that is a theorem (the FTC, p. 3). The definition is about area trapped between rectangles: chop [a, b] into strips, over-estimate with the sup on each strip, under-estimate with the inf, and force the gap to zero. 定积分不是“在两端点处取值的原函数” -- 那是一个定理(FTC,见第3页)。其定义关乎被夹在矩形之间的面积:把分成小 条,用每条上的上确界高估,用下确界低估,再迫使缺口趋于零。 UPPER & LOWER (DARBOUX) SUMS Partition P= {a = x0 <. . . < xn = b}, strip width Axk :
• Cauchy MVT 证明核心：构造辅助函数 $F(x)$，检查端点相等，套 Rolle 得 $F'(c)=0$，推出结论。[14]Source: asksia-bible-math1961-bilingual.pdfA PROOF . CENTREPIECE One auxiliary function does both 一个辅助函数同时搞定两者 Build the right helper, apply Rolle, and the MVT falls out as g(x) = x 构造出合适的辅助函数,应用 Rolle,MVT 便随之而出 The course proves the Mean Value Theorem as a corollary of the Cauchy MVT, and the Cauchy MVT itself by a single clever application of Rolle. This two-step descent is the most reproduced proof on the exam. 本课程把中值定理作为 Cauchy MVT 的推论来证明,而 Cauchy MVT 本身则通过对 Rolle 的一次巧妙应用得到。这一两步 下降是考试中被复现最多的证明。 CAUCHY MEAN VALUE THEOREM Cauchy MVT : f,g cts on [a,b], diff. on (a, b) >Ice (a,b) : [f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c) Proof idea - the auxiliary function 证明思路 -- 辅助函数 1 Build the helper F(x) = [f(b) - f(a)] [g(x) -g(a)] - [g(b) - g(a)] [f(x) - f(a)]. It is continuous on [a, b] and differentiable on (a, b) (sums/products of such f, g). 构造辅助函数。它在 上连续、在 上可微(此类函数的和/积)。 2 Check the endpoints. Substituting gives F(a) = 0 and - the design pays off - F(b) = 0 as well, so F(a) = F(b). 检查端点。代入得,同样 -- 设计奏效了 -- 于是。 3 Apply Rolle to F. There is c € (a, b) with F'(c) = 0. 对 应用 Rolle。存在 使。 4 Differentiate and read off: F'(c) = [f(b)-f(a)]g'(c) -[g(b)-g(a)]f'(c) = 0, which is exactly the claim. " 求导并读出:,这恰好就是所断言的。 MEAN VALUE THEOREM MVT: f cts on [a,b], diff. on (a,b) f(b) - f(a) > Ice (a,b) : f'(c) = b - a MVT as a one-line corollary MVT 作为一行推论 Put g(x) = x in the Cauchy MVT. Then g'(c) = 1 and g(b) - g(a) =b-a, so [f(b)-f(a)] · 1 = (b-a)f'(c). Divide by b - a: the MVT. (Equivalently: apply Rolle to f minus its chord L(x) - same idea, narrower. ) 在 Cauchy MVT 中令。则与,于是。除以:即得 MVT。 (等价地:对减去其弦线应用 Rolle -- 同样的思路,范 围更窄。) MATH1961 . Mathematics 1A (Advanced) C4 EX 3 MVT gives a clean inequality
• MVT 作为推论：在 Cauchy MVT 里取 $g(x)=x$。[14]Source: asksia-bible-math1961-bilingual.pdfA PROOF . CENTREPIECE One auxiliary function does both 一个辅助函数同时搞定两者 Build the right helper, apply Rolle, and the MVT falls out as g(x) = x 构造出合适的辅助函数,应用 Rolle,MVT 便随之而出 The course proves the Mean Value Theorem as a corollary of the Cauchy MVT, and the Cauchy MVT itself by a single clever application of Rolle. This two-step descent is the most reproduced proof on the exam. 本课程把中值定理作为 Cauchy MVT 的推论来证明,而 Cauchy MVT 本身则通过对 Rolle 的一次巧妙应用得到。这一两步 下降是考试中被复现最多的证明。 CAUCHY MEAN VALUE THEOREM Cauchy MVT : f,g cts on [a,b], diff. on (a, b) >Ice (a,b) : [f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c) Proof idea - the auxiliary function 证明思路 -- 辅助函数 1 Build the helper F(x) = [f(b) - f(a)] [g(x) -g(a)] - [g(b) - g(a)] [f(x) - f(a)]. It is continuous on [a, b] and differentiable on (a, b) (sums/products of such f, g). 构造辅助函数。它在 上连续、在 上可微(此类函数的和/积)。 2 Check the endpoints. Substituting gives F(a) = 0 and - the design pays off - F(b) = 0 as well, so F(a) = F(b). 检查端点。代入得,同样 -- 设计奏效了 -- 于是。 3 Apply Rolle to F. There is c € (a, b) with F'(c) = 0. 对 应用 Rolle。存在 使。 4 Differentiate and read off: F'(c) = [f(b)-f(a)]g'(c) -[g(b)-g(a)]f'(c) = 0, which is exactly the claim. " 求导并读出:,这恰好就是所断言的。 MEAN VALUE THEOREM MVT: f cts on [a,b], diff. on (a,b) f(b) - f(a) > Ice (a,b) : f'(c) = b - a MVT as a one-line corollary MVT 作为一行推论 Put g(x) = x in the Cauchy MVT. Then g'(c) = 1 and g(b) - g(a) =b-a, so [f(b)-f(a)] · 1 = (b-a)f'(c). Divide by b - a: the MVT. (Equivalently: apply Rolle to f minus its chord L(x) - same idea, narrower. ) 在 Cauchy MVT 中令。则与,于是。除以:即得 MVT。 (等价地:对减去其弦线应用 Rolle -- 同样的思路,范 围更窄。) MATH1961 . Mathematics 1A (Advanced) C4 EX 3 MVT gives a clean inequality

3.5 Taylor + Lagrange remainder：期末高价值证明点

• 拉格朗日余项形式（你要会写、会用来界误差）：
  $$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1},\quad c\text{ 在 }x_0\text{ 与 }x\text{ 之间}.$$
  证明思路是对特定 $F,G$ 用 Cauchy MVT，并注意“相消（telescoping）”步骤。[26]Source: asksia-cheatsheet-math1961.pdf7d . Diff => Cts, not conversely CLASSIC TRAP f(x)=|x| is continuous at 0 but not differentiable: the left difference quotient -> -1, the right -> +1, so f'(0) does not exist. Differentiable => continuous; continuous differentiable. 2nd-deriv test inconclusive case: f(x)=x4 at 0 has f'=f"=0 yet a min - fall back to the sign change of f. Rolle in one line: any polynomial of degree n has at most n real roots - between consecutive roots f'(c)=0 by Rolle, so the derivative (degree n-1) caps the root count by induction. The product/chain proofs run cleanest through the Carathéodory form: write each factor as f(x0)+m(x) (x-X0) and read off the derivative from m(x0). Singular critical points (f' undefined) count too: |x| at 0 is a minimum with no derivative - always check both stationary and singular points. 8 . Taylor + Lagrange HIGH -VALUE PROOF NTH TAYLOR POLY AT XO Τη (x) = Σ_{k=Θ}{n} f (k) (xo)/k !. (x-Xe)k Tn is the unique degree-≤n poly P with lim(x->x0) [f-P]/(x-x0)" = 0 - so build it by substitution / multiply / integrate, no repeated differentiation. LAGRANGE REMAINDER A Rn (x) = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 for some c between xe and X Proof: Cauchy MVT on F(t) =__ {k=0}^{n}f(k)(t)/k !· (x-t)k and G(t) =- (x-t)"+1; the sum telescopes to F'(t)=f(n+1) (t)/n !- (x-t)". ERROR BOUND (THE EXAM USE) |Rn (x) | ≤ max | f ( n+1) | / (n+1) ! . |x-Xe | n+ 1 e. g. In(1. 1) to <10-4: smallest n with 1/[(n+1)10"+1] < 10-4 = n=3. 8b . Standard Series 0 MEMORISE . ALL AT ex = Σ xk/k! · 1/(1-x) = Σ xk (|x|<1) sin x = Σ (-1)*x2k+ 1 / (2k+1)! COS x = Σ (-1)*x2* / (2k)! In (1+x) = Σ (-1) *** +1/(k+1) (|x|<1) Parity: sin/arctan (odd) -> odd powers; cos (even) > even powers. A Taylor poly always exists; the series represents f only if Rn >0 - must be shown (ex via Lagrange + |x|"+1/(n+1) !- >0). 8c . Build Without Differentiating METHODS[7]Source: asksia-bible-math1961-bilingual.pdfsin x x Taylor approximation of sin x at 0: the degree-3 polynomial P3 = x - x^3/6 hugs sin x near 0 and peels away as x grows. The poly matches f and its derivatives at the centre. sin x 在 0 处的泰勒逼近:3 次多项式 P3=x-x^3/6 在 0 附近紧贴 sin x,随 x 增大而剥离。该多项式在中 心处匹配 f 及其各阶导数。 MATH1961 . Mathematics 1A (Advanced) 对F 求导 -- 它会相消。每一项乘法法则所产生的部分都与 下一项相抵,只留下唯一的幸存者: 。 3 Read off the endpoints. F(x)=f(x), F(x0)=Tn(x); G(x)=O, G(x0) =- (x-x0)"+1. 读出端点。F(x)=f(x), F(x)=Tn(x); G(x)=0, G(x0) =- (x-xo)n+1。 4 Apply the Cauchy MVT to F,G on the interval: F(x)-F(x0) G(x)-G(I) - G'(C) FIC) for some interior c. 在该区间上对 F,G 应用柯西中值定理:对某个内点 c 成 立。 5 Substitute and simplify. The left side is Rn(x)/(x-x0)n+1; with G'(t)=(n+1)(x-t)" the right side collapses to f(n+1)(c)/(n+1) !. Rearranging gives the Lagrange remainder. 代入并化简。左边是 Rn(x)/(x-xo)n+1;由 G'(t)=(n+1) (x-t)”,右边坍缩为f(n+1)(c)/(n+1) !。 整理即得拉格朗日余 项。■ ★ The Cauchy MVT you are quoting 你所引用的 Cauchy MVT f,g continuous on [a,b], differentiable on (a,b) = = c: [f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c). Itself proved by Rolle on F(x)=[f(b)-f(a)][g(x)-g(a)]-[g(b)-g(a)][f(x)-f(a)]. Name the theorem you invoke - method marks live there. f,g 在[a,b]上连续、在(a,b)上可微⇒ 3c: [f(b) -f(a)]g'(c)=[g(b)-g(a)]f'(c)。它本身由对 F(x)=[f(b) -f(a)][g(x)-g(a)]-[g(b)-g(a)][f(x)-f(a)] 应用 Rolle 定理来证明。点明你所援引的定理 -- 方法分就在那 里。 ! Where students lose the proof 学生在证明中失分之处 The telescoping step (2) is the crux: write out F'(t) for n=1 or 2 by hand first so you see the cancellation before claiming it. And c is between x0 and x, unknown - you never solve for it, you bound f(n+1)(c). 相消步骤(2)是关键所在:先对 n=1 或 2 手工写出 F' (t),让你在断言相消之前先看见它。并且 c 是介于 xo 与 x之间的、未知的 -- 你永远不去解出它,而是去 界定f(n+1)(c)。 MATH1961 . Mathematics 1A (Advanced) AskSia Library EXAM BIBLE . ASKSIA SCHOOL OF MATHEMATICS & STATISTICS a±8 SEMESTER 1 . 2026 L+£ SE 3A sup A T THE COMPLETE EXAM BIBLE · ADVANCED / PROOF - BASED
• 误差界（考试用法）：
  $$|R_n(x)|\le \frac{\max|f^{(n+1)}|}{(n+1)!}|x-x_0|^{n+1}.$$
  用它来“给定精度，求最小 $n$”。[26]Source: asksia-cheatsheet-math1961.pdf7d . Diff => Cts, not conversely CLASSIC TRAP f(x)=|x| is continuous at 0 but not differentiable: the left difference quotient -> -1, the right -> +1, so f'(0) does not exist. Differentiable => continuous; continuous differentiable. 2nd-deriv test inconclusive case: f(x)=x4 at 0 has f'=f"=0 yet a min - fall back to the sign change of f. Rolle in one line: any polynomial of degree n has at most n real roots - between consecutive roots f'(c)=0 by Rolle, so the derivative (degree n-1) caps the root count by induction. The product/chain proofs run cleanest through the Carathéodory form: write each factor as f(x0)+m(x) (x-X0) and read off the derivative from m(x0). Singular critical points (f' undefined) count too: |x| at 0 is a minimum with no derivative - always check both stationary and singular points. 8 . Taylor + Lagrange HIGH -VALUE PROOF NTH TAYLOR POLY AT XO Τη (x) = Σ_{k=Θ}{n} f (k) (xo)/k !. (x-Xe)k Tn is the unique degree-≤n poly P with lim(x->x0) [f-P]/(x-x0)" = 0 - so build it by substitution / multiply / integrate, no repeated differentiation. LAGRANGE REMAINDER A Rn (x) = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 for some c between xe and X Proof: Cauchy MVT on F(t) =__ {k=0}^{n}f(k)(t)/k !· (x-t)k and G(t) =- (x-t)"+1; the sum telescopes to F'(t)=f(n+1) (t)/n !- (x-t)". ERROR BOUND (THE EXAM USE) |Rn (x) | ≤ max | f ( n+1) | / (n+1) ! . |x-Xe | n+ 1 e. g. In(1. 1) to <10-4: smallest n with 1/[(n+1)10"+1] < 10-4 = n=3. 8b . Standard Series 0 MEMORISE . ALL AT ex = Σ xk/k! · 1/(1-x) = Σ xk (|x|<1) sin x = Σ (-1)*x2k+ 1 / (2k+1)! COS x = Σ (-1)*x2* / (2k)! In (1+x) = Σ (-1) *** +1/(k+1) (|x|<1) Parity: sin/arctan (odd) -> odd powers; cos (even) > even powers. A Taylor poly always exists; the series represents f only if Rn >0 - must be shown (ex via Lagrange + |x|"+1/(n+1) !- >0). 8c . Build Without Differentiating METHODS[27]Source: asksia-cheatsheet-math1961.pdfERROR BOUND (THE EXAM USE) |Rn (x) | ≤ max | f ( n+1) | / (n+1) ! . |x-Xe | n+ 1 e. g. In(1. 1) to <10-4: smallest n with 1/[(n+1)10"+1] < 10-4 = n=3. 8b . Standard Series 0 MEMORISE . ALL AT ex = Σ xk/k! · 1/(1-x) = Σ xk (|x|<1) sin x = Σ (-1)*x2k+ 1 / (2k+1)! COS x = Σ (-1)*x2* / (2k)! In (1+x) = Σ (-1) *** +1/(k+1) (|x|<1) Parity: sin/arctan (odd) -> odd powers; cos (even) > even powers. A Taylor poly always exists; the series represents f only if Rn >0 - must be shown (ex via Lagrange + |x|"+1/(n+1) !- >0). 8c . Build Without Differentiating METHODS EXAMINED · Substitution: order-n poly of f(ax™) = Tn (ax™) (order mn). e. g. e4[x2] = Σ x2k/k !. · Integrate/differentiate: integrate 1/(1+t2)=>(-1)kt2k => arctan x = Σ (-1) x2k+ 1/(2k+1). Multiply: multiply two polys, keep terms up to order n. Uniqueness (the limit characterisation) guarantees any method gives the same Taylor polynomial. Also arctan x= Σ(-1)Kx2 ** 1/(2k+1) on|x|<1. 8d . Convergence RADIUS Geometric prototype: 1/(1-x) = lim(1+x+ . . . +x"), remainder x"+1/(1-x) -> 0 for |x|<1. Radius: factorial coefficients (e), e^{-x2}, sin, cos) converge Vx; geometric-type (1/(1+x2)) only on (-1,1). Representing f always needs Rn >0, proved case-by- case. 8e · Worked . Estimate ex LAGRANGE BOUND Estimate e^{0. 5} with T3 and bound the error. T3(0. 5) = 1 +0. 5 +0. 125 + 0. 0208 . . . = 1. 6458. Since f(4)=e` and e^{0. 5}<2,|R3|≤2/4 !· 0. 54=2/24 . 0. 0625 = 0. 0052. So e^{0. 5} =1. 646±0. 006 (true 1. 6487). Taylor poly of degree 2 at 0 for f(x)=(1+x): f(0)=1, f'(0)=1/2, f'(0) =- 1/4, so T2 = 1 + x/2 - x2/8. Limits of indeterminate forms can be read straight off the leading Taylor terms (a fast alternative to L'Hôpital). e. g. lim(x->0)(sin x-x)/x3: sin x=x-x3/6 + . . . , so the quotient -> -1/6. Reading the first non-vanishing term beats repeated L'Hôpital. A A function can be infinitely differentiable yet its Taylor series not represent it away from xo - convergence of the series and equality with f are separate facts. 9 . Riemann Integral DEF + DARBOUX

3.6 FTC + Riemann/Darboux：别把“定义”当“结论”

• 重点提醒：定积分不是“原函数两端相减”——那是 FTC 的结论，不是定义。定义来自 Darboux 上下和/夹逼面积。[9]Source: asksia-bible-math1961-bilingual.pdfEvaluate limx-+0 e™ - 1 - x 22 . 求。 At x = 0 top and bottom -> 0 (form 0/0): differentiate, 2x ea - 1 - still 0/0. 在分子分母(型):求导, -- 仍是。 2 Apply again: → e 1 2 2 再次应用 :。 3 So the limit is }. . 所以极限为。 ! Check the hypotheses BEFORE you differentiate 在求导之前先检查假设 L'Hôpital applies only to genuine 0/0 (or co/00) forms with g' # 0. On lima-+0 COS z - which is not 0/0 - differentiating gives the wrong answer. Verify the form first, every time. - L'Hopital 法则仅适用于真正的 (或)型且。在上 -- 它不是 -- 求导会给出错误答案。每一次都先核 实其形式。 - THEOREM (ROLLE) SEEDS IT ALL 定理(Rolle)是一切的种子 NAMED MVTS IN THE CHAIN 链条中各个命名的 MVT 0/0 THE FORM L'HÔPITAL NEEDS L'Hopital 所需的形式 ⚠ ALWAYS STATE HYPOTHESES 始终陈述假设 Rolle is the seed; Cauchy MVT is the trunk; the MVT, monotonicity and L'Hôpital are the branches. Prove the seed once and you can grow the whole tree on demand. Rolle 是种子;柯西中值定理是树干;MVT、单调性与洛必达法则是枝条。把种子证明一次,你就能随时长出整棵树。 MATH1961 . THE PROOF YOU REUSE ALL SEMESTER MATH1961 . Mathematics 1A (Advanced) x THE RIEMANN INTEGRAL . THE FTC - THE RIEMANN INTEGRAL C6 . PROOF -HEAVY Trap the area between two sums 把面积夹在两个和之间 Darboux upper & lower sums squeeze the integral from above and below Darboux 上和与下和从上下两侧夹逼积分 The definite integral is not "the antiderivative evaluated at the ends" - that is a theorem (the FTC, p. 3). The definition is about area trapped between rectangles: chop [a, b] into strips, over-estimate with the sup on each strip, under-estimate with the inf, and force the gap to zero. 定积分不是“在两端点处取值的原函数” -- 那是一个定理(FTC,见第3页)。其定义关乎被夹在矩形之间的面积:把分成小 条,用每条上的上确界高估,用下确界低估,再迫使缺口趋于零。 UPPER & LOWER (DARBOUX) SUMS Partition P= {a = x0 <. . . < xn = b}, strip width Axk :
• FTC I/II 的证明思路与常用结论（包括 Leibniz 变上限求导公式）在速记里给了“证明骨架+常用结果”。[19]Source: asksia-cheatsheet-math1961.pdfWhy unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +co, so U(f,P)=+co for every P and U-L can't be made < = not integrable (e. g. 1/x on (0,1]). 10 . FTC . both parts PROVED A F(x)=[ a" f is continuous when f integrable (|f|≤C + triangle inequality + squeeze). FTC I. f cts = F(x)=[ a" f is differentiable, F'(x)=f(x). Proof: [F(x+h)-F(x)]/h = (1/h)[x^{x+h} f; by EVT/IVT = f(¿), E=[x,x+h], and {->x as h->0 gives f(x). FTC II. G any primitive (G'=f) = [ ab f=G(b)-G(a). Proof: F-G has zero derivative = constant. LEIBNIZ (VARIABLE LIMITS) d/dx [_{g1 (x)}^{g2 (x)} f = f(g2)g2' - f(g1)g1' From FTC: parts /fg' = [fg] - [f'g; substitution [f(g)g' = J_{g(a)}^{g(b)} f(u)du; partial fractions for rationals ( dt/(1+t2)=arctan t). Worked Leibniz: H(x)=[_{x}^{x2} e^{t2} dt => H'(x) = e^{x4} ·2x- e^{x2]. 1. Improper: [. 1 dx/x diverges (unbounded at 0, handled as lim_{&->0+}[_€1). Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors -> A/(x-r) + (Bx+C)/(x2+px+q) + . . . . Recognise / dt/(1+t2)=arctan t and Jdt/t=In|t|. Integration by parts and substitution both descend from the product and chain rules via FTC II - so every technique is, at heart, a theorem you can prove. The integral function F(x)=[ a" f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II). Proof Recap SIDE 1 ε-δ: Ψε 30, 01x-al<δ = |f-1|<ε sup: (1) upper bound (2) least - BOTH Rolle-Cauchy MVT-MVT-L'Hôpital Rn = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 FTC: F'=f . Jabf = G(b) -G(a) Write proofs in full sentences ; state the definition, name the technique, justify each step. Quick proof-chain map: Completeness (LUB) = IVT & EVT => Rolle => Cauchy MVT => MVT = {monotonicity, L'Hôpital, Lagrange remainder}; EVT+IVT => FTC I = FTC II. Knowing the arrows lets you rebuild any proof from the one before it. Revision aid . check the current unit outline for assessment . @ 2026 flip - for side 2 . complex numbers, linear algebra & eigenvalues asksia. ai/cheatsheet/ usyd-math1961 . side 1/2 AskSia CHEATSHEET SERIES REVISION SHEET . ALL TOPICS Compiled by AskSia . mapped to the MATH1961 syllabus . asksia. ai/cheatsheet/usyd- math1961 -[25]Source: asksia-cheatsheet-math1961.pdfPartition P: a=x <. . . < Xn=b, AXk=Xk-Xk-1, norm |P|=max Δχκ. Riemann sum S=Σ f (xk*) ΔΧκ. INTEGRABLE bounded f is integrable if lim(IPII-0) S = A exists (same A V partitions & samples); Jab f := A Darboux 4 : with U(f,P)=Σ (sup f) Axk L(f,P)=Σ (inff)Axk, f integrable VE>0 3P: U-L<&; then L ≤ ] ≤U. Continuous => integrable; monotone => integrable. A Bounded = integrable (Dirichlet 1_Q: U=1, L =- 1 always). Unbounded => not integrable. From definition: for f(x)=x on [a,b], equidistant partition gives Un,Ln > (b2-a2)/2 using Ek = n(n+1)/2, so Jab x dx = (b2-a2)/2. 9b . Integral F,G Properties INTEGRABLE | [(kf+2g) = kff + {[g (linearity) Sab f = fac f + [_c^b f (additivity) f & g - [f & fg . |[f] > [|f| Symmetry tricks: on [-a,a] even => 21. ª, odd => 0; J . ^Tt X f(sin x)dx = (Tt/2)/0^Tt f(sin x)dx. Why unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +co, so U(f,P)=+co for every P and U-L can't be made < = not integrable (e. g. 1/x on (0,1]). 10 . FTC . both parts PROVED A F(x)=[ a" f is continuous when f integrable (|f|≤C + triangle inequality + squeeze). FTC I. f cts = F(x)=[ a" f is differentiable, F'(x)=f(x). Proof: [F(x+h)-F(x)]/h = (1/h)[x^{x+h} f; by EVT/IVT = f(¿), E=[x,x+h], and {->x as h->0 gives f(x). FTC II. G any primitive (G'=f) = [ ab f=G(b)-G(a). Proof: F-G has zero derivative = constant. LEIBNIZ (VARIABLE LIMITS) d/dx [_{g1 (x)}^{g2 (x)} f = f(g2)g2' - f(g1)g1' From FTC: parts /fg' = [fg] - [f'g; substitution [f(g)g' = J_{g(a)}^{g(b)} f(u)du; partial fractions for rationals ( dt/(1+t2)=arctan t). Worked Leibniz: H(x)=[_{x}^{x2} e^{t2} dt => H'(x) = e^{x4} ·2x- e^{x2]. 1. Improper: [. 1 dx/x diverges (unbounded at 0, handled as lim_{&->0+}[_€1). Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors -> A/(x-r) + (Bx+C)/(x2+px+q) + . . . . Recognise / dt/(1+t2)=arctan t and Jdt/t=In|t|. Integration by parts and substitution both descend from the product and chain rules via FTC II - so every technique is, at heart, a theorem you can prove. The integral function F(x)=[ a" f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II). Proof Recap SIDE 1 ε-δ: Ψε 30, 01x-al<δ = |f-1|<ε sup: (1) upper bound (2) least - BOTH Rolle-Cauchy MVT-MVT-L'Hôpital Rn = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 FTC: F'=f . Jabf = G(b) -G(a) Write proofs in full sentences ; state the definition, name the technique, justify each step.

4.1 L’Hôpital（洛必达）：先检查形式与条件

• 只对真正 $0/0$（或 $\infty/\infty$）形式可用，并且要检查导数条件（如 $g'\neq 0$ 等要求）；不是这个形式硬求导会错。[9]Source: asksia-bible-math1961-bilingual.pdfEvaluate limx-+0 e™ - 1 - x 22 . 求。 At x = 0 top and bottom -> 0 (form 0/0): differentiate, 2x ea - 1 - still 0/0. 在分子分母(型):求导, -- 仍是。 2 Apply again: → e 1 2 2 再次应用 :。 3 So the limit is }. . 所以极限为。 ! Check the hypotheses BEFORE you differentiate 在求导之前先检查假设 L'Hôpital applies only to genuine 0/0 (or co/00) forms with g' # 0. On lima-+0 COS z - which is not 0/0 - differentiating gives the wrong answer. Verify the form first, every time. - L'Hopital 法则仅适用于真正的 (或)型且。在上 -- 它不是 -- 求导会给出错误答案。每一次都先核 实其形式。 - THEOREM (ROLLE) SEEDS IT ALL 定理(Rolle)是一切的种子 NAMED MVTS IN THE CHAIN 链条中各个命名的 MVT 0/0 THE FORM L'HÔPITAL NEEDS L'Hopital 所需的形式 ⚠ ALWAYS STATE HYPOTHESES 始终陈述假设 Rolle is the seed; Cauchy MVT is the trunk; the MVT, monotonicity and L'Hôpital are the branches. Prove the seed once and you can grow the whole tree on demand. Rolle 是种子;柯西中值定理是树干;MVT、单调性与洛必达法则是枝条。把种子证明一次,你就能随时长出整棵树。 MATH1961 . THE PROOF YOU REUSE ALL SEMESTER MATH1961 . Mathematics 1A (Advanced) x THE RIEMANN INTEGRAL . THE FTC - THE RIEMANN INTEGRAL C6 . PROOF -HEAVY Trap the area between two sums 把面积夹在两个和之间 Darboux upper & lower sums squeeze the integral from above and below Darboux 上和与下和从上下两侧夹逼积分 The definite integral is not "the antiderivative evaluated at the ends" - that is a theorem (the FTC, p. 3). The definition is about area trapped between rectangles: chop [a, b] into strips, over-estimate with the sup on each strip, under-estimate with the inf, and force the gap to zero. 定积分不是“在两端点处取值的原函数” -- 那是一个定理(FTC,见第3页)。其定义关乎被夹在矩形之间的面积:把分成小 条,用每条上的上确界高估,用下确界低估,再迫使缺口趋于零。 UPPER & LOWER (DARBOUX) SUMS Partition P= {a = x0 <. . . < xn = b}, strip width Axk :

4.2 复数与 de Moivre + n 次根

• 极坐标/指数形式：$z=re^{i\theta}=r(\cos\theta+i\sin\theta)$，乘除时“模相乘/相除，幅角相加/相减”。[21]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Complex . linear algebra . prove it SIDE 2/2 Eigenvalues 11 . Complex Numbers CARTESIAN + POLAR z=x+iy, i2 =- 1. Argand plot (x,y). Conjugate z = x-iy; Re z = (z+z)/2, Im z = (z-z)/2i; |z|2 = z ż. POLAR / EXPONENTIAL Z = r(cos 0 + i sin 0) = r e^{i0} r = |z| = V(x2+y2), 0 = arg z arg z mod 21; principal Arg z € (-1,T]. Euler: e^{i0} = cos 0 + i sin 0. Product/quotient: multiply/divide moduli, add/subtract args. A Cartesian>> polar is where sign/quadrant errors cost marks - check which quadrant (x,y) sits in before fixing 0. Arithmetic: (a+ib)(c+id) = (ac-bd) + i(ad+bc); divide by multiplying by the conjugate, z/w = z·w/|w|2. Quadratics over C: usual formula; for real coefficients complex roots come in conjugate pairs. 11b . de Moivre & INDUCTION PROOF Roots DE MOIVRE, NEZ (r(cos 8 + i sin 0))" = rm(cos ne + i sin n0) i. e. (r e^{i0})" = rn e^{in0} Proof for nEN by induction: n=1 trivial; assume for k, then (e^{i0})^{k+1}= (e^{i0})^k. e^{i0]=e^{ik0}e^{i0}=e^{i(k+1)0} by the angle- sum identities. # (extend to Z via z-1. ) Use: multiple-angle formulas; expand (1+ic)" by binomial, split Re/Im. N-TH ROOTS OF Z = R E^{IO} ak = r^{1/n} e^{i(0+2mk)/n}, k=0, . . . , n-1 n-th roots of unity: e^{2Ttik/n} - n equally-spaced points, a regular n-gon on the unit circle. ! For z"=a give all n roots. Real-coeff polynomials: complex roots in conjugate pairs. 11c . Worked . Roots of Unity SOLVE Zn=A
• de Moivre：$(re^{i\theta})^n=r^n e^{in\theta}$，教材里还强调可以用归纳证明。[21]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Complex . linear algebra . prove it SIDE 2/2 Eigenvalues 11 . Complex Numbers CARTESIAN + POLAR z=x+iy, i2 =- 1. Argand plot (x,y). Conjugate z = x-iy; Re z = (z+z)/2, Im z = (z-z)/2i; |z|2 = z ż. POLAR / EXPONENTIAL Z = r(cos 0 + i sin 0) = r e^{i0} r = |z| = V(x2+y2), 0 = arg z arg z mod 21; principal Arg z € (-1,T]. Euler: e^{i0} = cos 0 + i sin 0. Product/quotient: multiply/divide moduli, add/subtract args. A Cartesian>> polar is where sign/quadrant errors cost marks - check which quadrant (x,y) sits in before fixing 0. Arithmetic: (a+ib)(c+id) = (ac-bd) + i(ad+bc); divide by multiplying by the conjugate, z/w = z·w/|w|2. Quadratics over C: usual formula; for real coefficients complex roots come in conjugate pairs. 11b . de Moivre & INDUCTION PROOF Roots DE MOIVRE, NEZ (r(cos 8 + i sin 0))" = rm(cos ne + i sin n0) i. e. (r e^{i0})" = rn e^{in0} Proof for nEN by induction: n=1 trivial; assume for k, then (e^{i0})^{k+1}= (e^{i0})^k. e^{i0]=e^{ik0}e^{i0}=e^{i(k+1)0} by the angle- sum identities. # (extend to Z via z-1. ) Use: multiple-angle formulas; expand (1+ic)" by binomial, split Re/Im. N-TH ROOTS OF Z = R E^{IO} ak = r^{1/n} e^{i(0+2mk)/n}, k=0, . . . , n-1 n-th roots of unity: e^{2Ttik/n} - n equally-spaced points, a regular n-gon on the unit circle. ! For z"=a give all n roots. Real-coeff polynomials: complex roots in conjugate pairs. 11c . Worked . Roots of Unity SOLVE Zn=A
• $z^n=a$ 要写出 全部 $n$ 个根：
  $$a_k=r^{1/n}e^{i(\theta+2\pi k)/n},\ k=0,\dots,n-1.$$
  常见扣分点：只写一个根就停。[21]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Complex . linear algebra . prove it SIDE 2/2 Eigenvalues 11 . Complex Numbers CARTESIAN + POLAR z=x+iy, i2 =- 1. Argand plot (x,y). Conjugate z = x-iy; Re z = (z+z)/2, Im z = (z-z)/2i; |z|2 = z ż. POLAR / EXPONENTIAL Z = r(cos 0 + i sin 0) = r e^{i0} r = |z| = V(x2+y2), 0 = arg z arg z mod 21; principal Arg z € (-1,T]. Euler: e^{i0} = cos 0 + i sin 0. Product/quotient: multiply/divide moduli, add/subtract args. A Cartesian>> polar is where sign/quadrant errors cost marks - check which quadrant (x,y) sits in before fixing 0. Arithmetic: (a+ib)(c+id) = (ac-bd) + i(ad+bc); divide by multiplying by the conjugate, z/w = z·w/|w|2. Quadratics over C: usual formula; for real coefficients complex roots come in conjugate pairs. 11b . de Moivre & INDUCTION PROOF Roots DE MOIVRE, NEZ (r(cos 8 + i sin 0))" = rm(cos ne + i sin n0) i. e. (r e^{i0})" = rn e^{in0} Proof for nEN by induction: n=1 trivial; assume for k, then (e^{i0})^{k+1}= (e^{i0})^k. e^{i0]=e^{ik0}e^{i0}=e^{i(k+1)0} by the angle- sum identities. # (extend to Z via z-1. ) Use: multiple-angle formulas; expand (1+ic)" by binomial, split Re/Im. N-TH ROOTS OF Z = R E^{IO} ak = r^{1/n} e^{i(0+2mk)/n}, k=0, . . . , n-1 n-th roots of unity: e^{2Ttik/n} - n equally-spaced points, a regular n-gon on the unit circle. ! For z"=a give all n roots. Real-coeff polynomials: complex roots in conjugate pairs. 11c . Worked . Roots of Unity SOLVE Zn=A[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors

4.3 特征值/对角化：考的是“推理”，不是报答案

• 必做顺序：先特征多项式 $\chi_A(\lambda)=\det(A-\lambda I)$，再找特征向量；并写清楚重数。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.
• 重根陷阱：代数重数 2 不代表可对角化，必须检查特征空间维数（几何重数）。示例 $A=\begin{pmatrix}2&1\0&2\end{pmatrix}$ 只有 1 个线性无关特征向量，所以不可对角化。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.
• 快速 sanity check：$\mathrm{tr}(A)=\sum\lambda$，$\det(A)=\prod\lambda$。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.

4.4 Rank–Nullity & 子空间/基

• 记住：$\mathrm{rank}(A)+\mathrm{nullity}(A)=n$（$n$ 是列数）。并能从 RREF 的 pivot 数读 rank，再得 nullity。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.[30]Source: asksia-cheatsheet-math1961.pdfFinding bases (R = RREF A): row space -> nonzero rows of R; column space > columns of A in R's pivot positions (col(A)#col(R) but same dim); null space -> solve Rx=0, separate parameters. RANK-NULLITY A rank (A) + nullity (A) = n rank = # leading 1s = dim row = dim col; nullity = dim null. Proof: the n columns split into pivot (-> rank) vs free (-> nullity) . Corollary: in dim-n space, n vectors are l. i. - they span. 20 . Linear Transformations MATRIX STANDARD T:Rn->R™ linear if T(u+v)=T(u)+T(v) and T(cu)=cT(u). Hence T(0)=0. Standard-matrix thm A : every linear T = L_A with A = [T(e1) . . . T(en)] (mxn, unique). Proof: x=Ex,e, + linearity. Composition: A_{S . T} = A_SA_T. A Prove T NOT linear: exhibit a failing case - absolute values, constants or squares break it (e. g. T with a |z| term fails T(cu)=cT(u) at c =- 1). T:R"->R" bijective - A_T invertible. 23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts.
• Column space 的陷阱：$\mathrm{col}(A)$ 要用 原矩阵 A 的 pivot 列，不是用 RREF 的列。[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors[30]Source: asksia-cheatsheet-math1961.pdfFinding bases (R = RREF A): row space -> nonzero rows of R; column space > columns of A in R's pivot positions (col(A)#col(R) but same dim); null space -> solve Rx=0, separate parameters. RANK-NULLITY A rank (A) + nullity (A) = n rank = # leading 1s = dim row = dim col; nullity = dim null. Proof: the n columns split into pivot (-> rank) vs free (-> nullity) . Corollary: in dim-n space, n vectors are l. i. - they span. 20 . Linear Transformations MATRIX STANDARD T:Rn->R™ linear if T(u+v)=T(u)+T(v) and T(cu)=cT(u). Hence T(0)=0. Standard-matrix thm A : every linear T = L_A with A = [T(e1) . . . T(en)] (mxn, unique). Proof: x=Ex,e, + linearity. Composition: A_{S . T} = A_SA_T. A Prove T NOT linear: exhibit a failing case - absolute values, constants or squares break it (e. g. T with a |z| term fails T(cu)=cT(u) at c =- 1). T:R"->R" bijective - A_T invertible. 23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts.

第 1 优先级：把“可背诵的证明”背下来（你会直接赚分）

• $\varepsilon$–$\delta$ 定义 + 证明一条典型极限（按模板写）[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF
• $\sup$ 的两子句证明法（特别是 clause (2) 的 $\varepsilon$-argument）[15]Source: asksia-bible-math1961-bilingual.pdf下确界是对偶概念:最大的下界。 i sup A need not lie in A 未必落在其中 For A = {1 - 1 : n ≥ 1} = {0, 2, 2, . . . } we have sup A = 1 ¢ A but inf A = 0 € A. When the sup is attained, sup A = max A. 对于 我们有 但。当 sup 确实被取到时,。 ★ The Least Upper Bound Axiom - completeness of R 最小上界公理––R的完备性 Every nonempty A C R that is bounded above has a supremum in R. (Dual: the Greatest Lower Bound Axiom gives inf A. ) This is the axiom Q fails - and it is the foundation every later existence proof stands on. 每一个非空且有上界的ACR在R 中都有上确界。 (对偶:最大下界公理给出。)这正是 所不具备的公 理 -- 也是后续每一个存在性证明所立足的根基。 How to PROVE sup A = '- both c clauses ‘ 如何证明“” -- 两个子句 1 Clause (1): c is an upper bound. Take an arbitrary a E A and show a ≤ c. 子句(1): c 是一个上界。取任意 并证明。 2 Clause (2): nothing smaller works. The standard move - fix any & > 0 and exhibit an a € A with a > c - 8. That proves c - & is not an upper bound, so no number below c can be one. 子句(2):没有更小的数行得通。标准做法 -- 固定任意 并 给出一个满足 的。这就证明了 不是上界,故任何小于 的 数都不可能是上界。 MATH1961 . Mathematics 1A (Advanced) ! BOTH clauses, every time 每一次都要写齐两个子句 A 'find / prove the sup' answer that shows only that c is an upper bound is half a proof. You must also kill everything below c - the s-argument in step 2. Forgetting clause (2) is the most common mark-loser here. 一个“求/证 sup”的答案如果只表明 是一个上界,那 只是半个证明。你还必须排除其下方的一切 -- 即步 骤2 中的 论证。漏掉子句(2)是这里最常见的失分原 因。 ✓ The bridge to limits: an approximating sequence 通往极限的桥梁:一个逼近数列 From clause (2) with a = - you get points an € A with c - - < an ≤ c, hence an -> sup A. This is precisely how completeness powers monotone convergence and the IVT in Ch 2. 由子句(2),取 可得到满足 的点 ,因此。这恰恰就 是完备性如何驱动第 2 章中的单调收敛与 IVT。 MATH1961 . Mathematics 1A (Advanced) E-A LIMIT 2 . LIMITS, CONTINUITY & THE REALS AHA - UNIT The precise &-o definition of a limit 极限的精确 8-ǒ 定义 The centrepiece of the Advanced unit - you must PROVE limits, not just apply laws 进阶单元的核心 -- 你必须证明极限,而不只是套用法则 Mainstream calculus says "f (x) gets close to L as x gets close to a. " That is a feeling, not mathematics. MATH1961 replaces it with a challenge-response game with a precise winning condition - the £-o definition, the single most important statement in the calculus stream.
• Rolle $\to$ Cauchy MVT $\to$ MVT（命名+假设+辅助函数）[14]Source: asksia-bible-math1961-bilingual.pdfA PROOF . CENTREPIECE One auxiliary function does both 一个辅助函数同时搞定两者 Build the right helper, apply Rolle, and the MVT falls out as g(x) = x 构造出合适的辅助函数,应用 Rolle,MVT 便随之而出 The course proves the Mean Value Theorem as a corollary of the Cauchy MVT, and the Cauchy MVT itself by a single clever application of Rolle. This two-step descent is the most reproduced proof on the exam. 本课程把中值定理作为 Cauchy MVT 的推论来证明,而 Cauchy MVT 本身则通过对 Rolle 的一次巧妙应用得到。这一两步 下降是考试中被复现最多的证明。 CAUCHY MEAN VALUE THEOREM Cauchy MVT : f,g cts on [a,b], diff. on (a, b) >Ice (a,b) : [f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c) Proof idea - the auxiliary function 证明思路 -- 辅助函数 1 Build the helper F(x) = [f(b) - f(a)] [g(x) -g(a)] - [g(b) - g(a)] [f(x) - f(a)]. It is continuous on [a, b] and differentiable on (a, b) (sums/products of such f, g). 构造辅助函数。它在 上连续、在 上可微(此类函数的和/积)。 2 Check the endpoints. Substituting gives F(a) = 0 and - the design pays off - F(b) = 0 as well, so F(a) = F(b). 检查端点。代入得,同样 -- 设计奏效了 -- 于是。 3 Apply Rolle to F. There is c € (a, b) with F'(c) = 0. 对 应用 Rolle。存在 使。 4 Differentiate and read off: F'(c) = [f(b)-f(a)]g'(c) -[g(b)-g(a)]f'(c) = 0, which is exactly the claim. " 求导并读出:,这恰好就是所断言的。 MEAN VALUE THEOREM MVT: f cts on [a,b], diff. on (a,b) f(b) - f(a) > Ice (a,b) : f'(c) = b - a MVT as a one-line corollary MVT 作为一行推论 Put g(x) = x in the Cauchy MVT. Then g'(c) = 1 and g(b) - g(a) =b-a, so [f(b)-f(a)] · 1 = (b-a)f'(c). Divide by b - a: the MVT. (Equivalently: apply Rolle to f minus its chord L(x) - same idea, narrower. ) 在 Cauchy MVT 中令。则与,于是。除以:即得 MVT。 (等价地:对减去其弦线应用 Rolle -- 同样的思路,范 围更窄。) MATH1961 . Mathematics 1A (Advanced) C4 EX 3 MVT gives a clean inequality[9]Source: asksia-bible-math1961-bilingual.pdfEvaluate limx-+0 e™ - 1 - x 22 . 求。 At x = 0 top and bottom -> 0 (form 0/0): differentiate, 2x ea - 1 - still 0/0. 在分子分母(型):求导, -- 仍是。 2 Apply again: → e 1 2 2 再次应用 :。 3 So the limit is }. . 所以极限为。 ! Check the hypotheses BEFORE you differentiate 在求导之前先检查假设 L'Hôpital applies only to genuine 0/0 (or co/00) forms with g' # 0. On lima-+0 COS z - which is not 0/0 - differentiating gives the wrong answer. Verify the form first, every time. - L'Hopital 法则仅适用于真正的 (或)型且。在上 -- 它不是 -- 求导会给出错误答案。每一次都先核 实其形式。 - THEOREM (ROLLE) SEEDS IT ALL 定理(Rolle)是一切的种子 NAMED MVTS IN THE CHAIN 链条中各个命名的 MVT 0/0 THE FORM L'HÔPITAL NEEDS L'Hopital 所需的形式 ⚠ ALWAYS STATE HYPOTHESES 始终陈述假设 Rolle is the seed; Cauchy MVT is the trunk; the MVT, monotonicity and L'Hôpital are the branches. Prove the seed once and you can grow the whole tree on demand. Rolle 是种子;柯西中值定理是树干;MVT、单调性与洛必达法则是枝条。把种子证明一次,你就能随时长出整棵树。 MATH1961 . THE PROOF YOU REUSE ALL SEMESTER MATH1961 . Mathematics 1A (Advanced) x THE RIEMANN INTEGRAL . THE FTC - THE RIEMANN INTEGRAL C6 . PROOF -HEAVY Trap the area between two sums 把面积夹在两个和之间 Darboux upper & lower sums squeeze the integral from above and below Darboux 上和与下和从上下两侧夹逼积分 The definite integral is not "the antiderivative evaluated at the ends" - that is a theorem (the FTC, p. 3). The definition is about area trapped between rectangles: chop [a, b] into strips, over-estimate with the sup on each strip, under-estimate with the inf, and force the gap to zero. 定积分不是“在两端点处取值的原函数” -- 那是一个定理(FTC,见第3页)。其定义关乎被夹在矩形之间的面积:把分成小 条,用每条上的上确界高估,用下确界低估,再迫使缺口趋于零。 UPPER & LOWER (DARBOUX) SUMS Partition P= {a = x0 <. . . < xn = b}, strip width Axk :
• Taylor remainder（拉格朗日余项公式 + 误差界怎么用）[26]Source: asksia-cheatsheet-math1961.pdf7d . Diff => Cts, not conversely CLASSIC TRAP f(x)=|x| is continuous at 0 but not differentiable: the left difference quotient -> -1, the right -> +1, so f'(0) does not exist. Differentiable => continuous; continuous differentiable. 2nd-deriv test inconclusive case: f(x)=x4 at 0 has f'=f"=0 yet a min - fall back to the sign change of f. Rolle in one line: any polynomial of degree n has at most n real roots - between consecutive roots f'(c)=0 by Rolle, so the derivative (degree n-1) caps the root count by induction. The product/chain proofs run cleanest through the Carathéodory form: write each factor as f(x0)+m(x) (x-X0) and read off the derivative from m(x0). Singular critical points (f' undefined) count too: |x| at 0 is a minimum with no derivative - always check both stationary and singular points. 8 . Taylor + Lagrange HIGH -VALUE PROOF NTH TAYLOR POLY AT XO Τη (x) = Σ_{k=Θ}{n} f (k) (xo)/k !. (x-Xe)k Tn is the unique degree-≤n poly P with lim(x->x0) [f-P]/(x-x0)" = 0 - so build it by substitution / multiply / integrate, no repeated differentiation. LAGRANGE REMAINDER A Rn (x) = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 for some c between xe and X Proof: Cauchy MVT on F(t) =__ {k=0}^{n}f(k)(t)/k !· (x-t)k and G(t) =- (x-t)"+1; the sum telescopes to F'(t)=f(n+1) (t)/n !- (x-t)". ERROR BOUND (THE EXAM USE) |Rn (x) | ≤ max | f ( n+1) | / (n+1) ! . |x-Xe | n+ 1 e. g. In(1. 1) to <10-4: smallest n with 1/[(n+1)10"+1] < 10-4 = n=3. 8b . Standard Series 0 MEMORISE . ALL AT ex = Σ xk/k! · 1/(1-x) = Σ xk (|x|<1) sin x = Σ (-1)*x2k+ 1 / (2k+1)! COS x = Σ (-1)*x2* / (2k)! In (1+x) = Σ (-1) *** +1/(k+1) (|x|<1) Parity: sin/arctan (odd) -> odd powers; cos (even) > even powers. A Taylor poly always exists; the series represents f only if Rn >0 - must be shown (ex via Lagrange + |x|"+1/(n+1) !- >0). 8c . Build Without Differentiating METHODS[27]Source: asksia-cheatsheet-math1961.pdfERROR BOUND (THE EXAM USE) |Rn (x) | ≤ max | f ( n+1) | / (n+1) ! . |x-Xe | n+ 1 e. g. In(1. 1) to <10-4: smallest n with 1/[(n+1)10"+1] < 10-4 = n=3. 8b . Standard Series 0 MEMORISE . ALL AT ex = Σ xk/k! · 1/(1-x) = Σ xk (|x|<1) sin x = Σ (-1)*x2k+ 1 / (2k+1)! COS x = Σ (-1)*x2* / (2k)! In (1+x) = Σ (-1) *** +1/(k+1) (|x|<1) Parity: sin/arctan (odd) -> odd powers; cos (even) > even powers. A Taylor poly always exists; the series represents f only if Rn >0 - must be shown (ex via Lagrange + |x|"+1/(n+1) !- >0). 8c . Build Without Differentiating METHODS EXAMINED · Substitution: order-n poly of f(ax™) = Tn (ax™) (order mn). e. g. e4[x2] = Σ x2k/k !. · Integrate/differentiate: integrate 1/(1+t2)=>(-1)kt2k => arctan x = Σ (-1) x2k+ 1/(2k+1). Multiply: multiply two polys, keep terms up to order n. Uniqueness (the limit characterisation) guarantees any method gives the same Taylor polynomial. Also arctan x= Σ(-1)Kx2 ** 1/(2k+1) on|x|<1. 8d . Convergence RADIUS Geometric prototype: 1/(1-x) = lim(1+x+ . . . +x"), remainder x"+1/(1-x) -> 0 for |x|<1. Radius: factorial coefficients (e), e^{-x2}, sin, cos) converge Vx; geometric-type (1/(1+x2)) only on (-1,1). Representing f always needs Rn >0, proved case-by- case. 8e · Worked . Estimate ex LAGRANGE BOUND Estimate e^{0. 5} with T3 and bound the error. T3(0. 5) = 1 +0. 5 +0. 125 + 0. 0208 . . . = 1. 6458. Since f(4)=e` and e^{0. 5}<2,|R3|≤2/4 !· 0. 54=2/24 . 0. 0625 = 0. 0052. So e^{0. 5} =1. 646±0. 006 (true 1. 6487). Taylor poly of degree 2 at 0 for f(x)=(1+x): f(0)=1, f'(0)=1/2, f'(0) =- 1/4, so T2 = 1 + x/2 - x2/8. Limits of indeterminate forms can be read straight off the leading Taylor terms (a fast alternative to L'Hôpital). e. g. lim(x->0)(sin x-x)/x3: sin x=x-x3/6 + . . . , so the quotient -> -1/6. Reading the first non-vanishing term beats repeated L'Hôpital. A A function can be infinitely differentiable yet its Taylor series not represent it away from xo - convergence of the series and equality with f are separate facts. 9 . Riemann Integral DEF + DARBOUX
• FTC I/II 的核心思路（至少会把结构写出来）[19]Source: asksia-cheatsheet-math1961.pdfWhy unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +co, so U(f,P)=+co for every P and U-L can't be made < = not integrable (e. g. 1/x on (0,1]). 10 . FTC . both parts PROVED A F(x)=[ a" f is continuous when f integrable (|f|≤C + triangle inequality + squeeze). FTC I. f cts = F(x)=[ a" f is differentiable, F'(x)=f(x). Proof: [F(x+h)-F(x)]/h = (1/h)[x^{x+h} f; by EVT/IVT = f(¿), E=[x,x+h], and {->x as h->0 gives f(x). FTC II. G any primitive (G'=f) = [ ab f=G(b)-G(a). Proof: F-G has zero derivative = constant. LEIBNIZ (VARIABLE LIMITS) d/dx [_{g1 (x)}^{g2 (x)} f = f(g2)g2' - f(g1)g1' From FTC: parts /fg' = [fg] - [f'g; substitution [f(g)g' = J_{g(a)}^{g(b)} f(u)du; partial fractions for rationals ( dt/(1+t2)=arctan t). Worked Leibniz: H(x)=[_{x}^{x2} e^{t2} dt => H'(x) = e^{x4} ·2x- e^{x2]. 1. Improper: [. 1 dx/x diverges (unbounded at 0, handled as lim_{&->0+}[_€1). Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors -> A/(x-r) + (Bx+C)/(x2+px+q) + . . . . Recognise / dt/(1+t2)=arctan t and Jdt/t=In|t|. Integration by parts and substitution both descend from the product and chain rules via FTC II - so every technique is, at heart, a theorem you can prove. The integral function F(x)=[ a" f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II). Proof Recap SIDE 1 ε-δ: Ψε 30, 01x-al<δ = |f-1|<ε sup: (1) upper bound (2) least - BOTH Rolle-Cauchy MVT-MVT-L'Hôpital Rn = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 FTC: F'=f . Jabf = G(b) -G(a) Write proofs in full sentences ; state the definition, name the technique, justify each step. Quick proof-chain map: Completeness (LUB) = IVT & EVT => Rolle => Cauchy MVT => MVT = {monotonicity, L'Hôpital, Lagrange remainder}; EVT+IVT => FTC I = FTC II. Knowing the arrows lets you rebuild any proof from the one before it. Revision aid . check the current unit outline for assessment . @ 2026 flip - for side 2 . complex numbers, linear algebra & eigenvalues asksia. ai/cheatsheet/ usyd-math1961 . side 1/2 AskSia CHEATSHEET SERIES REVISION SHEET . ALL TOPICS Compiled by AskSia . mapped to the MATH1961 syllabus . asksia. ai/cheatsheet/usyd- math1961 -[25]Source: asksia-cheatsheet-math1961.pdfPartition P: a=x <. . . < Xn=b, AXk=Xk-Xk-1, norm |P|=max Δχκ. Riemann sum S=Σ f (xk*) ΔΧκ. INTEGRABLE bounded f is integrable if lim(IPII-0) S = A exists (same A V partitions & samples); Jab f := A Darboux 4 : with U(f,P)=Σ (sup f) Axk L(f,P)=Σ (inff)Axk, f integrable VE>0 3P: U-L<&; then L ≤ ] ≤U. Continuous => integrable; monotone => integrable. A Bounded = integrable (Dirichlet 1_Q: U=1, L =- 1 always). Unbounded => not integrable. From definition: for f(x)=x on [a,b], equidistant partition gives Un,Ln > (b2-a2)/2 using Ek = n(n+1)/2, so Jab x dx = (b2-a2)/2. 9b . Integral F,G Properties INTEGRABLE | [(kf+2g) = kff + {[g (linearity) Sab f = fac f + [_c^b f (additivity) f & g - [f & fg . |[f] > [|f| Symmetry tricks: on [-a,a] even => 21. ª, odd => 0; J . ^Tt X f(sin x)dx = (Tt/2)/0^Tt f(sin x)dx. Why unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +co, so U(f,P)=+co for every P and U-L can't be made < = not integrable (e. g. 1/x on (0,1]). 10 . FTC . both parts PROVED A F(x)=[ a" f is continuous when f integrable (|f|≤C + triangle inequality + squeeze). FTC I. f cts = F(x)=[ a" f is differentiable, F'(x)=f(x). Proof: [F(x+h)-F(x)]/h = (1/h)[x^{x+h} f; by EVT/IVT = f(¿), E=[x,x+h], and {->x as h->0 gives f(x). FTC II. G any primitive (G'=f) = [ ab f=G(b)-G(a). Proof: F-G has zero derivative = constant. LEIBNIZ (VARIABLE LIMITS) d/dx [_{g1 (x)}^{g2 (x)} f = f(g2)g2' - f(g1)g1' From FTC: parts /fg' = [fg] - [f'g; substitution [f(g)g' = J_{g(a)}^{g(b)} f(u)du; partial fractions for rationals ( dt/(1+t2)=arctan t). Worked Leibniz: H(x)=[_{x}^{x2} e^{t2} dt => H'(x) = e^{x4} ·2x- e^{x2]. 1. Improper: [. 1 dx/x diverges (unbounded at 0, handled as lim_{&->0+}[_€1). Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors -> A/(x-r) + (Bx+C)/(x2+px+q) + . . . . Recognise / dt/(1+t2)=arctan t and Jdt/t=In|t|. Integration by parts and substitution both descend from the product and chain rules via FTC II - so every technique is, at heart, a theorem you can prove. The integral function F(x)=[ a" f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II). Proof Recap SIDE 1 ε-δ: Ψε 30, 01x-al<δ = |f-1|<ε sup: (1) upper bound (2) least - BOTH Rolle-Cauchy MVT-MVT-L'Hôpital Rn = f(n+1) (c)/(n+1) ! . (x-Xe)n+1 FTC: F'=f . Jabf = G(b) -G(a) Write proofs in full sentences ; state the definition, name the technique, justify each step.

第 2 优先级：把“高频陷阱”做成错题清单（考前背一遍）

• 量词顺序写错（尤其 $\forall\varepsilon,\exists\delta$）[5]Source: asksia-bible-math1961-bilingual.pdf为什么是这个顺序 The Advanced unit hangs everything on a rigour toolkit - logic, the proof methods, induction, and the completeness of R (sup / inf). We put that first (Chs 1-2) because the same machinery proves the IVT, the EVT, the MVT and the FTC later on. Master the spine and the rest of the course is 'apply the toolkit'. The three chapters in this build are Ch 1 Foundations and Ch 2 Limits & continuity, sitting under the front matter you are reading now. 高阶单元把一切都挂在一套严谨性工具箱上 -- 逻辑、各种证明方法、数学归纳法,以及R的完备性(sup/ inf)。我们 把它放在最前(第1-2章),因为同一套机制随后会用来证明 IVT、EVT、MVT 和 FTC。掌握这条主干,课程其余部分 便只是“套用工具箱”。本版本中的三章是 第1章 基础 与 第2 章 极限与连续性,置于你正在阅读的前言之下。 MATH1961 . Mathematics 1A (Advanced) LOGIC 1 . THE RIGOUR TOOLKIT LO1 . RIGOUR Logic, quantifiers & reading a statement exactly 逻辑、量词,以及精确读懂一个命题 Marks are lost for sloppy logic - the quantifier order is part of the answer 逻辑松散会丢分 -- 量词的顺序也是答案的一部分 Advanced mathematics is written in a precise language. Before any proof, you must read a statement exactly. what is being asserted, in what order, and over which set. The lecturer grades this - a misplaced quantifier or a confused implication arrow loses marks even when the underlying idea is right. 进阶数学以一种精确的语言书写。在任何证明之前,你必须精确地读懂一条命题:它断言了什么,以什么顺序,在哪个集合 上。讲师会对此评分 -- 即使底层想法正确,一个错位的量词或一个混淆的蕴含箭头都会丢分。 - 1. 1 Notation you must use correctly 1. 1 你必须正确使用的记号 Symbol Reads as €,¢ is / is not an element of = defined to equal (vs =, a deduced equality) >, <, >> implies . is implied by . iff (all different!) E'A for all . there exists NCZCQCRC C the number-system chain ! Quantifier ORDER changes the meaning 量词顺序会改变含义 "Valy : y > x" (true in R) is not "By Vx : y > x" (false - no single y beats every x). In the &-o definition the order Ve Ed is the whole point: o is allowed to depend on &. Swap them and you have written something false. (在 中为真)不同于“”(为假 -- 不存在单一的 y 胜过每一个 x)。在c-ǒ 定义中,顺序 VE38 才是关键 所在:ō 是允许依赖于c的。把它们对调,你写出的 便是错的。 1. 2 Negating quantified statements[18]Source: asksia-cheatsheet-math1961.pdfAssessment: exam 60% · mid-sem quiz 13% . 10 online quizzes 10% . 2 assignments 15% . tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper. Where marks are won: E-8 limit proofs . sup/inf completeness . induction (weak & strong) . the Rolle->MVT->Taylor->FTC chain . rank-nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do. Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers -> limits -> continuity -> differentiation -> Taylor -> integration; Linear Algebra (Brownlowe) runs vectors -> systems -> matrices -> subspaces -> transformations -> eigenvalues. Reproduce named proofs verbatim where you can they recur. SIA > Style is graded (LO2): write proofs as full English sentences mixing words + symbols. Pure symbol-soup OR pure prose both lose marks even with the right idea. State the definition first, name the technique, then argue. 1 . Proof Toolkit NAME THE METHOD Quantifiers & negation - get these exact or lose partial credit. - (vx P(x)) = 3x -P(x); - (3x P(x)) = Vx -P(x). Prove 3x:P(x) by exhibiting one witness; prove vx:P(x) by a generalised argument on arbitrary x; disprove V by a single counterexample . THREE GENERALISED PROOFS Direct - assume P, deduce Q. e. g. m=j2, n=k2 => mn=(jk)2. Notation marks: E "element of"; := "defined to be" (vs = a deduced equality); distinguish =, =, =. Number systems NcZcQcR CC. Set-builder {xEZ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right. 1b . Induction Template EXAMINED EVERY SEM Weak induction. P(n) for all nEN if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1). SKELETON 1. State P(n) precisely. 2. Basis: verify P(0). 3. Step: "Assume P(k). Then . . . " - P(k+1). 4. By induction P(n) Vn. Canonical: 3 | 4"-1, via 4^{k+1}-1 = 4(4^k-1)+3. Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0sisk then P(k+1). Use strong induction when P(k+1) needs several earlier cases e. g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)). 2 . The e-8 Limit CENTRAL DEFINITION f defined on an open interval around a (not necessarily at a). LIM(X->A) F(X) = L νε>θ δ=δ(ε)>0: 0 < |x-a| < & = |f(x)-L| < & Variants (all examinable): Lim(x-a) f = +% : VM 38>0, 0x-al<8 = f(x)>M Lim(x-+co) f = L : VE>0 3N, X>N => |f (x) -L|<8 plus one-sided (x->a+, a") and ±oo point/value combinations. The quantifier ORDER is the marked part .
• 只写 “$c$ 是上界” 不写 “没有更小的上界”（sup 少一半分）[15]Source: asksia-bible-math1961-bilingual.pdf下确界是对偶概念:最大的下界。 i sup A need not lie in A 未必落在其中 For A = {1 - 1 : n ≥ 1} = {0, 2, 2, . . . } we have sup A = 1 ¢ A but inf A = 0 € A. When the sup is attained, sup A = max A. 对于 我们有 但。当 sup 确实被取到时,。 ★ The Least Upper Bound Axiom - completeness of R 最小上界公理––R的完备性 Every nonempty A C R that is bounded above has a supremum in R. (Dual: the Greatest Lower Bound Axiom gives inf A. ) This is the axiom Q fails - and it is the foundation every later existence proof stands on. 每一个非空且有上界的ACR在R 中都有上确界。 (对偶:最大下界公理给出。)这正是 所不具备的公 理 -- 也是后续每一个存在性证明所立足的根基。 How to PROVE sup A = '- both c clauses ‘ 如何证明“” -- 两个子句 1 Clause (1): c is an upper bound. Take an arbitrary a E A and show a ≤ c. 子句(1): c 是一个上界。取任意 并证明。 2 Clause (2): nothing smaller works. The standard move - fix any & > 0 and exhibit an a € A with a > c - 8. That proves c - & is not an upper bound, so no number below c can be one. 子句(2):没有更小的数行得通。标准做法 -- 固定任意 并 给出一个满足 的。这就证明了 不是上界,故任何小于 的 数都不可能是上界。 MATH1961 . Mathematics 1A (Advanced) ! BOTH clauses, every time 每一次都要写齐两个子句 A 'find / prove the sup' answer that shows only that c is an upper bound is half a proof. You must also kill everything below c - the s-argument in step 2. Forgetting clause (2) is the most common mark-loser here. 一个“求/证 sup”的答案如果只表明 是一个上界,那 只是半个证明。你还必须排除其下方的一切 -- 即步 骤2 中的 论证。漏掉子句(2)是这里最常见的失分原 因。 ✓ The bridge to limits: an approximating sequence 通往极限的桥梁:一个逼近数列 From clause (2) with a = - you get points an € A with c - - < an ≤ c, hence an -> sup A. This is precisely how completeness powers monotone convergence and the IVT in Ch 2. 由子句(2),取 可得到满足 的点 ,因此。这恰恰就 是完备性如何驱动第 2 章中的单调收敛与 IVT。 MATH1961 . Mathematics 1A (Advanced) E-A LIMIT 2 . LIMITS, CONTINUITY & THE REALS AHA - UNIT The precise &-o definition of a limit 极限的精确 8-ǒ 定义 The centrepiece of the Advanced unit - you must PROVE limits, not just apply laws 进阶单元的核心 -- 你必须证明极限,而不只是套用法则 Mainstream calculus says "f (x) gets close to L as x gets close to a. " That is a feeling, not mathematics. MATH1961 replaces it with a challenge-response game with a precise winning condition - the £-o definition, the single most important statement in the calculus stream.
• EVT 使用不检查“闭+有界+连续”两个条件（少一个就可能错）[12]Source: asksia-bible-math1961-bilingual.pdf2 By the LUB Axiom, x1 = sup A exists in [a, b]. 由 LUB公理,在 中存在。 3 Continuity squeezes from both sides: approaching x1 from within A forces f (x1) ≤ l; from just above forces f(x1) > l. Hence f(x1) = l. 连续性从两侧夹逼:从内部趋近迫使;从恰好上方趋近迫 使。因此。 ★ Extreme Value Theorem - also via completeness 最值定理 -- 同样经由完备性 A continuous f on a CLOSED, BOUNDED [a, b] attains a global min and max: 3xm, IM E [a, b] : f(am) ≤ f(x) ≤ f(xM) Vx € [a, b]. So Range(f) = [f(Im), f(xM)]. The proof sets M = sup{f(x) : x € [a, b]} (finite by completeness) and builds a point where it is hit. 定义在闭、有界区间 [a,b] 上的连续函数 f 取到全局 最小值与最大值: 3xm, IM E [a, b] : f (Im) ≤ f(x) ≤ f(xM)Vx € [a, b]. 故。该证明设 (由完备性,它是有限的),并构造出 一个取到它的点。 MATH1961 . Mathematics 1A (Advanced) ✓ The exam use of IVT - roots & surjectivity IVT 在考试中的用法 -- 根与满射 "Show f(x) = 0 has a root": find a sign change f(p) < 0 < f(q) (or use limits at Loo), then IVT. To prove a continuous f : R -> R is surjective: show limz->- f =- o and lim-++o f =+00, then IVT hits every value in between. “证明 有一个根”:找出一个变号(或用 在 处的极 限),然后用 IVT。要证一个连续的 是满射:证明 且 然后由IVT 取到其间的每个值。 ， ! EVT needs BOTH hypotheses - drop one and it fails EVT 需要两个假设 -- 去掉一个它就失效 Closed and bounded interval AND continuity are both essential. On the open (0, 1), f(x) = x has no max (sup = 1 not attained). With a discontinuity, f(x) = 1/x on (0, 1] is unbounded. Quoting EVT without checking these hypotheses is a classic hand-waving error the Advanced exam punishes. 闭且有界的区间与连续性两者都不可或缺。在开区间 上,没有最大值(sup=1取不到)。带一个间断点 时,在 上无界。引用EVT 却不核查这些假设,正是 高阶考试要惩罚的典型含糊错误。 "State the definition with the right quantifier order, name the theorem, check its hypotheses, then build the argument - that is where the Advanced marks live. " “以正确的量词顺序陈述定义、点明定理、核查其假 设,然后搭建论证 -- 高阶班的分数就在那里。” MATH1961 . EXAM-STRATEGY SUMMARY MATH1961 . Mathematics 1A (Advanced) DIFFERENTIATION . THE MVT FAMILY DIFFERENTIATION C4 . PROOF -HEAVY The derivative, rigorously 严格意义下的导数 Continuity is not optional - in 1961 it is built into the definition 连续性不是可选项 -- 在 1961 中它已写进定义 In the Advanced stream the derivative is not "the slope you remember from school" - it is a limit of difference quotients, and you are expected to know why it forces continuity and how the rules are proved. A single picture - the secant line through two points collapsing onto the tangent - carries the whole idea. 在进阶线中,导数不是“你在中学记得的那个斜率” -- 它是差商的极限,并且你被要求知道它为什么迫使连续性、以及那些 法则如何被证明。一幅图 -- 过两点的割线坍缩到切线上 -- 承载了整个思想。[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors
• “极限不存在”不写两条数列（只说振荡不给分）[24]Source: asksia-cheatsheet-math1961.pdf2b . e-8 Proof Skeleton PROVE A LIMIT METHOD 1. Fix ε>0 (arbitrary). 2. Bound |f(x)-LI ≤ (expr in |x-al). 3. Choose o so that bound < E. 4. Verify 0<x-al<b >> |f(x)-L|<E. . Worked: lim(x->a) [x= Ja (a>0). |Vx-Val= |x-a|/(/x+/a) ≤ |x-a|//a. So take 8 = & /a : then |x-a| <8 = |/x-/a| ≤ 8//a=£. Squeeze law: f & h ≤ g near a and lim f = lim g = L => lim h = L. Kills x. cos(1/x)->0. Fundamental limit: lim(x->0) sin x/x=1. DNE by two sequences A : if xn > a, Xn -> a give lim f(Xn) # lim f(X'n), the limit does not exist. You must exhibit both sequences - don't just assert oscillation. 3 . Continuity LIMIT = VALUE f is continuous at a if lim(x-> a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x->a+) f = f(a). Continuous on A = cts at each point. A A limit can exist yet f be discontinuous because f(a) # lim. Piecewise "find a,b so f continuous": match one- sided limits to the value. Substitution (composition): f cts at m and lim(x->a) g = m => lim(x->a) f(g(x)) =f(m). Proof chains two 8-8 statements. Worked (DNE): lim(x->0+) cos(1//x) fails to exist. Take Xn = 1/(4n2m2) -> 0 with cos->1, and xn = 1/(Tt/2+2nTt)2 > 0 with cos->0. Two limits differ = no limit. 3b · Limit Laws USE AFTER PROVING ONCE Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f. lim g; lim(f/g)=lim f / lim g (denominator limit # 0). Polynomials & rationals are continuous on their domain, so limits = substitution there. A Laws presuppose each piece's limit exists - never split a limit you haven't shown exists (0. 00, 00-00 traps). Asymptotes from limits: lim(x-> a)f = +co => vertical asymptote x=a; lim(x-> too)f = L => horizontal asymptote y=L. e. g. arctan x has horizontals y=±Tt/2; 1/x has both. Worked squeeze: show lim(x->0) x2sin(1/x) = 0. Since -x2 ≤ x2sin(1/x) ≤ x2 and both bounds -> 0, the squeeze law gives 0. 1 (note sin(1/x) alone has no limit at 0 - the x2 factor is what forces it. ) Likewise lim(x->oo) cos x/x = 0 by squeeze between ±1/x. The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences - the same E-N machinery underlies both. 4 . Completeness . THE CORE OF[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors
• 重特征值不检查几何重数（误判可对角化）[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors

第 3 优先级：计算题要练到“写得出理由”

• 复数极坐标/辐角象限检查；de Moivre；$n$ 次根写全。[21]Source: asksia-cheatsheet-math1961.pdfMATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 2 OF 2 Complex . linear algebra . prove it SIDE 2/2 Eigenvalues 11 . Complex Numbers CARTESIAN + POLAR z=x+iy, i2 =- 1. Argand plot (x,y). Conjugate z = x-iy; Re z = (z+z)/2, Im z = (z-z)/2i; |z|2 = z ż. POLAR / EXPONENTIAL Z = r(cos 0 + i sin 0) = r e^{i0} r = |z| = V(x2+y2), 0 = arg z arg z mod 21; principal Arg z € (-1,T]. Euler: e^{i0} = cos 0 + i sin 0. Product/quotient: multiply/divide moduli, add/subtract args. A Cartesian>> polar is where sign/quadrant errors cost marks - check which quadrant (x,y) sits in before fixing 0. Arithmetic: (a+ib)(c+id) = (ac-bd) + i(ad+bc); divide by multiplying by the conjugate, z/w = z·w/|w|2. Quadratics over C: usual formula; for real coefficients complex roots come in conjugate pairs. 11b . de Moivre & INDUCTION PROOF Roots DE MOIVRE, NEZ (r(cos 8 + i sin 0))" = rm(cos ne + i sin n0) i. e. (r e^{i0})" = rn e^{in0} Proof for nEN by induction: n=1 trivial; assume for k, then (e^{i0})^{k+1}= (e^{i0})^k. e^{i0]=e^{ik0}e^{i0}=e^{i(k+1)0} by the angle- sum identities. # (extend to Z via z-1. ) Use: multiple-angle formulas; expand (1+ic)" by binomial, split Re/Im. N-TH ROOTS OF Z = R E^{IO} ak = r^{1/n} e^{i(0+2mk)/n}, k=0, . . . , n-1 n-th roots of unity: e^{2Ttik/n} - n equally-spaced points, a regular n-gon on the unit circle. ! For z"=a give all n roots. Real-coeff polynomials: complex roots in conjugate pairs. 11c . Worked . Roots of Unity SOLVE Zn=A[23]Source: asksia-cheatsheet-math1961.pdf23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts. * Repeated eigenvalue => check geom = alg mult. * col(A) uses A's columns, not RREF's. * Eigenvector must be nonzero; report a basis of E_N. · * z"=a has all n roots - don't stop at one. * Proofs in full sentences (LO2), not symbol-soup. . SIA -> Method marks survive a slipped number. State the definition, name the technique, write full sentences. That is the whole difference between 1961 and 1061. LA Recap SIDE 2 z = re^{i0} . (re^{i0})" = rme^{in0} |u . v | ≤ llull livIl · proj = (u. v)/llull2 . u rank + nullity = n . det=0 = invertible Ax=Àx . X_A(A)=det(A-AI)=0 diag = n l. i. eigenvectors
• 特征多项式 $\to$ 特征向量基；rank-nullity；列空间/零空间的标准提取。[17]Source: asksia-cheatsheet-math1961.pdfA is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 - 2 = 2 (two free variables). Column space dim 2 9 R 3; null space dim 2 S R 4. Sum 2+2 = 4 = n V. Standard matrices: rotation by 0 = [cos 0 -sin 0; sin @ cos 0]; RaRg = R_{a+B} recovers the angle-sum identities. Reflection, projection, scaling each have their own A. FAILS 21c . The Multiplicity WHEN DIAG Trap A=[2 1;02]: χ(λ)=(2-λ)2, so X=2 with algebraic mult 2. But A-21 = [0 1; 0 0] has nullity 1 => geometric mult 1 < 2. So A is NOT diagonalisable - only one independent eigenvector. Contrast A=[2 0; 0 2] (=21): same )=2 twice but every vector is an eigenvector, geometric mult 2 = already diagonal. Eigenvectors for distinct A are l. i. - the lemma behind "n distinct eigenvalues = diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum. A real matrix can have complex eigenvalues (a rotation [cos 0 -sin 0; sin 0 cos 0] has ) = e^{+i0}) - this is exactly where the complex-number stream feeds back into linear algebra. Always compute x_A(A) and factor it before hunting eigenvectors - the marks for "eigenvalue reasoning" live in showing det(A-NI)=0, not just stating the answers. Revision aid . check the current unit outline for assessment . @ 2026 good luck. prove, don't just compute. R3 THE KEY PROOF Elimination RANK STRUCTURE . Complex numbers . de Moivre & roots . Vectors & Cauchy-Schwarz . Gaussian elimination . Vector spaces . Rank-nullity . REVISION SHEET . ALL TOPICS MATH1961 Mathematics 1A (Advanced) UNIVERSITY OF SYDNEY . SCHOOL OF MATHEMATICS & STATISTICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Rigour & calculus . prove it SIDE 1/2 RIGOUR . Proof toolkit & induction . 8-o limits . Completeness & sup . Continuity . IVT/EVT . Rolle-MVT-Taylor . Riemann & FIC 0 . Exam Blueprint READ FIRST * PROVE, don't just compute. This is the Advanced unit (=4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct 8-8 argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning - not the final number alone.[30]Source: asksia-cheatsheet-math1961.pdfFinding bases (R = RREF A): row space -> nonzero rows of R; column space > columns of A in R's pivot positions (col(A)#col(R) but same dim); null space -> solve Rx=0, separate parameters. RANK-NULLITY A rank (A) + nullity (A) = n rank = # leading 1s = dim row = dim col; nullity = dim null. Proof: the n columns split into pivot (-> rank) vs free (-> nullity) . Corollary: in dim-n space, n vectors are l. i. - they span. 20 . Linear Transformations MATRIX STANDARD T:Rn->R™ linear if T(u+v)=T(u)+T(v) and T(cu)=cT(u). Hence T(0)=0. Standard-matrix thm A : every linear T = L_A with A = [T(e1) . . . T(en)] (mxn, unique). Proof: x=Ex,e, + linearity. Composition: A_{S . T} = A_SA_T. A Prove T NOT linear: exhibit a failing case - absolute values, constants or squares break it (e. g. T with a |z| term fails T(cu)=cT(u) at c =- 1). T:R"->R" bijective - A_T invertible. 23 . Proof Pattern Belt PICK THE RECIPE · "Prove V . . . " > arbitrary element OR induction (strong if order ≥2). · "Prove a limit" -> 8-8: fix &, bound, pick 8. · "Bound the error" > Lagrange remainder, solve for n. "'Is it a subspace / VS" > check 0 + closure / 8 axioms. · "L. i . ? " > trivial-solution-only of Ec;v=0. · "'Diagonalisable?" -> n l. i. eigenvectors / geom = alg mult. 24 . High-Yield Traps WHERE RIGOUR WINS sup proof needs both clauses, not just an upper bound. * Limit DNE needs two explicit sequences. · * EVT needs closed + bounded + cts.

• A $\varepsilon$–$\delta$ / 连续 / DNE 两数列
• B 完备性 $\sup/\inf$、IVT/EVT 证明
• C Rolle–MVT–Taylor–FTC 链条
• D 复数 + de Moivre + roots
• E 线代（rank-nullity / 子空间 / 线性变换）
• F eigenvalues / diagonalisation（重根陷阱）