ECON003

• 说明：这门课本质是经济学里的数学（calculus + algebra），不是统计（没有均值/概率/回归/假设检验）。[4]Source: asksia-bible-econ1003-bilingual.pdfMethod, not memorisation 方法,而非死记硬背 A formula sheet is provided - so marks live in the steps 考试提供公式表 -- 所以分数藏在步骤里 This is not a transcript of the lecture decks. It is a self-contained course in every technique ECON1003 tests - each rule stated plainly, each method shown on a worked example with real arithmetic, each common slip flagged. Because the exams hand you a formula sheet, the examiner cannot test whether you remember a formula - only whether you can apply it. That is exactly what these pages drill. 这不是讲义幻灯片的誊录稿,而是一门自成体系的课程,覆盖 ECON1003 所考查的每一项技巧 -- 每条法则直白陈述,每 种方法用带真实运算的例题演示,每个常见失误都被标出。由于考试会发给你公式表,考官无法考你是否记得某个公式 -- 只能考你能否应用它。而这正是这些页面要反复操练的。 A 1 . LEARN 1·学 You haven't seen the lecture yet. Read a chapter top to bottom. Every topic opens with a one-line TL;DR, then define - formula - worked example - exam trap. You can learn the method here cold. 你还没上课。把一章从头读到 尾。每个主题以一行 TL;DR 开 头,然后是定义→公式→ 例题 →考试陷阱。你完全可以在这里 从零学会方法。 B 2 . DRILL 2 · 练 You've done the lecture and tutorial. Cover the worked- example steps and re-derive each answer yourself. The aim is speed and accuracy on fresh numbers - that is what both exams reward. 你已经上完讲课和辅导课了。盖 住例题步骤,自己重新推导每个 答案。目标是在全新数字上做到 又快又准 -- 这正是两场考试 所奖励的。 C 3 . EXAM 3 · 考 It's revision week. The TL;DR strips and the recap box at each chapter's end are your map. The blueprint overleaf shows what the Midterm covers vs the cumulative Final. 到了复习周。各处的 TL;DR条 目和每章末尾的复习框就是你的 地图。背面的蓝图展示了期中考 查什么、对比累计期末考查什 么。 ! The single most important thing to understand about ECON1003 关于 ECON1003 最该理解的一件事 Despite the name, this unit is mathematics for economics, NOT statistics. There is no mean, no probability, no hypothesis testing, no regression. The seven taught topics are straight lines, non-linear functions, financial maths, differentiation, functions of several variables, integration and linear algebra - calculus and algebra, dressed in demand, cost, utility and production. If you came expecting stats, recalibrate now. 尽管名字如此,本单元是经济学数学,而非统计学。没有均值、没有概率、没有假设检验、没有回归。所教的七大主题 是直线、非线性函数、金融数学、求导、多变量函数、积分法和线性代数 -- 即微积分和代数,披着需求、成本、效用 和生产的外衣。如果你是奔着统计来的,现在就重新调整预期。 ECON1003 . Quantitative Methods in Economics i How this book was built - and the two-layer rule 本书如何构建 -- 以及两层规则 Standard mathematical rules and formulas are stated plainly (they are universal). The unit's own framing and example numbers are paraphrased, never copied verbatim from slides. The textbook followed closely is Bradley, Essential Mathematics for Economics & Business, 4th ed. (Wiley). Every arithmetic step here has been checked - but verify dates and weights against your own Canvas (canvas. sydney. edu. au), as details can shift between cohorts. 标准的数学法则和公式被直白陈述(它们是普适的)。本单元自身的表述和示例数字都经过改写,绝不逐字照抄幻灯片。 紧密遵循的教材是 Bradley, Essential Mathematics for Economics & Business,第4版(Wiley)。这里每一步运 算都经过核验 -- 但请对照你自己的 Canvas (canvas. sydney. edu. au)核实日期和权重,因为细节可能在不同届之间 有所变动。 ECON1003 . Quantitative Methods in Economics[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).
• 另外：两场考试都有公式表，所以真正拿分点在步骤、方法链条、别踩坑。[2]Source: asksia-bible-econ1003-bilingual.pdfTHE BLUEPRINT - THE EXAM BLUEPRINT MIDTERM 40% + FINAL 50% Two exams, one formula sheet 两场考试,一张公式表 Quizzes 10% . Midterm 40% · Final 50% (cumulative) 测验 10% · 期中 40% · 期末 50%(累计) Your mark is built from three pieces. The two written exams together are 90% of the grade - and both supply a formula sheet, so the examiner is testing method, not recall. 你的成绩由三部分构成。两场笔试合计占总评的 90% -- 且两场都提供公式表,所以考官考查的是方法,而非记忆。 10% ONLINE QUIZZES 在线测验 40% MIDTERM EXAM 期中考 50% FINAL (CUMULATIVE) 期末考(累计性) 7 topics ALL EXAMINABLE 全部可考 The three assessment pieces 三项考核构成 Component Weight When / detail Final examination - cumulative, all 7 topics 50% Final exam period Mid-semester exam - 60 min 40% 19 Apr, 15:40 Online quizzes (4 x on Canvas) 10% 1 wk each · no extensions Quizzes are posted across the semester (Mar-May), each open for a full week with no extensions under any circumstance - do them early. Q&A runs through Ed in Canvas, not email. 测验贯穿整个学期发布(3月至5月),每次开放整整一 周,任何情况下都不延期 -- 尽早完成。答疑通过 Canvas 中的 Ed 进行,而非邮件。[4]Source: asksia-bible-econ1003-bilingual.pdfMethod, not memorisation 方法,而非死记硬背 A formula sheet is provided - so marks live in the steps 考试提供公式表 -- 所以分数藏在步骤里 This is not a transcript of the lecture decks. It is a self-contained course in every technique ECON1003 tests - each rule stated plainly, each method shown on a worked example with real arithmetic, each common slip flagged. Because the exams hand you a formula sheet, the examiner cannot test whether you remember a formula - only whether you can apply it. That is exactly what these pages drill. 这不是讲义幻灯片的誊录稿,而是一门自成体系的课程,覆盖 ECON1003 所考查的每一项技巧 -- 每条法则直白陈述,每 种方法用带真实运算的例题演示,每个常见失误都被标出。由于考试会发给你公式表,考官无法考你是否记得某个公式 -- 只能考你能否应用它。而这正是这些页面要反复操练的。 A 1 . LEARN 1·学 You haven't seen the lecture yet. Read a chapter top to bottom. Every topic opens with a one-line TL;DR, then define - formula - worked example - exam trap. You can learn the method here cold. 你还没上课。把一章从头读到 尾。每个主题以一行 TL;DR 开 头,然后是定义→公式→ 例题 →考试陷阱。你完全可以在这里 从零学会方法。 B 2 . DRILL 2 · 练 You've done the lecture and tutorial. Cover the worked- example steps and re-derive each answer yourself. The aim is speed and accuracy on fresh numbers - that is what both exams reward. 你已经上完讲课和辅导课了。盖 住例题步骤,自己重新推导每个 答案。目标是在全新数字上做到 又快又准 -- 这正是两场考试 所奖励的。 C 3 . EXAM 3 · 考 It's revision week. The TL;DR strips and the recap box at each chapter's end are your map. The blueprint overleaf shows what the Midterm covers vs the cumulative Final. 到了复习周。各处的 TL;DR条 目和每章末尾的复习框就是你的 地图。背面的蓝图展示了期中考 查什么、对比累计期末考查什 么。 ! The single most important thing to understand about ECON1003 关于 ECON1003 最该理解的一件事 Despite the name, this unit is mathematics for economics, NOT statistics. There is no mean, no probability, no hypothesis testing, no regression. The seven taught topics are straight lines, non-linear functions, financial maths, differentiation, functions of several variables, integration and linear algebra - calculus and algebra, dressed in demand, cost, utility and production. If you came expecting stats, recalibrate now. 尽管名字如此,本单元是经济学数学,而非统计学。没有均值、没有概率、没有假设检验、没有回归。所教的七大主题 是直线、非线性函数、金融数学、求导、多变量函数、积分法和线性代数 -- 即微积分和代数,披着需求、成本、效用 和生产的外衣。如果你是奔着统计来的,现在就重新调整预期。 ECON1003 . Quantitative Methods in Economics i How this book was built - and the two-layer rule 本书如何构建 -- 以及两层规则 Standard mathematical rules and formulas are stated plainly (they are universal). The unit's own framing and example numbers are paraphrased, never copied verbatim from slides. The textbook followed closely is Bradley, Essential Mathematics for Economics & Business, 4th ed. (Wiley). Every arithmetic step here has been checked - but verify dates and weights against your own Canvas (canvas. sydney. edu. au), as details can shift between cohorts. 标准的数学法则和公式被直白陈述(它们是普适的)。本单元自身的表述和示例数字都经过改写,绝不逐字照抄幻灯片。 紧密遵循的教材是 Bradley, Essential Mathematics for Economics & Business,第4版(Wiley)。这里每一步运 算都经过核验 -- 但请对照你自己的 Canvas (canvas. sydney. edu. au)核实日期和权重,因为细节可能在不同届之间 有所变动。 ECON1003 . Quantitative Methods in Economics

1）你现在最需要知道的：Final 的“考试结构 + 出题风格”

考核占比（总评怎么来的）

• Online quizzes：10%（Canvas 上，通常每次开一周，不延期）[2]Source: asksia-bible-econ1003-bilingual.pdfTHE BLUEPRINT - THE EXAM BLUEPRINT MIDTERM 40% + FINAL 50% Two exams, one formula sheet 两场考试,一张公式表 Quizzes 10% . Midterm 40% · Final 50% (cumulative) 测验 10% · 期中 40% · 期末 50%(累计) Your mark is built from three pieces. The two written exams together are 90% of the grade - and both supply a formula sheet, so the examiner is testing method, not recall. 你的成绩由三部分构成。两场笔试合计占总评的 90% -- 且两场都提供公式表,所以考官考查的是方法,而非记忆。 10% ONLINE QUIZZES 在线测验 40% MIDTERM EXAM 期中考 50% FINAL (CUMULATIVE) 期末考(累计性) 7 topics ALL EXAMINABLE 全部可考 The three assessment pieces 三项考核构成 Component Weight When / detail Final examination - cumulative, all 7 topics 50% Final exam period Mid-semester exam - 60 min 40% 19 Apr, 15:40 Online quizzes (4 x on Canvas) 10% 1 wk each · no extensions Quizzes are posted across the semester (Mar-May), each open for a full week with no extensions under any circumstance - do them early. Q&A runs through Ed in Canvas, not email. 测验贯穿整个学期发布(3月至5月),每次开放整整一 周,任何情况下都不延期 -- 尽早完成。答疑通过 Canvas 中的 Ed 进行,而非邮件。[18]Source: asksia-cheatsheet-econ1003.pdf* Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356 · Break-even: TC=100+4Q, P=9 -> Q =20 · Turning pt of y=x2-8x+1 -> (4, -15) · J(6x2+2) dx -> 2x3 + 2x + C * Quiz & Logistics IT 10% . DON'T LOSE 4 online quizzes (10% total), 1 week each, NO extensions - do them early. Q&A via Ed, not email. Textbook: Bradley, Essential Mathematics for Economics & Business, 4th ed. (Chs 2,4,5,6,7,8,9). The unit is maths in the service of economics - every technique gets an economic reading (slope = marginal effect, A = shadow price, integral = surplus). maths for economics . check your current unit outline . @ 2026 flip - for side 2 . several variables, integration & matrices asksia. ai/cheatsheet/ usyd-econ1003 . side 1/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd-econ1003 CH2
• Midterm：40%（侧重 Topic 1–4 / Side 1）[2]Source: asksia-bible-econ1003-bilingual.pdfTHE BLUEPRINT - THE EXAM BLUEPRINT MIDTERM 40% + FINAL 50% Two exams, one formula sheet 两场考试,一张公式表 Quizzes 10% . Midterm 40% · Final 50% (cumulative) 测验 10% · 期中 40% · 期末 50%(累计) Your mark is built from three pieces. The two written exams together are 90% of the grade - and both supply a formula sheet, so the examiner is testing method, not recall. 你的成绩由三部分构成。两场笔试合计占总评的 90% -- 且两场都提供公式表,所以考官考查的是方法,而非记忆。 10% ONLINE QUIZZES 在线测验 40% MIDTERM EXAM 期中考 50% FINAL (CUMULATIVE) 期末考(累计性) 7 topics ALL EXAMINABLE 全部可考 The three assessment pieces 三项考核构成 Component Weight When / detail Final examination - cumulative, all 7 topics 50% Final exam period Mid-semester exam - 60 min 40% 19 Apr, 15:40 Online quizzes (4 x on Canvas) 10% 1 wk each · no extensions Quizzes are posted across the semester (Mar-May), each open for a full week with no extensions under any circumstance - do them early. Q&A runs through Ed in Canvas, not email. 测验贯穿整个学期发布(3月至5月),每次开放整整一 周,任何情况下都不延期 -- 尽早完成。答疑通过 Canvas 中的 Ed 进行,而非邮件。[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)
• Final：50%（累计性：7 个主题全考，Side 1 + Side 2 都回归）[2]Source: asksia-bible-econ1003-bilingual.pdfTHE BLUEPRINT - THE EXAM BLUEPRINT MIDTERM 40% + FINAL 50% Two exams, one formula sheet 两场考试,一张公式表 Quizzes 10% . Midterm 40% · Final 50% (cumulative) 测验 10% · 期中 40% · 期末 50%(累计) Your mark is built from three pieces. The two written exams together are 90% of the grade - and both supply a formula sheet, so the examiner is testing method, not recall. 你的成绩由三部分构成。两场笔试合计占总评的 90% -- 且两场都提供公式表,所以考官考查的是方法,而非记忆。 10% ONLINE QUIZZES 在线测验 40% MIDTERM EXAM 期中考 50% FINAL (CUMULATIVE) 期末考(累计性) 7 topics ALL EXAMINABLE 全部可考 The three assessment pieces 三项考核构成 Component Weight When / detail Final examination - cumulative, all 7 topics 50% Final exam period Mid-semester exam - 60 min 40% 19 Apr, 15:40 Online quizzes (4 x on Canvas) 10% 1 wk each · no extensions Quizzes are posted across the semester (Mar-May), each open for a full week with no extensions under any circumstance - do them early. Q&A runs through Ed in Canvas, not email. 测验贯穿整个学期发布(3月至5月),每次开放整整一 周,任何情况下都不延期 -- 尽早完成。答疑通过 Canvas 中的 Ed 进行,而非邮件。[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

Final 的出题“共同特征”（非常关键）

• 题目几乎都是程序化（procedural）：给你函数/系统 → 选对方法 → 算 → 再解释经济含义 → 按要求取整。[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)
• 方法分真实存在：写步骤比只写一个数字更值钱；只在最后一步再四舍五入。[1]Source: asksia-bible-econ1003-bilingual.pdf代数余子式 / 余子式 Minor = sub-determinant; cofactor = minor x the +-+ sign. ECON1003 . Quantitative Methods in Economics Term 术语 中文 One-line definition 释义 Inverse matrix (A-1) 逆矩阵 A-1 = CT/det(A); satisfies AA1 = I; solves X = A-1B. ✓ How to use the glossary 用法 如何使用术语表 用法 Cover the right column and recite the definition from the English term - then flip and recall the English from the +X. 〔遮住释义、由术语复述定义;再由中文回想英文。〕 The exam rewards the exact wording: "next-best forgone", "CT/det(A)", "signed area" - precision is marks. 遮住右栏,从英文术语复述出定义 -- 再反过来,由中文回忆英文。〔遮住释义、由术语复述定义;再由中文回想英文。〕 考试奖励精确的措辞:“next-best forgone”、“CT/det(A)”、“signed area" -- 精确就是分数。 ECON1003 . Quantitative Methods in Economics PRACTICE P1 -P2 - PRACTICE BANK DRILL TO EXAM STANDARD Drill both exams, with full solutions 两场考试齐刷,附完整解答 Fourteen AskSia-authored problems across all seven topics - worked end to end 横跨全部七大主题的14 道 AskSia 原创题 –––完整解答 These are exam-style problems written in the ECON1003 style - not copied from any real quiz or exam. They span lines, finance, differentiation, the Lagrangian, integration and matrices, so they serve both the Midterm (Topics 1-4) and the cumulative Final (all seven). Every number is verified. Cover the solution, work it on paper, then read the answer. 这些是按 ECON1003 风格编写的考试风格题目 -- 并非抄自任何真实测验或考试。它们覆盖直线、金融、求导、拉格朗日 函数、积分和矩阵,因此同时服务于期中(主题1-4)和累计的期末(全部七大主题)。每个数字都经过核验。盖住解答,在 纸上做完,再看答案。 ★ How to use this bank 如何使用这个题库 Cover the solution - recognition is not recall. Budget ~1 minute per mark. Show your work: method marks are real and partial credit is given, so a stated step beats a bare number. Round only the final answer to the stated decimals. Check every answer by substituting back - differentiate an integral, multiply an inverse, re-solve a system. 遮住答案 认得出不等于回忆得起。按每分约1分 钟来分配时间。写出你的过程:方法分是实打实的, 也会给部分分,所以一个写出来的步骤胜过一个孤零 零的数字。只对最终答案按规定小数位取整。对每个 答案都做检验,把它代回去 -- 对积分求导、把逆矩 阵乘回去、重新解一遍方程组。 i Formula sheet vs memory 公式表 vs 记忆 Both exams give a formula sheet (quadratic, series, interest, NPV, annuity, by-parts, 3×3 determinant, ax/log derivatives, quotient rule). You must memorise the rest: power / product / chain rules, substitution, the Lagrangian conditions, Cramer's rule and the inverse methods. The marks are in applying them cleanly.[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1
• 很多题都长得像“链条题”（下面第 2 部分我会给你“链条清单”）：
  • 直线 → 斜率/截距 → 弹性解释
  • 建 $TR=PQ$ → 求导得 $MR$ → 令 $MR=MC$ → 求 $Q^*$ 再回代求 $P^*$
  • 积分 $MC$ → 还原 $TC$（别丢 $+C$）
  • Lagrangian → 解内点+角点 → 解释 shadow price（$\lambda$）
  • 矩阵/线代 → 先看维度 → 再看 $\det(A)\neq 0$ → Cramer / inverse / 消元 → 代回验证
    这些都是宝典里明确点名的“反复考法”。[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

2）Final 最高频“核心链条”（你要练到闭眼都能写）

链条 A：利润最大化（Final/期中都爱考，但 Final 一定会回归）

必会步骤（按得分顺序）：[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)[20]Source: asksia-cheatsheet-econ1003.pdfWORKED MIN . F=X2-6X+5 f'=2x-6=0 = x=3 . f' ' =2>0 - MIN f(3) = - 4 WORKED MAX . F =- X2+4X f' =- 2x+4=0 = x=2 . f' ' =- 2<0 - MAX f(2) = 4 17 . Economic Applications CH6 . EXAM FAVE MARGINALS MR = dTR/dQ MC = dTC/dQ revenue max: MR = 0 profit max: MR = MC Non-linear elasticity: Ed = (dQ/dP) . (P/Q) with calculus for dQ/dP. WORKED . P=100-Q, TC=Q2 TR=PQ=100Q-Q2 = MR=100-2Q MC=2Q . MR=MC = Q=25, P=75 check: revenue max (MR=0) at Q=50 - different! TRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max[21]Source: asksia-cheatsheet-econ1003.pdfTRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max * Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356

• 先把需求写成 $P$ 关于 $Q$ 的形式（inverse demand），因为你要写 $TR(Q)$。
• $TR(Q)=P(Q)\cdot Q$
• $MR(Q)=\dfrac{dTR}{dQ}$
• $MC(Q)=\dfrac{dTC}{dQ}$（如果给的是 $TC$）
• 利润最大化条件：$MR=MC$，解出 $Q^*$
• 回代求 $P^*$，再求利润 $\pi=TR-TC$（若题目要）

超级常见陷阱：

• “求 $MR$ 时直接对 $P$ 求导”是错的：必须先写 $TR=P(Q)\cdot Q$ 再求导。[20]Source: asksia-cheatsheet-econ1003.pdfWORKED MIN . F=X2-6X+5 f'=2x-6=0 = x=3 . f' ' =2>0 - MIN f(3) = - 4 WORKED MAX . F =- X2+4X f' =- 2x+4=0 = x=2 . f' ' =- 2<0 - MAX f(2) = 4 17 . Economic Applications CH6 . EXAM FAVE MARGINALS MR = dTR/dQ MC = dTC/dQ revenue max: MR = 0 profit max: MR = MC Non-linear elasticity: Ed = (dQ/dP) . (P/Q) with calculus for dQ/dP. WORKED . P=100-Q, TC=Q2 TR=PQ=100Q-Q2 = MR=100-2Q MC=2Q . MR=MC = Q=25, P=75 check: revenue max (MR=0) at Q=50 - different! TRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max[21]Source: asksia-cheatsheet-econ1003.pdfTRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max * Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356

链条 B：直线 + 截距/斜率 + 弹性（Topic 1 经典）

直线标准形：$y=mx+c$（先整理再读斜率截距）[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1

• 斜率：$m=\Delta y/\Delta x$（rise/run）[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1
• 截距：$x=0$ 时的 $y$（基准值）[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1

需求 vs 反需求：

• $Q=f(P)$ 和 $P=f(Q)$ 的斜率互为倒数，不是相等；先看题目问哪一个。[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1

价格弹性（直线/非线性通用）：

• $$E_d=\left(\frac{dQ}{dP}\right)\left(\frac{P}{Q}\right)$$[5]Source: asksia-bible-econ1003-bilingual.pdf两场考试都提供公式表(二次方程、级数、利息、 NPV、年金、分部积分、3x3 行列式、ax/log 导数、 商法则)。其余的你必须背熟:幂/乘积/链式法则、 换元法、拉格朗日条件、Cramer's rule 以及逆矩阵方 法。分数就在于把它们干净利落地用出来。 P1 CALCULATION 6 marks . lines & break-even . Topic 1 A firm sells output at P = $10 per unit with total cost 某厂商以 P=$10 的单价出售产品,总成本为 TC = 600 + 4Q TC = 600 + 4Q (a) Find the break-even quantity. [3] (b) Find profit when Q = 150. [3] (a) 求盈亏平衡数量。[3](b)求Q=150 时的利润。[3] WORKED SOLUTION 1 (a) Break-even is profit = 0. Revenue = TC: 10Q = 600 + 4Q - 6Q = 600 -+ Q=100 units. (Fixed cost 600 is the TC intercept, not a marginal cost. ) ECON1003 . Quantitative Methods in Economics (a) 盈亏平衡即利润=0。收益 = TC: 10Q= 600+ 4Q→6Q= 600→Q =100单位。(固定成本 600是 TC的截距,不 是边际成本。) (b) Profit at Q = 150. TR = 10-150 = 1500; TC = 600 + 4. 150 = 1200; Profit = 1500 - 1200 = $300. 2 (b) Q =150 时的利润。TR= 10·150= 1500; TC = 600+ 4·150= 1200; Profit = 1500 - 1200= $300。 P2 CALCULATION 5 marks . linear elasticity . Topic 1 Inverse demand is P = 120 - 3Q. 反需求为 P= 120-3Q。 (a) Find the price elasticity of demand at Q = 20. [3] (b) Is demand elastic, inelastic or unit-elastic there? [2] (a) 求 Q=20 时的需求价格弹性。[3](b)在该点需求是富有弹性、缺乏弹性还是单位弹性?[2] WORKED SOLUTION 1 Get dQ/dP. Rearrange P = 120 - 3Q - Q = 40 - P/3, so dQ/dP = - 1/3. 求 dQ/dP。整理 P=120-3Q →Q=40-P/3,所以 dQ/dP =- 1/3。 2 Price at Q = 20. P = 120 - 3(20) = 60. Q=20 时的价格。P= 120-3(20)= 60。 3 (a) Apply E = (dQ/dP)(P/Q). E =(-1/3)(60/20) = (-1/3)(3) = - 1. (a) 套用 E=(dQ/dP)(P/Q)。E= (-1/3)(60/20) = (-1/3)(3)= - 1。 4 (b) |E| = 1 - unit-elastic. This is the midpoint of a linear demand curve (choke price 120 at Q = 0, intercept Q = 40 at P = 0; midpoint Q = 20). Elastic above it, inelastic below. (b)|E|=1→单位弹性。这是线性需求曲线的中点(Q=0 时窒息价格为120,P=0时截距 Q=40;中点 Q=20)。其上 方富有弹性,下方缺乏弹性。 ECON1003 . Quantitative Methods in Economics PRACTICE P3-P4 - PRACTICE BANK (CONT. ) Financial mathematics 金融数学 Compounding the right number of periods, and solving for an unknown rate 用正确的期数复利,并求解未知利率[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。
• 分类：$|E|>1$ 富有弹性；$|E|<1$ 缺乏弹性；$|E|=1$ 单位弹性。[14]Source: asksia-bible-econ1003-bilingual.pdfElastic / inelastic 富有弹性/ 缺乏弹性 |E| > 1 elastic (responsive); |E| < 1 inelastic; |E| = 1 unit-elastic. Topic 2 -Non-linear functions 非线性函数 Quadratic 二次函数 ax2 + bx + c; one turning point; a parabola, min if a > O. Discriminant 判别式 b2 - 4ac; >O two roots, =O one root, <O no real roots. Exponential function 指数函数 y = ax; constant base raised to a variable power; e ~ 2. 71828. Logarithm 对数 log (b) = the power a must be raised to for b; In = log base e. Turning point 转折点/ 极值点 Where a curve changes from rising to falling (or vice versa). ECON1003 . Quantitative Methods in Economics GLOSSARY 2/3 - GLOSSARY · 术语表 (CONT. ) Finance, differentiation 金融、求导 Compound interest, NPV, the derivative rules and their economic meaning 复利、NPV、求导法则及其经济含义 Term 术语 中文 One-line definition ##X Topic 3 - Financial mathematics 金融数学 Arithmetic series[5]Source: asksia-bible-econ1003-bilingual.pdf两场考试都提供公式表(二次方程、级数、利息、 NPV、年金、分部积分、3x3 行列式、ax/log 导数、 商法则)。其余的你必须背熟:幂/乘积/链式法则、 换元法、拉格朗日条件、Cramer's rule 以及逆矩阵方 法。分数就在于把它们干净利落地用出来。 P1 CALCULATION 6 marks . lines & break-even . Topic 1 A firm sells output at P = $10 per unit with total cost 某厂商以 P=$10 的单价出售产品,总成本为 TC = 600 + 4Q TC = 600 + 4Q (a) Find the break-even quantity. [3] (b) Find profit when Q = 150. [3] (a) 求盈亏平衡数量。[3](b)求Q=150 时的利润。[3] WORKED SOLUTION 1 (a) Break-even is profit = 0. Revenue = TC: 10Q = 600 + 4Q - 6Q = 600 -+ Q=100 units. (Fixed cost 600 is the TC intercept, not a marginal cost. ) ECON1003 . Quantitative Methods in Economics (a) 盈亏平衡即利润=0。收益 = TC: 10Q= 600+ 4Q→6Q= 600→Q =100单位。(固定成本 600是 TC的截距,不 是边际成本。) (b) Profit at Q = 150. TR = 10-150 = 1500; TC = 600 + 4. 150 = 1200; Profit = 1500 - 1200 = $300. 2 (b) Q =150 时的利润。TR= 10·150= 1500; TC = 600+ 4·150= 1200; Profit = 1500 - 1200= $300。 P2 CALCULATION 5 marks . linear elasticity . Topic 1 Inverse demand is P = 120 - 3Q. 反需求为 P= 120-3Q。 (a) Find the price elasticity of demand at Q = 20. [3] (b) Is demand elastic, inelastic or unit-elastic there? [2] (a) 求 Q=20 时的需求价格弹性。[3](b)在该点需求是富有弹性、缺乏弹性还是单位弹性?[2] WORKED SOLUTION 1 Get dQ/dP. Rearrange P = 120 - 3Q - Q = 40 - P/3, so dQ/dP = - 1/3. 求 dQ/dP。整理 P=120-3Q →Q=40-P/3,所以 dQ/dP =- 1/3。 2 Price at Q = 20. P = 120 - 3(20) = 60. Q=20 时的价格。P= 120-3(20)= 60。 3 (a) Apply E = (dQ/dP)(P/Q). E =(-1/3)(60/20) = (-1/3)(3) = - 1. (a) 套用 E=(dQ/dP)(P/Q)。E= (-1/3)(60/20) = (-1/3)(3)= - 1。 4 (b) |E| = 1 - unit-elastic. This is the midpoint of a linear demand curve (choke price 120 at Q = 0, intercept Q = 40 at P = 0; midpoint Q = 20). Elastic above it, inelastic below. (b)|E|=1→单位弹性。这是线性需求曲线的中点(Q=0 时窒息价格为120,P=0时截距 Q=40;中点 Q=20)。其上 方富有弹性,下方缺乏弹性。 ECON1003 . Quantitative Methods in Economics PRACTICE P3-P4 - PRACTICE BANK (CONT. ) Financial mathematics 金融数学 Compounding the right number of periods, and solving for an unknown rate 用正确的期数复利,并求解未知利率

终极考点（考官最爱抓）：

• “弹性不是斜率”：直线需求的斜率恒定，但弹性沿线变化；中点处 $E=-1$。[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。[5]Source: asksia-bible-econ1003-bilingual.pdf两场考试都提供公式表(二次方程、级数、利息、 NPV、年金、分部积分、3x3 行列式、ax/log 导数、 商法则)。其余的你必须背熟:幂/乘积/链式法则、 换元法、拉格朗日条件、Cramer's rule 以及逆矩阵方 法。分数就在于把它们干净利落地用出来。 P1 CALCULATION 6 marks . lines & break-even . Topic 1 A firm sells output at P = $10 per unit with total cost 某厂商以 P=$10 的单价出售产品,总成本为 TC = 600 + 4Q TC = 600 + 4Q (a) Find the break-even quantity. [3] (b) Find profit when Q = 150. [3] (a) 求盈亏平衡数量。[3](b)求Q=150 时的利润。[3] WORKED SOLUTION 1 (a) Break-even is profit = 0. Revenue = TC: 10Q = 600 + 4Q - 6Q = 600 -+ Q=100 units. (Fixed cost 600 is the TC intercept, not a marginal cost. ) ECON1003 . Quantitative Methods in Economics (a) 盈亏平衡即利润=0。收益 = TC: 10Q= 600+ 4Q→6Q= 600→Q =100单位。(固定成本 600是 TC的截距,不 是边际成本。) (b) Profit at Q = 150. TR = 10-150 = 1500; TC = 600 + 4. 150 = 1200; Profit = 1500 - 1200 = $300. 2 (b) Q =150 时的利润。TR= 10·150= 1500; TC = 600+ 4·150= 1200; Profit = 1500 - 1200= $300。 P2 CALCULATION 5 marks . linear elasticity . Topic 1 Inverse demand is P = 120 - 3Q. 反需求为 P= 120-3Q。 (a) Find the price elasticity of demand at Q = 20. [3] (b) Is demand elastic, inelastic or unit-elastic there? [2] (a) 求 Q=20 时的需求价格弹性。[3](b)在该点需求是富有弹性、缺乏弹性还是单位弹性?[2] WORKED SOLUTION 1 Get dQ/dP. Rearrange P = 120 - 3Q - Q = 40 - P/3, so dQ/dP = - 1/3. 求 dQ/dP。整理 P=120-3Q →Q=40-P/3,所以 dQ/dP =- 1/3。 2 Price at Q = 20. P = 120 - 3(20) = 60. Q=20 时的价格。P= 120-3(20)= 60。 3 (a) Apply E = (dQ/dP)(P/Q). E =(-1/3)(60/20) = (-1/3)(3) = - 1. (a) 套用 E=(dQ/dP)(P/Q)。E= (-1/3)(60/20) = (-1/3)(3)= - 1。 4 (b) |E| = 1 - unit-elastic. This is the midpoint of a linear demand curve (choke price 120 at Q = 0, intercept Q = 40 at P = 0; midpoint Q = 20). Elastic above it, inelastic below. (b)|E|=1→单位弹性。这是线性需求曲线的中点(Q=0 时窒息价格为120,P=0时截距 Q=40;中点 Q=20)。其上 方富有弹性,下方缺乏弹性。 ECON1003 . Quantitative Methods in Economics PRACTICE P3-P4 - PRACTICE BANK (CONT. ) Financial mathematics 金融数学 Compounding the right number of periods, and solving for an unknown rate 用正确的期数复利,并求解未知利率

链条 C：积分还原经济量（Final 专属高频）

从 $MC$ 还原 $TC$：

• $$TC(Q)=\int MC(Q),dQ + C$$
• 并用条件 $TC(0)=FC$（固定成本）来定出 $C$。[6]Source: asksia-bible-econ1003-bilingual.pdf1 Setup. MC = dTC/dQ = 3Q2 + 4, and fixed cost TC(O) = 50. 设定。MC = dTC/dQ=3Q2 +4,固定成本 TC(0) =50。 Integrate. TC =[(3Q2 + 4)dQ = Q3 +4Q+C. 积分。TC=S(3Q2 + 4)dQ= Q3+4Q + C。 3 Use the condition. TC(0) = 0 + 0 + C = 50 -+ C = 50 (the fixed cost is the constant). 使用条件。TC(0)=0+0+C=50→C=50(固定成本就是那个常数)。 4 Particular solution. TC = Q3 + 4Q + 50. The +C was the fixed cost all along - drop it and you lose the whole economic content. I 特解。TC=Q3+4Q+50。那个+C 自始至终就是固定成本 -- 丢掉它,你就丢掉了全部的经济内涵。 ECON1003 . Quantitative Methods in Economics EX 6. 8 A separable growth law separable . growth 1 Setup. dy/dx = ky (proportional growth), with y(O) = yo. 设定。dy/dx =ky(比例增长),且 y(0)= yo。 2 Separate. (1/y)dy = k dx. 分离变量。(1/y)dy=k dx。 3 Integrate both sides. [(1/y)dy = [k dx -> Inlyl = kx + C. 对两边积分。S(1/y)dy = Sk dx → Inlyl = kx+ C。 Isolate y. y = ekx+C = A. ekx (write A = eC). Apply y(O) = y. - A = yo, so y = yo ekx - the exponential growth law, derived from scratch. 解出 y。y = ekx+C = A·ekx(记 A= eC)。代入 y(0)= yo → A= yo,所以 y = yo ekx -- 从零开始推导出来的指数增 长律。 ! Solve for C, do not drop it; and don't forget the In step 解出 C,不要丢掉它;也别忘了 In 这一步 In a differential equation the constant of integration is the whole point - the initial condition exists to fix it. With a separable equation, integrating 1/y produces a ln|y|; you must then exponentiate to get y back, turning the additive +C into a multiplicative A = eC. 在微分方程中,积分常数才是关键所在 -- 初始条件存在就是为了确定它。对于可分离方程,对 1/y 积分会产生一个 Inlyl;接着你必须取指数才能把 y 还原回来,从而把加性的+C 变成乘性的 A= eC。 Integration is differentiation run backwards - so every answer can be checked by differentiating it. On the final, if you have a minute, differentiate your integral and watch it collapse back to the integrand. Free insurance on method marks. 积分是反向运行的求导 -- 所以每个答案都可以通过对它求导来检验。在期末考上,如果你还有一分钟,就对你的积分 求导,看着它塌缩回被积函数。这是方法分上的免费保险。 MARKER'S NOTE . ECON1003 FINAL ECON1003 . Quantitative Methods in Economics T7 . MATRICES - TOPIC 7 . LINEAR ALGEBRA FINAL EXAM TOPIC Matrices, arithmetic & multiplication[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS

你必须养成“$+C$ 习惯”：不写 $+C$ 是白送丢分，做 ODE/还原成本时甚至是致命错误。[6]Source: asksia-bible-econ1003-bilingual.pdf1 Setup. MC = dTC/dQ = 3Q2 + 4, and fixed cost TC(O) = 50. 设定。MC = dTC/dQ=3Q2 +4,固定成本 TC(0) =50。 Integrate. TC =[(3Q2 + 4)dQ = Q3 +4Q+C. 积分。TC=S(3Q2 + 4)dQ= Q3+4Q + C。 3 Use the condition. TC(0) = 0 + 0 + C = 50 -+ C = 50 (the fixed cost is the constant). 使用条件。TC(0)=0+0+C=50→C=50(固定成本就是那个常数)。 4 Particular solution. TC = Q3 + 4Q + 50. The +C was the fixed cost all along - drop it and you lose the whole economic content. I 特解。TC=Q3+4Q+50。那个+C 自始至终就是固定成本 -- 丢掉它,你就丢掉了全部的经济内涵。 ECON1003 . Quantitative Methods in Economics EX 6. 8 A separable growth law separable . growth 1 Setup. dy/dx = ky (proportional growth), with y(O) = yo. 设定。dy/dx =ky(比例增长),且 y(0)= yo。 2 Separate. (1/y)dy = k dx. 分离变量。(1/y)dy=k dx。 3 Integrate both sides. [(1/y)dy = [k dx -> Inlyl = kx + C. 对两边积分。S(1/y)dy = Sk dx → Inlyl = kx+ C。 Isolate y. y = ekx+C = A. ekx (write A = eC). Apply y(O) = y. - A = yo, so y = yo ekx - the exponential growth law, derived from scratch. 解出 y。y = ekx+C = A·ekx(记 A= eC)。代入 y(0)= yo → A= yo,所以 y = yo ekx -- 从零开始推导出来的指数增 长律。 ! Solve for C, do not drop it; and don't forget the In step 解出 C,不要丢掉它;也别忘了 In 这一步 In a differential equation the constant of integration is the whole point - the initial condition exists to fix it. With a separable equation, integrating 1/y produces a ln|y|; you must then exponentiate to get y back, turning the additive +C into a multiplicative A = eC. 在微分方程中,积分常数才是关键所在 -- 初始条件存在就是为了确定它。对于可分离方程,对 1/y 积分会产生一个 Inlyl;接着你必须取指数才能把 y 还原回来,从而把加性的+C 变成乘性的 A= eC。 Integration is differentiation run backwards - so every answer can be checked by differentiating it. On the final, if you have a minute, differentiate your integral and watch it collapse back to the integrand. Free insurance on method marks. 积分是反向运行的求导 -- 所以每个答案都可以通过对它求导来检验。在期末考上,如果你还有一分钟,就对你的积分 求导,看着它塌缩回被积函数。这是方法分上的免费保险。 MARKER'S NOTE . ECON1003 FINAL ECON1003 . Quantitative Methods in Economics T7 . MATRICES - TOPIC 7 . LINEAR ALGEBRA FINAL EXAM TOPIC Matrices, arithmetic & multiplication[12]Source: asksia-bible-econ1003-bilingual.pdf逆向求导:幂次加一,再除以新的幂次 TL;DR. Integration undoes differentiation. To integrate a power, raise the index by one and divide by the new index - the one exception is x", which integrates to In(x). Every indefinite integral carries a + C; never drop it, because in differential-equation questions you later solve for it. Definite integrals turn this machinery into area - and area is how this course measures consumer and producer surplus. 一句话总结。积分抵消求导。要对一个幂求积分,指数加一,再除以新的指数 -- 唯一例外是x-1,它积分为 In(x)。每个不 定积分都带一个+ C;绝不要丢掉它,因为在微分方程题里你之后要解出它。定积分把这套机制变成面积 -- 而本课程正是 用面积来度量消费者剩余和生产者剩余。 ★ What the exam asks here 考试在这里怎么问 Integration sits on the cumulative final (not the midterm). Expect: a routine indefinite integral testing the power / ln / e* rules; a substitution or by-parts integral; a definite integral read as an area or as consumer / producer surplus; and a differential equation (recovering total cost from marginal cost, or a separable growth law). The formula sheet gives you the by-parts formula - the rest you memorise. 积分出现在累计性期末考(而非期中考)上。预计会有:一道检验幂/ In/ex法则的常规不定积分;一道换元或分部积 分;一道被解读为面积或消费者剩余/生产者剩余的定积分;以及一道微分方程(由边际成本还原总成本,或一个可分 离的增长律)。公式表会给你分部积分公式 -- 其余的要背熟。 6. 1 The integral as an anti-derivative 6. 1 积分作为反导数 A function g is an anti-derivative (#) off if g'(x) = f(x). Because the derivative of a constant is zero, anti- derivatives are only pinned down up to a constant - hence the constant of integration C. 若 g'(x)= f(x),则函数 g 是 f 的反导数〔原函数〕。由于常数的导数为零,反导数只能确定到相差一个常数 -- 这就是积分 常数 C 的由来。 DEFINITION [ f(x) dx = g(x) + C where g' (x) = f(x) 6. 2 The power rule & the n = - 1 exception 6. 2 幂法则与 n =- 1 的例外 For every index except -1, push the index up by one and divide by the new index. At n = - 1 that would divide by zero - so that single case becomes the natural log. 对除 -1 以外的每个指数,把指数加一,再除以新的指数。在 n =- 1 时这样做会除以零 -- 所以这唯一一种情形变成了自 然对数。 ECON1003 . Quantitative Methods in Economics POWER RULE + FRIENDS [ x" dx = x"+1/ (n+1) + C (n = - 1) [ x-1 dx = [ (1/x) dx = In| x | + C (the n = - 1 case) [ K dx = Kx + C . § ex dx = ex + C . [ K. f dx = KJ f dx EX 6. 1 Power rule, three terms at once power rule 1 Rewrite roots & fractions as powers first. Integrate (3x2 + 4Vx - 5/x2)dx, i. e. f(3x2 + 4x1/2 - 5x-2)dx. 先把根式与分式改写成幂。对 S(3x2+4√x-5/x2)dx 积分,即 S(3x2+4x1/2-5x-2)dx。 2 Term by term. 3. x3/3 = x3; 4. x3/2/(3/2) = (8/3)x3/2; - 5. x-1/(-1) = +5x-1. 逐项处理。3·x3/3=x3; 4·x3/2/(3/2)= (8/3)×3/2; - 5·x-1/(-1) = +5x-1。 3 Assemble + C. Answer = x3 + (8/3)x3/2 + 5/x + C. Differentiate back to check: 3x2 + 4×1/2 - 5x-2. V 组装并加+ C。答案=x3+(8/3)×3/2+5/x+C。求导验证:3x2+4×1/2-5x-2。√ ! n = - 1 is the log, and the "+ C" is not optional n =- 1 对应对数,「+C」不是可选项 (1/x)dx is In|x| + C, never xº/O. And forgetting + C is a free dropped mark on an indefinite integral - worse, it is fatal in a differential equation, where the initial condition exists precisely to solve for that C. S(1/x)dx 是 In|xl+ C,绝不是 x°/0。而忘写+C 在不定积分中是白白丢一分 -- 更糟的是,它在微分方程中是致命 的,因为初始条件存在的意义恰恰就是解出那个C。 ECON1003 . Quantitative Methods in Economics[18]Source: asksia-cheatsheet-econ1003.pdf* Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356 · Break-even: TC=100+4Q, P=9 -> Q =20 · Turning pt of y=x2-8x+1 -> (4, -15) · J(6x2+2) dx -> 2x3 + 2x + C * Quiz & Logistics IT 10% . DON'T LOSE 4 online quizzes (10% total), 1 week each, NO extensions - do them early. Q&A via Ed, not email. Textbook: Bradley, Essential Mathematics for Economics & Business, 4th ed. (Chs 2,4,5,6,7,8,9). The unit is maths in the service of economics - every technique gets an economic reading (slope = marginal effect, A = shadow price, integral = surplus). maths for economics . check your current unit outline . @ 2026 flip - for side 2 . several variables, integration & matrices asksia. ai/cheatsheet/ usyd-econ1003 . side 1/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd-econ1003 CH2

链条 D：消费者/生产者剩余（用定积分表达“面积”）

先求均衡：先找 $Q^*$（它是积分上限），再算剩余。[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS

公式结构（按宝典写法）：[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS

• $$CS=\int_{0}^{Q^*}\big(D(Q)-P^*\big),dQ$$
• $$PS=\int_{0}^{Q^*}\big(P^*-S(Q)\big),dQ$$

易错点：没先求 $Q^*$ 就直接积分；或把上下函数顺序写反。[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

链条 E：拉格朗日（多变量最优化，Final 高频）

基本形式：

• 有约束最大化：$L=f(\cdot)-\lambda(\text{constraint})$（宝典写法就是 “$L=f-A(\text{constraint})$”）[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors

必须“列全四件套”（cheatsheet 明确点名）：[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5

• 所有一阶条件 $\partial L/\partial(\text{每个变量})=0$
• 约束要检查/满足
• 非负性/不等式约束要处理（含 complementary slackness：$\lambda(g-b)=0$ 这种结构）[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5
• 角点（corner cases） 也要算并比较（例如 $x=0$ 或 $y=0$）[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

shadow price（影子价格）定义：

• $\lambda$ 表示：约束放松 1 单位时，最优目标值改变多少（单位：目标/约束）。这是“写一句话就能拿分”的点。[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5

链条 F：矩阵/线代（Final 专属，且很容易丢“低级分”）

开做之前先做两件事：[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

• 先检查维度（dims first）
• 再检查 $\det(A)\neq 0$（否则没有逆矩阵，也不能用 $A^{-1}$ 方案）[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

逆矩阵（cofactor 法）核心公式：

• $$A^{-1}=\frac{C^{T}}{\det(A)}$$
• 其中 $C$ 是余子式/代数余子式矩阵（minor × “$+ - +$ 棋盘符号”）。[1]Source: asksia-bible-econ1003-bilingual.pdf代数余子式 / 余子式 Minor = sub-determinant; cofactor = minor x the +-+ sign. ECON1003 . Quantitative Methods in Economics Term 术语 中文 One-line definition 释义 Inverse matrix (A-1) 逆矩阵 A-1 = CT/det(A); satisfies AA1 = I; solves X = A-1B. ✓ How to use the glossary 用法 如何使用术语表 用法 Cover the right column and recite the definition from the English term - then flip and recall the English from the +X. 〔遮住释义、由术语复述定义;再由中文回想英文。〕 The exam rewards the exact wording: "next-best forgone", "CT/det(A)", "signed area" - precision is marks. 遮住右栏,从英文术语复述出定义 -- 再反过来,由中文回忆英文。〔遮住释义、由术语复述定义;再由中文回想英文。〕 考试奖励精确的措辞:“next-best forgone”、“CT/det(A)”、“signed area" -- 精确就是分数。 ECON1003 . Quantitative Methods in Economics PRACTICE P1 -P2 - PRACTICE BANK DRILL TO EXAM STANDARD Drill both exams, with full solutions 两场考试齐刷,附完整解答 Fourteen AskSia-authored problems across all seven topics - worked end to end 横跨全部七大主题的14 道 AskSia 原创题 –––完整解答 These are exam-style problems written in the ECON1003 style - not copied from any real quiz or exam. They span lines, finance, differentiation, the Lagrangian, integration and matrices, so they serve both the Midterm (Topics 1-4) and the cumulative Final (all seven). Every number is verified. Cover the solution, work it on paper, then read the answer. 这些是按 ECON1003 风格编写的考试风格题目 -- 并非抄自任何真实测验或考试。它们覆盖直线、金融、求导、拉格朗日 函数、积分和矩阵,因此同时服务于期中(主题1-4)和累计的期末(全部七大主题)。每个数字都经过核验。盖住解答,在 纸上做完,再看答案。 ★ How to use this bank 如何使用这个题库 Cover the solution - recognition is not recall. Budget ~1 minute per mark. Show your work: method marks are real and partial credit is given, so a stated step beats a bare number. Round only the final answer to the stated decimals. Check every answer by substituting back - differentiate an integral, multiply an inverse, re-solve a system. 遮住答案 认得出不等于回忆得起。按每分约1分 钟来分配时间。写出你的过程:方法分是实打实的, 也会给部分分,所以一个写出来的步骤胜过一个孤零 零的数字。只对最终答案按规定小数位取整。对每个 答案都做检验,把它代回去 -- 对积分求导、把逆矩 阵乘回去、重新解一遍方程组。 i Formula sheet vs memory 公式表 vs 记忆 Both exams give a formula sheet (quadratic, series, interest, NPV, annuity, by-parts, 3×3 determinant, ax/log derivatives, quotient rule). You must memorise the rest: power / product / chain rules, substitution, the Lagrangian conditions, Cramer's rule and the inverse methods. The marks are in applying them cleanly.[8]Source: asksia-bible-econ1003-bilingual.pdfC = matrix of cofactors (minor x the +-+ checkerboard sign). Transpose it, then divide by det(A). C = 余子式矩阵(小行列式 ×+-+棋盘格符号)。把它转 置,再除以det(A)。 THE 2x2 INVERSE a b 2×2 shortcut : A = - A-1 = (1/(ad-bc)) . cd -C a det) d -b (swap a&d, negate bac, ECON1003 . Quantitative Methods in Economics AskSia Library EXAM BIBLE . ASKSIA SCHOOL OF ECONOMICS SEMESTER 1 . 2026 THE COMPLETE EXAM BIBLE Quantitative Methods in Economics 经济学定量方法 IT SAYS "QUANTITATIVE"-IT MEANS THE MATHS ENGINE ROOM OF ECONOMICS. 悉尼大学 ECON1003 · 双语考试 bible · 期中 40%+期末 50% · 数学 (非统计) ECON1003 . UNIVERSITY OF SYDNEY 中英双语版 · BILINGUAL EDITION 英文主讲,中文随行 一考试要点与术语保留英文原词 Lines, logs, interest, calculus, multivariable optimisation and matrices - every method drilled on economic examples. A formula sheet is provided in the exams, so this book trains the one thing that earns marks: method, not memorisation. Two exams to clear - a 40% Midterm and a 50% cumulative Final. Independent study companion. Not affiliated with or endorsed by the University of Sydney. Corrections: takedowns@asksia. ai PREFACE - HOW TO USE THIS BOOK[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL
• 关键陷阱：别忘了 转置 $C^T$。[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors

$2\times 2$ 速算法：若 $A=\begin{pmatrix}a&b\c&d\end{pmatrix}$，

• $$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}$$（交换 $a,d$，$b,c$ 取负）。[8]Source: asksia-bible-econ1003-bilingual.pdfC = matrix of cofactors (minor x the +-+ checkerboard sign). Transpose it, then divide by det(A). C = 余子式矩阵(小行列式 ×+-+棋盘格符号)。把它转 置,再除以det(A)。 THE 2x2 INVERSE a b 2×2 shortcut : A = - A-1 = (1/(ad-bc)) . cd -C a det) d -b (swap a&d, negate bac, ECON1003 . Quantitative Methods in Economics AskSia Library EXAM BIBLE . ASKSIA SCHOOL OF ECONOMICS SEMESTER 1 . 2026 THE COMPLETE EXAM BIBLE Quantitative Methods in Economics 经济学定量方法 IT SAYS "QUANTITATIVE"-IT MEANS THE MATHS ENGINE ROOM OF ECONOMICS. 悉尼大学 ECON1003 · 双语考试 bible · 期中 40%+期末 50% · 数学 (非统计) ECON1003 . UNIVERSITY OF SYDNEY 中英双语版 · BILINGUAL EDITION 英文主讲,中文随行 一考试要点与术语保留英文原词 Lines, logs, interest, calculus, multivariable optimisation and matrices - every method drilled on economic examples. A formula sheet is provided in the exams, so this book trains the one thing that earns marks: method, not memorisation. Two exams to clear - a 40% Midterm and a 50% cumulative Final. Independent study companion. Not affiliated with or endorsed by the University of Sydney. Corrections: takedowns@asksia. ai PREFACE - HOW TO USE THIS BOOK[26]Source: asksia-cheatsheet-econ1003.pdf· Same A, many b's -> find A-1 once · Big / numeric system -> Gauss-Jordan 34 . The Inverse Matrix CH9 . MEMORISE DEFN - EXISTS IFF DET(A) # 0 A· A-1 = A-1. A = I COFACTOR METHOD A-1 = CT / det (A) C = cofactors (minor x +-+ sign) 2x2 SHORTCUT A-1 = (1/(ad-bc)) . [[d,-b], [-c,a]] WORKED . A = [2,1],[1,1]] det = 2-1 = 1 - A-1 = [ [1,-1], [-1,2]] SOLVE A SYSTEM AX = B = X = A-1B TRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions

解线性方程组：

• $AX=B \Rightarrow X=A^{-1}B$（可做，但先确认 $\det(A)\neq 0$）[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL[26]Source: asksia-cheatsheet-econ1003.pdf· Same A, many b's -> find A-1 once · Big / numeric system -> Gauss-Jordan 34 . The Inverse Matrix CH9 . MEMORISE DEFN - EXISTS IFF DET(A) # 0 A· A-1 = A-1. A = I COFACTOR METHOD A-1 = CT / det (A) C = cofactors (minor x +-+ sign) 2x2 SHORTCUT A-1 = (1/(ad-bc)) . [[d,-b], [-c,a]] WORKED . A = [2,1],[1,1]] det = 2-1 = 1 - A-1 = [ [1,-1], [-1,2]] SOLVE A SYSTEM AX = B = X = A-1B TRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions
• 单次求解常用消元/Cramer 更快；多次不同 $B$ 才更适合先算 $A^{-1}$ 复用。[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

3）公式表 vs 你必须背的：记忆清单（按宝典原意整理）

• 二次公式、数列求和、利息/NPV/年金、分部积分、$3\times 3$ 行列式、$a^x/\log$ 的导数、商法则等。[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

• 求导：幂/和/常数、乘积、链式（product & chain 特别爱组合考）[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'
• 积分：换元（substitution）、幂积分规则与 $n=-1$ 例外、以及永远写 $+C$ [12]Source: asksia-bible-econ1003-bilingual.pdf逆向求导:幂次加一,再除以新的幂次 TL;DR. Integration undoes differentiation. To integrate a power, raise the index by one and divide by the new index - the one exception is x", which integrates to In(x). Every indefinite integral carries a + C; never drop it, because in differential-equation questions you later solve for it. Definite integrals turn this machinery into area - and area is how this course measures consumer and producer surplus. 一句话总结。积分抵消求导。要对一个幂求积分,指数加一,再除以新的指数 -- 唯一例外是x-1,它积分为 In(x)。每个不 定积分都带一个+ C;绝不要丢掉它,因为在微分方程题里你之后要解出它。定积分把这套机制变成面积 -- 而本课程正是 用面积来度量消费者剩余和生产者剩余。 ★ What the exam asks here 考试在这里怎么问 Integration sits on the cumulative final (not the midterm). Expect: a routine indefinite integral testing the power / ln / e* rules; a substitution or by-parts integral; a definite integral read as an area or as consumer / producer surplus; and a differential equation (recovering total cost from marginal cost, or a separable growth law). The formula sheet gives you the by-parts formula - the rest you memorise. 积分出现在累计性期末考(而非期中考)上。预计会有:一道检验幂/ In/ex法则的常规不定积分;一道换元或分部积 分;一道被解读为面积或消费者剩余/生产者剩余的定积分;以及一道微分方程(由边际成本还原总成本,或一个可分 离的增长律)。公式表会给你分部积分公式 -- 其余的要背熟。 6. 1 The integral as an anti-derivative 6. 1 积分作为反导数 A function g is an anti-derivative (#) off if g'(x) = f(x). Because the derivative of a constant is zero, anti- derivatives are only pinned down up to a constant - hence the constant of integration C. 若 g'(x)= f(x),则函数 g 是 f 的反导数〔原函数〕。由于常数的导数为零,反导数只能确定到相差一个常数 -- 这就是积分 常数 C 的由来。 DEFINITION [ f(x) dx = g(x) + C where g' (x) = f(x) 6. 2 The power rule & the n = - 1 exception 6. 2 幂法则与 n =- 1 的例外 For every index except -1, push the index up by one and divide by the new index. At n = - 1 that would divide by zero - so that single case becomes the natural log. 对除 -1 以外的每个指数,把指数加一,再除以新的指数。在 n =- 1 时这样做会除以零 -- 所以这唯一一种情形变成了自 然对数。 ECON1003 . Quantitative Methods in Economics POWER RULE + FRIENDS [ x" dx = x"+1/ (n+1) + C (n = - 1) [ x-1 dx = [ (1/x) dx = In| x | + C (the n = - 1 case) [ K dx = Kx + C . § ex dx = ex + C . [ K. f dx = KJ f dx EX 6. 1 Power rule, three terms at once power rule 1 Rewrite roots & fractions as powers first. Integrate (3x2 + 4Vx - 5/x2)dx, i. e. f(3x2 + 4x1/2 - 5x-2)dx. 先把根式与分式改写成幂。对 S(3x2+4√x-5/x2)dx 积分,即 S(3x2+4x1/2-5x-2)dx。 2 Term by term. 3. x3/3 = x3; 4. x3/2/(3/2) = (8/3)x3/2; - 5. x-1/(-1) = +5x-1. 逐项处理。3·x3/3=x3; 4·x3/2/(3/2)= (8/3)×3/2; - 5·x-1/(-1) = +5x-1。 3 Assemble + C. Answer = x3 + (8/3)x3/2 + 5/x + C. Differentiate back to check: 3x2 + 4×1/2 - 5x-2. V 组装并加+ C。答案=x3+(8/3)×3/2+5/x+C。求导验证:3x2+4×1/2-5x-2。√ ! n = - 1 is the log, and the "+ C" is not optional n =- 1 对应对数,「+C」不是可选项 (1/x)dx is In|x| + C, never xº/O. And forgetting + C is a free dropped mark on an indefinite integral - worse, it is fatal in a differential equation, where the initial condition exists precisely to solve for that C. S(1/x)dx 是 In|xl+ C,绝不是 x°/0。而忘写+C 在不定积分中是白白丢一分 -- 更糟的是,它在微分方程中是致命 的,因为初始条件存在的意义恰恰就是解出那个C。 ECON1003 . Quantitative Methods in Economics[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors
• 拉格朗日：条件清单 + 角点 + complementary slackness [29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors
• 线代：Cramer、逆矩阵方法、cofactor 转置、$\det(A)\neq 0$ 的前置检查 [11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

4）逐主题复习重点（7 个主题你该怎么分配时间）

Topic 1：Lines / Elasticity / Break-even（基础但最容易“粗心送分”）

你要会：

• 非标准形式先整理成 $y=mx+c$ 或 $P=f(Q)$ 再读信息。[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1[18]Source: asksia-cheatsheet-econ1003.pdf* Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356 · Break-even: TC=100+4Q, P=9 -> Q =20 · Turning pt of y=x2-8x+1 -> (4, -15) · J(6x2+2) dx -> 2x3 + 2x + C * Quiz & Logistics IT 10% . DON'T LOSE 4 online quizzes (10% total), 1 week each, NO extensions - do them early. Q&A via Ed, not email. Textbook: Bradley, Essential Mathematics for Economics & Business, 4th ed. (Chs 2,4,5,6,7,8,9). The unit is maths in the service of economics - every technique gets an economic reading (slope = marginal effect, A = shadow price, integral = surplus). maths for economics . check your current unit outline . @ 2026 flip - for side 2 . several variables, integration & matrices asksia. ai/cheatsheet/ usyd-econ1003 . side 1/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd-econ1003 CH2
• 盈亏平衡：Profit $=0$；固定成本是 $TC$ 截距，不是边际成本。[5]Source: asksia-bible-econ1003-bilingual.pdf两场考试都提供公式表(二次方程、级数、利息、 NPV、年金、分部积分、3x3 行列式、ax/log 导数、 商法则)。其余的你必须背熟:幂/乘积/链式法则、 换元法、拉格朗日条件、Cramer's rule 以及逆矩阵方 法。分数就在于把它们干净利落地用出来。 P1 CALCULATION 6 marks . lines & break-even . Topic 1 A firm sells output at P = $10 per unit with total cost 某厂商以 P=$10 的单价出售产品,总成本为 TC = 600 + 4Q TC = 600 + 4Q (a) Find the break-even quantity. [3] (b) Find profit when Q = 150. [3] (a) 求盈亏平衡数量。[3](b)求Q=150 时的利润。[3] WORKED SOLUTION 1 (a) Break-even is profit = 0. Revenue = TC: 10Q = 600 + 4Q - 6Q = 600 -+ Q=100 units. (Fixed cost 600 is the TC intercept, not a marginal cost. ) ECON1003 . Quantitative Methods in Economics (a) 盈亏平衡即利润=0。收益 = TC: 10Q= 600+ 4Q→6Q= 600→Q =100单位。(固定成本 600是 TC的截距,不 是边际成本。) (b) Profit at Q = 150. TR = 10-150 = 1500; TC = 600 + 4. 150 = 1200; Profit = 1500 - 1200 = $300. 2 (b) Q =150 时的利润。TR= 10·150= 1500; TC = 600+ 4·150= 1200; Profit = 1500 - 1200= $300。 P2 CALCULATION 5 marks . linear elasticity . Topic 1 Inverse demand is P = 120 - 3Q. 反需求为 P= 120-3Q。 (a) Find the price elasticity of demand at Q = 20. [3] (b) Is demand elastic, inelastic or unit-elastic there? [2] (a) 求 Q=20 时的需求价格弹性。[3](b)在该点需求是富有弹性、缺乏弹性还是单位弹性?[2] WORKED SOLUTION 1 Get dQ/dP. Rearrange P = 120 - 3Q - Q = 40 - P/3, so dQ/dP = - 1/3. 求 dQ/dP。整理 P=120-3Q →Q=40-P/3,所以 dQ/dP =- 1/3。 2 Price at Q = 20. P = 120 - 3(20) = 60. Q=20 时的价格。P= 120-3(20)= 60。 3 (a) Apply E = (dQ/dP)(P/Q). E =(-1/3)(60/20) = (-1/3)(3) = - 1. (a) 套用 E=(dQ/dP)(P/Q)。E= (-1/3)(60/20) = (-1/3)(3)= - 1。 4 (b) |E| = 1 - unit-elastic. This is the midpoint of a linear demand curve (choke price 120 at Q = 0, intercept Q = 40 at P = 0; midpoint Q = 20). Elastic above it, inelastic below. (b)|E|=1→单位弹性。这是线性需求曲线的中点(Q=0 时窒息价格为120,P=0时截距 Q=40;中点 Q=20)。其上 方富有弹性,下方缺乏弹性。 ECON1003 . Quantitative Methods in Economics PRACTICE P3-P4 - PRACTICE BANK (CONT. ) Financial mathematics 金融数学 Compounding the right number of periods, and solving for an unknown rate 用正确的期数复利,并求解未知利率[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1
• 弹性公式 + “弹性≠斜率”解释。[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。[5]Source: asksia-bible-econ1003-bilingual.pdf两场考试都提供公式表(二次方程、级数、利息、 NPV、年金、分部积分、3x3 行列式、ax/log 导数、 商法则)。其余的你必须背熟:幂/乘积/链式法则、 换元法、拉格朗日条件、Cramer's rule 以及逆矩阵方 法。分数就在于把它们干净利落地用出来。 P1 CALCULATION 6 marks . lines & break-even . Topic 1 A firm sells output at P = $10 per unit with total cost 某厂商以 P=$10 的单价出售产品,总成本为 TC = 600 + 4Q TC = 600 + 4Q (a) Find the break-even quantity. [3] (b) Find profit when Q = 150. [3] (a) 求盈亏平衡数量。[3](b)求Q=150 时的利润。[3] WORKED SOLUTION 1 (a) Break-even is profit = 0. Revenue = TC: 10Q = 600 + 4Q - 6Q = 600 -+ Q=100 units. (Fixed cost 600 is the TC intercept, not a marginal cost. ) ECON1003 . Quantitative Methods in Economics (a) 盈亏平衡即利润=0。收益 = TC: 10Q= 600+ 4Q→6Q= 600→Q =100单位。(固定成本 600是 TC的截距,不 是边际成本。) (b) Profit at Q = 150. TR = 10-150 = 1500; TC = 600 + 4. 150 = 1200; Profit = 1500 - 1200 = $300. 2 (b) Q =150 时的利润。TR= 10·150= 1500; TC = 600+ 4·150= 1200; Profit = 1500 - 1200= $300。 P2 CALCULATION 5 marks . linear elasticity . Topic 1 Inverse demand is P = 120 - 3Q. 反需求为 P= 120-3Q。 (a) Find the price elasticity of demand at Q = 20. [3] (b) Is demand elastic, inelastic or unit-elastic there? [2] (a) 求 Q=20 时的需求价格弹性。[3](b)在该点需求是富有弹性、缺乏弹性还是单位弹性?[2] WORKED SOLUTION 1 Get dQ/dP. Rearrange P = 120 - 3Q - Q = 40 - P/3, so dQ/dP = - 1/3. 求 dQ/dP。整理 P=120-3Q →Q=40-P/3,所以 dQ/dP =- 1/3。 2 Price at Q = 20. P = 120 - 3(20) = 60. Q=20 时的价格。P= 120-3(20)= 60。 3 (a) Apply E = (dQ/dP)(P/Q). E =(-1/3)(60/20) = (-1/3)(3) = - 1. (a) 套用 E=(dQ/dP)(P/Q)。E= (-1/3)(60/20) = (-1/3)(3)= - 1。 4 (b) |E| = 1 - unit-elastic. This is the midpoint of a linear demand curve (choke price 120 at Q = 0, intercept Q = 40 at P = 0; midpoint Q = 20). Elastic above it, inelastic below. (b)|E|=1→单位弹性。这是线性需求曲线的中点(Q=0 时窒息价格为120,P=0时截距 Q=40;中点 Q=20)。其上 方富有弹性,下方缺乏弹性。 ECON1003 . Quantitative Methods in Economics PRACTICE P3-P4 - PRACTICE BANK (CONT. ) Financial mathematics 金融数学 Compounding the right number of periods, and solving for an unknown rate 用正确的期数复利,并求解未知利率

Topic 2：Non-linear（Quadratic / Exponential / Log）

你要会：

• 二次：判别式 $\Delta=b^2-4ac$ 判断根；驻点在两根中点；$a>0$ 最小、$a<0$ 最大。[7]Source: asksia-bible-econ1003-bilingual.pdf(1) Solve ax2+bx+c=O with the formula; check the discriminant A first (<0 = no real root). (2) Turning point sits midway between the roots; a>0 is a min, a<0 a max. (3) Transformations: outside = vertical (normal), inside = horizontal (reversed); (x-2)2 moves right. (4) Index rules combine powers; solve a *= bx by matching bases. (5) Logs undo exponentials: product/quotient/power rules, but no rule for log of a sum. (6) Growth P = P eft; for time, take In of both sides. - (1) 用公式解 ax2+bx+c=0;先检查判别式 △(<0表示无实根)。(2)驻点/极值点位于两根的正中间;a>0 为极小值, a<0 为极大值。(3)变换:外部= 竖直方向(正常),内部= 水平方向(相反);(x-2)2 向右移。(4)指数法则用于合并 幂次;解 ax=bx 时令底数相同。(5)对数抵消指数:乘积/商/幂法则,但和的对数没有法则。(6)增长 P= Poef t;求 时间时,对两边取 In。 - ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS - TOPIC 3 . FINANCIAL MATHEMATICS FORMULA-SHEET TOPIC Money over time: series, interest, present value 货币的时间价值:数列、利息、现值 Arithmetic & geometric series . simple / compound / continuous interest 等差与等比数列 · 单利/复利 /连续复利 Financial maths is the most procedural topic in the unit: every question is "plug the right numbers into the right formula and grind. " The good news - nearly every formula is on the provided sheet. The marks are lost on the small things: choosing compound vs continuous, getting the m (periods per year) right, and telling a future-value annuity from a present-value one. Get those decisions right and the rest is calculator work. 金融数学是本单元最程序化的主题:每道题都是「把正确的数字代入正确的公式,然后硬算」。好消息是 -- 几乎每个公式 都在所提供的公式表上。失分都在小细节上:选复利还是连续复利、把 m(每年期数)取对、以及区分终值年金与现值年 金。把这些判断做对,其余就是按计算器的活儿。 ★ What the exam does with this topic 考试对这个主题怎么考 Expect numeric-entry items: "find the value after t years," "find the rate / time," "compute the NPV," "find the monthly repayment. " You will pick a formula from the sheet, substitute, and round to the stated decimal places. Show your substitution - method marks are awarded even if the final arithmetic slips. 预计会出现数值填空题:“求t年后的数值”、“求利率/时间”、“计算 NPV”、“求每月还款额”。你需要从公式表中挑出 一个公式,代入数值,并按规定的小数位取整。把代入过程写出来 -- 即使最后的算术出了差错,方法分依然会给。 3. 1 Sequences & series 3. 1 数列与级数 A sequence is a list following a pattern; a series is the sum of its terms. Two patterns matter. An arithmetic sequence adds a constant d each step (a, a+d, a+2d, . . . ); a geometric sequence multiplies by a constant ratio r each step (a, ar, ar2 , . . . ). 数列是按某规律排列的一串数;级数是其各项之和。两种规律最重要。等差数列每步加上一个常数 d (a, a+d, a+2d, ARITHMETIC SERIES Arithmetic sum of first n terms : Sn = (n/2) . (2a + (n-1)d) a = first term . d = common difference GEOMETRIC SERIES Geometric sum of first n terms : Sn = a(1 - rn) / (1 - r) Infinite sum (only if |r| < 1): Sco = a / (1 - r) ECON1003 . Quantitative Methods in Economics[14]Source: asksia-bible-econ1003-bilingual.pdfElastic / inelastic 富有弹性/ 缺乏弹性 |E| > 1 elastic (responsive); |E| < 1 inelastic; |E| = 1 unit-elastic. Topic 2 -Non-linear functions 非线性函数 Quadratic 二次函数 ax2 + bx + c; one turning point; a parabola, min if a > O. Discriminant 判别式 b2 - 4ac; >O two roots, =O one root, <O no real roots. Exponential function 指数函数 y = ax; constant base raised to a variable power; e ~ 2. 71828. Logarithm 对数 log (b) = the power a must be raised to for b; In = log base e. Turning point 转折点/ 极值点 Where a curve changes from rising to falling (or vice versa). ECON1003 . Quantitative Methods in Economics GLOSSARY 2/3 - GLOSSARY · 术语表 (CONT. ) Finance, differentiation 金融、求导 Compound interest, NPV, the derivative rules and their economic meaning 复利、NPV、求导法则及其经济含义 Term 术语 中文 One-line definition ##X Topic 3 - Financial mathematics 金融数学 Arithmetic series
• 指数与对数：对数用来“把指数拽下来”解未知时间/未知幂；不能配底就两边取对数。[7]Source: asksia-bible-econ1003-bilingual.pdf(1) Solve ax2+bx+c=O with the formula; check the discriminant A first (<0 = no real root). (2) Turning point sits midway between the roots; a>0 is a min, a<0 a max. (3) Transformations: outside = vertical (normal), inside = horizontal (reversed); (x-2)2 moves right. (4) Index rules combine powers; solve a *= bx by matching bases. (5) Logs undo exponentials: product/quotient/power rules, but no rule for log of a sum. (6) Growth P = P eft; for time, take In of both sides. - (1) 用公式解 ax2+bx+c=0;先检查判别式 △(<0表示无实根)。(2)驻点/极值点位于两根的正中间;a>0 为极小值, a<0 为极大值。(3)变换:外部= 竖直方向(正常),内部= 水平方向(相反);(x-2)2 向右移。(4)指数法则用于合并 幂次;解 ax=bx 时令底数相同。(5)对数抵消指数:乘积/商/幂法则,但和的对数没有法则。(6)增长 P= Poef t;求 时间时,对两边取 In。 - ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS - TOPIC 3 . FINANCIAL MATHEMATICS FORMULA-SHEET TOPIC Money over time: series, interest, present value 货币的时间价值:数列、利息、现值 Arithmetic & geometric series . simple / compound / continuous interest 等差与等比数列 · 单利/复利 /连续复利 Financial maths is the most procedural topic in the unit: every question is "plug the right numbers into the right formula and grind. " The good news - nearly every formula is on the provided sheet. The marks are lost on the small things: choosing compound vs continuous, getting the m (periods per year) right, and telling a future-value annuity from a present-value one. Get those decisions right and the rest is calculator work. 金融数学是本单元最程序化的主题:每道题都是「把正确的数字代入正确的公式,然后硬算」。好消息是 -- 几乎每个公式 都在所提供的公式表上。失分都在小细节上:选复利还是连续复利、把 m(每年期数)取对、以及区分终值年金与现值年 金。把这些判断做对,其余就是按计算器的活儿。 ★ What the exam does with this topic 考试对这个主题怎么考 Expect numeric-entry items: "find the value after t years," "find the rate / time," "compute the NPV," "find the monthly repayment. " You will pick a formula from the sheet, substitute, and round to the stated decimal places. Show your substitution - method marks are awarded even if the final arithmetic slips. 预计会出现数值填空题:“求t年后的数值”、“求利率/时间”、“计算 NPV”、“求每月还款额”。你需要从公式表中挑出 一个公式,代入数值,并按规定的小数位取整。把代入过程写出来 -- 即使最后的算术出了差错,方法分依然会给。 3. 1 Sequences & series 3. 1 数列与级数 A sequence is a list following a pattern; a series is the sum of its terms. Two patterns matter. An arithmetic sequence adds a constant d each step (a, a+d, a+2d, . . . ); a geometric sequence multiplies by a constant ratio r each step (a, ar, ar2 , . . . ). 数列是按某规律排列的一串数;级数是其各项之和。两种规律最重要。等差数列每步加上一个常数 d (a, a+d, a+2d, ARITHMETIC SERIES Arithmetic sum of first n terms : Sn = (n/2) . (2a + (n-1)d) a = first term . d = common difference GEOMETRIC SERIES Geometric sum of first n terms : Sn = a(1 - rn) / (1 - r) Infinite sum (only if |r| < 1): Sco = a / (1 - r) ECON1003 . Quantitative Methods in Economics[27]Source: asksia-cheatsheet-econ1003.pdfWORKED . 3^{2X} = 81 81 = 34 - 2x = 4 = x = 2 TRAP > Can't match the base? Take logs of both sides and use log(ab)=b. log a to drop the exponent down. 8 . Logarithms CH4 Log_a (b) = power on a giving b. Log=log10, Ln=loge. LOG RULES Log(a)+Log(b) = log(ab) Log(a)-log(b) = log(a/b) Log(ab) = b · Log(a) change base: log_c(a) = Log(a)/Log(c) WORKED . 1750 = 753E^{0. 03T} Ln(1750/753) = 0. 03t = t = 28 TRAP > In(A+B) + InA + InB. Rules apply only to x/+ inside one log; argument must be >0. 9 · Sequences & Series N-TH TERM arithmetic: an = a + (n-1)d geometric: an = a. pn-1 ARITHMETIC SUM (+D) Sn = (n/2) (2a + (n-1)d) GEOMETRIC SUM (xR) Sn = a(1-rn)/(1-r) infinite (|r|<1): S = a/(1-r) WORKED . 2+6+18+ . . . (8 TERMS) a=2, r=3 - Sa = 2(38-1)/2 = 6560 TRAP > Want the n-th term or the sum? And the infinite sum needs |r|<1. 10 . Interest CH5 . ON SHEET GROWTH OF PO simple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate.
• 对数陷阱：$\ln(A+B)\neq \ln A+\ln B$；log 规则只对乘/除/幂，不对加法。[27]Source: asksia-cheatsheet-econ1003.pdfWORKED . 3^{2X} = 81 81 = 34 - 2x = 4 = x = 2 TRAP > Can't match the base? Take logs of both sides and use log(ab)=b. log a to drop the exponent down. 8 . Logarithms CH4 Log_a (b) = power on a giving b. Log=log10, Ln=loge. LOG RULES Log(a)+Log(b) = log(ab) Log(a)-log(b) = log(a/b) Log(ab) = b · Log(a) change base: log_c(a) = Log(a)/Log(c) WORKED . 1750 = 753E^{0. 03T} Ln(1750/753) = 0. 03t = t = 28 TRAP > In(A+B) + InA + InB. Rules apply only to x/+ inside one log; argument must be >0. 9 · Sequences & Series N-TH TERM arithmetic: an = a + (n-1)d geometric: an = a. pn-1 ARITHMETIC SUM (+D) Sn = (n/2) (2a + (n-1)d) GEOMETRIC SUM (xR) Sn = a(1-rn)/(1-r) infinite (|r|<1): S = a/(1-r) WORKED . 2+6+18+ . . . (8 TERMS) a=2, r=3 - Sa = 2(38-1)/2 = 6560 TRAP > Want the n-th term or the sum? And the infinite sum needs |r|<1. 10 . Interest CH5 . ON SHEET GROWTH OF PO simple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate.

Topic 3：Financial Maths（最程序化：会选模型就赢一半）

你要会做的决策：复利 vs 连续复利；$m$（每年期数）是否正确；FV 年金 vs PV/贷款年金不要混。[7]Source: asksia-bible-econ1003-bilingual.pdf(1) Solve ax2+bx+c=O with the formula; check the discriminant A first (<0 = no real root). (2) Turning point sits midway between the roots; a>0 is a min, a<0 a max. (3) Transformations: outside = vertical (normal), inside = horizontal (reversed); (x-2)2 moves right. (4) Index rules combine powers; solve a *= bx by matching bases. (5) Logs undo exponentials: product/quotient/power rules, but no rule for log of a sum. (6) Growth P = P eft; for time, take In of both sides. - (1) 用公式解 ax2+bx+c=0;先检查判别式 △(<0表示无实根)。(2)驻点/极值点位于两根的正中间;a>0 为极小值, a<0 为极大值。(3)变换:外部= 竖直方向(正常),内部= 水平方向(相反);(x-2)2 向右移。(4)指数法则用于合并 幂次;解 ax=bx 时令底数相同。(5)对数抵消指数:乘积/商/幂法则,但和的对数没有法则。(6)增长 P= Poef t;求 时间时,对两边取 In。 - ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS - TOPIC 3 . FINANCIAL MATHEMATICS FORMULA-SHEET TOPIC Money over time: series, interest, present value 货币的时间价值:数列、利息、现值 Arithmetic & geometric series . simple / compound / continuous interest 等差与等比数列 · 单利/复利 /连续复利 Financial maths is the most procedural topic in the unit: every question is "plug the right numbers into the right formula and grind. " The good news - nearly every formula is on the provided sheet. The marks are lost on the small things: choosing compound vs continuous, getting the m (periods per year) right, and telling a future-value annuity from a present-value one. Get those decisions right and the rest is calculator work. 金融数学是本单元最程序化的主题:每道题都是「把正确的数字代入正确的公式,然后硬算」。好消息是 -- 几乎每个公式 都在所提供的公式表上。失分都在小细节上:选复利还是连续复利、把 m(每年期数)取对、以及区分终值年金与现值年 金。把这些判断做对,其余就是按计算器的活儿。 ★ What the exam does with this topic 考试对这个主题怎么考 Expect numeric-entry items: "find the value after t years," "find the rate / time," "compute the NPV," "find the monthly repayment. " You will pick a formula from the sheet, substitute, and round to the stated decimal places. Show your substitution - method marks are awarded even if the final arithmetic slips. 预计会出现数值填空题:“求t年后的数值”、“求利率/时间”、“计算 NPV”、“求每月还款额”。你需要从公式表中挑出 一个公式,代入数值,并按规定的小数位取整。把代入过程写出来 -- 即使最后的算术出了差错,方法分依然会给。 3. 1 Sequences & series 3. 1 数列与级数 A sequence is a list following a pattern; a series is the sum of its terms. Two patterns matter. An arithmetic sequence adds a constant d each step (a, a+d, a+2d, . . . ); a geometric sequence multiplies by a constant ratio r each step (a, ar, ar2 , . . . ). 数列是按某规律排列的一串数;级数是其各项之和。两种规律最重要。等差数列每步加上一个常数 d (a, a+d, a+2d, ARITHMETIC SERIES Arithmetic sum of first n terms : Sn = (n/2) . (2a + (n-1)d) a = first term . d = common difference GEOMETRIC SERIES Geometric sum of first n terms : Sn = a(1 - rn) / (1 - r) Infinite sum (only if |r| < 1): Sco = a / (1 - r) ECON1003 . Quantitative Methods in Economics[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'[23]Source: asksia-cheatsheet-econ1003.pdfsimple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate. WORKED . $1000, QUARTERLY, 8%, 2YR P = 1000(1+0. 08/4)^{4. 2} = 1000(1. 02)8 = $1171. 66 WORKED NPV . - 1000 NOW, +600/YR x2, I=10% -1000 + 600/1. 1 + 600/1. 21 = - 1000 + 545. 5 + 495. 9 = +41. 3 = accept 12 . Annuities & CH5 . ON SHEET Loans FUTURE VALUE (DEPOSIT A. ) Vt = A0 . [(1+i/m)^{tm} - 1]/(i/m) PRESENT VALUE / LOAN L L = A0 . [1 - (1+i/m)^{-mt}]/(i/m) solve for Ao = each payment Total interest = (payment x no. payments) - L. WORKED . LOAN $20000, 6%/YR, 5YR ANN. L = A0 . [1-(1. 06)-5]/0. 06 20000 = A0 . 4. 2124 - Ao = $4747. 93/yr TRAP > FV uses ( . . . )^{tm) - 1; PV/loan uses 1 - ( . . . )^{-tm) (negative exp). Match m to payment frequency (monthly m=12). 13 . The Derivative - DEFINITION (NOT EXAMINED TO COMPUTE) f'(x) = lim_{k-0} [f(x+k)-f(x)]/k = instantaneous rate of change = slope of the tangent. Notation: f'(x), dy/dx, df/dx all mean the same thing. 14 . Differentiation Rules CH6 . MEMORISE

核心公式（常见形态）：

• 复利（每年计息 $m$ 次）：$$P_t=P_0\left(1+\frac{i}{m}\right)^{mt}$$[13]Source: asksia-bible-econ1003-bilingual.pdf等差数列求和 Sum of terms with a constant difference d; Sn = (n/2)(2a + (n-1)d). Geometric series 等比数列求和 Constant ratio r; Sn = a(1-r")/(1-r); infinite sum a/(1-r) if Ir|<1. Compound interest 复利 Pt = P. (1 + i/m)mt; interest earns interest each period. Continuous compounding 连续复利 Pt = Peit; the m->co limit of compounding. Present value 现值 Today's worth of a future sum: Po= Pt/(1+i)t. Depreciation 折旧 Fall in asset value; reducing-balance At = A. (1-i)t. Net present value (NPV) 净现值 E of discounted cash flows minus outlay; invest if NPV > O. Annuity 年金 A stream of equal periodic payments; has future- and present-value formulas. Topic 4 – Differentiation 微分 Derivative 导数 Instantaneous rate of change f'(x); slope of the tangent line. Power rule 幂法则 d/dx(x") = n·x™-1.[15]Source: asksia-bible-econ1003-bilingual.pdfEX 3A Two production-growth schemes - arithmetic vs geometric L · series A plant makes 1200 units in week 1. Scheme 1 adds 80 units each week (arithmetic, d = 80); Scheme 2 grows 5% each week (geometric, r = 1. 05). Find the total output over the first 20 weeks under each. 某工厂第1周生产1200件。方案1每周增加80件(等差,d=80);方案2每周增长5%(等比,r=1. 05)。求前 20 周在每种方案下的总产量。 1 Scheme 1 (arithmetic). a = 1200, d = 80, n = 20. S20 = (20/2)(2×1200 + 19×80) = 10x(2400+1520) = 39,200 units. 方案1(等差)。a =1200, d= 80, n = 20。S20 =(20/2)(2×1200+19×80)= 10×(2400+1520)= 39,200 单位。 2 Scheme 2 (geometric). a = 1200, r = 1. 05, n = 20. S20 = 1200(1 - 1. 0520)/(1 - 1. 05) ~ 39,679 units - the compounding 5% just edges ahead. 方案 2(等比)。a =1200, r=1. 05, n= 20。S20 = 1200(1-1. 0520)/(1-1. 05)~39,679 单位 -- 5% 的复利效应 恰好略微领先。 ! Term vs sum - read the question 单项 vs 求和 -- 看清题目 "Output in week 20" wants the 20th term (1200 + 19x80 = 2720, or 1200x1. 0519 ~ 3032); "total output over 20 weeks" wants the sum S20. And the infinite-sum formula is only valid when Irl < 1. “第20周当周的产量”要的是第 20项(1200+19×80=2720,或 1200×1. 0519~3032);“20周总产量”要的 是总和 S20。而无穷级数求和公式只在 Irl<1时才成立。 3. 2 Simple, compound & continuous interest 3. 2 单利、复利与连续复利 The single most important choice in the topic is which interest model applies. Simple interest pays only on the original principal; compound interest pays interest on accumulated interest too; continuous compounding is the limit as the number of conversion periods goes to infinity. 这个主题里最重要的一个选择是适用哪种利息模型。单利只对原始本金计息;复利还对累积的利息计息;连续复利是计息期数 趋于无穷时的极限。 INTEREST MODELS Simple : Pt = Po(1 + i. t) Compound (annual): Pt = P0(1 + i) t Compound m times/yr: Pt = P0(1 + i/m) mt Continuous : Pt = Po'eit ECON1003 . Quantitative Methods in Economics ! The m-trap - divide the rate AND multiply the periods m 陷阱–––利率要除,期数要乘 With m conversion periods per year you must do both: use rate i/m per period and exponent mt. The classic slip is dividing the rate but forgetting to raise the exponent to mt (or vice versa). Match m to the frequency: monthly - m = 12, quarterly -> m = 4, semi-annual -+ m = 2. 每年有 m 个计息周期时,你必须两件事都做:每周期使用利率 i/m,并且指数取 mt。典型的疏忽是除了利率却忘了把 指数提到 mt 次方(或反过来)。让 m 与频率匹配:按月→ m=12,按季→m=4,半年→m=2。 ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS 3. 3 Worked interest & present-value problems 3. 3 利息与现值例题精解 Four classic shapes. Each is "identify the model, rearrange for the unknown, substitute. " The unknown can be the future value, the principal (present value), the rate, or the time - the last two need logs. 四种经典题型。每种都是「识别模型,对未知量重排,代入」。未知量可以是终值、本金(现值)、利率或时间 -- 后两者需 要用对数。
• 连续复利：$$P_t=P_0e^{it}$$[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'[15]Source: asksia-bible-econ1003-bilingual.pdfEX 3A Two production-growth schemes - arithmetic vs geometric L · series A plant makes 1200 units in week 1. Scheme 1 adds 80 units each week (arithmetic, d = 80); Scheme 2 grows 5% each week (geometric, r = 1. 05). Find the total output over the first 20 weeks under each. 某工厂第1周生产1200件。方案1每周增加80件(等差,d=80);方案2每周增长5%(等比,r=1. 05)。求前 20 周在每种方案下的总产量。 1 Scheme 1 (arithmetic). a = 1200, d = 80, n = 20. S20 = (20/2)(2×1200 + 19×80) = 10x(2400+1520) = 39,200 units. 方案1(等差)。a =1200, d= 80, n = 20。S20 =(20/2)(2×1200+19×80)= 10×(2400+1520)= 39,200 单位。 2 Scheme 2 (geometric). a = 1200, r = 1. 05, n = 20. S20 = 1200(1 - 1. 0520)/(1 - 1. 05) ~ 39,679 units - the compounding 5% just edges ahead. 方案 2(等比)。a =1200, r=1. 05, n= 20。S20 = 1200(1-1. 0520)/(1-1. 05)~39,679 单位 -- 5% 的复利效应 恰好略微领先。 ! Term vs sum - read the question 单项 vs 求和 -- 看清题目 "Output in week 20" wants the 20th term (1200 + 19x80 = 2720, or 1200x1. 0519 ~ 3032); "total output over 20 weeks" wants the sum S20. And the infinite-sum formula is only valid when Irl < 1. “第20周当周的产量”要的是第 20项(1200+19×80=2720,或 1200×1. 0519~3032);“20周总产量”要的 是总和 S20。而无穷级数求和公式只在 Irl<1时才成立。 3. 2 Simple, compound & continuous interest 3. 2 单利、复利与连续复利 The single most important choice in the topic is which interest model applies. Simple interest pays only on the original principal; compound interest pays interest on accumulated interest too; continuous compounding is the limit as the number of conversion periods goes to infinity. 这个主题里最重要的一个选择是适用哪种利息模型。单利只对原始本金计息;复利还对累积的利息计息;连续复利是计息期数 趋于无穷时的极限。 INTEREST MODELS Simple : Pt = Po(1 + i. t) Compound (annual): Pt = P0(1 + i) t Compound m times/yr: Pt = P0(1 + i/m) mt Continuous : Pt = Po'eit ECON1003 . Quantitative Methods in Economics ! The m-trap - divide the rate AND multiply the periods m 陷阱–––利率要除,期数要乘 With m conversion periods per year you must do both: use rate i/m per period and exponent mt. The classic slip is dividing the rate but forgetting to raise the exponent to mt (or vice versa). Match m to the frequency: monthly - m = 12, quarterly -> m = 4, semi-annual -+ m = 2. 每年有 m 个计息周期时,你必须两件事都做:每周期使用利率 i/m,并且指数取 mt。典型的疏忽是除了利率却忘了把 指数提到 mt 次方(或反过来)。让 m 与频率匹配:按月→ m=12,按季→m=4,半年→m=2。 ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS 3. 3 Worked interest & present-value problems 3. 3 利息与现值例题精解 Four classic shapes. Each is "identify the model, rearrange for the unknown, substitute. " The unknown can be the future value, the principal (present value), the rate, or the time - the last two need logs. 四种经典题型。每种都是「识别模型,对未知量重排,代入」。未知量可以是终值、本金(现值)、利率或时间 -- 后两者需 要用对数。
• 现值：$$P_0=\frac{P_t}{(1+i)^t}$$[13]Source: asksia-bible-econ1003-bilingual.pdf等差数列求和 Sum of terms with a constant difference d; Sn = (n/2)(2a + (n-1)d). Geometric series 等比数列求和 Constant ratio r; Sn = a(1-r")/(1-r); infinite sum a/(1-r) if Ir|<1. Compound interest 复利 Pt = P. (1 + i/m)mt; interest earns interest each period. Continuous compounding 连续复利 Pt = Peit; the m->co limit of compounding. Present value 现值 Today's worth of a future sum: Po= Pt/(1+i)t. Depreciation 折旧 Fall in asset value; reducing-balance At = A. (1-i)t. Net present value (NPV) 净现值 E of discounted cash flows minus outlay; invest if NPV > O. Annuity 年金 A stream of equal periodic payments; has future- and present-value formulas. Topic 4 – Differentiation 微分 Derivative 导数 Instantaneous rate of change f'(x); slope of the tangent line. Power rule 幂法则 d/dx(x") = n·x™-1.
• NPV：把折现现金流相加，$t=0$ 的支出记为负；$NPV>0$ 接受项目。[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'[23]Source: asksia-cheatsheet-econ1003.pdfsimple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate. WORKED . $1000, QUARTERLY, 8%, 2YR P = 1000(1+0. 08/4)^{4. 2} = 1000(1. 02)8 = $1171. 66 WORKED NPV . - 1000 NOW, +600/YR x2, I=10% -1000 + 600/1. 1 + 600/1. 21 = - 1000 + 545. 5 + 495. 9 = +41. 3 = accept 12 . Annuities & CH5 . ON SHEET Loans FUTURE VALUE (DEPOSIT A. ) Vt = A0 . [(1+i/m)^{tm} - 1]/(i/m) PRESENT VALUE / LOAN L L = A0 . [1 - (1+i/m)^{-mt}]/(i/m) solve for Ao = each payment Total interest = (payment x no. payments) - L. WORKED . LOAN $20000, 6%/YR, 5YR ANN. L = A0 . [1-(1. 06)-5]/0. 06 20000 = A0 . 4. 2124 - Ao = $4747. 93/yr TRAP > FV uses ( . . . )^{tm) - 1; PV/loan uses 1 - ( . . . )^{-tm) (negative exp). Match m to payment frequency (monthly m=12). 13 . The Derivative - DEFINITION (NOT EXAMINED TO COMPUTE) f'(x) = lim_{k-0} [f(x+k)-f(x)]/k = instantaneous rate of change = slope of the tangent. Notation: f'(x), dy/dx, df/dx all mean the same thing. 14 . Differentiation Rules CH6 . MEMORISE
• 总利息：$$(\text{每期还款}\times \text{期数})-\text{本金}$$[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'[23]Source: asksia-cheatsheet-econ1003.pdfsimple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate. WORKED . $1000, QUARTERLY, 8%, 2YR P = 1000(1+0. 08/4)^{4. 2} = 1000(1. 02)8 = $1171. 66 WORKED NPV . - 1000 NOW, +600/YR x2, I=10% -1000 + 600/1. 1 + 600/1. 21 = - 1000 + 545. 5 + 495. 9 = +41. 3 = accept 12 . Annuities & CH5 . ON SHEET Loans FUTURE VALUE (DEPOSIT A. ) Vt = A0 . [(1+i/m)^{tm} - 1]/(i/m) PRESENT VALUE / LOAN L L = A0 . [1 - (1+i/m)^{-mt}]/(i/m) solve for Ao = each payment Total interest = (payment x no. payments) - L. WORKED . LOAN $20000, 6%/YR, 5YR ANN. L = A0 . [1-(1. 06)-5]/0. 06 20000 = A0 . 4. 2124 - Ao = $4747. 93/yr TRAP > FV uses ( . . . )^{tm) - 1; PV/loan uses 1 - ( . . . )^{-tm) (negative exp). Match m to payment frequency (monthly m=12). 13 . The Derivative - DEFINITION (NOT EXAMINED TO COMPUTE) f'(x) = lim_{k-0} [f(x+k)-f(x)]/k = instantaneous rate of change = slope of the tangent. Notation: f'(x), dy/dx, df/dx all mean the same thing. 14 . Differentiation Rules CH6 . MEMORISE

最常见陷阱（必背）：

• $m$ 陷阱：利率要除 $i/m$，期数指数要乘 $mt$，两件事必须同时做。[15]Source: asksia-bible-econ1003-bilingual.pdfEX 3A Two production-growth schemes - arithmetic vs geometric L · series A plant makes 1200 units in week 1. Scheme 1 adds 80 units each week (arithmetic, d = 80); Scheme 2 grows 5% each week (geometric, r = 1. 05). Find the total output over the first 20 weeks under each. 某工厂第1周生产1200件。方案1每周增加80件(等差,d=80);方案2每周增长5%(等比,r=1. 05)。求前 20 周在每种方案下的总产量。 1 Scheme 1 (arithmetic). a = 1200, d = 80, n = 20. S20 = (20/2)(2×1200 + 19×80) = 10x(2400+1520) = 39,200 units. 方案1(等差)。a =1200, d= 80, n = 20。S20 =(20/2)(2×1200+19×80)= 10×(2400+1520)= 39,200 单位。 2 Scheme 2 (geometric). a = 1200, r = 1. 05, n = 20. S20 = 1200(1 - 1. 0520)/(1 - 1. 05) ~ 39,679 units - the compounding 5% just edges ahead. 方案 2(等比)。a =1200, r=1. 05, n= 20。S20 = 1200(1-1. 0520)/(1-1. 05)~39,679 单位 -- 5% 的复利效应 恰好略微领先。 ! Term vs sum - read the question 单项 vs 求和 -- 看清题目 "Output in week 20" wants the 20th term (1200 + 19x80 = 2720, or 1200x1. 0519 ~ 3032); "total output over 20 weeks" wants the sum S20. And the infinite-sum formula is only valid when Irl < 1. “第20周当周的产量”要的是第 20项(1200+19×80=2720,或 1200×1. 0519~3032);“20周总产量”要的 是总和 S20。而无穷级数求和公式只在 Irl<1时才成立。 3. 2 Simple, compound & continuous interest 3. 2 单利、复利与连续复利 The single most important choice in the topic is which interest model applies. Simple interest pays only on the original principal; compound interest pays interest on accumulated interest too; continuous compounding is the limit as the number of conversion periods goes to infinity. 这个主题里最重要的一个选择是适用哪种利息模型。单利只对原始本金计息;复利还对累积的利息计息;连续复利是计息期数 趋于无穷时的极限。 INTEREST MODELS Simple : Pt = Po(1 + i. t) Compound (annual): Pt = P0(1 + i) t Compound m times/yr: Pt = P0(1 + i/m) mt Continuous : Pt = Po'eit ECON1003 . Quantitative Methods in Economics ! The m-trap - divide the rate AND multiply the periods m 陷阱–––利率要除,期数要乘 With m conversion periods per year you must do both: use rate i/m per period and exponent mt. The classic slip is dividing the rate but forgetting to raise the exponent to mt (or vice versa). Match m to the frequency: monthly - m = 12, quarterly -> m = 4, semi-annual -+ m = 2. 每年有 m 个计息周期时,你必须两件事都做:每周期使用利率 i/m,并且指数取 mt。典型的疏忽是除了利率却忘了把 指数提到 mt 次方(或反过来)。让 m 与频率匹配:按月→ m=12,按季→m=4,半年→m=2。 ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS 3. 3 Worked interest & present-value problems 3. 3 利息与现值例题精解 Four classic shapes. Each is "identify the model, rearrange for the unknown, substitute. " The unknown can be the future value, the principal (present value), the rate, or the time - the last two need logs. 四种经典题型。每种都是「识别模型,对未知量重排,代入」。未知量可以是终值、本金(现值)、利率或时间 -- 后两者需 要用对数。[23]Source: asksia-cheatsheet-econ1003.pdfsimple: Pt = Po (1+it) compound: Pt = Pe (1+i)t m/yr: Pt = Po(1+i/m)^{mt} continuous: Pt = Poe^{it} PV: Po = P:/(1+i)t WORKED . 10000-> 20000 IN 6YR 2=(1+i)6 = i = 2^{1/6}-1 = 0. 122 TRAP > With m periods/yr you must divide rate (i/m) AND multiply periods (mt) - the classic slip drops the mt. 11 . Depreciation & NPV CH5 Straight-line: subtract a fixed amount/yr. Reducing- balance: At = Ao (1-i) +; total depr = A0 - At. NET PRESENT VALUE NPV = Σ cashflow_t/(1+1)t outlay at t=0 enters NEGATIVE NPV > 0 => project beats the discount rate. WORKED . $1000, QUARTERLY, 8%, 2YR P = 1000(1+0. 08/4)^{4. 2} = 1000(1. 02)8 = $1171. 66 WORKED NPV . - 1000 NOW, +600/YR x2, I=10% -1000 + 600/1. 1 + 600/1. 21 = - 1000 + 545. 5 + 495. 9 = +41. 3 = accept 12 . Annuities & CH5 . ON SHEET Loans FUTURE VALUE (DEPOSIT A. ) Vt = A0 . [(1+i/m)^{tm} - 1]/(i/m) PRESENT VALUE / LOAN L L = A0 . [1 - (1+i/m)^{-mt}]/(i/m) solve for Ao = each payment Total interest = (payment x no. payments) - L. WORKED . LOAN $20000, 6%/YR, 5YR ANN. L = A0 . [1-(1. 06)-5]/0. 06 20000 = A0 . 4. 2124 - Ao = $4747. 93/yr TRAP > FV uses ( . . . )^{tm) - 1; PV/loan uses 1 - ( . . . )^{-tm) (negative exp). Match m to payment frequency (monthly m=12). 13 . The Derivative - DEFINITION (NOT EXAMINED TO COMPUTE) f'(x) = lim_{k-0} [f(x+k)-f(x)]/k = instantaneous rate of change = slope of the tangent. Notation: f'(x), dy/dx, df/dx all mean the same thing. 14 . Differentiation Rules CH6 . MEMORISE

Topic 4：Differentiation（整门课的脊梁）

你要会：

• 导数的意义：瞬时变化率/切线斜率；在经济学里叫“边际”。[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1
• 题型：求导（常链式×乘积/商）、判断增减/凹凸、极值点与拐点、二阶导检验、以及 $MR=MC$ 的应用。[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'
• 二阶导判别：$f''<0$ 极大；$f''>0$ 极小。[20]Source: asksia-cheatsheet-econ1003.pdfWORKED MIN . F=X2-6X+5 f'=2x-6=0 = x=3 . f' ' =2>0 - MIN f(3) = - 4 WORKED MAX . F =- X2+4X f' =- 2x+4=0 = x=2 . f' ' =- 2<0 - MAX f(2) = 4 17 . Economic Applications CH6 . EXAM FAVE MARGINALS MR = dTR/dQ MC = dTC/dQ revenue max: MR = 0 profit max: MR = MC Non-linear elasticity: Ed = (dQ/dP) . (P/Q) with calculus for dQ/dP. WORKED . P=100-Q, TC=Q2 TR=PQ=100Q-Q2 = MR=100-2Q MC=2Q . MR=MC = Q=25, P=75 check: revenue max (MR=0) at Q=50 - different! TRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max[10]Source: asksia-bible-econ1003-bilingual.pdf总利息。一笔$5,000 的贷款,利率20%,分4个季 度还清(m= 4,t=1),得 Ao =$1,410. 06;总支付 4×1410. 06 =$5,640. 24,故利息 = 5640. 24- 5000 $640. 24。 Total interest on a loan = (each payment x number of payments) - the amount borrowed. The repayment formula gives the payment; the interest is what you add on top. 贷款的总利息=(每期还款额 ×还款次数)-借款本金。还款公式给出每期还款额;利息就是你在本金之上额外多付的 部分。 L . DEBT REPAYMENT ECON1003 . Quantitative Methods in Economics i Recap - Financial Mathematics in seven lines 复习 -- 金融数学七行讲清 (1) Arithmetic series Sn = (n/2)(2a + (n-1)d); geometric Sn = a(1-rn)/(1-r); infinite sum a/(1-r) only if Ir|<1. (2) Compound m times/yr: rate i/m, exponent mt - do both. (3) Continuous: Poeit. (4) Unknown rate - take a root; unknown time - take a log. (5) NPV sums discounted cash flows, outlay negative at t = 0; accept if NPV > 0. (6) Reducing-balance depreciation A0(1-i)t. (7) FV annuity has ( . . . )tm - 1; PV/loan annuity has 1 - ( . . . )-tm - never swap them. (1) 等差级数 Sn = (n/2)(2a +(n-1)d);等比级数 Sn = a(1-rn)/(1-r);无穷求和 a/(1-r)仅在 |r/<1 时成立。(2)每年计 息m次:利率 i/m,指数 mt -- 两者都要做。(3)连续复利:Poeft。(4)利率未知→取根;时间未知→取对数。(5) NPV 是折现现金流之和,初始支出在 t= 0 时为负;当 NPV>0 时接受。(6)余额递减折旧 Ao(1-i)。(7)终值年金含 ( . . . )tm -1;现值/贷款年金含1-( . . . )-tm -- 切勿互换。 ECON1003 . Quantitative Methods in Economics T4 . DIFFERENTIATION - TOPIC 4 . DIFFERENTIATION MIDTERM + FINAL The derivative: slope, shape and the marginal idea 导数:斜率、形状与边际思想 Rules . higher derivatives . concavity . turning & inflection points 求导法则 · 高阶导数 · 凹凸性 · 驻点与拐点 Differentiation is the spine of the unit. A derivative is just a rate of change - the slope of the tangent - and economics calls that rate "marginal. " Master the rules, then read what the first and second derivatives say about a curve's shape, and you can locate maxima, minima and the profit-maximising output. This topic spans both exams: the Midterm covers up to the application of differentiation, the Final revisits all of it. 求导是本单元的脊梁。导数不过是变化率 -- 切线的斜率 -- 而经济学把这个率称为「边际」。掌握法则后,再去读一阶和 二阶导数对曲线形状的说明,你就能定位极大值、极小值和利润最大化的产量。这个主题横跨两场考试:期中考到求导的应用 为止,期末再全面复盘。 ★ What the exam asks here 考试在这里怎么问 Differentiate a given function (often needing chain x product or quotient together); find where it is increasing / decreasing / concave up / down; locate turning and inflection points; apply the second-derivative test; and the headline application - given a demand and cost function, find the profit-maximising Q where MR = MC. The limit definition is shown for understanding but the lecturer states he will not ask you to differentiate from the limit. 对给定函数进行求导(往往需要同时用到链式 ×乘积或商法则);判断它在何处递增/递减/上凹/下凹;定位驻点/极 值点和拐点;运用二阶导数检验;以及最重要的应用 -- 给定需求与成本函数,找出 MR = MC 时的利润最大化产量 Q。极限定义会展示出来以帮助理解,但讲师明确表示不会要求你用极限定义来求导。 4. 1 The derivative & the differentiation rules 4. 1 导数与求导法则 Formally f'(x) = limk-o [f(x+k) - f(x)] / k - the instantaneous rate of change. In practice you never use the limit; you use rules. The provided formula sheet gives you the ax, loga and quotient rules; you must memorise the power, constant, sum, chain and product rules. 形式上 f'(x)= limk→0 [f(x+k) - f(x)] / k -- 即瞬时变化率。实务中你从不用这个极限;你用的是法则。所提供的公式表给 了你 aX、loga 和商法则;幂、常数、和、链式和乘积法则你必须背下来。 BASIC RULES (MEMORISE) Power: (xn)' = n ·xn-1 Constant : (k)' = 0 Constx : (k . f)' = k . f'

Topic 5：Several Variables（偏导、MPL/MPK、MRS/MRTS、拉格朗日）

你要会（“公式带”那一组）：[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors

• $$MPL=\frac{\partial Q}{\partial L},\quad MPK=\frac{\partial Q}{\partial K}$$
• $$MRTS=-\frac{MPL}{MPK}$$
• $$MRS=-\frac{MU_y}{MU_x}$$
• 拉格朗日写法、shadow price 含义、complementary slackness、角点比较。[29]Source: asksia-cheatsheet-econ1003.pdfTHE WHOLE MOVE MAX U=XY S. T. 2X+Y=10 L = xy - A(2x+y-10) L_x: y - 2X = 0 = y = 2A L_y: x - X = 0 = X = A = y = 2x; sub: 2x+2x=10 = x=2. 5 x=2. 5, y=5, A=2. 5, U=12. 5 At the optimum MRS = price ratio : MUx/MUy = y/x = 2 = P1/P2. V Cost-min mirrors it: min C s. t. Q = f(L, K) gives the tangency MRTS = w/r (input-price ratio). 24 . The Multiplier 2 CH7 - SHADOW PRICE A = shadow price : how much the optimal objective changes if you relax that constraint by one unit. In utility-max s. t. budget: ) = marginal utility of income (above, A=2. 5 => +$1 income raises U by =2. 5). In cost-min s. t. output: ) = marginal cost of one more unit. Uses: utility maximisation & cost minimisation. SIA - If asked "what does À mean?", answer in units of the objective per unit of the constraint - that one sentence is worth marks. 24b . Lagrangian Checklist BEFORE YOU STOP · All ÒL/>(var) = 0; constraints checked · Inequality X's all ≥ 0 · Complementary slackness on each ≤ · Corner cases (x=0, y=0) evaluated . f compared across ALL candidates 25 . The Integral CH8 ANTI-DERIVATIVE [f(x)dx = g(x) + C, where g' (x)=f(x) ALWAYS add + C (indefinite) RULES [xm dx = xn+1/(n+1) + C (n = - 1) [x-1 dx = Ln(x) + C (the n =- 1 case) SK dx = Kx + C . fex dx = ex + C TRAP > Power rule breaks at n =- 1 (+0) - that case is the log. Dropping +C is fatal once you solve for C in an ODE. 26 · Substitution CH8 Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back. WORKED . J (5X-2)1º DX U=5x-2, dx=du/5[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors

Topic 6：Integration（Final 必考：规则 + 经济面积）

幂积分规则与 $n=-1$ 例外：[12]Source: asksia-bible-econ1003-bilingual.pdf逆向求导:幂次加一,再除以新的幂次 TL;DR. Integration undoes differentiation. To integrate a power, raise the index by one and divide by the new index - the one exception is x", which integrates to In(x). Every indefinite integral carries a + C; never drop it, because in differential-equation questions you later solve for it. Definite integrals turn this machinery into area - and area is how this course measures consumer and producer surplus. 一句话总结。积分抵消求导。要对一个幂求积分,指数加一,再除以新的指数 -- 唯一例外是x-1,它积分为 In(x)。每个不 定积分都带一个+ C;绝不要丢掉它,因为在微分方程题里你之后要解出它。定积分把这套机制变成面积 -- 而本课程正是 用面积来度量消费者剩余和生产者剩余。 ★ What the exam asks here 考试在这里怎么问 Integration sits on the cumulative final (not the midterm). Expect: a routine indefinite integral testing the power / ln / e* rules; a substitution or by-parts integral; a definite integral read as an area or as consumer / producer surplus; and a differential equation (recovering total cost from marginal cost, or a separable growth law). The formula sheet gives you the by-parts formula - the rest you memorise. 积分出现在累计性期末考(而非期中考)上。预计会有:一道检验幂/ In/ex法则的常规不定积分;一道换元或分部积 分;一道被解读为面积或消费者剩余/生产者剩余的定积分;以及一道微分方程(由边际成本还原总成本,或一个可分 离的增长律)。公式表会给你分部积分公式 -- 其余的要背熟。 6. 1 The integral as an anti-derivative 6. 1 积分作为反导数 A function g is an anti-derivative (#) off if g'(x) = f(x). Because the derivative of a constant is zero, anti- derivatives are only pinned down up to a constant - hence the constant of integration C. 若 g'(x)= f(x),则函数 g 是 f 的反导数〔原函数〕。由于常数的导数为零,反导数只能确定到相差一个常数 -- 这就是积分 常数 C 的由来。 DEFINITION [ f(x) dx = g(x) + C where g' (x) = f(x) 6. 2 The power rule & the n = - 1 exception 6. 2 幂法则与 n =- 1 的例外 For every index except -1, push the index up by one and divide by the new index. At n = - 1 that would divide by zero - so that single case becomes the natural log. 对除 -1 以外的每个指数,把指数加一,再除以新的指数。在 n =- 1 时这样做会除以零 -- 所以这唯一一种情形变成了自 然对数。 ECON1003 . Quantitative Methods in Economics POWER RULE + FRIENDS [ x" dx = x"+1/ (n+1) + C (n = - 1) [ x-1 dx = [ (1/x) dx = In| x | + C (the n = - 1 case) [ K dx = Kx + C . § ex dx = ex + C . [ K. f dx = KJ f dx EX 6. 1 Power rule, three terms at once power rule 1 Rewrite roots & fractions as powers first. Integrate (3x2 + 4Vx - 5/x2)dx, i. e. f(3x2 + 4x1/2 - 5x-2)dx. 先把根式与分式改写成幂。对 S(3x2+4√x-5/x2)dx 积分,即 S(3x2+4x1/2-5x-2)dx。 2 Term by term. 3. x3/3 = x3; 4. x3/2/(3/2) = (8/3)x3/2; - 5. x-1/(-1) = +5x-1. 逐项处理。3·x3/3=x3; 4·x3/2/(3/2)= (8/3)×3/2; - 5·x-1/(-1) = +5x-1。 3 Assemble + C. Answer = x3 + (8/3)x3/2 + 5/x + C. Differentiate back to check: 3x2 + 4×1/2 - 5x-2. V 组装并加+ C。答案=x3+(8/3)×3/2+5/x+C。求导验证:3x2+4×1/2-5x-2。√ ! n = - 1 is the log, and the "+ C" is not optional n =- 1 对应对数,「+C」不是可选项 (1/x)dx is In|x| + C, never xº/O. And forgetting + C is a free dropped mark on an indefinite integral - worse, it is fatal in a differential equation, where the initial condition exists precisely to solve for that C. S(1/x)dx 是 In|xl+ C,绝不是 x°/0。而忘写+C 在不定积分中是白白丢一分 -- 更糟的是,它在微分方程中是致命 的,因为初始条件存在的意义恰恰就是解出那个C。 ECON1003 . Quantitative Methods in Economics[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors

• $$\int x^n dx=\frac{x^{n+1}}{n+1}+C\quad (n\neq -1)$$
• $$\int \frac{1}{x},dx=\ln|x|+C$$

三类 Final 常见题：

• 不定积分（规则题）+ 换元/分部积分（分部积分公式在公式表上）[12]Source: asksia-bible-econ1003-bilingual.pdf逆向求导:幂次加一,再除以新的幂次 TL;DR. Integration undoes differentiation. To integrate a power, raise the index by one and divide by the new index - the one exception is x", which integrates to In(x). Every indefinite integral carries a + C; never drop it, because in differential-equation questions you later solve for it. Definite integrals turn this machinery into area - and area is how this course measures consumer and producer surplus. 一句话总结。积分抵消求导。要对一个幂求积分,指数加一,再除以新的指数 -- 唯一例外是x-1,它积分为 In(x)。每个不 定积分都带一个+ C;绝不要丢掉它,因为在微分方程题里你之后要解出它。定积分把这套机制变成面积 -- 而本课程正是 用面积来度量消费者剩余和生产者剩余。 ★ What the exam asks here 考试在这里怎么问 Integration sits on the cumulative final (not the midterm). Expect: a routine indefinite integral testing the power / ln / e* rules; a substitution or by-parts integral; a definite integral read as an area or as consumer / producer surplus; and a differential equation (recovering total cost from marginal cost, or a separable growth law). The formula sheet gives you the by-parts formula - the rest you memorise. 积分出现在累计性期末考(而非期中考)上。预计会有:一道检验幂/ In/ex法则的常规不定积分;一道换元或分部积 分;一道被解读为面积或消费者剩余/生产者剩余的定积分;以及一道微分方程(由边际成本还原总成本,或一个可分 离的增长律)。公式表会给你分部积分公式 -- 其余的要背熟。 6. 1 The integral as an anti-derivative 6. 1 积分作为反导数 A function g is an anti-derivative (#) off if g'(x) = f(x). Because the derivative of a constant is zero, anti- derivatives are only pinned down up to a constant - hence the constant of integration C. 若 g'(x)= f(x),则函数 g 是 f 的反导数〔原函数〕。由于常数的导数为零,反导数只能确定到相差一个常数 -- 这就是积分 常数 C 的由来。 DEFINITION [ f(x) dx = g(x) + C where g' (x) = f(x) 6. 2 The power rule & the n = - 1 exception 6. 2 幂法则与 n =- 1 的例外 For every index except -1, push the index up by one and divide by the new index. At n = - 1 that would divide by zero - so that single case becomes the natural log. 对除 -1 以外的每个指数,把指数加一,再除以新的指数。在 n =- 1 时这样做会除以零 -- 所以这唯一一种情形变成了自 然对数。 ECON1003 . Quantitative Methods in Economics POWER RULE + FRIENDS [ x" dx = x"+1/ (n+1) + C (n = - 1) [ x-1 dx = [ (1/x) dx = In| x | + C (the n = - 1 case) [ K dx = Kx + C . § ex dx = ex + C . [ K. f dx = KJ f dx EX 6. 1 Power rule, three terms at once power rule 1 Rewrite roots & fractions as powers first. Integrate (3x2 + 4Vx - 5/x2)dx, i. e. f(3x2 + 4x1/2 - 5x-2)dx. 先把根式与分式改写成幂。对 S(3x2+4√x-5/x2)dx 积分,即 S(3x2+4x1/2-5x-2)dx。 2 Term by term. 3. x3/3 = x3; 4. x3/2/(3/2) = (8/3)x3/2; - 5. x-1/(-1) = +5x-1. 逐项处理。3·x3/3=x3; 4·x3/2/(3/2)= (8/3)×3/2; - 5·x-1/(-1) = +5x-1。 3 Assemble + C. Answer = x3 + (8/3)x3/2 + 5/x + C. Differentiate back to check: 3x2 + 4×1/2 - 5x-2. V 组装并加+ C。答案=x3+(8/3)×3/2+5/x+C。求导验证:3x2+4×1/2-5x-2。√ ! n = - 1 is the log, and the "+ C" is not optional n =- 1 对应对数,「+C」不是可选项 (1/x)dx is In|x| + C, never xº/O. And forgetting + C is a free dropped mark on an indefinite integral - worse, it is fatal in a differential equation, where the initial condition exists precisely to solve for that C. S(1/x)dx 是 In|xl+ C,绝不是 x°/0。而忘写+C 在不定积分中是白白丢一分 -- 更糟的是,它在微分方程中是致命 的,因为初始条件存在的意义恰恰就是解出那个C。 ECON1003 . Quantitative Methods in Economics
• 定积分当面积（净面积会有正负，若要总面积要分段）[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS
• ODE/还原成本：$TC=\int MC,dQ + FC$（$+C$ 就是固定成本）[6]Source: asksia-bible-econ1003-bilingual.pdf1 Setup. MC = dTC/dQ = 3Q2 + 4, and fixed cost TC(O) = 50. 设定。MC = dTC/dQ=3Q2 +4,固定成本 TC(0) =50。 Integrate. TC =[(3Q2 + 4)dQ = Q3 +4Q+C. 积分。TC=S(3Q2 + 4)dQ= Q3+4Q + C。 3 Use the condition. TC(0) = 0 + 0 + C = 50 -+ C = 50 (the fixed cost is the constant). 使用条件。TC(0)=0+0+C=50→C=50(固定成本就是那个常数)。 4 Particular solution. TC = Q3 + 4Q + 50. The +C was the fixed cost all along - drop it and you lose the whole economic content. I 特解。TC=Q3+4Q+50。那个+C 自始至终就是固定成本 -- 丢掉它,你就丢掉了全部的经济内涵。 ECON1003 . Quantitative Methods in Economics EX 6. 8 A separable growth law separable . growth 1 Setup. dy/dx = ky (proportional growth), with y(O) = yo. 设定。dy/dx =ky(比例增长),且 y(0)= yo。 2 Separate. (1/y)dy = k dx. 分离变量。(1/y)dy=k dx。 3 Integrate both sides. [(1/y)dy = [k dx -> Inlyl = kx + C. 对两边积分。S(1/y)dy = Sk dx → Inlyl = kx+ C。 Isolate y. y = ekx+C = A. ekx (write A = eC). Apply y(O) = y. - A = yo, so y = yo ekx - the exponential growth law, derived from scratch. 解出 y。y = ekx+C = A·ekx(记 A= eC)。代入 y(0)= yo → A= yo,所以 y = yo ekx -- 从零开始推导出来的指数增 长律。 ! Solve for C, do not drop it; and don't forget the In step 解出 C,不要丢掉它;也别忘了 In 这一步 In a differential equation the constant of integration is the whole point - the initial condition exists to fix it. With a separable equation, integrating 1/y produces a ln|y|; you must then exponentiate to get y back, turning the additive +C into a multiplicative A = eC. 在微分方程中,积分常数才是关键所在 -- 初始条件存在就是为了确定它。对于可分离方程,对 1/y 积分会产生一个 Inlyl;接着你必须取指数才能把 y 还原回来,从而把加性的+C 变成乘性的 A= eC。 Integration is differentiation run backwards - so every answer can be checked by differentiating it. On the final, if you have a minute, differentiate your integral and watch it collapse back to the integrand. Free insurance on method marks. 积分是反向运行的求导 -- 所以每个答案都可以通过对它求导来检验。在期末考上,如果你还有一分钟,就对你的积分 求导,看着它塌缩回被积函数。这是方法分上的免费保险。 MARKER'S NOTE . ECON1003 FINAL ECON1003 . Quantitative Methods in Economics T7 . MATRICES - TOPIC 7 . LINEAR ALGEBRA FINAL EXAM TOPIC Matrices, arithmetic & multiplication[28]Source: asksia-cheatsheet-econ1003.pdf= (5x-2)11/55 + C WORKED . [2X. E^{X2} DX U=x2 => du=2x dx = Je" du = e^{x2}+C TRAP > The leftover x's must all cancel via du - if any remain, substitution isn't the right tool. 27 . By Parts CH8 . ON SHEET LECTURER'S FORM [f g dx = f. [g dx - [f' . (fg dx) dx WORKED . JLN X DX f=Ln x, g=1 = x ln x - x + C WORKED . JX EX DX f=x, g=ex = x ex - [ex dx = ex(x-1)+C TRAP + Pick f (to differentiate) so f' simplifies - logs & powers of x are good f. 28 · Definite Integral & Area CH8 NET SIGNED AREA Jab f dx = F(b) - F(a) between curves: fab [f - g] dx (f above g) WORKED . J,3 (2X+1) DX = [x2+x]13 = (9+3) - (1+1) = 10 TRAP > Gives NET area - below-axis counts negative (§. 33 x dx = 0). For "total area" split at crossings. 28b . TC from MC CH8 . ECON USE RECOVER TOTAL COST TC = [ MC dQ + FC the constant of integration = fixed cost WORKED . MC = 3Q2+2, FC=50 TC = Q3 + 2Q + 50 29 . Surplus & ODEs CONSUMER / PRODUCER SURPLUS CS = Jo^{Q*} [demand(Q) - P*] dQ PS = Jo^{Q*} [P* - supply (Q)] dQ Find equilibrium Q* (from P*) first - it's the upper limit. WORKED CS . D: P=20-Q, P *= 8 Q *= 12 - CS = Jo12 (20-Q-8) dQ = [12Q - Q2/2] @12 = 144 - 72 = 72 WORKED PS . S: P=2+Q, P *= 8 Q *= 6 = PS = Je6 (8-2-Q) dQ = [6Q - Q2/2] 06 = 36 - 18 = 18 DIFFERENTIAL EQUATIONS

Topic 7：Linear Algebra（矩阵、Cramer、逆矩阵、解方程组）

你要会：

• 先 $\det(A)\neq 0$ 再谈逆矩阵；cofactor 法必须转置。[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL[17]Source: asksia-cheatsheet-econ1003.pdfTRAP + Cofactor method: don't forget to TRANSPOSE (C') and divide by det(A). If det(A)=0 there is no inverse. * Formula Belt · Side 2 MEMORISE COLD SEVERAL VARIABLES MPL=OQ/aL . MPK=OQ/aK MRTS =- MPL/MPK . MRS =- MUy/MUx L = f - A(constraint) . A = shadow price compl. slackness A(g-b)=0 INTEGRATION [x^=xn+1/(n+1)+C (n =- 1) . [x-1=Ln x+C by parts: f[g - [f' (Jg) CS=Je^{Q*}(D-P*) . PS=Je^{Q*}(P *- S) LINEAR ALGEBRA det 2×2 = ad-bc Cramer X1 = det(Bi)/det(A) A-1 = CT/det (A) . X = A-1B ON THE SHEET YOU MEMORISE 3×3 determinant Cramer's rule a”, log_a, quotient rule inverse methods by-parts formula Lagrangian conditions * Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors
• $A^{-1}$ 的定义与用途：$AA^{-1}=I$，并用于 $X=A^{-1}B$。[1]Source: asksia-bible-econ1003-bilingual.pdf代数余子式 / 余子式 Minor = sub-determinant; cofactor = minor x the +-+ sign. ECON1003 . Quantitative Methods in Economics Term 术语 中文 One-line definition 释义 Inverse matrix (A-1) 逆矩阵 A-1 = CT/det(A); satisfies AA1 = I; solves X = A-1B. ✓ How to use the glossary 用法 如何使用术语表 用法 Cover the right column and recite the definition from the English term - then flip and recall the English from the +X. 〔遮住释义、由术语复述定义;再由中文回想英文。〕 The exam rewards the exact wording: "next-best forgone", "CT/det(A)", "signed area" - precision is marks. 遮住右栏,从英文术语复述出定义 -- 再反过来,由中文回忆英文。〔遮住释义、由术语复述定义;再由中文回想英文。〕 考试奖励精确的措辞:“next-best forgone”、“CT/det(A)”、“signed area" -- 精确就是分数。 ECON1003 . Quantitative Methods in Economics PRACTICE P1 -P2 - PRACTICE BANK DRILL TO EXAM STANDARD Drill both exams, with full solutions 两场考试齐刷,附完整解答 Fourteen AskSia-authored problems across all seven topics - worked end to end 横跨全部七大主题的14 道 AskSia 原创题 –––完整解答 These are exam-style problems written in the ECON1003 style - not copied from any real quiz or exam. They span lines, finance, differentiation, the Lagrangian, integration and matrices, so they serve both the Midterm (Topics 1-4) and the cumulative Final (all seven). Every number is verified. Cover the solution, work it on paper, then read the answer. 这些是按 ECON1003 风格编写的考试风格题目 -- 并非抄自任何真实测验或考试。它们覆盖直线、金融、求导、拉格朗日 函数、积分和矩阵,因此同时服务于期中(主题1-4)和累计的期末(全部七大主题)。每个数字都经过核验。盖住解答,在 纸上做完,再看答案。 ★ How to use this bank 如何使用这个题库 Cover the solution - recognition is not recall. Budget ~1 minute per mark. Show your work: method marks are real and partial credit is given, so a stated step beats a bare number. Round only the final answer to the stated decimals. Check every answer by substituting back - differentiate an integral, multiply an inverse, re-solve a system. 遮住答案 认得出不等于回忆得起。按每分约1分 钟来分配时间。写出你的过程:方法分是实打实的, 也会给部分分,所以一个写出来的步骤胜过一个孤零 零的数字。只对最终答案按规定小数位取整。对每个 答案都做检验,把它代回去 -- 对积分求导、把逆矩 阵乘回去、重新解一遍方程组。 i Formula sheet vs memory 公式表 vs 记忆 Both exams give a formula sheet (quadratic, series, interest, NPV, annuity, by-parts, 3×3 determinant, ax/log derivatives, quotient rule). You must memorise the rest: power / product / chain rules, substitution, the Lagrangian conditions, Cramer's rule and the inverse methods. The marks are in applying them cleanly.[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

5）Final 冲刺“最省命”的复习安排（你照做就行）

• 利润最大化链条、CS/PS 积分链条、拉格朗日四件套、矩阵 det→方法→验证。[3]Source: asksia-bible-econ1003-bilingual.pdfWhat the formula sheet means for you 公式表对你意味着什么 On the sheet (don't memorise) You must know cold Quadratic formula, series sums Power / sum / constant rules Compound & continuous interest, NPV, annuity Product, quotient & chain rules Elasticity, integration by parts Substitution; how to set up TR=PQ 3×3 determinant; d/dx of ax, log Lagrangian conditions; Cramer; inverse ECON1003 . Quantitative Methods in Economics i What the Midterm covers (19 Apr) 期中考查什么 (4月19日) The Midterm examines Weeks 1-4 plus the early differentiation material - in this book that is roughly Ch 1 (Lines), Ch 2 (Non-linear functions), Ch 3 (Financial maths) and the opening of Ch 4 (Differentiation: the rules and basic optimisation). Multivariable, integration and linear algebra are Final- only. 期中考查第 1-4 周外加早期求导内容––在本书中 大致是第1章(直线)、第2章(非线性函数)、第3 章(金融数学)以及第4 章开头(求导:法则与基础 最优化)。多变量、积分法和线性代数仅在期末考。 ★ The strategy this dictates 由此决定的策略 Every exam question is procedural: take a function or a system, apply the right technique, interpret the economic meaning, round as asked. The recurring chains are - line - slope/intercept - elasticity; set up TR - differentiate - MR=MC; integrate MC - recover TC; Lagrangian - solve cases - shadow price. Show every line - method marks and partial credit are real. Drill the chains and fresh numbers can't surprise you. 每道考题都是程序化的:拿到一个函数或一个方程 组,套用正确的技巧,解读其经济含义,并按要求取 整。反复出现的链条是 -- line→ slope/intercept -> elasticity; set up TR -> differentiate -> MR=MC; integrate MC -> recover TC; Lagrangian -> solve cases -> shadow price. - 步都要写出来 -- 方法分和部分分都是实打实的。把 这些链条练熟,再陌生的数字也无法让你措手不及。 ECON1003 . Quantitative Methods in Economics CONTENTS - CONTENTS Seven topics, in textbook order 七大主题,按教材顺序 Bradley Ch 2 - Ch 9, mapped to one exam-ready book Bradley 第2章→第9章,整合为一本备考书 Ch Topic Core methods Part 1 . Algebra & functions (Midterm core)[11]Source: asksia-cheatsheet-econ1003.pdf* Final Exam Discipline ALL 7 TOPICS · Cumulative - Side 1 maths returns; redo profit-max & finance Lagrangian - list ALL 4 conditions; check corners · Integration - n =- 1 -> In; +C; find Q* before surplus · Matrices - dims first; det#0 before Cramer/inverse; transpose cofactors asksia. ai/cheatsheet/ usyd-econ1003 . side 2/2 AskSia CHEAT SHEET SERIES MATHS, NOT STATS Compiled by AskSia . mapped to the ECON1003 syllabus . asksia. ai/cheatsheet/usyd- econ1003 CHE maths for economics . check your current unit outline . @ 2026 good Luck. show your work. ECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

• $m$ 陷阱（金融）[15]Source: asksia-bible-econ1003-bilingual.pdfEX 3A Two production-growth schemes - arithmetic vs geometric L · series A plant makes 1200 units in week 1. Scheme 1 adds 80 units each week (arithmetic, d = 80); Scheme 2 grows 5% each week (geometric, r = 1. 05). Find the total output over the first 20 weeks under each. 某工厂第1周生产1200件。方案1每周增加80件(等差,d=80);方案2每周增长5%(等比,r=1. 05)。求前 20 周在每种方案下的总产量。 1 Scheme 1 (arithmetic). a = 1200, d = 80, n = 20. S20 = (20/2)(2×1200 + 19×80) = 10x(2400+1520) = 39,200 units. 方案1(等差)。a =1200, d= 80, n = 20。S20 =(20/2)(2×1200+19×80)= 10×(2400+1520)= 39,200 单位。 2 Scheme 2 (geometric). a = 1200, r = 1. 05, n = 20. S20 = 1200(1 - 1. 0520)/(1 - 1. 05) ~ 39,679 units - the compounding 5% just edges ahead. 方案 2(等比)。a =1200, r=1. 05, n= 20。S20 = 1200(1-1. 0520)/(1-1. 05)~39,679 单位 -- 5% 的复利效应 恰好略微领先。 ! Term vs sum - read the question 单项 vs 求和 -- 看清题目 "Output in week 20" wants the 20th term (1200 + 19x80 = 2720, or 1200x1. 0519 ~ 3032); "total output over 20 weeks" wants the sum S20. And the infinite-sum formula is only valid when Irl < 1. “第20周当周的产量”要的是第 20项(1200+19×80=2720,或 1200×1. 0519~3032);“20周总产量”要的 是总和 S20。而无穷级数求和公式只在 Irl<1时才成立。 3. 2 Simple, compound & continuous interest 3. 2 单利、复利与连续复利 The single most important choice in the topic is which interest model applies. Simple interest pays only on the original principal; compound interest pays interest on accumulated interest too; continuous compounding is the limit as the number of conversion periods goes to infinity. 这个主题里最重要的一个选择是适用哪种利息模型。单利只对原始本金计息;复利还对累积的利息计息;连续复利是计息期数 趋于无穷时的极限。 INTEREST MODELS Simple : Pt = Po(1 + i. t) Compound (annual): Pt = P0(1 + i) t Compound m times/yr: Pt = P0(1 + i/m) mt Continuous : Pt = Po'eit ECON1003 . Quantitative Methods in Economics ! The m-trap - divide the rate AND multiply the periods m 陷阱–––利率要除,期数要乘 With m conversion periods per year you must do both: use rate i/m per period and exponent mt. The classic slip is dividing the rate but forgetting to raise the exponent to mt (or vice versa). Match m to the frequency: monthly - m = 12, quarterly -> m = 4, semi-annual -+ m = 2. 每年有 m 个计息周期时,你必须两件事都做:每周期使用利率 i/m,并且指数取 mt。典型的疏忽是除了利率却忘了把 指数提到 mt 次方(或反过来)。让 m 与频率匹配:按月→ m=12,按季→m=4,半年→m=2。 ECON1003 . Quantitative Methods in Economics T3 . FINANCIAL MATHS 3. 3 Worked interest & present-value problems 3. 3 利息与现值例题精解 Four classic shapes. Each is "identify the model, rearrange for the unknown, substitute. " The unknown can be the future value, the principal (present value), the rate, or the time - the last two need logs. 四种经典题型。每种都是「识别模型,对未知量重排,代入」。未知量可以是终值、本金(现值)、利率或时间 -- 后两者需 要用对数。
• 弹性≠斜率（直线）[9]Source: asksia-bible-econ1003-bilingual.pdfFig 1. 2 -- 线性的需求与供给在均衡处相交(这里 Q* = 2400, P* = 1200)。沿着同一条需求线,弹性 在顶端附近很大(P高、Q低),在底端附近很小 -- 斜率相同,弹性不同。 ECON1003 . Quantitative Methods in Economics ! Elasticity is NOT the slope - it changes along the line 弹性不是斜率 -- 它沿直线不断变化 The slope of a straight-line demand is constant, but elasticity varies: it is elastic at high prices (small Q), inelastic at low prices, and unit-elastic (E = - 1) exactly at the midpoint. Quoting 'the slope is the elasticity' is wrong and the examiner knows students do it. 直线型需求的斜率是恒定的,但弹性会变化:在高价处(Q小)富有弹性,在低价处缺乏弹性,并恰好在中点处呈单位 弹性(E =- 1)。说“斜率就是弹性”是错的,而考官清楚学生会这么犯错。 -3 Ed AT P = 1800 P = 1800 时的 Ed elastic SINCE |E| > 1 由于 |티| > 1 -0. 5 DP/DQ (THE SLOPE) dP/dQ (斜率) -2 DQ/DP (ITS RECIPROCAL) dQ/dP (其倒数) Read a slope as a marginal effect, an intercept as a baseline, and an elasticity as a responsiveness - the maths is only worth marks once you say what it means. 把斜率读作边际效应,把截距读作基准值,把弹性读作反应程度 -- 只有当你说出它的含义时,这套数学才值分。 ECON1003 . INTERPRET- THE-RESULT OUTCOME i Recap - Chapter 1 in six lines 复习 -- 第1章六行讲清 (1) Every line is y = mx + c; rearrange to that form before reading m and c. (2) Slope = rise/run = Ay/Ax. (3) Demand Q = f(P) and inverse demand P = f(Q) have reciprocal slopes - read which is asked. (4) TC = FC + VC; break-even sets Profit = 0; fixed cost is the intercept. (5) Budget line: price change pivots, income change shifts parallel. (6) E = (dQ/dP)(P/Q); elasticity is not the slope and varies along a linear curve. (1) 每条直线都是 y = mx+c;先整理成这种形式再读出 m 和 c。(2)斜率= rise/run = △y/△x。(3)需求 Q=f(P) 与 反需求 P=f(Q)的斜率互为倒数 -- 看清题目问的是哪一个。(4) TC= FC+VC;盈亏平衡令 Profit=0;固定成本就 是截距。(5)预算线:价格变化使其旋转,收入变化使其平行移动。(6) E=(dQ/dP)(P/Q);弹性不是斜率,且在一条线 性曲线上会变化。 ECON1003 . Quantitative Methods in Economics CH 2 . NON -LINEAR - CHAPTER 2 . NON- LINEAR FUNCTIONS MIDTERM CORE When the line bends: parabolas, growth and their inverse 当直线弯曲时:抛物线、增长及其反函数 Quadratics . transformations . exponentials & e . logarithms . growth 二次方程 · 函数变换 · 指数与 e · 对数 · 增长 i TL; DR 一句话总结 Three function families and one big idea. Quadratics - solve with the formula, read the shape from the discriminant. Exponentials - the natural engine of growth, with e at the centre. Logarithms - the inverse that undoes an exponential so you can solve for an unknown power. The formula sheet gives you the quadratic formula; the log and index rules you must wield fluently. - 三个函数族和一个核心理念。二次方程 -- 用公式求解,从判别式读出图形形状。指数 -- 增长的天然引擎,e 处于中 心。对数 -- 它是抵消指数的反函数,让你能解出未知的幂次。公式表会给你二次方程求解公式;而对数和指数运算法 则,你必须运用自如。
• $MR$ 必须从 $TR(Q)$ 求导（别对 $P$ 求导）[21]Source: asksia-cheatsheet-econ1003.pdfTRAP > To get MR, first write TR=P. Q with P in terms of Q, THEN differentiate. Never differentiate price directly. 17b · Worked Derivatives IN ACTION CHAIN . Y = (3X2+1)5 = 5(3x2+1)4. 6x = 30x(3x2+1)4 PRODUCT . Y = X2EX = 2x. ex + x2 . ex = ex(x2+2x) QUOTIENT . Y = (X+1)/(X-1) = [(x-1)-(x+1)]/(x-1)2 = - 2/(x-1) 2 18 . Midterm Blueprint 40% . 60 MIN Held 19 Apr; covers Wks 1-4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided. Drill: rearrange to y=mx+c . pick FV vs PV annuity . MR=MC profit . 2nd-deriv test. Show work; round at the end. Bring nothing you don't understand - the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the ax and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold. * Formula Belt · Side 1 MEMORISE COLD LINES & ELASTICITY y = mx + c · m = ΔΥ/ ΔΧ Profit = PQ - TC; break-even Profit=0 budget x2 = m/p2 - (p1/p2)x1 Ed = (dQ/dP) (P/Q) NON-LINEAR x = (-b±V(b2-4ac))/(2a) 4 = b2-4ac Log(ab)=log a+log b . log(ab)=b log a FINANCE Pt = Po (1+i)t . = Po(1+i/m)^{mt} continuous P: = Poe^{it} NPV = Σ CFt/(1+1) t annuity PV & 1-(1+i/m)^{-tm} CALCULUS xn-nxn-1 . ex-ex . Ln x-1/x chain (dy/du) (du/dx) · ax+axln a [fg]'=f'g+g'f . [f/g]'=(f'g-g'f)/g2 2nd-deriv: f''<0 max, f' '>0 min MR=MC at profit max . MR=0 at rev max * Exam Discipline MARKS DON'T LOSE EASY · Rearrange first - get y=mx+c or P=f(Q) before reading anything off · +C habit even on Side 1 (it matters on Side 2) CH6 . THE RULES SIA > The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic. * Quick Self-Check CAN YOU DO THESE? . Rearrange 3y-6x+9=0 -> y=2x-3 · $5000 at 8% cts for 3yr = 5000e^{0. 24} = $6356
• 积分 $+C$ 不能丢（尤其 TC/ODE）[12]Source: asksia-bible-econ1003-bilingual.pdf逆向求导:幂次加一,再除以新的幂次 TL;DR. Integration undoes differentiation. To integrate a power, raise the index by one and divide by the new index - the one exception is x", which integrates to In(x). Every indefinite integral carries a + C; never drop it, because in differential-equation questions you later solve for it. Definite integrals turn this machinery into area - and area is how this course measures consumer and producer surplus. 一句话总结。积分抵消求导。要对一个幂求积分,指数加一,再除以新的指数 -- 唯一例外是x-1,它积分为 In(x)。每个不 定积分都带一个+ C;绝不要丢掉它,因为在微分方程题里你之后要解出它。定积分把这套机制变成面积 -- 而本课程正是 用面积来度量消费者剩余和生产者剩余。 ★ What the exam asks here 考试在这里怎么问 Integration sits on the cumulative final (not the midterm). Expect: a routine indefinite integral testing the power / ln / e* rules; a substitution or by-parts integral; a definite integral read as an area or as consumer / producer surplus; and a differential equation (recovering total cost from marginal cost, or a separable growth law). The formula sheet gives you the by-parts formula - the rest you memorise. 积分出现在累计性期末考(而非期中考)上。预计会有:一道检验幂/ In/ex法则的常规不定积分;一道换元或分部积 分;一道被解读为面积或消费者剩余/生产者剩余的定积分;以及一道微分方程(由边际成本还原总成本,或一个可分 离的增长律)。公式表会给你分部积分公式 -- 其余的要背熟。 6. 1 The integral as an anti-derivative 6. 1 积分作为反导数 A function g is an anti-derivative (#) off if g'(x) = f(x). Because the derivative of a constant is zero, anti- derivatives are only pinned down up to a constant - hence the constant of integration C. 若 g'(x)= f(x),则函数 g 是 f 的反导数〔原函数〕。由于常数的导数为零,反导数只能确定到相差一个常数 -- 这就是积分 常数 C 的由来。 DEFINITION [ f(x) dx = g(x) + C where g' (x) = f(x) 6. 2 The power rule & the n = - 1 exception 6. 2 幂法则与 n =- 1 的例外 For every index except -1, push the index up by one and divide by the new index. At n = - 1 that would divide by zero - so that single case becomes the natural log. 对除 -1 以外的每个指数,把指数加一,再除以新的指数。在 n =- 1 时这样做会除以零 -- 所以这唯一一种情形变成了自 然对数。 ECON1003 . Quantitative Methods in Economics POWER RULE + FRIENDS [ x" dx = x"+1/ (n+1) + C (n = - 1) [ x-1 dx = [ (1/x) dx = In| x | + C (the n = - 1 case) [ K dx = Kx + C . § ex dx = ex + C . [ K. f dx = KJ f dx EX 6. 1 Power rule, three terms at once power rule 1 Rewrite roots & fractions as powers first. Integrate (3x2 + 4Vx - 5/x2)dx, i. e. f(3x2 + 4x1/2 - 5x-2)dx. 先把根式与分式改写成幂。对 S(3x2+4√x-5/x2)dx 积分,即 S(3x2+4x1/2-5x-2)dx。 2 Term by term. 3. x3/3 = x3; 4. x3/2/(3/2) = (8/3)x3/2; - 5. x-1/(-1) = +5x-1. 逐项处理。3·x3/3=x3; 4·x3/2/(3/2)= (8/3)×3/2; - 5·x-1/(-1) = +5x-1。 3 Assemble + C. Answer = x3 + (8/3)x3/2 + 5/x + C. Differentiate back to check: 3x2 + 4×1/2 - 5x-2. V 组装并加+ C。答案=x3+(8/3)×3/2+5/x+C。求导验证:3x2+4×1/2-5x-2。√ ! n = - 1 is the log, and the "+ C" is not optional n =- 1 对应对数,「+C」不是可选项 (1/x)dx is In|x| + C, never xº/O. And forgetting + C is a free dropped mark on an indefinite integral - worse, it is fatal in a differential equation, where the initial condition exists precisely to solve for that C. S(1/x)dx 是 In|xl+ C,绝不是 x°/0。而忘写+C 在不定积分中是白白丢一分 -- 更糟的是,它在微分方程中是致命 的,因为初始条件存在的意义恰恰就是解出那个C。 ECON1003 . Quantitative Methods in Economics[6]Source: asksia-bible-econ1003-bilingual.pdf1 Setup. MC = dTC/dQ = 3Q2 + 4, and fixed cost TC(O) = 50. 设定。MC = dTC/dQ=3Q2 +4,固定成本 TC(0) =50。 Integrate. TC =[(3Q2 + 4)dQ = Q3 +4Q+C. 积分。TC=S(3Q2 + 4)dQ= Q3+4Q + C。 3 Use the condition. TC(0) = 0 + 0 + C = 50 -+ C = 50 (the fixed cost is the constant). 使用条件。TC(0)=0+0+C=50→C=50(固定成本就是那个常数)。 4 Particular solution. TC = Q3 + 4Q + 50. The +C was the fixed cost all along - drop it and you lose the whole economic content. I 特解。TC=Q3+4Q+50。那个+C 自始至终就是固定成本 -- 丢掉它,你就丢掉了全部的经济内涵。 ECON1003 . Quantitative Methods in Economics EX 6. 8 A separable growth law separable . growth 1 Setup. dy/dx = ky (proportional growth), with y(O) = yo. 设定。dy/dx =ky(比例增长),且 y(0)= yo。 2 Separate. (1/y)dy = k dx. 分离变量。(1/y)dy=k dx。 3 Integrate both sides. [(1/y)dy = [k dx -> Inlyl = kx + C. 对两边积分。S(1/y)dy = Sk dx → Inlyl = kx+ C。 Isolate y. y = ekx+C = A. ekx (write A = eC). Apply y(O) = y. - A = yo, so y = yo ekx - the exponential growth law, derived from scratch. 解出 y。y = ekx+C = A·ekx(记 A= eC)。代入 y(0)= yo → A= yo,所以 y = yo ekx -- 从零开始推导出来的指数增 长律。 ! Solve for C, do not drop it; and don't forget the In step 解出 C,不要丢掉它;也别忘了 In 这一步 In a differential equation the constant of integration is the whole point - the initial condition exists to fix it. With a separable equation, integrating 1/y produces a ln|y|; you must then exponentiate to get y back, turning the additive +C into a multiplicative A = eC. 在微分方程中,积分常数才是关键所在 -- 初始条件存在就是为了确定它。对于可分离方程,对 1/y 积分会产生一个 Inlyl;接着你必须取指数才能把 y 还原回来,从而把加性的+C 变成乘性的 A= eC。 Integration is differentiation run backwards - so every answer can be checked by differentiating it. On the final, if you have a minute, differentiate your integral and watch it collapse back to the integrand. Free insurance on method marks. 积分是反向运行的求导 -- 所以每个答案都可以通过对它求导来检验。在期末考上,如果你还有一分钟,就对你的积分 求导,看着它塌缩回被积函数。这是方法分上的免费保险。 MARKER'S NOTE . ECON1003 FINAL ECON1003 . Quantitative Methods in Economics T7 . MATRICES - TOPIC 7 . LINEAR ALGEBRA FINAL EXAM TOPIC Matrices, arithmetic & multiplication
• 逆矩阵要 $C^T/\det(A)$，先查 det，别忘转置。[16]Source: asksia-bible-econ1003-bilingual.pdfECON1003 . Quantitative Methods in Economics X = A- 1B EX 7. 5 Invert a 2×2, then solve X = A-1B Solve 求解: 3 1 10 5 i. e. the system 3x + y = 5, 2x + 4y = 10. 2 4 X = 1 Determinant. det(A) = 3-4 - 1. 2 = 10 (# 0, so A-1 exists). 行列式。det(A)= 3·4-1·2=10(≠0,所以A-1存在)。 2 Inverse via the shortcut. swap 3&4, negate 1&2: A-1 = (1/10) -2 3 4 -1 用速算法求逆矩阵。交换3和 4,对1和 2 取负:A-1=(1/10) -2 4 -1 3 。 3 X = A-1B. x = (4. 5 - 1. 10)/10 = (20 -10)/10 = 1; y = (-2. 5 + 3. 10)/10 = (-10 + 30)/10 = 2. X = A-1B. x = (4. 5-1. 10)/10 = (20-10)/10 =1; y = (-2. 5+ 3. 10)/10 = (-10 +30)/10 = 2. 4 Check. 3(1) + 2 = 5 V; 2(1) + 4(2) = 10 V. 检验。3(1)+2=5√;2(1)+ 4(2)= 10√。 ! Don't forget the transpose - or the determinant test 别忘了转置 -- 也别忘了行列式检验 Cofactor method: it is C /det(A), not C/det(A) - the transpose is the step everyone drops. And before any of this, check det(A) # 0: if det(A) = O there is no inverse, and X = A-1B does not exist. 余子式法:它是 CT/det(A),不是 C/det(A) -- 转置正是人人都会漏掉的那一步。而在这一切之前,先检查 det(A)≠ 0:如果 det(A)=0,就没有逆矩阵,X=A-1B 也不存在。 ✓ Which method, when? 何时用哪种方法? For a one-off solve, elimination / Cramer is usually faster. Build the full inverse when you must solve AX = B for several different right-hand sides B - compute A-1 once, then reuse it. 对于一次性的求解,消元法/Cramer 通常更快。只有当你必须为多个不同的右端项 B 求解 AX= B时,才去构造完整 的逆矩阵 -- 先算出 A-1一次,然后反复使用它。 Always verify: multiply A by your A- and confirm you get the identity, or substitute (x, y) back into the original equations. Method marks reward a stated check - and it catches the sign slip before it costs you. 永远要验证:把 A乘以你算出的 A-,确认得到单位矩阵;或者把(x,y)代回原方程组。方法分奖励写出来的检验 而且它能在符号疏忽害你丢分之前就抓住它。 MARKER'S NOTE . ECON1003 FINAL

• 每种题型你都准备一套：
  • “先整理→写公式→代入→解→解释经济含义→最后取整/验证”。[1]Source: asksia-bible-econ1003-bilingual.pdf代数余子式 / 余子式 Minor = sub-determinant; cofactor = minor x the +-+ sign. ECON1003 . Quantitative Methods in Economics Term 术语 中文 One-line definition 释义 Inverse matrix (A-1) 逆矩阵 A-1 = CT/det(A); satisfies AA1 = I; solves X = A-1B. ✓ How to use the glossary 用法 如何使用术语表 用法 Cover the right column and recite the definition from the English term - then flip and recall the English from the +X. 〔遮住释义、由术语复述定义;再由中文回想英文。〕 The exam rewards the exact wording: "next-best forgone", "CT/det(A)", "signed area" - precision is marks. 遮住右栏,从英文术语复述出定义 -- 再反过来,由中文回忆英文。〔遮住释义、由术语复述定义;再由中文回想英文。〕 考试奖励精确的措辞:“next-best forgone”、“CT/det(A)”、“signed area" -- 精确就是分数。 ECON1003 . Quantitative Methods in Economics PRACTICE P1 -P2 - PRACTICE BANK DRILL TO EXAM STANDARD Drill both exams, with full solutions 两场考试齐刷,附完整解答 Fourteen AskSia-authored problems across all seven topics - worked end to end 横跨全部七大主题的14 道 AskSia 原创题 –––完整解答 These are exam-style problems written in the ECON1003 style - not copied from any real quiz or exam. They span lines, finance, differentiation, the Lagrangian, integration and matrices, so they serve both the Midterm (Topics 1-4) and the cumulative Final (all seven). Every number is verified. Cover the solution, work it on paper, then read the answer. 这些是按 ECON1003 风格编写的考试风格题目 -- 并非抄自任何真实测验或考试。它们覆盖直线、金融、求导、拉格朗日 函数、积分和矩阵,因此同时服务于期中(主题1-4)和累计的期末(全部七大主题)。每个数字都经过核验。盖住解答,在 纸上做完,再看答案。 ★ How to use this bank 如何使用这个题库 Cover the solution - recognition is not recall. Budget ~1 minute per mark. Show your work: method marks are real and partial credit is given, so a stated step beats a bare number. Round only the final answer to the stated decimals. Check every answer by substituting back - differentiate an integral, multiply an inverse, re-solve a system. 遮住答案 认得出不等于回忆得起。按每分约1分 钟来分配时间。写出你的过程:方法分是实打实的, 也会给部分分,所以一个写出来的步骤胜过一个孤零 零的数字。只对最终答案按规定小数位取整。对每个 答案都做检验,把它代回去 -- 对积分求导、把逆矩 阵乘回去、重新解一遍方程组。 i Formula sheet vs memory 公式表 vs 记忆 Both exams give a formula sheet (quadratic, series, interest, NPV, annuity, by-parts, 3×3 determinant, ax/log derivatives, quotient rule). You must memorise the rest: power / product / chain rules, substitution, the Lagrangian conditions, Cramer's rule and the inverse methods. The marks are in applying them cleanly.[24]Source: asksia-cheatsheet-econ1003.pdfECON1003 Quantitative Methods in Economics UNIVERSITY OF SYDNEY . SCHOOL OF ECONOMICS EXAM REVISION Sem 1 2026 . SIDE 1 OF 2 Midterm 40% + Final 50% · formula sheet provided SIDE 1/2 LINES - CALCULUS . Straight lines & elasticity . Quadratics/exponentials/logs . Financial maths . Differentiation & MR=MC 0 . How to Use This READ FIRST * ECON1003 is maths for economics - NOT statistics . No probability, distributions or hypothesis tests. Seven topics: lines . non-linear functions . financial maths . differentiation . several variables . integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9). Two exams: Midterm 40% (Wks 1-4, up to applications of differentiation - Side 1) . Final 50% (cumulative, all 7 - both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse). SIA > Method marks matter - show every step; partial credit is given. Round only the FINAL answer to the stated decimal places. 1 . The Straight Line LINE CH2 y = mx + c m = slope, c = y- intercept m = (y2-y1)/(x2-X1) = rise/run Slope m: "x up 1 = y changes by m. " Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope). WORKED . LINE THRU (2,1), M=1 1 = 1. 2 + c = c = - 1 = y = x - 1 TRAP > Non-standard form (e. g. 2y+4x-4=0) must be rearranged to y=mx+c before reading m,c: = >y = - 2x + 2. 2 . Demand & Supply CH2 Demand Q=f(P), e. g. Q=200-2P (-2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100-0. 5Q. The two slopes are reciprocals, not equal . COST . REVENUE . PROFIT TC = FC + VC (e. g. 800 + 1. 50) R = P. Q Profit = R - TC Break-even: set Profit = 0, solve Q WORKED . TC=800+1. 5Q, P=3. 5 Profit = 3. 5Q - (800+1. 5Q) = 20-800 =0 - break-even Q = 400 TRAP > Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted. 3 . Budget Line TWO GOODS, INCOME M CH2 m = P1X1 + P2X2 ×2 = m/p2 - (p1/p2)×1

6）你下一步给我一个信息，我就能把“详细复习重点”继续细化到题型清单

• 你这次 Final 最怕的是哪一块？（选一个或多个）
  • A 直线/弹性/盈亏平衡
  • B 金融数学（利息/NPV/年金）
  • C 求导 + $MR=MC$
  • D 积分（换元/分部/CS-PS/还原 TC）
  • E 拉格朗日（含角点/影子价格）
  • F 矩阵/逆矩阵/Cramer

• “必背定义一句话”（考试写得分那种）
• “公式 + 何时用”
• “最常见题型模板”
• “3 个最常见失误与纠正”