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187800
https://zh.khanacademy.org/math/statistics-probability/random-variables-stats-library/binomial-random-variables/e/calculating-binomial-probability
计算二项分布概率 (练习) | 二项随机变量 | 可汗学院 跳转到主内容 如果你看到这则信息,这表示下载可汗学院的外部资源时遇到困难. 如果您在网络过滤器后面，请确保.kastatic.org 和 .kasandbox.org 域未被阻止。 探索 数学 科学 计算机 艺术与人文 大学, 职业和更多 经济和金融 选择一个类别以查看它的课程 搜索 捐款登录注册 搜索科目, 技能, 教学视频 跳到课程内容 统计和概率 课程: 统计和概率>单元 9 课程 5: 二项随机变量 二项变量 识别二项变量 用于假设实验 "独立性" 的 10% 法则 识别二项变量 二项式分布 可视化二项式分布 二项式概率示例 推广至n投k进 罚球的二项概率分布 投篮二项式分布图的绘制 二项分布概率密度函数和二项累积分布函数 二项式概率（基础） 二项分布概率公式 计算二项分布概率 数学> 统计和概率> 随机变量> 二项随机变量 © 2025 Khan Academy 使用条款隐私政策Cookie 通知可访问性声明 计算二项分布概率 Google课堂 Microsoft Teams 您可能需要：计算器 问题 某种西红柿在从盆移植到花园时有 70%‍的几率存活.小娜移植了 3‍棵. 假设每棵植物的存活相互独立.设 X‍为存活的西红柿的数量. 在这 3‍ 棵西红柿中存活 2‍ 棵的几率是多少? 保留小数点到百分位. P(X=2)=‍ 你的答案是 一个精确的十进位小数，例如0.75 显示计算器 相关内容 视频10 分钟 24 秒 10:24 二项式概率示例 视频5 分钟 27 秒 5:27 二项分布概率密度函数和二项累积分布函数 视频4 分钟 13 秒 4:13 推广至n投k进 报告问题 做所有4道题 跳过 检查 Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices 计算器 RAD DEG sin⁻¹ sin del ac cos⁻¹ cos()tan⁻¹ tan π ans ln log eˣ EXP +-= 1 2 3 4 5 6 7 8 9 0. ×÷xʸ√
187801
https://fiveable.me/discrete-mathematics/unit-8/euler-hamiltonian-paths/study-guide/nxBy2f3LfpTk9HLo
Euler and Hamiltonian Paths | Discrete Mathematics Class Notes | Fiveable | Fiveable ap study content toolsprintables upgrade 🧩Discrete Mathematics Unit 8 Review 8.3 Euler and Hamiltonian Paths All Study Guides Discrete Mathematics Unit 8 – Graph Theory Topic: 8.3 🧩Discrete Mathematics Unit 8 Review 8.3 Euler and Hamiltonian Paths Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA 🧩Discrete Mathematics Unit & Topic Study Guides Introduction to Logic and Proofs Set Theory Functions and Relations Algorithms and Complexity Number Theory and Cryptography Induction and Recursion Counting and Probability Graph Theory 8.1 Graph Fundamentals and Representations 8.2 Graph Connectivity and Traversals 8.3 Euler and Hamiltonian Paths 8.4 Planar Graphs and Coloring Trees and Applications Boolean Algebra Modeling Computation Discrete Probability Recurrence Relations Generating Functions print guide report error Euler and Hamiltonian paths are key concepts in graph theory. They help us understand how to traverse graphs efficiently, visiting edges or vertices exactly once. These ideas have real-world applications in route planning, puzzle solving, and network design. Euler paths focus on edges, while Hamiltonian paths deal with vertices. Both have specific conditions for existence and algorithms for finding them. Understanding these paths is crucial for solving complex problems in various fields, from logistics to computer science. Euler Paths and Circuits Fundamental Concepts of Euler Paths and Circuits Euler path traverses every edge in a graph exactly once Euler circuit forms a closed loop traversing every edge exactly once Eulerian graph contains an Euler circuit Semi-Eulerian graph contains an Euler path but not an Euler circuit Conditions for Eulerian Graphs Necessary condition for Euler path requires all vertices except at most two have even degree Sufficient condition for Euler path ensures graph is connected with at most two odd-degree vertices Necessary and sufficient condition for Euler circuit mandates all vertices have even degree and graph is connected Fleury's algorithm finds Euler circuits in Eulerian graphs Practical Applications of Euler Paths Snow plow route optimization utilizes Euler paths to clear streets efficiently Mail delivery routes leverage Euler paths to minimize repeated travel Network design employs Euler circuits for efficient data transmission (token ring networks) Puzzle solving applies Euler paths in games like "draw without lifting your pencil" Hamiltonian Paths and Cycles Core Concepts of Hamiltonian Paths and Cycles Hamiltonian path visits every vertex in a graph exactly once Hamiltonian cycle forms a closed loop visiting every vertex exactly once Hamiltonian graph contains a Hamiltonian cycle Semi-Hamiltonian graph contains a Hamiltonian path but not a Hamiltonian cycle Conditions for Hamiltonian Graphs Necessary conditions for Hamiltonian graphs include connectivity and absence of articulation points Dirac's theorem provides sufficient condition: graph with n ≥ 3 vertices and minimum degree ≥ n/2 is Hamiltonian Ore's theorem offers another sufficient condition: sum of degrees of non-adjacent vertices ≥ n for graph with n ≥ 3 vertices Bondy-Chvátal theorem generalizes sufficient conditions for Hamiltonian graphs Complexity and Algorithms for Hamiltonian Problems Determining existence of Hamiltonian paths or cycles belongs to NP-complete class of problems Held-Karp algorithm solves Hamiltonian cycle problem in O(n 2 2 n)O(n^2 2^n)O(n 2 2 n) time Nearest neighbor algorithm provides heuristic approach for finding approximate Hamiltonian cycles Genetic algorithms offer alternative method for tackling Hamiltonian problems in large graphs Applications Optimization Problems in Graph Theory Traveling Salesman Problem seeks shortest Hamiltonian cycle in weighted graph Nearest Neighbor and 2-opt heuristics provide approximate solutions for Traveling Salesman Problem Branch and bound algorithm offers exact solution for smaller instances of Traveling Salesman Problem Vehicle routing problems extend Traveling Salesman Problem to multiple vehicles and constraints Historical and Practical Graph Traversal Problems Konigsberg Bridge Problem led to development of Euler path concept Seven Bridges of Konigsberg represented as multigraph with land masses as vertices and bridges as edges Euler proved impossibility of traversing all bridges exactly once, establishing foundations of graph theory Modern applications include optimizing routes for mail delivery and waste collection Graph Theory in Scientific and Engineering Applications Utility graph traversal problem involves connecting houses to utilities without crossing lines K3,3 graph represents utility problem, proving its impossibility on a plane Molecular structure analysis uses graph theory to represent chemical bonds and atomic arrangements Graph isomorphism helps identify structurally equivalent molecules in chemistry Study Content & Tools Study GuidesPractice QuestionsGlossaryScore Calculators Company Get $$ for referralsPricingTestimonialsFAQsEmail us Resources AP ClassesAP Classroom every AP exam is fiveable history 🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history social science ✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾‍⚖️ ap us government english & capstone ✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar arts 🎨 ap art & design🖼️ ap art history🎵 ap music theory science 🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics math & computer science 🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p world languages 🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature go beyond AP high school exams ✏️ PSAT🎓 Digital SAT🎒 ACT honors classes 🍬 honors algebra II🐇 honors biology👩🏽‍🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history college classes 👩🏽‍🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽‍🔬 science💶 social science RefundsTermsPrivacyCCPA © 2025 Fiveable Inc. 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187802
https://www.chemistrysteps.com/preparation-and-reaction-mechanism-of-carboxylic-anhydrides/
Preparation and Reaction Mechanisms of Carboxylic Anhydrides - Chemistry Steps Skip to content Chemistry Steps Menu Live Help Topics Study Guides Reagent Guide Reaction Maps Quizzes Practice CS Benefits Log In Register Menu Live Help Topics Study Guides Reagent Guide Reaction Maps Quizzes Practice CS Benefits Log In Register Carboxylic Acids and Their Derivatives Preparation and Reaction Mechanisms of Carboxylic Anhydrides Preparation of Anhydrides Anhydrides are prepared by dehydration of carboxylic acids either at extremely high temperatures (800 o C) or by using P 2 O 5 as a dehydrating agent: For most carboxylic acids are not compatible with excessive heating, and even using P 2 O 5 is not a practically applicable solution. Therefore, the most common method for preparing carboxylic anhydrides in the laboratory is the conversion of the acid to an acid chloride and a carboxylate salt, which react readily at room temperature: Reactions of Anhydrides Although slightly less reactive than acid chlorides, anhydrides react with all the nucleophiles in an identical mechanism. The only difference in these nucleophilic acyl substitutions is the change of the leaving group from chloride to a carboxylate. Hydrolysis of Anhydrides Anhydrides can be hydrolyzed to carboxylic acids. Adding a base would increase the rate of the conversion: Esters from Anhydrides Anhydrides can be converted into esters by using an alcohol with a base or, even better, an alkoxide which is a stronger nucleophile: In a similar reaction, thioesters can be prepared from anhydrides: Amides from Anhydrides Amines, being good nucleophiles, react readily with anhydrides to form primary, secondary or tertiary amides: Anhydrides in Grignard Reaction Just like the acid chlorides and esters, anhydrides react with excess Grignard reagent, forming a tertiary alcohol: Reduction of Anhydrides Anhydrides can be reduced to a primary alcohol using LiAlH 4: To stop the reduction at the aldehyde, a less powerful reducing agent such as lithium tri(t-butoxy) aluminum hydride LiAl(O t Bu)3 H can be used: Check Also Preparation of Carboxylic Acids Naming Carboxylic Acids Naming Nitriles Naming Esters Naming Carboxylic Acid Derivatives – Practice Problems The Addition-Elimination Mechanism Fischer Esterification Ester Hydrolysis by Acid and Base-Catalyzed Hydrolysis What is Transesterification? Esters Reaction with Amines – The Aminolysis Mechanism Ester Reactions Summary and Practice Problems Preparation of Acyl (Acid) Chlorides (ROCl) Reactions of Acid Chlorides (ROCl) with Nucleophiles R 2 CuLi Organocuprates – Gilman Reagent Reaction of Acyl Chlorides with Grignard and Gilman (Organocuprate) Reagents Reduction of Acyl Chlorides by LiAlH 4, NaBH4, and LiAl(OtBu)3 H Reduction of Carboxylic Acids and Their Derivatives Preparation and Reaction Mechanism of Carboxylic Anhydrides Amides – Structure and Reactivity Naming Amides Amides Hydrolysis: Acid and Base-Catalyzed Mechanism Amide Dehydration Mechanism by SOCl 2, POCl 3, and P 2 O 5 Amide Reduction Mechanism by LiAlH4 Reduction of Amides to Amines and Aldehydes Amides Preparation and Reactions Summary Amides from Carboxylic Acids-DCC and EDC Coupling The Mechanism of Nitrile Hydrolysis To Carboxylic Acid Nitrile Reduction Mechanism with LiAlH4 and DIBAL to Amine or Aldehyde The Mechanism of Grignard and Organolithium Reactions with Nitriles The Reactions of Nitriles Converting Nitriles to Amides Carboxylic Acids to Ketones Esters to Ketones Carboxylic Acids and Their Derivatives Practice Problems Carboxylic Acids and Their Derivatives Quiz Reactions Map of Carboxylic Acid Derivatives Share Your Thoughts, Ask that Question! Cancel reply Comment Name Email Website - [x] Notify me via e-mail if anyone answers my comment. Organic Chemistry Study Guides Structure and Bonding Lewis Structures in Organic Chemistry Valency and Formal Charges in Organic Chemistry How to Determine the Number of Lone Pairs Bonding Patterns in Organic Chemistry sp3, sp2, and sp Hybridization in Organic Chemistry with Practice Problems How to Quickly Determine The sp3, sp2, and sp Hybridization Bond Lengths and Bond Strengths VSEPR Theory – Molecular and Electron Geometry of Organic Molecules Dipole-dipole, London Dispersion and Hydrogen Bonding Interactions Dipole Moment and Molecular Polarity Boiling Point and Melting Point in Organic Chemistry Boiling Point and Melting Point Practice Problems Solubility of Organic Compounds General Chemistry Overview Quiz Molecular Representations Bond-Line or Skeletal Structures Functional Groups in Organic Chemistry with Practice Problems Bond-line, Lewis, and Condensed Structures with Practice Problems Curved Arrows with Practice Problems Resonance Structures in Organic Chemistry with Practice Problems Rules for Drawing Resonance Structures Major and Minor Resonance Contributor How to Choose the More Stable Resonance Structure Drawing Complex Patterns in Resonance Structures Localized and Delocalized Lone Pairs with Practice Problems Molecular Representations Quiz Acids and Bases Acids and Bases – General Chemistry Organic Acids and Bases Organic Acid-Base Mechanisms Acid Strength and pKa How to Determine the Position of Equilibrium for an Acid-Base Reaction Inductive and Resonance (Mesomeric) Effects Factors That Determine the pKa and Acid Strength How to Choose an Acid or a Base to Protonate or Deprotonate a Given Compound Lewis Acids and Bases Basicity of Amines Organic Acids and Bases Practice Problems Organic Acids and Bases Quiz Alkanes and Cycloalkanes Naming Alkanes by IUPAC Nomenclature Rules Practice Problems Naming Bicyclic Compounds Naming Bicyclic Compounds-Practice Problems How to Name a Compound with Multiple Functional Groups Primary, Secondary, and Tertiary Carbon Atoms in Organic Chemistry Constitutional or Structural Isomers with Practice Problems Degrees of Unsaturation or Index of Hydrogen Deficiency The Wedge and Dash Representation Sawhorse Projections Newman Projections with Practice Problems Staggered and Eclipsed Conformations Conformational Isomers of Propane Newman Projection and Conformational Analysis of Butane Newman Projection of Chair Conformation Gauche Conformation Gauche Conformation, Steric, Torsional Strain Energy Practice Problems Ring Strain Steric vs Torsional Strain Conformational Analysis Drawing the Chair Conformation of Cyclohexane Ring Flip: Drawing Both Chair Conformations with Practice Problems 1,3-Diaxial Interactions and A value for Cyclohexanes Ring-Flip: Comparing the Stability of Chair Conformations with Practice Problems Cis and Trans Decalin IUPAC Nomenclature Practice Problems IUPAC Nomenclature Summary Quiz Alkanes and Cycloalkanes Practice Quiz Stereochemistry How to Determine the R and S Configuration The R and S Configuration Practice Problems What is Nonsuperimposable in Organic Chemistry Chirality and Enantiomers Diastereomers-Introduction and Practice Problems Enantiomers vs Diastereomers Cis and Trans Isomers E and Z Alkene Configuration with Practice Problems Enantiomers, Diastereomers, the Same or Constitutional Isomers with Practice Problems Configurational Isomers Optical Activity Specific Rotation Racemic Mixtures Enantiomeric Excess (ee): Percentage of Enantiomers from Specific Rotation with Practice Problems Symmetry and Chirality. Meso Compounds Fischer Projections with Practice Problems R and S Configuration in the Fischer Projection R and S configuration on Newman projections R and S Configuration of Allenes Converting Bond-Line, Newman Projection, and Fischer Projections Resolution of Enantiomers: Separate Enantiomers by Converting to Diastereomers Stereochemistry Practice Problems Stereochemistry Practice Quiz Energy Changes In Organic Chemistry Energy and Organic Chemistry Reactions Bond Lengths and Bond Strengths Homolytic and Heterolytic Bond Cleavage The Heat of Reaction from Bond Dissociation Energies Nucleophilic Substitution Reactions Introduction to Alkyl Halides Nomenclature of Alkyl Halides Substitution and Elimination Reactions Nucleophilic Substitution Reactions – An Introduction All You Need to Know About the S N 2 Reaction Mechanism The S N 2 Mechanism: Kinetics, Thermodynamics, Curved Arrows, and Stereochemistry with Practice Problems The Stereochemistry of S N 2 Reactions Stability of Carbocations The S N 1 Nucleophilic Substitution Reaction Reactions of Alkyl Halides with Water The Stereochemistry of the S N 1 Reaction Mechanism The S N 1 Mechanism: Kinetics, Thermodynamics, Curved Arrows, and Stereochemistry with Practice Problems Steric Hindrance in S N 2 and S N 1 Reactions Carbocation Rearrangements in S N 1 Reactions with Practice Problems Ring Expansion Rearrangements Ring Contraction Rearrangements When Is the Mechanism S N 1 or S N 2? Reactions of Alcohols with HCl, HBr, and HI Acids SOCl 2 and PBr 3 for Conversion of Alcohols to Alkyl Halides Alcohols in S N 1 and S N 2 Reactions How to Choose Molecules for Doing S N 2 and S N 1 Synthesis-Practice Problems Exceptions in S N 2 and S N 1 Reactions Nucleophilic Substitution and Elimination Practice Quiz Reactions Map of Alkyl Halides Alkenes: Structure, Stability, and Nomenclature Alkenes: Structure and Stability Naming Alkenes by IUPAC Nomenclature Rules Cis and Trans Isomers E and Z Alkene Configuration with Practice Problems Elimination Reactions Substitution and Elimination Reactions The E2 Mechanism E2 Elimination Practice Problems Zaitsev's Rule - Regioselectivity of E2 Elimination Reactions The Hofmann Elimination of Amines and Alkyl Fluorides Stereoselectivity of E2 Elimination Reactions Stereospecificity of E2 Elimination Reactions S N 2 and E2 Rates of Cyclohexanes Elimination Reactions of Cyclohexanes with Practice Problems POCl 3 for Dehydration of Alcohols The E1 Mechanism with Practice Problems Regioselectivity of E1 Reactions Stereoselectivity of E1 Reactions How to tell if it is E2 or E1 Mechanism S N 1 vs E1 Reactions S N 2 vs E2 Reactions Dehydration of Alcohols by E1 and E2 Elimination Mesylates and Tosylates as Good Leaving Groups Mitsunobu Reaction S N 1 S N 2 E1 E2 – How to Choose the Mechanism Polar Protic and Polar Aprotic Solvents S N 1 S N 2 E1 or E2 - the Largest Collection of Practice Problems The Hammond Postulate The E1cB Elimination Mechanism Nucleophilic Substitution and Elimination Practice Quiz Reactions Map of Alkyl Halides Addition Reactions of Alkenes Electrophilic Addition Reactions to Alkenes Markovnikov's Rule Markovnikov's Rule with Practice Problems Addition of Water to Alkenes Acid-Catalyzed Hydration of Alkenes with Practice Problems Rearrangements in Alkene Addition Reactions Oxymercuration-Demercuration Addition of Alcohols to Alkenes Free-Radical Addition of HBr: Anti-Markovnikov Addition Hydroboration-Oxidation: The Mechanism Hydroboration-Oxidation of Alkenes: Regiochemistry and Stereochemistry with Practice Problems Halogenation of Alkenes and Halohydrin Formation The Regiochemistry of Alkene Addition Reactions The Stereochemistry of Alkene Addition Reactions Cis product in an anti-Addition Reaction of Alkenes Ozonolysis of Alkenes with Practice Problems Syn Dihydroxylation of Alkenes with KMnO 4 and OsO 4 Anti-Dihydroxylation of Alkenes with MCPBA and Other Peroxides with Practice Problems Oxidative Cleavage of Alkenes with KMno 4 and O 3 Alkene Reactions Practice Problems Changing the Position of a Double Bond Changing the Position of a Leaving Group Alkenes Multi-Step Synthesis Practice Problems Alkene Addition Reactions Practice Quiz Reactions Map of Alkenes Alkynes Introduction to Alkynes Naming Alkynes by IUPAC Nomenclature Rules - Practice Problems Preparation of Alkynes by Elimination Reactions Hydrohalogenation of Alkynes Addition of Water to Alkynes Acid-Catalyzed Hydration of Alkynes with Practice Problems Reduction of Alkynes Halogenation of Alkynes Hydroboration-Oxidation of Alkynes with Practice Problems Ozonolysis of Alkynes with Practice Problems Alkylation of Terminal Alkynes in Organic Synthesis with Practice Problems Reactions of Acetylide Ions Alkyne reactions summary practice problems Alkyne Synthesis Reactions Practice Problems Alkyne Naming and Reactions Practice Quiz Reactions Map of Alkynes Nuclear Magnetic Resonance (NMR) Spectroscopy NMR Spectroscopy – An Easy Introduction NMR Chemical Shift NMR Chemical Shift Range and Value Table NMR Number of Signals and Equivalent Protons Homotopic, Enantiotopic, Diastereotopic, and Heterotopic Homotopic Enantiotopic Diastereotopic Practice Problems Integration in NMR Spectroscopy Splitting and Multiplicity (N+1 rule) in NMR Spectroscopy NMR Signal Splitting N+1 Rule Multiplicity Practice Problems 13 C Carbon NMR DEPT NMR: Signals and Problem Solving NMR Spectroscopy-Carbon-Dept-IR Practice Problems Organic Structure Determination NMR Spectroscopy-Carbon-Dept-IR Practice Problems Infrared (IR) Spectroscopy Practice Problems Solving Mass Spectrometry Practice Problems Radical Reactions Free-Radical Addition of HBr: Anti-Markovnikov Addition Initiation, Propagation, Termination in Radical Reactions Selectivity in Radical Halogenation Stability of Radicals Resonance Structures of Radicals Stereochemistry of Radical Halogenation with Practice Problems Allylic Bromination by NBS with Practice Problems Radical Halogenation in Organic Synthesis Reactions of Alcohols Nomenclature of Alcohols: Naming Alcohols based on IUPAC Rules with Practice Problems Preparation of Alcohols via Substitution or Addition Reactions Reaction of Alcohols with HCl, HBr, and HI Acids Mesylates and Tosylates as Good Leaving Groups SOCl 2 and PBr 3 for Conversion of Alcohols to Alkyl Halides Alcohols in Substitution Reactions Practice Problems POCl 3 for Dehydration of Alcohols Dehydration of Alcohols by E1 and E2 Elimination The Oxidation States of Organic Compounds LiAlH4 and NaBH4 Carbonyl Reduction Mechanism Alcohols from Carbonyl Reductions - Practice Problems Grignard Reaction in Preparing Alcohols with Practice Problems Grignard Reaction in Organic Synthesis with Practice Problems Protecting Groups For Alcohols in Organic Synthesis Oxidation of Alcohols: PCC, PDC, CrO 3, DMP, Swern, and All of That Diols: Nomenclature, Preparation, and Reactions NaIO4 Oxidative Cleavage of Diols The Pinacol Rearrangement The Williamson Ether Synthesis Alcohol Reactions Practice Problems Naming Thiols and Sulfides Reactions of Thiols Alcohols Quiz – Naming, Preparation, and Reactions Reactions Map of Alcohols Ethers and Epoxides Preparation of Epoxides Ring-Opening Reactions of Epoxides Reactions of Epoxides under Acidic and Basic Conditions Reactions of Epoxides Practice Problems The Grignard Reaction of Epoxides Naming Ethers The Williamson Ether Synthesis Reactions of Ethers-Ether Cleavage Conjugated Systems Resonance and Conjugated Dienes Allylic Carbocations 1,2 and 1,4 Electrophilic Addition to Dienes Kinetic vs Thermodynamic Control of Electrophilic Addition to Dienes The Diels-Alder Reaction Diels-Alder Reaction: Dienes and Dienophiles Predict the Products of the Diels-Alder Reaction with Practice Problems Endo and Exo Products of Diels-Alder Reaction with Practice Problems Regiochemistry of the Diels–Alder Reaction with Practice Problems Identify the Diene and Dienophile of the Diels-Alder reaction with Practice Problems Diels-Alder Reaction in Organic Synthesis Practice Problems Aromatic Compounds Naming Aromatic Compounds Introduction to Aromatic Compounds Benzene – Aromatic Structure and Stability Aromaticity and Huckel's Rule The 4n+2 Rule Identify Aromatic, Antiaromatic, or Nonaromatic Compounds Frost Circle Annulenes Electrophilic Aromatic Substitution Electrophilic Aromatic Substitution – The Mechanism The Halogenation of Benzene The Nitration of Benzene The Sulfonation of Benzene Activating and Deactivating Groups Friedel-Crafts Alkylation with Practice Problems Friedel-Crafts Acylation with Practice Problems Vilsmeier-Haack Reaction The Alkylation of Benzene by Acylation-Reduction Ortho, Para, Meta in EAS with Practice Problems Ortho, Para, and Meta in Disubstituted Benzenes Why Are Halogens Ortho-, Para- Directors yet Deactivators? Is Phenyl an Ortho/Para or Meta Director? Limitations of Electrophilic Aromatic Substitution Reactions Orientation in Benzene Rings With More Than One Substituent Synthesis of Aromatic Compounds From Benzene Arenediazonium Salts in Electrophilic Aromatic Substitution Reactions at the Benzylic Position Benzylic Bromination Nucleophilic Aromatic Substitution Nucleophilic Aromatic Substitution Practice Problems Reactions of Phenols Reactions of Aniline Meta Substitution on Activated Aromatic Ring Electrophilic Aromatic Substitution Practice Problems Aromatic Compounds Quiz Reactions Map of Aromatic Compounds Aldehydes and Ketones Nomenclature of Aldehydes and Ketones How to Name a Compound with Multiple Functional Groups Preparation of Aldehydes and Ketones Nucleophilic Addition to Carbonyl Groups Reduction of Aldehydes and Ketones Reactions of Aldehydes and Ketones with Water Reactions of Aldehydes and Ketones with Alcohols: Acetals and Hemiacetals Acetals as Protecting Groups for Aldehydes and Ketones Formation and Reactions of Imines and Enamines Reductive Amination Acetal Hydrolysis Mechanism Imine and Enamine Hydrolysis Mechanism Hydrolysis of Acetals, Imines, and Enamines-Practice Problems Reaction of Aldehydes and Ketones with CN, Cyanohydrin Formation The Wittig Reaction: Examples and Mechanism The Wittig Reaction: Practice Problems Aldehydes and Ketones to Carboxylic Acids Reactions of Aldehydes and Ketones - Practice Problems Aldehydes and Ketones Reactions Practice Quiz Reactions Map of Aldehydes Reactions Map of Ketones Carboxylic Acids and Their Derivatives-Nucleophilic Acyl Substitution Preparation of Carboxylic Acids Naming Carboxylic Acids Naming Nitriles Naming Esters Naming Carboxylic Acid Derivatives – Practice Problems The Addition-Elimination Mechanism Fischer Esterification Ester Hydrolysis by Acid and Base-Catalyzed Hydrolysis What is Transesterification? 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187803
https://brilliant.org/wiki/inequalities-with-strange-equality-conditions/
Inequalities with Strange Equality Conditions Sign up with Facebook or Sign up manually Already have an account? Log in here. Daniel Liu, Calvin Lin, and Jimin Khim contributed This page serves to debunk the myth that An expression attains its maximum or minimum when all (some) of the variables are equal. If a,b are real numbers such that a+b=4, what is the minimum value of [(a−1)(b−1)]2? Clearly, since the expression is a perfect square, the minimum that it can attain is 0. This can indeed be attained, say with a=1,b=3. The minimum does not occur when a=b=24​, even though the expression is symmetric. □​ Some inequalities are less obvious and they do not have equality case when all variables are equal. If x,y,z≥0 satisfy x+y+z=3, find the maximum value of x2y+y2z+z2x. Setting x=y=z=1 gives x2y+y2z+z2x=3. However, is this the maximum value? In fact, setting x=0,y=2,z=1 gives x2y+y2z+z2x=4, which is the actual maximum value. □​ Try your hand at the list of problems below. Be warned that these problems are hard to (properly) solve, and could require a lot of ingenuity. Let a,b,c,d∈[21​,2]. Suppose abcd=1. Find the maximum possible value of (a+b1​)(b+c1​)(c+d1​)(d+a1​). The correct answer is: 25 Let a,b,c,d be positive integers such that a+b+c+d=120, and define S=ab+bd+da+bc+cd. Find the number of ordered quadruples (a,b,c,d) for which S attains its maximum. The correct answer is: 39 Let x,y,z be non-negative real numbers satisfying the condition x+y+z=1. The maximum possible value of x3y3+y3z3+z3x3 has the form ba​, where a and b are positive, coprime integers. What is the value of a+b? The correct answer is: 65 Without loss of generality, we may assume that x≥y≥z≥0. Since y and z are non-negative real numbers, x=1−y−z≤1. Furthermore, since squares are non-negative, (x−21​)2≥0⇒x2−x+41​≥0⇒41​≥x(1−x). We can cube both sides of the inequality since x(1−x)≥0, and both terms are positive. This gives 641​≥x3(1−x)3=x3(y+z)3=x3y3+3x3y2z+3x3yz2+x3z3. Observe that the right hand side is very close to the quantity that we want to maximize. In fact, since 0≤y,z≤x, we have 3x3yz2≥3(y2z)yz2=3y3z3≥y3z3. Thus, 641​≥x3y3+3x3y2z+3x3yz2+x3z3≥x3y3+y3z3+z3x3. In order for equality to hold throughout, we need x=21​ in the first inequality, and z=0 in the last inequality. We can check that the maximium 641​ is attained at (x,y,z)=(21​,21​,0). Hence, a+b=1+64=65. Note: This inequality is tricky because the maximum is attained at (21​,21​,0) and its permutations, hence standard approaches will tend to fail to work. Let x,y,z be real numbers such that x2+y2+z2+(x+y+z)2=9 and xyz≤3215​. To 2 decimal places, what is the greatest possible value of x? The correct answer is: 2.5 Suppose a, b, and c are non-negative real numbers with a+b+c=1. The largest possible value of the expression ab2+bc2+ca2 can be written as mn​, where n and m are coprime positive integers. What is the value of n+m? The correct answer is: 31 What is the smallest real number k (to 3 decimal places), such that for all ordered triples of non-negative reals (a,b,c) which satisfy a+b+c=1, we have 1−c​a​+1−a​b​+1−b​c​≤1+k? The correct answer is: 0.250 If a,b,c are non-negative real numbers, what is the maximum value of (a+b+c)3ab2+bc2+ca2​? The correct answer is: 0.148148148 1 0 0.25 6.25E-2 Consider all pairs of real numbers such that a+b=3. What is the minimum value of a2b2−2a2b−2ab2+a2+4ab+b2−2a−2b+1? The correct answer is: 0 Let a triangle with sides of length a,b,c have perimeter 2. What is the maximum value of k such that b1−a​+c1−b​+a1−c​≥k is always true? Prove your claim. The correct answer is: 1.000 Over all positive triples of real numbers, what is the largest value of k (to 2 decimal places) such that b+ca​​+c+ab​​+a+bc​​≥k? The correct answer is: 2.00 For real numbers a and b such that a>b>0, what is the minimum value of a+b(a−b)1​? Cite as: Inequalities with Strange Equality Conditions. Brilliant.org. Retrieved 17:05, August 23, 2025, from
187804
https://www.dictionary.com/browse/fanciful
Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement View synonyms for fanciful fanciful [fan-si-fuhl] adjective characterized by or showing fancy; capricious or whimsical in appearance. a fanciful design of butterflies and flowers. 2. suggested by fancy; imaginary; unreal. fanciful lands of romance. Synonyms: illusory, baseless, visionary 3. led by fancy rather than by reason and experience; whimsical. a fanciful mind. fanciful / ˈfænsɪfʊl/ adjective not based on fact; dubious or imaginary fanciful notions 2. made or designed in a curious, intricate, or imaginative way 3. indulging in or influenced by fancy; whimsical Discover More Other Word Forms fancifully adverb fancifulness noun overfanciful adjective overfancifully adverb overfancifulness noun unfanciful adjective Discover More Word History and Origins Origin offanciful1 First recorded in 1620–30; fancy + -ful Discover More Example Sentences Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com. The continent may call, where he could find a set-up that suits him, but the notion of a big Premier League post is fanciful in the extreme. FromBBC The possibility of having to turn, cap in hand, to the International Monetary Fund for a loan or to require intervention from the European Central Bank, is no longer fanciful. FromBBC In response to the petition, one Supreme Court judge on the two-judge bench called the allegations "fanciful ideas". FromBBC To think treasure hunters could be making the journey in the pitch-black sounds fanciful and yet this site has fallen victim to nighthawking multiple times. FromBBC The all-ages appeal of the museum is a testament to the everlasting approach of the couple’s narratives, which handle difficult life moments with a fanciful nature, but never hold your hand. FromLos Angeles Times Advertisement Discover More Related Words absurd bizarre extravagant fantastic fantastical www.thesaurus.com fictional imaginative offbeat preposterous unreal whimsical Advertisement Advertisement Advertisement fancierfancify
187805
https://general.chemistrysteps.com/the-effect-of-a-common-ion-on-solubility/
Skip to content Chemistry Steps General Chemistry ### Acid–Base and Solubility Equilibria The Effect of a Common Ion on Solubility In the previous post, we talked about the solubility and solubility product constant (Ksp) of ionic compounds with low solubility. For example, the dissolution equation and the Ksp for CaF2 are: CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) Ksp = [Ca2+][ F–]2 = 3.9 x 10-11 By assigning x mol/L as the concentration of CaF2 dissolved in a saturated solution, we were able to determine the molar solubility of CaF2 from the Ksp. Setting up an ICE table helps determine the concentrations correctly. x x 2x CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) | | | | --- | | [Ca2+] | [F–] | | Initial | 0 | 0 | | Change | +x | +2x | | Equil | x | 2x | So, the expression for Ksp can be written as: Ksp = [Ca2+][ F–]2 = (x)(2x) = 3.9 x 10-11 Therefore, (x)(2x)2 = 3.9 x 10-11 4x3 = 3.9 x 10-11 x = 2.1 x 10-4 2.1 x 10-4 mol/L is the concentration of dissolved CaF2 because it is in 1:1 ratio with Ca2+, and this is the molar solubility of CaF2. The Effect of a Common Ion on Solubility Let’s now assume that the solution of CaF2 contains Ca(NO3)2 with a concentration of 0.20 M. What is the molar solubility of CaF2 in this solution? Before doing the calculations, we can predict the solubility will go down because Ca(NO3)2is a strong electrolyte and completely dissociates into Ca2+ and NO3– ions. Importantly, the Ca2+ is also formed when CaF2 dissolves in water, so it is a common ion, and as we discussed earlier it pushes the equilibrium to the opposite direction according to the Le Chtelier’s principle. And now, let’s calculate the molar solubility of CaF2. It is going to be the same procedure, except the initial concentration of Ca2+ ions is not zero since the dissociation of Ca(NO3)2 produces an equivalent amount of the cation. 0.20 0.20 0.40Ca(NO3)2 (aq) ⇆ Ca2+(aq) + 2NO3–(aq) x x 2x CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) | | | | --- | | [Ca2+] | [F–] | | Initial | 0.20 | 0 | | Change | +x | +2x | | Equil | 0.20 + x | 2x | So, the expression for Ksp is: Ksp = [Ca2+][ F–]2 = (0.20 + x)(2x)2 = 3.9 x 10-11 Now, because the ionization of CaF2 is negligible compared to the one of Ca(NO3)2, we assume that 0.20 + x ≈ 0.20, and the simplified equation will be: (0.20)(2x)2 = 3.9 x 10-11 0.80x2 = 3.9 x 10-11 Therefore, x = 7.0 x 10-6 The x is very small compared to 0.20, and therefore, the approximation was valid, and 7.0 x 10-6 M is the molar solubility of CaF2 in the presence of 0.20 M Ca(NO3)2. As expected it is lower than the solubility of CaF2 in pure water (2.1 x 10-4). Let’s do another example, where the solution contains 0.15 M NaF. In this case, the common ion is F- and therefore, its initial concentration is going to be equal to 0.15 M since NaF is a strong electrolyte and completely dissociates in aqueous solutions. 0.15 0.15 0.15NaF(aq) ⇆ Na+(aq) + F–(aq) x x 2x CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) | | | | --- | | [Ca2+] | [F–] | | Initial | 0 | 0.15 | | Change | +x | +2x | | Equil | x | 0.15 + 2x | So, the expression for Ksp is: Ksp = [Ca2+][ F–]2 = (x)(0.15 + 2x)2 = 3.9 x 10-11 Now, because the ionization of CaF2 is negligible compared to the one of NaF, we assume that 0.15 + x ≈ 0.15, and the simplified equation will be: (x)(0.15)2 = 3.9 x 10-11 0.0225x = 3.9 x 10-11 Therefore, x = 1.7 x 10-9 The x is very small compared to 0.15, and therefore, the approximation was valid, and 1.7 x 10-9 M is the molar solubility of CaF2 in the presence of 0.15 M NaF. As expected it is lower than the solubility of CaF2 in pure water (2.1 x 10-4). So, to summarize, remember that in general, the solubility of a slightly soluble ionic compound decreases with the presence of a common ion in the solution. Check Also Buffer Solutions The Henderson–Hasselbalch Equation The pH of a Buffer Solution Preparing a Buffer with a Specific pH The Common Ion Effect The pH and pKa Relationship Strong Acid–Strong Base Titrations Titration of a Weak Acid by a Strong Base Titration of a Weak Base by a Strong Acid Titration of Polyprotic Acids Buffer Solutions Practice Problems Kspand Molar Solubility The Effect of pH on Solubility Will a Precipitate Form? Ksp and Q Kspand Molar Solubility Practice Problems Leave a Comment Cancel reply
187806
https://byjus.com/physics/isothermal-process/
Physics, studies systems and objects to measure their temperatures, motions and other physical characteristics. It can be applied to anything from single-celled organisms to mechanical systems to galaxies, stars and planets and the processes that govern them. In physics, thermodynamics is a branch that deals with the relationships between heat energy and other forms of energy. It describes how thermal energy is converted into other forms of energy and how it affects matter. An isothermal process is a thermodynamic process in which the temperature of a system remains constant. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. At a particular constant temperature, the change of a substance, object or system is known as the Isothermal Process. Usually, there are two phenomena under which this process can take place. If a system is in contact with a thermal reservoir from outside, then, to maintain thermal equilibrium, the system slowly adjusts itself with the temperature of the reservoir through heat exchange. In contrast, in another phenomenon, no heat transfer occurs between a system and its surrounding. In this process, the temperature of the system is changed in order to keep the heat constant. This process is known as the Adiabatic Process. Isothermal and Adiabatic Process DifferenceExamples of Isothermal ProcessWhat is Boyles Law Difference Between Isothermal and Adiabatic Process An isothermal process is a process that occurs under constant temperature but other parameters of the system can be changed accordingly. On the other hand, in an adiabatic process, heat transfer occurs to keep the temperature constant. The main difference between isothermal and adiabatic process is that the isothermal process occurs under constant temperature, while the adiabatic process occurs under varying temperature. The work done in an isothermal process is due to the change in the net heat content of the system. Meanwhile, the work done in an adiabatic process is due to the change in its internal energy. Examples of Isothermal Process An isothermal process occurs in systems that have some means of regulating the temperature. This process occurs in systems ranging from highly structured machines to living cells. A few examples of an isothermal process are given below. What is Boyle’s Law? An isothermal process is of special interest for ideal gasses. An ideal gas is a hypothetical gas whose molecules don’t interact and face an elastic collision with each other. Joule’s second law states that the internal energy of a fixed amount of an ideal gas only depends on the temperature. Thus, the internal energy of an ideal gas in an isothermal process is constant. In an isothermal condition, for an ideal gas, the product of Pressure and Volume (PV) is constant. This is known as Boyle’s law. Physicist and chemist Robert Boyle published this law in 1662. Boyle’s law is often termed as Boyle–Mariotte law, or Mariotte’s law because French physicist Edme Mariotte independently discovered the same law in 1679. Boyle’s Law Equation The absolute pressure exerted by an object of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system. There are a couple of ways in which the above-stated law can be expressed. The most basic way is given as follows: PV = k where P is the pressure, V is the volume and k is a constant. The law can also be used to find the volume and pressure of a system when the temperature is held constant in the system as follows: PiVi = Pf Vf where, The way people breathe and exhale air out of their lungs can be explained by Boyle’s Law. When the diaphragm contracts and expands, lung volume decreases and increases respectively, changing the air pressure inside them. The pressure difference between the interior of the lungs and the external air produces either inhalation or exhalation. Recommended Video When scientists study isothermal processes in systems, they examine heat and energy and their relation and also the mechanical energy it takes to change or maintain the temperature of a system. Such understanding helps biologists study the regulation of temperature in living organisms. It also comes into play in planetary science, space science, engineering, geology, and many other branches of science. Visit BYJU’S to learn more Physics concepts. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Physics related queries and study materials Your result is as below Request OTP on Voice Call Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Request OTP on Voice Call Website Post My Comment Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
187807
https://physics.stackexchange.com/questions/446658/regarding-the-mathbfe-%C3%97-mathbfb-drift-in-the-earths-magnetic-field
electromagnetism - Regarding the $\mathbf{E} × \mathbf{B} $ drift in the Earth's magnetic field - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Regarding the E×B E×B drift in the Earth's magnetic field Ask Question Asked 6 years, 9 months ago Modified4 months ago Viewed 891 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. So I have a burning question: The only reason that the E×B E×B drift doesn't generate an electric current is because both the electrons and the positive ions move towards the same direction (towards Earth's ionosphere) therefore a charge separation isn't formed? Are opposite velocities the deal breaker for charge separation or am I missing something else? electromagnetism atmospheric-science plasma-physics geomagnetism Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Mar 30, 2024 at 11:54 Sancol. 958 1 1 gold badge 6 6 silver badges 20 20 bronze badges asked Dec 11, 2018 at 17:36 Lysandros BafaloukosLysandros Bafaloukos 51 3 3 bronze badges 2 Can you explain what you mean by ExB drift? I know this as the momentum or equivalently the energy transport of the EM field.my2cts –my2cts 2018-12-11 18:40:02 +00:00 Commented Dec 11, 2018 at 18:40 1 The force that comes as the product of the electrical current of the magnetotail being perpendicular to Earth's magnetic field that results in the particles drifting towards Earth Lysandros Bafaloukos –Lysandros Bafaloukos 2018-12-11 18:43:04 +00:00 Commented Dec 11, 2018 at 18:43 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The only reason that the E x B drift doesn't generate an electric current is because both the electrons and the positive ions move towards the same direction... The ExB-drift is independent of charge, so yes, both ions and electrons will undergo the same ExB-drift velocity. ...therefore a charge separation isn't formed? Counter streaming particles of opposite charge do not a charge separation make. Rather, such a scenario leads to a current given by: j=∑s n s q s V s j=∑s n s q s V s where n s n s is the number density, q s q s is the total charge (including sign), and V s V s is the bulk velocity vector of species s s. As an example, if electrons and protons flow in the same direction both at V o V o, there would be zero net current (assuming quasi-neutrality) because the two species have the opposite charge. If they flow relative to each other, i.e., V e≠V i V e≠V i, then there will be a net current. Are opposite velocities the deal breaker for charge separation or am I missing something else? Charge separation in a plasma is generally difficult because it results in electric fields. Since a plasma is an ionized gas of freely moving electrons and ions, any electric field will quickly act to eliminate itself by doing work on the charged particles. Most plasmas have incredibly high conductivities, so the lifetime of charge separations are very short. I think you are confusing flow with charge separation, which are two different phenomena. Particle species in plasmas can flow relative to each other without generating any charge separation. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Dec 15, 2018 at 16:19 honeste_viverehoneste_vivere 16.2k 4 4 gold badges 44 44 silver badges 153 153 bronze badges Add a comment| This answer is useful 0 Save this answer. Show activity on this post. In uniform crossed magnetic and electric fields, the motion of a charge can be separated into a gyromotion and a constant drift. A constant motion implies zero force, so the charge doesn't affect the drift velocity. The drift velocity is derived in George K. Parks, Physics of Space Plasmas (1991) pages 93-97, and the derivation is a bit too long to reproduce here, so I'll just give the answer: v=E×B B 2.v=E×B B 2. Note that charge doesn't appear in this expression. The Lorentz force on a particle with charge q q is F=q E+q v×B.F=q E+q v×B. If you substitute the drift velocity in this equation and use (E×B)×B=(E⋅B)B−B 2 E(E×B)×B=(E⋅B)B−B 2 E), you can confirm that F=0.F=0. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Dec 13, 2018 at 17:16 A. NewellA. Newell 574 2 2 silver badges 10 10 bronze badges 2 But the E x B field actually acts on the particles and it propels them towards the ionosphere the fact that it doesn't distinguishes between the two charges is irrelevant Lysandros Bafaloukos –Lysandros Bafaloukos 2018-12-13 17:24:29 +00:00 Commented Dec 13, 2018 at 17:24 You wanted to know why there is no charge separation, and the answer is that the opposite charges are drifting in the same direction - on average. The E×B E×B drift applies to the center of the gyromagnetic motion, not the instantaneous motion of each charge.A. Newell –A. Newell 2018-12-14 17:24:35 +00:00 Commented Dec 14, 2018 at 17:24 Add a comment| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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187808
https://math.stackexchange.com/questions/4988288/proof-that-a2b2-2ab
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Proof that $a^2+b^2 > 2ab$ Ask Question Asked Modified 11 months ago Viewed 285 times 0 $\begingroup$ I'm going through Cunningham's A Logical Introduction to Proof. An exercise is asking to prove that $a^2+b^2 > 2ab$, by letting $a$ and $b$ be distinct numbers and using the fact $x^2 >0$ I understand $a^2+b^2 > 2ab$ is related to the extended form of $(a-b)^2$, but using something like $a^2+b^2 - 2ab > 0$ to start my proof feels like going back to what I'm being asked to prove. What I know so far tells me that I can take $a^2 > 0$, but if I wanted to get $a^2 + b^2$, wouldn't that turn my inequality in $a^2 + b^2 > b^2$ (by the law that states $a+c < b+c$)? Where does the $2ab$ derives from and how can I even start the proof? inequality proof-writing Share edited Oct 22, 2024 at 17:32 andalodandalod asked Oct 22, 2024 at 17:31 andalodandalod 3155 bronze badges $\endgroup$ 4 2 $\begingroup$ You're making it harder than it is. $(a-b)^2 \geq 0$. And thus...? $\endgroup$ David G. Stork – David G. Stork 2024-10-22 17:34:36 +00:00 Commented Oct 22, 2024 at 17:34 1 $\begingroup$ $a^2+b^2-2ab=(a-b)^2>0$ for $a\neq b$. For a different solution, see this post. $\endgroup$ Dietrich Burde – Dietrich Burde 2024-10-22 17:36:54 +00:00 Commented Oct 22, 2024 at 17:36 $\begingroup$ You don't know $a\neq0,$ so you can't say $a^2>0.$ You do know $a-b\neq0,$ however. $\endgroup$ Thomas Andrews – Thomas Andrews 2024-10-22 17:36:59 +00:00 Commented Oct 22, 2024 at 17:36 1 $\begingroup$ "...feels like going back to what I'm being asked to prove" : it is not circular reasoning. Instead, it is perfectly valid. $\endgroup$ user2661923 – user2661923 2024-10-22 18:36:45 +00:00 Commented Oct 22, 2024 at 18:36 Add a comment | 3 Answers 3 Reset to default 4 $\begingroup$ You may be experiencing the difference between discovering a proof and articulating it. This is possibly the result of a long-lived movement in mathematics education: the tendency to present results as though they were handed down from on high, without much motivation. From that perspective, there's no problem, because although it may seem arbitrary to start your proof with "$a \not= b$, therefore $(a-b)^2 > 0$"€”for some people, it's not clear why you would know to start with that€”it's not actually logically prior to your desired conclusion. You're not assuming that which you want to prove. However, for those people, it does seem epistemologically prior, in the sense that they don't see how to motivate starting that way. All I can offer in the way of guidance there is that this seems to me mostly a matter of experience: You see those terms, and your mind will eventually more or less immediately leap to that approach. There's no particular short cut to that state of mind, I'm afraid. Share edited Oct 22, 2024 at 18:01 answered Oct 22, 2024 at 17:40 Brian TungBrian Tung 35.6k33 gold badges4646 silver badges8282 bronze badges $\endgroup$ 0 Add a comment | 3 $\begingroup$ WARNING: possibly wildly overkill, but I literally don't even care (because I think it will help OP). Similar to Brian Tung's answer. There is a difference between doing scratch work and writing a logical proof. Scratch work is stuff you quickly scribble down to see what's going on and get ideas going to try to prove the statement/proposition. Often with scratch work for algebraic proofs like the one here, we start with the statement we are trying to prove and try to end up at a familiar true statement. Note that this does not give us a formal proof, however, since in a formal proof, we cannot start by assuming the thing we are trying to prove is true (as that's cheating!). However, before we even do scratch work, we need to be crystal clear of the precise statement we are trying to prove. Statement we are trying to prove (A.K.A. "Proposition"): If $a\neq b,$ then $a^2+b^2 > 2ab.$ Scratch work: $$a^2 +b^2 > 2ab$$ $$+(-2ab) \text{ to both sides: }\quad a^2 + b^2 -2ab > 0 $$ $$ (a-b)^2 > 0.$$ We know that last statement is true, so we have sort of got somewhere, maybe. However, this is not yet a logical proof. It is a bunch of (maybe) equivalent statements that haven't yet been joined up in a logical argument (which is all that a proof is at the end of the day). In our proof proper, we have to start with facts we know to be true end up with "Therefore it is true that $a^2 +b^2 > 2ab$ whenever $a\neq b$". Most people who are new to proofs would do well to write something like the following as a solution: $$ (a-b)^2 = a^2 + b^2 -2ab $$ $$ (a-b)^2 \text{ is always } > 0, \text{ therefore } a^2 + b^2 -2ab > 0,$$ $$ \text{ therefore, by adding } 2ab \text{ to both sides, } a^2 + b^2 > 2ab. $$ However, this is not perfect as it is missing details, like working/reasoning of some steps, and it's also missing all-important implication signs or logical indications like "therefore" or "because". Better is: We are trying to prove that, If $a\neq b,$ then $a^2+b^2 > 2ab.$ So suppose that $a,b\in\mathbb{R}$ such that $a\neq b.$ [We aim to show that $a^2+b^2 > 2ab.$] $$ \text{ By expanding brackets, for all } a\in\mathbb{R},\ b\in\mathbb{R},\quad (a-b)^2 = (a-b)(a-b) = a(a-b) -b(a-b)=a^2 -ab -ba+b^2= a^2 + b^2 -2ab. $$ $$ \text{In particular, for all } a\in\mathbb{R},\ b\in\mathbb{R}, \text{ with } a\neq b,\quad (a-b)^2 = a^2 + b^2 -2ab. $$ $$ \text{ Since } a\neq b,\ \text{ it follows that } a-b\neq 0, \text{ and so either } a-b <0 \text{ or } a-b>0. \text{ Either way, we have: } (a-b)^2=(a-b)(a-b),$$ $$ \text{ which is either two positive numbers multiplied together or two negative numbers multiplied together, and therefore is always } > 0.$$ $$\text{ In summary, } (a-b)^2 >0. \text{ Therefore, } a^2 + b^2 -2ab \text{ is always }> 0.$$ $$ \text{ Therefore, by adding } 2ab \text{ to both sides, } a^2 + b^2 > 2ab. $$ This is overkill, but it's better to be overkill with proofs and then later on reel them in by making them more concise, but only once you are confident that the steps you intend on omitting from the proof are correct. So you see that formulating algebraic proofs takes several stages. Clearly state the proposition you wish to prove. Jot down ideas and scratch work that hopefully sort-of prove the proposition (albeit not in a logical manner, but maybe it would be logical if the order of the steps are reversed). Try to arrange all your ideas in a logically deductive proof. Check to make sure your proof is valid and sound, and that you have started with statements you already know to be true like axioms, and then the proof finishes with the proposition being shown to be a true statement. Keep in mind that proofs are usually a work in progress and can always be improved in clarity and brevity. For example, my Proposition statement itself can be clarified to: "If $a,b\in\mathbb{R}$ and $a\neq b,$ then $a^2+b^2 > 2ab$", since one might wonder from the proposition I originally wrote if $a$ and $b$ could be allowed to be complex numbers or something else. As you get better at proofs, sometimes you can omit some of the details, although you should not omit the logical implication part of your proof, because, for example, $A\implies B$ is not the same as $B \implies A.$ Share edited Oct 23, 2024 at 12:06 answered Oct 22, 2024 at 18:58 Adam RubinsonAdam Rubinson 24.8k66 gold badges2929 silver badges6969 bronze badges $\endgroup$ 0 Add a comment | 1 $\begingroup$ Work with equivalence signs: $a^2+b^2>2ab \Leftrightarrow a^2-2ab+b^2>0 \Leftrightarrow (a-b)^2 > 0 \Leftrightarrow a \neq b$ Share answered Oct 22, 2024 at 18:03 VosoniVosoni 27111 silver badge88 bronze badges $\endgroup$ Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inequality proof-writing See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 3 Showing $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$? Related 1 Question about Logic Proof 2 Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ 1 How to show ¬ in terms of ˆ¨ and ˆ§ 4 Proof: $a < \sqrt{ab} < \frac{a+b}{2} < b$ 1 Is this proof style legitimate? Can anyone clarify the rules for $\forall$ intro and elimination, and $\exists$ intro and elimination? 4 How do you know when trying to find a proof/counterexample isn't worth it? 0 Check proof that any integer can be written as the difference of two positive integers 3 What is wrong with this proof by contradiction? Hot Network Questions Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator What meal can come next? 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187809
https://arxiv.org/pdf/2012.05198
Published Time: Mon, 23 Jan 2023 07:53:35 GMT On cyclic and nontransitive probabilities ∗ Pavle Vuksanovic † A.J. Hildebrand ‡ August 30, 2020 Abstract Motivated by classical nontransitivity paradoxes, we call an n-tuple ( x1, . . . , x n) ∈ [0 , 1] n cyclic if there exist independent random variables U1, . . . , U n with P (Ui = Uj ) = 0 for i 6 = j such that P (Ui+1 > U i) = xi for i = 1 , . . . , n − 1 and P (U1 > U n) = xn. We call the tuple ( x1, . . . , x n) nontransitive if it is cyclic and in addition satisfies xi > 1/2for all i.Let pn (resp. p∗ n ) denote the probability that a randomly chosen n-tuple ( x1, . . . , x n) ∈ [0 , 1] n is cyclic (resp. nontransitive). We determine p3 and p∗ 3 exactly, while for n ≥ 4we give upper and lower bounds for pn that show that pn converges to 1 as n → ∞ .We also determine the distribution of the smallest, middle, and largest elements in a cyclic triple. 1 Introduction A classic example of a nontransitive probability paradox is provided by the Efron dice , a set of four dice invented by Bradley Efron and popularized by Martin Gardner . The Efron dice are six-sided dice with face values given as follows: A = {0, 0, 4, 4, 4, 4}, B = {1, 1, 1, 5, 5, 5},C = {2, 2, 2, 2, 6, 6}, D = {3, 3, 3, 3, 3, 3}. (1.1) One can easily check that, with probability 2 /3 each, B beats A, C beats B, and D beats C while, at the same time, A beats D with probability 2 /3. In this sense, the dice A, B, C, D form a nontransitive cycle. More formally, if we let A, B, C, D denote independent discrete random variables that are uniformly distributed over the values listed in (1.1), then these variables satisfy (1.2) P (B > A ) = P (C > B ) = P (D > C ) = P (A > D ) = 23. ∗ Subject Classification: 60C05, 91A60 † University of Illinois, email pavlev2@illinois.edu ‡ University of Illinois, email ajh@illinois.edu (corresponding author) 1 arXiv:2012.05198v3 [math.PR] 24 Feb 2021 Another classic example of nontransitivity is given by the following set of three-sided dice which seems to have first appeared (in a different, but equivalent, context) in a paper by Moon and Moser : (1.3) A = {1, 5, 9}, B = {2, 6, 7}, C = {3, 4, 8}. The three-sided dice A, B, C defined by (1.3) form a nontransitive cycle with probabilities (1.4) P (B > A ) = P (C > B ) = P (A > C ) = 59. There now exists a large body of research motivated by such nontransitivity phenomena; see [2, 5, 7, 9, 10, 11, 13, 15] for some recent work on this subject. It is natural to ask what the most “extreme” level of nontransitivity is that can be achieved with constructions such as the Efron and Moon-Moser dice. Can one replace the probabilities 2 /3 and 5 /9 in (1.2) and (1.4) by even larger numbers? To formalize this question, one can consider, for each integer n ≥ 3, the quantity (1.5) πn = max min ( P (U2 > U 1), . . . , P (Un > U n−1), P (U1 > U n)) , where the maximum is taken over all sets of independent random variables U1, . . . , U n. The Efron and Moon-Moser dice constructions show that π4 ≥ 2/3 and π3 ≥ 5/9. The quantities πn were first investigated in the 1960s by Steinhaus and Trybula , Try-bula [21, 22], Chang , and Usiskin who showed that lim n→∞ πn = 3 /4 and determined the first few values of πn. It particular, the values of π3 and π4 are given by (1.6) π3 = √5 − 12 , π4 = 23. Thus the Efron dice construction is best-possible in the sense of achieving the value πn for n = 4; meanwhile, the Moon-Moser dice construction for n = 3 is not best-possible as 5/9 < (√5 − 1) /2. More recently (see, e.g., [3, 11, 17]) it was shown that, for any n ≥ 3, (1.7) πn = 1 − 14 cos 2(π/ (n + 2)) . Interestingly, the numbers πn defined by (1.5) have come up independently in very different contexts such as graph theory [4, 14, 17] and theoretical computer science [12, 20]. In this paper, we focus on another aspect of the nontransitivity phenomenon that has received less attention in the literature, namely the question of which n-tuples can be realized as a tuple of cyclic probabilities (P (U2 > U 1), . . . , P (Un > U n−1), P (U1 > U n)), and how common such tuples are among all n-tuples in [0 , 1] n. We introduce the following definitions: Definition 1.1 (Cyclic and nontransitive tuples) . Let n be an integer with n ≥ 3. (i) An n-tuple ( x1, . . . , x n) ∈ [0 , 1] n is called cyclic if there exist independent random variables U1, . . . , U n with P (Ui = Uj ) = 0 for i 6 = j such that (1.8) P (Ui+1 > U i) = xi (i = 1 , . . . , n − 1) and P (U1 > U n) = xn. 2(ii) An n-tuple ( x1, . . . , x n) ∈ [0 , 1] n is called nontransitive if it is cyclic and satisfies xi > 1/2 for all i.The dice examples (1.1) and (1.3) show that the 4-tuple (2 /3, 2/3, 2/3, 2/3) and the triple (5 /9, 5/9, 5/9) are nontransitive and, hence, also cyclic. The n-tuple (1 /2, 1/2, . . . , 1/2) is cyclic for any n ≥ 3 as can be seen by taking U1, . . . , U n to be independent random variables with a continuous common distribution. On the other hand, the tuple (1 , 1, . . . , 1) is not cyclic. Indeed, otherwise there would exist random variables Ui such that, with probability 1, both U1 < U 2 < · · · < U n and Un < U 1 hold. But the first of these relations implies that U1 < U n holds with probability 1, contradicting the second relation. A similar contradiction arises whenever the components of ( x1, . . . , x n) are sufficiently close to 1. The first non-trivial case of Definition 1.1 is the case n = 3, i.e., the case of cyclic triples . For this case, Trybula and, independently, Suck , gave necessary and sufficient conditions for a triple ( x, y, z ) to be cyclic (see Lemma 2.3 below). For n ≥ 4 a complete characterization of cyclic n-tuples is not known, though some partial results are known (see, for example, Trybula ). In this paper we consider the question of how common cyclic and nontransitive tuples are among all tuples in the n-dimensional unit cube [0 , 1] n. Let pn (resp. p∗ n ) be the probability that an n-tuple ( x1, . . . , x n) chosen randomly and uniformly from [0 , 1] n is cyclic (resp. non-transitive). These probabilities are given by the volumes of the regions Cn (resp. C ∗ n ) of cyclic (resp. nontransitive) n-tuples inside the unit cube [0 , 1] n; that is, we have (1.9) pn = vol( Cn), p∗ n = vol(C ∗ n ), where Cn = {(x1, . . . , x n) ∈ [0 , 1] n : (x1, . . . , x n) is cyclic },(1.10) C∗ n = {(x1, . . . , x n) ∈ [0 , 1] n : (x1, . . . , x n) is nontransitive }.(1.11) In our first theorem, we determine the probabilities p3 and p∗ 3 exactly. Theorem 1. (i) A random triple (x1, x 2, x 3) in [0 , 1] 3 is cyclic with probability p3 = 11 √54 − 17 4 − 6 ln( √5 − 1) ≈ 0.627575 . . . . (1.12) (ii) A random triple (x1, x 2, x 3) in [0 , 1] 3 is nontransitive with probability p∗ 3 = 11 √58 − 43 16 − 3 ln( √5 − 1) + 3 ln 2 8 ≈ 0.011217 . . . (1.13) In particular, the theorem shows that a random triple in [0 , 1] 3 is more likely than not to be cyclic, while only about 1% of all such triples are nontransitive. In our second theorem, we consider the case of general n ≥ 4. We derive upper and lower bounds for the probabilities pn that show that pn converges to 1 as n → ∞ .3Theorem 2. For any integer n ≥ 4 the probability pn that a random n-tuple (x1, . . . , x n) ∈ [0 , 1] n is cyclic satisfies (1.14) 1 − 3 ( 2 π )n ≤ pn ≤ 1 − 2 (14 )n . In particular, pn converges exponentially to 1 as n → ∞ . In our final result we consider the distribution of the smallest, middle, and largest elements of a cyclic triple. Let ( X1, X 2, X 3) be a random vector that is uniformly dis-tributed over the region C3 of all cyclic triples, and let ( X∗ 1 , X ∗ 2 , X ∗ 3 ) be the order statistics of ( X1, X 2, X 3). Thus, Xi is the i-th coordinate of a random cyclic triple, while X∗ i is the i-th smallest among the three coordinates of a random cyclic triple, where “random” is to be interpreted with respect to the usual Lebesgue measure on C3.In particular, X∗ 1 is the smallest element of a random cyclic triple, and the triple ( X1, X 2, X 3)is nontransitive if and only if X∗ 1 1/2. Thus it is of interest to determine the precise dis-tribution of the random variable X∗ 1 . More generally, in the following theorem we determine the density function fi(x) for each of the three random variables X∗ i , i = 1 , 2, 3. Theorem 3. Let f1(x), f2(x), and f3(x) denote, respectively, the density function of the smallest, middle, and largest element of a random cyclic triple; i.e., fi(x) is the density of the random variable X∗ i defined above. Then: f1(x) =  3 p3 (x3 − 3x2 + 1−x 2−x − (1 − x) ln(1 − x)) if 0 ≤ x ≤ 3−√52 , 3 p3 (x2 − 3x + 1 − (1 − x) ln(1 − x)) if 3−√52 < x ≤ 12 , 3 p3 (x2 + x − 1 + (1 − x) ln(1 − x) − 2(1 − x) ln x) if 12 < x ≤ √5−12 , 0 otherwise; (1.15) f2(x) =  3 p3 (3 x2 − x3) if 0 ≤ x ≤ 3−√52 , 6 p3 ( 3x − x2 − 12(1 −x) ) if 3−√52 < x ≤ 12 , f2(1 − x) if 12 < x ≤ 1, 0 otherwise; (1.16) f3(x) = f1(1 − x),(1.17) where p3 is given by (1.12) . 4Figure 1: The density functions f1(x) and f2(x) of the smallest, respectively middle, value in a random cyclic triple. The dark-shaded portion of the graph on the left corresponds to nontransitive triples. The densities f1(x) and f2(x) are shown in Figure 1. The dark-shaded portion of the graph of f1(x) is the portion of this density corresponding to nontransitive triples. As can be seen from the graph, the contribution of such triples to cyclic triples is quite small. This is consistent with the result of Theorem 1, which shows that nontransitive triples make up a proportion of only around 0 .627575 /0.011217 ≈ 1/55 of all cyclic triples. Formula (1.15) shows that the density f1(x) is supported on the interval [0 , (√5 − 1) /2]. Note that the right endpoint of this interval, ( √5 − 1) /2, is equal to the number π3 defined above (see (1.5) and (1.6)). Figure 1 shows that the density f1(x) is strictly positive on the entire interval [0 , (√5 − 1) /2), and that it has a unique mode (i.e., local maximum) located at around 0 .1. It is interesting to compare the distribution of X∗ 1 , the smallest element in a random cyclic triple in [0 , 1] 3, to that of X∗, the smallest element in an unrestricted random triple in [0 , 1] 3. An easy calculation shows that X∗ has density function f (x) = 3(1 − x)2, mean 1/4, and median 1 − 2−1/3 = 0 .206 . . . . In contrast to the density function f1(x), the latter density function is supported on the entire interval [0 , 1] and is strictly decreasing on this interval. Further statistics on the distributions f1(x) and f (x) are given in Table 1. Random Variable Expected Value Median Mode X∗ 1 0.211 . . . 0.197 . . . 0.107 . . . X∗ 0.25 0.206 . . . 0Table 1: Statistics on the distributions of X∗ 1 , the smallest element in a random cyclic triple, and X∗, the smallest element in a random triple in [0 , 1] 3.The graph on the right of Figure 1 shows the density of the middle value in a random cyclic triple. As can be seen from formula (1.16), this distribution is symmetric with respect to the line x = 1 /2, and it is supported on the full interval [0 , 1]. 5Formula (1.17) shows that the distribution of the largest value in a random cyclic triple, up to a reflection at the line x = 1 /2, is the same as the distribution of the smallest value. This is a consequence of the symmetry properties of cyclic triples (cf. Lemma 2.1 below). The remainder of this paper is organized as follows. In Section 2 we prove some ele-mentary properties of cyclic n-tuples, we present Trybula’s characterization of cyclic triples, and we derive a simplified form of this characterization under additional assumptions on the triples. Sections 3–5 contain the proofs of our main results, Theorems 1–3. We conclude in Section 6 with a discussion of some related questions and open problems suggested by our results. 2 Auxiliary results We begin by deriving some elementary properties of cyclic n-tuples. Here, and in the remain-der of the paper, we make the convention that subscripts of n-tuples are to be interpreted modulo n. Thus, for example, the definition (1.8) of a cyclic n-tuple ( x1, . . . , x n) can be written more concisely as P (Ui+1 > U i) = xi (i = 1 , . . . , n ). Given a real number t ∈ [0 , 1], we write (2.1) t = 1 − t. Lemma 2.1. Let (x1, . . . , x n) ∈ [0 , 1] n be a cyclic n-tuple. Then: (i) Any cyclic permutation of (x1, . . . , x n) is cyclic; that is, for any i ∈ { 1, 2, . . . , n }, the tuple (xi, x i+1 , . . . , x i+n) (with subscripts interpreted modulo n) is cyclic as well. (ii) The “reverse” tuple (xn, x n−1, . . . , x 1) is cyclic. (iii) The “complementary” tuple (x1, . . . , x n) is cyclic. Proof. Part (i) of the lemma follows immediately from the definition (1.8) of cyclic n-tuples. For parts (ii) and (iii) suppose ( x1, . . . , x n) ∈ [0 , 1] n is a cyclic n-tuple with associated random variables U1, . . . , U n satisfying (1.8). Setting U ∗ i = −Un+1 −i we have P (U ∗ i+1 U ∗ i ) = P (−Un−i > −Un+1 −i)= P (Un+1 −i > U n−i) = xn−i (i = 0 , 1, . . . , n − 1) , which shows that the tuple ( xn, x n−1, . . . , x 1) is cyclic. Similarly, the fact that ( x1, . . . , x n) is cyclic follows by letting U ∗ i = −Ui and observing that P (U ∗ i+1 U ∗ i ) = P (−Ui+1 > −Ui)= P (Ui > U i+1 ) = 1 − xi (i = 1 , . . . , n ). This completes the proof of the lemma. 6Lemma 2.2. If (x, y, z ) ∈ [0 , 1] 3 is a cyclic triple, then so is any permutation of (x, y, z ).Proof. Suppose ( x, y, z ) ∈ [0 , 1] 3 is cyclic. By part (i) of Lemma 2.1 the cyclic permutations (y, z, x ) and ( z, x, y ) are also cyclic. By part (ii) the reverse triple ( z, y, x ) is cyclic. Applying part (i) again to the triple ( z, y, x ), we obtain that ( y, x, z ) and ( x, z, y ) are cyclic as well. Hence all permutations of ( x, y, z ) are cyclic. The next result contains Trybula’s characterization of cyclic triples . We state this characterization in the slightly different—though equivalent—version given by Suck . Lemma 2.3 (Trybula [21, Theorem 1]; Suck [19, Theorems 2 and 3]) . A triple (x, y, z ) ∈ [0 , 1] 3 is cyclic if and only if it satisfies the following inequalities: min ( x + yz, y + zx, z + xy ) ≤ 1,(2.2) min ( x + y z, y + z x, z + x y ) ≤ 1.(2.3) The conditions (2.2) and (2.3) in this characterization are rather unwieldy to work with directly. However, by imposing additional constraints on the variables x, y, z, the conditions simplify significantly as the next lemma shows. Lemma 2.4. Let (x, y, z ) ∈ [0 , 1] 3.(i) If x ≤ y ≤ z, then the triple (x, y, z ) is cyclic if and only if it satisfies the following two inequalities: x + yz ≤ 1,(2.4) z + x y ≤ 1.(2.5) (ii) If x = min( x, y, z ) and y, z ≥ 1/2, then (x, y, z ) is cyclic if and only if (2.4) holds. Proof. We need to show that the conditions (2.2) and (2.3) reduce to (2.4) and (2.5) under the assumptions of part (i), and to (2.4) under the assumptions of part (ii). Suppose first that ( x, y, z ) satisfies the conditions of part (i), i.e., that x ≤ y ≤ z. Then x(1 −y) ≤ z(1 −y) and x(1 −z) ≤ y(1 −z), and therefore x+yz ≤ z +xy and x+yz ≤ y +zx .Thus, the minimum on the left of the inequality (2.2) is equal to x + yz , and the inequality therefore holds if and only if x + yz ≤ 1, Similarly, noting that x ≤ y ≤ z is equivalent to z ≤ y ≤ x, we see that condition (2.3) is equivalent to z + x y ≤ 1. This proves part (i) of the lemma. Next, suppose that y, z ≥ 1/2 and x = min( x, y, z ). Then either z = max( x, y, z ) or y = max( x, y, z ). In the first case we have x ≤ y ≤ z, so part (i) of the lemma applies and shows that the triple ( x, y, z ) is cyclic if and only if it satisfies (2.4) and (2.5). But since y, z ≥ 1/2, we have y, z ≤ 1/2 and therefore z + x y ≤ (1 /2) + 1 · (1 /2) = 1. Hence condition (2.5) holds trivially, so ( x, y, z ) is cyclic if and only if (2.4) holds. In the case where we have x ≤ z ≤ y, we note that, by Lemma 2.2, ( x, y, z ) is cyclic if and only if ( x, z, y ) is cyclic. Applying the above argument to ( x, z, y ) then yields the same conclusion. This completes the proof of part (ii). 73 Proof of Theorem 1 By (1.9) we have p3 = vol( C3) and p∗ 3 = vol(C ∗ 3 ), so computing these probabilities amounts to computing the volumes of the regions C3 and C ∗ 3 of cyclic and nontransitive triples. We begin by using the symmetry properties established in Lemmas 2.1 and 2.2 to reduce this computation to one involving simpler regions. Let C(I)3 = {(x, y, z ) ∈ C3 : 1 /2 < x ≤ 1 and x ≤ y, z ≤ 1},(3.1) C(II )3 = {(x, y, z ) ∈ C3 : 0 ≤ x < 1/2 and 1 /2 < y, z ≤ 1}.(3.2) Lemma 3.1. We have vol(C ∗ 3 ) = 3 vol( C(I)3 ),(3.3) vol( C3) = 6 vol( C(I)3 ) + 6 vol( C(II )3 ).(3.4) Proof. By definition, the set C ∗ 3 of nontransitive triples consists of those cyclic triples ( x, y, z )in which all coordinates are > 1/2. Since, by Lemma 2.2, the “cyclic triple” property is invariant with respect to taking permutations, the volume of this set is three times the volume of the set of those triples ( x, y, z ) in C ∗ 3 which satisfy x = min( x, y, z ), i.e., the set C(I)3 . This proves (3.3). To prove (3.4), note first that we may ignore triples ( x, y, z ) in which one of the coordi-nates is equal to 1 /2 as these do not contribute to the volume. We classify the remaining triples ( x, y, z ) in C3 into 8 mutually disjoint classes, according to their signature σ, defined as σ = (sign( x − 1/2) , sign( y − 1/2) , sign( z − 1/2)) , where sign( t) = 1 if t > 0 and sign( t) = −1 if t < 0. For example, a triple ( x, y, z ) with x < 1/2, y > 1/2, z > 1/2 has signature ( −1, 1, 1). Letting C σ 3 denote the set of cyclic triples ( x, y, z ) with signature σ, we then have (3.5) vol( C3) = ∑ σ=( ±1,±1,±1) vol( C σ 3 ), where the sum is over all 8 possible values of the signature σ. Now note that the cyclic triples with signature (1 , 1, 1) are exactly the nontransitive triples, and the cyclic triples with signature ( −1, 1, 1) are exactly those counted in the set C(II )3 . Thus we have vol( C(1 ,1,1) 3 ) = vol(C ∗ 3 ) = 3 vol( C(I)3 ),(3.6) vol( C(−1,1,1) 3 ) = vol( C(II )3 ).(3.7) Next, observe that if ( x, y, z ) has signature σ, then the complementary triple ( x, y, z ) = (1 − x, 1 − y, 1 − z) has signature −σ. Since, by Lemma 2.1(iii), a triple ( x, y, z ) is cyclic if and only if ( x, y, z ) is cyclic, it follows that (3.8) vol( C(−1,−1,−1) 3 ) = vol( C(1 ,1,1) 3 ) = vol(C ∗ 3 ). 8Finally, using Lemma 2.1(iii) along with Lemma 2.2, we see that vol( C(−1,−1,1) 3 ) = vol( C(1 ,−1,−1) 3 ) = vol( C(−1,1,−1) 3 )(3.9) = vol( C(1 ,1,−1) 3 ) = vol( C(1 ,−1,1) 3 ) = vol( C(−1,1,1) 3 )= vol( C(II )3 ). Combining (3.5)–(3.9) yields (3.4). Let (3.10) ω = √5 − 12 = 0 .618034 . . . and note that ω is the positive root of the quadratic equation (3.11) ω2 + ω = 1 . We remark that ω is also equal to the number π3 defined in (1.5) and (1.6), i.e., ω is the largest number for which there exists a cyclic triple ( x, y, z ) with min( x, y, z ) ≥ ω. We will, however, not use this fact in our proof. Lemma 3.2. (i) A triple (x, y, z ) belongs to the set C(I)3 if and only if it satisfies (3.12)  12 < x ≤ ωx ≤ y ≤ 1 − xxx ≤ z ≤ 1 − xy  . (ii) A triple (x, y, z ) belongs to the set C(II )3 if and only if it satisfies (3.13)  0 ≤ x < 1212 < y ≤ 1 − x 12 < z ≤ 1  or  0 ≤ x < 121 − x < y ≤ 112 < z ≤ 1 − xy  . Proof. By definition (see (3.1) and (3.2)) the sets C(I)3 and C(II )3 consist of those cyclic triples (x, y, z ) ∈ [0 , 1] 3 that satisfy 1 /2 < x ≤ y, z ≤ 1 and 0 ≤ x < 1/2 < y, z ≤ 1, respectively. By part (ii) of Lemma 2.4, under either of the latter two conditions, a triple ( x, y, z ) is cyclic if and only if it satisfies x + yz ≤ 1. Next, note that any triple ( x, y, z ) ∈ C(I)3 must satisfy x ≤ y and x ≤ z, so the inequality x + yz ≤ 1 can only hold if x + x2 ≤ 1, i.e., if x ≤ ω, where ω is defined by (3.10) and (3.11). 9For each x in the range 1 /2 < x ≤ ω, the set of pairs ( y, z ) with x ≤ y, z ≤ 1 satisfying x + yz ≤ 1 is nonempty and consists of exactly those pairs that satisfy x ≤ y ≤ (1 − x)/x and x ≤ z ≤ (1 − x)/y . It follows that a triple ( x, y, z ) belongs to C(I)3 if and only if it satisfies (3.12). This proves part (i) of the lemma. For the proof of part (ii), note that C(II )3 is the set of triples ( x, y, z ) satisfying 0 ≤ x < 1/2, 1 /2 < y, z ≤ 1, and x + yz ≤ 1. For any fixed pair ( x, y ) with 0 ≤ x < 1/2 and 1/2 < y ≤ 1, the set of values z with 1 /2 < z ≤ 1 for which x + yz ≤ 1 holds is exactly the interval (1 /2, min((1 − x)/y, 1)]. The latter interval is nonempty, and it reduces to (1 /2, 1] if y ≤ 1 − x, and to (1 /2, (1 − x)/y ] if 1 − x < y ≤ 1. The desired characterization (3.13) now follows. Lemma 3.3. We have (3.14) vol( C(II )3 ) = 316 − ln 2 8 . Proof. Using the characterization (3.13) of the set C(II )3 we obtain vol( C(II )3 ) = ∫ 1/20 ∫ 1−x 1/2 ∫ 11/2 dz dy dx + ∫ 1/20 ∫ 11−x ∫ 1−xy 1/2 dz dy dx = 12 ∫ 1/20 (12 − x ) dx + ∫ 1/20 ∫ 11−x (1 − xy − 12 ) dy dx = 12 ∫ 1/20 (12 − 2x ) dx − ∫ 1/20 (1 − x) ln(1 − x) dx = − ∫ 1/20 (1 − x) ln(1 − x) dx = − [(1 − x)2 4 (1 − 2 ln(1 − x)) ]1/20 = 316 − ln 2 8 . Lemma 3.4. We have (3.15) vol( C(I)3 ) = ln 2 8 − ln(2 ω) + 11 ω 12 − 716 , where ω = ( √5 − 1) /2 is defined as in (3.10) .Proof. Using the characterization (3.12) of the set C(I)3 we obtain, on noting that 1 > (1 − x)/x ≥ x for 1 /2 < x ≤ ω (since ω is the positive root of ω2 = 1 − ω) and 1 > (1 − x)/y ≥ x for x ≤ y ≤ (1 − x)/x ,vol( C(I)3 ) = ∫ ω 1/2 ∫ 1−xx x ∫ 1−xy x dz dy dx = ∫ ω 1/2 ∫ 1−xx x (1 − xy − x ) dy dx (3.16) = ∫ ω 1/2 (1 − x) ln (1 − xx2 ) dx − ∫ ω 1/2 x (1 − xx − x ) dx = I1 − I2, 10 say. The integrals I1 and I2 can be evaluated as follows, using the relations (see (3.11)) ω2 = 1 − ω and ω3 = ω − ω2 = 2 ω − 1: I1 = [−(1 − x)2 2 ln (1 − xx2 )] ω 1/2 + ∫ ω 1/2 (1 − x)2 2 ( − 11 − x − 2 x ) dx (3.17) = −(1 − ω)2 2 ln (1 − ωω2 ) ln 2 8 + ∫ ω 1/2 ( − 1 x + 32 − x 2 ) dx = ln 2 8 − ln(2 ω) + 3( ω − 1/2) 2 − ω2 − 1/44= ln 2 8 − ln (2 ω) + 7ω 4 − 15 16 ,I2 = ∫ ω 1/2 (1 − x − x2) dx = ω − 12 − ω2 − 1/42 − ω3 − 1/83 = 5ω 6 − 12.(3.18) Substituting (3.17) and (3.18) into (3.16) yields (3.15). Proof of Theorem 1. Combining (1.9) and Lemmas 3.1, 3.3, and 3.4, we obtain p∗ 3 = vol(C ∗ 3 ) = 3 vol( C(I)3 )= 3 (ln 2 8 − ln(2 ω) + 11 ω 12 − 716 ) = 3 ( ln 2 8 − ln( √5 − 1) + 11 √5 − 11 24 − 716 ) = 3 ln 2 8 − 3 ln( √5 − 1) + 11 √58 − 43 16 = 0 .01121759 . . . and p3 = vol( C3) = 6 vol( C(I)3 ) + 6 vol( C(II )3 )= 6 ( ln 2 8 − ln( √5 − 1) + 11 √5 − 11 24 − 716 ) 6 ( 316 − ln 2 8 ) = 11 √5 − 17 4 − 6 ln( √5 − 1) = 0 .6275748 . . . These are the desired formulas (1.13) and (1.12). 4 Proof of Theorem 2 Proof of Theorem 2, upper bound. For the upper bound in (1.14), note that, by (1.7), a cyclic tuple ( x1, . . . , x n) ∈ [0 , 1] n must satisfy min( x1, . . . , x n) ≤ πn = 1 − 14 cos 2(π/ (n + 2)) < 34. 11 Therefore any tuple ( x1, . . . , x n) ∈ [3 /4, 1] n is not cyclic. Moreover, since, by Lemma 2.1(iii), a tuple ( x1, . . . , x n) is cyclic if and only if the complementary tuple (1 − x1, . . . , 1 − xn) is cyclic, any tuple ( x1, . . . , x n) satisfying xi ∈ [0 , 1/4] for all i is also not cyclic. Thus the set Cn of cyclic tuples lies in the complement of the sets [3 /4, 1] n and [0 , 1/4] n. Hence we have pn = vol( Cn) ≤ 1 − vol ([3 /4, 1] n) − vol ([0 , 1/4] n) = 1 − 2 (14 )n , which is the desired upper bound. We next turn to the lower bound in (1.14). The argument is based on the following lemma, which gives a sufficient condition for an n-tuple to be cyclic. Recall our convention that subscripts in n-tuples ( x1, . . . , x n) are to be interpreted modulo n. Lemma 4.1. Let (x1, . . . , x n) ∈ [0 , 1] n, and suppose that there exists an index i ∈ { 1, . . . , n } such that (4.1) xi + xi+1 ≥ 1 and xi+2 + xi+3 ≤ 1. Then (x1, . . . , x n) is cyclic. Proof. Since, by Lemma 2.1(i), a cyclic permutation of a cyclic tuple is also cyclic, we may assume without loss of generality that the assumption (4.1) of the lemma holds with i = n−2, i.e., that (4.2) xn−2 + xn−1 ≥ 1 and xn + x1 ≤ 1. Let ( x1, . . . , x n) ∈ [0 , 1] n be an n-tuple satisfying (4.2). Define independent random variables U1, . . . , U n with values in {− n + 1 , −n + 2 , . . . , n + 1 , n + 2 } as follows: (4.3)  P (Un−1 = −n + 1) = 1 − xn−2,P (Ui = −i) = 1 − xi−1 (n − 2 ≥ i ≥ 3) ,P (U2 = −2) = 1 − x1 1 − xn ,P (U1 = 0) = 1 − xn,P (U2 = 2) = x1 1 − xn ,P (Ui = i) = xi−1 (3 ≤ i ≤ n − 2) ,P (Un−1 = n − 1) = xn−1 + xn−2 − 1,P (Un = n) = 1 ,P (U1 = n + 1) = xn,P (Un−1 = n + 2) = 1 − xn−1. The values of the random variables Ui are shown below: Un−1 −(n − 1) . . . Ui −i . . . U2 −2 U1 0 U2 . . Ui i . . . Un−1 n − 1 Un n U1 n + 1 Un−1 n + 2 12 The idea behind this construction is the following: If we let Ui be random variables with values ±i, then Ui+1 > U i holds if and only if Ui+1 = i + 1. Thus, if we require that P (Ui+1 = i + 1) = xi (and hence P (Ui+1 = −i − 1) = 1 − xi), then Ui+1 > U i holds with the desired probability xi. This is indeed how the variables Ui in (4.3) are defined for the “interior” indices i = 3 , . . . , n − 2, so the probabilities P (Ui+1 > U i) have the desired value xi if 2 ≤ i ≤ n − 3. For the remaining indices i = 1 , 2, n − 1, n , the definition of Ui has to be adjusted to ensure that the “wrap-around” probability P (Un > U 1) also has the desired value. The following calculations show that if U1, U 2, U n−1, U n are defined as in (4.3), then the remaining probabilities P (Ui+i > U i) also have the desired values: P (U2 > U 1) = P (U2 = 2) P (U1 = 0) = x1 1 − xn · (1 − xn) = x1,P (Un > U n−1) = 1 − P (Un−1 = n + 2) = 1 − (1 − xn−1) = xn−1,P (Un−1 > U n−2) = 1 − P (Un−1 = −n + 1) = 1 − (1 − xn−2) = xn−2,P (U1 > U n) = P (U1 = n + 1) = xn. The assumption (4.2) ensures that the numbers x1/(1 − xn) and xn−1 + xn−2 − 1 arising in the definition of U2 and Un−1 are contained in the interval [0 , 1] and thus can represent probabilities. Thus the tuple ( x1, . . . , x n) is indeed cyclic. Lemma 4.2. Suppose (x1, . . . , x n) ∈ [0 , 1] n satisfies neither of the conditions (4.4) xi + xi+1 < 1 (i = 1 , . . . , n ) and (4.5) xi + xi+1 > 1 (i = 1 , . . . , n ). Then (x1, . . . , x n) is cyclic. Proof. Let si = xi + xi+1 . By Lemma 4.1, it suffices to show that if neither (4.4) nor (4.5) hold, then there exists an index i ∈ { 1, . . . , n } such that (4.6) si ≥ 1 and si+2 ≤ 1. We split the argument into two cases based on the parity of n.If n is odd, consider the sequence of n numbers s1, s 3, . . . , s n, s 2, s 4, . . . , s n−1. If there is no i for which (4.6) holds, then the numbers in this sequence must be either all strictly greater than 1 or all strictly less than 1. But since this sequence is a permutation of s1, s 2, . . . , s n,it then follows that either (4.4) or (4.5) holds, contradicting the assumption of the lemma. If n is even, consider the two sequences A = {s1, s 3, . . . , s n−1} and B = {s2, s 4, . . . , s n}.If there is no i for which (4.6) holds, we conclude as before that, within each of these two sequences , either all elements are greater than 1 or all elements are less than 1. If the elements 13 of both sequences are all greater than 1 or all less than 1, then (4.4) or (4.5) follows, and we again have a contradiction. In the remaining case the elements of one sequence are all greater than 1 and those of the other sequence are all less than 1. Since each sequence has exactly n/ 2 elements, one of the two sums ∑ a∈A a and ∑ b∈B b must be strictly greater than n/ 2, while the other sum must be strictly less than n/ 2. On the other hand, the identity ∑ a∈A a = n/ 2 ∑ j=1 (x2j−1 + x2j ) = n ∑ i=1 xi = n/ 2 ∑ j=1 (x2j + x2j+1 ) = ∑ b∈B b shows that the two sums ∑ a∈A a and ∑ b∈B b are in fact equal. Thus, this case cannot occur and the proof of the lemma is complete. Define the sets D(I) n = {(x1, . . . , x n) ∈ [0 , 1] n : (x1, . . . , x n) satisfies (4.4) },(4.7) D(II ) n = {(x1, . . . , x n) ∈ [0 , 1] n : (x1, . . . , x n) satisfies (4.5) },(4.8) D∗ n = {(x1, . . . , x n) ∈ D(I) n : x1 = min( x1, . . . , x n)}.(4.9) Lemma 4.3. We have vol( D(I) n ) = vol( D(II ) n ),(4.10) vol( D(I) n ) = n vol( D∗ n ),(4.11) vol( Cn) ≥ 1 − 2n vol( D∗ n ).(4.12) Proof. First note that an n-tuple ( x1, . . . , x n) satisfies condition (4.5) if and only if the complementary n-tuple ( x1, . . . , x n) = (1 − x1, . . . , 1 − xn) satisfies (4.4). The transformation (x1, . . . , x n) → (x1, . . . , x n) then shows that the regions D(I) n and D(II ) n have the same volume. This yields (4.10). For the proof of (4.11), note that if a tuple ( x1, . . . , x n) satisfies condition (4.4), then so does any cyclic permutation of this tuple. Therefore the set D(I) n is invariant with respect to taking cyclic permutations. It follows that the volume of D(I) n is n times that of D∗ n , the set consisting of those tuples in D(I) n with smallest component x1. This proves (4.11). Finally, (4.12) follows on noting that, by Lemma 4.2, the set Cn of cyclic n-tuples contains the complement of the set D(I) n ∪ D(II ) n . Therefore, vol( Cn) ≥ 1 − vol( D(I) n ∪ D(II ) n ) ≥ 1 − vol( D(I) n ) − vol( D(II ) n )= 1 − 2 vol( D(I) n ) = 1 − 2n vol( D∗ n ), where in the last step we have used (4.10) and (4.11). To obtain the desired lower bound for the volume of Cn, it now remains to obtain an appropriate upper bound for the volume of the region D∗ n . The following lemma expresses this volume in terms of a combinatorial quantity counting alternating permutations. Here a permutation of a set of n distinct real numbers x1, . . . , x n is called up-down alternating if it satisfies x1 < x 2 > x 3 < . . . , i.e., if xi+1 − xi is positive for odd i and negative for even i; see Andr´ e for more about such permutations. 14 Lemma 4.4. We have (4.13) vol( D∗ n ) = An−1 (2 n)( n − 1)! , where An−1 is the number of up-down alternating permutations of length n − 1.Proof. By definition, D∗ n is the set of n-tuples ( x1, . . . , x n) ∈ [0 , 1] n satisfying the inequalities (4.4) and in addition x1 = min( x1, . . . , x n). Under the latter condition the last inequality in (4.4), xn + x1 < 1, is implied by the second-last inequality, xn−1 + xn < 1, and thus can be omitted. Moreover, the inequalities (4.4) can only hold if 0 ≤ x1 < 1/2. Thus, the set D∗ n is the set of tuples ( x1, . . . , x n) satisfying (4.14)  0 ≤ x1 < 1/2 x1 ≤ x2 < 1 − x1 ... x1 ≤ xn < 1 − xn−1  . Hence we have vol( D∗ n ) = ∫ 1/20 ∫ 1−x1 x1 ∫ 1−x2 x1 · · · ∫ 1−xn−1 x1 dx n · · · dx 3dx 2dx 1.(4.15) To evaluate the latter integral, we apply the change of variables (4.16) y0 = x1, yi = xi+1 − x1 1 − 2x1 (i = 1 , . . . , n − 1) . It is easily checked that the transformation (4.16) has Jacobian determinant ±(1 − 2y0)n−1 and maps the region (4.14) to the set of n-tuples ( y0, . . . , y n−1) satisfying (4.17) 0 ≤ y0 < 1/2, 0 ≤ y1 < 1, 0 ≤ yi < 1 − yi−1 (i = 2 , . . . , n − 1) . It follows that vol( D∗ n ) = ∫ 1/20 (1 − 2y0)n−1 dy 0 ∫ 10 ∫ 1−y1 0 · · · ∫ 1−yn−2 0 dy n−1 · · · dy 2dy 1(4.18) = 12n ∫ 10 ∫ 1−y1 0 · · · ∫ 1−yn−2 0 dy n−1 · · · dy 2dy 1 = 12n vol( En−1), where En−1 = { (y1, . . . , y n−1) ∈ [0 , 1] n−1 : 0 ≤ y1 < 1, 0 ≤ yi < 1 − yi−1 (i = 2 , . . . , n − 1) } . Now set, for i = 1 , . . . , n − 1, (4.19) ui = { yi if i is odd, 1 − yi if i is even. 15 The transformation (4.19) has Jacobian determinant ±1 and thus is volume-preserving. Moreover, noting that the condition yi < 1 − yi−1 is equivalent to ui < u i−1 when i is odd, and to ui−1 < u i when i is even, we see that this transformation maps the region En−1 to the region E∗ n−1 = {(u1, . . . , u n−1) ∈ [0 , 1] n−1 : u1 < u 2 > u 3 < . . . }. The set E∗ n−1 is, up to a set of volume 0, the set of tuples in the ( n − 1)-dimensional unit cube whose coordinates form an up-down alternating permutation of length n − 1. Since there are ( n − 1)! permutations of the coordinates and, by symmetry, each such permutation contributes an amount 1 /(n − 1)! to the volume of the unit cube, it follows that (4.20) vol( En−1) = vol( E∗ n−1 ) = An−1 (n − 1)! , where An−1 is the number of up-down alternating permutations of length n − 1. Combining (4.18) with (4.20) we obtain vol( D∗ n ) = 12n vol( En−1) = 12n vol( E∗ n−1 ) = An−1 (2 n)( n − 1)! . This is the desired formula (4.13). Lemma 4.5. The number An of alternating permutations satisfies (4.21) An n! ≤ 3 ( 2 π )n+1 (n ≥ 1) . Proof. Andr´ e [1, p. 23] showed that, for any integer m ≥ 1, A2m (2 m)! = 2 ( 2 π )2m+1 ( 1 − 132m+1 + 152m+1 − 172m+1 + . . . ) ,(4.22) A2m−1 (2 m − 1)! = 2 ( 2 π )2m ( 1 + 132m + 152m + 172m + . . . ) .(4.23) The desired bound (4.21) follows as the series in parentheses in (4.22) is bounded above by 1, while the series in (4.23) is at most ≤ 1 + ∞ ∑ n=3 1 n2 = −14 + π2 6 < 32. Proof of Theorem 2, lower bound. Combining Lemmas 4.3, 4.4, and 4.5 we obtain pn = vol( Cn) ≥ 1 − 2n vol( D∗ n ) = 1 − 2n An−1 (2 n)( n − 1)! = 1 − An−1 (n − 1)! ≥ 1 − 3 ( 2 π )n , which is the desired lower bound for pn.16 5 Proof of Theorem 3 Let (5.1) Co3 = {(x, y, z ) ∈ [0 , 1] 3 : ( x, y, z ) is cyclic and x ≤ y ≤ z} , i.e., Co3 is the set of triples in C3 with nondecreasing coordinates. The following lemma characterizes the set Co3 in terms of inequalities on x, y, and z.Recall (see (3.10) and (3.11)) that ω = ( √5 − 1) /2 is the positive root of ω2 + ω = 1. Lemma 5.1. We have: (i) A triple (x, y, z ) belongs to the set Co 3 if and only if it satisfies (5.2)  0 ≤x ≤ ωx ≤y ≤ √1 − x max( y, (1 − x)(1 − y)) ≤z ≤ min ( 1, 1 − xy ) . (ii) A triple (x, y, z ) belongs to the set Co 3 if and only if it satisfies (5.3)  0 ≤y ≤ 10 ≤x ≤ min (y, 1 − y2) max( y, (1 − x)(1 − y)) ≤z ≤ min ( 1, 1 − xy ) . Moreover, the intervals for the variables x, y, and z in (5.2) and (5.3) are always non-empty, i.e., the upper bounds on these variables are always greater or equal to the corresponding lower bounds. Proof. By Lemma 2.4(i), a triple ( x, y, z ) ∈ [0 , 1] 3 with x ≤ y ≤ z is cyclic if and only if it satisfies the inequalities (5.4) x + yz ≤ 1 and (1 − z) + (1 − y)(1 − x) ≤ 1. Hence Co3 is the set of triples ( x, y, z ) ∈ [0 , 1] 3 satisfying x ≤ y ≤ z and (5.4). Consider now a triple ( x, y, z ) ∈ [0 , 1] 3 satisfying these conditions. Then x+x2 ≤ x+yz ≤ 1, and hence x ≤ ω, with ω = ( √5 − 1) /2 defined as above. Thus, we must have (5.5) 0 ≤ x ≤ ω. Next, note that x + y2 ≤ x + yz ≤ 1 and hence y ≤ √1 − x. Since we also have x ≤ y ≤ z,it follows that (5.6) x ≤ y ≤ √1 − x. Note that since x ≤ ω, with ω being the positive root of ω2 + ω = 1, we have x2 + x ≤ 1 and hence x ≤ √1 − x. Thus, the interval for y in (5.6) is non-empty. 17 Next, we have trivially (5.7) 0 ≤ y ≤ 1, while the inequalities x ≤ y ≤ z and x + y2 ≤ x + yz ≤ 1 imply (5.8) 0 ≤ x ≤ min( y, 1 − y2). Since 0 ≤ y ≤ 1, the interval (5.8) is clearly non-empty. Finally, the inequalities (5.4) imply z ≤ (1 − x)/y and z ≥ (1 − y)(1 − x), which combined with 0 ≤ z ≤ 1 and x ≤ y ≤ z yields (5.9) max( y, (1 − x)(1 − y)) ≤ z ≤ min ( 1, 1 − xy ) . Again the interval for z here is non-empty since, by (5.6), (1 − x)/y ≥ y, while we also have (1 − x)/y ≥ (1 − x) ≥ (1 − x)(1 − y), 1 ≥ y, and 1 ≥ (1 − x)(1 − y) which follow from 0 ≤ x, y ≤ 1. Note that the inequalities (5.5), (5.6), and (5.9) are exactly those in the desired charac-terization (5.2) of Co3 , while (5.7), (5.8), and (5.9) are those in (5.3). Thus, we have shown that any triple ( x, y, z ) ∈ Co3 must satisfy (5.2) and (5.3) and that the intervals for x, y, and z in (5.2) and (5.3) are all non-trivial. Conversely, assume that ( x, y, z ) satisfies (5.2) or (5.3). Then 0 ≤ x ≤ y ≤ z ≤ 1, and z must satisfy (5.9). The upper bound for z in (5.9) implies z ≤ (1 −x)/y and hence x+yz ≤ 1, i.e., the first inequality in (5.4). The lower bound for z in (5.9) implies (1 − x)(1 − y) ≤ z and hence (1 − z) + (1 − x)(1 − y) ≤ 1, i.e., the second inequality in (5.4). Thus, ( x, y, z )must be an element of Co3 .This completes the proof of the lemma. Proof of Theorem 3. Since, by Lemma 2.2, the “cyclic triple” property is invariant with respect to taking permutations, we have (5.10) vol( Co3 ) = 13! vol( C3) = 16p3. Moreover, the distribution of the order statistics ( X∗ 1 , X ∗ 2 , X ∗ 3 ) of a random triple in C3 is the same as the distribution of a random vector ( X, Y, Z ) that is distributed uniformly over the region Co3 . It follows that the density functions f1, f2, and f3 in Theorem 3 are the marginal densities of the latter vector and thus can be obtained by integrating over appropriate slices of the region Co3 and dividing by the volume of this region. That is, we have (5.11) f1(x) = 1vol( Co3 ) ∫ ∫ A1(x) 1 dz dy = 6 p3 ∫ ∫ A1(x) 1 dz dy, where (5.12) A1(x) = {(y, z ) ∈ [0 , 1] 2 : ( x, y, z ) ∈ Co3 }, along with analogous formulas for f2 and f3.18 Using the first characterization of Co3 in Lemma 5.1, we see that f1(x) is supported on the interval [0 , ω ], and that for each x with 0 ≤ x ≤ ω we have f1(x) = 6 p3 ∫ √1−xx ∫ min (1, 1−xy ) max( y, (1 −x)(1 −y)) dz dy = 6 p3 ∫ √1−xx ( min ( 1, 1 − xy ) − max( y, (1 − x)(1 − y)) ) dy = 6 p3  12 (x3 − 3x2 + 1−x 2−x − (1 − x) ln(1 − x)) if 0 ≤ x ≤ 1 − ω, 12 (x2 − 3x + 1 − (1 − x) ln(1 − x)) if 1 − ω ≤ x ≤ 1/2, 12 (x2 + x − 1 + (1 − x) ln(1 − x) − 2(1 − x) ln x) if 1 /2 ≤ x ≤ ω.This yields the desired formula (1.15) on noting that ω = (1 − √5) /2 and (5.13) 1 − ω = 1 −√5 − 12 = 3 − √52 . Similarly, using the second characterization of Lemma 5.1, we obtain f2(y) = 6 p3 ∫ min (y, 1−y2) 0 ∫ min (1, 1−xy ) max( y, (1 −x)(1 −y)) dz dx = 6 p3 ∫ min (y, 1−y2) 0 ( min ( 1, 1 − xy ) − max( y, (1 − x)(1 − y)) ) dx = 6 p3  12 (3 y2 − y3) if 0 ≤ y ≤ 1 − ω,3y − y2 − 12(1 −y) if 1 − ω ≤ y ≤ 1/2, 2 − y − y2 − 12y if 1 /2 ≤ y ≤ ω, 12 (y3 2 − 3y 2 1 ) if ω ≤ y ≤ 1=  3 p3 (3 x2 − x3) if 0 ≤ x ≤ 1 − ω, 6 p3 ( 3x − x2 − 12(1 −x) ) if 1 − ω < x ≤ 1/2, f2(1 − x) if 1 /2 < x ≤ 1, This yields the desired formula (1.16) for f2(y). Finally, since, by Lemma 2.1(iii), a triple ( x1, x 2, x 3) is cyclic if and only if (1 − x1, 1 − x2, 1 − x3) is cyclic, the distribution of X∗ 3 , the largest element in a cyclic triple, is the same as that of 1 − X∗ 1 . Thus, we have f3(x) = f1(1 − x). This completes the proof of Theorem 3. 6 Concluding Remarks In this section we mention some related questions and conjectures suggested by our results. 19 Precise asymptotic behavior of pn. Theorem 2 implies that e−(1+ o(1)) c1n ≤ 1 − pn ≤ e−(1+ o(1)) c2n (n → ∞ )holds with c1 = ln 4 and c2 = ln( π/ 2). It seems reasonable to expect that there exists a single constant c, with c2 ≤ c ≤ c1, such that 1 − pn = e−(1+ o(1)) cn (n → ∞ ), i.e., that the limit c = lim n→∞ − ln(1 − pn) n exits. If true, the value of c must lie between the constants c2 = ln( π/ 2) = 0 .451 . . . and c1 = ln 4 = 1 .386 . . . . Prescribing all pairwise probabilities P (Ui > U j ). By definition, a cyclic n-tuple is an n-tuple of real numbers in [0 , 1] that can represent the n probabilities P (Ui+1 > U i), i =1, . . . , n , with a suitable choice of independent random variables Ui, i = 1 , . . . , n , satisfying P (Ui = Uj ) = 0 for i 6 = j.As a natural extension of this concept, one can ask which arrays xij , i, j = 1 , . . . , n , can represent probabilities of the form (6.1) xij = P (Uj > U i) (i, j = 1 , . . . , n ), under the same assumptions on the random variables Ui. This problem arises in the theory of Social Choice, where it is known under the term utility representations ; see Suck for background and references. Obviously, in order for (6.1) to hold, it is necessary that xii = 0 for all i. Moreover, the assumption that P (Ui = Uj ) = 0 for i 6 = j implies xij = P (Uj > U i) = 1 − P (Ui > U j ) = 1 − xji for i > j . Thus the array xij is determined by the (n 2 ) values xij , 1 ≤ i < j ≤ n.Considering these values as tuples in the (n 2 )-dimensional unit cube, one can then ask for the probability that a random tuple in this cube has a representation of the form (6.1). When n = 3, the values xij are determined by x12 = P (U2 > U 1), x23 = P (U3 > U 2), and x13 = P (U3 > U 1) = 1 − P (U1 > U 3). Thus we see that a representation (6.1) is possible if and only if the triple ( x12 , x 23 , 1 − x13 ) is cyclic, so the problem is equivalent to the problem considered in Theorem 1. In the general case, however, the two problems are different. Indeed, the question of characterizing arrays xij that have representations of the form (6.1) seems to be an ex-traordinarily hard problem that remains largely unsolved; we refer to Suck for further discussion and some partial results. Representations with dependent random variables Ui. In our definition of cyclic tuples we assumed the random variables Ui to be independent. This is a natural assumption that is made in most of the mathematical literature on nontransitivity phenomena. In the context of dice rolls the assumption reflects the fact that if n dice are rolled, then the values showing on the faces of these dice are indeed independent. 20 However, there also exist well-known nontransitivity paradoxes—especially in voting the-ory and social choice theory—in which the underlying random variables are dependent. Thus, it would be of interest to study the analogs of cyclic and nontransitive tuples if the indepen-dence assumption on Ui is dropped. We refer to Marengo et al. for some recent results in this direction, and to Suck for an in-depth discussion of the differences between dependent and independent representations of the form (6.1). References D´ esir´ e Andr´ e, Sur les permutations altern´ ees , J. Math. Pures Appl. 7 (1881), 167–184. Levi Angel and Matt Davis, Properties of sets of nontransitive dice with few sides ,Involve 11 (2018), no. 4, 643–659. MR 3778917 I.I. Bogdanov, Nontransitive Roulette , Mat. Prosveshch. 3 (2010), 240–255. Adrian Bondy, Jian Shen, St´ ephan Thomass´ e, and Carsten Thomassen, Density condi-tions for triangles in multipartite graphs , Combinatorica 26 (2006), no. 2, 121–131. MR 2223630 Joe Buhler, Ron Graham, and Al Hales, Maximally nontransitive dice , Amer. Math. Monthly 125 (2018), no. 5, 387–399. MR 3785874 Li-Chien Chang, On the maximin probability of cyclic random inequalities , Scientia Sinica 10 (1961), no. 5, 499. Brian Conrey, James Gabbard, Katie Grant, Andrew Liu, and Kent E. Morrison, In-transitive dice , Math. Mag. 89 (2016), no. 2, 133–143. MR 3510988 Martin Gardner, The paradox of nontransitive dice and the elusive principle of indiffer-ence , Scientific American 223 (1970), no. 6, 110–115. E. Gilson, C. Cooley, W. Ella, M. Follett, and L. Traldi, The Efron dice voting system ,Soc. Choice Welf. 39 (2012), no. 4, 931–959. MR 2983388 Artem Hulko and Mark Whitmeyer, A game of nontransitive dice , Math. Mag. 92 (2019), no. 5, 368–373. MR 4044971 Andrzej Komisarski, Nontransitive random variables and nontransitive dice , Amer. Math. Monthly (2021), to appear. Daniel Kr´ al, Locally satisfiable formulas , Proceedings of the Fifteenth Annual ACM-SIAM Symposium on Discrete Algorithms, ACM, New York, 2004, pp. 330–339. MR 2291069 A. V. Lebedev, The nontransitivity problem for three continuous random variables ,Autom. Remote Control 80 (2019), no. 6, 1058–1068. S0005117919060055 21 L. Lov´ asz and J. Pelik´ an, On the eigenvalues of trees , Period. Math. Hungar. 3 (1973), 175–182. MR 416964 James E. Marengo, Quinn T. Kolt, and David L. Farnsworth, An upper bound for a cyclic sum of probabilities , Statist. Probab. Lett. 165 (2020), 108861. MR 4118940 J. W. Moon and L. Moser, Generating oriented graphs by means of team comparisons ,Pacific J. Math. 21 (1967), 531–535. MR 216984 Zolt´ an L´ or´ ant Nagy, A multipartite version of the Tur´ an problem—density conditions and eigenvalues , Electron. J. Combin. 18 (2011), no. 1, Paper 46, 15. MR 2776822 Hugo Steinhaus and S. Trybula, On a paradox in applied probabilities , Bull. Acad. Polon. Sci 7 (1959), no. 67-69, 108. Reinhard Suck, Independent random utility representations , Math. Social Sci. 43 (2002), no. 3, 371–389. MR 2072963 Luca Trevisan, On local versus global satisfiability , SIAM J. Discrete Math. 17 (2004), no. 4, 541–547. MR 2085212 S. Trybula, On the paradox of three random variables , Zastos. Mat. 5 (1960/61), 321– 332. MR 126865 , On the paradox of n random variables , Zastos. Mat. 8 (1965), 143–156. MR 190970 Zalman Usiskin, Max-min probabilities in the voting paradox , Ann. Math. Statist. 35 (1964), 857–862. MR 161353 22
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361001200094 Department of Taxation and Finance Amended Resident Income Tax Return New York State • New York City • Yonkers • MCTMT IT-201-X c Single d Married filing joint return (enter spouse’s Social Security number above) e Married filing separate return (enter spouse’s Social Security number above) f Head of household (with qualifying person) g Qualifying widow(er) For the full year January 1, 2020, through December 31, 2020, or fiscal year beginning ... 20 and ending ... A Filing status (mark an X in one box): D1 Did you file an amended federal return? (see instructions) .................................................... Yes No D2 Were you required to report any nonqualified deferred compensation, as required by IRC § 457A, on your 2020 federal return? (see Form IT-201-I, page 15) Yes No E (1) Did you or your spouse maintain living quarters in NYC during 2020? ..................... 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see instructions) ................................................................................................ 60 .00 61 Total New York State, New York City, Yonkers, and sales or use taxes, MCTMT, and voluntary contributions (add lines 46, 58, 59, and 60) .............................................................. 61 .00 Tax computation, credits, and other taxes 38 Taxable income (from line 37 on page 3) ....................................................................................... 38 .00 39 NYS tax on line 38 amount .......................................................................................................... 39 .00 40 NYS household credit ........................................................ 40 .00 41 Resident credit .................................................................. 41 .00 42 Other NYS nonrefundable credits (Form IT-201-ATT, line 7) 42 .00 43 Add lines 40, 41, and 42 .............................................................................................................. 43 .00 44 Subtract line 43 from line 39 (if line 43 is more than line 39, leave blank) .......................................... 44 .00 45 Net other NYS taxes (Form IT-201-ATT, line 30) ............................................................................. 45 .00 46 Total New York State taxes (add lines 44 and 45) ........................................................................ 46 .00 47 NYC taxable income ........................................................ 47 .00 47a NYC resident tax on line 47 amount ................................ 47a .00 48 NYC household credit ...................................................... 48 .00 49 Subtract line 48 from line 47a (if line 48 is more than line 47a, leave blank) ........................................................ 49 .00 50 Part-year NYC resident tax (Form IT-360.1) ....................... 50 .00 51 Other NYC taxes (Form IT-201-ATT, line 34) ........................ 51 .00 52 Add lines 49, 50, and 51 .................................................. 52 .00 53 NYC nonrefundable credits (Form IT-201-ATT, line 10) ........ 53 .00 54 Subtract line 53 from line 52 (if line 53 is more than line 52, leave blank) ......................................................... 54 .00 54a MCTMT net earnings base .... 54a .00 54b MCTMT ............................................................................ 54b .00 55 Yonkers resident income tax surcharge ........................... 55 .00 56 Yonkers nonresident earnings tax (Form Y-203) ................ 56 .00 57 Part-year Yonkers resident income tax surcharge (Form IT-360.1) 57 .00 58 Total New York City and Yonkers taxes / surcharges and MCTMT (add lines 54 and 54b through 57) 58 .00 59 Sales or use tax as reported on your original return (see instructions. Do not leave line 59 blank.) 59 .00 New York City and Yonkers taxes, credits, and surcharges and MCTMT Page 4 of 6 IT-201-X (2020) Your Social Security number 361005200094 62 Enter amount from line 61 ........................................................................................................... 62 .00 Amount you owe 81 If line 79 is less than line 62, subtract line 79 from line 62 (see instructions) ............................... 81 .00 82 Account information for direct deposit or electronic funds withdrawal (see instructions) If the funds for your payment (or refund) would come from (or go to) an account outside the U.S., mark an X in this box (see instructions) .............................................................................................................................. See Important information in the instructions. Your refund Account information 80 If line 79 is more than line 62, subtract line 62 from line 79 and indicate how you want your refund direct (fill in lines 82 paper Mark one refund choice: deposit through 82c) - or - check ................................... 80 .00 Payments and refundable credits 63 Empire State child credit .................................................. 63 .00 64 NYS/NYC child and dependent care credit ...................... 64 .00 65 NYS earned income credit (EIC) .............................. 65 .00 66 NYS noncustodial parent EIC .......................................... 66 .00 67 Real property tax credit .................................................... 67 .00 68 College tuition credit ......................................................... 68 .00 69 NYC school tax credit (fixed amount) (also complete F on page 1) 69 .00 69a NYC school tax credit (rate reduction amount) ................. 69a .00 70 NYC earned income credit ....................................... 70 .00 70a This line intentionally left blank ........................................ 70a 71 Other refundable credits (Form IT-201-ATT, line 18) ............. 71 .00 72 Total New York State tax withheld ................................... 72 .00 73 Total New York City tax withheld ..................................... 73 .00 74 Total Yonkers tax withheld ............................................... 74 .00 75 Total estimated tax payments / Amount paid with Form IT-370 75 .00 76 Amount paid with original return, plus additional tax paid after your original return was filed (see instructions) ........ 76 .00 77 Total payments (add lines 63 through 76) ..................................................................................... 77 .00 78 Overpayment, if any, as shown on original return or previously adjusted by NY State (see instr.) ... 78 .00 78a Amount from original Form IT-201, line 79 (see instructions) 78a .00 79 Subtract line 78 from line 77 ....................................................................................................... 79 .00 IT-201-X (2020) Page 5 of 6 82a Account type: Personal checking - or - Personal savings - or - Business checking - or - Business savings 82b Routing number 82c Account number 82d Electronic funds withdrawal (see instructions) ............... Date Amount .00 Name(s) as shown on page 1 Your Social Security number To pay by electronic funds withdrawal, mark an X in the box and fill in lines 82 through 82d. If you pay by check or money order you must complete Form IT-201-V and mail it with your return. You must submit all required forms. Failure to do so will result in an adjustment to your return. 361006200094 Your signature Your occupation Spouse’s signature and occupation (if joint return) Date Daytime phone number Email: Name of partnership or S corporation Identifying number Principal business activity Address of partnership or S corporation See instructions for where to mail your return. 83a Federal audit change (complete lines 84 through 91 below) ................................................ 83b Worthless stock/securities .............. 83c Claim of right ............................... 83d Wages ........................................... 83e Military ............................................. 83f Court ruling .................................. 83g Workers’ compensation .................. 83h Treaties/visa .................................... 83i Tax shelter transaction ................ 83j Credit claim ..................................... 83k Protective claim (see instructions) ...... 83 Reason(s) for amending your return (mark an X in all applicable boxes; see instructions) Print designee’s name Designee’s phone number Personal identification ( ) number (PIN) Email: Third-party designee? Yes No ▼ Taxpayer(s) must sign here ▼ ( ) 83l Net operating loss (see instructions). Mark an X in the box .... and enter the year of the loss .... 83m Report Social Security number (SSN) Prior identification number Date SSN was issued 83n Other. Mark an X in the box ... and explain: 83o To report adjustments to partnership or S corporation income, gain, loss or deduction, provide the following information: Partnership S corporation 90 Federal credits disallowed ........ Earned income credit Amount disallowed Child care credit Amount disallowed 91 Federal penalties assessed 91a Fraud ............................................. 91b Negligence ........................ 91c Other (explain below) .......................... 84 Enter the date (mmddyyyy) of the 85 Do you concede the federal audit final federal determination changes (If No, explain below.) ......... Yes No (Explain) If you marked an X in box 83a above, you must complete lines 84 through 91 below. All others may skip lines 84 through 91 and go directly to the Third-party designee question. You must sign your amended return below. 86 List federal changes 86a 86a .00 86b 86b .00 86c 86c .00 86d 86d .00 86e 86e .00 87 Net federal changes (increase or decrease) ........................................................................... 87 .00 88 Federal taxable income (mark an X in one box) .... Per return Previously adjusted 88 .00 89 Corrected federal taxable income ............................................................................................ 89 .00 Page 6 of 6 IT-201-X (2020) Your Social Security number ▼ Paid preparer must complete ▼ (see instructions) Preparer’s NYTPRIN NYTPRIN excl. code Preparer’s signature Preparer’s printed name Firm’s name (or yours, if self-employed) Preparer’s PTIN or SSN Address Employer identification number Date Email: Department of Taxation and Finance Instructions for Form IT-201-X Amended Resident Income Tax Return New York State • New York City • Yonkers • MCTMT IT-201-X-I Important information Follow these steps to complete your amended Form IT-201-X: • Complete your Form IT-201-X as if you are filing your return for the first time. • Carefully review and follow the instructions below. You must enter the same amount of sales and use tax and voluntary contributions from your original return; you cannot change these amounts. • Do not submit a copy of your original Form IT-201, IT-203, or IT-195 with your amended Form IT-201-X. • Submit with your amended Form IT-201-X any: – amended Form IT-196; – amended credit claim form or other amended form (do not submit the original version); – new credit claim form or any other form that you are filing for the first time with your amended Form IT-201-X; and – original credit claim form(s) (for example, Forms IT-213, IT-215, and IT-216); withholding form(s) (for example, Form IT-2), and all other form(s) that you submitted with your original return and are not amending (for example, Forms IT-196, IT-201-ATT, and IT-227). If you do not submit all the necessary forms with your amended return, we will adjust your return and disallow the amounts claimed on the missing forms. General information You must file an amended 2020 New York State return if: • You made an error when you filed your original 2020 New York State income tax return. • The Internal Revenue Service (IRS) made changes to your 2020 federal return. • You need to file a protective claim for 2020. • You need to report an NOL carryback for 2020. See the instructions for 2020 Form IT-201 to determine which amended return to file (Form IT-201-X or IT-203-X). Do not file an amended return on Form IT-201-X to protest a paid assessment that was based on a statement of audit changes. If you receive an assessment from the Tax Department, do not file an amended return strictly to protest the assessment. Follow the instructions you receive with the assessment. To file an amended return, complete all six pages of Form IT-201-X, using your original return as a guide, and make any necessary changes to income, deductions, and credits. Use 2020 Form IT-201-I, Instructions for Form IT-201, and the specific instructions below to complete Form IT-201-X. Generally, Form IT-201-X must be filed within three years of the date the original return was filed or within two years of the date the tax was paid, whichever is later. (A return filed early is considered filed on the due date.) Do not file Form IT-201-X unless you have already filed your original return. If you file an amended federal return to make changes to your federal income, total taxable amount, capital gain or ordinary income portion of a lump-sum distribution, the amount of your earned income credit or credit for child and dependent care expenses, or the amount of your foreign tax credit affecting the computation of the resident credit for taxes paid to a province of Canada, you must also file an amended New York State return within 90 days of the date you amend your federal return. If the IRS changes any of these items, report these changes to the New York State Tax Department on an amended return within 90 days of the IRS final determination. If you do not agree with the IRS determination, you must still file an amended state return indicating your disagreement. To report changes for a tax year prior to 1988, use Form IT-115, Report of Federal Changes. If you file an amended return to report an NOL carryback, you must generally file Form IT-201-X within three years from the date the loss year return was due (including any extensions). Specific instructions Use the 2020 Form IT-201 instructions when completing Form IT-201-X, along with the following specific line instructions. If you are amending any credit claim form or other form, or are using any credit claim form or other form for the first time, write Amended across the top of that form and submit it with your amended return. Any other credit claim form or other form that you submitted with your original return (including Form IT-558, Form IT-196, or Form IT-227) must also be submitted with your amended return. Entering whole dollar amounts When entering amounts on this form, enter whole dollar amounts only (zeros have been preprinted). Do not write in dollar signs or commas when making entries. Use the following rounding rules when entering your amounts; drop amounts below 50 cents and increase amounts from 50 to 99 cents to the next dollar. For example, $1.39 becomes $1 and $2.50 becomes $3. Item D1 – Amended federal return You must mark an X in the Yes or No box. Item G – Special condition code If you entered a special condition code(s) on your original return, enter the same code(s). In addition, if you qualify for one or more of the special conditions below, enter the 2-character code(s). Code A6 Enter this code if you are filing Form IT-201-X to reduce your NYAGI for Build America Bond interest included in your recomputed federal AGI. Code C7 Enter this code if you now qualify for an extension of time to file and pay your tax due under the combat zone or contingency operation relief provisions. See Publication 361, New York State Income Tax Information for Military Personnel and Veterans. Code 56 Enter this code if you are filing Form IT-201-X to report a theft loss for a Ponzi-type fraudulent investment. Code P2 Enter this code if you are filing Form IT-201-X to file a protective claim. Also, be sure to mark an X in the line 83k box. Page 2 of 4 IT-201-X-I (2020) Code N3 Enter this code if you are filing Form IT-201-X to report an NOL. Also, be sure to mark an X in the line 83l box and complete the information requested for the loss year. For more information on claiming an NOL carryback, see the instructions for Form IT-201. Code M4 Enter this code if as a civilian spouse of a military servicemember you are making an election to use the same state of legal residence as the servicemember for state income tax purposes. For additional information, see TSB-M-19(3)I, Veterans Benefits and Transition Act of 2018, available on our website. Line 34 – Standard or itemized deduction Standard deduction: If you are claiming the standard deduction on your amended return, enter the appropriate amount, for your filing status, from the table on page 3 of Form IT-201-X. Itemized deduction: If you are claiming the New York itemized deduction on your original and amended return and you meet all three of the following conditions, submit a copy of your original Form IT-196: • You are not amending your New York State itemized deductions. • Your NYAGI on your original and amended returns is $100,000 or less. • You are not claiming the college tuition itemized deduction. If you do not meet all of the above conditions, you must recalculate your New York State itemized deduction using Form IT-196. If you are reporting an NOL carryback and you were subject to the New York itemized deduction adjustment on your original 2020 Form IT-196, you should recompute your New York itemized deduction adjustment to reflect the decrease in your NYAGI. Line 59 – Sales or use tax Enter the amount of New York State and local sales or use tax you reported on your original return. You cannot change the amount of sales or use tax you owe using Form IT-201-X. If you need to increase the amount of sales or use tax paid with your original return, you must file Form ST-140, Individual Purchaser’s Annual Report of Sales and Use Tax. If you are entitled to a refund of any amount you originally paid, you must file Form AU-11, Application for Credit or Refund of Sales or Use Tax. Line 60 – Voluntary contributions Enter the total amount of voluntary contributions you reported on your original return. This amount should be the same as the total reported on your original Form IT-227, New York State Voluntary Contributions. If the voluntary contributions you reported on your original Form IT-227 were previously adjusted by the Tax Department, enter the total adjusted amount on this line. You cannot change the amount of your contributions as reported (or adjusted) on your original return or original Form IT-227. You must submit your original Form IT-227 with your amended Form IT-201-X. Line 76 – Amount paid with original return, plus additional tax paid after your original return was filed From your original Form IT-201, line 80 (or Form IT-203, line 70). If you paid additional amounts since your original return was filed, also include these payments on line 76. If you did not pay the entire balance due shown on your original return, enter the actual amount that was paid. Do not include payments of interest or penalties. Line 78 – Overpayment, if any, as shown on original return From your original Form IT-201, line 77 (or Form IT-203, line 67). If the overpayment claimed on your original return was previously adjusted by the Tax Department, enter the adjusted overpayment on this line. Do not include interest you received on any refund. Line 78a – Amount from original return If you filed Form IT-203, enter the amount from Form IT-203, line 69. Line 80 – Refund If line 79 is more than line 62, subtract line 62 from line 79; this is your refund amount. You have two ways to receive your refund. You can choose direct deposit to have the funds deposited directly into your bank account (the fastest option for most filers), or you can choose to have a paper check mailed to you. Mark an X in one box to indicate your choice. Refund options Direct deposit – If you choose direct deposit, enter your account information on line 82 for a fast and secure direct deposit of your refund. If you do not enter complete and correct account information at line 82, we will mail you a paper check. Paper check refunds – We will mail your refund check to the mailing address on your return. Paper checks for joint filers will be issued with both names and must be signed by both spouses. Paper checks take weeks to be processed, printed, and mailed. If you do not have a bank account, you will likely be charged a fee to cash your check. Line 81 – Amount you owe Enter on line 81 the amount of tax you owe. Payment options By automatic bank withdrawal You may authorize the Tax Department to make an electronic funds withdrawal from your bank account either by completing line 82, or on our website. This payment option is not available if the funds for your payment would come from an account outside the U.S. (see Note below). If you choose to complete line 82 to pay by electronic funds withdrawal, mark an X in the box, enter your account information on lines 82a through 82c, and enter your electronic funds withdrawal information on line 82d. By check or money order If you owe more than one dollar, submit Form IT-201-V, Payment Voucher for Income Tax Returns, and full payment with your return. Make your check or money order payable in U.S. funds to New York State Income Tax, and write the last four digits of your Social Security number and 2020 Income Tax on it. Do not send cash. Interest – If a balance due is shown on your amended return, include the interest amount on line 81. Compute the interest by accessing our website or call 518-457-5181, and we will compute the interest for you. Include with your payment any interest computed. Fee for payments returned by banks – The law allows the Tax Department to charge a $50 fee when a check, money order, or electronic payment is returned by a bank for nonpayment. However, if an electronic payment is returned as a result of an error by the bank or the department, the department will not charge the fee. If your payment is returned, we will send a separate bill for $50 for each return or other tax document associated with the returned payment. Line 82 – Account information If you marked the box that indicates your payment (or refund) would come from (or go to) an account outside the U.S., stop. Do not complete lines 82a through 82d (see Note below). All others, supply the information requested. Note: Banking rules prohibit us from honoring requests for electronic funds withdrawal or direct deposit when the funds for your payment (or refund) would come from (or go to) an account outside the U.S. Therefore, if you marked this box, you must pay any amount you owe by check or money order (see above); or if you are requesting a refund, we will send your refund to the mailing address on your return. The following requirements apply to both direct deposit and electronic funds withdrawal: Use the sample image as a guide; enter your own information exactly as it appears on your own check or bank records. Do not enter the information from the sample check below. On line 82a, mark an X in the box for the type of account. On line 82b, enter your bank’s 9-digit routing number (refer to your check or contact your bank). The first two digits always begin with 01 through 12, or 21 through 32. On the sample check below, the routing number is 111111111. Note: If your check states that it is payable through a bank different from the one where you have your checking account, do not use the routing number on that check. Instead, contact your bank for the correct routing number to enter on line 82b. On line 82c, enter your account number. • If you marked personal or business checking on line 82a, enter the account number shown on your checks. • If you marked personal or business savings on line 82a, enter your savings account number from a preprinted savings account deposit slip, your passbook or other bank records, or from your bank. The account number can be up to 17 characters (both numbers and letters). Include hyphens (-) but omit spaces and special symbols. Enter the number from left to right. On the sample check below, the account number is 9999999999. Sample JOHN SMITH 999 Maple Street Someplace, NY 10000 Date 1234 15-0000/0000 Pay to the Order of Dollars SOME BANK Someplace, NY 10000 For Do not include the check number QQQQ9QQQQQ 1234 Note: The routing and account numbers may appear in different places on your check. $ 111111111 X X Sample routing number Sample account number Note: The routing and account numbers may appear in different places on your check. IT-201-X-I (2020) Page 3 of 4 Contact your bank if you need to verify routing and account numbers or confirm that it will accept your direct deposit or process your electronic funds withdrawal. If you encounter any problem with direct deposit to, or electronic withdrawal from, your account, call 518-457-5181. Allow six to eight weeks for processing your return. Line 82d – Electronic funds withdrawal Enter the date you want the Tax Department to make an electronic funds withdrawal from your bank account and the amount from line 81 you want electronically withdrawn. If you are amending your return prior to the original due date (generally April 15), enter a date that is on or before the due date of your return. If we receive your amended return after the original return due date or you do not enter a date, we will withdraw the funds on the day we accept your return. Your confirmation will be your bank statement that includes a NYS Tax Payment line item. We will only withdraw the amount that you authorize. If we determine that the amount you owe is different from the amount claimed on your return, we will issue you a refund for any amount overpaid or send you a bill for any additional amount owed, which may include penalty and interest. You may revoke your electronic funds withdrawal authorization only by contacting the Tax Department at least 5 business days before the payment date. If you complete the entries for electronic funds withdrawal, do not send a check or money order for the same amount due unless you receive a notice. Line 83k – Protective claim If you marked the Protective claim box, be sure you have entered code P2 at item G on the front of your Form IT-201-X. Complete your amended return in full assuming that the item(s) that is the subject of the protective claim is eligible for refund. A protective claim is a refund claim that is based on an unresolved issue(s) that involves the Tax Department or another taxing jurisdiction that may affect your New York tax(es). The purpose of filing a protective claim is to protect any potential overpayment for a tax year for which the statute of limitations is due to expire. Line 83l – Net operating loss For New York State income tax purposes, your NOL carryback is limited to the federal NOL carryback that would have been allowed using the rules in place prior to any changes made to the IRC after March 1, 2020. Therefore, there is no carryback of NOLs for New York State purposes; except for certain farming losses. If you marked the Net operating loss box, you must enter the year of the loss at line 83l and enter code N3 at item G on the front of your Form IT-201-X. You must file Form IT-201-X to claim an NOL carryback within three years from the date the loss year return was due (including any extensions). Submit all of the following with your Form IT-201-X: • A copy of your federal Form 1040 and Schedule A, if applicable, for the loss year. In addition, provide any schedules or statements that are related to your loss. If your NOL will have an effect on more than one tax year, this federal information must only be submitted with the amended return for the first carryback year. • A copy of your federal NOL computation, including federal Form 1045 and all related schedules. You do not have to include the alternative minimum tax NOL computation. Visit our website at www.tax.ny.gov • get information and manage your taxes online • check for new online services and features Telephone assistance Automated income tax refund status: 518-457-5149 Personal Income Tax Information Center: 518-457-5181 To order forms and publications: 518-457-5431 Text Telephone (TTY) or TDD Dial 7-1-1 for the equipment users New York Relay Service Need help? Page 4 of 4 IT-201-X-I (2020) • A copy of your original federal Form 1040 and Schedule A, if applicable, for the carryback year. No additional schedules/statements are required. • A copy of any federal documentation (if available) showing the IRS has accepted your NOL carryback claim. Line 83m – Report Social Security number If you filed your original return using either an individual taxpayer identification number (ITIN) or a New York State temporary identification number (with a TF prefix) and have received a Social Security number (SSN), then mark the box, enter the identification number used on your original return, and enter the date when the SSN was issued. If you received notification (Form TR-298) from the Tax Department that you were assigned a temporary identification number, follow the instructions in that notice to report your valid identification number (SSN or ITIN) to us. Do not file Form IT-201-X to report only your new identification number. Line 83n – Other If you marked the Other box, include an explanation of the change on the explanation line at line 83n (for example, you are changing your New York State dependent exemption amount). If you need additional room, submit a separate sheet with your explanation. Include your name and SSN on the additional sheet. Line 83o – Partnership or S corporation If you marked a box at line 83o, give the partnership’s or S corporation’s name, identifying number, principal business activity, and address. Lines 84 through 91 If you marked the line 83a box and are reporting changes made by the IRS, complete lines 84 through 91 by entering the information requested as it appears on your final federal report of examination changes. Use a minus sign to show any decreases. Important: Fully explain the changes you are making on Form IT-201-X. Submit any schedules or forms that apply, along with any available federal documentation. Documentation may include, but is not limited to, copies of: your federal Form 1040X; federal acceptance of your amended federal return (include copies of the refund check, if applicable); amended federal Schedule B, Schedule C, or Schedule D; and revised federal Schedule K-1. Failure to include this information when filing Form IT-201-X may delay the processing of your return or the issuance of your refund. Where to file If enclosing a payment (check or money order), mail your return and Form IT-201-V to: STATE PROCESSING CENTER PO BOX 15555 ALBANY NY 12212-5555 If not enclosing a payment, mail your return to: STATE PROCESSING CENTER PO BOX 61000 ALBANY NY 12261-0001 Private delivery services – If you are not submitting your form by U.S. Mail, be sure to consult Publication 55, Designated Private Delivery Services, for the address and other information. Paid preparer’s signature If you pay someone to prepare your return, the paid preparer must also sign it and fill in the other blanks in the paid preparer’s area of your return. A person who prepares your return and does not charge you should not fill in the paid preparer’s area. Paid preparer’s responsibilities – Under the law, all paid preparers must sign and complete the paid preparer section of the return. Paid preparers may be subject to civil and/or criminal sanctions if they fail to complete this section in full. When completing this section, enter your New York tax preparer registration identification number (NYTPRIN) if you are required to have one. If you are not required to have a NYTPRIN, enter in the NYTPRIN excl. code box one of the specified 2-digit codes listed below that indicates why you are exempt from the registration requirement. You must enter a NYTPRIN or an exclusion code. Also, you must enter your federal preparer tax identification number (PTIN) if you have one; if not, you must enter your Social Security number. Code Exemption type Code Exemption type 01 Attorney 02 Employee of attorney 03 CPA 04 Employee of CPA 05 PA (Public Accountant) 06 Employee of PA 07 Enrolled agent 08 Employee of enrolled agent 09 Volunteer tax preparer 10 Employee of business preparing that business’ return See our website for more information about the tax preparer registration requirements. Privacy notification New York State Law requires all government agencies that maintain a system of records to provide notification of the legal authority for any request for personal information, the principal purpose(s) for which the information is to be collected, and where it will be maintained. To view this information, visit our website, or, if you do not have Internet access, call and request Publication 54, Privacy Notification. See Need help? for the Web address and telephone number.
187811
https://www.reddit.com/r/learnmath/comments/tqziq5/possible_5_digit_codes_using_0123456789/
Possible 5 Digit codes using 0,1,2,3,4,5,6,7,8,9 : r/learnmath Skip to main contentPossible 5 Digit codes using 0,1,2,3,4,5,6,7,8,9 : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Get App Get the Reddit app Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 390K Members Online •3 yr. ago Hutchie2306 Possible 5 Digit codes using 0,1,2,3,4,5,6,7,8,9 RESOLVED Hey everyone I am running into some problems with this problem I invented for myself. Note: Just for simplicity sake, as an example 00140 is a 5 digit code even though it really is just 140 Possible code options without repeat = 10P5 = 30,240 Possible code options with repeat : 10^5 = 100,000 So that means there are 69,760 codes where a digit is repeated at least twice. As there are 10 digits to choose from, that means 6,976 where 9 is represented at least twice within the code Codes with 5 9s and 0 other numbers: 9,9,9,9,9 So total codes = 1 Codes with 4 9s and 1 other number: x,9,9,9,9 9,x,9,9,9 9,9,x,9,9 9,9,9,x,9 9,9,9,9,x where x can be 0,1,2,3,4,5,6,7,8 (9 options) So total codes = 5 x 9 = 45 Codes with 3 9s and 2 other numbers: x,x,9,9,9 x,9,x,9,9 x,9,9,x,9 x,9,9,9,x 9,x,x,9,9 9,x,9,x,9 9,x,9,9,x 9,9,x,x,9 9,9,x,9,x 9,9,9,x,x where x,x can be any 2 numbers from 0,1,2,3,4,5,6,7,8,9 so x,x has 9^2 options as repetitions are allowed Hence total codes = 10 x 81 = 810 Codes with 4 9s and 1 other number: 9,9,x,x,x 9,x,9,x,x 9,x,x,9,x 9,x,x,x,9 x,9,9,x,x x,9,x,9,x x,9,x,x,9 x,x,9,9,x x,x,9,x,9 x,x,x,9,9 where x,x,x can be any 3 numbers from 0,1,2,3,4,5,6,7,8,9 so x,x,x has 9^3 options as repetitions are allowed Hence total codes = 10 x 729 = 7290 But if we add together the total codes where 9 has at least 2 digits I run into the error: 7290 + 810 + 45 + 1 = 8146 but 8146 =! 6976 Any help is greatly appreciated. Thanks :) Read more Share Share Related Answers Section Related Answers Possible 5 digit codes using digits 0-9 List of all possible 5 digit combinations 0-9 Common 5 digit passcodes Popular codes for 5 digit combinations 5 digit number combinations without repeats New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts RedditreReddit: Top posts of March 29, 2022 RedditreReddit: Top posts of March 2022 RedditreReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation TOPICS Internet Culture (Viral) Amazing Animals & Pets Cringe & Facepalm Funny Interesting Memes Oddly Satisfying Reddit Meta Wholesome & Heartwarming Games Action Games Adventure Games Esports Gaming Consoles & Gear Gaming News & Discussion Mobile Games Other Games Role-Playing Games Simulation Games Sports & Racing Games Strategy Games Tabletop Games Q&As Q&As Stories & Confessions Technology 3D Printing Artificial Intelligence & Machine Learning Computers & Hardware Consumer Electronics DIY Electronics Programming Software & Apps Streaming Services Tech News & Discussion Virtual & Augmented Reality Pop Culture Celebrities Creators & Influencers Generations & Nostalgia Podcasts Streamers Tarot & Astrology Movies & TV Action Movies & Series Animated Movies & Series Comedy Movies & Series Crime, Mystery, & Thriller Movies & Series Documentary Movies & Series Drama Movies & Series Fantasy Movies & Series Horror Movies & Series Movie News & Discussion Reality TV Romance Movies & Series Sci-Fi Movies & Series Superhero Movies & Series TV News & Discussion RESOURCES About Reddit Advertise Reddit Pro BETA Help Blog Careers Press Communities Best of Reddit Top Translated Posts Topics
187812
https://k12.libretexts.org/Bookshelves/Mathematics/Precalculus/05%3A_Trigonometric_Functions/5.04%3A_Vertical_Shift_of_Sinusoidal_Functions
5.4: Vertical Shift of Sinusoidal Functions - K12 LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not availabl e. chrome_reader_mode Enter Reader Mode 5: Trigonometric Functions Precalculus { } { "5.01:_The_Unit_Circle" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.02:_The_Sinusoidal_Function_Family" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.03:_Amplitude_of_Sinusoidal_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.04:_Vertical_Shift_of_Sinusoidal_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.05:_Frequency_and_Period_of_Sinusoidal_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.06:_Phase_Shift_of_Sinusoidal_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.07:_Graphs_of_Other_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.08:_Graphs_of_Inverse_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Functions_and_Graphs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Polynomials_and_Rational_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Logs_and_Exponents" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Basic_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Analytic_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Vectors" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Systems_and_Matrices" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Conics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Polar_and_Parametric_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Complex_Numbers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Discrete_Math" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Finance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Concepts_of_Calculus" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Concepts_of_Statistics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Logic_and_Set_Theory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 13 Feb 2022 22:29:19 GMT 5.4: Vertical Shift of Sinusoidal Functions 981 981 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "program:ck12", "authorname:ck12", "license:ck12", "source@ ] [ "article:topic", "showtoc:no", "program:ck12", "authorname:ck12", "license:ck12", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Mathematics 4. Precalculus 5. 5: Trigonometric Functions 6. 5.4: Vertical Shift of Sinusoidal Functions Expand/collapse global location Precalculus Front Matter 1: Functions and Graphs 2: Polynomials and Rational Functions 3: Logs and Exponents 4: Basic Triangle Trigonometry 5: Trigonometric Functions 6: Analytic Trigonometry 7: Vectors 8: Systems and Matrices 9: Conics 10: Polar and Parametric Equations 11: Complex Numbers 12: Discrete Math 13: Finance 14: Concepts of Calculus 15: Concepts of Statistics 16: Logic and Set Theory Back Matter 5.4: Vertical Shift of Sinusoidal Functions Last updated Feb 13, 2022 Save as PDF 5.3: Amplitude of Sinusoidal Functions 5.5: Frequency and Period of Sinusoidal Functions Page ID 981 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Vertical Shift of Sinusoidal Functions 2. Examples 1. Example 1 2. Example 2 3. Example 3 4. Example 4 5. Example 5 Your knowledge of transformations, specifically vertical shift, apply directly to sinusoidal functions. In practice, sketching shifted sine and cosine functions requires greater attention to detail and more careful labeling than other functions. Can you describe the following transformation in words? f⁡(x)=sin⁡x→g⁡(x)=−3⁢sin⁡x−4 In what order do the reflection, stretch and shift occu r? Is there a difference? Vertical Shift of Sinusoidal Functions The general form of a sinusoidal function is: f⁡(x)=±a⋅sin⁡(b⁢(x+c))+d Recall that a controls amplitude and the ± controls reflection. Here you will see how d controls the vertical shift. The most straightforward way to think about vertical shift of sinusoidal functions is to focus on the sinusoidal axis, the horizontal line running through the middle of the sine or cosine wave. At the start of the problem identify the vertical shift and immediately draw the new sinusoidal axis. Then proceed to graph amplitude and reflection about that axis as opposed to the x axis. The graphs of the following three functions are shown below: f⁡(x)=sin⁡x+3 g⁡(x)=sin⁡x−2 h⁡(x)=sin⁡x+1 2 To draw these graphs, the new sinusoidal axis for each graph is drawn first. Then, a complete sine wave for each one is drawn. Note the five important points that separate each quadrant to help to get a clear sense of the graph. There are no reflections in these graphs and they all have an amplitude of 1. Right now every cycle starts at 0 and ends at 2⁢π but this will not always be the case. Watch the portions of the following video focused on vertical translations: Examples Example 1 Earlier, you were asked which order vertical shift and reflection should be performed in and if it matters. The following transformation can be described as follows. f⁡(x)=sin⁡x→g⁡(x)=−3⁢sin⁡x−4 Describe the stretching and reflecting first and then the vertical shift. This is the most logical way to discuss the transformation verbally because then the numbers like 3 and -4 can be explicitly identified in the graph. The order in describing the transformation matters. When describing vertical transformations it is most intuitive to simply describe the transformations in the same order as the order of operations. Example 2 Identify the equation of the following transformed cosine graph. since there is no sinusoidal axis given, you must determine the vertical shift, stretch and reflection. The peak occurs at (π,3) and the trough occurs at (0,-1) so the horizontal line directly between +3 and -1 is y=1. since the sinusoidal axis has been shifted up by one unit d=1. From this height, the graph goes two above and two below which means that the amplitude is 2 . since this cosine graph starts its cycle at (0,-1) which is a lowest point, it is a negative cosine. The function is f⁡(x)=−2⁢cos⁡x+1 Example 3 Transform the following sine graph in two ways. First, transform the sine graph by shifting it vertically up 1 unit and then stretching it vertically by a factor of 2 units. Second, transform the sine graph by stretching it vertically by a factor of 2 units and then shifting it vertically up 1 unit. When doing ordered transformations it is good to show where you start and where you end up so that you can effectively compare and contrast the outcomes. See how both transformations start with a regular sine wave. The two columns represent the sequence of transformations that produce different outcomes. Example 4 What equation models the following gra ph? f⁡(x)=3⋅sin⁡x−1 Example 5 Graph the following function: f⁡(x)=−2⋅cos⁡x+1 First draw the horizontal sinusoidal axis and identify the five main points for the cosine wave. Be careful to note that the amplitude is 2 and the cosine wave starts and ends at a low point because of the negative sign. Review Graph each of the following functions that have undergone a vertical stretch, reflection, and/or a vertical shift. f⁡(x)=−2⁢sin⁡x+4 g⁡(x)=1 2⁢cos⁡x−1 h⁡(x)=3⁢sin⁡x+2 j⁡(x)=−1.5⁢cos⁡x+1 2 k⁡(x)=2 3⁢sin⁡x−3 Find the minimum and maximum values of each of the following functions. f⁡(x)=−3⁢sin⁡x+1 g⁡(x)=2⁢cos⁡x−4 h⁡(x)=1 2⁢sin⁡x+1 j⁡(x)=−cos⁡x+5 k⁡(x)=sin⁡(x)−1 Give the equation of each function graphed below. 11. 12. 13. 14. 15. This page titled 5.4: Vertical Shift of Sinusoidal Functions is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. LICENSED UNDER Back to top 5.3: Amplitude of Sinusoidal Functions 5.5: Frequency and Period of Sinusoidal Functions Was this article helpful? Yes No Recommended articles 5.7: Graphs of Other Trigonometric Functions 5.8: Graphs of Inverse Trigonometric Functions 5.1: The Unit Circle 5.3: Amplitude of Sinusoidal Functions Article typeSection or PageAuthorCK12LicenseCK-12OER program or PublisherCK-12Show TOCno Tags source@ © Copyright 2025 K12 LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? 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https://docs.oracle.com/cd/E19205-01/819-5267/bkakn/index.html
14.3.1 Output Using iostream (Sun Studio 12: C++ User's Guide) Documentation Home>Sun Studio 12: C++ User's Guide>Part III Libraries>Chapter 14 Using the Classic iostream Library>14.3 Using the Classic iostream Library> 14.3.1 Output Using iostream Sun Studio 12: C++ User's Guide Previous: 14.2 Basic Structure of iostream Interaction Next: 14.3.2 Input Using iostream 14.3.1 Output Using iostream Output using iostream usually relies on the overloaded left-shift operator (<<) which, in the context of iostream, is called the insertion operator. To output a value to standard output, you insert the value in the predefined output stream cout. For example, given a value someValue, you send it to standard output with a statement like: cout << someValue; The insertion operator is overloaded for all built-in types, and the value represented by someValue is converted to its proper output representation. If, for example, someValue is a float value, the << operator converts the value to the proper sequence of digits with a decimal point. Where it inserts float values on the output stream, << is called the float inserter. In general, given a type X, << is called the X inserter. The format of output and how you can control it is discussed in the ios(3CC4) man page. The iostream library does not support user-defined types. If you define types that you want to output in your own way, you must define an inserter (that is, overload the << operator) to handle them correctly. The << operator can be applied repetitively. To insert two values on cout, you can use a statement like the one in the following example: cout << someValue << anotherValue; The output from the above example will show no space between the two values. So you may want to write the code this way: cout << someValue << " " << anotherValue; The << operator has the precedence of the left shift operator (its built-in meaning). As with other operators, you can always use parentheses to specify the order of action. It is often a good idea to use parentheses to avoid problems of precedence. Of the following four statements, the first two are equivalent, but the last two are not. cout << a+b; // + has higher precedence than << cout << (a+b); cout << (a&y); // << has precedence higher than & cout << a&y // probably an error: (cout << a) & y 14.3.1.1 Defining Your Own Insertion Operator The following example defines a string class: include include class string { private: char data; size_t size; public: // (functions not relevant here) friend ostream& operator<<(ostream&, const string&); friend istream& operator>>(istream&, string&); }; The insertion and extraction operators must in this case be defined as friends because the data part of the string class is private. ostream& operator<< (ostream& ostr, const string& output) { return ostr << output.data;} Here is the definition of operator<< overloaded for use with string s. cout << string1 << string2; operator<< takes ostream& (that is, a reference to an ostream) as its first argument and returns the same ostream, making it possible to combine insertions in one statement. 14.3.1.2 Handling Output Errors Generally, you don’t have to check for errors when you overload operator<< because the iostream library is arranged to propagate errors. When an error occurs, the iostream where it occurred enters an error state. Bits in the iostream’s state are set according to the general category of the error. The inserters defined in iostream ignore attempts to insert data into any stream that is in an error state, so such attempts do not change the iostream’s state. In general, the recommended way to handle errors is to periodically check the state of the output stream in some central place. If there is an error, you should handle it in some way. This chapter assumes that you define a function error, which takes a string and aborts the program. error is not a predefined function. See 14.3.9 Handling Input Errors for an example of an error function. You can examine the state of an iostream with the operator !,which returns a nonzero value if the iostream is in an error state. For example: if (!cout) error("output error"); There is another way to test for errors. The ios class defines operator void(), so it returns a NULL pointer when there is an error. You can use a statement like: if (cout << x) return; // return if successful You can also use the function good, a member of ios: if (cout.good()) return; // return if successful The error bits are declared in the enum: enum io_state {goodbit=0, eofbit=1, failbit=2, badbit=4, hardfail=0x80}; For details on the error functions, see the iostream man pages. 14.3.1.3 Flushing As with most I/O libraries, iostream often accumulates output and sends it on in larger and generally more efficient chunks. If you want to flush the buffer, you simply insert the special value flush. For example: cout << "This needs to get out immediately." << flush; flush is an example of a kind of object known as a manipulator, which is a value that can be inserted into an iostream to have some effect other than causing output of its value. It is really a function that takes an ostream& or istream& argument and returns its argument after performing some actions on it (see 14.7 Manipulators). 14.3.1.4 Binary Output To obtain output in the raw binary form of a value, use the member function write as shown in the following example. This example shows the output in the raw binary form of x. cout.write((char)&x, sizeof(x)); The previous example violates type discipline by converting &x to char. Doing so is normally harmless, but if the type of x is a class with pointers, virtual member functions, or one that requires nontrivial constructor actions, the value written by the above example cannot be read back in properly. Previous: 14.2 Basic Structure of iostream Interaction Next: 14.3.2 Input Using iostream © 2010, Oracle Corporation and/or its affiliates
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https://www.youtube.com/watch?v=VjQm6eHPJ-A
Thomas Calculus || Exercise 2.7 || Question 05 || Slopes and Tangent Lines || Allah Dad Allah Dad 6650 subscribers 76 likes Description 4508 views Posted: 23 Feb 2024 Thomas Calculus || Exercise 2.7 || Question 05 || Slopes and Tangent Lines || Allah Dad Thomas Calculus 11th edition chapter number 2 limits and continuity exercise 2.7 it means section 2.7 question number 5 to 10 slope and tangent line. (00:00) Introduction (01:26) Tangent to a Circle (02:40) Slope and Tangent Line (04:04) Method of finding the tangent line (04:55) Difference Quotient (05:20) Derivative (08:36) Question NO 05 Book.............👉👉👉.........Thomas Calculus Writer............👉👉👉............Thomas Finney Edition..........👉👉👉..................11th Edition Chapter.........👉👉👉........................No # 02 Name............👉...LIMITS AND CONTINUITY ( ..............Tangents and Derivatives............... ) Exercise 2.7....................................................... Slopes and Tangent Lines................................. Question .....................................................(05) thomascalculus slopes tangents allahdad Please Like, Share, Comments, Subscribe and press the bell icon on the YouTube app. Thank You Very Much.....😊🙂 13 comments Transcript: [संगीत] बिस्मिल्लाह रहमा रहीम देखें स्टूडेंट ये आपके पास थॉमस कैलकुलस 11थ एडिशन चैप्टर नंबर टू एक्ससाइज 2.7 है देखें स्लोप्स एंड टेंज लाइन इन एक्ससाइज 5 टू 10 फाइंड एन इक्वेशन फॉर द टेंज टू द कर्व एट द गिवन पॉइंट देन स्केच द कर्व एंड टेंज टुगेदर क्या कह रहा है कि हमने इक्वेशन एक्ससाइज में क्वेश्चन 5 टू 10 में जितने भी क्वेश्चन है एक इक्वेशन ऑफ टेंज लाइन फाइंड आउट करनी होगी ठीक है फिर हमारे पास जो फंक्शन है उस फंक्शन का जो ग्राफ बनेगा इस फंक्शन का जो ग्राफ बनेगा उस ग्राफ के ऊपर हमने इक्वेशन ऑफ टेंज लाइन को ड्रा करना होगा फाइंड करने के बाद देखि न क्या कह रहा है स्केच द कर्व एंड टेंज टूगेदर कि वो जो फंक्शन का फंक्शन का जो कर्व बनेगा उस कर्व के ऊपर हमने क्या करना होगा टेंज को ड्रा करना होगा एक साथ टूगेदर ठीक है ये बात बताया अब जो गिवन पॉइंट है ना वो यहां पर कह रहा है एट द गिवन पॉइंट कि दिए गए पॉइंट के ऊपर हमने क्या फाइंड आउट करनी है इक्वेशन ऑफ टेंज एट लाइन फाइंड आउट करनी है ठीक है इक्वेशन इस गिवन पॉइंट की मदद से इक्वेशन ट लाइन फाइंड कर करेंगे ना फिर उसके बाद जो हमारे पास इक्वेशन ऑफ टेंज लाइन आ जाएगी वो जो हमारे पास फंक्शन का कर्व बनेगा उस कर्व के ऊपर उस इक्वेशन ऑफ टेंज लाइन को ड्रा कर देंगे बस सिंपल हमारा यही काम है ठीक है तो सबसे पहले हम सीख लेते हैं हमारा जो टॉपिक है वो है स्लोप एंड टेंज लाइन का बेसिकली स्लोप होता क्या है टेंज लाइन क्या होता है इसकी मदद से हम एक्ससाइज के इस क्वेश्चन को सॉल्व करते तो सबसे पहले हम इसके थोरेट्स को समझ लेते हैं कि बेसिकली स्लोप एंड टेंज लाइन होती क्या है ठीक है स्टूडेंट देखें तो देखें स्टूडेंट टेंज टू अ सर्कल क्या होता है दोस्तो इसको सीख लेते हैं देखें लाइन एल इज द टेंज टू द सर्कल एट पॉइंट प कि पॉइंट प के ऊपर ठीक है पॉइंट प पर सर्कल की जो टेंज लाइन है वो कौन सी है एल है देखें कौन सा पॉइंट ये वाला पॉइंट अब ये आपके पास क्या है सर्कल है इस सर्कल में हमारे पास यहां पर भी पॉइंट यहां पर भी पॉइंट यहां पर भी हर जगह पर कोई ना कोई पॉइंट होगा ना तो ये उस पॉइंट को हमने यहां पर एक पॉइंट है इस पॉइंट को हमने क्या कहा हुआ है प कहा हुआ है ठीक है उस पॉइंट प के ऊपर सर्कल की सर्कल की टेंज लाइन है वो कौन सी है पॉइंट p के ऊपर सर्कल के पॉइंट प के ऊपर जो टेंज लाइन है वो कौन सी है l है किस तरह बनती है इफ इट पास थ्रू द पॉइंट p एंड परपेंडिकुलर टू द रेडियस o देखिए आपके पास ये सर्कल है इस सर्कल का ये क्या है सेंटर है जीरो का नाम दिया हुआ है इस सेंटर o से लेकर p तक आपके पास ये क्या है सर्कल का रेडियस है इस रेडियस के जो परपेंडिकुलर है इस रेडियस के क्या है ये देखें परपेंडिकुलर लाइन लगाई गई है ये लाइन क्या है टेंज लाइन है सर्कल के ऊपर क्योंकि ये देखें आपको यहां पर नजर भी आ रहा है क्योंकि ये सर्कल के साथ टच करके गुजर रही है तो ये कौन सी लाइन है लाइन है सर्कल के ऊपर किस पॉइंट के ऊपर p के ऊपर रेडियस के क्या ओ रेडियस के क्या है परपेंडिकुलर है यही बात बताई जा रही है इसके ऊपर आगे चलते हैं स्टूडेंट आगे क्या बात बताई जा रही है स्लोप एंड टेंज क्या होता है देखें स्टूडेंट अ स्लोप द स्लोप ऑफ द कर्व y इ इक्वल ट f f एक आपके पास कोई भी फंक्शन है इस फंक्शन में आपने ना कोई रियल नंबर वैल्यूज जैसे कि नेचुरल नंबर आपने इंटी जर पुट करने है ठीक है वो पुट करने के बाद आपके पास एक्सवा में कोई ग्राफ जनरेट होगा ना व वो बेसिकली क्या होगा कर्व होगा वो उसका कर्व के ऊपर स्लोप को हम डिफाइन कर रहे हैं उस कर्व के ऊपर स्लोप को किस तरह से डिफाइन करेंगे देखिए द स्लोप ऑफ द कर्व y = f x एट द पॉइंट कौन सा पॉइंट p x नॉ और ये कौन सा f x नॉ बेसिकली y नॉट है तो इस पॉइंट के ऊपर कर्व का जो स्लोप बनेगा वो कौन सा होगा स्लो का क्या फार्मूला है इज द नंबर m इक्वल टू लिमिट h अप्रोच टू 0 f x0 + h - f x0 डिवा h अगर इसकी लिमिट प्रोवाइडेड है एजिस्ट करती है तब फिर आपके पास क्या कहेंगे वो वो क्या होगा स्लोप होगा स्टूडेंट ठीक है अब बेसिकली ये डेरिवेटिव डेरिवेटिव की डेफिनेशन लिखी हुई है ठीक है मैं आपको आगे चलके बताऊंगा किस तरह से ये आगे क्या बता रहा है द टेंज लाइन टू द कर्व एट पॉइंट p कि देखिए p कोई भी पॉइंट है कर्व का उसके ऊपर अगर कोई टेंज एट लाइन हमने ड्रा करनी है तो वो किस तरह है इज द लाइन थ्रो p विद द स्लोप m तो स्लोप m और उस थ्रो पॉइंट p के थ्रू क्या होगी एक लाइन होगी टेंज लाइन क्या है बेसिकली एक लाइन है कर्व के ऊपर एक लाइन है जिसका स्लोप क्या रिप्रेजेंट कर रहा है m स्लोप क्या होता है m ठीक है अब देखें आगे क्या कह रहा है टेंज को किस तरह से हम फाइंड करेंगे स्टूडेंट फाइंड करने के लिए टेंज को देखें ये कुछ स्टेप है इस इन्हीं स्टेप को हमें फॉलो करना होगा कैलकुलेट सबसे पहले हम f x नॉ कैलकुलेट करेंगे ठीक है जो फंक्शन गिवन होगा उसमें आपने x नॉ पुट कर देना है तो फिर x नॉ में प्लस h ऐड करना है ये दोनों चीजें कैलकुलेट करनी है फिर कैलकुलेट स्लोप को कैलकुलेट करना है किस तरह से उसका फार्मूला स्लोप का आप याद रखें m = लिमिट h अप्रोच टू 0 f x नॉ + h - f x न डि ब बहुत ही आसान सा फार्मूला है स्लोप का आपने ये उसकी ये दोनों स्टेप करने के बाद ये फार्मूला में वैल्यूज को पुट करना है सिंपलीफाई करना है तो स्लोप m आ जाएगा फिर उसके बाद आपने क्या करना है अगर इसकी लिमिट ये वाली लिमिट एजिस्ट करती है न फिर हम नेक्स्ट स्टेप में चले जाएंगे फाइंड द टेंज लाइन इस तरह से फाइंड करेंगे देखिए ये फार्मूला है टेंज लाइन का y = y नॉ + m x - x नॉ ठीक है अब लास्ट एक छोटी सी डेफिनेशन है स्टूडेंट वो करते हैं फिर उसके बाद एक्सरसाइज के कुछ रिमेनिंग क्वेश्चन को सॉल्व करेंगे स्टूडेंट अब देखें आपके पास ये एक डेफिनेशन है डेफिनेशन क्या बोलते हैं स्टूडेंट देखें डिफरेंस क्वेश की डेफिनेशन है द एक्सप्रेशन f x न प् h - f x0 / h यहां पर पहले क्या था हमारे पास स्लो न टेंज में हमने पढ़ा था लिमिट यहां पर थी h अप्रोच टू 0 लेकिन वो लिमिट नहीं है लिमिट को छोड़ कर उसके अंदर वाला जो हिस्सा है ना ये वाला इसको क्या बोलते हैं डिफरेंस क्वेश बोलते हैं फंक्शन का ट पॉइंट x नॉ के ऊपर ठीक है विद इंक्रीमेंट h के ऊपर अब इसी तरह देखें डेरिवेटिव क्या है डेरिवेटिव का कांसेप्ट मैंने आपको बता रहा है देखिए यहां पर लिमिट इफ द लिमिट ऑफ द एक्सप्रेशन f x नॉ - f x नॉ डिवाइड बा h तो ये क्या है अगर मैं अगर यहां पर लगा दूं लिमिट h अप्रोच टू 0 और ये चीज ठीक है तो वो क्या बनता है हमारे पास स्लोप बनता है लेकिन अगर ना इसकी लिमिट यहां पर इसकी लिमिट एजिस्ट करती हो ठीक है लिमिट h अप्रोच टू 0 और ये वाला फंक्शन जैसे एजिस्ट करती हो जैसे ही हम लिमिट में x अप्रोच टू x नॉट अप्लाई करें ठीक है इसकी लिमिट एजिस्ट करती हो तब हम कहेंगे कि लिमिट इज कॉल्ड डेरिवेटिव ऑफ द फंक्शन f x एट x = x नॉ बेसिकली ये चीज किसका है ये ये क्या बता रहा है आपके पास डेरिवेटिव की डेफिनेशन बता रहा है डेरिवेटिव की डेफिनेशन क्या होती है लिमिट मैं यहां पर नीचे ख लिख रहा हूं देखें लिमिट h अप्रोचेबल निकलता है इसका जो भी आंसर निकलता है इट मीन आपने उस फंक्शन के डेरिवेटिव का आंसर निकाला हुआ है ठीक है ये डेरिवेटिव की डेफिनेशन है डेरिवेटिव का ही फार्मूला है वैसे तो हमें पता है डेरिवेटिव का फार्मूला होता है हमारे पास जैसा कि हमने फंक्शन का डेरिवेटिव लेना है वो किस तरह लेते हैं मान ले आपने डेरिवेटिव ना है वो पावर रूल की मदद से d / [संगीत] इससे भी आंसर वही निकलता है जो इससे निकलता है लेकिन हमने ये वाला यूज़ करना है ठीक है स्टूडेंट इसकी मदद से इसका जो आंसर आएगा वही स्लोप होगा ठीक है तो अब हम एक्ससाइज के क्वेश्चन को सॉल्व करते हैं स्टूडेंट स्टेप बाय स्टेप तो अब स्टूडेंट यहां पर आप क्वेश्चन के स्टेटमेंट पढ़े ये क्या बोल रहा है स्टूडेंट स्टेटमेंट के ऊपर ये देखें स्टूडेंट ये देखिए इन एक्सरसाइज 5 टू 10 फाइंड एन इक्वेशन फॉर द टेंज टेंज की इक्वेशन का फार्मूला क्यों कौन सा होता है स्टूडेंट अभी हमने पढ़ा है इक्वेशन ऑफ टेंज कहां हमें पता है हमारे पास फार्मूला होता है y = y नॉ + m x - x नॉ हमें ये तो पता है y नॉ और x नॉ को अगर हम इस पॉइंट में से चूज कर ले ठीक हो गया फिर m की m क्या होगा m होगा स्लोप ठीक है फिर हमने स्लोप भी तो फाइंड आउट करना है तो स्लोप का भी हमें फार्मूला पता है स्टूडेंट टेंज लाइन के लिए हमें लाजमी है इस स्लोप की वैल्यू को फाइंड करना इसके लिए अलग से फार्मूला होता है m इक्वल टू अभी हमने पढ़ा है स्लोप का फार्मूला क्या होता है लिमिट h डिवाइडेड बाय h इसकी लिमिट एजिस्ट करनी चाहिए तब जाकर इसका जो आंसर आएगा वो क्या होगा आपके पास स्लोप होगा ठीक है स्टूडेंट अब अगर आप गौर से देखे ना स्टूडेंट यहां पर तो हमें सबसे पहले इक्वेशन टेंज लाइ फाइंड आट करनी है लेकिन टेंज लाइट में हमने देखा हमारे पास m पड़ा हुआ है तो m के लिए हमारे पास ये फार्मूला होता है तो सबसे पहले हम ये फार्मूला लगाकर m की वैल्यू निकालते हैं m की फम वैल्यू यहां पर रखेंगे तो हमारे पास क्या आ जाएगी इक्वेशन ऑफ टेंज लाइन आ आ जाएगी फिर उस उसमें पॉइंट की मदद से हमने उसको कर्व ड्रा करना होगा कर्व के ऊपर फिर हमने इक्वेशन ऑ टेंज लाइन को क्या करना होगा स्केच करना होगा ठीक है तो अब मैं आपको ये स्टेप बाय स्टेप करके दिखाता हूं ठीक है स्टूडेंट तो देखिए आपके पास ये क्वेश्चन नंबर फाइव है स्टेप बाय स्टेप सॉल्व करते हैं स्टूडेंट तो देखें हमें क्या चाहिए इक्वेशन ऑ टेंज लाइन चाहिए तो इक्वेशन ऑ टेंज लाइन से पहले हमें यहां पर m पड़ा हुआ है तो m का फार्मूला ये होता है तो इसमें हमें सबसे पहले कौन सा चाहिए f x न चाहिए स्टूडेंट तो f x न देखें हमने y किसके इक्वल होता है y बेसिकली आपके पास f एक के इक्वल होता है ठीक है तो मैं यहां पर लिखता हूं गिवन है हमारे पास f ऑफ एक क्या है स्टूडेंट f एक है आपके पास 4 - x स् हियर तो अब हमें क्या चाहिए स्टूडेंट हमें चाहिए f x नॉ तो जहां पर x होगा उसकी जगह आपने रिप्लेस कर देना है x नॉ तो क्या बनेगा f x0 = 4 - x0 की पावर 2 ठीक है अब हमें हमारे पास ये चीज आ गई स्टूडेंट अब हमें और क्या चाहिए स्टूडेंट और हमें चाहिए f x0 + h यानी कि x में + h को ऐड कर देता हूं तो क्या बनेगा स्टूडेंट देखें यहां पर जहां पर भी आपके पास x नट होगा ना देखिए ना हमें ये चाहिए था ये मिल गया ठीक है अब हमें क्या चाहिए f x नॉ प् h चाहिए यानी कि x नॉ के अंदर आपने h को ऐड करना है तो क्या बनेगा यहां से देखें तो मैं ऐड करता हूं देखें f ऑफ x न इसमें प्लस h को ऐड कर दें तो मैं अगर h को ऐड करूं तो क्या बनेगा देखें 4 माइनस हम किसम ऐड कर रहे हैं x नॉ में h को ऐड कर ें हैं फोर में ऐड नहीं कर रहे तो फोर एज इट आएगा x नॉ में h कर रहे हैं तो x नॉ में h को आप ऐड कर रहे हैं तो ये उसका स्क्वायर एज इट है तो वो एज इट आएगा देखें हमें ये चीज भी मिल गई हमें ये चीज भी मिल गई अब हमने फाइनली लिमिट लगानी है तो अब हम लिमिट लगाते हैं देखें स्टूडेंट तो मैंने यहां पर लिखा है सिंस वी नो दैट फॉर स्लोप स्लोप आगे मैंने ब्रैकेट में लिखा है m तो उसका क्या फार्मूला है ये वाला फार्मूला तो देखिए m इ इक्वल टू आपके पास क्या है लिमिट h अप्रोचेबल निकाली है ये वाली वैल्यू यहां पर रख दें तो क्या बनेगा देखें 4 - x0 + h और उसकी पावर टू आगे क्या है आपके पास माइनस f x नॉ f x नॉ क्या अपने ये वैल्यू निकाल ली तो यहां पर रखें ब्रैकेट लगाए क्योंकि माइनस है 4 - x नॉ की पावर 2 डिवाइडेड बाय नीचे आपके पास क्या है स्टूडेंट नीचे देखें एच है एच को आपने एज इट रखना है वैसे ठीक है अब आप इसको सिंपलीफाई करें देखिए स्टूडेंट अगर आप गौर से देखें ना अगर मैं डायरेक्ट यहां पर h मैं पुट कर दूं रो तो समथिंग और जीरो इंफिनिट बन जाएगा वो लिमिट एजिस्ट नहीं करेगी तो लिमिट को एजिस्ट करवाने के लिए लाजमी है कि नीचे वाले एज को आप खत्म करें किसी ना किसी तरीके से तो नीचे वाले एच को खत्म करने के लिए ऊपर वाले मुझे स्क्वेयर ओपन करने पड़ेंगे तो मैं इसके स्क्वेयर को ओपन करता हूं देखें स्टूडेंट देखिए आप आपने इसके स्क्वायर को ओपन करना किस तरह से करेंगे देखि इसका स्क्वायर ओपन किस तरह होगा एक्स नॉ की पावर 2 प्लस ए की पावर टू प्स 2 ए एक्स नॉ इस तरह ओपन होगा ना फिर आप एज लिखें 4 माइन एक्स नॉ का स्क्वायर ठीक है नीचे डिवाइडेड बाय क्या है आपके पास एच एच को वैसे ही लिखा रहने दे ठीक है तो अब इसको मैं सॉल्व करता हूं देखि सॉल्व किस तरह से करेंगे आपके पास क्या है लिमिट ए मल्टीप्लाई करें तो क्या बनेगा 4 - x नॉ की पावर 2 - h का स्क्वा - 2hx न -4 और ये माइनस और ये माइनस मिलके पॉजिटिव हो जाएंगे x नॉ की पावर 2 डिवाइडेड बा आपके पास क्या बनेगा स्टूडेंट नीचे h बनेगा अब अगर आप गौर से देखें ना जो चीज कैंसिल आउट होती है स्टूडेंट उन्हें आप कैंसिल आउट करने की कोशिश करें अगर गौर से देखें ये देखें पॉजिटिव फोर है ये नेगेटिव फोर है कैंसिल आउट हो गया ये देखिए आपके पास क्या है - x नॉ का स्क्वायर है और ये क्या है पॉजिटिव x नॉट का स्क्वायर है कैंसिल आउट हो गया इन इन दोनों में से आपने h को कॉमन निकाल लेना है तो क्या बनेगा देखें लिमिट h अप्रोचेबल गे तो क्या बनेगा देखें यहां से मैं अगर h को कॉमन निकालूं तो अंदर क्या बनेगा - h -2 ठीक है h कॉमन आ गया अंतर आपके पास रह गया x नॉ डिवाइडेड बा नीचे आपके पास क्या है h अब देखें इस h से ये वाला एच कैंसिल आउट हो जाएगा अब नीचे तो कुछ है नहीं तो हम ऐसा जीरो तो पोट कर नहीं सकते क्योंकि नीचे h खत्म हो चु गया अब हमारे पास लिमिट का कोई ना कोई फाइना इट आंसर आएगा तो देखें उसके लिए लिमिट में h की जगह 0 रख द तो ये क्या है -0 -2 एक न तो आंसर क्या आया स्टूडेंट -2 और ये क्या है x नॉट तो x नॉट आपके पास पॉइंट कौन सा गिवन था स्टूडेंट मैं आपको पॉइंट दिखाता हूं देखें ये देखें ये वाला पॉइंट गिवन था ना क्वेश्चन की स्टेटमेंट में -1 और थ तो आपके पास ये कौन सा पॉइंट है स्टूडेंट ये आपके पास है x नॉट और ये आपके पास कौन सा है आप इसे कंसीडर करें y नॉट ठीक है तो ये x नॉट y नॉ तो x0 की वैल्यू कौन सी है -1 तो मैं -1 यहां से उठाकर यहां पर पुट कर देता हूं इसमें x नॉ की जगह क्या रख देता हूं -1 तो क्या बनेगा आपके पास पॉजिटिव 2 ठीक है अब अगर आप गौर से देखें ना स्टूडेंट तो इस फंक्शन का मैंने क्या लिया डेरिवेटिव लिया डेरिवेटिव को फिर स्लोप के इक्वल पुट करते हैं तो m की वैल्यू क्या आ गई टू आ गई ये देखिए ना स्लोप m की वैल्यू किसके इक्वल आ गई है ये टू के इक्वल अगर मैं डायरेक्ट डेरिवेटिव भी लेता ना इस फंक्शन का ये देखें इस फंक्शन का डेरिवेटिव लेता तो क्या बनता मेरे पास देखें डेरिवेटिव ऑफ f x f x का डेरिवेटिव d / dx2 = d / ठीक है डायरेक्ट जो जो हम आगे चैप्टर नंबर थ्री में पढ़ेंगे तो इसका इसका डेरिवेटिव क्या बनेगा आपके पास f का फर्स्ट डेरिवेटिव इज इक्वल टू फोर कांस्टेंट इसका डेरिवेटिव जीरो माइनस वैसे ठीक है आपके पास तो ये क्या है पावर टू है आपके पास तो अब हमने पावर को लगाना है इसमें तो क्या बनेगा आपके पास टू और ये क्या बनेगा x बनेगा ठीक है तो अब आपने क्या निकालना है स्लोप की वैल्यू निकालना है तो ये आपके पास डेरिवेटिव आया f का फर्स्ट डेरिवेटिव इनटू x आया तो इसमें आपने स्लोप की वैल्यू निकालनी है किस पॉइंट के ऊपर -1 के ऊपर x नॉट की वैल्यू कौन सी है या x = माइ 1 के ऊपर निकालेंगे तो -2 -1 तो आंसर इज इक्वल टू 2 तो ये डेरिवेटिव का डायरेक्ट डेरिवेटिव का क्या है आपके पास डायरेक्ट रूल है ये आपने यूज नहीं करना वैसे मैं समझाने के लिए कर रहा हूं आपने ये वाली डेफिनेशन को फॉलो करना है स्टूडेंट ठीक है तो अब हमने देखें हमने ये स्लोप निकाला f का फर्स्ट डेरिवेटिव लेकर उसमें आपने क्या निकालना है x की जगह -1 रखना है जो पॉइंट गिवन है आपके पास ये देखें ना यहां पर आपके पास पॉइंट कौन सा गिवन अब हम हमारे पास यहां पर कौन सा था x था ये देखें ये x है आपके पास ठीक है ये देखें ये x था तो अब आपने इसको x कंसीडर करना है ये इस वाले को अगर x नॉ है तो x न कंसीडर करें हमारे पास x था तो मैंने x x को -1 पुट कर दिया तो स्लोप आ गया टू ठीक है अब इसी तरह हमारे पास यहां पर क्योंकि x नॉट था स्टूडेंट ये देखिए यहां पर x नॉ था तो इस वजह से मैंने इसे कहा x नॉ ठीक है तो x नॉट की वैल्यू कौन सी आई -1 पुट करेंगे तो आंसर क्या आएगा टू तो अब ये ये क्या है आपके पास स्लोप की वैल्यू है स्टूडेंट ठीक है अब इसी तरह स्टूडेंट अब हमें ये चीज मिलेगी हम इक्वेशन ऑफ टेंज लाइन को इजली फाइंड कर सकते हैं ठीक है तो ये देखें इस इक्वेशन में पुट कर इक्वेशन ऑफ टेंज लाइन हमारा फाइनल काम कौन सा रह गया इक्वेशन ऑ टेंज लाइन को फाइंड आउट करना तो मैं आप लिखता हूं नो वी नो दैट फॉर टेंज तो मैं लिखता हूं टेंज के लिए स्टूडेंट हमें पता है हमारे पास इक्वेशन होती है फार्मूले के अकॉर्डिंग y = होता है y नॉ + m x - x0 स्टूडेंट अगर आप गौर से देखें ना आपके पास पॉइंट गिवन है कौन सा पॉइंट गिवन है स्टूडेंट -1 और 3 तो अगर आप गौर से देखें इस -1 को आप कहते x नॉट ठीक है और इस थी को आप कहते y नॉट इस पॉइंट के इक्वल करके यहां पर रख दें तो क्या बनेगा देखें y नॉ की वैल्यू कौन सी है स्टूडेंट थ्री के इक्वल है तो थी पुट करें m की वैल्यू आपने निकाली है स्टूडेंट टू ये देखें टू है तो आपने टू पुट करें x को आपने एज रखना है - x नॉ की वैल्यू आपने कौन सी लिखी है -1 तो माइनस और माइनस मिलक पॉजिटिव वन हो जाएंगे इसे आपने सॉल्व करना है स्टूडेंट 3 + 2x + 2 तो आंसर क्या बनेगा 2x + 5 तो टेंज लाइन की कौन सी इक्वेशन बनी है हमारे पास टेंज लाइन कौन सी है y = 2x + 5 इज द टेंज तो मैं लिखता हूं इज दी टेंज टू दी पॉइंट p -1 और 3 ये इक्वेशन किस पॉइंट के ऊपर टेंज एट होगी p -1 और 3 पॉइंट के ऊपर टेंज होगी बताऊंगा आगे मैं ग्राफ में जैसे जनरेट करूंगा ना कुछ चीजें आपको ग्राफ में क्लियर होंगी स्टूडेंट ठीक है अब हम इसके ग्राफ को जनरेट करेंगे तो ग्राफ को जनरेट करने के लिए सबसे पहले हमने उसमें इंटी जर रियल नंबर वैल्यूज पुट करेंगे ठीक है पहले फंक्शन का कर्व जनरेट करेंगे फंक्शन का कर्व जो आ जाएगा ना फिर उस फंक्शन के कर्व के ऊपर इस टेंज लाइन को ड्रा कर देंगे फाइनल हमारा काम हो जाएगा बस खत्म ठीक है तो अब हम इसको सॉल्व करें हैं देखें तो स्टूडेंट अब यहां पर आप देखें तो देखें स्टूडेंट आपके पास मैं आपको सिखा रहा हूं कि आपने फंक्शन का ग्राफ किस तरह से जनरेट करना है x प्लेन के ऊपर ठीक है तो आपके पास फंक्शन कौन सा गिवन था y = 4 - x स् था ना तो सबसे पहले आपने ना चूज करना है एट तो मैं लिखता हूं सबसे पहले ट x = 0 के ऊपर y का कौन सा आंसर बनेगा यहां से देखें y इ x को आपने रो पुट किया तो रो का स्क्वा 0 तो y का आंसर क्या बनता है स्टूडेंट फोर बनता है इट मीन आपके पास एक पॉइंट मिल गया हमें कौन सा पॉइंट मिल गया है जब x 0 है तब y का आंसर क्या है फोर है एक पॉइंट आपके पास ये है अब इसी तरह स्टूडेंट एक और पुट करें जब ट x = अब x को बढ़ाएं रो से लेकर वन फिर उसके बाद y का क्या आंसर बनेगा y इ टू जब x को आपने रखा वन तो वन का स्क्वायर वन तो y का आंसर क्या आया 4 - 1 3 तो y का आंसर आया थी तो एक और पॉइंट हमें मिला है p जब x1 है और y का आंसर क्या है आपके पास थ्री है ठीक है इसी तरह एक और पॉइंट भी लिख रहे हैं स्टूडेंट एट x = 2 तो y का आंसर क्या बनेगा देखिए जब आप x को टू पुट करेंगे 2 टू 4 - 4 0 तो y का आंसर क्या बन रहा है स्टूडेंट्स जीरो बन रहा है तो इसमें एक और पॉइंट मिल गया जब x2 है तो y क्या बनता है जीरो इसी तरह आपके पास और भी बहुत सारे पॉइंट मिलेंगे आपके पास ठीक है समझाने के लिए इतने पॉइंट मैंने आपको यहां पर लिखे हैं अब अब यहां पर देखें इक्वेशन ऑफ टेंज लाइन में आपने पॉइंट्स को देखें तो यहां पर जब आपके पास x क्या है जीरो है स्टूडेंट तोत y का आंसर क्या बनता है जब x 0 है तो y का आंसर क्या बनेगा फव बनेगा यहां से तो क्या बनता है देखें पॉइंट कौन सा मिलेगा आपके पास जब x जीरो है तो y का आंसर क्या बनता है फ बनता है कौन सा y टेंज लाइन टेंज लाइन का आंसर ठीक है इसी तरह स् आगे देखें जब x को आप बढ़ाए x = 1 रखें तो y का आंसर क्या बनेगा x को मैंने रख दिया वन तो y का आंसर क्या बनेगा 2 प् 4 7 तो इसका आंसर एक और पॉइंट मिल गया हमें p इन जब x1 है तब y का आंसर क्या है स्टूडेंट सेन है ठीक है अब इसी तरह देखें स्टूडेंट तो अब आप एक एलिमेंट पॉजिटिव रखा है एक एलिमेंट नेगेटिव रखें x = -1 के ऊपर y का आंसर क्या बनता है देखिए -1 रखेंगे तो ये क्या बनेगा -2 प् 5 तो बनता है क्या 3 तो एक पॉइंट हमें मिल गया p -1 और 3 इसी तरह अगर हम माइनस अब देखि अगर हम टू रख द अब x = 1 रख दिया फिर माइव भी रख के देखा अगर हम x = 2 रखें तो टू पर क्या बनेगा देखें y का आंसर क्या बनेगा 2 टू 4 प् 5 9 बनता है तो एक पॉइंट हमें मिल गया p इन कौन सा 2 और 9 का इसी तरह एक और पॉइंट रख क्या माइनस वाला तो क्या बनता है x इ ट अगर मैं -2 रखूं तो y का आंसर क्या बनेगा -2 -4 + 5 तो ये क्या बनता है वन बनता है तो एक पॉइंट हमें मिल गया p -2 और वन का पॉइंट मिल गया स्टूडेंट ठीक है अब ग्राफ जनरेट करते हैं देखें तो स्टूडेंट मैं यहां पर इसके ग्राफ को जनरेट करने के लिए यहां पर मैं एक एक्सवा प्लेन बना लेता हूं आपको समझाने के लिए इस तरह से ये देखें ये मैंने कोई प्लेन बनाया इस तरह से ठीक है अब जरा आपने बात को समझना है स्टूडेंट ये आपके पास इस क्वेश्चन का लास्ट एंड है बस ठीक है अब देखें यहां पर ये आपके पास क्या है जीरो है स्टूडेंट ठीक है ये आपके पास क्या है वन है ये आपके पास क्या है -1 है ये आपके पास क्या है स्टूडेंट टू होगा यहां पर क्या होगा -2 होगा इसी तरह यहां पर आपके पास क्या होगा स्टूडेंट थ्री होगा यहां पर आपके पास क्या होगा फोर होगा अप टू सो ऑन ठीक है अब इसी तरह यहां पर आप देखें स्टूडेंट मैं आपको यहां पर ये दिखा देता हूं ये देखें यहां पर क्या होगा आपके पास -3 होगा अच्छा ये वाला जो है वो ये वाले जो पॉइंट्स निकले हैं किस फंक्शन के लिए निकले हैं जब फंक्शन y इ इक्वल 4 - x किया था और ये किसके लिए निकले हैं इक्वेशन ऑफ टेंज के लिए y = 2x प् 5 तो मैंने नीचे अगेन लिख दिया क्योंकि ऊपर आपको नजर नहीं आ रहे थे तो मैंने एक साथ आपको समझाना है तो इसलिए मैंने अगेन लिख दिया तो ये आपके पास टेंट के लिए है टेंट के लिए पॉइंट्स निकल के आए हैं और ये आपके पास क्या है फंक्शन जो गिवन था उस पर जो आपके पास पॉइंट बन रहे हैं वो ये है ठीक है तो अब हमने हमने यहां पर देखें जीरो के बाद वन पुट किया टू पुट किया आप नेगेटिव भी पुट कर सकते हैं ठीक है पॉजिटिव वन नेगेटिव पुट करें फिर टू फिर नेगेटिव टू पुट करें जितना मैंने यहां पर किया वन फिर -1 फिर टू फिर -2 अप टू सो न इस तरह से ठीक है अब मैं आपको ग्राफ को समझा समझाने की कोशिश करता हूं स्टूडेंट ये देखें आपके पास ना सबसे पहले आप देखें पहले उसने उसने क्या कहा उसने कहा आपने ना फंक्शन का जो कर्व बनेगा वो और टेंज लाइन को टूगेदर एक साथ ड्रा करना है ये क्वेश्चन की स्टेटमेंट में हमसे पूछा गया था ना सबसे पहले हम ये ड्रा करते हैं इसके ऊपर ये इस टेंज लाइन को ड्रा कर देंगे फिर तो सबसे पहले हम इस फंक्शन के कर्व को जनरेट करते हैं तो कार को किस तरह से जनरेट करेंगे जब मैंने x को जीरो रखा ये देखें ये x वाली लाइन है ये कौन सी है y वाली लाइन है ठीक है वा वाली लाइन है ये वाली जब मैंने x को रखा जीरो तो y का आंसर क्या मिला मुझे फोर ये देखें x 0 है y 4 है जब मैंने x को रखा 0 y का आंसर क्या मिला मुझे यहां पर क्या होगा वन यहां पर क्या होगा टू यहां पर क्या होगा थ्री यहां पर क्या होगा फोर तो y का मुझे आंसर मिला फोर एक पॉइंट यहां पर लाई करता है ठीक है एक पॉइंट ये एक पॉइंट ये तो फर्स्ट पॉइंट मैंने ड्रा कर दिया अब नेक्स्ट पॉइंट को देखें जब x = 1 था तो y का आंसर क्या था 3 था जब x = 1 था तोत y का आंसर क्या था स्टूडेंट y का आंसर क्या था थ्री था ये देखें x1 है तो y 3 है तो ये पॉइंट कहां पर लाई करता है जब x देखें वन है y क्या है थ्री है 1 2 3 तो यहां पर लाई करता है पॉइंट देखें यहां पर ये पॉइंट यहां पर लाई करता है ठीक है दूसरा पॉइंट भी हमने ड्रा कर दिया जब x व है और y का आंसर क्या है थ्री तो ये यहां पर लाई करता है पॉइंट अब दूसरे पॉइंट की तरफ आते हैं जब x2 है तब y का आंसर क्या है स्टूडेंट जीरो है यहां पर देखिए जब x क्या है टू है जब x क्या है स्टूडेंट टू है तब वा का आंसर क्या है जीरो है इट मीन एक पॉइंट हमें यहां पर मिलता है स्टूडेंट ठीक है अब इसी तरह आगे आप नेक्स्ट आपने क्या करना है माइनस वाले पॉइंट को पुट करेंगे माइनस वाले तो माइनस वाले पॉइंट किस तरह आएंगे वो इस तरह आएंगे स्टूडेंट इस तरह ये देखें मैं यहां पर इसको लिखता हूं अब क्योंकि ये मैंने पॉजिटिव नंबर ऑफ एलिमेंट्स रखे तो पॉजिटिव एरिया ये देखिए ये वाला एरिया कौन सा है पॉजिटिव और ये वाला कौन सा है नेगेटिव तो मैंने कौन से पॉइंट रखे हुए हैं पॉजिटिव तो पॉजिटिव साइड प मुझे इसके पॉइंट्स का ग्राफ मिलेगा इस तरह ठीक है अब मैंने यहां पर भी मैंने नेगेटिव में रखे तो आप यहां पर भी नेगेटिव रखें तो आप नेगेटिव रखेंगे तो क्या बनेगा देखिए नेगेटिव की साइड पर आए जब x = -1 मैं पुट करता हूं इस फंक्शन में यहां पर इसमें देखिए x को मैं क्या पुट करता हूं -1 तो फिर y का आंसर क्या बनेगा जब मेरे पास पॉइंट कौन सा है यहां पर य मैं लिखि यहां पर लिखता हूं देखि x जब मैं -1 रखता हूं तो फिर y का आंसर क्या बनता है -1 का स्क्वा पॉजिटिव व तो माइनस तो आंसर क्या बनता है थ्री तो जो ये वाला पॉइंट कहां पर नहीं करेगा देखिए माइनस वाला मैंने नहीं रखा तो वो भी रख ले यहां पर तो क्या बनेगा जब x -1 है y का आंसर क्या है थ जब x -1 है तो y का आंसर क्या है थ्री 1 2 3 तो थ पॉइंट यहां पर ला करता है स्टूडेंट ठीक है अब इसी तरह जब x को आप -2 रखेंगे तब आपके पास क्या कौन सा ग्राफ मिलेगा आपको -2 0 मिलेगा तो फिर आपके पास ये ये ये यहां पर पॉइंट लाई करेगा तो अब आपने अगर गौर से देखें ना तो आपके पास ग्राफ किस तरह से जनरेट हो रहागा स्टूडेंट इस तरह से जनरेट हो रहा है देखिए इस तरह से ये वाले पॉइंट को आपने इस तरह से मिलाना है ठीक है यूं और फिर आपने इसको ये मुझे अच्छे अच्छी तरीके से नहीं बन रहा आपने अच्छी तरीके से बनाना है या आपके पास बहुत ही खूबसूरत सा बनेगा ये देखिए इस तरह से ये इस तरह से आपने इस पॉइंट को मिलाना है ठीक है ये इस तरह से आपके पास ग्राफ बनेगा ये ठीक है स्टूडेंट अब आपने ना ये इसका काम खत्म हो गया अब आपने टेंज क बना वो क्वेश्चन के स्टेटमेंट में ये कहा था कि आपके पास जो आंसर है ना उस आपके पास जो ग्राफ बनर है उस ग्राफ के ऊपर जो कर्व के ऊपर आपने इक्वेशन ऑफ टेंज लाइन को एक साथ ड्रा करने है अभी इक्वेशन टेंज लाइन के लिए देखें जब आपने पॉइंट x = 0 रखा स्टूडेंट तो y का आंसर क्या बना 0 और 5 बना तो कहां पर ये लाई करना है देखिए इसके पॉइंट को पहले आप ड्रा करें जब आपने x को रखा रो y का आंसर बना फ जब x क्या है 0 y का आंसर क्या है फाइव तो 1 2 3 4 5 तो फाइव यहां पर पॉइंट तो पॉइंट यहां पर बन रहा है एक टेंज का ठीक है अब नेक्स्ट आगे आए स्टूडेंट जब आपने x को रखा वन तो y का आंसर क्या आया 7 x को रखा वन तो y का क्या है सेन ये वन है y का y2 y3 y4 y5 ठीक है y6 y7 यहां पर होगा तो एक पॉइंट हमें यहां पर मिल मिल रहा है स्टूडेंट ठीक है ना जरा बात को समझे अब इसी तरह अ आगे हमने जब x को हमने माइनस मा रखा तो y का आंसर क्या बना जब x को हमने रखा -1 और y का आंसर क्या बना थ्र तो देखें x जब आपके पास क्या है यहां पर -1 यहां पर है ना जब x आपके पास -1 है तो y का आंसर क्या बन रहा हैथ बन रहा है तो देखें x जब -1 है तो y का आंसर क्या है 1 2 3 तो y का आंसर यहां पर बन रहा है एक पॉइंट यहां पर मि मिल गया हमें एक पॉइंट ये एक पॉइंट ये एक पॉइंट ये ये वाले तीन पॉइंट आपने याद रखने है स्टूडेंट ठीक है तो मैं इसको रेड कलर से डिनोट करता हूं आपको समझाने के लिए तो देखें ये मैं मैंने रेड कलर से लिखने की कोशिश कर तो आपको टेंज का समझ जाए तो ये कौन से वाले पॉइंट है टेंज एट के पॉइंट है स्टूडेंट ठीक है अब उसके बाद आगे देखें जब x को मैंने रखा टू y का आंसर क्या आया नाइन जब x को मैंने रखा टू y का आंसर क्या आया नाइन तो ये x को टू है y1 y2 y3 होगा y4 होगा y5 होगा y6 होगा y7 होगा y8 y9 तो ये यहां पर पॉइंट बनेगा इस तरह से ठीक है ये इस तरह से स्टूडेंट ठीक है अब इसी तरह आगे वाला पॉइंट यहां पर बनेगा फिर यहां पर फिर यहां पर इस तरह चलता जाएगा ऊपर अब नेगेटिव वाला देखते हैं जब x को मैंने रखा -2 y को फिर रखा वन y का आंसर क्या आया वन जब x -2 है तो y क्या है वन है ये वाला पॉइंट जब x क्या है आपके पास -2 है y का आंसर क्या है स्टूडेंट y का आंसर वन है तो इट मीन यहां पर बनता है आपके पास पॉइंट जब x क्या आपके पास - 2y का आंसर क्या है वन ये देखें ये है वन तो इट मीन इसी तरह एक और पॉइंट आएगा आपके पास वो यहां पर होगा इसी तरह एक और पॉइंट आएगा आपके पास वो यहां पर होगा ठीक है इसी तरह चलते जाएंगे एक पॉइंट यहां पर मिलता जाएगा आपके पास तो इसका मतलब ये है यहां से लेकर यहां तक ये आपके पास क्या है टेंज एट लाइन बन रही है आपके पास आपने इसके पॉइंट को यहां से ले यहां तक मिलाने हैं स्टूडेंट तो मैं मिलाता हूं देखें तो इक्वेशन ऑफ टेंज लाइन कौन सी बनी स्टूडेंट ये वाली बनी देखें वो यहीं पास बता रहा था ना कि ये आपके पास क्या है क्वेश्चन के स्टेटमेंट में फंक्शन गिवन था ये हमने इक्वेशन ऑफ टेंज लाइन निकाली थी ये आपके पास क्वेश्चन स्टेटमेंट में फंक्शन गिवन था इस फंक्शन का जो ग्राफ बना है इस ग्राफ के ऊपर हमने इक्वेशन ऑफ इस टेंज एट लाइन को ड्रा कर दिया है इस ग्राफ के ऊपर तो क्या बना कर्व और टेंज लाइन को टूगेदर हमने स्केच कर दिया है ठीक है स्टूडेंट ये कर्व है ये कर्व के ऊपर क्या बन रही है टेंज लाइन बन रही है फॉर एग्जांपल आपके पास ये सर्कल है तो टेंज लाइन किस तरह इस तरह से होती है फॉर एग्जांपल ये आपके पास कर्व है कर्व के ऊपर टेंज लाइन किस तरह से होती है इस तरह होती है या इस तरह से आ रही आ रही होती है उसको टच करके गुजर रही होती है ठीक है तो तो ये आपके पास क्या है ये आपके पास टेंज का कांसेप्ट है तो बेसिकली इस तरह से आपने डिनोट करना है ठीक है वैसे ये वाला पॉइंट आपने यहां पर लिख द तो ये आपके पास क्या है टू 3 4 अप टू सोन यहां पर -1 हो गया -2 हो गया ठीक है अप टू सो ऑन अब इस तर स्टूडेंट नेक्स्ट क्वेश्चन को मैं सॉल्व नहीं करने वाला आपको जस्ट कांसेप्ट बताऊंगा आपने उसी को सॉ खुद से सॉल्व करना है अब इसी तरह स्टूडेंट आप देखें ये आपके पास क्वेश्चन नंबर सिक्स है ठीक है आपके पास ये फंक्शन गिवन है y = x - 1 का स्क्वा प् 1 और आपके पास ये पॉइंट गिवन है ठीक है आपने इसको f एक एक् कहना है क्योंकि आपको पता है स्लोप का ये फार्मूला तो इसके लिए हम सबसे पहले ये चीज चाहिए तो हमने लिख x को x नॉट से रिप्लेस कर दे तो ये बनेगा फिर x नॉट में h को ऐड करेंगे तो ये बनेगा फिर इसमें पुट करेंगे तो m की वैल्यू आएगी ये देखिए मैंने वैल्यू को पुट किया हुआ है सिंपलीफाई किया हुआ है आप यहां से स्क्रीनशॉट भी ले लें तो अगर आप चाहे तो ये मेरे पास इसका पीडीएफ प्रंट में भी है ठीक है तो इसको सॉल्व करेंगे तो आंसर आएगा आपके पास जीरो स्लोप का ठीक है अब इस स्लोप का आंसर आ गया इसी तरह आपने निकालना है इक्वेशन ऑफ टेंज टेंज लाइन को ठीक है तो टेंज लाइन को आपने किस तरह से निकालना है देखिए मैं आपको यहां पर दिखाता हूं ये देखें टेंज लाइन का आपके पास ये फार्मूला है स्टूडेंट y = y नॉ + m x - x नॉ ठीक है तो अब सिंपली फा करना है उसमें ठीक है y नॉट और x नॉ का जो पॉइंट गिवन था वो पुट करना है m का जो आंसर आया था वो पुट करना है तो y = 1 आपके पास कौन सी लाइन है टेंट लाइन है तो अपने सबसे पहले क्या करना है तो मैंने यहां पर पॉइंट को पुट किया हुए देखें तो अब x = 0 पुट करें फिर -1 पुट करें फिर वन पुट करें फिर -2 पुट करें फिर टू पुट करें आप टू सोन चलते हैं इसी तरह y = 1 मेंथ से पहले आप x = 0 पुट करें फिर x = 1 पुट करें x = -1 पुट करें x = 2 पुट करें x = -2 पुट करें तो आपके पास ये ग्राफ बनते आएंगे ठीक है तो इसके बाद आपने ये इस तरह से इसके ग्राफ को जनरेट करना है स्टूडेंट ठीक है तो मैंने यहां पर देखिए पॉइंट को कंप्लीट पुट नहीं किया तो आपने कंप्लीट पुट करना है माइनस वाले भी और पॉजिटिव वाले भी ठीक है अब इसी तरह ये देखिए यहां पर इसकी जो टेंज लाइन बनता है व ये देखि इस कर्व को टच करके गुजर रही है ये टेंज लाइन ठीक है इस तरह से मजीद आप देख ले यहां से ठीक है अब इसी तरह क्वेश्चन नंबर सेवन को मैं आपको दिखाता हूं वो किस तरह से होगा ये देखिए स्टूडेंट आपके पास क्वेश्चन नंबर सेवन है y = 2 x x अंडररूट ठीक है आपके पास ये पॉइंट गिवन है ठीक है तो सबसे पहले आपने इस इसको f x कहना है फिर x को x नॉ से रिप्लेस कर देना है x नॉट में h को ऐड कर देना है फिर स्लोप के लिए आपके अपने ये फार्मूला लगाना है इसको सिंपलीफाई करना है नीचे आपने एच को खत्म करना है स्टूडेंट ठीक है देखिए स्टूडेंट नीचे आपने एच ना एच को किस तरह से आप खत्म करेंगे रेशन आइज करेंगे ठीक है रेशन आइज करने के लिए देखिए इसका स्क्वायर बनेगा इसका इसका h नटू इसको सिंपल नीचे अंदर मल्टीप्लाई हो जाएगा ठीक है इसका स्क्वायर ये अंडर रूट को स्क्वायर कैंसिल आट हो जाएगा फरर का स्क्वायर 2 का स्क्वायर फर बनेगा इसका रूट कैंसिल हो हो जाएगा 4 का स्क 2 तो बनता है आपके पास देखिए सॉरी 2 का स्क्वा 4 बनता है ना तो अब इसको सिंपलीफाई करें नीचे एच को खत्म करने के लिए ऊपर में से मैंने h को निकाला है h से h कैंसिल आउट हो जाएगा ऊपर फोर आ जाएगा यानी कि सिंपल तरीके से यहां पर ना कैलकुलेशन है और कुछ भी नहीं है स्टूडेंट ठीक है तो m का आंसर आ जाएगा वन इसी तरह टेंट लाइन क्या आपके फार्मूला है y = y नॉ + m x - x नॉ m की वैल्यू आपने निका ली है वो पुट करें x नॉ और y नॉट जो क्वेश्चन की स्टेटमेंट में जो पॉइंट गिवन थे वो क्या है x नॉ और y नॉ है वो यहां पर पुट करें तो y = 1 + x आपके पास क्या है चेंज लाइन है तो सबसे पहले आपने नेगेटिव वाले भी पॉइंट्स पुट करने हैं और पॉजिटिव वाले भी पॉइंट्स पुट करने हैं ये आपके पास क्वेश्चन की स्टेटमेंट में फंक्शन गिवन था और ये आपके पास आपने न्यू निकाला है टेंज लाइन का ठीक है तो ये देखें इसकी टेंज लाइन कौन सी बनती है ये वाली टेंज लाइन बनती है y = x + 1 और ये आपके पास कौन सा है अच्छा बेसिकली ना ये टेंज लाइन बन रही है ना तो आपने यहां पर इसके साथ ये भी लिखना है तो फर्स्ट क्वेश्चन में मैंने नहीं लिखा था तो वो आप लिख लें ये जो आपके पास कार् बन रहा है कौन सा फंक्शन का तो y = 2 x रूट ठीक है अब इस स्टूडेंट क्वेश्चन नंबर एट को देखें स्टेप बाय स्टेप हम क्वेश्चन नंबर एट को सॉल्व करते हैं स्टूडेंट ये आपके पास क्वेश्चन नंबर एट है y = 1 अप x स् -1 और वन आपके पास पॉइंट गिवन है ठीक है ये देखिए मजीद अगर आपको यहां पर समझ नहीं आ रही ना तो कमेंट्स बॉक्स में मुझे जरूर बताएं मैं इनका आपको पीडीएफ सेंड कर दूंगा या मैं आपको इनका इन एक्ससाइज से जो रिलेटेड क्वेश्चन है ना ये खुद से मैं आपको समझाऊ अलग से पेज प जिस तरह मैंने क्वेश्चन नंबर फाइव को समझाया अगर आपको समझ नहीं आ रही तो ठीक है तो ये देखें ये बिल्कुल सेम उसी मेथड से सॉल्व होंगे आप यहां से इसे देख रहे हैं स्टेप बाय स्टेप ठीक है इसका मैं आपको ग्राफ भी जनरेट करके दिखा देता हूं ये देखि ग्राफ इसका ये है ये इसका ग्राफ है ठीक है अब इसी तरह क्वेश्चन नंबर नाइन को देखें स्टूडेंट नाइन क्या है y = x की पावर 3 पॉइंट कौन से हैं -2 और -8 पॉइंट आपके पास गिवन हो अच्छा ये पॉइंट्स गिवन होंगे आपके पास ठीक है तो इस तरह आपने सबसे पहले स्लोप को निकालना है स्टूडेंट ये मैं आपको दिखा रहा हूं आपने इसको लिख लेना है अपने पास ठीक है उसके बाद फिर आपके पास कौन सा गिवन है ये देखें ये आपके पास टेंज लाइन आपने निकालनी है फिर पॉइंट्स आपने निकालने हैं ठीक है फंक्शन के पॉइंट्स निकालने हैं टेंज के भी पॉइंट्स आपने अलग-अलग निकालने है आगे देखें तो ये आपके पास फंक्शन के पॉइंट्स हैं ठीक है ये टेंट्स के हैं ठीक है तो इसकी मदद से आपने देखें ग्राफ को जनरेट करना है ग्राफ आपके पास कुछ इस तरह से बनेगा ठीक है ये देखें टेंज लाइन जो है वो ग्राफ को टच करके गुजरी है यहां से देखें अब इसी तरह क्वेश्चन नंबर 10 को देखें स्टूडेंट 10 आपके पास गिवन है y = 1 / x की पावर 3 -2 और - 1.8 आपके पास पॉइंट गिवन है ठीक है तो ये देखिए आप यहां से देख सकते हैं इसके ये देखिए मैंने स्टेप बाय स्टेप सॉल्व कर रखा है थाकि कोई भी बंदा देखे ना उस पहली दफा अगर कोई कोई भी बंदा इसे देखे तो उसे इजली समझ आ जाएगा ठीक है अब इस तरह आगे दिखाता हूं आपको तो यहां से टेंज लाइन के फार्मूला लगाएंगे m की वैल्यू पुट करेंगे सिंपलीफाई करेंगे तो ये पास क्या होगा टेंज लाइन होगा इसी तरह जो पॉइंट होंगे वो फंक्शन के ऊपर पॉइंट्स अलग से बनेंगे टेंज एट के पॉइंट्स अलग से बनेंगे ये देखें ये टेंज एट के पॉइंट्स ये फंक्शन के पॉइंट्स सिंपलीफाई करेंगे यहां पर रखेंगे तो आपके पास ये आंसर आएगा ये आपके पास क्या बनेगा ग्राफ बनेगा ठीक है तो आज की इस वीडियो में स्टूडेंट हमने क्वेश्चन नंबर फव टू 10 को स्टेप बाय स्टेप सॉल्व किया है अगर फिर भी आपके पास क्वेश्चन 6 7 8 9 और 10 में अगर आपको कोई प्रॉब्लम आ रही है तो कमेंट्स बॉक्स में मुझे अपने कंफ्यूजन को लाजमी बताएं ताकि मैं आपको नेक्स्ट वीडियो में उसकी क्लीयरेंस दे दूं ठीक है तो स्टूडेंट अगर आप हमारे चैनल पर नए हैं तो कै चैनल को जरूर सब्सक्राइब करें मिलते हैं नेक्स्ट वीडियो में तब तक के लिए अल्लाह हाफिज
187815
https://www.gauthmath.com/solution/1812326852679813/Length-of-a-chord-2rsin-frac-2
Solved: Length of a chord =2rsin ( θ /2 ) [Calculus] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Calculus Questions Question Length of a chord =2rsin ( θ /2 ) Gauth AI Solution 100%(1 rated) Answer The length of chord AB is 2 r sin(θ/2) Explanation Draw a circle with center O and radius r. Draw a chord AB with central angle θ. Draw a perpendicular from O to AB, intersecting AB at point C. Triangle OAC is a right triangle with hypotenuse OA = r and angle AOC = θ/2. Using the sine function in triangle OAC, we have sin(θ/2) = AC/OA = AC/r. Therefore, AC = rsin(θ/2). Since AC is half the length of chord AB, the length of chord AB is 2 AC = 2 rsin(θ/2) Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Geometry Show that the length d of a chord of a circle of radius is given by the formula d=2rsin frac θ 2 where θ is the central angle formed by the radii to the ends of the chord see the figure. Use this result to derive the fact that sin θ < θ where mured in radians. 100% (4 rated) Given AOB is a sector of a circle with centre O and a radius of r cm where angle AOB= θ rad. a Show that the length of the chord AB is 2rsin frac θ 2. [2 marks] b Given r=5cm and angle AOB=1.2 rad, find the difference in length between the arc length AB and the chord AB. [3 marks] Answer: 100% (4 rated) Fig.10.4 Let the length of the chord AB=x , then x=2rsin frac θ 2 r=6cm θ =60 ° ∴ frac rho 2=30 ° x=2 6 sin 30 =12 0.500 =6cm 100% (2 rated) Ungkapkan k dalam sebutan r dan . The chord MN of length k cm forms an angle of square MON= θ rad with O being the center of the circle. Given the radius of a circle is r cm. Express k in terms ofr and . 2rsin frac 90 ° θ π 100% (5 rated) Show that the length d of a chord of a circle of radius r is given by the formula d=2rsin frac θ 2 where 0 is the central angle formed by the radii to the ends of the chord. See the figure. Use this result to derive the fact that sin θ < θ , where θ >0 is measured in radians. While multiple methods can be used to prove that d=2rsin frac θ 2 , we will use the Law of Cosines for this problem Substitute the appropriate values into the formula for the Law of Cosines. c2=a2+b2-2abcos C d2=square 2+square 2-2square square cos square 100% (2 rated) art II: Match the ollowing statement in column “A“ with their corresponding formula column “B”. “A” “B” 11. Chord length A 2πr 12. Arc length B 2rsin frac θ 2 13. Area of Secto C frac π r θ 180 ° D 14. Circumference frac π r2 θ 360 ° 15. Area of circle E π r2 100% (4 rated) If I is the length of a circular arc, a is the length of the chord of the whole arc, and b is the length of the chord of half the arc, show that: a a=2rsin l/2r and b b=2rsin frac l4r' , where r is the radius of the circle. By expanding sin 1/2r and sin l/4r as series, show that I= 8b-a/3 approximately. Page 2 100% (5 rated) How long are the chords that are joining the radii together? The length of each chord can be found using the formula for the side length 8 of a regular polygon inscribed in a circle with radius r: s=2rsin frac π n Where n is the number of sides. Since this is a pentagon, n=5 and r=4cm . We can calculate the chord length from this formula. 100% (5 rated) 6.5 po= If central angle @ cuts off a chord of length € in a circle of radius f, then the relationship betwern θ, c, and # is given by 2rsin frac θ 2=c Find B,ifc=r. θ =60 ° ,300 ° θ =30 ° ,300 ° θ =60 ° ,330 ° e=30 ° ,330 ° B=60 ° ,180 ° 05 guints 100% (2 rated) Find the: a length of the minor arc AB _ _ b Perimeter of the sector OAB _ 2 Y _ c Area of the sector OAB _ _ d Area of the shaded segment _ 6. _ _ _ _ _ _ e find the exact length of the chord AB hint: chord length =2rsin [frac θ 2] _ a _ fiiraan 100% (1 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
187816
http://math.uchicago.edu/~may/REU2017/REUPapers/Zhang,Alec.pdf
POLYA’S ENUMERATION ALEC ZHANG Abstract. We explore Polya’s theory of counting from first principles, first building up the necessary algebra and group theory before proving Polya’s Enumeration Theorem (PET), a fundamental result in enumerative combi-natorics. We then discuss generalizations of PET, including the work of de Bruijn, and its broad applicability. Contents 1. Introduction 1 2. Basic Definitions and Properties 2 3. Supporting Theorems 4 3.1. Orbit-Stabilizer Theorem 4 3.2. Burnside’s Lemma 4 4. Polya’s Enumeration 5 4.1. Prerequisites 5 4.2. Theorem 7 5. Extensions 11 5.1. De Bruijn’s Theorem 12 6. Further Work 15 Acknowledgments 16 References 16 1. Introduction A common mathematical puzzle is finding the number of ways to arrange a necklace with n differently colored beads. Yet (n −1)! and (n−1)! 2 are both valid answers, since the question has not defined what it means for necklaces to be distinct. The former counts the number of distinct necklaces up to rotation, while the latter counts the number of distinct necklaces up to rotation and reflection. Questions like these become more complex when we consider “distinctness” up to arbitrary transformations and with objects of more elements and non-standard symmetries. The search for a general answer leads us to concepts in group theory and symmetry, and ultimately towards Polya’s enumeration, which we will explore below. 1 2 ALEC ZHANG 2. Basic Definitions and Properties We start with one of the most basic algebraic structures: Definition 2.1. Group. A group is a set G equipped with an operation ∗satisfying the properties of associativity, identity, and inverse: • Associativity: ∀a, b, c ∈G, (a ∗b) ∗c = a ∗(b ∗c). • Identity: ∃e ∈G|∀a ∈G, e ∗a = a ∗e = a. • Inverse: ∀a ∈G, ∃a−1 ∈G|a ∗a−1 = a−1 ∗a = e. Given group elements g, h in group G, we denote g ∗h as gh. Definition 2.2. Subgroup. A subgroup of a group G is a group under the same operation of G whose elements are all contained in G. If H is a subgroup of G, we write H ≤G. One of the most important groups is the symmetric group Sn, whose elements are all permutations of the set {1, ..., n}, and whose operation is composition. Indeed, permutations are associative under composition, there is an identity permutation, and all permutations have an inverse. Every finite set X also has an implied sym-metric group Sym(X), which simply involves all permutations of its elements. Groups can also “act” on sets in the following manner: Definition 2.3. Group action. Given a group G and a set X, a left group action is a function φ : G×X →X satisfying the properties of left identity and compatibility: • Left Identity: For the identity element e ∈G, for all x ∈X, φ(e, x) = x. • Left Compatibility: For all g, h ∈G, for all x ∈X, φ(gh, x) = φ(g, φ(h, x)). A right group action is similarly defined as a function φ : X × G →X satisfying right identity and compatibility: • Right Identity: For the identity element e ∈G, for all x ∈X, φ(x, e) = x. • Right Compatibility: For all g, h ∈G, for all x ∈X, φ(x, gh) = φ(φ(x, g), h). Group actions will be left group actions unless specified otherwise, but the defini-tions and properties below apply analogously to right group actions as well. Given a group element g in a group G, an element x in a set X, and a group action φ, we denote φ(g, x) as gx if φ is a left group action and φ(x, g) as xg if φ is a right group action. In any group action, the group also acts in a bijective manner on the set: Proposition 2.4. Given a group action φ of group G on a set X, the function fφ : x 7→φ(g, x) is bijective for all g ∈G. Proof. It suffices to find an inverse function. We see that hφ : x 7→g−1x is such an inverse, since fφ(hφ(x)) = fφ(g−1x) = g(g−1x) = (gg−1)x = x, hφ(fφ(x)) = hφ(gx) = g−1(gx) = (g−1g)x = x by compatibility of the group action. □ Thus, one may alternatively view the group action as associating a permutation POLYA’S ENUMERATION 3 pg ∈Sym(X) with every g ∈G, where gx for g ∈G and x ∈X is determined by pg(x), the image of x in pg. Formally, a group action φ is a homomorphism from G to Sym(X). If we actually consider G = Sym(X) as our group acting on X, then G naturally acts on X; that is, for pg ∈G, φ(pg, x) = pg(x) is the natural group action associated with G and X. Associated with the elements of the set in any group action are two important notions: Definition 2.5. Orbit. Given a group action φ of a group G on a set X, the orbit of a set element x ∈X is orb(x) = Ox = {gx : g ∈G} = {y ∈X|∃g ∈G : y = gx} . Definition 2.6. Stabilizer. Given a group action φ of a group G on a set X, the stabilizer of a set element x ∈X is stab(x) = Sxx = {g ∈G|gx = x} . Using the stabilizer notation, we can similarly define the transformer: Definition 2.7. Transformer. Given a group action φ of a group G on a set X, the transformer of two set elements x, y ∈X is trans(x, y) = Sxy = {g ∈G|gx = y} . Associated with a group action is the set of orbits, called the quotient: Definition 2.8. Quotient. Given a group action φ of a group G on a set X, the quotient of φ is defined as X/G = {Ox : x ∈X} . As it turns out, the orbits of a set partition it: Proposition 2.9. For any group action φ of a group G on a set X, X/G is a partition of X. Proof. It is well-known that equivalence classes of a set partition it. Then it suffices to show that the relation x∼y ⇐ ⇒x, y ∈Ox is an equivalence relation. We check the reflexive, symmetric, and transitive properties: • Reflexive: For all x ∈X, x∼x since ex = x ∈Ox for the identity element e ∈G. • Symmetric: For all x, y ∈X, x∼y clearly implies y∼x. • Transitive: For all x, y, z ∈X, if x∼y and y∼z, then x, y, z ∈Ox, so x∼z as well. □ It is also worth noting that the stabilizer of any element x ∈X forms a subgroup of G: Proposition 2.10. For any group action φ of a group G on a set X, Sxx ≤G for all x ∈X. Proof. Associativity is inherited from the group structure of G. We check the closure, identity, and inverse properties. For gi, gj ∈Sxx and x ∈X: • Closed: Clearly gi(gjx) = gix = x. But by the compatibility property of φ, we must also have (gigj)x = x ⇒gigj ∈Sxx. 4 ALEC ZHANG • Identity: The identity e ∈G is in Sxx since ex = x. • Inverse: Consider arbitrary gi ∈Sxx. Since gix = x, we also have g−1 i (gix) = g−1 i x ⇒g−1 i x = (g−1 i gi)x = ex = x by compatibility of φ, so g−1 i ∈Sxx. □ 3. Supporting Theorems 3.1. Orbit-Stabilizer Theorem. With our notions of orbits and stabilizers in hand, we prove the fundamental orbit-stabilizer theorem: Theorem 3.1. Orbit −Stabilizer Theorem. Given any group action φ of a group G on a set X, for all x ∈X, |G| = |Sxx||Ox|. Proof. Let g ∈G and x ∈X be arbitrary. We first prove the following lemma: Lemma 1. For all y ∈Ox, |Sxx| = |Sxy|. Proof. It suffices to show a bijection between Sxx and Sxy. Let gxx ∈Sxx and gxy ∈Sxy. Clearly gxygxxx = gxyx = y, so gxygxx ∈Sxy. In addition, by definition of Sxy, we have gxyx = y, so multiplying by g−1 xy gives us g−1 xy gxyx = g−1 xy y = ⇒ ex = x = g−1 xy y by compatibility; thus, g−1 xy ∈Syx and so g−1 xy gxy ∈Sxx. Consider any h ∈Sxy. Since gxygxx ∈Sxy, we may define χ : Sxx →Sxy : gxx 7→ hgxx. Since g−1 xy gxy ∈Sxx, we may define ψ : Sxy →Sxx : gxy 7→h−1gxy, which is also an inverse for χ: χ(ψ(gxy)) = χ(h−1gxy) = hh−1sxy = gxy ψ(χ(gxx)) = ψ(hgxx) = h−1hgxx = gxx. Thus χ is a bijection and |Sxx| = |Sxy|. □ By Lemma 1, we have |Sxy| = |Sxx| for all y ∈Ox. Now note that the sets Sxy : y ∈Ox must partition G; this follows from the definition of the orbit and the fact that the group action is a function.1 Thus |G| = |Sxx||Ox|, as desired. ■ 3.2. Burnside’s Lemma. We can now calculate the order of the group, the size of the stabilizer of an arbitrary set element, or the size of that element’s orbit given the other two quantities. However, one quantity of interest, the number of orbits, is still in complete question! The following theorem, now attributed to Cauchy in 1845, determines the number of orbits in terms of the order of the group and the number of fixed elements under the group action: Theorem 3.2. Burnside′s Lemma. Given a finite group G, a finite set X, and a group action φ of G on X, the number of distinct orbits is |X/G| = 1 |G| X g∈G |Xg|, where Xg = {x ∈X|gx = x}, the set of elements of X fixed by action by g. 1It is important to note the difference between this statement and Proposition 2.9. The propo-sition states that the different orbits partition the set, whereas here we state that given any one of those orbits, every group element acting on x gives exactly one element in Ox, and that all elements in Ox are covered. POLYA’S ENUMERATION 5 Proof. We first note that X g∈G |Xg| = |(g, x) ∈(G, X) : gx = x| = X x∈X |Sxx|, so we just need to show |X/G| = 1 |G| X x∈X |Sxx|. By the Orbit-Stabilizer Theorem, we have that |Sxx| = |G| |Ox|, so 1 |G| X x∈X |Sxx| = 1 |G| X x∈X |G| |Ox| = X x∈X 1 |Ox|. Since orbits partition X by Proposition 2.9, we can split up X into disjoint orbits of X/G. Thus, we can rewrite our sum, where A is an orbit in X: X x∈X 1 |Ox| = X A∈X/G X x∈A 1 |A| = X A∈X/G 1 = |X/G|, so |X/G| = 1 |G| P x∈X |Xg|, as desired. ■ 4. Polya’s Enumeration 4.1. Prerequisites. Polya’s enumeration introduces functions f from a finite set X to a new finite set Y . Notation-wise, Y X is the set of all functions f : X →Y , represented as a set of ordered pairs (xi, f(xi)) for xi ∈X. For instance, if Y is a set of colors, then f ∈Y X is a coloring of the elements in X, and Y X/G is the number of distinct colorings of X under some group action of G on Y X. From this perspective, an action φ of G on X induces a natural group action φ′ of G on Y X, namely: φ′ : (g, f) 7→f ′ = f ◦p−1 g = {(φ(g, x), f(x))|x ∈X} for f ∈Y X. Indeed, φ′ satisfies identity and compatibility: ef = {(ex, f(x))|x ∈X} = {(x, f(x))|x ∈X} = f, g1(g2f) = g1f ′ = g1({(g2x, f(x))|x ∈X}) = {(g1(g2x), f ′(g2x))|x ∈X} = {(g1g2)x, f(x)|x ∈X} = (g1g2)f. Throughout this section, we assume an implicit group action φ of a group G on a finite set X of size n, where φ is arbitrary. In addition, we assume Y is another finite set. To state Polya’s Enumeration Theorem, we introduce some more machinery: Definition 4.1. Type. Let p be a permutation on X. Then the type of p is the set {b1, ..., bn}, where bi is the number of cycles of length i in the cycle decomposition of p. 6 ALEC ZHANG Definition 4.2. Cycle index polynomial. The cycle index polynomial Zφ of the group action φ is defined as 2 Zφ(x1, ..., xn) = 1 |G| X g∈G n Y i=1 xbi(g) i , where bi(g) is the ith element of the type of the implied permutation pg ∈Sym(X). Definition 4.3. Function equivalence. Two functions f ∈Y X are said to be equivalent under the action of G (f1 ∼G f2) if they are in the same orbit of φ′, i.e. there exists g ∈G such that f2 = gf1. By the proof of Proposition 2.9, function equivalence is an equivalence relation, so Y X will have equivalence classes under function equivalence: Definition 4.4. Configuration. A configuration is an equivalence class of the equivalence relation ∼G on Y X. Analogously, we have that every configuration c is just an orbit of φ′, and that the set of configurations C is just Y X/G under φ′. To weight our functions differently, we can assign weights to elements in Y : Definition 4.5. Weight. Let w : Y →R be a weight assignment to each element in Y . 3 Then the weight of a function f ∈Y X is defined as W(f) = Y x∈X w(f(x)). It follows that all functions in a configuration c have the same common weight, which we call the weight of the configuration W(c): Proposition 4.6. All functions in a configuration have the same common weight. Proof. Consider arbitrary functions f1, f2 ∈Y X in configuration C. Since f1 ∼G f2, there exists g ∈G such that f1(gx) = f2(x). In addition, from Proposi-tion 2.4, we know every group element acting on a set permutes it, so W(f) = Q x∈X w(f(x)) = Q x∈X w(f(gx)) for any g ∈G. Thus, W(f1) = Y x∈X w(f1(x)) = Y x∈X w(f1(gx)) = Y x∈X w(f2(x)) = W(f2). □ Definition 4.7. Configuration Generating Function (CGF). Let C be the set of all configurations c. Then the CGF is defined as F(C) = X c∈C W(c). 2The standard notation is ZG, but here we use Zφ to explicitly show that the cycle index polynomial not only depends on the algebraic structure of the group G, but also its induced permutation on X through the group action φ. 3In general, we may replace R with any commutative ring. POLYA’S ENUMERATION 7 4.2. Theorem. We are now equipped to tackle Polya’s Enumeration Theorem (PET). We cover both the unweighted and weighted versions of the theorem; the first can be proved directly from Burnside’s Lemma: Theorem 4.8. Polya′s Enumeration Theorem (Unweighted). Let G be a group and X, Y be finite sets, where |X| = n. Then for any group action φ of G on X, the number of distinct configurations in Y X is |C| = 1 |G| X g∈G |Y |c(g), where c(g) denotes the number of cycles in the cycle decomposition of pg ∈Sym(X), the permutation of X associated with the action of g on X. Proof. Since configurations are orbits of φ′, we have |C| = |Y X/G| under φ′. We apply Burnside’s Lemma to the finite set Y X with group action φ′, which states that |Y X/G| = 1 |G| X g∈G |(Y X)g|. It remains to show that |(Y X)g| = |Y |c(g). But any function f ∈Y X will remain constant under the action of g if and only if all elements in X in each cycle are assigned the same set element in Y . There are thus |Y | choices of elements in Y for each of the c(g) cycles in the cycle decomposition, and the result follows. ■ We now state the weighted version of PET: Theorem 4.9. Polya′s Enumeration Theorem (Weighted). Let G be a group and X, Y be finite sets, where |X| = n. Let w be a weight function on Y . Then for any group action φ of G on X, the CGF is given by Zφ  X y∈Y w(y), X y∈Y w(y)2, ..., X y∈Y w(y)n  . Proof. We first prove the following lemma: Lemma 1. |C| = 1 |G| P g∈G  f ∈Y X|(∀x ∈X)(f(gx) = f(x)) . Proof. Let φ′ R be the right group action on Y X induced by φ: φ′ R : (f, g) 7→f ′ R = f ◦pg = {(x, f(φ(g, x)))|x ∈X} , where f ∈Y X and g ∈G. The result follows by applying Burnside’s Lemma to Y X under φ′ R, as in Theorem 4.8. □ We now take φ′ R to be our group action on Y X. Let A(ω) = {c ∈C|W(c) = ω} be the set of all configurations with common weight ω. Sgg =  f ∈Y X|f = fg is the set of all functions stabilized by g; let Sgg(ω) =  f ∈Y X|f = fg, W(f) = ω be the set of all functions stabilized by g with common weight ω. Then by Lemma 1, we have |A(ω)| = 1 |G| X g∈G |Sgg(ω)|. 8 ALEC ZHANG We can also group our CGF by weights: CGF = X c∈C W(c) = X ω ω|A(ω)| = 1 |G| X ω X g∈G ω|Sgg(ω)| by the above equality. Since our sum is finite, we can switch the order of summation: CGF = 1 |G| X g∈G X ω ω|Sgg(ω)| = 1 |G| X g∈G X f∈Sgg W(f). G permutes X through the group action, so the corresponding permutation pg for g ∈G has a cycle decomposition C1, ..., Ck, where k ≤n. It follows that if f ∈Sgg, then f(x) = f(gx) = f(g2x) = ... for all x ∈X, g ∈G and f is constant on each cycle Ci in the cycle decomposition. Then we have X f∈Sgg W(f) = X f∈Sgg Y x∈X w(f(x)) = X f∈Sgg k Y i=1 Y x∈Ci w(f(x)) = X f∈Sgg k Y i=1 w(f(xi))|Ci|, where xi ∈Ci. Let |Y | = m. Since we are summing over all f ∈Sgg, we need to cover all possible assignments of y ∈Y to cycles Ci, so our expression becomes X f∈Sgg W(f) = k Y i=1  w(y1)|Ci| + ... + w(ym) |Ci| = k Y i=1 X y∈Y w(y)|Ci|, and plugging this into the CGF expression gives us CGF = 1 |G| X g∈G   k Y i=1 X y∈Y w(y)|Ci|  . Regardless of cycle length, by definition of the type, there will be bj(g) cycles of length j, so our expression is CGF = 1 |G| X g∈G n Y j=1  X y∈Y w(y)j   bj(g) = Zφ  X y∈Y w(y), X y∈Y w(y)2, ..., X y∈Y w(y)n  . ■ Note that setting w(y) = 1 for all y ∈Y makes W(f) = 1 for all f ∈Y X and gives us a CGF of ZG(|Y |, ..., |Y |), so the unweighted version of PET immediately follows. We summarize the concepts in PET with a concrete example: Example 4.10. Classify the non-isomorphic multigraphs with n = 4 vertices and with up to m = 2 separate edges between two vertices allowed. Solution. We first clarify our sets, groups and actions. • Sets: Let V be the set of vertices {V1, ..., Vn}, X be the set of edges  E12, E13, ..., E(n−1)(n) of Kn, indicating all possible distinct edges, and Y be the set {y0, ..., ym}, indicating the number of possible edges between two vertices; let the weight of yi be w(yi) = wi. • Groups: We have SV = Sym(V ) associated with V ; let SX|V be the group of permutations on X induced by SV . Note that this is not the same as SX = Sym(X), since |SX|V | = n! but |SX| = n 2  !. POLYA’S ENUMERATION 9 • Actions: Let group SV act on set V with the natural group action φV . Then group SX|V acts on set X through an induced action φ, and acts on set Y X through an induced (right) action φ′ R as shown in the proof of PET. SV φV V SXjV φ X φ0 R Y X Then Y X represents all possible multigraphs, and |Y X/SX|V | is the number of multigraphs up to isomorphism. For instance, the multigraph V1 V2 V3 V4 is represented by the function f : {E12, E13, E14, E23, E24, E34} 7→{0, 2, 1, 1, 0, 0}. We first compute the cycle index polynomial Zφ′ R. To do so, we need to deter-mine the corresponding types to the elements of SX|V . SV acting on K4 leads to the following elements in SX|V : • The identity (V1)(V2)(V3)(V4) ∈SV leads to the corresponding identity (E12)(E13)(E14)(E23)(E24)(E34) ∈SX|V with type {6, 0, 0, 0, 0, 0}. This contributes an x6 1 term to the cycle index. • There are 4 2  = 6 elements of the form (VaVb)(Vc)(Vd) ∈SV leading to the corresponding element (Eab)(Ecd)(EacEbc)(EadEbd) ∈SX|V with type {2, 2, 0, 0, 0, 0}. Each of the 6 elements contributes an x2 1x2 2 term to the cycle index. • There are ( 4 2) 2 = 3 elements of the form (VaVb)(VcVd) ∈SV leading to the corresponding element (Eab)(Ecd)(EacEbd)(EadEbc) ∈SX|V with type {2, 2, 0, 0, 0, 0}. Each of the 3 elements contributes an x2 1x2 2 term to the cycle index. • There are 4 3  ∗2 = 8 elements of the form (VaVbVc)(Vd) ∈SV (note that (123)(4) is different from (132)(4)) leading to the corresponding element (EabEbcEac)(EadEbdEcd) ∈SX|V with type {0, 0, 2, 0, 0, 0}. Each of the 8 elements contributes an x2 3 term to the cycle index. • There are 3! = 6 elements of the form (VaVbVcVd) ∈SV leading to the corre-sponding element (EabEbcEcdEad)(EacEbd) ∈SX|V with type {0, 1, 0, 1, 0, 0}. Each of the 6 elements contributes an x2x4 term to the cycle index. 10 ALEC ZHANG Thus, the cycle index is Zφ′ R(x1, x2, x3, x4, x5, x6) = 1 24(x6 1 + 9x2 1x2 2 + 8x2 3 + 6x2x4). Now the weighted version of PET tells us that the CGF of Y X/G is Zφ′ R  X y∈Y w(y), ..., X y∈Y w(y)6  = Zφ′ R (w0 + w1 + w2), ..., (w6 0 + w6 1 + w6 2)  = 1 24 (w0 + w1 + w2)6 + 9(w0 + w1 + w2)2(w2 0 + w2 1 + w2 2)2 + 8(w3 0 + w3 1 + w3 2)2 + 6(w2 0 + w2 1 + w2 2)(w4 0 + w4 1 + w4 2)  = w6 0 + w5 0w1 + 2w4 0w2 1 + 3w3 0w3 1 + 2w2 0w4 1 + w0w5 1 + w6 1 + w5 0w2 + 2w4 0w1w2 + 4w3 0w2 1w2 + 4w2 0w3 1w2 + 2w0w4 1w2 + w5 1w2 + 2w4 0w2 2 + 4w3 0w1w2 2 + 6w2 0w2 1w2 2 + 4w0w3 1w2 2 + 2w4 1w2 2 + 3w3 0w3 2 + 4w2 0w1w3 2 + 4w0w2 1w3 2 + 3w3 1w3 2 + 2w2 0w4 2 + 2w0w1w4 2 + 2w2 1w4 2 + w0w5 2 + w1w5 2 + w6 2. The CGF then completely classifies the non-isomorphic multigraphs of degree n = 4 with up to m = 2 separate edges between two vertices allowed. For instance, the term 4w3 0w2 1w2 indicates that there are four non-isomorphic multigraphs with 3 absent edges, 2 edges, and 1 double-edge, namely the multigraphs below: If we set w0 = w1 = 1, w2 = 0, we just get the number of non-isomorphic graphs of 4 vertices: Zφ′ R(2, 2, 2, 2, 2, 2) = 11. If we set w0 = w1 = w2 = 1, we get the total number of non-isomorphic multigraphs: Zφ′ R(3, 3, 3, 3, 3, 3) = 66. If we set w0 = 0, w1 = 1, w2 = 2, we get the total number of edges among all non-isomorphic multigraphs: Zφ′ R(1 + 2, 1 + 22, ..., 1 + 26) = 163. Other quantities of interest may be found by substituting different values for wi. □ POLYA’S ENUMERATION 11 5. Extensions Up until now, we have considered a group action φ with group G acting on set X, inducing an action φ′ on the set Y X. However, recall that φ′ : (g, f) 7→f ′ = {(φ(g, x), f(x))} permutes both X and f(X) through φ. More generally, we can permute the set Y independently of φ; that is, we can consider an action ψ on set Y with another group H. We now define a more general equivalence between functions: Definition 5.1. Generalized function equivalence. Two functions f1, f2 ∈Y X are equivalent (f1 ∼gh f2) if ∃g ∈G, h ∈H such that for all x ∈X, f1(gx) = hf2(x). For this definition to be compatible with our definitions of configuration, CGF, etc., ∼gh must be an equivalence relation. Proving this is left as an exercise to the reader. Note that our previous proposition that equivalent functions have the same weight does not necessarily hold with generalized function equivalence; it is a require-ment on ψ. However, assuming that generalized-equivalent functions have the same weight, we then have analogous results to the ones in section 4: Theorem 5.2. Generalized PET (Weighted.) Let groups G, H act on the sets X, Y through the group actions φ and ψ, respectively. Let w : Y →R be a weight function for Y . Using generalized function equivalence, the CGF is CGF = 1 |G||H| X (g,h)∈G×H X f∈S(g,h) W(f), where S(g,h) is the set of functions f ∈Y X stabilized by (g, h), i.e. S(g,h) =  f ∈Y X|(∀x ∈X)(f(gx) = hf(x)) . Proof. The proof follows exactly the same way as in the proof of PET (Weighted.) Note that function equivalence can also be written in the following form: f1 ∼gh f2 ⇐ ⇒(∃(g, h) ∈G × H)(∀x ∈X)(h−1f1(gx) = f2(x)). Then we can define a right group action χ of group G×H on set Y X in the following manner: χ(f, (g, h)) = h−1fg, where g ∈G acts on f ∈Y X through the induced right group action φ′ R : (f, g) 7→ f ′ φR = {(x, f(φ(g, x)))} and h ∈H acts on f ∈Y X through the induced left group action ψ′ : (h, f) 7→f ′ ψ = {(x, ψ(h, f(x)))}. We then have that configurations are the orbits of f ∈Y X under χ: f2 = {(x, f2(x))} = f1(g, h) = h−1f1g = h−1({(x, f1(gx))}) =  (x, h−1f1(gx)) ⇐ ⇒f1 ∼gh f2. Taking χ as our group action on Y X, we again let A(ω) = {c ∈C|W(c) = ω} , S(g,h)(ω) =  f ∈Y X|f(g, h) = f, W(f) = ω . By Burnside’s Lemma with the group action χ, we have |A(ω)| = 1 |G × H| X (g,h)∈G×H |S(g,h)(ω)|. 12 ALEC ZHANG Finally, recall that the CGF can be grouped into weights: CGF = X c∈C W(c) = X ω ω|A(ω)| = 1 |G × H| X ω X (g,h)∈G×H ω|S(g,h)(ω)| = 1 |G||H| X (g,h)∈G×H X f∈S(g,h) W(f). ■ 5.1. De Bruijn’s Theorem. We finally arrive at the problem of counting the number of orbits, which involves weighting each function (up to generalized equiv-alence) with the weight 1. Before, we could simply substitute w(y) = 1 to get an answer of Zφ(|Y |, ..., |Y |). Here, however, we no longer have a concise expression for the answer in terms of the CGF; we are looking for the quantity P (g,h)∈G×H |f ∈Y X : f(g, h) = f| |G||H| , which simply follows from Burnside’s Lemma. We turn to de Bruijn’s theorem: Theorem 5.3. (de Bruijn.) Let group G act on finite set X through group action φ, and let group H act on finite set Y through group action ψ. Then the number of functions up to function equivalence is Zφ  ∂ ∂z1 , ∂ ∂z2 , ∂ ∂z3 , ...  Zψ  e P k zk, e2 P k z2k, e3 P k z3k, ...  {zi}=0. Proof. Let bi(g), cj(h) be the types of g ∈G, h ∈H, respectively, and let |X| = n, |Y | = m. For ease of notation, let bi(g) = 0, cj(h) = 0 for all g ∈G, h ∈H if i > n and j > m, respectively, where i, j are taken over the positive integers Z+. We first prove the following lemmas: Lemma 1. If f ∈S(g,h), then f(x) = y implies f(gix) = hiy for all i. Proof. Since fg = hf, we have fg2 = (fg)g = (hf)g = h(fg) = h(hf) = h2f. The result easily follows by induction on i: fgi−1 = hi−1f ⇒fgi = (fgi−1)g = hi−1fg = hi−1hf = hif. □ Lemma 2. If f ∈S(g,h), then each cycle CX in X is mapped by f to a cycle CY in Y where |CY | divides |CX|. Proof. Consider any f ∈S(g,h), where fg = hf for some g ∈G, h ∈H. Let x ∈X belong to cycle Cg,x of length j. Then we have Cg,x =  x, gx, ..., gj−1x , where gjx = x. By Lemma 1, we have f(gix) = hif(x) for all positive integers i. Thus, the image of Cg,x under f is f(Cg,x) =  f(x), hf(x), h2f(x), ..., hj−1f(x) , and we have hjf(x) = f(gjx) = f(x), so the cycle CY in Y containing f(x) must have a length that divides j. □ POLYA’S ENUMERATION 13 Lemma 3. The total number of functions stabilized by (g, h) is X f∈S(g,h) W(f) = Y i  X j|i jcj(h)   bi(g) . (Recall that we are setting W(f) = 1.) Proof. We count the number of functions using the condition in Lemma 2. Let f ∈S(g,h). For each cycle Cg,i in the cycle decomposition of pg ∈Sym(X), pick an arbitrary element xi ∈Cg,i. Since there are cj(h) cycles of length j in the cycle decomposition of ph ∈Sym(Y ), and Cg,x can only map to cycles whose length divides its own by Lemma 2, xi can map to y ∈Y under f in P j|i jcj(h) ways. But note that after the mapping xi 7→f(xi) has been determined, the rest of the mappings in Cg,i are determined as well, due to the condition in Lemma 1. Thus, the number of functions is the number of ways to choose mappings for each cycle in pg; since pg has bi(g) cycles of length i, the result follows. □ By Lemma 3, we have X f∈S(g,h) W(f) = (c1(h))b1(g) · (c1(h) + 2c2(h))b2(g) · (c1(h) + 3c3(h))b3(g)... But note that each of the terms of the form ab in this product can be written as the partial derivative expression ∂b ∂zb eaz z=0:  X j|i jcj(h)   bi(g) = ∂bi(g) ∂zbi(g) i e(P j|i jcj(h))zi zi=0. More generally, ab = ∂b ∂zb ea(P i zi) {zi}=0, so we can write our expression as X f∈S(g,h) W(f) = Y i ( ∂ ∂zi )bi(g) ! e(P j|i jcj(h))(P i zi) {zi}=0. We can also express the exponent solely in terms of j:  X j|i jcj(h)   X i zi ! =  X j jcj(h) ∞ X k=1 zkj   Let ri = ∂ ∂zi , sj = ej P k zkj. Then we have 1 |G| X g∈G Y i ( ∂ ∂zi )bi(g) = Zφ(r1, r2, ...), 1 |H| X h∈H e(P j|i jcj(h))(P i zi) = 1 |H| X h∈H e P j jcj(h) P k=1 zkj = 1 |H| X h∈H Y j (ejcj(h) P k zkj) = 1 |H| X h∈H Y j (ej P k zkj)cj(h) = Zψ(s1, s2, ...). 14 ALEC ZHANG Finally, by Generalized PET, we have CGF = 1 |G||H| X (g,h)∈G×H X f∈S(g,h) W(f) =  1 |G| X g∈G Y i ( ∂ ∂zi )bi(g)   1 |H| X h∈H e(P j|i jcj(h))(P i zi) ! = Zφ( ∂ ∂z1 , ∂ ∂z2 , ∂ ∂z3 , ...)Zψ(e P k zk, e2 P k z2k, e3 P k z3k, ...) {zi}=0. ■ Here is a simple example that highlights the differences between PET and its gen-eralizations: Example 5.4. Compute the number of distinct colorings of the vertices of a square with 3 colors, under the following equivalencies: • 1. Rotations are not distinct. • 2. Rotations and reflections are not distinct. • 3. Rotations, reflections, and color permutations are not distinct. Solution. The first two cases can be solved using Burnside’s Lemma and PET; the third case involves generalized PET and de Bruijn’s Theorem. Our sets are X = {V1, V2, V3, V4}, the set of vertices, and Y = {R, G, B} , our three colors. Let r = (V1V2V3V4) be a clockwise 90-degree rotation and s = (V1V2)(V3V4) be a reflection across the vertical axis. The corresponding groups for each case are • 1. G = C4 =  e, r, r2, r3 , the cyclic group on the vertices, • 2. G = D4 =  e, r, r2, r3, s, sr, sr2, sr3 , the dihedral group on the vertices, • 3. G = D4 and H = S3, the symmetric group on the colors, where the group actions are the natural group actions. For the sake of demonstration, we solve case 1 and 2 in two different ways. By Burnside’s Lemma, the answer to the first case is |Y X/G| = 1 |G| X g∈G (Y X)g = 1 4(34 + 31 + 32 + 31) = 24. For the second case, we compute the cycle index for the natural group action φ2 : D4 × X →X: Zφ2(x1, x2, x3, x4) = 1 |D4| X g∈D4 4 Y i=1 x(bi(g)) i = 1 8(x4 1+x4+x2 2+x4+x2 2+x2 1x2+x2 2+x2 1x2) = 1 8(x4 1 + 2x2 1x2 + 3x2 2 + 2x4). The number of distinct colorings is then equal to Zφ2(3, 3, 3, 3) = 1 8(34 + 2(32)(3) + 3(32) + 2(3)) = 21. The three coloring-pairs distinct under rotation but not rotation and reflection are shown below: POLYA’S ENUMERATION 15 Finally, for the last case, we use de Bruijn’s theorem. We have Zφ(x1, x2, x3, x4) = 1 4(x4 1 + x2 2 + 2x4) Zψ(x1, x2, x3) = 1 6(x3 1 + 3x1x2 + 2x3), so the number of distinct colorings is equal to Zφ( ∂ ∂z1 , ∂ ∂z2 , ∂ ∂z3 , ...)Zψ(e P i=1 zi, e2 P i=1 z2i, e3 P i=1 z3i, ...) {zi}=0 = 1 24( ∂4 ∂z4 1 + ∂2 ∂z2 2 + 2 ∂ ∂z4 )(e3(z1+z2+z3+z4) + 3ez1+z2+z3+z4e2z2+2z4 + 2e3z3) {zi}=0 = 1 24(34 + (3) + 0 + 32 + (3)(32) + 0 + (2)(3) + (2)(3)(3) + 0) = 6. □ The 6 distinct colorings are shown below: where all of the colorings below are equivalent to one another: rotation reflection permutation (RGB) 6. Further Work Multiple generalizations of Polya’s Enumeration Theorem exist, most coming from the work of de Bruijn, that are not fully addressed in this paper. As more generalized theorems of PET are developed, the most meaningful work on the subject may certainly come from clever substitutions or cases of these theorems. More modern applications of the theorem can be found in the fields of analytic combinatorics and random permutation statistics, among others. 16 ALEC ZHANG Acknowledgments. It is a pleasure to thank my mentor Nat Mayer, who pa-tiently guided me through unfamiliar concepts and sat through oft-disorganized presentations. Great thanks also go to Professor Laci Babai for a rigorous and engaging apprentice class, and to Peter May for hosting this REU and giving me the opportunity to meet many bright minds. References G. Polya. Kombinatorische Anzahlbestimmungen fr Gruppen, Graphen und chemische Verbindungen. Acta Math., 68. 1937. Pages 145-254. Matias von Bell. Polya’s Enumeration Theorem and its Applications. 2015. Pages 17-23. N. G. de Bruijn. Polya’s Theory of Counting. Applied Combinatorial Mathematics. Wiley, New York. 1964. Pages 144-184.
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https://www.cs.utep.edu/vladik/2018/tr18-14.pdf
Optimization of Quadratic Forms and t-norm Forms on Interval Domain and Computational Complexity Milan Hlad´ ık Charles University Faculty of Mathematics and Physics Department of Applied Mathematics Malostransk´ e n´ am. 25 11800, Prague, Czech Republic milan.hladik@matfyz.cz Michal ˇ Cern´ y Department of Econometrics University of Economics, Prague W. CHurchill’s sq. 4 13067 Prague, Czech Republic cernym@vse.cz Vladik Kreinovich Department of Computer Science University of Texas at El Paso 500 W. University El Paso, Texas 79968, USA vladik@utep.edu Abstract—We consider the problem of maximization of a quadratic form over a box. We identify the NP-hardness bound-ary for sparse quadratic forms: the problem is polynomially solvable for O(log n) nonzero entries, but it is NP-hard if the number of nonzero entries is of the order nε for an arbitrarily small ε > 0. Then we inspect further polymonially solvable cases. We define a sunflower graph over the quadratic form and study efficiently solvable cases according to the shape of this graph (e.g. the case with small sunflower leaves or the case with a restricted number of negative entries). Finally, we define a generalized quadratic form, called t-norm form, where the quadratic terms are replaced by t-norms. We prove that the optimization problem remains NP-hard with an arbitrary Lipschitz continuous t-norm. I. INTRODUCTION In this paper we elaborate the problems outlined in in more details. In that work we studied processing imprecise data from multiple sources which interact together. The inter-action among inputs x1, . . . , xn is formalized by a function f(x1, . . . , xn) which cannot be written in a separable form as ∑n i=1 fi(xi) (for some functions fi). An example is a quadratic form xT Ax with nonzero off-diagonal entries, which is studied in this paper. Then we consider a more general form of pairwise interactions: formally, we replace the bilinear terms xixj (i ̸= j) from xT Ax by so-called t-norms (which can be regarded as generalizations of the “AND” logical connective). The general question is: when the inputs x1, . . . , xn are imprecise but are known to be in given compact intervals x1 = [x1, x1], . . . , xn = [xn, xn], and we are given a function f : Rn →R, can we find tight bounds for f(x1, . . . , xn)? Formally, denoting x = x1 × · · · × xn, (1) the problem reduces to the computation of supx∈Rn{f(x) | x ∈x} and infx∈Rn{f(x) | x ∈x}. Here, the expression “to find the bounds” refers to computational complexity: We are to determine under which conditions the bounds can be evaluated in polynomial time and when the computation is NP-hard. Recall that in general, finding the tight bounds for a general function f need not be recursive. This is why various classes of functions of interest in data processing need to be studied separately. In this text, bold symbols—such as x—refer to n-dimensional intervals of the form (1). The real n-vectors of lower and upper bounds are denoted by x and x, respectively, and we write x = [x, x] for short. Basics in computations complexity and interval computing can be found e.g. in . II. QUADRATIC FORMS ON INTERVAL DOMAIN Consider a quadratic form f : Rn →R f(x) = bT x + xT Ax = n ∑ i=1 bixi + n ∑ i,j=1 aijxixj restricted to a given interval domain x = [x, x]. It is known that computing the range of f on x, i.e., f := min f(x) subject to x ∈x, f := max f(x) subject to x ∈x is an NP-hard problem. This is true even for A positive definite, in which case computing f is polynomial whereas computing f is NP-hard. Assumption. For simplicity of exposition, we focus only on computation of f in the sequel. We will also assume for the remainder of the paper that f(x) is convex (i.e., that A is positive semidefinite). III. SPARSE QUADRATIC FORMS Suppose that the matrix A is sparse, that is, most of the off-diagonal entries are zero. Becomes the problem of computing the range tractable? In this section it is sufficient to fix x = [0, 1]n. Proposition 1. The problem of computing f remains NP-hard even when the number of off-diagonal non-zeros in A is bounded by O(n1/k). Proof: Let f(x) = bT x + xT Ax be a quadratic function on Rn. Consider the quadratic form g(x, y) := f(x) + m ∑ i=1 (2yi −1)2. Then the maximum of g(x, y) on [0, 1]n+m is the same as the maximum of f(x), shifted by the amount of m. That is, g = f + m. (2) Putting m := n2k we get that the quadratic form g(x, y) of dimension d = n + m has O(d1/k), non-zero off-diagonal en-tries in the corresponding matrix. Since f(x) was an arbitrary quadratic form, computing the range of g(x, y) is NP-hard, too. Corollary (to the proof). Under the assumption of Proposi-tion 1, it is NP-hard to approximate f with a given (arbitrarily large) absolute error. This follows from the fact that the maximum of a quadratic form is known to be NP-hard to approximate with an absolute error , and (2) does not change the absolute error. On the other hand, approximating f with a relative error can be done efficiently via semidefinite ralaxation even for a nonconvex f(x), see . Proposition 2. The problem becomes polynomial if the num-ber of off-diagonal non-zeros in A is bounded by O(log n). Proof: Denote I := {i = 1, . . . , n | ∃j ̸= i : aij ̸= 0}. Now, f(x) can be expressed as f(x) = ∑ i̸∈I (bixi + aiix2 i ) + ∑ i,j∈I aijxixj, and its maximum as f = max x∈x  ∑ i̸∈I (bixi + aiix2 i )  +max x∈x  ∑ i,j∈I aijxixj  . (3) The first term in (3) is computed easily as max x∈x  ∑ i̸∈I (bixi + aiix2 i )  = ∑ i̸∈I max xi∈xi(bixi + aiix2 i ), and maximizing a univariate quadratic function is a trivial task. The second term in (3) requires maximizing a quadratic function on an interval domain in dimension O(log n). Hence, by brute force, we find the maximum in exponential time w.r.t. O(log n), which is polynomial w.r.t. n. IV. POLYNOMIAL CASES BASED ON SUNFLOWER GRAPHS Without loss of generality assume that A is upper triangular and that x = [0, 1]n. Consider the graph G = (V, E), where V = {x1, . . . , xn} and {xi, xj} is an edge of G if and only if aij ̸= 0. So we are in fact maximizing a quadratic form f(x) on the graph G (see Chapter 10 of ). Let D ⊆V be a vertex cut such that the graph G′ = (V \ D, E′) after removing the cut D consists of connected components of vertex size O(log n). Suppose further that the size of the cut is |D| = O(log n). (A graph with such cut is sometimes called sunflower graph, see Figure 1.) Then the cut is associated with |D| variables. Hence we can process all 0/1-assignments of these variables. There are at most 2|D| such assignments. For every such assignments, we resolve the problem by brute force in each of the components. Therefore, the overall time complexity is 2|D|(T1 + T2 + · · · + Tk) ≤2O(log n)(2O(log n) + 2O(log n) + · · · + 2O(log n)) ≤poly(n), where Ti is time complexity of maximization over ith com-ponent, k ≤n is the number of components, and poly(n) is a polynomial in n. x1 x3 x8 Component 1 Component 2 Component 3 Component 4 · · · · · · cut Fig. 1. A sunflower graph with a cut of size O(log n) and components of size O(log n). A problem. How to find a suitable cut? This is an open challenging question. Notice that minimum cut splitting graph G into two components can be found efficiently by means of linear programming. Nevertheless, incorporating restrictions on size of the components seems a hard problem. Special graphs The above reasoning can be extended even to the compo-nents larger than O(log n), but having a special structure. So, we will now discuss a few of special graphs possessing a suitable structure. For the sake of simplicity of exposition, we will illustrate it on the graph G = (V, E). Few negative coefficients: Provided that all coefficients are nonnegative, that is, bi ≥ 0 and aij ≥ 0 for all i, j ∈{1, . . . , n}, then the optimal solution is simply x = (1, . . . , 1)T . If it is not the case, we can still effectively compute an optimal solution as long as the number of negative coefficients bi and aij is small. Define a cut D to contain all variables incident with negative coefficients: D := {xi; bi < 0 or aij < 0 for some j}. If |D| = O(log n), then we are done since applying the cut we obtain a subproblem with nonnegative coefficients and the 0/1-variables in D can be tested brute-force in time 2|D| = poly(n). Other special graphs Assume now for simplicity in the remainder of this section that the domain of variables is x = [−1, 1]n. Further assume that bi = 0 for every i. Trees: If G is a tree, then maximizing the quadratic function on G is easy: Take an arbitrary vertex in xi ∈G as a root, and distinguish two assignments xi = ±1. For each assignment, the remaining variables associated with G have determined values. Sorting the vertices according to some tree search algorithm, we put xj := sgn(aijxi) when xi precedes xj. Planar graphs: The above class can be extended to planar graphs with O(log n) faces because by removing O(log n) vertices we obtain a tree. Bipartite graphs: Complete bipartite graphs Km,n and their subgraphs are also efficiently processed provided aij ≤0 for i ̸= j. The variables associated with the first set of vertices will be set as xi := 1, and the others xi := −1. If the assumption aij ≤0 for i ̸= j is not satisfied, then the bipartite graph is still efficiently processed as long as m = O(log n), in which case the vertex cut D is the smaller of those two subsets. Remark. For related results see . V. T-NORM FORMS Recall that a t-norm is a function T : [0, 1]2 →[0, 1] satisfying: • commutativity: T(a, b) = T(b, a), • monotonicity: a ≤c, b ≤d ⇒T(a, b) ≤T(c, d), • associativity: T(a, (T(b, c))) = T(T(a, b), c), • 1 is identity element: T(a, 1) = a. From the definition, we immediately have T(0, 0) = T(0, 1) = T(1, 0) = 0, T(1, 1) = 1. (4) Given t-norms Tij, the question is how easy is evaluation of the t-norm form fT (x) = n ∑ i=1 (bixi + aiix2 i ) + ∑ i̸=j aijTij(xi, xj) on a given interval domain x. Proposition 3. Maximizing a t-norm form on x = [0, 1]n is NP-hard even if we choose and fix for every Tij a Lipschitz continuous t-norm, that is, |Tij(x) −Tij(x′)| ≤α · ∥x −x′∥, where α is a Lipschitz constant and ∥· ∥is any vector norm. Proof: Let f(x) = bT x + xT Ax be a convex quadratic function on Rn. Consider the t-norm form fT (x) := bT x + n ∑ i=1 aiix2 i + β n ∑ i=1 (2xi −1)2 + ∑ i̸=j aijTij(xi, xj). By the Lipschitz continuity assumption, for sufficiently large β the function fT (x) is convex. Thus the maximum of fT (x) is attained in a vertex of x. However, on a set of vertices x ∈{0, 1}n, fT (x) = f(x) + βn since Tij(xi, xj) = xixj and (2xi −1)2 = 1. This means that the maximum of fT (x) is the same as the maximum of f(x), shifted by the amount of βn. Since maximizing f(x) on x is NP-hard, maximizing t-norm forms on x is NP-hard, too. Remark 1. It is interesting that the proof does not require all the axioms of a t-norm. Basically, we used (4) only. Thus the statement holds true for any Lipschitz continuous functions Tij satisfying (4). Notice that the commonly used t-norms satisfy the assump-tion of the proposition: • product t-norm T(x, y) = xy (in this case, the t-norm form is a quadratic form), • minimum t-norm T(x, y) = min{x, y}, • Łukasiewicz t-norm T(x, y) = max{0, x + y −1}, • nilpotent minimum t-norm T(x, y) = { min{x, y} if x + y > 1, 0 otherwise, • Hamacher product t-norm T(x, y) = { 0 if x = y = 0, xy x+y−xy otherwise. On the other hand, the drastic t-norm defined as T(x, y) = { min{x, y} if max{x, y} = 1, 0 otherwise does not satisfy the assumption. ACKNOWLEDGMENTS M. Hlad´ ık was supported by the Czech Science Foundation Grant P403-18-04735S. M. ˇ Cern´ y was supported by the Czech Science Foundation Grant P402/12/G097. V. Kreinovich was supported in part by the National Science Foundation Grant HRD-1242122 (Cyber-ShARE Center of Excellence). REFERENCES M. ˇ Cern´ y M and M. Hlad´ ık M. The complexity of computation and approximation of the t-ratio over one-dimensional interval data. Computational Statistics & Data Analysis 80, 2014, 26–43. J.-A. Ferrez, K. Fukuda, and T. Liebling. Solving the fixed rank convex quadratic maximization in binary variables by a parallel zonotope construction algorithm. European Journal of Operational Research 166, 2005, 35–50. B. G¨ artner and J. Matouˇ sek. Approximation Algorithms and Semidefi-nite Programming. Springer, 2012. M. Hlad´ ık, M. ˇ Cern´ y, and V. Kreinovich, “When Is Data Processing under Interval and Fuzzy Uncertainty Feasible: What If Few Inputs Interact? Does Feasibility Depend on How We Describe Interaction?” These Proceedings. Nesterov Yu. Semidefinite relaxation and nonconvex quadratic opti-mization. Optimization Methods & Software 9 (1–3), 1998, 141–160. V. Kreinovich, A. Lakayev, J. Rohn, and P. Kahl, Computational Com-plexity and Feasibility of Data Processing and Interval Computations, Kluwer, Dordrecht, 1998. Yu G.-D. (2014). Quadratic forms on graphs with applications to minimizing the least eigenvalue of signless Laplacian over bicyclic graphs. Electronic Journal of Linear Algebra 27, article 13.
187818
https://math.stackexchange.com/questions/4623654/define-if-two-planes-are-parallel
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams define if two planes are parallel Ask Question Asked Modified 2 years, 8 months ago Viewed 152 times 0 $\begingroup$ I am trying to develop a function to determine if 2 planes are parallel or not. I have their equations : P1 : a1.x + b1.y + c1.z + d1 = 0 P2 : a2.x + b2.y + c2.z + d1 = 0 I check : a1=0 & a2=0 => parallel b1=0 & b2=0 => parallel c1=0 & c2=0 => parallel else, if a1/a2 = b1/b2 = c1/c2 => parallel I have some big doubts about it, because I read on courses that none of parameters may be equal to 0, but if I have a plane on $(x,y)$, with $z=10$, the equation may be : $z-10 = 0$, so $a, b =0$, right? How could I check parallelism, taking in account each particular case? After thinking, I guess it may be something like that : (a1=0 & a2=0) => no need to check equality (b1=0 & b2=0) => no need to check equality (c1=0 & c2=0) => no need to check equality if (a1=0 & a2!=0) or (a1!=0 & a2=0) => not parallel if (b1=0 & b2!=0) or (b1!=0 & b2=0) => not parallel if (c1=0 & c2!=0) or (c1!=0 & c2=0) => not parallel Where != means different Then check if a1/a2 = b1/b2 = c1/c2 (if I need to check) Aside : How can I insert here some mathematical signs, I cannot find? geometry 3d plane-geometry Share edited Jan 22, 2023 at 16:32 whoisit 4,05911 gold badge99 silver badges2929 bronze badges asked Jan 22, 2023 at 15:22 Siegfried.VSiegfried.V 33633 silver badges1313 bronze badges $\endgroup$ 8 1 $\begingroup$ To avoid worrying about zero, you could check whether $(a1.b2=a2.b1) and (a1.c2=a2.c1) and (b1.c2=b2.c1)$ $\endgroup$ Empy2 – Empy2 2023-01-22 15:27:09 +00:00 Commented Jan 22, 2023 at 15:27 $\begingroup$ In fact, I just found a better course, that is giving the definition. In fact, I can have 1a=0, or b1=0, or c1=0, but never 3 of thm = 0, right? $\endgroup$ Siegfried.V – Siegfried.V 2023-01-22 15:32:14 +00:00 Commented Jan 22, 2023 at 15:32 $\begingroup$ Yes, check for that too $\endgroup$ Empy2 – Empy2 2023-01-22 16:25:57 +00:00 Commented Jan 22, 2023 at 16:25 $\begingroup$ You should check mathjax - where we write math within $...$ for typing out math signs. Math Meta Stackexchange has a mathjax tutorial, search for it $\endgroup$ whoisit – whoisit 2023-01-22 16:27:06 +00:00 Commented Jan 22, 2023 at 16:27 $\begingroup$ @Empty thanks that helped. if you write as answer, I could accept it, or may I accept jp bouchon's answer, as it is good explained too, don't know what may be the correct way. $\endgroup$ Siegfried.V – Siegfried.V 2023-01-22 16:29:38 +00:00 Commented Jan 22, 2023 at 16:29 | Show 3 more comments 1 Answer 1 Reset to default 2 $\begingroup$ If a plane $P$ has $ax+by+cz+d=0$ as cartesian equation €”with $(a,b,c)\ne(0,0,0)$€”, then the vector $\left(\begin{smallmatrix} a\ b\ c \end{smallmatrix}\right)$ is orthogonal to $P$. Now two planes $P_1$ and $P_2$ are parallel iff $\left(\begin{smallmatrix} a_1\ b_1\ c_1 \end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} a_2\ b_2\ c_2 \end{smallmatrix}\right)$ are €œproportional€, i.e., linearly dependent, which is in turn equivalent to $$\begin{vmatrix} a_1&a_2\ b_1&b_2 \end{vmatrix} = \begin{vmatrix} a_1&a_2\ c_1&c_2 \end{vmatrix} = \begin{vmatrix} b_1&b_2\ c_1&c_2 \end{vmatrix}=0.$$ That is: $a_1b_2-a_2b_1 =a_1c_2-a_2c_1 =b_1c_2-b_2c_1=0$. Share edited Jan 22, 2023 at 16:27 Apass.Jack 13.5k11 gold badge2222 silver badges3333 bronze badges answered Jan 22, 2023 at 16:15 jp boucheronjp boucheron 83622 silver badges77 bronze badges $\endgroup$ 3 $\begingroup$ you wrote with (a,b,c)‰ (0,0,0). Could you explain why? As I understood, a,b,c are not necessarly ‰ 0 ? $\endgroup$ Siegfried.V – Siegfried.V 2023-01-23 17:05:14 +00:00 Commented Jan 23, 2023 at 17:05 $\begingroup$ This is because the cartesian equation $0.x + 0.y + 0.z + d=0$ doesn't define a plane. According to the value of $d$, this equation defines $\emptyset$ (for $d‰ 0$) or $\Bbb{R}^3$ (for $d=0$). To define a plane at least one of the coefficients must be ${}\neq0$. Put otherwise: the affine plane P has for direction a vector space $\vec P$, which is defined as the kernel of a non-zero linear form $\ell$, and $P$ has $\ell(x,y,z)+d=0$ for equation (for some $d\in\Bbb{R}$. Such a linear form is of the type $\ell(u,v,w)=au+bv+cw$ for some $(a,b,c)\neq(0,0,0)$. $\endgroup$ jp boucheron – jp boucheron 2023-01-23 17:23:17 +00:00 Commented Jan 23, 2023 at 17:23 $\begingroup$ Sorry, I missunderstood the comment. I believed none of them may be equal to 0. Thanks for your answer, it helped a lot :). $\endgroup$ Siegfried.V – Siegfried.V 2023-01-23 18:16:29 +00:00 Commented Jan 23, 2023 at 18:16 Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry 3d plane-geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 0 Obtaning a quality intersection line in 3D and variance-covariances contribution Is there any good reason for a programmer to study geometry? 1 I have some problems with straight lines and planes 1 simple 3d space collision check algorithm How can I prove that 3 planes are arranged in a triangle-like shape without calculating their intersection lines? 1 Prerequisites for the book "Three-dimensional geometry and topology" Are there affine planes with no abelian group of translations? Hot Network Questions в ответе meaning in context Real structure on a complex torus How to locate a leak in an irrigation system? What "real mistakes" exist in the Messier catalog? 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187819
https://www.sea-astronomia.es/glosario/medio-interplanetario
medio interplanetario | Sociedad española de astronomía Pasar al contenido principal Acceso SEAMOS La SEA Quiénes somos Área de socios Reuniones Científicas Boletín Comisiones Eventos Sinergias SEA Investigación Centros de investigación Observatorios Estudios RRHH Publicaciones Tesis Enseñanza y divulgación Enseñanza Divulgación TERMINOLOGÍA Libros miembros SEA Astronomía y Cultura Mujeres y Astronomía Pro-Am Agenda Empleos / Becas Noticias Contacto medio interplanetario Material y campos magnéticos que pueblan el Sistema Solar en el espacio entre los cuerpos celestes (planetas, asteroides y cometas). Este medio está formado por polvo interplanetario, rayos cósmicos, plasma proveniente del viento solar y la combinación de los campos magnéticos del Sol y los planetas. La temperatura del medio interplanetario es aproximadamente de 100 000 K y su densidad muy baja, del orden de cinco partículas por centímetro cúbico en la vecindad de la Tierra. Esta densidad disminuye conforme aumenta la distancia al Sol (la relación es de proporcionalidad inversa del cuadrado de la distancia). El reflejo de la luz solar en las partículas pulverulentas del medio interplanetario da lugar a un resplandor difuso que recibe el nombre de luz zodiacal y luz antisolar (Gegenschein). Volver al índice Inicio Acceso web Política de cookies SÍGUENOS
187820
https://deutsch.heute-lernen.de/grammatik/der-die-das/verstaerker
Heißt es der, die oder das Verstärker? Direkt zum Inhalt Deutsch lernen „Der”, „die” oder „das”? Das Genus im Deutschen Verstärker Heißt es der, die oder das Verstärker? Das grammatikalische Geschlecht (Genus) von Verstärker ist maskulin Der Artikel im Nominativ ist deswegen der. Deutsche sagen also: der Verstärker. Der, die oder das? Welche Regeln gibt es? Du weißt jetzt, dass Verstärker maskulin ist. Aber wie ist es mit all den anderen deutschen Wörtern? Musst du den Artikel für jedes einzelne Wort lernen? Leider gibt es wirklich nicht sehr viele universelle Regeln, die dir bei dem Problem helfen. Denn das grammatikalische Geschlecht im Deutschen ist oft sehr unlogisch. Das natürliche Geschlecht hilft dir selten weiter. Mädchen sind zum Beispiel per Definition weiblich. Aber das Genus des Wortes Mädchen ist trotzdem neutral: es heißt das Mädchen. Ganz sicher kannst du beim Genus von einem Wort also nur sein, wenn du es auswendig gelernt hast. Teste dein Wissen mit unserem Quiz Wie gesagt: Logisch ist das Genus im Deutschen nicht. Aber ein paar Regeln gibt es zum Glück doch: Maskuline Wörter Sehr oft (aber leider nicht immer) maskulin sind Wörter, mit denen man über Zeit und Datum spricht, also zum Beispiel die Tageszeiten, die Wochentage, Monate und auch Jahreszeiten. Die vier Himmelsrichtungen sind maskulin. Auch Vokabeln, mit denen man das Wetter beschreibt, brauchen oft den Artikel der: der Wind, der Schnee, der Regen. Und auch wenn das berühmteste deutsche Getränk – das Bier – neutral ist; die meisten anderen Getränke mit Alkohol sind maskulin. Außerdem gibt es bestimmte Wortendungen, die man fast nur bei maskulinen Wörtern findet; zum Beispiel: -ig, -ich, -ling oder -en. Feminine Wörter Der Apfel ist die wichtigste Ausnahme. Aber fast alle anderen Obstsorten – die Kiwi, die Orange, die Traube – sind feminin. Namen für Schiffe und Motorräder benutzt man im Deutschen auch immer mit die. Es gibt auch einige Suffixe, die zeigen, dass ein Wort sehr wahrscheinlich feminin ist: -in, -keit, -heit, -ung, -schaft oder -ei. Neutrale Wörter Die Suffixe -ment, -tum und -chen sind typisch für Wörter, die den Artikel das brauchen. Außerdem benutzen Deutsche das, wenn sie über Farben (das Rot) oder Biersorten sprechen. Was ist der richtige unbestimmte Artikel zu Verstärker? Wenn du der Verstärker sagst, heißt das, dass dein Gesprächspartner wissen sollte, worüber genau du sprichst. Willst du weniger konkret über etwas sprechen, benutzt du stattdessen die unbestimmten Artikel ein und eine. Es gibt nur diese beiden Formen: eine für feminine Nomen und ein für maskuline und neutrale Wörter. Verstärker ist maskulin, die korrekte Form ist also: ein Verstärker. Was ist der Plural von Verstärker? der Verstärker=> die Verstärker ein Verstärker=> viele Verstärker Im Plural sind die deutschen Artikel sehr viel weniger problematisch. Der bestimmte Artikel ist im Plural immer die, egal ob der Begleiter im Singular der, die oder das heißt. Unbestimmte Artikel gibt es im Deutschen im Plural nicht. Du solltest dann also einfach die Pluralform ohne einen Artikel benutzen. Ein bisschen komplizierter sind die Pluralformen des Nomens. Bei der Bildung des Plurals gibt es nämlich auch einige Ausnahmen, die du lernen musst. Und wie dekliniert man Verstärker? Alles verstanden? Dann teste doch gleich dein Deutsch-Wissen: QUIZ Klicke auf den richtigen Artikel: React App Polizei der die das Sagt man der, die, oder das... Sonderheft Seenotretter Wiesnanstich Zeitungsjunge Handschlag Nachrichtenpodcast Wiesnhendl Internetauftritt Eingabefehler Dokumentarfilm Käserinde Coworker Gummi Grillhendl Baumschule Nach Artikeln suchen © ZEIT SPRACHEN GmbH Datenschutz Impressum AGB
187821
https://mathoverflow.net/questions/109582/seeking-a-geometric-proof-of-a-generalized-alternating-series-convergence
mathematics education - Seeking a Geometric Proof of a Generalized Alternating Series' Convergence - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Seeking a Geometric Proof of a Generalized Alternating Series' Convergence Ask Question Asked 12 years, 11 months ago Modified12 years, 9 months ago Viewed 2k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. Let z∈C∖{1}z∈C∖{1} with |z|=1|z|=1. We consider the following infinite series, which necessarily converges: S(z):=∑n=1∞z n n S(z):=∑n=1∞z n n Note that S(−1)S(−1) is the alternating harmonic series. A straightforward application of the Dirichlet Convergence Test proves any such series converges, but I feel this is a bit like killing a fly with a sledgehammer. (I realize some of you might not think this test is a sledgehammer; I wonder also whether this series is a fly.) In any event, I'm wondering whether there is a way to prove convergence using only a simple geometric argument (with some basic analysis). For example, we can think of S(i)S(i) as taking steps in the plane of length 1/n 1/n, but turning ninety degrees after each one. Then the partial sums correspond to a nested sequence of squares, where the area of the squares is clearly converging to 0 0. Thus, an argument using the Nested Interval Property (or really its corresponding 2 D 2 D version) indicates that the series converges. More generally, I'd think that because we are taking steps of size decreasing to 0 0 and rotating by the same amount after each step, there should be a general geometric argument for why S(z)S(z) will converge. Ideally, I'd like to have a proof that could be made accessible to early Calculus students, even if not every step is presented in fully rigorous form. For clarity's sake, I will directly state my question: How does one prove S(z)S(z) converges using a simple geometric argument that relies at most on basic analysis (e.g., makes no appeal to stronger theorems from Complex Analysis)? mathematics-education mg.metric-geometry cv.complex-variables real-analysis Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Dec 29, 2012 at 5:48 Benjamin DickmanBenjamin Dickman asked Oct 14, 2012 at 5:46 Benjamin DickmanBenjamin Dickman 8,096 2 2 gold badges 51 51 silver badges 96 96 bronze badges 2 2 Your question is interesting, but since Abel summation is the discrete-time analogue of integration by part, in a calculus course where both series and integrals are discussed I would rather use it to have one more link between the two theories.Benoît Kloeckner –Benoît Kloeckner 2012-10-14 07:48:15 +00:00 Commented Oct 14, 2012 at 7:48 I'm not suggesting a geometric proof for this particular problem be presented in lieu of introducing other topics; instead, I agree with your (apparent) fondness for linking different theories, so I enjoy when proofs can be carried out in multiple ways. That said, for better or worse, often a student's first Calculus course that covers Leibniz's Alternating Series Test will not get into Abel summation, Dirichlet series, etc.Benjamin Dickman –Benjamin Dickman 2012-10-14 08:02:12 +00:00 Commented Oct 14, 2012 at 8:02 Add a comment| 4 Answers 4 Sorted by: Reset to default This answer is useful 17 Save this answer. Show activity on this post. Here is what I think you are looking for: First, note that if you take steps of fixed length ℓ ℓ, and keep rotating by an angle of θ≠0 θ≠0, then you will stay inside a circle of radius r=ℓ 2 cos(θ/2)−1 r=ℓ 2 cos⁡(θ/2)−1. In fact, the steps will all land on the circle, and you can calculate its center as the point of distance r r from one of the steps, at an angle which bisects the angle θ θ. Now if at a certain step you decide to change the length of your steps to ℓ′<ℓ ℓ′<ℓ, the new circle will have radius r′=ℓ′2 cos(θ/2)−1 r′=ℓ′2 cos⁡(θ/2)−1, with center the point of distance r′r′ away from the current step in the same direction as the old center. From this description it's clear that the new circle is contained in the old circle. Now you can apply the nested interval property. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 14, 2012 at 8:17 Kevin VentulloKevin Ventullo 4,840 1 1 gold badge 32 32 silver badges 42 42 bronze badges 0 Add a comment| This answer is useful 5 Save this answer. Show activity on this post. This proof is not really much more than a rewriting of Dirichlet's test, but here goes: (z−1)∑n=1 N z n n=∑n=2 N+1 z n n−1−∑n=1 N z n n=z N+1 N−z 1+∑n=2 N z n n(n−1)(z−1)∑n=1 N z n n=∑n=2 N+1 z n n−1−∑n=1 N z n n=z N+1 N−z 1+∑n=2 N z n n(n−1) which converges as N→∞N→∞. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 14, 2012 at 7:15 Robert IsraelRobert Israel 54.7k 1 1 gold badge 79 79 silver badges 156 156 bronze badges 1 Isn't it more like convergence acceleration?ACL –ACL 2018-03-21 08:40:41 +00:00 Commented Mar 21, 2018 at 8:40 Add a comment| This answer is useful 3 Save this answer. Show activity on this post. Here's an idea. Group the series into blocks ∑n=d k d(k+1)−1 z n n∑n=d k d(k+1)−1 z n n where d d is fixed and large enough that the complex numbers 1,z,z 2,...z d−1 1,z,z 2,...z d−1 are approximately uniformly distributed across the unit circle. The terms in each block should then be approximately uniformly distributed in phase across the unit circle, and in particular each term should be pairable with a term approximately the negative of it, the two of them canceling to order O(1 n 2)O(1 n 2). But making this precise seems to me to require more effort than proving that Dirichlet's convergence test works. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 14, 2012 at 6:55 Qiaochu YuanQiaochu Yuan 124k 42 42 gold badges 466 466 silver badges 764 764 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. Think of an alternating series ∑n≥0(−1)n a n,a n>a n+1>0∑n≥0(−1)n a n,a n>a n+1>0 as describing the motion of a hesitating person, who starts at a 0 a 0, and goes alternatively, one step backward, one step forward. If we denote by S n S n its location after n n-steps S n=∑k=0 n(−1)k a k,S n=∑k=0 n(−1)k a k, then we observe that S 0>S 1>0 S 0>S 1>0 and we deduce inductively S 2 n>S 2 n+2>S 2 n+1>0 S 2 n>S 2 n+2>S 2 n+1>0 Thus during the travel, the person is always on the right-hand side of the origin, and every two steps he gets closer to the origin. Mathematically, this means that the subsequence S 2 n S 2 n is decreasing and positive thus it has a limit. The odd subsequence S 2 n+1 S 2 n+1 converges to the same limit since S 2 n−S 2 n+1=a 2 n+1→0.S 2 n−S 2 n+1=a 2 n+1→0. We can visualize this process by considering the continuous function S:[0,∞)→R S:[0,∞)→R which is linear on each of the intervals [n,n+1][n,n+1], n n nonegative integer, and such that S(n)=S n S(n)=S n. It can be visualized as a zig-zag, down-up-down-up, that stays above the x x-axis, while the peaks are getting shorter and shorter. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Oct 14, 2012 at 13:08 answered Oct 14, 2012 at 12:54 Liviu NicolaescuLiviu Nicolaescu 35.7k 2 2 gold badges 96 96 silver badges 172 172 bronze badges Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions mathematics-education mg.metric-geometry cv.complex-variables real-analysis See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 70Behaviour of power series on their circle of convergence 10Power series with funny behavior at the boundary Related 7completeness axiom for the real numbers 6Puiseux series expansion for space curves? 3Simple bound for generalized geometric series 1Abscissa of convergence for a very specific Dirichlet series / Euler product 26What is the origin/history of the following very short definition of the Lebesgue integral? 1Exponent of convergence of sequence of zeros of e P(z)+Q(z)e P(z)+Q(z) Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. 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187822
https://brainly.com/question/24222111
[FREE] b. Compare the similar triangle proof from question 3 with the inscribed square proof. How are they - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +33,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +13,1k Ace exams faster, with practice that adapts to you Practice Worksheets +5,6k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified b. Compare the similar triangle proof from question 3 with the inscribed square proof. How are they different? Which method was easier for you to understand? (1 point) 1 See answer Explain with Learning Companion NEW Asked by Bayabbay5180 • 07/12/2021 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 23216 people 23K 4.5 9 Upload your school material for a more relevant answer i might be wrong but this is what i put Explanation In question 3 it was comparing three triangles where now it is using the triangles to find the area of a square instead of proving that they are the same. Answered by 1036059 •51 answers•23.2K people helped Thanks 9 4.5 (10 votes) Expert-Verified⬈(opens in a new tab) This answer helped 23216 people 23K 4.4 8 Upload your school material for a more relevant answer The proof of similar triangles focuses on the equality of angles and proportionality of sides, while the inscribed square proof emphasizes area relations and geometric fitting. Understanding may vary, with some students finding triangle properties more straightforward. Both proofs highlight distinct geometric principles at play. Explanation To compare the proof of similar triangles with the proof of an inscribed square, we can follow a structured approach: Understanding the Concepts: Similar Triangles: These triangles have the same shape, which means that their corresponding angles are equal and the ratios of their corresponding sides are proportional. The proof often involves showing that these conditions hold true using various methods such as side ratios or angle comparisons. Inscribed Square: This involves positioning a square within a larger geometric shape (like a circle or triangle) and proving its dimensions in relation to that shape. For example, finding the area of the square relative to the circle's radius. Differences in Proofs: The proof for similar triangles centers around angle and side length comparisons, while the inscribed square proof is more about area relations and how the square fits within another shape. In the case of similar triangles, one might use the properties of triangles and angles, while for an inscribed square, calculations might involve using formulas related to area and geometry (like the area of a square being A = side²). Personal Understanding: Some may find the similar triangle proof easier to understand because it mainly deals with fundamental properties of triangles that are more straightforward. In contrast, the inscribed square proof might involve more steps or require visualizing the geometric arrangement more clearly, which could complicate understanding. To summarize, comparing these two proofs highlights the differences in geometry's approach to proving relationships between shapes. The triangle proof focuses on angle and side ratios while the square proof involves area and fitting shapes, which may determine ease of understanding based on a student's proficiency with spatial reasoning. Examples & Evidence For instance, similar triangles can be proven by showing that two triangles with one angle equal will lead to the remaining angles being equal and the sides being in proportion. An inscribed square could be demonstrated by showing how the square's area relates to the enclosing circle's radius. This comparison is based on established geometric principles, where similar triangles rely on angle and side relationships established by Euclidean geometry, and the inscribed square's area relations are derived from basic area formulas of square and shapes like circles. Thanks 8 4.4 (9 votes) Advertisement Bayabbay5180 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Solve for x: 3 2​x+5=3 Solve for d. −2 d−8=8+8+4 d Solve for w in the proportion. 42 w​=6 1​w= Find the sum. 5 2​+(−5 3​)=□ The equation of a circle is x 2+(y−10)2=16. The radius of the circle is □ units. The center of the circle is at □ . 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187823
https://www.youtube.com/watch?v=p0ABG7y9jXw
The Orthocenter of a Triangle Using Vectors Mike, the Mathematician 19000 subscribers 35 likes Description 2038 views Posted: 31 Mar 2023 We use vectors to find the representation of the orthocenter of a triangle. mikethemathematician, #mikedabkowski, #profdabkowski, #calc3 5 comments Transcript: hello students in this video we'll see how to find the orthocenter of a triangle using vectors if we're given a triangle a b c we choose the circumcenter to be the origin so in other words we have is this we have a triangle a b C and the center of the circumcircle is going to be the origin for our Vector configuration so with this choice of origin we know the distance the length of any Vector will be R so in other words the length of the vector B is equal to the length of the vector a is equal to the length of the vector C with the origin being the circum Center and that's equal to R the circum radius now we claim that we can find the orthocenter so our claim is at the orthocenter which is the intersection of the altitudes is given by H which is a plus b plus c okay so to show this is true it suffices to show so this suffice is a show that if I look at a h Point H so if I look at ah the vector a to H is perpendicular to BC and with all the other relationships the same so let's look at AA so if I look at a h minus a and I dot product that with C minus B what will we get well what is h i claim that H is a plus b plus C so if H is a plus b plus C then H minus H is just going to be B plus c I'm going to dot that with C minus B so what this will be is this is going to be B dot C minus C dot b Plus the length of c squared would you just see that c and then minus the length of B squared now B dot C and C dot b are the same so those are going to cancel out and the length of C and the length of B are both equal to R so this is r squared minus r squared and that's equal to zero so that says that the vector h a or a to H is perpendicular to CB in other words this point H is on the altitude that goes from a to H they go from a to the side like BC similarly we have that H minus B dot C minus a is equal to zero and H minus C dot B minus a is equal to zero so that shows me that this point H is on every altitude so H is not only on each altitude all the altitudes are concurrent at this point H so the orthocenter of a triangle is in Vector form is a plus b plus C if we choose the circumcenter to be the origin thank you very much
187824
https://en.wikipedia.org/wiki/Template:Logical_connectives
Template:Logical connectives - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 See also Template:Logical connectives [x] 19 languages العربية Azərbaycanca Eesti Emiliàn e rumagnòl فارسی Français 한국어 Հայերեն Italiano עברית Македонски 日本語 ਪੰਜਾਬੀ Português Русский Українська اردو Tiếng Việt 中文 Edit links Template Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia | hide v t e Common logical connectives | | Tautology/True⊤{\displaystyle \top } | | | Alternative denial(NAND gate)∧¯{\displaystyle {\overline {\wedge }}} Converse implication⇐{\displaystyle \Leftarrow } Implication(IMPLY gate)⇒{\displaystyle \Rightarrow } Disjunction(OR gate)∨{\displaystyle \lor } | | Negation(NOT gate)¬{\displaystyle \neg } Exclusive or(XOR gate)⊕{\displaystyle \oplus } Biconditional(XNOR gate)⊙{\displaystyle \odot } Statement(Digital buffer) | | Joint denial(NOR gate)∨¯{\displaystyle {\overline {\vee }}} Nonimplication(NIMPLY gate)⇏{\displaystyle \nRightarrow } Converse nonimplication⇍{\displaystyle \nLeftarrow } Conjunction(AND gate)∧{\displaystyle \land } | | Contradiction/False⊥{\displaystyle \bot } | | Philosophy portal | Template documentation[view] [edit] [history] [purge] ‹The template below (Collapsible option) is being considered for merging with Navbox documentation. See templates for discussion to help reach a consensus.› This template's initial visibility currently defaults to autocollapse, meaning that if there is another collapsible item on the page (a navbox, sidebar, or table with the collapsible attribute), it is hidden apart from its title bar; if not, it is fully visible. To change this template's initial visibility, the |state=parameter may be used: {{Logical connectives|state=collapsed}} will show the template collapsed, i.e. hidden apart from its title bar. {{Logical connectives|state=expanded}} will show the template expanded, i.e. fully visible. See also [edit] | show v t e Logic templates | | Types of logic | Classical logic Mathematical logic Metalogic Non-classical logic Philosophical logic | | Other templates | Common logical symbols Logical connectives Logical paradoxes Logical truth Set theory Transformation rules Normal forms Diagrams | | Logic navbar | The above documentation is transcluded from Template:Logical connectives/doc. (edit | history) Editors can experiment in this template's sandbox (create | mirror) and testcases (create) pages. Add categories to the /doc subpage. Subpages of this template. Retrieved from " Categories: Logic navigational boxes Philosophy and thinking navigational boxes WikiProject Logic This page was last edited on 17 April 2025, at 10:23(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Template:Logical connectives 19 languagesAdd topic
187825
https://www.lcgdbzz.org/custom/news/id/7440
如何根据变量类型选择合适的统计分析方法？ 中文｜English ISSN 1001-5256 (Print) ISSN 2097-3497 (Online) CN 22-1108/R 首页 关于本刊 本刊简介 编委会 审稿专家 编辑部风采 广告合作 联系我们 投稿指南 本刊稿约 本刊规范 开放获取声明 出版伦理规范 稿件出版流程 同行评议流程 网络首发 论文模板 在线期刊 网络首发 当期目录 过刊浏览 F5000论文 肝胆学院 国外期刊导读 行业资讯 视频课堂 指南共识 全部指南 国内指南 国外指南 指南翻译 指南解读 Menu 首页 关于本刊 本刊简介 编委会 审稿专家 编辑部风采 广告合作 联系我们 投稿指南 本刊稿约 本刊规范 开放获取声明 出版伦理规范 稿件出版流程 同行评议流程 网络首发 论文模板 在线期刊 网络首发 当期目录 过刊浏览 F5000论文 肝胆学院 国外期刊导读 行业资讯 视频课堂 指南共识 所有指南 国内指南 国外指南 指南翻译 指南解读 留言板 尊敬的读者、作者、审稿人, 关于本刊的投稿、审稿、编辑和出版的任何问题, 您可以本页添加留言。我们将尽快给您答复。谢谢您的支持! | 姓名 | | | 邮箱 | | | 手机号码 | | | 标题 | | | 留言内容 | | | 验证码 | | 如何根据变量类型选择合适的统计分析方法？ 分类 阅读次数：1698 分享到: 用微信扫码二维码 分享至好友和朋友圈 发布日期：2015-12-23 来源：MedSci 把握两个关键 1、抓住业务问题不放松。你想解决什么问题？这是核心，是方向。 2、全面理解数据。须把握三大关键：变量、数据分析方法、变量和方法的关联。 认识变量 认识数据分析方法 其一 其二 分享到: 用微信扫码二维码 分享至好友和朋友圈 阅读次数：1698 相关指南 医学中常用计数资料的统计描述 如何使用SPSS软件进行正态分布检验 SPSS软件如何实现t检验 浅谈医学科研中计数资料的统计学处理原则 【SPSS软件应用】两指标间的关系分析 如何根据变量类型选择合适的统计分析方法？ 第一类错误和第二类错误 单因素方差分析(ANOVA)：两两比较检验Post-Hoc选项详解 统计方法选用手册 计量资料的统计描述 关于卡方检验与单因素logistic回归的个人看法 全部指南 国内指南 国外指南 指南翻译 指南解读 1 病毒性肝炎 436 1.1 乙型肝炎 229 1.2 丙型肝炎 163 1.3 甲型肝炎 5 1.4 戊型肝炎 10 1.5 其他肝炎 29 2 肝硬化及并发症 157 3 酒精性肝病 40 4 代谢相关脂肪性肝病 91 5 肝衰竭/肝性脑病/人工肝 89 6 肝肿瘤 284 7 自身免疫性肝病 53 8 药物性肝病 46 9 肝移植 103 10 其他肝病 94 10.1 遗传及代谢性肝病 46 10.2 胆汁淤积性肝病 28 10.3 肝脏血管病 20 11 一般肝病/肝脏检查 146 12 胆道疾病 206 13 胰腺疾病 358 14 全身疾病与肝病/内镜 101 15 肝胆胰疾病相关评分系统汇总 18 推荐指南 2020年欧洲肝病学会推荐意见：丙型肝炎的治疗（最终更新版） 慢性乙型肝炎防治指南（2019年版） 中国门静脉高压经颈静脉肝内门体分流术临床实践指南（2019年版） 2019 LICAGE意见书：肝硬化患者围手术期止血管理 肝纤维化中西医结合诊疗指南（2019年版） 《2019年欧洲肝脏重症监护组专家共识：肝硬化患者围手术期止血管理》摘译 《2019年美国胃肠病学会临床实践更新：肝硬化患者凝血治疗》摘译 AI 助手 中国首个肝胆病专业杂志 主管单位：中华人民共和国教育部 主办单位：吉林大学 学术支持：中华医学会肝病学分会 通信地址：长春市新疆街461号 投稿咨询：0431-88782044 审稿咨询：0431-88783542 电子信箱：lcgdb@vip.163.com 关于本刊 本刊简介 编委会 审稿专家 编辑部风采 广告合作 联系我们 投稿指南 本刊稿约 本刊规范 开放获取声明 出版伦理规范 稿件出版流程 同行评议流程 网络首发 论文模板 在线期刊 网络首发 当期目录 过刊浏览 F5000论文 肝胆学院 国外期刊导读 行业资讯 视频课堂 指南共识 全部指南 国内指南 国外指南原文 指南翻译 指南解读 昨日IP昨日PV今日IP今日PV当前在线 网站设计 © 2020 《临床肝胆病杂志》编辑部 吉ICP备10000617号-1技术支持: 仁和软件 AI助手(RAG) Hi，AI搜索已经支持Deepseek 检索增强生成 （Retrieval-augmented Generation）
187826
https://en.wikipedia.org/wiki/Cycloalkane
Jump to content Cycloalkane Afrikaans العربية Azərbaycanca Беларуская (тарашкевіца) Български Bosanski Català Čeština Deutsch Eesti Ελληνικά Español Euskara فارسی Français 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Italiano עברית ქართული Қазақша Lietuvių Magyar Македонски Монгол Nederlands 日本語 Norsk bokmål Oʻzbekcha / ўзбекча Polski Português Romnă Русский Shqip Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Татарча / tatarça Türkçe Українська Tiếng Việt 粵語 中文 Edit links From Wikipedia, the free encyclopedia Saturated alicyclic hydrocarbon In organic chemistry, the cycloalkanes (also called naphthenes, but distinct from naphthalene) are the monocyclic saturated hydrocarbons. In other words, a cycloalkane consists only of hydrogen and carbon atoms arranged in a structure containing a single ring (possibly with side chains), and all of the carbon-carbon bonds are single. The larger cycloalkanes, with more than 20 carbon atoms are typically called cycloparaffins. All cycloalkanes are isomers of alkenes. The cycloalkanes without side chains (also known as monocycloalkanes) are classified as small (cyclopropane and cyclobutane), common (cyclopentane, cyclohexane, and cycloheptane), medium (cyclooctane through cyclotridecane), and large (all the rest). Besides this standard definition by the International Union of Pure and Applied Chemistry (IUPAC), in some authors' usage the term cycloalkane includes also those saturated hydrocarbons that are polycyclic. In any case, the general form of the chemical formula for cycloalkanes is CnH2(n+1−r), where n is the number of carbon atoms and r is the number of rings. The simpler form for cycloalkanes with only one ring is CnH2n. Examples [edit] Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cyclooctane Nomenclature [edit] See also: IUPAC nomenclature See also: Alkanes Unsubstituted cycloalkanes that contain a single ring in their molecular structure are typically named by adding the prefix "cyclo" to the name of the corresponding linear alkane with the same number of carbon atoms in its chain as the cycloalkane has in its ring. For example, the name of cyclopropane (C3H6) containing a three-membered ring is derived from propane (C3H8) - an alkane having three carbon atoms in the main chain. The naming of polycyclic alkanes such as bicyclic alkanes and spiro alkanes is more complex, with the base name indicating the number of carbons in the ring system, a prefix indicating the number of rings ( "bicyclo-" or "spiro-"), and a numeric prefix before that indicating the number of carbons in each part of each ring, exclusive of junctions. For instance, a bicyclooctane that consists of a six-membered ring and a four-membered ring, which share two adjacent carbon atoms that form a shared edge, is [4.2.0]-bicyclooctane. That part of the six-membered ring, exclusive of the shared edge has 4 carbons. That part of the four-membered ring, exclusive of the shared edge, has 2 carbons. The edge itself, exclusive of the two vertices that define it, has 0 carbons. There is more than one convention (method or nomenclature) for the naming of compounds, which can be confusing for those who are just learning, and inconvenient for those who are well-rehearsed in the older ways. For beginners, it is best to learn IUPAC nomenclature from a source that is up to date, because this system is constantly being revised. In the above example [4.2.0]-bicyclooctane would be written bicyclo[4.2.0]octane to fit the conventions for IUPAC naming. It then has room for an additional numerical prefix if there is the need to include details of other attachments to the molecule such as chlorine or a methyl group. Another convention for the naming of compounds is the common name, which is a shorter name and it gives less information about the compound. An example of a common name is terpineol, the name of which can tell us only that it is an alcohol (because the suffix "-ol" is in the name) and it should then have a hydroxyl group (–OH) attached to it. The IUPAC naming system for organic compounds can be demonstrated using the example provided in the adjacent image. The base name of the compound, indicating the total number of carbons in both rings (including the shared edge), is listed first. For instance, "heptane" denotes "hepta-", which refers to the seven carbons, and "-ane", indicating single bonding between carbons. Next, the numerical prefix is added in front of the base name, representing the number of carbons in each ring (excluding the shared carbons) and the number of carbons present in the bridge between the rings. In this example, there are two rings with two carbons each and a single bridge with one carbon, excluding the carbons shared by both the rings. The prefix consists of three numbers that are arranged in descending order, separated by dots: [2.2.1]. Before the numerical prefix is another prefix indicating the number of rings (e.g., "bicyclo+"). Thus, the name is bicyclo[2.2.1]heptane. Cycloalkanes as a group are also known as naphthenes, a term mainly used in the petroleum industry. Properties [edit] Containing only C–C and C–H bonds, cycloalkanes are similar to alkanes in their general properties. Cycloalkanes with high angle strain, such as cyclopropane, have weaker C–C bonds, promoting ring-opening reactions. Cycloalkanes have higher boiling points, melting points, and densities than alkanes. This is due to stronger London forces because the ring shape allows for a larger area of contact. Even-numbered cycloalkanes tend to have higher melting points than odd-numbered cycloalkanes. While variations in enthalpy and orientational entropy of the solid-phase crystal structure largely explain the odd-even alternation found in alkane melting points, conformational entropy of the solid and liquid phases has a large impact on cycloalkane melting points.: 98 For example, cycloundecane has a large number of accessible conformers near room temperature, giving it a low melting point,: 22 whereas cyclododecane adopts a single lowest-energy conformation: 25 (up to chirality) in both the liquid phase and solid phase (above 199 K),: 32–34 and has a high melting point. These trends are broken from cyclopentadecane onwards, due to increasing variation in solid-phase conformational mobility, though higher cycloalkanes continue to display large odd-even fluctuations in their plastic crystal transition temperatures.: 99–100 Sharp plastic crystal phase transitions disappear from C48H96 onwards, and sufficiently high molecular weight cycloalkanes, such as C288H576, have similar crystal lattices and melting points to high-density polyethylene.: 27, 37 Table of cycloalkanes [edit] | Alkane | Formula | Melting point [°C] | Boiling point [°C] | Liquid density [g·cm−3] (at 20 °C) | --- --- | Cyclopropane | C3H6 | −127.6: 27 | −33 | | | Cyclobutane | C4H8 | −90.7: 27 | 12.5 | 0.720 | | Cyclopentane | C5H10 | −93.4: 27 | 49.2 | 0.751 | | Cyclohexane | C6H12 | 6.7: 27 | 80.7 | 0.778 | | Cycloheptane | C7H14 | −8.0: 27 | 118.4 | 0.811 | | Cyclooctane | C8H16 | 14.5 | 151.2 | 0.840 | | Cyclononane | C9H18 | 10–11: 262 | 178: 265 | 0.8534 | | Cyclodecane | C10H20 | 9.9: 27 | 201 | 0.871 | | Cycloundecane | C11H22 | −7.2: 1613 | 179–181: 142 | 0.81: 142 | | Cyclododecane | C12H24 | 60.4 | 244.0 | 0.855 (extrapolated) | | Cyclotridecane | C13H26 | 24.5: 27 | | 0.861: 143[a] | | Cyclotetradecane | C14H28 | 56.2: 27 | | | | Cyclopentadecane | C15H30 | 63.5: 27 | | | | Cyclohexadecane | C16H32 | 60.6: 27 | 319 | | | Cycloheptadecane | C17H34 | 64–67 | | | | Cyclooctadecane | C18H36 | 74–75 | | | | Cyclononadecane | C19H38 | 79–82 | | | | Cycloeicosane | C20H40 | 49.9: 27[b] | | | Conformations and ring strain [edit] Main article: Strain (chemistry) In cycloalkanes, the carbon atoms are sp3 hybridized, which would imply an ideal tetrahedral bond angle of 109° 28′ whenever possible. Owing to evident geometrical reasons, rings with 3, 4, and (to a small extent) also 5 atoms can only afford narrower angles; the consequent deviation from the ideal tetrahedral bond angles causes an increase in potential energy and an overall destabilizing effect. Eclipsing of hydrogen atoms is an important destabilizing effect, as well. The strain energy of a cycloalkane is the increase in energy caused by the compound's geometry, and is calculated by comparing the experimental standard enthalpy change of combustion of the cycloalkane with the value calculated using average bond energies. Molecular mechanics calculations are well suited to identify the many conformations occurring particularly in medium rings.: 16–23 Ring strain is highest for cyclopropane, in which the carbon atoms form a triangle and therefore have 60° C–C–C bond angles. There are also three pairs of eclipsed hydrogens. The ring strain is calculated to be around 120 kJ mol−1. Cyclobutane has the carbon atoms in a puckered square with approximately 90° bond angles; "puckering" reduces the eclipsing interactions between hydrogen atoms. Its ring strain is therefore slightly less, at around 110 kJ mol−1. For a theoretical planar cyclopentane the C–C–C bond angles would be 108°, very close to the measure of the tetrahedral angle. Actual cyclopentane molecules are puckered, but this changes only the bond angles slightly so that angle strain is relatively small. The eclipsing interactions are also reduced, leaving a ring strain of about 25 kJ mol−1. In cyclohexane the ring strain and eclipsing interactions are negligible because the puckering of the ring allows ideal tetrahedral bond angles to be achieved. In the most stable chair form of cyclohexane, axial hydrogens on adjacent carbon atoms are pointed in opposite directions, virtually eliminating eclipsing strain. In medium-sized rings (7 to 13 carbon atoms) conformations in which the angle strain is minimised create transannular strain or Pitzer strain. At these ring sizes, one or more of these sources of strain must be present, resulting in an increase in strain energy, which peaks at 9 carbons (around 50 kJ mol−1). After that, strain energy slowly decreases until 12 carbon atoms, where it drops significantly; at 14, another significant drop occurs and the strain is on a level comparable with 10 kJ mol−1. At larger ring sizes there is little or no strain since there are many accessible conformations corresponding to a diamond lattice. Ring strain can be considerably higher in bicyclic systems. For example, bicyclobutane, C4H6, is noted for being one of the most strained compounds that is isolatable on a large scale; its strain energy is estimated at 267 kJ mol−1. Reactions [edit] Cycloalkanes, referred to as naphthenes, are a major substrate for the catalytic reforming process. In the presence of a catalyst and at temperatures of about 495 to 525 °C, naphthenes undergo dehydrogenation to give aromatic derivatives: The process provides a way to produce high octane gasoline. In another major industrial process, cyclohexanol is produced by the oxidation of cyclohexane in air, typically using cobalt catalysts: : 2 C6H12 + O2 → 2 C6H11OH This process coforms cyclohexanone, and this mixture ("KA oil" for ketone-alcohol oil) is the main feedstock for the production of adipic acid, used to make nylon. The small cycloalkanes – in particular, cyclopropane – have a lower stability due to Baeyer strain and ring strain. They react similarly to alkenes, though they do not react in electrophilic addition, but in nucleophilic aliphatic substitution. These reactions are ring-opening reactions or ring-cleavage reactions of alkyl cycloalkanes. Preparation [edit] Many simple cycloalkanes are obtained from petroleum. They can be produced by hydrogenation of unsaturated, even aromatic precursors. Numerous methods exist for preparing cycloalkanes by ring-closing reactions of difunctional precursors. For example, diesters are cyclized in the Dieckmann condensation: The acyloin condensation can be deployed similarly. For larger rings (macrocyclizations) more elaborate methods are required since intramolecular ring closure competes with intermolecular reactions. The Diels-Alder reaction, a [4+2] cycloaddition, provides a route to cyclohexenes: The corresponding [2+2] cycloaddition reactions, which usually require photochemical activation, result in cyclobutanes. See also [edit] Prelog strain Conformational isomerism Alkane Cycloalkene Cycloalkyne Notes [edit] ^ Liquid-phase density measured for a sample melting at 18 °C. The same sample was measured to have a solid density of 0.864 g·cm−3 at 16 °C. ^ Dale et al. (1963) instead gives a melting point of 61–62 °C. References [edit] ^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2014) "Cycloalkane". doi:10.1351/goldbook.C01497 ^ Jump up to: a b "Alkanes & Cycloalkanes". www2.chemistry.msu.edu. Retrieved 2022-02-20. ^ "Blue Book". iupac.qmul.ac.uk. Retrieved 2023-04-01. ^ Fahim, MA, et al. (2010). Fundamentals of Petroleum Refining. p. 14. doi:10.1016/C2009-0-16348-1. ISBN 978-0-444-52785-1. ^ Boese, Roland; Weiss, Hans-Christoph; Bläser, Dieter (1999-04-01). "The Melting Point Alternation in the Short-Chain n-Alkanes: Single-Crystal X-Ray Analyses of Propane at 30 K and of n-Butane to n-Nonane at 90 K". Angewandte Chemie International Edition. 38 (7): 988–992. doi:10.1002/(SICI)1521-3773(19990401)38:7<988::AID-ANIE988>3.0.CO;2-0. ISSN 1433-7851. ^ Brown, RJC; Brown, RFC (June 2000). "Melting Point and Molecular Symmetry". Journal of Chemical Education. 77 (6): 724. doi:10.1021/ed077p724. ^ Jump up to: a b Dale, Johannes (1963). "15. Macrocyclic compounds. Part III. Conformations of cycloalkanes and other flexible macrocycles". Journal of the Chemical Society (Resumed): 93–111. doi:10.1039/JR9630000093. ^ Jump up to: a b c d e f g h i j k l m n Wunderlich, Bernhard; Möller, Martin; Grebowicz, Janusz; Baur, Herbert (1988). "Condis crystals of cyclic alkanes, silanes and related compounds". Conformational Motion and Disorder in Low and High Molecular Mass Crystals. Berlin, Heidelberg: Springer-Verlag Springer e-books. pp. 26–44. doi:10.1007/BFb0008610. ISBN 978-3-540-38867-8. ^ Jump up to: a b c d Dragojlovic, Veljko (2015). "Conformational analysis of cycloalkanes" (PDF). Chemtexts. 1 (3): 14. Bibcode:2015ChTxt...1...14D. doi:10.1007/s40828-015-0014-0. S2CID 94348487. ^ "ECHA CHEM". chem.echa.europa.eu. ^ "ECHA CHEM". chem.echa.europa.eu. ^ "ECHA CHEM". chem.echa.europa.eu. ^ Jump up to: a b Kaarsemaker, Sj.; Coops, J. (January 1952). "Thermal quantities of some cycloparaffins. Part III. results of measurements". Recueil des Travaux Chimiques des Pays-Bas. 71 (3): 261–276. doi:10.1002/recl.19520710307. ^ Ruzicka, L; Plattner, PA; Wild, H (January 1946). "209. Zur Kenntnis des Kohlenstoffringes. (40. Mitteilung). Über die Schmelzpunkte in der Reihe der Polymethylen-Kohlenwasserstoffe von Cyclo-propan bis Cyclo-octadecan" [209. On carbon rings. (Part 40). On the melting points in the series of polymethylene hydrocarbons from cyclopropane to cyclooctadecane]. Helvetica Chimica Acta (in German). 29 (6): 1611–1615. doi:10.1002/hlca.19460290631. ^ Jump up to: a b c Egloff, Gustav (1940). Physical constants of hydrocarbons. Vol. 2 : Cyclanes, cyclenes, cyclynes, and other alicyclic hydrocarbons. Reinhold Publishing Corporation. ^ "ECHA CHEM". chem.echa.europa.eu. ^ "ECHA CHEM". chem.echa.europa.eu. ^ "ECHA CHEM". chem.echa.europa.eu. The density of cyclododecane was measured at 8 temperatures between 66 and 134 °C with a dilatometer. Extrapolation to 20 °C leads to 0.855 g·cm−3. ^ Jump up to: a b c d Dale, Johannes; Hubert, A. J.; King, G. S. D. (1963). "13. Macrocyclic compounds. Part I. Synthesis of macrocyclic polyynes: conformational effects in ring formation and in physical properties". Journal of the Chemical Society (Resumed): 77. doi:10.1039/JR9630000073. ^ McMurry, John (2000). Organic chemistry (5th ed.). Pacific Grove, CA: Brooks/Cole. p. 126. ISBN 0534373674. ^ Wiberg, K. B. (1968). "Small Ring Bicyclo[n.m.0]alkanes". In Hart, H.; Karabatsos, G. J. (eds.). Advances in Alicyclic Chemistry. Vol. 2. Academic Press. pp. 185–254. ISBN 9781483224213. ^ Wiberg, K. B.; Lampman, G. M.; Ciula, R. P.; Connor, D. S.; Schertler, P.; Lavanish, J. (1965). "Bicyclo[1.1.0]butane". Tetrahedron. 21 (10): 2749–2769. doi:10.1016/S0040-4020(01)98361-9. ^ Irion, Walther W.; Neuwirth, Otto S. (2000). "Oil Refining". Ullmann's Encyclopedia of Industrial Chemistry. doi:10.1002/14356007.a18_051. ISBN 3-527-30673-0. ^ Michael Tuttle Musser "Cyclohexanol and Cyclohexanone" in Ullmann's Encyclopedia of Industrial Chemistry, Wiley-VCH, Weinheim, 2005. IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (1995) "Cycloalkanes". doi:10.1351/goldbook.C01497 Organic Chemistry IUPAC Nomenclature. Rule A-23. Hydrogenated Compounds from Fused Polycyclic Hydrocarbons Organic Chemistry IUPAC Nomenclature.Rule A-31. Bridged Hydrocarbons: Bicyclic Systems. Organic Chemistry IUPAC Nomenclature.Rules A-41, A-42: Spiro Hydrocarbons Organic Chemistry IUPAC Nomenclature.Rules A-51, A-52, A-53, A-54:Hydrocarbon Ring Assemblies External links [edit] "Cycloalkanes" at the online Encyclopædia Britannica | v t e Hydrocarbons | | --- | | Saturated aliphatic hydrocarbons | | | | | | | | --- --- --- | | Alkanes CnH2n + 2 | | | | --- | | Linear alkanes | Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane | | Branched alkanes | Isobutane Isopentane 3-Methylpentane Neopentane Isohexane Isoheptane Isooctane Isononane Isodecane | | | Cycloalkanes | Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cyclooctane Cyclononane Cyclodecane | | Alkylcycloalkanes | Methylcyclopropane Methylcyclobutane Methylcyclopentane Methylcyclohexane Isopropylcyclohexane | | Bicycloalkanes | Housane (bicyclo[2.1.0]pentane) Norbornane (bicyclo[2.2.1]heptane) Decalin (bicyclo[4.4.0]decane) | | Polycycloalkanes | Adamantane Diamondoid Perhydrophenanthrene Sterane Cubane Prismane Dodecahedrane Basketane Churchane Pagodane Twistane | | Other | Spiroalkanes | | | Unsaturated aliphatic hydrocarbons | | | | | | | | --- --- --- | | Alkenes CnH2n | | | | --- | | Linear alkenes | Ethene Propene Butene Pentene Hexene Heptene Octene Nonene Decene | | Branched alkenes | Isobutene Isopentene Isohexene Isoheptene Isooctene Isononene Isodecene | | | Alkynes CnH2n − 2 | | | | --- | | Linear alkynes | Ethyne Propyne Butyne Pentyne Hexyne Heptyne Octyne Nonyne Decyne | | Branched alkynes | Isopentyne Isohexyne Isoheptyne Isooctyne Isononyne Isodecyne | | | Cycloalkenes | Cyclopropene Cyclobutene Cyclopentene Cyclohexene Cycloheptene Cyclooctene Cyclononene Cyclodecene | | Alkylcycloalkenes | Methylcyclopropene Methylcyclobutene Methylcyclopentene Methylcyclohexene Isopropylcyclohexene | | Bicycloalkenes | Norbornene | | Cycloalkynes | Cyclopropyne Cyclobutyne Cyclopentyne Cyclohexyne Cycloheptyne Cyclooctyne Cyclononyne Cyclodecyne | | Dienes | Propadiene Butadiene Pentadiene Hexadiene Heptadiene Octadiene Nonadiene Decadiene | | Other | Alkatriene Alkadiyne Cumulene Cyclooctatetraene Cyclododecatriene Enyne | | | Aromatic hydrocarbons | | | | | | | | --- --- --- | | PAHs | | | | --- | | Acenes | Naphthalene Anthracene Tetracene Pentacene Hexacene Heptacene Octacene | | Other | Azulene Fluorene Helicenes Circulenes Butalene Phenanthrene Chrysene Pyrene Corannulene Kekulene | | | Alkylbenzenes | Toluene | | | | | | | --- --- --- | | C2-Benzenes | | | | --- | | Xylenes | o-Xylene m-Xylene p-Xylene | | Other | | | | C3-Benzenes | | | | --- | | Trimethylbenzenes | Mesitylene Pseudocumene Hemellitene | | Other | Cumene n-Propylbenzene 4-Ethyltoluene | | | C4-Benzenes | | | | --- | | Cymenes | o-Cymene m-Cymene p-Cymene | | Tetramethylbenzenes | Durene Prehnitene Isodurene | | Other | n-Butylbenzene sec-Butylbenzene tert-Butylbenzene Isobutylbenzene | | | Other | Hexamethylbenzene 2-Phenylhexane 1,3,5-Triethylbenzene 1,3,5-Triheptylbenzene | | | Vinylbenzenes | Styrene Divinylbenzene 4-Vinyltoluene | | Other | Benzene Cyclopropenylidene Phenylacetylene trans-Propenylbenzene | | | Other | Annulenes Annulynes Alicyclic compounds Petroleum jelly | | v t e | | --- | | Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cyclooctane Cyclononane Cyclodecane Cycloundecane Cyclododecane Cyclotridecane Cyclotetradecane Cyclopentadecane Cyclohexadecane | | | Authority control databases | | --- | | National | Germany United States France BnF data Israel | | Other | Yale LUX | Retrieved from " Category: Cycloalkanes Hidden categories: CS1 German-language sources (de) Articles with short description Short description matches Wikidata
187827
https://www.youtube.com/watch?v=kECiI8D6j7k
3D to 2D Projection Formula: PROOF Math Easy Solutions 56800 subscribers 36 likes Description 3392 views Posted: 1 Nov 2023 In this video I derive the formula for projecting 3D coordinates onto a 2D screen using similar triangles. First I plot out the coordinates in 3D and then draw a straight line from the point where we are projecting the coordinates and onto the x = 0 plane. The point of projection can be thought of as the point at which our camera or eyes are located at, thus the resulting 2D projection maintains 3D perspective from that specific point. Drawing a second line passing through the z coordinates of the 3D line, we can see two similar triangles. Either of the similar triangles can be used to obtain identical formulas for the y and z projection coordinates. Note that in this projection derivation, I project the x coordinates to x = 0 and the camera or eyes location is at x = 1000. Epic stuff! The timestamps of key parts of the video are listed below: Projecting 3D coordinates to 2D coordinates: 0:00 Two Similar Triangles: 2:45 Determining the formula for the y and z projection: 4:41 3D to 2D projection formula: 8:53 This video was taken from my earlier video listed below: Laboratory Project: Putting 3D in Perspective: HIVE video notes: Video sections playlist: Related Videos: Sequences and Series playlist: . Become a MES Super Fan! DONATE! ʕ •ᴥ•ʔ SUBSCRIBE via EMAIL: MES Links: MES Truth: Official Website: Hive: Email me: contact@mes.fm Free Calculators: BMI Calculator: Grade Calculator: Mortgage Calculator: Percentage Calculator: Free Online Tools: iPhone and Android Apps: 4 comments Transcript: Projecting 3D coordinates to 2D coordinates yeah to determine what happens or how to project it first let's consider how the X Y and Zed coordinates get projected from 3D onto the 2D screen all right so let's say we have our camera here so here's our camera and this is going to be let's say it's at a distance X or let's move this uh down further all right so here's how they get projected so we have let's say this is this this is the distance X Al so this distance X here and there's the origin right here called the O and now what we're going to have is this is the screen so the screen looks something like this and then it goes like this and like that and then let's say we are projecting this point onto the screen so we're going to have to draw a perfect straight line from it to the screen so it has to connect it all the way to the screen all right so let's say uh instead of this so let's say we have this distance here to here is a th000 so this is uh the Thousand that we're looking at erase this little thingy all right so from here to here is a th000 but then from this point here let's say this coordinates of this point are uh it goes from um let's draw this all out like this uh this this coordinates has I'll put put a dot dot dot this way I'll go in the wide side up then the the then downwards like this or let's erase this so let's say from here to here is X then that's the x coordinate of it and let's just say this point is x y z then it gets to this point where this point is let's say it's projected as X prime or just x with a dash yeah X Prime y Prime and this is the projected coordinates of that point like that so we just projected through a straight line all right so that's the X this is the Zed and this is the Y so that's the Y is this setup like that all right and now the next setup is uh let's say that's the um projected line let now I mean this is the this is the point let's see the projected point now is going to be and also finish this off let's finish the screen off like that all right like this all right so finish that off and now this part right here uh this is going to be it's going to have a new Y and a new Zed this is z Prime and this is going to be y Prime like that actually I put the Y on top there all right so that is the projection so we Two Similar Triangles have we can have two similar triangles and we can see that here I'm going to draw this thin line across like that so now we have two similar triangles we have the Zed so the Zed projects up to the Z Prime and we have the y projects up to the Y Prime uh so I'll write this one as one this one as two so let's write these out so we have the first triangle so for the Z how it gets projected uh this goes from uh just like this all right right here let fix that up let's move this over so this the one and this The Zed uh parts so we have the point here this is at Zed and then this this distance is X and then this uh this height is just Z Prime and then this full length is a th000 all right and then similarly for the the Y we have the top triangle like that and this this goes from uh from here from y to Y Prime is the exact same thing but we're doing a with a Y so in fact I'll just copy this exact same thing here copy and then paste then move this here so we have the exact same thing then remove this with a y y and Y like this yes so Y and Y Prime let's go y Prime all right so Y and Y Prime and then lastly those are that's how the z and y get projected the X gets projected wherever the x is just goes to zero so we just have uh X goes to is to Zer all right three x goes to zero so X goes to zero like that so we have the the all the X's goes to zero Determining the formula for the y and z projection all right so now we can determine a formula for the projection of both the Y and Zed coordinates using similar triangles while the x coordinate just becomes equal to zero so X goes to zero and these ones goes to here so we could use similar triangles and let's solve that out let's just do it for the Zed and then will be the same for the uh uh y like this so say we have this set this up better so like this all right and this is going to be our uh Point here I'm going to write this dash dash like that or instead of dash line put a straight line like this whoops all right here fix that up anyway so this is going to be a zed and now what we'll do is we need a similar triangle set up and this is going to be our angle Theta this angle Theta and this is going to be a right angle this going be right angle um let's remove that so it's just common sense right angle and this is going to be this height here is our Zed Prime so then this height here is going to be Zed Prime minus Z and likewise this is going to be X and now this this distance here is going to be 1,000 - x that's because this whole thing is a th000 all right all right so thus we could apply similar triangles so similar triangles like this and for completeness we could even write just write this out as tan theta equals to well you could have similar triangle Z over 1 - x this also equals to well this angle you could use tan of this one so Zed Prime minus Z over uh this is going to be X like that all right so let's do some simplification um let's move this over to this side and then this over to the other side so we get a z x z x equals to now we have a zed uh like this or actually instead of that uh instead of this we'll move the we'll move the X over to the side and then this part and then this separately would keep the 1,00 overx so that's what we'll do that that's better setup there so we'll have a z x move this up top z x and then this is going to be divided by 1, - x then plus Z equal to Z Prime and I'll just move that over uh move the Z Prime further so that we can multiply the top and bottom by 1X like that and then simp move this even further equal Zed Prime all right so we do that so we don't change anything just we have the same common denominator and this is going to be now uh multiply that inside and then add those all up so Z X plus uh Z 1 1000us z x over 1,000 - x = 2 to Z Prime all right so what we have now is yeah what we have now is z x plus Z th000 well these ones cancel z x z x all right so we're just going to be left with zed time a th000 or th000 uh yeah let put it like that over 1,00 - x = to Z prime or what we can do is well I'll move this over yeah I'll move this over to the other side just put Z Prime equals and then just simplifications okay move this over to this side so we get a z Prime is equal to is equal to Z 100000 / 1,000 - x this can be times it by 1 over a th000 simplified and then div by 1 over a th000 so we get rid of the thousands so 3D to 2D projection formula then what we end up having is final equation Z Prime equal 2 well 1,000 can cancel we have a z on top over th000 th000 is 1us then X over a th000 so we just have 1,000 there as opposed to 2 THS there yes epic epic stuff there so there is our our projection and similarly I'll write down similarly for the Y exact same uh triangles so y Prime is equal to Y over 1us it's it's still going to be the same X so 1,000 x over 1,000 like that epic stuff
187828
https://beyondtheworksheet.com/effective-strategies-for-teaching-decimal-operations/
Skip to content Effective Strategies for Teaching Decimal Operations in Middle School Math Effective Strategies for Teaching Decimal Operations in Middle School Math Hey teachers! I’m excited to share some effective and engaging strategies for teaching decimal operations to your middle school math students. Decimal concepts can be tricky, but with the right approach, you can make them accessible and fun for your students. The Decimal Point Anchor Method One common challenge students face is aligning decimals correctly during operations like subtraction. They might align the numbers by their end digits, which can lead to errors. To combat this, I recommend using the “Decimal Point Anchor” method. In this approach, encourage students to view the decimal point as an anchor, stabilizing the numbers in the right place. Before students begin subtracting, have them draw vertical lines to connect the decimal points across the numbers. This visual aid ensures that every digit is correctly placed in its corresponding place value column. Practice this technique in class by solving examples together and correcting intentionally misaligned problems. This method not only helps students understand decimal alignment but also builds their confidence in handling decimal operations. Culinary Math Chef Challenge To make multiplying decimals more engaging, introduce the “Culinary Math Chef” challenge. Turn your classroom into a cooking show where recipes typically serve four people. Challenge your students to adjust the recipes to serve different numbers, like 10 or 5, using decimal multiplication. This real-world application of decimals makes the lesson memorable and shows the practical use of math in everyday life. After completing the challenge, have students explain their process as if they were on a cooking show. This reinforces their learning and helps them articulate their understanding of decimals. Decimal Detectives Activity Another hands-on activity is the “Decimal Detectives” activity. Place various items around the classroom, each tagged with a price that includes decimals. Assign each student a “case file” with a specific amount of money to spend. Their task is to choose items that collectively stay within their budget without going over. This activity teaches students to add and subtract decimals accurately and encourages them to apply these skills strategically. Following the activity, hold a “detective debrief” where students can explain their choices and discuss how they used decimal operations to stay within their budget. This debrief is a great opportunity for peer learning and helps reinforce mathematical concepts in a fun, interactive way. Avoiding Decimal Drudgery It’s important to remember that engagement is key in learning. Relying solely on worksheets full of decimal problems can quickly disengage students. Instead, incorporate decimals into activities and engaging, authentic scenarios that show their application in real life. By doing so, we not only help students understand the mechanics of decimals but also appreciate their relevance and importance. I’m all about making math both understandable and enjoyable. By integrating these strategies into your lessons, you can help your students master decimal operations and see the real-world applications of their math skills. Follow along on Instagram and/or Facebook for daily math content! Could you use FREE Resources? Provide your email address and instantly receive access to a free resource library and stay up to date with tips, products and insider information! Thanks! Now, head to your email to get access to the free resources! Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Click to email a link to a friend (Opens in new window) Email Click to share on Pinterest (Opens in new window) Pinterest Click to share on LinkedIn (Opens in new window) LinkedIn Related PrevPreviousBringing Ratios, Rates, and Proportions to Life in the Middle School Classroom NextGaps in Learning Identified in High School GeometryNext Hi, I'm Lindsay! I create ready to go resources for middle school math teachers, so they can get back what matters most – their time! Search By Topic Shop TPT INTERVENTION FULL CURRICULUM ACTIVITIES STATIONS Exclusive Freebie If you’d like a free spin and solve game, click the image below. BLOG EXCLUSIVE! Facebook-f Pinterest Instagram Linkedin Youtube Envelope COPYRIGHT © 2025 · Beyond the Worksheet Inc. Loading Comments...
187829
https://www.amazon.com/Books-Vasile-Cirtoaje/s?rh=n%3A283155%2Cp_27%3AVasile%2BCirtoaje
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187830
https://math.ipm.ac.ir/combin/IPMCCC2019/slides/Karasev.pdf
Fair partition of a convex planar pie Roman Karasev 1 joint work with Arseniy Akopyan 2 and Sergey Avvakumov 2 1Moscow Institute of Physics and Technology 2IST Austria Tehran, April, 2019 The problem statement Question (Nandakumar and Ramana Rao, 2008) Given a positive integer m and a convex body K in the plane, can we cut K into m convex pieces of equal areas and perimeters? Previously known results For m = 2 it can be done by a simple continuity argument. Split in two parts of equal area and rotate. At some point the perimeters must be equal. Previously known results For m = 2 it can be done by a simple continuity argument. Split in two parts of equal area and rotate. At some point the perimeters must be equal. A generalization of the continuity argument through an appropriate Borsuk–Ulam-type theorem yields a proof for m = pk with p prime. The topological tool was used previously by Viktor Vassiliev for a different problem (1989). The fair partition result for m = 2k was proved explicitly by Mikhail Gromov (2003). Previously known results For m = 2 it can be done by a simple continuity argument. Split in two parts of equal area and rotate. At some point the perimeters must be equal. A generalization of the continuity argument through an appropriate Borsuk–Ulam-type theorem yields a proof for m = pk with p prime. The topological tool was used previously by Viktor Vassiliev for a different problem (1989). The fair partition result for m = 2k was proved explicitly by Mikhail Gromov (2003). For m, which is not a prime power, this direct technique fails. A classical example: the ham sandwich theorem Theorem Any 3 sufficiently nice probability measures in R3 can be simultaneously equipartitioned by a plane. Scheme of proof The configuration space is the sphere S3, which naturally parametrizes (oriented) planes in R3. Scheme of proof The configuration space is the sphere S3, which naturally parametrizes (oriented) planes in R3. The test map f : S3 →R3 sends an oriented plane u ∈S3 to the point f (u) ∈R3 whose i-th coordinate is the difference of the values of the i-th measure on the two corresponding halfspaces. Scheme of proof The configuration space is the sphere S3, which naturally parametrizes (oriented) planes in R3. The test map f : S3 →R3 sends an oriented plane u ∈S3 to the point f (u) ∈R3 whose i-th coordinate is the difference of the values of the i-th measure on the two corresponding halfspaces. Solutions are in f −1(0). Scheme of proof The configuration space is the sphere S3, which naturally parametrizes (oriented) planes in R3. The test map f : S3 →R3 sends an oriented plane u ∈S3 to the point f (u) ∈R3 whose i-th coordinate is the difference of the values of the i-th measure on the two corresponding halfspaces. Solutions are in f −1(0). This map is Z2-equivariant, i.e., f (−u) = −f (u), and the classical Borsuk–Ulam theorem guarantees that any such map must have a zero, which yields the desired equipartition. Convex fair partitions for prime power Theorem (Karasev, Hubard, Aronov, Blagojevi´ c, Ziegler, 2014) If m is a power of a prime then any convex body K in the plane can be partitioned into m parts of equal area and perimeter. The case m = 3 was done first by B´ ar´ any, Blagojevi´ c, and Sz˝ ucs. In dimension n ≥3 a similar result with equal volumes and equal n −1 other continuous functions of m convex parts was also established for m = pk. Configuration space F(m) is the space of m-tuples of pairwise distinct points in R2. Given ¯ x ∈F(m) we can use Kantorovich theorem on optimal transportation to equipartition K into m parts of equal area. The partition is a weighted Voronoi diagram with centers in ¯ x. ¯ x ∈F(3). No need to consider partitions not in F(m) The space F(m) is smaller than the space E(m) of all equal are convex partitions. However, there is an Sm-equivariant map F(m) →E(m), given by the Kantorovich theorem, and an Sm-equivariant map E(m) →F(m), sending a partition into its centers of mass. The maps do not commute, but show that the spaces are equivalent from the points of view of plugging them into a Borsuk–Ulam-type theorems. Further simplification of F(m) The dimension of F(m) is 2m. We can further simplify it. Lemma (Blagojevi´ c and Ziegler, 2014) Space F(m) retracts Sm-equivariantly to its subpolyhedron P(m) ⊂F(m) with dim(P(m)) = m −1. This lemma allows to imagine how the solution changes if we consider a family of problems depending on a parameter. Cellular decomposition of F(3) A 6-cell. A 5-cell. A 4-cell. Equivariant map Let the map f : P(m) →Rm send a generalized Voronoi equal area partition into the perimeters of the m parts. The test map is Sm-equivariant, if Sm acts on Rm by permutations of the coordinates. A partition corresponding to u ∈P(m) solves the problem if f (u) ∈∆:= {(x, x, . . . , x) ∈Rm}. Homology of the solution set Theorem (Blagojevi´ c and Ziegler) If m = pk is a prime power and f : P(m) →Rm is an Sm-equivariant map in general position, then f −1(∆) is a non-trivial 0-dimensional cycle modulo p in homology with certain twisted coefficients. If m is not a prime power then there exists an Sm-equivariant map f : P(m) →Rm with f −1(∆) = ∅. Our main result Theorem (Akopyan, Avvakumov, K.) For any m ≥2 any convex body K in the plane can be partitioned into m parts of equal area and perimeter. Our main result Theorem (Akopyan, Avvakumov, K.) For any m ≥2 any convex body K in the plane can be partitioned into m parts of equal area and perimeter. When m is not a prime power, the theorem was unknown even for simplest K, e.g., for generic triangles. “Naive” continuity argument “Naive” argument for m = 6 (the smallest non-prime-power): “Naive” continuity argument “Naive” argument for m = 6 (the smallest non-prime-power): Pick a direction and a halving line in that direction. “Naive” continuity argument “Naive” argument for m = 6 (the smallest non-prime-power): Pick a direction and a halving line in that direction. Fair partition each half into 3 pieces. “Naive” continuity argument “Naive” argument for m = 6 (the smallest non-prime-power): Pick a direction and a halving line in that direction. Fair partition each half into 3 pieces. Rotate the direction. “Naive” continuity argument “Naive” argument for m = 6 (the smallest non-prime-power): Pick a direction and a halving line in that direction. Fair partition each half into 3 pieces. Rotate the direction. There are difficulties arguing this way, because the partitions in three parts may not depend continuously on parameters of the half subproblem. Proof sketch for m = 6 As we rotate the direction, plot the perimeters of all the solutions, the language of multivalued functions must be useful. Solid and dashed are perimeters on the left and right, resp. Solid/dashed intersections are fair partitions. Number of solutions In this particular example the number of solutions, with signs, is 0! Proof sketch for m = 6 Solid graph separates the bottom from the top, from the homology modulo 3 description of the solution set by Blagojevi´ c and Ziegler. Proof sketch for m = 6 After choosing an appropriate subgraph of the multivalued function, bold solid and bold dashed curves intersect at 1 point, modulo 2. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Parameterize the partitions on each stage by P(pi) and assume the areas equalized by the weighted Voronoi argument. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Parameterize the partitions on each stage by P(pi) and assume the areas equalized by the weighted Voronoi argument. Argue from bottom to top: Assume that the perimeters are equalized in all parts of every i-th stage region and thus form a multivalued function of the corresponding region. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Parameterize the partitions on each stage by P(pi) and assume the areas equalized by the weighted Voronoi argument. Argue from bottom to top: Assume that the perimeters are equalized in all parts of every i-th stage region and thus form a multivalued function of the corresponding region. Establish the top-from-bottom separation property for this multivalued function, using the modulo pi homology argument. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Parameterize the partitions on each stage by P(pi) and assume the areas equalized by the weighted Voronoi argument. Argue from bottom to top: Assume that the perimeters are equalized in all parts of every i-th stage region and thus form a multivalued function of the corresponding region. Establish the top-from-bottom separation property for this multivalued function, using the modulo pi homology argument. Choose a “nice” multivalued subfunction, for which a Borsuk–Ulam-type theorem holds. Plan of the proof for arbitrary m Decompose into primes m = p1 . . . pk, then consider iterated partitions, first cut into p1 parts, then cut every part into p2 parts, and so on. Parameterize the partitions on each stage by P(pi) and assume the areas equalized by the weighted Voronoi argument. Argue from bottom to top: Assume that the perimeters are equalized in all parts of every i-th stage region and thus form a multivalued function of the corresponding region. Establish the top-from-bottom separation property for this multivalued function, using the modulo pi homology argument. Choose a “nice” multivalued subfunction, for which a Borsuk–Ulam-type theorem holds. Step i →i −1 equializing the perimeter values in parts of (i −1)th stage region, keeping the separation property for the new multivalued function, the common value of the perimeter. Summary Generalizations: “Area” can be any finite Borel measure, zero on hyperplanes. “Perimeter” can be any Hausdorff-continuous function on convex bodies (e.g., diameter). Unknown, if we replace “area” with an arbitrary (i.e., non-additive) rigid-motion-invariant continuous function of convex bodies. If we want to equalize the volumes and two perimiter-like functions in R3, then it is possible for m = pk (K., Aronov, Hubard, Blagojevi´ c, Ziegler), but our current method does not work already for m = 2pk. Full version is arXiv:1804.03057. Summary Thank you for your attention! Full version is arXiv:1804.03057.
187831
https://mathmaine.com/2010/05/27/function-translations/
Skip to content Function Translations: How to recognize and analyze them For the approach I now prefer to this topic, using transformation equations, please follow this link: Function Transformations: Translation A function has been “translated” when it has been moved in a way that does not change its shape or rotate it in any way. A function can be translated either vertically, horizontally, or both. Other possible “transformations” of a function include dilation, reflection, and rotation. Imagine a graph drawn on tracing paper or a transparency, then placed over a separate set of axes. If you move the graph left or right in the direction of the horizontal axis, without rotating it, you are “translating” the graph horizontally. Move the graph straight up or down in the direction of the vertical axis, and you are translating the graph vertically. In the text that follows, we will explore how we know that the graph of a function like which is the blue curve on the graph above, can be described as a translation of the graph of the green curve above: Describing as a translation of a simpler-looking (and more familiar) function like makes it easier to understand and predict its behavior, and can make it easier to describe the behavior of complex-looking functions. Before you dive into the explanations below, you may wish to play around a bit with the green sliders for “h” and “k” in this Geogebra Applet to get a feel for what horizontal and vertical translations look like as they take place (the “a” slider dilates the function, as discussed in my Function Dilations post). Vertical Translation Consider the equation that describes the line that passes through the origin and has a slope of two: What happens to the graph of this line if every value of has three added to it? The function is defined as the result of with three added to each result. If we then substitute the definition of from above for , we get: Since produces the y-coordinate corresponding to x for every point on the original graph, adding 3 to each value moves every point on the graph up by 3. Adding “+3” to the definition of causes the entire function to be “translated vertically” by a positive three. This process works for any function, and is usually thought through in the reverse order: when looking at a more complex function, do you see a constant added or subtracted? If so, you can think of it as a vertical translation of the rest of the function: Another example: Horizontal Translation Consider the same function described at the beginning of the Vertical Translation section, which describes a line that passes through the origin with a slope of two: What happens to the graph of this equation if every “x” in the equation is replaced by a value that is 4 less? We can describe this algebraically by evaluating instead of , and let’s call this new function : Now let’s compare the behaviors of and : produces the same results as , but only when its input values are four greater than the input to . Comparing the graphs of the two functions, the graph of will have the same shape as , but that shape has been shifted four units to the right along the x-axis. A helpful way to think about the above (thanks to Michael Paul Goldenberg’s 2016 comment below) is to think of the independent variable “x” as measuring time in seconds. Therefore, “x-4” is 4 seconds earlier then “x”, and evaluating produces a result from 4 seconds earlier than time “x”. When we graph , all of the results will appear to be 4 seconds later (to the right) than those on the graph of . The fact that substituting “x-4” for “x” produces a horizontal translation of +4 (not -4) is a source of errors when people get horizontal and vertical translation behaviors confused. One way to address this is to use a procedural approach whenever you see a variable with a constant added or subtracted (often together in a set of parentheses). To find the direction of the translation, set the transformation expression equal to zero and solve: The result will always give your the magnitude and direction of the translation (see Keep Your Eye On The Variable). This process works for any function: so set and solve for x. The graph of is the same as that of translated horizontally by -3. Or for the graph of is the same as that of translated horizontally by -5. Note that requires everyinstance of “x” in to have (x+5) substituted for it. So a function like will only be a horizontal translation of if every instance of “x” has the same constant added or subtracted. The notation expresses this idea compactly and elegantly. One last example: so the graph of is the same as that of translated horizontally by . Reconciling Horizontal And Vertical Translations Let’s re-examine why translates a function in a positive vertical direction, yet translates the function in a negative horizontal direction. This apparent difference in the way we analyze horizontal and vertical translations can be reconciled by treating both independent and dependent variables in the same manner. If and we subtract 7 from both sides, it becomes: Since every instance of occurs as a , and every instance of “x” occurs as , you may treat both and “x” as having been translated relative to a parent function, and you may analyze them both in exactly the same manner: – what value of makes ? Positive 7. So the translation in the direction, along the vertical axis, is positive 7. – what value of “x” makes ? Negative 5. So the translation in the “x” direction, along the horizontal axis, is negative 5. Therefore, if we define as shown below, a can be created which is translated horizontally by -5 and vertically by +7 when compared to : Equivalent Translations In mathematics, it is often (but not always) possible to produce the same end result in different ways. When working with linear equations and using the approach described in the last section above, you may have wondered how to handle a situation such as: The above describes a horizontal translation by +4, but if we subtract 4 from both sides the equation becomes: which describes a vertical translation by -4. Are they both valid interpretations? Since both of the above are valid algebraic manipulations of the same equation, they must both have the same graph. Imagine the graph of , which will be a line with a slope of one that passes through the origin. Now translate the graph vertically by +4. This translation will also cause the x-intercept to move… four to its left. Equivalent translations do not always translate by the same distance. If the slope of the line is not 1, we need to translate by different amounts: The first representation of above is a horizontal translation of by +2, while the last one is a vertical translation by -4. Yet, they both describe the same graph. We could be even trickier if we wished to: So, we can choose to describe g(x) as either: – f(x) translated horizontally by +2 (1st line) – f(x) translated vertically by -4 (2nd line) – f(x) translated vertically by -2 and horizontally by +1 (4th line) Just as there are often multiple ways of describing something using English, a particular situation can often be described in more than one mathematically too. Share this: Click to print (Opens in new window) Print Click to email a link to a friend (Opens in new window) Email Tweet Save Click to share on Reddit (Opens in new window) Reddit Like Loading... Related Function Dilations: How to recognize and analyze themIn "Concepts" Function Transformations: TranslationIn "Concepts" Using Corresponding Points to Determine Dilation Factors and Translation AmountsIn "Concepts" By Whit Ford Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc. View all of Whit Ford's posts. 6 comments Interesting timing. I was reading about transformation of axes and matrix multiplication in Lillian R. Lieber’s book on Lattice Theory last night. Reply cool Reply 2. I am learning about this right now in Geometry Class. 🙂 Reply 3. Since reading this post and commenting on it, I’ve had an epiphany of sorts regarding horizontal transformations and why they seem to behave counterintuitively. Take as an example function f(x) = |x|. Compare the position of the vertex of the parent graph with that of g(x) = |x – 4|. As noted with other functions, the second graph is translated four units to the right. One way to think about this is to consider the independent variable, x, to be time. In that case, subtracting 4 from x DELAYS the outcome by 4 units of time. That means that things happen LATER than they would in the case of f(x), a shift of 4 units to the right. On the other hand, h(x) = |x + 4| would entail adding 4 units to x, ACCELERATING the outcome by 4 units of time (4 units EARLIER), and hence shifting the graph 4 units to the left. In the last couple of years, the vast majority of my students have found that a helpful metaphor that makes sense of the horizontal shifts of the graphs of various functions. Reply I have taken the occasion to rewrite (and hopefully further clarify) much of this post, and have included your approach with attribution as well. Thank you! Reply Glad you liked that idea. Leave a comment Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed. Comment Reblog Subscribe Subscribed MathMaine Already have a WordPress.com account? Log in now. MathMaine Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar
187832
https://www.quora.com/In-frogs-fertilisation-is-external-then-what-is-the-significance-of-copulation
In frogs, fertilisation is external then, what is the significance of copulation? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Biology Amphibians as Pets External Fertilization Copulation (zoology) Bell Frogs Animal Reproduction Mating Examples of Amphibians Frog Tadpoles 5 In frogs, fertilisation is external then, what is the significance of copulation? All related (32) Sort Recommended Suresh Katare BSC agri from Vasantrao Naik Marthwada Krishi Vidyapeeth Parbhani ·7y This process increases the metabolic activity and the rate of protein synthesis of the egg Fertilization provide new Genetic constitution to the zygote Upvote · 9 1 Related questions More answers below What is the reason behind external fertilization in frogs and fish? Why can’t sperm pass inside their ovaries? Why Do fish, frogs show external fertilisation? Why is the fertilization in frog called external fertilization? Which animal vagina is most similar to that of a human? What would happen if a female was filled with animal semen? David Kirshner Zoologist, wildlife interpreter, wildlife illustrator and photographer · Author has 903 answers and 3.3M answer views ·8y Related How do land frogs reproduce? One of the interesting things about frogs is that they have the most variation in reproductive modes among all of the back-boned, land-dwelling animals (i.e. tetrapod vertebrates). Most terrestrial (land-dwelling) frogs return to water to mate and lay eggs. However, many have evolved all sorts of weird and wonderful ways to get around this. Some frogs lay eggs on land which skip the tadpole stage and develop straight into tiny frogs. A number of different frogs have developed skin folds which act much like a marsupial’s pouch and the eggs (or tadpoles, depending on which type of frog) develop insi Continue Reading One of the interesting things about frogs is that they have the most variation in reproductive modes among all of the back-boned, land-dwelling animals (i.e. tetrapod vertebrates). Most terrestrial (land-dwelling) frogs return to water to mate and lay eggs. However, many have evolved all sorts of weird and wonderful ways to get around this. Some frogs lay eggs on land which skip the tadpole stage and develop straight into tiny frogs. A number of different frogs have developed skin folds which act much like a marsupial’s pouch and the eggs (or tadpoles, depending on which type of frog) develop inside those. Others carry the developing tadpoles in their vocal pouch (the skin on the throat which fills with air when they call). Two species of (now extinct) frogs in Australia, called gastric brooding frogs, used to swallow their eggs and have the young develop inside the stomach (which temporarily stopped producing acid). There are also many terrestrial (land) frogs whose eggs and/or tadpoles still require water, but the frogs still have an unusual way of getting them there. Many poison dart frogs, for example, carry their young on their back up into the canopy of a rainforest tree and deposit them in a tiny pools of water found in the centre of plants such as bromeliads. The female may even lay extra eggs as food for the developing tadpoles. On the other hand, many tree frogs lay their eggs on the underside of leaves overhanging a small pond so the hatching tadpoles drop into the water. These are just a small sampling of the huge variety of ways that land frogs deal with reproduction. Upvote · 9 8 9 1 9 1 PVishal Naidu Studied at Birla Institute of Technology and Science, Pilani · Author has 904 answers and 2.2M answer views ·7y Frogs don’t copulate. The male and female get into a mating posture called amplexus, in which the male climbs onto the female's back and clasps his forelegs around her middle. Then the female releases her eggs and the male releases his sperm at the same time. Upvote · 9 1 Abhisek Mallick Studied at National Institute of Technology, Raipur · Author has 242 answers and 552.1K answer views ·3y Related How do frogs reproduce? The vocal sacs are present only in male and during the breeding season a nuptial pad is developed at the base of first finger of the male frog. the vocal sacs increas e the pitch of sound while nuptial pads help in grasping the female during amplexus. germinal epithelium of seminiferous tubule produce sperms which are transferred to bidders canal via vasa efferentia from there sperms are carried to transverse collecting tubules, longitudinal tubules and then to urinogenital duct. the latter carries sperms to seminal vesicle where they are stored temporarily. then they are transferred to cloacc Continue Reading The vocal sacs are present only in male and during the breeding season a nuptial pad is developed at the base of first finger of the male frog. the vocal sacs increas e the pitch of sound while nuptial pads help in grasping the female during amplexus. germinal epithelium of seminiferous tubule produce sperms which are transferred to bidders canal via vasa efferentia from there sperms are carried to transverse collecting tubules, longitudinal tubules and then to urinogenital duct. the latter carries sperms to seminal vesicle where they are stored temporarily. then they are transferred to cloacca and then they are shed into female ova through amplexus Upvote · 9 4 Related questions More answers below Which other animals apart from human beings will do mating (have sexual intercourse) for pleasure and not for reproduction purposes? Is it possible for humans to have sex with frogs? Why is animal intercourse so rapid as opposed to humans? Do animals other than hominids experience any orgasm when they procreate? Do animals contract diseases through sex like humans? Bir Bahadur Professor Retd, working as Visiting Professor, at Kakatiya University (1970–present) · Author has 2K answers and 998.3K answer views ·4y Related Why Do fish, frogs show external fertilisation? According to Giese and Kanatani 1987 external fertilization is a common and widespread reproductive strategy in the aquatic environmental ecosystem, and is generally thought to be ancestral to internal modes of reproduction (Jägersten 1972; Parker 1984; Wray 1995; but see Rouse and Fitzhugh 1994). Therefore, estimates of male and female fertilization success in external fertilizers may provide not only information on sperm competition for the majority of animal phyla but also an insight into the evolution of sexual dimorphism and internal fertilization. Despite the need to understand the patter Continue Reading According to Giese and Kanatani 1987 external fertilization is a common and widespread reproductive strategy in the aquatic environmental ecosystem, and is generally thought to be ancestral to internal modes of reproduction (Jägersten 1972; Parker 1984; Wray 1995; but see Rouse and Fitzhugh 1994). Therefore, estimates of male and female fertilization success in external fertilizers may provide not only information on sperm competition for the majority of animal phyla but also an insight into the evolution of sexual dimorphism and internal fertilization. Despite the need to understand the patterns and consequences of variation in both male and female fertilization success, little is known about the fate of gametes released in aquatic environments. Historically, discussions about reproductive success in external fertilizers were based on speculation or laboratory studies (reviewed by Levitan 1995a). It has only been in the last decade that some of the practical obstacles associated with ‘chasing’ gametes in an aquatic medium have been overcome. Estimates of gametes concentration and fertilization have been made, but there is still no direct information on sperm competition and multiple paternity.. To be brief, External fertilization is when the sperm and eggs are released outside of the body and meet outside of the body. External fertilization is limited essentially to animals living in aquatic environments In contrast to most other organisms, the available evidence on external fertilizers suggests that sperm is limiting. Evidence from field experiments and natural observations of spawning demonstrate that the proportion of a female's eggs that are fertilized is often much less than 100%, and a majority of the variation in female fertilization success can be explained by male abundance, proximity, or synchrony. This somewhat different view of sexual selection has implications for the generality of Bateman's principle (Bateman 1948) and the evolution of sexual dimorphism in this presumptive ancestral reproductive strategy. Disadvantages of External Fertilization,A large quantity of gametes is wasted and left unfertilized. Chances of fertilization are diminished by environmental hazards and Predators. Eggs and sperms, essentially, may not come in contact. Among vertebrates, external fertilization is most common in amphibians and fish. Invertebrates utilizing external fertilization are mostly benthic, sessile, or both, including animals such as coral, sea anemones, and tube-dwelling polychaetes. Benthic marine plants also use external fertilization to reproduce External fertilization in an aquatic environment protects the desiccation of eggs. Broadcast spawning leads to higher genetic diversity due to a larger mixing of genes within a group. The chances of survival of the species also increase. For sessile aquatic organisms such as sponges, broadcast spawning is the only ... Internal fertilization has the advantage of protecting the fertilized egg from ... gonads to produce sperm and egg was a major step in the evolutionary process Upvote · Kartik Upadhyay Infrastructure Engineer at Sopra Steria Group (2016–present) ·11y Related How do frogs reproduce? In the male frog, the two testes are attached to the kidneys and semen passes into the kidneys through fine tubes called efferent ducts. It then travels on through the ureters, which are consequently known as urinogenital ducts. There is no penis, and sperm is ejected from the cloaca directly onto the eggs as the female lays them. The ovaries of the female frog are beside the kidneys and the eggs pass down a pair of oviducts and through the cloaca to the exterior. When frogs mate, the male climbs on the back of the female and wraps his fore limbs round her body, either behind the front legs or Continue Reading In the male frog, the two testes are attached to the kidneys and semen passes into the kidneys through fine tubes called efferent ducts. It then travels on through the ureters, which are consequently known as urinogenital ducts. There is no penis, and sperm is ejected from the cloaca directly onto the eggs as the female lays them. The ovaries of the female frog are beside the kidneys and the eggs pass down a pair of oviducts and through the cloaca to the exterior. When frogs mate, the male climbs on the back of the female and wraps his fore limbs round her body, either behind the front legs or just in front of the hind legs. This position is called amplexus and may be held for several days. The male frog has certain hormone-dependent secondary sexual characteristics. These include the development of special pads on his thumbs in the breeding season, to give him a firm hold.The grip of the male frog during amplexus stimulates the female to release eggs, usually wrapped in jelly, as spawn. In many species the male is smaller and slimmer than the female. Males have vocal cords and make a range of croaks, particularly in the breeding season, and in some species they also have vocal sacs to amplify the sound. Hope my answer satisfied the need. Upvote · 9 3 9 1 Lee Duer USA, Earth (1953–present) · Author has 13.8K answers and 5.1M answer views ·1y Related What is the reproduction of frogs? Q: What is the reproduction of frogs? Amphibians life cycle is tied to water. Their eggs must be wet and their larval forms (like tadpoles) live their part of the life cycle in the water. After they reach their adult stage they can stay in or near water (like many frogs and salamanders do) or they can live on land and return to water only to the lay their eggs (like most toads do). Frog Life Cycle Toad Life Cycle Salamander Life Cycle Continue Reading Q: What is the reproduction of frogs? Amphibians life cycle is tied to water. Their eggs must be wet and their larval forms (like tadpoles) live their part of the life cycle in the water. After they reach their adult stage they can stay in or near water (like many frogs and salamanders do) or they can live on land and return to water only to the lay their eggs (like most toads do). Frog Life Cycle Toad Life Cycle Salamander Life Cycle Upvote · Arthur Lawrence Bennett Retired teacher of English and European languages. · Author has 740 answers and 722.9K answer views ·6y Related Can two different species of frog procreate successfully together? The answer is probably not but possibly yes. There is a tendency for us not to notice the differences between strangers (the all Chinese look the same to me syndrome). Humans will tend to see all frogs including those of different species as more or less the same, just your basic Kermit jobbie. Amphibians are however an extremely ancient class of vertebrates and have had countless millions of years to develop differences in their species underneath their basic froggy exterior which would be barriers to interspecific hybridization. The Australian golden bell frog, the New Zealand Hamilton ‘s frog Continue Reading The answer is probably not but possibly yes. There is a tendency for us not to notice the differences between strangers (the all Chinese look the same to me syndrome). Humans will tend to see all frogs including those of different species as more or less the same, just your basic Kermit jobbie. Amphibians are however an extremely ancient class of vertebrates and have had countless millions of years to develop differences in their species underneath their basic froggy exterior which would be barriers to interspecific hybridization. The Australian golden bell frog, the New Zealand Hamilton ‘s frog, Canadian wood frog, European fire bellied toad, Brazilian arrow frog, European marsh frog - there is no way that any of the above could procreate together. You will probably discover that we have assigned each of the above to a different genus which is always a good guide to breeding incompatibility. If on the other hand you take frogs of the same genus but different species such as the various European frogs of the genus Rana, temporaria, arvalis, dalmatina, lataste, graeca, iberica, ridibunda, lessonae, esculenta, you might expect hybridization to be possible, yet this doesn't seem to be the case. This is especially remarkable because fertilization takes place outside the body. The male simply releases his sperm into the water where the female has just released her eggs. If several species were using the same pond you could well expect that any eggs could accidentally encounter sperm from another species. What this means is that however similar various species of brown or green European frogs might seem to us, the differences are already so anciently embedded that they are incompatible at a cellular or chemical level. This has to be so to maintain species integrity when fertilization takes place in an open pond. It would be interesting to investigate the different species of south American arrow frogs. The different species have violently different colour patterns. Although all the patterns advertise their toxicity, they are also very clear species markers. I do not know where fertilization takes place with arrow frogs. The female lays a single egg in a bromeliad leaf base. I do not know whether the male is at hand to deposit sperm there or whether the eggs have already been fertilized either on the female's body or even internally. I suspect that in the case of arrow frogs it is not a chemical barrier to hybridization but a cultural one based on the colour patterns to identify mates. The South American poison arrow frogs could be analogous to New Guinea birds of paradise. Upvote · 9 1 Marguerite Church Former Instructor in Biology Appalachian State University (1984–2008) · Author has 6.8K answers and 5M answer views ·6y Related Why is the fertilization in frog called external fertilization? With frog fertilization , the eggs are laid in a select spot in water . The male frog may or may not be close by or may even clasp the female as she lays the eggs. He releases sperm over the eggs and the adults leave. It is literally fertilization outside the frog’s body and the development of the eggs into tadpoles occur outside as well. My first experience of science was in the second grade when our teacher brought fertilized frog eggs into the class for us to watch them develop. She had us draw pictures of the developing tadpole and the resulting froglets. I just realized how formative for Continue Reading With frog fertilization , the eggs are laid in a select spot in water . The male frog may or may not be close by or may even clasp the female as she lays the eggs. He releases sperm over the eggs and the adults leave. It is literally fertilization outside the frog’s body and the development of the eggs into tadpoles occur outside as well. My first experience of science was in the second grade when our teacher brought fertilized frog eggs into the class for us to watch them develop. She had us draw pictures of the developing tadpole and the resulting froglets. I just realized how formative for me that experience was. Thank you Ms. Baxter!!! Upvote · Calvin Peck Evolutionary Biologist, Masters in Human Cognition and Learning · Author has 171 answers and 1.2M answer views ·9y Related Why is external fertilization used by aquatic animals? A better question would be why is internal fertilization used by land animals. As a living being we all started out in the sea. All reproduction must be conducted in an aquatic environment that can support the medium in which we live (again, water). Imagine trying to breath oxygen through water…can’t do it right? It’s called drowning because our bodies are only built to absorb oxygen in a specific medium…the air. The same thing is true in reverse for basically all other of our life’s processes; they require water. Digestion, circulation, excretion, hormonal response….all conducted in an aquatic Continue Reading A better question would be why is internal fertilization used by land animals. As a living being we all started out in the sea. All reproduction must be conducted in an aquatic environment that can support the medium in which we live (again, water). Imagine trying to breath oxygen through water…can’t do it right? It’s called drowning because our bodies are only built to absorb oxygen in a specific medium…the air. The same thing is true in reverse for basically all other of our life’s processes; they require water. Digestion, circulation, excretion, hormonal response….all conducted in an aquatic environment. All plants, bacteria, fungi, and animals live in the medium of water. The difference between you and a sea sponge is that you developed ways of simply carrying around all that water with you. That’s where 80%+ of your weight comes from. So the development characteristics that allowed creatures to use water for external fertilization precedes those that developed for us to live on land. The bottom line is: all reproduction requires a watery environment. Simply laying eggs then having them fertilized later is a pretty basic way of getting it done compared to what humans have to endure. Upvote · Mary Higgins Nutritionist and Yoga Teacher · Author has 4.4K answers and 2.3M answer views ·1y Related What is the reproductive process of frogs? Where are their eggs laid and where are the tadpole larvae found? The majority of frogs come about this way: males sit around the pond calling out for females. If a female frog is interested, she calls out in a much softer voice that she is interested. Usually there is a mad dash of males toward the female and the competition amongst the males can get quite rough. Some male frogs may be drowned by others as they rush to give her a hug around the waist. The male sits on her back in a position known as amplexus. At this point, the female who is full of eggs, gets squeezed by the male and the eggs exit from her body and the male frog squirts sperm over the eggs Continue Reading The majority of frogs come about this way: males sit around the pond calling out for females. If a female frog is interested, she calls out in a much softer voice that she is interested. Usually there is a mad dash of males toward the female and the competition amongst the males can get quite rough. Some male frogs may be drowned by others as they rush to give her a hug around the waist. The male sits on her back in a position known as amplexus. At this point, the female who is full of eggs, gets squeezed by the male and the eggs exit from her body and the male frog squirts sperm over the eggs. The eggs are surrounded in a jelly like matter and may stick to the rocks, to plants or to the surface of the pond. Other species of frogs such as the tree frogs, lay their eggs on land or on the branch of a tree located above water. The eggs can look like long strands, or clumps depending on the species of frog. The eggs hatch and the tadpole enters the water. Upvote · Alex Hirsekorn Former Aquarium Volunteer in Port Angeles, WA · Author has 3.2K answers and 5.1M answer views ·5y Related Unfortunately, amphibians have external fertilization. Why? Simply put, amphibians have evolved to use external fertilization because that method works and there has yet to be any evolutionary pressure to use other methods. In other words “If it works, don’t fix it”. Why do you use the word “Unfortunately”? Amphibians have been around, externally fertilizing their brains out for 370 million years! Please name an animal group that has exceeded that level of success using internal fertilization. Hint: There are a few such animals but I’m not going to tell you which ones. Upvote · 9 2 Richard Pierce Photographer, Aquarist, Marine Biologist, Philosopher · Author has 2.2K answers and 5.2M answer views ·7y Related In which species does external fertilisation take place? External fertilization is common in most fish and many aquatic invertebrates. Internal fertilization is the exception to the rule in fishes, although many different taxa have some examples of internal fertilization. The most common example of internal fertilization in fish is the Family Poeciliidae. These are the guppies, mollies, platies and swordtails. Internal fertilization also occurs in some sharks and rays, some catfish, other families within the order Cyprinodontidae, some tetras and a few lesser known families. Most fish, bivalves, corals, starfish, and other non-mobile invertebrates al Continue Reading External fertilization is common in most fish and many aquatic invertebrates. Internal fertilization is the exception to the rule in fishes, although many different taxa have some examples of internal fertilization. The most common example of internal fertilization in fish is the Family Poeciliidae. These are the guppies, mollies, platies and swordtails. Internal fertilization also occurs in some sharks and rays, some catfish, other families within the order Cyprinodontidae, some tetras and a few lesser known families. Most fish, bivalves, corals, starfish, and other non-mobile invertebrates also practice external fertilization. In order to maximize the chance of fertilization, the release of gametes is often synchronized, either by a chemical or environmental signal. Fish will often come in close proximity to maximize the chance of successful fertilization. The gametes have a limited lifespan outside of the body, and both will quickly die if fertilization does not occur quickly. External fertilization is much less common in terrestrial animals. No mammals, birds or reptiles practice external fertilization. Some invertebrates, both terrestrial and aquatic practice what might be considered a hybrid type of fertilization. Males may produce sperm packets called spermatophores. These packets are delivered or attached to the female. The spermatophores may be used for internal or external fertilization depending on the species, often long after the male has departed. Some species can store spermatophores for up to several years and can fertilize many batches of eggs. Plants are the only terrestrial organisms that practice external fertilization by broadcasting their gametes into the air or by using intermediaries. Pollen grains spread by the wind or through pollinators are the equivalent of sperm in animals. The pollen grains land on the receptor organs in the plant but the actual fertilization may be internal. Upvote · 9 3 9 1 Related questions What is the reason behind external fertilization in frogs and fish? Why can’t sperm pass inside their ovaries? Why Do fish, frogs show external fertilisation? Why is the fertilization in frog called external fertilization? Which animal vagina is most similar to that of a human? What would happen if a female was filled with animal semen? Which other animals apart from human beings will do mating (have sexual intercourse) for pleasure and not for reproduction purposes? Is it possible for humans to have sex with frogs? Why is animal intercourse so rapid as opposed to humans? Do animals other than hominids experience any orgasm when they procreate? Do animals contract diseases through sex like humans? How do land frogs reproduce? In evolutionary terms, why do male mammals have penises? Do all mammals (or only primates) eliminate urine and sperm through the penis? Is sperm an animal? How do frogs reproduce? Where does the fertilization of a frog take place? Related questions What is the reason behind external fertilization in frogs and fish? Why can’t sperm pass inside their ovaries? Why Do fish, frogs show external fertilisation? Why is the fertilization in frog called external fertilization? Which animal vagina is most similar to that of a human? What would happen if a female was filled with animal semen? Which other animals apart from human beings will do mating (have sexual intercourse) for pleasure and not for reproduction purposes? Is it possible for humans to have sex with frogs? Why is animal intercourse so rapid as opposed to humans? Do animals other than hominids experience any orgasm when they procreate? Do animals contract diseases through sex like humans? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.andrew.cmu.edu/user/ramesh/teaching/course/48-175/lectures/2.BasicsOfDescriptiveGeometry.pdf
2 Basic Concepts of Descriptive Geometry From this moment onwards we look at a particular branch of geometry—descriptive geometry—developed by Gaspard Monge in the late eighteenth century, who, incidentally, played an important role in Napoleonic war efforts, and which, now plays a major part of current architectural drawing practice. Gaspard Monge Gaspard Monge (1746-1818) discovered (or invented) the principles of descriptive geometry at the tender age of 18, working as a military engineer on the design of fortifications, which were made of stones accurately cut to fit one onto another so that a wall or turret so constructed was self-supporting and strong enough to withstand bombardment. Monge’s descriptive geometry system was declared classified and a military secret and it was not until many years later around 1790s (when Monge was a Professor at the Beaux Arts) that it became a part of French engineering and archi-tectural education and then adopted virtually universally. Descriptive geometry deals with physical space, the kind that you have been used since birth. Things you see around you and even things that you cannot see have geometry. All these things concern geometric objects almost always in relationship to one another that sometimes requires us to make sense of it all—in other words, when we try to solve geometric problems albeit in architecture, engineering, science. Descriptive geometry deals with manually solving problems in three-dimensional geometry by generating two-dimensional views. So …what is a view? 52 2.1 VIEWS A view is a two dimensional picture of geometric objects. Not any old picture, but, more precisely, a ‘projection’ of geometrical objects onto a planar surface. This notion is more familiar than some of us of may think. For example, whenever we see a movie on the silver screen, we are really seeing a ‘projection’ of a sequence of ‘moving pictures’ captured on transparent film through a cone of light rays emanating from a lamp so that each picture appears enlarged on a flat screen placed at a distance from the image. Each such picture is a view. 2-1 Movie projection Another example is the shadow cast by an object, say, a tree, on another object, say, a wall. In this example, the shadow cast by the tree can be viewed as being ‘projected’ on the wall by the rays of light emanating from the sun. 2-2 Projecting shadows Here, the rays are almost parallel, in contrast to the rays emanating from a single point source as in the movie example. Another difference is that a tree is a truly 3-dimensional object, while the picture on a piece of film is essentially flat. In either case the types of projection is a close physical model of the mathematical notion of a projection. So … what is a projection? 2.2 PROJECTIONS In geometry, projections are mappings of 2- or 3-dimensional figures onto planes or 3-dimensional surfaces. For our purpose, we consider a projection to be an association between points on an object and points on a plane, known as the picture plane. This association— between a geometric figure and its image—is established by lines from points on the figure to corresponding points on the image in the picture plane. These lines are referred to as projection lines. 53 The branch of geometry that investigates projections, including a study of the properties that are preserved under them, goes under the name of projective geometry. Descriptive geometry is really a subfield of projective geometry. Problems solved using descriptive geometry can be intricate. For example, the task may be to depict accurately in a drawing the shadow cast by a tree on a roof that may not be flat. Since this shadow is in itself the result of a projection, this tasks calls for depicting the projection of a projection. An understanding of projections is therefore essential not only for the generation of images, but also for an understanding of what goes on in the scenes depicted by these images. The present chapter introduces the principles of parallel projections to build a foundation for the specific techniques of descriptive geometry dealing with ‘orthographic views’, which are commonly represented in architecture by floor plans, sections and elevation drawings. 2.3 PARALLEL PROJECTION BETWEEN LINES Let us start simple … with lines. Definition 2-1: Family The set of all lines parallel to any given line is a family of parallel lines. When no misunderstandings are possible, a family of parallel lines is simply referred to as a line family. Being parallel is an equivalence relation for lines in the sense that if a line is parallel to another, which is also parallel to a third, then the first and third lines are also parallel. The relation ‘line family’ partitions the all lines into classes so that each line belongs to exactly one class, containing all the lines parallel to it. 2-3 A line family As parallel lines do not intersect: Property 2-2: Uniqueness For a line family and a given point, there is exactly one line in the family that passes through that point. Consider two coplanar lines, l and m, and a coplanar line family as shown in Figure 2-4. A parallel projection of l on m maps every point P of l to that point P' of m, where m meets the projection line that passes through P. P’ is called the image of P. 54 2-4 Projection between lines This projection establishes a one-to-one correspondence between the points on l and the points on m. We call this simply a projection between lines l and m. From elementary geometry, whenever parallel lines are intersected by a traversal (a line not parallel to the line), opposite interior angles formed at the intersection points are identical in measure (congruent). 2-5 Opposite angles along a transversal are identical Consider now a parallel projection of a line l on a line m and two distinct points, A and B, on l and their images on m, A’ and B’. There are two cases to consider: • l and m are parallel, in which case polygon ABB'A' is a parallelogram and consequently, AA' = BB'; that is, the projection preserves distances. • l and m intersect at a point, say P. In this case, P is fixed and triangle !PAA' is similar to !PBB'. Consequently, !"! !" = !"! !" = !"! !" = k That is, the projection multiplies distances by a constant factor k. 55 2-6 Parallel projections multiply distances by a constant factor 2.3.1 Between-ness and parallel projections A point B lies between two points, A and C, whenever AC = AB+BC. Since a parallel projection multiplies distance by a constant factor, k, (which may be identically equal to 1) it follows that image A'C'= kAC = kAB + kBC = A'B' + B'C', it must also preserve between-ness of points. That is: Property 2-3: Distance Preserving and Between-ness A parallel projection between two lines multiplies distances by a constant; if the lines are parallel, the constant is one. Moreover, a parallel projection preserves between-ness Applying properties of parallel projections to Figure 2-7, it is easy to see that the sum of the projections of segments of a polyline onto a line equals the projection of the segment between the first and last end-points of the polyline. Between-ness preserving entities An entity, which preserves between-ness, also preserves distances. Rays, lines, segments, conic sections in general geometric figures are preserved by between-ness preserving entities. m l C A A' B B' C' m l C P A A' B B' C' 56 2-7 Sum of the projections of segments of a polyline onto a line equals the projection of the segment between the first and last end-points of the polyline Construction 2-1 Dividing a given segment of arbitrary length in a given ratio We are going to revisit the constructible constructions albeit via a variation. Suppose 𝐴𝐵 is a given segment and suppose we are given a ratio, say m:n, where m and n are integers. From one of the end points, say A, draw a ray, r. Mark m units on a convenient unit of measure from A on the ray. Let the mark be M. From repeat this step with a measure of n units. Call this mark N. Draw a line –M– parallel to –NB– and let it meet 𝐴𝐵 at M', which divides the segment into the required ratio. 2-8 Dividing a segment into two segments in a given ratio This construction clearly employs a parallel projection from r to 𝐴𝐵. Note that construction also works if 𝐴𝐵 is to be extended in the ratio m:n; in this case, we draw a line –N– parallel to –MB– and let it meet –AB– at N’. Then, AN' extends AB in the required ratio. line projections polyline 57 EXAMPLE: Dividing and extending segments The above construction also works for any number of divisions or extensions by integer ratios and for any combination of divisions and extensions. For example, if AB is to be divided into three segments with ratios 4:2:3 and extended by two segments in the ratio 2:4, we mark off points on r after 4, 6 (= 4+2), 9 (= 4+2+3), 11(= 9+2) and 15 (= 9+2+4) units and draw the line l through the point at mark 9; the desired points on AB or its extension are the projections of the marks on r by lines parallel to l. This is illustrated in Figure 2-9. 2-9 Dividing and extending a segment into an arbitrary number of given ratios The reader may notice the relationship to constructions [1.16], which is interpreted in terms of similarity between lengths, whereas this construction is interpreted in terms of parallel projections. 2.1.1 Parallel projections between planes We can now extend the notion of a parallel projection to planes in space. Definition 2-4: Parallel Projection between Planes Let p and p' be two distinct planes and ƒ a line family not parallel to either plane. A parallel projection of p onto p’ maps every point P of p onto point P' of p' where p' meets the line in ƒ that passes through P. Because there is exactly one line in f that passes through P, this type of mapping establishes a 1-1 correspondence between points of p and p’, and we often, simply, call this a projection between the points of the planes. Consider a line l on p. The lines in ƒ that pass through l form a plane distinct from p’ that intersects p’ at a line l’ which is the image of l under the projection. It thus maps lines on lines. Furthermore, l’ is the image of l under a coplanar parallel projection. Because a parallel projection between two lines multiplies distances by a constant factor 58 k, which equals 1 when p and p’ are parallel (see Figure 2-10i), in which case it is constant for the entire projection. Otherwise, k may vary for different lines (see Figure 2-10ii). The projection thus preserves between-ness and the properties that depend on it, but does not always multiply distances by a constant factor. i ii 2-10 Illustrating a parallel projection between planes We summarize these observations in the following property: Property 2-5: Between-ness Preserving A parallel projection between two planes is a 1-1 mapping between points on planes, which preserves between-ness between points and parallelism, concurrence and ratio of division between lines. Distances are preserved only if the planes are parallel. 2.1.2 Not all points of an object have to be projected Since a parallel projection preserves between-ness, it maps not only lines on lines, but also segments on segments, rays on rays etc.; that is, it maps linear figures on linear figures of the same type. Notice that not all points of an object have to be projected. Because of Property 2-5 on preserving between-ness, we can project just distinguished points such as the endpoints of lines. The projection of the line is the line joining the projected endpoints. In a similar fashion the projection of a surface is constructed from the projection of its ‘boundary’ lines, each, in turn, constructed from the projection of their endpoints. f p' p l' l f a p' p l' l 59 If we apply this to a piece of architecture, we see that a projection of a planar facade on a plane parallel to it shows all features of the facade in true size (see Figure 2-11). This fact explains why elevations are so important in building design; plans and sections are important for similar reasons. But note also in the figure that the roof, which is not projected on a plane parallel to it, appears in the projection not in true size. The next section will get back to the role of parallel projections in architecture in greater detail. 2-11 Orthographic projection of a facade The method works even for curved objects. In fact, we can show (although we do not do so here) the following property, which will be important for subsequent chapters: Property 2-6: Type Preserving A parallel projection between two planes maps parabolas onto parabolas, hyperbolas onto hyperbolas, circles or ellipses onto circles or ellipses, and, more generally, curves of degree n onto curves of degree n. The ‘boundary’ of the surface corresponds to points on the object at which the projection lines are tangential to the surface of the object. 2.1.3 Parallel projections of general figures The notion of a parallel projection that underlies Definition 2-4 can be extended to projections of general 3-dimensional objects by means of a line family on a plane, even a non-planar surface in space. For example, to model geometrically the shadow cast by a cylindrical tower on a domed roof under parallel light rays like the ones generated by the sun, one would use a parallel projection of a cylinder on part of a sphere under a line family whose lines are parallel to the rays of the sun. More straightforward is the projection of a spatial figure on a plane. The image created on that plane can be viewed as a 2-dimensional representation of the figure; descriptive 60 geometry deals with precisely the generation of such images. Since objects of interest in architecture and related fields are often composed of linear shapes, such images can often be pieced together by a series of parallel projections between planes as defined— Figure 2-11 illustrates this as well. However, that scenes thus depicted may involve images of more complicated projections such as the shadows described above. The handling of ‘shades and shadows’ and similar projections constitutes indeed a special subfield of descriptive geometry, which we will consider later in the course. We conclude this section with a special case of particular significance for parallel projections used in descriptive geometry. Consider a plane p' and a line family ƒ not parallel to p'. Any two distinct lines in ƒ define a plane, p, which must intersect p’ at a line, l. Consider the image of p projected by the lines in ƒ on p', that is, the set of points where a line in ƒ through a point in p meets p'. Every point on p is projected by a line in p, which intersects p' at a, and every point P' on l is the image of infinitely many points on p (namely the points on the line in ƒ that passes through P'). l is thus the image of p under the projection (see Figure 2-12). Observe that l is also the image of p' on p under a parallel projection by a line family parallel to p'. 2-12 The image of a plane onto another is the line where the two planes meet The following property states this result for further reference. p p' l 61 Property 2-7: Image of plane onto another If ƒ is a line family, p a plane parallel to ƒ and p' a plane not parallel to ƒ, the image of p projected on p' by ƒ is the line l where p and p' meet. Conversely, the image of p’ projected on p by a line family is the line l where p and p' meet. 2.4 ORTHOGRAPHIC VIEWS When you draw a floor plan, section or elevation of a building, you are (consciously or unconsciously) using a special form of parallel projection, which is introduced in the following definition: Definition 2-8: Orthographic Projection and Views Let ƒ be a family of lines and p a plane not parallel to ƒ. The parallel projection of a figure on p by ƒ is an orthographic projection if the lines in ƒ are normal (perpendicular) to p. If g is an orthographic projection on a plane p, and h is a distance-preserving mapping of p, the image of hg is called an orthographic view of the figure. In principle, there are two consecutive mappings involved in the generation of an orthographic view. Figure 2-13 illustrates this for a single point, X. The plane p on which X is mapped is called the picture plane. A particular line, l, in ƒ maps X onto a point Xp of p; l is called the projection line through X. The mapping h maps this image on a point X' in the Euclidean plane—which, for us, will be a sheet of paper. 2-13 Specifying an orthographic view of a point on a sheet a paper It is important to realize that the mapping from a spatial to a 2-dimensional plane can be done in one of two ways, where each way implies a specific point of view or viewer projection line l h picture plane p sheet of paper Xp X X' 62 position on one or the other side of the spatial plane (see Figure 2-14). The view from one side is the mirror image of the other view. 2-14 Two ways of mapping onto a picture plane We normally assume that the picture plane is between the viewer and the object being viewed. 2-15 Normally the picture (projection) plane is between viewer and object We call the view generated by a horizontal picture plane that is placed above an object and viewed from above a top view of the object. Similarly we call a view generated by a horizontal picture plane place below an object and viewed from below a bottom view of the object. A view generated similarly by a vertical picture plane placed in front, of an object and viewed from the front is called respectively, a front view of the object. Likewise we have back or side views of an object. As locations, and therefore measurements, are involved in descriptive geometry we devise a method by means of which accurate perpendicular measurements are represented on a sheet of paper. We can apply an orthographic view (Definition 2-8) by imagining the horizontal plane to be hinged to the frontal plane and likewise the side or profile plane to be hinged to the frontal plane so that all three planes are represented on the same sheet of paper. The viewer’s lines of sight remains perpendicular to the respective picture planes. Even after the planes have been swung into place as illustrated in Figure 2-17, the observer considers the horizontal picture plane to be perpendicular to the frontal picture plane. viewpoint 1 viewpoint 2 PROJECTION PLANE Projector - a line from a point in space perpendicular to a plane surface called a projection plane Observer's line of sight is perpendicular to the projection plane Point in space 63 2-16 Three orthogonal views: top, front and side (profile) 2-17 Unfolding the picture planes onto a sheet of paper Observer's line of sight to see frontal projection of points Projection of the point on frontal plane (similar to a wall in a room) Projection of the point on horizontal plane (similar to a ceiling in a room) Projectors Observer's line of sight to see horizontal projection of points FRONTAL PLANE HORIZONTAL PLANE Point in space Observer's line of sight Observer's line of sight Projection of the point on frontal plane Projection of the point on horizontal plane Projection of the point on left profile plane Projectors Observer's line of sight LEFT PROFILE PLANE FRONTAL PLANE HORIZONTAL PLANE Point in space Horizontal plane Frontal projection of point Profile projection of point Horizontal projection of point Line of sight after the projection planes are in the plane of the drawing surface Projection planes before being swung into the plane of the drawing surface Profile plane Frontal plane Plane of the drawing surface Point The six orthographic views Top and Bottom views Front and Back views Left and Right Side view 64 Thus, when the viewer looks at the horizontal picture plane, he/she sees the frontal picture plane as an edge. Likewise when the viewer sees the frontal picture plane, he/she sees the horizontal and profile planes as edges. 2-18 Visualizing a picture plane in the other picture planes The viewer’s line of sight would appear as a point in each projection plane. This can be visualized by standing right behind the lines of sight in each view of a point in space. Thus, when the viewer sees the frontal plane, he/she sees the point at distance below the horizontal plane. When viewing the horizontal plane, the viewer sees the point at a distance behind the frontal plane. When he/she views the side (profile) plane, he/she stills sees the point behind the front plane. See Figure 2-19. 2-19 Individual views Horizontal and frontal planes seen as edges Horizontal and profile planes seen as edges Frontal and profile planes seen as edges Edge of the profile plane Edge of the profile plane Edge of the frontal plane Line of sight is perpendicular to the profile projection plane Line of sight is perpendicular to the frontal projection plane Edge of the frontal plane Edge of the horizontal plane Edge of the horizontal plane Line of sight is perpendicular to the horizontal projection plane PROFILE PLANE FRONTAL PLANE HORIZONTAL PLANE HORIZONTAL PLANE PROFILE PLANE FRONTAL PLANE Distance behind frontal projection plane Distance below horizontal projection plane Distance behind profile projection plane 65 2.1.4 Notational convention 2-20 Only the reference (hinge or folding) lines are important 2.5 ADJACENT VIEWS Figure 2-19 illustrates an important concept in descriptive geometry. When dealing with orthographic views, descriptive geometry always assumes that the figure under consideration is given in at least two adjacent views, which is captured by the following definition: 3 2 2 1 Distance behind frontal projection plane Distance below horizontal projection plane Distance behind profile projection plane P1 P2 P3 3 2 2 1 Folding, reference or hinge line 1|2 Line of sight for hotizontal plane Line of sight for left profile plane Line of sight for frontal plane Folding, reference or hinge line 2|3 LEFT PROFILE PLANE FRONTAL PLANE HORIZONTAL PLANE P3 P P1 P2 66 Definition 2-9: Adjacent Views Two intersecting planes are perpendicular if the line of intersection is the image of one plane in the other plane under an orthographic projection. Two views obtained from two perpendicular picture planes are called adjacent. Figure 2-21 illustrates perpendicular planes. In architecture you will find countless examples of adjacent views for instance, plan and elevation, an elevation and a second elevation in a picture plane perpendicular to the one used in the first elevation, an elevation and section with picture planes that have the same relationship with each other etc. 2-21 Perpendicular planes The line where the picture planes of two adjacent views, t and f, meet is called the folding line between t and f and denoted by t | f. The folding line is the image of each view in the adjacent view. We denote the view of a point or object, X, in a view, t, by Xt. Sometimes views are numbered instead of a letter. The same naming conventions apply. 2-22 A point in two adjacent views X Xtop Xfront folding line projection line top front Xtop Xfront 67 The folding line is also known as the reference or hinge line 2.1.5 Literal versus normal renderings of orthographic views A literal rendering of the orthographic view of an object would treat each point of the object equally; for a cylinder, this could result in a drawing like the one shown at the top of Figure 2-23, which depicts the cylinder as an unstructured mass without distinguishing between parts that are more important than others. This is hardly ever done. The normal way of rendering the orthographic view of an object is demonstrated at the bottom of Figure 2-23. The visible outline of the figure is always drawn in its entirety because it separates the figure from the background or from other figures. In addition, such a rendering tends to emphasize important other features; in the example shown, the upper rim of the cylinder is shown in its entirety, including the part that does not lie on the outline; this implies that the upper surface is visible from the particular direction from which the object is viewed, while the bottom is not visible and therefore not shown in its entirety. 2-23 Two renderings of an orthographic view of a cylinder Orthographic views, in architecture or other fields, are generated for a purpose, and the selection of the features to be shown may vary with that purpose. In general, one shows, aside from the outline, at least the boundary of each surface that can be seen from the direction of view; hidden boundaries are normally deleted if showing them would confuse the image; otherwise, they are shown with dashed lines (see Figure 2-24) or by similar means that distinguish them from the visible lines. folding line f t Xt Xf 68 2-24 A tetrahedron with a hidden edge Note also that different parts of an object may lie on the outline in different views. For example, the boundary of the cylinder is delineated in Figure 2-23 by two segments, which correspond to two specific segments on the cylinder’s boundary; two different segments would fall on the outline if we were to shift the direction of view (by choosing a different picture plane). 2.6 ORTHOGRAPHIC VIEWS IN ARCHITECTURE: FLOOR PLANS Floor plans, sections and elevations are for the most part orthographic views of a building or of portions thereof. When we produce such a drawing, we are often not even aware of the underlying principles because the problems involved can be solved intuitively. But in cases when it is not immediately clear how to draw a certain part of a design, we have to go back to the underlying principles in order to resolve the issue at hand. Take, for example, a floor plan. It shows in the majority of cases the walls and partitions on a floor that separate spaces from each other and the outside and indicates the position and width of doors, windows and other openings, along with other important objects. When the walls and openings are vertical, this can be done without complications because the sides of these objects are on planes perpendicular to the picture plane and project on line segments. But how should we draw the floor plan of an attic under a pitched roof, where the main parts are not vertical? This problem can be resolved when we consider what a floor plan really is. Definition 2-10: Floor Plan A floor plan of a building is the top view of a portion of a building below a picture plane cutting horizontally through the building. It shows the parts of the building underneath this plane as seen when we view the picture plane from above. The solid parts that intersect the picture plane (also called the cutting plane) are the most important features depicted and are shown, normally, in outline; openings appear as gaps in this outline. In order to make such a plan as informative as possible, the height of the picture plane must be chosen carefully; especially, it should cut through as many openings as possible (in practice, some ‘cheating’ may be tolerated here if it increases the clarity of the plan; but in case of doubt, it is advisable to stick to a literal interpretation of this process). 69 Objects such as steps, furniture, floor patterns etc. that are below the cutting plane can be projected into that plane and are then shown as in any other orthographic view. Important features above the cutting plane that are visible from it in reality can also be projected into the cutting plane, possibly with dashed lines to distinguish them from other objects. If the sole purpose of a plan is to show ceiling pattern in this way, the plan is usually called a reflected ceiling plan. Many of these devices are illustrated in the first floor plan shown in Figure 2-25. 2-25 Standard drawings of a house in the Tudor style (Top) Third floor plan; and (Bottom) First floor plan It is important to understand that all of the projections that come into play in such a drawing use the same family of projection lines normal to the cutting plane (and this 70 family is unique). It is up to the designer to decide which of the many features projected by these lines are actually drawn. The normal case illustrated by the first floor plan in Figure 2-25 depicts vertical walls, that is, walls perpendicular to the picture plane, whose sides consequently project into lines. This is the single most important feature that makes floor plans so easy to construct. But consider the third floor plan shown in Figure 2-25. The rooms on this floor are enclosed by vertical walls only up to a certain height above which the underside of the roof becomes the boundary of the room towards the outside (see the section in the same figure). If one understands that a floor plan is an orthographic projection into a horizontal cutting plane, viewed from above, it becomes apparent how this situation can be drawn. The plan shown in Figure 2-25 uses a cutting plane slightly below the point where the roof meets the vertical portions of the walls. The plan depicts the mass of the roof as it is cut by the picture plane; the parts of the roof that extend from this plane outwards and down to the eaves appear in top view, while the parts that rise above the plane are indicated by dashed lines outlining the different sides of the roof that are visible from below. Details such as the elaborate articulation of the gables are drawn following the same principles. A section is developed in the same way using a vertical cutting plane that cuts through the building. Elevations are developed with vertical picture planes that do not intersect the building. Features of the building beyond the picture plane are again shown as orthographic views to the degree of detail desired. See Figures 2-27 and 2-27. 2-26 Standard drawings of a house in the Tudor style: Longitudinal section; 71 2-27 Standard drawings of a house in the Tudor style: Front elevation In these constructions, the given adjacent views are always assumed to be placed relative to each other as they would appear when one of the picture planes is ‘folded’ at the folding line to become co-planar with the other picture plane as demonstrated in Figures 2-21 and 2-22 (the folding hinge line receives its name from this convention). In such an arrangement, any point in one view appears in the other view on a line perpendicular to the folding line; this line is, in fact, the image of a projection line projecting the point into the picture plane of the adjacent view; it appears as a point in the other view, mapping every point on it on that point. Figure 2-28 shows two adjacent views of a building, a floor plan and a side elevation, generated from two perpendicular picture planes and arranged as two adjacent views in the way described. It is important to realize that in order to resolve all details in any one of these standard views, some of the more advanced techniques may have to be used. The core of descriptive geometry consists of a collection of constructions that can be used to depict spatial objects in orthographic views and to determine geometric properties of such objects from these views; in many cases, additional views must be generated in order to solve a certain problem. But these additional views can in most cases be obtained by purely 2-dimensional constructions if at least two adjacent views are given. This assumption underlies most constructions. 72 2-28 Two adjacent views of a building 2.7 VISIBILITY OF LINES We stated already that a projected object is hardly ever shown in its entirety—that is, by showing every point that belongs to the object—because this would lead to a completely unstructured, incomprehensible image. Most prominent among the features shown are the edges that form the boundary of the object (or of its parts) in the view under consideration. The edges appear as line segments or as curves in the view and are often simply called lines. If all of these lines are displayed, the view is called a wire-frame image. For all but the simplest objects, however, this type of view is still too confusing to be informative. Wire-frame views are therefore usually edited in order to make them more comprehensible. Most commonly, they show which lines are visible or hidden in a particular view. Hidden lines are either dashed (this is usual in many engineering fields) or omitted altogether (this is usual in architectural design). 73 Construction 2-2 Visibility/Intersection Test Given two lines in two adjacent views, neither line perpendicular to the folding line, that meet at a point, X, in at least one view, t, determine which line is in front of the other (relative to t) at the intersection point. (Right) The problem 1 Determine which line is in front (or above) of the other There are two steps. 1. Draw the projection line a through Xt into view f. There are two possibilities. 1.1 If the lines meet also at a point on a in view f, the lines truly intersect. (Right) The lines truly intersect 1.2 Otherwise, the lines do not intersect. Determine the spatial relation between the lines at Xt from the relative positions of their intersections with a in f: the line that intersects a at a point closer to the t f X a t f X 74 folding line than the other line is closer to the picture plane of f at that point; consequently, it is in front of the other line at Xt in t. For example, in the figure below, line l satisfies this condition. (Right) The lines do not intersect Project from the (apparent) point of intersection onto the other view The same test is applied if the two lines also intersect in view f. See Figure 2-29. (Left) Relative visibility of two lines 2-29 Intersection test – l is in front in view t and behind in view The obstructed line is often shown interrupted at the intersection point (when it is visible everywhere except at that point) or dashed (when an entire portion is obstructed) as the example below illustrates. f t X a l l t f X 75 2.1.6 Worked example – Visibility Determine visibility of a three-sided pyramid in given top and front views, t and f Consider the front and side views of the pyramid as shown in Figure 2-30. 2-30 Front and side views of a pyramid The outside edges of the pyramid are visible in both views since the points of intersection coincide in both views. However, edges l and m intersect in both views, but not at the same point. Applying the visibility test to the intersection point in the top view t, we find that the corresponding point on l is closer to the folding line than the corresponding point on m in the front view f. Similarly, applying the visibility test to the intersection point in the front view f, we find that the corresponding point on m is closer to the folding line than the corresponding point on l in the top view ƒ. Consequently, we show m dashed in view t and l dashed in view f as shown on the right side in Figure 2-31. m l m l front top 76 2-31 Visibility test for a pyramid 2.8 PRINCIPAL VIEWS (OR PROJECTIONS) See Figure 2-32 2-32 Principal views m l m l front top m l m l front top Line of sight for top view Line of sight for bottom view Line of sight for left side elevation view LEFT SIDE FRONT TOP Line of sight for back elevation view Line of sight for front elevation view Line of sight for right side elevation view 77 It is important to note: • Elevation views are always perpendicular to the horizontal view. That is, perpendicular distances below or above the horizontal plane are always seen in the elevation views. • Lines of sight that are perpendicular to an elevation view are always horizontal. Figure 2-33 showing the view swung into the plane of a drawing paper. 2-33 Principal views unfolded onto paper 5 1 6 4 2 3 2 2 1 Bottom View Top View Right Side Elevation Right Side Elevation Front Elevation Back Elevation 78 2.9 AUXILIARY VIEWS Definition 2-11: Auxiliary View If p and q are two adjacent views, a view using a picture plane perpendicular to the picture plane used in p or q is an auxiliary view. A primary auxiliary view is a view using a picture plane perpendicular to one of the coordinate planes and inclined to the other two coordinate planes. A secondary auxiliary view is an auxiliary perpendicular to a primary auxiliary view. For each of the three different picture planes parallel to the coordinate planes, there are three types of primary auxiliary views, depending on the coordinate plane to which their picture planes are perpendicular; possible picture planes for these views are shown in Figure 2-34. 2-34 Picture planes for principal auxiliary views for different coordinate planes 2.1.7 Auxiliary elevation views Consider the left-most figure in Figure 2-34. It shows a plane perpendicular to the horizontal plane. As shown in Figure 2-35 we can consider of number of auxiliary views based on different lines of sight, each parallel to the horizontal plane (and hence, horizontal) and each perpendicular its picture plane. 2-35 Auxiliary views Planes 2,3,4,5 and 6 are all elevations as each is perpendicular to the horizontal projection planes Observer's line of sight remains horizontal when viewing elevations 6 5 4 3 1 2 Auxiiary elevation Auxiiary elevation Frontal projection plane Front view Horizontal projection plane Top view 79 2-35(continued) Unfolded auxiliary views When solving descriptive geometry problems we eliminate the outlines of the projection planes and keep just the folding, hinge or reference lines as shown in Figure 2-36. Note that point X is located at a distance H below the horizontal picture plane as indicated by the projector from X1 to X2 in the top and front views. For the other auxiliary views 3, 4, 5 and 6, draw projectors from X1 perpendicular to the folding lines 1|k, k > 2, and transfer the distance H to points Xk. 2-36 Auxiliary elevation views 2.1.8 Auxiliary inclined views The same idea applies when the line of sight is neither horizontal nor vertical. See Figure 2-37, where the viewer’s line of sight is inclined at some angle and the picture plane perpendicular to the line of sight is inclined with respect to the top and front views. 3 1 5 6 1 4 2 1 Aux Elevation Aux Elevation Aux Elevation Aux Elevation Front Elevation Top view H = distance below horizontal plane H H H H 3 1 5 6 1 4 2 1 X3 X6 X4 X5 X1 X2 80 2-37 Auxiliary inclined view Figure 2-38 shows the top, front, and auxiliary projection planes. Note that as the top picture plane is perpendicular to the front plane, the viewer will see point X at a distance F behind the front plane. Likewise as the auxiliary picture plane is perpendicular to the front plane the point will be seen at the same distance F in the auxiliary view. 2-38 Auxiliary inclined view 2.10 TRANSFER DISTANCE An auxiliary view of an object can be constructed from two adjacent views by use of the construction below. This construction reiterates the notion of ‘transfer distance’ introduced above, of transferring the distance from an adjacent view to an auxiliary view. It is important that you thoroughly understand the construction, especially the role of transfer distance. No other construction is more fundamental in descriptive geometry. Horizontal projection plane Frontal projection plane Inclined line of sight – horizontal plane (top view) appears as an edge Aux inclined projection plane Top view Front elevation 3 2 2 1 F F Auxiliary inclined plane Frontal projection plane Horizontal projection plane 3 2 2 1 X3 X1 X2 81 Construction 2-3 Transfer Distance Given a point, X, in two adjacent views, t and f, construct an auxiliary view of X using a picture plane perpendicular to the picture plane of t. (Right) The problem There are three steps. 1. Call the auxiliary view a, and select a folding line, t | a, in t (any convenient line other than t | f will do). 2. Draw the projection line, lX, through Xt perpendicular to t | a. 3. Let dx be the distance of Xf from folding line t | f. Xa (that is, the view of X in a) is the point on lx that has distance dx from the folding line t | a. The distance dx is called a transfer distance. The construction is illustrated in Figure 2-39. 2-39 Constructing an auxiliary view from two adjacent views (Left) Constructing an auxiliary view of the point; (Right) Visualization of the construction t f Xt Xf dX dX lX t a f t Xa T Xt Xf dX dX dX a f t Xf Xt X Xa 82 2.1.9 More on transfer distances Figure 2-40 illustrates five consecutively adjacent projection planes in which two are always perpendicular to the third. Note that point P is located at a distance H below the horizontal view, F behind the front view and E behind the auxiliary elevation. 2-40 An example with five projection planes and its unfolding #5 Auxiliary inclined projection plane #4 Auxiliary inclined projection plane #3 Auxiliary elevation plane #2 Frontal projection plane #1 Horizontal projection plane Inclined line of sight #4 -auxiliary elevation plane appears as an edge Inclined line of sight #5 -frontal plane appears as an edge Level line of sight #3 -horizontal projection plane #1 and auxiliary inclined projection plane #4 appear as edges Level line of sight #2 -horizontal projection plane always appears as an edge Vertical line of sight #1 - frontal projection plane and all other elevation planes appear as edges P5 P2 P4 P P3 P1 1 2 5 3 4 P1 P2 P5 P3 P4 83 2-41 Unfolding 2-40 and the method of transfer distances 2.1.10 Examples of auxiliary view from given top and front elevation views A typical problem situation is that the top and front views are given along with an auxiliary axis specified as shown on the right. (Right) The problem F = distance behind frontal plane H = distance below horizontal plane F E = distance behind aux elevation #3 H E view #4 - aux. inclined projection plane view #3 - aux. elevation view #1 - top view view #2 - front elevation view #5 - aux. inclined projection plane 5 2 1 2 1 3 3 4 P4 p3 P5 P1 P2 1 f f t 60° 60° 84 When constructing an auxiliary especially if it is your first time at it, or a truly dense drawing it is often wise to number the points. See Figure 2-42(top left), where points 3 and 7, 4 and 6 overlap in top view and points 1 and 3 overlap in front view. We can now proceed as per Construction 2-3. See Figure 2-42(right). Hidden lines are shown dashed by Construction 2-2 (see page 73) for visibility of lines. 2-42 Constructing an auxiliary view (Top Left) The problem with points numbered (Right) Auxiliary view for the specified folding line with hidden lines shown dashed (Bottom Left) Rendered view 1 f f t 1,3 7 4 7,3 6,4 2 5,1 2 6 5 1 f f t 60° 60° 1 5 3 7 6 4 2 1,3 7 4 7,3 6,4 2 5,1 2 6 5 85 Figures 2-43 and 2-44 give two further examples of given top and front views and the location of the folding lines for the auxiliary view(s) although the points are numbered. These are left as exercises. 2-43 Example top, front and constructed auxiliary views transfer distances from view t transfer distances from view f 2 f 1 t f t 86 2-44 Another example top, front and constructed auxiliary views ∠45° lines not visible in any particular view are normally shown dashed (or dotted) in that view 1 f f t
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https://www.quora.com/What-conditions-determine-that-two-lines-in-3-D-space-are-parallel
Something went wrong. Wait a moment and try again. Straight Line in Space Vectors (geometry) Geometric Lines Analytic Geometry 3-d Geometry Lines of Latitude Geometric Analysis Vector Algebra 5 What conditions determine that two lines in 3-D space are parallel? · In 3-D space, two lines are considered parallel if they meet the following conditions: Direction Vectors : The direction vectors of the two lines must be scalar multiples of each other. If line L 1 has a direction vector d 1 and line L 2 has a direction vector d 2 , then L 1 and L 2 are parallel if there exists a scalar k such that: d 1 = k d 2 or d 2 = k d 1 2. Non-Intersecting : If the lines do not intersect, they are either parallel or skew. Skew lines are not parallel and do not lie in the same plane. 3. Same Plane Condition : For two lines In 3-D space, two lines are considered parallel if they meet the following conditions: Direction Vectors: The direction vectors of the two lines must be scalar multiples of each other. If line L1 has a direction vector d1 and line L2 has a direction vector d2, then L1 and L2 are parallel if there exists a scalar k such that: d1=kd2 or d2=kd1 Non-Intersecting: If the lines do not intersect, they are either parallel or skew. Skew lines are not parallel and do not lie in the same plane. Same Plane Condition: For two lines to be parallel, they must either lie in the same plane or be parallel in three-dimensional space. If the lines are coplanar and their direction vectors are parallel, they are parallel lines. For two lines in 3-D space to be parallel: - Their direction vectors must be scalar multiples of each other. - They can either be distinct parallel lines or coincident lines (the same line). Michael Paglia Former Journeyman Wireman IBEW · Author has 33.3K answers and 5.3M answer views · 1y Originally Answered: What is the condition for two lines being parallel in 3D? · Not much on 3d But 2d slopes need to be the same And their distance needs to be the same I'd say same slope And same plane I'm not getting how the intercepts work there Sorry Hey how's about that No Alzheimer's yet Lost both mom.and mom in law to it SoI know Not much on 3d But 2d slopes need to be the same And their distance needs to be the same I'd say same slope And same plane I'm not getting how the intercepts work there Sorry Hey how's about that No Alzheimer's yet Lost both mom.and mom in law to it SoI know Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Donald Hartig PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views · 6y Originally Answered: How can you check if 2 lines are perpendicular in 3D space? · Dot their direction vectors. The lines are perpendicular if and only if the dot product is zero. Related questions How can you determine if two lines are parallel without finding their equations? Are two lines parallel if they are the same line? How can we determine which side of a line a point is on, if the two given lines are parallel? Do parallel lines in 3-D space have the same slope? What is the side view of a straight line parallel to both plane? Jeff Koch Physics / Math Professor · Author has 11.1K answers and 1.7M answer views · 2y The two lines must be co planar to be parallel. Any set of parallel lines must be in the same 2 dimensional plane. And of course, they must be non intersecting. Another words, they must have the same slope in their plane. If two lines are not in the same plane, they are called skew lines. Alexandru Carausu Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010) · Author has 3K answers and 875.1K answer views · Updated 1y Related How can you determine if two lines are parallel? What happens when these lines intersect? If the direction vectors of the two lines are (respectively) v_ 1 & v _ 2 then ( L _1 ) || ( L _2 ) <==> v_ 1 || v _ 2 . (1) This means that the two vectors are collinear, which means that the two free vectors can be represented by two lines segments that lie on the same line or parallel lines. If the two vectors have known Cartesian coordinates in the orthonormal system ( O ; i , j , k ) , then relation (1) is equivalent to the proportionality of their respective coordinates. In the 3D space, v_ 1 ( x _1 , y _1 , z _1 ) , v _2 ( x _2 , y _2 , z _2 ) & (1) <==> x _1 / x _2 = y _1 / y _2 = z _1 / z _2 . (2) The common v If the direction vectors of the two lines are (respectively) v_1 & v_2 then ( L_1 ) || ( L_2 ) <==> v_1 || v_2 . (1) This means that the two vectors are collinear, which means that the two free vectors can be represented by two lines segments that lie on the same line or parallel lines. If the two vectors have known Cartesian coordinates in the orthonormal system ( O ; i , j , k ) , then relation (1) is equivalent to the proportionality of their respective coordinates. In the 3D space, v_1 ( x_1 , y_1 , z_1 ) , v_2 ( x_2 , y_2 , z_2 ) & (1) <==> x_1 / x_2 = y_1 / y_2 = z_1 / z_2 . (2) The common value of the three ratios in (2) can be = λ , hence v_1 = λ v_2 : the two direction vectors are proportional, with the multiplication of vectors by scalars. As regards the second question, this is more than strange ! Two parallel lines do not intersect one another at any point(s) : ( L_1 ) ∩ ( L_2 ) = ∅ , the empty set ! Note: A few (minor) corrections operated on Thu, January 18, 9:39 AM (RO time) Related questions How can we determine if two lines are parallel using only their slopes (no other information)? How can I prove that two lines in 3D are parallel? How do you tell if two lines are parallel? How can you determine if two lines are parallel using only two points on each line, without knowing the equations of the lines? How do you find the midpoint of two parallel lines? Michael Paglia Former Journeyman Wireman IBEW · Author has 33.3K answers and 5.3M answer views · 1y Related How can you determine if two lines are parallel? What happens when these lines intersect? I can tell you 3 ways But I learned you needed to prove anything you used And it seems some proofs are definitions these days If 2 lines are intersected by a 3rd And the opposite angles are congruent they are parallel I'd have to have that actual theorem number in my school in 1973 But that's one way Perpendicular lines are another Any 2 lines that never meet are parallel Lines have no length or width Line segments do A point has no dimension Logia den Drase Agent of the Alternate New South Wales Education Review · Author has 456 answers and 881.1K answer views · 9y Related Given the components of two three dimensional vectors, how do you determine if the two are parallel to each other? For each vector, take the largest component, and divide the other components by the largest component. (“Normalise the largest component to 1.”) If these are parallel vectors, the normalised vectors will be identical. Buddha Buck Studied at University at Buffalo · Author has 5.8K answers and 16.9M answer views · 9y Related Given the components of two three dimensional vectors, how do you determine if the two are parallel to each other? Take their wedge product. If vectors →a,→b are parallel, then →a∧→b=0. If you just have their components, →a=a1→e1+a2→e2+a3→e3,→a=b1→e1+b2→e2+b3→e3, then by remembering that the wedge product is antisymmetric and distributes over vector addition, you get →a∧→b=(a1→e1+a2→e2+a3→e3)∧(→a=b1→e1+b2→e2+b3→e3)=(a1→e1)∧(b1→e1+(a1→e1)∧(b2→e2+(a1→e1)∧(b3→e3+(a1→e2)\we Take their wedge product. If vectors →a,→b are parallel, then →a∧→b=0. If you just have their components, →a=a1→e1+a2→e2+a3→e3,→a=b1→e1+b2→e2+b3→e3, then by remembering that the wedge product is antisymmetric and distributes over vector addition, you get Unable to parse this math expression.\vec{a}\wedge\vec{b} = (a_1\vec{e_1} + a_2\vec{e_2} + a_3\vec{e_3})\wedge(\vec{a} = b_1\vec{e_1} + b_2\vec{e_2} + b_3\vec{e_3}) = (a_1\vec{e_1})\wedge(b_1\vec{e_1} + (a_1\vec{e_1})\wedge(b_2\vec{e_2} +(a_1\vec{e_1})\wedge(b_3\vec{e_3} +(a_1\vec{e_2})\wedge(b_1\vec{e_1} +(a_2\vec{e_2})\wedge(b_2\vec{e_2} +(a_2\vec{e_2})\wedge(b_3\vec{e_3} +(a_3\vec{e_3})\wedge(b_1\vec{e_1} +(a_3\vec{e_3})\wedge(b_2\vec{e_2} +(a_3\vec{e_3})\wedge(b_3\vec{e_3} = a_1b_2\vec{e_1}\wedge\vec{e_2} +a_1b_3\vec{e_1}\wedge\vec{e_3} +a_2b_1\vec{e_2}\wedge\vec{e_1} +a_2b_3\vec{e_2}\wedge\vec{e_3} +a_3b_1\vec{e_3}\wedge\vec{e_1} +a_3b_2\vec{e_3}\wedge\vec{e_2} = (a_1b_2-a_2b_1)\vec{e_1}\wedge\vec{e_2} + (a_1b_3-a_3b_1)\vec{e_1}\wedge\vec{e_3} + (a_2b_3-a_3b_2)\vec{e_3}\wedge\vec{e_2} In order for that to be zero, all three components of the wedge product (e.g, a1b2−b2a1) must be zero. Andrea Baisero Does not compute. · Author has 140 answers and 329.3K answer views · 11y Related Given the components of two three dimensional vectors, how do you determine if the two are parallel to each other? Given x and y the above mentioned vectors, check that ⟨ x , y ⟩ 2 == ⟨ x , x ⟩ ⟨ y , y ⟩ . Ajay Sreenivas Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010) · Author has 5.3K answers and 1.7M answer views · 1y Related How can you determine if two lines are parallel? What happens when these lines intersect? Draw two lines that are perpendicular to one of the lines at two different points and compare the distance from these points to the points of intersection with the second line. If the lines are parallel the distances will be equal else the lines will intersect at some point. Calvin L. 18, Mathematics & Statistics major · Author has 10K answers and 2.2M answer views · 2y Related How do you find the condition of parallel lines? Two lines are parallel if their gradients/slopes are the same. Parallel lines will never touch each other. e.g. y=7x−3 and y=7x+14 are parallel since their gradient, 7 is the same. You'll realize that there is no value of x and y where these lines touch. y=13x−6 and y=19 are not parallel since they have different gradients (—13 and 0). You'll notice that these lines will touch at some point in x and when y is 19. Justin Rising PhD in statistics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 12.1K answers and 26.5M answer views · 5y Related Are two lines parallel if they are the same line? Do you agree with the following statements? If a is parallel to b , then b is parallel to a . 2. If a is parallel to b and b is parallel to c , then a is parallel to c . 3. For every line a , there is some line b such that a is parallel to b . If so, you are forced to conclude that every line is parallel to itself. Alexis Sullivan Senior Product Manager · 11mo Related How do you tell if two lines are parallel? Method 1 of 3:Comparing the Slopes of Each Line The slope of a line is defined by (Y2 - Y1)/(X2 - X1) where X and Y are the horizontal and vertical coordinates of points on the line. You must define two points on the line to calculate this formula. The point closer to the bottom of the line is (X1, Y1) and the point higher on the line, above the first point, is (X2, Y2). This formula can be restated as the rise over the run. It is the change in vertical difference over the change in horizontal difference, or the steepness of the line. If a line points upwards to the right, it will have a positive Method 1 of 3:Comparing the Slopes of Each Line The slope of a line is defined by (Y2 - Y1)/(X2 - X1) where X and Y are the horizontal and vertical coordinates of points on the line. You must define two points on the line to calculate this formula. The point closer to the bottom of the line is (X1, Y1) and the point higher on the line, above the first point, is (X2, Y2). This formula can be restated as the rise over the run. It is the change in vertical difference over the change in horizontal difference, or the steepness of the line. If a line points upwards to the right, it will have a positive slope. If the line is downwards to the right, it will have a negative slope. Identify the X and Y coordinates of two points on each line. A point on a line is given by the coordinate (X, Y) where X is the location on the horizontal axis and Y is the location on the vertical axis. To calculate the slope, you need to identify two points on each of the lines in question. Points are easily determined when you have a line drawn on graphing paper. To define a point, draw a dashed line up from the horizontal axis until it intersects the line. The position that you started the line on the horizontal axis is the X coordinate, while the Y coordinate is where the dashed line intersects the line on the vertical axis. For example: line l has the points (1, 5) and (-2, 4) while line r has the points (3, 3) and (1, -4). Plug the points for each line into the slope formula. To actually calculate the slope, simply plug in the numbers, subtract, and then divide. Take care to plug in the coordinates to the proper X and Y value in the formula. To calculate the slope of line l: slope = (5 – (-4))/(1 – (-2)) Subtract: slope = 9/3 Divide: slope = 3 The slope of line r is: slope = (3 – (-4))/(3 - 1) = 7/2 Compare the slopes of each line. Remember, two lines are parallel only if they have identical slopes. Lines may look parallel on paper and may even be very close to parallel, but if their slopes are not exactly the same, they aren’t parallel. In this example, 3 is not equal to 7/2, therefore, these two lines are not parallel. Method 2 of 3:Using the Slope-Intercept Formula 1. Define the slope-intercept formula of a line. The formula of a line in slope-intercept form is y = mx + b, where m is the slope, b is the y-intercept, and x and y are variables that represent coordinates on the line; generally, you will see them remain as x and y in the equation. In this form, you can easily determine the slope of the line as the variable "m". For example. Rewrite 4y - 12x = 20 and y = 3x -1. The equation 4y - 12x = 20 needs to be rewritten with algebra while y = 3x -1 is already in slope-intercept form and does not need to be rearranged. Rewrite the formula of the line in slope-intercept form. Oftentimes, the formula of the line you are given will not be in slope-intercept form. It only takes a little math and rearranging of variables to get it into slope-intercept. For example: Rewrite line 4y-12x=20 into slope-intercept form. Add 12x to both sides of the equation: 4y – 12x + 12x = 20 + 12x Divide each side by 4 to get y on its own: 4y/4 = 12x/4 +20/4 Slope-intercept form: y = 3x + 5. Compare the slopes of each line. Remember, when two lines are parallel to each other, they will have the exact same slope. Using the equation y = mx + b where m is the slope of the line, you can identify and compare the slopes of two lines. In our example, the first line has an equation of y = 3x + 5, therefore it’s slope is 3. The other line has an equation of y = 3x – 1 which also has a slope of 3. Since the slopes are identical, these two lines are parallel. Note that if these equations had the same y-intercept, they would be the same line instead of parallel. Method 3 of 3:Defining a Parallel Line with the Point-Slope Equation 1. Define the point-slope equation. Point-slope form allows you to write the equation of a line when you know its slope and have an (x, y) coordinate. You would use this formula when you want to define a second parallel line to an already given line with a defined slope. The formula is y – y1= m(x – x1) where m is the slope of the line, x1 is the x coordinate of a point given on the line and y1 is the y coordinate of that point. As in the slope-intercept equation, x and y are variables that represent coordinates on the line; generally, you will see them remain as x and y in the equation. The following steps will work through this example: Write the equation of a line parallel to the line y = -4x + 3 that goes through point (1, -2). Determine the slope of the first line. When writing the equation of a new line, you must first identify the slope of the line you want to draw yours parallel to. Make sure the equation of the original line is in slope-intercept form and then you know the slope (m). The line we want to draw parallel to is y = -4x + 3. In this equation, -4 represents the variable m and therefore, is the slope of the line. This equation only works if you have a coordinate that passes through the new line. Make sure you don’t choose a coordinate that is on the original line. If your final equations have the same y-intercept, they are not parallel, but the same line. In our example, we will use the coordinate (1, -2). Write the equation of the new line with the point-slope form. Remember the formula is y – y1= m(x – x1). Plug in the slope and coordinates of your point to write the equation of your new line that is parallel to the first. Using our example with slope (m) -4 and (x, y) coordinate (1, -2): y – (-2) = -4(x – 1) Simplify the equation. After you have plugged in the numbers, the equation can be simplified into the more common slope-intercept form. This equation's line, if graphed on a coordinate plane, would be parallel to the given equation. For example: y – (-2) = -4(x – 1) Two negatives make a positive: y + 2 = -4(x -1) Distribute the -4 to x and -1: y + 2 = -4x + 4. Subtract -2 from both side: y + 2 – 2 = -4x + 4 – 2 Simplified equation: y = -4x + 2 Related questions How can you determine if two lines are parallel without finding their equations? Are two lines parallel if they are the same line? How can we determine which side of a line a point is on, if the two given lines are parallel? Do parallel lines in 3-D space have the same slope? What is the side view of a straight line parallel to both plane? How can we determine if two lines are parallel using only their slopes (no other information)? How can I prove that two lines in 3D are parallel? How do you tell if two lines are parallel? How can you determine if two lines are parallel using only two points on each line, without knowing the equations of the lines? How do you find the midpoint of two parallel lines? How can you determine the distance between two parallel lines? What lines are parallel to y = − 4 x + 3 ? What is the difference between parallel lines and non-parallel lines? Why do parallel lines have a slope of 0? How do you determine if a point lies between two parallel lines on a graph? How can two lines be identified as parallel without using any instruments? Is there a specific condition or rule that determines parallelism? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
187835
https://psu.pb.unizin.org/chem112maluz4/
Chemistry 112- Chapters 12-17 of OpenStax General Chemistry – Simple Book Publishing Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Book Title: Chemistry 112- Chapters 12-17 of OpenStax General Chemistry by OpenStax Book Description: The textbook provides an important opportunity for students to learn the core concepts of chemistry and understand how those concepts apply to their lives and the world around them. The book also includes a number of innovative features, including interactive exercises and real-world applications, designed to enhance student learning. License: Creative Commons Attribution Read Book Contents Show All Contents Hide All Contents Book Contents Navigation Preface Chapter 12. Kinetics Introduction 12.1 Chemical Reaction Rates 12.2 Factors Affecting Reaction Rates 12.3 Rate Laws 12.4 Integrated Rate Laws 12.5 Collision Theory 12.6 Reaction Mechanisms 12.7 Catalysis Chapter 13. Fundamental Equilibrium Concepts Introduction 13.1 Chemical Equilibria 13.2 Equilibrium Constants 13.3 Shifting Equilibria: Le Châtelier’s Principle 13.4 Equilibrium Calculations Chapter 14. Acid-Base Equilibria Introduction 14.1 Brønsted-Lowry Acids and Bases 14.2 pH and pOH 14.3 Relative Strengths of Acids and Bases 14.4 Hydrolysis of Salt Solutions 14.5 Polyprotic Acids 14.6 Buffers 14.7 Acid-Base Titrations Chapter 15. Equilibria of Other Reaction Classes Introduction 15.1 Precipitation and Dissolution 15.2 Lewis Acids and Bases 15.3 Multiple Equilibria Chapter 16. Thermodynamics Introduction 16.1 Spontaneity 16.2 Entropy 16.3 The Second and Third Laws of Thermodynamics 16.4 Free Energy Chapter 17. Electrochemistry Introduction 17.1 Balancing Oxidation-Reduction Reactions 17.2 Galvanic Cells 17.3 Standard Reduction Potentials 17.4 The Nernst Equation 17.5 Batteries and Fuel Cells 17.6 Corrosion 17.7 Electrolysis Appendix Appendix A: The Periodic Table Appendix B: Essential Mathematics Appendix C: Units and Conversion Factors Appendix D: Fundamental Physical Constants Appendix E: Water Properties Appendix F: Composition of Commercial Acids and Bases Appendix G: Standard Thermodynamic Properties for Selected Substances Appendix H: Ionization Constants of Weak Acids Appendix I: Ionization Constants of Weak Bases Appendix J: Solubility Products Appendix K: Formation Constants for Complex Ions Appendix L: Standard Electrode (Half-Cell) Potentials Appendix M: Half-Lives for Several Radioactive Isotopes Book Information Author OpenStax License Chemistry 112- Chapters 12-17 of OpenStax General Chemistry Copyright © 2016 by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. Subject Chemistry Metadata Title Chemistry 112- Chapters 12-17 of OpenStax General Chemistry Author OpenStax License Chemistry 112- Chapters 12-17 of OpenStax General Chemistry Copyright © 2016 by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. © Jun 20, 2016 OpenStax.Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. Under this license, any user of this textbook or the textbook contents herein must provide proper attribution as follows: The OpenStax College name, OpenStax College logo, OpenStax College book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University.For questions regarding this license, please contact partners@openstaxcollege.org. If you use this textbook as a bibliographic reference, then you should cite it as follows:OpenStax, Chemistry. OpenStax CNX. Jun 20, 2016 If you redistribute this textbook in a print format, then you must include on every physical page the following attribution: “Download for free at If you redistribute part of this textbook, then you must retain in every digital format page view (including but not limited to EPUB, PDF, and HTML) and on every physical printed page the following attribution: “Download for free at Primary Subject Chemistry Pressbooks Powered by Pressbooks Pressbooks User Guide |Pressbooks Directory Pressbooks on YouTubePressbooks on LinkedIn
187836
https://web.stanford.edu/~lindrew/8.044.pdf
8.044: Statistical Physics I Lecturer: Professor Nikta Fakhri Notes by: Andrew Lin Spring 2019 My recitations for this class were taught by Professor Wolfgang Ketterle. 1 February 5, 2019 This class’s recitation teachers are Professor Jeremy England and Professor Wolfgang Ketterle, and Nicolas Romeo is the graduate TA. We’re encouraged to talk to the teaching team about their research – Professor Fakhri and Professor England work in biophysics and nonequilibrium systems, and Professor Ketterle works in experimental atomic and molecular physics. 1.1 Course information We can read the online syllabus for most of this information. Lectures will be in 6-120 from 11 to 12:30, and a 5-minute break will usually be given after about 50 minutes of class. The class’s LMOD website will have lecture notes and problem sets posted – unlike some other classes, all pset solutions should be uploaded to the website, because the TAs can grade our homework online. This way, we never lose a pset and don’t have to go to the drop boxes. There are two textbooks for this class: Schroeder’s “An Introduction to Thermal Physics” and Jaffe’s “The Physics of Energy.” We’ll have a reading list that explains which sections correspond to each lecture. Exam-wise, there are two midterms on March 12 and April 18, which take place during class and contribute 20 percent each to our grade. There is also a final that is 30 percent of our grade (during finals week). The remaining 30 percent of our grade comes from 11 or 12 problem sets (lowest grade dropped). Office hours haven’t been posted yet; they will also be posted on the website once schedules are sorted out. 1.2 Why be excited about 8.044? One of the driving principles behind this class is the phrase “More is different.” We can check the course website for the reading “More is Different” by P.W. Anderson. Definition 1 Thermodynamics is a branch of physics that provides phenomenological descriptions for properties of macroscopic systems in thermal equilibrium. Throughout this class, we’ll define each of the words in the definition above, and more generally, we’re going to learn about the physics of energy and matter as we experience it at normal, everyday time and length scales. The 1 most important feature is that we’re dealing with the physics of many particles at once – in fact, we’re going to be doing a statistical description of about 1024 particles at once. It would be very hard and basically useless to try to use ordinary equations of motion to describe the behavior of each particle. Fact 2 Because thermodynamics is a study of global properties, like magnetism or hardness, the largeness of our systems will often actually be an advantage in calculations. The concept of time asymmetry will also come up in this class. In Newton’s laws, Maxwell’s equations, or the Schrodinger equation, there is no real evidence that time needs to travel in a certain direction for the physics to be valid. But the “arrow of time” is dependent on some of the ideas we’ll discuss in this class. Two more ideas that will repeatedly come up are temperature and entropy. We’ll spend a lot of time precisely understanding those concepts, and we’ll understand that it doesn’t make sense to talk about the temperature of an individual particle – it only does to define temperature with regards to a larger system. Meanwhile, entropy is possibly the most influential concept coming from statistical mechanics: it was originally understood as a thermodynamic property of heat engines, which is where much of this field originated. But now, entropy is science’s fundamental measure of disorder and information, and it can quantify ideas from image compression to the heat death of the Universe. Here’s a list of some of the questions we’ll be asking in this class: • What is the difference between a solid, liquid, and gas? • What makes a material an insulator or a conductor? • How do we understand other properties of materials, like magnets, superfluids, superconductors, white dwarfs, neutron stars, stretchiness of rubber, and physics of living systems? None of these are immediately apparent from the laws of Newton, Maxwell, or Schrodinger. Instead, we’re going to need to develop a theoretical framework with two main parts: • Thermodynamics: this is the machinery that describes macroscopic quantities such as entropy, temperature, magnetization, and their relationship. • Statistical mechanics: this is the statistical machinery at the microscopic level. What are each of the degrees of freedom doing in our system? These concepts have been incorporated into different other STEM fields: for example, they come up in Monte-Carlo methods, descriptions of ensembles, understanding phases, nucleation, fluctuations, bioinformatics, and (now the foundation of most of physics) quantum statistical mechanics. 1.3 An example from biology Many living systems perform processes that are irreversible, and the behavior of these processes can be quantified in terms of how much entropy is produced by them. Statistical physics and information theory help us do this! As a teaser, imagine we have a biological system where movement of particles is influenced by both thermal motion and motor proteins. By watching a video, we can track each individual particle, and looking at the trajectory forward and backward, and we can construct a relative entropy ⟨˙ S⟩ kB ≡D[pforward||pbackward] = X pf ln pf pb 2 which compares the probability distributions of forward and backward motion, and the point is that this relates to the entropy production rate of the system! But it’ll take us a lot of work to get to that kind of result, so we’ll start with some definitions and important concepts. To summarize this general overview, there’s two complementary paths going on here: Thermodynamics = ⇒global properties = ⇒(temperature, entropy, magnetization, etc.), and Statistical physics = ⇒(microscopic world to macroscopic world). We’ll also spend time on two “diversions:” quantum mechanic will help us construct the important states that we will end up “counting” in statistical physics, and basic probability theory will give us a statistical description of the properties we’re trying to describe (since entropy itself is an information theory metric)! To fully discuss these topics, we’re going to need some mathematics, particularly multivariable calculus. 1.4 Definitions We’ll start by talking about the basic concepts of heat, internal energy, thermal energy, and temperature. Definition 3 (Tentative) Thermal energy is the collective energy contained in the relative motion of a large number of particles that compose a macroscopic system. Heat is the transfer of that thermal energy. (We’ll try to be careful in distinguishing between energy and the transfer of that energy throughout this class.) Definition 4 Internal energy, often denoted U, is the sum of all contributions to the energy of a system as an isolated whole. This internal energy U is usually made up of a sum of different contributions: • Kinetic energy of molecular motion, including translational, vibrational, and rotational motion, • Potential energy due to interactions between particles in the system, and • Molecular, atomic, and nuclear binding energies. Notably, this does not include the energy of an external field or the kinetic and potential energy of the system as a whole, because we care about behavior that is internal to our system of study. Example 5 Consider a glass of water on a table, and compare it to the same glass at a higher height. This doesn’t change the internal energy, even though the glass has gained some overall gravitational potential energy. Definition 6 (Tentative) Temperature is what we measure on a thermometer. 3 As a general rule, if we remove some internal energy from a system, the temperature will decrease. But there are cases where it will plateau as well! For example, if we plot temperature as a function of the internal energy, it is linear for each phase state (solid, liquid, vapor), but plateaus during phase changes, because it takes some energy to transform ice to water to vapor. And now we’re ready to make some slightly more precise definitions: Definition 7 Let U0 be the energy of a system at tempreature T = 0. Thermal energy is the part of the internal energy of a system above U = U0. Notice that with this definition, the binding energy does not contribute to thermal energy (because that’s present even at T = 0, U = U0), but the other sources of internal energy (kinetic energy, potential energy) will still contribute. Definition 8 Heat is the transfer of thermal energy from a system to another system. This is not a property of the system: instead, it’s energy in motion! And heat transfer can occur as heat conduction or radiation, a change in temperature, or other things that occur at the microscopic level. 1.5 States and state variables The next discussion is a little bit more subtle – we want to know what it means for our system to be in a particular state. In classical mechanics, a state is specified by the position and velocity of all objects at time t. So if we’re given the two numbers {xi(t), ˙ xi(t)} for each i (that is, for every particle in our system), we have specified everything we might want to know. Meanwhile, in quantum mechanics, the state of a system is specified by quantum numbers: for example, |n1, · · · , nM⟩(for some nonnegative integers ni) is one way we might describe the system. But we have a completely different definition of “state” now that we’re in a macroscopic system: Definition 9 A system which has settled down is in a state of thermodynamic equilibrium or thermodynamic macrostate. Here are some of the characteristics of a system at thermal equilibrium: • The temperature of the system is uniform throughout the space. • More generally, at the macroscopic scale, no perceptible changes are occurring, though there are still changes at the microscopic level (like atomic or molecular movement). • The system is dynamic (meaning that the system continues to evolve and change at the microscopic level). It’s important to note that other properties of the system (that are not temperature) can be non-uniform at equilibrium! For example, if we mix water and oil, there will obviously be some differences in different parts of the system no matter how long we wait for the atoms to mix. Definition 10 State functions and state variables are properties that we can use to characterize an equilibrium state. 4 Some examples are pressure P, temperature T, volume T, the number of particles N, and the internal energy U. Note that some quantities are only defined for systems at thermal equilibrium, such as pressure and temperature, while others are defined for more general systems, such as volume. Another big part of this class is coming up with equations that relate these state functions: Example 11 The most famous equation of state, the ideal gas law (PV = NkBT), dictates the behavior of an ideal gas. Definition 12 A macrostate is a particular set of values for the state variables of a system. Meanwhile, a microstate tells us more at the particle level, specifying the state of each individual particle. For example, if we have a glass of water, we could (in principle) track each particle, writing down a microstate and describing our system at the microscopic level. But there are many different configurations that give a specific pressure and temperature, so a vast number of microstates can be consistent with a given macrostate. Definition 13 A macrostate’s multiplicity is determined by the number of microstates consistent with that macrostate, an ensemble is defined as the set of all possible microstates that are consistent with a given macrostate. This class will develop methods for describing such ensembles corresponding to a specific macrostate, and in particular one important consideration is that each microstate in the ensemble occurs with some probability. So we’ll be developing some probability theory in the next few lectures, and that will help us approach the physics more precisely. 2 February 6, 2019 (Recitation) 2.1 Introduction Usually, Professor Ketterle introduces himself in the first class, and this time he notices some people from 8.03. We’ll start today by giving a spiel for 8.044, and this class is particularly exciting to Professor Ketterle because he feels a connection to his research! Professor Fakhri does statistical physics in the classical world (in cells in aqueous solution), while Professor Ketterle examines systems at cold temperatures, which require quantum mechanics to describe. But it doesn’t always matter whether the microscopic picture is quantum or classical! In this class, we’re going to learn a framework for systems that have many degrees of freedom, but where we only know the macrostate (such as a liter of water at a certain pressure and temperature). Fact 14 Professor Ketterle’s research can be described as taking temperature T →0. His lab has gotten to temperatures of 450 picoKelvin, which is “pretty cold.” (In fact, according to Wikipedia, it’s the coldest temperature ever achieved.) For comparison, the background temperature in interstellar space is about 2.5 Kelvin, which is more than a billion times warmer than what’s been achieved in labs. 5 Low-temperature developments have opened up fields in physics today, because when we cool gas down to that regime, quantum mechanics becomes more apparent. Basically, when atoms have higher energy (at higher tempera-tures), they behave like classical particles that collide. But in reality, atoms should be thought of as de Broglie waves – it’s just that in the classical situations, the de Broglie wavelength λ = h mv does not play any role in the collisions, because λ is shorter than the size of the nucleus. But when we have an atom, which has low m, and cool it down so that v is small, the de Broglie wavelength can increase to the order of a micron or millimeter, which is large at atomic scales. So if we have a bunch of gas particles, and each particle’s wave is localized, new forms of matter can form with completely new properties, and that’s what makes low-temperature physics interesting. Remark 15. When Professor Ketterle was an undergraduate student, he found statistical mechanics fascinating. He found it attractive that we can predict so much in physics given so few pieces of information: just using statistics and ideas of “randomness” (so that we have isotopy of velocity), we can make many predictions: find limits on efficiency of engines, derive gas laws, and so on. So we’ll get a taste of that through this class! Professor Ketterle wants recitations to be interactive, so we should bring questions to the sessions. (He does not want to copy to the blackboard what we can just read from solutions.) Concepts are important: it’s important to address all of the important key points so that we can understand derivations. However, to become a physicist, we do need to do problems as well. So in order to compensate for the focus on problems in lecture, usually we will not have material “added.” So Professor Ketterle will prepare topics that help us get a deeper understanding or general overview of concepts, but we can also bring questions about homework if we have any. 2.2 Review of lecture material There were many buzzwords covered during the first lecture. We started thinking about what it means to have a state (microstates and macrostates, both classical and quantum), ensemble, energy (thermal energy Q, internal energy U, heat, and work), and temperature (which is particularly interesting in the limit T →0 or T →∞). Basically, we have introduced certain words, and it will take some practice to get fluent in those ideas. Question 16. How do we explain to a non-physicist why there is an “absolute zero” lowest temperature? (This is a good question to ask a physicist.) In fact, how is temperature actually defined? We still don’t have a formal definition of temperature, but let’s look at an example of a real-life situation in which temperature is very relevant: an ideal gas. Fact 17 Here’s an important related fact: starting on May 20, 2019, the kilogram will be defined in terms of the fixed constant ℏ, and along with that, the definition of a Kelvin will also change to depend on the Boltzmann constant kB rather than the triple point of water. So the Boltzmann constant will soon specify our units – it’s some constant that is approximately 1.38 · · ·×10−23J/K (and will come up frequently in the rest of this class), and now temperature is related to a measure of energy: kBT = [J]. 6 In an ideal gas (in which we neglect weak interactions between particles), we’ll make the definition E = 1 2mv 2 ≡3 2kBT, where v 2 will be defined next lecture. If we take all of this for granted, the lowest possible temperature must occur when there is no kinetic energy: if v 2 →0, T →0. And this also explains why there is no negative temperature in the ideal gas – we can’t have less than zero energy. So absolute zero is essentially created by definition: it’s a situation in which there is no kinetic energy for any of our particles. For an analogous situation, we can measure the pressure in a vacuum chamber, which is proportional to the density of particles. And thus the lowest pressure is zero, since we can’t have negative particles. On the other hand, what’s the highest temperature we can achieve? In principle, there is no upper limit: we can make kinetic energy per particle arbitrarily large. We can add some entertainment to the situation as well: even though the velocity v is upper bounded by c, we do have a divergent expression for relativistic energy which is not just 1 2mv 2: KE = 1 2 m0c2 q 1 −v 2 c2 . In other words, we can keep adding energy to a particle, and it will just get heavier (without going over the speed of light). And in the lab, the highest temperatures we’ve ever achieved are around 2 × 1012 K in particle accelerators. In general, if we take two nuclei and smash them together when moving near the speed of light, temperature changes happen when during the actual collision. Then energy is converted into particles, and we have a piece of matter in which we are almost at thermal equilibrium. But what really happens at 1012 Kelvin is interesting on its own. As our temperature rises, a solid melts, a liquid evaporates, molecules dissociate, atoms ionize into electrons and ions, and then ions lose more and more electrons until they are bare nuclei. If we go hotter, the nuclei dissociate as well! All of this behavior does actually happen in white dwarfs and neutron stars, but if we go even hotter, the protons and neutrons will dissociate into quarks, and we get a soup of quarks and gluons. Overall, it’s interesting that we’ve actually achieved room temperature times 1010, and we’ve also achieved room temperature divided by 1010, and there is physical behavior going on at both temperatures. And now we can talk about what “negative temperature” actually means – we’ll have a more rigorous discussion when we study spin systems, but it’s good for us to know that there are some magnetic systems that can reach infinite temperatures at finite energies. What’s basically happening is that we’re traversing 1 T from ∞to 0: when we cross over 0, we can get into negative temperatures, and thus negative temperatures are in some way “even hotter” than infinite temperatures! Expressions of the form e−E/(kBT ) are going to show up frequently in this class, so we will actually get 1 T reasonably often in our calculations. And we’ll understand why this happens as probability theory comes into the picture! 3 February 7, 2019 The first lecture was a glossary of the terms we will see in this class; we’ll be slowly building up those concepts over the semester. The first problem set will be posted today, and the deadline will generally be the following Friday at 9pm. Lecture notes have been uploaded, and they will generally be uploaded after each class (as will some recitation materials) under the Materials tab of the LMOD website. For example, starting next Monday in recitation, there will be a review of partial derivatives, and the notes for those are posted already. Also, we have two graduate TAs this semester! Pearson (from last semester’s 8.03 team) will also be available to 7 help. General office hours will also be posted by tomorrow, but in general, we should use email instead of Piazza. 3.1 Overview We will always begin lecture with a small overview of what will be covered. Statistical physics and thermodynamics are for bringing together the macroscopic and microscopic world, and we’re going to start by defining state functions like pressure and temperature using a tractable, simply-modeled system and working from first principles. Essentially, we will use a monatomic ideal gas to define temperature and pressure, and then we will derive the ideal gas law. From there, we’ll see how to make empirical corrections to have a more realistic understanding of a system (for example, a van der Waals gas). We’ll also briefly talk about the equipartition theorem, which lets us connect temperature to energy, as well as the first law of thermodynamics, which is basically a restatement of conservation of energy. 3.2 The simplest model Definition 18 A monatomic ideal gas is a system with N molecules, each of which is a single atom with no internal dynamics (such as rotation or vibration). The molecules collide elastically as point particles (and take up no space), and the only energy in the system is kinetic energy. So putting in our definitions, the kinetic energy, thermal energy, and internal energy are all essentially the same thing, and they are all equal to U = N X i=1 Ei = m 2 N X i=1 v 2 ix + v 2 iy + v 2 iz if all molecules have the same mass. Now assuming that we have an isotropic system, we can assume the three coordinates have equal averages, and we can define an average (squared) velocity v 2 ≡⟨v 2 ix⟩= ⟨v 2 iy⟩= ⟨v 2 iz⟩. Plugging this in, the average internal energy is then Uavg = N⟨E⟩= 3 2Nmv 2. Definition 19 The temperature is a measure of the thermal energy of the system, given by mv 2 ≡kBT where kB is a proportionality constant and T has units of Kelvin (in degrees above absolute zero). The Boltzmann constant kB has units of energy per temperature, and it is experimentally about 1.381×10−23J/K. Plugging this in, we find that the internal energy of an ideal gas is U = 3 2NkBT. 8 Fact 20 Usually chemists write this slightly differently: N is defined to be NAn, where NA is Avogadro’s number ≈ 6.023 × 1023mol−1, and n is the number of moles of the gas. Then the ideal gas constant is defined to be R ≡NAkB ≈8.314J/mol K, and our equation can be written as U = 3 2NkBT = 3 2nRT. The idea is that each of the three dimensions is contributing an equal amount to the energy in the system. We’ve used the fact that particles have energy to define a temperature, and now we’ll similarly use the fact that particles have momentum as well to define pressure. Consider a container shaped like a box with a piston as one of the walls (in the x-direction). We know that by Newton’s law, the force can be described as Fx = dpx dt , where p = mvx is the momentum in the x-direction for one of the particles. Since the piston will barely move, we can just say that the particle will reverse x-momentum when it bounces off, but has no change in the other two directions: ∆px = 2mvx, ∆py = ∆pz = 0. If we let the cross-sectional area of the piston-wall be A and the length of the box in the x-direction be ℓ, then the time between two collisions with the piston is ∆t = 2ℓ vx (since it must hit the opposite wall and bounce back). So now plugging in our values, the average force from this one molecule is Fx = ∆px ∆t = mv 2 x ℓ . Assuming no internal molecular collisions (since we have an ideal gas), the total force on the piston for this system is then Fx = N X i=1 Fxi = Nm ℓv 2 = N ℓkbT by our definition of temperature. So now the pressure on the piston, defined to be force per unit area, is P = Fx A = NkBT ℓA = NkBT V Now we’re making some assumptions: if we say that collisions and interactions with the wall don’t matter, and that the shape of the container does not matter (an argument using small volume elements), we can rearrange this as PV = NkBT , which is our first equation of state for the class. (As a sidenote, the shape of the container will matter if our system is out of equilibrium, though.) So pressure, volume, and temperature are not independent: knowing two of them defines the third in our ideal gas system, and we’re beginning to find a way to relate our state functions! 9 3.3 An empirical correction Let’s start modifying our equation of state now – we’re going to use the chemistry version of the ideal gas law, where n is measured in moles. One key assumption that we have right now is that the particles have no volume: we have to make some corrections if we don’t have point particles any more. So we change our volume: letting b be some measure of how much volume is taken up by the particles, we replace V →V −nb. Also, some particles may have attractive intermolecular forces, and to introduce this, we claim the pressure will change as P = nRT V −a  n V 2 . The constants a and b are empirically measured in a lab, but the point is that these modifications give us the van der Waals equation  P + a n2 V 2  (V −nb) = nRT. This means the effective volume for a real gas is smaller than an ideal gas, but the pressure can be larger or smaller than an ideal gas because we could have attractive or repulsive molecular interactions. 3.4 The equipartition theorem If we take another look at the equation for internal energy U = 3 2NkBT = 3N 1 2kBT  , notice that our system has 3N degrees of freedom: one in each of the x, y, and z coordinates for each of the particles. Proposition 21 (Classical equipartition theorem) At thermal equilibrium at temperature T, each quadratic degree of freedom contributes 1 2kBT to the total internal energy U of the system. This is important for being able to consistently define temperature! Unfortunately, this is only true in classical limits at high temperatures. And we should make sure we’re precise with our language: Definition 22 A degree of freedom is a quadratic term in a single particle’s energy (or Hamiltonian). Examples include: • translational (in each coordinate) about the center of mass: 1 2m v 2 x + v 2 y + v 2 z  , • rotational (in each axis): 1 2 ℓ2 x Ix and so on, and • vibrational: 1 2m ˙ x2 + 1 2kx2, imagining a molecule with two atoms in simple harmonic oscillation. Example 23 Let’s try writing down the different degrees of freedom for molecules of a diatomic gas. 10 In such a system, we have 3 translational degrees of freedom (looking at the center of mass), 2 rotational degrees of freedom (we don’t have the third because there’s no moment of inertia about the axis connecting the two atoms), and 2 vibrational degrees of freedom (coming from the simple harmonic oscillator of the two atoms stretching). Thus, by the equipartition theorem, we already know that we’ll have U = 7N 1 2kBT  , and that’s the power of the equipartition theorem: it allows us to have a general method for relating energy and temperature. And as an exercise, we should try to figure out why a simple crystalline solid has internal energy U = 3NkBT, using the same kinds of argument. 3.5 The first law of thermodynamics We now have a relationship between thermal energy and temperature, and our next step is to think about how the energy of a system can change. We’ll start by trying to define a relationship between work and pressure. Let’s go back to the piston wall that we used to derive the ideal gas law: the differential work being done on the piston by a particle is dW = F dx = PA dx = P dV. We will use the convention in this class where whenever dV < 0, work is being done on the system. So the change in the internal energy of the system is the mechanical work done, and dU = −P dV (with the negative sign because work is being done by, rather than on, the system.) But there are also other ways to transfer energy, particularly through heat (which we denote with the letter Q). We’ll use the convention is that heat flow into the system is positive, so dU = dQ . So we can write these together to get a total change in internal energy dU = dQ + dW. In other words, the internal energy of the system changes if we add infinitesimal heat to it, or if the system does work externally. We can also add particles to the system to further modify this first law: we’ll see later on that we sometimes also get a contribution of the form dU = µ dN where µ is the chemical potential, then dU = dQ + dW + µ dN. Explicitly, the whole point of this kind of statement is to have an energy conservation law, but implicitly, we also have that U, the internal energy of the system, is also a state function. But we should keep in mind that work and heat are path-dependent quantities, so the expressions dQ and dW are inexact differentials. And in particular, W, Q are not state functions! Example 24 Consider an evolution of a system as follows. Start in state 1, with N1 particles, a temperature of T1, and a volume of V1. This tells us that we have some internal energy U(N1, T1, V1). We can take this to state 2 by adding some heat, so now we have N2 particles, a temperature of T2, and a volume of V2, giving us a new internal energy. Finally, perform some work on the system to take us back to state 1. The idea is that U is an exact differential, so the internal energy doesn’t depend on the path taken. But dW and dQ are both inexact and path-dependent quantities – the example above showed us that dU = 0, but the work and 11 heat done depend on what state 2 is. So from here on out, we will use the notation ¯ dW and ¯ dQ, and now we write the first law as dU = ¯ dQ + ¯ dW . So calculus tells us that dU can be obtained from differentiation (of some other function), while ¯ dQ and ¯ dW cannot. In general, state functions can be divided into generalized displacements and generalized forces, which will be denoted {x} and {j}, respectively. These xis and jis are conjugate variables, and the idea is to write our differential work as ¯ dW = X i ji dxi. Here are some sample state functions that we’ll be working with throughout this class Forces (ji) Displacements (xi) Pressure P Volume V Tension F Length L Surface tension σ Area A Magnetic field H Magnetization M Chemical potential µ Number of particles N The quantities under “displacements” are generally extensive, while the quantities under “force” are intensive. We’ll use these words more as the class progresses, but the basic point is that scaling the system up changes our generalized displacements, but not the generalized forces. 4 February 11, 2019 (Recitation) 4.1 Questions and review We’ll begin with a few ideas from the homework. If we want to calculate the work done by moving from one state to another, we can integrate along the path taken and compute W = I final initial P dV. Many such problems are solved using PV diagrams, which plot each state based on their pressure (on the y-axis) and volume (on the x-axis), and knowing P and V tells us the temperature T (through the equation of state) as well. And if we have an ideal gas, lines PV = NkBT on the PV diagram are where the temperature is constant. For example, if we have an isothermal expansion along one of these lines, we find that Z P dV = Z NkBT V dV = NkBT(ln Vf −ln Vi). There are other kinds of work that can be specified: for instance, isobaric compression is done under constant pressure, so the work is just P∆V . But we’ll come back to all of this later! Remark 25. State functions are all the parameters that characterize a system. So for an ideal gas, pressure, volume, and temperature are enough – if we are asked to find the final state, we just need to find the values of those parameters. 12 Fact 26 The word adiabatic (referring to a process in which a system evolves) has two different definitions in statistical physics. In one definition, no heat is exchanged, meaning that ¯ dQ = 0 throughout the process. (This can happen if our system has insulating walls, so no heat is transferred in or out of the system, or if a process proceeds quickly enough so that heat can’t move.) But in the other definition, our process is slow enough so that the system is always in equilibrium. And the main feature of this definition (which isn’t true in the other) is that entropy is conserved. Adiabaticity in quantum mechanics means that a particle doesn’t deviate from its quantum state, because the process happens very gradually. (If we try to evolve too slowly, though, we will get noise that also interferes with the system.) The bottom line is that adiabaticity forces our system to act “not too slow and not too fast,” so that we get the desired constraints. 4.2 Energy Let’s start from the first law of thermodynamics, stated as dU = ¯ dW + ¯ dQ. As physicists, we don’t have access to absolute truth: we do our best to find better and better approximations. So Professor Ketterle doesn’t always like it when we call things “laws:” for example, why are we trying to test Coulomb’s law to the tenth decimal place, and why do we do it for two electrons that are very close? Regardless, the “law” above is a statement about energy. Question 27. What’s a form of energy that is internal energy but not thermal energy? Two possible answer are “the binding energy of the nucleus” or “the mass energy,” but these aren’t exactly correct. Thermal energy is supposed to “come and go” as our system heats up, so let’s think about a system of water molecules at increasing temperature. At first, our molecules gain kinetic energy, and then after we continue to heat, chemical energy will change through dissociation. So in this system, the binding energy of hydrogen is “reaction enthalpy,” which is indeed considered in thermal energy. And similarly, if we increase temperature so that the kinetic energy is comparable to rest mass energy, we get issues with relativity. If two such particles collide, they can create a particle-antiparticle pair, and in this regime, even the rest mass becomes part of a dynamic process. Therefore, that rest mass can become thermal energy as well. One takeaway we can have is that U = Ethermal + U0 for some constant U0, and we basically always only care about differences in energy anyway. So this distinction between “thermal energy” and “internal energy” isn’t really that important in Professor Ketterle’s eyes. 4.3 Scale of systems Question 28. Can a single particle have a temperature? That is, can we have an ensemble consisting of one particle? Normally, we are given some constant P, V, T, U, and a microstate is specified by a set of positions and momenta {xi, pi : 1 ≤i ≤N}, where N is a large number of particles. It turns out, though, that even single particles can be 13 thermal ensembles! For example, if we connect a particle in a box to a certain thermal reservoir at temperature T, we can find a “Boltzmann probability distribution” ∝e−E/(kBT ) for being at state E (this is a point which we’ll study later). So having an ensemble just means we have many copies of a system that are equally prepared macroscopically, regardless of how many particles this system has, as long as we follow all of the important laws. And in particular, remember that in an ideal gas, we’ve assumed the particles are not interacting! So it’s perfectly fine to take N →1 for an ideal gas; rephrased, an ideal gas is just N copies of a single-particle system. Remark 29. Schrodinger once said that the Schrodinger equation only describes ensembles when measurements are applied many times. He made the claim that the equation would not apply to just one particle, but recently, single photons, atoms, and ions were observed repeatedly, and it was shown that the quantum mechanical ideas applied there too. So we may think it’s nonsense that statistics can apply to a single particle, but we can often study a complicated system by simplifying it into multiple copies of a simple one. 4.4 Partial derivatives Consider the two equations dz dx = dz dy dy dx , dz dx = −dz dy dy dx . We can ask ourselves “which one is correct?”. One general rule to keep in mind is that in each field of physics, we need to learn some mathematics. In particular, two tools we’ll need to learn for this class are partial derivatives and statistics. In the handout posted online, we’ll see the second statement, but of course we need to ask ourselves why we don’t cancel the dys like we have done in ordinary calculus. The key point is that we use the two equations in different situations. The first equation is valid when z is a function of x, but we have it written as an implicit function y. Rephrased, if we have a function z(y(x)), such as exp(sin x), then indeed the chain rule indeed tells us that dz dx = dz dy dy dx . This is true for a function that depends on a single independent parameter. But on the other hand, suppose we have a function where x and y are independent variables: that is, we have z(x, y)? (For example, pressure is a function of volume and temperature.) Now z can change by the multivariable chain rule, and we can say that dz = ∂z ∂x y dx + ∂z ∂y x dy. Often we’ll have a situation where we want to keep z constant: for example, we might be keeping the pressure constant as we heat up our system, and we’re thinking about pressure as a function of other state variables. In a situation like that, we have 0 = ∂z ∂x y dx + ∂z ∂y x dy, so now x and y must be changed in a certain ratio: dy dx = − ∂z ∂x y ∂z ∂y x 14 And the left hand side of this equation, in more rigorous language, is just ∂y ∂x as we keep z fixed. So the second equation is now true, and the moral of the story is that we need to be very careful about what variables are being kept constant. We’ll do much more study of this in the following weeks! 5 February 12, 2019 We’ll start with some housekeeping: our first problem set is due on Friday, and there will be office hours for any questions we have. If we click on the instructor names on the course website, we can see what times office hours are being held. (Hopefully we all have access to the online class materials: if there are any problems, we should send an email to the staff.) Last lecture, we introduced thermodynamics with a simple and tractable model (the ideal gas). Once we defined pressure and temperature, we derived an equation of state, and we learned that we can empirically modify the ideal gas law to capture real-life situations more accurately. Next, we introduced the first law of thermodynamics, which is essentially conservation of energy. We learned that there are many ways to do work, and we can write the infinitesimal work as the product of a “force” and its corresponding “conjugated variable.” This led us to the generalized first law dU = ¯ dQ + ¯ dW, ¯ dW = X i ji dxi, where each ji is a generalized force and xi is its conjugated generalized displacement. In recitation, we reviewed some material about partial derivatives, and there will be a “zoo” of them in this class. Any macroscopic quantity can be found by taking derivatives of “free energy” (which will be defined later as well). For example, we can take derivatives of energy to find temperature and pressure, which will be helpful since we can use statistical physics to find general quantities like the free energy or entropy. But again, this is all a preview for what will become more rigorous later. 5.1 Experimental properties Definition 30 A response functions are quantities that change when parameters of a system are adjusted, and they are used to characterize the macroscopic properties of that system. Basically, we introduce a perturbation to a system, and we can then observe the response in our measurements. Example 31 (Heat capacity) Suppose we add some heat to a system, and we want to keep track of what happens to the temperature. We need to be careful, because the system can change while keeping pressure or volume constant? Both are useful quantities, and we’ll define the heat capacities CV ≡¯ dQ dT V , CP ≡¯ dQ dT P . (These can be thought of as variables for a gas on which we perform experiments.) 15 Example 32 (Force constant) Suppose we apply a force F on our system, and we want to see the displacement x that results from this external force. (This is a generalization of a spring constant.) We can therefore define an effective force constant via the equation dx dF ≡1 k . For example, we can define the isothermal compressibility of a gas via KT ≡−1 V ∂V ∂P T . This is, again, something we can measure experimentally. Example 33 (Thermal responses) The expansivity of a system is defined as αP = 1 V ∂V ∂T P . And finally, if we have an equation of the form dU = ¯ dQ + X i ji dxi, it makes sense to try to write ¯ dQ in a similar way as well. And it turns out that if we treat T, the temperature, as a force, we can find a conjugate displacement variable S (called the entropy)! So for a reversible process (which we will discuss later), we can write ¯ dQ = T dS, and now all of our contributions to the internal energy are in the same form. 5.2 Experimental techniques Now that we’ve discussed the abstract quantities that we care about measuring, we should ask how we actually measure pressure, volume, and temperature for a given system. We generally deal with quasi-equilibrium processes, in which the process is performed sufficiently slowly that the system is always (basically) at equilibrium at any given time. This means that thermodynamic state functions do actually exist throughout our evolution, so we can always calculate well-defined values of P, V, T, and other state functions. And the work done on the system (which is the negative of the work done by the system) is related to changes in thermodynamic quantities, as we wrote above. Example 34 Let’s say we want to measure the potential energy of a rubber band experimentally, and we do this by stretching the rubber band and applying a force. The idea is that if we are performing the stretching slowly enough, at any point, the force that we apply is basically the same as the internal force experienced by the system. That means that we can indeed take the force we’re applying to the rubber band, and we will find that U = R F dℓ. 16 5.3 PV diagrams Since our state functions are only defined in equilibrium states, all derivatives are also only described in the space of equilibrium states. Definition 35 In a PV diagram, the pressure P is plotted on the y-axis, the volume V is plotted on the x-axis, and every equilibrium state lies somewhere on the graph. The work done by or on the system as it transitions from a state I to a state II is defined via Wby = −Won = Z II I P dV. The idea here is that (as we discussed with the ideal gas) pressure is force per area, and volume is length times area, so this integral is basically computing R F dx. There are ways to go from one state to another without being in equilibrium states along the way as well: for example, if we have sudden free expansion, there is no heat being exchanged and no work done on or by the system, so ∆Q = ∆W = 0 = ⇒UA = UB. This tells us that U is a function of only the temperature of the gas (because it doesn’t change even when we change our pressure and volume). Example 36 (Isothermal expansion) Consider a situation in which a gas moves along an isotherm in the PV diagram: for an ideal gas, the equation of this isotherm is just PV = NkBT. As the name suggests, we’re keeping the temperature of the system constant while we compress the ideal gas. If we start with a volume V1 and pressure P1, and we end up at volume V2 and pressure P2, then the work done on the system is − Z V2 V1 P dV = Z V1 V2 NkBT V dV = NkBT(ln V2 −ln V1) = NkBT ln V1 V2 , since N, kB, and V are all constants independent of T. If we define r = V1 V2 , and we want to know how much heat is required for this process, we must have 0 = dU = ⇒ Q = −W = −NkBT ln r , because the internal energy U(T) does not change if T is fixed. Example 37 (Adiabatic compression) In an adiabatic process, there is no heat added or removed from the system (for example, this happens if we have an isolated container). Since ¯ dQ = 0, the first law of thermodynamics tells us that dU = ¯ dW = −P dV. 17 Since we have an ideal gas, we also know that PV = NkBT, U = f 2NkBT, where f is the number of active degrees of freedom for the molecules of the gas. We’ll now manipulate these equations a bit: in differential form, we have (using the product rule) that P dV + V dP = NkBdT, dU = f 2NkBdT. Combining these equations, we have a relation between changes in internal energy and the state variables: dU = f 2(P dV + V dP). So now since dU = −PdV from above, −P dV = f 2(P dV + V dP) = ⇒(f + 2)P dV + f V dP = 0. Definition 38 The adiabatic index for a gas with f degrees of freedom is given by γ = f + 2 f . Now if we rearrange the equation and integrate, 0 = γP dV + V dP = ⇒dP P = −γ dV V = ⇒ P P1 = V1 V γ . This tells us that PV γ is constant, and equivalently that TV γ−1 and T γP 1−γ are constant as well. So now the rest is just integration: the work done on the system is W = − Z V2 V1 P dV = −P1V γ 1 Z V2 V1 dV V γ = −P1V γ 1 (V 1−γ 2 −V 1−γ 1 ) . Plugging in our definition of r, W = NkBT1 γ −1 (r γ−1 −1). This quantity depends on the number of degrees of freedom in the system, but we can see that in general, the work done for an isothermal process is less than for an adiabatic process! (This is true because the PV curves for PV γ = c are “steeper” in a PV diagram than those for PV = c, so we are also increasing the temperature, meaning more work is required for us to get the same change in volume.) Fact 39 It’s hard to design an adiabatic experiment, since it’s hard to insulate a system completely from its surroundings. Example 40 (Isometric heating) In an isometric heating, we keep the volume of a system constant while we increase the heat (temperature goes up). 18 In a PV diagram, this corresponds to moving vertically up. Since there is no change in volume, there is no work done on the system, and thus dU = ¯ dQ: any internal energy change is due to addition of heat. But this is a measurable quantity: we can write ¯ dQ dT V = ∂U ∂T V = CV , and we can measure the change in internal energy to find the specific heat capacity CV experimentally. Since the energy is dependent only on temperature, for an ideal gas, U = f 2NkBT = ⇒ CV = f 2NkB . We can also define ˆ cV (per molecule) as CV N = f 2kB. Finally, we have one more important process of change along PV diagrams: Example 41 (Isobaric heating) This time, we keep the pressure constant and move horizonally in our PV diagram. Let’s differentiate the first law of thermodynamics with respect to T: since ¯ dW = −PdV , ¯ dQ = dU −¯ dW = ⇒ ¯ dQ dT P = ∂U ∂T P + P ∂V ∂T P . Since pressure is constant, our internal energy U is a function of T and V , and taking differentials, dU = ∂U ∂T V dT + ∂U ∂V T dV. Dividing through by a temperature differential to make this look more like the equation we had above, ∂U ∂T P = ∂U ∂T V + ∂U ∂V T ∂V ∂T P . Combining our equations by substituting the second equation into the first, ¯ dQ dT P = CV +  P + ∂U ∂V T  ∂V ∂T P . But the left side is CP , so we now have our general relation between CP and CV CP = CV +  P + ∂U ∂V T  ∂V ∂T P . Example 42 Let’s consider this equation when we have an ideal gas. Then U is only a function of T, so ∂U ∂V while keeping T constant is zero. This leaves CP + CV + P ∂V ∂T P 19 and since we have an ideal gas where PV = NkBT, ∂V ∂T at constant P is just NkB P , and we have CP = CV + NkB. Example 43 What if we have an incompressible system, like in solids or liquids? Then the volume does not change with respect to temperature noticeably, so CP = CV + P ∂V ∂T P . Defining αP = 1 V ∂V ∂T P , CP = CV + PV αP . For ideal gasses, αP = 1 T , and at room temperature this is about 1 300K−1. For solids and liquids, the numbers are smaller: αP = 10−6K−1 for quartz and αP = 2 × 10−4K−1 for water. This essentially means CP ≈CV for solids and liquids! Let’s look back again at isometric heating. We found that we had a state function U, which is exactly the amount of heat we added to the system. So in this case, we can directly measure dU = ¯ dQ. Are there any new state functions such that the change in heat for isobaric heating is the same as the change in that state function? That is, is there some quantity H such that ¯ dQ = dH? The answer is yes, and we’ll discuss this next time! It’s enthalpy, and it is H ≡U + PV . 6 February 13, 2019 (Recitation) Today, we have a few interesting questions, and we’ll be using clickers! We can see how we will respond to seemingly simple questions, because professor Ketterle likes to give us twists. 6.1 Questions Let’s start by filling in the details of “adiabatic” processes. In thermodynamics, there are two definitions of “adiabatic” in different contexts. Fact 44 “dia” in the word means “to pass through,” much like “diamagnetic.” In fact, in Germany, slide transparencies are called “dia”s as well. So “adia” means nothing passes through: a system in a perfectly insulated container does not allow transfer of heat. Adiabatic will mean thermally isolated in general! On the other hand, we have adiabatic compression (which we discussed in class), in which we have an equation of the form PV γ = c. But what is adiabatic expansion? It sounds like it should just be a decompression: perhaps it is just a reverse of the adiabatic compression process. But this isn’t quite right. In compression, we do the compression slowly: we assume the equation of state for the ideal gas is always valid, so we are always at equilibrium. Indeed, there also exists an adiabatic expansion that is very slow. But in problems 20 like the pset, we can have sudden changes: a sudden, free expansion is not in equilibrium all the time, and it is not reversible! Fact 45 Adiabatic compression increases the temperature, and slow adiabatic expansion does the opposite. But in sudden free expansion, the internal energy of the system is 0 (as ¯ dW = ¯ dQ = 0). So the temperature is constant in free expansion. All three processes have ¯ dQ = 0, since there is no heat transfer. The point is to be careful about whether we have reversible processes, since different textbooks may have different interpretations! We’ll talk about entropy later, but the key idea is that the slow adiabatic compression and expansion are isentropic: dS = 0. Fact 46 For example, if we change the frequency of our harmonic oscillator in quantum mechanics slowly, so that the energy levels of our system does not jump, that’s an adiabatic process in quantum mechanics. Example 47 Given a PV diagram, what is the graph of sudden, free expansion? We start and end on the same isotherm, since the temperature is the same throughout the process. But we can’t describe the gas as a simple equation of state! In fact, we’re not at equilibrium throughout the process, so there is no curve on the PV diagram. After all, the work done W = R PdV has to be zero. In other words, be careful! 6.2 Clickers Let’s talk about the idea of “degrees of freedom.” Molecules can look very different: they can be monatomic, diatomic, or much larger. The degrees of freedom can be broken up into • center of mass motion • rotational motion • vibrational motion. There are always 3 center of mass degrees of freedom, and let’s try to fill in the rest of the table! (We did this using clickers.) COM ROT VIB total atom 3 0 0 3 diatomic 3 2 1 6 CO2 3 2 4 9 H2O 3 3 3 9 polyatomic 3 3 3N −6 3N Some important notes that come out of this: • There are 2 rotational degrees of freedom for diatomic and straight triatomic molecules: both axes that are not along the line connecting the atoms work. As long as we can distinguish the three directions, though, there are 3 rotational degrees of freedom. 21 • Here, we count degrees of freedom as normal modes (which is different from 8.223). Recall that in 8.03, we distinguished translational from oscillatory normal modes. • Water is a triatomic molecule with three modes: the “bending” mode, the “symmetric” stretch, and the “anti-symmetric” stretch. • Carbon dioxide has 4 modes: the symmetric stretch, the asymmetric stretch, and two bending modes (in both perpendicular axes). In classical mechanics, if we’re given one particle, we can write 3 differential equations for it: each coordinate gets a Newton’s second law. That’s why we have 3 total degrees of freedom. Similarly, with two particles, we have 6 total degrees of freedom, and the numbers should add up to 3N in general. This lets us make sure we don’t forget any vibrational modes! 6.3 Wait a second... Notice that this definition of “degrees of freedom” is different from what is mentioned in lecture. Thermodynamic degrees of freedom are a whole different story! Now let’s change to using f , the thermodynamic degrees of freedom. Recall that we define γ = f +2 f . f measures the number of quadratic terms in the Hamiltonian, and as we will later rigorously derive, if the modes are thermally populated, the energy of each degree of freedom is kBT 2 . But the vibrations count twice, since they have both kinetic and potential energy! We’ll also rigorously show this later. So it’s time to add another column to our table: COM ROT VIB total thermodynamic atom 3 0 0 3 3 diatomic 3 2 1 6 7 CO2 3 2 4 9 13 H2O 3 3 3 9 12 polyatomic 3 3 3N −6 3N 6N −6 and as we derived in lecture, E = f kBT 2 , CV = f kB 2 N. However, keep in mind that this concept breaks down when we add too much energy and stop having well-defined molecular structure. Finally, let’s talk a bit about adiabatic and isothermal compression. Example 48 Let’s say we do an isothermal compression at T1, versus doing an adiabatic compression to temperature T2. We measure the work it takes to go from an initial volume V1 to a final volume V2 under both compressions. The total work is larger for the adiabatic process, since the “area under the curve is larger,” but why is this true intuitively? One way to phrase this is that we press harder, and that means there is more resistance against the work done. So now let’s prepare a box and do the experiment. We find that there is now no difference: why? (Eliminate the answer of bad isolation.) • The gas was monatomic with no rotational or vibrational degrees of freedom. 22 • Large molecules were put in with many degrees of freedom. • The gas of particles had a huge mass. • This is impossible. This is because γ = f +2 f ≈1 if f is large! Intuitively, the gas is absorbing all the work in its vibrational degrees of freedom instead of actually heating up. 7 February 14, 2019 Remember the pset is due tomorrow night at 9pm. As a reminder, the instructors are only accepting psets on the website: make a pdf file and submit them on LMOD. This minimizes psets getting lost, and it lets TAs and graders make comments directly. Fact 49 Don’t use the pset boxes. I’m not really sure who put them there. The solutions will become available soon after. Today, we’re starting probability theory. The professor uploaded a file with some relevant information, and delta functions (a mathematical tool) will be covered later on as well. Also, go to the TA’s office hours! 7.1 Review from last lecture We’ve been studying thermodynamic systems: we derived an ideal gas law by defining a pressure, temperature, and internal energy of a system. We looked at different processes that allow us to move from one point in phase space (in terms of P, V ) to another point. Thermodynamics came about by combining such motions to form engines, and the question was about efficiency! First of all, let’s review the ideas of specific heat for volume and pressure: • Remember that we discussed an isometric heating idea, where the volume stays constant. We could show that dV = 0 = ⇒¯ dW = 0 = ⇒dU = ¯ dQ, which means we can actually get access to the change in internal energy (which we normally cannot do). We also found that ¯ dQ dT V = dU dT V = CV . • When we have constant pressure (an isobaric process), we don’t quite have dU = ¯ dQ, but we wanted to ask the question of whether there exists a quantity H such that dH = ¯ dQ|P . The idea is that d(PV ) = V dP + PdV, ¯ dQ|P = dU|P + PdV |P and this last expression is just dU|P + d(PV )p since P is constant. So combining all of these, d(U + PV )|P = ¯ dQ|P = ⇒H ≡U + PV. 23 Definition 50 H is known as the enthalpy. It is useful in the sense that ¯ dQ dT P = ∂H ∂T P = CP . This is seen a lot in chemistry, since many experiments are done at constant pressure! We can write a general expression that combines those two: CP = CV +  P + ∂U ∂V T  ∂V ∂T P and for an ideal gas, this simplifies nicely to CP = CV + NkB. Last lecture, we also found that CP CV = f + 2 f ≡γ, which is the adiabatic index. For a monatomic ideal gas, f = 3 = ⇒ γ = 5 3, and for a diatomic ideal gas, f = 7 = ⇒γ = 9 7. 7.2 Moving on Fact 51 If you plot heat capacity CV per molecule as a function of temperature, low temperatures have CV ≈3 2 (only translational modes), corresponding to γ = 5 3, but this jumps to CV ≈5 2 = ⇒γ = 7 5 for temperatures between 200 to 1000 Kelvin. Hotter than that, vibrational modes start to come in, and CV increases while γ approaches 1. In a Carnot engine, we trace out a path along the PV diagram. How can we increase the efficiency? There are two important principles here: energy is conserved, but entropy is always increasing. Definition 52 (Unclear) Define the entropy as ∆S = Q T . But this doesn’t give very much physical intuition of what entropy really is: it’s supposed to be some measure of an “ignorance” of our system. Statistical physics is going to help us give an information theoretic definition later: S = −kB⟨ln Pi⟩, which will make sense as we learn about probability theory in the next three or four lectures! 7.3 Why do we need probability? Almost all laws of thermodynamics are based on observations of macroscopic systems: we’re measuring thermal properties like pressure and temperature, but any system is still inherently made up of atoms and molecules, so the motion is described by more fundamental laws, either classical or quantum. 24 So we care about likelihoods: how likely is it that particles will be in a particular microscopic state? 7.4 Fundamentals Definition 53 A random variable x has a set of possible outcomes S = {x1, x2, · · · , }. (This set is not necessarily countable, but I think this is clear from the later discussion.) This random variable can be either discrete or continuous. Example 54 (Discrete) When we toss a coin, there are two possible outcomes: Scoin = {H, T}. When we throw a die, Sdie = {1, 2, 3, 4, 5, 6}. Example 55 (Continuous) We can have some velocity of a particle in a gas dictated by S = {−∞< vx, vy, vz < ∞}. Definition 56 An event is a subset of some outcomes, and every event is assigned a probability. Example 57 When we roll a die, here are some probabilities: Pdie({1}) = 1 6, Pdie({1, 3}) = 1 3. Probabilities satisfy three important conditions: • positivity: any event has a nonnegative real probability P(E) ≥0. • additivity: Given two events A and B, P(A ∪B) = P(A) + P(B) −P(A ∩B) where A ∪B means “A or B” and A ∩B means “A and B”. • normalization: P(S) = 1, where S is the set of all outcomes. In other words, all random variables have some outcome. There are two ways to find probabilities: 25 • Objective approach: given a random variable, do many trials and measure the result each time. After N of them, we have probabilities NA for each event A: this is just the number of times A occurs, divided by N. In particular, as we repeat this sufficiently many times, P(A) = lim N→∞ NA N . • Subjective approach: We assign probabilities due to our uncertainty of knowledge about the system. For example, with a die, we know all six outcomes are possible, and in the absence of any prior knowledge, they should all be equally probable. Thus, P({1}) = 1 6. We’ll basically do the latter: we’ll start with very little knowledge and add constraints like “knowledge of the internal energy of the system.” 7.5 Continuous random variables Fact 58 We’ll mostly be dealing with these from now on, since they’re are what we’ll mostly encounter in models. Let’s say we have a random variable x which is real-valued: in other words, SX = {−∞< x < ∞}. Definition 59 The cumulative probability function for a random variable X, denoted FX(x), is defined as the probability that the outcome is less than or equal to x: FX(x) = Pr[E ⊂[−∞, x]]. Note that FX(−∞) = 0 and FX(∞) = 1, since x is always between −∞and ∞. Definition 60 The probability density function for a random variable X is defined by pX(x) ≡dFX dx In particular, pX(x)dx = Pr [E ⊂[x, x + dx]] . It’s important to understand that Z ∞ −∞ pX(x)dx = 1, since this is essentially the probability over all x. Fact 61 The units or dimension of pX(X) is the reciprocal of the units of X. 26 Note that there is no upper bound on pX; it can even be infinity as long as p is still integrable. Definition 62 Let the expected value of any function f (x) of a random variable x be ⟨F(x)⟩= Z ∞ −∞ F(x)p(x)dx. As a motivating example, the expected value of a discrete event is just ⟨X⟩= X i pixi, so this integral is just an “infinite sum” in that sent. 7.6 More statistics Definition 63 Define the mean of a random variable x to be ⟨x⟩= Z ∞ −∞ xp(x)dx. For example, note that ⟨x −⟨x⟩⟩= 0. In other words, the difference from the average is 0 on average, which should make sense. But we can make this concept into something useful: Definition 64 Define the variance of a random variable x to be var(x) = ⟨(x −⟨x⟩)2⟩. This tells something about the spread of the variable: basically, how far away from the mean are we? Note that we can expand the variance as var(x) = ⟨x2 −2x⟨x⟩+ ⟨x⟩2⟩= ⟨x2⟩−⟨x⟩2. Fact 65 (Sidenote) The reason we square instead of using an absolute value is that the mean is actually the value that minimizes the sum of the squares, while the median minimizes the sum of the absolute values. The absolute value version is called “mean absolute deviation” and is less useful in general. We’re going to use this idea of variance to define other physical quantities like diffusion later! 27 Definition 66 Define the standard deviation as σ(x) = p var(x). With this, define the skewness as a dimensionless metric of asymmetry: (x −⟨x⟩)3 σ3 , and define the kurtosis as a dimensionless measure of shape (for a given variance). (x −⟨x⟩)4 σ4 , Let’s look at a particle physics experiemnt to get an idea of what’s going on: e+e−→µ+µ−. Due to quantum mechanical effects, there is some probability distribution for θ, the angle of deflection: p(θ) = c sin θ(1 + cos2 θ), 0 ≤θ ≤π. To find the constant, we normalize with an integral over the range of θ: 1 = c Z π 0 sin θ(1 + cos2 θ)dθ. We will solve this with a u-substitution: letting x = cos θ, 1 = c Z 1 −1 (1 + x2)dx = ⇒1 = 8 3c = ⇒c = 3 8 So our probability density function is p(θ) = 3 8 sin θ(1 + cos2 θ) Fact 67 This has two peaks and is symmetric around π 2 . Thus, the mean value of θ is ⟨θ⟩= π 2 , and σ is approximately the distance to the peak. We can calculate the standard deviation exactly: ⟨θ2 ⟩= 3 8 Z π 0 (1 + cos2 θ) sin θ · θ2 dθ = π2 4 −17 9 , and therefore var(θ) = ⟨θ2⟩−⟨θ⟩2 ≈0.579 = ⇒σ ≈0.76. Finally, let’s compute the cumulative probability function: F(θ) = Z θ 0 3 8 sin θ(1 + cos2 θ)dθ = 1 8(4 −3 cos θ −cos3 θ). This has value 0 at θ = 0, 1 2 at θ = π 2 , and 1 at θ = π. 28 Next time, we will talk about discrete examples and start combining discrete and continuous probability. We’ll also start seeing Gaussian, Poisson, and binomial distributions! 8 February 19, 2019 (Recitation) A guest professor is teaching this recitation. We’re going to discuss delta functions as a mathematical tool, following the supplementary notes on the website. There are multiple different ways we can represent delta functions, but let’s start by considering the following: Definition 68 Define δε(x) = 1 √ 2πε exp  −x2 2ε2  . This is a Gaussian (bell curve) distribution with a peak at x = 0 and an inflection point at ±ε. It also has the important property Z ∞ −∞ δε(x) = 1, so it is already normalized. This can be shown by using the fact that I = Z ∞ −∞ dxe−αx2 = ⇒I2 = Z ∞ −∞ Z ∞ −∞ dxdye−α(x2+y 2) and now switch to polar coordinates: since dxdy = rdrdθ, I2 = Z ∞ 0 drr Z 2π 0 e−αr 2 = π α using a u-substitution. So δε is a function with area 1 regardless of the choice of ε. However, ε controls the width of our function! So if ε goes down, the peak at x = 0 will get larger and larger: in particular, δε(0) = 1 √ 2πε goes to ∞ as ε →0. So we have a family of such functions, and our real question is now what we can do with integration? What’s Z ∞ −∞ dxδε(x)f (x)? For a specific function and value of ε, this may not be a question you can answer easily. But the point is that if we put an arbitrary function in for f , we don’t necessarily know how to do the integration. What can we do? Well, let’s think about taking ε →0. Far away from x = 0, δε(x)f (x) is essentially zero. If we make δε extremely narrow, we get a sharp peak at x = 0: zooming in, f is essentially constant on that peak, so we’re basically dealing with Z ε −ε dxf (0)e−x2 2ε2 1 √ 2πε = f (0) · 1 = f (0). So the idea is that we start with a particular family of functions and take ε →0, and this means that δε is a pretty good first attempt of a “sharp peak.” 29 Definition 69 Let the Dirac delta function δ satisfy the conditions • δ(x −x0) = 0 for all x ̸= x0. • R ∞ −∞dxδ(x −x0) = 1, where the integral can be over any range containing x0. • R dxδ(x −x0)f (x) = f (x0), again as an integral over any range containing x0. This seems pretty silly: if we already know f (x), why do we need its evaluation at a specific point by integrating? We’re just evaluating the function at x = 0. It’s not at all clear why this is even useful. Well, the idea is that it’s often easier to write down integrals in terms of the delta function, and we’ll see examples of how it’s useful later on. For now, let’s keep looking at some more complicated applications of the delta function. What if we have something like Z dxδ(g(x))f (x)? We know formally what it means to replace δ(x) with δ(x −x0), but if we have a function g with multiple zeros, we could have many peaks: what does that really mean, and how tall are the peaks here? This is useful because we could find the probability that g(x, y, z) = c by integrating Z p(x, y, z)δ(g(x, y, z) −c)dxdydz and this answer is not quite obvious yet. So we’re going to have to build this up step by step. Let’s start in a simpler case. What’s Z dxf (x)δ(cx)? We can do a change of variables, but let’s not rush to that. Note that δε(−x) = δε(x), and similarly δ(x) = δ(−x): this is an even function. So replacing y = cx, = Z dy 1 |c|f y c  δ(y) = 1 |c|f (0). So we get back f (0), just with some extra constant factor. Be careful with the changing integration limits, both in this example and in general: that’s why we have the absolute value in the denominator. In general, the delta function “counts” things, so we have to make sure we don’t make bad mistakes with the sign! Similar to the above, we can deduce that linear functions give nice results of the form Z dxδ(cx −a)f (x) = 1 |c|f a c  . But this is all we need! Remember that we only care about δ when the value is very close to 0. So often, we can just make a linear approximation! f (x) looks linear in the vicinity of x0, and there’s a δ peak at x0. So if we make the Taylor expansion f (x) ≈f (x0) + f ′(x0)(x −x0), we have found everything relevant to the function that we need. Note that by definition, δ(g(x)) = 0 whenever g(x) ̸= 0. Meanwhile, if g(xi) = 0, g(x) ≈g(xi) + g′(xi)(x −xi) = g′(xi)(x −xi). So that means we can treat δ(g(x)) = X i δ(g′(xi)(x −xi)) where we are summing over all zeros of the function! 30 Fact 70 Remember that δ is an everywhere-positive function, so δ(g(x)) cannot be negative either. Well, we just figured out how to deal with δ(g(x)) where g is linear! So Z dxf (x)δ(g(x)) = X i 1 |g′(xi)|f (xi). So at each point xi where g is 0, we just take f (xi) and modify it by a constant. Now this function is starting to look a lot less nontrivial, and we’ll use it to do a lot of calculations over the next few weeks. Example 71 Let’s say you want to do a “semi-classical density of states calculation” to find the number of ways to have a particle at a certain energy level. Normally, we’d do a discrete summation, but what if we’re lazy? Then in the classical case, if u is the velocity, we have an expression of the form f (E) = Z p(u)δ  E −mu2 2  du. To evaluate this, note that the δ function is zero at u = ± q 2E m , and the derivative g′(u) = −mu = ⇒|g′(u±)| = √ 2mE, so the expression is just equal to f (E) = 1 √ 2mE (p(u+) + p(u−)). This is currently a one-dimensional problem, so there’s only two values of u. In highest dimensions, we might be looking at something like Z d3⃗ up(⃗ u)δ  E −mu2 x + u2 y + u2 z 2  . Now the zeroes lie on a sphere, and now we have to integrate over a whole surface! By the way, there are different ways to formulate the delta function. There also exists a Fourier representation δ(x) = Z ∞ −∞ dkeikx. It’s not obvious why this behaves like the delta function, but remember eikx is a complex number of unit magnitude. Really, we care about Z δ(x)f (x)dx, and the point is that for any choice of x other than 0, we just get a spinning unit arrow that gives net zero contribution. But if x = 0, eikx is just 1, so this starts to blow up just like the δε function. There also exist a Lorentzian representation δε = 1 π ε x2 + ε2 and an exponential representation δε = 1 2εe|x|/ε. 31 The point is that there are many different families of functions to capture the intended effect (integrates to 1), but as all of them get sharper, they end up having very similar properties for the important purposes of the delta function. 9 February 20, 2019 (Recitation) It is a good idea to talk about the concept of an exact differential again, and also to look over delta functions. 9.1 Questions We often specify a system (like a box) with a temperature, volume, and pressure. We do work on the system when the volume is reduced, so dW = −PdV . When we have a dielectric with separated charges, we can orient the dipoles and get a dipole energy ∝⃗ E · ⃗ p, where ⃗ p is the dipole moment. We now have to be careful if we want to call this potential energy: are we talking about the energy of the whole system, the external field, or something else? Well, the differential energy can be written as dW = EdP: how can we interpret this? Much like with −PdV , when the polarization of the material changes, the electrostatic potential energy changes as well. So now, what’s the equation of state for an electrostatic system? Can we find an equation like PV = nRT to have E(T, P)? Importantly, note that some analogies to break down: the electric field E is sort of necessary (from the outside) to get a polarization P. So if we consider an exact differential, and we’re given ∂E ∂P T = T AT + B , ∂E ∂T P = BP (AT + B)2 , we know everything we could want to know about the system. First, we should show that these do define an equation of state: is dE = T AT + B dP + BP (AT + B)2 dT an exact differential? Well, we just check whether ∂ ∂T T AT + B = ∂ ∂P BP (AT + B)2 . Once we do this, we can integrate ∂E ∂P with respect to P to get E up to a function of T, and then differentiate with respect to T to find that unknown function T. Next, let’s talk a bit more about mean and variance. We can either have a set of possible (enumerable, discrete) events {pi}, or we could have a probability density p(x). The idea with the density function is that p(x)dx = Pr[in the range [x, x + dx]]. Remember that probabilities must follow a normalization, which means that X pn = 1 or Z p(x)dx = 1. 32 What does it mean to have an average value? In the discrete case, we find the average as ⟨n⟩= X i ipi, since an outcome of i with probability pi should be counted pi of the time. Similarly, the continuous case just uses an integral: ⟨x⟩= Z xp(x)dx. Note that we can replace n and x with any arbitrary functions of n and x. Powers of n and x are called moments, so the mean value is the “first moment.” (This is a lot like taking the second “moment of inertia” R r 2ρ(r)dr.) Then the variance is an average value of ⟨(n −⟨n⟩)2⟩= ⟨n2 −2n⟨n⟩+ ⟨n⟩2⟩= ⟨n2⟩−⟨n⟩2. Let’s look at another situation where we have a probability density dP dw = p(w). 9.2 Delta functions Here’s a question: what is the derivative δ′(x)? Well, let’s start with some related ideas: δ(10x) = 1 10δ(x), and this might make us cringe a bit since δ is mostly infinite, but it works for all purposes that we are using it. The important idea is to not always think about infinity: we could consider the delta function to be a rectangle of width ε and height 1 ε. This makes it seem more like a real function. So now, if we take any function f (x), f (10x) is just a function that keeps the maximum constant but shrinks the width by a factor of 10. Well, if we integrate over f (10x), we’ll get a factor of 1 10 less, and that’s how we should understand δ(10x). Is δ′(x) defined, then? Let’s say we have a triangle function with peak 1 ε and width from −ε to ε. The derivative of this function is not defined at x = 0! So δ′(x) doesn’t necessarily need to be defined. The idea is that we can take ε →0 using any representation of a real function, and a derivative would have to be well-defined across all representations: that just doesn’t happen here. Curiously, though, if we used the triangle function, we can actually represent the derivative as a delta function itself, because the derivative is a large number from −ε to 0 and from 0 to ε: δ′(x) = 1 εδ  −ε 2  −1 εδ ε 2  , where the 1 ε factor is just for normalization, since the area of the rectangle for the derivative is ε · 1 ε2 . So it seems that δ′(x) = −1 ε  δ ε 2  −δ  −ε 2  . It’s okay, though: delta functions always appear under an integral. (So continuity is important, but not necessarily 33 differentiability.) This means that if we’re integrating this with a function f (x), Z f (x)δ′(x)dx = −1 ε Z  δ ε 2  −δ  −ε 2  f (x)dx = −1 εf ε 2  −f  −ε 2  = −f ′(0). But the idea is that we want to be faster at manipulating such things. What if we integrated by parts? Then Z f (x)δ′(x) = − Z f ′(x)δ(x)dx + f (x)δ(x)|∞ −∞. The delta function is mathematically zero, so the boundary term disappears (unless we have a bad Lorentzian or other description of the delta function). But now this just gives − Z f ′(x)δ(x)dx = −f ′(0). This is maybe how we should use delta functions, but it’s still important to have confidence that what we’re doing is correct! Finally, let’s ask one more question. We can store energy by pressing air into an underground cave, and we can do that in two ways: adiabatic and isothermal. If we compare the two situations, where does the energy go? In the isothermal case, the internal energy is the same. So isothermal compression is just transferring the energy as heat to the surrounding ground! Is there a way to retrieve it? (Hint: the process may be reversible if we do it slow enough!) 10 February 21, 2019 Today we’re going to continue learning about probability. As a reminder, we’re learning probability because there’s a bottom-up and top-down approach to statistical physics: thermodynamics gives us state functions that tell us physical properties of the world around us, and we can connect those with microscopic atoms and molecules that actually form the system. Probability allows us to not just follow every particle: we can just think about general distributions instead! We’ll discuss some important distributions today and start our connections between probability distributions and physical quantities. We’ll eventually get to the Central Limit Theorem! 10.1 A discrete random variable Consider a weighted coin such that P(H) = 5 8, P(T) = 3 8. This is a “biased distribution.” Let’s say that every time we get a head, we gain $1, and every time we get a tail, we lose $1. Letting x be our net money, our discrete probability distribution P(x) satisfies P(1) = 5 8, P(−1) = 3 8. Here are some important statistics: • P(1) + P(−1) = 1. • ⟨x⟩= P(1) · 1 + P(−1) · (−1) = 5 8 −3 8 = 1 4. • ⟨x2⟩= P(1)12 + P(−1)12 = 1, so the variance of this distribution is var(x) = ⟨x2⟩−⟨x⟩2 = 15 16. 34 This is a large variance, since most events are far away from the mean! • To find the probability density function p(x), we can use delta functions: p(x) = 5 8δ(x −1) + 3 8δ(x + 1) Note that if we integrate p(x)dx over any interval containing 1 but not −1, we get a probability of 5 8, which is what we want. In general, if we have a probability density function that has both discrete and continuous parts, we can write it as p(x) = f (x) + M X j=1 piδ(x −xj). See the Weibull distribution, as well as the Xenon lamp spectrum! 10.2 Important probability distributions We’re going to discuss the Gaussian, Poisson, and Binomial distributions. The idea is that the limit of a Binomial distribution will converge to a Poisson distribution, which will then converge to a Gaussian. We’ll see the second part today as part of the Central Limit Theorem! 10.3 Gaussian distribution Definition 72 The probability density function for a Gaussian with standard deviation σ is p(x) = 1 √ 2πσ exp  −(x −a)2 2σ2  . This distribution has mean a and variance σ2; let’s check for normalization to make sure this is indeed a valid distribution. So Z ∞ −∞ p(x)dx = 1 √ 2πσ Z ∞ −∞ exp  −(x −a)2 2σ2  , and now defining y ≡x −a √ 2σ = ⇒dy = 1 √ 2σ dx. Substituting in, we end up with 1 √ 2πσ · √ 2σ Z ∞ −∞ e−y 2dy. But the integral is known to be √π (by the polar substitution trick), and therefore our integral of p(x) is 1, and indeed we have a normalized distribution. So the cumulative distribution function is F(x) = Z x −∞ dξ 1 √ 2πσ exp  −(ξ −a)2 2σ2  . 35 Denoting ξ−a √ 2σ = z as before, we can substitute again to find F(x) = 1 √π Z (x−a)/( √ 2σ) −∞ dze−z2. Since our probability function is normalized, we can also write this as F(x) = 1 − 1 √π Z ∞ (x−a)/( √ 2σ) dze−z2 Let erfc, the complementary error function, be defined as erfc(η) = 2 √π Z ∞ η dze−z2. So that means our cumulative distribution can be written as F(x) = 1 −1 2 erfc x −a √ 2σ  . It’s important to note that the shape of p(x) is the familiar “bell curve” shape. When a = 0 (so the curve is centered at x = 0), p(0) = 1 √ 2πσ and we can also compute that P(σ) ≈0.61P(0), P(2σ) ≈0.135P(0). What exactly is the significance of σ? Well, for any a and σ in the Gaussian, Z a+σ a−σ p(x)dx ≈0.68, meaning that about 68% of the time, a random sample from the Gaussian distribution will be within σ of the mean. Similarly, Z a+2σ a−2σ p(x)dx ≈0.95, and that means that 95 percent of all measurements are within 2σ of the mean. Example 73 Consider a measurement of the magnetic moment of a muon m = geℏ 2Mµc . At first, we expect that g ≈2 theoretically. However, after many measurements, we get a Gaussian distribution for (g −2)µ: (g −2)µ = (116591802 ± 49) × 10−11, where the first term is the mean and the part after the ± is σ. Theoretical calculations actually end up giving (g −2)µ = (116592089 ± 63) × 10−11, and these distributions are actually significantly different: the discrepancy is still a point of ongoing research! The idea is that this measurement of σ allows us to compare two different distributions. 36 10.4 Poisson distribution For random variables X and Y , if they have probability distributions p(x) and p(y), then X, Y statistically independent = ⇒p(x, y) = p(x)p(y). Let’s start with an example. Given a random student, the probability that a student is born in May is 31 365.25 ≈0.0849. Meanwhile, the probability of being born between 9 and 10 in the morning is 1 24 ≈0.0417. So the probability of being born between 9 and 10 in the morning in May is 0.0849 × 0.0417 = 3.54 × 10−3. We need this to introduce the idea of a Poisson distribution! This is important for rare events with low probability. Here are two important ideas: • The probability of an event happening exactly once in the interval [t, t + dt] is proportional to dt as dt →0: dp = λdt for some λ. • The probability of events in different events are independent of each other. If we have these two conditions satisfied, the idea is that we can subdivide a time interval of length T into small intervals of length dt. In each interval, the probability that we observe an event is equal and independent to all the other ones! Definition 74 Then the probability that we observe a total of exactly n events in an interval time T is given by the Poisson distribution Pn(T). Let’s try to compute pn. We break T into N bins of length dt, so dt = T N , in such a way that the probability of getting two events in the same bin (small time interval) is negligible. Then dP = λdt = λT N ≪1. To compute the probability of finding n events, first we find the probability of computing no events, then 1 event, and so on. Note that the probability of observing no event in an interval dt is 1 −λT N , so the probability of observing no events overall is lim N→∞  1 −λT N N since we must have no event observed in all N intervals. By definition of an exponential, this is just P0(T) = e−λT . Next, find the probability of observing exactly 1 event in time interval T. There are N different places in which this one event can happen, and the probability that it happens is λT N . Then the other N −1 intervals must have no event happen, so this is P1(T) = lim N→∞N · λT N  1 −λT N N−1 = λT lim N→∞  1 −λT N N−1 . This gives, again by the definition of an exponential, P1(T) = λTe−λT . 37 Let’s do another example: what about two events? We pick which two intervals are chosen, and then Pk(T) = lim N→∞ N 2  · λT N 2  1 −λT N N−2 = (λT)2 2 e−λT . In general, the probability of k events happening is going to be Pk(T) = lim N→∞ N k  · λT N k  1 −λT N N−k = ⇒Pk(t) = (λT)k k! e−λT . It’s important to note that this is a discrete distribution! Let’s check some statistics for our probability distribution function. First of all, is it normalized? Well, ∞ X n=0 pn(T) = e−λT ∞ X n=0 (λT)n n! = e−λT eλT = 1, so the Poisson distribution is indeed normalized. Next, let’s find the mean: ⟨n⟩= ∞ X n=0 npn(T) = e−λT ∞ X n=0 n(λT)n n! Denoting Z ≡λT, this expression can be written as ⟨n⟩= e−Z ∞ X n=0 nZn n! = Ze−Z ∞ X n=1 zn−1 (n −1)! = e−ZZ · eZ = Z. So the mean of the Poisson distribution is ⟨n⟩= λT, which shouldn’t be that surprising: it’s saying that if events have a probability 1 N of happening, they happen on average once per N. Finally, let’s find the variance: we’ll leave this as an exercise, but the idea is to start by computing ⟨n(n −1)⟩= ⟨n2 −n⟩. It turns out the variance is also λT, and this is an interesting relationship between the mean and variance! We’ll introduce a dimensionless quantity σ(n) ⟨n⟩ which meausures the width of the distribution. Well, note that as T →∞for the Poisson distribution, this goes to 0, so the distribution becomes more and more spiked around λT. It turns out that this approaches a Gaussian distribution! How can we check that? Taking T →∞, λT ≫1 = ⇒n ≫1, and we want to find the probability Pn(λT). Denoting λT ≡Z, we expand around the maximum, and we’re going to look at the log of the function. By Stirling’s approximation, ln n! ∼n ln n −n + ln(2πn) as n →∞, so Pn(Z) = Zn n! e−Z = ⇒ln Pn(Z) = n ln Z −Z −ln n! ≈n ln Z −Z −n ln n + n −1 2 ln 2πn. The maximum of this function can be found by taking the derivative, ∂ ∂n ln Pn = ln Z −ln n −1 2n, 38 and ignoring the 1 2n term, we can say that n0 = Z at the maximum. Doing a Taylor expansion about n0, ln Pn(Z) = ln pn0(z) + (n −n0)2 2 ∂2 ∂n2 Pn(z) n=n0 , and taking the exponential of both sides, we find that Pn(z) = 1 √ 2πz exp  −(n −z)2 2z  which is a Gaussian with standard deviation √z and mean z, as desired! This is our first instance of the Central Limit Theorem, a powerful tool to deal with large numbers. 11 February 25, 2019 (Recitation) Professor Ketterle was writing a paper on the new definition of the kilogram. Let’s spend a few minutes talking about that! 11.1 The kilogram There’s a law that is taking effect in May. Currently, the kilogram is a physical object in France: it’s what weighs as much as the “original Paris kilogram.” There are copies around the world. But artifacts like this have discrepancies! There might be diffusion of atoms or fingerprints, so at the microgram level, there are still deviations. This wasn’t a problem, but now people have determined Planck’s constant with an error of ∆h h ≈10−8, and the error is limited from the mass deviation! So this is pretty inconvenient. Fact 75 So instead, why not define h to be 6.62 · · · × 10−34 Joule-seconds? Now we’ve turned it around: mass is now determined in terms of h, instead of the other way around! Question 76. Why exactly does h determine mass? Since E = hν, and we can consider the frequency of transition in Cesium = 9.1 · · · × 109 Hertz (which actually now defines units of time and frequency). With this, now we can measure energy (as either kinetic or rest energy). The idea is that all constants are now defined in terms of c, the speed of light, and h, Planck’s constant! So more precisely, we start with the frequency of Cesium, and we define a second to be 9.1 · · · GHz. But this means that if we take our value h, the mass of a photon will be m = hνCS c2 . How do we measure the mass of a photon? Well, a Cesium atom in the upper state is slightly heavier than a Cesium atom in the lower state! That gives us the change in mass ∆m, which is on the order of 10−40 kilograms. Fact 77 When a system loses energy, it often loses mass! 39 It’s a tiny effect, but it’s important for special relativity. So we can now set up a balance, where 1040 Cesium atoms are set up with spins in the “upper state.” On the other side, we have the same number of Cesium atoms, but the spins are in the ground state. Then any substance that balances the scale is exactly one kilogram! 11.2 Poisson distribution Think of the Poisson distribution as modeling a bunch of atoms radiating with some decay rate. If N is the number of particles, λ is the decay rate, and we observe for some rate dt, we have an expectation ⟨n⟩= Nλdt for the number of observed events in time dt. We can do combinatorics to find that pn = ⟨n⟩n n! e−⟨n⟩ This is a prettier way to write the Poisson distribution, and it shows that the whole distribution is based entirely on the value of ⟨n⟩. Based on this, let’s consider the concept of shot noise, which comes from us trying to count something that is random. When we have a random event like raindrops falling on a roof, you hear a random noise with fluctuations. This is because rain doesn’t come as a stream: it’s big droplets. So the shot noise comes from the fact that we have a stream of mass (or radioactively decaying particles) that are quantized. So sometimes we have a few more or a few less than expected. So if we are trying to observe 100 events, the expectation value is 100 ± 10. This is because the variance var(n) = ⟨n⟩, and therefore σ, the standard deviation, is p ⟨n⟩. So it’s important to remember that ±√n idea! Basically, the Poisson distribution looks almost normal, and the inflection point occurs around p ⟨n⟩from the mean. This leads to the next question: as you observe larger expectations, is the shot noise larger or smaller? Well, if we’re doing repeated measurements to determine some quantity, our precision goes as σ(n) ⟨n⟩∝ 1 √n. So if we measure 10 times longer, we are √ 10 times more accurate. But this is often the best we can do! Much along the same line, we want to do experiments with large counting rate to get higher expectations. So if we measure the noise in an electric current, the number of electrons that pass through gives shot noise as well. We’ll find that iδt = nee, where ne is the number of electrons and e is the charge of electron. Well, ne fluctuates by a factor of √ne, and we can then experimentally measure the charge of an electron! But people did experiments with superconductors, which can also carry current. This has no dissipation, and then the noise for ne was different. This is because q is now 2e (superconductivity happens when electrons combine into pairs). So when the current carriers are Cooper pairs instead of electrons, we only need half the number to get the same current, and this means our fluctuation is larger! This was the first proof that superconductors did not have electron charge carriers. But here’s a real life MIT example for shot noise and how to measure it. Let’s say we have 106 people on Killian Court, and let’s say there are a few exits that people randomly leave through. Through one exit, there’s 104 people who leave per unit time, and by shot noise, there’s a fluctuation of ±100 people. 40 But now let’s say that people leave in groups of 100 instead: now there’s 102 groups that leave per unit time, so the shot noise is just 10 groups. This means the actual absolute fluctuation is 103, which is larger than the original! So if the carrier of charge, or the unit size of people, is increased by a factor of N, the shot noise is increased by a factor of √ N. 11.3 Stirling’s formula How do you memorize the formula? If you do n! = n · (n −1) · · · · 2 · 1, we know nn is a bad estimate, and n 2 n would be a bit closer. And also because we have logarithms, e is a pretty good number to use. So instead n! ≈ n e n . This gives log n! ≈n(log n −log e) ≈n ln n −n. There’s another term of ln √ 2πn, but when n is large (as it is in this class), that’s several orders of magnitude smaller than n! So we can neglect it. 12 February 26, 2019 Remember that we have been introducing probability distributions. We found that the Poisson distribution converges to a Gaussian as the frequency of events becomes larger, and this was an important example of the Central Limit Theorem. Today, we’ll talk about the binomial distribution and connect it to the idea of diffusion. Finally, we’ll discuss conditional probability and figure out ideas like “energy given some other knowledge about the system.” 12.1 Binomial distribution Consider a random variable with two possible outcomes A and B, occurring with probability pA and pB = 1 −pA. Our goal is to find out how many times A occurs if we repeat the random variable N times. Then this can be calculated as PN(NA) =  N NA  pNA A (1 −pA)N−NA The first factor of  N NA  = N! NA!(N −NA)! comes from the number of ways in which we can choose which of the events are A, and the rest is just the probabilities of A and B multiplied the relevant number of times. Is this normalized? By the Binomial theorem, N X NA=0 PN(NA) = N X NA=0  N NA  pNA A (1 −pA)N−NA = (pA + (1 −pA))N = 1, 41 as desired. We can also find some other statistics by doing mathematical manipulation: ⟨NA⟩= N · pA, var(NA) = NpApB. Then, the ratio of the standard deviation to the mean of NA is σ(NA) ⟨NA⟩= p NpA(1 −pA) NpA = 1 √ N r 1 −pA pA , so as N →∞, the distribution becomes narrower and narrower. One question we should be asking ourselves: how is it possible for physics to have simple equations that explain the complexity of the world? This shows the beauty of statistical mechanics: we can explain the world by just using probability. 12.2 An application Fact 78 We’re going to use the binomial distribution to derive a diffusion equation. A random process is sometimes a random walk, and we can use the diffusion equation to understand that random walk! Definition 79 A random walk is a path of successive steps in random directions in some space. These describe many physical phenomena, like collisions of particles, shapes of polymers, and so on. For example, DNA is often curled up in a coil, and its shape is usually described by a random variable with mean zero! There are two kinds of emergent behavior for random walks. • Given any individual random walk, after a large number of steps, it becomes a fractal (scale invariant). We won’t be talking much about this in class though. • The endpoint of a random walk has a probability distribution that obeys a simple continuum law, which leads to the diffusion equation! The idea is that these phenomena are global, so they are independent of the microscopic details of the system. Example 80 Consider a random walk in one dimension: this is also known as Brownian motion. Let’s say that it moves left or right along a line with step size ℓ, and the probability is P(+ℓ) = P(−ℓ) = 1 2. First of all, we want to find the average displacement after N steps. Well, ⟨∆xi⟩= 1 2(ℓ) + 1 2(−ℓ) = 0, so the average is always 0 after N steps. On the other hand, we can consider the mean squared displacement: Then ⟨∆x2 i ⟩= 1 2(+ℓ)2 + 1 2(−ℓ)2 = ℓ2, 42 and the mean square displacement after N steps is ⟨∆x2⟩= N X i=1 N X j=1 ⟨∆xi∆xj⟩ and now all cross terms ∆xi∆xj with i ̸= j contribute 0 by independence. This means that ⟨∆x2⟩= N X i=1 ⟨∆x2 i ⟩= Nℓ2. Fact 81 We could also use the fact that variances of independent variables add! So since each step has variance ℓ2, the total sum has variance Nℓ2. So if successive jumps happen every δt, the number of jumps in a time t is N = t δt = ⇒⟨∆x2⟩= ℓ2 δt t. This is important: the variance scales linearly with time! In comparison, if our random walk has some average velocity ∆x(t) = vt = ⇒⟨∆x2⟩= ⟨v 2⟩t2, which is called ballistic motion. In more advanced statistics, this is the setup for the fluctuation-dissipation theorem! But what’s the main physics of what we’re working on: where is the randomness of our process coming from? • The existence of a randomly fluctuating force will push a particle in random directions. • There is some inertia of the system, as well as a viscous drag. Our goal is to compute the probability distribution of finding a particle x away from the original position after N steps. If we denote NL to be the number of steps to the left and NR the number of steps to the right (so N = NL+NR), then the net displacement of the walk is x = ℓ(NR −NL). Question 82. How many distinct walks are possible if we give ourselves N steps, NL of which are to the left and NR of which are to the right? This is just  N NL  = N! NL!NR!. In total, since each move can be to the left or to the right, there are 2N distinct ways to form a walk of N steps, and the probability of any sequence is 1 2N . Fact 83 It’s important to note that sequences each have equal probability, but x, the net distance, is not uniformly distributed. So the probability of having a walk of net length x is p(x, N) = N! NR!NL! 1 2 N , which is the number of sequences times the probability of any given sequence. 43 12.3 In the limiting case We claim that this becomes a Gaussian as N becomes large. Indeed, note that we’ve define x = ℓ(NR −NL), so NL = N −x/ℓ 2 , NR = N + x/ℓ 2 . Substituting these in, we can then use Stirling’s approximation ln n! ≈n ln n −n + 1 2 ln(2πn). This yields p(x, N) ∝ r 2 πN exp  −x2 2Nℓ2  This is a Gaussian symmetric about its mean 0, which tells us that we’re mostly likely to have a net displacement of x = 0. Fact 84 This explains why in polymers, most of the time there are blobs rather than straighter lines! It’s much more probable to be close to the mean. If we compute the variance, ⟨x2⟩= Nℓ2 as expected, and if we say that our events are equally spaced by some time δt, the variance is again ⟨x2⟩= ℓ2 δt t ∝t. Definition 85 Define the diffusion constant D such that ⟨x2⟩= 2Dt. This has various applications! 12.4 Multiple random variables Let’s say we have two variables x and y. Definition 86 The joint probability distribution function p(x0, y0) = d2F dxdy x=x0,y=y0 = ⇒d2F = p(x, y)dxdy is the probability of x occurring between x0 and x0 + dx and y between y0 and y0 + dy. Then F(x, y) = Z x −∞ dξ Z y −∞ dηp(ξ, η). We can also define some other quantities like we did in the one-dimensional case: 44 Definition 87 The expectation value of a function f (x, y) is Z ∞ −∞ dx Z ∞ −∞ dyp(x, y)f (x, y). We can also integrate one variable out of the equation to find the probability distribution for the other variable: for example, p(x) = Z ∞ −∞ dyp(x, y). With this in mind, let’s try to relate our variables. Can we answer questions like Question 88. What is the probability that X lies between x and x + dx given that Y is certain, denoted P(X|Y )? Note that p(x|y) should be proportional to p(x, y): we’ll talk more about this next time! 13 February 27, 2019 (Recitation) Some of Professor Ketterle’s colleagues said that photons don’t have mass, which is true in the basic sense. But there’s a relativistic mass-energy equivalence, so writing the equation for a photon E = m0c2 does actually make sense and has a nonzero m0 for photons. But there’s a question beyond semantics of “real mass” versus “relativistic mass” here: Question 89. If we take a cavity and put many photons inside bouncing back and forth, does the mass increase? Does it have more inertia? The answer is yes, since we have the photons “basically at a standstill!” But the whole point is to be careful what we mean by “mass.” 13.1 Small aside for surface tension Given a water droplet on a surface, there’s three different surface tensions, corresponding to the three pairs out of {surface, air, water}. The concept of surface tension is that surfaces want to shrink, and this creates net forces at interfaces. Water droplets stop existing when the forces cannot be balanced anymore, so an equilibrium state cannot exist. This point is called critical wetting, and anything beyond that point results in water coating the whole surface! 13.2 Delta functions and probability distributions Example 90 Let’s say we have a harmonic oscillator in a dark room, and there is a lightbulb on the oscillator. If we take a long-exposure photo, what does the light distribution look like? 45 If we let x = sin ωt, all phases φ are equally probable, since φ is proportional to time. So we want to go from a probability distribution of φ to one of x: how could we do that? Well, we know that the velocity is slower at the ends, so we expect more values of x on the ends than the middle. (This is explained more rigorously in the problem set.) The punchline is that the probability distribution is going to be proportional to 1 v(x), and now we can proceed with mathematics. So in this case, we have time as our random variable, and we need 1 = Z T 0 p(t)dt for a period of length T. So our starting distribution p(t) = 1 T , and now we want to turn this into a distribution p(x). Well, p(x) = Z T 0 p(t)δ(x −x(t))dt. since we ask for the moments in time where x(t) = x. In this case, p(t) is constant, so this is 1 T X i 1 |f ′(ti)| = 1 T 1 |v(t)| where f (t) = x −x(t) and ti are the roots. (Notice that this gives us our probability normalization for free!) So now dx dt = ω cos ωt = ω p 1 −sin2 ωt, which we can write in terms of x as dx dt = ω p 1 −x2 = ⇒p(x) = 1 ωT 1 √ 1 −x2 . But wait! We haven’t been careful enough, because there’s two different points where x(t) = x. The slopes are negative of each other, so even though dx dt are the same, we need to count the two roots separately. Thus the actual number we want is 2 ωT 1 √ 1 −x2 = 1 π √ 1 −x2 . The idea in general is that if we have a probability distribution p(x), and x = f (y) = ⇒y = g(x) is some function, we can find the probability distribution p(y) by p(y) = Z p(x)δ(y −g(x))dx. Basically, we want to take all values where g(x) = y. But here’s another way to see it: the probability differential p(x)dx should correspond to another probability differential p(y)dy, so p(y) = p(x)dx dy . We’d just have to be careful about multiple roots, which the delta function does a good job of. 13.3 Poisson radioactive decay When sampling decay, scientists often take a small time interval ∆t such that λ∆t = ⟨n⟩is very small. This is the limit ⟨n⟩≪1: the distribution is also correct in this limit. In general, the probability to get one count is ⟨n⟩= p: our question is to find the probability of two counts in that small interval. Is it p2? 46 Example 91 Consider a die with N faces. We throw the dice twice (n = 2): what is the probability we get two 1s? It’s 1 N2 . On the other hand, what’s the probability we get exactly one 1? It’ll be 2N−2 N2 ; as N →∞, this goes to 2 N . So notice that the probability of two 1s is actually not the square of the probability of one 1! In fact, it’s p2 4 . But back to the radioactive decay case. Does the same argument work here? Well, the Poisson distribution is pn = ⟨n⟩n n! e−⟨n⟩. Taking this to the limit where ⟨n⟩≪1, we can neglect the exponential term, and pn = pn n! . This isn’t the same as the p2 4 , because throwing a die is not Poisson - it’s binomial! To modify the distribution into one that’s more Poisson, we have to make N, the number of sides, go to infinity, but we also need to take n, the number of throws, to infinity. We’ll do this more formally next time, but if we take n, N →∞while keeping the expected number of events the same, then n N should be constant. This will indeed get us the desired Poisson distribution! Question 92. Let’s say we have a count of N for radioactive decay: what is σ? This is shot noise: it’s just √ N. Question 93. Let’s say we do a coin toss and flip it some number of times, getting N heads. What’s σ for the number of heads that appear? Binomial variance works differently: since σ2 = np(1 −p), σ = √ N/2! Question 94. What if the probability for a head is 0.999? In this case, σ2 is much less than N, and we’ll have basically no fluctuation relative to √ N. So binomial distributions work in the opposite direction! On the other hand, taking probability of a head to be 0.001 will give basically √ N. So that’s the idea of taking the binomial distribution an(1 −a)N−n N n  with mean Na and variance Na(1 −a). If 1 −a is very small, this yields similar statistics to the Poisson distribution! 14 February 28, 2019 There is an exam on March 12th, so here is some information on it! It will be at 11-12:30 (during regular class hours), and there will be four questions on the exam. This year, Professor Fakhri will post the past five years’ worth of exams, and three of the four problems will be from the past five years. They will be posted in the next few days, and we’ll have about 10 days to work through those. (One question will be new.) The material will cover things up to next Thursday. The next two lectures will talk about Shannon entropy, and those are the last lectures that will be on the exam. There will also be an optional review session held by the TAs next Thursday. Next week, the American Physical Society meeting will be taking place, so if we want extra office hours, we should send over an email! 47 14.1 Overview We started talking about conditional probability last time, which will be helpful in talking about canonical ensemble properties. We thought about the probability distribution of a sum of random variables, particularly thinking about doing repeated measurements of some quantity. The idea is that we’ll get closer and closer on average to the actual quantity. The idea was that we started with a two-variable probability distribution, and we wanted to find the probability that X lies between x and x + dx given a fixed value of y. This is denoted p(X|Y ). 14.2 Conditional probability Claim 95. p(x, y) is proportional to p(x|y). We know that Z p(x|y)dx = 1, since with y held fixed, we expect to find x somewhere (we’re just limiting ourselves to a one-dimensional probability distribution). In addition, Z p(x, y)dx = p(y), since this is “all possible values of x” for a given y. Thus, we can see that (essentially removing the integrals), p(x|y) = p(x, y) p(y) . This is the Bayesian conditional probability formula! Fact 96 We plot p(x|y) using “contour plots.” Example 97 Let’s say that the probability of an event happening is uniform inside a circle of radius 1 and 0 everywhere else. We can write this mathematically using the Heaviside step function: Definition 98 Define the Heaviside step function θ(x) = Z 0 −∞ δ(s)ds. This can be written as θ(x) =    1 x > 0 0 x < 0 and it is unclear what θ(0) is. Then the example above has a probability distribution of p(x, y) = 1 π θ(1 −x2 −y 2), 48 since we only count the points with 1 −x2 −y 2 ≥0 and the normalization factor comes from the area of the circle (π). Then if we want to find the probabilities p(x), p(y), p(y|x), we can do some integration: • To find p(x), integrate out the ys: p(x) = Z dy 1 π θ(1 −x2 −y 2) = 1 π Z √ 1−x2 √ 1−x2 dy = 2 π p 1 −x2 since we take the limits to be the zeros of the argument of θ. This holds for all |x| < 1 (the probability is 0 otherwise). • Similarly, we find p(y) = 2 π p 1 −y 2 for |y| < 1. • Finally, to find the conditional probability, p(y|x) = p(x, y) p(x) = 1 πθ(1 −x2 −y 2) 2 π √ 1 −x2 = θ(1 −x2 −y 2) 2 √ 1 −x2 , which is 1 2 √ 1−x2 for |y| < √ 1 −x2 and 0 everywhere else. The idea is that we were initially choosing points randomly in the circle, so the distribution for a given x should also be uniform in y. Definition 99 Given two random variables X and Y , define them to be statistically independent if and only if p(x, y) = p(x)p(y) This means knowing y tells us nothing about x and vice versa: in other words, it’s a corollary that p(x|y) = p(x), p(y|x) = p(y). Data analysis uses Bayes’ Theorem often, so we should read up on it if we’re curious! Also, see Greytak’s probability notes page 27 to 34 on jointly Gaussian random variables. 14.3 Functions of random variables Suppose x is a random variable with probability distribution dFx dx = p(x) [recall F is the cumulative distribution function]. Let y = f (x) be a function of a random variable x: what is the probability distribution dFy dx = p(y)? Example 100 Consider a one-dimensional velocity distribution pV (v) = dFV dv = ce−mv 2/(2kBT ). Given that E = mv 2 2 , is there any way we can find the probability distribution p(E)? The naive approach is to use the chain rule: just say dFE dE = dFE dv dv dE . But if we compute this, dE dv = mv = √ 2mE, 49 and we can plug this in to find p(E) = ce−E/kt 1 √ 2mE . Unfortunately, this is not normalized: instead, let’s use delta functions to try to get to the right answer! We can write dFx dx = p(x) = Z ∞ −∞ dξp(ξ)δ(x −ξ) where the delta function only plucks out the term where x = ξ. So dFy dy = Z ∞ −∞ dξp(ξ)δ(y −f (ξ)) (basically, we only select the values of ξ where f (ξ) = y). Using the important property of delta functions, this is dFy dy = X ξi p(ξi) df dξ ξi . So now if we have dFv dv = p(v) = ce−mv 2/(2kBT ) as before, we can write dFE dE = Z ∞ −∞ p(u)δ  E −mu2 2  du since we want all values of E equal to mu2 2 . Notice this happens at ui = ± q 2E m : there’s two roots, so that results in dFE dE = 2 √ 2mE ce−E/kBT which (can be checked) is normalized in the same way. The chain rule method misses the multiple roots! 14.4 Sums of random variables Let’s say we have x1, · · · , xn random variables with probability density functions pj(xj) for all 1 ≤j ≤n. Assume xjs are all statistically independent: then p(x1, x2, · · · , xn) = n Y i=1 pj(xj). For simplicity, let’s say that all probability distributions pj are the same function p(x). Then we can also write this as p(x1, · · · , xn) = n Y i=1 dyjp(yj)δ(xj −yj) (the delta function notation will make our lives easier later on). Our goal is to find the mean, average, variance, and other statistics for x1 + · · · + xn. Fact 101 This is applicable for experiments that repeatedly try to measure a quantity x. The average measured value x should have a probability distribution that grows narrower and narrower! 50 Using the notation Sn = n X j=1 xj, xn = Sn n , our goal is to find the uncertain in our measurement after n trials. More specifically, we want to find the probability distribution of x. Proposition 102 The variance of the average is proportional to 1 n, and as n →∞, this becomes a Gaussian. Why is this? The probability distribution for xn, much like the examples above, is p(xn) = n Y j=1 Z p(yj)δ xn −1 n n X k=1 yk ! dyj and the mean of xn is Z p(xn)xndxn = n Y i=1 Z dyjp(yj)1 n n X k=1 yk. Switching the sum and product, this is 1 n n X k=1 n Y j=1 Z p(yj)ykdyj = 1 nn⟨x⟩= ⟨x⟩, and this is just a convoluted way of saying that the average expected measurement is just the average of x. Next, let’s find the variance of the averages: we can first compute ⟨xn 2⟩= Z xn 2p(xn)dxn, which expands out to Z dxnxn 2 n Y j=1 Z dyjp(yj)δ xn −1 n n X k=1 yk ! and simplifying this by evaluating the delta functions, this becomes 1 n2 n Y j=1 Z p(yj)dyj n X k=1 yk !2 = 1 n2 n Y j=1 Z p(yj)dyj n X k=1 y 2 k + 2 X k>ℓ ykyℓ ! So this yields 1 n2 n⟨x2⟩+ n(n −1)⟨x⟩2 , so since we are trying to find the variance, var(xn) = ⟨xn 2⟩−⟨xn⟩2 = 1 n⟨x2⟩−1 n⟨x⟩2 = var(x) n , as desired. So σ(xn) = 1 √nσ(x), and this means that the standard deviation of the average gets smaller relative to the standard deviation of x as we make more measurements! 51 Proposition 103 So the distribution of the sum of random variables is dFSn dSn = pSn(Sn) = n Y j=1 Z dyjp(yj)δ Sn − n X k=1 yk ! . For example, if we have two random variables, p(S2) = ZZ dxdyp1(x)p2(y)δ(s2 −x −y) = Z dxp1(x)p2(s −x) which is the convolution of p1 and p2, denoted P1 ⊗P2. Fact 104 The sum of Gaussian random variables is another Gaussian with mean and variance equal to the sums of the means and variances of the two original Gaussians. Similarly, the sum of Poissons (λT)n n! e−λT is a Poisson distribution with ⟨SN⟩= N⟨n⟩and variance var(SN) = N var n = NλT. 15 March 4, 2019 (Recitation) We’ll go through some ideas from the problem set. 15.1 Probability distributions and brightness We can basically think of this problem as a sphere with fireflies emitting light. What do we see? Well, we know that the light intensity depends on ρ, where ρ = p x2 + y 2. So we basically want to integrate out the z-direction. The idea is that we should go from p(x, y, z) to different coordinates. Any probability distribution always integrates to 1, so we are interested in only picking out the values p(ρ) = Z p(x, y, z)δ(ρ − p x2 + y 2)dxdydz where ρ2 = x2 + y 2. Question 105. Is p(ρ) the brightness? Not quite! p(ρ)dρ gives the “probability” or number of stars in a narrow strip from ρ to ρ + dρ. So to find the brightness, we need to divide through by 2πρ, since we need to divide by the area. This means B(ρ) = p(ρ)dρ 2πρdρ = p(ρ) 2πρ . When we calculate this and plot it, p(ρ) looks linear for small ρ, so B(ρ) starts offapproximately constant close to the center! This makes sense: the “thickness” at each point is about the same near the center. 52 15.2 Change of random variables Let’s say we have some function (think potential energy) E = f (x) with inverse function x = g(E). Let’s say we’re given some spatial distribution p(x), but there is an energy E at each point: how can we find the probability density function p(E)? We can use the cumulative function F(x), defined by d dE F(E) = p(E). By the chain rule, we can regard E = f (x), so dF dx = dF df (x) df (x) dx = ⇒p(x) = p(E) df dx . But it is possible that our function is multi-valued: for example, what if E = f (x) = x2 is quadratic? Then the function is not one-to-one: in this case, our cumulative distribution should not look like F(E) = Z g(E)= √ E −∞ p(x)dx, but rather F(E) = Z √ E − √ E p(x)dx so that we get all values x such that x2 ≤E. Another example of this is that if we have E = x3, we just have F(E) = Z 3 √ E −∞ p(x)dx, since again we want all values of x that make x3 ≤E. The idea is that being “cumulative” in x doesn’t necessarily mean the corresponding values are any kind of “cumulative” function in f (x), so we need to be careful! This is part of the reason why we like to use delta functions instead: the central idea is that we need to look at all the roots of E = p(x). So the rest is now a differentiation: p(E) = d dE F(E) = d dE Z g(E) −∞ p(x)dx, where g is a root of the equation f (x) = E = ⇒x = f −1(E) (there could be multiple such roots, which means we’d have to split into multiple integrals). So by the first fundamental theorem of calculus, this is p(E) = p(g(E)) dg dE = ⇒p(E) = p(x) 1 df /dx as before! In general, if our expression looks like p(E) = d dE Z xi+1 xi p(x)dx + Z xi+3 xi+2 p(x)dx + · · ·  and this gives the sum X i p(xi) dg dE f (xi)=E , where the absolute value comes from the fact that the lower limits have a different sign of slope from the upper limits, which cancels out with the negative sign from the first fundamental theorem. Here’s one more method that we will all like! Add a theta function θ(E −f (x)): we want to integrate over all 53 values f (x) ≤E, where θ is the step function, and this is just F(E) = Z p(x)dxθ(E −f (x)) (think of this as “including” the parts that have a small enough value of f (x)). But now taking the derivative, p(E) = d dE F(E) = Z p(x)θ′(E −f (x))dx and the derivative of the theta function is the delta function p(E) = Z p(x)δ(E −f (x))dx. We’ve done this before! We just care about all roots where f (x) = E, and this is X i p(xi) 1 |f ′(xi)|, and indeed this is the same result as we’ve gotten through other methods. Fact 106 Key takeaway: sum over roots and correct with a factor of the derivative! Now if we have a function E as a function of multiple variables f (x, y, z), we can just pick out the “correct values” via p(E) = ZZZ p(x, y, z)δ(E −f (x, y, z))dxdydz. How do we evaluate the derivatives here? It’s possible that δ in general could be a product of three delta functions: for example, think of a point charge in electrostatics. But in this case, we’re dealing with a one-dimensional delta function. We need to solve the equation E = f (xi, y, z) = 0: we may have two roots x1, x2, so now we have p(E) = ZZ 2 X i=1 p(xi, y, z) · 1 ∂P ∂x (x, y, z) i dydz The point is that the delta function eliminates one variable, so integrate one variable at a time! Alternatively, there may be some condition on y and z (for example, if E = x2 + y 2 + z2, then y 2 + z2 are forced to be within some range), and that just means we have to add an additional condition: either an integration or a theta function. 16 March 5, 2019 Fact 107 The optical trap, a focused beam of light, can trap particles of size approximately 1 micron. People can play Tetris with those particles! Once the optical traps are turned off, though, particles begin to diffuse, and this can be explained by statistical physics. The American Physical Society meeting was held this week: 90 percent of the awards this year were given to statistical physics applied to different fields. The material covered was a bit more challenging than areas we’ve explored so far, but there are many applications of what we’ve discussed! 54 16.1 Entropy and information: an overview Let’s start by reviewing why we’ve been talking about all the topics of this class so far. We used the first law dU = ¯ dQ + ¯ dW, where ¯ dW is a product of an intensive and extensive quantity such as −PdV (generalized force and displacement). We found that ¯ dQ is not a state function: it does depend on the path we take to get to a specific state. Entropy is an old idea, and it comes from thermodynamics as a way to keep track of heat flow! It turns out that this S is indeed a state function, and it can characterize a macroscopic state. We’re going to use the ideas from probability theory to write down an “entropy of a probability distribution” and measure how much “space” is needed to write down a description of the states. One definition we’ll see later is the probabilistic S = − X pj log pj. Consider the following thought experiment: let’s say we have some particles of gas in a box, and we have the experimental tools to measure all of those positions and velocities. We write them down in a file, and we want to compress the files to the smallest possible length. If we compress efficiently, the length of the file tells us something about the entropy of the distribution. For example, if all of the particles behave “simply,” it will be easy to efficiently compress the data, but if the system looks more “random” or “variable,” the compression will be less effective. Fact 108 In general, if we heat up the room by 10 degrees and repeat this process, the compressed file will have a longer length. Generally, we want to see the change in length of the file per temperature or heat added! This is a connection between two different ideas: an abstract length of a computer program and a tangible heat. 16.2 Shannon entropy and probability Let’s say we have a set S = {s1, · · · , sN} whose outcomes have probability p1, · · · , pN. An example of a “well-peaked” distribution is p1 = 1, pj = 0 ∀j ̸= 1. If we see an event from this probability, we are “not surprised,” since we knew everything about the system from the beginning. On the other hand, if we have all pj approximately equal, pj = 1 N ∀j. The amount of surprise for any particular event is about as high as it can be in this case! So we’re really looking at the amount of randomness we have. Claude Shannon published a paper that was basically the birth of information theory: 55 Proposition 109 (Shannon, 1948) What is the minimum number of binary bits σ needed on average to reproduce the precise value of a symbol from the given bits? This turns out to be σ = − N X j=1 pj log2 pj. Example 110 Let’s say we have a fair coin that can come up heads or tails with PH = PT = 1 2. If we have a string of events like HTHHTHTTHTTT, we can represent this in a binary string by H →1, T →0. Clearly, we do need 1 bit to represent each coin flip. Our “symbol” here is an individual “head” or “tail” event, and the minimum number of bits needed is − 1 2 log2 1 2 + 1 2 log2 1 2  = 1. So our “coding scheme” sending heads and tails to 0 and 1 is “maximally efficient.” Example 111 Let’s say we have four symbols A, B, C, D, all equally likely to come up with probability 1 4. We can represent this via A →00, B →01, C →10, D →11. The Shannon entropy of this system is indeed − 4 X 1 1 4 log2 1 4 = 2, so we need at least 2 bits to encode each symbol. Example 112 Let’s say we have three symbols with probability PA = 1 2, PB = 1 4, PC = 1 4. Naively, we can represent A = 00, B = 01, C = 10, so we need 2 bits per symbol. But there is a better code? Yes, because the Shannon entropy − 1 2 log2 1 2 + 1 4 log2 1 4 + 1 4 log2 1 4  = 3 2, so there should be a way to code each symbol in 1.5 bits on average! Here’s a better coding scheme: use A = 0, B = 10, C = 11, which gives an average of 1 2 · 1 + 1 4 · 2 + 1 4 · 2 = 3 2 Now given a string 00101000111101001010, 56 we can reconstruct the original symbol: if we see a 0, then pull it out as an A, if we see a 1, pull it and the next number out to form a B or C, and rinse and repeat until we reach the end! In general, the idea is to group symbols to form composites, and associate high probability with shortest bit strings. In the case above, we had a high chance of having A, so we made sure it didn’t require too many bits whenever it appeared. Example 113 If we have a biased coin with probability 3 4 of heads (A) and 1 4 of tails (B), the Shannon entropy is − 1 4 log2 1 4 + 3 4 log2 3 4  ≈0.811. So there should be a way to represent the heads-tails method in less than 1 character per flip! We can group symbols into composites with probabilities                AA : 9 16 AB : 3 16 BA : 3 16 BB : 1 16 . These are all fairly close to powers of 2, so let’s represent A as 0, AB as 10, BA as 110, and BB as 111 (this is not perfect, but it works pretty well). Then on average, we need 9 16 · 1 + 3 16 · 2 + 3 16 · 3 + 1 16 · 3 ≈1.688 bits to represent two symbols, for an average of less than 1 bit per symbol! This is better than the version where we just use 1 for heads and 0 for tails. Fact 114 If we instead group 3 symbols, we may be able to get an even better coding scheme! We do have to make sure we can umambiguously decode, though. By the way, for our purposes from now on, we’ll be using natural log instead of base 2 log, since we have continuous systems instead. Note that log2 X = ln X ln 2 , so the Shannon entropy is σ = −1 ln 2 X n pn log pn, and if we instead have a continuous probability distribution, we can integrate instead: σ = − Z p(n) log2 p(n)dn, where we normalize such that R p(n)dn = 1. 57 16.3 Entropy of a physical system Now that we have some intuition for “representing” a system, let’s shift to some different examples. Consider the physical quantity S = −kB X i pi ln pi. Note that all terms here are nonnegative, so the minimum possible value is S = 0: this occurs when there is only one event with probability 1 and no other possibility. This is called a delta function distribution. On the other hand, the maximum possible value occurs with a uniform distribution: where all pis are the same. If there are M events each with probability 1 M , this evaluates to −kB X i 1 M ln M = kB ln M. (By the way, the kB is a way of converting from Joules to Kelvin for our measure of temperature.) Proposition 115 This means S is a measure of “dispersion” or “disorder” of the distribution! So this gives an estimate of our probability distribution, or at least its general shape. For example, if we have no information about our system, we expect it to be uniform. This yields the maximum possible value of S (or entropy), and this is the best unbiased estimate of our distribution. Once we obtain additional information, our unbiased estimate is obtained by maximizing the entropy given our new constraints. Fact 116 This is done using Lagrange multipliers! If we have some new information ⟨F(x)⟩= f (we measure the value of some function F(x)), we want to maximize S(α, β, {pj}) = − X i pi ln pi −α( X (pi) −1) −β( X (piF(xi)) −f ) Our constraints are that our distribution must be normalized and that we want ⟨F(x)⟩−f to be close to 0 as well. It turns out this gives a Boltzmann distribution pi = α exp(−βF(xi)). Here β is fixed by our constraints, and α is our normalization factor! For example, we could find β by knowing the average energy of particles. We’ll see this a bit later on in the course. 16.4 Entropy in statistical physics Recall that we specify a system by stating a thermodynamic equilibrium macrostate: for example, we give the internal energy, pressure, temperature, and volume of a system. This is specifying an ensemble. On the other hand, we can look at the microstates of our system: they can be specified in quantum systems by numbers {nj, 1 ≤j ≤N} or in the classical systems by positions and velocities {xi, vi, 1 ≤i ≤N}. We can set up a distinction here between information theory and statistical mechanics. In information theory, our ensembles look very simple: we have usually a small number of possible outcomes, but the probability distributions can look very complicated. On the other hand, ensembles in statistical mechanics are often much more complicated 58 (lots of different possible microstates), but our probability distributions are much more simple. The idea is that S, our entropy, will be the maximum value of S = −kB X pi ln pi across all probability distributions {pi}. But what are the distributions given our constraints? That’s what we’ll be looking at in the rest of this class! 17 March 6, 2019 (Recitation) Let’s start with a concrete example of the discussion from last recitation. Let’s say we have a probability distribution that is uniform inside a circle of radius R: p(x, y) = 1 πR2 for x2 + y 2 ≤R2 and 0 outside. We’re going to find the probability distribution p(r) in three different ways. 17.1 The messy way First of all, if we use Cartesian coordinates, we can directly write this in terms of a delta function p(r) = Z R −R Z p R2−y 2 − p R2−y 2 p(x, y)δ(r − p x2 + y 2)dxdy Let’s take care of the delta function as a function of x. We know the delta function δ(f (x)) is δ(x), divided by f ′(x) at a zero of the function, so let’s compute the roots! f (x) = r − p x2 + y 2 = ⇒x± = ± p r 2 −y 2. The absolute value of the derivative is equal at both roots: f ′(x) = − x p x2 + y 2 = ⇒|f ′(x±)| = p r 2 −y 2 r = r 1 − y r 2 . So now, we can evaluate our boxed expression above. The delta function is integrated out (except that we gain a factor of |f ′| in the denominator, and we replace x with the root xi wherever it appears. But here the probability distribution is uniform (does not depend on x explicitly), and the two roots have equal |f ′(xi)|, so we get a factor of 2. This simplifies to = X xi roots Z R −R Z p R2−y 2 − p R2−y 2 δ(x −xi) πR2 · 1 q 1 − y r 2 dx dy = Z R −R 1 πR2 · 2 q 1 − y r 2 dy(?) where the boxed terms integrate out to 1, since xi = ± p r 2 −y 2 is always in the range h − p R2 −y 2, p R2 −y 2 i . But we must be careful: if |y| > |r|, or if |r| > R, we don’t actually have these two roots! So we put in some constraints in the form of theta (step) functions: they force R > r and r 2 > y 2: = θ(R −r) 2 πR2 Z R −R θ(r 2 −y 2) q 1 − y r 2 dy (where we have r 2 and y 2 in the second theta function to deal with potentially negative values of y). What does that 59 θ function mean? The inner one just means we integrate across a different range of y: θ(R −r) 2 πR2 Z r −r 1 q 1 − y r 2 dy and now we can integrate this: substituting u = y r , this is θ(R −r) 2r πR2 Z 1 −1 1 √ 1 −u2 du and the integral is sin−1(u)|1 −1 = π, resulting in a final answer of p(r) = θ(R −r) 2r R2 =    2r R2 r < R 0 r ≥R It’s stupid to use Cartesian coordinates here, but this shows many of the steps needed! 17.2 Polar coordinates Here’s a faster way: the probability distribution x = ρ cos θ, y = ρ sin θ becomes p(x, y) = 1 πR2 for ρ < R. So now we can write our boxed double integral above in our new coordinates: Z R 0 Z 2π 0 p(x, y)δ(r −ρ)ρdθdρ. The integration over dθ gives a factor of 2π, and p is uniform, which simplifies this to 2π πR2 Z R 0 δ(r −ρ)ρdρ. The delta function has only the root ρ = r: since we’re integrating over [0, R], this is p(r) = 2π πR2 r · θ(R −r) = 2r R2 θ(R −r), which is identical to what we had before. 17.3 Without delta functions We can use cumulative density distributions instead! What is the cumulative probability F(r) = Z r 0 Z p r 2−y 2 − p r 2−y 2 p(x, y)dxdy? This is the probability over all x2 + y 2 ≤r 2. This integral is just p(x, y) (which is constant) times the area of a circle with radius r, which is πr 2 πR2 = r 2 R2 60 as long as r < R. So p(r) is just the derivative of F(r): p(r) = dF(r) dr = 2r R2 , and we’re done! We could have fixed up the edge case of r > R by adding a theta function θ(r −ρ) inside the original integrand. Then the derivative of the theta function is the delta function, which gives the same delta function as in our first method. To summarize, we can generally avoid delta functions with cumulative densities. 17.4 Parity violation We once assumed that if we flip our coordinates x →−x, y →−y, and so on, there is no difference in our laws. Basically, everything in a mirror would also obey the laws of physics. But the Wu experiment proved this to be false! This is called parity (P) violation. But there’s something more interesting: people managed to include charge (C) conjugation, changing matter and antimatter, and then CP conservation seemed to be true. But it was found that even this is violated! Fact 117 So if you want to tell your friends in the alien world what the right side is, you can say to run a current through a coil of wire. Put in Cobalt-60, and the magnetic field from the coil will have more electrons coming out from the top than the bottom if our coil is counterclockwise. This is a right-handed current! But if we want to explain that our heart is on the left-hand side, we can put ourselves in a solenoid. Run a current through the solenoid, and you can say that the Cobalt-60 emission goes up. Then the current now flows across our chest from the right to the left! But they’ll be made of antimatter, so they might hold out the wrong hand when they shake hands. 18 March 7, 2019 Just a reminder: there is an exam on Tuesday in this classroom. It will go from 11 to 12:30 (usual class time). The past exams have been uploaded: note that three of the four problems on this year’s exam will be reused from previous exams posted on Stellar. If we are able to do the problem sets and past exams, we have a good mastery of what’s going on. By the way, the concept of density of states has been moved: it will come later, and it will not come up on the exam. There will be an optional review session from 5 to 7 today, and Professor Fakhri will also hold additional office hours. Material is everything from class until today, though entropy will be more about concepts than specific examples. The next problem set will not be due on Friday. 18.1 Quick review and overview We’ve been learning about macroscopic quantities and connecting them to microscopic systems, and this led us to the idea of entropy. This was a concept that Boltzmann introduced before information theory: basically, we care about how “confused” we are about a system. 61 Today, we’re going to expand on the concept of thermodynamic entropy and introduce the second law of thermo-dynamics, which claims that entropy is nondecreasing with time. This is an emergent phenomenon! Remember that for a thermodynamic ensemble, we defined our entropy to be S = kB X j pj ln pj, where we’re summing over all possible microstates j that occur with a probability of pj. Note that this is also ≡(kB ln 2)σ, where σ is the Shannon entropy. 18.2 Looking more at entropy Proposition 118 Thermodynamic equilibrium occurs at the (unbiased) probability distribution which maximizes entropy: S = max pj −kb X j pj ln pj. Recall from last time that an example of such an unbiased probability distribution is the uniform distribution: all states occur with equal probability. If we have no prior information, this is the best “guess” we can have for what our system looks like. In this case, the distribution looks like pj = 1 Γ, where Γ is the total number of consistent microstates. (Γ is known as the multiplicity.) Plugging this in, S = −kb Γ X j=1 1 Γ ln 1 Γ = kb ln Γ . Fact 119 On Boltzmann’s tombstone in Vienna, S = k log W is written. This equation is kind of the foundation of statistical physics! Note that S is a measure of the macroworld, while W or Γ is a measure of the microworld, so this is a good relationship between the two. Here’s some additional facts about our entropy S. • S is a state function of P and V . In other words, it is independent of the path we took, so we can compute S from other macrostates. For example, we can write the entropy S of a gas in a box in terms of the volume, number of molecules, and internal energy S = S(U, V, N). On the other hand, if we have a magnet, the state function depends on the magnetization ⃗ M. 62 • The proportionality constant of kB arises because we chose to use units of temperature. In particular, we could have units of temperature in Joules if we just let kB = 1. 18.3 The second law Recall the first law of thermodynamics, which tells us about conservation of energy: dU = ¯ dQ −PdV. Proposition 120 (Second Law of Thermodynamics) Entropy of an isolated system cannot decrease. From an information theory perspective, this is saying that our ignorance of a system only increases with time. Let’s look at an example by time-evolving a system! Proposition 121 In both classical and quantum systems, the time-evolution of a microstate is both causal and time-reversal invariant. What do those words mean? Causality says that each microstate at some time t1 evolves into a unique, specific microstate at time t2 > t1. So causality says that we can’t have two different microstates at t2 that both originated from t1: if we had 100 microstates at time t1, we can’t have more than that at a later time t2. Meanwhile, the concept of time-reversal invariance is that both laws of classical and quantum physics are reversible if we switch t →−t. For instance, any wavefunction |ψ(t)⟩or classical ⃗ x(t), ⃗ v(t) that is a valid also gives a valid |ψ(−t)⟩and ⃗ x(−t), ⃗ v(−t). So if we think about this, it means we cannot have two microstates at time t1 that converge into one at a later time t2 either. So our ignorance about the system cannot decrease! But can the entropy increase? 18.4 A thought experiment Example 122 Consider a box with a partition, and one half of the box is filled with a gass with a known U, V, N at some time t < 0. (The other part is filled with a vacuum.) At time t = 0, the partition is removed. Now the gas fills a volume of 2V , and U and N are still the same, so there are many more microstates that are possible. This increases our entropy! The kinetic energy of the particles has not changed, but our ignorance of the system has increased. There are many more possible values for the initial position and momentum of every particle. What’s the change in the number of microstates Γ? If we assign a binary variable to each particle, which tells us whether the particle is on the left or right side of the box, after t > 0, we now need an extra binary bit to tell us about the system. Thus, with N particles, our change in Shannon entropy is ∆σ = N. Thus ∆S = kb ln 2∆σ = ⇒∆S = Nkb ln 2 and since S ∼log Γ, we get a factor of 2N more possible microstates! 63 Fact 123 This doesn’t break causality or time-reversal. The idea is that every microstate before t < 0 goes to exactly one microstate at t > 0, but we don’t know which one it is: the probability distribution is still uniform, just with a larger range of possibilities. Notice that there is some time from our initial state (U, V, N) to our final state (U, 2V, N) to reach equilibrium again (so that we can define our state functions). We can think of this as “mixing” states and making the probability distribution more uniform! There is a whole different field called ergodicity. Fact 124 In any (real) macroscopic system, regardless of the initial configuration, over a long time, the system will uniformly sample over all microstates. Basically, over a long time, the sampling of a probability distribution will yield all microstates with equal probability. For example, instead of preparing many initial configurations, we can prepare one particle and sample it many times. Fact 125 (Ergodic hypothesis) We can compute the properties of an equilibrium macrostate by averaging over the ensemble. If there are microstates Si that occur with probability pi, and we have some function f (Si) of microstates, we can compute a property ⟨f ⟩= X si f (si)p(si). But instead, we can sample our system and average: ⟨f ⟩= 1 T Z T 0 f (t)dt. This time T may be large though! Fact 126 (Systems out of equilibrium) There are some systems that have a slow relaxation time, so they’ll never reach equilibrium within a reasonable amount of time! An example is the universe. In the rest of this class, we’ll come up with ensembles, and find unbiased probability distributions consistent with a macrostate. We’ll try to see what conditions we can impose to define thermodynamic quantities! 18.5 Moving on We’ll be talking about different kinds of ensembles (collections of microstates) in this class. A microcanonical ensemble is mechanically and adiabatically isolated, so its volume V and number of particles N is constant. In such a system, we can define a temperature! After that, we will discuss the canonical ensemble, which trades fixed U for fixed T. We can then look at grand canonical ensembles, which are systems at fixed chemical potential. Recall that S = S(U, V, N) is a state function on equilibrium states, and ∆S > 0 for isolated systems. We also know that it is an extensive quantity (it is additive) like N, V , and U: it turns out the conjugated quantity (generalized force) here is temperature T. 64 How can we show additivity of entropy? Lemma 127 Given two independent non-interacting systems A and B, the entropy SAB = SA + SB. Proof. A has NA possible microstates with probability Pα,A, so SA = −kB X α Pα,A ln Pα,A and similar for B. By statistical independence, PA,B = PA · PB, so (here Pα,β refers to the probability Pα,APβ,B for brevity) SAB = −kB X α,β Pα,β ln Pα,β = −kB X α,β Pα,APβ,B ln(Pα,APβ,B) = −kB X α,β Pα,APβ,B (ln Pα + ln Pβ)) which can be written as SAB −kB X β Pβ,B X α Pα,A ln Pα,A −kB X α pα,A X β Pβ,B ln Pβ,B and as the boxed terms are 1, this is just SA + SB as desired. 19 March 11, 2019 (Recitation) We’ll cover some short questions and then relate Poisson, Binomial, and Gaussian distributions to each other. As a quick refresher, if we have a probability distribution p(x, y) and we want to find it in terms of another variable z = f (x, y), then p(z) = Z p(x, y)δ(z −f (x, y))dxdy will pick out the correct values of z. The rest is mathematics: find the roots, magnitudes of derivatives, and make the relevant substitutions. 19.1 Different probability distributions Question 128. What is the counting distribution for radioactive decay? Basically, measure the number of particles that decay / do something else in some interval T: if we do this multiple times, what’s the distribution going to look like? Remember that we’ve discussed three kinds of probability distributions here: binomial, Poisson, and Gaussian. We’re always looking for a count rate: can we distinguish anything between these three kinds? Well, in a binomial distribution, we have some finite number of trials N, so the possible range of n, our count, is always between 0 and N. But for the Poisson distribution, n is any nonnegative integer, and the Gaussian can be any real. The idea is that if our binomial distribution’s tail is sufficiently flat on the positive end, because our probability p is small or our number of trials N is large enough, then we can extend it to ∞and treat it similar to a Poisson 65 distribution. But on the other hand, if our tail is sufficiently flat on the negative end, we can also extend it to −∞ and treat it like a Gaussian distribution! So the answer to the question is “yes, Poisson is correct,” but not quite! There is indeed a maximum count rate: N, the number of total atoms in our radioactive material. So this is sort of binomial, but those events are so unlikely that we can neglect them completely. How do we rigorize this? Remember that our binomial distribution for N trials of an event of probability a is p(n, N) = an(1 −a)N−n N n  . If we let N →∞, but we keep our mean Na = ⟨n⟩constant, then a = ⟨n⟩ N , and our distribution becomes p(n, N) = ⟨n⟩n Nn  1 −⟨n⟩n Nn  N n  . We can neglect the −n in the second exponent, since n ≪N, and this middle term now approaches e−⟨n⟩. What’s more, N n  = N(N−1)···(N−n+1) n! is essentially Nn n! , and now we’re left with p(n, N) ≈⟨n⟩n Nn e−⟨n⟩Nn n! = ⟨n⟩n n! e−⟨n⟩ which is the Poisson distribution as we wanted! Fact 129 So letting N go to infinity but adjusting the probability accordingly, we get a Poisson distribution. On the other hand (or as a subsequent step), if we make ⟨n⟩larger and larger, this gives us a Gaussian distribution by using Stirling’s approximation and using a Taylor expansion. This will yield C exp  −(n −⟨n⟩)2 2⟨n⟩  . The idea is that all that matters is the values of n and ⟨n⟩. But here’s an alternative way to go from binomial to Gaussian: keep a constant, and let N get larger. This now yields C′ exp  −(n −⟨n⟩)2 2⟨n⟩  · (1 −a). So in this case, the variance is not ⟨n⟩= Na but Na(1 −a) = ⟨n⟩(1 −a). The reason this is different is because when we went to the Poisson as an intermediate step, we forced a to be small, which meant we could neglect the 1 −a term! So now let’s look at a = 1 2, which is the case of a random walk. So our variance is ⟨n⟩ 2 , and let’s say the step size of our random walk is 1 (so we move to the right or to the left by 1 unit each time). Fact 130 Notice that, for example, if we have 10 steps, we expect to move to the right 5 times. But if we move to the right 6 times instead, our net walk is 6−4 = 2: in general, if our number of right moves is k more than the mean, the net walk is 2k. So if we substitute in for a step size of our random walk x, n −⟨n⟩= x 2, and ⟨n⟩= N 2 . Rewriting our Gaussian, we will just get Brownian motion p(x) ∝e−x2/2N. 66 In general, measuring N atoms at a rate of λ for time t just yields ⟨n⟩= Nλt. But the concept behind all the counting is to track n, the number of counts, relative to ⟨n⟩, the average number of counts. Fact 131 If we have a probability distribution pn, the values of ⟨n⟩and var(n) are telling us data about one trial or sample from the distribution. But if we want to calculate n, which is the average of N measurements, the variance changes by a factor of 1 N (since variances add, so the variance of our sum is N var(n), and then we divide our sum by N, which divides our variance by N2). By the way, the formula for a normalized Gaussian distribution is on the equation sheet, so we don’t have to worry about it too much. Let’s think about the stars problem from a past exam: we have stars distributed with density ρ stars per light-year cubed. What is the probability that there are no stars within r of a given star? Basically, we take a volume V , and we want the probability no other star is in that given volume. We can think of this as taking small pieces of volume, where each one is independent, and where there is a finite, consistent value of ⟨n⟩: average number of stars in each piece of volume. This is a Poisson distribution! So our expectation value is ⟨n⟩= ρ · V = 4πρ 3 r 3, and p(n, N) = ⟨n⟩n n! e−⟨n⟩. But why do we take n = 0 instead of n = 1? The first star just gives us a volume to look at, so we can completely ignore it. Fact 132 (Clearly false) If we go on an airplane, we should bring an explosive to be safe, because the probability of there being two explosives on the plane is almost zero. The central idea here is independence! If we guarantee that we have one star, the other stars don’t care that the first star is there. 20 March 13, 2019 (Recitation) 20.1 Geometric versus binomial distribution Let’s quickly look at the tunneling problem from the quiz: we can think of having an alpha particle inside a nucleus that eventually escapes. We know the probability that it does exactly n bounces is pn = an(1 −a). 67 This is a geometric distribution! On a related note, if we have n + 1 atoms, what is the probability that exactly 1 of them decays? Well, this looks very similar to what we have above, but we get an extra combinatorial factor (because any of the atoms can decay): P1 = p(1 −p)n n + 1 1  , and this turns the geometric distribution into a binomial one! Fact 133 Here’s another example of a geometric distribution: let’s say a patient needs a kidney transplant, but we need to screen donors to see if there is a match. Given a random blood test, the probability of a match is p: then the number of people we need to screen is pn = p(1 −p)n−1. and we can replace a = 1 −p to get something similar to the tunneling problem. So the distinction is that you try again and again until success in a geometric distribution, but there’s a finite number of trials in a binomial one. 20.2 Entropy of a probability distribution Let’s say we have an average count rate of ⟨n⟩in a Poisson distribution, meaning the variance is ⟨n⟩as well. (Think of the standard deviation as being about at 0.6 times the maximum value in a distribution that is about Gaussian.) So what does it mean to have an entropy of a distribution? Example 134 Consider a coin toss or letters in a book: can we measure the entropy of that somehow? The idea is to have N random events pulled from our probability distribution: what is the number of bits needed to represent that information on average? It’s kind of like image compression: using much less space to display the same data, but we don’t want any loss of resolution. In a coin toss, we need a 1 or 0 for each toss, since all events are randomly likely. So N random events must come from N bits, and indeed, the Shannon entropy for one event S = − X pi log2 pi = 1, Let’s go to the extreme: let’s assume we have a coin which comes up heads 99 percent of the time. How many bits of information do we need to communicate the random series 0000 · · · 010 · · · 01? Fact 135 We can just send a number that counts the number of 0s between 1s! So instead of needing about 100 bits to represent each group between 1s, we can use on average ≈log2(100) bits. More rigorously, let’s say the probability of having a 1 (corresponding to a tail) is small: ε ≈ 1 100. If we have N coin tosses, we will need an expected N · ε differences between the 1s. Each difference is about 1 ε, and we need log2 1 ε 68 bits for each one. So the expected entropy here is Nε log 1 ε = −Nε log ε, while the theoretical Shannon entropy yields S = −N (ε log ε + (1 −ε) log(1 −ε)) and as ε →0, the second term dies out to first order! This confirms the result we had before. Fact 136 One way to think of this is that the entropy gets smaller (we need less bits) as our sequence becomes less random and more predictable. 20.3 Entropy of a Poisson distribution If we think in terms of information, we are looking at a random sequence: we want to think of coding the resulting data. Well, what’s the data that we’re trying to code? If we set up our system and repeatedly count, we’ll get a series of numbers pulled from the distribution. So our question is how we can code this? How concisely can we represent the stream of random numbers? Example 137 First of all, let’s say we have a uniform distribution from 91 to 110, so there are 20 equally likely outcomes. What’s the entropy of this system? Well, we need log2(20) bits on average to represent the data! As a sanity check, if we go to the formula S = − X pi log pi and we have W equally probable options, then this Shannon entropy is just S = −W 1 W log 1 W = log W. Fact 138 Ludwig Boltzmann has S = kB ln W written on his tombstone - notice that this is just kB ln 2 times the quantity we’ve been thinking about! The multiplicative factor is just a convention to connect different fields and stick to historical reasons. So back to the probability distributions. We can perform N measurements, each of which can be one of W possibilities (for example, lottery numbers or pieces of colored paper). How many bits do we need to convey the random series that comes up? Again, we need log2 W bits for each number. But how would we encode these large numbers? In general, we want to index them: if there are 20 possible numbers, we should send them out as 0, 1, · · · , 19, not as their actual values. So looking at a Poisson distribution, we care much more about the events that occur more frequently. Looking at the p log p term, as p →0, this number approaches 0. So the improbable wings of the Poisson distribution are not important: we really care about those values within a few standard deviations! 69 So the number of numbers that dominate the sequence is around p ⟨n⟩for the Poisson distribution, and thus we estimate the entropy to be log  c p ⟨n⟩  ∼log2 c + 1 2 log2(⟨n⟩) where c is some order 1 number (from 1 to 10). Converting this to statistical mechanics, this gives us a 1 2 ln 2 term in leading order! Indeed, if we actually plug in S = − X pi log pi for the Poisson distribution (as we do in our homework), and in fact the actual number looks like (with x = ⟨n⟩) S(x) = 1 2 ln 2(1 + ln 2πx) = 1 2 ln 2 ln x + 1 2 ln 2(1 + ln 2π). So our handwaving gave us the correct result asymptotically by replacing the Poisson with a Gaussian. That second factor is approximately 4.1, so indeed we have our order 1 number. 21 March 14, 2019 Happy Pi Day! Read the paper posted on the course website. We did well on the exam; some of them haven’t been graded and will be done by the end of today. They will be brought to class on Tuesday; we can also email Professor Fakhri. If we feel like our score doesn’t reflect our understanding, also send an email. There is a problem set due on Monday (instead of Friday) at 7pm. There will be another pset for the Friday before break, but it is short (only 2 problems). Finally, there will be an interesting problem in the problem set after that about statistics of the Supreme Court. Last time, we introduced Shannon and thermodynamic entropy. Near the end of class, we found that the total entropy of a system is the sum of its independent parts, so entropy is an extensive quantity. Now we’ll relax this assumption and allow heat and work to be exchanged as well, and we’ll see what the new entropy becomes! This allows us to define temperature, and then we can connect those microscopic pictures to the macroscopic world. After that, we’ll talk about reversibility and quasi-equilibrium and how we can compute changes in entropy based on the initial and final state. 21.1 Entropy and thermal equilibrium Recall from last time that if we have independent, non-interacting systems A and B, then SAB = SA + SB. This time, let’s say that we still have A and B isolated from the outside, but there is a partition between A and B. This means that NA, NB are fixed, and so are VA, VB, but heat can flow between the two systems. Our goal is to somehow define a temperature based on SAB. We’ll let the system go to equilibrium: at that point, our entropy is maximized. Then if A has some entropy SA and internal energy UA, and B has some entropy SB and internal energy UB, we claim that SAB = SA(UA) + SB(UB) for functions SA and SB that depend on UA and UB, respectively. This is a good assumption even if we have small fluctuations. Now let’s say some infinitesimal heat ¯ dQ passes from A to B. Then dUA = −¯ dQ and dUB = ¯ dQ. But the change 70 in SAB should be zero, since we have a maximum entropy at this point! Expanding out the differential, dSAB = ∂SA ∂UA VA,NA dUA + ∂SB ∂UB VB,NB dUB. Now plugging in dSAB = 0 and using the fact that dUA = −dUB, we have at thermal equilibrium that ∂SA ∂UA VA,NA = ∂SB ∂UB VB,NB . So we want to define a state function that is equal at these two different points (in two systems in thermal equilibrium)! When we bring two systems together, the temperatures should become equal, which motivates the following definition: Definition 139 Define the temperature T of a system via ∂S ∂U V,N = 1 T . There’s no constant of proportionality here, because we used kB in our definition of entropy. Here are some important facts: • ∂S ∂U V,N only applies to systems at thermal equilibrium, but it’s a “hot” area of research to think about non-equilibrium states as well. • Temperature is transitive: if A and B are at thermal equilibrium, and so are B and C, then A and C are at thermal equilibrium. This is the zeroth law of thermodynamics. • There are other notions of equilibrium (e.g. mechanical) as well, but for now we’re ignoring ideas like partitions being able to move due to pressure. 21.2 Particles in binary states Let’s put together everything we’ve learned so far with an example! Example 140 Let’s say we have N particles, each of which can have 2 states (for example, a bit taking on values 0 or 1, or a particle with spins in a magnetic field). One way to represent this is by placing them along a number line and representing each one with an up or down arrow. Spin up gives an energy of ε0 and spin down gives an energy of ε1: let’s say n0 is the number of particles in the lower energy state ε0; without loss of generality we let ε0 = 0. Similarly, define n1 to be the number of particles in the upper energy state ε1 = ε. Note that n0 + n1 = N, and we can write this as a frequency: n1 = f N = ⇒n0 = (1 −f )N. The total internal energy of this system is ε0 · n0 + ε1 · n1 = f εN. Let’s compute the entropy of this system. If we can count the number of microstates Γ (also called the multiplicity), all such states should be equally probable, and then we can compute the entropy from there: S = kB ln Γ. 71 If we have f N particles in the upper energy state, and we have (1 −f )N particles in the lower energy state, Γ = N! n0!n1!. Now by the Stirling approximation (since N is large), S = kb ln Γ = kb(N ln N −N −(n0 ln n0 −n0) −(n1 ln n1 −n1)) and since N = n0 + n1, this simplifies to S = kB(N ln n −n0 ln n0 −n1 ln n1) = kB(n0(ln n −ln n0) + n1(ln n −ln n1)) = −NkB(f ln f + (1 −f ) ln(1 −f )). Notice that S is indeed extensive: it depends on N, the number of particles that we have. We can also find our temperature: 1 T = ∂S ∂U N = ∂S ∂f ∂f ∂U = (−N ln f + N ln(1 −f )) kB εN (here volume is not well-defined and not relevant). Moving terms around, ε kBT = ln(1 −f ) −ln f and this can be represented as comparing two kinds of energy: the difference in energy ε between states and kBT, the thermal energy. That’s the kind of comparison we’ll be doing a lot in this class, since the ratio tells us a lot about the macroscopic quantities of the system! So now defining β = 1 kBT , we can write 1 −f f = eβε. Then f = n1 N = 1 1 + eβε , 1 −f = n0 N = 1 1 + e−βε , which also means we can rewrite n1 = Ne−βε 1 + e−βε , n0 = N 1 + e−βε . So now we can compute our internal energy: U = f εN = Nε 1 + eβε . If we plot n0 N and n1 N as functions of kBT = 1 β , the number of excited particles in the upper state increases to an asymptotic limit. Similarly, U εN approaches 1 2 as kBT →∞. Finally, let’s plot the heat capacity C = ∂U ∂T N: C = Nε2 kBT 2 eε/(kBT ) 1 + eε/(kBT )2 . We’ll often look at these when T →0, T →∞: as T →0, C → 1 T 2 e−ε/(kBT ), and as T →∞, C →0. There will be physical explanations for each of these behaviors as well! Finally, remember that we computed the entropy: we’re going to change our variables: S = S(U, N) = S(T, N) = ⇒S = −NkB(f ln f + (1 −f ) ln f ) 72 Using the fact that f = 1 1+eεβ , we now have S(N, T) = Nkb ln(1 + e−βε) + Nε T  1 1 + eβε  and notice that we can split this up in a familiar way: = U(N, T) T + Nkb ln(1 + e−βε). As T →0, S →0, and T →∞, S →Nkb ln 2. This gives us a good interpretation of information entropy: high temperature gives high uncertainty! 21.3 Back to macroscopic systems Let’s look back at our equation 1 T = ∂S ∂U V . This is important because S is computed from the number of microstates in our system! So let’s go back to our system with two parts and an immovable thermal partition (so again, we have fixed volume). Let a small amount of heat be transferred from one part to the other. No work is done, so ¯ dQ = dU, which means that dS = ∂S ∂U V,N dU = 1 T dU. Here dU is our thermal energy, and since we add a small amount of heat, our systems remain close to thermal equilibrium: thus U and S remain well-defined. Finally, it’s important to note that this can be reversed without any effect on our system, since dU is infinitesimal. Fact 141 Those three conditions are what dictate a reversible heat transfer: dU = ¯ dQ, so dS = dQrev T . However, this is only an equality when we have reversible quantities: in general, we have the inequality dS ≥¯ dQ T . This is the first time we write our heat as the product of an extensive and intensive quantity. It’s important that S is a state function, so we can compute the change dS by looking at any path! Example 142 Consider a system with a partition: the left half has volume V , temperature TA, and pressure PA, and the right half is empty. At time t = 0, we remove the partition, and now we have a new volume 2V , TB, PB. Since the system is isolated, TB = TA (as no heat is transferred, and an ideal gas’s energy only depends on temperature). By the ideal gas law, then, PB = PA 2 . This is an irreducible process, so we can’t follow a specific path on the PV diagram. But S is a state function, so we can pick any path! Let’s say we go from A to C (reversibly add heat at constant pressure P until volume 2V ), and then from C to B (reversibly remove heat at constant volume until the pressure drops to P 2 ). Then dQrev along the first part of our path is CP (T)dT, and dQrev along the second part is CV (T)dT, so our total change in entropy is SB = SA = Z TC TA dT ′ T ′ CP (T ′) − Z TC T dT T ′ CV (T ′) 73 and these integrals can be combined to Z TC T dT ′ T ′ (CP (T ′) −CV (T ′)) = NkB ln TC T Since we have an ideal gas, TC = 2T, and therefore ∆S = SB −SA = NkB ln 2, as we expect! 22 March 18, 2019 (Recitation) 22.1 Questions Let’s discuss the entropy of a Poisson ensemble from the problem set. We’ve been given a Poisson distribution with mean ⟨n⟩: then pn = ⟨n⟩n n! e−⟨n⟩. By definition, the Shannon entropy is − X n pn log2 pn. Mathematically, this is pretty straightforward, and we discussed last Wednesday what an entropy of a distribution actually means. For example, if ⟨n⟩≈100, the entropy lets us know how many bits we generally need to represent the random samples. In particular, if we measure 100, 90, 105, 98, and so on, then we can encode this by taking differences from the mean. We expect σ ≈√n, so we can find a coding that only uses about log2(c√n) bits! Entropy, then, is the number of bits needed for an optimal encoding. Fact 143 In most cases, the entropy is basically log2 of a few times the standard deviation. This means that we expect S = 1 2 log2 n + log2 c for some constant c. Next, let’s take a look again at a biased coin: the information-theory entropy is S = −p log2 p −(1 −p) log2(1 −p). This makes sense to be maximized at 1 2: near 0 and 1, it’s easy to expect or predict the outcome of the coin flip. So the information-theoretic limit is the Shannon entropy. How do we construct a better coding scheme than 1 bit each? We can block our flips into larger groups and encode them one by one. We’ll find that often, we do better than 1 bit each, but not better than the Shannon theoretical limit! 74 22.2 Entropy We know that the entropy of a macrostate is S = kb ln Γ, where Γ is the number of microstates. It’s a powerful concept, but it can also be confusing. The key concept is that entropy must always get larger. We can only lose information by not carefully controlling our experiment! The second law of thermodynamics can be formulated in many ways: dS dt > 0 is one example, but we need to be careful with that. If we have a particle in a box, the particle can be anywhere, and we can describe the entropy of the particle. (We’ll look at this more in the future.) If the particle is in the left part of the box only, though, its entropy will decrease. Since the number of microstates is a factor of 1 2 smaller due to entropy, we lose a factor of kb ln 2. What if we have a delocalized particle that is then measured? We can “catch” the particle in a box of smaller volume. But this is done through a measurement, and through the process, the observer increases the entropy! So the second law applied to isolated systems, not subsystems. Let’s say we have a reservoir that is connected to a system, and we cool down the system (for example, if we put an object in a refrigerator). The entropy flows in the same direction as the heat: the object that cools down will lose entropy as well. Essentially, our question is “what happens to a system when we add energy?” It will have more microstates, unless we have a pathological system. Intuitively, a system (like a harmonic oscillator) has more energy for larger velocities, because we have a larger sphere of possibilities. This has to do with density of states! Fact 144 So a system that is cold and heats up will gain entropy: entropy flows in the same direction as heat. But maybe there are 1010 microstates in the reservoir and only 20 microstates in the small system. Let’s say we heat up the small system so it now has 100 microstates, and at the same time, the reservoir reduces its microstates to 5 × 109. Is this possible? Let’s look at the total entropy of the reservoir plus the system! Remember that the second law of thermodynamics applies to complete systems: we should consider S = log2 Γ, where Γ is the total number of microstates. Fact 145 So what we really care about is the product of the number of microstates, because we have multiplicity! Since 20 · 1010 < 100 · 5 × 109, this process is indeed allowed in nature. Entropy extends beyond energy conservation, though: it also tells us when energetically allowed processes will not happen in nature. For example, is it possible for us to take a reservoir of heat, extract work, with only the consequence that the temperature cools down? Also, is it possible for two objects at the same temperature T to develop a temperature differential? No, because these things violate entropy! Reversibility is an important concept here: free expansion of a gas greatly increases the number of microstates. On the other hand, we know the time-reversibility of the system means the actual number of microstates must be constant, so what does it mean for entropy to increase? How can we go from a small number of microstates to a large number? People have done many studies of various systems and how they behave in time. These are called billiard problems: have particles bouncing offa wall, and if we have a spherical system, it’s possible that we may never fill the full region 75 of microstates! On the other hand, more irregular systems may allow much more randomness in phase space. So the important assumption in entropy is that all microstates are equally likely. We don’t have particles kept on certain trajectories, mostly due to our inability to measure perfectly. Loss of knowledge when going from highly deterministic systems to complete randomness is the central idea of entropy increase. It is in principle possible for small deviations at the microscopic level to happen (fluctuation), but it’s often immeasurably small. 23 March 19, 2019 Class is very quiet - maybe everyone is ready for spring break. Our exams are at the front of the room; the problems and solutions will be uploaded tomorrow, because there are two people still taking the exam. There is a problem set (2 problems) due on Friday. 23.1 Overview and review As a quick review, we started by reviewing Shannon entropy and thinking of entropy as a thermodynamic quantity connecting the microscopic world with macroscopic problems. Last time, we looked at a two-level system, going from counting states to computing entropy, temperature, and heat capacity. It’s interesting, because it’s the first system where quantum effects manifest themselves in the macroscopic world. One thing to think about is the unimodal heat capacity - there are important quantum effects there! We’ll define some terms that are useful for describing systems, and we’re going to keep thinking about changes in entropy in the PV plane. We have all the tools necessary into understanding fundamental systems! Next lecture, we’ll also look at some more quantum systems. Fact 146 Recall that a two-level system has N particles, each of which can be in a lower or higher energy level. If we plot n N versus kBT, which is the number of particles in the higher energy state versus energy, the number of particles in the higher energy state increases to 1 2. If we plot U/(εN), where ε is the energy level of the higher energy state, this also saturates at 1 2. Finally, heat capacity c/(NkB) increases and then decreases. Remember that we found the entropy of this system: as T →0, S →0 [this will be the third law of thermodynamics], and as T →∞, S →NkB ln 2. Let’s try to justify what we see and physically explain each of these graphs! • At lower temperatures / energies, all energy is in the lower level. But if we add energy to our system, we have particles evenly distributed among all states. • This is also why internal energy saturates at 1 2εN: half of the particles will be in the ε energy state. (We take groud state to be 0.) • The heat capacity is harder to explain. At high temperatures, we expect entropy to read its maximum value of NkB ln 2, and we have evenly distributed particles. So changing the temperature a little bit does not change the configuration very much, so the heat capacity is very low. This is a saturation phenomenon! Meanwhile, at very low temperatures, we have to overcome the gap of energy ε to actually change the state of the system. This gapped behavior is also an important characteristic: we should have a vanishing of order exp h − ε kBT i . 76 • The maximum value of C (heat capacity) is coming from the scale temperature Tε = ε kB . This is where we start exciting more and more particles. This is where we have the highest disorder. By the way, the saturation effect means that we can’t have a system where most particles are at the higher energy level. However, we can prepare our system in a way that gives more particles than we normally have at the high energy level! It’s called a metastable state, and it’s interesting for “negative temperature.” 23.2 A closer look at the First Law Remember the First Law of Thermodynamics, which relates an exact differential to inexact differentials: dU = ¯ dQ + ¯ dW. We know that if work is done in a quasi-equilibrium process (so that we’re pretty close to being in an equilibrium state), then pressure is defined throughout the process, so we can write ¯ dW = −PdV . Meanwhile, if heat is added in a quasi-equilibrum process, temperature is defined throughout, so we can write ¯ dQ = TdS. This means we can write the First Law in a bit more detail: dU = TdS −PdV + µdN + σdA + · · · . Recall that we defined (for a system at thermal equilibrium) ∂S ∂U V,N = 1 T . This also tells us ∂U ∂V S,N = −P (if we plug into dU = TdS −PdV , the TdS term goes away), and similarly ∂S ∂V U,N = P T which is “volume equilibrium.” Remember that we had a fixed partition that only allowed heat transfer between two parts of a system: now, let’s see what happens with a movable partition! 23.3 Deriving some important cases Let’s say we have a system with A and B separated by a partition. Now, let’s say that A and B can exchange volume V , but they cannot exchange N, the number of particles, or U, the internal energy. If A and B are at thermal equilibrium, then the entropy is maximized. What changes when we move the partition a bit? volume is being exchanged, so dV = dVA = −dVB. Writing out changes in entropy in terms of partial derivatives like we did last time, since we’re at a maximum entropy, ∂SA ∂VA UA,NA dVA + ∂SB ∂VB UB,NB dVB = 0, 77 and this means we want ∂SA ∂VA UA,NA = ∂SB ∂VB UB,NB. Plugging in, this means PA TA = PB TB = ⇒PA = PB (since we’re at thermal equilibrium). So this means that pressures are equal at equilibrium! Similarly, if we have a partition where we can exchange only particles but not internal energy or volume, we find that ∂S ∂N U,V = −µ T is constant. 23.4 Entropy’s role in thermodynamic processes Remember that an adiabatic process has no heat transfer (because it moves too quickly, for example): dQ = 0. Definition 147 An isoentropic process has no change in entropy: ∆S = 0. These two are not interchangeable! It’s possible to have zero change in heat but a nonzero change in entropy. For example, free expansion is adiabatic if it happens fast enough and is isolated, but the entropy does increase. Let’s also be a bit more specific about some other words we’ve discussed: Definition 148 A quasi-equilibrium process is one where the state is always near thermodynamic equilibrium, so that state variables are defined throughout the process, meaning we can use the first law. A reversible process is a quasi-equilibrium process in which the direction of heat flow and work can be changed by infinitesimal changes in external parameters. It’s not necessarily true that quasi-equilibrium processes are reversible though. For example, processes with friction have some dissipation of energy, which means that we can’t do them in reverse and get back to our original state. However, if we do them sufficiently slowly, state functions can be consistently defined. 23.5 Looking at change in entropy in the PV plane Example 149 Consider the isometric heating of a gas: in other words, we hold the volume constant and dV = 0. This means dU = ¯ dQ = TdS = ⇒dS = dU T . Therefore, ∂U ∂T V = T ∂S ∂T V = ⇒ ∂S ∂T V = CV (T) T = ⇒S(T, V ) −S(T0, V0) = Z T T0 dT ′ T ′ CV (T ′) where CV (T) is some function of T. Looking at a simple system like the monatomic ideal gas, we have CV = 3 2NkB, so ∆S = 3 2NkB ln T T0 . 78 Example 150 Now let’s look at the isobaric heating of a gas: keep pressure constant and dP = 0. Remember from our problem set and other discussions that we can define the enthalpy H = U + PV, and in our system here, dH = dU + d(PV ) = TdS −PdV + PdV + V dP = TdS + V dP. Now, because we have an isobaric process, dH = TdS = ⇒∂H ∂T P = T ∂S ∂T P . Rearranging this, we have that CP (T) = T ∂S ∂T P , so S(T, P) −S(T0, P) = Z T T0 dT ′ T ′ CP (T ′), so again for a monatomic ideal gas, CP = 5 2NkB, and we have ∆S = 5 2NkB ln T T0 . Example 151 Finally, let’s think about isothermal expansion (from a volume V0 to a final volume V ): dT = 0. Let’s rewrite dU = TdS −PdV as dS = 1 T (dU + PdV ) = 1 T  ∂U ∂T V dT + ∂U ∂V T dV + PdV  (since U is a function of V and T), and this simplifies to dS = 1 T  CV dT +  ∂U ∂V T + P  dV  . Since we have an isothermal expansion, the dT term goes away, and thus dS = 1 T  P + ∂U ∂V T  dV. To compute this more easily, we’ll expand by volume first: S(T, V ) −S(T, V0) = Z V V0 dV ′ T  P + ∂U ∂V T  For an ideal gas, U is only a function of T, so ∂U ∂V T = 0, and pressure P = NkBT V , so S(T, V ) −S(T, V0) = NkB ln V V0 . 79 Fact 152 Draw a cycle using all three of these processes: • Start at pressure P0 and volume V0, and heat it isometrically to pressure P. • Isothermally expand the gas from volume V0 to V . • Finally, do an isobaric compression from volume V back to volume V0. Notice that V V0 = T T0 in an ideal gas, and in this case, the change in entropy over the whole cycle is zero! So as long as we have a well-defined pressure, temperature, and internal energy at all points in our process, we are at thermal equilibrium, and our entropy is a state function. In general, this means we can use any path to calculate our entropy! 24 March 20, 2019 (Recitation) A lot of interesting new material was covered in class. 24.1 Reviewing the two-level system Let’s look at the two-level system, where we have N particles that can either be at low energy (0) or high energy (ε). The lowest possible energy of the system is 0, and the highest possible is Nε. We’ve picked three special states: 0, Nε 2 , Nε. Notice that 0 and Nε have minimum entropy, since there is only 1 possible microstate. In general, the number of microstates for having n0 objects in the lower state and n1 in the higher state is Γ(n0, n1) = n! n1!n0!. This is N n1  : we’re choosing n1 of the N states. This is consistent with the edge cases of 0 and Nε. Well, what’s the entropy when the energy is half of the maximum? We know that if we have N fair coins, we have entropy of N bits, which corresponds to Nkb ln 2. So we don’t need any mathematics to understand the entropy of the system! What about the temperature of the system? We can argue that it’s usually zero at the ground state, because we’ve “taken out all of the energy.” We claim that the temperature at the middle state is ∞and that the energy at the high state is negative zero! How do we argue this? We can plot entropy versus energy: it peaks at the middle and starts and ends at 0. So now 1 T = ∂S ∂U . so the temperature is infinite at Nε 2 , because the slope of the S versus E graph is 0. It turns out we have the Boltzmann factor n1 n0 = e−(E1−E0)/(kbT ); the idea is that if two groups are equally populated, and the energy states are different, then the ratio on the left hand side is 1, the exponent must be 0, and a finite energy different means T must be infinite. The idea is that at infinite temperature, energy doesn’t matter, because it is free! 80 Well, notice that the graph has positive slope for E < Nε 2 and negative slope for E > Nε 2 . So the temperature is positive at first but then becomes negative! So let’s start plotting other quantities versus temperature. Let’s plot only in the range from T = 0 to T = ∞: this means we only care about energies between 0 and Nε 2 . If we plot E versus T, we start at 0 and then asymptotically approach Nε 2 . Fact 153 By the way, note that T, U, E are all parametrized together, so ∂S ∂U keeps N constant, but there’s no other dependencies to worry about. So what’s the specific heat of the system? Specific heat measures how much energy we need to change the temperature, and as T →∞, the energy is not changing anymore: it’s saturated to Nε 2 . So at high temperatures, the energy saturates, and C →0. Fact 154 Here, we have a finite number of possible states and a bounded energy system. This is very different from a gas that can have infinite kinetic energy! We’ll also find later that there are usually more energy states for ordinary systems: the spin system is an exception. What if T is very small? The system wants to be in the ground state, and for very low (infinitesimal) temperature, we only care about the ground state or the first excited state: there’s only a possibility of one of the N particles increasing its energy by ε. In the Boltzmann distribution, we have a factor e−E/(kBT ), so the probability of the first energy distribution is proportional to e−ε/(kBT ). This is something that characterizes any system with a ground state and an excited state with an energy gap of ε! Then we just use the physics of combinatorics: we’ll discuss this as partition functions later, but this exponential factor is universal across all systems like this. So systems at low temperature show exponential behavior, and if we plot energy versus temperature, we’ll start offwith an exponential behavior. This is what we call gapped behavior. Example 155 What’s an example of a system without gapped behavior? In other words, when can the excited state have arbitrarily small energy? We want to say a classical gas, where the kinetic energy 1 2mv 2: the velocity v can be arbitrarily small. But not really: remember that a particle in a box has a smallest energy ℏ2 L2 , so we do need to take the classical limit or use a very big box. Then we’ll see that the heat capacity does not have that exponential curve near T →0. 24.2 Negative temperature? Let’s look back at T being negative for high energy (in our system, where E > Nε 2 . We’ve always heard that we can’t cool down and get to negative temperature: there’s an absolute zero. The idea is that there’s a lowest energy state, and there’s no way to get less energy that that lowest state. 81 So negative temperature doesn’t mean we can keep cooling to negative Kelvin. In fact, this system tells us that negative temperatures appear in a different way! We increase the energy of the system, and the temperature increases to ∞and then goes negative. So somehow ∞temperature and −∞temperature are very close to each other! This is because we inherently defined our temperature as a slope being equal to 1 T , and often, Boltzmann factors give 1 T as well. So if we get confused, think about 1 T instead! Plotting energy versus −1 T (so that we start offwith positive temperature on the left), we have our energy increasing from 0 to Nε 2 at −1 T = 0, and then it further increases from Nε 2 to Nε! This is a continuous curve, and everything is smooth between positive and negative temperatures. In other words, the connection between positive and negative temperatures does not happen at 0: it happens at ∞. This is only possible because we have a maximum energy state! If we were to do this with a system like a classical gas, the energy would go to ∞as T →∞. So there wouldn’t be a connection to negative temperature there. 25 March 21, 2019 Today’s physics colloquium is about turbulence, which is a very challenging problem. Part of Professor Fakhri’s group is looking at biological signals, and it turns out the dynamics of the signaling molecules in the cell follow a class of turbulence phenomena that is connected to quantum superfluid turbulence. Remember there’s a problem set due tomorrow, and there will be another one posted soon. 25.1 Overview Today, we’re going to start discussing some general systems where we can do counting of states. The idea is to do thermodynamics and statistical physics in systems like a two-level system! We start by counting states to find multiplicity, and from there we can compute entropy, temperature, and heat capacity. In particular, we’ll look at a particle in a box, as well as the quantum harmonic oscillator. We’ll introduce an idea called the density of states, which will tell us important information about the internal energy of the system! This is not a lecture on quantum mechanics, but we’ll get some descriptions that are related. By the way, there is a way to count a density of states in a classical system too, even though the states are continuous! We’ll treat them in a semi-classical way. 25.2 Quantum mechanical systems: the qubit Recall that in classical mechanics, the number of states is infinite: for example, a system of N particles depends on the position and momentum (⃗ xi, ⃗ pi), and these quantities (which define states) are continuous variables. However, in quantum mechanical systems, states are quantized and finite, so they are countable! We’ll learn how to label and count some simple examples: the qubit, quantum harmonic oscillator, and the quantum particle in a box. A qubit is a system with only two states: |0⟩= |+⟩= |↑⟩and |1⟩= |−⟩= |↓⟩. Here, |⟩refers to a set of states. The higher energy state |+⟩has energy E+, and the lower energy state |−⟩has energy E−. The simplest example of a qubit is a spin 1 2 particle: for example, an electron in a magnetic field. We know (or are being told) that the magnetic moment ⃗ µ = −g  e 2mc  ⃗ s, 82 where g ≈2 for an electron, e is the charge of the electron, m is the mass of the electron, and c is the speed of light. Then when a particle is placed in a magnetic field ⃗ H, the energy is E = −⃗ µ · ⃗ H = e mc ⃗ s · ⃗ H. In particular, the different energy states are E+ = eℏ 2mc H, E−= −eℏ 2mc H, where ℏis Planck’s constant h 2π ≈1.054 × 10−34J · s. There are many systems that behave like this at low energy! Fact 156 These are interesting derivations, but for those of us who haven’t taken 8.04, we shouldn’t worry too much about it. In general, if we have a quantum particle with spin j in a magnetic field, the different possible energy states are Em = eℏg 2Mc mH, where m = j, j −1, · · · , −j. This has to do with systems exhibiting paramagnetic properties! 25.3 Quantum harmonic oscillator Let’s start by looking at a classical example. We have a harmonic oscillator with spring k and mass m, where the potential energy V (x) = 1 2kx2 = 1 2mω2x2 (if the natural frequency is ω = q k m. This is used to explain vibrations of solids/liquids/gases, as well as other material properties. The energy (or Hamiltonian) of this system can be written classically as H = p2 2m + 1 2kx2, and we find that x(t) = A cos(ωt + φ), and the energy of this system is a continuous quantity mω2A2. On the other hand, in the quantum version, we can label our quantum states as |n⟩, for n = 0, 1, 2, · · · , and there’s only a few allowed energies. They are equally spaced in units of ℏω: En =  n + 1 2  ℏω, where the lower allowed energy 1 2ℏω is the zero point energy. We can also have a bunch of non-interacting particles in the same harmonic oscillator potential: then we just add the individual energies. Fact 157 The energy of a set of non-interacting particles moving in a harmonic oscillator potential is (for example) E(n1, n2, n3) = ℏω  n1 + n2 + n3 + 3 2  . The states here are denoted as |n1, n2, n3⟩, where each nj = 0, 1, 2, · · · . It’s pretty easy to see how this generalizes. Well, there’s a one-to-one correspondence between a one-dimensional system with N particles and an N-dimensional 83 harmonic oscillator for one particle: the energy states look the same! EN(n1, · · · , nN) = ℏω  n1 + · · · + nN + N 2  . 25.4 Particle in a box This system has applications to an ideal gas in a box, as well as to black-body radiation and Fermi and Bose gases. Basically, we care about the limit of a large box. Consider a particle with mass M in a three-dimensional cubic box of length L. Quantum energy states can be labeled |n1, n2, n3⟩by three integers! We can show that including boundary conditions, E(n1, n2, n3) = π2ℏ2 2ML2 (n2 1 + n2 2 + n2 3). These energy levels are very closely spaced! Let’s compute this for an oxygen molecule. If the box measures 1 mm on a side, E ≈ 10 × 10−68 2 × (32 × 1.67 × 10−27) · 10−6 ≈10−36J ≈6 × 10−16eV. These are very small numbers: compare this to kBT at room temperature kBT = 1 40eV. So most of the time, we can assume the energy levels are almost smooth (since we often compare energy to kBT in our calculations), which makes our work a lot easier! 25.5 Counting states and finding density Let’s look at our particle in a box. Fact 158 The math will be more complicated from this point on. We should review theta and delta functions! Let’s say we want to find the number of energy states N(E) in a box of volume V with energy less than E. Once we find N(E), the number of states, we can differentiate it to find dN dE , which will give the number of states between E and E+dE (this is similar to how we differentiated a cumulative distribution to get a probability distribution function!). Note that in our cumulative distribution N(E) will be stepwise (since we have some finite number of states for each energy level), and we will normalize it by dividing by π2ℏ2 2ML2 . But as we go to higher temperatures, the steps are very small compared to kBT, so we can do a smooth interpolation! The idea is that dN dE is a sum of delta functions (because N is a bunch of step functions), but we can approximate those with a smooth curve as well. This is called a delta comb, by the way. So now N(E) = X n1,n2,n3 θ  E −π2ℏ2 2ML2 (n2 1 + n2 2 + n2 3)  , since we only count a state if the total energy is at most E. Doing our interpolation, the sum becomes an integral: since n1, n2, n3 can take on positive values, N(E) = Z ∞ 0 dn1 Z ∞ 0 dn2 Z ∞ 0 dn3θ c2 −n2 1 −n2 2 −n2 3  , 84 where c2 = 2ML2E π2ℏ2 , just to make the calculations a bit easier. To evaluate this integral, let’s let nj = cyj: N(E) = c3 Z ∞ 0 dy1 Z ∞ 0 dy2 Z ∞ 0 dy3θ(1 −y 2 1 −y 2 2 −y 2 3 ). Notice that this is one-eighth of the volume of a unit sphere! This is because we want y1, y2, y3 to be positive and y 2 1 + y 2 2 + y 2 3 ≤1. So the volume is just π 6 , and N(E) = π 6 c3 = π 6 2ML2E π2ℏ2 3/2 and since L3 = V , we get a factor of V out as well, and we can simplify further. For any quantum system, if E = E(n1, · · · , nN), then N(E) = X {nj} θ(E −E({nj})). Then we can differentiate dN dE = X nj δ (E −E({nj})) . So in this case, since N(E) = π 6  2M π2ℏ2 3/2 V E3/2, we have dN dE = π 4  2M π2ℏ2 3/2 V E1/2. Fact 159 In general, if Ek occurs some gk number of times, this is called a degeneracy factor. Since we want to count all the states, we get a gk factor in front of our theta function! This is better explained with examples, though. Example 160 Let’s count the density of states in a quantum harmonic oscillator. Then En = n + 1 2  ℏω, and the number of states N(E) is a step function that increases by 1 at 1 2ℏω, 3 2ℏω, and so on. So N(E) = ∞ X n=0 θ  E −ℏω  n + 1 2  If E ≫ℏω, we can again make this into an integral: N(E) = Z ∞ 0 θ  E hω −n −1 2  dn ≈E ℏω , so we have dN dE = 1 ℏω. Next time, we’ll talk about how classical systems can still be counted! This is the idea of a semi-classical description. 26 April 1, 2019 (Recitation) Today’s recitation is being taught by Professor England. 85 Let’s go back to discussing the ideas of entropy! Remember that we started offby discussing information entropy, but now we want to start developing entropy more as a thermodynamic and statistical quantity. 26.1 Rewriting the first law Earlier in the class, we found that we could describe our energy dU as a product of the generalized force and generalized displacement: dU = X i ∂U ∂xi  dxi. Here, we referred to TdS as “heat” dQ, and PdV , as well as other terms, as “work” dW. But our dS = ¯ dQ T could be interpreted differently: if we define Ωas a function of state variables U, V, N, · · · as the number of different microstates consistent with our data, we could also define S = kB ln Ω. There’s a lot of plausibility arguments that can support our theory working this way! For example, two objects in contact have maximum entropy when they have the same temperature, and indeed we want entropy to increase as time evolves. So let’s run with this, and let’s see if we can rewrite the first law in a more sensible way. Since S is proportional to ln Ω, we have dS = kBd(ln Ω): plugging this into the first law, d(ln Ω) = dU kBT + PdV kBT −µdN kBT · · · This gives us another way to think of how to “count” our states! We can ask questions like “how does a flow of heat contribute to our entropy?” or “how does entropy change if I increase the volume?” So the right hand side is a bunch of levers that we can pull: now we can see how our statistical quantity changes when I adjust my other terms! 26.2 Applications to other calculations We can go back to our ideal gas model, and we’ll see that the calculations fall out pretty easily! Let’s say we have N particles in a box of volume V , and we want to think about the number of microstates we can have. Fact 161 Scaling is very useful; we don’t have to be too exact with our calculations. Microstates are dictated by two variables: the position and the momentum of the individual particles. It’s true that if we tried to count the number of discrete “float numbers” that would work, we’d have infinitely many possibilities for ⃗ x. But there’s still a sense in which having twice as much volume gives twice as many possible positions, so the number of states here is basically proportional to V ! Similarly, momentum, which is independently assigned to particles of position, has some function a(U) in terms of the internal energy. Since we have N particles, the number of microstates here is proportional to Ω= (cV · a(U))N. 86 Fact 162 One other way to patch this up is with the uncertainty principle: dpdx ∼ℏ: we do need discrete states in that case. To get to entropy from here, we take a logarithm and multiply by kB (which is just a units conversion factor): kB ln Ω= NkB(ln V + ln a(U) + c). Notice that the constant term c comes from the fact that we only care about differentials and differences in the first law! For example, pressure is really just P kBT = ∂(ln Ω) ∂V  U,N,··· . Let’s try plugging that into our expression for ln Ω: this yields P kBT = N V , which rearranges to the ideal gas as desired! Fact 163 We can always take a derivative (in principle) fixing all the other variables. It might be experimentally difficult to do this, though. So another way to say this is that the ideal gas is just related to the scaling of states with volume. But there’s something wrong with the entropy we’re using: we wanted the change in entropy to be an extensive quantity. U is extensive: putting two copies of the system next to each other doubles the energy. Since U and ln Ω play similar roles in the first law, we want them to both be extensive. But ln Ω= N ln V , but doubling the size of the system doubles both N and V , and this doesn’t give exact extensivity. What’s going on here? Fact 164 This is called Gibbs’ paradox! The idea is that we treated our particles as different: if particles 1 and 2 are on the boundary and 3 is in the middle, this is different from 1 and 3 on the boundary and 2 in the middle. But there’s really no way for us to be able to distinguish our particles! So it seems that there are N! permutations, and we want to divide through by N!. But this isn’t actually an exact answer! Remember that we have quantized states, so we’re not overcounting particle states when many particles are close to each other. So we’d run into problems where dividing by N! is artificially penalizing particles nearby - we’ll return to those ideas for now. But we’ll deal with dilute gases, since we’re doing ideal gas calculations anyway, and then it is exact to say that Ω= (V · a(U))N N! . With Stirling’s approximation, this gives ln Ω= N ln V −N ln N + N = N ln V N + N. Now this is an extensive quantity! V N is an intensive quantity: it’s related to the density, and now scaling the system by a factor of 2 does scale ln Ωby a factor of 2 as well. 87 Example 165 If we have two independent identical systems, they have the same number of possible microstates Ω, so the total system has Ω2 possible states: this does yield twice the entropy. On the other hand, if we put the two systems to together, what happens to the entropy? Doesn’t it increase again? This is again an issue of Gibbs’ paradox! The particles in the two systems are indistinguishable, so we should really think a bit more about those N!-type terms. Keep this in mind when we start thinking about quantum systems (particularly Bose-Einstein condensates), though. 26.3 Relating this to temperature Let’s think a bit more about our ideal gas system. The internal energy of this system is U = 3N X i=1 p2 i 2m, since each particle has 3 degrees of freedom. So if we have some given energy U, how many possible arrangements Ω(U, V, N) are there? Basically, how does Ωscale with U? Let’s do the cheap thing: we have a bunch of independent coordinates pi whose squares add to some constant, which means that the momentum coordinates in phase space are on a hypersphere of radius √ 2mU. Really, we care about scaling: how does the size of a sphereical shell change with volume? Because the volume of the whole sphere goes as r 3N, the boundary goes as r 3N−1, but we can ignore the 1 if N is really big. So that means Ω(U, V, N) ∼V N N! √ 2mU 3N ∼V N N! U3N/2. Going back to our first law equation, let’s look at the part d(ln Ω) = dU kBT + · · · and differentiate with respect to U. This yields ∂(ln Ω) ∂U  V,N = 1 kBT = ∂(ln U) ∂U  3N 2 = ⇒U = 3NkBT 2 . This gives us the temperature definition out of our calculation of microstates as well! 26.4 Increasing entropy Finally, let’s do another look at the second law of thermodynamics. This law is initially encountered empirically: given a process that we design and take a system through, we can (for example) draw a loop in our PV diagram. In these cases, the total entropy will always increase: ∆S of the environment plus ∆S of our system cannot get smaller. So this is a very empirical law, but our new description gives us a way to think about this more clearly! (By the way, it’s important to note that this is always an average law.) So suppose we have a surface of states in U, V, N space, and we follow our system by doing dynamics with Hamilton’s equations. What does it really mean for entropy to increase? There’s two notions here: one is to count the number of states on our surface and take the logarithm, but that doesn’t really tell us anything in this case. 88 Instead, we can think of some quantity R: maybe this is “the number of particles on the left side” or “some measure of correlation with a 3-dimensional figure.” The point is that we can assign a value of R to all microstates, and now we can group our microstates together by values of R: this is called “coarse-graining.” So what’s Ω(R)? Much like with Ω, it’s the number of microstates consistent with some value of R, and we can define an “entropy” kB ln Ω(R) along with it. The concept is that R will eventually stabilize to some quantity - for example, particles are not likely to unfill a region. That means S(R) will increase generally - it may fluctuate, just as R does, but S is an average quantity. One way to think of this is that as we move in our state space, we’ll generally get stuck in the biggest volumes of R: it’s much more likely that we’ll end up there on average! Analogously, two systems at thermal equilibrium can go out of equilibrium, but they’re likely to return back into equilibrium (because it is thermodynamically favorable to do so). So entropy can decrease, but that is usually nothing more than a statistical fluke. 27 April 2, 2019 The energy level of the class is higher after the break! There’s a problem set due this Friday; we should make sure we understand the material from the required reading. There will be an exam on April 17th, which is similar to last exam. We’ll start hearing announcements about it soon. Recall that we started by introducing information and thermodynamic entropy. We found that counting the number of microstates consistent with a macrostate gives us a measure of entropy as well, and then we could calculate the temperature, heat capacity, and many equations of state. This then led us to introducing a few different systems, where we could actually count states. We discussed a two-level qubit, particle in a box, and quantum harmonic oscillator, because these all have discrete states. Today, we’ll talk about how to extend this to classical systems and give them this kind of treatment as well. The idea is to count states to find the density of states as well, which gives us a probability distribution. 27.1 Returning to the particle in a box As we calculated last time, the density of states depends on the energy E of the system: dN dE = 1 4π2ℏ3 (2M)3/2V E1/2. Our goal is to find a connection here to a classical system: let’s say we have some volume dV that is infinitesimal on the classical scale but large enough to contain many quantum states. Then for our box, dV = dxdydz = d3x (for simplicity). If we rewrite E in terms of momentum as p2 2m, then dE = pdp m ; substitute this into our equation above, and we can pull out a factor of 4π: dN = 4π 8πℏ3 p2dpd3x Now notice that the 4πp2dp looks a lot like spherical coordinates: d3p = p2dp sin θdθdφ, and integrating out θ and φ gives the result we want. So that means (because ℏ= h 2π), dN = 1 h3 d3xd3p. 89 This means that dx and dp, which are classical degrees of freedom, yield something about our states dN, just with a normalization factor 1 h3 ! So looking back at dN dE = 1 4π2ℏ2 (2M)3/2V E1/2, we can think of 1 h3 as a “volume” that contains one state. This means that we can go from classical systems to semi-classical descriptions: our “volume” in phase space is h3 = ∆x∆px∆y∆py∆z∆pz. In other words, we can think of our volume h3 as contain 1 state in our phase space. Example 166 Let’s go back to the density of states calculation for a quantum harmonic oscillator, but we’ll start with a 1-dimensional classical harmonic oscillator and do this trick - let’s see if we end up with the same result. The number of states with an energy less than E is N(E) = Z dxdp 2πℏθ  E −p2 2m −1 2kx2  (where h = 2πℏis the normalization factor as before). Writing 1 2kx2 = ξ2, p2 2m = η2, our number of states can be written as N(E) = 1 πℏω Z dηdξθ(E −η2 −ξ2) and since the integral is the area of a circle with radius √ E, this is just N(E) = πE πℏω = E ℏω , which is the same result that we got with the quantum harmonic oscillator! Then we can find dN dE , which will give the number of states with energy between E and E + dE. We can think of our dxdp 2πℏidea as a “volume normalization.” 27.2 A useful application Now, we’re going to use this counting-states argument to look at thermodynamics of an ideal gas. Remember that we could count our states in our two-level system if we were given the energy of our overall system, so let’s try to think about a microcanonical ensemble: systems that are mechanically and adiabatically isolated with a fixed energy. We’ll discuss Sackur-Tetrode entropy along the way as well. First of all, our goal is to find the multiplicity: how many states Γ(U, V, N) are consistent with a given U, V, N? Once we know this, we’ll write our entropy S(U) = kB ln Γ (at thermal equilibrium), and then we can calculate our temperature with the familiar equation ∂S ∂U N,V = 1 T to find our energy U in terms of V, T, N (and by extension the heat capacity). Finally, we’ll find the equation of state by taking some derivatives. Let’s use the model where we have N molecules of a monatomic gas in a box of side length L. Then the energy 90 of our system (as a particle in a box) is E(nx, ny, nz) = π2ℏ2 2ML2 (n2 x + n2 y + n2 z). To count the number of states, let’s assign a vector ⃗ nj to each particle, where 1 ≤j ≤N. Every particle contributes to the state, but since the particles are indistinguishable, we overcount by a factor of N!. (We’ll return to this idea later.) So the multiplicity is Γ(U, V, N) = 1 N! X ⃗ n1 X ⃗ n2 · · · X ⃗ nN δ 2ML2U π2ℏ2 − X ⃗ nj 2  since we want to only pick out the states with a certain energy. We’ll find that we have very closely spaced states, so we can approximate this sum as an integral from 0 to ∞. Let’s then replace this as half of the integral from −∞ to ∞by symmetry: since each of N particles has 3 degrees of freedom, and there are 3 integrals, this means we now have an integral over all degrees of freedom Γ(U, V, N) = 1 23NN! Z d3n1 · · · d3nNδ(R2 − N X j=1 ⃗ nj 2), where R2 = 2ML2U π2ℏ2 . So we now have 3N variables (basically, we want the surface area of a 3N-dimensional sphere of radius R): let’s replace our ⃗ nis with the 3N variables ξ1, ξ2, · · · , ξ3N. Our integral is now Γ = 1 23NN! Z d3Nξδ(R2 −⃗ ξ2) and normalizing by letting ξj = Rzi, this becomes Γ = 1 23NN!R3N−2 Z d3Nzδ(⃗ z2 −1). (The 3N −2 comes from δ(cx) = 1 c δ(x).) Since N is very large, we can approximate 3N −2 as 3N, and so now we want to deal with the integral Z d3Nzδ(⃗ z2 −1) = 1 2 Z d3Nδ(|⃗ z| −1). The integral is now the surface area of a 3N-dimensional sphere S3N = 3N 2 π3N/2 3N 2  ! so we can plug that in: this yields Γ = R3N 23N−1N!S3N and we’re done with our first step! 27.3 Calculating the entropy of the system Now, let’s start calculating entropy and our other relevant quantities. Since S = kB ln Γ, we can plug in the values we have: we’ll approximate 23N−1 as 23N, and by Stirling’s approximation, S = kB  3N ln R 2 −N ln N + N + 3N 2 ln π −3N 2 ln 3N 2 + 3N 2  . 91 We’ll factor out an N and combine some other terms as well: S = NkB  3 ln R 2 −ln N + 3 2 ln π −3 2 ln 3N 2 + 5 2  . We’ll now substitute in our value of R = √ 2mL2U πℏ : S = NkB 3 2 ln U + 3 ln L + 3 2 ln M 2π2ℏ2 + 3 2 ln π −5 2 ln N −3 2 ln 3 2 + 5 2  . We’ll now combine the terms with a 3 2 in front: since 3 ln L = ln L3 = ln V , S = kBN 3 2 ln U N + ln V N + 3 2 ln M 3πℏ2 + 5 2  and now we’ve found our entropy in terms of the variables we care about! This is called the Sackur-Tetrode entropy. 27.4 A better analysis: looking at temperature So let’s try to think about the consequences of having all of the different parts here. If we didn’t have the ℏterm in our entropy, imagine we take ℏto 0. Then our entropy S goes to infinity, and we know that this isn’t supposed to happen. The idea then is that there is some length scale where quantum effects become obvious! Also, entropy is an extensive quantity: we have the N in front of our other terms, and that’s why the N! correction term was important for us to include. Notice that if we double our volume, ∆S = NkB ln 2 (exercise), so this is consistent with what we want! So now let’s calculate temperature: ∂S ∂U V,N = 1 T = 3NkB 2U = ⇒U = 3 2NkBT which is the same result we had before: this is consistent with the equipartition theorem! Finally, since TdS = PdV +c (where c is other terms that are not relevant to S and V ), P T = ∂S ∂V U,N = NkB V = ⇒PV = NkBT, which is the equation of state for an ideal gas! Similarly, we know that −µdN = TdS + c′ (where c′ is other terms that aren’t relevant to N and S), −µ T = S N −5 2kB = ⇒5 2NkBT = TS + µN, and this can be rewritten as 3 2NkBT + NkBT = U + PV = TS + µN , which is known as Euler’s equation. 27.5 Microcanonical ensembles 92 Definition 167 A microcanonical ensemble is a mechanically and adiabatically isolated system, so the total energy of the system is specified. This means that in phase space, all possible microstates - that is, all members of the ensemble - must be located on the surface H(µ) = E. At thermal equilibrium, all states are equally likely, so the probability of any given state is PE(µ) =    1 Γ(E) H(µ) = E 0 otherwise. Next class, we’ll comment a bit more about uncertainty in this surface (to create volumes)! 28 April 3, 2019 (Recitation) 28.1 The diffusion equation Let’s start with a concept from the problem set. We’ve always been discussing thermodynamic entropy as S = −kB X pi ln pi = kb ln Γ (where Γ is the number of microstates) if all states are equally likely. But now when we look at the diffusion equation, we’re given a slightly different equation for a gas in one dimension: S = −kB Z ρ(x, t) ln ρ(x, t)dx. What’s the connection between the two? If we take some small volume ∆V , then the probability that we find a molecule in that volume at position x is p(x) = ρ(x)∆V. So this equation is just the probability distribution entropy, ignoring the normalization constant! Next question: let’s look at the diffusion equation more carefully: ∂ρ ∂t = D ∂2ρ ∂x2 . Can we argue why this always increases the diffusion of the gas from general principles? If we spread our probability over more possibilities, then entropy goes up: the larger Γ is, the more microstates we have, which means all individual probabilities are small. Well, diffusion makes our peaks in ρ go down! So the probability distribution is getting wider and flatter, and this is a lot like making all of our probabilities go to 1 n: we’re moving toward “equilibrium.” 28.2 Black holes There’s a lot of physics going on here! We have a relation between relativistic mass and energy from special relativity: E = Mc2, and we have the Schwarzchild radius from general relativity; Rs = 2GM c2 . Black holes are a good place to do research, because they are a system where we can make new discoveries: there are big open questions about combining quantum physics with general relativity. 93 Here’s one interesting idea: if there’s nothing to say about a black hole other than its angular momentum and mass, then matter with entropy sucked into that black hole has disappeared: doesn’t this mean entropy has decreased? We can never combine two microstates into one, so what’s going on here? Turns out black holes aren’t completely black! They are actually at a “temperature” Th = ℏc3 8πGkBM . This is called “Hawking radiation,” and this is a way for black holes to communicate with the outside world (using electromagnetic radiation). This means that we can actually have an entropy for black holes, and the contradiction is gone! Fact 168 (Sort of handwavy) This is a bit past standard explanations, but the idea is that the Schwarzchild radius is an “infinite potential:” no particles can get past the event horizon. But adding quantum fluctuation (due to quantum field theory “spontaneously” producing particle-antiparticle pairs), we can create virtual photons at the boundary: one with positive energy can escape, and another with negative energy gets sucked into the black hole. So we can think of this as having a black-body radiation spectrum, where there is a peak at the temperature ℏω = kBTBB. So now you can integrate to find the entropy of a black hole: we can find that it is related to the surface area of a sphere with Schwarzchild radius! This is relevant to understanding complicated materials: for example, the entropy of most systems (like an ideal gas) is dependent on volume. Black holes tell us a different story: entropy doesn’t necessarily scale with volume. After all, 1 M , so the derivative ∂S ∂E is proportional to E. So S is proportional to E2, which is proportional to R2 s ! So the fact is that we have a complicated system for which only the surface matters: this scales much slower than volume. Thus, describing superconductors and other materials is motivated by this discussion of a black hole! This is called the “holographic principle.” 28.3 Negative temperatures Let’s start by talking about heat transfer: it’s about energy conservation and increase in entropy. Usually, we have systems coupled to a reservoir: those reservoirs have a lot of energy, but we don’t want to go against what nature wants! So then a “deal” is made: if the reservoir gives an energy ∆E = T∆S, then the entropy of the reservoir decreases by ∆SR = ∆E T . So the system just needs to make sure that the entropy created by the system is larger: |∆SS| ≥|∆SR|, and then all the energy in the world can be transferred. This has to do with free energy, which we’ll talk about later! So let’s say that we transfer the energy from our reservoir R to our system S, and this just heats up our system S. Then the increase in entropy of the system is ∆Ss = ∆E TS : then if we want total entropy to increase, clearly this only happens if TS ≤TR. So it’s only thermodynamically favorable to transfer heat ∆Q from higher temperature to lower temperature. But what happens when negative temperatures are involved? 94 Remember that when we drew our graph of S versus E for the two-level system: the temperature is 1 slope. The slope goes from positive to 0 and then 0 to some negative value, which avoids the singularity of ∞temperature! So now we want 1 TS > 1 TR : regardless of whether we have two systems at positive temperature or two systems at negative temperature, they will want to move “towards” each other. But if we look at the total entropy, the equation we care about is that heat is transferred from R to S if and only if 1 TR < 1 TS . But now the question: if we bring a temperature with negative temperature in contact with one of positive temperature, what will happen? 1 T is continuous in this case, and two sysetms always want to equilibriate! So what is going on here is that systems at negative temperature gain entropy when you lose energy! The reservoir pays for both the energy and the entropy, so it will willingly transfer energy ∆E: then the entropy of both the reservoir and the system both go up. So what happens is that if a system with negative temperature is in contact with an ideal gas with some positive temperature, then the gas will always gain more energy! 29 April 4, 2019 This is a reminder that the problem set is due tomorrow! 29.1 Brief overview Last class, we used our knowledge of quantum systems (namely the particle in a box) to determine thermodynamics of an ideal gas. The idea is that the internal energy of the system is fixed, so we can count the number of possible states: then because at thermal equilibrium, all microstates are equally likely, we can just use the formula S = kB ln Γ to find the entropy. From there, we could find the temperature of the system, as well as the general equation of state. This then led us to the definition of a microcanonical ensemble. Today, we’re going to go back to how thermodynamic started: using heat engines! We’ll see how to extract work from those engines, and we’ll find a bound on the efficiency on such engines. From there, we’ll start looking beyond having “isolated systems:” we’ll see how to do statistical physics with just parts of systems, which will lead us to the idea of free energy. By the way, the question of N! in our calculations from last week is an important idea: this will be a problem on our next problem set. 29.2 The third law of thermodynamics We’ve talked quite a bit about entropy as information and also in terms of heat transfer, but let’s start looking beyond just those ideas! 95 Theorem 169 (Third law of thermodynamics) The following are equivalent statements: • As temperature T →0, the entropy of a perfect crystal goes to 0. • As temperature T →0, the heat capacity C(T) →0. • It is impossible to cool a system to T = 0 in any finite number of steps. Let’s go through each of these concepts one by one. The idea with the first statement is that a perfect crystal should always be at the ground state (in quantum mechanics) when all thermal energy is taken away (T = 0), so there should only be Γ = 1 microstate! Thus, S = kb ln Γ = 0. However, systems do have imperfections: for example, some atoms in a piece of glass may not be in this perfect crystal shape. So it will take a very long time for the system to “relax:” this is a kind of residual entropy, and there are interesting ways to deal with this. Thinking about the second statement, let’s write entropy in terms of the heat capacity. Lemma 170 Entropy for a given pressure and temperature is S(T, P) = S0 + Z T 0 dT T CP (T). where S0 is some “residual entropy” mentioned above. Proof. We start from the first law and formula for enthalpy: dU = TdS −PdV = ⇒dH = TdS + V dP. At constant pressure, ∂H ∂T P = T ∂S ∂T P = CP (T), where the third term disappears due to dP = 0, and now integrating both sides of ∂S ∂T P = CP (T ) T yields the result above. But now if we take T →0, the integral will diverge unless CP (T) goes to zero! So we must have CP (T) →0. Finally, let’s think about the third statement: how can we think about cooling a system in this way? 29.3 Engines and efficiency Definition 171 A heat engine is a machine that executes a closed path in the (phase) space of thermodynamic states by absorbing heat and doing work. Since internal energy is a state function, a closed loop will not change the internal energy of our system! We can think of this alternatively as assigning a U to each point in our phase space. This can be written as 0 = I dU = I TdS − I PdV. 96 Note that H PdV > 0 = ⇒our engine does work, and H TdS > 0 = ⇒heat is added to the system. If we draw a clockwise loop in PV -space, the engine will do work on the surroundings, and we can make an analogous statement about TS-space. So our goal is to construct such a closed loop and find the efficiency: how much work do we get for the amount of heat we are putting in? Definition 172 A Carnot engine is a special engine where all heat is absorbed at a fixed temperature T + and expelled at a fixed T −. This is an ideal engine: in other words, we are assuming that this is a reversible process and that all heat exchange only takes place between a source T + and sink T −. We can think of this as having our system between a source and sink, absorbing some heat Q+, expelling some heat Q−, and doing some work W. By the first law, since ∆U = 0, Q+ = Q−+ W. If we have a reversible transfer of energy between the system and the source, then the change in entropy S+ = Q+ T + . (If our system is not ideal, we may have additional entropy gain, since we lose additional knowledge about our system. So in general, we’d have S+ ≥Q+ T + .) Similarly, we also have a reversible transfer of energy between the source and sink, so the entropy expelled to the environment is S− = Q− T −. (In an irreversible process, some of the entropy remains as residual entropy inside our engine, so generally we have S−≤Q− T −. Definition 173 Define the efficiency of an engine to be η = W Q+ = Q+ −Q− Q+ = 1 −Q− Q+ to be the ratio of work we can do with the engine to the heat that we put in. We know that in our cycle, the entropy in and out of our system should also be a state function. So ∆S = 0, and S+ = S−= ⇒Q+ T + ≤Q− T −= ⇒Q− Q+ ≥T − T + . Plugging this in to our definition, we have the following: Proposition 174 (Carnot’s bound on efficiency of a heat engine) The maximum efficiency of a Carnot engine between a source of temperature T + and sink of temperature T −is η = 1 −T − T + . 97 Example 175 The maximum efficiency of an engine between 327◦Celsius and 27◦Celsius is 600−300 300 = 1 2 (convert to Kelvin). This Carnot efficiency is an absolute thermodynamic bound, and it’s saturated when all heat is added at some temperature T + and expelled at some temperature T −. If we plot this process in the TS plane, we trace out a rectangle and have to do the following steps: • Isothermally compress the gas at temperature T −: this requires expelling Q−heat and doing an equal amount of work. • Put in work to adiabatically compress our gas from T −to T +. • Isothermally expand the gas at temperature T +: this requires putting in Q+ heat and expelling that amount of work. • Let the gas adiabatically expand back to temperature T −(and get work out of it). In the PV plane, the first and third steps (for an ideal gas) follow the isotherms PV = c, and the second and fourth steps follow PV γ = c. The total work done here is the area enclosed by our curve in the PV plane! For an ideal gas, we have the following: If we say that our steps lead us between states 1, 2, 3, 4, then we can make the following table: Step Process Work by Heat 1 →2 isothermal compression −nRT ln V1 V2 −nRT −ln V1 V2 2 →3 adiabatic compression not needed 0 3 →4 isothermal expansion nRT + ln V4 V3 nRT + ln V4 V3 4 →1 adiabatic expansion not needed 0 where the “not needed” steps will cancel out. So the total heat added is Q+ = nRT + ln V4 V3 , while the work done is Q+ −Q−= nRT + ln V4 V3 −nRT −ln V1 V2 . We can show that V4 V3 = V1 V2 by using the fact that PV γ = c for adiabatic compression and expansion, and therefore we indeed have indeed η = W Q+ = T + −T − T + saturates the Carnot efficiency! Does our system change when we deal with small numbers and more fluctuation? The question of “whether we can beat the Carnot efficiency” is still a good question of research today at small scales! Next session, we will introduce an interesting engine called Stirling’s engine, whose work per cycle is better than the Carnot theoretical efficiency! 30 April 8, 2019 (Recitation) This recitation was taught by Professor England. 98 We often consider a situation where a piston is pushing on a gas. One assumption that is usually made is that we have a “quasi-equilibrium” state: the system is compressed slowly, so that we’re always close to an equilibrium state (for example on our PV diagram). In this scenario, we do actually use exact differentials in a way that can be measured. Then if we apply some force F over an infinitesimal distance dx, we have our first law dU = ¯ dW + ¯ dQ = −PdV + TdS. But in real life, we usually push our piston down quickly, and the compression is done rapidly. Now our question is: what’s the amount of work we do if our gas isn’t in equilibrium? (Basically, our system is not given enough time to relax.) Example 176 Let’s say our gas is placed in a bath at temperature T. When we compress an ideal gas very slowly, dU = 0 (since the temperature stays constant). This means ¯ dQ = ¯ dW = ⇒PdV = TdS. On the other hand, what’s the change in entropy for this process? The entropy change over the universe is ∆Suniverse = ∆Ssystem + ∆Sbath. In a quasi-static process, ∆Suniverse = 0: we have a reversible process, because the change of entropy at a constant temperature is ∆S = ¯ dQ T ; since all heat transferred out of the system is transferred into the bath at the same rate! So the total change in a quasi-static process is ∆S = ¯ dQ T −¯ dQ T = 0. Fact 177 Also, we can think about the fact that the entropy is proportional to −N ln ρ, where ρ is the density of our ideal gas. Indeed, this process is reversible: if we push the piston back out slowly enough, the heat will flow back into the system, and we’ll again have equal and opposite changes in entropy. But now let’s say we push the piston the same amount, but we do it fast enough that heat doesn’t transfer as smoothly: for example, we can imagine doing an adiabatic process, and then (once that’s finished) let the heat transfer with the bath happen. This is significantly different: in an isothermal process, PV is constant, while in an adiabatic constant, PV γ is constant: these yield different amounts of work for the same compression ∆V , because (mathematically) the integral of PdV is different or (physically) adiabatic compression means the particles will push back more, so the work we have to do is larger! So W > Z isothermal PdV ; and this argument works even when the process isn’t adiabatic. Ultimately, we are putting in some extra energy, and 99 that energy will eventually exit into the heat bath, since U = 3 2NkBT only depends on the temperature of our gas. Ultimately, then we know that the change in entropy of our system (ideal gas) doesn’t depend on our final state: it will be ∆Ssystem = −kB∆(N ln ρ) , which is the same as what we had before in the slow case. But the difference comes from the change ∆S in the environment: because the fast work we did is larger than the isothermal work, some heat is pushed into the heat bath! So ∆Q T , entropy change in the surrounding bath, is greater than the original R ¯ dQ T in the isothermal case, which means our process is now irreversible: ∆S > 0. Fact 178 So in summary, a fast process means we do work that is larger than the slow work we needed to do, which increases heat flow and therefore generates entropy. That’s where the second law of thermodynamics is coming from: ∆Sfast = 0. But it would also be nice to understand this in a microscopic scale as well. Let’s say we have a bunch of particles in our ideal gas that are flying around in this case: there will be times where (by chance) there’s some vacuum near the piston, so we can get lucky and compress the gas without doing any work (as fast as we want)! So we’ve then decreased the entropy of the gas without doing any work at all, and that seems to violate the second law of thermodynamics: how do we deal with this? This gets us into a more current topic: let’s say we have our system and we follow some path from the initial to final state (for example, we have some fast compression h(t) which dictates the height of our piston). Now we can define a state function F = U −TS, the Helmholtz free energy. We know that dU = −PdV + µdN + · · · + TdS, so taking differentials of the free energy (by the product rule), the TdS cancels out: dF = −SdT + ¯ dW So at constant temperature, dF = ¯ dW. That means that holding the temperature constant will give a total change in free energy equal to the work that we’ve done! So it’s important to build here because a first law that holds for quasi-static processes is deeply connected to it. To deal with the second law violation, we have to start thinking about the statistical properties of the microstates. A slow process always averages over all possibilities: a bunch of states are getting visited. But a really fast piston will visit the system in some microstate, which explains why our work is statistically varying. So we get some distribution p(W), and this will lead us to a fact: Proposition 179 (Jarzynski, 1997) We have Z dWp(W)e−W/(kBT ) = e−∆F/(kBT ) We’ll be able to prove this later on! What does this tell us? Inherently, the right hand side is a change in a state function, and the left hand side measures our statistical fluctuations: this is a limitation on the shape of our distribution. For example, p(W) = δ(W −∆F) in a quasi-static process, and indeed this checks out with our integral. 100 This can alternatively be written as a statement about expectations: D e−W/(kBT )E = e−∆F/(kBT ). So now we can define a quantity Wd = W −∆F which tells us how much work is dissipated. A quasi-static process has ∆F = W, but in general we can have some difference based on statistical variance. But dividing our statement above by e−∆F/(kBT ), we have D e−(W −∆F )/(kBT )E = D e−Wd/(kBT )E = 1. Now recall that the second law is about doing extra work that we didn’t need to! Dividing our ∆F definition through by T, ∆F T = ∆U T −∆S we can think of the right side as −∆Senv −∆Ssys, which is −∆Stotal. But by convexity (specifically, Jensen’s inequality), ⟨ex⟩≥e⟨x⟩ So plugging that in, ⟨Wd⟩≥0, so we always expect to do a nonnegative amount of dissipated work! 31 April 9, 2019 Some announcements about the exam: it will take place next Thursday, April 18. It’ll cover material after quiz 1 -it won’t have any problems on probability theory, but we will need to understand those concepts to do (for example) problems on a microcanonical ensemble! Again, previous exams will be posted online, and 3 of the 4 problems will be from previous years’ exams. There will be an optional review session as well. Notice that there won’t be any classes next week: Tuesday has no classes, and Thursday will be the exam. 31.1 A quick overview Last time, we talked about extracting work from heat engines, and we found ways to combine different processes in the PV plane to make a cycle and extract work. We found a theoretical bound, called the Carnot efficiency, and we constructed an example that saturates that efficiency. Fact 180 Note that it’s not possible to actually achieve this efficiency in real life because of dissipation and other processes! Now we’re going to try to introduce some other cycles as well, such as the Stirling engine. We’ll also discuss statistical physics for systems that are actually connected to the outside world using free energy! 101 31.2 A new heat engine Recall that in a Carnot cycle, we trace a clockwise cycle in our PV plane bounded by curves of the form PV = c and PV γ = C. We derived that the efficiency here is the theoretical bound η = W Q+ = T + −T − T + . But if we replace our adiabatic processes with isometric properties, this allows us to do more work in a single cycle! This is called a Stirling engine. In the TS diagram, we now no longer trace out a rectangle, since we no longer have adiabatic steps that keep entropy constant. Instead, for a monoatomic ideal gas, the entropy change is S = NkB 3 2 ln T + · · ·  . So in the TS-plane, our process is now bounded by horizontal lines (at constant temperature) and exponential curves (T ∝exp( 2 3 S NkB )). So our question now is how we construct such an engine? Again, let’s say our process takes us between states 1, 2, 3, 4. Step Process Work by engine Heat added 1 →2 Isothermal compression −nRT −ln V1 V2 −nRT −ln V1 V2 2 →3 Isometric heating 0 CV (T + −T −) 3 →4 Isothermal expansion nRT + ln V1 V2 nRT + ln V1 V2 4 →1 Isometric cooling 0 −CV (T + −T −) So the efficiency here is η = W Q+ = nR(T + −T −ln(V1/V2) nRT + ln(V1/V2) + CV (T + −T −). So the idea here is that the CV (T + −T −) is what limits us from reaching the Carnot efficiency - how do we get around this? Proposition 181 Consider a chamber with two pistons: the left piston is for “expansion” and the right piston is for “compression.” Keep the portion of the system to the left of the chamber at temperature T +, and keep the portion of the system to the right at temperature T −. In between, we have an ideal gas (equivalently a fluid), but put a material with high heat capacity in the middle as well (called the regenerator). So that amount of heat can be reused: it is stored and then reused in the next cycle! So now this effectively eliminates the CV (T + −T −) term, and we do indeed reach the Carnot efficiency T +−T − T + as desired. How much work do we get for each cycle? We can find that WStirling WCarnot = 1 1 −ln(T +/T −) (γ−1) ln r where r is the ratio of volumes. Plugging in T + T −= 2, r = 10, γ = 1.4, we find that a Stirling engine gives 4 times as much work as a Carnot engine for the same ratio of volumes and temperatures! 31.3 Brief aside about the third law Remember that one of our statements of the third law was that it is impossible to cool any system to absolute zero in a finite amount of time. One example of this is by switching between two different pressures: low to high 102 with an adiabatic process and high to low with an isothermal process. This can be done using nuclear magnetization! But both P curves end at (0, 0): that’s the main point of the third law. If we try to do any process, following some curve in the TS-plane, we will never actually be able to get to the origin. However, we’ve gotten pretty good at getting close to absolute zero: ask professor Ketterle about this! 31.4 Free energy So as a basic summary of what we’ve been talking about: we introduced S, our entropy, as a function of the number of microstates. We’ve made some statements about general properties of S, but we’re going to start looking at smaller parts of the system (and treat the rest as being a heat bath with constant temperature). So now our system exchanges energy, volume, and particles with an outside world in thermal equilibrium, so we have our temperature fixed (rather than the total internal energy). There’s many different notions of free energy: they will be used in different constraints. Remember that in an isolated system, we need ∆S ≥0 to have a spontaneous change of an isolated system (though this isn’t necessary sufficient). Let’s now imagine that our system is in contact with a heat bath at a fixed T: now the total entropy change is (at constant temperature) ∆Stotal = ∆Ssystem + ∆Sbath. If our system does no work (because it is at a fixed volume), ∆Sbath = ∆Q T = −∆Usystem T , and we can plug this in to find ∆Stotal = ∆Ssystem −∆Usystem T This means that we have a new necessary condition: Fact 182 We must have ∆Ssys −∆Usystem T ≥0 for a spontaneous change of our system in contact with a heat bath. This motivates us to make a new definition: Definition 183 Define the Helmholtz free energy F of a system to be the state function F = U −TS. We now require ∆F ≤0 to make a spontaneous change in a system at thermal equilibrium with fixed V and T. It seems that because S is a function of U, V, N in our system, F should also be a function of the four variables T, V, N, U. But the free energy is not actually a function of the internal energy! Indeed, F = U −TbSsystem(US) = ⇒ ∂F ∂US = 1 −Tb ∂SS ∂US = 1 −Tb Ts , and at thermal equilibrium (which is the only situation where we use F to represent the system), Tb = Ts. So the derivative is 0, so US is not an independent variable here! 103 Fact 184 This is an example of a Legendre transformation, which helps us change from one type of energy or free energy to another by using a different set of independent variables. For example, we go from (U, V, N) to (T, V, N). As another example, we can change our variables in another way: considering a system at constant temperature and pressure (this happens a lot in biology and chemistry). Then recall that dU = dQ −PdV = ⇒dH = dQ + V dP = ⇒∆Q = ∆H at constant pressure, and now we can play the same trick! The heat emitted into the environment at constant pressure is −∆H, so the total change in entropy ∆Stotal = ∆Ssystem + ∆Sbath = ∆Ssystem −∆H T , and by the second law, we must have this quantity be at least zero. Definition 185 This motivates the definition of the Gibbs free energy ∆G = H −TS and for a spontaneous change in our system, we must have ∆G ≤0: free energy needs to decrease! It’s important to note that this wasn’t all just to introduce a new state function: here’s a preview of what’s coming. In some systems, we have a fixed energy and can calculate the temperature of that system: in a microcanonical ensemble, we consider all possible microstates that are consistent with it. Then we can take some subset of our system, where the temperature T is still fixed (though the energy is not), and that’s what a canonical ensemble deals with! 31.5 Natural variables and Maxwell relations The idea we’ve been developing in this class so far is to go from various constraints on our system to some mathematical relationship between thermodynamic variables. For example, the first law tells us that if we define our internal energy U in terms of S and V , then dU = TdS −PdV = ⇒∂U ∂S V = T, ∂U ∂V S = −P. Because we have a state function U, the mixed second partial derivatives should be the same: this means ∂ ∂V  ∂U ∂S V  S = ∂ ∂S  ∂U ∂V S  V and that gives us the relation ∂T ∂V S = −∂P ∂S V . This applies to other state functions we have too - we can apply the same logic with our free energy definitions as well! The idea is that free energies are often easier to measure experimentally under certain conditions. For example, 104 since H = U + PV , dH = TdS + V dP, so writing H as a function of S and P, ∂H ∂S P = T, ∂H ∂P S = V, and doing the mixed partial derivatives yields ∂T ∂P S = ∂V ∂S P . Finally, since F = U −TS, dF = dU −TdS −SdT, which is also dF = −PdV −SdT. So this time, it’s natural to write F as a function of V and T. Now ∂F ∂T V = −S, ∂F ∂V T = −P = ⇒ ∂S ∂V T = ∂P ∂T V . These partial derivative equations often relate a quantity that is easy to measure with something that is generally less experimental in nature! We’ll talk more about these concepts next time. 32 April 10, 2019 (Recitation) 32.1 Exchanging energy Let’s start with the Helmholtz free energy F = E −TS. It’s easy to say what energy or momentum “does,” but what about free energy? One way to think about this is the “balance” between energy and entropy. Any system that interacts with a reservoir at temperature T (for example, room temperature in the world around us) cares about “free energy” to consider favorability of a reaction. The main idea is basically that the second law must hold! If we create an entropy ∆S, then we can gain an energy of E = T∆S from the reservoir. Another way to think about this is connected to the problem set: if we have a system with (non-normalized) occupation numbers na ∝e−βEa, notice that all Boltzmann factors are equal if we have infinite temperature, and that means all states are equally populated! On the other hand, with zero temperature, only the ground state is occupied, because any higher state has exponentially smaller occupation numbers. 32.2 Looking at the Sackur-Tetrode entropy Recall the equation that was derived by these two researchers from Germany back in 1913: S(U, T, N) = kBN  ln U N + ln V N + 3 2 ln 4πm 3h2 + 5 2  Note that the constants after the first two terms are just prefactors. Interestingly, though, Sackur got 3 2 instead of 5 2: this comes from ln n! = n(ln n −1), and he didn’t include the −1 in his approximation. But does this really matter? In almost all calculations, we ignore the entropy and only consider ∆S. But what would we actually get wrong? Notice that in the correct form of Sackur-Tetrode entropy, the entropy S goes to 0 as the temperature T goes to 105 0. It turns out we can measure absolute entropy via S(T) = Z T 0 dQ T = Z T 0 cP (T)dT T . (You also have to add latent heat to go from solid to liquid state, and so on. This can be written as a delta function, but the details aren’t that important.) Well, some researchers followed mercury across different temperatures: knowing CP , they integrated from 0, and eventually mercury became an ideal gas at high enough temperature! So through measurements, they found agreement with the theoretical answer. (They did assume 5 2.) Fact 186 This allowed them to find an experimental value of h, Planck’s constant, and they did this to 1 percent precision! It’s pretty amazing that we can determine this by measuring heat - this is really adding “entropy.” But if we replace 5 2 with 3 2 and try to get that experimental result for h, we actually change it by a factor of e−1/3 ≈0.72. So that would give you an incorrect experimental result by 30 percent! This is why absolute entropy does occasionally matter. 32.3 Multiple ground states In quantum mechanics, we have multiple energy levels: it’s possible that (for example at zero magnetic field) we have two low-energy states. Then we have a special symmetry in our system! For example, if we have N distinguishable particles and each can be in 2 degenerate states, we have an extra contribution S0 to the entropy. This was actually mentioned in Sackur-Tetrode: mercury has multiple isotopes, so we have to be careful. There’s other ways quasi-degeneracy could come up as well. 32.4 Fluctuations in a microcanonical ensemble Let’s try to think about a system with two subsystems, but instead of exchanging energy, let’s think about exchanging particles. If the system is divided into two symmetric parts A and B, and we have NA + NB = N particles, we expect there to generally be N 2 particles in both halves. The total number of microstates ΓAB is then multiplicative, because we essentially pick a microstate from both A and B. But then taking logarithms, the entropy SAB is now additive! But let’s think a bit more about the number fluctuation. Is it true that the number of microstates for the whole system, is ΓN,A+B = Γ N 2 ,A · Γ N 2 ,B? Not quite! Some arrangements of the system A + B don’t actually have the number of particles or the energy split exactly: there are fluctuations! So we basically have to add over all possible number of particles in system A: ΓN,A+B = X k Γk,AΓN−k,B. But in principle, the extreme values of k are not likely - they contribute very little to the entropy. Specifically, the distribution of probabilities NA is centered around N 2 , and now we have a sharp distribution that’s basically Gaussian with standard deviation ∝ √ N. 106 That means that allowing fluctuations, ΓN,A+B = Γ N 2 ,A · Γ N 2 ,B has an approximate “width” of √ N: this gives us an approximate answer of ΓN,A+B = √ NΓ N 2 ,A · Γ N 2 ,B. But this kind of fluctuation is very small: if we have, for example, 1026 particles, we have a precision of 10−13: this is unmeasurably small! So we wouldn’t be able to just count them. Now note that our counting of microstates is not quite multiplicative with this approximation: we have an entropy S ∝S N 2 ,A + S N 2 ,B + c ln N. So we should always be careful! The first two terms are extensive quantities, while the last term is not. Luckily, that is almost negligible compared to the other terms. 33 April 11, 2019 33.1 Overview and review This class is about studying the connection between microscopic and macroscopic descriptions of a system. What we’ve been doing recently is imposing various constraints on our system: for example, setting a fixed energy U gives us a microcanonical ensemble, and we’ll find that setting a fixed temperature T will give us what is called a canonical ensemble. Last time, we started discussing more relationships between our thermodynamic variables. If we have a first law condition dU = TdS −PdV, then we can define our energy U in terms of S and V : then because U is a state function, the mixed partial derivatives with respect to S and V gives us ∂U ∂S V = T, ∂U ∂V S = −P = ⇒∂T ∂V S = −∂P ∂S V . With the same kind of argument, we can also derive values from the enthalpy H = U + PV : this yields ∂H ∂S P = T, ∂H ∂P S = V = ⇒∂T ∂P S = ∂V ∂S P . With the Helmholtz free energy F = U −TS, ∂F ∂T V = −S, ∂F ∂V T = −P = ⇒ ∂S ∂V T = ∂P ∂T V , and finally with the Gibbs free energy G = F + PV , ∂G ∂T P = −S, ∂G ∂P T = V = ⇒ ∂S ∂P T = −∂V ∂T P . These four relations between our state variables are known as Maxwell’s equations: we’ll soon see why they’re important. For example, if we want to see change in entropy per volume at constant temperature, there’s really no way to measure that quantity directly, since it’s really hard to measure entropy. On the other hand, we can measure ∂P ∂T V a lot more easily: just change our temperature and measure the pressure inside some fixed volume! 107 Fact 187 Don’t memorize these equations: we can always derive these from the first law directly. More generally, we can make a table between intensive variables X and extensive conjugate variables Y : X Y −P V σ A H M F L E P µ N and in general, we can always write down our internal energy U as a function of S and Y , H as a function of S and X, F as a function of T and Y , and G as a function of T and X! We’ve mostly only been focusing on a three-dimensional gas, which is why we’ve been using P and V , but we could replace this with other pairs of variables. 33.2 An application of Maxwell’s relations Let’s try to compute heat capacity of an arbitrary material: in general, the formula CV = ∂U ∂T V measures how much internal energy of a system depends on its temperature. For an ideal gas, we know that internal energy only depends on temperature, but we may want to measure other heat capacities as well. Example 188 What is a good way to find ∂U ∂V T ? Let’s do this systematically so that it’s easy to understand how to do related problems! Start with the first law, dU = TdS −PdV. This means that ∂U ∂V S = −P. Let’s write U as a function of T and V , so dU = ∂U ∂T V dT + ∂U ∂V T dV. This includes the term we want! Going back to the first law, writing S as a function of T and V as well, ∂U ∂T V dT + ∂U ∂V T dV = dU = TdS −PdV = T  ∂S ∂T V dT + ∂S ∂V T dV  −PdV Now all differentials are in terms of dV and dT, so we can gather terms of dT and dV and divide to find ∂U ∂V T = T ∂S ∂V T −P. 108 If we want to make this a measurable quantity, though, we should replace the ∂S ∂V T term: this can be done with Maxwell’s relations! Since ∂S ∂V T = ∂P ∂T V , we just have ∂U ∂V T = T ∂P ∂T V −P . So if we just have the equation of state, we have enough to find the quantity we desired! Example 189 What is the V -dependence of CV ; that is, what is ∂CV ∂V T ? Recall that CV is itself a partial derivative! Specifically, ∂ ∂T  ∂U ∂V T = ∂ ∂V  ∂U ∂T T = ∂CV ∂V T . Since we just found the leftmost quantity, we can plug that in: ∂CV ∂V T = ∂ ∂T  T ∂P ∂T V −P  = T ∂2P ∂T 2 V . Example 190 Can we find a relationship in general between CP and CV ? Remember that we defined a quantity H = U + PV to help with this kind of problem: ∂H ∂T P = CP = ∂U ∂T P + P ∂V ∂T P . (This came from expanding out the derivatives on both sides.) Expanding out the first term on the right hand side, ∂U ∂T P = ∂U ∂T V + ∂U ∂V T ∂V ∂T P Note that ∂U ∂T V = CV , and we also computed earlier that ∂U ∂V T = T ∂P ∂T V −P, so plugging everything in, CP = CV +  ∂U ∂V T + P  ∂V ∂T P = ⇒ CP = CV + T ∂P ∂T V ∂V ∂T P . We defined response functions earlier in the class: we had variables like expansivity that are known for certain materials! So if we’re doing a problem in that realm, we can rewrite our equation in terms of coefficients like thermal expansion: α = 1 V ∂V ∂T P = ⇒CP = CV + TV α ∂P ∂T V . So now can we say anything about our function ∂P ∂T V ? Note that we have the partial derivative identity ∂P ∂T V ∂T ∂V P ∂V ∂P T = −1 = ⇒∂P ∂T V = − 1 ∂T ∂V P ∂V ∂P T = −V α ∂V ∂P T 109 where β = −1 V ∂V ∂P T is the coefficient of isothermal compressibility. This lets us just write down ∂P ∂T V = α β : plugging this in, we get the relation CP = CV + TV α2 β . This is basically a game of thinking like an experimental physicist: we know what’s easy to measure, so we just try to write our derivatives in terms of those quantities! 33.3 The Joule-Thomson effect This phenomenon is also known as the throttling process! Example 191 What is the temperature change of a real gas when it is forced through a valve, if the container is isolated from the environment (so no heat exchange)? In a real gas, there is actually interaction between the particles, so the temperature will actually change. Specifically, we then have a potential energy associated with our system as well! To describe our system, imagine having two containers A and B connected with a small valve in the middle: let’s say there is a pressure P0 on the left container A and a pressure P1 in the right container 2. We force some volume V0 through the valve from P0 to P1: then the work done by the piston for container A is WA = P0V0, and the work done by the piston for container B is WB = −P1V1. where V1 is the volume the gas takes up in the new container. Since there is no heat exchange, by the first law, U1 −U0 = WA + WB = P0V0 −P1V1 = ⇒U0 + P0V0 = U1 + P1V1. This means that this is a constant enthalpy process! So does the gas that is pushed through get cooler or warmer? We can define the Joule-Thomson coefficient µJT = ∂T ∂P H . Since we have expansion, ∂P < 0. This means that µJT < 0 = ⇒dT > 0, so the gas warms up, and µJT > 0 = ⇒ dT < 0, so the gas cools. This actually has to do with liquifying gases at room temperature! So how can we say things about µJT ? Starting from the enthalpy equation, if H is written as a function of T and P, since dH = 0, dH = ∂H ∂T P dT + ∂H ∂P T dP = ⇒µJT = − ∂H ∂P T ∂H ∂T P = −1 CP ∂H ∂P T . This ∂H ∂P T term is analogous to the ∂U ∂V T term from earlier! Specifically, we can go through the analogous derivations to find µJT = −1 CP  V −T ∂V ∂T P  . For an ideal gas, PV = nRT = ⇒ ∂V ∂T P = V T , so µJT = 0: we can’t see any effect on temperature. But for a van der 110 Waals gas, we have constants a and b so that (V −nb)  P + a n2 V 2  = nRT; if we expand out the left side, since a and b are generally very small, the −nb  an2 V 2  term is negligible, and thus PV −Pnb + an2 V = nRT; taking differentials, PdV −an2 V 2 dV = nRdT, and substituting in for P and doing the relevant calculations, µJT ≈n CP  2a RT −b  . So the temperature where µJT changes sign is the inversion temperature Tinversion = 2a Rb. If our temperature is larger than the inversion temperature, µJH is negative, so changing the pressure will increase our temperature. Otherwise, if T is smaller, then temperature will decrease! 34 April 17, 2019 (Recitation) There is a quiz tomorrow! 34.1 Types of free energy Last time, we mentioned the Helmholtz free energy and Gibbs free energy F = U −TS G = U + PV −TS, where both of these are a kind of balance between energy and entropy. We can think of this by looking at the Boltzmann distribution as T →0 and T →∞. Question 192. What does free energy mean? Let’s say we have our system S connected to a reservoir R fixed at temperature T. If we want to see if a certain reaction can happen, we must have the total ∆Stotal > 0 to have a favorable reaction (by the second law). Information-theoretically, we can go from more precise knowledge to less precise knowledge. What is the change in entropy here? The change in internal energy of the reservoir is ∆U = ∆Q (since no work is being done), which is −T∆Ssys in a reversible process. This means that the change in entropy ∆Stotal = ∆Ssys + ∆Sres = ∆Ssys −∆Usys T = −1 T (∆U −T∆Ssys) ≥0. Defining F = U −TS, any favorable reaction has to go to lower free energy! 111 Fact 193 Think of a reservoir as a swimming pool and our system as a piece of cold chalk being thrown into it. The change in temperature of the swimming pool can be assumed to be almost zero, and if not in a specific case, we can always scale the system. This is the limit we take in a canonical ensemble! On the other hand, when we have a system at constant pressure, we often use the Gibbs free energy G = F + PV , much like how we use enthalpy for some calculations instead of internal energy. But why is it called free energy? We can think of this as “energy that can be converted.” Looking again at a system with T and P constant (so we want to look at G), let’s say that we are part of the system, and we want to transform some internal energy U into something useful. For example, can we turn the energy gained from 2H2 + O2 →2H2O into something useful by the laws of thermodynamics? Is the energy actually available? Well, let’s say 1 mole of a large molecule dissociates into 5 moles of smaller constituents: then there are more molecules, so to conserve the pressure P, we must increase the volume V , which does work ∆U = −P∆V. This works in the water reaction as well! Since we now have less molecules, we have to reduce our volume, which does work on the system. This is where the PV term comes from in our equation, and we can finally account for the +T∆S from the change in entropy in our system (since that needs to be transferred to our reservoir). So we change the internal energy by some ∆U, and this now gives us an associated work P∆V and heat transfer −T∆S. This is indeed the G = U + PV −TS that we desire! Fact 194 In summary, we do “boundary condition” corrections for the variables we are fixing constant: in this case P and T. 34.2 Ice packs Some of these work without needing a frozen material: break something in a plastic pack, and it gets cold. How does this happen? It turns out that it requires energy ∆U > 0 to dissolve ammonium nitrate in water. So if this is in contact with something warmer (like an arm), it will grab energy and cool down the environment. Now think of the system as the plastic pack and our body as the environment! This is allowed because it creates some entropy ∆S > 0. So now F = U −TS < 0 is indeed true because the temperature T is sufficiently large (so the entropy overcompensates for the change in internal energy). But what happens if the ice pack were not in contact with anything? Then the water and ammonium nitrate would be completely isolated, and now this cannot dissolve! 112 34.3 Heat engines We have a theoretical bound on the efficiency of a heat engine ηC = T + −T − T + . Let’s say someone managed to create some ˜ η > ηC: what would this look like? We know that a reservoir T + provides heat Q+, and the sink T −gains heat Q+ −W. Well, let’s imagine doing this in reverse: pull heat Q+ −W from the T −sink and output heat Q+ to the T + reservoir. (For example, this is a refrigerator!) We can scale this in such a way that the Q+ here is the same as the Q+ in the theoretical other engine! Then putting the two engines together, the work we’ve created is ˜ W −W, and the heat we’ve extracted is −( ˜ W −W): we’ve removed the upper reservoir from the problem when we run the Carnot engine in reverse. But then we’d create work by extracting energy from a single temperature reservoir: this doesn’t make sense thermodynamically! We can’t get work and do a refrigeration at the same time, or at least it hasn’t been observed. Thus, the Carnot efficiency is actually the best possible bound on our heat engine! What’s more, any reversible engine has to work at exactly the Carnot efficiency, or else we could run it in reverse and combine it with the Carnot cycle forward, and we’d get the exact same contradiction. 35 April 22, 2019 (Recitation) 35.1 Engines Since we’ve been talking about engines, Professor Ketterle wanted to show us a Stirling engine! Most engines we have in real life have some combustible material, so we have an open system with fuel and some exhaust material (and this turns out to be more efficient in general). In principle, though, we should be able to operate an engine by just having one heat source (like in our Carnot cycle). The example at the front of class is just an alcohol burner, which acts as a hot reservoir, together with a cold reservoir at room temperature. Then we can run a motor in reverse, which is an electrical generator! 113 How does a heat engine work, generally? In principle, we can just say that we have a gas that compresses and contracts with a piston. But in this case, when we heat up the gas in the hot reservoir, a piston system actually moves the gas to the cold reservoir, where it cools down! We have pistons 90 degrees out of phase, and that lets our motor run. Recall that last time, we showed that if we ever had an engine with efficiency η > ηrev, we could couple it together with a Carnot engine (running one in reverse) to create work for free by extracting heat from a single reservoir. This would be very convenient, but it violates the second law of thermodynamics. Fact 195 This also told us that all reversible engines between reservoirs at T1 and T2 must have the same η efficiency. This is the maximum efficiency! The Carnot engine is just special because it’s an example of an easily described engine with that efficiency η = T2 −T1 T2 . 35.2 Free energy Recall that F = U −TS, the free energy, tells us something about whether a system’s entropy is going up or down: ∆F < 0 = ⇒∆Stotal = ∆Ssys + ∆Sres > 0 at constant temperature. Basically, defining new “energies” like F, G, H includes information about the reservoir as well: for example, G describes our environment on earth (at constant pressure and constant temperature), which is why we use G = U −TS + PV. Basically, TS and PV tell us that we need to do some work or transfer some heat to satisfy the environment conditions! So F and G tell us what is “left,” and that’s why their sign tells us about whether processes can happen spontaneously. Fact 196 We can think of “spontaneous” processes as those that can give us work, and that’s how to reconcile thoughts like “spontaneous combustion.” 35.3 Looking at free expansion of a gas again Let’s say we have a gas that expands from a volume V into a volume of 2V , while keeping the internal energy U and temperature T the same. (For example, consider an ideal gas). Then F = U −TS goes down, and free energy going down means that we can extract work out of this! That means we “missed our opportunity:” we’ve increased our entropy without getting the work out of it. But we can also just keep that whole system at constant temperature T: if we’re being formal about it, with a quasi-static process, we should actually have a piston that is isothermally expanding at temperature T. We’ll then find that the work done does satisfy |∆W| = |∆F|. If the internal energy and temperature stay the same, where is that work coming from? Well, the gas is losing some kinetic energy when the piston is moving back! It is the heat added by the environment at constant T that keeps the 114 internal energy constant, and that’s basically coming from the TS term in the free energy. So we create energy by reducing our internal energy, but the PV and TS terms are “taxed” by our reservoir and conditions. In our case here, we actually get a bonus from the environment to get work! Proposition 197 Use F when we have constant temperature, and use G when we also additionally have a requirement that P is constant. 35.4 Partial derivatives We have a bunch of thermodynamic variables: P, V, S, T, H, U, F, G. This is a system with only PV work. Well, that already gives us eight variables: we can now play in eight-dimensional space and write expressions of the form ∂a ∂b c . But how many of these eight variables are independent? For example, in 3-dimensional space, we can use r, θ, φ versus x, y, z and do lots of coordinate transformations, but no matter what, we have three independent variables. How many do we have in this situation. Well, let’s think about an ideal gas, where N is constant. Defining volume and temperature gives us pressure (by the ideal gas law), entropy (by the Sackur-Tetrode equation), and then we can get all types of energy. This means that we always have two independent variables! Often, we students will write things like ∂S ∂P V,T , which doesn’t actually make any sense, since fixing V and T tells us exactly what our other variables are! It’s important to not get lost in the jungle of variables: we can only keep one constant at a time (of course, unless one of the two is N). So Maxwell’s relations come about because P, V, S, T can be written as first derivatives of the thermodynamic potentials H, U, F, G. Specifically, for a function X (which is one of U, H, F, G), we can think of ∂2x ∂a∂b = ∂2x ∂b∂a. For example, in problem 4 of last week’s exam, we were told to compare ∂V ∂T , ∂S ∂P . The relevant differentials here are dU = TdS −PdV, dG = −SdT + V dP. This means U is written most naturally as a function of V and S, but G is written as a function of T and P. We can always convert between our variables, but this is the way we generally want to do it! We are supposed to say what we keep constant in the two partial derivatives above, but we should ask the question: should we use the Maxwell’s relation for U or for G? 115 We do have V and S in the numerators, and we have T and P in the denominators, so either would work at this point. But note that we were told to actually deal with ∂V ∂T S , ∂S ∂P V . So now looking at the V and S, it’s most natural to work with U! But G gives the answer as well, and this is what Professor Ketterle did initially! Let’s work out some of the steps: dG = −SdT + V dP = ⇒−S = ∂G ∂T P , V = ∂G ∂P T . So now ∂2G ∂T∂P = −∂S ∂P T = ∂V ∂T P . This isn’t actually what we want, though: we have the wrong variables held constant! There is a way to get around this though: just like we can transform CP and CV into each other, let’s look at how to relate ∂S ∂P T to ∂S ∂P V . We’re going to have to label our variables carefully: writing S as a function of P and V (since this is what we want in the end), where V is actually a function of P and T. This means we’re representing our entropy as S(P, T) = S(P, V (P, T)) = ⇒ ∂S ∂P T = ∂S ∂P V + ∂S ∂V P ∂V ∂P T by the chain rule. So it’s possible to work from there, but it is very messy! Alternatively, we could go back and do the smarter thing: looking at U as the thermodynamic potential instead, we end up with dU = TdS −PdV = ⇒∂T ∂V S = −∂P ∂S V . This is exactly the reciprocal of what we wanted: we end up with ∂V ∂T S = −∂S ∂P V . 36 April 23, 2019 All exam grades will be posted by the end of today. We can always email the professor to schedule a meeting if we want to discuss! Also, drop date is very soon. There is no pset due this week, because we just started some new material. Instead, we can focus on other homework assignments we might have! 36.1 Overview It’s time to start a new section in our class - we’re going to talk about a new kind of ensemble. We found that for a system that is well-defined in a certain way (mechanically and adiabatically isolated) can be described by a microcanonical ensemble, which relates to the probability distribution of the individual microstates. The key idea there was that the energy was held constant, and in these cases, all microstates are equally probable. The temperature T is then a natural consequence! We often deal with systems that aren’t adiabatically isolated, though. Then the microstates may have different probabilities: that’s what we’ll discuss today, and we’ll use the canonical ensemble as a tool! We’ll find that there 116 really isn’t much of a difference between the properties of a microcanonical and canonical ensemble in some aspects, but we don’t have to compute the actual number of microstates anymore. 36.2 The canonical ensemble Let’s start by making a distinction: Fact 198 In a microcanonical ensemble, we specify the internal energy U, and at thermal equilibrium we can deduce the temperature T. On the other hand, in a canonical ensemble, we specify the temperature T and use this to deduce the internal energy U. Then our macrostates M are specified by our temperature T, as well as (potentially) other variables. In this scenario, we’re allowing heat to be inputted, but no external work can be done on or by the system. Proposition 199 We can have our system maintained at a fixed temperature T if it is in contact with a reservoir or heat bath at temperature T. We make the assumption that the heat bath is sufficiently large that it has basically negligible change in temperature! For example, a glass of boiling water in a large room will cool down to room temperature, while the room’s temperature is basically fixed. Question 200. How do we find the probability of any given microstate PT (µ)? Let’s say our reservoir has some microstates µR and energies associated to them HR(µR). Similarly, our system has some microstates µS and energies HS(µs). We can think of the reservoir and system as one larger system R ⊕S: this is now mechanically and adiabatically isolated, so we can think of this as a microcanonical ensemble! Our total energy here is Etotal, and we know that (because our system is much smaller than the reservoir), we have Esys ≪Etotal. Then the probability of some microstate (µs ⊕µR) is P(µS ⊕µR) = 1 ΓS⊕R(Etotal) if our total energy HS(µS) + HR(µR) = Etotal), and 0 otherwise. This is a joint probability distribution, so if we want a specific µS for our system, we “integrate” or “sum out” all values of R: P(µS) = X {µR} P(µS ⊕µR). Also recall that we can write the conditional probability P(S|R) = P(S, R) P(R) , so (because probability is 1 Γ, the multiplicity), this expression can also be written as P(µS) = ΓR(Etotal −HS(µS)) ΓS⊕R(Etotal) 117 since we sum over the multiplicity of R given a specific energy. Notice that the denominator here is independent of µS: it’s a constant, so we can rewrite this in terms of the entropy of our reservoir: given that S = kb ln Γ, P(µs) ∝exp  1 kB SR(Etotal −HS(µS)  . In the limit where the energy of the system is significantly smaller than the total energy of the system and reservoir, we can do a Taylor expansion of this entropy to simplify this further! Then (treating HS(µS) as our variable), because ER = 1 −HS(µS), SR(Etotal −HS(µS)) ≈SR(Etotal) −HS(µS)∂SR ∂ER . Dropping the S subscripts for simplicity and plugging into the probability distribution, since β = 1 kBT , and exp[SR] is some constant that doesn’t depend on our microstate (to first order), we have the following fact: Theorem 201 (Canonical ensemble) For a fixed temperature, the probability of our microstate is P(µ) = e−βH(µ) Z . This is a probability distribution, which means Z is our normalization factor! Z is taken from the German word “zustandssumme,” which means “sum over states.” That explains why it’s our normalization factor here: if we add up P(µ) for all states, we’ll get Z Z = 1. Definition 202 Z here is called a partition function. In this case, Z = X µ e−βH(µ). 36.3 Looking more at the partition function Why are partition functions useful? Basically, many macroscopic quantities can be described in terms of our function Z. Remember that the multiplicity Γ was important in our microcanonical ensemble for finding S: now Z takes its place, and it’s a lot easier to write this down! Fact 203 The exponential term here is called a Boltzmann factor. So now notice that the energy is no longer fixed: it’s some random variable H, and we want to find its mean, variance, and other information. (Spoiler: the fluctuations are small, so in the thermodynamic limit, this will look a lot like the microcanonical ensemble!) First of all, we can write the probability of a given energy H as a sum P(H) = X µ P(µ)δ(H(µ) −H), 118 since we want to add up the states with our given energy. Since all probabilities in this sum are equal, this evaluates to e−βH Z X µ δ(H(µ) −H) The sum is the number of microstates with a given energy H′, which is just our multiplicity Γ(H). Thus, this can be rewritten as P(H) = Γ(H)e−βH Z . We’ll talk more about this later, but we can write Γ in terms of our entropy: this will yield an expression of the form 1 Z exp S(H) kB − H kBT  , and since this is a constant times H −TS, this is related to our Helmholtz free energy! 36.4 Evaluating the statistics Note that as H increases, Γ rapidly increases, while e−βH rapidly decreases. Therefore, if we plot this, we’ll find that the distribution is sharply peaked around some energy U. What’s the mean and variance of this distribution? Well, the average here is ⟨H⟩= X µ H(µ)e−βH(µ) Z which can be written as a derivative −1 Z ∂ ∂β X µ e−βH(µ). But the sum here is just Z, so ⟨H⟩= −1 Z ∂Z ∂β = −∂(ln Z) ∂β . How sharp is this distribution - that is, how narrow is it? We can compute the variance var(H) by first considering −∂Z ∂β = X µ He−βH. Taking another derivative with respect to β, −∂2Z ∂β2 = X µ H2e−βH, but this looks a lot like the average value of H2! We then find ⟨H2⟩= 1 Z ∂2Z ∂β2 . Therefore, var(H) = ⟨H2⟩−⟨H⟩2 = 1 Z ∂2Z ∂β2 −  1 Z ∂Z ∂β 2 . This can be rewritten as 1 Z ∂ ∂β ∂Z ∂β  + ∂ ∂β  1 Z  ∂Z ∂β 119 which is actually just ∂ ∂β  1 Z ∂Z ∂β  by the product rule! If we look back, this tells us that var(H) = −∂ ∂β ⟨H⟩, where ⟨H⟩= U is our “average energy.” Since β = 1 kBT , this means we can rewrite this as var(H) = kBT 2 ∂U ∂T . Since CV , our heat capacity, is defined to be ∂U ∂T V,N, if we define ˆ CV = CV N (the heat capacity per particle), var(H) = NkBT 2 ˆ CV . If we think of fractional fluctuations, we want to look at the ratio of our standard deviation to the mean. The mean is proportional to N, the number of particles, since it is an extensive quantity, but the standard deviation of H is proportional to √ N. This means p var(H) mean(H) ∝ 1 √ N which is very small on the thermodynamic scale, and thus our variable is highly concentrated! This means that we can basically think of the system almost as being a microcanonical ensemble with fixed energy ⟨H⟩: it was just easier to get to this point, since all we need to do is compute our partition function Z. 36.5 Writing macroscopic quantities in terms of the partition function We found earlier that ⟨H⟩≡U = −∂ ∂β ln Z. Since β = 1 kBT , ∂ ∂β = ∂ ∂T ∂T ∂β  = ⇒ ∂ ∂β = −kBT 2 ∂ ∂T = ⇒U = kBT 2 ∂ ∂T ln Z. Let’s next look at the entropy: by definition, S = −kB X j pj ln pj. Plugging in our probability distribution, S = −kB X e−βEj Z (−βEj −ln Z). Rewriting this in terms of the internal energy U, S = kBβU + kB ln Z = U T + kB ln Z . From this, we can see that −kBT ln Z = U −TS = F, 120 which is the Helmholtz free energy! Recall that we have (by the differential dF = −SdT −PdV , ∂F ∂T V = −S, ∂F ∂V T = −P, so we can then get our other quantities by taking partial derivatives of F. This can always yield our equation of state! 37 April 24, 2019 (Recitation) Let’s start by discussing a topic from last problem set. 37.1 Mixing entropy Consider two gases (red and white) that are mixed in a box with volume V . How can we compare this situation to one where the two are in separated boxes (each with volume V )? Specifically, we have three scenarios: in A, red and white are in the same box of volume V , in B, red and white are in two different boxes each with volume V , and in C, red and white are in two different boxes each with volume V 2 . What can we say about the entropy? First of all, any scenario in C can occur in scenario A, so the entropy SA > SC. But let’s go ahead and do the mixing derivation again! Recall the Sackur-Tetrode equation S = NkB  · · · + ln V N  . (Everything else - masses, temperatures, and so on - are constant.) Usually, A and C are compared in textbooks: there is a partition in a box that is then removed. If this partition breaks our volume V into two parts with N1 = c1N and N2 = c2N of the total (for example, 80 percent Nitrogen on one side and 20 percent on the other), we can apply the Sackur-Tetrode equation: V and N are proportional if our system is at equal pressure on both sides: SC = kB  c1N ln c1V c1N  + kB  c2N ln c2V c2N  = NkB ln N V (before we mix). After we mix, though, we sum instead SA = kB  c1N ln V c1N  + kB  c2N ln V c2N  because each component now has the full volume instead of only a fraction! Note that we can write this as SA = kB  c1N ln V N −ln c1  + kB  c2N ln V N −ln c2  = SC −kBN(c1 ln c1 + c2 ln c2) , and this last term is called the mixing entropy. Fact 204 This generalizes to more than two components as well. Here’s a second derivation that is even simpler! We can ignore the color of the gas particles at first and go through the derivation for the Sackur-Tetrode equation in both cases. But then we have N = N1 + N2 particles, and we need to label N1 of them red and N2 of them white. In situation C, we must label all of the ones on the left red and all of 121 the ones on the right white, but we’re free to do this arbitrarily in situation A! So we get an extra multiplicity of N N1  , and that will contribute an extra ∆S = kB ln Γ = kB N! N1!N2! = −kBN N1 N ln N1 N + N2 N ln N2 N  , as expected. But notice that situations A and B are now essentially the same! When doing the Sackur-Tetrode equation, we made the explicit assumption that we could treat the red and white gases separately. So that actually tells us that SA = SB. Proposition 205 So is there a way for us to separate the two gases from situation A into situation B without doing any work? We can just construct a (one-directional) special membrane that only allows one of them to pass through! For example, if one of them is large and one is small, we could have small pores - we don’t violate any real laws of physics this way. Fact 206 Specifically, think of a red and white membrane that allow only red and white particles to pass through, respectively. Then we can enclose our red and yellow particles in boxes and “translate” the red box until it is side-by-side with the white box! So to get from situation A to situation B, we can place two boxes next to each other (each with volume V ), separated by a membrane. No work is done this way - the translation doesn’t do any work or allow for any transfer of heat! To get from situation B to situation C, then, we can compress the containers of situation B: if we do this isothermally, the work that we do creates an equivalent amount of heat (since ∆U = ∆Q + ∆W = 0), and the heat ∆Q is actually just T∆S, the change in entropy of mixing. 37.2 Looking at the canonical ensemble Recall the setup: we have a system connected to a reservoir at constant temperature T. What’s the probability of our system being at an energy E? Basically, we think of the reservoir plus the system as a microcanonical ensemble: then the probability is just proportional to the number of microstates of the system at energy E, which is the same as the reservoir being at energy U −E (where the total energy of reservoir plus system is U). Then writing this in terms of entropy, p(E) = ceSR(U−E)/kB ∝ΓR(U −E). But now assuming E ≪U, we can do the Taylor expansion = ceSR(U)/kBe ∂SR ∂U ·−E/kB 122 and the first two terms here can be encapsulated as one constant: remembering that ∂S ∂U = 1 T , = e−E/(kBT ) Z for some constant Z. As a normalization factor, we now have our “partition function” Z = X j e−Ej/(kBT ). Fact 207 Probably the most important idea in this derivation is thinking of the reservoir and the system as one big system, since it allows us to think about energy! As a question: Taylor expansions are first order, so what’s the range of validity for the assumptions that we made? Specifically, remember that the second derivative, which is related to the “curvature,” tells us how much the first derivative (which is related to temperature) is changing when we vary our parameter. So we neglect the change in temperature as we change our energy E! So this error term only comes about when the reservoir is too small: this means that with small systems (or theoretical “canonical ensembles,” really), our derivation gives an exact result. Recall that we did some derivations related here earlier on: on an earlier problem set, we had na particles in states of energy εa, and we had specific limits on P na = N and P naεa = U. We found then that na = N e−βεa P j e−βεj ; this was found by looking at the combinatorics of distributing particles over energy states and incorporating Lagrange parameters for the constraints on N and U. Notice that we can now translate this: we have N systems, each at some energy level, and now na N tells us the probability that a random system has energy εa! How do we make this into the language of a “canonical ensemble?” Any given system is coupled to the combined reservoir of N −1 other systems: as long as 1 ≪N, and we are at thermal equilibrium, the temperature of the reservoir will not change very much, and we can use our canonical ensemble formula. 38 April 25, 2019 We’ll discuss the canonical ensemble some more today! We’ll also start looking at a few more examples - the main idea is that the canonical ensemble will result in significantly simpler calculations, since microcanonical and canonical ensembles predict the same results in the thermodynamic limit. 38.1 Review and remarks on the canonical ensemble Here’s a helpful table: “macrostate” tells us which macrostates are fixed or given to us as constraints. Ensemble Macrostate p(µ) Normalization factor Microcanonical (U, x) δ(H(µ)−U) Γ S(U, x) = kB ln Γ Canonical (T, x) exp(−βH(µ)) Z F(T, x) = −kBT ln Z 123 In other words, the free energy in the canonical ensemble and the entropy in the microcanonical ensemble play similar roles. Note that the logarithm of the partition function also gives other quantities. For example, CV = Nˆ cV = 1 kBT 2 σ2 x where σ2 x is the variance of the energy of our system at thermal equilibrium. This shows up in more sophisticated discussions of statistical physics! Since our heat capacity is a response function CV = ∂Q ∂T V , it describes how much heat is transferred given some perturbation of our temperature. But instead of having to do that perturbation, this equations tells us that observing fluctuations at thermal equilibrium works as well! Fact 208 This is a special case of something called the fluctuation-dissipation theorem. As another comment: we’ve mostly been summing over states to calculate the partition function, but we’re often instead given the energies as opposed to states. For example, if we’re given that our energies are {Ei|1 ≤i ≤M}, we can write our partition function as Z(T, N, V ) = X i gie−βEi, where gi is the degeneracy of the energy level (letting us know how many microstates are at that given energy Ei). In the continuous limit, we can approximate this instead as Z(T, N, V ) = Z dE dN dE e−βE, where dN dE is the density of states we discussed earlier in class. One more comment: since F = −kBT ln Z, we know that Z = exp(−βF). This looks a lot like the Boltzmann factor exp(−βEi) for a particular microstate! Thus we can think of exp(−βEi) as being a volume or weight in phase space for a particular state: Z is then the weight contributed by all states for a given N and V . (For a canonical ensemble, since F = U −TS is the “free energy,” it can be thought of as the amount of energy available for the system to do useful work.) 38.2 Looking at the two-state system again Example 209 Let’s go back and say that we have a system where a particle can be in two states: E↑= ε 2 and E↓= −ε 2. The partition function here is then Z = eβε/2 + e−βε/2 = 2 cosh βε 2  , and now the internal energy of our system U = −d dβ ln Z = − ∂Z ∂β Z = −ε 2 tanh βε 2  124 goes from −ε 2 to 0 as our temperature T gets larger (which looks a lot like our microcanonical ensemble!). Then the probability of being in each of our two states is P↑= e−ε/(2kBT ) 2 cosh ε 2kBT + 1 = 1 1 + eε/(kBT ) , P↓= 1 −P↑. Indeed, P↑goes from 0 to 1 2, and P↓goes from 1 to 1 2 as our temperature T gets larger. Next, let’s calculate the entropy: from our free energy F, we find that S = U T + kB ln Z = −ε 2T tanh ε 2kBT + kB ln  2 cosh ε 2kBT  . Looking at the limits, this goes to −ε 2T +kB ln exp ε 2kBT →0 as T →0. This is consistent with what we already know: there’s only one possible ground state! On the other hand, as T →∞, this goes to kB ln 2, which is consistent with the microcanonical ensemble and the intuition of our system. We can also calculate our heat capacity: C = ∂U ∂T = −ε 2 ∂ ∂T tanh ε 2kBT = kB  ε 2kBT 2 1 cosh2 ε 2kBT . Plotting this as a function of thermal energy kBT, the distribution is (just like in the microcanonical ensemble) unimodal. Think of this in terms of fluctuations! We can calculate the standard deviation of our energy at a given temperature: σ(U) = kBT r CV kB = ε/2 cosh ε kBT , and the behavior models the microcanonical ensemble as T →0 and as T →∞: it is 38.3 Systems with a large number of particles Let’s now think about how to compute the partition function in larger systems! Specifically, how can we do this for distinguishable versus indistinguishable particles - more specifically, how many microstates are there with a given number distribution (n1, · · · , nM)? Assume our particles are distinguishable. If we have N particles, there are N n1  ways to pick n1 of them to go in the first energy level, then N−n1 n2  ways for the next energy level, and so on: this gives the multinomial coefficient N! n1!n2! · · · nm!. This is the degeneracy factor for a given energy that we want! Let’s look at us having N distinct harmonic oscillators, where there are M distinct energy levels ε1, · · · , εM. Then nk denotes the number of distinct harmonic oscillators that are excited at our energy level εk. Now we can write our partition function Z = X n1+···+nm=N N! Qm j=1 nj! exp " −β M X k=1 εknk # . where the first term is the degeneracy factor, and the PM k=1 εknk is the total energy of our state. Let’s see if we can simplify this: first of all, we can expand the sum inside the exponential as a product: Z = X n1+···+nm=N N! QM j=1 nj! M Y k=1 exp[−βεk]nk. 125 Now remember the binomial theorem: (a1 + a2)N = X n1+n2=N N! n1!n2!an1 1 an2 2 which generalizes to the multinomial theorem (a1 + · · · + aM)N = X n1+···+nM=N N! n1! · · · nM!an1 1 · · · anM M . This is exactly what we want here! By the multinomial theorem, we can rewrite our Z as Z = (exp[−βε1] + exp[−βε2] + · · · + exp[−βεM])N , and since the sum inside the parentheses is actually the partition function for an individual harmonic oscillator Z1, we can write Z = ZN 1 . Proposition 210 This means that the partition function for identical distinguishable systems is multiplicative! It’s important here that the individual oscillators here were distinguishable to get the multinomial coefficient. But if we want to do the same exercise with indistinguishable particles, the degeneracy factor becomes just 1. Then calculations in general are a lot uglier, but we can look at a special case: let’s say we have high temperatures, which is basically looking at our harmonic oscillators in the classical limit! Then our states are basically uniquely defined by just the list of energy levels (and we rarely need to deal with overcounting issues), so we just end up with Z →1 N!ZN 1 . Talk to Professor Ketterle when we’re not looking at the classical limit, though! Example 211 Let’s explicitly compute the partition function for N distinct harmonic oscillators and work with it! Then we have (using facts from quantum physics) Z1 = ∞ X n=0 exp  −βℏω  n + 1 2  = e−βℏω/2 ∞ X n=0 e−βh¯ ωn = e−βℏω/2 1 −e−βℏω by the geometric series formula, and Z is just ZN 1 . Now that we have our partition function, we can compute our other macroscopic variables: U = −∂ ∂B ln Z = −N ∂ ∂β ln Z1 which evaluates out to U = Nℏω  1 e−βℏω −1 + 1 2  . As we take our temperature T →0, we only occupy the ground energy level, which is 1 2Nℏω. On the other hand, if our energy is large, kBT ≫ 126 hbarω = ⇒T ≫ℏω kB , which is the high-temperature classical limit: think about what happens this way! It’s also good to think about how to calculate the heat capacity and entropy of this system from here. One comment: we’ll deal with a lot of series in these kinds of calculations, so we should become comfortable with the relevant techniques! 39 April 29, 2019 (Recitation) Today’s recitation is being taught by the TA. Recently we’ve been discussing the canonical ensemble - as a first question, when do we use it? Basically, whenever we have a system, we have some different states: for example, we can consider them by energy level. In a microcanonical ensemble, we consider a system that is cut offfrom all of its surroundings: no mass or heat transfer, so we have conservation of energy. But in the real world, most systems are connected to the environment, even if we have a fixed number of particles. A canonical ensemble is just the most basic example of this!The main idea is that the probability of finding the system in an energy εi is pi = e−βεi Z , where Z is the partition function. How do we derive this? In general, our system needs to maximize entropy S S({pi}) kB = − X pi ln pi. We have the constraints P pi = 1, and also we can say that the “average” energy of our system is constant: writing that in terms of our probabilities, we have P piεi = U. Fact 212 An idea here is that thermodyanmics and statistical mechanics are connected by the idea of “averaging.” Now that we have our constraints, we use Lagrange multipliers! Basically, given a function f that we want to maximize and a bunch of constraints of the form {gi(x) = 0}, it’s equivalent to minimizing over x and λ (the Lagrange multipliers) F(x) = f (x) − X λigi(x). To minimize a vector function, consider the gradient: we want ∂F ∂xi = ∂F ∂λj = 0 for all i, j. (The latter, by the way, is just saying that each gi(x) = 0.) So let’s apply this to our problem! We have G({pi}) = − X i pi ln pi −λ( X i pi −1) −β( X piεi −U) Taking the derivative with respect to pi, ∂G ∂pi = −ln pi −1 −λ −βEi = 0 = ⇒pi = e−1−λ−βEi. Meanwhile, taking the derivative with respect to λ just yields P pi = 1: this means we have to normalize our probability distribution: this yields pi = e−βεi P a e−βεa , 127 as before! So we can actually define temperature T directly in terms of our Lagranage multiplier: β = 1 kBT . Our partition function looks important, but can we do anything with it? First of all, we can write it in two different ways: Z = X states i e−βεi = X energies {εn} gne−βεn. (The gn degeneracy factor is pretty important here!) Let’s try to write this in terms of some quantities that we already know about: since our energy U = ⟨E⟩= X i piεi, X i εie−βεi Z , note that −1 Z ∂Z ∂β = 1 Z X i εie−βεi = −∂ln Z ∂B = U. Now that we have energy, we can think about this in certain kinds of systems. A real magnet is a bunch of small atoms that are magnetic dipoles: microscopically, those dipoles are oriented in magnetic domains, and this order breaks down if the magnetic heats up! (Basically, there’s no clear direction for the dipoles to point.) On the other hand, magnets at very low temperature have a defined order: there’s a quantity “magnetization” which is the average |µ| N , and it decreases to 0 at some Curie temperature TC. We might see later on that the decay is actually square-root up until that point! Example 213 So if we have a system with a bunch of magnetic spins, we have a total energy (in a two-state system) E = −( X µi)B. Knowing B and knowing the temperature T, we want to think about the average magnetization m = ⟨P µi⟩ N . We can think about this in another way: consider X {µi} (P µj)eβ P i µi)B Z = 1 β ∂ln Z ∂B = Nm, and now we have what we want: we’ve found m in terms of the partition function! So looking at probability distribution, we want the fluctuation in U (∆U)2 = ⟨E2⟩−⟨E⟩2. The second term here is already known: it is  − ∂ln Z |partialβ 2 , and then how can we find the first term? Take more derivatives! Note that ∂2 ln Z ∂β2 = ∂ ∂β  −1 Z X εie−βεi  , and now by the product rule, = 1 Z X ε2 i e−βεi  −1 Z2 ∂Z ∂β 2 . The first term is now the expected value of ⟨E2⟩, and therefore the fluctuations (∆U)2 are connected to partial derivatives ∂2 ln Z ∂β2 ! 128 We can think of this in terms of other situations as well: remember that when we have our heat capacity C = ∂U ∂T , we can also think about the relationship between ∂ ∂T and ∂ ∂β = ∂ ∂T ∂T ∂B, and now ∂B ∂T = − 1 kBT 2 by direct substitution. Fact 214 It will be important to switch between β and T in the future! So then C = ∂U ∂T = − 1 kBT 2 ∂U ∂β = 1 kBT 2 ∂2 ln Z ∂β2 = ⇒ (∆U)2 = kBT 2C . C is a response function: we can generally do something to our system and see how this changes the properties. The equation, then, connects theoretical quantities to measurable experimental quantities! As an exercise, we can prove that if χ = ∂⟨M⟩ ∂B , we have a similar relation: then (∆M)2 = kBT 2χ. Finally, is there a way for us to compute the entropy from our partition function? We have S = −kB X pi ln pi = −kB X pi(−ln Z −βεi). This can be rewritten as = −kBβ X piεi + kB ln Z X pi. The sum of the pis is 1, and the sum of the piεis is our energy ⟨E⟩, so we have S = 1 T ⟨E⟩+ kB ln Z. This should look familiar: since the Helmholtz free energy F = U −TS, we can rewrite F = kBT ln Z = 1 β ln Z. But now knowing F and U, we are able to find S, and then we can get whatever else we want! In fact, we can also use this to prove the first and second law. Example 215 Let’s go back to our two-state system. We can say that our energy is E = −B X µi = −( X σi)µB, where each σi is ±1. Intuitively, we should expect independence, so Z should be multiplicative. Indeed, our collection of spins {σi} = (±1, · · · , ±1): this is 2N configurations, and we can then write this as X {σi} e−βµB(P σi) and now we can expand this out as (since P aibj = P ai P bj) X {σi} eβµBσ1eβµBσ2 · · · = Y i X σi=±1 eβµBσi ! But we always have the same B, µ, β, σ, so each sum is equal: this means we have Z = e−βµB + eβµBN , = 2N cosh(βµB)N 129 and we’ve found Z explicitly. Finally, looking at the magnetization one more time, ⟨M⟩= 1 β ∂ln Z ∂B = Nµ tanh(βµB). So m, the average magnetization, goes from 1 to −1: a strong enough magnetic field will align all the spins in the same direction! We also can expand the curve around B = 0 to figure out the linear relationship between m and B at small B. 40 April 30, 2019 There is a problem set due this week on Friday - it covers a lot of concepts about the canonical ensemble, and each one is a good way to put what we have learned into practice! Today, we’ll look at more examples of canonical ensembles and see the applications to black-body radiation and other topics. (This is because we can think of black-body radiation as a bunch of oscillators, and we can learn about their equation of state to deduce further information.) 40.1 Back to the two-level system Let’s again imagine that we have N particles, each with two energy levels. This can be thought of as assigning an up or down arrow to each of N spots on a lattice - the particles are then distinguishable, because we can identify them by their position. Let’s say the up and down spin have energy levels of ε 2 and −ε 2, respectively. We found last time that the partition function for one particle was Z1 = e−εβ/2 + eεβ/2 = 2 cosh βε 2 , and then the partition function for all N particles, by independence, is just Z = ZN 1 =  2 cosh βε 2 N . We can now calculate the thermodynamic properties of the system: first of all, ln Z = N ln  2 cosh βε 2  This actually gives us the other quantities that we want: we can calculate U = −Nε 2 tanh ε 2kBT , S = N  − ε 2kBT tanh ε 2kBT + kB ln  2 cosh ε 2kBT  . We’ll also find that the heat capacity CV is just N times the heat capacity of a single particle, and the fluctuation standard deviation of U is σ(U) = kBT r CV kB = √ Nε/2 cosh ε 2kBT . Since U ∝N and σ ∝N, we have concentration of the energy! 130 40.2 Another look at the harmonic oscillators Remember that for a similar system we examined (the quantum harmonic oscillator), we have that Z1 = ∞ X n=0 exp(−βℏω(n + 1 2)), and then again by independence, Z = ZN 1 = ⇒U = −∂(ln Z) ∂β = Nℏω  1 eβℏω −1 + 1 2  . A good question here: what do we know about temperature in the limits? As T →0, β →∞, and that means U = Nℏω 2 . This is the ground state: all harmonic oscillators are in their ground state of ℏω 2 . Meanwhile, when T ≫ℏω kB , so the temperature is large enough for βℏω to go to 0, U becomes large. We can define βℏω = x: now doing a Laurent expansion because our quantity blows up at 0, 1 ex −1 = 1 x + x2 2 + x3 6 + · · · = 1 x · 1 1 + x 2 + x2 2 + · · · . Now we can neglect higher order terms and this simplifies to 1 x  1 −x 2 + O(x2)  = 1 x −1 2 + O(x). Plugging this back in, as T →∞, we have U →Nℏω  1 βℏω −1 2 + 1 2  = NkBT, and this is the famous result from the equipartition theorem! (Harmonic oscillators contribute a potential and kinetic quadratic term to the Hamiltonian.) So now, how can we determine the probability distribution for a single oscillator? We have p(E) = 1 Z1 e−E/(kBT ) (as a property of the canonical ensemble in general), and plugging in our specific value of Z1, p(E) = (1 −e−βℏω)eβℏω/2e−β n+ 1 2  ℏω, and this is actually a probability distribution over our ns: p(E) = p(n) = (1 −e−βℏω)e−nβℏω. This is actually a geometric distribution: if we take a = e−βℏω, the probability p(n) = (1 −a)an. This is indeed normalized: P n p(n) = 1, and the average value of n, called the average occupation number of a harmonic oscillator, is ⟨n⟩= 1 eℏω/(kBT ) −1. 131 Finally, let’s compute the heat capacity: C −∂U ∂T = Nℏω ∂ ∂T  1 eℏω/(kBT ) −1 + 1 2  which simplifies to = NkB  ℏω kBT 2 eℏω/(kBT ) (eℏω/(kBT ) −1)2 . Again, let’s look at the limits: when T →∞, all ℏω kBT terms go to 0 (define that fraction to be y). Specifically, (ey −1)2 ∼(y + O(y 2))2 ∼y 2, ey ∼1 + · · · ∼1, and thus as T →∞, we have C →NkB. This is a familiar result as well - at high temperatures, we’re expecting U = NkBT, so it makes sense for C = NkB. Remember also that because we have a gap between the ground state and next lowest energy level, we have gapped behavior where C becomes exponentially small at low temperature T. The distinguishing factor here from a two-level system is that there are always higher and higher energy levels! 40.3 Deriving the ideal gas law again Let’s try to do this derivation without needing to count states! We have our N indistinguishable particles, so we found last time that the partition function (by overcounting arguments) is Z = 1 N!(Z1)N. To calculate Z1, remember that we need to do our density of states argument: by the semi-classical density of states argument, Z1 = X j eEj/(kBT ) → Z d3xd3p (2πℏ)3 e−p2/(2mkBT ). Evaluating this integral, integrating out the d3x gives us a volume, and writing d3p = 4πp2dp by spherical symmetry, Z1 = V 4π (2πℏ)3 Z ∞ 0 dpp2e−p2/(2mkBT ). Using the change of variables y 2 = p2 2mkBT , the integral now becomes Z1 = V 4π 2mkBT 4π2ℏ2 3/2 Z ∞ 0 dyy 2e−y 2 and the known integral has value √π 4 . This gives us a final partition function for one particle of Z1 = mkBT 2π2ℏ2 3/2 V. Notice that the units of the partition function are dimensionless! That means that the mkBT 2πℏ2 term should have units of 1 length2 , and this helps us define a new length scale λD = s 2πℏ2 mkBT = 2πℏ √2πmkBT = h √2mEthermal . 132 This is called the thermal de Broglie wavelength: how do we compare it to others? Thinking of our ideal gas as noninteracting point particles, we have an average inter-particle distance V N 1/3 . The main point is this: if we have V N 1/3 ≫λD, we have a classical system, and we can use Maxwell-Boltzmann statistics to evaluate our system. But if V N 1/3 ≪λD (for example, when we start lowering our temperature), quantum mechanical effects are dominant, and we describe the system either with Bose-Einstein statistics or Fermi-Dirac statistics, based on whether we have distinguishability. Example 216 Consider an electron at room temperature. Then we have λD = h √2πmekBT ≈4.5nm, which is pretty small. This means that in our ideal gas situation, we should be using Maxwell-Boltzmann statistics! So now we have Z1 = V λ3 D = ⇒Z = ZN 1 N! , and now let’s try to calculate our thermodynamic quantities. By Stirling’s approximation, U = −∂ ∂β ln Z = −∂ ∂β (N ln Z1 −N ln N + N), and if we work this out, we’ll derive the well-known U = 3 2 N β = 3 2NkBT. Calculating the entropy for this system, S = U T + kB ln Z = 3 2NkB + kB ln ZN 1 N!  . If we use Stirling’s approximation again, we’ll find that S = NkB 3 2 + ln Z1 −ln N + 1  , and we can substitute in to find S = NkB 3 2 ln mkBT 2πℏ2  + ln V N + 5 2  which is the familiar Sackur-Tetrode equation! 40.4 The Maxwell-Boltzmann equation Finally, we want to look at the probability distribution for the kinetic energy of a specific molecule in an ideal gas. Using the semi-classical limit argument again, now that we know the partition function Z1, p(E) = 1 Z1 Z d3xd3p (2πℏ)3 e−p2/(2mkBT )δ  E −p2 2m  , 133 since we want to pick out those energies that are equal to p2 2m specifically. Making the same simplifications and plugging in Z1, =  2πℏ2 mkBT 3/2 1 V 1 (2πℏ)3 V · 4π Z ∞ 0 dpp2e−p2/(2mkBT )δ  E −p2 2m  =  2π mkBT 3/2 1 2π2 2mEe−E/kBT  m |p|  . Now since we can write p = p m 2E in terms of kinetic energy, this all simplifies to p(E) = 2π  1 πkBT 3/2 √ Ee−E/(kBT ) . The √ E factor essentially tells us about the momentum: the density of states is proportional to √ E. This is known as the Maxwell-Boltzmann distribution for the kinetic energy of a molecule in an ideal gas situation, and it works whenever we have V N 1/3 ≫λD. We can find the average energy here: ⟨E⟩= Z Ep(E)dE = 3 2kBT = ⟨1 2mv 2⟩, which gives us the root-mean-square of the velocity: vrms = p ⟨v 2⟩= r 3kBT m . This also gives us the velocity distribution p(v) = Z ∞ 0 dEp(E)δ v − r 2E m ! and through some routine algebra, we get the Maxwell-Boltzmann velocity p(v) = 4π  m 2πkBT 3/2 v 2e−mv 2/(2kBT ) . Next time, we’ll start talking about black-body radiation. Look at the problem set to examine some of these systems more carefully! 41 May 1, 2019 (Recitation) 41.1 General comments Let’s start with some general remarks about the partition function. We’re in a situation where we know about states, entropy, and the general behavior of nature: now, we’ll try to tie that in to the canonical ensemble using our function Z = X i e−βEi. 134 This was derived in class: in particular, the probability of a state i is e−βEi Z , and then the average value of E is ⟨E⟩= X i Ei e−βEi Z which is connected to a β-derivative. We also know that S = kB P pi ln pi can be written in terms of the derivatives, as can free energy and all our other variables! So it basically has all the knowledge we need in our system. So in relation to the problem set, we always “sum over states,” even if some of them may look different from others! What makes the partition function easy? The idea is that having N indistinguishable particles gives a partition function Z = 1 N!ZN 1 = ⇒ln Z = N ln Z1 −ln N!, and it’s often much easier to calculate Z1 for one particle. This wasn’t true in the microcanonical ensemble: since we had a fixed total U, we didn’t have independence between the energy states of our particles there, so our calculations were more complicated! We don’t have to make the ugly approximation for the 3N-dimensional sphere’s surface area, as we did when deriving Sackur-Tetrode with the microcanonical ensemble. 41.2 Particles in a potential There’s a lot of different ways we can work with a particle with a potential and kinetic energy. The potential energy can be constant (a box), linear (for example, due to gravity on Earth), or quadratic (a harmonic oscillator) in terms of x, and the kinetic energy can be quadratic or linear in terms of p. Finally, we can pick how many degrees of freedom d we have for our particle. The idea is that many problems are just some combinations of these parameters! We’re going to generalize so that we can see the big picture. First of all, if our energy H(⃗ x, ⃗ p) can be written in terms of these phase variables, our equation for the partition function Z1 (for one particle) is just (semi-classically) Z = Z d3xd3p h3 e−βH(⃗ p,⃗ x). The first term’s 3 can be replaced with the number of degrees of freedom we have. So now we just insert our expressions for kinetic and potential energy into H, but let’s not integrate yet. What are we trying to work with? For example, if we’re trying to find the internal energy, we don’t need Z: instead, we want U = −∂ ∂β (ln Z). So all the normalization factors in Z can be ignored if all we want is U: all of the Gaussian integrals give multiplicative factors, and ln Z turns those into constants, which become zero under β. Given this, let’s rederive the relevant part of this calculation for the ideal gas. The d3x integration gives a volume, and we don’t care about the h3 either, so we’re left with Z1 ∝ Z ∞ −∞ d3pe−β(p2 x +p2 y +p2 z )/(2m). This is essentially just three independent integrals! To work with this, let’s define a new variable ξ2 = βp2 x 2m : then dξ = βdpx · C - here we only care about the β-dependence, since we take the derivative with respect to β later on. 135 This gives us a scale factor of β−1/2 in each direction! So we’re left with Z1 =  β−1/2 Z dξe−ξ23 = β−3/2 · C, where C has the volume, the h3 term, and so on - those aren’t relevant right now! Now ln Z is some constant plus −3 2 ln β, and now U = −∂ ∂β  −3 2 ln β  = 3 2β = 3 2kBT, as we expect! (This also gives us things like CV = 3 2kB for free.) Fact 217 If we need something like the entropy, then we do need the constants we’ve tossed out along the way, but the internal energy doesn’t depend on them! So we’ve now looked at one example of a potential: let’s now think about the relativistic gas, where the kinetic energy is c|p|. Most of our derivation stays the same, except that now Z1 = Z ∞ −∞ d3pe−βc|p|/2m. Defining ⃗ ξ = −βc⃗ p, this converts us to an integral proportional to β−3 Z d3ξe−|ξ| (with the point being that we want to convert to generic, normalized integrals which just give us fixed constants!). Then almost everything stays the same: we just have U = 3kBT and CV = 3kB now. In both of these cases, we’ve found that our partition function Z1 = (c)β−x, where x = d 2 is half the number of degrees of freedom for the quadratic kinetic energy and x = d is the number of degrees of freedom for the linear kinetic energy. But we’ve only been dealing with cases where our potential is constant: what if we have a linear or quadratic potential? Well, the situation is exactly symmetric! We can make exactly the same arguments and substitutions, so in general, we’ll actually have Z1 = (c)β−x+y, where y = 1 2d if we have a harmonic oscillator (quadratic) potential and d if we have a linear potential! Fact 218 The whole point is that this is why statistics is useful: we can just look at how everything scales with β. This gives us (at least in the semi-classical limit) the internal energy and many other quantities. So now if we want the specific heat for a relativistic gas in two dimensions which lives in a potential V (⃗ x) = α|x|10, 136 we know that the relativistic gas integration gives us (since d = 2), x + y = 1 · 2 + 1 10 · 2 = 11 5 . So the partition function is just going to be proportional to β−11/5. That gives us a specific heat of 11 5 kBT with very few calculations! Fact 219 This is a generalization of the equipartition theorem, which tells us that each quadratic term gives a β−1/2 in the partition theorem, which gives a specific heat of 1 2kB. 42 May 2, 2019 In terms of exam regrades, everything will be updated later today. The problem set is due tomorrow! We’re going to discuss black-body radiation today: the main idea is that being able to absorb light at all frequencies leads to interesting behavior in terms of the spectrum of colors. In particular, we’ll start understanding ideas like why we humans can see in the visible light range! (In particular, things would be very different if the sun was at a different temperature.) Looking ahead, we’ll look at photons as particles of a gas, and we’ll consider why the classical model of this gas isn’t good enough to understand that spectrum, particularly in the UV range! This was part of the birth of quantum mechanics, and it’ll let us actually plot the energy density with respect to our frequency ω correctly. 42.1 Black-body radiation As a reminder, quantum mechanical particles can be described by a wavefunction, and the energy eigenstates (for a particle in a box) are given by ψk = 1 √ V ei⃗ k⃗ x. In three dimensions, we have ⃗ k = (k1, k2, k3), where each ki = 2πni L is in terms of the mode numbers ni ∈Z. It turns out this can also describe the quanta of the electromagnetic field! Each mode can be thought of associated with a quantum harmonic oscillator, and we need all plane waves to travel at the speed of light, so we have ωk = c|k|. (Looking at the polarization, there are two modes per wavevector ⃗ k.) So we’ll treat our black-body system as a gas of photons, and let’s see what happens as our temperature T increases! We’ll start by essentially counting states: how many states are there available to a single photon whose energy is between E and E + dE? We’ll calculate this in terms of frequency: dN dω dω = (4πk2)  dk dω dω   2V (2π)3  , where k2 = ω2 c2 and dk dω = 1 c . What do each of these terms mean? The first term 4πk2 is the surface area of a sphere of radius k, the second term is the thickness of a spherical shell for dω, and the last term is the density of a single particle plane-wave eigenstate wavevector (where the factor of 2 comes from the fact that there are 2 directions of polarization). So now simplifying, dN dω dω = V ω2 π2c2 dω, 137 We want to calculate the partition function, so let’s look at a particular frequency ω for our photons and then integrate later. This is essentially the partition function for a harmonic oscillator, but we’ll disregard the zero point energy: it’s a constant, and in practice, we only care about energy differences (unless we’re talking about dark energy and so on). This means that our partition function simplifies to Zω = 1 + e−βℏω + e−2βℏω + · · · = 1 1 −e−βℏω . We want to sum this over all possible frequencies - theoretically, we don’t have any upper limit (though in solids, we do have some physical limit due to the properties). So now we can integrate out ln Z = Z ∞ 0 dω dN dω ln Zω, which simplifies to ln Z = −V π2c3 Z ∞ 0 dωω2 ln(1 −e−βℏω). We’ll leave the partition function like this, but we can already calculate some quantities. First of all, the average internal energy can be found via ⟨E⟩= U = −∂ ∂β ln Z = V ℏ2 π2c3 Z ∞ 0 ω3 eβℏω −1dω, (since we can swap the integral and derivative). In differential form, this says that E(ω)dω = V ℏ2 π2c3 ω3 eβℏω −1dω, which is the amount of energy carried by photons with frequency between ω and ω+dω. This is actually Planck’s distribution! If we plot the integrand with respect to ω, this has a single peak: at low frequency, the distribution is consistent with the equipartition theorem, but then the energy gaps at higher frequencies bring I back to 0 instead of going offto infinity. Fact 220 Calculating dE(ω) dω gives Wein’s radiation law, which is a good exercise. By the way, when the temperature is around 6000 Kelvin, which is the temperature of the Sun, the peak is mostly in the visible light range, which makes sense! Fact 221 This is known as a “radiation gas,” which has different properties from the normal gasses that we’ve been dealing with. 42.2 Doing some derivations Looking again at our internal energy integral, we have U = Z ∞ 0 dω  ω2V π2c3  ℏω  1 eβℏω −1  . Here, the first term  ω2V π2c3  is the density of modes per frequency. The ℏω is a kind of energy unit excitation, and combined with the last term, this gives the mean thermal excitation energy per mode. Substituting in x = βℏω, we 138 actually want to calculate U = V π2c3 kBT ℏ3 4 Z ∞ 0 x3dx ex −1. That last integral is actually known because there is no upper limit on the frequency - otherwise, it’s only possible to numerically approximate it! It turns out that this is = Γ(4)ζ(4) = 6 · π4 90 = π4 15, and therefore we can think about the energy density of our gas of photons U V = ξ = π2k4 B 15ℏ3c3 T 4. In other words, we can just write down this law as U V ∝T 4 . Example 222 So we can imagine having a box of photons with a hole: how can we find the energy flux? Flux is the rate of energy transfer, and it will just be ξc 4 , where c is the speed of light: the 1 4 factor comes from us not having a point source, and where the size of the hole is larger than the wavelength of the photons. This can then be written as σT 4 , and this proportionality is called the Stefan-Boltzmann law. Here, σ = π2k4 B 60ℏ3c2 . Example 223 Let’s see if we can find the pressure of this gas. (It will be interesting, because the pressure only depends on the temperature!) First of all, let’s calculate the Helmholtz free energy F = −kBT ln Z = V kBT π2c3 Z ∞ 0 dωω2 ln(1 −e−βℏω). Again letting x = βℏω and doing similar simplifications, F = V (kBT)4 πc3ℏ3 Z ∞ 0 dxx2 ln(1 −e−x), and now by integration by parts, the integral simplifies to Z ∞ 0 dx 1 3 d dx x3  ln(1 −e−x) = 1 3x3 ln(1 −e−x ∞ 0 −1 3 Z ∞ 0 dx x3 ex −1. We’ve seen the second term before: it evaluates to π4 45, and therefore F = −π2(kBT)4V 45ℏ3c3 = −1 3U. 139 Now we can find our pressure: P = −∂F ∂V T = −U 3V = 4σ 3c T 4. This equation of state tells us that interestingly, this gas’s pressure only depends on the temperature! In particular, the pressure P is 1 3 of the energy density, and this is an interesting fact in cosmology. 42.3 Continuing on Let’s try calculating some other thermodynamic quantities: the entropy of our gas is S = −∂F ∂T V = 16V σ 3c T 3, the specific heat CV = ∂U ∂T V = 16V σ c T 3, and to find the number of photons, dN dωdV = ω2 π2c3 1 eβℏω −1, and integrating out V and ω, we find that N = V π2c3 Z ∞ 0 dω ω2 eβℏω −1 ≈1.48σT 3V kBc . We can also now calculate the fluctuations in energy: normalizing the standard deviation with respect to U, σ(E) U = kBT 4 c σT 4V s 16σT 3V kBc = r kBc 6T 3V = r 1.48 N . These fluctuations are very small as N grows large! So we can basically treat this from the microcanonical point of view, and the thermodynamic behavior would look basically the same. Question 224. So what does all of this mean? First of all, this shows that classical mechanics isn’t enough to describe our systems. If we take ℏ→0 - that is, we don’t assume that we have quantization - then d2U dωdV = ℏω3 π2c3 1 eℏωβ−1 → ω2 π2c3 kBT as ℏ→0. This quadratic growth is consistent for ℏω ≪kBT with the equipartition theorem, but then quantum mechanical effects give an exponential decay eventually, which classical mechanics can’t predict. Also, this gas is very interesting, because local properties only depend on temperature. U V , N V , S V , and P are all dependent on T and not V ! Finally, this was a gas of photons, but we can also think about vibrations of solids in terms of phonons instead. The same derivations can be made, but we just have some different quantities: c is now the speed of sound (since we care about sound modes), and also, we have a bounded integral instead of integrating from 0 to ∞. Basically, the wavelength can’t be smaller than the inter-atom distance! That makes it interesting to think about how CV of a solid changes as we decrease the temperature - this will lead us to Einstein’s law. An exercise: what is the chemical potential µ for this radiation gas? It will turn out to be zero, and we’ll see a connection when we discuss bosons later on! 140 43 May 6, 2019 (Recitation) 43.1 Homework questions In one of the problems of the problem set, we have particles oriented with a dipole moment ⃗ mu which gives us an energy −⃗ E · ⃗ µ = −|E||µ| cos θ. We want to find the partition function, which is an integral over states: = Z ”dxdp” h e−βH, where H, the Hamiltonian, is the kinetic plus potential energy that we have if we use the semi-classical approximation. The main differences here are that we have two-dimensional space (and therefore two-dimensional momentum space as well), so we really want Z1 = Z d2xd2p h2 e−βH. What’s special is that we are integrating over angles θ and π, and we also have the canonical momenta pθ and pφ as a result. But now we can integrate our cos θ out through our polar coordinates, and everything works out! (We will get a cos θ in the exponent.) In a different problem, when we give a gravitational potential energy −mgh to each particle, that just adds a e−βmgh to each term. This is generally solved more easily with the grand canonical ensemble (where we are actually allowed to exchange particles), but if we are to solve the problem in the way that we know, the e−βmgh factor comes out of the z-integral. Now we just get an additive constant in F (because we take ln Z) and the rest of the problem looks like an ideal gas! Finally, considering the polymer chain, the main concept is to treat each individual monomer separately! Since we have a classical system, we can just compute Z1 for one monomer and take the Nth power. We basically have a bunch of two-level systems, for which the partition function is easy to calculate! Now all the degeneracies of the form N n  are accounted for in the binomial expansion of the two-level system 1 + e−β∆, raised to the Nth power. We don’t even need to do the combinatorics from the microcanonical ensemble! 43.2 Clicker questions Question 225. Canonical versus microcanonical ensembles: do both ensembles have the same properties? The answer is yes! We’ll unfold more details about this soon. Question 226. Do both ensembles have the same number of microstates? No! In the microcanonical ensemble, we’re given a total energy E for which all our states must reside, but in the canonical ensemble, we’re given more freedom. Here’s a way to think about it: let’s say we divide a system into two halves with N 2 particles. Then the total number of microstates is Γtotal = ΓLeftΓRight, but this isn’t quite correct in general because we might not have exactly N 2 on each side! (Think of “fixing the number of particles on the left side” as “fixing the energy of our system” as in the microcanonical ensemble.) It turns out in 141 general that the right order of magnitude is to deal with the Gaussian fluctuation Γ2V = (ΓV )2√ N. In this case, it is easy to count microstates, but it may be more difficult in other versions! Well, the reason we don’t worry about it too much is because taking log often makes the √ N negligible. We really only care about those microstates with non-negligible probability, so the Binomial distribution with mean N 2 and standard deviation on the order of √ N can be viewed as basically Gaussian, and we only care about those values within a few standard deviations. So what is the number of microstates in a microcanonical distribution? We often have a density of states ∂N(E) ∂E , but that’s a classical description and therefore can’t be used directly for counting. So we need to have some energy width δE, and we often didn’t specify what that was! Fact 227 Usually we don’t need to because the δE cancels out later, or because it only adds a negligible logarithmic term. There was one example where it didn’t need to be specified: given a harmonic oscillator, there are only specific states at width ℏω apart, so perfect, identical harmonic oscillators do actually have some “exact number of states.” So in our equation Γ(E) = ∂N(E) ∂E δE, we can think of “δE = ℏω′′ as an effective width between our energies. Question 228. So what’s the energy gap that we’ve assumed in the canonical ensemble? We have a Boltzmann factor e−βE that exponentially decreases with E, and we have a density of states that increases very rapidly. Putting these together, we get a sharp peak at the average energy ⟨E⟩, which is why we say that the microcanonical and canonical ensembles are equivalent for lots of purposes. So how broad is the distribution? Well, we calculated this in class: we can find ⟨E⟩and ⟨E2⟩, which allow us to find σE = kBT r C kB where C is the specific heat. Thinking about this for an ideal gas: C ∝NkB, so this is proportional to √ N. This is the same factor that’s been popping up in all the other situations, and that’s because that’s the right factor for fluctuations! Now it’s the same idea: √ N is small compared to N, so it can be neglected. 44 May 7, 2019 Some preliminary announcements: homework will be graded soon, and we should check the website to make sure everything is accurate. The last graded problem set is due on Friday! The next one (on photons, phonons, gases) is not graded but is good practice for the final. Finally, some past finals will be posted soon - there will be more problems on the final because it is longer (3 hours), but some fraction will be taken from past exams. 44.1 Overview Last time, we went over black-body radiation by thinking of photons as quanta of the electromagnetic field. We find that raising the temperature of a black-body gives a spectrum of radiation, and we can figure out the thermodynamic quantities of this body using the canonical ensemble! 142 We’re essentially going to go over the physics of solids today, examining quantities like the heat capacity with different models. We’ll also consider what happens when we look at the extremes of temperature: for example, what if the de Broglie wavelength is comparable in size to the inter-atom distance? That’s where the grand canonical ensemble comes in! 44.2 Phonons Our goal, ultimately, is to consider the heat capacity C of a solid. A phonon is essentially a quanta of vibrational mechanical energy! Fact 229 We can discretize sound waves in solids using these quasi-particles. The energy of a phonon is E = ℏω = ℏkcs, where cs is the speed of sound. We’ll be working with a perfect crystal of atoms as our system for convenience. The first thing we want to do is to consider the density of states of our phonons dN dω dω = 3V 2π2c2 s ω2dω. where the 3 in the numerator comes from having a multiplicity of polarization. Fact 230 The factor of 3 is really from having one longitudinal mode (compression) and two transverse modes (shear). Since we have N atoms in our crystal, this is 3N different normal modes, meaning that there are 3N different types of phonons with frequency ω1, · · · , ω3N. As a result of this, we have another difference between phonons and photons: the frequency spectrum for light waves is unbounded (it can go arbitrarily high), but the sound waves have a minimum wavelength λ = 2πcs ω , where λ is the spacing between atoms (since something needs to be able to propagate the wave). This means we have a maximum frequency corresponding to our λD ∼ 3 q V N , meaning that our maximum frequency ωD ∼ N V 1/3 cs. To find the proportionality constant in front, note that we can count single-phonon states as Z ωD 0 dω dN dω = V ω3 D 2π2c2 s (by direct integration). One way to deal with this (with solid-state physics) is to deal with a “primitive cell,” but instead we’ll argue that this number is just 3N, the number of degrees of freedom. Thus we can find our maximum allowed 143 frequency: 3N = V ω3 D 2π2c3 s = ⇒ ωd = 6π2N V 1/3 cs . We’ll associate ωD with a new temperature scale now: define the Debye temperature TD = ℏω kB . 44.3 Using the canonical ensemble We’ll now calculate the partition function for our system: for a fixed frequency, Zω = 1 + e−βℏω + e−2βℏω + · · · = 1 1 −e−βℏω . Assuming the frequencies are independent of each other, the partition function is just the product over all ω Z = Y ω Zω, and now taking the log, we can approximate the sum as an integral Z = Z ωD 0 dω dN dω ln Zω. (Note that we now have an upper frequency ωD instead of ∞.) Now the energy of our system is E = Z ∞ 0 dω dN dω ℏω eβℏω−1 , which simplifies to = 3V ℏ 2πc3 s Z ωD 0 dω ω3 eβℏω −1. There isn’t an analytic expression for this anymore, so we’ll instead just look at the low- and high-temperature limits. Letting x = βℏω, we can rewrite our energy as E = 3V 2π2(ℏcs)3 (kBT)4 Z TD/T 0 dx x3 ex −1. Example 231 One extreme is where T ≪TD, so our upper limit goes to ∞. Then we’ve seen our integral before: it evaluates to π4 15. So in this case, we can calculate CV = ∂E ∂T = 2π2V k4 B 5ℏ3c3 s T 3 = NkB 12π4 5  T TD 3 . In other words, at low temperatures, our heat capacity is proportional to T 3. This does line up with experimental observations! In particular, CV →0 as T →0., which is consistent with our third law of thermodynamics. 144 Fact 232 Einstein has a different description of this behavior with a different kind of calculation - Debye just extended that model by allowing many different frequencies. Example 233 On the other hand, can we recover the results we expect at high (classical) temperatures T ≫TD? Then we integrate over very small values of x, so we can do a Taylor expansion. In these cases, Z TD/T 0 dx x3 ex −1 = Z TD/T 0 dx(x2 + O(x3)) = 1 3 TD T 3 + · · · , and now we find that our heat capacity reaches a constant CV = V k4 BT 3 D 2π2ℏ3c3 s = 3NkB, which is consistent with the Dulong-Petit law! The data doesn’t perfectly fit with this, and that’s because we have a flaw in the calculation: our dispersion relation ω = kcs isn’t quite accurate, but that can be left for a physics of solids class. 44.4 Back to the monatomic ideal gas Remember that when we talked about the ideal gas for a canonical ensemble, we treated the particles quantum mechanically in a box: we put the system at fixed temperature, and we used the partition function to pull out the thermodyanmics of our system. There, we introduced the de Broglie wavelength (or length scale), and we mentioned that we should relate that to the inter-particle distance to see whether or not quantum effects are important. Well, when we’re close to the de Broglie wavelength, we can’t use the Z = 1 N!ZN 1 formula for our partition function anymore: we need a better way to describe our system. That’s where the grand canonical ensemble comes in: we relax our constraint on having a fixed number of particles, but we fix the chemical potential. Remember that this length scale λ = s 2πℏ2 mkBT increases as our temperature decreases. Eventually, this becomes comparable to our interparticle spacing ∝ V N 1/3, which is where quantum effects start to have an effect. Remember that in a microcanonical ensemble, we fix V, U, N, and in a canonical ensemble, we fix V, T, N. In both of these cases, we can calculate the energy, and the macroscopic properties are basically the same here because the fluctuations are so small. So in a grand canonical ensemble, we fix V, T, µ instead: it’s a pressure-like term that adds particles. Specifically, our internal energy in differential form can be written as dU = TdS −PdV + µdN = ⇒ ∂U ∂N S,V = µ. It’s usually difficult to keep S and V constant, so we instead work with the Gibbs free energy: there, we can instead 145 calculate ∂G ∂N T,P = µ, which is generally easier to work with experimentally! We can show that in a system at thermal equilibrium, ∂S ∂N U,V = −µ T , so two systems brought together at thermal equilibrium will also have the same chemical potential. Proposition 234 This means we can think of our system as being connected to a large reservoir, so that both are at some fixed T, µ. Now the total system and reservoir are a microcanonical ensemble! So we want the probability distribution of a given state sj for our system: then p(Sj) = ΓR ΓR⊕S , and similar to our canonical ensemble derivation, we can write the Γs in terms of our entropy: ΓR(sj) = exp(SR(Sj)/kB) = ⇒pj(sj) = 1 Z exp [SR(U −Ej, M −Nj)/kB] where Z is our normalization factor, analogous to the partition function. Now ∂pj ∂Ej = −1 kB ∂SR ∂U U−Ej pj = −1 kBT pj, and ∂pj ∂N = −1 kB ∂S ∂M M−Nj pj = µ kBT pj. Thus, we can write our probability of any state pj(Sj) = 1 Z exp  −Ej kBT + µNj kBT  = 1 Z exp [(µNj −Ej)/(kBT)] . We can now write our expression for our grand partition function Z = X j exp [(µNj −Ej)/(kBT)] . 44.5 Thermodynamics of the grand canonical ensemble How can we do physics with this new quantity? Let’s look at a similar quantity as in the canonical ensemble ∂ ∂β ln Z = 1 Z X j (µNj −Ej)e(µNj−Ej)β, which can be rewritten as = X j pj(sj)(µNj −Ej). Thus, we actually have the average value of µNj −Ej, which is ∂ ∂β ln Z = µ⟨N⟩−⟨E⟩= µN −U. 146 We can also consider ∂ ∂µ ln Z = 1 Z X j βNje(µNj−Ej)β; again, we can extract out our probability term to get = β⟨N⟩= βN. This means we can find the expected number of particles N, and in the thermodynamic limit, fluctuations are much smaller (in order of magnitude) than N, so this is pretty accurate almost all the time! Is it possible for us to find the entropy of our system? We know that S = −kB X pj ln pj = −kB X j eβ(µNj−Ej) Z · (β(µNj −Ej) −ln Z) , which can be arranged as S = −1 T (µN −U) + kB ln Z = ⇒ U −TS −µN = kBT ln Z . So the central theme is that ln Z basically gives us an energy term! This lets us create a new free energy Ω= U −TS −µN = −kBT ln Z, which is the grand potential: this is essentially a sum over all the different states! In differential form, the grand potential can be written out as dΩ= dU −TdS −SdT −µdN −Ndµ, and writing out dU = TdS −PdV + µdN, this simplifies as dΩ= −SdT −PdV −Ndµ. This means we can take derivatives again: we can find our entropy, pressure, and number of particles via ∂Ω ∂T V,µ = −S, ∂Ω ∂V T,µ = −P, ∂Ω ∂µ T,V = −N. Next time, we’ll look at bosons and fermions, and we’ll try to recover the Boltzmann statistics at high temperatures! 45 May 8, 2019 (Recitation) 45.1 More on density of states Let’s start by summarizing some concepts from last recitation. We have a density of states dN dE (which usually goes as ENA if we have a mole of atoms), which is very steep. On the other hand, we have the Boltzmann factor e−E/(kBT ), which usually goes as e−N. Multiplying together these rapidly growing and decaying distributions, we get a very sharp peak! Almost all of the action occurs around that sharp peak, because the probability of being found far away from ⟨E⟩ is very small. More quantitatively, note that the distribution is a delta function at ⟨E⟩for a microcanonical ensemble, since we fix the energy. Really, the only way to do this is to pick a bunch of harmonic oscillators with quantized energy 147 states, but there are still some perturbations there - in other words, our discrete energy states are broadened a bit. Fact 235 To account for this, we can think of the microcanonical ensemble as having an energy width of ℏω if our energy states of the oscillators differ by ℏω. So when we derived the Sackur-Tetrode equation, remember that we used quantized box numbers ni such that the sum of the squares of those quantities is fixed: this ended up being the surface area of a sphere, which helped us find the multiplicity of a given energy. Specifically, if we say that X i n2 xi + n2 yi + n2 zi is constant, we can integrate to find the number of states: since we have a microcanonical ensemble, we pick out a specific energy U to find that Γ ∝ Z d3Nni · δ X n2 i −U  . But there’s one thing we did not do: we never specified that we use a spherical shell with some energy width δE. Surface areas are not volumes, so we have a density of states instead of a number of states - how can we introduce an energy uncertainty? Well, remember that we replaced r 3N−1 with r 3N due to N ≫1: this means that we’re essentially allowing for the whole volume instead of the surface area, so our energy width is actually from 0 to our energy E. This also tells us that in an N-dimensional sphere, almost all of the volume is concentrated towards the surface area: in fact, half the volume is concentrated within an O r N  distance from the surface area! That’s why it’s okay for us to also count the volume of states inside: it’s negligible compared to that part near the surface. Fact 236 So going back to the canonical ensemble, remember that we can find the uncertainty σE = kBT r CV kB ∝ √ N ∝ E √ N . Thus, the width of the energy peak is proportional to √ N, which is precise (proportionally) up to experimental accuracy! But when we made the approximation from r 3N−1 to r 3N, we actually get a fluctuation on the order of E N . So it’s often okay to make what looks like a crude approximation! 45.2 Quick note about the different distributions Fact 237 As a result, we almost never use the microcanonical ensemble. An awkward situation is that if we divide a system in two, a microcanonical ensemble doesn’t stay microcanonical (because we can exchange energy between the two halves)! On the other hand, the partition function is simple: we can just multiply Z = Z1 · Z2, and that makes many calculations much easier. 148 On the other hand, allowing particle exchange means we have to use the grand canonical distribution. This helps us with quantum mechanical descriptions of particles - in particular, because of quantum indistinguishability, we no longer have independent particles, but we do have independent states, and we can take our partition function and factor it Z = Y Zi over states! We’ll probably go over this a bit more in class soon, but it’s interesting that non-interacting particles still affect independence for each other. (For example, we have the Pauli exclusion principle for fermions!) We’re going to find that some derivations are easier for the grand canonical ensemble than in the microcanonical or canonical ensemble as well. Le’ L 45.3 One last conceptual question We know that we have a Boltzmann factor e−E/(kBT ) in our probability distribution, so any energy state, no matter how huge, has some positive probability of happening. Is that an artifact of the approximations we have made, or is it physical? One assumption we make is that our reservoir (that our system is connected with) is large enough to sustain the fluctuations of energy! Since we treat the reservoir and system as a microcanonical ensemble with some total energy U, we can’t actually have energies E > U. But other than that, this is indeed how nature works! Let’s assume that there is indeed some point where the energy can’t get any larger. and then we have some small probability p of getting to an energy above that. Then the entropy and energy change are approximately ∆= −kBp ln p, ∆E = pE = ⇒∆F = ∆E −T∆S = p(E −TkB ln p) So transferring some probability to a higher state changes the free energy F: ∆F < 0 if p < e−E/(kBT ), and that means that it is always favorable to allow that energy state! This means the canonical ensemble probability distribution is indeed a stable equilibrium, and any collision between atoms (or other perturbation) of the system moves us towards that exact distribution. 46 May 9, 2019 46.1 Overview Today, we’re going to continue talking about the grand canonical ensemble, which is a useful framework for trying to understand quantum effects in different kinds of gases! As a quick reminder, this is a system where instead of fixing energy or temperature (as in the microcanonical and canonical ensembles) along with the number of particles, we have to make a more careful argument. At the quantum level, we can no longer assume that all of our particles are likely to be in different states, so we can’t have a simple overcounting term like 1 N!. In addition, we now allow our system to be open. The number of particles can now fluctuate, but we’ll find that in the thermodynamic limit again, the fluctuations are small. So this will also give similar results to the other formulations! We’ll see some systems where we can use this model today. 149 46.2 Types of quantum particles The main idea is that in relativistic quantum field theory, there are two types of particles. Definition 238 Bosons are particles that have an intrinsic spin of 0, 1, 2, · · · (in multiples of ℏω). The main concept here is that we have indistinguishable particles: the quantum states are symmetric until particle exchange, which can be written as ψ(⃗ r1, ⃗ r2) = ψ(⃗ r2, ⃗ r1). Examples of bosons include photons, gluons, Helium-4, as well as quasi-particles like phonons and plasmons. Definition 239 On the other hand, fermions are particles that have an intrinsic spin of 1 2, 3 2, · · · (in multiples of ℏω). Examples here include electrons in metals at low temperature, liquid Helium-3, white dwarfs, and neutron stars. This time, quantum states are antisymmetric: we have ψ(⃗ r1, ⃗ r2) = −ψ(⃗ r2, ⃗ r1). Fact 240 Bosons can have infinitely many particles per state, but by the Pauli exclusion principle, there can only be 0 or 1 fermion in each quantum state. We’re going to label our states by occupation numbers {nr}, often as |n1, n2, · · · , nr⟩. At high temperatures, they will be the same as in our Boltzmann statistics. Fact 241 By the way, these are three-dimensional particles: 2-D systems are completely different! Remember that we have our canonical ensemble partition function Z = X {nr } e−βnr εr , where we sum over all ways of partitioning N particles into sets {nr}, with the constraint P nr = N. But we have indistinguishable particles here, and it’s hard to account for the fact that we can only have 1 particle in each energy state for fermions. 46.3 Using the grand canonical ensemble So our first step is to look at an open system: let’s fix our chemical potential µ and let N fluctuate! (We’ll also put our system in contact with a heat bath at temperature T.) Then for a given state, we have the variables Nj = X nr, Ej = X r nrεr 150 which are allowed to vary, and now our grand partition function Z = X j exp [(µNj −Ej)β] = X {nr } exp " (µ X r nr − X r nrεr)β # . We can rewrite the sum of the exponents as a product: = X {nr } Y r exp [(µ −εr)nrβ] = Y r X {nr } exp [(µ −εr)nrβ] by swapping the sum and product. Now our path diverges for bosons and fermions: in one case, the nrs can be anything, and in the other, the nrs must be 0 or 1. Proposition 242 For fermions, since nrs are all 0 or 1, we find that ZF D = Y r (1 −e(µ−εr )β). This is referred to as Fermi-Dirac. Proposition 243 Meanwhile, for bosons, since nrs can b, anything, we have an infinite sum ZBE = Y r ∞ X n=0 e(mu−εr )β !n = Y r 1 1 −e(µ−εr )β . This is referred to as Bose-Einstein. In both cases, we often want to deal with the logarithm of the partition function: then the product becomes a sum, and we have the following expressions: ln ZF D = X r ln(1 + e(µ−εr )β), ln ZBE = X r −ln(1 −e(µ−εr )β), It’s important to note here that P r is the sum over a single state! So if we want the ensemble average occupation number, we take (as was derived previously) N = kBT ∂ ∂µ ln Z. For fermions, N = X r e(µ−εr )β 1 + e(µ−εr )β = X r 1 1 + e(εr −µ)β Each term here can be thought of as ⟨nr⟩, the average occupation number of state r! That means that for a given state, ⟨nr⟩= 1 e(εr −µ)β , 151 and this is known as the Fermi-Dirac occupation number. In particular, we have the Fermi function f (tr) = 1 1 + eβ(tr −µ) . Note that µ is allowed to be both positive and negative here. If we take low temperature T, β →∞, and then chemical potential µ essentially separates our filled and empty states! The probability of having a state εr > µ is almost zero, and the probability of having a state for εr < µ is almost 1. As we increase temperature T, the jump from probability 0 to 1 becomes less steep, and we’ll see how that works in a minute. Meanwhile, for bosons, N = X r − e(µ−εr )β 1 −e(µ−εr )β = X r 1 e(εr −µ)β −1. Analogously, this can be thought os as a sum P r⟨nr⟩, so we know that the average occupation ⟨nr⟩= 1 eβ(tr −µ)−1 . Notice that the main difference here from the Fermi-Dirac occupation number is the −1 instead of the +1! (By the way, remember that this is exactly the expression we saw with photons if we set µ = 0.) One important idea with this system, though, is that ⟨nr⟩≥0 for non-interacting particles. If µ →εr, the denominator goes to 0, and thus ⟨nr⟩goes to infinity. In fact, µ > εr means our occupation number becomes negative! This isn’t allowed, so that actually sets a bound on our allowed µ in relation to our energy states εr. Proposition 244 The allowed values of µ for a boson system are µ ≤εrmin, where εrmin is the lowest energy state for a single particle. By the way, Professor Ketterle may talk about Bose-Einstein condensates next Thursday! 46.4 Looking more carefully at the occupation numbers Let’s take ⟨nr⟩to high temperature in both cases, so εr −µ ≫kBT. Then we should expect that our results agree with the canonical ensemble answer of ⟨nr⟩= e−β(εr −µ) for classical, identical particles. Indeed, this is what happens in the limiting case for both FD and BE statistics, since the exponential terms dominate the 1 in the denominator! So our new model is really mostly useful at lower temperatures. This means we really care about having our inter-atom distance on the order of of our length scale meaning that V N ∼  h √2πmekBT 3 . 152 Example 245 Electrons at room temperature have a length scale of around 4.5 nanometers, and copper has a density of 9×103 kg/m3, which means there are 8.5 × 1028 atoms per cubic meter. So if we have one “conduction electron” per atom, the volume we have to work with is about 1 8.5 × 1028 m3 per atom. This is about (0.23nm)3, and that means the inter-atom distance is small enough for us to want to use Fermi-Dirac statistics! But looking at copper atoms instead of the electrons themselves, the relevant length scale is around 0.012 nm. In other words, copper is a bunch of non-degenerate atoms immersed in a degenerate electron gas. 46.5 Fermions at low temperature We should think of “degenerate” as having density large enough for quantum effects to be important. This is because two particles being too close gives overlapping wavefunctions. So for the remainder of this class, we’ll be looking at degenerate Fermi systems (where we use Fermi-Dirac statistics). Examples include electrons in a conduction band of metals, white dwarf stars, neutron stars, and heavy metals. Here our occupation numbers follow nj = 1 1 + e(εj−µ)β . Note that when µ ≫kBT, levels with ε < µ have nj →1, and levels with ε > µ have nj →0. We get complete degeneracy at T = 0, where we have essentially a sharp change from filled to empty states. Let’s think a little more about such a degenerate Fermi gas: now we essentially have nj = 1 1 + e(εj−µ)∞=    1 εj < µ 0 εj > µ . This is essentially a step function! All states with low energy are filled, and all others are empty. If we increase our temperature a little, the ∞in the exponent becomes a large number, and we get a little bit of wiggle room. Then in some interval ∆ε ∼kBT, we can have partially filled states. Fact 246 The filled levels are known as the “Fermi sea,” and the set of states with ε = µ is known as the “Fermi surface.” We can then define a “Fermi momentum” pF = p 2mεF where εF = µ. In three dimensions, we can calculate the density of states of our Fermi gas: dN = g d3xd3p (2πℏ)3 , where g is our degeneracy factor (2 for an electron). This can then be written as N = g Z d3xd3p (2πℏ)3 1 e(p2/2m−µ)β + 1, 153 and we can do similar tricks with the spherical integration as before: this evaluates to = 4πV g (2πℏ)3 Z ∞ 0 p2dp e(p2/2m−µ)β + 1. We’ll finish this and calculate some of the thermodynamic properties here next time! 47 May 13, 2019 (Recitation) Professor Ketterle showed us another Stirling engine today! Basically, we have a displacer which displaces gas between a hot reservoir (hot water) and cold reservoir (room temperature air). Whenever the piston moves down towards the hot reservoir, the air above gets colder, so there is a temperature differential. This causes a pressure modulation, which allows another piston to drive the wheel forward. 47.1 Entropic force Often, we think of energy as the cause of force. But in statistics, we can think of entropy causing some kind of force as well! Recall the differential formula dU = TdS −PdV ; we can use this to find the pressure at constant S or U: P = T ∂S ∂V U = −∂U ∂V S . On the other hand, we also know that we have the free energy F = U −TS = ⇒dF = −SdT −PdV, which leads us to P in terms of isothermal derivatives: P = −∂F ∂V T = ∂(U −TS) ∂V T = −∂U ∂V T + T ∂S ∂V T . But the pressure is always the same, regardless of what process we’re using to get to this point (because we have a state function)! So does pressure come from internal energy or from entropy? There’s actually contributions from both an internal energy change and an entropy change! Applying this to the ideal gas, because ∂U ∂V T = 0, we actually have P = T ∂S ∂V T , and pressure is determined by entropy alone! Fact 247 It’s important to think about isothermal versus adiabatic compression here: in the former case, compression leads to no change in internal energy (which is only dependent on T), so we get an output of heat. On the other hand, adiabatic compression leads to an increase in the kinetic energy of the particles, which means the temperature does increase. 154 47.2 Applying this to the chain model Let’s go back to the question where we have N chains of length ℓ, each of which is either flipped to the left or to the right in a one-dimensional system. Here, assume that the system doesn’t have any internal energy, so all states are equally probable. There are 2N possible configurations of our chain (since each chain can be flipped to the left or right independently), and this is essentially a random walk! We can describe such a system by a binomial distribution, which can be approximated as a Gaussian for large N: since the variance of each individual chain is ℓ2, p(x) = p0 exp  −1 2 x2 ℓ2N  . From this, we know that our multiplicity Γ(x) = Γ0 exp  −1 2 x2 ℓ2N  , so the entropy S = kB ln Γ = kB  c −1 2 x2 ℓ2N  : (since the Gaussian exponential nicely cancels out with the logarithm!) But now, we can calculate the force exerted by the chain: since P and V were conjugate variables in our boxed equation above, we can also replace them with F and x. Notably, if ∂U ∂x T = 0 here (notice that −PdV is work, and so is +Fdx, so we gain a negative sign), F = −T ∂S ∂x T = ⇒ F = kBT x ℓ2N . So force is proportional to temperature, and it’s also a Hooke’s law relation! This is a “pure entropy” situation, where we just needed to count the number of microstates to find dependence of S on the length x of the chain. In other words, we put energy into the chain when we stretch it, but there’s no way to store internal energy in this chain! So the energy of pulling will be transferred as heat, as that’s the only way we can move the chain to a lower entropy state by the second law. Question 248. What causes the restoration here? Let’s imagine that our chain is now vertical, and there is a mass hanging from the end. At a given temperature, this gives us the length x of the chain where we have equilibrium. But if we increase the temperature of our reservoir, what happens? Because our weight is constant, the force in the above equation is constant, so Tx is constant. In other words, when temperature goes up, the chain gets shorter! This is a lot like if we put a weight on a piston sitting on top of an ideal gas: increasing the temperature of the gas increases the pressure, which makes the piston move higher up. How exactly can such a massless chain with no internal energy even hold up a mass? Remember that the chain is always connected to a reservoir at temperature T! For entropic reasons, this nudges some of the chains upward. There’s always a process to transfer force and energy between any reservoirs that exist and the system at hand. 47.3 Blackbody radiation and the Debye model The former deals with a “gas” of photons, and the latter (which deals with specific heat of a solid) deals with a “gas” of phonons, the quantized excitations of sound. 155 In both cases, we can say that we have a bunch of energy states: then what’s the expectation value for the average occupation number of a harmonic oscillator? It’s given by the expression 1 eℏω/(kBT ) −1. If we want to know the total energy of a bunch of harmonic oscillator quanta like this, we multiply by the energy ℏω (for each harmonic oscillator), and then we need a density of states to know how many harmonic oscillators we have per interval. So U = Z dωℏω 1 eℏω/(kBT ) −1 dNω dω , and this is essentially just a dimensional analysis argument! Integrating this out from 0 frequency to some maximum ωmax, we have found our internal energy. But how are the two different in their descriptions? The main difference is that we have dispersion relations ω = csk and ω = ck, and in both cases, we find that dN dω is quadratic in ω2. The difference is just the factor of the speed of photons versus phonons! In addition, there is also a factor of 3 2 from the polarization factors. But most importantly, the maximum frequency is upper bounded for the Debye model because of the limitations of the solid, while the maximum frequency is unbounded for blackbody radiation. Each of these concepts is important to understand! 48 May 14, 2019 Today’s lecture is being given by Professor Ketterle. 40 minutes ago, Professor Ketterle sent a press release about the kilogram changing next Monday! It will no longer be defined by the Paris artifact but by Planck’s constant. We should celebrate this, because Planck’s constant is much more beautiful than a chunk of metal. But he wants to mention that explaining physics should be simple: it should not just be for “the physicists and the nerds.” Today’s class will go in three parts: an introduction with the essence of quantum statistics, using the Bose-Einstein distribution to describe phase transition, and how to create the first Bose-Einstein condensate in a gas. 48.1 Bose-Einstein condensate: an introduction Bose-Einstein condensate allows us to create a matter wave by having many particles in one quantum state. This is based on quantum statistics, and we can use situations from class to understand it! Example 249 Let’s say we have 3 particles with some total energy 3. This is a microcanonical ensemble: what’s the classical description of this? There’s 1 way to have all three particles in energy level 1, 3! = 6 ways to have 3 particles in energy levels 2, 1, 0, and 3 ways to have one in energy 3 and the others in energy 0. But if we have indistinguishable particles, our distribution is different. Now we don’t have the multiplicity of 6 and 3; in particular, the distribution of particles in energy levels for these bosons is different from the classical model. This means bosons actually have a tendency to clump together! Finally, there’s one more problem: if we have fermions, we’re forced into a specific case, because you can’t have two different particles in the same state! 156 Fact 250 Notice that the difference for occupancy level iis just a −1 or +1 in the denominator of 1 e(ε−µ)/(kBT ) . In particular, photons are bosons, but they can be freely created! This means there is no “chemical potential,” and we can set µ = 0 here. This gives us the Planck blackbody spectrum n(ε) = 1 eε/(kBT ) −1. Notably, Bose rederived Planck’s formula with a new method, and Einstein used that method to add in the µ chemical potential term! That became the “Bose-Einstein distribution.” Question 251. How do we describe Bose-Einstein condensation, then? The idea is that the classical distribution just shifts to be skinnier and taller when we adjust our temperature T. But then at a certain critical temperature T < TC, corresponding to a critical wavelength, the population per energy state goes to infinity. This specific singularity is described in Einstein’s paper. But back in 1924, Einstein said (after describing the theory) “The theory is pretty, but is there also some truth to it?” The idea is that “mathematical singularities” may not need to be part of actual physics, so there was lots of skepticism. It was only in 1938 that Fritz London realized that Bose-Einstein condensation is indeed physical and observable! 48.2 Why use the formalism that we do? We’re going to look more at the equations and understand the singularity mathematically. First of all, note that we can formulate things in many ways, but smarter choices (for example, in terms of coordinates) make our job easier. We know that for atoms and molecules, energy is conserved, and so is the number of particles. But there is a problem here: distributing energies is not independent, and this means one of the particles needs to “pay the price.” We like to assume in classical physics that each particle is an independent entity, and fluctuations shouldn’t affect each other! That’s why we use the canonical ensemble: we can then say that N particles have a partition function Zn = ZN 1 or 1 N!ZN 1 . But then Einstein’s 1924 paper seemed to cause some problems: the particles turned out to not be independent anymore under the canonical distribution. But indeed, descriptions under quantum physics no longer have particle independence (for example, the Pauli exclusion principle)! Instead, our independence shifts to the quantum states themselves. By allowing each quantum state to run through each of its possible occupation numbers, we also allow for our total number of particles N to fluctuate, and now we use the grand canonical ensemble Z = Y i Zi. That’s the beauty of the grand canonical ensemble - we get our independence back again! 48.3 Mathematical derivation We’ve derived in class before that the occupation number under the Bose-Einstein distribution nj = 1 e(εj−µ)/(kBT ) −1. 157 Question 252. What is the population n0 (corresponding to zero momentum and lowest energy)? We use the fact that as T →0, the chemical potential is negative but approaches 0−. Then the occupation of the zero energy state is n0 →kBT −µ →∞. Notably, this means the chemical potential can’t be positive, or we’d have a singularity at some non-ground state! Question 253. What is the number of particles in an excited state (all j ̸= 0)? This is just the sum over all nonzero states N = X εj>δ 1 eβ(εj−µ) −1 where we’ll make δ →0. If we only care about having an upper bound, we can ignore the small negative value of chemical potential: N ≤ X εj>δ 1 eβεj −1. Remember that when we sum over all states, we can do this in a semi-classical manner instead: = Z d3xd3p h3 1 eβεj −1. The position integral just becomes V , the volume of our system, and then we can use the spherical integration trick again: = V 4π h3 Z dpp2 1 eβεj −1. We can now replace p2dp with √εdε with a constant factor; C√ε here is our density of states! This ends up giving us (replacing h with ℏfor simplicity) an upper bound Nmax = Z ∞ δ N(ε) eβε−1 , N(ε) = V 2m ℏ 3/2 √ε 4π2 . We’ll remove most of the ugly constants by introducing a thermal de Broglie wavelength and dimensionless variable λ = s 2πℏ2 mkBT , x = βε. Remember that our integral is still counting the number of particles in non-ground states: this gives us = V λ3 2 √π Z ∞ δβ √xdx ex −1. But now we can actually replace δβ with 0, since the integral converges, and we get an estimate of Nmax ≈2.612 V λ3 . Notice that this is a fixed number dependent on T! So if we put more particles into our system at a fixed temperature, they don’t go into excited states! So we have some absolute limit on the number of excited particles in terms of V and T: all others go into the ground state. 158 Proposition 254 In other words, eventually we have a saturation of the quantum gas: when we have too many particles, the whole system must condense! This is similar to the way in which eventually water vapor at 100 degrees Celsius must start forming water droplets when the pressure is too large. So if we lower the temperature enough, and Nmax reaches a point comparable to N, every subsequent increase in the number of particles or decrease in temperature will create a Bose-Einstein condensate. Fact 255 Notably, if we set the boxed Nmax = N, we can find our critical temperature kBTC =  N 2.612 2/3 ℏ22π L2m , where V = L3 if we assume we have a box. Notably, the N2/3 L2 is proportional to n2/3, where n is the density of our gas! There’s two things we should discuss here. First of all, the density we can get to realistically (when we work with individual atoms) is n = 1014, which is a factor of 105 smaller than room temperature gases. Then we find that the TC here is 100 nanoKelvin: that means verifying this experimentally requires us to get to very cold temperatures! Question 256. If we want almost all atoms in the ground state, what’s the temperature scale that we are allowed to have? Remember that there is a first excited state: we have gapped behavior here. In our classical model, an energy lower than that must put all the particles in the ground state! On the other hand, though, the Bose-Einstein model has an extra factor of N2/3 in it. Because bosons are indistinguishable, we get this extra factor that actually helps us: everything goes to the ground state much faster! 48.4 How do we do this experimentally? Fact 257 In a paper that Schrodinger wrote in 1952, he expressed the opinion that van der Waals corrections (attractions and repulsions between molecules) and effects of Bose-Einstein condensation are almost impossible to separate. He thinks that most systems will become liquids before any quantum effects can be seen. But it turns out that there exist cooling methods that can get us to very cold atomic clouds! Atoms cannot “keep photons,” so if we shine a laser beam at an atom, the photons that hit the atom must be emitted out by fluorescence. But now we can blue-shift the emitted radiation, every time an atom absorbs and emits a photon, it radiates away some of its energy! This is hard to implement in the laboratory, but this is the way laser cooling works. Fact 258 Atoms are a special system: they are almost completely immune to black-body radiation, so we don’t need to do the same shielding from cameras and beams and other materials as in more complicated systems. 159 Well, this gets us to the microKelvin level: what can we do to get closer to the 100 nanoKelvin level that we want? It turns out evaporative cooling is very easy to understand. In a thermos, steam molecules (which have the highest energy) escape, and the lower-energy molecules stay behind. In our atomic system, we have a container made by magnetic or electric fields, and then use “radio frequency spin flips” to select the particles with highest energy (This is the same as “blowing on a cup of coffee”). In other words, if we remove our electric/magnetic field container, the gas will expand at a thermal velocity. Since the kinetic energy mv 2 2 = kBT 2 by equipartition, this is a way for us to figure out the temperature without needing to explicitly put something like a thermometer in contact with it! Fact 259 By the way, right now the temperature of 1 Kelvin is 1 273.15 times the triple point of water. Soon it’ll be defined in terms of the Boltzmann constant instead! Basically, when we “blow” with our radio waves, we should expect smaller clouds, and indeed, this is what happens! There’s an elliptical object in the middle that stays put - that’s the Bose-Einstein condensate that we can observe. Remember that a thermal gas is isotropic: the shadow should always be perfectly circular. So why is the condensate elliptical? It’s because the ground state of an elliptical container is elliptical! Fact 260 Finally, how do we prove that the atoms act as one single wave? The key idea is interference! Two waves that collide form a standing wave, and it turns out that we can do the same kind of interference by taking a Bose-Einstein condensate and cutting it in half. The interesting thing here is that we have now accomplished interference of matter waves, since the positive and negative part of the wavefunction add up to zero! 49 May 15, 2019 (Recitation) 49.1 Drawing more connections Let’s start by trying to understand the last concept we introduced in this class. Exchanging energy in the canonical ensemble is very similar to exchanging particles in a grand canonical ensemble! In the former, we’re connected to a temperature reservoir, and in the second, we’re connected to some chemical potential reservoir. What are the transitions here? We fix the internal energy U in our microcanonical ensemble, but in a canonical ensemble, we essentially tweak our energy as a function of T, our temperature. In particular, we don’t clamp down the energy: one thing we can do is to maximize entropy as a function of β (using Lagrange multipliers). The mathematics here is that if we want to maximize F(x, y, z) subject to some constraint E(x, y, z) = ε0, we can introduce a new parameter β which also varies independently: F(x, y, z) −βE = 0 and then all derivatives with respect to x, y, z, β must be zero. So the microcanonical condition is enforced with the Lagrange parameter, and ultimately that parameter is adjusted (aka, the temperature is our independent variable) to reach our energy E. 160 Fact 261 So temperature plays a dual role: temperature affects internal energy, but we can also use it as a Lagrange parameter to ask questions like “how to maximize entropy.” E and β depend each other. Well, this same dependence happens with N and µ! If we have a reservoir with some chemical potential, µ controls the number of particles N, just like the temperature T controls our energy E. We now want to maximize the free energy F, and we use a similar Lagrange parameter: this time, it is µ. So now µ can also be our “knob” that controls N. Example 262 Consider a semiconductor piece of metal. A battery is then a source of chemical potential: increasing the voltage charges up the system and introduces more electrons! So the battery becomes the “energetic cost” of delivering another particle. Our grand partition function Z = X j e−β(εj−njµ now takes into account that each particle being present changes our energy somehow. Remember where this all comes from: we treat our system as being connected to a reservoir, and we think of this whole system-reservoir as being a microcanonical ensemble. Then the probability of any specific microstate µj is Pr(µj) ∝ΓRes(E −εj, N −nj), since E and N are fixed across the whole system-reservoir entity. This is then proportional to eSRes, and expanding out S to first order in our Taylor expansion yields the result that we want: = ∂S ∂N (nj) + ∂S ∂E (−εj) = βµnj −βεj and that yields the −β(εj −µnj) that we want! This Taylor expansion basically tells us “how much energy it costs” to give away a particle or give away some energy. 49.2 Going to the lab Most particles in nature are fermions (quarks, electrons, and so on), rather than bosons. Fact 263 By the way, the main difference is the spin (inherent angular momentum) of the particle, and whether it’s an integer or a half-integer. Two indistinguishable particles must satisfy ψ(x1, x2) = ±ψ(x2, x1), where the ± comes from us only observing the square of the wave function. Turns out −corresponds to fermions, and + corresponds to bosons: now the Pauli exclusion principle comes from the “spin statistic theorem!” 50 May 16, 2019 We’re going to finish talking about Fermi gas thermodynamics, and we’ll finish by talking about what lies beyond this class! 161 Fact 264 By the way, the grade cutoffs for A/B, B/C, and C/D last year were 85, 65, and 55. This fluctuates year to year, though. We should check this weekend that everything is graded! The deadline for all grading-related things is tentatively Sunday, but everything will definitely be done by Wednesday morning. Exam 1 and 2 solutions will be posted soon, and past exams have been posted for reviewing for the final as well. 50.1 Back to Fermi systems: a gas of electrons Remember that fermions are a gas of spin 1 2. We found that because of the Pauli exclusion principle, we have an interesting result for the occupation number at zero temperature for a degenerate Fermi gas: nj(ε) = 1 1 + e(ε−µ)β this is basically a step function at ε = µ. (If we increase the temperature to some T, we get partially filled states with a width of around ∆ε ≈kBT.) At temperature T = 0, define µ = Ef to be the Fermi energy: we can then also define a Fermi momentum pF = √2mEf , which has applications in condensed matter. The terms Fermi sea and Fermi surface then refer to the filled states where E < Ef and the set of states where E = Ef , respectively. We’re going to try to derive the thermodynamics of a 3D Fermi gas now! We start with a density of states calculation: we have dN = g d3xd3p (2πℏ)3 , where g, the degeneracy factor, is 2s + 1 (where s is the spin of our particle). Since electrons have spin 1 2, g = 2 in this case! Integrating out, N = g Z d3xd3p (2πℏ)3 1 e(p2/2m−µ)β + 1. We’ll do the same tricks we keep doing: integrating out d3x and using spherical coordinates, this simplifies to a single integral N = 4πV g (2πℏ)3 Z ∞ 0 p2dp e(p2/2m−µ) + 1. As we take temperature to 0, we have some upper limit equal to our Fermi momentum (since the denominator is 1 for p < pF and ∞for p > pF ), so this simplifies to (combining some constants) N = gV 2π2ℏ3 Z pF 0 p2dp = gV p3 F 6π2ℏ3 ; substituting back for our Fermi energy yields Ef = ℏ2 2m 6π2 g N V 2/3 . The idea is that this is the energy of the last filled state, because fermions that enter our system successively fill energy levels from lowest to highest! This gives us a Fermi wavenumber pF ℏ= kf = ⇒N = g 6π2 V k3 F . 162 We can also define a Fermi temperature Tf = Ef kB ; surprisingly, for electrons in metal, this temperature is around 104 Kelvin, and for electrons in a white dwarf, this is around 107 Kelvin! Since these numbers are so large, the description of Fermi gases is very good for deriving material properties of solids, as well as other fermion systems. So now if we want to calculate our energy, we integrate U = g Z d3xd3p (2πℏ)3 p2/2m e(p2/2m−µ)β + 1; If we take our temperature T →0, the same behavior with the denominator happens, and we’re left with U = gV 2π2ℏ3 Z pF 0 dpp2 p2 2m = gV p5 F 20mπ2ℏ3 . Again, we can rearrange to write U in terms of N, our number of particles, and Ef , our Fermi energy: this yields U = 3 5NEF . Finally, how do we derive the equation of state? If U is written as a function of N, V, T = 0 (in our limiting case), then P = −∂U ∂V N,T =0 . We can rewrite our expression above for U: it turns out that we have U = constant · N5/3 V 2/3 , so taking the derivative, P = 2 3 U V = gp5 F 30π2mℏ3 , which can be rewritten in terms of Fermi energy as PV = 2 5NEf . So even at temperature 0, there is some residual pressure! This is known as “degenerate pressure,” and it occurs because of the Pauli exclusion principle - this has stabilizing effects in certain systems. The only problem is that we’ve assumed our electrons are free particles, but we know that this isn’t true - they’re bound to nuclei! So we need to start adding correction terms to account for interactions like this. 50.2 What’s next? There’s a lot of cool and exciting areas that we can study after this class! Here’s some of them: • Phase transitions. These can be observed in real life, and they also have applications in biophysics and other areas like particle physics! • Non-equilibrium physics: how does a system out of equilibrium relax into equilibrium? There’s something called “linear response theory” here, as well as a notion of “non-equilibrium steady states.” • Dynamical processes. 163 • Thermodynamics of small systems - we’ve been using large N to simplify a lot of our calculations, but there’s an exciting area of theoretical development where we start caring more about fluctuations! There’s notions of “work-fluctuation” and other strange phenomena. We should all continue with statistical physics after this point! This class is a basis for doing other, more exciting things. 164
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Stanford Engineering Everywhere | EE364B - Convex Optimization II | Lecture 7 - Example: Piecewise Linear Minimization Stanford Engineering Everywhere Home Courses Using SEE Survey Contact Us Stanford University stanford engineering S tanford E ngineering E verywhere S EE Menu Home Courses Using SEE Survey Contact Us EE364B - Convex Optimization II Lecture 7 - Example: Piecewise Linear Minimization To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration-:- Loaded: 0% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time--:- 1x Playback Rate Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. 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Expand/Collapse Video Bookmarks Playlist Info Bookmarks 00:00:34 Recap: ACCPM Algorithm 00:02:57 Example: Piecewise Linear Minimization 00:17:07 ACCPM With Constraint Dropping 00:26:33 Epigraph ACCPM 00:28:00 Ellipsoid Method 00:33:42 Motivation (For Ellipsoid Method) 00:36:52 Ellipsoid Algorithm For Minimizing Convex Function 00:47:21 Properties Of Ellipsoid Method 00:49:58 Example (Using Ellipsoid Method) 00:51:16 Updating The Ellipsoid 01:07:57 Simple Stopping Criterion 01:11:14 Basic Ellipsoid Algorithm 01:12:39 Interpretation (Of Basic Ellipsoid Algorithm) 01:12:47 Example (Of Ellipsoid Method) Lectures 1 - Course Logistics 2 - Recap: Subgradients 3 - Convergence Proof 4 - Project Subgradient For Dual Problem 5 - Stochastic Programming 6 - Addendum: Hit-And-Run CG Algorithm 7 - Example: Piecewise Linear Minimization 8 - Recap: Ellipsoid Method 9 - Comments: Latex Typesetting Style 10 - Decomposition Applications 11 - Sequential Convex Programming 12 - Recap: 'Difference Of Convex' Programming 13 - Recap: Conjugate Gradient Method 14 - Methods (Truncated Newton Method) 15 - Recap: Example: Minimum Cardinality Problem 16 - Model Predictive Control 17 - Stochastic Model Predictive Control 18 - Announcements About the Lecture TITLE:Lecture 7 - Example: Piecewise Linear Minimization DURATION:1 hr 14 min TOPICS:Example: Piecewise Linear Minimization ACCPM With Constraint Dropping Epigraph ACCPM Motivation (For Ellipsoid Method) Ellipsoid Algorithm For Minimizing Convex Function Properties Of Ellipsoid Method Example (Using Ellipsoid Method) Updating The Ellipsoid Simple Stopping Criterion Basic Ellipsoid Algorithm Interpretation (Of Basic Ellipsoid Algorithm) Example (Of Ellipsoid Method) Course Details Show All Course Description Continuation of Convex Optimization I. Subgradient, cutting-plane, and ellipsoid methods. Decentralized convex optimization via primal and dual decomposition. Alternating projections. Exploiting problem structure in implementation. Convex relaxations of hard problems, and global optimization via branch & bound. Robust optimization. Selected applications in areas such as control, circuit design, signal processing, and communications. Course requirements include a substantial project. Prerequisites: Convex Optimization I Syllabus DOWNLOAD All Course Materials Instructor Boyd, Stephen Stephen P. Boyd is the Samsung Professor of Engineering, and Professor of Electrical Engineering in the Information Systems Laboratory at Stanford University. His current research focus is on convex optimization applications in control, signal processing, and circuit design. Professor Boyd received an AB degree in Mathematics, summa cum laude, from Harvard University in 1980, and a PhD in EECS from U. C. Berkeley in 1985. In 1985 he joined the faculty of Stanford’s Electrical Engineering Department. He has held visiting Professor positions at Katholieke University (Leuven), McGill University (Montreal), Ecole Polytechnique Federale (Lausanne), Qinghua University (Beijing), Universite Paul Sabatier (Toulouse), Royal Institute of Technology (Stockholm), Kyoto University, and Harbin Institute of Technology. He holds an honorary doctorate from Royal Institute of Technology (KTH), Stockholm. Professor Boyd is the author of many research articles and three books: Linear Controller Design: Limits of Performance (with Craig Barratt, 1991), Linear Matrix Inequalities in System and Control Theory (with L. El Ghaoui, E. Feron, and V. Balakrishnan, 1994), and Convex Optimization (with Lieven Vandenberghe, 2004). Professor Boyd has received many awards and honors for his research in control systems engineering and optimization, including an ONR Young Investigator Award, a Presidential Young Investigator Award, and an IBM faculty development award. In 1992 he received the AACC Donald P. Eckman Award, which is given annually for the greatest contribution to the field of control engineering by someone under the age of 35. In 1993 he was elected Distinguished Lecturer of the IEEE Control Systems Society, and in 1999, he was elected Fellow of the IEEE, with citation: “For contributions to the design and analysis of control systems using convex optimization based CAD tools.” He has been invited to deliver more than 30 plenary and keynote lectures at major conferences in both control and optimization. In addition to teaching large graduate courses on Linear Dynamical Systems, Nonlinear Feedback Systems, and Convex Optimization, Professor Boyd has regularly taught introductory undergraduate Electrical Engineering courses on Circuits, Signals and Systems, Digital Signal Processing, and Automatic Control. In 1994 he received the Perrin Award for Outstanding Undergraduate Teaching in the School of Engineering, and in 1991, an ASSU Graduate Teaching Award. In 2003, he received the AACC Ragazzini Education award, for contributions to control education, with citation: “For excellence in classroom teaching, textbook and monograph preparation, and undergraduate and graduate mentoring of students in the area of systems, control, and optimization.” Handouts Lecture Materials SubgradientsLecture SlidesLecture Notes Subgradient MethodsLecture SlidesLecture NotesMatlab Files Subgradient Methods for Constrained ProblemsLecture Slides Stochastic Subgradient MethodLecture SlidesLecture NotesMatlab Files Localization and Cutting-plane MethodsLecture SlidesLecture Notes Analytic Center Cutting-plane MethodLecture SlidesLecture NotesMatlab Files Ellipsoid MethodLecture SlidesMatlab Files Ellipsoid Method Part IILecture SlidesMatlab Files Primal and Dual DecompositionLecture SlidesLecture NotesMatlab Files Decomposition ApplicationsLecture Slides Sequential Convex ProgrammingLecture SlidesMatlab Files Conjugate-gradient MethodLecture SlidesMatlab Files Truncated Newton MethodsLecture SlidesMatlab Files Methods for Convex-cardinality ProblemsLecture SlidesMatlab Files Methods for Convex-cardinality Problems, Part IILecture SlidesMatlab Files Model Predictive ControlLecture SlidesMatlab Files Stochastic Model Predictive ControlLecture Slides Branch-and-Bound MethodsLecture SlidesLecture NotesPython Files Additional Lecture Notes Notes on relaxation and randomized methods for nonconvex QCQP Notes on convex-concave games and minimax Numerical linear algebra software Resources This page contains links to various interesting and useful sites that relate in some way to convex optimization. It goes without saying that you’ll be periodically checking things using google and wikipedia. The wikipedia entry on convex optimization (and related topics) could be improved or extended. Stephen Boyd’s research page. There’s a lot of material there, and you don’t have to know every detail in every paper, but you should certainly take an hour or more to browse through these papers. EE364a web page. We expect you to know what’s in these pages. The Convex Optimization book. You’re expected to know pretty well the material in this book. Unless you have a really good memory, you should be browsing through this. Lieven Vandenberghe’s ee236a and ee236b course pages. Athena Scientific books on optimization. You can also check the MIT courses that use some of these books. CVX. Be sure to check out the every extensive library of examples. (Indeed, feel free to add to it.) CVXOPT, which also includes an extensive library of examples, and CVXMOD YALMIP, a Matlab toolbox for optimization modeling. SOSTOOLS, a toolbox for formulating and solving sums of squares (SOS) optimization problems. Assignments Homework Assignments Assignments may require Matlab files, see Software below. | Assignment | Solutions | Due Date | --- | Assignment 1 | Solutions | Lecture 4 | | Assignment 2 | Solutions | Lecture 5 | | Assignment 3 | Solutions | Lecture 7 | | Assignment 4 | Solutions | Lecture 11 | | Assignment 5 | Solutions | Lecture 14 | | Assignment 6 | Solutions | Lecture 17 | | Assignment 7 | Solutions | Lecture 18 | Final Project Convex Optimization II requires an extensive project. Here are the project guidelines. Here are the project deadlines: Initial proposal, due Lecture 7 Revised proposal, due Lecture 12 Midterm progress report, due Lecture 14 Final report, due Lecture 18 Here is some example Latex code you can use for a template. Project proposals, reports, and posters must use Latex, with either our template or a good alternative. | Matlab files for homework problems: | | camera_data.m | | flowgray.png | | illum_data.m | | log_normcdf.m | | log_opt_invest.m | | nonlin_meas_data.m | | ps_data.m | | pwl_fit_data.m | | sep3way_data.m | | sp_ln_sp_data.m | | team_data.m | | thrusters_data.m | | tv_img_interp.m | Software Matlab Files | Matlab files for homework problems: | | bicommodity_data.m | | ex_blockprecond.m | | l1_heuristic_portfolio_data.m | | quad2_min.m | | sp_bayesnet_data.m | Course Sessions (18): Show All Lecture 1 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 2 min Watch Online:Topics: Course Logistics, Course Organization, Course Topics, Subgradients, Basic Inequality, Subgradient Of A Function, Subdifferential, Subgradient Calculus, Some Basic Rules (For Subgradient Calculus), Pointwise Supremum, Weak Rule For Pointwise Supremum, Expectation, Minimization, Composition, Subgradients And Sublevel Sets, Quasigradients Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 2 min Topics: Course Logistics, Course Organization, Course Topics, Subgradients, Basic Inequality, Subgradient Of A Function, Subdifferential, Subgradient Calculus, Some Basic Rules (For Subgradient Calculus), Pointwise Supremum, Weak Rule For Pointwise Supremum, Expectation, Minimization, Composition, Subgradients And Sublevel Sets, Quasigradients Transcripts HTML PDF Lecture 2 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 7 min Watch Online:Topics: Recap: Subgradients, Subgradients And Sublevel Sets, Quasigradients, Optimality Conditions – Unconstrained, Example: Piecewise Linear Minimization, Optimality Conditions – Constrained, Directional Derivative And Subdifferential, Descent Directions, Subgradients And Distance To Sublevel Sets, Descent Directions And Optimality, Subgradient Method, Step Size Rules, Assumptions, Convergence Results, Aside: Example: Applying Subgradient Method To Abs(X) Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 7 min Topics: Recap: Subgradients, Subgradients And Sublevel Sets, Quasigradients, Optimality Conditions – Unconstrained, Example: Piecewise Linear Minimization, Optimality Conditions – Constrained, Directional Derivative And Subdifferential, Descent Directions, Subgradients And Distance To Sublevel Sets, Descent Directions And Optimality, Subgradient Method, Step Size Rules, Assumptions, Convergence Results, Aside: Example: Applying Subgradient Method To Abs(X) Transcripts HTML PDF Lecture 3 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 15 min Watch Online:Topics: Convergence Proof, Stopping Criterion, Example: Piecewise Linear Minimization, Optimal Step Size When F Is Known, Finding A Point In The Intersection Of Convex Sets, Alternating Projections, Example: Positive Semidefinite Matrix Completion, Speeding Up Subgradient Methods, A Couple Of Speedup Algorithms, Subgradient Methods For Constrained Problems, Projected Subgradient Method, Linear Equality Constraints, Example: Least L_1-Norm Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 15 min Topics: Convergence Proof, Stopping Criterion, Example: Piecewise Linear Minimization, Optimal Step Size When F Is Known, Finding A Point In The Intersection Of Convex Sets, Alternating Projections, Example: Positive Semidefinite Matrix Completion, Speeding Up Subgradient Methods, A Couple Of Speedup Algorithms, Subgradient Methods For Constrained Problems, Projected Subgradient Method, Linear Equality Constraints, Example: Least L_1-Norm Transcripts HTML PDF Lecture 4 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 19 min Watch Online:Topics: Project Subgradient For Dual Problem, Subgradient Of Negative Dual Function, Example (Strictly Convex Quadratic Function Over Unit Box), Subgradient Method For Constrained Optimization, Convergence, Example: Inequality Form LP, Stochastic Subgradient Method, Noisy Unbiased Subgradient, Stochastic Subgradient Method, Assumptions, Convergence Results, Convergence Proof, Stochastic Programming Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 19 min Topics: Project Subgradient For Dual Problem, Subgradient Of Negative Dual Function, Example (Strictly Convex Quadratic Function Over Unit Box), Subgradient Method For Constrained Optimization, Convergence, Example: Inequality Form LP, Stochastic Subgradient Method, Noisy Unbiased Subgradient, Stochastic Subgradient Method, Assumptions, Convergence Results, Convergence Proof, Stochastic Programming Transcripts HTML PDF Lecture 5 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 16 min Watch Online:Topics: Stochastic Programming, Variations (Of Stochastic Programming), Expected Value Of A Convex Function, Example: Expected Value Of Piecewise Linear Function, On-Line Learning And Adaptive Signal Processing, Example: Mean-Absolute Error Minimization, Localization And Cutting-Plane Methods, Cutting-Plane Oracle, Neutral And Deep Cuts, Unconstrained Minimization, Deep Cut For Unconstrained Minimization, Feasibility Problem, Inequality Constrained Problem, Localization Algorithm, Example: Bisection On R, Specific Cutting-Plane Methods, Center Of Gravity Algorithm, Convergence Of CG Cutting-Plane Method Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 16 min Topics: Stochastic Programming, Variations (Of Stochastic Programming), Expected Value Of A Convex Function, Example: Expected Value Of Piecewise Linear Function, On-Line Learning And Adaptive Signal Processing, Example: Mean-Absolute Error Minimization, Localization And Cutting-Plane Methods, Cutting-Plane Oracle, Neutral And Deep Cuts, Unconstrained Minimization, Deep Cut For Unconstrained Minimization, Feasibility Problem, Inequality Constrained Problem, Localization Algorithm, Example: Bisection On R, Specific Cutting-Plane Methods, Center Of Gravity Algorithm, Convergence Of CG Cutting-Plane Method Transcripts HTML PDF Lecture 6 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 12 min Watch Online:Topics: Addendum: Hit-And-Run CG Algorithm, Maximum Volume Ellipsoid Method, Chebyshev Center Method, Analytic Center Cutting-Plane Method, Extensions (Of Cutting-Plane Methods), Dropping Constraints, Epigraph Cutting-Plane Method, PWL Lower Bound On Convex Function, Lower Bound, Analytic Center Cutting-Plane Method, ACCPM Algorithm, Constructing Cutting-Planes, Computing The Analytic Center, Infeasible Start Newton Method Algorithm, Properties (Of Infeasible Start Newton Method Algorithm), Pruning Constraints, PWL Lower Bound On Convex Function, Lower Bound In ACCPM, Stopping Criterion, Example: Piecewise Linear Minimization Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 12 min Topics: Addendum: Hit-And-Run CG Algorithm, Maximum Volume Ellipsoid Method, Chebyshev Center Method, Analytic Center Cutting-Plane Method, Extensions (Of Cutting-Plane Methods), Dropping Constraints, Epigraph Cutting-Plane Method, PWL Lower Bound On Convex Function, Lower Bound, Analytic Center Cutting-Plane Method, ACCPM Algorithm, Constructing Cutting-Planes, Computing The Analytic Center, Infeasible Start Newton Method Algorithm, Properties (Of Infeasible Start Newton Method Algorithm), Pruning Constraints, PWL Lower Bound On Convex Function, Lower Bound In ACCPM, Stopping Criterion, Example: Piecewise Linear Minimization Transcripts HTML PDF Lecture 7 Watch Online:Now Playing Download: Right Click, and Save AsDownload Duration:1 hr 14 min Watch Online:Topics: Example: Piecewise Linear Minimization, ACCPM With Constraint Dropping, Epigraph ACCPM, Motivation (For Ellipsoid Method), Ellipsoid Algorithm For Minimizing Convex Function, Properties Of Ellipsoid Method, Example (Using Ellipsoid Method), Updating The Ellipsoid, Simple Stopping Criterion, Basic Ellipsoid Algorithm, Interpretation (Of Basic Ellipsoid Algorithm), Example (Of Ellipsoid Method) Watch Online:Download: Right Click, and Save As Duration: Now PlayingDownload1 hr 14 min Topics: Example: Piecewise Linear Minimization, ACCPM With Constraint Dropping, Epigraph ACCPM, Motivation (For Ellipsoid Method), Ellipsoid Algorithm For Minimizing Convex Function, Properties Of Ellipsoid Method, Example (Using Ellipsoid Method), Updating The Ellipsoid, Simple Stopping Criterion, Basic Ellipsoid Algorithm, Interpretation (Of Basic Ellipsoid Algorithm), Example (Of Ellipsoid Method) Transcripts HTML PDF Lecture 8 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 11 min Watch Online:Topics: Recap: Ellipsoid Method, Improvements (To Ellipsoid Method), Proof Of Convergence, Interpretation Of Complexity, Deep Cut Ellipsoid Method, Ellipsoid Method With Deep Objective Cuts, Inequality Constrained Problems, Stopping Criterion, Epigraph Ellipsoid Method, Epigraph Ellipsoid Example, Summary: Methods For Handling, Nondifferentiable Convex Optimization Problems Directly, Decomposition Methods, Separable Problem, Complicating Variable, Primal Decomposition, Primal Decomposition Algorithm, Example (Using Primal Decomposition), Aside: Newton's Method With A Complicating Variable, Dual Decomposition, Dual Decomposition Algorithm Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 11 min Topics: Recap: Ellipsoid Method, Improvements (To Ellipsoid Method), Proof Of Convergence, Interpretation Of Complexity, Deep Cut Ellipsoid Method, Ellipsoid Method With Deep Objective Cuts, Inequality Constrained Problems, Stopping Criterion, Epigraph Ellipsoid Method, Epigraph Ellipsoid Example, Summary: Methods For Handling, Nondifferentiable Convex Optimization Problems Directly, Decomposition Methods, Separable Problem, Complicating Variable, Primal Decomposition, Primal Decomposition Algorithm, Example (Using Primal Decomposition), Aside: Newton's Method With A Complicating Variable, Dual Decomposition, Dual Decomposition Algorithm Transcripts HTML PDF Lecture 9 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 10 min Watch Online:Topics: Comments: Latex Typesetting Style, Recap: Primal Decomposition, Dual Decomposition, Dual Decomposition Algorithm, Finding Feasible Iterates, Interpretation, Decomposition With Constraints, Primal Decomposition (With Constraints) Algorithm, Example (Primal Decomposition With Constraints), Dual Decomposition (With Constraints), Dual Decomposition (With Constraints) Algorithm, General Decomposition Structures, General Form, Primal Decomposition (General Structures), Dual Decomposition (General Structures), A More Complex Example, Aside: Pictorial Representation Of Primal And Dual Decomposition Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 10 min Topics: Comments: Latex Typesetting Style, Recap: Primal Decomposition, Dual Decomposition, Dual Decomposition Algorithm, Finding Feasible Iterates, Interpretation, Decomposition With Constraints, Primal Decomposition (With Constraints) Algorithm, Example (Primal Decomposition With Constraints), Dual Decomposition (With Constraints), Dual Decomposition (With Constraints) Algorithm, General Decomposition Structures, General Form, Primal Decomposition (General Structures), Dual Decomposition (General Structures), A More Complex Example, Aside: Pictorial Representation Of Primal And Dual Decomposition Transcripts HTML PDF Lecture 10 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 17 min Watch Online:Topics: Decomposition Applications, Rate Control Setup, Rate Control Problem, Rate Control Lagrangian, Aside: Utility Functions, Rate Control Dual, Dual Decomposition Rate Control Algorithm, Generating Feasible Flows, Convergence Of Primal And Dual Objectives, Maximum Capacity Violation, Single Commodity Network Flow Setup, Network Flow Problem, Network Flow Lagrangian, Network Flow Dual, Recovering Primal From Dual, Dual Decomposition Network Flow Algorithm, Electrical Network Analogy, Example: Minimum Queueing Delay, Optimal Flow, Convergence Of Dual Function, Convergence Of Primal Residual, Convergence Of Dual Variables, Aside: More Complicated Problems Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 17 min Topics: Decomposition Applications, Rate Control Setup, Rate Control Problem, Rate Control Lagrangian, Aside: Utility Functions, Rate Control Dual, Dual Decomposition Rate Control Algorithm, Generating Feasible Flows, Convergence Of Primal And Dual Objectives, Maximum Capacity Violation, Single Commodity Network Flow Setup, Network Flow Problem, Network Flow Lagrangian, Network Flow Dual, Recovering Primal From Dual, Dual Decomposition Network Flow Algorithm, Electrical Network Analogy, Example: Minimum Queueing Delay, Optimal Flow, Convergence Of Dual Function, Convergence Of Primal Residual, Convergence Of Dual Variables, Aside: More Complicated Problems Transcripts HTML PDF Lecture 11 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 16 min Watch Online:Topics: Sequential Convex Programming, Methods For Nonconvex Optimization Problems, Sequential Convex Programming (SCP), Basic Idea Of SCP, Trust Region, Affine And Convex Approximations Via Taylor Expansions, Particle Method, Fitting Affine Or Quadratic Functions To Data, Quasi-Linearization, Example (Nonconvex QP), Lower Bound Via Lagrange Dual, Exact Penalty Formulation, Trust Region Update, Nonlinear Optimal Control, Discretization, SCP Progress, Convergence Of J And Torque Residuals, Predicted And Actual Decreases In Phi, Trajectory Plan, 'Difference Of Convex' Programming, Convex-Concave Procedure Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 16 min Topics: Sequential Convex Programming, Methods For Nonconvex Optimization Problems, Sequential Convex Programming (SCP), Basic Idea Of SCP, Trust Region, Affine And Convex Approximations Via Taylor Expansions, Particle Method, Fitting Affine Or Quadratic Functions To Data, Quasi-Linearization, Example (Nonconvex QP), Lower Bound Via Lagrange Dual, Exact Penalty Formulation, Trust Region Update, Nonlinear Optimal Control, Discretization, SCP Progress, Convergence Of J And Torque Residuals, Predicted And Actual Decreases In Phi, Trajectory Plan, 'Difference Of Convex' Programming, Convex-Concave Procedure Transcripts HTML PDF Lecture 12 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 13 min Watch Online:Topics: Recap: 'Difference Of Convex' Programming, Alternating Convex Optimization, Nonnegative Matrix Factorization, Comment: Nonconvex Methods, Conjugate Gradient Method, Three Classes Of Methods For Linear Equations, Symmetric Positive Definite Linear Systems, CG Overview, Solution And Error, Residual, Krylov Subspace, Properties Of Krylov Sequence, Cayley-Hamilton Theorem, Spectral Analysis Of Krylov Sequence Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 13 min Topics: Recap: 'Difference Of Convex' Programming, Alternating Convex Optimization, Nonnegative Matrix Factorization, Comment: Nonconvex Methods, Conjugate Gradient Method, Three Classes Of Methods For Linear Equations, Symmetric Positive Definite Linear Systems, CG Overview, Solution And Error, Residual, Krylov Subspace, Properties Of Krylov Sequence, Cayley-Hamilton Theorem, Spectral Analysis Of Krylov Sequence Transcripts HTML PDF Lecture 13 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 15 min Watch Online:Topics: Recap: Conjugate Gradient Method, Recap: Krylov Subspace, Spectral Analysis Of Krylov Sequence, A Bound On Convergence Rate, Convergence, Residual Convergence, CG Algorithm, Efficient Matrix-Vector Multiply, Shifting, Preconditioned Conjugate Gradient Algorithm, Choice Of Preconditioner, CG Summary, Truncated Newton Method, Approximate Or Inexact Newton Methods, CG Initialization, Hessian And Gradient, Methods, Convergence Versus Iterations, Convergence Versus Cumulative CG Steps, Truncated PCG Newton Method, Extensions Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 15 min Topics: Recap: Conjugate Gradient Method, Recap: Krylov Subspace, Spectral Analysis Of Krylov Sequence, A Bound On Convergence Rate, Convergence, Residual Convergence, CG Algorithm, Efficient Matrix-Vector Multiply, Shifting, Preconditioned Conjugate Gradient Algorithm, Choice Of Preconditioner, CG Summary, Truncated Newton Method, Approximate Or Inexact Newton Methods, CG Initialization, Hessian And Gradient, Methods, Convergence Versus Iterations, Convergence Versus Cumulative CG Steps, Truncated PCG Newton Method, Extensions Transcripts HTML PDF Lecture 14 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 13 min Watch Online:Topics: Methods (Truncated Newton Method), Convergence Versus Iterations, Convergence Versus Cumulative CG Steps, Truncated PCG Newton Method, Truncated Newton Interior-Point Methods, Network Rate Control, Dual Rate Control Problem, Primal-Dual Search Direction (BV Section 11.7), Truncated Netwon Primal-Dual Algorithm, Primal And Dual Objective Evolution, Relative Duality Gap Evolution, Relative Duality Gap Evolution (N = 10^6), L_1-Norm Methods For Convex-Cardinality Problems, L_1-Norm Heuristics For Cardinality Problems, Cardinality, General Convex-Cardinality Problems, Solving Convex-Cardinality Problems, Boolean LP As Convex-Cardinality Problem, Sparse Design, Sparse Modeling / Regressor Selection, Estimation With Outliers, Minimum Number Of Violations, Linear Classifier With Fewest Errors, Smallest Set Of Mutually Infeasible Inequalities, Portfolio Investment With Linear And Fixed Costs, Piecewise Constant Fitting, Piecewise Linear Fitting, L_1-Norm Heuristic, Example: Minimum Cardinality Problem, Polishing, Regressor Selection Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 13 min Topics: Methods (Truncated Newton Method), Convergence Versus Iterations, Convergence Versus Cumulative CG Steps, Truncated PCG Newton Method, Truncated Newton Interior-Point Methods, Network Rate Control, Dual Rate Control Problem, Primal-Dual Search Direction (BV Section 11.7), Truncated Netwon Primal-Dual Algorithm, Primal And Dual Objective Evolution, Relative Duality Gap Evolution, Relative Duality Gap Evolution (N = 10^6), L_1-Norm Methods For Convex-Cardinality Problems, L_1-Norm Heuristics For Cardinality Problems, Cardinality, General Convex-Cardinality Problems, Solving Convex-Cardinality Problems, Boolean LP As Convex-Cardinality Problem, Sparse Design, Sparse Modeling / Regressor Selection, Estimation With Outliers, Minimum Number Of Violations, Linear Classifier With Fewest Errors, Smallest Set Of Mutually Infeasible Inequalities, Portfolio Investment With Linear And Fixed Costs, Piecewise Constant Fitting, Piecewise Linear Fitting, L_1-Norm Heuristic, Example: Minimum Cardinality Problem, Polishing, Regressor Selection Transcripts HTML PDF Lecture 15 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 3 min Watch Online:Topics: Recap: Example: Minimum Cardinality Problem, Interpretation As Convex Relaxation, Interpretation Via Convex Envelope, Weighted And Asymmetric L_1 Heuristics, Regressor Selection, Sparse Signal Reconstruction, L_1-Norm Methods For Convex-Cardinality Problems Part II, Total Variation Reconstruction, Total Variation Reconstruction, TV Reconstruction, L_2 Reconstruction, Iterated Weighted L_1 Heuristic, Sparse Solution Of Linear Inequalities, Detecting Changes In Time Series Model, Time Series And True Coefficients, TV Heuristic And Iterated TV Heuristic, Extension To Matrices, Factor Modeling, Trace Approximation Results, Summary: L_1-Norm Methods Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 3 min Topics: Recap: Example: Minimum Cardinality Problem, Interpretation As Convex Relaxation, Interpretation Via Convex Envelope, Weighted And Asymmetric L_1 Heuristics, Regressor Selection, Sparse Signal Reconstruction, L_1-Norm Methods For Convex-Cardinality Problems Part II, Total Variation Reconstruction, Total Variation Reconstruction, TV Reconstruction, L_2 Reconstruction, Iterated Weighted L_1 Heuristic, Sparse Solution Of Linear Inequalities, Detecting Changes In Time Series Model, Time Series And True Coefficients, TV Heuristic And Iterated TV Heuristic, Extension To Matrices, Factor Modeling, Trace Approximation Results, Summary: L_1-Norm Methods Transcripts HTML PDF Lecture 16 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 19 min Watch Online:Topics: Model Predictive Control, Linear Time-Invariant Convex Optimal Control, Greedy Control, 'Solution' Via Dynamic Programming, Linear Quadratic Regulator, Finite Horizon Approximation, Cost Versus Horizon, Trajectories, Model Predictive Control (MPC), MPC Performance Versus Horizon, MPC Trajectories, Variations On MPC, Explicit MPC, MPC Problem Structure, Fast MPC, Supply Chain Management, Constraints And Objective, MPC And Optimal Trajectories, Variations On Optimal Control Problem Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 19 min Topics: Model Predictive Control, Linear Time-Invariant Convex Optimal Control, Greedy Control, 'Solution' Via Dynamic Programming, Linear Quadratic Regulator, Finite Horizon Approximation, Cost Versus Horizon, Trajectories, Model Predictive Control (MPC), MPC Performance Versus Horizon, MPC Trajectories, Variations On MPC, Explicit MPC, MPC Problem Structure, Fast MPC, Supply Chain Management, Constraints And Objective, MPC And Optimal Trajectories, Variations On Optimal Control Problem Transcripts HTML PDF Lecture 17 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 17 min Watch Online:Topics: Stochastic Model Predictive Control, Causal State-Feedback Control, Stochastic Finite Horizon Control, 'Solution' Via Dynamic Programming, Independent Process Noise, Linear Quadratic Stochastic Control, Certainty Equivalent Model Predictive Control, Stochastic MPC: Sample Trajectory, Cost Histogram, Simple Lower Bound For Quadratic Stochastic Control, Branch And Bound Methods, Methods For Nonconvex Optimization Problems, Branch And Bound Algorithms, Comment: Example Problem Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 17 min Topics: Stochastic Model Predictive Control, Causal State-Feedback Control, Stochastic Finite Horizon Control, 'Solution' Via Dynamic Programming, Independent Process Noise, Linear Quadratic Stochastic Control, Certainty Equivalent Model Predictive Control, Stochastic MPC: Sample Trajectory, Cost Histogram, Simple Lower Bound For Quadratic Stochastic Control, Branch And Bound Methods, Methods For Nonconvex Optimization Problems, Branch And Bound Algorithms, Comment: Example Problem Transcripts HTML PDF Lecture 18 Watch Online:Watch Now Download: Right Click, and Save AsDownload Duration:1 hr 19 min Watch Online:Topics: Announcements, Recap: Branch And Bound Methods, Basic Idea, Unconstrained, Nonconvex Minimization, Lower And Upper Bound Functions, Branch And Bound Algorithm, Comment: Picture Of Branch And Bound Algorithm In R^2, Comment: Binary Tree, Example, Pruning, Convergence Analysis, Bounding Condition Number, Small Volume Implies Small Size, Mixed Boolean-Convex Problem, Solution Methods, Lower Bound Via Convex Relaxation, Upper Bounds, Branching, New Bounds From Subproblems, Branch And Bound Algorithm (Mixed Boolean-Convex Problem), Minimum Cardinality Example, Bounding X, Relaxation Problem, Algorithm Progress, Global Lower And Upper Bounds, Portion Of Non-Pruned Sparsity Patterns, Number Of Active Leaves In Tree, Global Lower And Upper Bounds, Watch Online:Download: Right Click, and Save As Duration: Watch NowDownload1 hr 19 min Topics: Announcements, Recap: Branch And Bound Methods, Basic Idea, Unconstrained, Nonconvex Minimization, Lower And Upper Bound Functions, Branch And Bound Algorithm, Comment: Picture Of Branch And Bound Algorithm In R^2, Comment: Binary Tree, Example, Pruning, Convergence Analysis, Bounding Condition Number, Small Volume Implies Small Size, Mixed Boolean-Convex Problem, Solution Methods, Lower Bound Via Convex Relaxation, Upper Bounds, Branching, New Bounds From Subproblems, Branch And Bound Algorithm (Mixed Boolean-Convex Problem), Minimum Cardinality Example, Bounding X, Relaxation Problem, Algorithm Progress, Global Lower And Upper Bounds, Portion Of Non-Pruned Sparsity Patterns, Number Of 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187838
https://en.wikipedia.org/wiki/Reaction_rate_constant
Jump to content Search Contents (Top) 1 Elementary steps 2 Relationship to other parameters 3 Dependence on temperature 3.1 Comparison of models 4 Units 5 Plasma and gases 6 Rate constant calculations 7 Divided saddle theory 8 See also 9 References Reaction rate constant العربية Català Čeština Deutsch Eesti Español فارسی Français 한국어 Bahasa Indonesia Italiano עברית Magyar Nederlands 日本語 Norsk bokmål Polski Romnă Русский Српски / srpski Srpskohrvatski / српскохрватски Suomi தமிழ் Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Coefficient of rate of a chemical reaction In chemical kinetics, a reaction rate constant or reaction rate coefficient (⁠⁠) is a proportionality constant which quantifies the rate and direction of a chemical reaction by relating it with the concentration of reactants. For a reaction between reactants A and B to form a product C, a A + b B → c C where : A and B are reactants : C is a product : a, b, and c are stoichiometric coefficients, the reaction rate is often found to have the form: Here ⁠⁠ is the reaction rate constant that depends on temperature, and [A] and [B] are the molar concentrations of substances A and B in moles per unit volume of solution, assuming the reaction is taking place throughout the volume of the solution. (For a reaction taking place at a boundary, one would use moles of A or B per unit area instead.) The exponents m and n are called partial orders of reaction and are not generally equal to the stoichiometric coefficients a and b. Instead they depend on the reaction mechanism and can be determined experimentally. Sum of m and n, that is, (m + n) is called the overall order of reaction. Elementary steps [edit] For an elementary step, there is a relationship between stoichiometry and rate law, as determined by the law of mass action. Almost all elementary steps are either unimolecular or bimolecular. For a unimolecular step A → P the reaction rate is described by , where is a unimolecular rate constant. Since a reaction requires a change in molecular geometry, unimolecular rate constants cannot be larger than the frequency of a molecular vibration. Thus, in general, a unimolecular rate constant has an upper limit of k1 ≤ ~1013 s−1. For a bimolecular step A + B → P the reaction rate is described by , where is a bimolecular rate constant. Bimolecular rate constants have an upper limit that is determined by how frequently molecules can collide, and the fastest such processes are limited by diffusion. Thus, in general, a bimolecular rate constant has an upper limit of k2 ≤ ~1010 M−1s−1. For a termolecular step A + B + C → P the reaction rate is described by , where is a termolecular rate constant. There are few examples of elementary steps that are termolecular or higher order, due to the low probability of three or more molecules colliding in their reactive conformations and in the right orientation relative to each other to reach a particular transition state. There are, however, some termolecular examples in the gas phase. Most involve the recombination of two atoms or small radicals or molecules in the presence of an inert third body which carries off excess energy, such as O + O2 + N2 → O3 + N2. One well-established example is the termolecular step 2 I + H2 → 2 HI in the hydrogen-iodine reaction. In cases where a termolecular step might plausibly be proposed, one of the reactants is generally present in high concentration (e.g., as a solvent or diluent gas). Relationship to other parameters [edit] For a first-order reaction (including a unimolecular one-step process), there is a direct relationship between the unimolecular rate constant and the half-life of the reaction: . Transition state theory gives a relationship between the rate constant and the Gibbs free energy of activation , a quantity that can be regarded as the free energy change needed to reach the transition state. In particular, this energy barrier incorporates both enthalpic () and entropic () changes that need to be achieved for the reaction to take place: The result from transition state theory is , where h is the Planck constant and R the molar gas constant. As useful rules of thumb, a first-order reaction with a rate constant of 10−4 s−1 will have a half-life (t1/2) of approximately 2 hours. For a one-step process taking place at room temperature, the corresponding Gibbs free energy of activation (ΔG‡) is approximately 23 kcal/mol. Dependence on temperature [edit] The Arrhenius equation is an elementary treatment that gives the quantitative basis of the relationship between the activation energy and the reaction rate at which a reaction proceeds. The rate constant as a function of thermodynamic temperature is then given by: The reaction rate is given by: where Ea is the activation energy, and R is the gas constant, and m and n are experimentally determined partial orders in [A] and [B], respectively. Since at temperature T the molecules have energies according to a Boltzmann distribution, one can expect the proportion of collisions with energy greater than Ea to vary with e−Ea⁄RT. The constant of proportionality A is the pre-exponential factor, or frequency factor (not to be confused here with the reactant A) takes into consideration the frequency at which reactant molecules are colliding and the likelihood that a collision leads to a successful reaction. Here, A has the same dimensions as an (m + n)-order rate constant (see Units below). Another popular model that is derived using more sophisticated statistical mechanical considerations is the Eyring equation from transition state theory: where ΔG‡ is the free energy of activation, a parameter that incorporates both the enthalpy and entropy change needed to reach the transition state. The temperature dependence of ΔG‡ is used to compute these parameters, the enthalpy of activation ΔH‡ and the entropy of activation ΔS‡, based on the defining formula ΔG‡ = ΔH‡ − TΔS‡. In effect, the free energy of activation takes into account both the activation energy and the likelihood of successful collision, while the factor kBT/h gives the frequency of molecular collision. The factor (c⊖)1-M ensures the dimensional correctness of the rate constant when the transition state in question is bimolecular or higher. Here, c⊖ is the standard concentration, generally chosen based on the unit of concentration used (usually c⊖ = 1 mol L−1 = 1 M), and M is the molecularity of the transition state. Lastly, κ, usually set to unity, is known as the transmission coefficient, a parameter which essentially serves as a "fudge factor" for transition state theory. The biggest difference between the two theories is that Arrhenius theory attempts to model the reaction (single- or multi-step) as a whole, while transition state theory models the individual elementary steps involved. Thus, they are not directly comparable, unless the reaction in question involves only a single elementary step. Finally, in the past, collision theory, in which reactants are viewed as hard spheres with a particular cross-section, provided yet another common way to rationalize and model the temperature dependence of the rate constant, although this approach has gradually fallen into disuse. The equation for the rate constant is similar in functional form to both the Arrhenius and Eyring equations: where P is the steric (or probability) factor and Z is the collision frequency, and ΔE is energy input required to overcome the activation barrier. Of note, , making the temperature dependence of k different from both the Arrhenius and Eyring models. Comparison of models [edit] All three theories model the temperature dependence of k using an equation of the form for some constant C, where α = 0, 1⁄2, and 1 give Arrhenius theory, collision theory, and transition state theory, respectively, although the imprecise notion of ΔE, the energy needed to overcome the activation barrier, has a slightly different meaning in each theory. In practice, experimental data does not generally allow a determination to be made as to which is "correct" in terms of best fit. Hence, all three are conceptual frameworks that make numerous assumptions, both realistic and unrealistic, in their derivations. As a result, they are capable of providing different insights into a system. Units [edit] The units of the rate constant depend on the overall order of reaction. If concentration is measured in units of mol·L−1 (sometimes abbreviated as M), then For order (m + n), the rate constant has units of mol1−(m+n)·L(m+n)−1·s−1 (or M1−(m+n)·s−1) For order zero, the rate constant has units of mol·L−1·s−1 (or M·s−1) For order one, the rate constant has units of s−1 For order two, the rate constant has units of L·mol−1·s−1 (or M−1·s−1) For order three, the rate constant has units of L2·mol−2·s−1 (or M−2·s−1) For order four, the rate constant has units of L3·mol−3·s−1 (or M−3·s−1) Plasma and gases [edit] Calculation of rate constants of the processes of generation and relaxation of electronically and vibrationally excited particles are of significant importance. It is used, for example, in the computer simulation of processes in plasma chemistry or microelectronics. First-principle based models should be used for such calculation. It can be done with the help of computer simulation software. Rate constant calculations [edit] Rate constant can be calculated for elementary reactions by molecular dynamics simulations. One possible approach is to calculate the mean residence time of the molecule in the reactant state. Although this is feasible for small systems with short residence times, this approach is not widely applicable as reactions are often rare events on molecular scale. One simple approach to overcome this problem is Divided Saddle Theory. Such other methods as the Bennett Chandler procedure, and Milestoning have also been developed for rate constant calculations. Divided saddle theory [edit] The theory is based on the assumption that the reaction can be described by a reaction coordinate, and that we can apply Boltzmann distribution at least in the reactant state. A new, especially reactive segment of the reactant, called the saddle domain, is introduced, and the rate constant is factored: where αSDRS is the conversion factor between the reactant state and saddle domain, while kSD is the rate constant from the saddle domain. The first can be simply calculated from the free energy surface, the latter is easily accessible from short molecular dynamics simulations See also [edit] Reaction rate Equilibrium constant Molecularity References [edit] ^ "Chemical Kinetics Notes". www.chem.arizona.edu. Archived from the original on 31 March 2012. Retrieved 5 May 2018. ^ Lowry, Thomas H. (1987). Mechanism and theory in organic chemistry. Richardson, Kathleen Schueller (3rd ed.). New York: Harper & Row. ISBN 978-0060440848. OCLC 14214254. ^ Moore, John W.; Pearson, Ralph G. (1981). Kinetics and Mechanism (3rd ed.). John Wiley. pp. 226–7. ISBN 978-0-471-03558-9. ^ The reactions of nitric oxide with the diatomic molecules Cl2, Br2 or O2 (e.g., 2 NO + Cl2 → 2 NOCl, etc.) have also been suggested as examples of termolecular elementary processes. However, other authors favor a two-step process, each of which is bimolecular: (NO + Cl2 ⇄ NOCl2, NOCl2 + NO → 2 NOCl). See: Compton, R.G.; Bamford, C. H.; Tipper, C.F.H., eds. (2014) . "5. Reactions of the Oxides of Nitrogen §5.5 Reactions with Chlorine". Reactions of Non-metallic Inorganic Compounds. Comprehensive Chemical Kinetics. Vol. 6. Elsevier. p. 174. ISBN 978-0-08-086801-1. ^ Sullivan, John H. (1967-01-01). "Mechanism of the Bimolecular Hydrogen—Iodine Reaction". The Journal of Chemical Physics. 46 (1): 73–78. Bibcode:1967JChPh..46...73S. doi:10.1063/1.1840433. ISSN 0021-9606. ^ Kotz, John C. (2009). Chemistry & chemical reactivity. Treichel, Paul., Townsend, John R. (7th ed.). Belmont, Calif.: Thomson Brooks/ Cole. p. 703. ISBN 9780495387039. OCLC 220756597. ^ Laidler, Keith J. (1987). Chemical Kinetics (3rd ed.). Harper & Row. p. 113. ISBN 0-06-043862-2. ^ Steinfeld, Jeffrey I.; Francisco, Joseph S.; Hase, William L. (1999). Chemical Kinetics and Dynamics (2nd ed.). Prentice Hall. p. 301. ISBN 0-13-737123-3. ^ Carpenter, Barry K. (1984). Determination of organic reaction mechanisms. New York: Wiley. ISBN 978-0471893691. OCLC 9894996. ^ Blauch, David. "Differential Rate Laws". Chemical Kinetics. ^ a b Daru, János; Stirling, András (2014). "Divided Saddle Theory: A New Idea for Rate Constant Calculation" (PDF). J. Chem. Theory Comput. 10 (3): 1121–1127. doi:10.1021/ct400970y. PMID 26580187. ^ Chandler, David (1978). "Statistical mechanics of isomerization dynamics in liquids and the transition state approximation". J. Chem. Phys. 68 (6): 2959–2970. Bibcode:1978JChPh..68.2959C. doi:10.1063/1.436049. ^ Bennett, C. H. (1977). Christofferson, R. (ed.). Algorithms for Chemical Computations, ACS Symposium Series No. 46. Washington, D.C.: American Chemical Society. ISBN 978-0-8412-0371-6. ^ West, Anthony M.A.; Elber, Ron; Shalloway, David (2007). "Extending molecular dynamics time scales with milestoning: Example of complex kinetics in a solvated peptide". The Journal of Chemical Physics. 126 (14): 145104. Bibcode:2007JChPh.126n5104W. doi:10.1063/1.2716389. PMID 17444753. Retrieved from " Category: Chemical kinetics Hidden categories: Articles with short description Short description is different from Wikidata Reaction rate constant Add topic
187839
https://en.wikipedia.org/wiki/Direction_cosine
Jump to content Direction cosine বাংলা Català Deutsch Español فارسی Français Italiano Қазақша Norsk nynorsk Polski Romnă Svenska தமிழ் Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Cosines of the angles between a vector and the coordinate axes | | | --- | | | This article includes a list of references, related reading, or external links, but its sources remain unclear because it lacks inline citations. Please help improve this article by introducing more precise citations. (January 2017) (Learn how and when to remove this message) | In analytic geometry, the direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three positive coordinate axes. Equivalently, they are the contributions of each component of the basis to a unit vector in that direction. Three-dimensional Cartesian coordinates [edit] Further information: Cartesian coordinates If v is a Euclidean vector in three-dimensional Euclidean space, ⁠⁠ where ex, ey, ez are the standard basis in Cartesian notation, then the direction cosines are It follows that by squaring each equation and adding the results Here α, β, γ are the direction cosines and the Cartesian coordinates of the unit vector and a, b, c are the direction angles of the vector v. The direction angles a, b, c are acute or obtuse angles, i.e., 0 ≤ a ≤ π, 0 ≤ b ≤ π and 0 ≤ c ≤ π, and they denote the angles formed between v and the unit basis vectors ex, ey, ez. General meaning [edit] More generally, direction cosine refers to the cosine of the angle between any two vectors. They are useful for forming direction cosine matrices that express one set of orthonormal basis vectors in terms of another set, or for expressing a known vector in a different basis. Simply put, direction cosines provide an easy method of representing the direction of a vector in a Cartesian coordinate system. Applications [edit] Determining angles between two vectors [edit] If vectors u and v have direction cosines (αu, βu, γu) and (αv, βv, γv) respectively, with an angle θ between them, their unit vectors are Taking the dot product of these two unit vectors yield, where θ is the angle between the two unit vectors, and is also the angle between u and v. Since θ is a geometric angle, and is never negative. Therefore only the positive value of the dot product is taken, yielding us the final result, See also [edit] Cartesian tensor Euler angles References [edit] Kay, D. C. (1988). Tensor Calculus. Schaum’s Outlines. McGraw Hill. pp. 18–19. ISBN 0-07-033484-6. Spiegel, M. R.; Lipschutz, S.; Spellman, D. (2009). Vector analysis. Schaum’s Outlines (2nd ed.). McGraw Hill. pp. 15, 25. ISBN 978-0-07-161545-7. Tyldesley, J. R. (1975). An introduction to tensor analysis for engineers and applied scientists. Longman. p. 5. ISBN 0-582-44355-5. Tang, K. T. (2006). Mathematical Methods for Engineers and Scientists. Vol. 2. Springer. p. 13. ISBN 3-540-30268-9. Weisstein, Eric W. "Direction Cosine". MathWorld. Retrieved from " Categories: Algebraic geometry Vectors (mathematics and physics) Geometry stubs Hidden categories: Articles with short description Short description is different from Wikidata Articles lacking in-text citations from January 2017 All articles lacking in-text citations All stub articles
187840
https://www.onlinemathlearning.com/compare-unit-fractions-worksheet.html
Compare Unit Fractions Worksheets (answers, printable, online, grade 3) OnlineMathLearning.com - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. Personal Data Processing Opt Outs CONFIRM × Compare Unit Fractions Worksheets Related Pages Math Worksheets Lessons for Third Grade Free Printable Worksheets Printable Math Worksheets Discover more Mathematics math Math College prep math Assessments Calculators graphing scientific Video math lessons Math puzzles games Math problem solving Tutoring services Share this page to Google Classroom Printable “Fraction” worksheets: Equal Parts Introduction to Fractions Compare Unit Fractions Compare Fractions with Same Numerator Fractions on the Number Line Compare Fractions on the Number Line Compare Fractions Order Fractions Printable Math Worksheets Compare Unit Fractions Worksheets In these free math worksheets, you will learn how to represent and compare unit fractions. How to compare unit fractions? Comparing unit fractions involves determining which of two or more unit fractions is greater or smaller. A unit fraction is a fraction where the numerator (top number) is 1. We can use visual aids such as fraction strips, fraction bars, circles, or number lines to represent the unit fractions. For example, we will be able to see that 1/3 is greater than 1/4. Imagine a pizza cut into 3 slices and another same size pizza cut into 4 slices. We can visualize that the slice of pizza that was cut into 3 parts would be bigger than the slice of pizza that was cut into 4 parts. The fraction with the larger denominator has smaller pieces, so it represents a smaller fraction of the whole. In this case, 1/4 is smaller than 1/3. Printable Math Worksheets Algebra Textbooks 2. We can also compare the denominators of the unit fractions. A smaller denominator means larger pieces, indicating a bigger fraction. For example, 1/4 is greater than 1/6. Have a look at this video if you need to learn how to compare unit fractions. Compare Unit Fractions Click on the following worksheet to get a printable pdf document. Scroll down the page for more Compare Unit Fractions Worksheets. Learning Software Printable Math Worksheets More Compare Unit Fractions Worksheets Printable (Answers on the second page.) Compare Unit Fractions Worksheet #1 Compare Unit Fractions Worksheet #2 Compare Unit Fractions Worksheet #3 Compare Unit Fractions Worksheet #4 Related Lessons & Worksheets Compare Unit Fractions Equivalent Fractions Reduce Proper Fractions Simplify Proper & Improper Fractions Improper Fractions to Mixed Numbers Mixed Numbers to Improper Fractions Printable Math Worksheets More Printable Worksheets Discover more Mathematics Math math School supplies Assessments Purchase GRE preparation materials Purchase physics textbook Classroom Tools Math Lesson Plans Printable Math Worksheets Discover more Math math Mathematics Educational assessment Education Learning Software Buy TOEFL test prep Purchase calculus online course Math Lesson Plans Educational Games Try out our new and fun Fraction Concoction Game. Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level. × We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. × Back to Top | Interactive Zone | Home Copyright © 2005, 2025 - OnlineMathLearning.com. 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187841
https://www.elephango.com/index.cfm/pg/k12learning/lcid/10961/Prime_Factorization
Home Tour Elephango for Families Elephango for Schools Discover Pricing Standards Search About Login Sign Up Family Sign-Up School Account Sign Up Student Parent Educator -1) { var s = source.indexOf("", e); // Add to scripts array scripts.push(source.substring(s_e+1, e)); // Strip from source source = source.substring(0, s) + source.substring(e_e+1); } // Loop through every script collected and eval it for(var i=0; i FAVORITE Like Prime Factorization Lesson ID: 10961 How do prime numbers "factor" into your life? Numbers have unique fingerprints, and you will learn about them by watching a video, playing online games, online practice, and, of course, your computer! categories Pre-Algebra, Whole Numbers and Operations subject Math learning style Visual personality style Lion Grade Level Middle School (6-8) Lesson Type Skill Sharpener Lesson Plan - Get It! Audio: Did you know that, just like you have a unique fingerprint, numbers also have a unique fingerprint? It is called their prime factorization. Other than zero (0) and one (1), all numbers have a prime factorization. Prime factorization is when numbers are broken down into factors of prime numbers. A prime number is any number that cannot be divided down into smaller numbers. If you need to review prime numbers, check out our lesson found under Additional Resources in the right-hand sidebar. Look at this example of prime factorization: Next, watch this great Math Antics - Prime Factorization video: What is the definition of prime factorization? If you said, "The set of prime numbers multiplied together to get another number," you are correct! If you said, "The action of finding the set of prime numbers multiplied to get a number," you are also correct! Keep going in the Got It? section! Resources and Extras Supplies printer paper scissors Resources Referenced in the Lesson Factors of Numbers Game Additional Resources Prime Suspects Suggested Lessons Skip Counting by Sixes Math | Intermediate (3-5) Solving One-Step Equation Mysteries Math | Middle School (6-8) Greater Than and Less Than Math | PreK/K, Primary (K-2) Mr. D Math - The Distributive Property in Action! Math | Middle School (6-8) Facebook Twitter Pinterest Email Copy Link By Subject College & Career English / Language Arts Fine Arts Geography Government History Life Skills Math Reading Science Social Studies Teaching Tips & Tools Technology By Grade PreK/K Primary (K-2) Intermediate (3-5) Middle School (6-8) High School (9-12) Adult Learning Parent Resources By Time Less than 30 minutes 30 minutes to an hour 1-2 hours 3-4 hours 5+ hours By Content Type Interactive Video By Lesson Type Dig Deeper Quick Query Skill Sharpener Auditory Kinesthetic Visual Beaver Golden Retriever Lion Otter Elephango © 2025 | Contact Us | About | Terms & Conditions | Privacy Policy | Acceptable Use Policy
187842
https://mathoverflow.net/questions/30156/demystifying-complex-numbers
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Demystifying complex numbers Ask Question Asked Modified 1 year, 4 months ago Viewed 22k times $\begingroup$ At the end of this month I start teaching complex analysis to 2nd year undergraduates, mostly from engineering but some from science and maths. The main applications for them in future studies are contour integrals and Laplace transform, but this should be a "real" complex analysis course which I could later refer to in honours courses. I am now confident (after this discussion, especially Gauss€™s complaints given in Keith€™s comment) that the name "complex" is discouraging to average students. Why do we need to study numbers which do not belong to the real world? We all know that the thesis is wrong and I have in mind some examples where the use of complex variable functions simplify solving considerably (I give two below). The drawback is that all them assume some knowledge from students already. So I would be happy to learn elementary examples which may convince students that complex numbers and functions of a complex variable are useful. As this question runs in the community wiki mode, I would be glad to see one example per answer. Thank you in advance! Here are the two promised examples. I was reminded of the second one by several answers and comments about trigonometric functions (and also by the notification that "the bounty on your question Trigonometry related to Rogers€“Ramanujan identities expires within three days"; it seems to be harder than I expected). Example 1. What is the Fourier expansion of the (unbounded) periodic function $$ f(x)=\ln\Bigl\lvert\sin\frac x2\Bigr\rvert\ ? $$ Solution. The function $f(x)$ is periodic with period $2\pi$ and has poles at the points $2\pi k$, $k\in\mathbb Z$. Consider the function on the interval $x\in[\varepsilon,2\pi-\varepsilon]$. The series $$ \sum_{n=1}^\infty\frac{z^n}n, \qquad z=e^{ix}, $$ converges for all values $x$ from the interval. Since $$ \Bigl\lvert\sin\frac x2\Bigr\rvert=\sqrt{\frac{1-\cos x}2} $$ and $\operatorname{Re}\ln w=\ln\lvert w\rvert$, where we choose $w=\frac12(1-z)$, we deduce that $$ \operatorname{Re}\Bigl(\ln\frac{1-z}2\Bigr)=\ln\sqrt{\frac{1-\cos x}2} =\ln\Bigl\lvert\sin\frac x2\Bigr\rvert. $$ Thus, $$ \ln\Bigl\lvert\sin\frac x2\Bigr\rvert =-\ln2-\operatorname{Re}\sum_{n=1}^\infty\frac{z^n}n =-\ln2-\sum_{n=1}^\infty\frac{\cos nx}n. $$ As $\varepsilon>0$ can be taken arbitrarily small, the result remains valid for all $x\ne2\pi k$. Example 2. Let $p$ be an odd prime number. $\newcommand\Legendre{\genfrac(){}{}}$For an integer $a$ relatively prime to $p$, the Legendre symbol $\Legendre ap$ is $+1$ or $-1$ depending on whether the congruence $x^2\equiv a\pmod{p}$ is solvable or not. Using the elementary result (a consequence of Fermat's little theorem) that $$ \Legendre ap \equiv a^{(p-1)/2}\pmod p, \tag{}\label{star} $$ show that $$ \Legendre 2p=(-1)^{(p^2-1)/8}. $$ Solution. In the ring $\mathbb Z+\mathbb Zi=\Bbb Z[i]$, the binomial formula implies $$ (1+i)^p\equiv1+i^p\pmod p. $$ On the other hand, $$ (1+i)^p =\bigl(\sqrt2e^{\pi i/4}\bigr)^p =2^{p/2}\biggl(\cos\frac{\pi p}4+i\sin\frac{\pi p}4\biggr) $$ and $$ 1+i^p =1+(e^{\pi i/2})^p =1+\cos\frac{\pi p}2+i\sin\frac{\pi p}2 =1+i\sin\frac{\pi p}2. $$ Comparing the real parts implies that $$ 2^{p/2}\cos\frac{\pi p}4\equiv1\pmod p, $$ hence from $\sqrt2\cos(\pi p/4)\in{\pm1}$ we conclude that $$ 2^{(p-1)/2}\equiv\sqrt2\cos\frac{\pi p}4\pmod p. $$ Then using the elementary result \eqref{star}: $$ \Legendre2p \equiv2^{(p-1)/2} \equiv\sqrt2\cos\frac{\pi p}4 =\begin{cases} 1 & \text{if } p\equiv\pm1\pmod8, \cr -1 & \text{if } p\equiv\pm3\pmod8, \end{cases} $$ which is exactly the required formula. soft-question cv.complex-variables teaching Improve this question edited Jan 4, 2022 at 18:50 community wiki 8 revs, 5 users 62%Wadim Zudilin $\endgroup$ 14 $\begingroup$ Maybe an option is to have them understand that real numbers also do not belong to the real world, that all sort of numbers are simply abstractions. $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2010-07-01 14:50:45 +00:00 Commented Jul 1, 2010 at 14:50 $\begingroup$ Probably your electrical engineering students understand better than you do that complex numbers (in polar form) are used to represent amplitude and frequency in their area of study. $\endgroup$ Gerald Edgar – Gerald Edgar 2010-07-01 15:36:55 +00:00 Commented Jul 1, 2010 at 15:36 $\begingroup$ Not an answer, but some suggestions: try reading the beginning of Needham's Visual Complex Analysis (usf.usfca.edu/vca/) and the end of Levi's The Mathematical Mechanic (amazon.com/Mathematical-Mechanic-Physical-Reasoning-Problems/dp/…). $\endgroup$ Qiaochu Yuan – Qiaochu Yuan 2010-07-01 17:05:29 +00:00 Commented Jul 1, 2010 at 17:05 $\begingroup$ Your example has a hidden assumption that a student actually admits the importance of calculating F.S. of $\ln\left|\sin{x\over 2}\right|$, which I find dubious. The examples with an oscillator's ODE is more convincing, IMO. $\endgroup$ Paul Yuryev – Paul Yuryev 2010-07-02 03:02:50 +00:00 Commented Jul 2, 2010 at 3:02 $\begingroup$ @Mariano, Gerald and Qiaochu: Thanks for the ideas! Visual Complex Analysis sounds indeed great, and I'll follow Levi's book as soon as I reach the uni library. @Paul: I give the example (which I personally like) and explain that I do not consider it elementary enough for the students. It's a matter of taste! I've never used Fourier series in my own research but it doesn't imply that I doubt of their importance. We all (including students) have different criteria for measuring such things. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 05:06:23 +00:00 Commented Jul 2, 2010 at 5:06 | Show 9 more comments 44 Answers 44 Reset to default 2 Next $\begingroup$ The nicest elementary illustration I know of the relevance of complex numbers to calculus is its link to radius of convergence, which student learn how to compute by various tests, but more mechanically than conceptually. The series for $1/(1-x)$, $\log(1+x)$, and $\sqrt{1+x}$ have radius of convergence 1 and we can see why: there's a problem at one of the endpoints of the interval of convergence (the function blows up or it's not differentiable). However, the function $1/(1+x^2)$ is nice and smooth on the whole real line with no apparent problems, but its radius of convergence at the origin is 1. From the viewpoint of real analysis this is strange: why does the series stop converging? Well, if you look at distance 1 in the complex plane... More generally, you can tell them that for any rational function $p(x)/q(x)$, in reduced form, the radius of convergence of this function at a number $a$ (on the real line) is precisely the distance from $a$ to the nearest zero of the denominator, even if that nearest zero is not real. In other words, to really understand the radius of convergence in a general sense you have to work over the complex numbers. (Yes, there are subtle distinctions between smoothness and analyticity which are relevant here, but you don't have to discuss that to get across the idea.) Similarly, the function $x/(e^x-1)$ on the real line is smooth, but its power series at $x = 0$ has finite radius of convergence $2\pi$ (not sure if you can make this numerically apparent). Again, on the real line the reason for this is not visible, but in the complex plane there is a good explanation. If someone is not happy about that function looking initially problematic at $x = 0$, where its value is $1$, use $x/(e^x+1)$ instead and the radius of convergence of its power series at $x = 0$ has radius of convergence $\pi$ rather than $2\pi$. Share Improve this answer edited May 4, 2024 at 23:48 community wiki 2 revsKConrad $\endgroup$ 5 1 $\begingroup$ Thanks, Keith! That's a nice point which I always mention for real analysis students as well. The structure of singularities of a linear differential equation (under some mild conditions) fully determines the convergence of the series solving the DE. The generating series for Bernoulli numbers does not produce sufficiently good approximations to $2\pi$, but it's just beautiful by itself. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 05:14:35 +00:00 Commented Jul 2, 2010 at 5:14 $\begingroup$ Ī̲ wouldn€™t say it demystifies anything. $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 04:24:23 +00:00 Commented Apr 14, 2021 at 4:24 3 $\begingroup$ @IncnisMrsi my answer was posted almost 11 years ago, so I had to remind myself what the OP's original request was. It was not about the title, but about the following, taken from the OP's post: "to learn elementary examples which may convince students in usefulness of complex numbers and functions in complex variable." Судя по этой просьбе, я думаю, что все в порядке $\endgroup$ KConrad – KConrad 2021-04-14 05:18:39 +00:00 Commented Apr 14, 2021 at 5:18 $\begingroup$ My guess is that the main example you cite is the sort of thing that is actually appealing to the already-convinced, i.e. mathematicians who are reading this question, and yet doesn't necessarily appeal to someone who feels strong and comfortable with some real analysis but doesn't want to do complex analysis. i.e. it is not at all "strange" from the point of view of real analysis as to why the series doesn't converge: The tests that a student has learned for convergence of basic series will show it easily. $\endgroup$ SBK – SBK 2024-06-11 14:41:34 +00:00 Commented Jun 11, 2024 at 14:41 1 $\begingroup$ @SBK I of course am already convinced, but I still think an appropriately thoughtful student might wonder why the real analysis convergence tests are saying the radius of convergence is a value $R$ when nothing appears to be going wrong at $x = \pm R$ (a contrast to the series for $1/(1-x)$ and $\log(1+x)$ at $x = 0$, where $R = 1$). $\endgroup$ KConrad – KConrad 2024-06-11 23:57:12 +00:00 Commented Jun 11, 2024 at 23:57 Add a comment | 71 $\begingroup$ You can solve the differential equation $y''+y=0$ using complex numbers. Just write $$(\partial^2 + 1) y = (\partial +i)(\partial -i) y$$ and you are now dealing with two order one differential equations that are easily solved $$(\partial +i) z =0,\qquad (\partial -i)y =z.$$ The multivariate case is a bit harder and uses quaternions or Clifford algebras. This was done by Dirac for the Schrodinger equation ($-\Delta \psi = i\partial_t \psi$), and that led him to the prediction of the existence of antiparticles (and to the Nobel prize). Share Improve this answer edited May 5, 2024 at 4:33 community wiki 2 revs, 2 users 83%coudy $\endgroup$ 5 $\begingroup$ Thanks, coudy! This is really nice (Pietro's answer only hints a possible use of 2nd order DEs). $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 12:24:00 +00:00 Commented Jul 1, 2010 at 12:24 2 $\begingroup$ I was taught a similar trick as a second-year honours undergraduate (same target that the OP suggests), and I don't find it helpful at all. The problem is that you'll probably need to spend a full lecture trying to formalize it and explain why it works, otherwise it just looks like a magical unmotivated trick that abuses notation. $\endgroup$ Federico Poloni – Federico Poloni 2014-01-27 11:30:51 +00:00 Commented Jan 27, 2014 at 11:30 1 $\begingroup$ @Frederico: why do you say it abuses notation? That would be true in higher dimensional case without introducing Clifford algebras. But here it's perfectly valid. You just need to know what complex numbers are. $\endgroup$ Marek – Marek 2014-01-27 15:09:25 +00:00 Commented Jan 27, 2014 at 15:09 1 $\begingroup$ How is Dirac theory based on (parabolic) Schrödinger equation? We know that a solution to Dirac equation is a solution to Klein€“Gordon equation which is hyperbolic. $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 04:29:30 +00:00 Commented Apr 14, 2021 at 4:29 11 $\begingroup$ @Marek I see only now this old comment but I'll answer anyway just in case. The 'abuse of notation' is that it at that point students do not know that $\partial$ belongs to an operator algebra and that it makes sense to factor out $\partial^2 y +y = (\partial^2 +1)y$, let alone decompose it as $(\partial + i)(\partial -i)$. The fact that the derivative notation $\partial^2 y$ reads like a square followed by a product by $y$ at that point is just notation, so this seems to belong to the same category as simplifying out $dx$ to change variables in integrals. $\endgroup$ Federico Poloni – Federico Poloni 2021-04-14 06:43:56 +00:00 Commented Apr 14, 2021 at 6:43 Add a comment | 63 $\begingroup$ Students usually find the connection of trigonometric identities like $\sin(a+b)=\sin a\cos b+\cos a\sin b$ to multiplication of complex numbers striking. Share Improve this answer edited Jul 1, 2010 at 14:12 community wiki 2 revisions $\endgroup$ 6 25 $\begingroup$ Not sure about the students, but I do. :-) $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 12:21:09 +00:00 Commented Jul 1, 2010 at 12:21 $\begingroup$ Well, I am not claiming this about all students, but whenever I mentioned this (or same things in terms of rotation matrices) in class, I always get excited feedback at least from some of them. $\endgroup$ Yuri Bakhtin – Yuri Bakhtin 2010-07-01 12:24:53 +00:00 Commented Jul 1, 2010 at 12:24 10 $\begingroup$ This is an excellent suggestion. I can never remember these identities off the top of my head. Whenever I need one of them, the simplest way (faster than googling) is to read them off from $(a+ib)(c+id)=(ac-bd) + i(ad+bc)$. $\endgroup$ alex – alex 2010-07-01 20:35:54 +00:00 Commented Jul 1, 2010 at 20:35 15 $\begingroup$ When I first started teaching calculus in the US, I was surprised that many students didn't remember addition formulas for trig functions. As the years went by, it's gotten worse: now the whole idea of using an identity like that to solve a problem is alien to them, e.g. even if they may look it up doing the homework, they "get stuck" on the problem and "don't get it". What is there to blame: calculators? standard tests that neglect it? teachers who never understood it themselves? Anyway, it's a very bad omen. $\endgroup$ Victor Protsak – Victor Protsak 2010-07-02 01:43:42 +00:00 Commented Jul 2, 2010 at 1:43 8 $\begingroup$ @Victor: It can be worse... When I taught Calc I at U of Toronto to engineering students, I was approached by some students who claimed they had heard words "sine" and "cosine" but were not quite sure what they meant. $\endgroup$ Yuri Bakhtin – Yuri Bakhtin 2010-07-02 08:51:34 +00:00 Commented Jul 2, 2010 at 8:51 | Show 1 more comment 45 $\begingroup$ From "Birds and Frogs" by Freeman Dyson [Notices of Amer. Math. Soc. 56 (2009) 212--223]: One of the most profound jokes of nature is the square root of minus one that the physicist Erwin Schrödinger put into his wave equation when he invented wave mechanics in 1926. Schrödinger was a bird who started from the idea of unifying mechanics with optics. A hundred years earlier, Hamilton had unified classical mechanics with ray optics, using the same mathematics to describe optical rays and classical particle trajectories. Schrödinger€™s idea was to extend this unification to wave optics and wave mechanics. Wave optics already existed, but wave mechanics did not. Schrödinger had to invent wave mechanics to complete the unification. Starting from wave optics as a model, he wrote down a differential equation for a mechanical particle, but the equation made no sense. The equation looked like the equation of conduction of heat in a continuous medium. Heat conduction has no visible relevance to particle mechanics. Schrödinger€™s idea seemed to be going nowhere. But then came the surprise. Schrödinger put the square root of minus one into the equation, and suddenly it made sense. Suddenly it became a wave equation instead of a heat conduction equation. And Schrödinger found to his delight that the equation has solutions corresponding to the quantized orbits in the Bohr model of the atom. It turns out that the Schrödinger equation describes correctly everything we know about the behavior of atoms. It is the basis of all of chemistry and most of physics. And that square root of minus one means that nature works with complex numbers and not with real numbers. This discovery came as a complete surprise, to Schrödinger as well as to everybody else. According to Schrödinger, his fourteen-year-old girl friend Itha Junger said to him at the time, "Hey, you never even thought when you began that so much sensible stuff would come out of it." All through the nineteenth century, mathematicians from Abel to Riemann and Weierstrass had been creating a magnificent theory of functions of complex variables. They had discovered that the theory of functions became far deeper and more powerful when it was extended from real to complex numbers. But they always thought of complex numbers as an artificial construction, invented by human mathematicians as a useful and elegant abstraction from real life. It never entered their heads that this artificial number system that they had invented was in fact the ground on which atoms move. They never imagined that nature had got there first. Share Improve this answer answered Jul 9, 2010 at 12:12 community wiki Wadim Zudilin $\endgroup$ 1 3 $\begingroup$ As discussed at physics.stackexchange.com/questions/428033/… putting in an imaginary diffusion coefficient into the heat equation changes the equation to become invariant against time reversal $t \rightarrow -t$ like a wave equation. This is what Dyson means with "it suddenly made sense". $\endgroup$ user4503 – user4503 2018-10-07 19:57:12 +00:00 Commented Oct 7, 2018 at 19:57 Add a comment | 33 $\begingroup$ If the students have had a first course in differential equations, tell them to solve the system $$x'(t) = -y(t)$$ $$y'(t) = x(t).$$ This is the equation of motion for a particle whose velocity vector is always perpendicular to its displacement. Explain why this is the same thing as $$(x(t) + iy(t))' = i(x(t) + iy(t))$$ hence that, with the right initial conditions, the solution is $$x(t) + iy(t) = e^{it}.$$ On the other hand, a particle whose velocity vector is always perpendicular to its displacement travels in a circle. Hence, again with the right initial conditions, $x(t) = \cos t, y(t) = \sin t$. (At this point you might reiterate that complex numbers are real $2 \times 2$ matrices, assuming they have seen this method for solving systems of differential equations.) Share Improve this answer answered Jul 1, 2010 at 17:11 community wiki Qiaochu Yuan $\endgroup$ 2 1 $\begingroup$ Thanks, Qiaochu! They have some background in DEs, and this is a very good way to get DEs, trig identities and matrices at the same time. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 05:43:39 +00:00 Commented Jul 2, 2010 at 5:43 $\begingroup$ I just used basically this in a first course in differential equations to prove Euler's formula for them. One of us thought it was really cool, anyway... $\endgroup$ Ryan Reich – Ryan Reich 2013-03-22 14:10:32 +00:00 Commented Mar 22, 2013 at 14:10 Add a comment | 31 $\begingroup$ Here are two simple uses of complex numbers that I use to try to convince students that complex numbers are "cool" and worth learning. (Number Theory) Use complex numbers to derive Brahmagupta's identity expressing $(a^2+b^2)(c^2+d^2)$ as the sum of two squares, for integers $a,b,c,d$. (Euclidean geometry) Use complex numbers to explain Ptolemy's Theorem. For a cyclic quadrilateral with vertices $A,B,C,D$ we have $$\overline{AC}\cdot \overline{BD}=\overline{AB}\cdot \overline{CD} +\overline{BC}\cdot \overline{AD}$$ Share Improve this answer edited Jul 1, 2010 at 16:15 community wiki 2 revisions $\endgroup$ 3 27 $\begingroup$ And even more amazingly, one can completely solve the diophantine equation $x^2+y^2=z^n$ for any $n$ as follows: $$x+yi=(a+bi)^n, \ z=a^2+b^2.$$ I learned this from a popular math book while in elementary school, many years before studying calculus. $\endgroup$ Victor Protsak – Victor Protsak 2010-07-02 01:21:18 +00:00 Commented Jul 2, 2010 at 1:21 $\begingroup$ @Byron: Thanks! 2 examples in one answer: I can vote only once. :-( @Victor: I am indeed amazed! This elementary knowledge is new to me. I probably spent too much on complex DEs... $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 05:21:43 +00:00 Commented Jul 2, 2010 at 5:21 $\begingroup$ How does the vector calculus identity explain complex numbers? It is about an inner product space, true in every dimension (and not necessary for a positive-definite scalar product). There is dot product in every $\mathbb{R}^n, n\ge 1$, but only for $n = 1, 2, 4$ multiplication exists (and only for $n = 1, 2$ is it commutative). Downvote. $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 04:49:02 +00:00 Commented Apr 14, 2021 at 4:49 Add a comment | 23 $\begingroup$ One cannot over-emphasize that passing to complex numbers often permits a great simplification by linearizing what would otherwise be more complex nonlinear phenomena. One example familiar to any calculus student is the fact that integration of rational functions is much simpler over $\mathbb C$ (vs. $\mathbb R$) since partial fraction decompositions involve at most linear (vs quadratic) polynomials in the denominator. Similarly one reduces higher-order constant coefficient differential and difference equations to linear (first-order) equations by factoring the linear operators over $\mathbb C$. More generally one might argue that such simplification by linearization was at the heart of the development of abstract algebra. Namely, Dedekind, by abstracting out the essential linear structures (ideals and modules) in number theory, greatly simplified the prior nonlinear theory based on quadratic forms. This enabled him to exploit to the hilt the power of linear algebra. Examples abound of the revolutionary power that this brought to number theory and algebra - e.g. for one little-known gem see my recent post explaining how Dedekind's notion of conductor ideal beautifully encapsulates the essence of elementary irrationality proofs of n'th roots. Share Improve this answer edited Jun 15, 2020 at 7:27 community wiki 4 revsBill Dubuque $\endgroup$ Add a comment | 20 $\begingroup$ If you really want to "demystify" complex numbers, I'd suggest teaching what complex multiplication looks like with the following picture, as opposed to a matrix representation: If you want to visualize the product "z w", start with '0' and 'w' in the complex plane, then make a new complex plane where '0' sits above '0' and '1' sits above 'w'. If you look for 'z' up above, you see that 'z' sits above something you name 'z w'. You could teach this picture for just the real numbers or integers first -- the idea of using the rest of the points of the plane to do the same thing is a natural extension. You can use this picture to visually "demystify" a lot of things: Why is a negative times a negative a positive? --- I know some people who lost hope in understanding math as soon as they were told this fact i^2 = -1 (zw)t = z(wt) --- I think this is a better explanation than a matrix representation as to why the product is associative |zw| = |z| |w| (z + w)v = zv + wv The Pythagorean Theorem: draw (1-it)(1+it) = 1 + t^2 etc. One thing that's not so easy to see this way is the commutativity (for good reasons). After everyone has a grasp on how complex multiplication looks, you can get into the differential equation: $\frac{dz}{dt} = i z , z(0) = 1$ which Qiaochu noted travels counterclockwise in a unit circle at unit speed. You can use it to give a good definition for sine and cosine -- in particular, you get to define $\pi$ as the smallest positive solution to $e^{i \pi} = -1$. It's then physically obvious (as long as you understand the multiplication) that $e^{i(x+y)} = e^{ix} e^{iy}$, and your students get to actually understand all those hard/impossible to remember facts about trig functions (like angle addition and derivatives) that they were forced to memorize earlier in their lives. It may also be fun to discuss how the picture for $(1 + \frac{z}{n})^n$ turns into a picture of that differential equation in the "compound interest" limit as $n \to \infty$; doing so provides a bridge to power series, and gives an opportunity to understand the basic properties of the real exponential function more intuitively as well. But this stuff is less demystifying complex numbers and more... demystifying other stuff using complex numbers. Here's a link to some Feynman lectures on Quantum Electrodynamics (somehow prepared for a general audience) if you really need some flat out real-world complex numbers Share Improve this answer answered Jul 2, 2010 at 2:50 community wiki Phil Isett $\endgroup$ 0 Add a comment | 19 $\begingroup$ One of my favourite elementary applications of complex analysis is the evaluation of infinite sums of the form $$\sum_{n\geq 0} \frac{p(n)}{q(n)}$$ where $p,q$ are polynomials and $\deg q > 1 + \deg p$, by using residues. Share Improve this answer answered Jul 1, 2010 at 9:25 community wiki José Figueroa-O'Farrill $\endgroup$ 1 $\begingroup$ Thanks, José! This was not my list (even I use this very often in serious problems). I only wonder whether it is possible to start a course with such example. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 10:01:24 +00:00 Commented Jul 1, 2010 at 10:01 Add a comment | 13 $\begingroup$ They're useful just for doing ordinary geometry when programming. A common pattern I have seen in a great many computer programs is to start with a bunch of numbers that are really ratios of distances. Theses numbers get converted to angles with inverse trig functions. Then some simple functions are applied to the angles and the trig functions are used on the results. Trig and inverse trig functions are expensive to compute on a computer. In high performance code you want to eliminate them if possible. Quite often, for the above case, you can eliminate the trig functions. For example $\cos(2\cos^{-1} x) = 2x^2-1$ (for $x$ in a suitable range) but the version on the right runs much faster. The catch is remembering all those trig formulae. It'd be nice to make the compiler do all the work. A solution is to use complex numbers. Instead of storing $\theta$ we store $(\cos\theta,\sin\theta)$. We can add angles by using complex multiplication, multiply angles by integers and rational numbers using powers and roots and so on. As long as you don't actually need the numerical value of the angle in radians you need never use trig functions. Obviously there comes a point where the work of doing operations on complex numbers may outweigh the saving of avoiding trig. But often in real code the complex number route is faster. (Of course it's analogous to using quaternions for rotations in 3D. I guess it's somewhat in the spirit of rational trigonometry except I think it's easier to work with complex numbers.) Share Improve this answer answered Jul 2, 2010 at 4:23 community wiki Dan Piponi $\endgroup$ 1 $\begingroup$ Thanks! Your answer and Victor's comments reminded me about a related deduction in number theory (I'll try to add one more example to the original question). $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 08:46:25 +00:00 Commented Jul 2, 2010 at 8:46 Add a comment | 13 $\begingroup$ This answer is an expansion of the answer of Yuri Bakhtin. Here is a kind of mime show. Silently write the formulas for $\cos(2x)$ and $\sin(2x)$ lined up on the board, something like this: $$\cos(2x) = \cos^2(x) \hphantom{+ 2 \cos(x) \sin(x)} - \sin^2(x) $$ $$\sin(2x) = \hphantom{\cos^2(x)} + 2 \cos(x) \sin(x) \hphantom{- \sin^2(x)} $$ Do the same for the formulas for $\cos(3x)$ and $\sin(3x)$, and however far you want to go: $$\cos(3x) = \cos^3(x) \hphantom{+ 3 \cos^2(x) \sin(x)} - 3 \cos(x) \sin^2(x) \hphantom{- \sin^3(x)} $$ $$\sin(3x) = \hphantom{\cos^3(x)} + 3 \cos^2(x) \sin(x) \hphantom{- 3 \cos(x) \sin^2(x)} - \sin^3(x) $$ Maybe then let out a loud noise like "hmmmmmmmmm... I recognize those numbers..." Then, on a parallel board, write out Pascal's triangle, and parallel to that write the application of Pascal's triangle to the binomial expansions $(x+y)^n$. Make some more puzzling sounds regarding those pesky plus and minus signs. Then maybe it's time to actually say something: "Eureka! We can tie this all together by use of an imaginary number $i = \sqrt{-1}$". Then write out the binomial expansion of $$(\cos(x) + i\,\sin(x))^n $$ break it into its real and imaginary parts, and demonstrate equality with $$\cos(nx) + i\, \sin(nx). $$ Share Improve this answer edited Aug 7, 2013 at 23:24 community wiki 2 revs, 2 users 86%Lee Mosher $\endgroup$ Add a comment | 12 $\begingroup$ Several motivating physical applications are listed on wikipedia Why do we need to study numbers which do not belong to the real world? You may want to stoke the students' imagination by disseminating the deeper truth - that the world is neither real, complex nor p-adic (these are just completions of Q). Here is a nice quote by Yuri Manin picked from here On the fundamental level our world is neither real nor p-adic; it is adelic. For some reasons, reflecting the physical nature of our kind of living matter (e.g. the fact that we are built of massive particles), we tend to project the adelic picture onto its real side. We can equally well spiritually project it upon its non-Archimediean side and calculate most important things arithmetically. The relations between "real" and "arithmetical" pictures of the world is that of complementarity, like the relation between conjugate observables in quantum mechanics. (Y. Manin, in Conformal Invariance and String Theory, (Academic Press, 1989) 293-303 ) Share Improve this answer answered Jul 1, 2010 at 12:52 community wiki SandeepJ $\endgroup$ 2 3 $\begingroup$ Thanks for the tip! I'll better not cite Yuri Ivanovich to my electrical engineers; this will hardly encourage them to do complex analysis. :-) $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 13:24:07 +00:00 Commented Jul 1, 2010 at 13:24 $\begingroup$ There is some hype about alleged physical relevance of p-adic numbers (in fact, Ī̲ know in person several researchers who built their career upon it), but€”as for $\mathbb{Q}_p$€”these are either overreaching generalizations or barren conjectures, whereas complex numbers are justified by quantum mechanics alone. $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 05:04:24 +00:00 Commented Apr 14, 2021 at 5:04 Add a comment | 12 $\begingroup$ If they have a suitable background in linear algebra, I would not omit the interpretation of complex numbers in terms of conformal matrices of order 2 (with nonnegative determinant), translating all operations on complex numbers (sum, product, conjugate, modulus, inverse) in the context of matrices: with special emphasis on their multiplicative action on the plane (in particular, "real" gives "homotety" and "modulus 1" gives "rotation"). The complex exponential, defined initially as limit of $(1+z/n)^n$, should be a good application of the above geometrical ideas. In particular, for $z=it$, one can give a nice interpretation of the (too often covered with mystery) equation $e^{i\pi}=-1$ in terms of the length of the curve $e^{it}$ (defined as classical total variation). A brief discussion on (scalar) linear ordinary differential equations of order 2, with constant coefficients, also provides a good motivation (and with some historical truth). Related to the preceding point, and especially because they are from engineering, it should be worth recalling all the useful complex formalism used in Electricity. Not on the side of "real world" interpretation, but rather on the side of "useful abstraction" a brief account of the history of the third degree algebraic equation, with the embarrassing "casus impossibilis" (three real solutions, and the solution formula gives none, if taken in terms of "real" radicals!) should be very instructive. Here is also the source of such terms as "imaginary". Share Improve this answer edited Jul 1, 2010 at 20:09 community wiki 3 revisions $\endgroup$ 4 $\begingroup$ Thanks a lot, Pietro, for so many fruitful suggestions! Except for the point on $(1+z/n)^n$ (unfortunately, this is not the usual way to define the exponential), I can use all them. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 11:52:10 +00:00 Commented Jul 1, 2010 at 11:52 $\begingroup$ In fact the equivalence with the definition of exp(z) by the exponential series may be a nice exercise about dominated convergence for series. BTW, I realized only now that you asked for one suggestion for answer...sorry :-) $\endgroup$ Pietro Majer – Pietro Majer 2010-07-01 12:49:22 +00:00 Commented Jul 1, 2010 at 12:49 $\begingroup$ No worries, Pietro, this is a standard requirement for wiki community questions. Yes, it's a nice exercise, but most probably not for the level I get. (Last year I taught ODEs and, in particular, the linear systems where I needed to compute the exponential of matrices. The limit definition was mentioned as equivalent form of the series, and it was needed for proving $e^{\operatorname{tr}(A)}=\operatorname{det}(e^A)$. But that wasn't really accepted... :-( ) $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 13:00:30 +00:00 Commented Jul 1, 2010 at 13:00 3 $\begingroup$ @Wadim: The $(1+z/n)^n$ definition of the exponential is exactly what you get by applying Euler's method to the defining diff Eq of the exponential function, if you travel along the straight line from 0 to z in the domain, and use n equal partitions. $\endgroup$ Steven Gubkin – Steven Gubkin 2012-08-27 13:24:12 +00:00 Commented Aug 27, 2012 at 13:24 Add a comment | 12 $\begingroup$ This is a specific example where complex numbers aid a task in elementary real analysis; I haven't thought about the extent to which it generalizes. In my first year, I was given the task of formally proving that the Taylor series for arctan is $$ \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots, $$ where "formally" meant not simply integrating the series for $1/(1+x^2)$ termwise, since we hadn't yet seen any theorems that said you could do that. We had, however, seen Taylor's theorem. Hence the problem was to determine the values of all derivatives of $f(x)=\arctan(x)$, or of $f'(x)=1/(1+x^2)$, at $x=0$. However, it's not so easy to find a closed-form expression for the $n$-th derivative of $1/(1+x^2)$, unless you write it as $$ f'(x) = \frac{1}{2i} \left( \frac{1}{x-i}-\frac{1}{x+i} \right), $$ which then immediately yields $$ f^{(n)}(x) = \frac{(-1)^{n-1} (n-1)!}{2i} \cdot \left( \frac{1}{(x-i)^n}-\frac{1}{(x+i)^n} \right) $$ for $n>0$, which then gives the answers $f^{(2n)}(0)=0$ and $f^{(2n+1)}(0)=(-1)^n \cdot (2n)!$. Combining this with Taylor's theorem gives the desired series. I still think this is pretty neat. There really isn't any obvious way to cut the complex numbers and still have as painless a calculation as the one above. Share Improve this answer answered Aug 28, 2016 at 12:10 community wiki R.P. $\endgroup$ Add a comment | 11 $\begingroup$ In answer to "Why do we need to study numbers which do not belong to the real world?" you might simply state that quantum mechanics tells us that complex numbers arise naturally in the correct description of probability theory as it occurs in our (quantum) universe. I think a good explanation of this is in Chapter 3 of the third volume of the Feynman lectures of physics, although I don't have a copy handy to check. (In particular, similar to probability theory with real numbers, the complex amplitude of one of two independent events A or B occuring is just the sum of the amplitude of A and the amplitude of B. Furthermore, the complex amplitude of A followed by B is just the product of the amplitudes. After all intermediate calculations one just takes the magnitude of the complex number squared to get the usual (real number) probability.) Share Improve this answer answered Jul 1, 2010 at 22:06 community wiki Jon $\endgroup$ 3 1 $\begingroup$ Perhaps you are referring to Feynman's book QED? $\endgroup$ S. Carnahan – S. Carnahan ♦ 2010-07-02 04:41:02 +00:00 Commented Jul 2, 2010 at 4:41 $\begingroup$ Yes, I think it was there. (The "strange theory of light and matter" book.) $\endgroup$ Jon – Jon 2010-07-02 16:19:37 +00:00 Commented Jul 2, 2010 at 16:19 $\begingroup$ This is a conceptual mess. Basic probability theory absolutely doesn€™t need complex numbers but€”applied to the real world€”it is a simplification just like any other theory, QM included. Surely we can (and do) reduce complex amplitudes to dumb probability measures using |·|², but it has nothing to do with €œcorrect description of probability theory€. It is indeed about real-world randomness which extends far beyond Kolmogorov-style probability. Can anybody rewrite, please? $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 05:22:46 +00:00 Commented Apr 14, 2021 at 5:22 Add a comment | 10 $\begingroup$ I never took a precalculus class because every identity I've ever needed involving sines and cosines I could derive by evaluating a complex exponential in two different ways. Perhaps you could tell them that if they ever forget a trig identity, they can rederive it using this method? Share Improve this answer answered Jul 1, 2010 at 16:05 community wiki Dylan Wilson $\endgroup$ 2 2 $\begingroup$ I especially like the complex derivation of $\cos^n x$ and $\sin^n x$ in terms of trig functions of multiple angle, which is very useful if you need to integrate them. $\endgroup$ Victor Protsak – Victor Protsak 2010-07-02 01:32:59 +00:00 Commented Jul 2, 2010 at 1:32 $\begingroup$ @Dylan, thanks! You expand Yuri's answer. @Victor: aren't these standard problems for 1st year in algebra? ;-) $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 05:38:09 +00:00 Commented Jul 2, 2010 at 5:38 Add a comment | 10 $\begingroup$ Tristan Needham's book Visual Complex Analysis is full of these sorts of gems. One of my favorites is the proof using complex numbers that if you put squares on the sides of a quadralateral, the lines connecting opposite centers will be perpendicular and of the same length. After proving this with complex numbers, he outlines a proof without them that is much longer. The relevant pages are on Google books: Share Improve this answer answered Jul 2, 2010 at 15:12 community wiki Eric O. Korman $\endgroup$ 1 $\begingroup$ The same as for mathoverflow.net/a/30185/56921 €” vector calculus is not complex calculus! Complex numbers certainly are a good example of an inner product space, and to teach vectors you can use the example of „‚, but complex multiplication and (especially) division go beyond the vector-based intuition. $\endgroup$ Incnis Mrsi – Incnis Mrsi 2021-04-14 05:38:50 +00:00 Commented Apr 14, 2021 at 5:38 Add a comment | 9 $\begingroup$ How about how the holomorphicity of a function $f(z)=x+yi$ relates to, e.g., the curl of the vector $(x,y)\in\mathbb{R}^2$? This relates nicely to why we can solve problems in two dimensional electromagnetism (or 3d with the right symmetries) very nicely using "conformal methods." It would be very easy to start a course with something like this to motivate complex analytic methods. Share Improve this answer answered Jul 1, 2010 at 10:21 community wiki jeremy $\endgroup$ 7 $\begingroup$ Jeremy, can you expand your answer or provide a reference where your example is worked in details? $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 10:30:04 +00:00 Commented Jul 1, 2010 at 10:30 $\begingroup$ This is in a number of advanced math methods for physics, and graduate electromagnetism texts, but one offhand I know it's in is "Electrodynamics of Continuous Media" by Landau and Lifshitz, chapter 1 section 3, on methods of solving electrostatic problems. But there are other books out there with more involved and more sophisticated treatments, but offhand I don't know any titles. If you search amazon's "search inside this book" for "the method of conformal mapping" you can find part of the discussion. But the basics of it are elementary enough that that book should be sufficient. $\endgroup$ jeremy – jeremy 2010-07-01 11:20:27 +00:00 Commented Jul 1, 2010 at 11:20 1 $\begingroup$ Thanks, Jeremy! I'll definitely do the search, the magic word "the method of conformal mapping" is really important here. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 11:45:20 +00:00 Commented Jul 1, 2010 at 11:45 4 $\begingroup$ I think most older Russian textbooks on complex analysis (e.g. Lavrentiev and Shabat or Markushevich) had examples from 2D hydrodynamics (Euler-D'Alambert equations $\iff$ Cauchy-Riemann equations). Also, of course, the Zhukovsky function and airwing profile. They serve more as applications of theory than motivations, since nontrivial mathematical work is required to get there. $\endgroup$ Victor Protsak – Victor Protsak 2010-07-02 02:04:02 +00:00 Commented Jul 2, 2010 at 2:04 $\begingroup$ Yes, those examples are great! I was thinking about those examples, too, when I said "conformal methods," but they are a little less basic than the E&M example. There are many more examples, too, though, such as classical gravity, or just about anything that can be described with a potential. They would make excellent topics to visit after developing some of the formalism more carefully, since they can lead to a lot of intuition about why things are constructed like they are, and why they're useful! $\endgroup$ jeremy – jeremy 2010-07-02 03:02:02 +00:00 Commented Jul 2, 2010 at 3:02 | Show 2 more comments 9 $\begingroup$ From the perspective of complex analysis, the theory of Fourier series has a very natural explanation. I take it that the students had seen Fourier series first, of course. I had mentioned this elsewhere too. I hope the students also know about Taylor theorem and Taylor series. Then one could talk also of the Laurent series in concrete terms, and argue that the Fourier series is studied most naturally in this setting. First, instead of cos and sin, define the Fourier series using complex exponential. Then, let $f(z)$ be a complex analytic function in the complex plane, with period $1$. Then write the substitution $q = e^{2\pi i z}$. This way the analytic function $f$ actually becomes a meromorphic function of $q$ around zero, and $z = i \infty$ corresponds to $q = 0$. The Fourier expansion of $f(z)$ is then nothing but the Laurent expansion of $f(q)$ at $q = 0$. Thus we have made use of a very natural function in complex analysis, the exponential function, to see the periodic function in another domain. And in that domain, the Fourier expansion is nothing but the Laurent expansion, which is a most natural thing to consider in complex analysis. I am am electrical engineer; I have an idea what they all study; so I can safely override any objections that this won't be accessible to electrical engineers. Moreover, the above will reduce their surprise later in their studies when they study signal processing and wavelet analysis. Share Improve this answer edited Apr 13, 2017 at 12:58 community wiki 2 revsAnweshi $\endgroup$ Add a comment | 8 $\begingroup$ This answer doesn't show how the complex numbers are useful, but I think it might demystify them for students. Most are probably already familiar with its content, but it might be useful to state it again. Since the question was asked two months ago and Professor Zudilin started teaching a month ago, it's likely this answer is also too late. If they have already taken a class in abstract algebra, one can remind them of the basic theory of field extensions with emphasis on the example of $\mathbb C \cong \mathbb R[x]/(x^2+1).$ It seems that most introductions give complex numbers as a way of writing non-real roots of polynomials and go on to show that if multiplication and addition are defined a certain way, then we can work with them, that this is consistent with handling them like vectors in the plane, and that they are extremely useful in solving problems in various settings. This certainly clarifies how to use them and demonstrates how useful they are, but it still doesn't demystify them. A complex number still seems like a magical, ad hoc construction that we accept because it works. If I remember correctly, and has probably already been discussed, this is why they were called imaginary numbers. If introduced after one has some experience with abstract algebra as a field extension, one can see clearly that the complex numbers are not a contrivance that might eventually lead to trouble. Beginning students might be thinking this and consequently, resist them, or require them to have faith in them or their teachers, which might already be the case. Rather, one can see that they are the result of a natural operation. That is, taking the quotient of a polynomial ring over a field and an ideal generated by an irreducible polynomial, whose roots we are searching for. Multiplication, addition, and its 2-dimensional vector space structure over the reals are then consequences of the quotient construction $\mathbb R[x]/(x^2+1).$ The root $\theta,$ which we can then relabel to $i,$ is also automatically consistent with familiar operations with polynomials, which are not ad hoc or magical. The students should also be able to see that the field extension $\mathbb C = \mathbb R(i)$ is only one example, although a special and important one, of many possible quotients of polynomial rings and maximal ideals, which should dispel ideas of absolute uniqueness and put it in an accessible context. Finally, if they think that complex numbers are imaginary, that should be corrected when they understand that they are one example of things naturally constructed from other things they are already familiar with and accept. Reference: Dummit & Foote: Abstract Algebra, 13.1 Share Improve this answer edited Sep 5, 2010 at 19:46 community wiki 2 revisions $\endgroup$ 1 $\begingroup$ Thanks for this point, Anthony! Although "my" students struggle abstract algebra as well, I would incorporate these ideas in my next year lectures (I should teach complex analysis again). $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-10-10 22:05:10 +00:00 Commented Oct 10, 2010 at 22:05 Add a comment | 8 $\begingroup$ This is not exactly an answer to the question, but it is the simplest thing I know to help students appreciate complex numbers. (I got the idea somewhere else, but I forgot exactly where.) It's something even much younger students can appreciate. Recall that on the real number line, multiplying a number by -1 "flips" it, that is, it rotates the point 180 degrees about the origin. Introduce the imaginary number line (perpendicular to the real number line) then introduce multiplication by i as a rotation by 90 degrees. I think most students would appreciate operations on complex numbers if they visualize them as movements of points on the complex plane. Share Improve this answer answered Jan 20, 2011 at 14:57 community wiki JRN $\endgroup$ 1 1 $\begingroup$ I now remember where I got the idea: mathoverflow.net/questions/47214/… $\endgroup$ JRN – JRN 2011-02-03 13:48:36 +00:00 Commented Feb 3, 2011 at 13:48 Add a comment | 7 $\begingroup$ Another classic I haven't seen mentioned yet is the proof of the Machin formula $$ \frac{\pi}{4} = 4\arctan \frac{1}{5}-\arctan \frac{1}{239}. $$ I honestly don't know a proof of this that avoids complex numbers, but surely it can be nowhere near as simple as the elementary computation needed to prove the identity $$ 2+2i = \frac{(5+i)^4}{239+i}. $$ Taking $\arg$ on both sides yields Machin's formula. Share Improve this answer answered Aug 28, 2016 at 12:47 community wiki R.P. $\endgroup$ 3 8 $\begingroup$ It follows from the arctan summation formula: $\operatorname{arctan}(x) + \operatorname{arctan}(y) = \operatorname{arctan}\frac{x+y}{1-xy}$, which in turn can be derived from the sine angle addition formula. $\endgroup$ S. Carnahan – S. Carnahan ♦ 2016-08-29 01:38:04 +00:00 Commented Aug 29, 2016 at 1:38 1 $\begingroup$ But surely $e^{i(x+y)}=e^{ix}e^{iy}$ is easier to remember than the addition formulae for sine and cosine. $\endgroup$ Michael Renardy – Michael Renardy 2017-09-28 15:51:30 +00:00 Commented Sep 28, 2017 at 15:51 1 $\begingroup$ A simpler formula like $\pi/4=2\arctan(1/3)+\arctan(1/7)$ can be easily shown without words: Draw the integer coordinate points $A:=(9,0)$, $B:=(9,3)$, $C:=(8,6)$, $D:=(7,7)$ and argue on the angles $\hat {AOB}$, $\hat{BOC}$, $\hat{COD}$, $\hat{AOD}$ by similarity of rectangle triangles. An analogous picture shows Machin formula (although one need to start from $(625,0)$ to get all required points with integer coordinates). Of course this is just the elementary geometry version of the complex number computation. $\endgroup$ Pietro Majer – Pietro Majer 2019-04-23 19:49:54 +00:00 Commented Apr 23, 2019 at 19:49 Add a comment | 6 $\begingroup$ Motivating complex analysis The physics aspect of motivation should be the strongest for engineering students. No complex numbers, no quantum mechanics, no solid state physics, no lasers, no electrical or electronic engineering (starting with impedance), no radio, TV, acoustics, no good simple way of understanding of the mechanical analogues of RLC circuits, resonance, etc., etc. Then the "mystery" of it all. Complex numbers as the consequence of roots, square, cubic, etc., unfolding until one gets the complex plane, radii of convergence, poles of stability, all everyday engineering. Then the romance of it all, the "self secret knowledge", discovered over hundreds of years, a new language which even helps our thinking in general. Then the wider view of say Smale/Hirsch on higher dimensional differential equations, chaos etc. They should see the point pretty quickly. This is a narrow door, almost accidentally discovered, through which we see and understand entire new realms, which have become our best current, albeit imperfect,descriptions of how to understand and manipulate a kind of "inner essence of what is" for practical human ends, i.e. engineering. (True, a little over the top, but then pedagogical and motivational). For them to say that they just want to learn a few computational tricks is a little like a student saying, "don't teach me about fire, just about lighting matches". It's up to them I suppose, but they will always be limited. There might be some computer software engineer who needs a little more, but then I suppose there is also modern combinatorics. :-) Share Improve this answer answered Jul 2, 2010 at 3:18 community wiki sigoldberg1 $\endgroup$ 2 $\begingroup$ Thanks! I'll borrow some of your wording for my lecture notes. (If you don't object.) $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 08:54:21 +00:00 Commented Jul 2, 2010 at 8:54 $\begingroup$ Absolutely, Sure. $\endgroup$ sigoldberg1 – sigoldberg1 2010-07-02 15:43:27 +00:00 Commented Jul 2, 2010 at 15:43 Add a comment | 6 $\begingroup$ Maybe artificial, but a nice example (I think) demonstrating analytic continuation (NOT just the usual $\mathrm{Re}(e^{i \theta})$ method!) I don't know any reasonable way of doing this by real methods. As a fun exercise, calculate $$ I(\omega) = \int_0^\infty e^{-x} \cos (\omega x) \frac{dx}{\sqrt{x}}, \qquad \omega \in \mathbb{R} $$ from the real part of $F(1+i \omega)$, where $$ F(k) = \int_0^\infty e^{-kx} \frac{dx}{\sqrt{x}}, \qquad \mathrm{Re}(k)>0 $$ (which is easily obtained for $k>0$ by a real substitution) and using analytic continuation to justify the same formula with $k=1+i \omega$. You need care with square roots, branch cuts, etc.; but this can be avoided by considering $F(k)^2$, $I(\omega)^2$. Of course all the standard integrals provide endless fun examples! (But the books don't have many requiring genuine analytic continuation like this!) Share Improve this answer answered Jul 8, 2010 at 0:52 community wiki Zen Harper $\endgroup$ 4 4 $\begingroup$ I rather suspect analytic continuation is a conceptual step above what the class in question coud cope with... $\endgroup$ Yemon Choi – Yemon Choi 2010-07-08 01:22:41 +00:00 Commented Jul 8, 2010 at 1:22 $\begingroup$ Thanks, Zen! Although this seems to be hard for the class, I keep your complexified example for my personal collection. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-08 03:47:30 +00:00 Commented Jul 8, 2010 at 3:47 $\begingroup$ That is a very nice example. $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2010-07-08 03:59:08 +00:00 Commented Jul 8, 2010 at 3:59 $\begingroup$ Thanks, and hi again to my friend Yemon! I agree that this is too hard for most classes, but I couldn't resist showing it! It needs a clear understanding of analytic continuation to do it, so it's only appropriate for courses which actually cover that rigorously (or at semi-rigorously). $\endgroup$ Zen Harper – Zen Harper 2010-07-10 12:33:20 +00:00 Commented Jul 10, 2010 at 12:33 Add a comment | 6 $\begingroup$ Consider the function f(x)=1/(1+x^2) on the real line. Using the geometric progression formula, you can expand f(x)=1-x^2+... . This series converges for |x|<1 but diverges for all other x. Why this is so? The function looks nice and smooth everywhere on the real line. This example is taken from the Introduction of the textbook by B. V. Shabat. Share Improve this answer answered Jul 28, 2010 at 8:30 community wiki Alex Eremenko $\endgroup$ 1 $\begingroup$ Alex, thanks for your contribution. But it's already here: see Keith Conrad's answer above, mathoverflow.net/questions/30156/demystifying-complex-numbers/…. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-28 08:57:50 +00:00 Commented Jul 28, 2010 at 8:57 Add a comment | 5 $\begingroup$ Try this: compare the problems of finding the points equidistant in the plane from (-1, 0) and (1, 0), which is easy, with finding the points at twice the distance from (-1, 0) that they are from (1, 0). The idea that "real" concepts are the only ones of use in the "real world" is of course a fallacy. I suppose it is more than a century since electrical engineers admitted that complex numbers are useful. Share Improve this answer answered Jul 1, 2010 at 10:17 community wiki Charles Matthews $\endgroup$ 2 $\begingroup$ Charles, you easily guessed that "my" engineers are electrical! The problem is (and I clearly see it after discussions with colleagues from EE) that they needed it one century ago but hardly need (in a reasonable generality) nowadays. Only concepts and rules of manipulation are required. :-( +1 As for your example, I don't see serious benefits for changing $\mathbb R^2$ by $\mathbb C$ (what happens if I am later asked about the same or similar problem in $\mathbb R^3$?). $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-01 10:28:09 +00:00 Commented Jul 1, 2010 at 10:28 2 $\begingroup$ I do see an undeniable benefit. If you are later asked about it in $\mathbb{R}^3$ then you use vectors and dot product. The historical way would have been to use quaternions; indeed, this is how the notion of dot product crystallized in the work of Gibbs, and more relevantly for your EE students, Oliver Heaviside. $\endgroup$ Victor Protsak – Victor Protsak 2010-07-02 01:26:03 +00:00 Commented Jul 2, 2010 at 1:26 Add a comment | 5 $\begingroup$ I always like to use complex dynamics to illustrate that complex numbers are "real" (i.e., they are not just a useful abstract concept, but in fact something that very much exist, and closing our eyes to them would leave us not only devoid of useful tools, but also of a deeper understanding of phenomena involving real numbers.) Of course I am a complex dynamicist so I am particularly partial to this approach! Start with the study of the logistic map $x\mapsto \lambda x(1-x)$ as a dynamical system (easy to motivate e.g. as a simple model of population dynamics). Do some experiments that illustrate some of the behaviour in this family (using e.g. web diagrams and the Feigenbaum diagram), such as: The period-doubling bifurcation The appearance of periodic points of various periods The occurrence of "period windows" everywhere in the Feigenbaum diagram. Then let x and lambda be complex, and investigate the structure both in the dynamical and parameter plane, observing The occurence of beautiful and very "natural"-looking objects in the form of Julia sets and the (double) Mandelbrot set; The explanation of period-doubling as the collision of a real fixed point with a complex point of period 2, and the transition points occuring as points of tangency between interior components of the Mandelbrot set; Period windows corresponding to little copies of the Mandelbrot set. Finally, mention that density of period windows in the Feigenbaum diagram - a purely real result, established only in the mid-1990s - could never have been achieved without complex methods. There are two downsides to this approach: It requires a certain investment of time; even if done on a superficial level (as I sometimes do in popular maths lectures for an interested general audience) it requires the better part of a lecture It is likely to appeal more to those that are mathematically minded than engineers who could be more impressed by useful tools for calculations such as those mentioned elsewhere on this thread. However, I personally think there are few demonstrations of the "reality" of the complex numbers that are more striking. In fact, I have sometimes toyed with the idea of writing an introductory text on complex numbers which uses this as a primary motivation. Share Improve this answer answered Jul 28, 2010 at 9:34 community wiki Lasse Rempe $\endgroup$ 5 $\begingroup$ Thanks a lot, Lasse, for the nice example. Can you give a link to a place where the dynamics is discussed in the complex case as well as the density of period windows in the Feigenbaum diagrams is discussed? $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-28 09:42:34 +00:00 Commented Jul 28, 2010 at 9:42 $\begingroup$ Hm, I am not sure what the best place is - I guess you don't want me to reference a textbook on holomorphic dynamics! Devaney's "A first course in chaotic dynamical systems" has some things on Julia sets and the Mandelbrot set; I'll have a closer look if I get a chance. His "An introduction to chaotic dynamical systems", which is on a slightly more advanced level, also does, according to the table of contents - I don't have it handy. There are many other references, but I'll have to think a little bit whether I can come up with one that makes it easy for you to find what you need. $\endgroup$ Lasse Rempe – Lasse Rempe 2010-07-28 10:54:16 +00:00 Commented Jul 28, 2010 at 10:54 $\begingroup$ Density of period windows in the Feigenbaum diagram is the celebrated result on "density of hyperbolicity" (or "density of Axiom A") in the quadratic family. It was established independently by Lyubich and by Graczyk and Swiatek in the 90s. Lyubich had an article on the quadratic family in the AMS Notices a while back, I think on his stronger "Regular or Stochastic" theorem. $\endgroup$ Lasse Rempe – Lasse Rempe 2010-07-28 10:56:00 +00:00 Commented Jul 28, 2010 at 10:56 $\begingroup$ If you look for even more recent results in the same vein - all of which use complex methods - density of hyperbolicity was established for real polynomials by Kozlovski, Shen and van Strien a few years ago (appeared in the Annals of Mathematics). Even more recently, Sebastian van Strien and I extended this to some families of real transcendental entire functions such as $x\mapsto a\sin(x)+b\cos(x)$. In this proof, we, in particular, consider complex points that tend to $\infty$ under iteration by such a map - even though the map is bounded on the real line. (Available on the arXiv.) $\endgroup$ Lasse Rempe – Lasse Rempe 2010-07-28 11:00:11 +00:00 Commented Jul 28, 2010 at 11:00 $\begingroup$ Thanks for the tips, Lasse! I promise to follow them, although slowly, without bothering myself about the students, more educating myself. Thank you very much. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-28 22:54:42 +00:00 Commented Jul 28, 2010 at 22:54 Add a comment | 5 $\begingroup$ From the point of view of enginieers, the most obvious application of complex numbers is computing alternating currents. Consider first direct current. If you have a network of resistors, and want to compute the current in this network, or the potential of a node, then Kirchhoff's rules reduce this problem to a system of linear equations. Kirchhoff's rules are obvious, essentially saying that ellectric current cannot just disappear. If you have alternating current, you have capacities and inductions in addition to the resistors, but if you consider them as imaginary resistance depending on the frequency, the computations are exactly the same as in the direct case, just over another field. The alternative would be computing the phase shift separately from the current, which is much more effort and only works for very simple networks, e.g. oscillators. Once you learned Fourier analysis, this approach immediately tells you how a filter works, and whether a given network acts as a filter. Share Improve this answer answered Mar 16, 2019 at 9:24 community wiki Jan-Christoph Schlage-Puchta $\endgroup$ Add a comment | 5 $\begingroup$ "Why do we need to study numbers which do not belong to the real world?" I don't think you can answer this in a single class. The best answer I can come up with is to show how complicated calculus problems can be solved easily using complex analysis. As an example, I bet most of your students hated solving the problem $\int e^{-x}\cos(x) dx$. Solve it for them the way they learned it in calculus, by repeated integration by parts and then by $\int e^{-x}\cos(x) dx\ \ =\ \ \Re \int e^{-x(1-i)}dx$. They should notice how much easier it was to use complex analysis. If you do this enough they might come to appreciate numbers that do not belong to the real world. Share Improve this answer edited Apr 14, 2021 at 6:07 community wiki 2 revs, 2 users 86%Daniel Parry $\endgroup$ 1 2 $\begingroup$ That idea to compute the integral does not use complex analysis :) $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2011-01-20 02:07:59 +00:00 Commented Jan 20, 2011 at 2:07 Add a comment | 4 $\begingroup$ As an example to demonstrate the usefulness of complex analysis in mechanics (which may seem counterintuitive to engineering students, since mechanics is introduced on the reals), one may consider the simple problem of the one dimensional harmonic oscillator, whose Hamiltonian equations of motion are diagonalized in the complex representation, equivalently one needs to integrate a single (holomorphic) first order ODE instead of a single second order or two first order ODEs. Share Improve this answer answered Jul 1, 2010 at 15:08 community wiki David Bar Moshe $\endgroup$ 1 $\begingroup$ Thanks, David! It goes in line with some other answers but it's interesting to see a different point of view on the linearised DE for the harmonic oscillator. $\endgroup$ Wadim Zudilin – Wadim Zudilin 2010-07-02 08:49:13 +00:00 Commented Jul 2, 2010 at 8:49 Add a comment | 1 2 Next You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions soft-question cv.complex-variables teaching See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 10 Why the unreasonable applicability of complex numbers in physics/engineering? How to present mathematics to non-mathematicians? 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https://www.chegg.com/homework-help/questions-and-answers/3-goal-exercise-prove-exponential-map-exp-r-0-00-ex-continuous-strictly-increasing-bijecti-q87045530
Solved 3. The goal of this exercise is to prove that the | Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Advanced Math Advanced Math questions and answers 3. The goal of this exercise is to prove that the exponential map exp : R + (0, +00) х ex is continuous, strictly increasing and bijective, and its inverse function is continuous. The inverse function of exp will be denoted by log : (0, +00) + R. Recall that in a previous homework we defined exp(x) as n exp(x):= lim 1+- n 5)" n-> = = for any x E R. We Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 3. The goal of this exercise is to prove that the exponential map exp : R + (0, +00) х ex is continuous, strictly increasing and bijective, and its inverse function is continuous. The inverse function of exp will be denoted by log : (0, +00) + R. Recall that in a previous homework we defined exp(x) as n exp(x):= lim 1+- n 5)" n-> = = for any x E R. We also Show transcribed image text There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 A series expansion is often e... View the full answer Step 2 UnlockAnswer Unlock Previous question Transcribed image text: The goal of this exercise is to prove that the exponential map exp : R + (0, +00) х ex is continuous, strictly increasing and bijective, and its inverse function is continuous. The inverse function of exp will be denoted by log : (0, +00) + R. Recall that in a previous homework we defined exp(x) as n exp(x):= lim 1+- n 5)" n-> = = for any x E R. We also proved the following properties: i) if x > -1, then the sequence an(x) (1 + )" is strictly increasing (during the discussion from week 5, you showed that (an(x))n is increasing when n > -x, which in particular tells us that it is globally increasing when x > -1); ii) exp(0) = 1 (it follows immediately from the definition), and exp(x) > 0 for every x ER (this follows from the fact that 1+ > 0 for n sufficiently large, and the sequence (an(x))n is increasing when n > -x); iii) for any M E N and for any x < M, we have en SeM (this follows from the argument < provided to see that the sequence (an(x))n is bounded); iv) for every x € R, exp(-x) = exp(x)-1; v) for every x,y e R, exp(x + y) = exp(x) exp(y). n = We will break the proof into small steps: 1. Show that, for every x > -1, we have exp(x) – 1 > x; 2. Prove that the function x H exp(x) is strictly increasing, i. e. for every x < x' we have exp(x) < exp(x'); 3. Show that, for every x E R and n E N \ {0}, exp(x/n)" = exp(x), and deduce that limnyoo exp(1/n) = limn70 ve = 1 (we saw in an exercise the fact that, for every a > 0, "a tends to 1 as n goes to 0o). Similarly we have that limnto exp(-1/n) = 1; - = = limn–100 The = 2 4. Prove that the function x H exp(x) is continuous at x = 0 (Hint: by the previous point, for every e > 0 there exists a large natural number n = ne EN such that 1-€ < e-1/n el/n <1+ ε. Now use the fact that exp is increasing to conclude); 5. Using property v) and the continuity at x = 0, deduce the continuity of exp at every point x eR; 6. Prove that = sup{exp(x) | X E R} = too and inf{exp(x) | X E R} = 0, х and deduce that exp : R + (0, +00) is bijective. = . Finally, combining this with the exercise seen in the last discussion, we can deduce that the inverse log exp-1: (0, +00) +R is continuous, bijective and strictly increasing. Here are a few consequences of this fact: • for any a > 0 and b E R, we can now define ab exp(blog a) > 0. The function bH a satisfies a 6+b' a'ab' and a-b = 1/(a”). If a > 1, then b + a' is strictly increasing, and if a < 1, it is strictly decreasing. In both cases, b H a' is bijective as a function from R to (0, +). The logarithm function loga : (0, +00) + R can then be defined as the inverse of b + a', and it is continuous for every a > 0. since ea > 0 for every x E R, the inequality e – 1 > x is true for every x € R. This implies that for every t > -1, we have log(1+t) 00 n lim log( Vn) = log(1) = 0, noo which implies that limg_0+ x log x = 0. Not the question you’re looking for? Post any question and get expert help quickly. 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https://www.youtube.com/watch?v=8zO0algdgNM
Primary productivity in ecosystems| Matter and Energy Flow| AP Environmental Science| Khan Academy Khan Academy 9090000 subscribers 744 likes Description 53484 views Posted: 13 Jan 2021 Keep going! Check out the next lesson and practice what you’re learning: Primary productivity is the rate at which solar energy (sunlight) is converted into organic compounds via photosynthesis over a unit of time. Net primary productivity is the rate of energy storage by photosynthesizers in a given area, after subtracting the energy lost to respiration. Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today! Donate here: Volunteer here: 60 comments Transcript: Intro in this video we're going to talk about energy and in particular we're going to talk about the energy of life the energy that i need to live and all of us need to live the energy you need to think the energy i'm using to make this video right now and some of you might already guess where this energy is coming from the surface of our planet is constantly being bombarded with light energy from the sun and you might know that there's certain organisms on our planet that are capable of taking that light energy and then storing it as chemical energy and those things and there's many types but the ones that we see most often in our day-to-day life are plants and so let's imagine a plant here and what it's doing is it's using that light energy in conjunction with water typically from the soil that maybe it's getting through its roots and carbon dioxide in the air and it's using that light energy to actually stick or you could say fix the carbon to construct itself and in its own tissue it's storing that energy and then if it were to break down that tissue it can release that energy in various forms now as it does this you might also be familiar that these photosynthesizers or these primary producers or these autotrophs they're also releasing molecular oxygen now if we were to describe this in in chemical terms or chemistry terms we would describe this process of photosynthesis as taking carbon dioxide from the air in conjunction with water from the soil and what's i guess you could say fueling all of this is light energy usually from the sun and what that is yielding is the tissue of the plant that is actually storing that energy is chemical energy in an organic form and the primary the primary way that this is done is through glucose which is c6h12o6 i know what you're thinking all plants don't taste sweet well if you take chains of sugars and put them together you get carbohydrates and you adapt them a little bit you get things like starches and that's what most of the plant tissue is so some variations of this or thing variations of this linked together but this is where the energy is stored so energy stored in the actual plant tissue and then of course it releases that molecular oxygen Photosynthesis and this is the process of photosynthesis and even if you look at the word photosynthesis and what the parts of it mean photo is referring to light and synthesis is referring to putting something together synthesizing something so photosynthesis you're using light to put together these essentially fix the carbon together to store energy now you might say all right that's nice i'm storing the energy this way how do i actually use the energy and that's something that all of us are doing all living systems have to do and that process is respiration and you could already guess what at least the chemical the chemical reaction for respiration will look like you're going to start with our stored energy our glucose cac6h12o6 in the presence of oxygen and since we're respiring all the time this is why we need to breathe oxygen and this is going to yield carbon dioxide and that's why we exhale more carbon dioxide than we inhale it's also going to release water and it's going to release and this is the whole point of it cellular energy and in other videos that you'll see in a biology class we'll talk about how this form of stored energy gets converted to other forms and then how that's used by the various machinery and cells to actually live to reproduce to move in many cases Measurement now an interesting question is how do you measure how much photosynthesis is going on how much primary productivity is going on well one way to think about it is find an ecosystem and take a certain area of the surface of that ecosystem and it could be a terrestrial ecosystem on land it could be a marine ecosystem and then say for this area in a given period of time often times a year how much stuff is growing so this is the stuff that grows and obviously or it would seem that the more stuff that is growing that the more photosynthesis that is taking place and the way that they measure how much how much is growing you can either measure it in terms of grams of biomass so grams of biomass and biomass is just a fancy way of saying that the mass of biological stuff that's just growing on this area and usually they'll take the water out so they get a consistent measurement or you can convert this to calories and it's usually measured in thousands of calories kilocalories and when you see calories on a packaging food label what we most of us think of calories those are actually kilocalories when we think about it in scientific terms and i know what you're thinking you're like wait mass and kilocalories calories that's just a form of energy well those two things you can go between because usually a certain type of biomass a gram of a certain type of biomass will have a certain amount of energy stored in it not energy that necessarily all animals could use or that we could use but it does have energy in it Net Primary Productivity now when we talk about this primary productivity you might already be thinking about well doesn't don't the plants need to use some of the energy that they are producing themselves to live and my answer to you is of course they need it in fact that's probably the most important reason why they need to photosynthesize is because they need to do respiration in order for them to grow and metabolize and live and reproduce and so when you see how much has been produced in a given area in a given year you're actually seeing the net primary productivity this is the you could think about it how much photosynthesis they did minus how much respiration they did so if you think of how much photosynthesis they did as gross primary productivity so that's the total amount of photosynthesis and then you subtract out the amount of energy chemical energy or cellular energy they needed for respiration that would then give you the net primary productivity and as i mentioned just to make things a little bit tangible if you took a very productive ecosystem let's say something like a rain forest that i have here in the background a very productive ecosystem like this if you were to take on average a square meter of this it produces in a year about 2000 grams of biomass so here we would say that the net primary productivity of this rainforest that you see in the background here would be approximately 2000 grams per square meter per year and if you wanted to think about this in terms of kilocalories you just have to say well each gram of biomass is how many kilocalories and it depends on the type of biomass but let's say that we have four kilocalories per gram of biomass so then we could also say that this net primary productivity is equal to 2000 grams per square meter per year times 4 kilocalories per gram the grams cancel out and then you multiply 4 times 2 000 that's going to be 8000 kilocalories kilocalories per square meter per year that would be the net primary productivity because that's after the plants have been doing respiration now how would you figure out gross primary productivity well you're not going to be able to do it directly but you can figure that out by figuring out the rate of respiration if you took some plants in that ecosystem and then you put them in a dark room with no light and then if you saw how much oxygen they are absorbing or they're having to use then that gives you a sense of how much respiration they are doing and there's ways that you can look at the ratios of the oxygens and the carbons to figure out exactly how much respiration is going on and then if you know the net primary productivity and the rate of respiration then you could figure out the gross primary productivity but i will leave you there because these are really useful measurements well one it's really useful to think about where all of the energy that allows us to live comes from but it's also useful for ecologists to think about how productive a system is or what's making it more productive or less productive and as we'll see these numbers here these are for sure on the high end of net primary productivity if we were in a desert type of ecosystem this number might be in the low hundreds and not in the 8 000 range
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https://mangolanguages.com/resources/learn/grammar/english/participles-in-english-what-are-they-and-how-are-they-used
English Articles Participles in English: What are they and how are they used? By: Isabel McKay Fri Sep 13 2024 English Verbs, Adjectives, Grammar Tips A participle is a word made from a verbVerbs are words used to describe an action, state, or occurrence, like “to run,” “to exist,” “to happen.” and used as an adjectiveAdjectives are words that describe nouns (things, people, ideas, places, etc.). or as part of a compound tenseA compound tense is formed using at least two verbs, as in "I am running.". There are two kinds of participles in English: the present participle: verb + -ing Used for: the continuous tenses (e.g. He is running) describing a noun that performs or does an action (e.g. a sleeping baby) the past participle: verb + -ed or an irregularIrregular is a word we use in grammar to describe a word form that isn’t formed according to the usual grammar rules. form Used for: the perfect aspect (e.g. He has walked) passives (e.g. She was seen) describing a noun that underwent or experienced an action (e.g. a burnt pancake) In this post we’ll go through how to form and use both kinds of participles in detail, then we’ll go through all of their major uses. Then, for more advanced learners, we’ll look at how to form different kinds of participle phrases in English. Now that I’ve saidthat, let’s get down to a concentratedexplanation of the entertainingworld of participles in English! Having read that introduction, you can bet that lots of information will be covered! Table of Contents What are the types of participles in English? How to form the present participle? How to form the past participle? How to use participles in English? When are participles used in verb tenses? Using the present participle in the past, present, and future continuous Using the past participle in the past, present, and future perfect How are participles used as adjectives? Past participles vs. present participles as adjectives How to use the past participle in the passive voice? An advanced use of participles: What is a participle phrase? What is a “classic” participle phrase? What is a perfect participle? How to use participles in reduced relative clauses? What is a dangling participle? Summary What are the types of participles in English? The words we call participles are actually two verb forms in English: the present participle (also called the “active participle” or “imperfect participle”) being, hoping, going, having… the past participle (also called the “passive participle”) root ... or an irregularIrregular is a word we use in grammar to describe a word form that isn’t formed according to the usual grammar rules. form been, hoped, gone, had, walked... Please don’t be confused by the words “present” and “past”! We use these participles in all three tenses: past, present, and future! A better way to think of these is as “finished” (past) and “unfinished” (present) participles. Want to learn why? Keep reading! Let’s get down to brass tacks and look at them one at a time! How to form the present participle? To form the present participle, just take the basic rootNo definition set for rootLorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. of the verb and add -ing! It’s easy! For example: | | Present participle | --- | | | dancing | | | cooking | | | studying | | | cutting | | find | finding | | speak | speaking | Tip Though present participles are completely regular in speech, there are some special spelling rules to learn. For example: make → making → e at the end is deleted cut → cutting → some final consonants are doubled Luckily these rules do not affect pronunciation and they are pretty simple to learn. Have a look at our reference sheet for spelling words with suffixes in English to learn more. How to form the past participle? There are two kinds of past participles in English: regular and irregular Regular participles have a predictable form: root + -ed: | Root | Past participle | --- | | dance | danced | | look | looked | | study | studied | ###### Tip When we add -ed to the end of an English word, there are some spelling rules you will have to follow. For example: study → studied → y at the end becomes i + dance → danced → e at the end is dropped + stop → stopped → certain final consonants are doubled Luckily these rules do not affect pronunciation and they are pretty simple to learn. Have a look at our reference sheet for spelling words with suffixes in English to learn more. Irregular participles are a little trickier because they have forms that must be memorized. There are about 200 irregular verbs in English, and you can find their participles in the third column of most irregular verb lists: | Root | Simple past | Past participle | --- | be | was / were | been | | cut | cut | cut | | have | had | had | | go | went | gone | | speak | spoke | spoken | ###### Tip Have a look at this chart of irregular verbs in English! We’ve given you two versions: A version that sorts the verbs by skill level (A1-C2) A version that sorts the verbs into categories to help you see the different patterns How to use participles in English? You will probably first use participles to form the continuous and the perfect aspectNo definition set for aspectLorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. in each tense: participle Continuous: I am teaching you participles right now. Perfect: You have seen participles used as verb forms! But we also use participles as adjectives, that is, words that modify nouns: participle noun Use our learning guides to help you understand. Please clean up the broken glass! Grammar term watch! When we use a verb as a noun, we call this a gerund in English! A gerund looks exactly like a present participle (root + -ing), but because it’s used as a noun, we call it something else. gerund Speaking two languages is important in today’s world Your reading has improved a lot since last September. Past participles are also used to form the passive voice in English – sentences where the subjectNo definition set for subjectLorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. has an action done to them: Active voice: Pedro ate a pear. → subject does the action Passive voice: A pear was eaten (by Pedro). → action is done to the subject Tip In this post we’re looking at the main uses of participles in English, but you will find them in a few other cases as well! Now, let’s look at each of these uses in detail! When are participles used in verb tenses? In all three tenses (past, present, and future) participles are used to form verbs in the continuous and perfect aspects. To do this, each participle is used with a specific auxiliary (helping) verb: the present participle goes with auxiliary be and the past participle goes with the auxiliary have. Continous aspect: be + present participle Perfect aspect: have + past participle Take a look at how this works in all three tenses! Using the present participle in the past, present, and future continuous To use the present participle in the past, present, or future continuous, just change the tense of the auxiliary verb be to the past, present, or future! The present participle does not change. Past continuous: Mark was playing the piano in the studio yesterday. We were playing... Present continuous: I am reading an interesting book right now. You are reading… She is reading… Future continuous: Patty will be entertaining guests at her house tonight. The continuous aspect is usually used for actions that continue for a period of time, but its use is a little different in each tense. Check out our posts on the present, past, and future continuous to learn more! Using the past participle in the past, present, and future perfect To use the past participle in the past, present, or future perfect, just change the tense of the auxiliary verb haveto the past, present, or future. The past participle does not change. Past perfect: Peter had seen that movie many times before. Present perfect: They have finished their test now. She has finished… Future perfect: Muriel will have made twelve apple pies before the end of the week. The perfect aspect is usually used to talk about something that finishes before an action or time that is the focus of a story. I have seen this movie before… → before "now" …so I know that it is good. ...so I do not need to see it again. → "now": the focus of the story. We use the perfect when the fact that an earlier action happened impacts how we understand the event or situation we are focused on. Read more in our posts on past, present, and future perfect! Tip The perfect continuous tenses in English use both a past participle and a present participle! I had been wondering about that! How are participles used as adjectives? Both present participles and past participles are commonly used as adjectives. That means that a participle can modify a noun (or noun phrase!), adding more information about it Just like any adjective, participles as adjectives are usually placed before the noun they modify: adjective noun Patty doesn’t like raw fish. There was a large cat in the window. participle as adjective noun Patty likes fried fish. There was a sleeping cat in the window. Like other adjectives, past participles, are often placed after a linking verb like be or seem: linking verb adjective That fish is raw. That cat looks sleepy. linking verb past participle This fish is fried. This cat looks tired. But present participles usually do not follow a verb (except for be in a continous tense!) linking verb present participle (as adjective) ❌ This cat looks sleeping. ❌ My eggs seemed burning. ✅ My eggs were burning. → This is the past continuous tense! Past participles vs. present participles as adjectives Past participles are used to describe nouns that underwent a finished action, but present participles describe nouns that are doing the action. For example: Past participles: This is fried fish. → Someone fried the fish. This is a forgotten toy. → Someone forgot the toy. This is a grown man. → The man is finished growing. This is a painted house. → Someone painted the house. Present participles: This is a sleeping baby → The baby is sleeping There is a leaking sink. → The sink is leaking. This is a running horse. → The horse is running This is a growing boy. → The boy is growing Tip Gerunds (verbs used as nouns) can be used to describe the purpose of a noun. Because gerunds look just like present participles (verb + -ing), you’ll need to use context to decide the meaning: a cleaning rag: ✅ a rag for cleaning (gerund) ❌ the rag is cleaning (participle) a growing boy: ❌ a boy for growing (gerund) ✅ the boy is growing (participle) How to use the past participle in the passive voice? Use the past participle after the verb be to form the English passive voice: be past participle The subject of a passive verb is the noun that undergoes an action, not the noun that does the action: The ball was kicked. The book will be purchased for John. Mary is being given a present. Notice that the past participle does not change when you change the tense of be! In English, we use the passive voice when the object of an action is the topic of a sentence (the noun a sentence is “about”). For example: I was talking to Mary’s mom today, and she said that Mary was caught using her cell phone in class so she got punished. → Mary is the topic of conversation, so even though she is the object of the “catching” action, she is the subject of the verb. An advanced use of participles: What is a participle phrase? A participle phrase is a phrase that begins with either a past or present participle followed by some associated words. There are actually two types of participle phrases in English: “Classic” participle phrases (surrounded by commas) Remembering to bring his lunch, Tom left the house. Reduced relative clauses (not surrounded by commas) Rita visited a portrait painted by da Vinci. Let’s talk about each type, and then we’ll talk about “dangling participles,” which are a common writing error that happens when you confuse the two! What is a “classic” participle phrase? A “classic” participle phrase is a participle phrase that is surrounded by commas and can only describe the subject of a sentence. These are often used in writing, but are less common in speech. A classic participle phrase can come in three main places: At the beginning of the clause: participle phrase subject Exhausted by the hike, Tom fell onto the sofa and groaned. Remembering to bring his lunch, Tom left the house. After the verb and object / at the end of the clause: subject participle phrase Tom fell onto the sofa, exhausted by the hike. Tom left the house, remembering to bring his lunch. After the subject (uncommon): subject participle phrase Tom, exhausted by the hike, fell onto the sofa. Tom, remembering to bring his lunch, left the house. Notice that these participle phrases are always separated out by commas, and they always describe the subject. What is a perfect participle? A perfect participle is a type of “classic” participle phrase that is formed from a verb in the perfect aspect. So they look like this: having past participle having slept all night... having forgotten his teddy bear.. The perfect participle tells us about something that the subject of the sentence completed before the main action. Look: past participle Having finished her work, Mary turned off her computer and left the office. → Mary finished her work before she turned off the computer and left. How to use participles in reduced relative clauses? A reduced relative clause can also begin with a participle, but these follow different grammar rules. This is because they are actually just English relative clauses with some words deleted. For example: | Full relative clause | Reduced relative clause | --- | | Rita watched her friend who was cutting the grass. | Rita watched her friend cutting the grass. | | Rita bought a table that wasmade of wood. | Rita bought a table made of wood. | Recall that relative clauses in English are clauses (with a subject and a verb) that describe a noun. If the verb in the relative clause is be and it is followed by a participle, you can delete the relative pronounNo definition set for relative pronounLorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. and the verb be. This is how we create reduced relative clauses that begin with participles! But because reduced relative clauses are still relative clauses, they follow the rules that relative clauses follow. For example: Reduced relative clauses can describe any noun in the sentence, not just the subject: Rita watched her neighbor cutting the grass. → The neighbor cut the grass. + Rita bought a table made from wood. → The table is made from wood. Reduced relative clauses always come right after the noun phrase they describe, but they do not need to be separated by commas. ❌ Cutting the grass, Rita watched her neighbor. → This is only allowed if it is a classic participle phrase, describing Rita like the examples we saw above. What is a dangling participle? A dangling participle is a common writing error in English that happens when it is unclear which noun a participle phrase describes. Sometimes, this is because the noun the participle phrase describes is completely missing from the sentence: ❌ Walking in the garden, weeds sprouted everywhere. → The weeds are walking in the garden?!? Fixing this is easy! Just make sure there is a subject in the sentence! ✅ Walking in the garden, Mary saw weeds everywhere. → Now we know who was walking in the garden! Other times it happens because you are using a “classic” participle phrase without the proper punctuation (no commas), so it looks like it is a reduced relative clause! ❌ Tom watched the zebra sitting on the porch. → Intended meaning: Tom was sitting on the porch. Tom saw a zebra. → Actual meaning: Tom saw a zebra. The zebra was sitting on the porch. This second kind of dangling participle is easy to fix in writing (just add the commas!) but it can be confusing in speech, because we cannot see the commas! Even in writing, it is better to rephrase a sentence like this, because they can confuse readers. If you want to be very clear, it is best to put a “classic” participle phrase close to the noun it describes, if you can! ✅ Sitting on the porch, Tom watched the zebra. A third kind of dangling participle comes about because a few English prepositions come from present participles. For example: Rita watched her neighbor using binoculars. This sentence is correct, but has two different meanings: If we read the word using as a participle, this is a reduced relative clauses, so the sentence means: Rita watched her neighbor who was using binoculars. + But if we read the word using as a prepositionNo definition set for prepositionLorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum., then the phrase using binoculars tells us how the watching happened instead. So the sentence means: Rita watched her neighbor by means of binoculars. To clarify sentences like these, it is best to just rephrase what you are trying to say! Good for you! You’ve discovered a lot more about the participle in English! Having read this post, you’ve seen that: there are two types of participles in English: the present participle (root + -ing) the past participle (root + -ed / irregular form) participles are used to form… the continuous tenses (be + present participle) the perfect tenses (have + past participle) participles are also used as adjectives to modify nouns participles can be combined with other words to make participle phrases, which fall into two main groups: “classic” participle phrases → separated by commas, describe the subject reduced relative clauses → no commas, follow the noun they describe That was a bunch of information there! If you’re ready to practice, have a look at our English participle activities. We’ve also got a useful chart that covers comma use with participles and participle phrases. To embark on your next language adventure, join Mango on social! Ready to take the next step? 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https://www.sciencedirect.com/science/article/abs/pii/S0387760498000229
CT and MR imaging of cerebral tuberous sclerosis - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (57) Cited by (66) Brain and Development Volume 20, Issue 4, April 1998, Pages 209-221 Review article CT and MR imaging of cerebral tuberous sclerosis Author links open overlay panel Yuichi Inoue a, Yutaka Nemoto a, Ryuusuke Murata b, Takahiko Tashiro a, Miyuki Shakudo a, Kinuko Kohno a, Osamu Matsuoka c, Kunizo Mochizuki a Show more Add to Mendeley Share Cite rights and content Abstract Tuberous sclerosis is a heredofamilial neurocutaneous syndrome, or phakomatosis, with multisystem involvement including the brain, kidney, skin, retina, heart, lung, and bone. The brain is the most frequently affected organ in tuberous sclerosis. Brain lesions in tuberous sclerosis are of three kinds; cortical tubers, white matter abnormalities, and subependymal nodules. We review the computed tomography (CT) and magnetic resonance (MR) features of the brain lesions in patients with tuberous sclerosis. CT clearly demonstrates calcified subependymal nodules. MR imaging demonstrates more clearly cortical, and white matter lesions than CT, since MR imaging shows excellent image contrast between various normal structures and high sensitivity in detecting pathological states due to intrinsic differences in proton density and in particular, in proton relaxation times of tissues. Possible pathogenesis of this disorder is also discussed Introduction Tuberous sclerosis is a heredofamilial neurocutaneous syndrome, or phakomatosis, with multisystem involvement including the brain, kidney, skin, retina, heart, lung and bone. The brain is the most frequently affected organ in tuberous sclerosis. Although the earliest report of a patient with tuberous sclerosis is said to have been by von Recklinghausen (1862) in a neonate with cardiac rhabdomyomata, the name tuberous sclerosis was introduced by Bourneville (1880) in a 3-year-old girl with mental retardation, seizures and facial angiofibromas (cited in Ref. ). In 1890, Pringle described the facial nevi of adenoma sebaceum. Vogt later emphasized the classic triad of seizures, mental retardation, and sebaceous adenoma. The eponym Pringle disease is used when there are only dermatological findings, and Bonneville disease when the nervous system is affected. A recent Swedish study found the prevalence to be at least one in 6800 among children aged 11–15 years and one in 12 900 in a population aged 0–20 years . No racial or sexual predilection has been detected. Familial cases are inherited in an autosomal dominant fashion, but most cases probably arise sporadically; the rate of spontaneous mutation approaches 60–75% 2, 3, 4, 5. The tumor-like growth in different organs may include cells of more than one type, and examples of these are fibroblasts, cardiac myoblasts and angioblasts or glioblasts and neuroblasts 2, 3. Due to multiple organ involvement and genetic heterogeneity, the disorder is called `tuberous sclerosis complex' (TSC). Linkage of tuberous sclerosis to markers on chromosome 9q was first reported in 1987 and in further details in the recent literature . It soon became clear that there is genetic heterogeneity. The gene on chromosome 9 is called TSC1. Another group has revealed an important tuberous sclerosis locus on chromosome 16p13 10, 11, which was confirmed later on and designated as TSC2. No significant phenotypical differences have been discovered between TSC1 and TSC2. The traditional clinical triad of adenoma sebaceum, seizures, and mental retardation has been refined with appreciation of the complexity of the disorder. Recently, diagnostic criteria have been established by the National Tuberous Sclerosis Association (Table 1). A definitive diagnosis is established when one of the following are present: (1) cortical tubers or subependymal nodules, (2) multiple retinal astrocytomas, (3) facial or truncal adenoma sebaceum (angofibromas) or ungual angiofibromas, or (4) subependymal giant cell astrocytomas. In the absence of a primary finding, the diagnosis can be made if any two of the following are present: hypopigmented macules, shagreen patches, a single retinal tuber, multiple renal hamartomas (angiomyolipomas), cardiac rhabdomyoma, and tuberous sclerosis in a first-degree relative. Many of the clinical manifestations such as facial angiofibromas, ungual fibromas, and subependymal nodules, may not be apparent in infancy and childhood. Mental retardation and seizures are common in this disorder, but because of the lack of specificity they are not included in the diagnostic criteria. As described earlier, the diagnosis of tuberous sclerosis has been based on the classic triad of characteristic cutaneous lesions, seizures, and mental retardation. Because intellectual deficits are seen in only 40% of patients and the cutaneous and intellectual deficits are not clinically apparent in the first 2–3 years of life, the radiologic manifestations of tuberous sclerosis have taken on considerable clinical importance in the diagnosis of this disorder in infants and young children. Plain radiographs of the skull demonstrate periventricular calcification infrequently (Fig. 1). Historically, pneumoencephalography demonstrated the characteristic `candle guttering' appearance of nodular subependymal masses. The cortical tubers that dominate the gross pathologic appearance of this disorder, however, were generally not discernible on these plain radiography and pneumoencephalography. Computed tomography (CT) and magnetic resonance (MR) imaging revolutionized the role of radiologic evaluation of this disorder. CT clearly demonstrates calcified nodules, but it fails to demonstrate cortical tubers in spite of the presumed presence of such lesions in most patients. MR imaging demonstrates cortical tubers and white matter lesions above a certain size, but it frequently fails to depict subependymal nodules. MR imaging utilizes programmed sequences of pulses to produce different kinds of images such as T1-weighted, T2-weighted, and proton-weighted images. In this article, unless stated otherwise, T1-weighted images indicate spin-echo T1-weighted images (TR: 500–600 ms, TE: 15–20 ms), and T2-weighted images indicate either spin-echo T2-weighted images (TR: 2000–3000 ms. TE: 90–100 ms) or fast spin-echo T2-weighted images (TR: 4000–6000 ms, Teff: 80–100 ms). The cerebral lesions in tuberous sclerosis are of three kinds: cortical tubers, white matter abnormalities, and subependymal nodules. They are almost always benign hamartomas. Bender and Yunis have suggested that the same cellular components are present in all the brain lesions in tuberous sclerosis, and that they represent a combination of both neuronal and astrocytic features. Cortical tubers are the most characteristic lesions of tuberous sclerosis at pathologic examination. Varying in size from millimeters to several centimeters, tubers are rounded or wart-like protrusions of single or adjacent gyri, very firm to touch and pale in color 1, 5, 13. They may be wide and flat or dimpled. Tubers expand the gyri and blur the margin between the gray and white matter 1, 13. They may be present in the depths of sulci. Histologically, tubers are characterized as atypical giant astrocytes, abnormal malorientated neurons or many indeterminate cells, and lack the normal six-layered lamination of the cortical gray matter. Numerous glial processes and fibers, especially in the subpial layers, make the tissue abnormally firm or sclerotic on palpation 5, 6. They also have gliosis and abnormal myelination. The gliosis and absence of myelin in cortical tubers may extend into the underlying subcortical white matter. Tubers are occasionally present in the cerebellum; disorganized cortex, abnormal astrocytes and Purkinje cells, and calcification are the main features 1, 13. On MR imaging, tubers are found in 95% of patients with tuberous sclerosis 14, 15. They show somewhat thick cortical gray matter and a less distinctive gray-white matter junction compared with the normal cortex 15, 16, 17(Fig. 2). The peripheral component of tubers is isointense to mildly hyperintense to normal gray matter on both T1- and T2-weighted images, and the inner core of tubers is isointense to hypointense to gray matter on T1-weighted images and hyperintense on T2-weighted images 15, 16, 17, 18(Fig. 2, Fig. 3). These signal intensity changes reflect increased interstitial water in the subcortical area due to absence of myelin and looseness of tissue structures. The inner core of tubers may show a signal that is isointense to cerebrospinal fluid on both T1- and T2-weighted images 18, 19(Fig. 3). Although uncommon, part of the inner core may be hyperintense to gray matter on T1-weighted images 18, 19. Cerebrospinal fluid in cerebral sulci or over the brain surface shows a hyperintense signal on T2-weighted images and may produce a partial volume effect, which may obscure small cortical tubers. Recently developed pulse sequence, fluid attenuated inversion recovery (FLAIR) sequence, which produces T2-weighted images with cerebrospinal fluid appearing hypointense , appears to delineate cortical lesions as well as small white matter lesions clearly, and may be helpful occasionally 21, 22(Fig. 3). Magnetic transfer imaging, another new technique to suppress signals from protein-bound water protons, seems an effective modality improving the detectability and signal noise ratio for all intracranial lesions of tuberous sclerosis . In infants less than 1 year old, and particularly in those less than 6 months old, the appearance of cortical tubers is different compared with the children older than 2 years, when myelination of white matter comes close to the adult pattern . In the brain of young infants, there is less myelination, and a high water content in white matter is present, which results in hyperintensity on T2-weighted images . Thus, the inner core of cortical tubers in very young children is hyperintense to premyelinated white matter on T1-weighted images and is hypointense to premyelinated white matter on T2-weighted images 14, 25, 26(Fig. 4). In adult patients with tuberous sclerosis, high signal intensity in tubers on T2-weighted images is less often seen than in children. In our patients who had follow-up MR studies, the high signal on T2-weighted images seen at a younger age became less distinct and often disappeared (Fig. 5). These suggest that dysmyelinated white matter subjacent to tubers may gain myelination with increasing age. Enhancement of tubers following intravenous administration of gadopentate meglumine (GdDTPA) has been reported to occur in less than 5% of patients . Some cortical tubers may distort the adjacent gyrus but do not demonstrate high signal intensity on T2-weighted images (Fig. 2, Fig. 8). Visual inspection of the cortical areas on T2-weighted images only fails to identify tubers that do not show hyperintensity. Therefore, it is important to look for gyral deformity (usually expanded), abnormal thickening of the cortical gray matter, and/or blurring of the gray-white matter junction. When only the latter findings are observed, cortical dysplasia must be considered 27, 28. It should be kept in mind that formes frustes of tuberous sclerosis are not uncommon; a solitary cortical tuber unassociated with other features of tuberous sclerosis should not be misdiagnosed as a brain tumor 28, 29. On CT scans, cortical tubers can occasionally be detected as a localized low-density area 30, 31, which probably reflects the inner core of the tuber (Fig. 3). Calcification in cortical tubers has been reported as high as 54% . The cortical calcification may be gyriform, simulating the appearance of Sturge–Weber syndrome . In young infants, cortical tubers may be radiologically occult (Fig. 6). With increasing age, the low-density area of the inner core of tubers diminishes and becomes less distinctive, and calcifications may appear . Subependymal nodules are found in 95% of patients with tuberous sclerosis 14, 15. Subependymal nodules may occur in the third and fourth ventricular walls, but most are found in the lateral ventricular walls, near the sulcus terminalis, with their deeper parts embedded in the caudate or thalamus. They are firm, or stony hard due to calcification, and form round or elongated protrusions into the ventricles 1, 13. Calcification is common with increasing age. Multiple adjacent subependymal nodules resemble dripping wax, hence the nickname `candle guttering' for the pneumoencephalographic appearance of these lesions. Histologically the nodules are composed of giant cells with glial and neural features and may have many vascular elements to account for the tumor stain, while in neonates there may be scattered neuroblasts 1, 4. They are covered by an intact layer of ependyma. On CT scans, subependymal nodules are easily detected because of calcification. They are rarely calcified in the first year of life; the number of calcifications increases with the age of the patient . Calcification may be globular, partial, or ring-like in appearance 17, 19(Fig. 3, Fig. 7). MR imaging is less sensitive in depicting calcified nodules compared with CT. Gradient echo pulse sequences are known to be more sensitive for detection of calcium, iron, or other metals. Gradient echo T2-weighted images (T2 images) have been reported to be useful in detecting calcified nodules in patients with tuberous sclerosis . Routine clinical use of T2 pulse sequences, we believe, is not necessary to study patients with tuberous sclerosis because CT is widely available. Subependymal nodules compared with cortical gray matter, are commonly isointense to hyperintense on T1-weighted images and isointense to hypointense on T2-weighted images (Fig. 2, Fig. 3 and 10) 16, 17, 18. Hypointensity on T2-weighted images is a reflection of the calcification in nodules. Relative hyperintensity on T1-weighted images seems characteristic of subependymal nodules but may partly reflect mild calcification itself : calcification of less than 30% has been shown to cause hyperintensity on T1-weighted images . On Gd-DTPA MR, 30–80% of subependymal nodules show enhancement 15, 36, 37with a nodular, heterogeneous, or ring-like appearance. Enhancement of nodules does not necessarily denote neoplastic transformation . Periodic follow-up studies of enhancing lesions near the foramen of Monro are indicated because subependymal giant cell tumors occur almost exclusively at this site. Grobus et al. were the first to report the association of subependymal giant cell tumor with tuberous sclerosis (cited in Ref. ). The average age at the time of diagnosis is about 13 years old, and most commonly between 5 and 10 years. Subependymal giant cell tumors differ from the subependymal nodules by their size and tendency to enlarge. Their incidence in tuberous sclerosis is approximately 10–15% 4, 15, 33. Prominent blood vessels are frequently present in the tumors . It is believed that subependymal giant cell tumors originate from subependymal nodules of tuberous sclerosis 39, 40. On CT scans, subependymal giant cell tumors are identified near the foramen of Monro as a partially calcified, solid mass with either ipsilateral or bilateral ventricular enlargement, and they enhance to various degrees after intravenous contrast injection (Fig. 7, Fig. 8). On MR imaging, those tumors are usually heterogeneous but can be homogeneous and are isointense to slightly hypointense on T1-weighted images and hyperintense on T2-weighted images, and show marked enhancement with Gd-DTPA . On rare occasions, they may have a cystic internal component (Fig. 8). Serpentine signal voids which represent dilated vessels are occasionally seen in the tumors 15. Cerebral angiography may demonstrate tumor vessels as well as a tumor stain . Although uncommon, subependymal giant cell tumors may bleed, resulting in intraventricular hemorrhage 42. Why subependymal giant cell tumors develop only in the area of the foramen of Monro has not been determined. Uncommonly, other cerebral tumors also occur in patients with tuberous sclerosis. These tumors seem to be an incidental occurrence and not specific to tuberous sclerosis. These include pilocytic astrocytoma, fibrillary astrocytoma, and diffuse gliomatosis of the cerebral hemispheres . Islets consisting of a group of heterotopic clustered cells are invariably present in the white matter of patients with tuberous sclerosis 1, 13. Histologically the cells are often bizarre and gigantic, with characteristics of both neurons and glia . They are associated with areas of hypomyelination similar to those seen in the cortical tubers. Many of these lesions are microscopic and therefore do not appear on imaging studies . Low power microscopic examination of heterotopic clusters suggests that they are distributed along the most direct line between the ependymal wall and the tubers . This distribution corresponds to the normal migratory path of spongioblasts during embryogenesis, and it is during this migration that they differentiate 13, 43. On CT, those lesions large enough to be seen demonstrate decreased density in the white matter without enhancement after intravenous contrast administration. Calcification may occur in white matter lesions (Fig. 9). On MR, these show similar signal intensity to the cortical tubers: isointense to hypointense on T1-weighted images and hyperintense on T2-weighted images. They demonstrate three distinctive patterns: straight or curvilinear bands 44, 15that extend from the ventricular wall through the cerebrum toward the cortex (Fig. 10); wedge-shaped lesions; non-specific conglomerate foci (Fig. 3, Fig. 5); and cerebellar radial bands . Approximately 12% of these white matter lesions have been reported to show enhancement after contrast administration . The most interesting among those are straight or curvilinear bands which may reflect a possible pathogenesis of tuberous sclerosis, which will be discussed later. The other type of abnormal signal intensity in the white matter is a hyperintense linear band on both T1-weighted and T2-weighted images, which has been reported previously but not widely discussed, to our knowledge. These lesions are uncommon and some of them appear to enhance after contrast administration (Fig. 10, Fig. 11). On CT, these lesions are not visualized. Corresponding histologic investigation has not been done, but the band appears to correspond to lesions described by Donegami et al. have described heterotopic clusters in white matter, some of which were distributed in a pattern similar to the normal migratory path of primitive neuroepithelial cells during embryogenesis . The reason why these lesions are hyperintense on T1-weighted images is not understood and microcalcification has been suggested . Hypermyelination is another possible cause for the T1 hyperintensity, but pathologically has not been proven. Abnormal enhancement is probably related to the absence or incompleteness of the blood brain barrier in these lesions. To maintain an intact blood brain barrier, astrocytes require normal structure and function . Abnormal enhancement on contrast MR images reflects the presence of abnormal astrocytes (bizarre-shaped astrocytes), which may not function as normal astrocytes or may not produce the necessary protein to maintain tight junctions (cited in Ref. ). In neonates or young infants, white matter lesions look somewhat different, and are hyperintense to premyelinated white matter on T1-weighted images and hypointense to premyelinated white matter on T2-weighted images. Cerebellar lesions have been described in tuberous sclerosis, but they are uncommon, occurring in approximately 10% of patients 33, 46. Lesions of the cerebellum are similar to those seen in the cerebral hemispheres, consisting of cortical tubers, heterotopic clusters in the white matter and subependymal nodules 1, 4, 12. On microscopic examination, bizarre giant cell clusters are the main feature of cortical tubers, white matter lesions, and subependymal nodules. Much discussion has focused on the ontogenesis of these bizarre giant cells in tuberous sclerosis. Classic morphologic studies by light microscopy and electron microscopy have demonstrated astrocytic features . Immunohistochemical studies have shown neuronal features . Investigations using light microscopy, electron microscopy, and immunohistochemical studies, when considered together, have indicated that these bizarre giant cells have both astrocytic and neuronal features in cortical tubers, subependymal nodules, white matter lesions, and even subependymal giant cell tumors . These findings suggest that the giant cell in tuberous sclerosis is the product of a dysgenetic event in early development, resulting in incomplete expression of astrocyte or neuronal differentiation in that cell 12, 49. At histologic examination, the white matter lesions of tuberous sclerosis show clusters of giant cells that characteristically align in rows that appear to follow the path of neuronal migration, streaming radially along the most direct line from the ependyma to the cortical tuber 12, 13. Although straight or curvilinear lines that extend from the ventricular wall toward the cortex seen on MR seem to represent a migration anomaly, it may not be the primary abnormality of tuberous sclerosis but the secondary change due to the presence of bizarre-shaped giant cells. At present, it is reasonable to speculate as follows concerning brain lesions of tuberous sclerosis. The primary brain abnormality in tuberous sclerosis appears to be in some of the germ cells in the germinal matrix 50, 24, probably controlled by genes. Some dysgenetic giant cells in the ventricular wall may completely migrate to the cortex, producing cortical tubers; some may incompletely migrate, producing white matter lesions; some may not migrate at all, producing subependymal nodules 12, 15, 24. There has been much attention paid to types of seizures in tuberous sclerosis patients, particularly infantile spasms, and their possible relation to the anatomic location of tubers. A statistical analysis by Shephard et al. has shown that the occurrence of infantile spasms rather than any other type of seizure was related to the total number of tubers and was not related to the number in any one area of the brain . Several investigations have been unable to demonstrate a link between intelligence and the number of subependymal nodules 18, 52. In our experience with a limited number of patients, severely mentally retarded patients tended to have a higher number of tubers; no correlation was noted between the severity of mental retardation and either the number of subependymal nodules or the degree of ventricular dilatation 18, 52. Shephard et al. have reported that all tuberous sclerosis patients with mental retardation had or have been had some type of seizure, usually generalized and very often infantile spasms. and that none of the patients who never had seizures were mentally disabled. Seizure-free tuberous sclerosis patients had a larger number of cortical tubers than the patients with partial seizures. Ocular findings are common and present in approximately 50% of patients with tuberous sclerosis 13, 53. The most common of these is retinal hamartoma, an astrocytic proliferation without necrosis, hemorrhage, or dissemination 13, 54, 55. Retinal hamartomas are usually present in both eyes and are often multiple 13, 54. They may present in children as leukocoria. They must be differentiated from non-neoplastic causes of leukocoria as well as from neoplastic masses, notably retinoblastoma . When retinal hamartomas calcify, they can be seen on CT images as small calcifications in the region of the optic nerve head . When they do not calcify, they are difficult to detect on either CT or MRI, even after contrast administration. There have been several case reports of patients with tuberous sclerosis associated with intracranial aneurysms 56, 57, 19. Although these intracranial aneurysms may be merely a coincidental finding, these reports suggest some association of intracranial aneurysms with tuberous sclerosis. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Conclusion This article presents CT and MR findings of cortical tubers, white matter lesions, subependymal tubers, and subependymal giant cell tumors. New findings on MR images are particularly emphasized. Many of cortical tubers expand cerebral gyri, and demonstrate thickening of the cortex and blurring of the gray and white matter junction. The inner core of cortical tubers is hypointense to nearly isointense to the gray matter on T1-weighted images, and nearly isointense to hyperintense on T2-weighted Recommended articles References (57) A.E. Fryer et al. Connor, et al. Evidence that the gene for tuberous sclerosis is on chromosome 9 Lancet (1987) Harding BN, Malformations of the central nervous system. In: Adams JH, Duchen LW, editors. Greanfield's Neuropathology.... Gold AP. Tuberous sclerosis. In: Rowland LP, editor. Merritt's Textbook of Neurology. ninth ed. Baltimore, MD: Williams... Adams RD, Victor M. Tuberous sclerosis. In: Principles of Neurology. fifth ed. New York: McGraw–Hill,... J.G. Smirniotopoulos et al. The phakomatoses AJNR Am J Neuroradol (1992) R.E. Scully et al. Case records of the Massachusetts General Hospital: case 411986 N Engl J Med (1986) D.W. Webb et al. On the incidence of fits and mental retardation in tuberous sclerosis J Med Genet (1991) G. Ahlsen et al. Tuberous sclerosis in Western Sweden. A population study of cases with early childhood onset Arch Neurol (1994) M. van Slegtenhorst et al. Identification of the tuberous sclerosis gene TSC1 on chromosome 9q34 Cancer (1997) R.S. Kandt et al. Linkage of an important gene locus for tuberous sclerosis to a chromosome 16 marker for polycystic kidney disease Nature Genet (1992) E.S. Roach et al. Diagnostic criteria: tuberous sclerosis complex. Report of the Diagnostic Criteria Committee of the National Tuberous Sclerosis Association J Child Neurol (1992) B.L. Bender et al. Central nervous system pathology of tuberous sclerosis in children Ultrastruct Pathol (1980) Donegani G, Grattarola FR, Wildi E. Tuberous sclerosis. Bouneville disease. In: Vinken PJ, Bruyyn GW editors. Handbook... N.R. Altman et al. Tuberous sclerosis: characteristics at CT and MR imaging Radiology (1988) B.H. Braffman et al. MR imaging of tuberous sclerosis: pathogenesis of this phakomatosis, use of gadopentetate dimeglumine, and literature review Radiology (1992) S.K. McMurdo et al. MR imaging of intracranial tuberous sclerosis Am J Roentgenol (1987) J.R. Nixon et al. Cerebral tuberous sclerosis: MR imaging Radiology (1989) Y. Inoue et al. Magnetic resonance images of tuberous sclerosis Further observation and clinical correlations. Neuroradiology (1988) N. Martin et al. MRI evaluation of tuberous sclerosis Neuroradiology (1987) J.V. Hajnal et al. Use of fluid attenuated inversion recovery (FLAIR) pulse sequences in MRI of the brain J Comput Asist Tomogr (1992) J. Takanashi et al. MR evaluation of tuberous sclerosis: increased sensitivity with fluid-attenuated inversion recovery and relation to severity of seizures and mental retardation AJNR Am J Neuroradiol (1995) T. Kato et al. Improved detection of cortical and subcortical tubers in tuberous sclerosis by fluid-attenuated inversion recovery MRI Neuroradiology (1997) M.G. Jeong et al. Application of magnetization transfer imaging for intracranial lesions of tuberous sclerosis J Comput Assist Tomogr (1997) A.J. Barkovich et al. Normal maturation of the neonatal and infant brain: MR imaging at 1.5 T Radiology (1988) C. Christophe et al. Neonatal tuberous sclerosis. US, CT, and MR diagnosis of brain and cardiac lesions Pediatric Radiol (1989) Barkovich AJ. Tuberous sclerosis. In: Pediatric Neuroimaging. second ed. New York: Raven Press,... R. Kuzniecky et al. Cortical dysplasia in temporal lobe epilepsy: magnetic resonance imaging correlations Ann Neurol (1991) A. Yagishita et al. Focal cortical dysplasia: appearance on MR images Radiology (1997) View more references Cited by (66) Updated International Tuberous Sclerosis Complex Diagnostic Criteria and Surveillance and Management Recommendations 2021, Pediatric Neurology Citation Excerpt : All individuals suspected of having TSC, regardless of age, should undergo MRI of the brain to assess for the presence of cortical or subcortical tubers, other types of neuronal migration defects, subependymal nodules, and subependymal giant cell astrocytomas (SEGAs). If MRI is not available or cannot be performed, CT or head ultrasound (in neonates or infants when fontanels are open) may be used, although these frequently will not detect all abnormalities revealed by MRI23,24 (Category 1). Recently, gadolinium deposition in the brain has drawn attention,25 although newer macrocyclic gadolinium agents lower the risk of depositions compared with older linear gadolinium agents26,27 and the long-term clinical implications are unknown.25,28 Show abstract Tuberous sclerosis complex (TSC) is an autosomal dominant genetic disease affecting multiple body systems with wide variability in presentation. In 2013, Pediatric Neurology published articles outlining updated diagnostic criteria and recommendations for surveillance and management of disease manifestations. Advances in knowledge and approvals of new therapies necessitated a revision of those criteria and recommendations. Chairs and working group cochairs from the 2012 International TSC Consensus Group were invited to meet face-to-face over two days at the 2018 World TSC Conference on July 25 and 26 in Dallas, TX, USA. Before the meeting, working group cochairs worked with group members via e-mail and telephone to (1) review TSC literature since the 2013 publication, (2) confirm or amend prior recommendations, and (3) provide new recommendations as required. Only two changes were made to clinical diagnostic criteria reported in 2013: “multiple cortical tubers and/or radial migration lines” replaced the more general term “cortical dysplasias,” and sclerotic bone lesions were reinstated as a minor criterion. Genetic diagnostic criteria were reaffirmed, including highlighting recent findings that some individuals with TSC are genetically mosaic for variants in TSC1 or TSC2. Changes to surveillance and management criteria largely reflected increased emphasis on early screening for electroencephalographic abnormalities, enhanced surveillance and management of TSC-associated neuropsychiatric disorders, and new medication approvals. Updated TSC diagnostic criteria and surveillance and management recommendations presented here should provide an improved framework for optimal care of those living with TSC and their families. ### Tuberous sclerosis complex surveillance and management: Recommendations of the 2012 international tuberous sclerosis complex consensus conference 2013, Pediatric Neurology Show abstract Tuberous sclerosis complex is a genetic disorder affecting every organ system, but disease manifestations vary significantly among affected individuals. The diverse and varied presentations and progression can be life-threatening with significant impact on cost and quality of life. Current surveillance and management practices are highly variable among region and country, reflective of the fact that last consensus recommendations occurred in 1998 and an updated, comprehensive standard is lacking that incorporates the latest scientific evidence and current best clinical practices. The 2012 International Tuberous Sclerosis Complex Consensus Group, comprising 79 specialists from 14 countries, was organized into 12 separate subcommittees, each led by a clinician with advanced expertise in tuberous sclerosis complex and the relevant medical subspecialty. Each subcommittee focused on a specific disease area with important clinical management implications and was charged with formulating key clinical questions to address within its focus area, reviewing relevant literature, evaluating the strength of data, and providing a recommendation accordingly. The updated consensus recommendations for clinical surveillance and management in tuberous sclerosis complex are summarized here. The recommendations are relevant to the entire lifespan of the patient, from infancy to adulthood, including both individuals where the diagnosis is newly made as well as individuals where the diagnosis already is established. The 2012 International Tuberous Sclerosis Complex Consensus Recommendations provide an evidence-based, standardized approach for optimal clinical care provided for individuals with tuberous sclerosis complex. ### The tuberous sclerosis 2000 study: Presentation initial assessments and implications for diagnosis and management 2011, Archives of Disease in Childhood ### Neurology of the Newborn 2008, Neurology of the Newborn ### Brain abnormalities in tuberous sclerosis complex 2004, Journal of Child Neurology ### Usefulness of diagnostic criteria of tuberous sclerosis complex in pediatric patients 2000, Journal of Child Neurology View all citing articles on Scopus View full text Copyright © 1998 Elsevier Science B.V. All rights reserved. Recommended articles Implications of the local hemodynamic forces on the formation and destabilization of neoatherosclerotic lesions International Journal of Cardiology, Volume 272, 2018, pp. 7-12 Ryo Torii, …, Christos V.Bourantas ### The evaluation of uterine artery embolization as a nonsurgical treatment option for adenomyosis International Journal of Gynecology & Obstetrics, Volume 133, Issue 2, 2016, pp. 202-205 Shaoguang Wang, …, Yaozhong Dong ### Results of celiac trunk stenting during fenestrated or branched aortic endografting Journal of Vascular Surgery, Volume 64, Issue 6, 2016, pp. 1595-1601 Hélène Wattez, …, Stéphan Haulon ### Ectopic arachnoid granulations Cerebrospinal Fluid and Subarachnoid Space, 2023, pp. 169-173 Uduak-Obong I.Ekanem, R. Shane Tubbs ### Thyroid Arterial Embolization for the Management of Benign and Malignant Thyroid Disease: A Systematic Review Endocrine Practice, 2025 Hannelore Iris Coerts, …, Tessa Malaika van Ginhoven ### Can we quantify the risk of embolization for a free-floating thrombus? The Journal of Thoracic and Cardiovascular Surgery, Volume 153, Issue 4, 2017, pp. 804-805 Ruggero De Paulis, Luca Weltert Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. 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https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17?srsltid=AfmBOooQDSNfbz6hyGzjiVlG720AK879wuAOaLFQguZ7uU3gesnvxvjQ
Art of Problem Solving 2020 AMC 12A Problems/Problem 17 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2020 AMC 12A Problems/Problem 17 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2020 AMC 12A Problems/Problem 17 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 5 Video Solution by TheBeautyofMath 6 See Also Problem The vertices of a quadrilateral lie on the graph of , and the -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is . What is the -coordinate of the leftmost vertex? Solution 1 Let the coordinates of the quadrilateral be . We have by shoelace's theorem, that the area is We know that the numerator must have a factor of , so given the answer choices, is either or . If , the expression does not evaluate to , but if , the expression evaluates to . Hence, our answer is . Solution 2 Like above, use the shoelace formula to find that the area of the quadrilateral is equal to . Because the final area we are looking for is , the numerator factors into and , which one of and has to be a multiple of and the other has to be a multiple of . Clearly, the only choice for that is ~Solution by IronicNinja Solution 3 How is a concave function, then: Therefore , all quadrilaterals of side right are trapezius ~Solution by AsdrúbalBeltrán Video Solution by TheBeautyofMath Another example of shoelace theorem included earlier in the video ~IceMatrix See Also 2020 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 16Followed by Problem 18 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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https://www.youtube.com/watch?v=pNkzuxQ0Cb4
Geometry of a Tetrahedron: Question 2: Volume of a Tetrahedron in Vector Coordinate Form Math Easy Solutions 56700 subscribers 11 likes Description 1026 views Posted: 14 Mar 2023 In this video I go over Question 2 of the Discovery Project: Geometry of a Tetrahedron. This question involves determining the volume of a tetrahedron but in vector coordinate form. The formula of the volume of a tetrahedron is V = 1/3·A·h. To get this formula in vector coordinate form, we can determine the area A of the triangular base as being half the cross product. For the height h, we first have to determine the equation of the plane that contains the area A and then calculate the distance from the opposite vertex to the plane. After going over the equation of a plane and the equation for the distance from a point to the plane, we can thus calculate the volume of a tetrahedron for any set of coordinates. I also graph the tetrahedron using the amazing GeoGebra 3D graphing calculator which also calculates its volume as well which I use as a double check. The timestamps of key parts of the video are listed below: Question 2: 0:00 Solution to Part (a): 1:11 The Equation of a Plane: 6:21 The Distance from a Point to a Plane: 10:33 Solution to Part (b): 17:31 Graphing 3D Tetrahedron in GeoGebra: 29:06 This video was taken from my earlier video listed below: Discovery Project: The Geometry of a Tetrahedron: Video notes: Playlist: . Become a MES Super Fan! DONATE! ʕ •ᴥ•ʔ SUBSCRIBE via EMAIL: MES Links: MES Truth: Official Website: Hive: Email me: contact@mes.fm Free Calculators: BMI Calculator: Grade Calculator: Mortgage Calculator: Percentage Calculator: Free Online Tools: iPhone and Android Apps: 1 comments Transcript: Question 2 all right and now let's go over to question two and uh just a note on question two I actually uh finished filming the video and then I realized that uh the actual calculus book had a mistake in it and then I thought it might as well uh redo this and add more clarity even though the final answer is still the same so question two uh the volume v of a tetrahedron is one-third the distance from a Vertex to the opposite face and here's the correction that means correction this is actually the height of the of the tetrahedron so it's not actually the distance from a Vertex to opposite face but it's the height of that tetrahedron from this uh opposite face and I'll uh illustrate that in a bit so times that the area of that face and again I will look to provide a proof in a later video and now look at a question a question a States a fine or part A of question two states find a formula for the volume of a tetrahedron in terms of the coordinates of its vertices P Q R and S and then Part B States find the volume of a tetrahedron whose vertices are p111 and then has Q is one two three then R is one one two and then s is three negative one and two so let's take a look at Solution to Part (a) solution to part A so after after finishing the filming of this video I realized that the formula for a tetrahedron as stated in question two is actually wrong so that's not actually the formula the actual formula in here added a link to the Wikipedia article involves the height of the tetrahedron and not the distance from vertex to opposite face and this is evident from a skewed tetrahedron and here I'll draw it out so let's say we had a tetrahedron like this something that's uh not symmetric like this and uh yeah so let's say that there and then inside there's the Point like this so there's the inside Triangle and goes there's the three triangles around like that and let's say there's a area here in this base there and so the opposite vertices is somewhere here so the distance from this face to that is going to be well something like that but that's not actually what we want the actual formula is involves the height h like that and the volume of the tetrahedron is equal to V 1 3 the height times by that area so this is the volume yeah so it's equal to uh yes one-third the distance is actually it's actually the height so one third the height times the area of that phase so from this area up to there so that's the actual formula all right and going further I think what the discovery project actually meant was the distance to the plane that contains the opposite triangle face which is the same thing as stating the height of the tetrade so this distance right here is the same as saying that the distance from this point to the opposite face or to the plane that consists of this opposite face plane like that um just have it in 3D like that and this this triangle is inside this plane so the distance from that plane that contains a triangle uh to that point I think that is what they were considering let's erase all of this all right so yes going further so thus we can write the volume of a tetrahedron in coordinate form as follows and uh first what I'm going to do is I'm going to actually draw that skewed uh tetration against it so it's more uh clear um so that yeah we're not dealing with yeah so we're dealing with a height so let's just draw this out I'm going to draw it like this and there is our triangle and then inside as a point say this point like this and I'm going to call the point inside here I'm going to call it P actually instead of the S I'm going to say this point s and it has coordinates S1 S2 S3 and then let's say and then the height here or the distance of the plane that contains this this face right there this is going to be our h like that and then this point right here is let's say this is our r says R1 R2 R3 and then this point right here is going to be let's call this q and this coordinates are q1 lowercase uh q1 yeah q1 Q2 Etc and Q3 and then at this point right here we will call this point p and I'll put this like this so that is our p and the coordinates are P1 P2 and P3 like that all right so now that we have this set up so we could write the volume as so we could write V equals one third H times it by the area of this opposite face which is a p the area of this pqr triangle and now what this equals to well let's write down our terms and then the height is let's call this uh let's write this in terms of the distance to the plane just so it's generic so distance called d from the planes we're going to go from this point right here so we'll go from S1 as 2 S3 and then that distance all the way up to the plane that contains pqr like that and then the area of this triangle we know that already that's just this Vector right here p p uh is PQ and this PR and let's put this p as well there and the coordinates our P1 P2 P3 Etc and so we have this I'm going to shift this over like this and then uh now this one here is going to be multiplied by the area of this and that we already saw that earlier is one half the length of the cross product PQ cross PR like that all right and now I'm going to box this in and the solutions that I find online I can find the solution manual the solution I find online just stopped at this point but I'm going to keep going further and get the exact stuff they're asking for which is in coordinate form because this one just says well the distance and so on we don't have a formula yet for the distance all right and now that we have this let's just keep going further and we're The Equation of a Plane going to look at now the equation of a plane because here we're distance from this point s uh for you from the point s to the to the plain pqr or that contains this triangle a pqr so let's write the equation of a plane so note that the equation of a plane can be derived based on the fact that the dot product is equal to zero for perpendicular vectors and yeah I haven't covered a equation of a plane before so it's a good good time to start now what I'm going to do is let's just draw a plane like this and then when we talk about playing just basically a rectangular in 3D uh rectangle Etc and now we're going to do is I'm going to put the X and Y axis of it like this so that's Y and there is our X and there's a this is our Z there's the origin zero and let's say we had a point here I'm going to call this point uh over here is at X zero y 0 Z 0 and let's say there is a just Vector like this and this goes to the X Y Z and a point right here so it's a vector like that and this could be anywhere so that's a whole idea of a plane so anywhere around this uh perfect rectangular sheet right there and now and then this can extend to Infinity Etc if you're if it's not bounded so now what we'll do is we're going to use the fact that the dot product is equal to zero for perpendicular vectors so if we take a a vector perpendicular to this plane like this so this is let's say this is perpendicular and we'll call this n or normal to it or perpendicular to it and let's say this has the components for this Vector a capital a capital B and C like that like that yeah so thus the dot product of this permanent of these two perpendular vectors where we can get the position Vector over from here to Here by subtracting so what we'll get is well uh we'll go a capital a capital b capital c Dot and then the position Vector of of this Vector right here position Vector all we do is shift it over to here Etc and likewise for this one there yes and this Vector right here uh this is actually the position Vector so this is the same thing as uh over here and so on so but these ones are the coordinates so then we could get the position vector and let's just write that down and that's going to be well just the difference x minus x0 and then y minus y 0 and then Z minus Z 0 and then recall from our DOT product video if they're perpendicular that's equal to zero so and now we can just do the dot product remember that we just multiply this by this B by the middle see that this one add them all so we just multiply the components and add so what we get is a x minus X zero plus b y minus y zero plus c z minus zero and this equals to all zero and you can stop right here or continue further I'm going to continue further and simplify it so that's one way of writing the equation of a plane the next one is you can expand this out this equals to well a times x then and then then minus a times x 0 but we'll do that after so a times x and the next one is B times y plus c times n B times y plus c times Z and then minus when we do this all together and add the add those up we get this ax 0 plus then this B times y 0 and then Z times I mean that's c times zero like that this equals to zero all right so now that we have this set up and what I'm going to do is I'm going to call this lowercase D and then just write this out as uh ax plus b y plus c z minus lowercase D equals to zero so this is the equation of a plane that's usually given all right so now that we have the equation of a plane here now the next step is well we need to find a The Distance from a Point to a Plane distance from a point to a plane because remember this distance from the point s to the to the plane pqr that contains the triangle pqr so on the equation for the distance from a point to a plane plane which I'll prove in a later video is given by yeah just bit out of the scope for this video right now and let's just put this down here let's do some drawings again just make it clear let's say we had a plane like this uh this this angle like that so this is plain and this has the equation ax plus b y plus c z plus c z minus b equals to zero so uh let's say we had like that and let's say we had this is our y this is our X and this our Z like that zero or a point let's say we had a point P let's say the point p is here and this point is um has X or instead of just putting a point because we already have P for someone something else let's say this is X1 uh y1 and Z1 like that all right so now the distance from here all the way to this uh part right there it's going to be perpendicular to it so let's just draw this a bit uh better so let's write this plane um here and let's say this this point like this goes across and say it goes all the way across here and it's gonna be at a right angle to it because it's uh that's the closest distance to it and this point right here we'll call this D so from there or from this distance like that let's call this D like that so that is the distance D and D is given by the formula which I'll prove in a future video a x one plus b x 2 I mean now B oh by1 Plus c z one minus lowercase D divided by square root a squared plus b squared plus c squared like that that is the distance B and let's hold alt or box all this in all right so that's putting this all together so that's putting this all together the volume of a tetrahedron in coordinate form is we're going to have V equals to wall one-third the distance of the plane or the height and I'll put a giant bracket like this a and instead of this X1 y1 Z1 we're going from the point S1 S2 S3 Etc to the plane pqr and there is the area of that triangle so that's going to be S1 plus b S2 plus C S 3 and then minus D like that divide this by this is going to be a squared plus b squared plus c squared and then bracket and then this is the area the triangle is going to be one half the length of the cross product PQ cross p r like that and yeah that's what we have let's just go look at the formula we had so there's a distance and there's a one-half PQ times up or across p r which is quite fascinating stuff indeed I'll move this into Center and uh yeah this is some epic stuff right here and I'm going to do is I'm going to box this whole thing in or actually box entire volume in like this all right and then I'm going to write here because we need this a b c Etc where uh the ABC well that is again that's going to be yeah that's going to be from a vector perpendicular to this plane because remember this equation right here perpendicular to the plane in other words um in other words we go back up here as well we have a perpendicular plane right there that's where n is our PQ yeah PR so in other words it's a cross product so this is going to be where uh PQ cross PR is going to be oh this this is a vector equals 2 well just the vector a b c and and we're also given uh yes we're that and we and we know that d this D is equal to well if we go back up with what D is B is a x 0 plus b y zero plus C zero and again if you go back up here so we have this part is there n so this is going to be our P there this is going to be our X zeros y zeros Etc so that's where n in other words uh that's going to be our this point right here that's just the point p and that's where we take a cross product from so then so then we replace uh we replaces x0 with the um with the P coordinates so that's going to be where D is equal to well a um this is going to be X1 I'm in P1 I mean P1 P1 Plus for our case you have B uh P2 plus c P3 like that so that is our d and we can illustrate this by um and actually before that I'm going to do uh where I'm going to box this in this is so this is part of this equation so this throws it all inside and then note just to illustrate that draw our skewed uh triangle our tetrahedron is like this this is a point s Point p is inside goes dashed line across and then this is going to be a dashed line across and this is our R this is our q and then our cross product goes up like this and goes straight across it like that this is our n like that and there's our Point p and this n is again equals two let's put this up in like this let's move this down and move this down and write this back again so n um and that is our perpendicular vector and that's just going to be the PQ so p q from this angle I'm from this cross I'm this Vector to this Vector across those and we get um cross PR like that so yeah that's what it is all right so now that we have this and uh yeah this is more detail than the solutions that I Solution to Part (b) was able to find online but they weren't the official ones but anyways it's pretty uh in-depth you could even throw this in a computer program and solve it all all right so that's what we have so let's go back to well question um the part B of it now so question two part B says find the volume of a tetrahedron whose vertices are p111 and q123 and r112 and S three minus one two so let's write these down and start solving it so solution to Part B so the first step is to derive the position vectors p q and PR and then compute their cross product and cross product length so so well let's write the point p is one one and then the next point is Q is well one two three and then R is one one two and uh we'll do deal with the S later so let's just double check that and double check it is one one one two three one one two all right so that's what we have and uh the position position Vector PQ is just a difference uh Q minus P so we're going to have a write this in Vector form the square bracket or the triangle bracket so one minus one two minus one and then three minus one like that and now this equals to well one minus one is zero two minus one is one three minus one is two that's position Vector for PQ the next one is PR all right PR is equal to well one minus one and then we'll have this one minus one then two minus one and now this equals two well one minus one is zero one minus one is zero then two minus one is one all right all right now that we have uh this right here let's solve the cross product and then do the cross product length afterwards because again we'll need the cross product for this ABC and then the cross product length that we need that for this area there so let's write this down so p q cross e R is equal to and then write this in indeterminate form and let's move this down a bit because we're going to write this the giant determined I J K like that the next one's going to be 0 1 2 and 0 0 or zero zero one like that all right this equals two well let's just uh solve this out so again as always you cross this cross this and then you multiply this by this and subtract by this by oh two times zero so we're going to have a 1 minus I mean one times one minus two times zero zero this is going to be the I component all right erase this erase this now the next one is here so we'll do a minus the middle one's always minus zero times one is zero minus two times zero is zero J and the next one's plus erase this cross cross this next one's going to be zero times zero minus 1 times 0 is also zero this would be K like that all right that's what we have all right so that's what we have and now we can simplify this uh further get rid of all the zeros this is going to be well I so 1 times 1 is I mean we have one times I just or just I or one I and all the others vanish and that's all we're getting is it's just the I component in other words and writing this in the other component set up like this this just equals to one zero zero like that and where again this cross product is going to be we need it for our equation that's just our a b C in other words all the B and C are zero and then a is one so that's what we have and now the area area p q r remember this is just going to be one half the cross product length one half cross product length like that PR and this is going to be equals two well one half of the length of this um this Vector I which is well again this is just going to be one half square root and then 1 squared all the components squared 0 squared well zero Square these old vanished just it just becomes one squared this is one this is equals to one half so the area is one half like that all right so now that we have this the next setup is to find a distance from the point at s which has coordinates three negative one two to the plane uh right to the plane uh p q r so let's write this down so we have our points uh this s right here um this is a three uh negative one two in other words this is uh the coordinates S1 S2 that's three that we were looking up and that is for the equation so there's S1 S2 S3 and we only know what a b c is ABC is this right here yeah so let's just write that down so we know that um the normal Vector of the plane is equal to uh in in our case p q cross p r and this equals 2 1 0 0 and then we have it this equals to well a b c like that yeah in other words we have a equals to one and b equals to C which equals to zero so just Circle these like that that's what we have and then we also have uh this right here and then the next setup is what we need to solve our D and this is a point p and the number of the coordinates of p is just one one so let's write this down here so we also have let's just write that down so P this is P1 P2 E3 this equals two P one one [Music] all right so we have that so all those are equal to uh yeah one one one so the so then this means that the lowercase D is equal to a A1 Plus B V2 plus c p three and a is again just one so one times one plus and then B times will be zero this is equals to one like that all right so that D is equal to one so in other words uh we'll have this right here uh the distance D is equal to uh that distance is just uh this whole part right here and we already know that D is ETC so let's just write that all down a S1 plus b s two plus C as 3 minus B uh divide Us by square root a squared plus b squared plus c squared this equals 2 and then a is just one times S1 is three minus L plus and then a b is 0 plus C zero minus d uh D is just one like that and then divide this by square root a squared is one yeah I mean a yeah a squares is 1 squared plus all the others are zero squared 0 squared so in other words all we get is well this becomes three minus one over one because this is just one squared and then square root this equals to two so the length is two all right and uh that's so the distance from the point to the plane is two but we can do a double check right here let's do a double check uh that's using the equation but first we gotta find the equation of a plane so we have the plane is just going to be equal if you just remember it uh a X plus b uh y plus c z minus D equals to zero and uh D is just one uh then a is just uh yeah we already know a is just one everything else zero this becomes one times x minus one equals zero so notice we have x minus one equals to zero or x equals two is shift to one over the other side x equals to one so this is the equation of a plane in three dimensions all right so let's graph that out so let's say we had the plane like this I'm going to draw this plane like this all right so that's the plane and let's say we had uh this point goes out like this this is the x-axis it goes through and over here and this is the Y and this is that I mean and at this point right here is one so that's x equals to one so this is the x equals to one plane and this is our y like that and let's say our point was the three or our Point s is going to be well three let's say this two let's say this is three now so this is 2 here and then it goes negative one so here like that negative one and then it goes up to uh two uh somewhere around here and this is our Point s I'll put Point s Point s which has coordinates three negative one two like this and basically what this is saying is Well the distance from here to here perpendicular to it is just going to be directly across on this x-axis because this is a perfectly straight with this x-axis that's going to be this three minus two I mean three minus one equals to two so we have double checked that that's exactly it all right now now that we double checked that right it's a bit better like this and now the next step is well let's just put it all together so that's the volume of the tetrahedron is and write this out so V equals to one third times D the distance from well the point s to the plane let's put this all down uh p q r and then Times by the area of the pqr area p q r this equals two well one third times the distance can be two Times by the area number is one half and yeah just a double check so there's the um this is the distance two area is one half and these two cancels cancels and this equals to one over three epic stuff all right and now let's go even further and uh as a double check we Graphing 3D Tetrahedron in GeoGebra can graph the tetrahedron and calculate its area using the uh gogebra or just uh using geogebra which is an amazing 3D graphing calculator and other kind of calculator here so here's what I've graphed so here's the points P yeah P Q R and S and then I made a pyramid pqrs in other words just a triangular pyramid is a tetrahedron and and then we get something that looks like this and actually calculus area two area is equal to 0.33 in other words uh one over three so we got that right and then when you shift this around and you see this perfectly in line with the x equals one plane as the x-axis this is z there's the Y Etc and uh also note that again if you have this plane here notice the distance to the point s or the perpendicular distance or the height from the plane is exactly three minus two so this three this one so this this distance is just two yeah so that's just two so yes epic epic stuff here and and now if we uh go to this link so here's a link you can play around with it and so on and it looks like this and you can move it around and so on which is quite amazing and you can adjust it as well you can move this I think you could also drag and drop so you could change it around and you see the area automatically changing and so on and let's do a control Z back there yes epic seven and again notice perfectly in line with the uh with this right here with this x equals one line and there's a distance three minus one is two so that's the height this epic epic stuff
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https://www.reddit.com/r/learnmath/comments/15pceug/multiplying_exponents_with_different_bases_and/
multiplying exponents with different bases and powers - help : r/learnmath Skip to main contentmultiplying exponents with different bases and powers - help : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •2 yr. ago [deleted] multiplying exponents with different bases and powers - help TOPIC I’ve been watching YouTube videos about this, but I don’t find them very clear and comprehensive. I understand you have to make the base or the power the same, I just don’t know how to go about that. can someone please show me how that works? and how do you choose whether you make the base or exponent the same? I will be taking a quantitative reasoning class in college in the next coming weeks. I want to go back to concepts I never really understood, so I can address them and actually have a chance of succeeding in my class. I would appreciate a demonstration or any help. thank you! Read more Share Related Answers Section Related Answers steps to multiply powers with different bases how to multiply exponents in algebra multiplication of exponents explained when to add exponents in math dividing variables with exponents rules New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of August 12, 2023 Reddit reReddit: Top posts of August 2023 Reddit reReddit: Top posts of 2023 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
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https://www.themathdoctors.org/order-of-operations-common-misunderstandings/
Order of Operations: Common Misunderstandings – The Math Doctors Typesetting math: 100% Skip to content Main Menu Home Ask a Question Blog FAQ About The Math Doctors Contact Search Search for: Order of Operations: Common Misunderstandings September 19, 2019 December 22, 2023 / Algebra / Ambiguity, Mistakes, PEMDAS / By Dave Peterson Last time I started a series looking at the Order of Operations from various perspectives. This time I want to consider several kinds of misunderstandings we often see. Multiplication before division? Here is a question from 2005 from a teacher, “WRW”: Confusion over Interpretation of PEMDAS In telling students to "do multiplication and division IN THE ORDER THEY APPEAR," it seems they want to always do multiplication first. I think they follow the PEMDAS rule BY THE LETTER, so they want to multiply before dividing. When doing multiplication first, 8 / 2 4 = 8 / 8 = 1 When doing multiplication and division from left to right, 8 / 2 4 = 4 4 = 16 I agreed with him: If you think that students have a tendency to misinterpret the rule, you're probably right; but I think the reason is that PEMDAS is a poorly stated version of the rule, and it is easy to misunderstand it as meaning you do Multiplication, then Division, then Addition, then Subtraction. That's not what the rule is supposed to mean, but many students don't get past the letters and see the meaning! It's really wiser to think of subtraction as addition of the opposite, and division as multiplication by the reciprocal, and just leave D and S out of PEMDAS entirely, rather than try to fit them into the rules. But we make the rules for people who aren't ready to see things in a mathematically mature way! (I myself prefer to avoid PEMDAS altogether, and teach the "rules" in a more natural way that leads into this mature perspective.) Where some people memorize the rule as “Please Excuse My Dear Aunt Sally”, if I use a mnemonic at all, I make it PEMA: “Please Excuse My Attitude”. It’s just Exponent-stuff, Multiplication-stuff, and Addition-stuff, with Parentheses acting as traffic cop, telling you when to do something other than what the signs say. I’ll include my own way of introducing the concept in a later post on why we need the rules. But, continuing with an example: Translating these ideas into the case of multiplication and division, when we write 8 / 2 4 we really mean 8 1/2 4 which we can do in any order, since multiplication is commutative; clearly, however you do it, it comes out to 16, not 1. The problem here is that people tend to see this as if it said 8 2 4 which means something different. Note particularly that if we did the multiplication first in my example, then instead of 8×1 2×4=16, we would be doing 8×1 2×4=1. Interpreting it correctly, the only number we divide by is the 2. Also, seeing it this way allows me to rearrange the expression at will (since multiplication, unlike division, is associative and commutative). If I had, for example, 7÷3×15, I could think of it as 7×1 3×15=7×15×1 3=7×(15÷3)=7×5=35 In effect, I’m dragging the division sign around with the number following it! “WRW” answered, Thanks so much! I really like the idea of thinking of division as multiplying by the reciprocal and turning the whole multiplication/division portion into just multiplication. I'll try that out with my students and see if it helps. Thanks again! I didn’t mention there the fact that students outside of America are taught mnemonics like BODMAS, and students there sometimes think Division has to be done before Multiplication. Get them together, and you might have quite an argument! Addition before subtraction? Here’s a question from another teacher, Monica, the next year: Incorrect Application of PEMDAS and Order of Operations I was working with students on the order of operations today and explained that multiplication and division are done from left to right, as are addition and subtraction. Apparently, they believe they were taught in the past to do all addition and then all subtraction. I tried to show examples of why that wouldn't work, but they simply did the problem their way, obtained a different answer and asked why it was incorrect. Are there any examples or explanations that would clearly explain why they must be done from left to right? The same reasoning I gave for multiplication applies here, as I explained: It's impossible to show that they MUST be done from left to right; that is nothing more than a convention we all agree on. Your class has shown that it makes a difference which order you use; that proves that we MUST make some choice that we can all follow. What that choice is, is not so definite. But it makes a lot of sense to go left to right, for the following reason. You can’t prove that a particular grammar is “correct”, as if nature forced us to use it; every language has a different grammar, and each is correct for its own speakers. What makes a grammar correct is only that it is the same grammar used by other speakers of the language. So you’d have to prove that addition and subtraction are done left to right by showing that all the books do it that way. But we can see why it was a good idea. It’s the same thing I said about multiplication and division: We define subtraction this way: a - b = a + -b This allows us to think of any subtraction as an addition; we essentially just attach the negative sign to the number following it, rather than taking it as a different operation. The subtraction requires no extra rules, just the rules we already have for addition. If we do this, then 2 - 3 + 4 = 2 + -3 + 4 = 3 That is the same result we get if we do the operations from left to right (and it doesn't depend on whether we do the ADDITIONS from left to right, since addition is commutative!). If we did the addition first, we would get 2 - 3 + 4 = 2 - (3 + 4) = 2 + -(3 + 4) = 2 + -7 = -5 Note that this time, the negative sign ended up applying to ALL the following numbers, rather than just to the one after it. So doing additions first would mean we are really subtracting everything after a subtraction sign. One benefit of replacing subtraction with addition of a negative, as in the multiplication case above, is to be able to move things around. For example, a common trick to evaluate a string of additions and subtractions more easily is to move all the subtractions to the end: 5+3−6+2−8+1=5+3+2+1−6−8=(5+3+2+1)−(6+8)=11–14=−3. Without left-to-right operations, we couldn’t do this; we would have to look at the whole expression before rewriting any one subtraction as addition of the negative.So doing additions and subtractions from left to right makes it easier to transform an expression into one involving only addition; and since addition is commutative and associative, it is MUCH nicer to work with! The rule, therefore, arises from the wish to make expressions easier to handle. Without it, a lot of algebra would turn out to be a lot harder. So your students should thank whoever first made this choice! I closed by referring to the MD question above: Now, your student's misunderstanding of the rule very likely comes from the use of PEMDAS or something equivalent, which is meant to be only a summary of the rules. It sounds as if A comes before S, but that twists the intended meaning of the mnemonic. See this page for another thought: Confusion over Interpretation of PEMDAS That says essentially the same thing I just said, but about multiplication and division, which is an even bigger problem. (Did you know that in other countries they use BODMAS instead of PEMDAS, so students often think division should be done first?) For another interesting take on left-to-right operations, see Left Associativity Where do negatives fit in? One of the most common difficulties in evaluating expressions is the mixing of negation with exponents. We have had many questions on this; in fact, I could have included this in the series Frequently Questioned Answers. I’ve chosen to use this question from 2002, whose answer covers most of the ideas we bring up (and refers to several other answers): Negative Squared, or Squared Negative? After reading your answer in Exponents and Negative numbers it seems to me that you're ignoring an important fact: -3 isn't just -13, but a number in its own right, i.e., the number 3 units to the left of zero. If that's the case, then shouldn't -3^2 have the value -3-3, or 9? If -3 was intended to mean -13, then shouldn't it be written that way and not implied? Thank you for your time. The answer he referred to is an early one from 1997 where Doctor Ken stated that −3 2=−9, because it means −(3 2), not (−3)2. Now, if the convention is that negation is done after exponentiation, then that’s all we need to say. But Tom is arguing from the fact that −3 is itself a number, so it has to be kept together. Does that require us to do the negation first? He has a strong argument (and a common one). I took the question: I do recognize that it is possible to disagree on -3^2. Dr. Rick's answer to a similar question, Squaring Negative Numbers mentions this disagreement. Like you, he notes that if you think of -3 as a single number, it makes sense for the negation to bind more tightly to the 3 than any operation. That reasoning makes some sense, though I think other arguments are stronger. But I do agree that since there is some reason to read it either way, it is prudent always to include parentheses one way or the other, to clarify your intent, i.e., to write either -(3^2) or (-3)^2. Minimally, we can say that it is wisest to avoid this form, either because it is easy to read wrongly, or because all the books teach it wrong (depending on your opinion)! In fact, I find that the books I most respect never show such a form (making it hard to find examples to point to!), while others, on the contrary, emphasize it because students always get it wrong unless it is drilled into them. Occasionally people will try to argue the point based on the behaviors of particular calculators or spreadsheet programs. However, these are really irrelevant, since they all define their own input formats, and programmers (of which I am one) are notorious for choosing what's easiest for them, rather than what is most appropriate for the user. I've noted in several answers in our archives that some calculators, and Excel, use non-standard orders of operation without apology. But calculators in particular just don't use standard algebraic notation in the first place. I’ll be including a link to one of these discussions at the bottom. But the main point is simply that calculators have to follow a convention that suits the way you enter expressions on them, which is different from print. (As calculators have come to display expressions more like type, however, they have been forced to follow conventions more closely.) We also get questions from people who claim they learned long ago something different from what their children or grandchildren are learning (either the whole PEMDAS business, or some part of it like this one): There also seems to be a generational difference, with older people (including some teachers) claiming that they were taught to interpret -3^2 as (-3)^2. I suspect that what has changed is not the rules governing "order of operation" (operation precedence), but that schools are introducing the issue earlier, before students get into algebra proper. That means that they start by looking at expressions for which it is less clear why the rules make sense. I think you will rarely find examples of "-3^2" in practice, because there is no need for mathematicians to write it. You will find "-x^2" frequently. Conventions of algebra apply primarily where we have variables; in arithmetic you don’t normally write long expressions with numbers, but just evaluate as you go. Basic four-function calculators were designed for such use, and don’t follow the order of operations. And the conventions make more sense on their home turf (with variables than with numbers): If you approach the idea starting with numerical expressions like -3^2, you are thinking of -3 as a number and assuming that the expression says to square it. If you approach it first using variables, having first discovered that "-" in a negative number is actually an operator, then it is easier to see why -x^2 should be taken as the negative of the square. So I'll start with the latter, and then it becomes natural to treat numbers the same way we treat variables. The point here is that in arithmetic, we see the negative sign as part of the number (“the number I’m attached to is negative”); in algebra, we see it primarily as an operator (“negate the thing that follows”). And the convention arises in the latter context. It isn’t primarily meant for use with numbers, but with variables. Negation as multiplication or addition Now, in an expression like -x, clearly "-" is a (unary) operator, which takes a value "x" and converts it to its opposite, or negative. The expression "-x" is not just a single symbol, but a statement that something is to be done to a value. As soon as we start combining symbols like this, as in -x^2 or -xy, we have to decide what order to use in evaluating them. The trouble is that the "order of operations" rules as commonly taught (PEMDAS) don't mention negatives. So if we are going to go by the rules, we have to figure out how a negative relates to them. Well, there are two ways to express a negative in terms of binary operations. There is no N in PEMDAS, or even in many fuller explanations of the convention. To see where it fits, we need to think about how its meaning relates to the other operations. How do mathematicians think of negation? One is as multiplication by -1: -x = -1 x Treating it this way, clearly -x^2 = -1 x^2 = -(x^2) That is, since -x means a product, we have to do the exponentiation first. So if we think of negation as a kind of multiplication, it belongs right in there with MD. And, if you think about it, you’ll realize it doesn’t matter whether we think of it as being done first, last, or left-to-right: If you negate a factor before multiplying or dividing, you get the same answer as if you negate after multiplying or dividing: (−x)⋅y=−(x⋅y) The other way to talk about negation is as the additive inverse, subtracting x from 0: -x = 0 - x (This is why the "-" sign is used for both negation and subtraction.) Using this view, we see that -x^2 = 0 - x^2 = -(x^2) In particular, we would like to be able to replace subtraction with negation wherever we find it, and not mess things up: x−y 2=x+−y 2, which would not be true if the latter meant x+(−y)2. This is why the subtraction idea is applicable even when we are not actually subtracting from 0. So both views of negation produce the same interpretation, which does exponents first, and it is logical to put negation here in the order of precedence. So if PEMDAS really means PE(MD)(AS), with operations in parentheses being done together, we can extend it as PE(NMD)(AS), where negation is definitely done after exponents and before addition. But the fact is that there is no authority decreeing these rules; just as in the grammar of English, we get the "rules" by observing how the language is actually used, not by deducing them from some first principles. The order of operations is just the grammar of algebra. So the real question is, how do mathematicians really interpret negatives and exponents combined in an expression? If you look in books, you will rarely find "-3^2" written out, but you will often find polynomials with negative coefficients. And you will find that -x^2 + 3x - 2 is read as the negative of the square of x, plus three times x, minus 2. So, even though there is not a lot of evidence of usage with numbers, usage in polynomials (with variables) is clearly on the side of negation-after-exponentiation, and we want to be consistent. I have come to believe that the order of operations is what it is largely so that polynomials can be written efficiently. If "-x^2" meant the square of -x, then we would have to write this as -(x^2) + 3x - 2 to make it mean what we intend. Since powers are the core of a polynomial, we ensure that powers are evaluated first, followed by products and negatives (the two ways to write a coefficient) and then sums (adding the terms). Since we can easily see that this is how -x^2 is universally interpreted, it makes sense to treat -3^2 the same way. Addition comes last so that a polynomial is a sum of terms; negation goes with multiplication in order to have the same base in every term, without having to use parentheses to avoid accidentally changing a base to −x. For some other discussions of this issue that haven’t already been mentioned, see Precedence of Unary Operators Negative Numbers Combined with Exponentials Negative vs. Subtraction in Order of Operations And for a long discussion with a programmer, see Order of Operations and Negation in Excel That is the discussion I referred to above, about how calculators and programs have different needs, as well as whether -3 should be thought of as a unitary entity. Post navigation ← Previous Post Next Post → 26 thoughts on “Order of Operations: Common Misunderstandings” Pingback: Order of Operations: Subtle Distinctions – The Math Doctors Amanda November 3, 2021 at 7:20 pm My mathematically minded friend told me that when there are no brackets you complete the equation left to right. If there are brackets you use BEDMAS. Is this true? Reply 1. Dave Peterson November 4, 2021 at 10:42 am No. In fact, “BEDMAS” is mostly about what to do in the absence of brackets (parentheses)! The role of brackets is just to intentionally change the order of evaluation from the default, which is multiplication and divisions before addition and subtraction. And if we used brackets everywhere, we wouldn’t need the “EDMAS” part at all. See the previous post, Order of Operations: The Basics. Reply Helge February 3, 2022 at 10:10 am I am discussing this expression on a Facebook group: 8÷2(2+2) = ? My answer is 16. My opponent says 1. She insists that the multiplication must be done first, because it has a special connection with the parentheses. Who is right? Reply 1. Dave Peterson February 3, 2022 at 10:46 am Hi, Helge. You’re both wrong … to be arguing over this. The fact is that different conventions are taught about this sort of expression, so each of you would be right according to some teachers. (For which reason, it’s best never to write such an expression, if you want to be understood.) For details, see the post Order of Operations: Implicit Multiplication?, and its follow-up, Order of Operations: Historical Caveats. If you have further questions about this, you should ask us directly, via Ask a Question. Reply Pingback: Why Properties Matter: Beyond Addition and Multiplication – The Math Doctors David April 24, 2023 at 1:30 pm Full disclosure, I’m a PEMDAS fanatic. (Promoting PEMDAS to the level of “1+1=2”, and there can be only one correct answer or answer set.) Amen. I would, without any crisis of faith, accept BODMAS as the one true rule. Processing mixed operations left to right ignores the fact that not all cultures process their written information that way. Near the beginning of this page, an equation was rewritten by substituting “/ 2” with “ 1/2” resulting in an equation where the commutative property of multiplication can be used. That substitution requires a huge assumption: The operation 2 4 could not be performed before the 8 / 2. [Otherwise, the rewritten equation would be “8 (1 /(2 4))” which essentially makes the commutative property of multiplication a moot point.] The example actually highlights the importance of operator order in rewriting equations for manipulation. An inaccurate substitution condemns any further manipulations to produce errant results. Reply 1. Dave Peterson April 24, 2023 at 2:42 pm Hi, David. I’m not quite sure what you mean by “PEMDAS”, and therefore what interpretation you are defending or refuting. You seem to think multiplications and divisions should not be done from left to right; but that is what we mean by PEMDAS, as I explained here, and also in Order of Operations: The Basics. If you are taking PEMDAS to mean “Multiplication, then Division”, why do you also approve of BODMAS, which in the same way would be taken to mean “Division, then Multiplication”? That is not what they mean; they mean what I described here. This post explains why we choose to define the order of operations in this way. This is a choice; it is not based on any particular culture, or on an assumption that all cultures read left to right. You’re right that we need to take the meaning of an expression into account when we make a substitution, and not mechanically replace one set of characters with another. That is the cause of several common errors. But what we did here is not an error. The assumption I am making is that mathematicians do read expressions this way, which you can easily confirm. Reply Ronald Alfonso Becerra Gil May 14, 2023 at 10:25 am Sorry to reopen the question about -3^2, but I was arguing about this on YouTube, and many people seem to be convinced that as a standalone expression it should be considered as 9, while others say it must be ambiguous, because there is not an universally accepted answer. The point is that many calculators give 9 as the result, including the one in Excel. It is also the answer in many scientific calculators, but that is because they don’t let an expression to start with the binary minus but only with the unary one, that for them has more precedence than the rest of operations. Therefore, typing -3^2 ends meaning 9. I read your previous post that what calculators say shouldn’t be considered the rules of what we do on paper, and also that it is hard to find examples of expressions like -3^2 in math texts, because when we write something like it is because we were dealing with variables. But my question is: since the actual appearances of standalone expressions like -3^2 end occurring mostly when typing in calculators, not in math texts, can we teach that the actual answer for it must be -9 despite that is not what people will find in “practice”? Or at least not always in practice, as the answer tends to differ: in Google and WolframAlpha, it means -9. I have also put as an argument that a standalone expression as -3^2 can come from a more complex one, and thinking about it as 9 can lead to some problems. For example, if we are solving this equation: 5 – 3^2 = 5 + F(x) no one doubts that the left side means 5 – (3^2). But we realize we can subtract 5 in both sides, leaving: -3^2 = F(x) If we were to interpret -3^2 as (-3)^2 when it appears alone, we would need to add here new parentheses to keep the original meaning, add a factor (-1), or be forced to write a zero: -(3)^2 = F(x) (-1) 3^2 = F(x) 0 – 3^2 = F(x) because otherwise we would have no longer an equivalent equation to the original. Instead, if our convention is that -3^2 is always interpreted as -(3^2) regardless of the context, then we can just remove the redundant term 5 without the need of modifying the structure of the rest because of that. I don’t know if the argument is wrong or some people didn’t understand the implied problem, but they insist that if the expression didn’t reflect a subtraction from the beginning, the minus sign should be part of the number. Reply 1. Dave Peterson May 14, 2023 at 8:33 pm Hi, Ronald. I think your arguments in favor of taking -3^2 to mean -9 agree with ours. (I hope you do see that we are saying it is -9, not 9.) We simply have to recognize that there are many people who don’t see things the way mathematicians do, and therefore we should “write defensively” and avoid that form. There’s no need (for either them or us) to argue about it. They can merely be told that, however irrational they think it is, it is the convention among mathematicians that -3^2 = -9. In exactly the same way, we who know that convention can recognize that, however irrational we know it is that Excel says -3^2 = 9, that is their convention and we have to live with it. As for calculators, I think (most?) scientific calculators distinguish the “-” (subtract) button from the “(-)” (negate) button, though they could probably use one button, just as computer languages typically use one symbol. My guess is that they do this in part so that they can distinguish between starting an expression with a negation, and continuing a previous result by subtracting from it. But they typically give both subtraction and negation a lower precedence than exponentiation. That has not always been true, and it is not true in Excel (which is stuck with their rule for the sake of backward compatibility). Which calculators do you find give the answer as 9? I find the link at the end, Order of Operations and Negation in Excel, a particularly interesting read. Reply 1. Ronald Alfonso Becerra Gil May 15, 2023 at 12:38 pm Hello. For sure, I was aware that the article intended that it is -9 and not 9. I was just going to type that my CASIO fx-82MS scientific calculator returned 9 when using negation, and to my suprise, when I tested it, the result was indeed -9. Maybe it was a Mandela Effect? Maybe I took that word from someone else and assumed it to be true, since we usually don’t have the need to write expressions like that, so I wasn’t used to knowing what to expect from the calculator. Despite this, I could not ensure that all versions of a scientific CASIO give the same result. In the YouTube video in which I was arguing this, someone mentioned that in some old math texts they interpreted expressions like -3^2 as -3 -3, but didn’t give references. I suspect it was a Mandela Effect too. I think that finding that the results in software programs are most of the time the same as those agreed upon in mathematics is an additional argument to say that the interpretation to learn should definitely be one and not the other, and Excel should be taught as a separate case to be careful with. And thanks for sharing that article, which I found interesting! Seriously, before finding that YouTube video, I didn’t suspect there was much debate about it. Reply 1. Dave Peterson May 15, 2023 at 9:20 pm Yes, I think it is appropriate to explicitly teach that -3^2 = -9 in written mathematics, even though some software or calculators may not interpret it that way, and some humans may not as well. The former fits most usage they will see (including, I think, most current calculators), while the latter is a warning to be careful in reading what others write, and in writing for others. And, yes, the reason that calculators and the like seem to tend in the direction of agreeing with what teachers say is because … that’s what teachers say! They want to be right, and to be recommended by teachers. Reply Andrew December 21, 2023 at 11:22 pm I feel like this is a fundamental problem of our notation using the minus sign for multiple operations. What do you take of the case of a negative exponent? If we follow the negation-after-exponentiation example proposed then a^-b becomes a much different equation than convention dictates. Most would assume a^-b is a^(-b) as written, but if we follow negation-after-exponentiation it would be a^-1b. So negation in a base should conventionally be after exponentiation, but negation in a exponent should be conventionally before exponentiation. That to me sounds like a poor/ambiguous convention. Reply 1. Dave Peterson December 22, 2023 at 11:34 am This is a very interesting question, which I don’t think I’ve ever heard before. It raises several issues that are worth addressing. First, I has to be noted that in traditional formatting, the question you raise is moot, because when written as a−b, the entire exponent is written as a superscript, which inherently groups it; you have to evaluate the entire exponent before using it. As a result, when this is written in in-line format using the “^” (caret) symbol, you would write it as “a^(-b)” in order to express that grouping. Either way, there is never a question as to which operation is done first. Second, restricting ourselves to the latter format, I claim that it is unambiguous to write a−b as a^-b, because there is still only one way to interpret it, regardless of the order of operations. That is because the caret says to use the immediately following number as an exponent, and since “-” itself is not a number, we can only take all of “-b” as the exponent. As a unary operation, “-” acts only on the number on its right, so in this position, it is not “in competition” with exponentiation in the order of operations. Third, your argument, as I understand it, is based on what I said about “-x^2 = -1 x^2 = -(x^2)”, where I replaced negation with multiplication by -1. But you can’t just replace an expression anywhere with an expression that would be equivalent when standing alone, without first considering the order of operations. (I discussed this idea in this more recent post, under “Substitution requires it”.) To do your substitution, you would properly write, not “a^-1b”, but “a^(-1b)”, because of my second point (that “-b” must be treated as a unit, so that parentheses are appropriate to maintain that). Also, note that if you instead used my “-x^2 = 0 – x^2 = -(x^2)” as a model, you would get “a^-b = a^0 – b = 1 – b”. This further illustrates the error of blind substitution. (My two statements are not presented as proofs that the negation on the left is done after the exponentiation, but as two experiments to see what makes most sense; it is their agreement that provides support for the choice to do exponents (on the left) before negation. As I just showed, attempting the same experiment with negation of the exponent shows that that leads to in consistency.) Fourth, if the rule as stated results in an interpretation different from the very convention that the rule is supposed to state, then clearly we need to change our statement of the rule, not the convention! If you were right that “negation after exponentiation” would entail taking a^-b as (a^-1)b, and we know very well that it is taken as a^(-b), then we would want to change our wording, not its meaning. I think that’s what you are saying. In summary, I’m not sure whether your point is that we simply shouldn’t use “-” for negation (an idea discussed here), or that the convention has to be stated in your longer form in two parts (which I think unnecessary because of my second point), or something else. I think what we have is perfectly consistent. Reply Pingback: Order of Operations: Fractions, Evaluating, and Simplifying – The Math Doctors Scott Stocking May 15, 2024 at 10:45 pm On the negative numbers: One expression I use to try to confound the Facebook OOO crowd is the following: 6 ÷ -(1 + 2) Is it 6 ÷ -3 or -6 3? Surprisingly, those who think 6 ÷ 2(1 + 2) = 9 would say that that the answer to my problem is -2! I was taught that a negative sign in front of parentheses without a coefficient is the same as multiplying what is inside the parentheses by -1, so I would have guessed they’d say -18. I believe the answer should be -2, however, just like I believe 6 ÷ 2(1 + 2) = 1. Reply 1. Dave Peterson May 20, 2024 at 5:19 pm Hi, Scott. I don’t see why anyone would say that 6 ÷ -(1 + 2) = -18. The only way to get that is to blindly replace -(1+2) with -1(1+2), ignoring how it would change the order of operations in the surrounding expression. I discussed such unthinking substitutions (in a very similar context) in Implied Multiplication 3: You Can’t Prove It. But there is no reason anyone would do that here, because it is already clear what to do. As I said above in response to Andrew, “you can’t just replace an expression anywhere with an expression that would be equivalent when standing alone, without first considering the order of operations.” By the way, looking at your site, I find numerous places where I would disagree with you; but I see no value in arguing about this or “trying to confound” people who disagree with me. My goal is peace and mutual understanding, not proving myself correct. Acknowledging ambiguity (even merely potential ambiguity) is a central part of that. Reply Doug Hendrie February 10, 2025 at 2:03 pm Dave, Near the top of the page, you stated “Translating these ideas into the case of multiplication and division, when we write 8 / 2 4 we really mean 8 1/2 4 which we can do in any order, since multiplication is commutative; clearly, however you do it, it comes out to 16, not 1. The problem here is that people tend to see this as if it said 8 —– 2 4 which means something different.” It doesn’t mean something different, and your reciprocal calculation is wrong. Not ambiguous. Just wrong. TERMS are reciprocated. Not factors. So your reciprocal will be 1 ——— 24 ———— Expression Reciprocal 1 2x —— 2x ———————————————- 3 x —— —— x 3 ——————————————— 2x-3 (x+5) ———- ———- (x+5) 2x-3 Reciprocal in Algebra ————— Reciprocals. Which of the following is the reciprocal of 21 × 5? 1 Correct answer is: ———- 21×5 ————- The proof of 1/(ab) = (1/a)(1/b) consists in showing that, (1/a)(1/b) acts as the reciprocal of ab, and the product of ab(1/a)(1/(b) = 1 Mary Dolciani Teachers Manual Page 17 Book 1 Modern Algebra Structure and Method Reply 1. Dave Peterson February 10, 2025 at 4:44 pm Doug, I’m sorry, but as in your previous comments, this is simply wrong. You are assuming, contrary to convention, that all multiplications are done before any divisions. Some authors have taught that, but even the sources you quote here teach otherwise: MathsIsFun says, “Divide and Multiply rank equally (and go left to right),” and gives the example Similarly, SplashLearn says, “Multiplication and Division: Next, moving from left to right, multiply and/or divide, whichever comes first,” and gives the example Dolciani also teaches left-to-right, at least for explicit multiplication. (In some books, she taught that implicit multiplication is done first, but that is not relevant here.) I showed that in my previous response. As a result of this misunderstanding, you seem to assume that “/” means “take the reciprocal of the product that follows”, which it does not; your examples of reciprocals don’t deal with the symbolic form in which you are claiming to prove me wrong, but only with words; so they are irrelevant. The fact that the reciprocal of 21×5 is 1 21×5 does not imply that 1/21 × 5 means the reciprocal of 21 × 5. Rather, it means “the reciprocal of 21, times 5”, or 1 21×5. [I tried to edit your comment to insert images of what you quoted; here they are: ] I hope this is clear. Reply 1. Doug Hendrie February 11, 2025 at 5:03 pm Hi Dave, I’m not assuming every multiplication should be done first however, for these expressions, the multiplications do come first. For instance 2b÷2b= 1, NOT b² Now 215÷215 = 1 and 1215=215 Now taking the reciprocal of 215 which is 1/(215) which you already have the example, gives 1/(215)215= 1 Now previously , you questioned my use of the Commutative Law, ab=ba. 8÷24=8÷42 can only equal 1 Commutative Law states, factors can be presented in any order, without changing the result of the expression. Now this Law contradicts the convention that is PEMDAS. Do we use a convention, or do we use an actual Mathematical Law. You are promoting PEMDAS as the ONLY way to simplify expressions, and that is most definitely NOT the case. Simplify Expressions Using the Commutative and Associative Properties. When we have to simplify algebraic expressions, we can often make the work easier by applying the Commutative or Associative Property first instead of automatically following the order of operations. I’ll stick with the Laws of Maths. Regards Doug Reply 1. Dave Peterson February 11, 2025 at 7:17 pm Thanks for condensing your argument, so I can answer it more easily. I hope this time you will understand what I have been saying. Let’s take a look at what you quoted, from an online textbook: (First, let me point out that, like last time, you included a link only to the publisher of the snippet you quoted; I had to search to locate it specifically, which will make it easier to answer, by seeing the context, so I edited your comment to improve the link.) What they are saying here is not that you should apply properties instead of following the order of operations, as if these would give different results. They are saying that, whereas the order of operations tells you what an expression means, you don’t have to evaluate the expression in that order, but can apply properties to simplify the work. In the example, they observe that two of the terms cancel one another, so it is easier to add them first, rather than to do everything left to right. The order of operations rules tell you that it means “this, plus this, plus this”, but the commutative property tells you that you will get the same result if you change the order. Other examples on that page show the same concept. Now, you say that you aren’t claiming that all multiplications are done before divisions, but that in these examples, they are. Yet the only grounds you give for that claim is that, as you see them, the multiplication are done first. On that assumption, you then apply the commutative property to that product. But the standard rules (as taught by Lumen as well as the others I’ve pointed out), 21×5÷21×5 does not mean (21×5)÷(21×5)=1, but ((21×5)÷21)×5=25. Here is what they teach: If you were right, the first example would equal 1, not 4. But they follow the order of operations – the same rules they mention in the section on simplifying, and do not violate; they just rewrite the expression before evaluating. Similarly, all current textbooks I know (though not all, historically) would say that 8÷2×4 means (8÷2)×4=16, while 8÷4×2 means (8÷4)×2=4. And, again, the commutative property must be applied to the correct meaning of the expression, according to those rules, and does not override the rules. If two factors are not meant to be multiplied together, then you can’t change their order. A law can’t contradict a convention! The latter takes precedence, and determines when the law can be applied. Reply 2. Dave Peterson February 14, 2025 at 12:23 pm Doug, If you want to continue debating this, please don’t just keep repeating the same claims; rather, try to understand and answer my arguments. In particular, explain why every source you quote agrees with me that an expression like 18÷9×2 means (18÷9)×2, not 18÷(9×2), as you take it. Are you the only person in the world who understands mathematics correctly? Or is there some chance that you might be wrong? The reality is that the order of operations convention does have to be applied first, in order to know what an expression is intended to mean, before you can correctly apply properties to rewrite it. Reply 1. Doug Hendrie February 14, 2025 at 12:42 pm Then why does Khan Academy show the following- Khan Academy. When do we NOT follow the order of operations? Lots of times, actually! While the order of operations gives us one way to evaluate an expression, the properties of addition and multiplication allow us to be more flexible. The distributive property says that we can multiply a value to each term inside of the parentheses instead of adding or subtracting inside the parentheses first. The commutative property of multiplication says that we can multiply the factors in any order instead of only left to right. Once we learn more about reciprocals, we’ll be able to rewrite expressions with multiplication in place of the division. The order of operations, or the convention of performing arithmetic calculations in a certain sequence, is not a universal or natural law, but a human invention that developed over time and across cultures. Exponents and order of operations FAQ (article) | Khan Academy 2. Image 30 Dave Peterson February 14, 2025 at 3:19 pm The answer is the same as what I explained before. You are quoting from here in Khan Academy: By “not following”, they don’t mean “disregarding” the convention, disobeying it, but rather going beyond it: recognizing, once we understand the meaning of an expression (according to the order of operations), we can use properties as a tool to change how we choose to carry out the work of evaluating it. As I’ve shown for your other sources, Khan, too, disagrees with you about expressions of the sort you are interested in. For example, at the end of the video here, he does this: He does not do the 3×2 first, but does everything from left to right. I have not often seen students confused in this way by the term “order of operations”, though I can see how it can happen. It doesn’t mean “you must always do the operations in this order”, but just “what is written means that it could be done in this order, but you can rewrite it, if you want, before you do all the operations”. That’s what this site, and the book you used last time, are saying. Once students have learned the basics, I usually explain that they have a choice between doing “exactly what an expression says” (which is what the order of operations indicates), and “whatever is the easiest way to evaluate it” (which the properties allow you to do). Some even get in the habit of automatically distributing, even when that actually makes the work harder; others never get beyond taking an expression literally. You, on the other hand, seem to be unaware of your own misinterpretation, because somehow doing multiplication first feels so natural to you that you aren’t even aware that you are interpreting at all, much less that you are doing so wrongly. Again: Before you can apply the commutative property, you must determine what numbers are to be multiplied. That is what the order of operations does: It says, before you do this addition, for example, you will have to multiply these numbers. Once you understand what the expression means, then you can decide to rewrite it as an equivalent expression (with the same value) by rearranging parts of it, if you wish. But that doesn’t mean that you can swap any numbers that are separated by a multiplication sign without first considering whether those numbers are to be multiplied. A different example may help. We also have the commutative property of addition. But when I see the expression 2×5+4×3, I can’t swap the 5 and 4 to get 2×4+5×3; that gives a different value, because the 5 and 4 are not meant to be added. We have to see that they have to be multiplied by the 2 and the 3, respectively, before we do the addition. So the numbers actually being added are 2×5 and 4×3; if we were to apply the commutative property of addition, we would actually get 4×3+2×5. In the examples we have been discussing, like 8÷2×4, the numbers you claim can be commuted are not meant to be multiplied! And we know that because the order of operations tells us that, nominally, the division will be done before the multiplication. Yes, we can rewrite it if we want, but that might be as 8×4÷2, or just as 8 2×4. Doug Hendrie February 15, 2025 at 9:03 am Hi Dave Thank you for the prompt reply, So from the above 8/24 Does it follow that ab÷ab = b² Regards, Doug Reply 1. Dave Peterson February 15, 2025 at 10:03 am Ah! That’s the $64,000 question! I thought you’d never get around to it. But it’s a separate issue. You’ve been talking about explicit multiplication. where there is hardly any question (but see here, and other references here to what I call “AMF”). When you take out the multiplication sign, you get to the issue of implicit multiplication, which people fight over. Since there are different opinions, and it can be easily misread even if you have a strong opinion, I recommend just never writing expressions like ab÷ab. I think there is a very good case for treating that as (ab)÷(ab) = 1. But some teachers, at least, insist on treating it the same as a×b÷a×b = ((a×b)÷a)×b = b², and there is some reason to believe that replacing one notation for multiplication with another should not change the meaning of an expression. But that’s the subject of several other pages on this site, starting with Order of Operations: Implicit Multiplication?. In fact, you’ve previously commented on a more recent page on that topic, though you used explicit multiplication, making your comment irrelevant there. Reply Leave a Comment Cancel Reply Your email address will not be published.Required fields are marked Type here.. Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Have a question? Ask it here! We are a group of experienced volunteers whose main goal is to help you by answering your questions about math. To ask anything, just click here. 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https://www.youtube.com/watch?v=_QmIL9n613E
How do you solve an exponential equation with e as the base Brian McLogan 1590000 subscribers 3184 likes Description 490556 views Posted: 4 Oct 2013 👉 Learn how to solve exponential equations in base e. An exponential equation is an equation in which a variable occurs as an exponent. e is a mathematical constant approximately equal to 2.71828. e^x is a special type of exponential function called the (natural) exponential function To solve a natural exponential equation, we use the properties of exponents to isolate the (natural) exponential functions. Then we take the natural log of both sides. Note that the natural log cancels out the (natural) exponential function (e), leaving out only the exponent. 👏SUBSCRIBE to my channel here: ❤️Support my channel by becoming a member: 🙋‍♂️Have questions? Ask here: 🎉Follow the Community: Organized Videos: ✅Solve Exponential Equations ✅Solve Exponential Equations | Learn About ✅Solve Exponential Equations with a Calculator ✅Solve Exponential Equations with Fractions ✅Solve Exponential Equations | Easy ✅Solve Exponential Equations with e ✅Solve Exponential Equations with Logarithms ✅Solve Exponential Equations without a Calculator 🗂️ Organized playlists by classes here: 🌐 My Website - 🎯Survive Math Class Checklist: Ten Steps to a Better Year: Connect with me: ⚡️Facebook - ⚡️Instagram - ⚡️Twitter - ⚡️Linkedin - 👨‍🏫 Current Courses on Udemy: 👨‍👩‍👧‍👧 About Me: I make short, to-the-point online math tutorials. I struggled with math growing up and have been able to use those experiences to help students improve in math through practical applications and tips. Find more here: brianmclogan #exponential 79 comments Transcript: so anyways Let Me Maybe explain it this way and I'll and I'll do this two different ways first of all when we learn now we have a single logarithm right logarithm it's in logarith or logarithmic or exponential form sorry so one way if I want to solve for this you can't solve for the X when it's raised up into the power but if I rewrite this in exponential form this is now going to be log base e of 12 = 3x now remember we write log base e of 12 as Ln of 12 = 3x so now to solve for x you divide by 3 so X = Ln base 12 of 3 okay now that's one way you guys can go ahead and do this the another way that you can look at this is Again by taking the log of both sides again ladies and gentlemen when we're looking at this and let's go to our properties of logs now of course ladies and gentlemen I don't want to take the log of both sides because I need to choose a base remember I kept on telling you the base has to be e right you always want to have the base of your logarithm be the same base of your exponent well log base e is what we call what Ln so let's just take the Ln of Both Sides Now by applying the properties of logarithms maybe this will make a little bit more sense for some of you if by applying the properties logarithms we can take this exponent and put it where in front right and write it as the product so this is 3x Ln of e = Ln of 12 now let's evaluate Ln of e so what that says is e raised to what power equals e e raised to what power equals e one 1 so guess what it's just 3x 1 = Ln of 12 or 3x = Ln of 12 IDE 3 IDE 3 x = Ln of 12 / 3 you're getting the exact same answer but now what we need to do is solve this so we go in our calculator and we type in Ln of 12 and then we divide that by three so therefore xal 83 approximately because your okay anybody have any questions on that question you
187852
https://convertermaniacs.com/deg-to-rad/0/what-is-67.5-degrees-in-radians.html
67.5 degrees in radians (67.5 deg to rad) Converter Maniacs 67.5 degrees in radians Here is how to calculate and covert 67.5 degrees (deg) to radians (rad). We will show you the degrees to radians formula, the math to convert 67.5 degrees to radians, and we will illustrate 67.5 degrees in radians on a circle. To convert degrees to radians, we multiply degrees by π and then divide the product by 180. Here is the formula to convert degrees to radians: (degrees × π) ÷ 180 = radians When we enter 67.5 degrees into our formula, we get 67.5 degrees in radians as follows: (degrees × π) ÷ 180 = radians (67.5 × π) ÷ 180 = 3π/8 67.5 degrees = 3π/8 radians Discover more Calculator KJ to Calories calculators Since the answer above includes a Pi (π), which is an irrational number, 67.5 degrees in radians in terms of Pi is the only way to give the exact answer. However, we can divide the numerator by the denominator in the answer above and get an approximate decimal answer to 67.5 degrees to radians, like this: 3π ÷ 8 ≈ 1.17809724509617 67.5 degrees ≈ 1.1781 radians To illustrate 67.5 degrees in radians on a circle, we first drew a circle with our compass and then outlined 67.5 degrees with our protractor. The counterclockwise area between the blue and the orange line is 67.5 degrees. The counterclockwise distance from a to b along the red perimeter is 3π/8 or approximately 1.1781 if the radius is 1. If the radius is not 1, simply multiply 3π/8 (or 1.1781) by the radius to get the distance between a and b. Degrees to Radians Calculator Submit another number of degrees for us to convert to radians for you. What is degrees in radians? 67.51 degrees in radians Here is a similar degrees to radians calculation you may find interesting. Copyright|Privacy Policy|Disclaimer|Contact
187853
http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC8zLzBiNWRjYTdiMjRiMzkwZGU5MzQwYTI3NTcyMDI0YzkwZTZjYmM1LnBkZg==&rn=QmlnT2xpdHRsZW8ucGRm
Big-O and Little-o The Landau symbolic notation – Big-O and little-o – can be one of the most useful notations in analysis. Learn it well and you become more flexible, quicker, and more sure of yourself in a wide variety of analytic settings. One concept associated with limits that we weave through this is the notion of “eventually.” To understand this, let P(x) be a statement depending on the variable x that can be either true or false. The statement “as x Æ0, eventually P(x)” means that $ d > 0 such that for |x| < d, P(x) is true. The statement “as n Æ ∞, eventually P(n)” means that $ N such that for n > N, P(n) is true. Similar statements can be made for all limit situations. As an example: lim xÆa ƒ( x) = L iff " e > 0, eventually |ƒ( x) – L| < e, where in this case, “eventually” means “$ d > 0 such that for | x – a| < d.” With the concept of “eventually” understood, we proceed to the definitions of Big-O and little-o. We can define these concepts in any limit situation. In what follows, the function g(x) that we are comparing ought be eventually monotone , nonzero, and should be “reasonably simple”, a concept that we won’t define. Definition: As x Æ ?, ƒ( x) = o(g(x)) iff lim xÆ? ƒ( x) g(x) = 0. Definition: As x Æ ?, ƒ( x) = O(g(x)) iff $ C such that eventually |ƒ( x)| ≤ C|g(x)|. The second definition could be rephrased as saying that the fraction ƒ( x) g(x) is eventually bounded. Some examples of this notation: If m < n, as x Æ ∞, xm = o(xn). If m < n, as x Æ 0, xn = o(xm). If n > 0, as x Æ ∞, ln( x) = o(xn). If n < 0, as x Æ 0+, ln( x) = o(xn). As x Æ ∞, sin x = O(1). As x Æ 0, sin x = O(x). As x Æ 0, sin x = x + O(x3). As x Æ 0, sin x = x – x3 6 + O(x5). As n Æ ∞, n2#+#1 = O(n). As n Æ ∞, n2#+#1 = n + OË ÁÊ ¯ ˜ˆ 1 n# . This notation is very closely tied up with differentiation and with Taylor polynomials. For instance, we can prove that the function ƒ is differentiable at x iff $ a number which we call ƒ¢(x) such that as h Æ 0, ƒ( x + h) = ƒ( x) + ƒ¢(x)h + o(h). That is the definition of the derivative right there. The thing we’re calling o(h) is ƒ( x + h) – ƒ( x) – ƒ¢(x)h and calling it o(h) is saying that lim hÆ0 ƒ( x#+#h)#–#ƒ( x)#–#ƒ¢(x)h h = lim hÆ0 ƒ( x#+#h)#–#ƒ( x) h – ƒ¢(x) = 0 That’s what we get if ƒ is differentiable at x. If we have something stronger, namely that ƒ has a bounded second derivative in a neighborhood of x, then we can say something stronger: ƒ( x + h) = ƒ( x) + ƒ¢(x)h + O(h2). In general, if ƒ has n bounded derivatives in a neighborhood of x, Taylor’s Theorem says ƒ( x + h) =  k=0 n–1 ƒkhk + O(hn) . You’ll notice that this last statement is slightly weaker than the full statement of Taylor’s Theorem. We have merely said that the error is no worse than some constant times hn. The full statement of the theorem gives us some tools for estimating how big that constant is; using big-O notation essentially announces that we don’t care how big the constant is, as long as it’s constant. The reason big-O and little-o is a powerful an flexible notation is that more often than not, we don’t care what the constant is. Manipulating this notation: Most of what we have here is common sense. Some useful ideas: O(g1(x)) + O(g2(x)) = O(max( g1(x), g2(x)) O(g1(x)) ⋅O(g2(x)) = O(g1(x)⋅g2(x)) We’d like to be able to say that for reasonable functions w, w(O(g(x))) = O(w(g(x)). There’s no problem with saying ( O(h)) 2 = O(h2), but eO(ln x) isn’t well defined. You have to be careful there. You should avoid dividing by big-O or little-o. However, one can make sense of something like 1 2+ O(x) by long division: 1 2+ O(x) = 1 2 + O(x) as x Æ 0. Here’s a messy calculus problem which can be used to illustrate the ideas. You should take a close look at each step of this and figure out why I’m doing what I’m doing: Example: Find lim nÆ∞ Î ÍÈ ˚ ˙˘ 1 π#ı Û 0 π Ë ÁÊ ¯ ˜ˆ 1#+## cos #x n –1 dx n . This is a 1 ∞ case, so we start by taking the logarithm. For small u, (1 + u)–1 = 1 – u + u2 + O(u3). (Geometric series.) Thus, Ë ÁÊ ¯ ˜ˆ 1#+## cos #x n –1 = 1 – cos #x n +cos 2x n + O(n–3/2 ) 1 π ı Û 0 π Ë ÁÊ ¯ ˜ˆ 1#+## cos #x n –1 dx = 1 π ı Û 0 π 1 – cos #x n +cos 2x n + O(n–3/2 ) = 1 + 0 + 1 2n + O(n–3/2 ) For small u, ln (1 + u) = u + O(u2). Thus, n ln Î ÍÈ ˚ ˙˘1 π#ı Û 0 π Ë ÁÊ ¯ ˜ˆ 1#+## cos #x n –1 dx = n ln Ë ÁÊ ¯ ˜ˆ 1#+#0#+# 1 2n#+#O(n–3/2 ) = n Ë ÁÊ ¯ ˜ˆ1 2n## +#O(n–3/2 )#+#O(n–2 ) = 1 2 + O(n–1/2 ) . This goes to 1 2 , so the limit is e1/2 = e . Exercises: Does the series  n=1 ∞# n3[sin(arctan( n–1 )) – arctan(sin( n–1 ))] converge or diverge? Suppose that ƒ is defined on an interval and " y in that interval, as x Æ y, ƒ( x) – ƒ( y) = o(| x – y|). Prove that ƒ is constant. In all parts of this problem, assume that ƒ Œ C∞ (the infinitely differentiable functions.) a. Show that ƒ( x#+#h)#–#f(x) h = ƒ¢(x) + O(h) as h Æ 0. b. Show that ƒ( x#+#h)#–#f(x#–#h) 2h = ƒ¢(x) + O(h2) as h Æ 0. c. Show that ƒ( x#+#h)#–#2ƒ( x)#+#ƒ( x#–#h) h2 = ƒ ¢¢ (x) + O(h2) as h Æ 0.
187854
https://en.wikipedia.org/wiki/Neural_crest
Jump to content Search Contents 1 History 2 Induction 2.1 Inductive signals 2.2 Neural plate border specifiers 2.3 Neural crest specifiers 2.4 Neural crest effector genes 3 Migration 3.1 Delamination 3.2 Migration 4 Clinical significance 4.1 Waardenburg syndrome 4.2 Hirschsprung's disease 4.3 Fetal alcohol spectrum disorder 4.4 DiGeorge syndrome 4.5 Treacher Collins syndrome 5 Cell lineages 5.1 Cranial neural crest 5.2 Trunk neural crest 5.3 Vagal and sacral neural crest 5.4 Cardiac neural crest 6 Evolution 7 Neural crest derivatives 8 See also 9 References 10 External links Neural crest العربية Català Čeština Deutsch Eesti Español Esperanto فارسی Français Galego 한국어 Bahasa Indonesia Italiano עברית Nederlands 日本語 Polski Português Русский Suomi Svenska Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Pluripotent embyronic cell group giving rise to diverse cell lineages | Neural crest | | The formation of neural crest during the process of neurulation. Neural crest is first induced in the region of the neural plate border. After neural tube closure, neural crest cells delaminate from the region between the dorsal neural tube and overlying ectoderm and migrate out towards the periphery. | | Identifiers | | MeSH | D009432 | | TE | crest_by_E5.0.2.1.0.0.2 E5.0.2.1.0.0.2 | | FMA | 86666 | | Anatomical terminology [edit on Wikidata] | The neural crest is a ridge-like structure that is formed transiently between the epidermal ectoderm and neural plate during vertebrate development. Neural crest cells originate from this structure through the epithelial-mesenchymal transition, and in turn give rise to a diverse cell lineage—including melanocytes, craniofacial cartilage and bone, smooth muscle, dentin, peripheral and enteric neurons, adrenal medulla and glia. After gastrulation, the neural crest is specified at the border of the neural plate and the non-neural ectoderm. During neurulation, the borders of the neural plate, also known as the neural folds, converge at the dorsal midline to form the neural tube. Subsequently, neural crest cells from the roof plate of the neural tube undergo an epithelial to mesenchymal transition, delaminating from the neuroepithelium and migrating through the periphery, where they differentiate into varied cell types. The emergence of the neural crest was important in vertebrate evolution because many of its structural derivatives are defining features of the vertebrate clade. Underlying the development of the neural crest is a gene regulatory network, described as a set of interacting signals, transcription factors, and downstream effector genes, that confer cell characteristics such as multipotency and migratory capabilities. Understanding the molecular mechanisms of neural crest formation is important for our knowledge of human disease because of its contributions to multiple cell lineages. Abnormalities in neural crest development cause neurocristopathies, which include conditions such as frontonasal dysplasia, Waardenburg–Shah syndrome, and DiGeorge syndrome. Defining the mechanisms of neural crest development may reveal key insights into vertebrate evolution and neurocristopathies. History [edit] The neural crest was first described in the chick embryo by Wilhelm His Sr. in 1868 as "the cord in between" (Zwischenstrang) because of its origin between the neural plate and non-neural ectoderm. He named the tissue "ganglionic crest," since its final destination was each lateral side of the neural tube, where it differentiated into spinal ganglia. During the first half of the 20th century, the majority of research on the neural crest was done using amphibian embryos which was reviewed by Hörstadius (1950) in a well known monograph. Cell labeling techniques advanced research into the neural crest because they allowed researchers to visualize the migration of the tissue throughout the developing embryos. In the 1960s, Weston and Chibon utilized radioisotopic labeling of the nucleus with tritiated thymidine in chick and amphibian embryo respectively. However, this method suffers from drawbacks of stability, since every time the labeled cell divides the signal is diluted. Modern cell labeling techniques such as rhodamine-lysinated dextran and the vital dye diI have also been developed to transiently mark neural crest lineages. The quail-chick marking system, devised by Nicole Le Douarin in 1969, was another instrumental technique used to track neural crest cells. Chimeras, generated through transplantation, enabled researchers to distinguish neural crest cells of one species from the surrounding tissue of another species. With this technique, generations of scientists were able to reliably mark and study the ontogeny of neural crest cells. Induction [edit] A molecular cascade of events is involved in establishing the migratory and multipotent characteristics of neural crest cells. This gene regulatory network can be subdivided into the following four sub-networks described below. Inductive signals [edit] First, extracellular signaling molecules, secreted from the adjacent epidermis and underlying mesoderm such as Wnts, BMPs and Fgfs separate the non-neural ectoderm (epidermis) from the neural plate during neural induction. Wnt signaling has been demonstrated in neural crest induction in several species through gain-of-function and loss-of-function experiments. In coherence with this observation, the promoter region of slug (a neural-crest-specific gene) contains a binding site for transcription factors involved in the activation of Wnt-dependent target genes, suggestive of a direct role of Wnt signaling in neural crest specification. The current role of BMP in neural crest formation is associated with the induction of the neural plate. BMP antagonists diffusing from the ectoderm generates a gradient of BMP activity. In this manner, the neural crest lineage forms from intermediate levels of BMP signaling required for the development of the neural plate (low BMP) and epidermis (high BMP). Fgf from the paraxial mesoderm has been suggested as a source of neural crest inductive signal. Researchers have demonstrated that the expression of dominate-negative Fgf receptor in ectoderm explants blocks neural crest induction when recombined with paraxial mesoderm. The understanding of the role of BMP, Wnt, and Fgf pathways on neural crest specifier expression remains incomplete. Neural plate border specifiers [edit] Signaling events that establish the neural plate border lead to the expression of a set of transcription factors delineated here as neural plate border specifiers. These molecules include Zic factors, Pax3/7, Dlx5, Msx1/2 which may mediate the influence of Wnts, BMPs, and Fgfs. These genes are expressed broadly at the neural plate border region and precede the expression of bona fide neural crest markers. Experimental evidence places these transcription factors upstream of neural crest specifiers. For example, in Xenopus Msx1 is necessary and sufficient for the expression of Slug, Snail, and FoxD3. Furthermore, Pax3 is essential for FoxD3 expression in mouse embryos. Neural crest specifiers [edit] Following the expression of neural plate border specifiers is a collection of genes including Slug/Snail, FoxD3, Sox10, Sox9, AP-2 and c-Myc. This suite of genes, designated here as neural crest specifiers, are activated in emergent neural crest cells. At least in Xenopus, every neural crest specifier is necessary and/or sufficient for the expression of all other specifiers, demonstrating the existence of extensive cross-regulation. Moreover, this model organism was instrumental in the elucidation of the role of the Hedgehog signaling pathway in the specification of the neural crest, with the transcription factor Gli2 playing a key role. Outside of the tightly regulated network of neural crest specifiers are two other transcription factors Twist and Id. Twist, a bHLH transcription factor, is required for mesenchyme differentiation of the pharyngeal arch structures. Id is a direct target of c-Myc and is known to be important for the maintenance of neural crest stem cells. Neural crest effector genes [edit] Finally, neural crest specifiers turn on the expression of effector genes, which confer certain properties such as migration and multipotency. Two neural crest effectors, Rho GTPases and cadherins, function in delamination by regulating cell morphology and adhesive properties. Sox9 and Sox10 regulate neural crest differentiation by activating many cell-type-specific effectors including Mitf, P0, Cx32, Trp and cKit. Migration [edit] Further information: Collective cell migration The migration of neural crest cells involves a highly coordinated cascade of events that begins with closure of the dorsal neural tube. Delamination [edit] After fusion of the neural folds to create the neural tube, cells originally located in the neural plate border become neural crest cells. For migration to begin, neural crest cells must undergo a process called delamination that involves a full or partial epithelial–mesenchymal transition (EMT). Delamination is defined as the separation of tissue into different populations, in this case neural crest cells separating from the surrounding tissue. Conversely, EMT is a series of events coordinating a change from an epithelial to mesenchymal phenotype. For example, delamination in chick embryos is triggered by a BMP/Wnt cascade that induces the expression of EMT promoting transcription factors such as SNAI2 and FOXD3. Although all neural crest cells undergo EMT, the timing of delamination occurs at different stages in different organisms: in Xenopus laevis embryos there is a massive delamination that occurs when the neural plate is not entirely fused, whereas delamination in the chick embryo occurs during fusion of the neural fold. Prior to delamination, presumptive neural crest cells are initially anchored to neighboring cells by tight junction proteins such as occludin and cell adhesion molecules such as NCAM and N-Cadherin. Dorsally expressed BMPs initiate delamination by inducing the expression of the zinc finger protein transcription factors snail, slug, and twist. These factors play a direct role in inducing the epithelial-mesenchymal transition by reducing expression of occludin and N-Cadherin in addition to promoting modification of NCAMs with polysialic acid residues to decrease adhesiveness. Neural crest cells also begin expressing proteases capable of degrading cadherins such as ADAM10 and secreting matrix metalloproteinases (MMPs) that degrade the overlying basal lamina of the neural tube to allow neural crest cells to escape. Additionally, neural crest cells begin expressing integrins that associate with extracellular matrix proteins, including collagen, fibronectin, and laminin, during migration. Once the basal lamina becomes permeable, neural crest cells can begin migrating throughout the embryo. Migration [edit] Neural crest cell migration occurs in a rostral to caudal direction without the need of a neuronal scaffold such as along a radial glial cell. For this reason the crest cell migration process is termed "free migration". Instead of scaffolding on progenitor cells, neural crest migration is the result of repulsive guidance via EphB/EphrinB and semaphorin/neuropilin signaling, interactions with the extracellular matrix, and contact inhibition with one another. While Ephrin and Eph proteins have the capacity to undergo bi-directional signaling, neural crest cell repulsion employs predominantly forward signaling to initiate a response within the receptor bearing neural crest cell. Burgeoning neural crest cells express EphB, a receptor tyrosine kinase, which binds the EphrinB transmembrane ligand expressed in the caudal half of each somite. When these two domains interact it causes receptor tyrosine phosphorylation, activation of rhoGTPases, and eventual cytoskeletal rearrangements within the crest cells inducing them to repel. This phenomenon allows neural crest cells to funnel through the rostral portion of each somite. Semaphorin-neuropilin repulsive signaling works synergistically with EphB signaling to guide neural crest cells down the rostral half of somites in mice. In chick embryos, semaphorin acts in the cephalic region to guide neural crest cells through the pharyngeal arches. On top of repulsive repulsive signaling, neural crest cells express β1and α4 integrins which allows for binding and guided interaction with collagen, laminin, and fibronectin of the extracellular matrix as they travel. Additionally, crest cells have intrinsic contact inhibition with one another while freely invading tissues of different origin such as mesoderm. Neural crest cells that migrate through the rostral half of somites differentiate into sensory and sympathetic neurons of the peripheral nervous system. The other main route neural crest cells take is dorsolaterally between the epidermis and the dermamyotome. Cells migrating through this path differentiate into pigment cells of the dermis. Further neural crest cell differentiation and specification into their final cell type is biased by their spatiotemporal subjection to morphogenic cues such as BMP, Wnt, FGF, Hox, and Notch. Clinical significance [edit] Neurocristopathies result from the abnormal specification, migration, differentiation or death of neural crest cells throughout embryonic development. This group of diseases comprises a wide spectrum of congenital malformations affecting many newborns. Additionally, they arise because of genetic defects affecting the formation of the neural crest and because of the action of teratogens Waardenburg syndrome [edit] Waardenburg syndrome is a neurocristopathy that results from defective neural crest cell migration. The condition's main characteristics include piebaldism and congenital deafness. In the case of piebaldism, the colorless skin areas are caused by a total absence of neural crest-derived pigment-producing melanocytes. There are four different types of Waardenburg syndrome, each with distinct genetic and physiological features. Types I and II are distinguished based on whether or not family members of the affected individual have dystopia canthorum. Type III gives rise to upper limb abnormalities. Lastly, type IV is also known as Waardenburg-Shah syndrome, and afflicted individuals display both Waardenburg's syndrome and Hirschsprung's disease. Types I and III are inherited in an autosomal dominant fashion, while II and IV exhibit an autosomal recessive pattern of inheritance. Overall, Waardenburg's syndrome is rare, with an incidence of ~ 2/100,000 people in the United States. All races and sexes are equally affected. There is no current cure or treatment for Waardenburg's syndrome. Hirschsprung's disease [edit] Also implicated in defects related to neural crest cell development and migration is Hirschsprung's disease, characterized by a lack of innervation in regions of the intestine. This lack of innervation can lead to further physiological abnormalities like an enlarged colon (megacolon), obstruction of the bowels, or even slowed growth. In healthy development, neural crest cells migrate into the gut and form the enteric ganglia. Genes playing a role in the healthy migration of these neural crest cells to the gut include RET, GDNF, GFRα, EDN3, and EDNRB. RET, a receptor tyrosine kinase (RTK), forms a complex with GDNF and GFRα. EDN3 and EDNRB are then implicated in the same signaling network. When this signaling is disrupted in mice, aganglionosis, or the lack of these enteric ganglia occurs. Fetal alcohol spectrum disorder [edit] Fetal alcohol spectrum disorder is among the most common causes of developmental defects. Depending on the extent of the exposure and the severity of the resulting abnormalities, patients are diagnosed within a continuum of disorders broadly labeled fetal alcohol spectrum disorder (FASD). Severe FASD can impair neural crest migration, as evidenced by characteristic craniofacial abnormalities including short palpebral fissures, an elongated upper lip, and a smoothened philtrum. However, due to the promiscuous nature of ethanol binding, the mechanisms by which these abnormalities arise is still unclear. Cell culture explants of neural crest cells as well as in vivo developing zebrafish embryos exposed to ethanol show a decreased number of migratory cells and decreased distances travelled by migrating neural crest cells. The mechanisms behind these changes are not well understood, but evidence suggests PAE can increase apoptosis due to increased cytosolic calcium levels caused by IP3-mediated release of calcium from intracellular stores. It has also been proposed that the decreased viability of ethanol-exposed neural crest cells is caused by increased oxidative stress. Despite these, and other advances much remains to be discovered about how ethanol affects neural crest development. For example, it appears that ethanol differentially affects certain neural crest cells over others; that is, while craniofacial abnormalities are common in PAE, neural crest-derived pigment cells appear to be minimally affected. DiGeorge syndrome [edit] DiGeorge syndrome is associated with deletions or translocations of a small segment in the human chromosome 22. This deletion may disrupt rostral neural crest cell migration or development. Some defects observed are linked to the pharyngeal pouch system, which receives contribution from rostral migratory crest cells. The symptoms of DiGeorge syndrome include congenital heart defects, facial defects, and some neurological and learning disabilities. Patients with 22q11 deletions have also been reported to have higher incidence of schizophrenia and bipolar disorder. Treacher Collins syndrome [edit] Treacher Collins syndrome (TCS) results from the compromised development of the first and second pharyngeal arches during the early embryonic stage, which ultimately leads to mid and lower face abnormalities. TCS is caused by the missense mutation of the TCOF1 gene, which causes neural crest cells to undergo apoptosis during embryogenesis. Although mutations of the TCOF1 gene are among the best characterized in their role in TCS, mutations in POLR1C and POLR1D genes have also been linked to the pathogenesis of TCS. Cell lineages [edit] Neural crest cells originating from different positions along the anterior-posterior axis develop into various tissues. These regions of the neural crest can be divided into four main functional domains, which include the cranial neural crest, trunk neural crest, vagal and sacral neural crest, and cardiac neural crest. Cranial neural crest [edit] Main article: cranial neural crest The cranial neural crest migrates dorsolaterally to form the craniofacial mesenchyme that differentiates into various cranial ganglia and craniofacial cartilages and bones. These cells enter the pharyngeal pouches and arches where they contribute to the thymus, bones of the middle ear and jaw and the odontoblasts of the tooth primordia. Trunk neural crest [edit] Main article: trunk neural crest The trunk neural crest gives rise to two populations of cells. One group of cells fated to become melanocytes migrates dorsolaterally into the ectoderm towards the ventral midline. A second group of cells migrates ventrolaterally through the anterior portion of each sclerotome. The cells that stay in the sclerotome form the dorsal root ganglia, whereas those that continue more ventrally form the sympathetic ganglia, adrenal medulla, and the nerves surrounding the aorta. Vagal and sacral neural crest [edit] Vagal and sacral neural crest cells develop into the ganglia of the enteric nervous system and the parasympathetic ganglia. Cardiac neural crest [edit] Main article: cardiac neural crest Cardiac neural crest develops into melanocytes, cartilage, connective tissue and neurons of some pharyngeal arches. Also, this domain gives rise to regions of the heart such as the musculo-connective tissue of the large arteries, and part of the septum, which divides the pulmonary circulation from the aorta. The semilunar valves of the heart are associated with neural crest cells according to new research. Evolution [edit] Several structures that distinguish the vertebrates from other chordates are formed from the derivatives of neural crest cells. In their "New head" theory, Gans and Northcut argue that the presence of neural crest was the basis for vertebrate specific features, such as sensory ganglia and cranial skeleton. Furthermore, the appearance of these features was pivotal in vertebrate evolution because it enabled a predatory lifestyle. However, considering the neural crest a vertebrate innovation does not mean that it arose de novo. Instead, new structures often arise through modification of existing developmental regulatory programs. For example, regulatory programs may be changed by the co-option of new upstream regulators or by the employment of new downstream gene targets, thus placing existing networks in a novel context. This idea is supported by in situ hybridization data that shows the conservation of the neural plate border specifiers in protochordates, which suggest that part of the neural crest precursor network was present in a common ancestor to the chordates. In some non-vertebrate chordates such as tunicates a lineage of cells (melanocytes) has been identified, which are similar to neural crest cells in vertebrates. This implies that a rudimentary neural crest existed in a common ancestor of vertebrates and tunicates. Neural crest derivatives [edit] Ectomesenchyme (also known as mesectoderm): odontoblasts, dental papillae, the chondrocranium (nasal capsule, Meckel's cartilage, scleral ossicles, quadrate, articular, hyoid and columella), tracheal and laryngeal cartilage, the dermatocranium (membranous bones), dorsal fins and the turtle plastron (lower vertebrates), pericytes and smooth muscle of branchial arteries and veins, tendons of ocular and masticatory muscles, connective tissue of head and neck glands (pituitary, salivary, lachrymal, thymus, thyroid) dermis and adipose tissue of calvaria, ventral neck and face Endocrine cells: chromaffin cells of the adrenal medulla, glomus cells type I/II. Peripheral nervous system: Sensory neurons and glia of the dorsal root ganglia, cephalic ganglia (VII and in part, V, IX, and X), Rohon-Beard cells, some Merkel cells in the whisker, Satellite glial cells of all autonomic and sensory ganglia, Schwann cells of all peripheral nerves. Enteric cells: Enterochromaffin cells. Melanocytes, iris muscle and pigment cells, and even associated with some tumors (such as melanotic neuroectodermal tumor of infancy). See also [edit] First arch syndrome DGCR2—may control neural crest cell migration List of human cell types derived from the germ layers References [edit] ^ a b c d e f Huang, X.; Saint-Jeannet, J.P. (2004). "Induction of the neural crest and the opportunities of life on the edge". Dev. Biol. 275 (1): 1–11. doi:10.1016/j.ydbio.2004.07.033. PMID 15464568. ^ Shakhova, Olga; Sommer, Lukas (2008). "Neural crest-derived stem cells". StemBook. Harvard Stem Cell Institute. doi:10.3824/stembook.1.51.1. PMID 20614636. Retrieved 27 December 2019. ^ Brooker, R.J. 2014, Biology, 3rd edn, McGraw-Hill, New York, NY, 1084 ^ a b c d e Meulemans, D.; Bronner-Fraser, M. (2004). "Gene-regulatory interactions in neural crest evolution and development". Dev Cell. 7 (3): 291–9. doi:10.1016/j.devcel.2004.08.007. 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External links [edit] Embryology at UNSW Notes/ncrest ancil-445 at NeuroNames Diagram at University of Michigan Hox domains in chicks | v t e Human embryonic development in the first three weeks | | Week 1 | Fertilization Oocyte activation Zygote Cleavage Blastomere Morula Cavitation Blastocoel Blastocyst Inner cell mass Trophoblast | | Week 2 (Bilaminar) | Hypoblast Epiblast | | Week 3 (Trilaminar) | | | | --- | | Germ layers | Archenteron/Primitive streak + Primitive pit + Primitive node/Blastopore + Primitive groove Gastrula + Gastrulation Regional specification | | Ectoderm | Surface ectoderm Neuroectoderm Somatopleuric mesenchyme Neurulation Neural crest | | Endoderm | Splanchnopleuric mesenchyme | | Mesoderm | Axial mesoderm Paraxial + Somite + Somitomere Intermediate Lateral plate + Intraembryonic coelom + Splanchnopleuric mesenchyme + Somatopleuric mesenchyme | | | v t e Development of the nervous system | | Neurogenesis | | | | --- | | General | Neural development Neurulation Neurula Notochord Neuroectoderm Neural plate + Neural fold + Neural groove Neuropoiesis Adult neurogenesis | | Neural crest | Cranial neural crest + Cardiac neural crest complex Truncal neural crest | | Neural tube | Rostral neuropore Neuromere / Rhombomere Cephalic flexure Cervical flexure Pontine flexure Alar plate Basal plate Glioblast Neuroblast Germinal matrix | | | Eye | Neural tube Optic vesicle Optic stalk Optic cup Surface ectoderm + Lens placode | | Ear | Otic placode + Otic pit + Otic vesicle | Retrieved from " Categories: Embryology of nervous system Ectoderm Chordate anatomy Animal nervous system Hidden categories: Articles with short description Short description is different from Wikidata Neural crest Add topic
187855
https://www.ck12.org/flexi/math-grade-8/slope-intercept-form-of-linear-equations/define-the-slope-intercept-form./
Flexi answers - Define the slope-intercept form. | CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Math Grade 8 Deriving the Equation y = mx+b Question Define the slope-intercept form. Flexi Says: Linear expressions can be written in slope-intercept form:y=m x+b, where m represents the slope of the line, and b is the y coordinate where the line crosses the y axis (a location called the y-intercept). The slope,m,is written in the form change in y change in x or r i s e r u n and whole-number slope values are read as having a denominator of 1. Let’s look at an example: Graph the linear equation: y=3 x+2. i Analogy / Example Try Asking: How can you derive the equation of a graph?What are m and b in the linear equation y = 26x + 12?Determine the equation of the line passing through the point (−3, −10), with a slope of m=4. Put your answer in slope-intercept form. How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
187856
https://www.liverpool.ac.uk/~pjgiblin/papers/zigzag-final.pdf
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https://www.quora.com/Fluid-Mechanics-What-is-the-difference-between-vapour-pressure-and-saturation-pressure
Fluid Mechanics: What is the difference between vapour pressure and saturation pressure? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Saturation Point Vapor Pressure Engineering Fluid Mechanics Thermodyn... Chemistry Science Physics Thermodynamics Fluid Dynamics 5 Fluid Mechanics: What is the difference between vapour pressure and saturation pressure? All related (38) Sort Recommended Assistant Bot · 11mo Vapor pressure and saturation pressure are closely related concepts in fluid mechanics and thermodynamics, especially in the context of phase changes between liquid and gas. Here's a breakdown of their differences: Vapor Pressure Definition: Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. Conditions: It can occur at any temperature and is specific to the substance in question. As temperature increases, vapor pressure generally increases. Measurement: Vapor pressure can be measured at various temperatures, reflecting how much Continue Reading Vapor pressure and saturation pressure are closely related concepts in fluid mechanics and thermodynamics, especially in the context of phase changes between liquid and gas. Here's a breakdown of their differences: Vapor Pressure Definition: Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. Conditions: It can occur at any temperature and is specific to the substance in question. As temperature increases, vapor pressure generally increases. Measurement: Vapor pressure can be measured at various temperatures, reflecting how much of the substance is present in the vapor phase. Saturation Pressure Definition: Saturation pressure is the specific vapor pressure at which a liquid (or solid) is in equilibrium with its vapor phase at a given temperature. It is the maximum pressure that the vapor can exert before condensation occurs. Conditions: Saturation pressure is always associated with a specific temperature; it increases with temperature. Context: It is particularly important in understanding phase changes, such as boiling, where the vapor pressure of a liquid equals the ambient pressure. Summary Relationship: Saturation pressure is a specific type of vapor pressure at equilibrium. When a liquid reaches its saturation pressure, it begins to boil if additional heat is supplied. Usage: Vapor pressure can be considered a broader term that applies to any vapor above a liquid or solid, while saturation pressure specifically refers to the pressure at which a phase change occurs. In practical applications, understanding these pressures helps in processes like distillation, refrigeration, and understanding weather phenomena. Upvote · Mrinmoy Dey Mechanical Engineer ·11y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · I assume that we know the textbook defination of Vapour Pressure and Saturation Pressure. Still I will try to explain it in the manner i felt it easy to understand. So Vapour Pressure can be defined as the pressure exerted by a gas(more appropiately, vopour) which is in an equiblirium condition with its condensed phase(liquid) in a closed system at a particular temperature(say T). And Saturation Pressure can be defined as the pressure corresponding to a temperature called saturated temperature (say T0) at which the vapour pressure of the mixture equals the atmospheric pressure. (The system i Continue Reading I assume that we know the textbook defination of Vapour Pressure and Saturation Pressure. Still I will try to explain it in the manner i felt it easy to understand. So Vapour Pressure can be defined as the pressure exerted by a gas(more appropiately, vopour) which is in an equiblirium condition with its condensed phase(liquid) in a closed system at a particular temperature(say T). And Saturation Pressure can be defined as the pressure corresponding to a temperature called saturated temperature (say T0) at which the vapour pressure of the mixture equals the atmospheric pressure. (The system is said to have reached its boiling point.) Lets, understand it from the graph above(from wiki). Let us consider the red line(graph of 'diethyl ether' corresponding to 'vapour pressure vs temperature' ignore all other lines). okay, so take a particular temp T( anything between -30 to 30 degree cent.). and see the coressponding pressure to the temp., this pressure is the vapour pressure( Yaxis), but if we increase the temperature from T to T0(30.2 or 30.3), the vapour pressure equals the atmospheric pressure and the corresponding pressure gives the Saturated pressure. So in a nut shell, vapour pressure of any fluid corresponding to the saturated temp of that fluid gives the saturated pressure..This hopefully fulfils your difference between the two. Upvote · 99 80 9 5 Sponsored by Grammarly Review your English writing for mistakes. Write clearly in English with no mistakes. Try our free writing app! Download 5.4K 5.4K Ayush Kumar Studied at National Institute of Technology, Jamshedpur (Graduated 2020) ·7y Originally Answered: Fluid Mechanics: What is the difference between Vapour pressure and saturation pressure? · Both are same but the way we define these two are different. for example vapour pressure:-at equilibrium condition between liquid and their vapour ,the pressure exerted by vapour on the surface of liquid or container. Saturation pressure:-it is the pressure just below which water started to evaporate. Numerical value will be nearly same. Upvote · 99 13 Related questions More answers below What are the differences between vapour pressure and saturated vapour pressure? What is the difference between Vapour pressure and saturation pressure? What is the difference between saturated temperature and saturated pressure? What is the saturation pressure of fluid? Is the equilibrium vapour pressure always equal to the saturated vapour pressure? Name Keep it simple · Author has 118 answers and 338.7K answer views ·7y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · There is NO difference between vapour pressure and saturation pressure. ... Vapour pressure is the pressure exerted by the vapour on its condensed phase at a give temperature in a closed system. Saturation pressure is the pressure at which a liquid satrts to boil at a given temperature in a closed system. Conceptually: Vapour pressure is the partial pressure of the vapour in the system. Saturation pressure is the maximum pressure at which liquid will boil at a predefined temperature. Saturation vapour pressure is the vapour pressure of the vapour in a saturated vapour phase. Think of it as the par Continue Reading There is NO difference between vapour pressure and saturation pressure. ... Vapour pressure is the pressure exerted by the vapour on its condensed phase at a give temperature in a closed system. Saturation pressure is the pressure at which a liquid satrts to boil at a given temperature in a closed system. Conceptually: Vapour pressure is the partial pressure of the vapour in the system. Saturation pressure is the maximum pressure at which liquid will boil at a predefined temperature. Saturation vapour pressure is the vapour pressure of the vapour in a saturated vapour phase. Think of it as the partial pressure of the vapour when the air is saturated with it. This occurs at equilibrium with the liquid. Upvote · 9 6 Benoit Cushman-Roisin Fascinated by fluid flows around us and the math of it all. · Author has 375 answers and 1.4M answer views ·9y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · Same difference as between 12 and a dozen, that is, none. The better expression is vapor pressure, with saturation (equilibrium between liquid and its vapor) being understood. Things get a little complicated in meteorology because meteorologists need to reckon with humidity in the air. Sometimes they refer to the presence of water vapor below saturation (less than 100% relative humidity) inappropriately as 'water vapor'. So, when relative humidity reaches 100% and condensation begins, creating saturation between vapor and liquid water droplets, they use the expression 'saturation vapor pressur Continue Reading Same difference as between 12 and a dozen, that is, none. The better expression is vapor pressure, with saturation (equilibrium between liquid and its vapor) being understood. Things get a little complicated in meteorology because meteorologists need to reckon with humidity in the air. Sometimes they refer to the presence of water vapor below saturation (less than 100% relative humidity) inappropriately as 'water vapor'. So, when relative humidity reaches 100% and condensation begins, creating saturation between vapor and liquid water droplets, they use the expression 'saturation vapor pressure'. The text below is cut and pasted from Wikipedia's article on vapor pressure: In meteorology, the term vapor pressure is used to mean the partial pressure of water vapor in the atmosphere, even if it is not in equilibrium, and the equilibrium vapor pressure is specified otherwise. Meteorologists also use the term saturation vapor pressure to refer to the equilibrium vapor pressure of water or brine above a flat surface, to distinguish it from equilibrium vapor pressure, which takes into account the shape and size of water droplets and particulates in the atmosphere. Upvote · 99 12 Sponsored by JetBrains DataGrip, a powerful GUI tool for SQL. Smart code completion, on-the-fly analysis, quick-fixes, refactorings that work in SQL files, and more. Download 999 678 Sanjay Sharma Studied at St. John's College, Agra ·2y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · Well there is no difference between vapour pressure and saturation pressure. Vapour pressure is the pressure that is exerted by the vapour on its condensed phase at a given temperature in a closed system. Saturation pressure is the pressure in which a liquid starts boiling in a closed system at a given temperature. In a system Vapour pressure is the partial pressure of the vapour. Infect Saturation pressure is a unique case of the vapour pressure. Commonly this is seen at the point of transition to the liquid state from the gaseous state. Upvote · Related questions More answers below What is saturated vapour pressure? What is the difference between partial pressure and saturated vapour pressure? What do you mean by saturated vapour pressure at a particular temperature? What is vapour pressure? What is the difference between vapour pressure and total vapour pressure? Igor Polozov author of "aneasycalc" program (calculator and LaTeX string generator) · Author has 1.1K answers and 1.2M answer views ·Updated 8y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · Saturation pressure is a special case of the vapor pressure. It is perfectly acceptable to say, that the saturation pressure is a such a vapor pressure which corresponds to the phase transition temperature. Usually, this is the point of transition from the gaseous to the liquid state (and backwards of course). But it is not always the case. Each temperature corresponds exactly one saturation pressure, if it exists. And a line graph showing the dependence is continuous and bounded curve. It has the beginning and the end. The beginning is the beginnig of the temperature scale. There is the gas p Continue Reading Saturation pressure is a special case of the vapor pressure. It is perfectly acceptable to say, that the saturation pressure is a such a vapor pressure which corresponds to the phase transition temperature. Usually, this is the point of transition from the gaseous to the liquid state (and backwards of course). But it is not always the case. Each temperature corresponds exactly one saturation pressure, if it exists. And a line graph showing the dependence is continuous and bounded curve. It has the beginning and the end. The beginning is the beginnig of the temperature scale. There is the gas phase - rigid phase balance within the interval of absolute zero to the so called triple point and this point is remarkable by the fact that the balance is achieved in three phases: - liquid, solid and gaseous in that point. The interval of liquid - gas balance starts from it. And the end of this line is the critical point. As far as approaching on the saturation line Ts (P) to the critical point, vapor properties differ from the fluids properties the less and less. And in the critical point that difference disappears. All said is true for only equilibrium states of the 2 or 3-phase system. A common property of all these intervals is the fact that for points higher than the temperature Ts (P) can be only gaseous state. That means, if we have a gas at T> Ts, we can cool it to Ts and it will stay a gas. Upvote · 9 4 Sponsored by CUHK Business Masters Kick-start your career with CUHK Business Master’s programmes. A transformative learning experience combining academic excellence with practical skills. Sign Up 999 196 Paolo Cuzzato Biochemistry has been my first love: lost, but never completely forgotten · Author has 8K answers and 6.9M answer views ·9y Originally Answered: What is the difference between Vapour pressure and saturation pressure? · The first is the actual pressure of the vapours of a given substance here and now (or there, of course……); the second is the pressure at which, at any given temperature, the vapours are in equilibrium with their liquid - that is, the max possible pressure at that temperature, since any exceeding amount of vapour would condense. Upvote · 9 5 Rob Hooft PhD in structural chemistry using molecular modeling and X-ray diffraction. · Author has 5K answers and 9.1M answer views ·7y Related What is saturated vapour pressure? The same numeric value as the vapour pressure. The only difference is that it is measured as a partial pressure with other gases present. Taking water as an example: if we have a closed container at room temperature with some water and water vapor in it, and no air, the pressure of the vapor will be about 2 kPa. This is the vapor pressure. If we have water at room temperature in air, the vapor pressure is still 2 kPa, but the total pressure of the air with the water will still be 100 kPa. The 2kPa is the maximum amount of water that can be contained in air at room temperature. That is the parti Continue Reading The same numeric value as the vapour pressure. The only difference is that it is measured as a partial pressure with other gases present. Taking water as an example: if we have a closed container at room temperature with some water and water vapor in it, and no air, the pressure of the vapor will be about 2 kPa. This is the vapor pressure. If we have water at room temperature in air, the vapor pressure is still 2 kPa, but the total pressure of the air with the water will still be 100 kPa. The 2kPa is the maximum amount of water that can be contained in air at room temperature. That is the partial water pressure of water saturated air. You call this the saturated vapour pressure. Upvote · 9 3 Sponsored by RedHat Know what your AI knows, with open source models. The only secrets your AI should keep are yours. Learn More Cristobal Cortes Professor, Thermal Engineering at University of Zaragoza (1989–present) · Author has 2.1K answers and 2.7M answer views ·2y Related Is there a difference between vapour (vapor) pressure and bubble point pressure of a liquid (also is there a difference between these terms and saturation vapour pressure)? Not very much. Those three terms designate the (equilibrium) pressure of pure liquid water in contact with pure water vapor or steam, which is an excellent approximation to what happens in reality most of the time. Only if you enter into the details of the phenomena do differences appear. For instance, the equilibrium assumed excludes surface tension, and thus a flat liquid-vapor interface. In ordinary boiling water at a pressure, small bubbles of steam appear and then their inner pressure is slightly above, so you have saturated steam at a slightly higher pressure than saturation, and also pur Continue Reading Not very much. Those three terms designate the (equilibrium) pressure of pure liquid water in contact with pure water vapor or steam, which is an excellent approximation to what happens in reality most of the time. Only if you enter into the details of the phenomena do differences appear. For instance, the equilibrium assumed excludes surface tension, and thus a flat liquid-vapor interface. In ordinary boiling water at a pressure, small bubbles of steam appear and then their inner pressure is slightly above, so you have saturated steam at a slightly higher pressure than saturation, and also pure liquid where it should be two-phase (superheated, metastable state). This is, of course rather difficult to explain in simple words (see Ref.), and its effect on most practical systems is very small indeed. Cf. J. H. Lienhard IV, J. H. Lienhard V. A heat transfer textbook. 4th Ed. Cambridge, MA: Phlogiston Press (2017), Ch. 9, Fig. 9.2, p. 564 and ss. Upvote · SUMEET KINI M.B.B.S. from Lokmanya Tilak Municipal Medical College and General Hospital ·9y Related What is vapour pressure? All liquids exhibit tendency for evaporation. Evaporation takes place at the surface of liquid. If the kinetic energy of liquid molecules overcomes the intermolecular force of attraction in the liquid state then the molecules from the surface of liquid escape into space above surface. The process is called 'evaporation'. If evaporation is carried out in a closed container system then the vapours of liquid remains in contact with surface of liquid. Like gas molecules vapour of molecules also execute continuous random motion. During this motions, molecules collide with each other and also with t Continue Reading All liquids exhibit tendency for evaporation. Evaporation takes place at the surface of liquid. If the kinetic energy of liquid molecules overcomes the intermolecular force of attraction in the liquid state then the molecules from the surface of liquid escape into space above surface. The process is called 'evaporation'. If evaporation is carried out in a closed container system then the vapours of liquid remains in contact with surface of liquid. Like gas molecules vapour of molecules also execute continuous random motion. During this motions, molecules collide with each other and also with the walls of the container, losses their energy and returns back to liquid state. This process is called as 'condensation'. Evaporation and condensation are continues processes. Hence, after some time an equilibrium is established, at constant temperature between evaporation and condensation. At equilibrium number of molecules in vapour state remains constant at constant temperature. "The pressure exerted by vapours of liquid on the surface of liquid when equilibrium is established between liquid and it's vapour is called VAPOUR PRESSURE of liquid." The vapour pressure of the liquid depends on the nature of the liquid and temperature. With increase of intermolecular force of attraction vapour pressure of liquid decrease and with rise of temperature vapour pressure of liquid increases. Mercury manometer may be used to determine vapour pressure of liquid. Upvote · 999 206 99 10 9 3 Subbanna Vadlamudy Ph. D. from Saugar University, Sagar (Graduated 1961) · Author has 6.6K answers and 1.5M answer views ·3y Related How can you explain the differences between partial pressure, vapour pressure, and saturated vapor pressure? All gases exert pressure. If you have a mixture of gasses, they all individually exert pressure. The total gases’ pressure, P is equal to the sum of each gas pressure, called partial pressure, in the mixture. P = p1+p2+p3… Vapour pressure: Substances exist in solid, liquid, and gaseous states. Water exists as ice (s), water (l) and vapor/steam (g). Under certain temperature and pressure conditions all three can exist together. Under normal conditions ice alone can have water in the form of vapor on its surface. Similarly liquid water can have water vapor (gas) too. The pressure exerted by the v Continue Reading All gases exert pressure. If you have a mixture of gasses, they all individually exert pressure. The total gases’ pressure, P is equal to the sum of each gas pressure, called partial pressure, in the mixture. P = p1+p2+p3… Vapour pressure: Substances exist in solid, liquid, and gaseous states. Water exists as ice (s), water (l) and vapor/steam (g). Under certain temperature and pressure conditions all three can exist together. Under normal conditions ice alone can have water in the form of vapor on its surface. Similarly liquid water can have water vapor (gas) too. The pressure exerted by the vapor is called vapor pressure. This applied to all substances, however, liquids have the highest vapor pressures. When the liquid has the maximum amount of vapor on its surface in equilibrium with its liquid the pressure exerted by it is called saturated vapor pressure at that temperature. Upvote · Guy Clentsmith Inorganic chemist ...M. Sc., Ph. D., PGCE · Author has 26.5K answers and 19.4M answer views ·2y Related How are saturated solutions and vapour pressure different? Well, a saturated solution saturated solution is a solution that contains the SAME amount of solute as would be in equilibrium with UNDISSOLVED solute … and this is a worthwhile definition to learn, since I have seen so many wrong answers that state that …. ..... the solution holds ALL the solute that it can.....An ERRONEOUS definition of a saturated solution..... the solution holds ALL the solute that it can.....⏟An ERRONEOUS definition of a saturated solution And thus … Dissolved solute solvent at given temperature⇌Undissolved soluteA CORRECT definition of a saturated solution Dissolved solute⇌solvent at given temperature Undissolved solute⏟A CORRECT definition of a saturated solution Saturation is thus an equilibr Continue Reading Well, a saturated solution saturated solution is a solution that contains the SAME amount of solute as would be in equilibrium with UNDISSOLVED solute … and this is a worthwhile definition to learn, since I have seen so many wrong answers that state that …. ..... the solution holds ALL the solute that it can.....An ERRONEOUS definition of a saturated solution..... the solution holds ALL the solute that it can.....⏟An ERRONEOUS definition of a saturated solution And thus … Dissolved solute solvent at given temperature⇌Undissolved soluteA CORRECT definition of a saturated solution Dissolved solute⇌solvent at given temperature Undissolved solute⏟A CORRECT definition of a saturated solution Saturation is thus an equilibrium scenario. And a temperature with respect to the solution is usually specified, given that a hot solvent can usually solvate more solute than a cold one… A supersaturated solution supersaturated solution is a solution that CONTAINS a greater quantity of solute than would be in equilibrium with undissolved solute.... And the vapour pressure is the partial pressure exerted by a condensed phase to the gaseous phase… At 298∙K 298•K, water exerts a vapour pressure of 23.8∙m m∙H g 23.8•m m•H g … What vapour pressure does water exert at 373∙K 373•K? Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 1 Related questions What are the differences between vapour pressure and saturated vapour pressure? What is the difference between Vapour pressure and saturation pressure? What is the difference between saturated temperature and saturated pressure? What is the saturation pressure of fluid? Is the equilibrium vapour pressure always equal to the saturated vapour pressure? What is saturated vapour pressure? What is the difference between partial pressure and saturated vapour pressure? What do you mean by saturated vapour pressure at a particular temperature? What is vapour pressure? What is the difference between vapour pressure and total vapour pressure? Why is the saturated vapour pressure of water 2.5 m of water in fluid mechanics? What is saturated pressure? What is the relationship between boiling point and vapour pressure? What is vapour pressure and cavitation? What is the difference between equilibrium vapor pressure and saturated vapor pressure? Related questions What are the differences between vapour pressure and saturated vapour pressure? What is the difference between Vapour pressure and saturation pressure? What is the difference between saturated temperature and saturated pressure? What is the saturation pressure of fluid? Is the equilibrium vapour pressure always equal to the saturated vapour pressure? What is saturated vapour pressure? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
187858
https://www.chemteam.info/Equilibrium/calc-Ksp-from-titration-data.html
ChemTeam: Calculate Ksp when Given Titration Data Calculate the K sp of a Saturated Solution When Given Titration Data Back to Equilibrium Menu Example #1: A solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl 2. Once equilibrated, some solid Ca(OH)2 remains undissolved. The solution is filtered and a 25.00 mL sample requires 22.50 mL of 0.0250 M HCl to neutralize it. Calculate the value for K sp of Ca(OH)2 from this data. Solution: 1) The chemical equation: Ca(OH)2⇌ Ca2++ 2OH¯ 2) The K sp expression: Ksp= [Ca2+] [OH¯]2 3) Use titration data to determine moles of OH¯ in the 25.0 mL sample (Remember, every one H+ neutralizes one OH¯.): molarity = moles ÷ volume (in liters) 0.0250 mol/L = x ÷ 0.02250 L x = 0.0005625 mol 4) Use moles of OH¯ and sample volume to determine [OH¯]: 0.0005625 mol / 0.02500 L = 0.0225 mol/L 5) Determine [Ca 2+]: it is exactly half the [OH¯]. This is because of the 1:2 molar ratio from the balanced equation. 6) Calculate the K sp for Ca(OH)2 Ksp= (0.01125) (0.0225)2= 5.70 x 10¯6 Example #2: 25.00 mL of saturated calcium hydroxide solution was titrated. It was found that it reacted completely with 8.13 mL of 0.102 mol L HCl. (a) Determine the solubility of Ca(OH)2 in grams per liter. (b) Determine the K sp of Ca(OH)2 Solution to (a): 1) Determine moles of HCl used: moles HCl = (0.102 mol / L) (0.00813 L) = 0.00082926 mol 2) Determine moles Ca(OH)2 titrated: 0.00082926 mol / 2 = 0.00041463 mol Remember, every one Ca(OH)2 titrated requires 2 H+ 3) Convert moles Ca(OH)2 to grams Ca(OH)2: 0.00041463 mol times 74.0918 g/mol = 0.030721 g This is grams per 25.0 mL 3) Convert to grams per liter: 0.030721 g / 0.0250 L = 1.23 g/L (to three sig fig) Solution to (b): 1) 0.00041463 mol of Ca(OH)2 in 25.0 mL means: [Ca2+] = 0.00041463 mol / 0.0250 L = 0.0165852 M [OH¯] = 0.0165852 M times 2 = 0.0331704 M 2) Calculate the K sp: Ksp= (0.0165852) (0.0331704)2= 1.82 x 10¯5 This value is in error. The book value is 5.02 x 10¯6. The main reason for the error is that the above is actual data gathered by a high school student doing this experiment for the first time. However, the calculational technique is correct. Example #3: A saturated solution of Pb(OH)2 is filtered and 25.00 mL of this solution is titrated with 0.000050 M HCl. The volume required to reach the equivalence point of this solution is 6.70 mL. Calculate the concentration of OH¯, Pb 2+ and the K sp of this satured solution. Solution: 1) Determine moles of HCl used: moles HCl = (0.000050 mol / L) (0.00670 L) = 0.000000335 mol 2) Determine moles Pb(OH)2 titrated: 0.000000335 mol / 2 = 0.0000001675 mol Remember, every one Pb(OH)2 titrated requires 2 H+ 3) 0.0000001675 mol of Pb(OH)2 in 25.0 mL means: [Pb2+] = 0.0000001675 mol / 0.0250 L = 0.0000067 M [OH¯] = 0.0000067 M times 2 = 0.0000134 M 4) Calculate the K sp: Ksp= (0.0000067) (0.0000134)2= 1.20 x 10¯15 Back to Equilibrium Menu
187859
https://en.wikipedia.org/wiki/Sunset_(disambiguation)
Jump to content Search Contents (Top) 1 Arts and entertainment 1.1 Film and television 1.2 Gaming 1.3 Literature 1.4 Music 1.4.1 Groups and labels 1.4.2 Albums 1.4.3 Songs 1.5 Visual arts 2 Places 2.1 Australia 2.2 Canada 2.3 United Kingdom 2.4 United States 3 Transportation 4 Other uses 5 See also Sunset (disambiguation) Cebuano Čeština Deutsch Español Français 한국어 Italiano עברית Nederlands Polski Português Русский Српски / srpski Srpskohrvatski / српскохрватски Svenska Türkçe Українська Tiếng Việt Volapük 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikisource Wikidata item Appearance From Wikipedia, the free encyclopedia Look up sunset in Wiktionary, the free dictionary. Sunset is the time when the Sun disappears below the horizon in the west. Sunset, Sunsets, The Sunset or Sun Set may also refer to: Arts and entertainment [edit] Film and television [edit] Sunset (1988 film), a 1988 western directed by Blake Edwards Sunset (1990 film), a 1990 Russian film about fictional gangster Benya Krik based in Isaac Babel's stories Sunset (2018 film), a 2018 Hungarian film directed by László Nemes "Sunset" (Breaking Bad), a season three episode of Breaking Bad "Sunset" (True Blood), an episode of the HBO TV series True Blood Sunset Productions, an American television syndication company Sunset Shimmer, a character in the My Little Pony: Equestria Girls franchise Gaming [edit] Sunset (video game), a 2015 art game by Belgium developer Tale of Tales Literature [edit] Sunset (magazine), an American lifestyle magazine Sunset (novel), a 2006 Warriors: The New Prophecy novel by Erin Hunter Sunset (play), a 1927 play novel by Isaac Babel "Sunsets" (poem), ("Soleils couchants"), a set of six poems, or a six-part poem, by Victor Hugo Music [edit] Groups and labels [edit] Shakin' Stevens and the Sunsets, a Welsh rock band Sunset Records, an American record label Albums [edit] Sunset (EP), by Pete Yorn, 2000 Sun Set, a box set by Klaatu, 2005 Sunset (Michel Teló album), 2013 Sunset (Superbus album), 2013 Sunset, half of the Coldplay album Everyday Life, 2019 Sunsets, by Chris Watson, 1994 Sunsets (DVD), by Powderfinger, 2004 Songs [edit] "Sunset" (bugle call), a military bugle call in the UK and British Commonwealth "Sunset (Bird of Prey)", by Fatboy Slim, 2000 "Sunset", by Air Supply from Air Supply, 1985 "Sunset", by Boyd Rice (NON) from Easy Listening for Iron Youth, 1991 "Sunset", by Caroline Polachek from Desire, I Want to Turn Into You, 2023 "Sunset", by Demi Lovato from Dancing with the Devil... the Art of Starting Over, 2021 "Sunset", by Farruko Visionary, 2015 "Sunset", by Jackie Lomax from Is This What You Want?, 1969 "Sunset", by J. Cole from Revenge of the Dreamers III, 2019 "Sunset", by Kate Bush from Aerial, 2005 "Sunset", by Kim Dong-han from D-Day, 2018 "Sunset", by Marques Houston from Mr. Houston, 2009 "Sunset", by Mike Oldfield from Light + Shade, 2005 "Sunset", by Roxy Music from Stranded, 1973 "Sunset", by Stevie Wonder from Tribute to Uncle Ray, 1962 "Sunsets" (song), by Powderfinger from Vulture Street, 2004 "A Sunset", by Pulp from More, 2025 Visual arts [edit] Sunset (Delacroix), a mid 19th century drawing by Eugene Delacroix Sunset (Friedrich), an 1830–1835 oil on canvas by Caspar David Friedrich Sunset (Rylov painting), a 1917 oil on canvas by Arkady Rylov Places [edit] Australia [edit] Sunset, Queensland, a suburb of Mount Isa Sunset, Victoria Canada [edit] Sunset, Vancouver, a neighbourhood United Kingdom [edit] Sunset, Herefordshire United States [edit] Sunset, Arizona Sunset, Washington County, Arkansas Sunset, Arkansas Sunset, California Sunset, Florida Sunset, Louisiana Sunset, Missouri Sunset, Maine Sunset, South Carolina Sunset, Montague County, Texas Sunset, Starr County, Texas Sunset, Utah Sunset, Texas (disambiguation) Sunset, West Virginia Sunset, Wisconsin Bankhead House (Jasper, Alabama), also known as Sunset Sunset 4A Region, Nevada Sunset Crater, Arizona, U.S. Sunset District, San Francisco, California, U.S. Sunset Drive, the major roadway through downtown South Miami, Florida, U.S. Transportation [edit] Sunset station (New York), in Brighton, New York, U.S. 17th Street/SMC station, in Los Angeles, U.S.; formerly named "Sunset" Other uses [edit] Sunset (apple), an apple cultivar Sunset (mango), a mango cultivar Sunset (color), a pale tint of orange Sunset (computing), the planned discontinuation of a server, service, software feature, etc. Sunset provision, a clause in a statute that terminates the law after a certain date unless further legislative action is taken See also [edit] Search for "sunset" , "sunsets", or "sunset's" on Wikipedia. Sunset Bay (disambiguation) Sunset Beach (disambiguation) Sunset Boulevard (disambiguation) Sunset Grill (disambiguation) Sunset High School (disambiguation) Sunset Park (disambiguation) Sunset station (disambiguation) Sunset town, all-white municipalities that prohibit black people from being in the town after sunset Into the Sunset (disambiguation) Sunset and Sunrise (disambiguation) Sunrise (disambiguation) All pages with titles beginning with Sunset All pages with titles containing Sunset Topics referred to by the same term This disambiguation page lists articles associated with the title Sunset.If an internal link led you here, you may wish to change the link to point directly to the intended article. Retrieved from " Categories: Disambiguation pages Place name disambiguation pages Hidden categories: Short description is different from Wikidata All article disambiguation pages All disambiguation pages Sunset (disambiguation) Add topic
187860
https://www.pbslearningmedia.org/resource/our20-math-755/representing-subtraction/
Representing Subtraction | PBS LearningMedia FOR TEACHERS Is Alabama Public Television your local station?Yes No, change Sign in Sign up FOR TEACHERS Subjects Grades Student site Illustrative Mathematics 6-8 Math and Algebra 1 Share to Google ClassroomShare link with studentsBuild a lessonSocial share Favorite Representing Subtraction Video Grades: 6-8 Collection: Illustrative Mathematics 6-8 Math and Algebra 1 Open Up Resources: 6-8 Math To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video Video Player is loading. Play Video Play Seek back 10 seconds Mute Current Time 0:00 / Duration 0:00 Loaded: 0% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.75x 1.5x 1.25x 1x, selected 0.75x 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off English Captions, selected Audio Track Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Download About Standards In this video lesson, students subtract signed numbers on a number line by relating it to an addition equation with a missing addend. Students see that a subtraction equation like -8 – 3 = ? can be thought of as the related addition equation 3 + ? = 8. After repeatedly calculating differences this way (MP8), students recognize that the answer to each subtraction problem is the same number they would get by adding the opposite of the number. For example, -8 – 3 = ? can also be thought of as -8 + -3 = ? Grade 7, Episode 10: Unit 5, Lesson 5 | Illustrative Math Permitted use Stream, Download and Share Accessibility Caption Credits Alabama State Department of Education Want to see state standards for this resource? Sign In Nationwide Common Core State Standards (7) Grades 6-8 CCSS.Math.Content.6.NS.B.3 See anchor statement Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.All “CCSS.Math.Content.6.NS.B.3” resources See all (7) Grades 6-8 standards College and Career Readiness Standards for Adult Education (7) Grades 6-8 NS.A.2 See anchor statement Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.All “NS.A.2” resources See all (7) Grades 6-8 standards Next: Standards Support Materials for Use with Students ACTIVITY Unit 5, Lesson 5: Representing SubtractionSpanish Student Activity: Unit 5, Lesson 5: Representing Subtraction HANDOUTS Unit 5, Lesson 5: Practice ProblemsSpanish Handout: Unit 5, Lesson 5: Practice Problems You May Also Like 26:46 ##### Changing Elevation Video 26:46 ##### Subtracting Rational Numbers Video 26:46 ##### Tape Diagrams and Equations Video 26:46 ##### Linear Functions Video Explore related topics MathematicsK-8 MathematicsThe Number SystemOperations for Fractions With Rational Numbers More from the Illustrative Mathematics 6-8 Math and Algebra 1 Collection Collection 96 26:46 ##### Subtracting Rational Numbers Video 26:46 ##### Position, Speed, and Direction Video 26:46 ##### Multiplying Rational Numbers Video 26:46 ##### Dividing Rational Numbers Video 26:46 ##### Expressions with Rational Numbers and Solving Problems with Rational Numbers Video See the Collection Made Possible Through producer contributor contributor funder funder Unlock the Power of PBS LearningMedia Create a free account to gain full access to the website Save & Organize Resources See State Standards Manage Classes & Assignments Sync with Google Classroom Create Lessons Customized Dashboard Get More Features Free Student site In partnership with Connect With Us Sign up for our weekly PreK-12 newsletter for the latest classroom resources, news, and more. Sign up! Learn More About Teachers’ Lounge Blog Educator Recognition Contact Us Newsletters Help © 2025 PBS & WGBH Educational Foundation. All rights reserved. Privacy Policy|Terms of Use Top
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https://www.youtube.com/watch?v=aT8L6EvhQgo
What Are Magic Squares and How Can I Use Them in Maths? Twinkl Teaches KS2 26900 subscribers 21 likes Description 4742 views Posted: 3 Oct 2022 In this video, Twinkl Teacher Lloyd showcases magic squares which are brilliant problem-solving activities. Magic squares are square grids, often in a 3x3 pattern that are filled with numbers, in such a way that each row, each column, and the two diagonals add up to the same number. Children try to work out the missing numbers. Check them out here: ➡️ 0:00 Are you looking to engage children with fun math activities? 0:10 What are magic squares? How do they work? 0:30 Magic Square to 100 0:50 Where can I find more magic squares? KS2 #magicsquare #mathsactivity #primaryschool #problemsolving #logicalreasoning #mathsforkids #mathisfun Transcript: Are you looking to engage children with fun math activities? are you looking to bring a bit of magic math to learning then look no further and check out these fantastic magic squares abracadabra What are magic squares? How do they work? what are magic squares these grids are filled with numbers in such a way that each row each column and two diagonals add up to the same number by giving some of the numbers they should be able to complete the first row and then complete the rest check out this magic square addition to 100 worksheet this magic square Magic Square to 100 worksheet is great for reinforcing addition by having some alternative solutions to the same outcome it enhances not just addition but also problem solving skills and develops logical reasoning skills see if your children can work out the missing numbers to complete the equations on our magic squares worksheet it's easily printable or editable online Where can I find more magic squares? if you want to save on paper it's a perfect resource for teaching a key stage 2 maths class there are plenty of magic square activity puzzles on the twink website so make sure you search for more we'd love to know how you got on with these resources and we're always happy to hear your ideas and suggestions too feel free to get in touch or drop a review via the star ratings and review box below thanks for watching and bye for now foreign [Music]
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https://brainly.com/question/40750306
[FREE] If you fold a piece of paper 50 times, how many layers will there be? - brainly.com 6 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +63,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +32,2k Ace exams faster, with practice that adapts to you Practice Worksheets +6k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified If you fold a piece of paper 50 times, how many layers will there be? 1 See answer Explain with Learning Companion NEW Asked by madmonee2447 • 10/25/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 11653630 people 11M 0.0 0 Upload your school material for a more relevant answer If you fold a piece of paper 50 times, there will be approximately 1.13 quadrillion layers of paper. Explanation When you fold a piece of paper in half, it doubles the number of layers. So, if you fold a piece of paper in half 50 times, the number of layers will be 250. Using exponential notation, this can be written as 2 × 2 × 2 × ... (50 times) = 250. Calculating this, we find that there will be approximately 1.12589991 × 1015 or 1.13 quadrillion layers of paper. Theoretically, if you fold a piece of paper 50 times, following an exponential growth formula, you would end up with 2^50 layers. However, due to physical limitations, paper can't actually be folded more than 7 to 8 times. If you fold a piece of paper, each fold doubles the number of layers. So, if you're folding it 50 times, the math involved is actually an application of exponential growth. For instance, if you make one fold, you have 2 layers. If you then make another fold, that number doubles to 4 layers. So with each new fold, the total number of layers is 2 raised to the power of the number of folds you've made. Therefore, if you fold a piece of paper 50 times, theoretically, you would end up with 2^50 layers, assuming you could physically accomplish it. However, due to the physical limitation of paper (i.e., it cannot actually be folded more than about 7 to 8 times), this is only a theoretical explanation. Learn more about Exponential Growth here: brainly.com/question/12490064 SPJ11 Answered by monur5195 •40.9K answers•11.7M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 11653630 people 11M 0.0 0 Physics - Paul Peter Urone, Roger Hinrichs Microbiology - Nina Parker, Mark Schneegurt, Anh-Hue Thi Tu, Philip Lister, Brian M. Forster Physics for AP® Courses 2e - Kenneth Podolak, Henry Smith Upload your school material for a more relevant answer Folding a piece of paper 50 times theoretically results in about 1.13 quadrillion layers due to exponential growth principles. However, practically, a piece of paper cannot be folded more than 7 to 8 times due to physical limitations. Thus, while the formula suggests vast layers, it's not achievable in reality. Explanation If you fold a piece of paper 50 times, the theoretical number of layers would be calculated using the formula for exponential growth. Each time you fold the paper in half, the number of layers doubles. This can be expressed mathematically as: Layers=2 n where n is the number of folds. In this case, n=50, so: Layers=2 50 Calculating this gives: 2 50≈1.12589991×1 0 15 or approximately 1.13 quadrillion layers. It’s important to note that this result is purely theoretical. In reality, due to physical constraints, it is impossible to fold a standard piece of paper more than about 7 to 8 times. When you try to fold the paper beyond this point, it simply becomes too thick and difficult to manipulate. Examples & Evidence An example to consider is that after the first fold, you have 2 layers; after the second fold, you have 4 layers; this pattern continues doubling with each fold. So the number of layers after the 3rd fold would be 8, then 16 after the 4th fold, and so on, illustrating the exponential growth clearly. The concept of exponential growth and its mathematical representation 2 n can be found in basic algebra textbooks and is commonly discussed in mathematics courses around the world. Thanks 0 0.0 (0 votes) Advertisement madmonee2447 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Find the area bounded by the lines y=2 3​x−4 and y=−2 x+7 and the x axis. Find the vertex for the parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function: y=0.05 x 2+0.5 x−130. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. 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https://medium.com/@michaelclion/understanding-p-np-np-complete-and-np-hard-problems-f09a2b09cbf1
Sitemap Open in app Sign in Sign in Understanding P, NP, NP-complete, and NP-hard problems An intuitive overview with examples Chuyao Wang 8 min readMar 6, 2025 Preface The thing is, I have been studying bioinformatics for a while, but I never understood what P, NP, NP-complete, and NP-hard problems are — I just stow them into a corner in my mind called “computational science concepts too hard-core to understand”, and only invoke a vague picture of them when necessary, but this approach is becoming insufficient, so I decided to get a real, good understanding of them. Here I will share what I have learned and hopefully it will help others eager to learn about the subject. We will go through them in the order of P, NP, NP-complete, and NP-hard problems, and finish with a Venn diagram of them to solidify the concepts. We will also discuss what does it mean to prove a problem is NP-hard, why prove it, and why is P=NP one of the biggest open questions in computer science. Let’s dive right in! A polynomial time algorithm The concepts of P, NP, NP-complete, and NP-hard problems all have to do with an algorithm being polynomial time. It means that for an input of length n, the algorithm computes the solution in time O(n^k) for some constant k. For example, we are interested in finding the maximum number in a list of numbers. One approach would be: Start by assuming the first number is the maximum. Go through each number in the list one by one. Compare the current number with your current maximum. Update the maximum if the current number is larger. Finish when you’ve checked all the numbers. For a sequence of 5 numbers, we make at most 4 comparisons. Since the algorithm run time is bounded by n^k, where k=1, we say that this algorithm runs in polynomial time. P problems P problems are exactly those problems that can be solved in deterministic polynomial time. The P stands for polynomial. Here determinstic means that the algorithm follows fixed and well defined steps and always gives the same output given the same input, like the maximum finding algorithm we just introduced. In contrast, a non-deterministic algorithm involves some randomness and guessing. We will get back to them later. Also, the solution to a P problem can be verified in deterministic polynomial time. In practice, a P problem is considered efficiently solvable or “tractable”. NP problems There are 2 equivalent definitions of NP problems. We will go through each and them together. Definition 1: NP problems can be solved in non-deterministic polynomial time. To understand what non-deterministic means, consider the (perhaps not very computationally sound) example below: You need to pass 3 sets of doors (red, blue, and yellow), and you have 3 keys. Each key can open only 1 door in each set of doors. There is no information on the doors to indicate which door is the correct door. In the luckiest case, you can guess all 3 correct doors in a row, and you will pass the doors in 3 steps. However, in the unluckiest case, you will have to try all the keys in all the doors. Here, for a given sets of doors, the steps it takes to pass through the doors depend on your luck, and this makes it a non-deterministic process. Back to the NP problems. It means that if your algorithm can magically guess all the right steps to take at each step, it is solvable in polynomial time. However, in reality computation is deterministic, meaning non-deterministic algorithms are only theoretical constructs, like an ideal line with 0 thickness. A deterministic algorithm cannot always make perfect guesses, and all the possibilities need to be checked (efficiently or inefficiently). This makes the run time slower than polynomial time. Definition 2: the solutions to NP problems can be verified in polynomial time It means that if a solution is given for a NP problem, it can be verified in deterministic polynomial time. Note that since the solutions to P problems can also be verified in polynomial time, by this definition, P problems are a subset of NP problems. Finally, note that the 2 definitions are equivalent, and you cannot have one without the other: If a problem can be solved by a nondeterministic polynomial-time algorithm, then its solution can be verified in polynomial time. This is by the very definition of NP. The nondeterministic algorithm “guesses” a solution, and if it exists, it can verify it within polynomial time. If a solution can be verified in polynomial time, then there is a nondeterministic polynomial-time algorithm to find the solution. The nondeterministic algorithm can nondeterministically “guess” the certificate and then check it deterministically in polynomial time. Deterministic vs. non-deterministic polynomial time Usually, when people talk about a “polynomial time algorithm” without specifying whether it is determinsitic or not, as in The nondeterministic algorithm “guesses” a solution, and if it exists, it can verify it within polynomial time. It refers to deterministic polynomial time only. Different difficulty levels of NP problems Although all NP problems share the trait that their solutions can be verified in polynomial time, the difficulty of finding their solutions vary from problem to problem. P problems are the easiest NP problems as solving them takes, by definition, deterministic polynomial time. Get Chuyao Wang’s stories in your inbox Join Medium for free to get updates from this writer. NP-complete problems are the most difficult NP problems, explained next. NP-complete problems As mentioned above, NP-complete problems are the most difficult NP problems. However, even if some NP problems are more difficult than others, why make them into a separate class? The reason is that this specific group of difficult problems interesects with NP-hard problems. NP-hardness is related to the tractability of a problem and to proving P=NP, one of the biggest open questions in computer science. To understand what NP-hard problems are, we must first introduce a concept called “problem reduction”. Problem reduction Three key points about reduction: A reduction function is a function that takes an instance of one problem (for example, A in NP) and transform it into an instance of another problem (for example, B in NP-hard). In the conversion, the answer to the problem is conserved such that B has the same answer as A. An algorithm that solves B can be composed with the reduction function to also solve A. The term reduction is a little misleading here because the connotation is that the reduced problem (B) is easier than the original problem (A). In contrast, B is at least as difficult as A. The purpose of reduction is not to make the problem easier but to connect the difficulty of the two problems such that an efficient solution for B maps to an efficient solution for A. For this reason, the reduction function must run in polynomial time. If the reduction itself adds significant computational overhead, say it runs exponential time, then an efficient solution for B would not mean an efficient solution for A, defeating the purpose of relating the complexities of the two problems. NP-hard problems A systematic way (a polynomial time reduction) exists to reduce an NP problem into an NP-hard problem. In fact, a problem is considered NP-hard if every problem in NP can be transformed (reduced) to it in polynomial time. This is the definition of NP-hardness. It means that an NP-hard problem is at least as difficult as the hardest problems in NP, which are the NP-complete problems. To clarify, even though some NP-hard problems (for example, some optimization problems) are not in NP (their solutions cannot be verified in polynomial time), they can still be reduced to from NP problems by definition. Still, a question remains: why do we want to reduce NP problems to NP-hard problems, if it does not make the problem easier to solve? As mentioned before, when we reduce NP problems to NP-hard problems, having an efficient solution for the NP-hard problem means having an efficient solution for the NP problem. This has critical implication in proving P=NP. P=NP? We have covered how P is a subset of NP: they both have polynomial time to verify a solution, but differs in the time it takes to find a solution. P problems take deterministic polynomial time, while NP problems take non-deterministic polynomial time. The statement that P=NP means that all NP problems also take deterministic polynomial time to find a solution. This has tremedous implications if proven. For a rough understanding, consider if the time it takes to verify your password was on the same scale with the time it takes to crack your password, your accounts would be in great jeopardy! To prove P=NP, we take the following route: if an NP-hard problem, which is at least as hard as the hardest problems in NP, can be solved in polynomial time, then by the definition of reduction, all NP problems can also be solved in polynomial time. Then it effectively means NP and P are the same, proving P=NP. Currently, no known ploynomial time algorithm exists for NP-hard problems. Hence, P=NP is not proven. Proving a problem is NP-hard Besides the elusive proof to P=NP, there are practical implications to reducing an NP problem to an NP-hard one. A often heard concept is “proving a problem is NP-hard”. It means to construct a conversion function, and show that it converts a known NP problem to the problem being proved. Proving NP-hardness of a problem proves the problem is at least as hard as the hardest problems in NP. Since P=NP is not proven yet, it means P≠NP, so no polynomial time solution exists for NP-hard problems. Hence, NP-hard problems are considered intractable for large inputs, in contrast to P problems, which are tractable. In summary, the proof of NP-hardness have several practical implications: understanding computational limits: a NP-hard problem likely does not have a polynomial time solution; algorithm design and practical approaches: a more practical approach to design an algorithm for an NP-hard problem would be to seek approximate solutions, use heuristics, or find special-case solutions that work well enough for practical applications; theoretical importance: contribution to our understanding of the boundaries between tractable and intractable problems. Summary: a Venn diagram We use a Venn diagram to summarize the key points mentioned in this article: P problems can be solved and their solution verified in deterministic polynomial time. NP problems can be solved in non-deterministic polynomial time and their solutions verified in polynomial time. Hence, P is a subset of NP. All NP problems can be converted to NP-hard problems, and NP-hard problems are at least as difficult as NP-complete problems. Not all NP-hard problems are NP. Those problems that are both NP and NP-hard are considered NP-complete. Solving a NP-hard problem in deterministic polynomial time = solving a NP-complete problem in deterministic polynomial time = solving a NP problem in deterministic polynomial time = proving P=NP (not proven yet!) Algorithms Computer Science Bioinformatics Data Science Machine Learning ## Written by Chuyao Wang 0 followers ·2 following Learning about AI/ML and bioinformatics No responses yet Write a response What are your thoughts? Recommended from Medium Mohith ## Linear Regression Explained: From Theory to Real-World Implementation Understanding the math, assumptions, and practical steps to predict continuous outcomes with confidence Aug 9 168 2 In Philosophy Today by Pedro Barbalho ## Have We Been Tricked by Gödel? According to Kreisel, yes. Aug 23 872 24 Maxwell's Demon ## Towards Hamiltonian Mechanics 002 — The Legendre Transformation An important step towards Hamiltonian mechanics Mar 5 10 1 In Science Spectrum by Cole Frederick ## What, Exactly, is Bayes’ Theorem? How to Make Better Decisions Aug 24 978 12 AncientEncoder ## Optical Prime Pair Solver for Goldbach’s Conjecture Using Spectral WaveGenesis Original paper: May 21 87 In Women in Technology by Sara A. Metwalli ## A Quantum Computing Dictionary: Understanding the Basics A dictionary of the basic quantum concepts Mar 17 256 4 See more recommendations Text to speech
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https://bayanbox.ir/view/4018456844032510992/104-Number-Theory-Problems.pdf
داﻧﻠﻮد از ﺳﺎﻳﺖ رﻳﺎﺿﻲ ﺳﺮا www.riazisara.ir About the Authors Titu Andreescu received his Ph.D. from the West University of Timisoara, Ro-mania. The topic of his dissertation was “Research on Diophantine Analysis and Applications.” Professor Andreescu currently teaches at The University of Texas at Dallas. He is past chairman of the USA Mathematical Olympiad, served as di-rector of the MAA American Mathematics Competitions (1998–2003), coach of the USA International Mathematical Olympiad Team (IMO) for 10 years (1993– 2002), director of the Mathematical Olympiad Summer Program (1995–2002), and leader of the USA IMO Team (1995–2002). In 2002 Titu was elected member of the IMO Advisory Board, the governing body of the world’s most prestigious mathematics competition. Titu co-founded in 2006 and continues as director of the AwesomeMath Summer Program (AMSP). He received the Edyth May Sliffe Award for Distinguished High School Mathematics Teaching from the MAA in 1994 and a “Certificate of Appreciation” from the president of the MAA in 1995 for his outstanding service as coach of the Mathematical Olympiad Summer Pro-gram in preparing the US team for its perfect performance in Hong Kong at the 1994 IMO. Titu’s contributions to numerous textbooks and problem books are recognized worldwide. Dorin Andrica received his Ph.D in 1992 from “Babes ¸-Bolyai” University in Cluj-Napoca, Romania; his thesis treated critical points and applications to the geometry of differentiable submanifolds. Professor Andrica has been chairman of the Department of Geometry at “Babes ¸-Bolyai” since 1995. He has written and contributed to numerous mathematics textbooks, problem books, articles and sci-entific papers at various levels. He is an invited lecturer at university conferences around the world: Austria, Bulgaria, Czech Republic, Egypt, France, Germany, Greece, Italy, the Netherlands, Portugal, Serbia, Turkey, and the USA. Dorin is a member of the Romanian Committee for the Mathematics Olympiad and is a member on the editorial boards of several international journals. Also, he is well known for his conjecture about consecutive primes called “Andrica’s Conjecture.” He has been a regular faculty member at the Canada–USA Mathcamps between 2001–2005 and at the AwesomeMath Summer Program (AMSP) since 2006. Zuming Feng received his Ph.D. from Johns Hopkins University with emphasis on Algebraic Number Theory and Elliptic Curves. He teaches at Phillips Exeter Academy. Zuming also served as a coach of the USA IMO team (1997–2006), was the deputy leader of the USA IMO Team (2000–2002), and an assistant director of the USA Mathematical Olympiad Summer Program (1999–2002). He has been a member of the USA Mathematical Olympiad Committee since 1999, and has been the leader of the USA IMO team and the academic director of the USA Mathe-matical Olympiad Summer Program since 2003. Zuming is also co-founder and academic director of the AwesomeMath Summer Program (AMSP) since 2006. He received the Edyth May Sliffe Award for Distinguished High School Mathe-matics Teaching from the MAA in 1996 and 2002. www.riazisara.ir Titu Andreescu Dorin Andrica Zuming Feng 104 Number Theory Problems From the Training of the USA IMO Team Birkh¨ auser Boston • Basel • Berlin داﻧﻠﻮد از ﺳﺎﻳﺖ رﻳﺎﺿﻲ ﺳﺮا www.riazisara.ir Titu Andreescu The University of Texas at Dallas Department of Science/Mathematics Education Richardson, TX 75083 U.S.A. titu.andreescu@utdallas.edu Dorin Andrica “Babes ¸-Bolyai” University Faculty of Mathematics 3400 Cluj-Napoca Romania dorinandrica@yahoo.com Zuming Feng Phillips Exeter Academy Department of Mathematics Exeter, NH 03833 U.S.A. zfeng@exeter.edu Cover design by Mary Burgess. Mathematics Subject Classification (2000): 00A05, 00A07, 11-00, 11-XX, 11Axx, 11Bxx, 11D04 Library of Congress Control Number: 2006935812 ISBN-10: 0-8176-4527-6 e-ISBN-10: 0-8176-4561-6 ISBN-13: 978-0-8176-4527-4 e-ISBN-13: 978-0-8176-4561-8 Printed on acid-free paper. c ⃝2007 Birkh¨ auser Boston All rights reserved. This work may not be translated or copied in whole or in part without the writ-ten permission of the publisher (Birkh¨ auser Boston, c/o Springer Science+Business Media LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter de-veloped is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 9 8 7 6 5 4 3 2 1 www.birkhauser.com (EB) www.riazisara.ir 104 Number Theory Problems Titu Andreescu, Dorin Andrica, Zuming Feng October 25, 2006 داﻧﻠﻮد از ﺳﺎﻳﺖ رﻳﺎﺿﻲ ﺳﺮا www.riazisara.ir Contents Preface vii Acknowledgments ix Abbreviations and Notation xi 1 Foundations of Number Theory 1 Divisibility 1 Division Algorithm 4 Primes 5 The Fundamental Theorem of Arithmetic 7 G.C.D. 11 Euclidean Algorithm 12 B´ ezout’s Identity 13 L.C.M. 16 The Number of Divisors 17 The Sum of Divisors 18 Modular Arithmetics 19 Residue Classes 24 Fermat’s Little Theorem and Euler’s Theorem 27 Euler’s Totient Function 33 Multiplicative Function 36 Linear Diophantine Equations 38 Numerical Systems 40 Divisibility Criteria in the Decimal System 46 Floor Function 52 Legendre’s Function 65 Fermat Numbers 70 Mersenne Numbers 71 Perfect Numbers 72 www.riazisara.ir vi Contents 2 Introductory Problems 75 3 Advanced Problems 83 4 Solutions to Introductory Problems 91 5 Solutions to Advanced Problems 131 Glossary 189 Further Reading 197 Index 203 www.riazisara.ir Preface This book contains 104 of the best problems used in the training and testing of the U.S. International Mathematical Olympiad (IMO) team. It is not a collection of very difficult, and impenetrable questions. Rather, the book gradually builds students’ number-theoretic skills and techniques. The first chapter provides a comprehensive introduction to number theory and its mathematical structures. This chapter can serve as a textbook for a short course in number theory. This work aims to broaden students’ view of mathematics and better prepare them for possible participation in various mathematical competitions. It provides in-depth enrichment in important areas of number theory by reorganizing and enhancing students’ problem-solving tactics and strategies. The book further stimulates stu-dents’ interest for the future study of mathematics. In the United States of America, the selection process leading to participation in the International Mathematical Olympiad (IMO) consists of a series of national contests called the American Mathematics Contest 10 (AMC 10), the American Mathematics Contest 12 (AMC 12), the American Invitational Mathematics Ex-amination (AIME), and the United States of America Mathematical Olympiad (USAMO). Participation in the AIME and the USAMO is by invitation only, based on performance in the preceding exams of the sequence. The Mathematical Olympiad Summer Program (MOSP) is a four-week intensive training program for approximately fifty very promising students who have risen to the top in the American Mathematics Competitions. The six students representing the United States of America in the IMO are selected on the basis of their USAMO scores and further testing that takes place during MOSP. Throughout MOSP, full days of classes and extensive problem sets give students thorough preparation in several important areas of mathematics. These topics include combinatorial arguments and identities, generating functions, graph theory, recursive relations, sums and products, probability, number theory, polynomials, functional equations, complex numbers in geometry, algorithmic proofs, combinatorial and advanced geometry, functional equations, and classical inequalities. Olympiad-style exams consist of several challenging essay problems. Correct solutions often require deep analysis and careful argument. Olympiad questions www.riazisara.ir viii Preface can seem impenetrable to the novice, yet most can be solved with elementary high school mathematics techniques, when cleverly applied. Here is some advice for students who attempt the problems that follow. • Take your time! Very few contestants can solve all the given problems. • Try to make connections between problems. An important theme of this work is that all important techniques and ideas featured in the book appear more than once! • Olympiad problems don’t “crack” immediately. Be patient. Try different approaches. Experiment with simple cases. In some cases, working back-ward from the desired result is helpful. • Even if you can solve a problem, do read the solutions. They may con-tain some ideas that did not occur in your solutions, and they may discuss strategic and tactical approaches that can be used elsewhere. The solutions are also models of elegant presentation that you should emulate, but they often obscure the tortuous process of investigation, false starts, inspiration, and attention to detail that led to them. When you read the solutions, try to reconstruct the thinking that went into them. Ask yourself, “What were the key ideas? How can I apply these ideas further?” • Go back to the original problem later, and see whether you can solve it in a different way. Many of the problems have multiple solutions, but not all are outlined here. • Meaningful problem solving takes practice. Don’t get discouraged if you have trouble at first. For additional practice, use the books on the reading list. Titu Andreescu Dorin Andrica Zuming Feng October 2006 www.riazisara.ir Acknowledgments Thanks to Sara Campbell, Yingyu (Dan) Gao, Sherry Gong, Koene Hon, Ryan Ko, Kevin Medzelewski, Garry Ri, and Kijun (Larry) Seo. They were the members of Zuming’s number theory class at Phillips Exeter Academy. They worked on the first draft of the book. They helped proofread the original manuscript, raised critical questions, and provided acute mathematical ideas. Their contribution im-proved the flavor and the structure of this book. We thank Gabriel Dospinescu (Dospi) for many remarks and corrections to the first draft of the book. Some ma-terials are adapted from , , , and . We also thank those students who helped Titu and Zuming edit those books. Many problems are either inspired by or adapted from mathematical contests in different countries and from the following journals: • The American Mathematical Monthly, United States of America • Crux, Canada • High School Mathematics, China • Mathematics Magazine, United States of America • Revista Matematicˇ a Timis ¸oara, Romania We did our best to cite all the original sources of the problems in the solution section. We express our deepest appreciation to the original proposers of the problems. www.riazisara.ir Abbreviations and Notation Abbreviations AHSME American High School Mathematics Examination AIME American Invitational Mathematics Examination AMC10 American Mathematics Contest 10 AMC12 American Mathematics Contest 12, which replaces AHSME APMC Austrian–Polish Mathematics Competition ARML American Regional Mathematics League Balkan Balkan Mathematical Olympiad Baltic Baltic Way Mathematical Team Contest HMMT Harvard–MIT Math Tournament IMO International Mathematical Olympiad USAMO United States of America Mathematical Olympiad MOSP Mathematical Olympiad Summer Program Putnam The William Lowell Putnam Mathematical Competition St. Petersburg St. Petersburg (Leningrad) Mathematical Olympiad Notation for Numerical Sets and Fields Z the set of integers Zn the set of integers modulo n N the set of positive integers N0 the set of nonnegative integers Q the set of rational numbers Q+ the set of positive rational numbers Q0 the set of nonnegative rational numbers Qn the set of n-tuples of rational numbers R the set of real numbers R+ the set of positive real numbers R0 the set of nonnegative real numbers Rn the set of n-tuples of real numbers C the set of complex numbers xn the coefficient of the term xn in the polynomial p(x) www.riazisara.ir xii Abbreviations and Notation Notation for Sets, Logic, and Number Theory |A| the number of elements in the set A A ⊂B A is a proper subset of B A ⊆B A is a subset of B A \ B A without B (set difference) A ∩B the intersection of sets A and B A ∪B the union of sets A and B a ∈A the element a belongs to the set A n | m n divides m gcd(m, n) the greatest common divisor of m, n lcm(m, n) the least common multiple of m, n π(n) the number of primes ≤n τ(n) number of divisors of n σ(n) sum of positive divisors of n a ≡b (mod m) a and b are congruent modulo m ϕ Euler’s totient function ordm(a) order of a modulo m µ M¨ obius function akak−1 . . . a0(b) base-b representation S(n) the sum of digits of n ( f1, f2, . . . , fm) factorial base expansion ⌊x⌋ floor of x ⌈x⌉ celling of x {x} fractional part of x ep Legendre’s function pk∥n pk fully divides n fn Fermat number Mn Mersenne number www.riazisara.ir 1 Foundations of Number Theory Divisibility Back in elementary school, we learned four fundamental operations on numbers (integers), namely, addition (+), subtraction (−), multiplication (× or ·), and di-vision (÷ or / or c). For any two integers a and b, their sum a + b, differences a −b and b −a, and product ab are all integers, while their quotients a ÷ b (or a/b or a b) and b ÷ a are not necessarily integers. For an integer m and a nonzero integer n, we say that m is divisible by n or n divides m if there is an integer k such that m = kn; that is, m n is an integer. We denote this by n | m. If m is divisible by n, then m is called a multiple of n; and n is called a divisor (or factor) of m. Because 0 = 0 · n, it follows that n | 0 for all integers n. For a fixed integer n, the multiples of n are 0, ±n, ±, 2n, . . . . Hence it is not difficult to see that there is a multiple of n among every n consecutive integers. If m is not divisible by n, then we write n ∤m. (Note that 0 ∤m for all nonzero integers m, since m ̸= 0 = k · 0 for all integers k.) Proposition 1.1. Let x, y, and z be integers. We have the following basic prop-erties: (a) x | x (reflexivity property); (b) If x | y and y | z, then x | z (transitivity property); (c) If x | y and y ̸= 0, then |x| ≤|y|; (d) If x | y and x | z, then x | αy + βz for any integers α and β; (e) If x | y and x | y ± z, then x | z; (f) If x | y and y | x, then |x| = |y|; www.riazisara.ir 2 104 Number Theory Problems (g) If x | y and y ̸= 0, then y x | y; (h) for z ̸= 0, x | y if and only if xz | yz. The proofs of the above properties are rather straightforward from the defini-tion. We present these proofs only to give the reader some relevant examples of writing proofs. Proof: For (a), we note that x = 1 · x. In (b) to (h), the condition x | y is given; that is, y = kx for some integer k. For (b), we have y | z; that is, z = k1y for some integer k1. Then z = (kk1)x, or x | z. For (c), we note that if y ̸= 0, then |k| ≥1, and so |y| = |k| · |x| ≥|x|. For (d), we further assume that z = k2x. Then αy + βz = (kα + k2β)x. For (e), we obtain y ± z = k3x, or ±z = k3x −y = (k3 −k)x. It follows that z = ±(k −k3)x. For (f), because x | y and y | x, it follows that x ̸= 0 and y ̸= 0. By (c), we have |y| ≥|x| and |x| ≥|y|. Hence |x| = |y|. For (g), y x = k ̸= 0 is an integer. Since y = x · k, k | y. For (h), since z ̸= 0, x ̸= 0 if and only if xz ̸= 0. Note that y = kx if and only if yz = kxz. □ The property (g) is simple but rather helpful. For a nonzero integer n, there is an even number of positive divisors of n unless n is a perfect square; that is, n = m2 for some integer m. (If an integer is not divisible by any perfect square, then it is called square free. If n = m3 for some integer m, then n is called a perfect cube. In general, if n = ms for integers m and s with s ≥2, then n is called a perfect power.) This is because all the divisors of y appear in pairs, namely, x and y x (observe that x ̸= y x if y is not a perfect square). Here is a classic brain teaser: Example 1.1. Twenty bored students take turns walking down a hall that con-tains a row of closed lockers, numbered 1 to 20. The first student opens all the lockers; the second student closes all the lockers numbered 2, 4, 6, 8, 10, 12, 14, 16, 18, 20; the third student operates on the lockers numbered 3, 6, 9, 12, 15, 18: if a locker was closed, he opens it, and if a locker was open, he closes it; and so on. For the ith student, he works on the lockers numbered by multiples of i: if a locker was closed, he opens it, and if a locker was open, he closes it. What is the number of the lockers that remain open after all the students finish their walks? Solution: Note that the ith locker will be operated by student j if and only if j | i. By property (g), this can happen if and only if the locker will also be operated by student i j . Thus, only the lockers numbered 1 = 12, 4 = 22, 9 = 32, www.riazisara.ir 1. Foundations of Number Theory 3 and 16 = 42 will be operated on an odd number of times, and these are the lockers that will be left open after all the operations. Hence the answer is 4. □ The set of integers, denoted by Z, can be partitioned into two subsets, the set of odd integers and the set of even integers: {±1, ±3, ±5, . . . } and {0, ±2, ±4, . . . }, respectively. Although the concepts of odd and even integers appear straightfor-ward, they come handy in tackling various number-theoretic problems. Here are some basic ideas: (1) an odd number is of the form 2k + 1, for some integer k; (2) an even number is of the form 2m, for some integer m; (3) the sum of two odd numbers is an even number; (4) the sum of two even numbers is an even number; (5) the sum of an odd and even number is an odd number; (6) the product of two odd numbers is an odd number; (7) a product of integers is even if and only if at least one of its factors is even. Example 1.2. Let n be an integer greater than 1. Prove that (a) 2n is the sum of two odd consecutive integers; (b) 3n is the sum of three consecutive integers. Proof: For (a), the relation 2n = (2k −1) + (2k + 1) implies k = 2n−2 and we obtain 2n = (2n−1 −1) + (2n−1 + 1). For (b), the relation 3n = (s −1) + s + (s + 1) implies s = 3n−1 and we obtain the representation 3n = (3n−1 −1) + 3n−1 + (3n−1 + 1). □ Example 1.3. Let k be an even number. Is it possible to write 1 as the sum of the reciprocals of k odd integers? Solution: The answer is negative. We approach indirectly. Assume that 1 = 1 n1 + · · · + 1 nk for some odd integers n1, . . . , nk; then clearing denominators we obtain www.riazisara.ir 4 104 Number Theory Problems n1 · · · nk = s1 + · · · + sk, where si are all odd. But this is impossible since the left-hand side is odd and the right-hand side is even. □ If k is odd, such representations are possible. Here is one example for k = 9 and n1, . . . , n9 are distinct odd positive integers: 1 = 1 3 + 1 5 + 1 7 + 1 9 + 1 11 + 1 15 + 1 35 + 1 45 + 1 231. Example 1.4. [HMMT 2004] Zach has chosen five numbers from the set {1, 2, 3, 4, 5, 6, 7}. If he told Claudia what the product of the chosen numbers was, that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd. What is the product of the chosen numbers? Solution: The answer is 420. Providing the product of the chosen numbers is equivalent to telling the prod-uct of the two unchosen numbers. The only possible products that are achieved by more than one pair of numbers are 12 ({3, 4} and {2, 6}) and 6 ({1, 6} and {2, 3}). But in the second case, the sum of the two (unchosen) numbers is odd (and so the five chosen numbers have odd sum too). Therefore, the first must hold, and the product of the five chosen numbers is equal to 1 · 2 · 3 · · · 7 12 = 420. □ Division Algorithm The following result is called the division algorithm, and it plays an important role in number theory: Theorem 1.2a. For any positive integers a and b there exists a unique pair (q,r) of nonnegative integers such that b = aq + r and r < a. We say that q is the quotient and r the remainder when b is divided by a. To prove this result, we need to consider two parts: the existence of such a pair and its uniqueness. Proof: To show the existence, we consider three cases. (1) In this case, we assume that a > b. We can set q = 0 and r = b < a; that is, (q,r) = (0, b). (2) Suppose that a = b. We can set q = 1 and r = 0 < a; that is, (q,r) = (1, 0). (3) Finally, assume that a < b. There exist positive integers n such that na > b. Let q be the least positive integer for which (q + 1)a > b. Then qa ≤b. Let r = b −aq. It follows that b = aq + r and 0 ≤r < a. www.riazisara.ir 1. Foundations of Number Theory 5 Combining the three cases, we have established the existence. For uniqueness, assume that b = aq′+r′, where q′ and r′ are also nonnegative integers satisfying 0 ≤r′ < a. Then aq + r = aq′ + r′, implying a(q −q′) = r′ −r, and so a | r′ −r. Hence |r′ −r| ≥a or |r′ −r| = 0. Because 0 ≤r, r′ < a yields |r′ −r| < a, we are left with |r′ −r| = 0, implying r′ = r, and consequently, q′ = q. □ Example 1.5. Let n be a positive integer. Prove that 32n + 1 is divisible by 2, but not by 4. Proof: Clearly, 32n is odd and 32n + 1 is even. Note that 32n = (32)2n−1 = 92n−1 = (8 + 1)2n−1. Recall the Binomial theorem (x + y)m = xm + m 1 xm−1y + m 2 xm−2y2 + · · · +  n n −1 xym−1 + ym. Setting x = 8, y = 1, and m = 2n−1 in the above equation, we see that each summand besides the last (that is, ym = 1) is a multiple of 8 (which is a multiple of 4). Hence the remainder of 32n on dividing by 4 is equal to 1, and the remainder of 32n + 1 on dividing by 4 is equal to 2. □ The above argument can be simplified in the notation of congruence modulo 4. Congruence is an important part of number theory. We will discuss it extensively. The division algorithm can be extended for integers: Theorem 1.2b. For any integers a and b, a ̸= 0, there exists a unique pair (q,r) of integers such that b = aq + r and 0 ≤r < |a|. We leave the proof of this extended version to the reader. Primes The integer p > 1 is called a prime (or a prime number) if there is no integer d with d > 1 and d ̸= p such that d | p. Any integer n > 1 has at least one prime divisor. If n is a prime, then that prime divisor is n itself. If n is not a prime, then let a > 1 be its least divisor. Then n = ab, where 1 < a ≤b. If a were not a prime, then a = a1a2 with 1 < a1 ≤a2 < a and a1 | n, contradicting the minimality of a. An integer n > 1 that is not a prime is called composite. If n is a composite integer, then it has a prime divisor p not exceeding √n. Indeed, as above, n = ab, where 1 < a ≤b and a is the least divisor of n. Then n ≥a2; hence a ≤√n. This idea belongs to the ancient Greek mathematician Eratosthenes (250 BCE). Note that all positive even numbers greater than 2 are composite. In other words, 2 is the only even (and the smallest) prime. All other primes are odd; that www.riazisara.ir 6 104 Number Theory Problems is, they are not divisible by 2. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. How many primes are there? Are we really sure that there are infinitely many primes? Please see Theorem 1.3 below. A comparison between the number of elements in two infinite sets might be vague, but it is obvious that there are more (in the sense of density) composite numbers than primes. We see that 2 and 3 are the only consecutive primes. Odd consecutive primes such as 3 and 5, 5 and 7, 41 and 43, are called twin primes. It is still an open question whether there are infinitely many twin primes. Brun has shown that even if there are infinitely many twin primes, the sum of their inverses converges. The proof is however extremely difficult. Example 1.6. Find all positive integers n for which 3n −4, 4n −5, and 5n −3 are all prime numbers. Solution: The sum of the three numbers is an even number, so at least one of them is even. The only even prime number is 2. Only 3n −4 and 5n −3 can be even. Solving the equations 3n −4 = 2 and 5n −3 = 2 yields n = 2 and n = 1, respectively. It is trivial to check that n = 2 does make all three given numbers prime. □ Example 1.7. [AHSME 1976] If p and q are primes and x2 −px + q = 0 has distinct positive integral roots, find p and q. Solution: Let x1 and x2, with x1 < x2, be the two distinct positive integer roots. Then x2 −px + q = (x −x1)(x −x2), implying that p = x1 + x2 and q = x1x2. Since q is prime, x1 = 1. Thus, q = x2 and p = x2 + 1 are two consecutive primes; that is, q = 2 and p = 3. □ Example 1.8. Find 20 consecutive composite numbers. Solution: Numbers 20! + 2, 20! + 3, . . . , 20! + 21 will do the trick. □ The following result by Euclid has been known for more than 2000 years: Theorem 1.3a. There are infinitely many primes. Proof: Assume by way of contradiction that there are only a finite number of primes: p1 < p2 < · · · < pm. Consider the number P = p1 p2 · · · pm + 1. If P is a prime, then P > pm, contradicting the maximality of pm. Hence P is composite, and consequently, it has a prime divisor p > 1, which is one of the primes p1, p2, . . . , pm, say pk. It follows that pk divides p1 · · · pk · · · pm + 1. This, together with the fact that pk divides p1 · · · pk · · · pm, implies pk divides 1, a contradiction. □ Even though there are infinitely many primes, there are no particular formulas to find them. Theorem 1.3b in the next section will reveal part of the reasoning. www.riazisara.ir 1. Foundations of Number Theory 7 The Fundamental Theorem of Arithmetic The fundamental result in arithmetic (i.e., number theory) pertains to the prime factorization of integers: Theorem 1.4. [The Fundamental Theorem of Arithmetic] Any integer n greater than 1 has a unique representation (up to a permutation) as a product of primes. Proof: The existence of such a representation can be obtained as follows: Let p1 be a prime divisor of n. If p1 = n, then n = p1 is a prime factorization of n. If p1 < n, then n = p1r1, where r1 > 1. If r1 is a prime, then n = p1 p2, where p2 = r1 is the desired factorization of n. If r1 is composite, then r1 = p2r2, where p2 is a prime, r2 > 1, and so n = p1 p2r2. If r2 is a prime, then n = p1 p2 p3, where r2 = p3 and r3 = 1, and we are done. If r2 is composite, then we continue this algorithm, obtaining a sequence of integers r1 > r2 > · · · ≥1. After a finite number of steps, we reach rk+1 = 1, that is, n = p1 p2 · · · pk. For uniqueness, let us assume that there is at least one positive integer n that has two distinct prime factorizations; that is, n = p1 p2 · · · pk = q1q2 · · · qh where p1, p2, . . . , pk, q1, q2, . . . , qh are primes with p1 ≤p2 ≤· · · pk and q1 ≤ q2 · · · qh such that the k-tuple (p1, p2, . . . , pk) is not the same as the h-tuple (q1, q2, . . . , qh). It is clear that k ≥2 and h ≥2. Let n be the minimal such integer. We will derive a contradiction by finding a smaller positive integer that also has two distinct prime factorizations. We claim that pi ̸= q j for any i = 1, 2, . . . , k, j = 1, 2, . . . , h. If, for example, pk = qh = p, then n′ = n/p = p1 · · · pk−1 = q1 · · · qh−1 and 1 < n′ < n, contradicting the minimality of n. Assume without loss of generality that p1 ≤q1; that is, p1 is the least prime factor of n in the above representations. By applying the division algorithm it follows that q1 = p1c1 + r1, q2 = p1c2 + r2, . . . qh = p1ch + rh, where 1 ≤ri < p1, i = 1, . . . , h. We have n = q1q2 · · · qh = (p1c1 + r1)(p1c2 + r2) · · · (p1ch + rh). Expanding the last product we obtain n = mp1 +r1r2 · · ·rh for some positive integer m. Setting n′ = r1r2 · · ·rh we have n = p1 p2 · · · pk = mp1 + n′. It www.riazisara.ir 8 104 Number Theory Problems follows that p1 | n′ and n′ = p1s. As we have shown, s can be written as a product of primes. We write s = s1s2 · · · si, where s1, s2, . . . , si are primes. On the other hand, using the factorization of r1,r2, . . . ,rh into primes, all their factors are less than ri < p1. From n′ = r1r2 · · ·rh, it follows that n′ has a factorization into primes of the form n′ = t1t2 · · · t j, where ts < p1, s = 1, 2, . . . , j. This factorization is different from n′ = p1s1s2 · · · si. But n′ < n, contradicting the minimality of n. □ From the above theorem it follows that any integer n > 1 can be written uniquely in the form n = pα1 1 · · · pαk k , where p1, . . . , pk are distinct primes and α1, . . . , αk are positive integers. This representation is called the canonical factorization (or factorization) of n. It is not difficult to see that the canonical factorization of the product of two integers is the product of the canonical factorizations of the two integers. This factorization allows us to establish the following fundamental property of primes. Corollary 1.5. Let a and b be integers. If a prime p divides ab, then p divides either a or b. Proof: Because p divides ab, p must appear in the canonical factorization of ab. The canonical factorizations of a, b, and ab are unique, and the canonical factorization of ab is the product of the canonical factorizations of a and b. Thus p must appear in at least one of the canonical factorizations of a and b, implying the desired result. □ Another immediate application of the prime factorization theorem is an alter-native way of proving that there are infinitely many primes. As in the proof of Theorem 1.3, assume that there are only finitely many primes: p1 < p2 < · · · < pm. Let N = m i=1  1 + 1 pi + 1 p2 i + · · ·  = m i=1 1 1 −1 pi . On the other hand, by expanding and by using the canonical factorization of pos-itive integers, we obtain N = 1 + 1 2 + 1 3 + · · · , yielding m i=1 pi pi −1 = ∞, a contradiction. We have used the well-known facts: www.riazisara.ir 1. Foundations of Number Theory 9 (a) the harmonic series 1 + 1 2 + 1 3 + · · · diverges; (b) the expansion formula 1 1 −x = 1 + x + x2 + · · · holds for real numbers x with |x| < 1. This expansion formula can also be interpreted as the summation formula for the infinite geometric progression 1, x, x2, . . . . From the formula ∞ i=1 pi pi −1 = ∞, using the inequality 1 + t ≤et, t ∈R, we can easily derive ∞  i=1 1 pi = ∞. For a prime p we say that pk fully divides n and write pk∥n if k is the greatest positive integer such that pk|n. Example 1.9. [ARML 2003] Find the largest divisor of 1001001001 that does not exceed 10000. Solution: We have 1001001001 = 1001 · 106 + 1001 = 1001 · (106 + 1) = 7 · 11 · 13 · (106 + 1). Note that x6 + 1 = (x2)3 + 1 = (x2 + 1)(x4 −x2 + 1). We conclude that 106 + 1 = 101 · 9901, and so 1001001001 = 7 · 11 · 13 · 101 · 9901. It is not difficult to check that no combination of 7, 11, 13, and 101 can generate a product greater than 9901 but less than 10000, so the answer is 9901. □ Example 1.10. Find n such that 2n∥31024 −1. Solution: The answer is 12. Note that 210 = 1024 and x2 −y2 = (x + y)(x −y). We have 3210 −1 = (329 + 1)(329 −1) = (329 + 1)(328 + 1)(328 −1) = · · · = (329 + 1)(328 + 1)(327 + 1) · · · (321 + 1)(320 + 1)(3 −1). By Example 1.5, 2∥32k + 1, for positive integers k. Thus the answer is 9 + 2 + 1 = 12. □ www.riazisara.ir 10 104 Number Theory Problems Theorem 1.4 indicates that all integers are generated (productively) by primes. Because of the importance of primes, many people have tried to find (explicit) formulas to generate primes. So far, all the efforts are incomplete. On the other hand, there are many negative results. The following is a typical one, due to Goldbach: Theorem 1.3b. For any given integer m, there is no polynomial p(x) with inte-ger coefficients such that p(n) is prime for all integers n with n ≥m. Proof: For the sake of contradiction, assume that there is such a polynomial p(x) = akxk + ak−1xk−1 + · · · + a1x + a0 with ak, ak−1, . . . , a0 being integers and ak ̸= 0. If p(m) is composite, then our assumption was wrong. If not, assume that p(m) = p is a prime. Then p(m) = akmk + ak−1mk−1 + · · · + a1m + a0 and for positive integers i, p(m + pi) = ak(m + pi)k + ak−1(m + pi)k−1 + · · · + a1(m + pi) + a0. Note that (m + pi) j = m j +  j i m j−1(pi) +  j 2 m j−2(pi)2 + · · · +  j j −1 m(pi) j−1 + (pi) j. Hence (m + pj) j −m j is a multiple of p. It follows that p(m + pi) −p(m) is a multiple of p. Because p(m) = p, p(m + pi) is a multiple of p. By our assumption, p(m + pi) is also prime. Thus, the possible values of p(m + pi) are 0, p, and −p for all positive integers i. On the other hand, the equations p(x) = 0, p(x) = p, and p(x) = −p can have at most 3k roots. Therefore, there exist (infinitely many) i such that m + pi is not a solution of any of the equations p(x) = 0, p(x) = p, and p(x) = −p. We obtain a contradiction. Hence our assumption was wrong. Therefore, such polynomials do not exist. □ Even though there are no definitive ways to find primes, the density of primes (that is, the average appearance of primes among integers) has been known for about 100 years. This was a remarkable result in the mathematical field of analytic number theory showing that lim n→∞ π(n) n/log n = 1, where π(n) denotes the number of primes ≤n. The relation above is known as the prime number theorem. It was proved by Hadamard and de la Vall´ ee Poussin in 1896. An elementary but difficult proof was given by Erd¨ os and Selberg. www.riazisara.ir 1. Foundations of Number Theory 11 G.C.D. For a positive integer k we denote by Dk the set of all its positive divisors. It is clear that Dk is a finite set. For positive integers m and n the maximal element in the set Dm ∩Dn is called the greatest common divisor (or G.C.D.) of m and n and is denoted by gcd(m, n). In the case Dm ∩Dn = {1}, we have gcd(m, n) = 1 and we say that m and n are relatively prime (or coprime). The following are some basic properties of G.C.D. Proposition 1.6. (a) if p is a prime, then gcd(p, m) = p or gcd(p, m) = 1. (b) If d = gcd(m, n), m = dm′, n = dn′, then gcd(m′, n′) = 1. (c) If d = gcd(m, n), m = d′m′′, n = d′n′′, gcd(m′′, n′′) = 1, then d′ = d. (d) If d′ is a common divisor of m and n, then d′ divides gcd(m, n). (e) If px∥m and py∥n, then pmin x,y∥gcd(m, n). Furthermore, if m = pα1 1 · · · pαk k and n = pβ1 1 · · · pβk k , αi, βi ≥0, i = 1, . . . , k, then gcd(m, n) = pmin(α1,β1) 1 · · · pmin(αk,βk) k . (f) If m = nq + r, then gcd(m, n) = gcd(n,r). Proof: The proofs of these properties are rather straightforward from the def-inition. We present only the proof property (f). Set d = gcd(m, n) and d′ = gcd(n,r). Because d | m and d | n it follows that d | r. Hence d | d′. Con-versely, from d′ | n and d′ | r it follows that d′ | m, so d′ | d. Thus d = d′. □ The definition of G.C.D. can easily be extended to more than two numbers. For given integers a1, a2, . . . , an, gcd(a1, a2, . . . , an) is the common greatest di-visor of all the numbers a1, a2, . . . , an. We can define the greatest common divi-sor of a1, a2, . . . , an by considering d1 = gcd(a1, a2), d2 = gcd(d1, a3), . . . , dn−1 = gcd(dn−2, an). We leave to the reader to convince himself that dn−1 = gcd(a1, . . . , an). We also leave the simple proofs of the following properties to the reader. Proposition 1.6. (Continuation) (g) gcd(gcd(m, n), p) = gcd(m, gcd(n, p)); proving that gcd(m, n, p) is well-defined; (h) If d | ai, i = 1, . . . , s, then d | gcd(a1, . . . , as); www.riazisara.ir 12 104 Number Theory Problems (i) If ai = pα1i 1 · · · pαki k , i = 1, . . . , s, then gcd(a1, . . . , as) = pmin(α11,...,α1k) 1 · · · pmin(αk1,...,αkk) k . We say that a1, a2, . . . , an are relatively prime if their greatest common divi-sor is equal to 1. Note that gcd(a1, a2, . . . , an) = 1 does not imply that gcd(ai, a j) = 1 for 1 ≤i < j ≤n. (For example, we can set a1 = 2, a2 = 3, and a3 = 6.) If a1, a2, . . . , an are such that gcd(ai, a j) = 1 for 1 ≤i < j ≤n, we say that these numbers are pairwise relatively prime (or coprime). Euclidean Algorithm Canonical factorizations help us to determine the greatest common divisors of integers. But it is not easy to factor numbers, especially large numbers. (This is why we need to study divisibility of numbers.) A useful algorithm for finding the greatest common divisor of two positive integers m and n is the Euclidean algorithm. It consists of repeated application of the division algorithm: m = nq1 + r1, 1 ≤r1 < n, n = r1q2 + r2, 1 ≤r2 < r1, . . . rk−2 = rk−1qk + rk, 1 ≤rk < rk−1, rk−1 = rkqk+1 + rk+1, rk+1 = 0. This chain of equalities is finite because n > r1 > r2 > · · · > rk. The last nonzero remainder, rk, is the greatest common divisor of m and n. Indeed, by applying successively property (f) above we obtain gcd(m, n) = gcd(n,r1) = gcd(r1,r2) = · · · = gcd(rk−1,rk) = rk. Example 1.11. [HMMT 2002] If a positive integer multiple of 864 is chosen randomly, with each multiple having the same probability of being chosen, what is the probability that it is divisible by 1944? First Solution: The probability that a multiple of 864 = 25 · 33 is divisible by 1944 = 23 · 35 is the same as the probability that a multiple of 22 = 4 is divisible by 32 = 9. Since 4 and 9 are relatively prime, the probability is 1 9. □ Second Solution: By the Euclidean algorithm, we have gcd(1944, 864) = gcd(1080, 864) = gcd(864, 216) = 216. Hence 1944 = 9·216 and 864 = 4·216. We can finish as in the first solution. □ www.riazisara.ir 1. Foundations of Number Theory 13 Example 1.12. [HMMT 2002] Compute gcd(2002 + 2, 20022 + 2, 20023 + 2, . . . ). Solution: Let g denote the desired greatest common divisor. Note that 20022 + 2 = 2002(2000 + 2) + 2 = 2000(2002 + 2) + 6. By the Euclidean algorithm, we have gcd(2002 + 2, 20022 + 2) = gcd(2004, 6) = 6. Hence g | gcd(2002 + 2, 20022 + 2) = 6. On the other hand, every number in the sequence 2002 + 2, 20022 + 2, . . . is divisible by 2. Furthermore, since 2002 = 2001 + 1 = 667 · 3 + 1, for all positive integers k, 2002k = 3ak + 1 for some integer ak. Thus 2002k + 2 is divisible by 3. Because 2 and 3 are relatively prime, every number in the sequence is divisible by 6. Therefore, g = 6. □ B´ ezout’s Identity Let’s start with two classic brain teasers. Example 1.13. In a special football game, a team scores 7 points for a touch-down and 3 points for a field goal. Determine the largest mathematically unreach-able number of points scored by a team in an (infinitely long) game. Solution: The answer is 11. It’s not difficult to check that we cannot obtain 11 points. Note that 12 = 3 + 3 + 3 + 3, 13 = 7 + 3 + 3, and 14 = 7 + 7. For all integers n greater than 11, the possible remainders when n is divided by 3 are 0, 1, and 2. If n has remainder 0, we can clearly obtain n points by scoring enough field goals; if n has remainder 1, then n −13 has remainder 0, and we can obtain n points by scoring one touchdown and enough field goals; if n has remainder 2, then n −14 has remainder 0, and we can obtain n points by scoring two touchdowns and enough field goals. In short, all integers n greater than 11 can be written in the form n = 7a + 3b for some nonnegative integers a and b. □ Example 1.14 There is an ample supply of milk in a milk tank. Mr. Fat is given a 5-liter (unmarked) container and a 9-liter (unmarked) container. How can he measure out 2 liters of milk? Solution: Let T, L5, and L9 denote the milk tank, the 5-liter container, and the 9-liter container, respectively. We can use the following table to achieve the desired result. www.riazisara.ir 14 104 Number Theory Problems T L5 L9 x 0 0 x −5 5 0 x −5 0 5 x −10 5 5 x −10 1 9 x −1 1 0 x −1 0 1 x −6 5 1 x −6 0 6 x −11 5 6 x −11 2 9 □ The key is to make the connection between 2 = 4 × 5 −2 × 9. We leave it to the reader to use the equation 2 = 3×9−5×5 to set up another process. For given integers a1, a2, . . . , an, we call α1a1 + α2a2 + · · · + αnan, where α1, α2, . . . , αn are arbitrary integers, linear combinations of a1, a2, . . . , an. Examples 1.13 and 1.14 are seemingly unrelated problems. But they both involve linear combinations of two given integers. What if we replace (7, 3) by (6, 3) in Example 1.13, and (5, 9) by (6, 9) in Example 1.14? We have the following general result. Theorem 1.7. [B´ ezout] For positive integers m and n, there exist integers x and y such that mx + ny = gcd(m, n). Proof: From the Euclidean algorithm it follows that r1 = m −nq1, r2 = −mq2 + n(1 + q1q2), . . . . In general, ri = mαi + nβi, for i = 1, . . . , k. Because ri+1 = ri−1 −riqi+1, it follows that αi+1 = αi−1 −qi+1αi, βi+1 = βi−1 −qi+1βi, for i = 2, . . . , k −1. Finally, we obtain gcd(m, n) = rk = αkm + βkn. □ Note that gcd(a, b) divides ax + by. In view of B´ ezout’s identity, for given integers a, b, and c, the equation ax + by = c is solvable for integers (x, y) if and only if gcd(a, b) divides c. In algebra, we solve systems of equations. In number theory, we usually try to find special solutions for systems of equations, namely, integer solutions, rational solutions, and so on. Hence most of the these systems have more variables than the number of equations in the system. These are called Diophantine equations, attributed to the ancient Greek mathematician Diophantus, which will be studied extensively in the sequel to this book: 105 Dio-phantine Equations and Integer Function Problems. For fixed integers a, b, and c, ax + by = c is a two-variable linear Diophantine equation. www.riazisara.ir 1. Foundations of Number Theory 15 Corollary 1.8. If a | bc and gcd(a, b) = 1, then a | c. Proof: If c = 0, the assertion is clearly true. Assume that c ̸= 0. Since gcd(a, b) = 1, by B´ ezout’s identity, ax + by = 1 for some integers x and y. Hence acx+bcy = c. Because a divides acx and bcy, a divides c, as desired. □ Corollary 1.9. Let a and b be two coprime numbers. If c is an integer such that a | c and b | c, then ab | c. Proof: Because a | c, we have c = ax for some integer x. Hence b divides ax. Because gcd(a, b) = 1, b | x, and by Corollary 1.8, it follows that x = by for some integer y, and so c = aby, or ab | c. □ Corollary 1.10. Let p be a prime, and let k be an integer with 1 ≤k < p. Then p | p k  . Proof: Note that from relation k p k = p p −1 k −1 it follows that p divides k p k  . Because gcd(p, k) = 1, the relation p divides p k  is obtained via Corollary 1.8. □ Example 1.15. [Russia 2001] Let a and b be distinct positive integers such that ab(a + b) is divisible by a2 + ab + b2. Prove that |a −b| > 3 √ ab. Proof: Set g = gcd(a, b) and write a = xg and b = yg with gcd(x, y) = 1. Then ab(a + b) a2 + ab + b2 = xy(x + y)g x2 + xy + y2 is an integer. Note that gcd(x2 + xy + y2, x) = gcd(y2, x) = 1. Similarly, gcd(x2 + xy + y2, y) = 1. Because gcd(x + y, y) = 1, we have gcd(x2 + xy + y2, x + y) = gcd(y2, x + y) = 1. By Corollary 1.9, x2 + xy + y2 | g, implying that g ≥x2 + xy + y2. Therefore, |a −b|3 = |g(x −y)|3 = g2|x −y|3 · g ≥g2 · 1 · (x2 + xy + y2) > g2xy = ab. It follows that |a −b| > 3 √ ab. □ Note that the key step x2 + xy + y2 divides g can also be obtained by clever algebraic manipulations such as a3 = (a2 + ab + b2)a −ab(a + b). www.riazisara.ir 16 104 Number Theory Problems L.C.M. For a positive integer k we denote by Mk the set of all multiples of k. As opposed to the set Dk defined earlier in this section, Mk is an infinite set. For positive integers s and t the minimal element of the set Ms ∩Mt is called the least common multiple of s and t and is denoted by lcm(s, t) or [s, t]. Proposition 1.11. (a) If lcm(s, t) = m, m = ss′ = tt′, then gcd(s′, t′) = 1. (b) If m′ is a common multiple of s and t and m′ = ss′ = tt′, gcd(s′, t′) = 1, then m′ = m. (c) If m′ is a common multiple of s and t, then m | m′. (d) If m | s and n | s, then lcm(m, n) | s. (e) If n is an integer, n lcm(s, t) = lcm(ns, nt). (f) If s = pα1 1 · · · pαk k and t = pβ1 1 · · · pβk k , αi, bi ≥0, i = 1, . . . , k, then lcm(s, t) = pmax(α1,β1) 1 · · · pmax(αk,βk) k . The properties in Proposition 1.11 are easily obtained from the definition of L.C.M., and we leave their proofs to the reader. The following property establishes an important connection between G.C.D. and L.C.M. Proposition 1.12. For any positive integers m and n the following relation holds: mn = gcd(m, n) · lcm(m, n). Proof: Let m = pα1 1 · · · pαk k , n = pβ1 1 · · · pβk k , αi, βi ≥0, i = 1, . . . , k. From Properties 1.6 (e) and 1.11 (f) we have gcd(m, n) lcm(m, n) = pmin(α1,β1)+max(α1,β1) 1 · · · pmin(αk,βk)+max(αk,βk) k = pα1+β1 1 · · · pαk+βk k = mn. □ Let a1, a2, . . . , an be positive integers. The least common multiple of a1, a2, . . . , an, denoted by lcm(a1, a2, . . . , an), is the smallest positive integer that is a multiple of all of a1, a2, . . . , an. Note that Proposition 1.12 cannot be easily generalized. For example, it is not true that gcd(a, b, c) lcm(a, b, c) = abc. We leave it to the reader to find interesting counterexamples. www.riazisara.ir 1. Foundations of Number Theory 17 The Number of Divisors We start with three examples. Example 1.16. [AIME 1988] Compute the probability that a randomly chosen positive divisor of 1099 is an integer multiple of 1088. Solution: What are the divisors of 1099? Is 3 a divisor? Is 220 a divisor? We consider the prime factorization of 1099, which is 299 · 599. The divisors of 1099 are of the form 2a · 5b, where a and b are integers with 0 ≤a, b ≤99. Because there are 100 choices for each of a and b, 1099 has 100 · 100 positive integer divisors. Of these, the multiples of 1088 = 288 · 588 must satisfy the inequalities 88 ≤a, b ≤99. Thus there are 12 choices for each of a and b; that is, 12 · 12 of the 100 · 100 divisors of 1099 are multiples of 1088. Consequently, the desired probability is 12·12 100·100 = 9 625. □ Example 1.17. Determine the number of ordered pairs of positive integers (a, b) such that the least common multiple of a and b is 23571113. Solution: Both a and b are divisors of 23571113, and so a = 2x5y11z and b = 2s5t11u for some nonnegative integers x, y, z, s, t, u. Because 23571113 is the least common multiple, max{x, s} = 3, max{y, t} = 7, and max{z, u} = 13. Hence (x, s) can be equal to (0, 3), (1, 3), (2, 3), (3, 3), (3, 2), (3, 1), or (3, 0), so there are 7 choices for (x, s). Similarly, there are 15 and 27 choices for (y, t) and (z, u), respectively. By the multiplication principle, there are 7×15×27 = 2835 ordered pairs of positive integers (a, b) having 23571113 as their least common multiple. □ Example 1.18. Determine the product of distinct positive integer divisors of n = 4204. Solution: Because n = (22 · 3 · 5 · 7)4, d is a divisor of n if and only if d can be written in the form 2a · 3b · 5c · 7d, where 0 ≤a ≤8, 0 ≤b ≤4, 0 ≤c ≤4, and 0 ≤d ≤4. Hence there are 9, 5, 5, and 5 possible values for a, b, c, and d, respectively. It follows that n has 9·5·5·5 = 1125 positive divisors. If d ̸= 4202, then 4204 d is also a divisor, and the product of these two divisors is 4204. We can thus partition 1124 divisors of n (excluding 4202) into 562 pairs of divisors of the form  d, n d  , and the product of the two divisors in each pair is 4204. Hence the answer is 4204·562 · 4202 = 4202250. □ Putting the last three examples together gives two interesting results in number theory. For a positive integer n denote by τ(n) the number of its divisors. It is www.riazisara.ir 18 104 Number Theory Problems clear that τ(n) =  d|n 1. Writing τ in this summation form allows us later to discuss it as an example of a multiplicative arithmetic function. Proposition 1.13. If n = pa1 1 pa2 2 · · · pak k is a prime decomposition of n, then n has τ(n) = (a1 + 1)(a2 + 1) · · · (ak + 1) divisors. Corollary 1.14. If n = pa1 1 pa2 2 · · · pak k is a prime decomposition of n, then there are (2a1 + 1)(2a2 + 1) · · · (2ak + 1) distinct pairs of ordered positive integers (a, b) with lcm(a, b) = n. Corollary 1.15. For any positive integer n, d|n d = n τ(n) 2 . The proofs of these three propositions are identical to those of Examples 1.16, 1.17, and 1.18. It is interesting to note that these three results can be generalized to the case that the powers of the primes in the prime decomposition are nonnegative (because if ai = 0 for some 1 ≤i ≤k, then ai + 1 = 2ai + 1 = 1, which does not affect the products). Corollary 1.16. For any positive integer n, τ(n) ≤2√n. Proof: Let d1 < d2 < · · · < dk be the divisors of n not exceeding √n. The remaining divisors are n d1 , n d2 , . . . , n dk . It follows that τ(n) ≤2k ≤2√n. □ The Sum of Divisors For a positive integer n denote by σ(n) the sum of its positive divisors, including 1 and n itself. It is clear that σ(n) =  d|n d. This representation will help us to show that σ is multiplicative. www.riazisara.ir 1. Foundations of Number Theory 19 Proposition 1.17. If n = pα1 1 · · · pαk k is the prime factorization of n, then σ(n) = pα1+1 1 −1 p1 −1 · · · pαk+1 k −1 pk −1 . Proof: The divisors of n can be written in the form pa1 1 · · · pak k , where a1, . . . , ak are integers with 0 ≤a1 ≤α1, . . . , 0 ≤ak ≤αk. Each divisor of n appears exactly once as a summand in the expansion of the product (1 + p1 + · · · + pα1 1 ) · · · (1 + pk + · · · + pαk k ), from which the desired result follows, by also noting the formula for the sum of a finite geometric progression: rk+1 −1 r −1 = 1 + r + r2 + · · · + rk. □ Example 1.19. Find the sum of even positive divisors of 10000. Solution: The even divisors of 10000 can be written in the form of 2a5b, where a and b are integers with 1 ≤a ≤5 and 0 ≤b ≤5. Each even divisor of 10000 appears exactly once as a summand in the expansion of the product (2 + 22 + 23 + 24 + 25)(1 + 5 + 52 + 53 + 54 + 55) = 62 · 56 −1 5 −1 = 242172. □ Modular Arithmetic Let a, b, and m be integers, with m ̸= 0. We say that a and b are congruent modulo m if m divides a −b. We denote this by a ≡b (mod m). The relation “≡” on the set Z of integers is called the congruence relation. If m does not divide a −b, then we say that integers a and b are not congruent modulo m and we write a ̸≡b (mod m). Proposition 1.18. (a) a ≡a (mod m) (reflexivity). (b) If a ≡b (mod m) and b ≡c (mod m), then a ≡c (mod m) (transitivity). (c) If a ≡b (mod m), then b ≡a (mod m). www.riazisara.ir 20 104 Number Theory Problems (d) If a ≡b (mod m) and c ≡d (mod m), then a + c ≡b + d (mod m) and a −c ≡b −d (mod m). (e) If a ≡b (mod m), then for any integer k, ka ≡kb (mod m). (f) If a ≡b (mod m) and c ≡d (mod m), then ac ≡bd (mod m). In general, if ai ≡bi (mod m), i = 1, . . . , k, then a1 · · · ak ≡b1 · · · bk (mod m). In particular, if a ≡b (mod m), then for any positive integer k, ak ≡bk (mod m). (g) We have a ≡b (mod mi), i = 1, . . . , k, if and only if a ≡b (mod lcm(m1, . . . , mk)). In particular, if m1, . . . , mk are pairwise relatively prime, then a ≡b (mod mi), i = 1, . . . , k, if and only if a ≡b (mod m1 · · · mk). Proof: The proofs are straightforward. We present the proof of (g) and leave the rest to the reader. From a ≡b (mod mi), i = 1, . . . , k, it follows that mi | (a −b), i = 1, . . . , k. Hence a −b is a common multiple of m1, . . . , mk, and so lcm(m1, . . . , mk) | (a −b). That is, a ≡b (mod lcm(m1, . . . , mk)). Conversely, from a ≡b (mod lcm(m1, . . . , mk)) and the fact that each mi divides lcm(m1, . . . , mk) we obtain a ≡b (mod mi), i = 1, . . . , k. □ Proposition 1.19. Let a, b, n be integers, n ̸= 0, such that a = nq1 + r1, b = nq2 + r2, 0 ≤r1,r2 < |n|. Then a ≡b (mod n) if and only if r1 = r2. Proof: Because a −b = n(q1 −q2) + (r1 −r2), it follows that n | (a −b) if and only if n | (r1 −r2). Taking into account that |r1 −r2| < |n|, we have n | (r1 −r2) if and only if r1 = r2. □ Example 1.20. Prove that there are infinitely many primes of the form 4k −1; that is, congruent to 3 modulo 4. Proof: We first note that there is at least one prime p with p ≡3 (mod 4) (simply set p = 3). Suppose there were only finitely many primes congruent to 3 modulo 4. Let p1, p2, . . . , pk be those primes, and let P = p1 p2 · · · pk denote their product. We have 4P −1 ≡3 (mod 4). If all the prime divisors of 4P −1 were congruent to 1 modulo 4, then 4P −1 would be congruent to 1 modulo 4 (by Proposition 1.18 (g)). Thus, some prime divisor p of 4P −1 would be congruent to 3 modulo 4. On the other hand, gcd(4P −1, pi) = 1 for all i with 1 ≤i ≤k, and so we find another prime that is congruent to 3 modulo 4, a contradiction to our assumption. Hence there are infinitely many primes of the form 4k −1. □ www.riazisara.ir 1. Foundations of Number Theory 21 In exactly the same way, we can show that there are infinitely many primes of the form 6k −1. We can view congruency as (part of) an arithmetic progression. For example, we can rewrite the last two results as follows: There are infinitely many primes in the arithmetic progression {−1 + ka}∞ k=1 with a = 4 or a = 6. These are the special cases of a famous result of Dirichlet: There are infinitely many primes in any arithmetic progression of integers for which the common difference is relatively prime to the terms. In other words, if a and m be relatively prime positive integers, then there are infinitely many primes p such that p ≡a (mod m). Dirichlet was also able to compute the density (in simpler terms, a certain kind of frequency of such primes) of these prime numbers in the set of all primes. This was another milestone in analytic number theory. The proof of this work is beyond the scope of this book. We present a more detailed form of this result in the glossary section of this book. Some problems in this book become easy if we apply this theorem directly. But all of these problems can also be solved in different ways, and we strongly encourage the reader to look for these different approaches, which will enhance the reader’s problem-solving abilities. In Example 1.20, it is very natural to work modulo 4. Many times, such a choice is not obvious. Taking the proper modulus holds the key to many problems. Example 1.21. [Russia 2001] Find all primes p and q such that p + q = (p −q)3. Solution: The only such primes are p = 5 and q = 3. Because (p −q)3 = p + q ̸= 0, p and q are distinct and hence relatively prime. Because p −q ≡2p (mod p + q), taking the given equation modulo p + q gives 0 ≡8p3 (mod p + q). Because p and q are relatively prime, so are p and p + q. Thus, 0 ≡8 (mod p + q); that is, p + q divides 8. Likewise, taking the given equation modulo p−q gives 2p ≡0 (mod p−q). Because p and q are relatively prime, so are p and p −q. We conclude that 2 ≡0 (mod p −q), or p −q divides 2. It easily follows that (p, q) is equal to (3, 5) or (5, 3); only the latter satisfies the given equation. □ There is another approach to the last problem: setting p −q = a leads to p + q = a3. Hence p = a3+a 2 and q = a3−a 2 . This kind of substitution is a very common technique in solving Diophantine equations. Example 1.22. [Baltic 2001] Let a be an odd integer. Prove that a2n + 22n and a2m + 22m are relatively prime for all positive integers n and m with n ̸= m. www.riazisara.ir 22 104 Number Theory Problems Proof: Without loss of generality, assume that m > n. For any prime p dividing a2n + 22n, we have a2n ≡−22n (mod p). We square both sides of the equation m −n times to obtain a2m ≡22m (mod p). Because a is odd, we have p ̸= 2. Thus, 22m + 22m = 22m+1 ̸≡0 (mod p), so that a2m ≡22m ̸≡−22m (mod p). Therefore, p ∤(a2m + 22m), proving the desired result. □ Setting a = 1 in the last example leads to a property of the Fermat numbers, which will soon be discussed. Example 1.23. Determine whether there exist infinitely many even positive in-tegers k such that for every prime p the number p2 + k is composite. Solution: The answer is positive. First note that for p = 2, p2 + k is always composite for all even positive integers k. Next we note that if p > 3, then p2 ≡1 (mod 3). Hence if k is an even positive integer with k ≡2 (mod 3), then p2 + k is composite for all all primes p > 3 (p2 + k is greater than 3 and is divisible by 3). Finally, we note that 32 + k ≡0 (mod 5) if k ≡1 (mod 5). Putting the above arguments together, we conclude that all positive integers k with ⎧ ⎨ ⎩ k ≡0 (mod 2), k ≡2 (mod 3), k ≡1 (mod 5), (∗) satisfy the conditions of the problem. By Proposition 1.18 (g), we consider (mod lcm(2, 3, 5)) = (mod 30). It is not difficult to check that all positive integers k with k = 26 (mod 30) satisfy the system, and hence the conditions of the problem. □ The system (∗) is a linear congruence system, and each of the three equa-tions in the system is a linear congruence equation. We will study the solutions of the linear congruence systems when we study the Chinese Remainder Theo-rem in the sequel to this book: 105 Diophantine Equations and Integer Function www.riazisara.ir 1. Foundations of Number Theory 23 Problems. The major difference between solving an equation and solving a con-gruence equation is the limitation of division in the latter situation. For example, in algebra, 4x = 4y implies that x = y. In modular arithmetic, 4x ≡4y (mod 6) does not necessarily imply that x ≡y (mod 6). (Why?) On the other hand, 4x ≡4y (mod 15) does imply that x ≡y (mod 15). (Why?) Proposition 1.18 (g) plays a key role in this difference. In algebra, xy = 0 implies that either x = 0 or y = 0 or both. But in modular arithmetic, xy ≡0 (mod m) does not imply x ≡0 (mod m) or y ≡0 (mod m). (For example, 3·5 ≡0 (mod 15), but 3 ̸≡0 (mod 15) and 5 ̸≡0 (mod 15). We will discuss this topic in a detailed fashion when we talk about linear congruence equations. For a little preview, we rewrite Corollary 1.5 in the language of modular arithmetic. Corollary 1.20. Let p be a prime. If x and y are integers such that xy ≡0 (mod p), then either x ≡0 (mod p) or y ≡0 (mod p) or both. This is an example of interchanging the faces of a common idea in number theory: p | xy (divisibility notation), xy ≡0 (mod p) (modular and congru-ence notation), and p = kxy (Diophantine equation forms). Simple applications Corollaries 1.8 and 1.9 also lead to the following properties. Corollary 1.21. Let m be a positive integer, and let a, b, and c be integers with c ̸= 0. If ac ≡bc (mod m), then a ≡b  mod m gcd(c, m) . Corollary 1.22. Let m be a positive integer. Let a be an integer relatively prime to m. If a1 and a2 are integers such that a1 ̸≡a2 (mod m), then a1a ̸≡a2a (mod m). The following property is useful in reducing the power in congruency rela-tions. Corollary 1.23. Let m be a positive integer, and let a and b be integers relatively prime to m. If x and y are integers such that ax ≡bx (mod m) and ay ≡by (mod m), then agcd(x,y) ≡bgcd(x,y) (mod m). Proof: By B´ ezout’s identity, there are nonnegative integers u and v such that gcd(x, y) = ux −vy. By the given conditions, we have aux ≡bux (mod m) and bvy ≡avy (mod m), www.riazisara.ir 24 104 Number Theory Problems implying that auxbvy ≡avybux (mod m). Since gcd(a, m) = gcd(b, m) = 1, by Corollary 1.21, we have agcd(x,y) ≡aux−vy ≡bux−vy ≡bgcd(x,y) (mod m). □ Residue Classes By Proposition 1.18 (a), (b), and (c), we conclude that for any given positive integer m, we can classify integers into a unique class according to their remainder on division by m. Clearly, there are m such classes. A set S of integers is also called a complete set of residue classes modulo n if for each 0 ≤i ≤n−1, there is an element s ∈S such that i ≡s (mod n). Clearly, {a, a + 1, a + 2, . . . , a + m −1} is a complete set of residue classes modulo m for any integer a. In particular, for a = 0, {0, 1, . . . , m −1} is the minimal nonnegative complete set of residue classes. Also, it is common to consider the complete set of residue classes {0, ±1, ±2, . . . , ±k} for m = 2k + 1 and {0, ±1, ±2, . . . , ±(k −1), k} for m = 2k. Example 1.24. Let n be an integer. Then (1) n2 ≡0 or 1 (mod 3); (2) n2 ≡0 or ± 1 (mod 5); (3) n2 ≡0 or 1 or 4 (mod 8); (4) n3 ≡0 or ± 1 (mod 9); (5) n4 ≡0 or 1 (mod 16); All the proofs can be done by checking complete sets of residue classes. We leave them to the reader. We also encourage the reader to review these relations after finishing studying Euler’s theorem. Example 1.25. [Romania 2003] Consider the prime numbers n1 < n2 < · · · < n31. Prove that if 30 divides n4 1 + n4 2 + · · · + n4 31, then among these numbers one can find three consecutive primes. Solution: Let s = n4 1 + n4 2 + · · · + n4 31. Firstly, we claim that n1 = 2. Otherwise, all numbers ni, 1 ≤i ≤31, are odd, and consequently s is odd, a contradiction. Secondly, we claim that n2 = 3. Otherwise, we have n4 i ≡1 (mod 3) for all 1 ≤i ≤31. It follows that s ≡31 ≡1 (mod 3), a contradiction. Finally, we prove that n3 = 5. Indeed, if not, then n2 i ≡±1 (mod 5) and n4 i ≡1 (mod 5) for all 1 ≤i ≤31. Thus, s ≡31 ≡1 (mod 5), a contradiction. We conclude that three consecutive primes, namely, 2, 3, and 5, appear in the given prime numbers. □ www.riazisara.ir 1. Foundations of Number Theory 25 Example 1.26. Let m be an even positive integer. Assume that {a1, a2, . . . , am} and {b1, b2, . . . , bm} are two complete sets of residue classes modulo m. Prove that {a1 + b1, a2 + b2, . . . , am + bm} is not a complete set of residue classes. Proof: We approach indirectly by assuming that it is. Then we have 1 + 2 + · · · + n ≡(a1 + b1) + (a2 + b2) + · · · + (am + bm) ≡(a1 + a2 + · · · + am) + (b1 + b2 + · · · + bm) ≡2(1 + 2 + · · · + m) (mod m), implying that 1 + 2 + · · · + m ≡0 (mod m), or m | m(m+1) 2 , which is not true for even integers m. Hence our assumption was wrong. □ Example 1.27. Let a be a positive integer. Determine all the positive integers m such that {a · 1, a · 2, a · 3, . . . , a · m} is a set of complete residue classes modulo m. Solution: The answer is the set of positive integers m that are relatively prime to a. Let Sm denote the given set. First we show that Sm is a complete set of residue classes if gcd(a, m) = 1. Because this set has exactly m elements, it suffices to show that elements in the set are not congruent to each other modulo m. Assume to the contrary that ai ≡aj (mod m) for some 1 ≤i < j ≤n. Because gcd(a, m) = 1, by Corollary 1.20, we have i ≡j (mod m), which is impossible since |i −j| < m. Hence our assumption was wrong and Sm is a complete set of residue classes modulo m. On the other hand, if g = gcd(a, m) > 1, then a = a1g and m = m1g, where m1 is a positive integer less than n. We have am1 ≡a1m1g ≡a1m ≡am ≡0 (mod m). Hence Sm is not a complete set of residue classes. □ Similarly, we can show the following result. Proposition 1.24. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. Assume that S is a complete set of residue classes modulo m. The set T = aS + b = {as + b | s ∈S} is also a complete set of residue classes modulo n. www.riazisara.ir 26 104 Number Theory Problems Now we are better equipped to discuss linear congruence equations a bit fur-ther. Proposition 1.25. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. There exist integers x such that ax ≡b (mod m), and all these integers form exactly one residue class modulo m. Proof: Let {c1, c2, . . . , cm} be a complete set of residue classes modulo m. By Proposition 1.24, {ac1 −b, ac2 −b, . . . , acm −b} is also a complete set of residue classes. Hence there exists ci such that ac1 −b ≡ 0 (mod m), or c1 is a solution to the congruence equation ax ≡b (mod m). It is easy to see that all the numbers congruent to c1 modulo m also satisfy the congruence equation. On the other hand, if both x and x′ satisfy the equation, we have ax ≡ax′ (mod m). By Corollary 1.20, we have x ≡x′ (mod m). □ In particular, setting b = 1 in Proposition 1.25 shows that if gcd(a, m) = 1, then there is x such that ax ≡1 (mod m). We call such x the inverse of a modulo m, denoted by a−1 or 1 a (mod m). Because all such numbers form exactly one residue class modulo m, the inverse of a is uniquely determined (or well defined) modulo m for all integers relatively prime to m. Now we are ready to prove Wilson’s theorem. Theorem 1.26. [Wilson’s Theorem] For any prime p, (p−1)! ≡−1 (mod p). Proof: The property holds for p = 2 and p = 3, so we may assume that p ≥5. Let S = {2, 3, . . . , p −2}. Because p is prime, for any s in S, s has a unique inverse s′ ∈{1, 2, . . . , p −1}. Moreover, s′ ̸= 1 and s′ ̸= p −1; hence s′ ∈S. In addition, s′ ̸= s; otherwise, s2 ≡1 (mod p), implying p | (s −1) or p | (s + 1), which is not possible, since s + 1 < p. It follows that we can group the elements of S in p−3 2 distinct pairs (s, s′) such that ss′ ≡1 (mod p). Multiplying these congruences gives (p −2)! ≡1 (mod p) and the conclusion follows. □ Note that the converse of Wilson’s theorem is true, that is, if (n −1)! ≡−1 (mod n) for an integer n ≥2, then n is a prime. Indeed, if n were equal to n1n2 for some integers n1, n2 ≥2, we would have n1 | 1 · 2 · · · n1 · · · (n −1) + 1, which is not possible. This provides us a new way to determine whether a number is prime. (However, this is not a very practical way, since for large n, (n −1)! is huge!) In most situations, there are no major differences in picking a particular com-plete set of residue classes to solve a particular problem. Here is a distinct exam-ple. www.riazisara.ir 1. Foundations of Number Theory 27 Example 1.28. [MOSP 2005, Melanie Wood] At each corner of a cube, an in-teger is written. A legal transition of the cube consists in picking any corner of the cube and adding the value written at that corner to the value written at some adjacent corner (that is, pick a corner with some value x written at it, and an ad-jacent corner with some value y written at it, and replace y by x + y). Prove that there is a finite sequence of legal transitions of the given cube such that the eight integers written are all the same modulo 2005. We present two solutions. Notice that if we take a legal transition and perform it 2004 times, then modulo 2005, this is the same as replacing y by y −x. Call such a repetition of a single legal transition 2004 times a super transition. First Solution: Look at the integers modulo 2005, and replace them with residue classes 1, 2, . . . , 2005. If all of the residue classes are the same, then we need no transitions. Otherwise, there is an edge with residue classes N and M with 1 ≤N < M ≤2005. Performing a super transition, we can replace M by M −N, which is a residue class, since 1 ≤M −N ≤2005. Since N ≥1, this reduces the sum of the residue classes by at least 1. Because the sum of the residue classes is always at least 8, by repeating this process, we will eventually get to a state in which all of the residue classes are the same. □ Note that the proof would not work well if we replaced the numbers with residue classes 0, 1, . . . , 2004. As in the case N = 0, the sum of the residue classes is not decreased. Second Solution: Look at the integers all modulo 2005. They are congruent to some set of positive integers modulo 2005. Performing a super transition on an edge is the same (modulo 2005) as performing a step of the Euclidean algorithm on the two numbers of the edge. Performing the Euclidean algorithm on a pair of positive integers will make them equal to the greatest common divisor of the two original integers after a finite number of steps. Thus, we can make two numbers of an edge congruent modulo 2005 after a finite number of super transitions. First we do this on all edges going in one direction, then on all the edges going in another direction, and then on all the edges going in the third direction. After this, we see that all the integers written at corners are congruent modulo 2005. □ Fermat’s Little Theorem and Euler’s Theorem From the last few results, we note that for a given positive integer m, it is useful to consider the congruence classes that are relatively prime to m. For any positive integer m we denote by ϕ(m) the number of all positive integers n less than m that are relatively prime to m. The function ϕ is called Euler’s totient function. It is clear that ϕ(1) = 1 and for any prime p, ϕ(p) = p −1. Moreover, if n is a positive integer such that ϕ(n) = n −1, then n is a prime. www.riazisara.ir 28 104 Number Theory Problems A set S of integers is also called a reduced complete set of residue classes modulo m if for each i with 0 ≤i ≤n −1 and gcd(i, m) = 1, there is an element s ∈S such that i ≡s (mod m). It is clear that a reduced complete set of residue classes modulo m consists of ϕ(m) elements. Proposition 1.27. Let m be a positive integer. Let a be an integer relatively prime to m. Assume that S is a reduced complete set of residue classes modulo m. Set T = aS = {as | s ∈S}, which is also a reduced complete set of residue classes modulo n. The proof is similar to that of Proposition 1.24, and we leave it to the reader. Proposition 1.27 allows us to establish two of the most famous theorems in num-ber theory. Theorem 1.28. [Euler’s Theorem] Let a and m be relatively prime positive integers. Then aϕ(m) ≡1 (mod m). Proof: Consider the set S = {a1, a2, . . . , aϕ(m)} consisting of all positive inte-gers less than m that are relatively prime to m. Because gcd(a, n) = 1, it follows from Proposition 1.26 that {aa1, aa2, . . . , aaϕ(m)} is another reduced complete set of residue classes modulo n. Then (aa1)(aa2) · · · (aaϕ(n)) ≡a1a2 · · · aϕ(n) (mod m). Using that gcd(ak, n) = 1, k = 1, 2, . . . , ϕ(n), the conclusion now follows. □ Setting m = p as prime, Euler’s theorem becomes Fermat’s little theorem. Theorem 1.29. [Fermat’s Little Theorem] Let a be a positive integer and let p be a prime. Then a p ≡a (mod p). Proof: We present an alternative proof independent of Euler’s theorem. We induct on a. For a = 1 everything is clear. Assume that p | (a p −a). Then (a + 1)p −(a + 1) = (a p −a) + p−1  k=1 p k ak. Using the fact that p | p k  for 1 ≤k ≤p −1 (Corollary 1.10) and the inductive hypothesis, it follows that p divides (a +1)p −(a +1); that is, (a +1)p ≡(a +1) (mod p). □ www.riazisara.ir 1. Foundations of Number Theory 29 Clearly, Fermat’s little theorem is a special case of Euler’s theorem. But with a few more properties on the Euler function ϕ that we will develop, we can derive Euler’s theorem from Fermat’s little theorem Note also another form of Fermat’s little theorem: Let a be a positive integer relatively prime to prime p. Then a p−1 ≡1 (mod p). Next, we present a few examples involving these two important theorems. Example 1.29. Let p be a prime. Prove that p divides abp −ba p for all integers a and b. Proof: Note that abp −ba p = ab(bp−1 −a p−1). If p | ab, then p | abp −ba p; if p ∤ab, then gcd(p, a) = gcd(p, b) = 1, and so bp−1 ≡a p−1 ≡1 (mod p), by Fermat’s little theorem. Hence p | bp−1 −a p−1, implying that p | abp −ba p. Therefore, p | abp −ba p for all p. □ Example 1.30. Let p ≥7 be a prime. Prove that the number 11 . . . 1   p−1 1’s is divisible by p. Proof: We have 11 . . . 1   p−1 1’s = 10p−1 −1 9 , and the conclusion follows from Fermat’s little theorem. (Note also that gcd(10, p) = 1.) □ Example 1.31. Let p be a prime with p > 5. Prove that p8 ≡1 (mod 240). Proof: Note that 240 = 24 · 3 · 5. By Fermat’s little theorem, we have p2 ≡1 (mod 3) and p4 ≡1 (mod 5). Because a positive integer is relatively prime to 24 if and only if it is odd, ϕ(24) = 23. By Euler’s theorem, we have p8 ≡1 (mod 16). Therefore, p8 ≡1 (mod m) for m = 3, 5, and 16, implying that p8 ≡1 (mod 240). □ Note that this solution indicates that we can establish Euler’s theorem by Fer-mat’s little theorem. Further, it is not difficult to check that n4 ≡1 (mod 16) for n ≡±1, ±3, ±5, ±7 (mod 16) (see Example 1.24 (5)). Hence we can improve the result to p4 ≡1 (mod 240) for all primes p > 5. www.riazisara.ir 30 104 Number Theory Problems Example 1.32. Prove that for any even positive integer n, n2 −1 divides 2n! −1. Proof: Let m = n + 1. We need to prove that m(m −2) divides 2(m−1)! −1. Because ϕ(m) divides (m −1)! we have (2ϕ(m) −1) | (2(m−1)! −1) and from Euler’s theorem, m | (2ϕ(m) −1). It follows that m | (2(m−1)! −1). Similarly, (m −2) | (2(m−1)! −1). Because m is odd, gcd(m, m −2) = 1 and the conclusion follows. □ For a given positive integer m, let {a1, a2, . . . , aϕ(m)} be a reduced complete set of residue classes modulo m. By the existence and uniqueness of inverses, it is not difficult to see that the set of their inverses, denoted by {a−1 1 , a−1 2 , . . . , a−1 ϕ(m)} or  1 a1 , 1 a2 , . . . , 1 aϕ(m)  , is also a reduced complete set of residue classes modulo m. One might attempt to generalize Wilson’s theorem by pairing residue classes that are inverses of each other. This approach fails, since there are residue classes other than 1 and −1 (or m −1) that are inverses of themselves. (In the proof of Wilson’s theorem, there are only two possible values for s, namely s = 1 or s = p −1, such that s2 ≡1 (mod p).) For example, 62 ≡1 (mod 35) for m = 35. Let m be a positive integer, and let a be an integer relatively prime to m. Assume that b = na is a multiple of a; that is, n = b a is an integer. From a−1a ≡1 (mod p), we have n ≡a−1an ≡a−1b (mod m). This means that n = a b under the usual arithmetic meaning identifies with n ≡1 a · b (mod m). This allows us to choose the order of operations to our advantage. Example 1.33. [IMO 2005] Consider the sequence a1, a2, . . . defined by an = 2n + 3n + 6n −1 for all positive integers n. Determine all positive integers that are relatively prime to every term of the sequence. First Solution: The answer is 1. It suffices to show that every prime p divides an for some positive integer n. Note that both p = 2 and p = 3 divide a2 = 22 + 32 + 62 −1 = 48. Assume now that p ≥5. By Fermat’s little theorem, we have 2p−1 ≡3p−1 ≡ 6p−1 ≡1 (mod p). Then 3 · 2p−1 + 2 · 3p−1 + 6p−1 ≡3 + 2 + 1 ≡6 (mod 6), or 6(2p−2 + 3p−2 + 6p−2 −1) ≡0 (mod p); that is, 6ap−2 is divisible by p. Because p is relatively prime to 6, ap−2 is divisible by p, as desired. www.riazisara.ir 1. Foundations of Number Theory 31 Second Solution: If we use the notation of inverse, the proof can be written as 6ap−2 ≡6(2p−2 + 3p−2 + 6p−2 −1) ≡6 1 2 + 1 3 + 1 6 −1 ≡0 (mod p), for every prime p greater than 5. □ Example 1.34. Find an infinite nonconstant arithmetic progression of positive integers such that each term is not a sum of two perfect cubes. Solution: Assume that the desired arithmetic progression is {a, a + d, a + 2d, . . . }. We are basically considering all integers in the residue class a modulo d. We want to limit the number of residue classes that are cubes modulo d. In this way, we limit the number of residue classes that can be written as the sum of cubes modulo d. We first look for d such that a3 ≡1 (mod d) for all integers a. Fermat’s little theorem states that a p−1 ≡1 (mod p) for prime p and integers a relatively prime to p. If we set p −1 = 3, we have p = 4, which is not a prime. Hence we cannot apply Fermat’s little theorem directly. On the other hand, if we set p = 7, then a6 ≡1 (mod 7) for integers relatively prime to 7. It not difficult to check that the possible residue classes for a3 modulo 7 are 0, 1, −1 (or 6). Hence, modulo 7, the possible residue classes for a3 + b3 are 0, 1, −1, 2, −2. Therefore, {3, 3+7, 3+2·7, . . . } and {4, 4+7, 4+2·7, . . . } are two sequences satisfying the conditions of the problem. □ By noting that ϕ(9) = 6, we can also find sequences of the form {a, a + 9, a + 2 · 9, . . . }. We leave the details to the reader. (Compare to Example 1.24 (4).) Example 1.35. [IMO 2003 shortlist] Determine the smallest positive integer k such that there exist integers x1, x2, . . . , xk with x3 1 + x3 2 + · · · + x3 k = 20022002. Solution: The answer is k = 4. We first show that 20022002 is not a sum of three cubes. To restrict the number of cubes modulo n, we would like to have ϕ(n) to be a multiple of 3. Again, consider n = 7. But adding three cubes modulo 7 gives too many residue classes (since 7 is too small). We then consider n = 9 with ϕ(9) = 6. Because 2002 ≡4 (mod 9) and 20023 ≡43 ≡1 (mod 9), it follows that 20022002 ≡(20023)667 · 2004 ≡4 (mod 9). On the other hand, x3 ≡0, ±1 (mod 9) for integers x. We see that x3 1 +x3 2 +x3 3 ̸≡ 4 (mod 9). www.riazisara.ir 32 104 Number Theory Problems It remains to show that 20022002 is a sum of four cubes. Starting with 2002 = 103 + 103 + 13 + 13 and using 2002 = 667 · 3 + 1 once again, we find that 20022002 = 2002 · (2002667)3 = (10 · 2002667)3 + (10 · 2002667)3 + (2002667)3 + (2002667)3. □ Fermat’s little theorem provides a good criterion to determine whether a num-ber is composite. But the converse is not true. For example, 3 · 11 · 17 divides a3·11·17 −a, since 3, 11, 17 each divide a3·11·17 −a (for instance, if 11 did not divide a, then from Fermat’s little theorem, we have 11 | (a10 −1); hence 11 | (a10·56 −1), i.e., 11 | (a561 −a) and 561 = 3 · 11 · 17). The composite integers n satisfying an ≡a (mod n) for any integer a are called Carmichael numbers. There are also even Carmichael numbers, for ex-ample n = 2 · 73 · 1103. Let a and m be relatively prime positive integers. Setting b = 1 in Corollary 1.23 leads to an interesting result. By Euler’s theorem, there exist positive integers x such that ax ≡1 (mod m). We say that a has order d modulo m, denoted by ordm(a) = d, if d is the smallest positive integer such that ad ≡1 (mod m). By Euler’s theorem, ordm(a) = d ≤ϕ(m). If x is a positive integer such that ax ≡1 (mod m), then by Corollary 1.23, agcd(x,d) ≡1 (mod m). Since gcd(x, d) ≤d, by the minimality of d, we must have gcd(x, d) = d. Hence d divides x. We have established the following property. Proposition 1.30. A positive integer x is such that ax ≡1 (mod m) if and only if x is a multiple of the order of a modulo m. For a pair of relatively prime positive integers a and m, it is not true that there always exists a positive integer s such that as ≡−1 (mod n). (For example, a = 2 and m = 7.) Assume that there exists a perfect power of a that is congruent to −1 modulo m, and assume that s is the least such integer. We have ordm(a) = 2s. Indeed, a2s ≡1 (mod m), so d divides 2s. If d < 2s, then 22s−d ≡−1 (mod m) violates the minimality assumption on s. Furthermore, if t is an integer such that at ≡−1 (mod m), then t is a multiple of s. Because a2t ≡1 (mod m), it follows that d = 2s divides 2t, and so s divides t. It is then clear that t must an odd multiple of s; that is, at ≡ −1 if t is an odd mulptiple of s; 1 if t is an even mulptiple of s. Example 1.36. [AIME 2001] How many positive integer multiples of 1001 can be expressed in the form 10 j −10i, where i and j are integers and 0 ≤i < j ≤99? www.riazisara.ir 1. Foundations of Number Theory 33 Solution: Because 10 j −10i = 10i(10 j−i −1) and 1001 = 7 · 11 · 13 is relatively prime to 10i, it is necessary to find i and j such that the 10 j−i −1 is divisible by the primes 7, 11, and 13. Notice that 1001 = 7 · 11 · 13; that is, 103 ≡−1 (mod 1001). It is easy to check that ord1001(10) = 6. By Proposition 1.30, 10i(10 j−i −1) is divisible by 1001 if and only if j −i = 6n for some positive integer n. Thus it is necessary to count the number of integer solutions to i + 6n = j, where j ≤99, i ≥0, and n > 0. For each n = 1, 2, 3, . . . , 15, there are 100−6n suitable values of i (and j), so the number of solutions is 94 + 88 + 82 + · · · + 4 = 784. □ Euler’s Totient Function We discuss some useful properties of Euler’s totient function ϕ. First of all, it is not difficult to see the following: Proposition 1.31. Let p be a prime, and let a be a positive integer. Then ϕ(pa) = pa −pa−1. Next, we show that ϕ is multiplicative: Proposition 1.32. Let a and b be two relatively prime positive integers. Then ϕ(ab) = ϕ(a)ϕ(b). Proof: : Arrange the integers 1, 2, . . . , ab into an a × b array as follows: 1 2 · · · a a + 1 a + 2 · · · 2a . . . . . . . . . . . . a(b −1) + 1 a(b −1) + 2 · · · ab. Clearly, there are ϕ(ab) numbers in the above table that are relatively prime to ab. On the other hand, there are ϕ(a) columns containing those elements in the table relatively prime to a. Each of those columns is a complete set of residues modulo b, by Proposition 1.24. Hence there are exactly ϕ(b) elements in each of those columns that are relatively prime to b. Therefore, there are ϕ(a)ϕ(b) numbers in the table that are relatively prime to ab. Hence ϕ(ab) = ϕ(a)ϕ(b) for relatively prime integers ab. □ www.riazisara.ir 34 104 Number Theory Problems Theorem 1.33. If n = pα1 1 · · · pαk k is the prime factorization of n > 1, then ϕ(n) = n  1 −1 p1 · · ·  1 −1 pk . First Proof: This follows directly from Propositions 1.31 and 1.32. □ Second Proof: We employ the inclusion and exclusion principle. Set Ti = {d : d ≤n and pi|d}, for i = 1, . . . , k. It follows that T1 ∪· · · ∪Tk = {m : m ≤n and gcd(m, n) > 1}. Hence ϕ(n) = n −|T1 ∪· · · ∪Tk| = n − k  i=1 |Ti| +  1≤i< j≤k |Ti ∩Tj| −· · · + (−1)k|T1 ∩· · · ∩Tk|. We have |Ti| = n pi , |Ti ∩Tj| = n pi p j , . . . , |T1 ∩· · · ∩Tk| = n p1 · · · pk . Finally, ϕ(n) = n  1 − n  i=1 1 pi +  1≤i< j≤k 1 pi p j −· · · + (−1)k 1 p1 · · · pk  = n  1 −1 p1 · · ·  1 −1 pk . □ Based on Theorem 1.33, we can establish Euler’s theorem from Fermat’s lit-tle theorem. Indeed, let n = pα1 1 · · · pαk k be the prime factorization of n. We have a pi−1 ≡1 (mod pi), hence a pi(pi−1) ≡1 (mod p2 i ), a p2 i (pi−1) ≡1 (mod p3 i ), . . . , a pαi −1 i (pi−1) ≡1 (mod pαi i ). That is, aϕ(pαi i ) ≡1 (mod pαi i ), i = 1, . . . , k. Applying this property to each prime factor, the conclusion fol-lows. Theorem 1.34. [Gauss] For any positive integer n,  d|n ϕ(d) = n. www.riazisara.ir 1. Foundations of Number Theory 35 Proof: We consider the rational numbers 1 n , 2 n , . . . , n n . Clearly, there are n numbers in the list. We obtain a new list by reducing each number in the above list to the lowest terms; that is, express each fraction as a quotient of relatively prime integers. The denominators of the numbers in the new list will all be divisors of n. If d | n, exactly ϕ(d) of the numbers in the list will have d as their denominator. (This is the meaning of lowest terms!) Hence, there are  d|n ϕ(d) in the new list. Because the two lists have the same number of terms, we obtain the desired result. □ Example 1.37. Let n be a positive integer. (1) Find the sum of all positive integers less than n and relatively prime to n. (2) Find the sum of all positive integers less than 2n and relatively prime to n. Solution: The answers are nϕ(n) 2 and 2nϕ(n), respectively. Let S1 =  d<n gcd(d,n)=1 d and S2 =  d<2n gcd(d,n)=1 d. Let d1 < d2 < · · · < dϕ(n) be the numbers less than n and relatively prime to n. Note that gcd(d, n) = 1 if and only if gcd(n −d, n) = 1. We deduce that d1 + dϕ(n) = n, d2 + dϕ(n)−1 = n, . . . , dϕ(n) + d1 = n, implying that S1 = nϕ(n) 2 . On the other hand,  n<d<2n gcd(d,n)=1 d =  d 1, (−1)k if n = p1 · · · pk, where p1, . . . , pk are distinct primes. For example, µ(2) = −1, µ(6) = 1, µ(12) = µ(22 · 3) = 0. Theorem 1.35. The M¨ obius function µ is multiplicative. Proof: Let m, n be positive integers such that gcd(m, n) = 1. If p2 | m for some p > 1, then p2 | mn and so µ(m) = µ(mn) = 0 and we are done. Consider now m = p1 · · · pk, n = q1 · · · qh, where p1, . . . , pk, q1, . . . , qh are distinct primes. Then µ(m) = (−1)k, µ(n) = (−1)h, and mn = p1 · · · pkq1 · · · qh. It follows that µ(mn) = (−1)k+h = (−1)k(−1)h = µ(m)µ(n). □ For an arithmetic function f we define its summation function F by F(n) =  d|n f (d). The connection between f and F is given by the following result. Theorem 1.36. If f is multiplicative, then so is its summation function F. Proof: Let m and n be positive relatively prime integers and let d be a divisor of mn. Then d can be uniquely represented as d = kh, where k | m and h | n. www.riazisara.ir 1. Foundations of Number Theory 37 Because gcd(m, n) = 1, we have gcd(k, h) = 1, so f (kh) = f (k) f (h). Hence F(mn) =  d|mn f (d) =  k|m h|n f (k) f (h) =  k|m f (k)   h|h f (h)  = F(m)F(n). □ Note that if f is a multiplicative function and n = pα1 1 · · · pαk k , then  d|n µ(d) f (d) = (1 −f (p1)) · · · (1 −f (pk)). Indeed, the function g(n) = µ(n) f (n) is multiplicative; hence from Theorem 1.36, so is its summation function G. Then G(n) = G(pα1 1 ) · · · G(pαk k ) and G(pαi i ) =  d|pαi i µ(d) f (d) = µ(1) f (1) + µ(pi) f (pi) = 1 −f (pi), and the conclusion follows. Theorem 1.37. [M¨ obius inversion formula] Let f be an arithmetic function and let F be its summation function. Then f (n) =  d|n µ(d)F n d  . Proof: We have  d|n µ(d)F n d  =  d|n µ(d) ⎛ ⎝ c| n d f (c) ⎞ ⎠=  d|n ⎛ ⎝ c| n d µ(d) f (c) ⎞ ⎠ =  c|n ⎛ ⎝ d| n c µ(d) f (c) ⎞ ⎠=  c|n f (c) ⎛ ⎝ d| n c µ(d) ⎞ ⎠= f (n), since for n c > 1 we have  d| n c µ(d) = 0. We have used the fact that  (d, c)| d|n and c|n d  =  (d, c)| c|n and d|n c  . □ Theorem 1.38. Let f be an arithmetic function and let F be its summation function. If F is multiplicative, then so is f . www.riazisara.ir 38 104 Number Theory Problems Proof: Let m, n be positive integers such that gcd(m, n) = 1 and let d be a divisor of mn. Then d = kh, where k | m, h | n, and gcd(k, h) = 1. Applying the M¨ obius inversion formula it follows F(mn) =  d|mn µ(d)F mn d  =  k|m h|n µ(kh)F mn kh =  k|m h|n µ(k)µ(h)F m k  F n h =  k|m µ(k)F m k   h|n µ(h)F n h  = f (m) f (n). □ We leave it to the reader to show that functions τ, σ, and ϕ are indeed mul-tiplicative. We also encourage the reader to redevelop some properties of these functions by the general results we developed in this section. Linear Diophantine Equations An equation of the form a1x1 + · · · + anxn = b, (∗) where a1, a2, . . . , an, b are fixed integers, is called a linear Diophantine equa-tion. We assume that n ≥1 and that coefficients a1, . . . , an are all different from zero. The main result concerning linear Diophantine equations is the following gen-eralization of Theorem 1.7. (B´ ezout’s identity). Theorem 1.39. The equation (∗) is solvable if and only if gcd(a1, . . . , an) | b. In case of solvability, all integer solutions to (∗) can be expressed in terms of n−1 integral parameters. Proof: Let d = gcd(a1, . . . , an). If b is not divisible by d, then (∗) is not solvable, since for any integers x1, . . . , xn the left-hand side of (∗) is divisible by d and the right-hand side is not. If d | b, then we obtain the equivalent equation a′ 1x1 + · · · + a′ nxn = b′, www.riazisara.ir 1. Foundations of Number Theory 39 where a′ i = ai/d for i = 1, . . . , n and b′ = b/d. Clearly, we have gcd(a′ 1, . . . , a′ n) = 1. We use induction on the number n of the variables. In the case n = 1 the equation has the form x1 = b or −x1 = b, and thus the unique solution does not depend on any parameter. We now assume that n ≥2 and that the solvability property holds for all linear equations in n −1 variables. Our goal is to prove the solvability of equations in n variables. Set dn−1 = gcd(a1, . . . , an−1). Then any solution (x1, . . . , xn) of (1) satisfies the congruence a1x1 + a2x2 + · · · + anxn ≡b (mod dn−1), which is equivalent to anxn ≡b (mod dn−1). (†) Multiplying both sides of (†) by aϕ(dn−1)−1 n and taking into account that aϕ(dn−1) n ≡1 (mod dn−1), we obtain xn ≡c (mod dn−1), where c = aϕ(dn−1)−1 n b. It follows that xn = c + dn−1tn−1 for some integer tn−1. Substituting in (∗) and rearranging yields the equation in (n −1) variables a1x1 + · · · + an−1xn−1 = b −anc −an−1dn−1tn−1. It remains to show that dn−1 | (b −anc −an−1dn−1tn−1), which is equivalent to anc ≡b (mod dn−1). The last relation is true because of the choice of c. Therefore we can divide the last equation by dn−1, and obtain a′ 1x1 + · · · + a′ n−1xn−1 = b′, (‡) where a′ i = ai/dn−1 for i = 1, . . . , n −1 and b′ = (b −anc)/dn−1 −antn−1. Because gcd(a′ 1, . . . , a′ n−1) = 1, by the induction hypothesis the equation (‡) is solvable for each integer tn−1 and its solutions can be written in terms of n −2 integral parameters. If we add to these solutions xn = c + dn−1tn−1, we obtain solutions to (∗) in terms of n −1 parameters. □ Corollary 1.40. Let a1, a2 be relatively prime integers. If (x0 1, x0 2) is a solution to the equation a1x1 + a2x2 = b, then all of its solutions are given by  x1 = x0 1 + a2t, x2 = x0 2 −a1t, for every integer t. www.riazisara.ir 40 104 Number Theory Problems Example 1.38. Determine all triples (x, y, z) of integers satisfying the equation 3x + 4y + 5z = 6. Solution: We have 3x + 4y ≡1 (mod 5); hence 3x + 4y = 1 + 5s for some integer s. A solution to this equation is x = −1 + 3s, y = 1 −s. Applying Corollary 1.40, we obtain x = −1 + 3s + 4t and y = 1 −s −3t, for some integer t, and substituting back into the original equation yields z = 1 −s. Hence all solutions are (x, y, z) = (−1 + 3s + 4t, 1 −s −3t, 1 −s), for all pairs of integers s and t, □ Example 1.39. Let n be a positive integer. Suppose that there are 666 ordered triples (x, y, z) of positive integers satisfying the equation x + 8y + 8z = n. Find the maximum value of n. Solution: The answer is 303. Write n = 8a + b, where a and b are integers with 0 ≤b < 8. Since x ≡n ≡b (mod 8), the possible values of x are b, 8 + b, . . . , 8(a −1) + b. For x = b + 8i, where 0 ≤i ≤a −1, 8(y + z) = 8(a −i) or y + z = a −i, which admits a −i −1 ordered pairs (y, z) of positive integer solutions, namely, (1, a −i −1), . . . , (a −i −1, 1). Hence there are a−1  i=0 (a −i −1) = a−1  i=0 i = a(a −1) 2 ordered triples satisfying the conditions of the problem. Solving a(a−1) 2 = 666 gives a = 37. Therefore, the maximum value for n is equal to 37 · 8 + 7 = 303, obtained by setting b = 7. □ Numerical Systems The fundamental result in this subsection is given by the following theorem: Theorem 1.41. Let b be an integer greater than 1. For any integer n ≥1 there is a unique system (k, a0, a1, . . . , ak) of integers such that 0 ≤ai ≤b −1, i = 0, 1, . . . , k, ak ̸= 0, and n = akbk + ak−1bk−1 + · · · + a1b + a0. (∗) www.riazisara.ir 1. Foundations of Number Theory 41 Proof: For the existence, we apply repeatedly the division algorithm: n = q1b + r1, 0 ≤r1 ≤b −1; q1 = q2b + r2, 0 ≤r2 ≤b −1; . . . qk−1 = qkb + rk, 0 ≤rk ≤b −1; where qk is the last nonzero quotient. Let q0 = n, a0 = n −q1b, a1 = q1 −q2b, . . . , ak−1 = qk−1 −qkb, ak = qk. Then k  i=0 aibi = k−1  i=0 (qi −qi+1b)bi + qkbk = q0 + k  i=1 qibi − k  i=1 qibi = q0 = n. For uniqueness, assume that n = c0 + c1b + · · · + cnbn is another such represen-tation. If h ̸= k, for example h > k, then n ≥bh ≥bk+1. But n = a0 + a1b + · · · + akbk ≤(b −1)(1 + b + · · · + bk) = bk+1 −1 < bk+1, a contradiction. If h = k, then a0 + a1b + · · · + akbk = c0 + c1b + · · · + ckbk, and so b | (a0 −c0). On the other hand, |a0 −c0| < b; hence a0 = c0, Therefore a1 + a2b + · · · + akbk−1 = c1 + c2b + · · · + ckbk−1. By repeating the above procedure, it follows that a1 = c1, a2 = c2, . . . , and ak = ck. □ Relation (∗) is called the base-b representation of n and is denoted by n = akak−1 . . . a0(b). The usual decimal representation corresponds to b = 10 and we write only n = akak−1 . . . a0 instead. (For example, 4567 = 4567(10).) Example 1.40. Let xy and yx be two 2-digit integers. Prove that their sum is composite. Proof: Since xy = 10x + y and yx = 10y+x, their sum is equal to 11x +11y = 11(x + y), a composite number. □ www.riazisara.ir 42 104 Number Theory Problems Example 1.41. [AHSME 1973] In the following equation, each of the letters represents uniquely a different digit in base ten: (Y E) · (M E) = T T T. Determine the sum E + M + T + Y. Solution: Because T T T = T ·111 = T ·3·37, one of Y E and M E is 37, implying that E = 7. But T is a digit and T ·3 is a two-digit number ending with 7, and so it follows that T = 9 and T T T = 999 = 27·37, and so E+M+T +Y = 2+3+7+9 = 21. □ Example 1.42. [AIME 2001] Find the sum of all positive two-digit integers that are divisible by each of their digits. Solution: Let ab denote an integer with the required property. Then 10a + b must be divisible by both a and b. It follows that b must be divisible by a, and that 10a must be divisible by b. The former condition requires that b = ka for some positive integer k, and the latter condition implies that k = 1 or k = 2 or k = 5. Thus the requested two-digit numbers are 11, 22, . . . , 99, 12, 24, 36, 48, and 15. Their sum is 11 · 45 + 12 · 10 + 15 = 630. □ Example 1.43. [AMC12A 2002] Some sets of prime numbers, such as {7, 83, 421, 659}, use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes can have? Solution: The answer is 207. Note that digits 4, 6, and 8 cannot appear in the units digit. Hence the sum is at least 40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207. On the other hand, this value can be obtained with the set {2, 5, 7, 43, 61, 89}. □ Example 1.44. Write 101011(2) in base 10, and write 1211 in base 3. Solution: We have 1010011(2) = 1 · 26 + 0 · 25 + 1 · 24 + 0 · 23 + 0 · 22 + 1 · 2 + 1 = 64 + 16 + 2 + 1 = 83. Dividing by 3 successively, the remainders give the digits of the base-3 represen-tation, beginning with the last. The first digit is the last nonzero quotient. We can www.riazisara.ir 1. Foundations of Number Theory 43 arrange the computations as follows: 1211 3 1209 403 3 2 402 134 3 1 132 44 3 2 42 14 3 2 12 4 3 2 3 1 1 Hence 1211 = 1122212(3). □ Example 1.45. The product of seven and the six-digit number abcdef is equal to the product of six and the six-digit number def abc. Find these two six-digit numbers. Solution: Let x and y denote the three-digit numbers abc and def , respectively. Then abcdef = 1000x+y and def abc = 1000y+x. By the given conditions, we have 7(1000x+y) = 6(1000y+x), or 6994x = 5993y. Since gcd(6994, 5993) = gcd(5993, 1001) = gcd(1001, 13) = 13, we have 538x = 461y, and so the two numbers are 461538 and 538461. □ Example 1.46. [AMC12A 2005] A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. For example, after traveling one mile the odometer changed from 000039 to 000050. If the odometer now reads 002005, how many miles has the car actually traveled? Solution: Because the odometer uses only 9 digits, it records mileage in base-9 numerals, except that its digits 5, 6, 7, 8, and 9 represent the base-9 digits 4, 5, 6, 7, and 8. Therefore the mileage is 2004(9) = 2 · 93 + 4 = 2 · 729 + 4 = 1462. □ Example 1.47. Prove that the number 11 . . . 1(9) in base 9 is triangular; that is, it is the sum of the first k positive integers for some positive integer k. Proof: Indeed 11 . . . 1   n 1’s (9) = 9n−1 + 9n−2 + · · · + 9 + 1 = 9n −1 9 −1 = 1 2 · 3n −1 2 · 3n + 1 2 = 1 + 2 + · · · + 3n −1 2 . Thus it is a triangular number. □ www.riazisara.ir 44 104 Number Theory Problems Example 1.48. Determine all positive integers n such that 11111(n) is a perfect square. Solution: The answer is n = 3. We have 11111(n) = n4 + n3 + n2 + n + 1. If n is even, then n2 + n 2 and n2 + n 2 +1 are two consecutive integers. We have  n2 + n 2 2 = n4 + n3 + n2 4 < n4 + n3 + n2 + n + 1 <  n2 + n 2 + 1 2 . Hence 11111(n) is not a perfect square for even positive integers n. If n is odd, then n2 + n 2 −1 2 and n2 + n 2 + 1 2 are integers. Clearly, we have  n2 + n 2 −1 2 2 < n4 + n3 + n2 + n + 1. Note that  n2 + n 2 + 1 2 2 = n4 + n3 + 5n2 4 + n 2 + 1 4 = n4 + n3 + n2 + n + 1 + n2 −2n −3 4 = n4 + n3 + n2 + n + 1 + (n −3)(n + 1) 4 . For odd integers n greater than 3, 11111(n) is strictly between two consecutive perfect squares, namely,  n2 + n 2 −1 2 2 and  n2 + n 2 + 1 2 2 . Hence 11111(n) is not a perfect square for any positive integers other than 3. For n = 3, we have 11111(3) = 121 = 112. □ In the last example, we showed that an integer is not a perfect square by plac-ing the integer between two consecutive perfect squares. This method works be-cause integers are discrete. Such methods will hardly work for real numbers, since there are no holes in between real numbers. This is a very useful method in solving Diophantine equations. In certain numerical systems, the base does not have to be constant. Here are two examples. www.riazisara.ir 1. Foundations of Number Theory 45 Proposition 1.42. Every positive integer k has a unique factorial base expan-sion ( f1, f2, f3, . . . , fm), meaning that k = 1! · f1 + 2! · f2 + 3! · f3 + · · · + m! · fm, where each fi is an integer, 0 ≤fi ≤i, and fm > 0. Proof: Note that there exists a unique positive integer m1 such that m1! ≤k < (m1 + 1)!. By the division algorithm, we can write k = m1! fm1 + r1 for some positive integer fm1 and some integer r1 with 0 ≤r1 < m1!. Because k < (m + 1)! = m! · (m + 1), it follows that fm1 ≤m. Repeating this process, we can then write r1 = m2! fm2 + r2, with m2 the unique positive integer with m2! ≤r1 < (m2 + 1)!, 1 ≤fm2 ≤m2, and 0 ≤r2 < m2!. Keeping this process on r2, and so on, we obtain a unique factorial base expansion of k. □ Proposition 1.43. Let F0 = 1, F1 = 1, and Fn+1 = Fn + Fn−1 for every posi-tive integer n. (This sequence is called the Fibonacci sequence, and its terms are called Fibonacci numbers.) Each nonnegative integer n can be uniquely written as a sum of nonconsecutive positive Fibonacci numbers; that is, each nonnegative integer n can be written uniquely in the form n = ∞  k=0 αk Fk, where αk ∈{0, 1} and (αk, αk+1) ̸= (1, 1) for each k. This expression for n is called its Zeckendorf representation. The proof of Proposition 1.43 is similar to that of Proposition 1.42, and we leave the details to the reader. Example 1.49. [AIME2 2000] Given that ( f1, f2, f3, . . . , f j) is the factorial base expansion of 16! −32! + 48! −64! + · · · + 1968! −1984! + 2000!, find the value of f1 −f2 + f3 −f4 + · · · + (−1) j−1 f j. www.riazisara.ir 46 104 Number Theory Problems Solution: Because (n + 1)! −n! = n!(n + 1) −n! = n!n, it follows that (n + 16)! −n! = (n + 16)! −(n + 15)! + (n + 15)! −(n + 14)! + · · · + (n + 1)! −n! = (n + 15)!(n + 15) + (n + 14)!(n + 14) + · · · + (n + 1)!(n + 1) + n!n. This shows that the factorial base expansion of (n + 16)! −n! is (0, 0, . . . , 0, n, n + 1, . . . , n + 14, n + 15), which begins with a block of n −1 zeros. The factorial base expansion of 16! is (0, 0, . . . , 0, 1), so the requested expansion is (0, 0, . . . , 0; 1; 0, . . . , 0; 32, 33, . . . , 47; 0, . . . , 0; 64, . . . , 79; . . . ; 1984, . . . , 1999). Notice that starting in position thirty-two, the expansion contains groups of six-teen nonzero numbers alternating with groups of sixteen zeros. With the exception of f16 = 1, each nonzero fi is i. Each of the 62 groups of sixteen nonzero num-bers contributes 8 to the alternating sum, and f16 contributes −1, so the requested value is 8 · 62 −1 = 495. □ Divisibility Criteria in the Decimal System We will prove some divisibility criteria for integers in decimal representation. Proposition 1.44. Let n = ahah−1 . . . a0 be a positive integer. (a) Let s(n) = a0 + a1 + · · · + ah denote the sum of its digits. Then n ≡s(n) (mod 3). In particular, n is divisible by 3 if and only if the sum S(n) of its digits is divisible by 3. (b) We can replace 3 by 9 in (1); that is, n ≡S(n) (mod 9). In particular, n is divisible by 9 if and only if the sum S(n) of its digits is divisible by 9. (c) Let s′(n) = a0 −a1 + · · · + (−1)hah (alternating sum). Then n is divisible by 11 if and only if s′(n) is divisible by 11. (d) n is divisible by 7, 11, or 13 if and only if ahah−1 . . . a3 −a2a1a0 has this property. (e) n is divisible by 27 or 37 if and only if ahah−1 . . . a3 + a2a1a0 has this property. www.riazisara.ir 1. Foundations of Number Theory 47 (f) n is divisible by 2k or 5k (k ≤h) if and only if ak−1 . . . a0 has this property. Proof: For (a) and (b), since 10k = (9 + 1)k, it follows that 10k ≡1 (mod 9). Hence n ≡h k=0 ak10k = h k=0 ak ≡S(n) (mod 9). For (c), we note that 10k = (11 −1)k. Hence 10k ≡(−1)k (mod 11), and so n ≡ h  k=0 ak10k ≡ h  k=0 ak · (−1)k ≡s′(n) (mod 11), from which the conclusion follows. For (d), the conclusion follows by the facts 1001 = 7 · 11 · 13 and n = ahah−1 . . . a3 · 1000 + a2a1a0 = ahah−1 . . . a3 · (1001 −1) + a2a1a0. For (e), the conclusion follows by the facts 999 = 27 · 37 and n = ahah−1 . . . a3 · 1000 + a2a1a0 = ahah−1 . . . a3 · (999 + 1) + a2a1a0. For (f), we note that 10k ≡0 (mod m) for m = 2k or m = 5k. We have n = ah . . . ak · 10k + ak−1 . . . a0, from which the conclusion follows. □ Example 1.50. Perfect squares or not? (1) Determine all positive integers k such that the k-digit number 11 . . . 1 is not a perfect square. (2) Can a 5-digit number consisting only of distinct even digits be a perfect square? (3) Determine whether 20 . . . 04   2004 is a perfect square. Solution: The answers are mostly negative for all these questions. (1) Clearly, k = 1 works. We claim that there are no other answers. Since 11 . . . 1   k 1’s ≡11 ≡3 (mod 4), 11 . . . 1   k 1’s is not a perfect square. (Example 1.24 (3)). (2) The answer is no. If n is a 5-digit number consisting only of distinct even digits, then the sum of its digits is equal to 0 + 2 + 4 + 6 + 8 = 20, which is congruent to 2 modulo 9; hence it is not a perfect square. (Example 1.24 (4).) www.riazisara.ir 48 104 Number Theory Problems (3) The given number is not a perfect square because the sum of its digits is 6, a multiple of 3 but not of 9. (Example 1.24 (4).) □ Example 1.51. [AIME 1984] The integer n is the smallest positive multiple of 15 such that every digit of n is either 0 or 8. Find n. Solution: An integer n is divisible by 15 if and only if it is divisible by both 3 and 5. By Proposition 1.44 (a) and (f), the answer is n = 8880. □ Example 1.52. Determine the number of five-digit positive integers abcde (a, b, c, d, and e not necessarily distinct) such that the sum of the three-digit number abc and the two-digit number de is divisible by 11. Solution: The answer is 8181. Note that abcde = abc × 100 + de = abc + de + 99 × abc. Hence abc + de is divisible by 11 if and only if abcde is divisible by 11. Note that 99990 is the greatest 5-digit number that is divisible by 11 and that 9999 is the greatest 4-digit number that is divisible by 11. Hence there are 99990 11 = 9090 multiples of 11 that have at most 5 digits, and there are 9999 11 = 999 multiples of 11 that have at most 4 digits. Therefore, there are exactly 9090 −999 = 8181 multiples of 11 that have exactly 5 digits. □ Example 1.53. [USAMO 2003] Prove that for every positive integer n there exists an n-digit number divisible by 5n all of whose digits are odd. First Solution: We proceed by induction. The property is clearly true for n = 1. Assume that N = a1a2 . . . an is divisible by 5n and has only odd digits. Consider the numbers N1 = 1a1a2 . . . an = 1 · 10n + 5n M = 5n(1 · 2n + M), N2 = 3a1a2 . . . an = 3 · 10n + 5n M = 5n(3 · 2n + M), N3 = 5a1a2 . . . an = 5 · 10n + 5n M = 5n(5 · 2n + M), N4 = 7a1a2 . . . an = 7 · 10n + 5n M = 5n(7 · 2n + M), N5 = 9a1a2 . . . an = 9 · 10n + 5n M = 5n(9 · 2n + M). The numbers 1 · 2n + M, 3 · 2n + M, 5 · 2n + M, 7 · 2n + M, 9 · 2n + M give distinct remainders when divided by 5. Otherwise, the difference of some two of them would be a multiple of 5, which is impossible, because 2n is not a multiple of 5, nor is the difference of any two of the numbers 1, 3, 5, 7, 9. It follows that one of the numbers N1, N2, N3, N4, N5 is divisible by 5n · 5, and the induction is complete. □ www.riazisara.ir 1. Foundations of Number Theory 49 Second Solution: For an m digit number a, where m ≥n, let ℓ(a) denote the m −n leftmost digits of a. (That is, we consider ℓ(a) as an (m −n)-digit number.) It is clear that we can choose a large odd number k such that a0 = 5n · k has at least n digits. Assume that a0 has m0 digits, where m0 ≥n. Note that a0 is an odd multiple of 5. Hence the units digit of a0 is 5. If the n rightmost digits of a0 are all odd, then the number b0 = a0−ℓ(a0)·10n satisfies the conditions of the problem, because b0 has only odd digits (the same as the n leftmost digits of a0) and that b0 is the difference of two multiples of 5n. If there is an even digit among the n rightmost digits of a0, assume that i1 is the smallest positive integer such that the i1th rightmost digit of a0 is even. Then the number a1 = a0 + 5n · 10i1−1 is a multiple of 5n with at least n digits. The (i −1)th rightmost digit is the same as that of a0 and the i1th rightmost digit of a1 is odd. If the n rightmost digits of a1 are all odd, then b1 = a1 −ℓ(a1) · 10n satisfies the conditions of the problem. If there is an even digit among the n rightmost digits of a1, assume that i2 is the smallest positive integer such that the i2th rightmost digit of a1 is even. Then i2 > i1. Set a2 = a1 +5n ·10i2−1. We can repeat the above process of checking the rightmost digits of a2 and eliminate the rightmost even digits of a2, if there is such a digit among the n rightmost digits of a2. This process can be repeated at most n −1 times because the units digit of a0 is 5. Thus, we can obtain a number ak, for some nonnegative integer k, such that ak is a multiple of 5n with its n rightmost digits all odd. Then bk = ak −ℓ(ak)·10n is a number that satisfies the conditions of the problem. □ We can replace the condition of odd digits by any collection of 5 digits that forms a complete set of residue classes modulo 5. In exactly the same way, we can show that for every positive integer n there exists an n-digit number divisible by 2n all of whose digits form a complete set of residue classes modulo 5. We close this section with some more discussion on S(n), the sum of the digits of a positive integer n. Proposition 1.45. Let n be a positive integer, and let S(n) denote the sum of its digits. Then (a) 9|S(n) −n; (b) S(n1 + n2) ≤S(n1) + S(n2) (subadditivity property); (c) S(n1n2) ≤min(n1S(n2), n2S(n1)); (d) S(n1n2) ≤S(n1)S(n2) (submultiplicativity property). Proof: Part (a) is simply Proposition 1.44 (b). Let us prove (b), (c), and (d). Consider n1 = akak−1 . . . a0, n2 = bhbh−1 . . . b0, and n1 + n2 = cscs−1 . . . c0. www.riazisara.ir 50 104 Number Theory Problems In order to prove (b), we choose the least t such that ai +bi < 10 for all i < t. Then at + bt ≥10; hence ct = at + bt −10 and ct+1 ≤at+1 + bt+1 + 1. We obtain t+1  i=1 ci ≤ t+1  i=1 ai + t+1  i=1 bi. Continuing this procedure, the conclusion follows. Because of the symmetry, in order to prove (c) it suffices to prove that S(n1n2) ≤n1S(n2). The last inequality follows by applying the subadditivity property (b) repeatedly. Indeed, S(2n2) = S(n2 + n2) ≤S(n2) + S(n2) = 2S(n2), and after n1 steps we obtain S(n1n2) = S(n2 + n2 + · · · + n2   n1 times ) ≤S(n2) + S(n2) + · · · + S(n2)   n1 times = n1S(n2). To establish (d), we observe that by (b) and (c), S(n1n2) = S  n1 h  i=0 bi10i  = S  h  i=0 n1bi10i  ≤ h  i=0 S(n1bi10i) = h  i=0 S(n1bi) ≤ h  i=0 bi S(n1) = S(n1) h  i=0 bi = S(n1)S(n2). as desired. □ From the proof of Proposition 1.45, we note that it is very important to deal with carryings in working with problems related to the sum of the digits. Example 1.54. [Russia 1999] In the decimal expansion of n, each digit (except the first digit) is greater than the digit to its left. What is S(9n)? Solution: Write n = akak−1 . . . a0. By performing the subtraction ak ak−1 . . . a1 a0 0 − ak . . . a2 a1 a0 we find that the digits of 9n = 10n −n are ak, ak−1 −ak, . . . , a1 −a2, a0 −a1 −1, 10 −a0. These digits sum to 10 −1 = 9. □ www.riazisara.ir 1. Foundations of Number Theory 51 Example 1.55. [Ireland 1996] Find a positive integer n such that S(n) = 1996S(3n). Solution: Consider n = 1 33 . . . 3   5986 3’s 5. Then 3n = 4 00 . . . 0   5986 0’s 5. We have S(n) = 3·5986+1+5 = 17964 = 1996·9 = 1996S(n), as desired. □ Example 1.56. Determine whether there is any perfect square that ends in 10 distinct digits. Solution: The answer is yes. We note that 1 1 1 1 × 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 3 2 1 Likewise, it is not difficult to see that 111111111112 = 12345678900987654321 is a number that satisfies the conditions of the problem. □ Example 1.57. [IMO 1976] When 44444444 is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B. Solution: The answer is 7. Let a = 44444444. By our notation, we have A = S(a) and B = S(A), and we want to compute S(B). First we will show that the sum of the digits of B is fairly small. Note that 4444 < 10000 = 104. Hence a = 44444444 < 104·4444 = 1017776, and so a cannot have more than 17776 digits. Because each digit is at most a 9, A = S(a) ≤17776 · 9 = 159984. Of the natural numbers less than or equal to www.riazisara.ir 52 104 Number Theory Problems 159984, the number with the largest digit sum is 99999, and so B = S(A) ≤45. Of the natural numbers less than or equal to 45, the number with the largest digit sum is 39. Hence S(B) ≤12. By Proposition 1.45 (a), we have S(B) ≡B ≡S(A) ≡A ≡S(a) ≡a ≡44444444 (mod 9). It suffices to show that 44444444 ≡7 (mod 9). Indeed, we have 44444444 ≡(4 + 4 + 4 + 4)4444 ≡164444 ≡(−2)4444 ≡(−2)3·1481+1 ≡((−2)3)1481 · (−2) ≡(−8)1481 · (−2) ≡1 · (−2) ≡7 (mod 9). □ Floor Function For a real number x there is a unique integer n such that n ≤x < n + 1. We say that n is the greatest integer less than or equal to x, or the floor of x. We write n = ⌊x⌋. The difference x −⌊x⌋is called the fractional part of x and is denoted by {x}. The least integer greater than or equal to x is called the ceiling of x and is denoted by ⌈x⌉. If x is an integer, then ⌊x⌋= ⌈x⌉and {x} = 0; if x is not an integer, then ⌈x⌉= ⌊x⌋+ 1. We start with four (algebraic) examples to get familiar with these functions. Example 1.58. [Australia 1999] Solve the following system of equations: x + ⌊y⌋+ {z} = 200.0, {x} + y + ⌊z⌋= 190.1, ⌊x⌋+ {y} + z = 178.8. Solution: Because x = ⌊x⌋+ {x} for all real numbers x, adding the three equations gives 2x + 2y + 2z = 568.9, or x + y + z = 284.45. Subtracting each of the three given equations from the last equation gives {y} + ⌊z⌋= 84.45, ⌊x⌋+ {z} = 94.35, {x} + ⌊y⌋= 105.65. Therefore 84 = ⌊84.45⌋= ⌊⌊z⌋+ {y}⌋= ⌊z⌋, and thus ⌊z⌋= 84 and {y} = 0.45. In the same way we obtain ⌊y⌋= 105, and so y = 105.45. Similarly, x = 94.65 and z = 84.35. □ www.riazisara.ir 1. Foundations of Number Theory 53 Example 1.59. Determine the distinct numbers in the sequence  12 2005 ,  22 2005 , . . . ,  20052 2005 . Solution: For 1 ≤i ≤2005, let ai =  i2 2005 . Because 442 = 1936 < 2005 < 2025 = 452, a1 = a2 = · · · = a44 = 0. For integers m with m ≥1002, since (m + 1)2 2005 −m2 2005 = 2m + 1 2005 ≥1, it follows that am < am+1. Hence, a1002, a1003, . . . , a2005 take distinct values. For positive integers m with m < 1002, since (m + 1)2 2005 −m2 2005 = 2m + 1 2005 < 1, it follows that am+1 ≤am + 1. Note that this sequence is clearly nondecreasing. We conclude that all the integer values less than a1001 have been taken. Finally, we compute a1001 = 499 and a1002 = 500. Therefore, the answer of the problem is 500 + 1004 = 1504 (namely, values 0, 1, . . . , 499, a1002, a1003, . . . , a2005). □ Example 1.60. [ARML 2003] Find the positive integer n such that 1 n is closest to { √ 123456789}. Solution: As shown in the Example 1.56, we have 11111.112 = 123456765.4321 < 123456789 < 123456789.87654321 = 11111.11112. Hence √ 123456789 = 11111 and 1 10 < 0.11 < { √ 123456789} < 0.1111 < 1 9. □ Example 1.61 [AIME 1997] Suppose that a is positive, {a−1} = {a2}, and 2 < a2 < 3. Find the value of a12 −144a−1. www.riazisara.ir 54 104 Number Theory Problems Solution: Notice first that the given hypothesis implies that {a−1} = a−1 (since 1 < a and 0 < a−1 < 1) and {a2} = a2 −2. Hence a must satisfy the equation a−1 = a2 −2, or a3 −2a −1 = 0. This factors as (a + 1)(a2 −a −1) = 0, whose only positive root is a = 1+ √ 5 2 . Now use the relations a2 = a + 1 and a3 = 2a + 1 to calculate a6 = 8a + 5, a12 = 144a + 89, and a13 = 233a + 144, from which it follows that a12 −144a−1 = a13 −144 a = 233. □ Note: By the relation a2 = a + 1, we can show easily that an = Fn−1a + Fn−2, where {Fn}∞ n=0 is the Fibonacci sequence with F0 = F1 = 1 and Fn+1 = Fn + Fn−1 for every positive integer n. This is not surprising if we note that a2 = a +1 is the characteristic equation of the Fibonacci sequence. For more details on this, the reader also can look at chapter five of . Example 1.62. Find all real solutions to the equation 4x2 −40⌊x⌋+ 51 = 0. Solution: Note that (2x −3)(2x −17) = 4x2 −40x + 51 ≤4x2 −40⌊x⌋+ 51 = 0, which gives 3 2 ≤x ≤17 2 and 1 ≤⌊x⌋≤8. Then x = √40⌊x⌋−51 2 , so it is necessary to have ⌊x⌋= √40⌊x⌋−51 2 . Testing ⌊x⌋∈{1, 2, 3, . . . , 8} in this equation, we find that ⌊x⌋can equal only 2, 6, 7, or 8. Thus the only solutions for x are √ 29 2 , √ 189 2 , √ 229 2 , and √ 269 2 . A quick check confirms that these values work. □ www.riazisara.ir 1. Foundations of Number Theory 55 Proposition 1.46. We have the following properties for the floor and the ceiling functions. (a) If a and b are integers with b > 0, and q is the quotient and r is the remainder when a is divided by b, then q = ! a b " and r = # a b $ · b. (b) For any real number x and any integer n, ⌊x + n⌋= ⌊x⌋+n and ⌈x + n⌉= ⌈x⌉+ n. (c) If x is an integer, then ⌊x⌋+ ⌊−x⌋= 0; if x is not an integer, ⌊x⌋+ ⌊−x⌋= 1. (d) The floor function is nondecreasing; that is, for x ≤y, ⌊x⌋≤⌊y⌋. (e) x + 1 2 rounds x to its nearest integer. (f) ⌊x⌋+ ⌊y⌋≤⌊x + y⌋≤⌊x⌋+ ⌊y⌋+ 1. (g) ⌊x⌋· ⌊y⌋≤⌊xy⌋for nonnegative real numbers x and y. (h) For any positive real number x and any positive integer n the number of positive multiples of n not exceeding x is ! x n " . (i) For any real number x and any positive integer n, ⌊x⌋ n = x n . Proof: The proofs of (a) to (d) are straightforward. We present only the proof of (e) to (i). For (e) note that if {x} < 1 2, then x + 1 2 = ⌊x⌋, which is the integer closest to x; if {x} > 1 2, then x + 1 2 = ⌈x⌉, which is the integer closest to x. This is a very simple but useful trick in computer programming. For (f), we write x = ⌊x⌋+ {x} and y = ⌊y⌋+ {y}. The desired result reduces to 0 ≤⌊{x} + {y}⌋≤1, which is clear since 0 ≤{x}, {y} < 1. For (g), we write again x = ⌊x⌋+{x} and y = ⌊y⌋+{y}. Then ⌊x⌋, ⌊y⌋, {x}, {y} are all nonnegative. It is clear that ⌊xy⌋= ⌊(⌊x⌋+ {x}) (⌊y⌋+ {y})⌋ = ⌊⌊x⌋⌊y⌋+ ⌊x⌋{y} + ⌊y⌋{x} + {x}{y}⌋≥⌊x⌋⌊y⌋. www.riazisara.ir 56 104 Number Theory Problems For (h), we consider all multiples 1·n, 2·n, . . . , k ·n, where k ·n ≤x < (k +1)n. That is, k ≤x n < k + 1, and the conclusion follows. Note that (i) follows immediately from (f), since the multiples of an integer n are integers. □ Furthermore, we extend Proposition 1.46 (f) to the following. Example 1.63. For real numbers x and y, prove that ⌊2x⌋+ ⌊2y⌋≥⌊x⌋+ ⌊y⌋+ ⌊x + y⌋. Proof: Write x = ⌊x⌋+ {x} and y = ⌊y⌋+ {y}. Then ⌊2x⌋+ ⌊2y⌋= 2⌊x⌋+ ⌊2{x}⌋+ 2⌊y⌋+ ⌊2{y}⌋ and ⌊x + y⌋= ⌊x⌋+ ⌊y⌋+ ⌊{x} + {y}⌋. It suffices to show that ⌊2{x}⌋+ ⌊2{y}⌋≥⌊{x} + {y}⌋. By symmetry, we may assume that {x} ≥{y}. Note that {x} is nonnegative. We have ⌊2{x}⌋+ ⌊2{y}⌋≥⌊2{x}⌋≥⌊{x} + {y}⌋, by Proposition 1.46 (e) (since 2{x} ≥{x} + {y}). □ Proposition 1.46 (e) also has different forms for special values of the variable. Example 1.64. For a given positive integer n, show that √n + 1 2 = % n −3 4 + 1 2 . Proof: Suppose that √n + 1 2 = k and % n −3 4 + 1 2 = m. Then we have k ≤√n + 1 2 < k + 1, or k −1 2 ≤√n < k + 1 2. Squaring both sides of the last inequality gives k2 −k + 1 4 ≤n < k2 + k + 1 4. www.riazisara.ir 1. Foundations of Number Theory 57 Since n is an integer, we have k2 −k + 1 ≤n ≤k2 + k. Likewise, we have m ≤ & n −3 4 + 1 2 < m + 1, implying that m2 −m + 1 4 ≤n −3 4 < m2 + m + 1 4. Because n is an integer, we again have m2 −m + 1 ≤n ≤m2 + m. Combining the above, we conclude that m = k, as desired. □ The graphs of the functions y = ⌊x⌋and y = ⌈x⌉are typical step functions. Their unique properties allow us to describe some very special sequences. Example 1.65. [AIME 1985] How many of the first 1000 positive integers can be expressed in the form ⌊2x⌋+ ⌊4x⌋+ ⌊6x⌋+ ⌊8x⌋, where x is a real number? Solution: Define the function f (x) = ⌊2x⌋+ ⌊4x⌋+ ⌊6x⌋+ ⌊8x⌋, and observe that if n is a positive integer, then f (x + n) = f (x) + 20n. In particular, this means that if an integer k can be expressed in the form f (x0) for some real number x0, then for n = 1, 2, 3, . . . one can express k + 20n similarly; that is, k + 20n = f (x0) + 20n = f (x0 + n). In view of this, one may restrict attention to determining which of the first 20 positive integers are generated by f (x) as x ranges through the half-open interval (0, 1]. Next observe that as x increases, the value of f (x) changes only when either 2x, 4x, 6x, or 8x attains an integral value, and that the change in f (x) is always to a new, higher value. In the interval (0, 1] such changes occur precisely when x is of the form m/n, where l ≤m ≤n and n = 2, 4, 6, or 8. There are 12 such fractions; in increasing order they are 1 8, 1 6, 1 4, 1 3, 3 8, 1 2, 5 8, 2 3, 3 4, 5 6, 7 8, and 1. Therefore, only 12 of the first 20 positive integers can be represented in the desired form. Since 1000 = 50·20, there are 50·12 = 600 positive integers of the desired form. □ Example 1.66. [Gauss] Let p and q be relatively prime integers. Prove that p q + 2p q + · · · + (q −1)p q = (p −1)(q −1) 2 . www.riazisara.ir 58 104 Number Theory Problems Solution: Since gcd(p, q) = 1, ip q is not an integer. By Proposition 1.46 (c), it follows that ip q + (q −i)p q = p + ip q + −ip q = p −1 for 1 ≤i ≤q −1. Therefore, 2  p q + 2p q + · · · + (q −1)p q =  p q + (q −1)p q + · · · + (q −1)p q + p q = (p −1)(q −1), from which the desired result follows. □ We can also interpret the above result as the number of lattice points lying inside the triangle bounded by the lines y = 0, x = p, and y = qx p . (A point in the coordinate plane is a lattice point if it has integer coordinates.) Example 1.67. The sequence {an}∞ n=1 = {2, 3, 5, 6, 7, 8, 10, . . . } consists of all the positive integers that are not perfect squares. Prove that an = n + √n + 1 2 . First Proof: We claim that √n + 1 2 2 < n + √n + 1 2 < √n + 1 2 + 1 2 . (†) With our claim, it is clear that among the integers 1, 2, . . . , n + √n + 1 2 , there are exactly √n + 1 2 perfect squares, namely, 12, 22, . . . , √n + 1 2 2 . Hence n + √n + 1 2 www.riazisara.ir 1. Foundations of Number Theory 59 is the nth number in the sequence after all perfect squares have been deleted; that is, an = n + √n + 1 2 . Now we prove our claim. Note that √n is either an integer or an irrational number. Hence {√n} ̸= 1 2. We consider two cases. In the first case, we assume that {√n} < 1 2. Set k = !√n " . Then k2 ≤n <  k + 1 2 2 , or k2 < n < k2 + k + 1 4. Then √n + 1 2 = !√n " = k, and the inequality (†) becomes k2 < n + k < (k + 1)2 = k2 + 2k + 1, which is evident. In the second case, we assume that {√n} > 1 2. Again, set k = !√n " . Then  k + 1 2 2 < n < (k + 1)2, or k2 + k + 1 4 < n < k2 + 2k + 1. Then √n + 1 2 = !√n " + 1 = k + 1, and the inequality (†) becomes (k + 1)2 < n + k + 1 < (k + 2)2 = k2 + 4k + 4, which is also evident. Combining the last two cases, we have shown that our claim is always true, and our proof is complete. □ The second proof reveals the origin of this closed form of an. Second Proof: Consider the sequence {bn}∞ n=1 = {1, 1; 2, 2, 2, 2; 3, 3, 3, 3, 3, 3; . . . }. We note that an −bn = n www.riazisara.ir 60 104 Number Theory Problems for all positive integers n. It is clear that there are are exactly (n + 1)2 −n2 −1 = 2n non-perfect squares strictly between two consecutive perfect squares n2 and (n + 1)2. It suffices to show that bn = √n + 1 2 . If bn = k, it is in the kth group and is preceded by at least k −1 groups containing 2 + 4 + · · · + 2(k −1) terms. Considering also the fact that there are n −1 terms before bn, we conclude that 2 + 4 + · · · + 2(bn −1) ≤n −1. Moreover, bn is the largest integer satisfying this inequality. Thus bn is the largest integer satisfying the inequality bn(bn −1) ≤n −1; that is, bn =  1 + √ 4n −3 2 = % n −3 4 + 1 2 = √n + 1 2 , by Example 1.64. □ Theorem 1.47. [Beatty’s Theorem] Let α and β be two positive irrational real numbers such that 1 α + 1 β = 1. The sets {an}∞ n=1 = {⌊α⌋, ⌊2α⌋, ⌊3α⌋, . . . } and {bn}∞ n=1 = {⌊β⌋, ⌊2β⌋, ⌊3β⌋, . . . } form a partition of the set of positive integers; that is, {an}∞ n=1 and {bn}∞ n=1 are nonintersecting sets with their union equal to the set of all positive integers. Proof: We first show that they are nonintersecting. We proceed indirectly by assuming the contrary, that is, we assume that there are indices i and j such that k = ai = b j = ⌊iα⌋= ⌊jβ⌋. Since both iα and jβ are irrational, it follows that k < iα < k + 1, k < jβ < k + 1, or i k + 1 < 1 α < i k and j k + 1 < 1 β < j k . Adding these two inequalities gives i + j k + 1 < 1 α + 1 β = 1 < i + j k , www.riazisara.ir 1. Foundations of Number Theory 61 or k < i + j < k + 1, which is impossible. Hence our assumption was wrong and these two sequences do not intersect. We next prove that every positive integer appears in one of the two sequences. We again approach indirectly by assuming that there is a positive integer k that does not appear in these two sequences. It follows that there are indices i and j such that iα < k, (i + 1)α > k + 1, jβ < k, ( j + 1)β > k + 1, or i k < 1 α < i + 1 k + 1 and j k < 1 β < j + 1 k + 1. Adding the last two inequalities gives i + j k < 1 α + 1 β = 1 < i + j + 2 k + 1 , implying that i + j < k and k + 1 < i + j + 2, and so i + j < k < i + j + 1, which is again impossible. Hence our assumption was wrong and every positive integer appears in exactly one of the two sequences. □ Example 1.68a. [USAMO 1981] For a positive number x, prove that ⌊x⌋+ ⌊2x⌋ 2 + ⌊3x⌋ 3 + · · · + ⌊nx⌋ n ≤⌊nx⌋. Indeed, we have a more general result. By Proposition 1.46 (f), Example 1.68a is a special case of Example 1.68b by setting ai = −⌊ix⌋. Example 1.68b. [APMO 1999] Let a1, a2, . . . be a sequence of real numbers satisfying ai+ j ≤ai + a j for all i, j = 1, 2, . . . . Prove that a1 + a2 2 + a3 3 + · · · + an n ≥an for all positive integers n. First Proof: We use strong induction. The base cases for n = 1 and 2 are trivial. Now assume that the statement is true for n ≤k for some positive integer k ≥2. www.riazisara.ir 62 104 Number Theory Problems That is, a1 ≥a1, a1 + a2 2 ≥a2, . . . a1 + a2 2 + · · · + ak k ≥ak. Adding all the inequalities gives ka1 + (k −1)a2 2 + · · · + ak k ≥a1 + a2 + · · · + ak. Adding (a1 + a2 + · · · + ak) to both sides of the last inequality yields (k + 1)  a1 + a2 2 + · · · + ak k  ≥(a1 + ak) + (a2 + ak−1) + · · · + (ak + a1) ≥kak+1. Dividing both sides of the last inequality by (k + 1) gives a1 + a2 2 + · · · + ak k ≥kak+1 k + 1 , or a1 + a2 2 + · · · + ak k + ak+1 k + 1 ≥ak+1. This completes the induction and we are done. □ Second Proof: [By Andreas Kaseorg] We can extend the condition by induction to ai1+i2+···+ik ≤ai1 + ai2 + · · · + aik. We apply a combinatorial argument. A permutation is a change in position within a collection. More precisely, if S is a set, then a permutation of S is a one-to-one function π that maps S onto itself. If S = {x1, x2, . . . , xn} is a finite set, then we may denote a permutation π of S by (y1, y2, . . . , yn), where yk = π(xk). An ordered k-tuple (xi1, xi2, . . . , xik) is a k-cycle of π if π(xi1) = xi2, π(xi2) = xi3, . . . , and π(xik) = x1. Let Sn denote the set of permutations of n elements. For an element π in Sn, define f (π, k) to be the number of k-cycles in π. Clearly, 1 · f (π, 1) + 2 · f (π, 2) + · · · + n · f (π, n) = n, since both sides count the number of elements in the permutation π. Note also that  π∈Sn f (π, k) is the total number of k-cycles in all permutations on n elements, www.riazisara.ir 1. Foundations of Number Theory 63 which is n k  (k −1)!(n −k)! = n! k ; that is,  π∈Sn f (π, k) = n k (k −1)!(n −k)! = n! k . (∗) This is because (a) we have n k  ways to choose k elements as the elements of a k-cycle; (b) we have (k −1)! ways to form a k-cycle using the k chosen elements; (c) we have (n −k)! ways to permute the n −k unchosen elements to complete the permutation of all n elements. Therefore, by (∗), we have a1 + a2 2 + a3 3 + · · · + an n = 1 n!  π∈Sn [ f (π, 1)a1 + f (π, 2)a2 + · · · + f (π, n)an] ≥1 n!  π∈Sn a1· f (π,1)+2· f (π,2)+···+n· f (π,n) = 1 n!  π∈Sn an = an, because there are exactly n! elements in Sn (since there are n! permutations of n elements). □ As shown in Examples 1.68a and 1.68b, many interesting and challenging problems related to the floor and ceiling functions have close ties to their special functional properties. We leave most of them to the sequel of this book: 105 Diophantine Equations and Integer Function Problems. We close this section by introducing the well-known Hermite identity. Proposition 1.48. [Hermite Identity] Let x be a real number, and let n be a positive integer. Then ⌊x⌋+ x + 1 n + x + 2 n + · · · + x + n −1 n = ⌊nx⌋. Proof: If x is an integer, then the result is clearly true. We assume that x is not an integer; that is, 0 < {x} < 1. Then there exists 1 ≤i ≤n −1 such that {x} + i −1 n < 1 and {x} + i n ≥1, (∗) www.riazisara.ir 64 104 Number Theory Problems that is, n −i n ≤{x} < n −i + 1 n . (∗∗) By (∗), we have ⌊x⌋= x + 1 n = · · · = x + i −1 n and x + i n = · · · = x + n −1 n = ⌊x⌋+ 1, and so ⌊x⌋+ x + 1 n + x + 2 n + · · · + x + n −1 n = i⌊x⌋+ (n −i) (⌊x⌋+ 1) = n⌊x⌋+ n −i. On the other hand, by (∗∗), we obtain n⌊x⌋+ n −i ≤n⌊x⌋+ n{x} = nx < n⌊x⌋+ n −i + 1, implying that ⌊nx⌋= n⌊x⌋+ n −i. Combining the above observations, we have ⌊x⌋+ x + 1 n + x + 2 n + · · · + x + n −1 n = n⌊x⌋+ n −i = ⌊nx⌋. □ Example 1.69. [AIME 1991] Suppose that r is a real number for which r + 19 100 + r + 20 100 + · · · + r + 91 100 = 546. Find ⌊100r⌋. Solution: The given sum has 91 −19 + 1 = 73 terms, each of which equals either ⌊r⌋or ⌊r⌋+ 1. But 73 · 7 < 546 < 73 · 8, and so it follows that ⌊x⌋= 7. Because 546 = 73·7+35, the first 38 terms take the value 7 and the last 35 terms take the value 8; that is, r + 56 100 = 7 and r + 57 100 = 8. It follows that 7.43 ≤r < 7.44 and hence that ⌊100r⌋= 743. □ www.riazisara.ir 1. Foundations of Number Theory 65 Example 1.70. [IMO 1968] Let x be a real number. Prove that ∞  k=0 x + 2k 2k+1 = ⌊x⌋. Solution: Setting n = 2 in Hermite’s identity gives ⌊x⌋+ x + 1 2 = ⌊2x⌋, or x + 1 2 = ⌊2x⌋−⌊x⌋. Repeatedly applying the last identity gives ∞  k=0 x + 2k 2k+1 = ∞  k=0 x 2k+1 + 1 2 = ∞  k=0  x 2k − x 2k+1  = ⌊x⌋, as desired. □ Legendre’s Function We use Proposition 1.46 (h) to develop some interesting results. Let p be a prime. For any positive integer n, let ep(n) be the exponent of p in the prime factorization of n!. The arithmetic function ep is called the Legendre function associated with the prime p. The following result gives a formula for the computation of ep(n). Proposition 1.49. [Legendre’s Formula] For any prime p and any positive inte-ger n, ep(n) =  i≥1 n pi = n p + n p2 + n p3 + · · · . We note that this sum is a finite one, because for large m, n < pm+1 and n pm+1 = 0. Let m be the least positive integer such that n < pm+1; that is, m = ln n ln p . It suffices to show that ep(n) = n p + n p2 + · · · + n pm . www.riazisara.ir 66 104 Number Theory Problems We present two closely related proofs. The first is written in the language of number theory, the second in the language of combinatorics. First Proof: For n < p it is clear that ep(n) = 0. If n ≥p, then in order to determine ep(n) we need to consider only the multiples of p in the product n! = 1 · 2 · · · n; that is, (1 · p)(2 · p) · · · (kp) = pkk!, where k = n p by Proposition 1.46 (h). Hence ep(n) = n p + ep  n p . Replacing n by n p and taking into account Proposition 1.46 (i), we obtain ep  n p = ⎢ ⎢ ⎢ ⎣ n p p ⎥ ⎥ ⎥ ⎦+ ep ⎛ ⎝ ⎢ ⎢ ⎢ ⎣ n p p ⎥ ⎥ ⎥ ⎦ ⎞ ⎠= n p2 + ep  n p2 . Continuing this procedure we get ep  n p2 = n p3 + ep  n p3 , . . . ep  n pm−1 = n pm + ep  n pm = n pm . Summing up the relations above yields the desired result. □ Second Proof: For each positive integer i, define ti such that pti ∥i. Because p is prime, we have pt1+t2+···+tn∥n!, or t = tn! = t1 +t2 +· · ·+tn. On the other hand, n pk counts all multiples of pk that are less than or equal to n exactly once. Thus the number i = pti · a (with a and p relatively prime) is counted ti times in the sum n p + n p2 + · · · + n pm , namely, in the terms n p , n p2 , . . . , n pi . Therefore, for each 1 ≤i ≤n, the number i contributes ti in both n p + n p2 + · · · + n pm www.riazisara.ir 1. Foundations of Number Theory 67 and t1 + t2 + · · · + tn. Hence t = t1 + t2 + · · · + tn = n p + n p2 + · · · + n pm . In a more formal language, consider the matrix M = (xi, j) with m rows and n columns, where m is the smallest integer such that pm > n. We define xi, j =  1 if pi divides j, 0 otherwise. Then the number of 1’s in the jth column of the matrix M is t j, implying that the column sums of M are t1, t2, . . . , tn. Hence the sum of all of the entries in M is t. On the other hand, the 1’s in the ith row denote the numbers that are multiples of pi. Consequently, the sum of the entries in the ith row is n pi . Thus the sum of the entries in M is also m i=1 n pi . It follows that t = n p + n p2 + · · · + n pm , as desired. □ Example 1.71. Let s and t be positive integers such that 7s∥400! and 3t∥((3!)!)!. Compute s + t. Solution: The answer is 422. Note that ((3!)!)! = (6!)! = 720!. Applying Legendre’s formula, we have s = e7(400) = 400 7 + 400 72 + 400 73 = 57 + 8 + 1 = 66 and t = e3(720) = 720 3 + 720 32 + 720 33 + 720 34 + 720 35 = 240 + 80 + 26 + 8 + 2 = 356, and so s + t = 356 + 66 = 422. □ www.riazisara.ir 68 104 Number Theory Problems Example 1.72. The decimal representation of 2005! ends in m zeros. Find m. Solution: It is equivalent to compute m such that 10m∥2005!. Since 10m = 2m5m, we have m = min{e2(2005!), e5(2005!)}. Because 2 < 5, we have m = e5(2005!) = 2005 5 + 2005 25 + 2005 125 + 2005 625 = 500. The answer is 500. □ Example 1.73. [HMMT 2003] Find the smallest n such that n! ends in 290 zeros. Solution: As shown in the solution of Example 1.72, we need to find the smallest n such that 290 = e5(n) = n 5 + n 52 + n 53 + · · · , which is roughly a geometric series (by taking away the floor function) whose sum is represented approximately by n/5 1−1/5. Solving 290 ≈ n 5 1 −1 5 , we estimate n = 1160, and this gives us e5(1160) = 288. Adding 10 to the value of n = 1160 gives the necessary two additional factors of 5 (from 1165 and 1170), and so the answer is 1170. □ Example 1.74. Let m and n be positive integers. Prove that (1) m! · (n!)m divides (mn)!. (2) m!n!(m + n)! divides (2m)!(2n)!. Proof: We present a common technique in this proof. (1) Let p be a prime. Let x and y be nonnegative integers such that px∥m! · (n!)m and py∥(mn)!. It suffices to show that x ≤y. Note that x = ep(m)+ mep(n) and y = ep(mn). It suffices to show that ∞  i=1 mn pi ≥ ∞  i=1 m pi + m ∞  i=1 n pi . If p > n, then the second summand on the right-hand side is 0 and the inequality is clearly true. We assume that p ≤n. Let s be the positive www.riazisara.ir 1. Foundations of Number Theory 69 integer such that ps ≤n < ps+1. By Proposition 1.46 (g), we have ∞  i=1 mn pi = s  i=1 m · n pi + ∞  i=1 m pi · n ps ≥m s  i=1 n pi + ∞  i=1 m pi n ps ≥m ∞  i=1 n pi + ∞  i=1 m pi , as desired. (2) The proof is very similar to that of (1). We leave it to the reader. □ Note: It is also to find combinatorial proofs for these facts. For example, there are (mn)! m!(n!)m ways to split mn people into m groups of n, implying (1). Example 1.75. Let k and n be positive integers. Prove that (k!)kn+kn−1+···+k+1 | (kn+1)!. Proof: For every i with 0 ≤i ≤n, setting (n, m) = (k, ki) in Example 1.74 (1) gives k! | k!, k!(k!)k | (k2)!, (k2)!(k!)k2 | (k3)!, . . . , (kn)!(k!)kn | (kn+1)!. Multiplying this together gives k!k!(k2!)(k3!) · · · (kn)!k!k+k2+···+kn | k!(k2!)(k3!) · · · (kn+1)!, from which the desired result follows. □ Example 1.76. Let n > 2 be a composite number. Prove that not all of the terms in the sequence n 1 , n 2 , . . . ,  n n −1 are divisible by n. www.riazisara.ir 70 104 Number Theory Problems Proof: Let p be a prime divisor of n, and let s be the integer such that ps ≤n < ps+1. We show that n ∤  n ps = n! (ps)!(n −ps)!. Since p | n, it suffices to show that p ∤  n ps  . Suppose that pk∥  n ps  . Then k = ep(n) −ep(ps) −ep(n −ps). It suffices to show that k = 0. By Legendre’s formula, we have k =  i≥1 n pi −  i≥1 ps pi −  i≥1 n −ps pi = s  i=1 n pi − s  i=1 ps pi − s  i=1 n −ps pi = s  i=1 n pi − s  i=1 ps pi − s  i=1 n pi + s  i=1 ps pi = 0, since ps pi are integers for each 1 ≤i ≤s. □ Legendre’s formula is a great tool in combinatorial number theory. It helps to establish two important theorems of Lucas and Kummer. We will discuss them in detail in the sequel to this book – 107 Combinatorial Number Theory Problems. The reader can also look at chapter three of . Fermat Numbers Trying to find all primes of the form 2m + 1, Fermat noticed that m must be a power of 2. Indeed, if m were equal to k · h with k an odd integer greater than 1, then 2m + 1 = (2h)k + 1 = (2h + 1)(2h(k−1) −2h(k−2) + · · · −2h + 1), and so 2m + 1 would not be a prime. The integers fn = 22n + 1, n ≥0, are called Fermat numbers. We have f0 = 3, f1 = 5, f2 = 17, f3 = 257, f4 = 65537, and f5 = 4294967297. After checking that these five numbers are primes, Fermat conjectured that fn is a prime for all n. But Euler proved that 641 | f5. His argument was the following: f5 = 232 + 1 = 228(54 + 24) −(5 · 27)4 + 1 = 228 · 641 −(6404 −1) = 641(228 −639(6402 + 1)). www.riazisara.ir 1. Foundations of Number Theory 71 It is still unknown whether there are infinitely many prime Fermat numbers (Fer-mat primes). The answer to this question is important because Gauss proved that a regular polygon Q1Q2 . . . Qn can be constructed using a straightedge and compass if and only if n = 2h p0 · · · pk, where k ≥0, p0 = 1, and p1, . . . , pk are distinct Fermat primes. Gauss was the first to construct such a polygon for n = 17. It is also unknown whether there are infinitely many composite Fermat numbers. (Well, the good thing is that the answer to one of these two questions must be positive ¨ ⌣.) Example 1.77. For positive integers m and n with m > n, fn divides fm −2. Proof: By repeatedly applying the difference of squares formula a2 −b2 = (a −b)(a + b), it is not difficult to show that fm −2 = fm−1 fm−2 · · · f1 f0, from which the desired result follows. □ Example 1.78. For distinct positive integers m and n, fm and fn are relatively prime. Proof: By the first example, we have gcd( fm, fn) = gcd( fn, 2) = 1. □ This result also is a special case of Example 1.22. Example 1.79. Prove that for all positive integers n, fn divides 2 fn −2. Proof: We have 2 fn −2 = 2  222n −1  = 2 + 22n22n−n −1 , . Clearly, 22n−n is even. Note that for an even positive integer 2m, x2m −1 is divisible by x + 1. Hence x + 1 divides x22n−n −1. Setting x = 22n leads to the desired conclusion. □ The result in Example 1.79 shows that 2 fn ≡2 (mod fn), which gives an-other counterexample to the converse of Fermat’s little theorem. That is, 2 f5 ≡2 (mod f5) but f5 is not a prime. Mersenne Numbers The integers Mn = 2n −1, n ≥1, are called Mersenne numbers. It is clear that if n is composite, then so is Mn. Hence Mk is a prime only if k is a prime. Moreover, if n = ab, where a and b are integers greater than 1, then Ma and Mb both divide Mn. But there are primes n for which Mn is composite. For example, 47 | M23, 167 | M83, 263 | M13, and so on. www.riazisara.ir 72 104 Number Theory Problems Theorem 1.50. Let p be an odd prime and let q be a prime divisor of Mp. Then q = 2kp + 1 for some positive integer k. Proof: From the congruence 2p ≡1 (mod q) and from the fact that p is a prime, by Proposition 1.30, it follows that p is the least positive integer satisfying this property. By using Fermat’s little theorem, we have 2q−1 ≡1 (mod q), hence p | (q−1), by Proposition 1.30 again. But q−1 is an even integer, so q−1 = 2kp and the conclusion follows. □ Perfect Numbers An integer n ≥2 is called perfect if the sum of its divisors is equal to 2n; that is, σ(n) = 2n. For example, the numbers 6, 28, 496 are perfect. The perfect numbers are closely related to Mersenne numbers. We first introduce a famous result on even perfect numbers. The “if” part belongs to Euclid and the “only if” part is due to Euler. Theorem 1.51. An even positive integer n is perfect if and only if n = 2k−1Mk for some positive integer k for which Mk is a prime. Proof: First we show the “if” part. Assume that n = 2k−1(2k −1), where Mk = 2k −1 is prime. Because gcd(2k−1, 2k −1) = 1 and the fact that σ is a multiplicative function, it follows that σ(n) = σ(2k−1)σ(2k −1) = (2k −1) · 2k = 2n; that is, n is perfect. Second we show the “only if” part. Assume that n is an even perfect number. Let n = 2tu, where t ≥0 and u is odd. Because n is perfect, we have σ(n) = 2n; hence σ(2tu) = 2t+1u. Using again that σ is multiplicative, we get 2t+1u = σ(2tu) = σ(2t)σ(u) = (2t+1 −1)σ(u). Because gcd(2t+1−1, 2t+1) = 1, it follows that 2t+1 | σ(u); hence σ(u) = 2t+1v for some positive integer v. We obtain u = (2t+1 −1)v. The next step is to show that v = 1. Assume to the contrary that v > 1. Then σ(u) ≥1 + v + 2t+1 −1 + v(2t+1 −1) = (v + 1)2t+1 > v · 2t+1 = σ(u), a contradiction. We get v = 1, hence u = 2t+1 −1 = Mt+1 and σ(u) = 2t+1. If Mt+1 is not a prime, then σ(u) > 2t+1, which is impossible. Finally, n = 2k−1Mk, where k = t + 1. □ Since Mk is a prime only if k is a prime, we can reword Theorem 1.51 as www.riazisara.ir 1. Foundations of Number Theory 73 An even positive integer n is perfect if and only if n = 2k−1Mk for some prime p for which Mp is a prime. Theorem 1.51 sets up a one-to-one correspondence between the prime Mer-senne numbers and the even perfect numbers. The following are two simple re-sults related to odd perfect numbers. Theorem 1.52. If n is an odd perfect number, then the prime factorization of n is of the form n = paq2b1 1 q2b2 2 · · · q2bt t , where both a and p are congruent to 1 modulo 4 and t ≥2. Proof: Let n = pa1 1 pa2 2 · · · pak k be the canonical prime factorization of n. Since n is perfect, we have k i=1 (1 + pi + p2 i + · · · + pai i ) = 2pa1 1 pa2 2 · · · pak k . Since n is odd, there is exactly one i, 1 ≤i ≤k, such that 1 + pi + p2 i + · · · + pai i ≡2 (mod 4). Then ai must be odd. Write ai = 2x + 1 for some integer x. Since p2 i ≡1 (mod 4), we can rewrite the above congruence equation as (x + 1)(pi + 1) ≡2 (mod 4), implying that pi ≡1 (mod 4) and x is even, and so ai ≡1 (mod 4). For j ̸= i with 1 ≤j ≤k, we have 1 + p j + p2 j + · · · + p a j j ≡1 (mod 2), and so j must be even. It follows that n = paq2b1 1 q2b2 2 · · · q2bt t , where both a and p are congruent to 1 modulo 4. It remains to show that t ≥2. Assume to the contrary that t = 1. We have (1 + p + p2 + · · · + pa 1)(1 + q + q2 + · · · + p2b 2 ) = 2paq2b, or pa+1 −1 p −1 · q2b+1 −1 q −1 = 2paq2b. It follows that 2 = p −1 pa p −1 · q − 1 q2b q −1 < p p −1 · q q −1 ≤5 4 · 3 2 = 15 8 , which is not true. Hence our assumption was wrong and t ≥2. □ www.riazisara.ir 74 104 Number Theory Problems In 1980, Hagis proved that t ≥7 and n > 1050. The existence of odd perfect numbers still remains one of the most challenging problems in number theory. www.riazisara.ir 2 Introductory Problems 1. Let 1, 4, . . . and 9, 16, . . . be two arithmetic progressions. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S? 2. Given a sequence of six strictly increasing positive integers such that each number (besides the first) is a multiple of the one before it and the sum of all six numbers is 79, what is the largest number in the sequence? 3. What is the largest positive integer n for which n3 + 100 is divisible by n + 10? 4. Those irreducible fractions! (1) Let n be an integer greater than 2. Prove that among the fractions 1 n , 2 n , . . . , n −1 n , an even number are irreducible. (2) Show that the fraction 12n + 1 30n + 2 is irreducible for all positive integers n. 5. A positive integer is written on each face of a cube. Each vertex is then assigned the product of the numbers written on the three faces intersecting the vertex. The sum of the numbers assigned to all the vertices is equal to 1001. Find the sum of the numbers written on the faces of the cube. www.riazisara.ir 76 104 Number Theory Problems 6. Call a number prime looking if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000? 7. A positive integer k greater than 1 is given. Prove that there exist a prime p and a strictly increasing sequence of positive integers a1, a2, . . . , an, . . . such that the terms of the sequence p + ka1, p + ka2, . . . , p + kan, . . . are all primes. 8. Given a positive integer n, let p(n) be the product of the nonzero digits of n. (If n has only one digit, then p(n) is equal to that digit.) Let S = p(1) + p(2) + · · · + p(999). What is the largest prime factor of S? 9. Let m and n be positive integers such that lcm(m, n) + gcd(m, n) = m + n. Prove that one of the two numbers is divisible by the other. 10. Let n = 231319. How many positive integer divisors of n2 are less than n but do not divide n? 11. Show that for any positive integers a and b, the number (36a + b)(a + 36b) cannot be a power of 2. 12. Compute the sum of the greatest odd divisor of each of the numbers 2006, 2007, . . . , 4012. 13. Compute the sum of all numbers of the form a/b, where a and b are rela-tively prime positive divisors of 27000. www.riazisara.ir 2. Introductory Problems 77 14. L.C.M of three numbers. (1) Find the number of ordered triples (a, b, c) of positive integers for which lcm(a, b) = 1000, lcm(b, c) = 2000, and lcm(c, a) = 2000. (2) Let a, b, and c be integers. Prove that lcm(a, b, c)2 lcm(a, b) lcm(b, c) lcm(c, a) = gcd(a, b, c)2 gcd(a, b) gcd(b, c) gcd(c, a). 15. Let x, y, z be positive integers such that 1 x −1 y = 1 z . Let h be the greatest common divisor of x, y, z. Prove that hxyz and h(y −x) are perfect squares. 16. Let p be a prime of the form 3k + 2 that divides a2 + ab + b2 for some integers a and b. Prove that a and b are both divisible by p. 17. The number 27000001 has exactly four prime factors. Find their sum. 18. Find all positive integers n for which n! + 5 is a perfect cube. 19. Find all primes p such that the number p2 + 11 has exactly six different divisors (including 1 and the number itself). 20. Call a positive integer N a 7-10 double if the digits of the base-7 represen-tation of N form a base-10 number that is twice N. For example, 51 is a 7-10 double because its base-7 representation is 102. What is the largest 7-10 double? 21. If a ≡b (mod n), show that an ≡bn (mod n2). Is the converse true? 22. Let p be a prime, and let 1 ≤k ≤p −1 be an integer. Prove that p −1 k ≡(−1)k (mod p). www.riazisara.ir 78 104 Number Theory Problems 23. Let p be a prime. Show that there are infinitely many positive integers n such that p divides 2n −n. 24. Let n be an integer greater than three. Prove that 1! + 2! + · · · + n! cannot be a perfect power. 25. Let k be an odd positive integer. Prove that (1 + 2 + · · · + n) | (1k + 2k + · · · + nk) for all positive integers n. 26. Let p be a prime greater than 5. Prove that p−4 cannot be the fourth power of an integer. 27. For a positive integer n, prove that σ(1) + σ(2) + · · · + σ(n) ≤n2. 28. Determine all finite nonempty sets S of positive integers satisfying i + j gcd(i, j) is an element of S for all i and j (not necessarily distinct) in S. 29. Knowing that 229 is a nine-digit number all of whose digits are distinct, without computing the actual number determine which of the ten digits is missing. Justify your answer. 30. Prove that for any integer n greater than 1, the number n5 + n4 + 1 is composite. 31. The product of a few primes is ten times as much as the sum of the primes. What are these (not necessarily distinct) primes? 32. A 10-digit number is said to be interesting if its digits are all distinct and it is a multiple of 11111. How many interesting integers are there? www.riazisara.ir 2. Introductory Problems 79 33. Do there exist 19 distinct positive integers that add up to 1999 and have the same sum of digits? 34. Find all prime numbers p and q such that pq divides the product (5p −2p)(5q −2q). 35. Prove that there are infinitely many numbers not containing the digit 0 that are divisible by the sum of their digits. 36. Prove that any number consisting of 2n identical digits has at least n distinct prime factors. 37. Let a and b be two relatively prime positive integers, and consider the arith-metic progression a, a + b, a + 2b, a + 3b, . . . . (1) Prove that there are infinitely many terms in the arithmetic progression that have the same prime divisors. (2) Prove that there are infinitely many pairwise relatively prime terms in the arithmetic progression. 38. Let n be a positive integer. (1) Evaluate gcd(n! + 1, (n + 1)! + 1). (2) Let a and b be positive integers. Prove that gcd(na −1, nb −1) = ngcd(a,b) −1. (3) Let a and b be positive integers. Prove that gcd(na +1, nb+1) divides ngcd(a,b) + 1. (4) Let m be a positive integer with gcd(m, n) = 1. Express gcd(5m + 7m, 5n + 7n) in terms of m and n. www.riazisara.ir 80 104 Number Theory Problems 39. Bases? What bases? (1) Determine whether it is possible to find a cube and a plane such that the distances from the vertices of the cube to the plane are 0, 1, 2, . . . , 7. (2) The increasing sequence 1, 3, 4, 9, 10, 12, 13, . . . consists of all those positive integers that are powers of 3 or sums of distinct powers of 3. Find the 100th term of this sequence (where 1 is the 1st term, 3 is the 2nd term, and so on). 40. Fractions in modular arithmetic. (1) Let a be the integer such that 1 + 1 2 + 1 3 + · · · + 1 23 = a 23!. Compute the remainder when a is divided by 13. (2) Let p > 3 be a prime, and let m and n be relatively prime integers such that m n = 1 12 + 1 22 + · · · + 1 (p −1)2 . Prove that m is divisible by p. (3) Let p > 3 be a prime. Prove that Let p > 3 be a prime. Prove that p2 | (p −1)!  1 + 1 2 + · · · + 1 p −1 . 41. Find all pairs (x, y) of nonnegative integers such that x2 + 3y and y2 + 3x are simultaneously perfect squares. 42. First digit? Not the last digit? Are your sure? (1) Given that 22004 is a 604-digit number with leading digit 1, determine the number of elements in the set {20, 21, 22, . . . , 22003} with leading digit 4. (2) Let k be a positive integer and let n = n(k) be a positive integer such that in decimal representation 2n and 5n begin with the same k digits. What are these digits? www.riazisara.ir 2. Introductory Problems 81 43. What are those missing digits? (1) Determine the respective last digit (unit digit) of the numbers 3100171002131003 and 7777...7   1001 7’s . (2) Determine the last three digits of the number 200320022001. (3) The binomial coefficient 99 19  is a 21-digit number: 107,196,674,080,761,936,xyz. Find the three-digit number xyz. (4) Find the smallest positive integer whose cube ends in 888. 44. Let p ≥3 be a prime, and let {a1, a2, . . . , ap−1} and {b1, b2, . . . , bp−1} be two sets of complete residue classes modulo p. Prove that {a1b1, a2b2, . . . , ap−1bp−1} is not a complete set of residue classes modulo p. 45. Let p ≥3 be a prime. Determine whether there exists a permutation (a1, a2, . . . , ap−1) of (1, 2, . . . , p −1) such that the sequence {iai}p−1 i=1 contains p −2 distinct congruence classes modulo p. 46. Prove that any positive integer less than n! can be represented as a sum of no more than n positive integer divisors of n!. 47. Let n > 1 be an odd integer. Prove that n does not divide 3n + 1. www.riazisara.ir 82 104 Number Theory Problems 48. Let a and b be positive integers. Prove that the number of solutions (x, y, z) in nonnegative integers to the equation ax + by + z = ab is 1 2[(a + 1)(b + 1) + gcd(a, b) + 1]. 49. Order! Order, please! (1) Let p be an odd prime, and let q and r be primes such that p divides qr + 1. Prove that either 2r | p −1 or p | q2 −1. (2) Let a > 1 and n be given positive integers. If p is a prime divisor of a2n + 1, prove that p −1 is divisible by 2n+1. 50. Prove that (n −1)! n(n + 1) is even for every positive integer n. 51. Determine all the positive integers m each of which satisfies the following property: there exists a unique positive integer n such that there exist rect-angles that can be divided into n congruent squares and also into n + m congruent squares. 52. Determine all positive integers n such that n has a multiple whose digits are nonzero. www.riazisara.ir 3 Advanced Problems 1. (a) Prove that the sum of the squares of 3, 4, 5, or 6 consecutive integers is not a perfect square. (b) Give an example of 11 consecutive positive integers the sum of whose squares is a perfect square. 2. Let S(x) be the sum of the digits of the positive integer x in its decimal representation. (a) Prove that for every positive integer x, S(x) S(2x) ≤5. Can this bound be improved? (b) Prove that S(x) S(3x) is not bounded. 3. Most positive integers can be expressed as a sum of two or more consecu-tive positive integers. For example, 24 = 7 + 8 + 9 and 51 = 25 + 26. A positive integer that cannot be expressed as a sum of two or more consec-utive positive integers is therefore interesting. What are all the interesting integers? 4. Set S = {105, 106, . . . , 210}. Determine the minimum value of n such that any n-element subset T of S contains at least two non-relatively prime elements. 5. The number 99 . . . 99   1997 9’s www.riazisara.ir 84 104 Number Theory Problems is written on a blackboard. Each minute, one number written on the black-board is factored into two factors and erased, each factor is (independently) increased or diminished by 2, and the resulting two numbers are written. Is it possible that at some point (after the first minute) all of the numbers on the blackboard equal 9? 6. Let d be any positive integer not equal to 2, 5, or 13. Show that one can find distinct a, b in the set {2, 5, 13, d} such that ab −1 is not a perfect square. 7. A heap of balls consists of one thousand 10-gram balls and one thousand 9.9-gram balls. We wish to pick out two heaps of balls with equal numbers of balls in them but different total weights. What is the minimal number of weighings needed to do this? (The balance scale reports the weight of the objects in the left pan minus the weight of the objects in the right pan.) 8. We are given three integers a, b, and c such that a, b, c, a +b−c, a +c−b, b + c −a, and a + b + c are seven distinct primes. Let d be the difference between the largest and smallest of these seven primes. Suppose that 800 is an element in the set {a+b, b+c, c+a}. Determine the maximum possible value of d. 9. Prove that the sum S(m, n) = 1 m + 1 m + 1 + · · · + 1 m + n is not an integer for any given positive integers m and n. 10. For all positive integers m > n, prove that lcm(m, n) + lcm(m + 1, n + 1) > 2mn √m −n . 11. Prove that each nonnegative integer can be represented in the form a2 + b2 −c2, where a, b, and c are positive integers with a < b < c. 12. Determine whether there exists a sequence of strictly increasing positive integers {ak}∞ k=1 such that the sequence {ak + a}∞ k=1 contains only finitely many primes for all integers a. www.riazisara.ir 3. Advanced Problems 85 13. Prove that for different choices of signs + and −the expression ±1 ± 2 ± 3 ± · · · ± (4n + 1) yields all odd positive integers less than or equal to (2n + 1)(4n + 1). 14. Let a and b be relatively prime positive integers. Show that ax + by = n has nonnegative integer solutions (x, y) for all integers n > ab −a −b. What if n = ab −a −b? 15. The sides of a triangle have integer lengths k, m, and n. Assume that k > m > n and  3k 104  =  3m 104  =  3n 104  . Determine the minimum value of the perimeter of the triangle. 16. Consider the following two-person game. A number of pebbles are lying on a table. Two players make their moves alternately. A move consists in taking off the table x pebbles, where x is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the player who goes second has a winning strategy? 17. Prove that the sequence 1, 11, 111, . . . contains an infinite subsequence whose terms are pairwise relatively prime. 18. Let m and n be integers greater than 1 such that gcd(m, n −1) = gcd(m, n) = 1. Prove that the first m −1 terms of the sequence n1, n2, . . . , where n1 = mn + 1 and nk+1 = n · nk + 1, k ≥1, cannot all be primes. 19. Find all positive integers m such that the fourth power of the number of positive divisors of m equals m. www.riazisara.ir 86 104 Number Theory Problems 20. (1) Show that it is possible to choose one number out of any 39 consecu-tive positive integers having the sum of its digits divisible by 11. (2) Find the first 38 consecutive positive integers none of which has the sum of its digits divisible by 11. 21. Find the largest integer n such that n is divisible by all positive integers less than 3 √n. 22. Show that for any fixed positive integer n, the sequence 2, 22, 222, 2222 , . . . (mod n) is eventually constant. (The tower of exponents is defined by a1 = 2 and ai+1 = 2ai for every positive integer i.) 23. Prove that for n ≥5, fn + fn−1 −1 has at least n + 1 prime factors, where fn = 22n + 1. 24. Prove that any integer can be written as the sum of the cubes of five integers, not necessarily distinct. 25. Integer or fractional parts? (1) Find all real numbers x such that x⌊x⌊x⌊x⌋⌋⌋= 88. (2) Show that the equation {x3} + {y3} = {z3} has infinitely many rational noninteger solutions. 26. Let n be a given positive integer. If p is a prime divisor of the Fermat number fn, prove that p −1 is divisible by 2n+2. 27. The sequence {an}∞ n=1 = {1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, . . . } of positive integers is formed by taking one odd integer, then two even integers, then three odd integers, etc. Express an in closed form. www.riazisara.ir 3. Advanced Problems 87 28. Prove that for each n ≥2, there is a set S of n integers such that (a −b)2 divides ab for every distinct a, b ∈S. 29. Show that there exist infinitely many positive integers n such that the largest prime divisor of n4 + 1 is greater than 2n. 30. For a positive integer k, let p(k) denote the greatest odd divisor of k. Prove that for every positive integer n, 2n 3 < p(1) 1 + p(2) 2 + · · · + p(n) n < 2(n + 1) 3 . 31. If pt is an odd prime power and m is an integer relatively prime to both p and p −1, then for any a and b relatively prime to p, am ≡bm (mod pt) if and only if a ≡b (mod pt). 32. Prove that for each prime p ≥7, there exists a positive integer n and inte-gers x1, . . . , xn, y1, . . . , yn not divisible by p such that ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ x2 1 + y2 1 ≡x2 2 (mod p), x2 2 + y2 2 ≡x2 3 (mod p), . . . x2 n + y2 n ≡x2 1 (mod p). 33. For every positive integer n, prove that σ(1) 1 + σ(2) 2 + · · · + σ(n) n ≤2n. 34. Prove that the system x6 + x3 + x3y + y = 147157, x3 + x3y + y2 + y + z9 = 157147, has no solutions in integers x, y, and z. www.riazisara.ir 88 104 Number Theory Problems 35. What is the smallest number of weighings on a balance scale needed to identify the individual weights of a set of objects known to weigh 1, 3, 32, . . . , 326 in some order? (The balance scale reports the weight of the objects in the left pan minus the weight of the objects in the right pan.) 36. Let λ be the positive root of the equation t2 −1998t −1 = 0. Define the sequence x0, x1, . . . by setting x0 = 1, xn+1 = ⌊λxn⌋ (n ≥0). Find the remainder when x1998 is divided by 1998. 37. Determine (with proof) whether there is a subset X of the integers with the following property: for any integer n there is exactly one solution of a + 2b = n with a, b ∈X. 38. The number xn is defined as the last digit in the decimal representation of the integer √ 2 n (n = 1, 2, . . . ). Determine whether the sequence x1, x2, . . . , xn, . . . is periodic. 39. Prove that every integer n can be represented in infinitely many ways as n = ±12 ± 22 ± · · · ± k2 for a convenient k and a suitable choice of the signs + and −. 40. Let n be a given integer with n ≥4. For a positive integer m, let Sm denote the set {m, m + 1, . . . , m + n −1}. Determine the minimum value of f (n) such that every f (n)-element subset of Sm (for every m) contains at least three pairwise relatively prime elements. 41. Find the least positive integer r such that for any positive integers a, b, c, d, ((abcd)!)r is divisible by the product of (a!)bcd+1, (b!)acd+1, (c!)abd+1, (d!)abc+1, ((ab)!)cd+1, ((bc)!)ad+1, ((cd)!)ab+1, ((ac)!)bd+1, ((bd)!)ac+1, ((ad)!)bc+1, ((abc)!)d+1, ((abd)!)c+1, ((acd)!)b+1, ((bcd)!)a+1. www.riazisara.ir 3. Advanced Problems 89 42. Two classics on L.C.M. (1) Let a0 < a1 < a2 < · · · < an be positive integers. Prove that 1 lcm(a0, a1) + 1 lcm(a1, a2) + · · · + 1 lcm(an−1, an) ≤1 −1 2n . (2) Several positive integers are given not exceeding a fixed integer con-stant m. Prove that if every positive integer less than or equal to m is not divisible by any pair of the given numbers, then the sum of the reciprocals of these numbers is less than 3 2. 43. For a positive integer n, let r(n) denote the sum of the remainders of n divided by 1, 2, . . . , n. Prove that there are infinitely many n such that r(n) = r(n −1). 44. Two related IMO problems. (1) A wobbly number is a positive integer whose digits are alternately nonzero and zero with the units digit being nonzero. Determine all positive integers that do not divide any wobbly numbers. (2) A positive integer is called alternating if among any two consecutive digits in its decimal representation, one is even and the other is odd. Find all positive integers n such that n has a multiple that is alternat-ing. 45. Let p be an odd prime. The sequence (an)n≥0 is defined as follows: a0 = 0, a1 = 1, . . . , ap−2 = p −2, and for all n ≥p −1, an is the least positive integer that does not form an arithmetic sequence of length p with any of the preceding terms. Prove that for all n, an is the number obtained by writing n in base p −1 and reading the result in base p. 46. Determine whether there exists a positive integer n such that n is divisible by exactly 2000 different prime numbers, and 2n + 1 is divisible by n. 47. Two cyclic symmetric divisibility relations. (1) [Russia 2000] Determine whether there exist pairwise relatively prime integers a, b, and c with a, b, c > 1 such that b | 2a + 1, c | 2b + 1, a | 2c + 1. www.riazisara.ir 90 104 Number Theory Problems (2) [TST 2003, by Reid Barton] Find all ordered triples of primes (p, q,r) such that p | qr + 1, q | r p + 1, r | pq + 1. 48. Let n be a positive integer, and let p1, p2, . . . , pn be distinct primes greater than 3. Prove that 2p1 p2···pn + 1 has at least 4n divisors. 49. Let p be a prime, and let {ak}∞ k=0 be a sequence of integers such that a0 = 0, a1 = 1, and ak+2 = 2ak+1 −pak for k = 0, 1, 2, . . . . Suppose that −1 appears in the sequence. Find all possible values of p. 50. Let F be a set of subsets of the set {1, 2, . . . , n} such that (1) if A is an element of F, then A contains exactly three elements; (2) if A and B are two distinct elements in F, A and B share at most one common element. Let f (n) denote the maximum number of elements in F. Prove that (n −1)(n −2) 6 ≤f (n) ≤(n −1)n 6 . 51. Determine all positive integers k such that τ(n2) τ(n) = k, for some n. 52. Let n be a positive integer greater than two. Prove that the Fermat number fn has a prime divisor greater than 2n+2(n + 1). www.riazisara.ir 4 Solutions to Introductory Problems 1. [AMC10B 2004] Let 1, 4, . . . and 9, 16, . . . be two arithmetic progres-sions. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S? Solution: The smallest number that appears in both sequences is 16. Since the least common multiple of 3 and 7 (the two common differences of the progressions) is 21, numbers appear in both sequences only if they are in the form 16+21k, where k is a nonnegative integer. The largest k such that 7k + 9 ≤2004 is k = 285. Hence there are 286 numbers each of which appears in both progressions. Thus the answer is 4008 −286 = 3722. 2. [HMMT 2004] Given a sequence of six strictly increasing positive integers such that each number (besides the first) is a multiple of the one before it and the sum of all six numbers is 79, what is the largest number in the sequence? Solution: Let a1 < a2 < · · · < a6 be the six numbers. If a4 ≥12, then a5 ≥2a4 ≥24 and a6 ≥2a5 ≥48, implying that a4 +a5 +a6 ≥84, which violates the conditions of the problem. Hence a4 < 12. Then the only way we can have the required divisibilities among the first four numbers is if they are a1 = 1, a2 = 2, a3 = 4, and a4 = 8. We write a5 = ma4 = 8m and a6 = na5 = 8mn for integers m and n with m, n ≥2. We get 8m + 8mn = 79 −(1 + 2 + 4 + 8) = 64, or m(1 + n) = 8. This leads to the unique solution m = 2 and n = 3. Hence the answer is a6 = 48. 3. [AIME 1986] What is the largest positive integer n for which n3 + 100 is divisible by n + 10? www.riazisara.ir 92 104 Number Theory Problems Solution: By division we find that n3 + 100 = (n + 10)(n2 −10n + 100) −900. Thus, if n + 10 divides n3 + 100, then it must also divide 900. Moreover, since n is maximized whenever n + 10 is, and since the largest divisor of 900 is 900, we must have n + 10 = 900. Therefore, n = 890. 4. Those irreducible fractions! (1) Let n be an integer greater than 2. Prove that among the fractions 1 n , 2 n , . . . , n −1 n , an even number of them are irreducible. (2) Show that the fraction 12n + 1 30n + 2 is irreducible for all positive integers n. Proof: We prove part (1) via a parity argument, and we establish part (2) applying the Euclidean algorithm. (1) The fraction k n is irreducible if and only if the fraction n−k n is irre-ducible, because gcd(k, n) = gcd(n −k, n). If the fractions k n and n−k n are distinct for all k, then pairing up yields an even number of irreducible fractions. If k n = n−k n for some k, then n = 2k and so k n = k 2k = 1 2 is reducible and the problem reduces to the previous case. (2) Note that gcd(30n + 2, 12n + 1) = gcd(6n, 12n + 1) = gcd(6n, 1) = 1, from which the desired result follows. 5. A positive integer is written on each face of a cube. Each vertex is then assigned the product of the numbers written on the three faces intersecting the vertex. The sum of the numbers assigned to all the vertices is equal to 1001. Find the sum of the numbers written on the faces of the cube. www.riazisara.ir 4. Solutions to Introductory Problems 93 Solution: Let a, b, c, d, e, and f be the numbers written on the faces, with a and f , b and d, c and e written on opposite faces. We are given that 1001 = abc + abe + acd + ade + bcf + bef + cd f + def = (a + f )(b + d)(c + e). (We can realize this factorization by noticing that the product xyz appears exactly once if and only if x and y, y and z, z and x are not written on the opposite faces.) Since 1001 = 7 · 11 · 13 and each of a + f, b + d, and c + e are greater than 1, it follows that {a + f, b + d, c + e} = {7, 11, 13}, implying that the answer is a + b + c + d + e + f = 7 + 11 + 13 = 31. 6. [AMC12A 2005] Call a number prime looking if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000? Solution: Of the numbers less than 1000, 999 2 = 499 of them are divisi-ble by two, 999 3 = 333 are divisible by 3, and 999 5 = 199 are divisible by 5. There are 999 6 = 166 multiples of 6, 999 10 = 99 multiples of 10, and 999 15 = 66 multiples of 15. Finally, there are 999 30 = 33 multiples of 30. By the inclusion and exclusion principle there are 499 + 333 + 199 −166 −99 −66 + 33 = 733 numbers that are divisible by at least one of 2, 3, and 5. Of the remaining 999 −733 = 266 numbers, 165 are primes other than 2, 3, or 5. Note that 1 is neither prime nor composite. This leaves exactly 100 prime-looking numbers. 7. A positive integer k greater than 1 is given. Prove that there exist a prime p and a strictly increasing sequence of positive integers a1, a2, . . . , an, . . . such that the terms of the sequence p + ka1, p + ka2, . . . , p + kan, . . . are all primes. Proof: The pigeonhole principle provides an elegant solution. There is nothing to be afraid of, just infinitely many pigeons in finitely many pi-geonholes. www.riazisara.ir 94 104 Number Theory Problems For each i = 1, 2, . . . , k −1 denote by Pi the set of all primes congruent to i modulo k. Each prime (except possibly k itself) is contained in exactly one of the sets P1, P2, . . . , Pk−1. Because there are infinitely many primes, at least one of these sets is infinite, say Pi. Let p = x1 < x2 < · · · < xn < · · · be its elements arranged in increasing order, and an = xn+1 −p k for every positive integer n. Then the p + kan simply run through the members of Pi, beginning at x2. The numbers an are positive integers. The prime p and the strictly increasing sequence a1, a2, . . . , an, . . . have the desired properties. 8. [AIME 1994] Given a positive integer n, let p(n) be the product of the nonzero digits of n. (If n has only one digit, then p(n) is equal to that digit.) Let S = p(1) + p(2) + · · · + p(999). What is the largest prime factor of S? Solution: Consider each positive integer less than 1000 to be a three-digit number by prefixing 0’s to numbers with fewer than three digits. The sum of the products of the digits of all such positive numbers is (0 · 0 · 0 + 0 · 0 · 1 + · · · + 9 · 9 · 9) −0 · 0 · 0 = (0 + 1 + · · · + 9)3 −0. However, p(n) is the product of nonzero digits of n. The sum of these products can be found by replacing 0 by 1 in the above expression, since ignoring 0’s is equivalent to thinking of them as 1’s in the products. (Note that the final 0 in the above expression becomes a 1 and compensates for the contribution of 000 after it is changed to 111.) Hence S = 463 −1 = (46 −1)(462 + 46 + 1) = 33 · 5 · 7 · 103, and the largest prime factor is 103. 9. [Russia 1995] Let m and n be positive integers such that lcm(m, n) + gcd(m, n) = m + n. Prove that one of the two numbers is divisible by the other. www.riazisara.ir 4. Solutions to Introductory Problems 95 First Proof: Let d = gcd(m, n). We write m = ad and n = bd. Then gcd(a, b) = 1 and lcm(m, n) = mn gcd(m, n) = abd. The given equation becomes abd +d = ad +bd, or ab −a −b +1 = 0. It follows that (a −1)(b −1) = 0, implying that either a = 1 or b = 1; that is, either m = d, n = bd = bm or n = d, m = an. Second Proof: Because lcm(m, n) · gcd(m, n) = mn, it follows that lcm(m, n) and gcd(m, n) as well as m, n are roots of x2−(m +n)x +mn = 0. Hence {lcm(m, n), gcd(m, n)} = {m, n} and the conclusion follows. 10. [AIME 1995] Let n = 231319. How many positive integer divisors of n2 are less than n but do not divide n? First Solution: Let n = prqs, where p and q are distinct primes. Then n2 = p2rq2s, so n2 has (2r + 1)(2s + 1) factors. For each factor less than n, there is a corresponding factor greater than n. By excluding the factor n, we see that there must be (2r + 1)(2s + 1) −1 2 = 2rs + r + s factors of n2 that are less than n. Because n has (r + 1)(s + 1) factors (including n itself), and because every factor of n is also a factor of n2, there are 2rs + r + s −[(r + 1)(s + 1) −1] = rs factors of n2 that are less than n but not factors of n. When r = 31 and s = 19, there are rs = 589 such factors. Second Solution: (By Chengde Feng) A positive integer divisor d of n2 is less than n but does not divide n if and only if d = 231+a319−b if 2a < 3b, 231−a319+b if 2a > 3b, where a and b are integers such that 1 ≤a ≤31 and 1 ≤b ≤19. Since 2a ̸= 3b for positive integers a and b, there are 19×31 = 589 such divisors. www.riazisara.ir 96 104 Number Theory Problems 11. [APMO 1998] Show that for any positive integers a and b, the number (36a + b)(a + 36b) cannot be a power of 2. Proof: Write a = 2c · p, b = 2d · q, with p and q odd. Assume without loss of generality that c ≥d. Then 36a + b = 36 · 2c p + 2dq = 2d(36 · 2c−d p + q). Consequently, (36a + b)(36b + a) = 2d(36 · 2c−d p + q)(36b + a) has the nontrivial odd factor 36 · 2c−d p + q, and thus is not a power of 2. 12. Compute sum of the greatest odd divisor of each of the numbers 2006, 2007, . . . , 4012. Solution: For a positive integer n, let p(n) denote its greatest odd divisor. We can write n = 2k · p(n) for some nonnegative integer k. If two positive integers n1 and n2 are such that p(n1) = p(n2), then one is at least twice the other. Because no number from 2007, 2008, . . . , 4012 is twice another such num-ber, p(2007), p(2008), . . . , p(4012) are 2006 distinct odd positive integers. Also note that these odd numbers belong to the set {1, 3, 5, . . . , 4011}, which also consists of exactly 2006 elements. It follows that {p(2007), p(2008), . . . , p(4012)} = {1, 3, . . . , 4011}. Hence the desired sum is equal to p(2006) + 1 + 3 + · · · + 4011 = 1003 + 20062 = 1003 · 4013 = 4025039. 13. Compute the sum of all numbers of the form a/b, where a and b are rela-tively prime positive divisors of 27000. www.riazisara.ir 4. Solutions to Introductory Problems 97 Solution: Because 27000 = 233353, each a/b can be written in the form of 2a3b5c, where a, b, c are integers in the interval [−3, 3]. It follows that each a/b appears exactly once in the expansion of (2−3 + 2−2 + · · · + 23)(3−3 + 3−2 + · · · + 33)(5−3 + 5−2 + · · · + 53). It follows that the desired sum is equal to 1 233353 · 27 −1 2 −1 · 37 −1 3 −1 · 57 −1 5 −1 = (27 −1)(37 −1)(57 −1) 263353 . 14. L.C.M of three numbers. (1) [AIME 1987] Find the number of ordered triples (a, b, c) of posi-tive integers for which lcm(a, b) = 1000, lcm(b, c) = 2000, and lcm(c, a) = 2000. (2) Let a, b, and c be integers. Prove that lcm(a, b, c)2 lcm(a, b) lcm(b, c) lcm(c, a) = gcd(a, b, c)2 gcd(a, b) gcd(b, c) gcd(c, a). Solution: We have two different approaches to these two parts. For part (1), we deal with the L.C.M. and G.C.D. of three integers via pairwise L.C.M. and G.C.D. of two integers. For part (2), we use prime factoriza-tions. (1) Because both 1000 and 2000 are of the form 2m5n, the numbers a, b, and c must also be of this form. We set a = 2m15n1, b = 2m25n2, c = 2m35n3, where the mi and ni are nonnegative integers for i = 1, 2, 3. Then the following equalities must hold: max{m1, m2} = 3, max{m2, m3} = 4, max{m3, m1} = 4 (∗) and max{n1, n2} = 3, max{n2, n3} = 3, max{n3, n1} = 3. (∗∗) From (∗), we must have m3 = 4, and either m1 or m2 must be 3, while the other one can take the values of 0, 1, 2, or 3. There are 7 such ordered triples, namely (0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 0, 4), (3, 1, 4), (3, 2, 4), and (3, 3, 4). www.riazisara.ir 98 104 Number Theory Problems To satisfy (∗∗), two of n1, n2, and n3 must be 3, while the third one ranges through the values of 0, 1, 2, and 3. The number of such ordered triples is 10; they are (3, 3, 0), (3, 3, 1), (3, 3, 2), (3, 0, 3), (3, 1, 3), (3, 2, 3), (0, 3, 3), (1, 3, 3), (2, 3, 3), and (3, 3, 3). Because the choice of (m1, m2, m3) is independent of the choice of (n1, n2, n3), they can be chosen in 7 · 10 = 70 different ways. This is the number of ordered triples (a, b, c) satisfying the given conditions. (2) Let a = pα1 1 · · · pαn n , b = pβ1 1 · · · pβn n , and c = pγ1 1 · · · pγn n , where p1, . . . , pn are distinct primes, and a1, . . . , an, b1, . . . , bn, c1, . . . , cn are nonzero integers. Then lcm(a, b, c)2 lcm(a, b) lcm(b, c) lcm(c, a) = .n i=1 p2 max{αi,βi,γi} i .n i=1 pmax{αi,βi} i .n i=1 pmax{βi,γi} i .n i=1 pmax{γi,αi} i = n i=1 p2 max{αi,βi,γi}−max{αi,βi}−max{βi,γi}−max{γi,αi} i and gcd(a, b, c)2 gcd(a, b) gcd(b, c) gcd(c, a) = .n i=1 p2 min{αi,βi,γi} i .n i=1 pmin{αi,βi} i .n i=1 pmin{βi,γi} i .n i=1 pmin{γi,αi} i = n i=1 p2 min{αi,βi,γi}−min{αi,βi}−min{βi,γi}−min{γi,αi} i It suffices to show that for each nonnegative numbers α, β, and γ , 2 max{α, β, γ } −max{α, β} −max{β, γ } −max{γ, α} = 2 min{α, β, γ } −min{α, β} −min{β, γ } −min{γ, α}. By symmetry, we may assume that α ≤β ≤γ . It is not difficult to deduce that both sides are equal −β, completing our proof. As a side result from the proof of (2), we note that lcm(a, b) lcm(b, c) lcm(c, a) lcm(a, b, c)2 and gcd(a, b) gcd(b, c) gcd(c, a) gcd(a, b, c)2 are equal integers. www.riazisara.ir 4. Solutions to Introductory Problems 99 15. [UK 1998] Let x, y, z be positive integers such that 1 x −1 y = 1 z . Let h be the greatest common divisor of x, y, z. Prove that hxyz and h(y −x) are perfect squares. Proof: Let x = ha, y = hb, z = hc. Then a, b, c are positive integers such that gcd(a, b, c) = 1. Let gcd(a, b) = g. So a = ga′, b = gb′ and a′ and b′ are positive integers such that gcd(a′, b′) = gcd(a′ −b′, b′) = gcd(a′, a′ −b′) = 1. We have 1 a −1 b = 1 c ⇐ ⇒c(b −a) = ab ⇐ ⇒c(b′ −a′) = a′b′g. So g|c and gcd(a, b, c) = g = 1. Therefore gcd(a, b) = 1 and gcd(b − a, ab) = 1. Thus b −a = 1 and c = ab. Now hxyz = h4abc = (h2ab)2 and h(y −x) = h2 are both perfect squares, as desired. 16. Let p be a prime of the form 3k + 2 that divides a2 + ab + b2 for some integers a and b. Prove that a and b are both divisible by p. Proof: We approach indirectly by assuming that p does not divide a. Because p divides a2 + ab + b2, it also divides a3 −b3 = (a −b)(a2 + ab + b2), so a3 ≡b3 (mod p). Hence a3k ≡b3k (mod p). Hence p does not divide b either. Applying Fermat’s little theorem yields a p−1 ≡bp−1 ≡1 (mod p), or a3k+1 ≡b3k+1 (mod p). Because p is relatively prime to a, we conclude that a ≡b (mod p). This, combined with a2 + ab + b2 ≡0 (mod p), implies 3a2 ≡0 (mod p). Because p ̸= 3, it turns out that p divides a, which is a contradiction. www.riazisara.ir 100 104 Number Theory Problems 17. [HMMT 2005] The number 27000001 has exactly four prime factors. Find their sum. Solution: Since x3+1 = (x +1)(x2−x +1) and x2−y2 = (x +y)(x −y), it follows that 27000001 = 3003 + 1 = (300 + 1)(3002 −300 + 1) = 301(3002 + 2 · 300 + 1 −900) = 301[(300 + 1)2 −900] = 301(3012 −302) = 301 · 331 · 271 = 7 · 43 · 271 · 331. Hence the answer is 7 + 43 + 271 + 331 = 652. 18. Find all positive integers n for which n! + 5 is a perfect cube. First Solution: The only answer is n = 5. One checks directly that n! + 5 is not a perfect cube for n = 1, 2, 3, 4, 6, 7, 8, 9 and that 5! + 5 is a perfect cube. If n! + 5 were a perfect cube for n > 9, then, since it is a multiple of 5, n! + 5 would be a multiple of 125. However, this is not true, since n! is a multiple of 125 for n > 9, but 5 is not. Thus the only positive integer with the desired property is n = 5. Second Solution: Again, we check the cases n = 1, 2, . . . , 6 directly. For n ≥7, n! + 5 ≡5 (mod 7), which is not a cubic residue class modulo 7. (The only cubic residue classes modulo 7 are 0 and ±1.) 19. [Russia 1995] Find all primes p such that the number p2 + 11 has exactly six different divisors (including 1 and the number itself). Solution: For p ̸= 3, p2 ≡1 (mod 3), and so 3 | (p2 + 11). Similarly, for p ̸= 2, p2 ≡1 (mod 4) and so 4 | (p2 + 11). Except in these two cases, then, 12 | (p2 + 11); since 12 itself has 6 divisors (1, 2, 3, 4, 6, 12) and p2 + 11 > 12 for p > 1, p2 + 11 must have more than 6 divisors. The only cases to check are p = 2 and p = 3. If p = 2, then p2 + 11 = 15, which has only 4 divisors (1, 3, 5, 15), while if p = 3, then p2 + 11 = 20, which indeed has 6 divisors {1, 2, 4, 5, 10, 20}. Hence p = 3 is the only solution. www.riazisara.ir 4. Solutions to Introductory Problems 101 20. [AIME 2001] Call a positive integer N a 7-10 double if the digits of the base-7 representation of N form a base-10 number that is twice N. For example, 51 is a 7-10 double because its base-7 representation is 102. What is the largest 7-10 double? Solution: Suppose that ak7k +ak−17k−1 +· · ·+a272 +a17+a0 is a 7-10 double, with ak ̸= 0. In other words, ak10k + ak−110k−1 + · · · + a2102 + a110 + a0 is twice as large, so that ak(10k −2 · 7k) + ak−1(10k−1 −2 · 7k−1) + · · · + a1(10 −2 · 7) + ao(1 −2) = 0. Because the coefficient of ai in this equation is negative only when i = 0 and i = 1, and no ai is negative, it follows that k is at least 2. Because the coefficient of ai is at least 314 when i > 2, and because no ai exceeds 6, it follows that k = 2 and 2a2 = 4a1 + a0. To obtain the largest possible 7-10 double, first try a2 = 6. Then the equation 12 = 4a1 + a0 has a1 = 3 and a0 = 0 as the solution with the greatest possible value of a1. The largest 7-10 double is therefore 6 · 49 + 3 · 7 = 315. 21. If a ≡b (mod n), show that an ≡bn (mod n2). Is the converse true? Proof: From a ≡b (mod n) it follows that a = b + qn for some integer q. By the binomial theorem we obtain an −bn = (b + qn)n −bn = n 1 bn−1qn + n 2 bn−2q2n2 + · · · + n n qnnn = n2  bn−1q + n 2 bn−2q2 + · · · + n n qnnn−2 , implying that an ≡bn (mod n2). The converse is not true because, for instance, 34 ≡14 (mod 42) but 3 ̸≡1 (mod 4). 22. Let p be a prime, and let 1 ≤k ≤p −1 be an integer. Prove that p −1 k ≡(−1)k (mod p). www.riazisara.ir 102 104 Number Theory Problems First Proof: We induct on k. The conclusion is clearly true for k = 1, since p −1 1 = p −1 ≡−1 (mod p). Assume that the conclusion is true for k = i, where 1 ≤i ≤p −2. It is well known (and easy to check by direct computation) that p −1 i + p −1 i −1 = p i . By Corollary 1.10, we have p −1 i + p −1 i −1 ≡0 (mod p). By the induction hypothesis, we have p −1 i ≡− p −1 i −1 ≡−(−1)i−1 ≡(−1)i (mod p), completing our induction. Second Proof: Because p −1 k = (p −1)(p −2) · · · (p −k) k! is an integer and gcd(k!, p) = 1, it suffices to show that (p −1)(p −2) · · · (p −k) ≡(−1)k · k! (mod p), which is evident. 23. Let p be a prime. Show that there are infinitely many positive integers n such that p divides 2n −n. Proof: If p = 2, p divides 2n −n for every even positive integer n. We assume that p is odd. By Fermat’s little theorem, 2p−1 ≡1 (mod p). It follows that 2(p−1)2k ≡1 ≡(p −1)2k (mod p); that is, p divides 2n −n for n = (p −1)2k. www.riazisara.ir 4. Solutions to Introductory Problems 103 24. Let n be an integer greater than three. Prove that 1! + 2! + · · · + n! cannot be a perfect power. Proof: For n = 4, we have 1! + 2! + 3! + 4! = 33, which is not a perfect power. For k ≥5, k! ≡0 (mod 10). It follows that for n ≥5, 1! + 2! + 3! + 4! + · · · + n! ≡3 (mod 10), so it cannot be a perfect square, or an even power, for this reason. For odd powers, the following argument settles all cases: one checks the claim for n < 9 directly; for k ≥9, k! is a multiple of 27, while 1! + 2! + · · · + 8! is a multiple of 9, but not 27. Hence 1! + 2! + · · · + n! cannot be a cube or higher power. 25. Let k be an odd positive integer. Prove that (1 + 2 + · · · + n) | (1k + 2k + · · · + nk) for all positive integers n. Proof: We consider two cases. In the first case, we assume that n is odd and write n = 2m + 1. Then 1 + 2 + · · · + n = (m + 1)(2m + 1). We have 1k + 2k + · · · + nk = 1k + 2k + · · · + (2m + 1)k = [1k + (2m + 1)k] + [2k + (2m)k] + · · · + [mk + (m + 2)k] + (m + 1)k. Since k is odd, x + y is a factor of xk + yk. Hence 2m + 2 divides ik + (2m + 2 −i)k for i = 1, 2, . . . , m. Consequently, m + 1 divides 1k + 2k + · · · + nk. Likewise, we have 1k + 2k + · · · + nk = 1k + 2k + · · · + (2m + 1)k = [1k + (2m)k] + [2k + (2m −1)k] + · · · + [mk + (m + 1)k] + (2m + 1)k. Hence 2m+1 divides ik +(2m+1−i)k for i = 1, 2, . . . , m. Consequently, 2m + 1 divides 1k + 2k + · · · + nk. We have shown that each of m + 1 and 2m + 1 divides 1k + 2k + · · · + nk. Since gcd(m + 1, 2m + 1) = 1, we conclude that (m + 1)(2m + 1) divides 1k + 2k + · · · + nk. In the second case, we assume that n is even. The proof is similar to that of the first case. We leave it to the reader. www.riazisara.ir 104 104 Number Theory Problems 26. Let p be a prime greater than 5. Prove that p−4 cannot be the fourth power of an integer. Proof: Assume that p −4 = q4 for some positive integer q. Then p = q4 + 4 and q > 1. We obtain p = q4 + 4q2 + 4 −4q2 = (q2 + 2)2 −(2q)2 = (q2 −2q + 2)(q2 + 2q + 2), a product of two integers greater than 1, contradicting the fact that p is a prime. (Note that for p > 5, q > 1, and so (q −1)2 = q2 −2q + 1 > 0, or q2 −2q + 2 > 1.) 27. For a positive integer n, prove that σ(1) + σ(2) + · · · + σ(n) ≤n2. Proof: The ith summand on the left-hand side is the sum of all the divisors of i. If we write out all these summands on the left-hand side explicitly, each number d, with 1 ≤d ≤n, appears ! n d " times, once for each multiple of d that is less than or equal to n. Hence the left-hand side of the desired inequality is equal to 1 · n 1 + 2 · n 2 + 3 · n 3 + · · · + n · n n ≤1 · n 1 + 2 · n 2 + 3 · n 3 + · · · + n · n n = n2. 28. [APMO 2004] Determine all finite nonempty sets S of positive integers satisfying i + j gcd(i, j) is an element of S for all i and j (not necessarily distinct) in S. Solution: The answer is S = {2}. First of all, taking i = j in the given condition shows that i+ j gcd(i, j) = 2i i = 2 is in S. We claim that there is no other element in S. Assume to the contrary that S contains elements other than 2. Let s be the smallest element in S that is not equal to 2. www.riazisara.ir 4. Solutions to Introductory Problems 105 If s is odd, then s+2 gcd(s,2) = s+2 is another odd element in S. In this way, we will have infinitely many odd numbers in S, contradicting to the fact that S is a finite set. Hence s must be even, and so s > 2. Then s+2 gcd(s,2) = s 2 + 1 is in S. For s > 2, 2 < s 2 + 1 < s, contradicting the minimality assumption of s. Note: What if i and j are distinct in the given condition? Kevin Mod-zelewski showed that the answer is all the sets in the form of {a + 1, a(a + 1)}, where a is a positive integer. We leave the proof to the reader. 29. Knowing that 229 is a nine-digit number all of whose digits are distinct, without computing the actual number determine which of the ten digits is missing. Justify your answer. Solution: Note that 23 ≡−1 (mod 9), and hence 229 ≡(23)9 · 22 ≡ −4 ≡5 (mod 9). The ten-digit number containing all digits 0 through 9 is a multiple of 9, because the sum of its digits has this property. So, in our nine-digit number, 4 is missing. (Indeed, 229 = 536870912.) 30. Prove that for any integer n great than 1, the number n5 + n4 + 1 is not prime. Proof: The given expression factors as n5 + n4 + 1 = n5 + n4 + n3 −n3 −n2 −n + n2 + n + 1 = (n2 + n + 1)(n3 −n + 1). Hence for n > 1, it is the product of two integers greater than 1. One senses the lack of motivation for this factoring. Indeed, with a little bit knowledge of complex numbers, we can present solid algebraic and number-theoretic reasoning for this factoring. We know that x = 1, ω, ω2, where ω = −1 2 + √ 3i 2 = cis 120◦, are the three roots of the equation x3 −1 = (x −1)(x2 + x + 1) = 0. More precisely, ω and ω2 are the two roots of x2 −x + 1 = 0. Since ω3 = 1, ω5 + ω4 + 1 = ω2 + ω + 1 = 0, it follows that ω and ω2 are roots of x5 + x4 + 1 = 0. We conclude that n2 + n + 1 must be a factor of n5 + n4 + 1. In the light of this argument, we can replace 4 and 5 in the problem statement by any pair of positive integers congruent to 1 and 2 modulo 3. www.riazisara.ir 106 104 Number Theory Problems 31. [Hungary 1995] The product of a few primes is ten times as much as the sum of the primes. What are these (not necessarily distinct) primes? Solution: Obviously 2 and 5 must be among the primes, and there must be at least one more. Let p1 < p2 < · · · < pn be the remaining primes. By the given conditions, we deduce that p1 + p2 + · · · + pn + 7 = p1 p2 · · · pn. (∗) The product of any collection of numbers, each at least 2, must be at least as large as their sum. For two numbers x and y this follows because 0 ≤(x −1)(y −1) −1 = xy −x −y. The general result follows by applying this fact repeatedly as x1x2 · · · xk ≥x1x2 · · · xk−1 + xk ≥· · · ≥x1 + x2 + · · · + xk. In this problem, we have p1 + p2 + · · · + pn + 7 = p1 p2 · · · pn ≥(p1 + p2 · · · + pn−1)pn. Setting s = p1+· · ·+ pn−1, the last equation can be written as s+ pn +7 ≥ spn, or (s −1)(pn −1) ≤8. We can have s = 0 only if there are no primes left, in which case equation (∗) becomes pn + 7 = pn, a contradiction. Hence s ≥2 and so we must have pn −1 ≤8. This leaves pn = 2, 3, 5 as the only options. If pn = 2, equation (∗) becomes 2n + 7 = 2n, which is impossible modulo 2. If pn = 3, then pn −1 = 2, and so s −1 ≤4. Then {p1, p2, . . . , pn−1} can equal only {2}, {3}, or {2, 2}, {2, 3}. We check easily that none of these sets satisfies the equation (∗). If pn = 5, then pn −1 = 4, and so s −1 ≤2, and so the remaining primes must be either a single 2 or a single 3. We check easily that only the latter case gives a solution. Hence the primes in the collection are {2, 3, 5, 5}. 32. [Russia 1998] A 10-digit number is said to be interesting if its digits are all distinct and it is a multiple of 11111. How many interesting integers are there? www.riazisara.ir 4. Solutions to Introductory Problems 107 Solution: There are 3456 such integers. Let n = abcdef ghi j be a 10-digit interesting number. The digits of n must be 0,1, . . . ,9, so modulo 9, n ≡a + b + c + d + e + f + g + h + i + j ≡0 + 1 + 2 + · · · + 9 ≡0; that is, 9 divides n. Because gcd(9, 11111) = 1, it follows that 99999 = 9·11111 divides n. Let x = abcde and y = f ghi j be two 5-digit numbers. We have n = 105x + y. Thus 0 ≡n ≡105x + y ≡x + y (mod 99999). But 0 < x + y < 2·99999, so n is interesting if and only if x + y = 99999, that is, if a + f = · · · = e + j = 9. There are 5! = 120 ways to distribute the pairs (0, 9), (1, 8), . . . , (4, 5) among (a, f ), (b, g), . . . , (e, j), and for each pair we can swap the order of the digits: for example, (b, g) could be (0, 9) or (9, 0). This gives 25 = 32 more choices for a total of 32 · 120 numbers. However, one-tenth of these numbers have a = 0, which is not allowed. So, there are 9 10 ·32·120 = 3456 interesting numbers, as claimed. 33. [Russia 1999] Do there exist 19 distinct positive integers that add up to 1999 and have the same sum of digits? Solution: The answer is negative. Suppose, by way of contradiction, that such integers did exist. The average of the numbers is 1999 19 < 106, so one number is at most 105 and has digit sum at most 18 (for number 99). Every number is congruent to its digit sum modulo 9, so all the numbers and their digit sums are congruent modulo 9, say congruent to k. Then k ≡19k ≡1999 ≡1 (mod 9), so the common digit sum is either 1 or 10. If it is 1, then all the numbers are equal to 1, 10, 100, or 1000, so that some two are equal. This is not allowed. Thus the common digit sum is 10. Note that the twenty smallest numbers with digit sum 10 are 19, 28, 37, . . . , 91, 109, 118, 127, . . . , 190, 208. The sum of the first nine numbers is (10 + 20 + · · · + 90) + (9 + 8 + · · · + 1) = 450 + 45 = 495, www.riazisara.ir 108 104 Number Theory Problems while the sum of the next nine numbers is (900) + (10 + 20 + · · · + 80) + (9 + 8 + 7 + · · · + 1) = 900 + 360 + 45 = 1305. Hence the first eighteen numbers add up to 1800. Because 1800 + 190 ̸= 1999, the largest number among the nineteen must be at least 208. Hence the smallest eighteen numbers add up to at least 1800, giving a total sum of at least 2028 > 1999, a contradiction. 34. [Bulgaria 1995] Find all prime numbers p and q for which pq divides the product (5p −2p)(5q −2q). Solution: The solutions are (p, q) = (3, 3), (3, 13), or (13, 3). It is easy to check that these are solutions. Now we show that they are the only solutions. By symmetry, we may assume that p ≤q. Since (5p −2p)(5q −2q) is odd, we have q ≤p ≤3. We observe that if a prime k divides 5k−2k, then by Fermat’s little theorem, we have 3 ≡5 −2 ≡5k −2k (mod k), or k = 3. Assume that p > 3. By our observation, we have that p divides 5q −2q, or 5q ≡2q (mod p). By Fermat’s little theorem, we have 5p−1 ≡2p−1 (mod p). By Corollary 1.23, 5gcd(p−1,q) ≡2gcd(p−1,q) (mod p). Because q ≥p, gcd(p −1, q) = 1. The last congruence relation now reads 5 ≡2 (mod p), implying that p = 3, a contradiction. Hence p = 3. If q > 3, by our observation, q must divide 5p −2p = 53 −23 = 9 · 13, and so q = 13, leading to the solution (p, q) = (3, 13). 35. Prove that there are infinitely many numbers not containing the digit 0 that are divisible by the sum of their digits. Proof: For a positive integer n, let an = 11 . . . 1   3n . It suffices to show that for all positive integers n, an is divisible by the sum of its digits; that is, an is divisible by 3n. www.riazisara.ir 4. Solutions to Introductory Problems 109 We induct on n. For n = 1, it is clear that an = 111, which is divisible by 3. Assume that an is divisible by 3n, for some positive integer n = k. We consider ak+1. Note that ak+1 = 11 . . . 1   3k+1 = 11 . . . 1   3k·3 = 11 . . . 1   3k 11 . . . 1   3k 11 . . . 1   3k = 11 . . . 1   3k  102·3k + 103k + 1  = ak · 1 0 . . . 0   3k−1 1 0 . . . 0   3k−1 1. Because 3 divides 1 0 . . . 0   3k−1 1 0 . . . 0   3k−1 1 and 3k divides ak, it follows that 3k+1 divides ak+1. This completes our induction. 36. Prove that any number consisting of 2n identical digits has at least n distinct prime factors. Proof: Such a number N can be written as N = k · 102n −1 10 −1 = k(10 + 1)(102 + 1) · · · (102n−1 + 1). The desired conclusion follows from the fact that the n factors 102h + 1, h = 0, 1, . . . , n −1, are pairwise relatively prime. Indeed, for h1 > h2, 102h2 + 1 | 102h1 −1 = 9 · (10 + 1)(102 + 1) · · · (102h2 + 1) · · · (102h1−1 + 1), so gcd(102h2 + 1, 102h1 + 1) = gcd(102h1 −1, 102h1 + 1) = gcd(2, 102h1 + 1) = 1. Note: There is another way to see that gcd(102h1 + 1, 102h2 + 1) = 1. If p divides 102h2 + 1, then p must be odd. Since 102h2 ≡−1 (mod p), it follows that 102h1 ≡  102h22h1−h2 ≡(−1)2h1−h2 ≡1 (mod p), implying that p divides 102h1 −1. Since p is odd, p does not divide 102h1 + 1. www.riazisara.ir 110 104 Number Theory Problems 37. Let a and b be two relatively prime positive integers, and consider the arith-metic progression a, a + b, a + 2b, a + 3b, . . . . (1) [G. Polya] Prove that there are infinitely many terms in the arithmetic progression that have the same prime divisors. (2) Prove that there are infinitely many pairwise relatively prime terms in the arithmetic progression. First Proof: In this approach, we apply properties of linear congruences. (1) Since gcd(a, b) = 1, a has an inverse modulo b. Let x be a positive integer such that ax ≡1 (mod b). For every positive integer n, let sn = (a + b)(ax)n. Then sn ≡a (mod b); that is, sn is a term in the arithmetic progression. It is clear that these terms have the same prime divisors, namely, the divisors of a, x, and a + b. (2) We construct these terms inductively with the additional condition that these terms are relatively prime to a. Let t1 = a + b. Then gcd(t1, t2) = 1 and gcd(t1, a) = 1. Assume that terms t1, . . . , tk have been chosen such that gcd(ti, t j) = 1 and gcd(a, ti) = 1 for 1 ≤i < j ≤k. Set tk+1 = t1 · · · tkb + a. Clearly, tk+1 is a term in the arithmetic progression. Because t1, . . . , tk are distinct integers greater than 1, it is not difficult to see that tk+1 > ti, 1 ≤i ≤k. It is also not difficult to see that gcd(tk+1, a) = 1 by the induction hypothesis and the given condition gcd(a, b) = 1. It remains to show that gcd(tk+1, ti) = 1 for 1 ≤i ≤k, which follows from gcd(tk+1, ti) = gcd(t1t2 · · · tkb + a, ti) = gcd(a, ti) = 1, again by the induction hypothesis. Our induction is thus complete. Second Proof: (By Sherry Gong) In this approach, we apply Euler’s the-orem. (1) The terms xn = (a + b)nϕ(b)+1 satisfy the conditions of the problem. We note that these terms share the same prime divisors, namely, the divisors of a + b. It remains to show that xn appears in the arithmetic progression for each large integer n. By Euler’s theorem, we have xn ≡anϕ(b)+1 ≡anϕ(b) · a ≡a (mod b). www.riazisara.ir 4. Solutions to Introductory Problems 111 Hence xn = a + kb. For large n, xn must appears in the given arith-metic progression. (2) Let y1 = a and y2 = a + b. Clearly, gcd(y1, y2) = 1. Assume that we have pairwise relatively prime terms y1 < y2 < · · · < yk in the sequence. We set yk+1 = y1y2 · · · ykazk+1ϕ(b)−k+1 + b, where zk+1 is some large integer such that yk+1 > yk. We claim that yk+1 is a term in the arithmetic progression that is relatively prime to each of y1, y2, . . . , yk. In this way, we can construct one new term at a time inductively to produce a subsequence of the arithmetic pro-gression satisfying the conditions of the problem. Now we prove our claim. We note that yk+1 ≡akazk+1ϕ(b)−k+1 ≡a (mod b), implying that yk+1 is a term in the arithmetic progression. For each 1 ≤i ≤k, we also have gcd(yk+1, yi) = gcd(b, yi) = gcd(b, a) = 1, by noting that yi is a term in the arithmetic progression. Our proof is thus complete. Note: We can slightly modify our proof so that the conclusions hold for all relatively prime integers a and b. In part (1), gcd(a, b) does not even need to be 1. Because by factoring out gcd(a, b) from each term in the progression we reduce to the current part (1). 38. Let n be a positive integer. (1) Evaluate gcd(n! + 1, (n + 1)! + 1). (2) Let a and b be positive integers. Prove that gcd(na −1, nb −1) = ngcd(a,b) −1. (3) Let a and b be positive integers. Prove that gcd(na +1, nb+1) divides ngcd(a,b) + 1. (4) Let m be a positive integer with gcd(m, n) = 1. Express gcd(5m + 7m, 5n + 7n) in terms of m and n. www.riazisara.ir 112 104 Number Theory Problems Proof: We apply the Euclidean algorithm and Corollary 1.23 to this prob-lem. (1) By the Euclidean algorithm, we have gcd(n! + 1, (n + 1)! + 1) = gcd(n! + 1, (n + 1)! + 1 −(n + 1)(n! + 1)) = gcd(n! + 1, n) = 1. (2) Without loss of generality, we assume that a ≥b. Then gcd(na −1, nb −1) = gcd(na −1 −na−b(nb −1), nb −1) = gcd(na−b −1, nb −1). Recall the process of finding gcd(a, b) = gcd(a −b, b). We see that the process of computing gcd(na−1, nb−1) is the same as the process of computing gcd(a, b) as the exponents, from which the conclusion follows. Alternatively, we can also approach the problem the following way. Since gcd(a, b) divides both a and b, the polynomial xgcd(a,b) −1 divides both xa −1 and xb −1. Hence ngcd(a,b) −1 divides both na −1 and nb −1, implying that ngcd(a,b) −1 | gcd(na −1, nb −1). On the other hand, assume that m divides both na −1 and nb −1; that is, na ≡1 ≡1a (mod m) and nb ≡1 ≡1b (mod m) (clearly, m and n are relatively prime to each other). By Corollary 1.23, we have ngcd(a,b) ≡1 (mod m); that is, m divides ngcd(a,b) −1. It follows that gcd(na −1, nb −1) | ngcd(a,b) −1. we conclude that ngcd(a,b) −1 = gcd(na −1, nb −1). (3) Assume that m divides both 2a + 1 and 2b + 1. Note that m is odd. It suffices to show that m divides 2gcd(a,b) + 1. Since 2a ≡2b ≡−1 (mod m), we have 22a ≡1 (mod m) and 22b ≡1 (mod m). By Corollary 1.23, it follows that 2gcd(2a,2b) ≡1 (mod m); that is, m divides 2gcd(2a,ab) −1 = 22 gcd(a,b) −1, or m | (2gcd(a,b) −1)(2gcd(a,b) + 1). www.riazisara.ir 4. Solutions to Introductory Problems 113 If m divides 2gcd(a,b) +1, we are done. Assume that m does not divide 2gcd(a,b) + 1. Since gcd(2gcd(a,b) −1, 2gcd(a,b) + 1) = gcd(2, 2gcd(a,b) −1) = 1, m must divide 2gcd(a,b)−1, which divides 2a −1 (as we showed in the proof of (3)). But m divides 2a+1 by our original assumption. Thus m divides gcd(2a +1, 2a −1) = 2. Since m is odd, m = 1, contradicting the assumption that m does not divide 2gcd(a,b) + 1. Thus, m must divide 2gcd(a,b) + 1, completing our proof. (4) Let sn = 5n + 7n. If n ≥2m, note that sn = smsn−m −5m7msn−2m, so gcd(sm, sn) = gcd(sm, sn−2m). Similarly, if m < n < 2m, we have sn = smsn−m −5n−m7n−ms2m−n, so gcd(sm, sn) = gcd(sm, s2m−n). Thus by the Euclidean algorithm, we conclude that if m + n is even, then gcd(sm, sn) = gcd(s1, s1) = 12, and if m + n is odd, then gcd(sm, sn) = gcd(s0, s1) = 2. Note: The interested reader might want to generalize part (3), considering the relation between gcd(na + 1, nb + 1) and ngcd(a,b) + 1. 39. Bases? What bases? (1) Determine whether it is possible to find a cube and a plane such that the distances from the vertices of the cube to the plane are 0, 1, 2, . . . , 7. (2) [AIME 1986] The increasing sequence 1, 3, 4, 9, 10, 12, 13, . . . con-sists of all those positive integers that are powers of 3 or sums of distinct powers of 3. Find the 100th term of this sequence (where 1 is the 1st term, 3 is the 2nd term, and so on). Solution: In this problem, we apply base 2 (binary representation) and base 4. (1) The answer is positive. We consider a unit cube S with vertices (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), and (1, 1, 1). We note that these www.riazisara.ir 114 104 Number Theory Problems coordinates match the binary representation of 0, 1, 2, 3, 4, 5, 6, 7. This motivates us to consider the plane x +2y+4z = 0. The distances from the vertices of S to the plane are 0, 1 √ 21 , 2 √ 21 , 3 √ 21 , 4 √ 21 , 5 √ 21 , 6 √ 21 , 7 √ 21 . By a simple scaling, we can find a cube satisfying the conditions of the problem. Indeed, we dilate S via the origin with a ratio of √ 21 to obtain cube T . Point (a, b, c) maps to point ( √ 21a, √ 21b, √ 21c). Then cube T and plane x + 2y + 4z = 0 satisfy the conditions of the problem. (2) Note that a positive integer is a term of this sequence if and only if its base-3 representation consists only of 0’s and 1’s. Therefore, we can set up a one-to-one correspondence between the positive integers and the terms of this sequence by representing both with binary digits (0’s and 1’s), first in base 2 and then in base 3: 1 = 1(2) ⇐ ⇒1(3) = 1, 2 = 10(2) ⇐ ⇒10(3) = 3, 3 = 11(2) ⇐ ⇒11(3) = 4, 4 = 100(2) ⇐ ⇒100(3) = 9, 5 = 101(2) ⇐ ⇒101(3) = 10, . . . This is a correspondence between the two sequences in the order given, that is, the kth positive integer is made to correspond to the kth sum (in increasing order) of distinct powers of 3. This is because when the binary numbers are written in increasing order, they are still in increasing order when interpreted in any other base. Therefore, to find the 100th term of the sequence, we need only look at the 100th line of the above correspondence: 100 = 1100100(2) ⇐ ⇒1100100(3) = 981. Note: The key facts in the solution of part (3) are the following: (a) for integers a and b in X, no carries appear (among the digits) in the addition a + 2b; (b) each digit in base 4 (namely, 0, 1, 2, 3) can be uniquely written in the form of a + 2b, where a and b are equal to either 0 or 1. In other words, we can uniquely write a base-4 digit in base 2: base 2 00(2) 01(2) 10(2) 11(2) base 4 0(4) 1(4) 2(4) 3(4) www.riazisara.ir 4. Solutions to Introductory Problems 115 Based on this, for each nonnegative integer n = nknk−1 . . . a0(4), we can find the solution a + 2b = n in this way: write each digit ni in base 2 according the above table to obtain the binary representation of n. The digits in odd positions from the left form the base-4 representation of a, and the digits in the even positions from the left form the base-4 representation of b. For example, 123210(4) = 01, 10, 11, 10, 01, 00(2) = 101010(4) + 2 · 011100(4) = 101010(4) + 2 · 11100(4). 40. Fractions in modular arithmetic. (1) [ARML 2002] Let a be the integer such that 1 + 1 2 + 1 3 + · · · + 1 23 = a 23!. Compute the remainder when a is divided by 13. (2) Let p > 3 be a prime, and let m and n be relatively prime integers such that m n = 1 12 + 1 22 + · · · + 1 (p −1)2 . Prove that m is divisible by p. (3) [Wolstenholme’s Theorem] Let p > 3 be a prime. Prove that p2 | (p −1)!  1 + 1 2 + · · · + 1 p −1 . Solution: (1) Note that a = 23! + 23! 2 + · · · + 23! 23 . Besides 23! 13 , each summand on the right-hand side is an integer divis-ible by 13. Hence, by Wilson’s theorem, we have a ≡23! 13 ≡12! · 14 · 15 · · · 23 ≡12!10! ≡(12!)2 11 · 12 ≡1 2 ≡7 (mod 13). www.riazisara.ir 116 104 Number Theory Problems (2) Note that ((p −1)!)2 m n = ((p −1)!)2  1 12 + 1 22 + · · · + 1 (p −1)2 is an integer. Note also that 1 1, 1 2, . . . , 1 p −1  is a reduced complete set of residue classes modulo p. By Proposition 1.18 (h) and Wilson’s theorem, we have ((p −1)!)2  1 12 + 1 22 + · · · + 1 (p −1)2 ≡(−1)2[12 + 22 + · · · + (p −1)2] ≡(p −1)p(2p −3) 6 ≡0 (mod p), since p ≥5 and so gcd(6, p) = 1. Hence p divides the integer ((p−1)!)2m n . Since gcd((p −1)!, p) = 1, we must have p | m, as desired. (3) Set S = (p −1)!  1 + 1 2 + · · · + 1 p −1 . Then 2S = (p −1)! p−1  i=1 /1 i + 1 p −i 0 = (p −1)! p−1  i=1 p i(p −i) = p · T, where T = (p −1)! p−1  i=1 1 i(p −i). Since 2S is an integer and p is relatively prime to the numerators of the summands in T , T must itself be an integer. Since p > 3, gcd(p, 2) = 1 and p must divide S. It suffices to show that p also divides T . By (2), we have T ≡(p −1)! p−1  i=1 −1 i2 ≡(p −1)!m n ≡0 (mod p), since p | m and gcd(m, n) = 1. www.riazisara.ir 4. Solutions to Introductory Problems 117 41. Find all pairs (x, y) of positive integers such that x2 + 3y and y2 + 3x are simultaneously perfect squares. Solution: The answers are (x, y) = (1, 1), (11, 16), or (16, 11). It is easy to check that they are solutions. We show that they are the only answers. The inequalities x2 + 3y ≥(x + 2)2 and y2 + 3x ≥(y + 2)2 cannot hold simultaneously because summing them up yields 0 ≥x+y+8, which is false. Hence at least one of x2 + 3y < (x + 2)2 and y2 + 3x < (y +2)2 is true. Without loss of generality assume that x2 +3y < (x +2)2. From x2 < x2 + 3y < (x + 2)2 we derive x2 + 3y = (x + 1)2; hence 3y = 2x + 1. Then x = 3k + 1 and y = 2k + 1 for some nonnegative integer k. Consequently, we have y2 + 3x = 4k2 + 13k + 4. If k > 5, then (2k + 3)2 < 4k2 + 13k + 4 < (2k + 4)2, and so y2 + 3x cannot be a square. It is not difficult to check that for k ∈{1, 2, 3, 4}, y2 + 3x is not a perfect square and that for k = 0, y2 + 3x = 4 = 22 and for k = 5, y2 + 3x = 132. For these two values of k, x2 + 3y is equal to 22 or 172, leading to solutions (x, y) = (1, 1) and (x, y) = (16, 11). 42. First digit? Not the last digit? Are your sure? (1) [AMC12B 2004] Given that 22004 is a 604-digit number with leading digit 1, determine the number of elements in the set {20, 21, 22, . . . , 22003} with leading digit 4. (2) Let k be a positive integer and let n = n(k) be a positive integer such that in decimal representation 2n and 5n begin with the same k digits. What are these digits? Solution: We present two solutions for part (1). (1) • First approach. The smallest power of 2 with a given number of digits has a first digit (most left digit) of 1, and there are ele-ments of S with n digits for each integer n ≤603, so there are 603 elements of S whose first digit is 1. Furthermore, if the first digit of 2k is 1, then the first digit of 2k+1 is either 2 or 3, and the www.riazisara.ir 118 104 Number Theory Problems first digit of 2k+2 is either 4, 5, 6, or 7. Therefore there are 603 elements of S whose first digit is 2 or 3, 603 elements whose first digits is 4, 5, 6, or 7, and 2004 −3(603) = 195 elements whose first digit is 8 or 9. Finally, note that the first digit of 2k is 8 or 9 if and only if the first digit of 2k−1 is 4, so there are 195 elements of S whose first digit is 4. • Second approach. We partition the set S into the following blocks: {20, 21, 22, 23; 24, 25, 26; , 27, . . . , 22003}, where the leading term in each block has first digit 1. Because 22004 has first digit 1, S has been partitioned into complete blocks. As we showed in the first approach, there are exactly 603 elements in S whose first digit is 1. Hence there are 603 blocks in S. Note that a block can have either 3 or 4 elements. If a block has 3 elements 2k, 2k+1, and 2k+2, the their first digits are 1, 2 or 3, 5 or 6 or 7; if a block has 4 elements 2k, 2k+1, 2k+2, and 2k+3, then their first digits are 1, 2, 4, 8, or 9. Thus the num-ber of elements in S having first digit 4 is equal to the number of 4-element blocks. Suppose that there are x 3-element blocks and y 4-element blocks. We have 3x + 4y = 2004 (since there is a total of 2004 elements in S) and x + y = 603 (since there are 603 complete blocks). Solving the equations gives x = 408 and y = 195. (2) Let s and t be unique positive integers such that 10s < 2n < 10s+1 and 10t < 5n < 10t+1. Set a = 2n 10s and b = 5n 10t . Clearly, 1 < a < 10, 1 < b < 10, and ab = 10n−s−t. Hence ab is a power of 10 and since 1 < ab < 102, the only possibility is ab = 10. We obtain min(a, b) < √ ab = √ 10 < max(a, b), implying that the first common k digits are the first k digits of √ 10. (For k = 1, 25 = 32 and 55 = 3125 have the same leading digit, the first digit of √ 10 = 3.1 . . . .) 43. What are those missing digits? (1) Determine the respective last digit (unit digit) of the numbers 3100171002131003 and 7777...7   1001 7’s . www.riazisara.ir 4. Solutions to Introductory Problems 119 (2) [Canada 2003] Determine the last three digits of the number 200320022001. (3) The binomial coefficient 99 19  is a 21-digit number: 107,196,674,080,761,936,xyz. Find the three-digit number xyz. (4) Find the smallest positive integer whose cube ends in 888. First Solution: The key values in this problem are ϕ(10) = 4 and ϕ(1000) = 400. We repeatedly apply Euler’s theorem. (1) The answers are 9 and 3, respectively. Note that 3100171002131003 ≡31000911002 · 3 · 13 ≡81250911002 · 39 ≡9 (mod 10). Since 74 ≡1 (mod 10), we obtain 7777...7   1000 7’s ≡3 (mod 4) by noting that 72k ≡1 (mod 4) and 72k+1 ≡3 (mod 4). Hence 7777...7   1001 7’s ≡73 ≡3 (mod 10). (2) The answer is 241. Since ϕ(1000) = 400 and 200320022001 ≡320022001 (mod 1000), we need to compute 20022001 modulo 400, or 22001 modulo 400. Since 400 = 16 · 25 and 16 clearly divides 22001, 22001 ≡16k (mod 400) for some positive integer k. By Corollary 1.21, we deduce 21997 ≡k (mod 25). Since ϕ(25) = 20, k ≡21997 ≡22000 23 ≡1 8 ≡22 (mod 25), www.riazisara.ir 120 104 Number Theory Problems or k = 22. It follows that 20022001 ≡22001 ≡16k ≡352 (mod 400), and so 200320022001 ≡320022001 ≡3352 ≡9176 ≡(10 −1)176 (mod 1000). By the binomial theorem, we have (10 −1)176 ≡ 176 2 · 102 − 176 1 · 10 + 1176 ≡0 −760 + 1 ≡241 (mod 1000). (3) The answer is 594. We have 99 19 = 99! 19!80! = 99 · 98 · · · 81 19! . Since 1000 = 8 · 125, we need to compute 99 19  modulo 8 and 125, respectively. Instead, we first compute 99 19  modulo 4 and 25, since 99 is very close to 100. (That is, we compute the y and z first.) We note that 99 · 98 · · · 81 19! = 99 · 98 · · · 96 · 95 · 94 · · · 91 · 90 · 89 · · · 86 · 85 · 84 · · · 81 4! · 5 · 6 · · · 9 · 10 · 11 · · · 14 · 15 · 16 · · · 19 = 19 · 18 · 17 · 99 · · · 96 · 94 · · · 91 · 89 · · · 86 · 84 · · · 81 3!4! · 6 · · · 9 · 11 · · · 14 · 16 · · · 19 . Consequently, 99 · 98 · · · 81 19! ≡19 · 18 · 17 3! ≡19 (mod 25). In a similar fashion, we can compute 99 19  modulo 4. Note that ∞  n=1 99 2n − ∞  n=1 19 2n − ∞  n=1 80 2n = 95 −16 −78 = 1, from which it follows that 99 19  ≡2 (mod 4). Combining the above, we conclude that 99 19  ≡94 (mod 100); hence that y = 9 and z = 4. www.riazisara.ir 4. Solutions to Introductory Problems 121 We also compute e3 99 19 = ∞  n=1 99 3n − ∞  n=1 19 3n − ∞  n=1 80 3n = 48 −8 −36 = 4, from which it follows that 99 19  ≡0 (mod 9). Hence, modulo 9, we have 1 + 0 + 7 + 1 + 9 + 6 + 6 + 7 + 4 + 0 + 8 + 0 + 7 + 6 + 1 + 9 + 3 + 6 + x + 9 + 4 ≡0, or x ≡5 (mod 9). Because x is a digit in decimal representation, x = 5. (4) The answer is 192. If the cube of an integer ends in 8, then the integer itself must end in 2; that is, it must be of the form 10k + 2. Therefore, n3 = (10k + 2)3 = 1000k3 + 600k2 + 120k + 8, where the penultimate term, 120k, determines the penultimate digit (tenth digit) of n3, which must also be 8. In other words, 88 ≡n3 ≡120k + 8 (mod 100), or 80 ≡120k (mod 100). In view of this, by Corollary 1.21, 8 ≡12k (mod 10), or 4 ≡6k (mod 5). Consequently, 4 ≡k (mod 5), or k = 5m + 4. Thus, modulo 1000, we have 888 ≡n3 ≡600(5m + 4)2 + 120(5m + 4) + 8 ≡9600 + 600m + 488, or 800 ≡600m (mod 1000). Consequently, by Corollary 1.21 again, we have 8 ≡6m (mod 10), or 4 ≡3m (mod 5). This leads to m ≡3 (mod 5). The smallest m that will ensure this is m = 3, implying that k = 5·3+4 = 19, and n = 10·19+2 = 192. (Indeed, 1923 = 7077888.) Second Solution: We present another approach for part (3). Similar to what we have shown in the first solution, it is not difficult to prove that 11 | 99 19  and 7 | 99 19  . Applying Proposition 1.44 (b), (c), and (d) leads to x + y + z ≡0 (mod 9), x −y + z ≡0 (mod 11), xyz + 1 ≡0 (mod 7), www.riazisara.ir 122 104 Number Theory Problems or x + y + z ≡0 (mod 9), x −y + z ≡0 (mod 11), 2x + 3y + z + 1 ≡0 (mod 7). Because x, y, and z are digits, the first equation leads to x + y + z = 9, or 18, or 27 (with x = y = z = 9); and the second equation leads to x −y + z = 0 or 11. It is not difficult to see that (x + y + z, x −y + z) = (18, 0), or (x +z, y) = (9, 9). Substituting this into the third equation gives 0 ≡x + 3y + (x + z) + 1 ≡x + 2 (mod 7), implying that x = 5, and so z = 4 and xyz = 594. Note: A common mistake in solving part (3) goes as follows: 99 19 = 99 · 98 · · · 81 19! ≡19 · 18 · · · 1 19! ≡1 (mod 8). Because 19! is not relatively prime to 8, we cannot operate division in this congruence. (Please see the discussion leading to Corollary 1.21.) 44. Let p ≥3 be a prime, and let {a1, a2, . . . , ap−1} and {b1, b2, . . . , bp−1} be two sets of complete residue classes modulo p. Prove that {a1b1, a2b2, . . . , ap−1bp−1} is not a complete set of residue classes modulo p. Proof: By Wilson’s theorem, a1a2 · · · ap−1 ≡b1b2 · · · bp−1 ≡(p −1)! ≡−1 (mod p). It follows that (a1b1)(a2b2) · · · (ap−1bp−1) ≡a1a2 · · · ap−1b1b2 · · · bp−1 ≡(−1)2 ≡1 (mod p −1), and so {a1b1, a2b2, . . . , ap−1bp−1} is not a complete set of residue classes modulo p, by Wilson’s theorem again. www.riazisara.ir 4. Solutions to Introductory Problems 123 45. Let p ≥3 be a prime. Determine whether there exists a permutation (a1, a2, . . . , ap−1) of (1, 2, . . . , p −1) such that the sequence {iai}p−1 i=1 contains p −2 distinct congruence classes modulo p. Solution: The answer is positive. For each 1 ≤i ≤p −2, since gcd(i, p) = 1, i is invertible modulo p, and so ix ≡i + 1 (mod p) has a unique solution (modulo p). Let ai be the unique integer with 1 ≤ai ≤p −1 such that iai ≡i + 1 (mod p). It remains to show that for 1 ≤i < j ≤p −2, ai ̸= a j. Assume to the contrary that ai = a j = a for 1 ≤i < j ≤p −2. Because iai ≡i + 1 (mod p) and ja j ≡j + 1 (mod p), it follows that 0 ≡a( j −i) ≡ja j −iai ≡j −i (mod p), which is impossible since 0 < j −i < p −2. Note: By Problem 43, {a1, 2a2, 3a3, . . . , (p −1)ap−1} is not a complete set of residue classes. By Problem 44, we conclude that the maximum num-ber of distinct congruence classes in the sequence {a1, 2a2, 3a3, . . . , (p − 1)ap−1} is p −2. 46. [Paul Erd¨ os] Prove that any positive integer less than n! can be represented as a sum of no more than n positive integer divisors of n!. First Proof: for each k = 1, 2, . . . , n, let ak = n! k!. Suppose we have some number m with ak ≤m < ak−1, where 2 ≤k ≤n. Then consider the number d = ak⌊m ak ⌋. We have 0 ≤m −d < ak; furthermore, because s = ⌊m ak ⌋< ak−1 ak = k, we know that n! d = akk! aks = k! s is an integer. Thus from m we can subtract d, a factor of n!, to obtain a number less than ak. Then if we start with any positive integer m < n! = a1, then by subtracting at most one factor of n! from m we can obtain an integer less than a2; by www.riazisara.ir 124 104 Number Theory Problems subtracting at most one more factor of n! we can obtain an integer less than a3; and so on, so that we can represent m as the sum of at most n−1 positive integer divisors of n!. Second Proof: We proceed by induction. For n = 3 the claim is true. Assume that the hypothesis holds for n −1. Let 1 < k < n! and let k′ and q be the quotient and the remainder when k is divided by n; that is, k = k′n + q, 0 ≤q < n, and 0 ≤k′ < k n < n! n = (n −1)!. From the inductive hypothesis, there are integers d′ 1 < d′ 2 < · · · < d′ s, s ≤n−1, such that d′ i | (n−1)!, i = 1, 2, . . . , s and kk′ = d′ 1+d′ 2+· · ·+d′ s. Hence k = nd′ 1 +nd′ 2 +· · ·+nd′ s +q. If q = 0, then k = d1 +d2 +· · ·+ds, where di = nd′ i, i = 1, 2, . . . , s, are distinct divisors of n!. If q ̸= 0, then k = d1 + d2 + · · · + ds+1, where di = nd′ i, i = 1, 2, . . . , s, and ds+1 = q. It is clear that di | n!, i = 1, 2, . . . , s, and ds+1 | n!, since q < n. On the other hand, ds+1 < d1 < d2 < · · · < ds, because ds+1 = q < n ≤nd′ 1 = d1. Therefore k can be written as a sum of at most n distinct divisors of n!, as claimed. 47. Let n > 1 be an odd integer. Prove that n does not divide 3n + 1. Proof: Assume to the contrary that there is a positive odd integer n that divides 3n + 1. Let p be the smallest prime divisor of n. Then p divides 3n + 1; that is, 3n ≡−1 (mod p), implying that 32n ≡1 (mod p). By Fermat’s little theorem, we also have 3p−1 ≡1 (mod p). By Corollary 1.23, 3gcd(2n,p−1) ≡1 (mod p). Because p is the smallest prime divisor of n, gcd(n, p −1) = 1. Because n is odd, p −1 is even. Hence gcd(2n, p −1) = 2. It follows that 32 ≡1 (mod p), or p divides 8, which is impossible (since p is odd). 48. Let a and b be positive integers. Prove that the number of solutions (x, y, z) in nonnegative integers to the equation ax + by + z = ab is 1 2[(a + 1)(b + 1) + gcd(a, b) + 1]. www.riazisara.ir 4. Solutions to Introductory Problems 125 Proof: It is clear that for each solution (x, y, z) in nonnegative integers to ax + by + z = ab we have the solution (x, y) in nonnegative integers to ax + by ≤ab. Conversely, for each solution (x, y) to ax + by ≤ab we have the solution (x, y, ab −ax −by) to the given equation. Hence it suffices to count the number of solutions (x, y) in nonnegative integers to ax + by ≤ab. Clearly, these solutions correspond to points of integer coordinates in the rectangle [0, b] × [0, a]. The number of lattice points (that is, points with integer coordinates) in this rectangle is (a+1)(b+1). The condition ax+by ≤ab means that the point (x, y) is situated under or on the diagonal AB. Because of the symmetry, the desired number of points (x, y) is 1 2(a + 1)(b + 1) + d 2 , where d is the number of such points on the diagonal AB. In order to find d, note that ax + by = ab is equivalent to y = a −a b x. The number of integers in the array 1 · a b , 2 · a b , . . . , b · a b is gcd(a, b). We also need to count the point A(0, a); hence d = gcd(a, b) + 1 and the conclusion follows. 49. Order! Order, please! (1) Let p be an odd prime, and let q and r be primes such that p divides qr + 1. Prove that either 2r | p −1 or p | q2 −1. www.riazisara.ir 126 104 Number Theory Problems (2) Let a > 1 and n be given positive integers. If p is a prime divisor of a2n + 1, prove that p −1 is divisible by 2n+1. Proof: In this problem, we apply Propositions 1.30 repeatedly. (1) Let d = ordp(q) (the order of q modulo p). Since p | qr + 1 and p > 2, we have qr ≡−1 ̸≡1 (mod p), and so q2r ≡(−1)2 ≡1 (mod p). From the above congruences, d divides 2r but not r. Since r is prime, the only possibilities are d = 2 and d = 2r. If d = 2r, then 2r | p−1 because d | p −1, by Fermat’s little theorem and Proposition 1.30. If d = 2, then q2 ≡1 (mod p), and so p | q2 −1. (2) The proof is similar to that of Theorem 1.50. From the congruence a2n ≡−1 (mod p), we have a2n+1 =  a2n2 ≡1 (mod p). By Proposition 1.30, ordp(a) divides 2n+1. Since 22n ≡ −1 (mod p), ordp(a) = 2n+1. Clearly, gcd(a, p) = 1. By Fermat’s little theorem, we have a p−1 ≡1 (mod p). By Proposition 1.30, we conclude that 2n+1 divides p −1. Note: Setting a = 2 in (2) shows that if p is a prime divisor of the Fermat number fn, then p −1 is divisible by 2n+1. 50. [APMO 2004] Prove that (n −1)! n(n + 1) is even for every positive integer n. Proof: One checks directly that the conclusion holds for n = 1, 2, . . . , 6. Now we assume that n ≥6. We consider three cases. In the first case, we assume that n = p is prime. Then n + 1 = p + 1 is even. Hence n + 1 = 2 · p+1 2 divides (n −1)! = (p −1)! and k = (n −1)! n + 1 = (p −1)! p + 1 www.riazisara.ir 4. Solutions to Introductory Problems 127 is an even integer, and so k + 1 is an odd integer. By Wilson’s theorem, k + 1 ≡(p −1)! p + 1 + 1 ≡−1 1 + 1 ≡0 (mod p), implying that k+1 p is an odd integer; that is, (p−1)! p+1 + 1 p = (p −1)! p(p + 1) + 1 p is an odd integer. Hence (n −1)! n(n + 1) = (p −1)! p(p + 1) is even. In the second case, we assume that n + 1 = p is prime. Then n = p −1 is even. Hence n = 2 · p−1 2 divides (n −1)! = (p −2)! and k′ = (n −1)! n = (p −2)! p −1 is an even integer, and so k′ + 1 is an odd integer. By Wilson’s theorem, k′ + 1 ≡(p −2)! p −1 + 1 ≡(p −1)! (p −1)2 + 1 ≡−1 + 1 ≡0 (mod p), implying that k′+1 p is an odd integer; that is, (p−2)! p−1 + 1 p = (p −2)! p(p −1) + 1 p is an odd integer. Hence (n −1)! n(n + 1) = (p −2)! p(p −1) is even. In the third case, we assume that both n and n + 1 are composite. It is not difficult to show that both n and n+1 divide (n−1)!. Since gcd(n, n+1) = 1, we conclude that n(n + 1) divides (n −1)!; that is, (n −1)! n(n + 1) is an integer. Also, using Legendre’s function, it is easy to see that this integer is also even. www.riazisara.ir 128 104 Number Theory Problems 51. [ARML 2002] Determine all the positive integers m each of which satisfies the following property: there exists a unique positive integer n such that there exist rectangles that can be divided into n congruent squares and also into n + m congruent squares. Solution: The integer m satisfies the conditions of the problem if and only if m is in the set S = {8, p, 2p, and 4p, where p is an odd prime}. Without loss of generality, consider a rectangle ABC D that can be divided into n + m squares of side 1 and m larger squares of side x. Because the sides of ABC D have integer lengths, x must be a rational number. Write x = a b, where a and b are relatively prime integers. Because x > 1, a > b. The area of ABC D is (n + m) · 1 = n · a b 2 . Solving the above equation for n gives n = mb2 a2 −b2 = mb2 (a + b)(a −b). Because gcd(a, b) = 1, gcd(b, a + b) = gcd(a, a −b) = 1, and so (a + b)(a −b) divides m. Note also that a + b and a −b have the same parity. If m has two odd factors each of which is greater than 1, write m = i jk, where j > 1 and k > 1 are odd integers. Then (a + b, a −b) = ( j, k) and (a + b, a −b) = ( jk, 1) lead to two distinct values for n, namely, n = i( j−k)2 4 and n = i( jk−1)2 4 , contradicting the uniqueness of n. Hence m has at most one odd factor greater than 1; that is, m = 2c or 2c · p for some prime p. We consider these two cases separately. In the first case, we assume that m = 2c. It is not difficult to check that there is no solution for n when c = 1 and 2. If c > 3, then (a −b, a +b) = (2, 4) and (a + b, a −b) = (2, 8) lead to two distinct values for n, namely, n = 2c−3 and n = 2c−4, contradicting the uniqueness of n. For c = 3 (and m = 8), we must have (a, b) = (2, 4) and n = 1. In the second case, we assume that m = 2c · p. Similar to the first case, we can show that c ≤2 (by also considering (a + b, a −b) = (1, p)). Hence www.riazisara.ir 4. Solutions to Introductory Problems 129 m = p, 2p, or 4p. The following table shows that all these values work. m (a + b, a −b) (a, b) n p (p, 1)  p+1 2 , p−1 2  (p−1)2 4 2p (p, 1)  p+1 2 , p−1 2  (p−1)2 2 4p (p, 1) or (2p, 2)  p+1 2 , p−1 2  or (p + 1, p −1) (p −1)2 52. Determine all positive integers n such that n has a multiple whose digits are nonzero. Solution: We claim that an integer n satisfies the conditions of the problem if and only if n is not a positive multiple of 10. We call a number good if it satisfies the conditions of the problem. Clearly, multiples of 10 are not good since their multiples always end in the digit 0. We will show that all the other positive integers are good. Let n be a positive integer not divisible by 10. We consider a few cases. In the first case, we assume that n = 5k or n = 2k for some positive integer k. As we have shown in Example 1.49, there exist k-digit multiples of n whose digits are nonzero, implying that n is good. In the second case, we assume that n is relatively prime to 10. We claim that n has a multiple whose digits are all equal to 1. We take 11 . . . 1   ϕ(9n) = 10ϕ(9n) −1 9 , which is divisible by n, from Euler’s theorem. In the third case, we assume that n = as · m, where a = 2 or 5 and m is relatively prime to 10. As we have discussed in the first case, there is an s-digit multiple of as whose digits are nonzero. Let t = as−1as−2 . . . a0 be that number. We consider the sequence as−1as−2 . . . a0, as−1as−2 . . . a0as−1as−2 . . . a0, . . . ; that is, the kth number in the sequence is the concatenation (the number obtained by writing them one after another) of k t’s. As we have shown in the second case, two terms, say the ith and jth terms (i < j) in the sequence, are congruent to each other modulo m. It follows that as−1as−2 . . . a0 . . . as−1as−2 . . . a0   j−i as−1as−2 . . . a0’s 00 . . . 0   ( j−i)s 0’s ≡0 (mod m). www.riazisara.ir 130 104 Number Theory Problems Since gcd(m, 10) = gcd(m, as) = 1 and as divides t = as−1as−2 . . . a0, it follows that as−1as−2 . . . a0 . . . as−1as−2 . . . a0   j−i as−1as−2 . . . a0’s is a multiple of n = as · m whose digits are nonzero. www.riazisara.ir 5 Solutions to Advanced Problems 1. [MOSP 1998] (a) Prove that the sum of the squares of 3, 4, 5, or 6 consecutive integers is not a perfect square. (b) Give an example of 11 consecutive positive integers the sum of whose squares is a perfect square. Proof: Define s(n, k) = n2 + (n −1)2 + · · · + (n + k −1)2 as the sum of squares of k consecutive integers, the least of which is n. (1) Note that s(n −1, 3) = (n −1)2 + n2 + (n + 1)2 = 3n2 + 2. Since s(n−1, 3) ≡2 (mod 3), s(n−1, 3) is not a perfect square; that is, the sum of the squares of 3 consecutive integers is not a perfect square. Note that s(n, 4) = 4(n2 + 3n + 3) + 2. Since s(n, 4) ≡2 (mod 4), s(n, 4) is not a perfect square; that is, the sum of the squares of 4 consecutive integers is not a perfect square. Note that s(n −2, 5) = 5(n2 + 2). Since s(n −2, 5) ≡n2 + 2 ≡2 or 3 modulo 4, s(n −2, 5) is not a perfect square; that is, the sum of the squares of 5 consecutive integers is not a perfect square. Note that s(n −2, 6) = 6n2 + 6n + 19. Since n2 + n = n(n + 1) is even, s(n −2, 6) ≡6n(n + 1) + 19 ≡3 (mod 4), and so s(n −2, 6) is not a perfect square; that is, the sum of the squares of 6 consecutive integers is not a perfect square. (2) We have s(n −5, 11) = 11(n2 + 10). It remains to find n such that 11(n2 + 10) is a perfect square. Hence 11 must divide n2 + 10, or n2 −1 ≡n2 + 10 (mod 11). Consequently, n −1 ≡0 (mod 11) or n + 1 ≡0 (mod 11), or n = 11m ± 1 for some integer k. It follows that s(n −5, 11) = 11[(11m ± 1)2 + 10] = 112(11m2 ± 2m + 1) = www.riazisara.ir 132 104 Number Theory Problems 112[10m2+(m±1)2]. We observe that for m = 2, 10m2+(m+1)2 = 49 = 72, which leads to an example so s(18, 11) = 772. 2. [MOSP 1998] Let S(x) be the sum of the digits of the positive integer x in its decimal representation. (a) Prove that for every positive integer x, S(x) S(2x) ≤5. Can this bound be improved? (b) Prove that S(x) S(3x) is not bounded. Proof: (a) The maximum carry is 1. This implies that the only carries in 2x are the ones accounted for in S(2d) for each digit d in the decimal representation of x. Hence S(2x) =  S(2d), where the sum is taken over all the digits of x. It is clear that S(d)/S(2d) ≤5 for every decimal digit d ̸= 0. Thus S(x) S(2x) =  S(d)  S(2d) ≤5. This bound cannot be improved, since S(5) = 5S(10). One can also apply Proposition 1.45 (d) to obtain S(x) = S(10x) ≤ S(5)S(2x) = 5S(2x). (b) Let pk = 33 . . . 3   k 4. Then 3pk = 3(33 . . . 3   k+1 +1) = 99 . . . 9   k+1 +3 = 1 00 . . . 0   k 2. Thus S(pk) S(3pk) = 3k + 4 3 , which is unbounded. This completes our proof. www.riazisara.ir 5. Solutions to Advanced Problems 133 3. Most positive integers can be expressed as a sum of two or more consecu-tive positive integers. For example, 24 = 7 + 8 + 9 and 51 = 25 + 26. A positive integer that cannot be expressed as a sum of two or more consec-utive positive integers is therefore interesting. What are all the interesting integers? Solution: A number n is interesting if and only if n is power of 2; that is, n = 2k for some nonnegative integer k. Assume that n is not interesting. We can write n = m + (m + 1) + · · · + (m + k) = (k + 1)(2m + k) 2 (∗) for some positive integers m and k. Since k + 1 and 2m + k are of different parity, one of them is an odd integer greater than 3, and so n must have an odd divisor greater than 3. It follows that 2k (for every positive integer k) is interesting. It remains to show that all other positive integers n are not interesting. We write n = 2h ·ℓwhere h is nonnegative and ℓis an odd number greater than 1. (Note that 2h+1 ̸= ℓ.) If 2h+1 < ℓ, n is not interesting since we can set k = 2h+1 −1 and m = ℓ−k 2 = ℓ+ 1 −2h+1 2 in (∗); if 2h+1 > ℓ, n is not interesting since we can set k = ℓ−1 and m = 2h+1 −k 2 = 2h+1 + 1 −ℓ 2 in (∗). Hence, in any case, n is not interesting if n has a odd divisor greater than 1, completing our proof. 4. Set S = {105, 106, . . . , 210}. Determine the minimum value of n such that any n-element subset T of S contains at least two non-relatively prime elements. Solution: The minimum value of n is 26. Our first step is to compute the number of prime numbers in the set S. For any positive integer k, let Ak denote the subset of multiples of k in S, and let P = {2, 3, 5, 7, 11}. We compute the cardinality of the subset of S consisting of numbers divisible by one or more of 2, 3, 5, 7, or 11: A = 1 k∈P Ak = A2 ∪A3 ∪A5 ∪A7 ∪A11, www.riazisara.ir 134 104 Number Theory Problems using the inclusion and exclusion principle, as follows: |A| =  k∈P |Ak| −  i< j∈P |Ai ∩A j| +  i< j<k∈P |Ai ∩A j ∩Ak| −  i< j 210. Therefore S consists of 87 composite numbers and 19 primes. We can now prove that given any 26 numbers in S, there exist two of them that are not relatively prime. By the pigeonhole principle, since there are 19 primes in S, at least 7 of the 26 numbers we have chosen are composite, which means that at least 6 of the numbers are in A. But this means, again by the pigeonhole principle, that two of them belong to the same set Ak, for some k ∈P. Thus they share a common factor (namely k), and hence are not relatively prime. Finally, we can construct a subset of S with 25 elements in which every pair of elements is relatively prime. Let P denote the set of all primes in S; then we can see that the set P ∪{112, 53, 27, 32 · 17, 132, 7 · 29} = P ∪{121, 125, 128, 153, 169, 203} is a set of 25 numbers that are all mutually relatively prime. 5. [St. Petersburg 1997] The number 99 . . . 99   1997 9’s is written on a blackboard. Each minute, one number written on the black-board is factored into two factors and erased, each factor is (independently) increased or diminished by 2, and the resulting two numbers are written. Is it possible that at some point (after the first minute) all of the numbers on the blackboard equal 9? www.riazisara.ir 5. Solutions to Advanced Problems 135 Solution: The answer is negative. Adding or subtracting 2 from a number motivates us to consider arithmetic modulo 4. Since a + 2 ≡a −2 (mod 4) for any integer a, adding or subtracting 2 becomes the same operation modulo 4. Note first that the original number is congruent to 3 modulo 4. We claim that there is always a number congruent to 3 modulo 4: factoring such a number gives one factor congruent to 1 modulo 4, and changing that by 2 in either direction gives a number congruent to 3 modulo 4. On the other hand, 9 is congruent to 1 modulo 4, and so we cannot have all 9’s written on the board at any moment. 6. IMO 1986] Let d be any positive integer not equal to 2, 5, or 13. Show that one can find distinct a, b in the set {2, 5, 13, d} such that ab −1 is not a perfect square. First Proof: Since 2 · 5 −1 = 32, 2 · 13 −1 = 52, and 5 · 13 −1 = 82, we will look for a non-perfect square in the set {2d −1, 5d −1, 13d −1}. Assume to the contrary that all these numbers are perfect squares; that is, 2d −1 = a2, 5d −1 = b2, and 13d −1 = c2, where a, b, and c are integers. Then a is an odd number, say, a = 2x + 1 and d = 2x(x +1)+1. Since x(x +1) is always even, it follows that d ≡1 (mod 4), and so b and c are even. Assume that b = 2y and c = 2z. From 5d = b2 + 1 and 13d = c2 + 1, we have 8d = c2 −b2. Thus, d = 4y2 + 1 5 = 4z2 + 1 13 = 4z2 −4y2 8 = z2 −y2 2 . It follows that z and y are of equal parity. In this case, z2−y2 ≡0 (mod 4), while d ≡1 (mod 4). Thus, we get a contradiction. Second Proof: We operate modulo 16. We first calculate n2 modulo 16 for n = 0, 1, . . . , 7, 8 to see that the possible residues modulo 16 are 0, 1, 4, 9. If 2d−1 is not a perfect square, we are done. Assume that 2d−1 is a perfect square. Then 2d −1 is congruent to 0, 1, 4, or 9 modulo 16. Since 2d is even, 2d is congruent to 2 or 10 modulo 16, implying that d is congruent to www.riazisara.ir 136 104 Number Theory Problems 1, 5, 9, or 13 modulo 16. Thus, we have the following table (modulo 16): d 5d −1 13d −1 1 4 12 5 8 4 9 12 4 13 4 8 Note that all the boldfaced numbers are not perfect squares, and there is such a number in each row. Thus, for all possible values of d that make 2d −1 a perfect square, at least one of 5d −1 and 13d −1 is a not perfect square, and we are done. 7. [Russia 2001] A heap of balls consists of one thousand 10-gram balls and one thousand 9.9-gram balls. We wish to pick out two heaps of balls with equal numbers of balls in them but different total weights. What is the minimal number of weighings needed to do this? (The balance scale reports the weight of the objects in the left pan minus the weight of the objects in the right pan.) Proof: It is clear that one has to use at least one weighing. We claim that it is also enough. Split the two thousand balls into three heaps H1, H2, H3 of 667, 667, and 666 balls, respectively. Weigh heaps H1 and H2 against each other. If the total weights are not equal, we are done. Otherwise, discard one ball from H1 to form a new heap H′ 1 of 666 balls. We claim that H′ 1 and H3 have different weights. If not, then they have the same number of 10-gram balls, say, n. Then H1 and H2 either each had n 10-gram balls or each had n + 1 10-gram balls. This would imply that 1000 equals 3n or 3n + 2, which is impossible. 8. [China 2001] We are given three integers a, b, and c such that a, b, c, a + b −c, a + c −b, b + c −a, and a + b + c are seven distinct primes. Let d be the difference between the largest and smallest of these seven primes. Suppose that 800 is an element in the set {a + b, b + c, c + a}. Determine the maximum possible value of d. First Solution: The answer is 1594. First, observe that a, b, and c must all be odd primes; this follows from the assumption that the seven quantities listed are distinct primes and the fact that there is only one even prime, 2. (If, say, a is even, then b and c must www.riazisara.ir 5. Solutions to Advanced Problems 137 be odd. Then a + b −c and a + c −b must both be even, and so equal to 2.) Therefore, the smallest of the seven primes is at least 3. Second, assume without loss of generality that a + b = 800. Because a + b −c > 0, we must have c < 800. We also know that c is prime; therefore, since 799 = 17 · 47, we have c ≤797. It follows that the largest prime, a + b + c, is no more than 1597. Combining these two bounds, we can bound d by d ≤1597 −3 = 1594. It remains to observe that we can choose a = 13, b = 787, c = 797 to achieve this bound. The other four primes are then 3, 23, 1571, and 1597. Second Solution: assume without loss of generality that a + b = 800. (Clearly, both a and b are odd.) Then c, a+b+c = 800+c, and a+b−c = 800 −c are primes. Consider c, 800 + c, and 800 −c modulo 3. It is not difficult to see that exactly one of them is congruent to 0 modulo 3; that is, one of them is equal to 3. Consequently, we have either c = 3 or 800 −c = 3 (and c = 797). If c = 3, d < a + b + c = 803. If c = 797, then d ≤a +b+c−3 = 1594. We can finish as we did in the first solution. 9. Prove that the sum S(m, n) = 1 m + 1 m + 1 + · · · + 1 m + n is not an integer for any given positive integers m and n. Proof: Assume to the contrary that S(m, n) is an integer for some pos-itive integers m and n. Clearly, n ≥1. Consequently, there are even integers among the numbers m, m + 1, . . . , m + n. Let ℓdenote lcm(m, m + 1, . . . , m + n). Then ℓis even. We have ℓS(m, n) = ℓ m + ℓ m + 1 + · · · + ℓ m + n . (∗) By our assumption, the left-hand side of the above identity is even. We will reach a contradiction by showing that the right-hand side is odd. For every integer i with 0 ≤i ≤n, assume that 2ai fully divides m + i. Let m = max{a0, a1, . . . , an}. It follows that 2m∥ℓ. Assume that a j = m (where 0 ≤j ≤n). We claim that j is unique. Assume to the contrary that a j = a j1 with 0 ≤j < j1 ≤n. Then m + j = 2a j · k and m + j1 = 2a j1 · k1, where k and k1 are odd positive integers. Hence k + 1 is an even integer between k and k1, and so m + j < 2a j · (k + 1) < 2a j · k1 = 2a j1 · k1 = m + j1, www.riazisara.ir 138 104 Number Theory Problems implying that 2a j+1 divides 2a j · (k + 1), contradicting the maximality of a j = m. Thus, such a j is unique. It follows that ℓ m+i is an even integer for all 0 ≤i ≤n with i ̸= j, and ℓ m+ j is odd. Hence all but one summand on the right-hand side of (∗) are even, implying that the right-hand side of (∗) is odd, contradicting the fact that the left-hand side of (∗) is even. Hence our original assumption was wrong and S(m, n) is not an integer. 10. [St. Petersburg 2001] For all positive integers m > n, prove that lcm(m, n) + lcm(m + 1, n + 1) > 2mn √m −n . Proof: Let m = n + k. Then lcm(m, n) + lcm(m + 1, n + 1) = mn gcd(m, n) + (m + 1)(n + 1) gcd(m + 1, n + 1) > mn gcd(n + k, n) + mn gcd(m + 1, n + 1) = mn gcd(k, n) + mn gcd(n + k + 1, n + 1) = mn gcd(k, n) + mn gcd(k, n + 1). Now, gcd(k, n) | k, and gcd(k, n + 1) | k. We conclude that gcd(k, n) has no common prime factor with gcd(k, n + 1), because if it did, n + 1 would have a common prime factor with n, which is impossible. Since both divide k, so does their product, implying that gcd(k, n) gcd(k, n + 1) ≤k. Consequently, lcm(m, n) + lcm(m + 1, n + 1) > mn gcd(k, n) + mn gcd(k, n + 1) ≥2mn 4 1 gcd(k, n) gcd(k, n + 1) ≥2mn % 1 k = 2mn √m −n by the AM-GM inequality. 11. Prove that each nonnegative integer can be represented in the form a2 + b2 −c2, where a, b, and c are positive integers with a < b < c. www.riazisara.ir 5. Solutions to Advanced Problems 139 First Proof: Let k be a nonnegative integer. If k is even, say k = 2n, the conclusion follows from the identity 2n = (3n)2 + (4n −1)2 −(5n −1)2 and the simple algebraic facts 3n < 4n −1 < 5n −1 for n > 1, 0 = 32 + 42 −52, and 2 = 52 + 112 −122. If k is odd, we use the identity 2n + 3 = (3n + 2)2 + (4n)2 −(5n + 1)2, where for n > 2, 3n + 2 < 4n < 5n + 1. Since 1 = 42 + 72 −82, 3 = 42 + 62 −72, 5 = 42 + 52 −62, and 7 = 62 + 142 −152, we have exhausted the case k odd as well. Second Proof: We present a more general approach for this problem. The key fact is that the positive differences between consecutive perfect squares are linearly increasing. For every nonnegative integer k, we choose a large positive integer a with different parity from that of k. We then set c = b+1. Then k = a2 + b2 −c2 = a2 −(2b + 1). Since k and a has different parity, a2 −k is odd, and so b = a2 −k −1 2 is a positive integer. Since the left-hand side of the above equation is a quadratic in a, its value is greater than a for large a, and so the condition a < b < c = b + 1 is satisfied, and we are done. 12. Determine whether there exists a sequence of strictly increasing positive integers {ak}∞ k=1 such that the sequence {ak + a}∞ k=1 contains only finitely many primes for all integers a. Note: One easily thinks about ak = k!. But it is then difficult to deal with a = 1 or a = −1. We present two ways to modify this sequence. www.riazisara.ir 140 104 Number Theory Problems First Solution: The answer is positive. There exists such a sequence. Indeed, for every positive integer k, let ak = (k!)3. If a = ±1, then ak+a = (k!)3 ± 1 is composite, since polynomials x3 + 1 and x3 −1 factor into (x + 1)(x2 −x + 1) and (x −1)(x2 + x + 1), respectively. If |a| > 1, then a divides k! for k ≥|a|, implying that a also divides ak + a for k ≥|a|. Second Solution: (By Kevin Modzelewski) We set ak = (2k)! + k. For all integers k ≥|a| and k ≥2 −a, we have 2 ≤k + a ≤2k. Therefore, ak + a is divisible by k + a, and thus composite, for all such k. 13. Prove that for different choices of signs + and −the expression ±1 ± 2 ± 3 ± · · · ± (4n + 1) yields all odd positive integers less than or equal to (2n + 1)(4n + 1). Proof: We induct on n. For n = 1, from ±1 ± 2 ± 3 ± 4 ± 5 we obtain all odd positive integers less than or equal to (2 + 1)(4 + 1) = 15: +1 −2 + 3 + 4 −5 = 1, −1 + 2 + 3 + 4 −5 = 3, −1 + 2 + 3 −4 + 5 = 5, −1 + 2 −3 + 4 + 5 = 7, −1 −2 + 3 + 4 + 5 = 9, +1 −2 + 3 + 4 + 5 = 11, −1 + 2 + 3 + 4 + 5 = 13, +1 + 2 + 3 + 4 + 5 = 15. Assume that the conclusion is true for n = k, where k is some positive integer; that is, from ±1 ± 2 ± · · · ± (4k + 1) with suitable choices of signs + and −we obtain all odd positive integers less than or equal to (2k + 1)(4k + 1). Now we assume that n = k + 1. Observe that −(4k + 2) + (4k + 3) + (4k + 4) −(4k + 5) = 0. Hence from ±1 ± 2 ± · · · ± (4k + 5) for suitable choices of signs + and −we obtain all odd positive numbers less than or equal to (2k + 1)(4k + 1). It suffices to obtain all odd integers m such that (2k + 1)(4k + 1) < m ≤(2k + 3)(4k + 5) = (2n + 1)(4n + 1). (∗) There are (2k + 3)(4k + 5) −(2k + 1)(4k + 1) 2 = 8k + 7 such odd integers m. Each of these integers can be written in exactly one of the following forms: (2n + 3)(4n + 5) = +1 + 2 + · · · + (4n + 5), www.riazisara.ir 5. Solutions to Advanced Problems 141 or (2n + 3)(4n + 5) −2k = +1 + 2 + · · · + (k −1) −k + (k + 1) + · · · + (4n + 4) + (4n + 5), for k = 1, 2, . . . , 4n + 5, or (2n + 1)(4n + 5) −2ℓ = +1 + 2 + · · · + (ℓ−1) −ℓ+ (ℓ+ 1) + · · · + (4n + 4) −(4n + 5), for ℓ= 1, 2, . . . , 4n + 1. Hence all numbers m from (∗) are obtained, completing the inductive step. 14. Let a and b be relatively prime positive integers. Show that ax + by = n has nonnegative integer solutions (x, y) for integers n > ab −a −b. What if n = ab −a −b? First Proof: We call an integer n representable if there are nonnegative integers x and y such that n = ax + by. First we show that n = ab −a −b is not representable. Assume to the contrary that ab−a−b = ax +by, where x and y are nonnegative integers. Taking the last equation modulo a and then modulo b leads to −b ≡by (mod a) and −a ≡ax (mod a). Since gcd(a, b) = 1, it follows that y ≡−1 (mod a) and x ≡−1 (mod b). Since x and y are nonnegative, y ≥a −1 and x ≥b −1. Hence ab −a −b = n = ax + by ≥a(b −1) + b(a −1) = 2ab −a −b, which is impossible for positive integers a and b. Therefore, our assump-tion was wrong and n = ab −a −b is not representable. Second we show that n > ab−a −b is representable. Since gcd(a, b) = 1, by Proposition 1.24, {n, n −b, n −2b, . . . , n −(a −1)b} is a complete set of residue classes modulo a. Hence there exists exactly one y, with 0 ≤y ≤a −1, such that n −yb ≡0 (mod a), or n −yb = ax for some integer x. If x ≥0, we are done. If x < 0, then x ≤−1, and so n −(a −1)b ≤n −yb = ax ≤−a, www.riazisara.ir 142 104 Number Theory Problems or n ≤ab −a −b, contradicts the condition n > ab −a −b. Hence both x and y are nonnegative, and so n > ab −a −b is representable. Second Proof: We prove the following claim: If m and n are integers with m + n = ab −a −b, then exactly one of m and n is representable. If n > ab −a −b, then m must be negative, which is clearly not repre-sentable. Hence by our claim, n is representable. If n = ab −a −b, then since m = 0 is clearly representable (with x = y = 0), n = ab −a −b is not representable, again by our claim. It remains to prove our claim. By B´ ezout’s identity, there exist pairs (x, y) of integers such that ax+by = n. Since ax+by = a(x−bt)+b(y+bt), we can always reduce or increase x by a multiple of b. Thus, we can always assume that 0 ≤x ≤b −1. Furthermore, a number n = ax + by is representable if and only if it is representable under the additional condition that 0 ≤x ≤b −1. Assume that n = ax + by and m = as + bt, where x, y, s, and t are integers with both x and s nonnegative integers less than b; that is, 0 ≤x, x ≤b −1. Then ax + by + as + bt = m + n = ab −a −b, or ab −(x + s + 1)a −(y + t + 1)b = 0. (∗) Since gcd(a, b) = 1, equation (∗) indicates that b must divide x + s + 1. Note that 1 ≤x + s + 1 ≤2b −1. Hence x + s + 1 = b, and the equations (∗) becomes (y +t +1)b = 0, or y +t +1 = 0. It is easy to see that exactly one of y and t is nonnegative, and exactly one of them is negative; that is, one of them is representable and the other is equal to 1. Note: Can you generalize this result for three pairwise relatively prime numbers a, b, and c? 15. [China 2003] The sides of a triangle have integer lengths k, m, and n. As-sume that k > m > n and  3k 104  =  3m 104  =  3n 104  . Determine the minimum value of the perimeter of the triangle. www.riazisara.ir 5. Solutions to Advanced Problems 143 Solution: It suffices to find positive integers k, m, and n with k > m > n and k < m + n such that 3k ≡3m ≡3n (mod 104), or 3k ≡3m ≡3n (mod 24) and 3k ≡3m ≡3n (mod 54). (∗) Let d1 = ord24(3), d2 = ord54(3), and d = gcd(d1, d2). By Proposition 1.30, d divides both of k −m and m −n. It is easy to check that d1 = 4. We note that d2 divides ϕ(54) = 54 −53 = 500, by Proposition 1.30 again. We claim that d2 = 500. If d2 < 500, it must be a divisor of either 250 = 500 2 or 100 = 500 5 . It suffices to show that (a) 3250 ̸≡1 (mod 54) and (b) 3100 ̸≡1 (mod 54). We establish (a) by noting that 3250 ≡32 ≡−1 (mod 5), since ϕ(5) = 4. By the binomial theorem, we have 3100 ≡(10 −1)50 ≡ 50 48 · 102 − 50 49 · 10 + 1 ̸≡1 (mod 54), establishing (b). It follows that d2 = 500 and d = 500. Condition (∗) is satisfied if and only if both of k −m and m −n are multiples d = 500. We set m = 500s + n and k = 500t + m = 500(s + t) + n for positive integers s and t. The perimeter of the triangle is equal to k + m + n = 500(2s + t) + 3n. Condition k < m + n now reads 500t < n. Therefore, the minimum value of the perimeter is equal to 500 · 3 + 3 · 501 = 3003, obtained when s = t = 1 and n = 501. 16. [Baltic 1996] Consider the following two-person game. A number of peb-bles are lying on a table. Two players make their moves alternately. A move consists in taking off the table x pebbles, where x is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the player who goes second has a winning strategy. Proof: Assume to the contrary that there are only finitely many initial situations in which the player who goes second has a winning strategy. Under our assumption, there exists a positive integer N such that if there www.riazisara.ir 144 104 Number Theory Problems are n > N pebbles on the table, the player who goes first at the moment has a winning strategy. Consider the initial situation with (N + 1)2 −1 pebbles on the table. Let P1 and P2 denote the players who go first and second, respectively. By our assumption, P1 has a winning strategy, which requires P1 to remove x pebbles on his first move. It is clear that x ̸= N 2, and P2 is left with at least (N + 1)2 −1 −N 2 = 2N > N pebbles to make the first move. By our assumption, at this moment, P2 has a winning strategy. But it is impossible for both players to have a winning strategy for the same initial situation. Hence our original assumption was wrong and there are infinitely many initial situations in which the player who goes second has a winning strategy. 17. [MOSP 1997] Prove that the sequence 1, 11, 111, . . . contains an infinite subsequence whose terms are pairwise relatively prime. First Proof: Let xn denote the nth term in the sequence. Then xn+1 −10xn = 1, implying that gcd(xn+1, xn) = 1. To prove that there is an infinite subsequence of numbers any of two of which are relatively prime, it suffices to prove that no matter how many terms the subsequence contains, it can always contain at least one more term. To do this, note that xn divides xmn. Let p be the product (or least common multiple) of all the indices already in the subsequence. Then any number in the subsequence divides x p. Hence x p+1 can be added to the subsequence and we are done. Second Proof: We maintain the same notation as in the first proof. Note that xn = 10n−1 9 . By introductory problem 38 (2), we have gcd(xm, xn) = gcd(10m −1, 10n −1) 9 = 10gcd(m,n) −1 9 = 1 for integers m and n with gcd(m, n) = 1. Hence the subsequence {x p} where p are primes satisfies the conditions of the problem. Note: Euler’s proof of the existence of infinitely many primes reveals the connection between these two proofs. 18. Let m and n be integers greater than 1 such that gcd(m, n −1) = gcd(m, n) = 1. Prove that the first m −1 terms of the sequence n1, n2, . . . , where n1 = mn + 1 and nk+1 = n · nk + 1, k ≥1, cannot all be primes. www.riazisara.ir 5. Solutions to Advanced Problems 145 Proof: It is straightforward to show that nk = nkm + nk−1 + · · · + n + 1 = nkm + nk −1 n −1 for every positive integer k. Hence nϕ(m) = nϕ(m) · m + nϕ(m) −1 n −1 . From Euler’s theorem, m | (nϕ(m) −1), and since gcd(m, n −1) = 1, it follows that m 2 2 2 2 nϕ(m) −1 n −1 . Consequently, m divides nϕ(m). Because ϕ(m) ≤m −1, nϕ(m) is not a prime, and we are done. 19. [Ireland 1999] Find all positive integers m such that the fourth power of the number of positive divisors of m equals m. Solution: If the given condition holds for some integer m, then m must be a perfect fourth power and we may write its prime factorization as m = 24a234a354a574a7 · · · for nonnegative integers a2, a3, a5, a7, . . . . The num-ber of positive divisors of m equals (4a2 + 1)(4a3 + 1)(4a5 + 1)(4a7 + 1) · · · . This number is odd, so m is odd and a2 = 0. Thus, 1 = 4a3 + 1 3a3 · 4a5 + 1 5a5 · 4a7 + 1 7a7 · · · = x3x5x7 · · · , where we write x p = 4ap+1 pap for each p. We proceed to examine x p through three cases: p = 3, p = 5, and p > 5. When a3 = 1, x3 = 5 3; when a3 = 0 or 2, x3 = 1. When a3 > 2, by Bernoulli’s inequality we have 3a3 = (8 + 1)a3/2 > 8(a3/2) + 1 = 4a3 + 1, so that x3 < 1. When a5 = 0 or 1, x5 = 1; when a5 ≥2, by Bernoulli’s inequality we have 5a5 = (24 + 1) a5 2 ≥24 · a5 2 + 1 = 12a5 + 1, www.riazisara.ir 146 104 Number Theory Problems so that x5 ≤4a5 + 1 12a5 + 1 ≤9 25. Finally, for any p > 5, when ap = 0 we have x p = 1; when ap = 1 we have pap = p > 5 = 4ap + 1, so that x p < 1; and when ap > 1 then again by Bernoulli’s inequality we have pap > 5ap > 12ap + 1, so that as above, x p < 9 25. Now if a3 ̸= 1 then we have x p ≤1 for all p. Because 1 = x2x3x5 · · · we must actually have x p = 1 for all p. This means that a3 ∈{0, 2}, a5 ∈{0, 1}, and a7 = a11 = · · · = 0. Hence m = 14, (32)4, 54, or (32 · 5)4. Otherwise, if a3 = 1 then 3 divides m = 54(4a5 + 1)4(4a7 + 1)4 · · · . Then for some prime p′ ≥5, 3 | (4ap′ + 1), so that ap′ ≥2; from above, we have x p′ ≤9 25. Then x3x5x7 · · · ≤5 3 · 9 25 < 1, which is a contradiction. Thus, the only such integers m are 1, 54, 38, and 38 · 54, and it is easily verified that these integers work. 20. [Romania 1999] (1) Show that it is possible to choose one number out of any 39 consecu-tive positive integers having the sum of its digits divisible by 11. (2) Find the first 38 consecutive positive integers none of which has the sum of its digits divisible by 11. Proof: Call an integer deadly if its sum of digits is divisible by 11, and let d(n) equal the sum of the digits of a positive integer n. We have the following observations: (a) If n ends in a 0, then the numbers n, n + 1, . . . , n + 9 differ only in their units digits, which range from 0 to 9. Hence d(n), d(n + 1), . . . , d(n+9) is an arithmetic progression with common difference 1. Thus if d(n) ̸≡1 (mod 11), then one of these numbers is deadly. (b) Next suppose that n ends in k ≥0 nines. Then d(n + 1) = d(n) + 1 −9k: the last k digits of n + 1 are 0’s instead of 9’s, and the next digit to the left is 1 greater than the corresponding digit in n. www.riazisara.ir 5. Solutions to Advanced Problems 147 (c) Finally, suppose that n ends in a 0 and that d(n) ≡d(n + 10) ≡1 (mod 11). Because d(n) ≡1 (mod 11), we must have d(n+9) ≡10 (mod 11). If n +9 ends in k 9’s, then we have 2 ≡d(n +10)−d(n + 9) ≡1 −9k (mod 11), implying that k ≡6 (mod 11). (1) Suppose we had 39 consecutive integers, none of them deadly. One of the first ten must end in a 0: call it n. Because none of n, n+1, . . . , n+ 9 are deadly, we must have d(n) ≡1 (mod 11), by (a) above. Simi-larly, d(n + 10) ≡1 (mod 11) and d(n + 20) ≡1 (mod 11). From (c) above, this implies that both n + 9 and n + 19 must end in at least six 9’s. This is impossible, because n + 10 and n + 20 can’t both be multiples of one million! (2) Suppose we have 38 consecutive numbers N, N+1, . . . , N+37, none of which is deadly. By an analysis similar to that in part (1), none of the first nine can end in a 0. Hence, N + 9 must end in a 0, as must N + 19 and N + 29. Then we must have d(N + 9) ≡d(N + 19) ≡ 1 (mod 11). Therefore d(N + 18) ≡10 (mod 11). Furthermore, if N + 18 ends in k 9’s we must have k ≡6 (mod 11). The smallest possible such number is 999999, yielding the 38 consec-utive numbers 999981, 999982, . . . , 1000018. Indeed, none of these numbers is deadly: their sums of digits are congruent to 1, 2, . . . , 10, 1, 2, . . . , 10, 1, 2, . . . , 10, 2, 3, . . . , 9, and 10 (mod 11), respec-tively. 21. [APMO 1998] Find the largest integer n such that n is divisible by all posi-tive integers less than 3 √n. Solution: The answer is 420, which satisfies the condition since 7 < 3 √ 420 < 8 and 420 = lcm{1, 2, 3, 4, 5, 6, 7}. Suppose n > 420 is an integer such that every positive integer less than 3 √n divides n. Then 3 √n > 7, so 420 = lcm(1, 2, 3, 4, 5, 6, 7) divides n; thus n ≥840 and 3 √n > 0. Thus 2520 = lcm(1, 2, . . . , 9) divides n and 3 √n > 13. Now let m be the largest positive integer less than 3 √n; that is, m < 3 √n ≤m + 1. We have m ≥13 and lcm(1, 2, . . . , m) divides n. But lcm(m −3, m −2, m −1, m) ≥m(m −1)(m −2)(m −3) 6 , (†) since 2 and 3 are the only possible common divisors of these four numbers. Thus m(m −1)(m −2)(m −3) 6 ≤n ≤(m + 1)3, www.riazisara.ir 148 104 Number Theory Problems implying that m ≤6  1 + 2 m −1  1 + 3 m −2  1 + 4 m −3 . The left-hand side of the inequality is an increasing function of m, and the right-hand side is a decreasing function of m. But for m = 13, we have 13 · 12 · 11 · 10 = 17160 > 16464 = 6 · 143, so this inequality is false for all m ≥13. Thus no n > 420 satisfies the given condition. Note: Ryan Ko pointed out that the inequality (†) can be improved to lcm(m −3, m −2, m −1, m) ≥(m −1)(m −2)(m −3)(m −4) 2 . Why? 22. [USAMO 1991] Show that for any fixed positive integer n, the sequence 2, 22, 222, 2222 , . . . (mod n) is eventually constant. (The tower of exponents is defined by a1 = 2 and ai+1 = 2ai for every positive integer i.) Proof: We apply strong induction on n. The base case n = 1 is clearly true. Assume that the conclusion is true for n ≤k, where k is some positive integer. We consider the case n = k + 1. If n = k + 1 is odd, 2ϕ(n) ≡1 (mod n) by Euler’s theorem. Because ϕ(n) < n, by the induction hypothesis, the sequence a1, a2, . . . is eventu-ally constant modulo ϕ(n); that is, ai ≡c (mod ϕ(n)) for large i. Conse-quently, ai+1 ≡2ai ≡2c (mod n) is constant, completing the inductive step for this case. If n = k + 1 is even, we write n = k + 1 = 2q · m for some positive integer k and odd positive integer m. By the induction hypothesis, the sequence a1, a2, . . . is eventually constant modulo m. Clearly, ai ≡0 (mod 2q) for all sufficiently large i. Because 2q and m are relatively prime, each of 2q and m divides ai+1 −ai, which implies that n = 2q · m divides ai+1 −ai; that is, the sequence a1, a2, . . . is eventually constant modulo n = k + 1, completing the inductive step for this case, and our induction is complete. www.riazisara.ir 5. Solutions to Advanced Problems 149 23. Prove that for n ≥5, fn + fn−1 −1 has at least n + 1 prime factors. Proof: For each k ≥1, we have fk+1 + fk −1 = 22k+1 + 22k + 1 = (22k + 1)2 −(22k−1)2 = (22k + 1 −22k−1)(22k + 1 + 22k−1). Hence fk+1 + fk −1 = ak( fk + fk−1 −1), (∗) where ak = fk −fk−1 + 1. We proceed by induction. We have f5 + f4 −1 = 3 · 7 · 13 · 97 · 241 · 673, and the property holds. Assume that for some k ≥5, fk + fk−1 −1 has at least k + 1 prime factors. Using (∗) and the fact that gcd( fk + fk−1 −1, ak) = gcd( fk + fk−1 −1, fk −fk−1 + 1) = gcd( fk −fk−1 + 1, 2 · 22k−1) = 1, we conclude that fk+1 + fk −1 has at least k + 2 prime factors, and we are done. 24. Prove that any integer can be written as the sum of the cubes of five integers, not necessarily distinct. Proof: We use the identity 6k = (k + 1)3 + (k −1)3 −k3 −k3 for k = n3 −n 6 = n(n −1)(n + 1) 6 , which is an integer for all n. We obtain n3 −n =  n3 −n 6 + 1  3 +  n3 −n 6 −1  3 −  n3 −n 6  3 −  n3 −n 6  3 . Therefore, n is equal to the sum (−n)3 +  n3 −n 6  3 +  n3 −n 6  3 +  n −n3 6 −1  3 +  n −n3 6 + 1  3 . Remark: One can prove that any rational number is the sum of the cubes of three rational numbers. www.riazisara.ir 150 104 Number Theory Problems 25. Integer or fractional parts? (1) [Czech and Slovak 1998] Find all real numbers x such that x⌊x⌊x⌊x⌋⌋⌋= 88. (2) [Belarus 1999] Show that the equation {x3} + {y3} = {z3} has infinitely many rational noninteger solutions. Solution: (1) Let f (x) = x⌊x⌊x⌊x⌋⌋⌋. We claim that if a and b are real numbers with the same sign and |a| > |b| ≥1, then | f (a)| > | f (b)|. We notice that |⌊a⌋| ≥|⌊b⌋| ≥ 1. Multiplying this by |a| > |b| ≥1, we have |a⌊a⌋| > |b⌊b⌋| ≥ 1. Notice that a⌊a⌋and a⌊a⌊a⌋⌋have the same signs as b⌊b⌋and b⌊b⌊b⌋⌋respectively. In a similar manner, |a⌊a⌊a⌋⌋| > |b⌊b⌊b⌋⌋| ≥1, |⌊a⌊a⌊a⌋⌋⌋| ≥|⌊b⌊b⌊b⌋⌋⌋| ≥1, and | f (a)| > | f (b)|, establishing our claim. We have f (x) = 0 for |x| < 1, f (1) = f (−1) = 1. Suppose that f (x) = 88. So |x| > 1, and we consider the following two cases. In the first case,we assume that x ≥1. It is easy to check that f  22 7  = 88. From our claim, we know that f (x) is increasing for x > 1. So x = 22 7 is the unique solution on this interval. In the second case, we assume that x ≤−1. From our claim, we know that f (x) is decreasing for x < 1. Since | f (−3)| = 81 < f (x) = 88 < 2 2 2 2 f 112 37 2 2 2 2 = 112, −3 > x > −112 37 and ⌊x⌊x⌊x⌋⌋⌋= −37. But then x = −88 37 > −3, a contradiction. Thus there is no solution on this interval. Therefore, x = 22 7 is the only solution. Finally, we note that 22 7 and −112 37 are found by finding ⌊x⌋, ⌊x⌊x⌋⌋, and ⌊x⌊x⌊x⌋⌋⌋in that order. For example, for x ≥1, f (3) < 88 < f (4), and so 3 < x < 4. Then ⌊x⌋= 3 and x⌊x⌊3x⌋⌋= 88. Then f (3) < 88 < f (10/3), so ⌊x⌊x⌋⌋= 9, and so on. www.riazisara.ir 5. Solutions to Advanced Problems 151 (2) Let x = 3 5 · (125k + 1), y = 4 5 · (125k + 1), and z = 6 5 · (125k + 1) for every integer k. These are never integers because 5 does not divide 125k + 1. Moreover, we note that 125x3 = 33(125k + 1)3 ≡33 (mod 125). Hence, 125 divides 125x3 −33 and x3 −  3 5 3 is an integer. Thus, {x3} = 27 125. Similarly, {y3} = 64 125 and {z3} = 216 125 −1 = 91 125 = 27 125 + 64 125, implying that {x3} + {y3} = {z3}. 26. Let n be a given positive integer greater than 1. If p is a prime divisor of the Fermat number fn, prove that p −1 is divisible by 2n+2. Proof: Since n > 1, fn−1 = 22n−1 + 1 is defined. Note that ( fn−1)2n+1 =  22n−1 + 1 2n+1 =  22n + 1 + 22n−1+12n =  fn + 22n−1+12n . By the binomial theorem, we obtain ( fn−1)2n+1 ≡  fn + 22n−1+12n ≡  22n−1+12n ≡  22n2n−1+1 ≡( fn −1)2n−1+1 ≡(−1)2n−1+1 ≡−1 (mod fn), implying that fn divides ( fn−1)2n+1 + 1. Since p divides fn, p divides ( fn−1)2n+1 + 1, from which the desired conclusion follows by setting a = fn−1 in introductory problem 1.49 (2). www.riazisara.ir 152 104 Number Theory Problems 27. [USAMO 1999 proposal, by Gerald Heuer] The sequence {an}∞ n=1 = {1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, . . . } of positive integers is formed by taking one odd integer, then two even integers, then three odd integers, etc. Express an in closed form. Solution: The solution is similar to the second proof of Example 1.70. We claim that an = 2n −  1 + √ 8n −7 2 for every positive integer n. We rewrite the given sequence in blocks as {an}∞ n=1 = {1; 2, 4; 5, 7, 9; 10, 12, 14, 16; 17, . . . }. Consider the sequence {bn}∞ n=1 = {1; 2, 2; 3, 3, 3; 4, 4, 4, 4; 5, . . . }. We show that an + bn = 2n (∗) for all positive integers n. This is clear for n = 1 and n = 2. Within each block in each sequence, an+1 = an +2 and bn+1 = bn, so if the relation (∗) holds for the first integer of a block, it holds for all integers in that block. If it is true for the last integer of a block, then it is true for the first integer of the next block because an and bn each increase by 1. By induction, relation (∗) holds for every positive integer n. It suffices to show that bn =  1 + √ 8n −7 2 . (†) If bn = k, it is in the kth group and is preceded by at least k −1 groups containing 1 + 2 + · · · + (k −1) terms. Considering also the fact that there are n −1 terms before bn, we conclude that 1 + 2 + · · · + (bn −1) ≤n −1, or bn(bn −1) 2 ≤n −1. www.riazisara.ir 5. Solutions to Advanced Problems 153 Solving the above quadratic inequality for bn gives bn ≤1 + √ 8n + 7 2 , from which (†) follows, by noting that bn is the largest integer satisfying this inequality. 28. [USAMO 1998, by Bjorn Poonen] Prove that for each n ≥2, there is a set S of n integers such that (a −b)2 divides ab for every distinct a, b ∈S. Proof: We will prove the assertion by induction on n that we can find such a set, all of whose elements are nonnegative. For n = 2, we may take S = {0, 1}. Now suppose that for some n ≥2, the desired set Sn of n nonnegative integers exists. Let L be the least common multiple of (a −b)2 and ab, with (a, b) ranging over pairs of distinct elements from Sn. Define Sn+1 = {L + a : a ∈S} ∪{0}. Then Sn+1 consists of n + 1 nonnegative integers, since L > 0. If α, β ∈ Sn+1 and either α of β is zero, then (α −β)2 divides αβ. If L + a, L + b ∈ Sn+1, with a, b distinct elements of Sn, then (L + a)(L + b) ≡ab ≡0 (mod (a −b)2)), so [(L + a) −(L + b)]2 divides (L + a)(L + b), completing the inductive step. 29. [St. Petersburg 2001] Show that there exist infinitely many positive integers n such that the largest prime divisor of n4 + 1 is greater than 2n. Proof: We claim first that there are infinitely many numbers that are prime divisors of m4 + 1 for some m. Suppose to the contrary that there is only a finite number of such primes. Let p1, p2, . . . , pk be all of them. Let p be any prime divisor of (p1 p2 · · · pk)4 + 1. This number cannot equal any pi. This contradicts our assumption, and establishes the claim. Let P be the set of all numbers that are prime divisors of m4 + 1 for some m. Pick any p from P and any integer m such that p divides m4 + 1. Let r be the residue of m modulo p. It follows that r < p and p divides both r4 + 1 and (p −r)4 + 1. Let n be the minimum of r and p −r. It follows that n < p 2 or p > 2n. If n can be obtained using the construction above, www.riazisara.ir 154 104 Number Theory Problems then it satisfies the desired condition. If it is constructed using the prime p, then p divides n4 + 1. Thus, any such number n can be constructed with only a finite number of primes p. Since the set P is infinite, and for each integer m such a number n can be constructed, there is an infinite number of integers n satisfying the desired condition. Note: The interested reader might want to solve the following more chal-lenging problem which appeared in USAMO in 2006. For integral m, let p(m) be the greatest prime divisor of m. By convention, we set p(±1) = 1 and p(0) = ∞. Find all polynomials f with integer coefficients such that the sequence {p( f (n2)) −2n}n≥0 is bounded above. (In particular, this re-quires f (n2) ̸= 0 for n ≥0.) 30. [Hungary 2003] For a positive integer k, let p(k) denote the greatest odd divisor of k. Prove that for every positive integer n, 2n 3 < p(1) 1 + p(2) 2 + · · · + p(n) n < 2(n + 1) 3 . Proof: Let s(n) = p(1) 1 + p(2) 2 + · · · + p(n) n . We need to show that 2n 3 < s(n) < 2(n + 1) 3 . (∗) We apply strong induction on n. The statement (∗) is true for n = 1 and n = 2, since 2 · 1 3 = 2 3 < s(1) = 1 < 2(1 + 1) 3 = 4 3 and 2 · 2 3 = 4 3 < s(2) = 1 + 1 2 = 3 2 < 2(2 + 1) 3 = 2. Assume that the statement (∗) is true for all integers n less than k, where k is some positive integer. We will show that the statement (∗) is true for integers n = k + 1. The key fact is that p(2k) = p(k). We consider two cases. www.riazisara.ir 5. Solutions to Advanced Problems 155 In the first case, we assume that k is even. We write k = 2m, where m is a positive integer less than k. For n = k + 1 = 2m + 1, we note that s(2m + 1) =  p(1) 1 + p(3) 3 + · · · + p(2m + 1) 2m + 1 +  p(2) 2 + p(4) 4 + · · · + p(2m) 2m = (m + 1) +  p(1) 2 + p(2) 4 + · · · + p(m) 2m = (m + 1) + 1 2  p(1) 1 + p(2) 2 + · · · + p(m) m = (m + 1) + s(m) 2 . By the induction hypothesis, we have (m + 1) + m 3 < (m + 1) + s(m) 2 = s(2m + 1) < (m + 1) + (m + 1) 3 . Since 2(2m+1) 3 = 4m+2 3 < 4m+3 3 = (m + 1) + m 3 and (m + 1) + (m+1) 3 = 4(m+1) 3 = 2(2m+1+1) 3 , it follows that 2(2m + 1) 3 < s(2m + 1) < 2(2m + 1 + 1) 3 , which is (∗) for n = 2m + 1. In the second case, we assume that k is odd. We write k = 2m + 1 and n = k + 1 = 2m + 2. Similar to the first case, we can show that s(2m + 2) = (m + 1) + s(m + 1) 2 . By the induction hypothesis, it is not difficult to show that the statement (∗) is also true for n = 2m + 2, which completes our induction. 31. If pt is an odd prime power and m is an integer relatively prime to both p and p −1, then for any a and b relatively prime to p, am ≡bm (mod pt) if and only if a ≡b (mod pt). Proof: Since (a −b) divides (am −bm), if pt divides (a −b) then pt divides (am −bm). www.riazisara.ir 156 104 Number Theory Problems Conversely, suppose a and b are relatively prime to p and am ≡bm (mod pt). Since m is relatively prime to both p and p −1, m is rela-tively prime to (p −1)pt−1 = ϕ(pt), so there exists a positive integer k such that mk ≡1 (mod ϕ(pt)). Then a ≡amk = (am)k ≡(bm)k = bmk ≡b (mod pt), as desired. Note: We can view this as an additional property for Proposition 1.18. In Proposition 1.18 (f), if we have a ≡b (mod m), then for any positive integer k, ak ≡bk (mod m). This problem allows us to take roots for congruence relations under certain relations. 32. [Turkey 1997] Prove that for each prime p ≥7, there exists a positive integer n and integers x1, . . . , xn, y1, . . . , yn not divisible by p such that x2 1 + y2 1 ≡x2 2 (mod p), x2 2 + y2 2 ≡x2 3 (mod p), . . . x2 n + y2 n ≡x2 1 (mod p). Proof: We claim that n = p −1 satisfies the conditions of the problem. We first consider a system of equations x2 1 + y2 1 = x2 2, x2 2 + y2 2 = x2 3, . . . x2 n + y2 n = x2 n+1. We repeatedly use the most well-known Pythagorean triple 32 + 42 = 52 to obtain the following equalities (3n)2 + (3n−1 · 4)2 = (3n−1 · 5)2, (3n−1 · 5)2 + (3n−2 · 5 · 4)2 = (3n−2 · 52)2, (3n−2 · 52)2 + (3n−3 · 52 · 4)2 = (3n−3 · 53)2, . . . (3n+1−i · 5i−1)2 + (3n−i · 5i−1 · 4)2 = (3n−i · 5i)2, . . . (3 · 5n−1)2 + (5n−1 · 4)2 = (5n)2. www.riazisara.ir 5. Solutions to Advanced Problems 157 Indeed, we set xi = 3n+1−i · 5i−1, yi = 4 · 3n−i · 5i−1, for every i = 1, . . . , n, and xn+1 = 5n. To finish our proof, we only need to note that by Fermat’s little theorem, we have x2 n+1 −x2 1 ≡52n −32n ≡25p−1 −9p−1 ≡0 (mod p). Note: There are infinitely many such n, for instance all multiples of p −1. 33. [HMMT 2004] For every positive integer n, prove that σ(1) 1 + σ(2) 2 + · · · + σ(n) n ≤2n. Proof: If d is a divisor of i, then so is i d , and i/d i = 1 d . Summing over all divisors d of i (which is σ(i)), we see that σ(i) i is the sum of all the reciprocals of the divisors of i; that is, σ(i) i =  d|i 1 d for every positive integer i. Consequently, the desired inequality becomes  d|1 1 d +  d|2 1 d + · · · +  d|n 1 d ≤2n. As we have shown in the solution of introductory problem 27, if we write out all these summands on the left-hand side explicitly, each number 1 d , with 1 ≤d ≤n, appears ! n d " times, once for each multiple of d that is less than or equal to n. Hence the desired inequality becomes 1 1 n 1 + 1 2 n 2 + 1 3 n 3 + · · · + 1 n n n < 2n. For each positive integer i, we have 1 i ! n i " < 1 i · n i = n i2 . Hence it suffices to show that n 12 + n 22 + · · · + n n2 < 2n, www.riazisara.ir 158 104 Number Theory Problems or 1 22 + 1 32 + · · · + 1 n2 < 1, which follows from 1 22 + 1 32 + · · · + 1 n2 < 1 1 · 2 + 1 2 · 3 + · · · + 1 n(n −1) = 1 1 −1 2 + 1 2 −1 3 + · · · +  1 n −1 −1 n = 1 −1 n < 1. Note: From calculus, we also know that 1 12 + 1 22 + · · · = π2 6 < 2. 34. [USAMO 2005, by R˘ azvan Gelca] Prove that the system x6 + x3 + x3y + y = 147157, x3 + x3y + y2 + y + z9 = 157147, has no solutions in integers x, y, and z. First Proof: Add the two equations; then add 1 to each side to obtain (x3 + y + 1)2 + z9 = 147157 + 157147 + 1. We prove that the two sides of this expression cannot be congruent modulo 19. We choose 19 because the least common multiple of the exponents 2 and 9 is 18, and by Fermat’s little theorem, a18 ≡1 (mod 19) when a is not a multiple of 19. In particular, (z9)2 ≡0 or 1 (mod 19), and it follows that the possible remainders when z9 is divided by 19 are −1, 0, 1. Next calculate n2 modulo 19 for n = 0, 1, . . . , 9 to see that the possible residues modulo 19 are −8, −3, −2, 0, 1, 4, 5, 6, 7, 9. www.riazisara.ir 5. Solutions to Advanced Problems 159 Consequently, adding a number from the last two lists gives the possible residues modulo 19 for (x3 + y + 1)2 + z9: −8 −3 −2 0 1 4 5 6 7 9 −1 −9 −4 −3 −1 0 3 4 5 6 8 0 −8 −3 −2 0 1 4 5 6 7 9 1 −7 −2 −1 1 2 5 6 7 8 10 Finally, apply Fermat’s little theorem to see that 147157 + 157147 + 1 ≡14 (mod 19). Because we cannot obtain 14 (or −5), which does not appear in the table above, the system has no solution in integers x, y, and z. Second Proof: We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain (x3 + y + 1)2 + z9 = 147157 + 157147 + 1. By Fermat’s little theorem, a12 ≡1 (mod 13) when a is not a multiple of 13. Hence we compute 147157 ≡41 ≡4 (mod 13) and 157147 ≡13 ≡1 (mod 13). Thus (x3 + y + 1)2 + z9 ≡6 (mod 13). The cubes modulo 13 are 0, ±1, and ±5. Writing the first given equation as (x3 + 1)(x3 + y) ≡4 (mod 13), we see that there is no solution in the case x3 ≡−1 (mod 13) and for x3 congruent to 0, 1, 5, −5 modulo 13. Correspondingly, x3 + y must be congruent to 4, 2, 5, −1. Hence (x3 + y + 1)2 ≡12, 9, 10, or 0 (mod 13). Also, z9 is a cube; hence z9 must be 0, 1, 5, 8, or 12 modulo 13. The following table shows that 6 modulo 13 is not obtained by adding one of 0, 9, 10, 12 to one of 0, 1, 5, 8, 12: 0 1 5 8 12 0 0 1 5 8 12 9 9 10 1 4 8 10 10 11 2 5 9 12 12 0 4 7 11 Hence the system has no solutions in integers. www.riazisara.ir 160 104 Number Theory Problems Note: This argument shows that there is no solution even if z9 is replaced by z3. 35. [St. Petersburg 2000] What is the smallest number of weighings on a bal-ance scale needed to identify the individual weights of a set of objects known to weigh 1, 3, 32, . . . , 326 in some order? (The balance scale reports the weight of the objects in the left pan minus the weight of the objects in the right pan.) Solution: At least three weighings are necessary: each of the first two weighings divides the weights into three categories (the weights in the left pan, the weights in the right pan, and the weights remaining off the scale). Because 27 > 3 · 3, some two weights must fall into the same category on both weighings, implying that these weights cannot be distinguished. We now show that three weighings indeed suffice. Label the 27 weights using the three-letter words made up of the letters L, R, O. In the ith weighing, put the weights whose ith letter is L on the left pan and the weights whose ith letter is R on the right pan. The difference between the total weight of the objects in the left pan and the total weight of the objects in the right pan equals ϵ030 + ϵ131 + · · · + ϵ26326, where ϵ j equals 1, −1, or 0 if 3 j is in the left pan, in the right pan, or off the scale, respectively. The value of the above sum uniquely determines all of the ϵ j: the value of the sum modulo 3 determines ϵ0, then the value of the sum modulo 9 determines ϵ1; and so on. Thus, for j = 0, . . . , 26, the ith weighing determines the ith letter of the weight that measures 3 j. After three weighings, we thus know exactly which weight measures 3 j, as desired. Note: This is a case of a more general result, that each integer has a unique representation in base 3 using the digits −1, 0, 1. Clearly, this works for numbers n with 0 ≤n < 31 (since 0 = 0, 1 = 1, and 2 = 3 −1). Assume that this works for numbers n with 0 ≤n < 3k for some positive integer k. We consider n with 3k ≤n < 3k+1. If 3k ≤n < 2 · 3k, it works because n = 3k + n1, where 0 ≤n1 < 3k; if 2 · 3k ≤n < 3k+1, it works because n = 3k+1 −3k + n1 with 0 ≤n1 < 3k. It is not difficult to see that it works for negative numbers and the representation is unique for every integer. Indeed, we can convert a regular base-3 representation easily to www.riazisara.ir 5. Solutions to Advanced Problems 161 this new base-3 representation. For example, 49 = 1211(3) = 33 + 2 · 32 + 3 + 1 = 2 · 33 −32 + 3 + 1 = 34 −33 −32 + 3 + 1. 36. [Iberoamerican 1998] Let λ be the positive root of the equation t2 −1998t −1 = 0. Define the sequence x0, x1, . . . by setting x0 = 1, xn+1 = ⌊λxn⌋ (n ≥0). Find the remainder when x1998 is divided by 1998. Solution: We have 1998 < λ = 1998 + √ 19982 + 4 2 = 999 + 5 9992 + 1 < 1999, x1 = 1998, x2 = 19982. Since λ2 −1998λ −1 = 0, λ = 1998 + 1 λ and xλ = 1998x + x λ for all real numbers x. Since xn = ⌊xn−1λ⌋and xn−1 is an integer and λ is irrational, we have xn < xn−1λ < xn + 1, or xn λ < xn−1 < xn + 1 λ . Since λ > 1998, ! xn λ " = xn−1 −1. Therefore, xn+1 = ⌊xnλ⌋= 1998xn + xn λ = 1998xn + xn−1 −1, that is, xn+1 ≡xn−1 −1 (mod 1998). Therefore by induction x1998 ≡ x0 −999 ≡1000 (mod 1998). 37. [USAMO 1996, by Richard Stong] Determine (with proof) whether there is a subset X of the integers with the following property: for any integer n there is exactly one solution of a + 2b = n with a, b ∈X. First Proof: Yes, there is such a subset. As shown in introductory problem 39 (3), if the problem is restricted to the nonnegative integers, then the set of integers whose representations in base-4 contain only the digits 0 and 1 satisfies the desired property. To accommodate the negative integers as www.riazisara.ir 162 104 Number Theory Problems well, we switch to “base-(−4).” That is, we represent every integer in the form k i=0 ci(−4)i, with ci ∈{0, 1, 2, 3} for all i and ck ̸= 0, and let X be the set of numbers whose representations use only the digits 0 and 1. This X will again have the desired property, once we show that every integer has a unique representation in this fashion. To show that base-(−4) representations are unique, let {ci} and {di} be two distinct finite sequences of elements of {0, 1, 2, 3}, and let j be the smallest integer such that c j ̸= d j. Then k  i=0 ci(−4)i ̸≡ k  i=0 di(−4)i (mod 4 j), so in particular the two numbers represented by {ci} and {di} are distinct. On the other hand, to show that n admits a base-(−4) representation, find an integer k such that 1 + 42 + · · · + 42k ≥n and express n + 4 + · · · + 42k−1 = 2k  i=0 ci4i. Now set d2i = c2i and d2i−1 = 3−c2i−1, and note that n = 2k i=0 di(−4)i. Second Proof: For any S of integers, let S∗= {a + 2b| a, b ∈S}. Call a finite set of integers S = {a1, a2, . . . , am} good if |S∗| = |S|2; that is, if the values ai +2a j (1 ≤i, j ≤m) are distinct. We first prove that given a good set and an integer n, we can always find a good superset T of S such that n is an element in T ∗. If n is in S∗already, take T = S. Otherwise, take T = S ∪{k, n −2k}, where k is to be chosen. Then put T ∗= S∗∪Q ∪R, where Q = {3k, 3(n −2k), k + 2(n −2k), (n −2k) + 2k} and R = {k + 2ai, (n −2k) + 2ai, ai + 2k, ai + 2(n −2k)| 1 ≤i ≤m}. Note that for any choice of k, we have n = (n −2k) + 2k in Q, which is a subset of T ∗. Except for n, the new values are distinct nonconstant linear forms in k, so if k is sufficiently large, they will all be distinct from each other and from the elements of S∗. This proves that T ∗is good. www.riazisara.ir 5. Solutions to Advanced Problems 163 Starting with the good set X0 = {0}, we thus obtain a sequence of sets X1, X2, X3, . . . such that for each positive integer j, X j is a good superset of X j−1 and X∗ j contains the jth term of the sequence 1, −1, 2, −2, 3, −3, . . . . It follows that X = ∞ 1 j=0 X j has the desired property. 38. The number xn is defined as the last digit in the decimal representation of the integer √ 2 n (n = 1, 2, . . . ). Determine whether the sequence x1, x2, . . . , xn, . . . is periodic. Solution: The answer is negative. Set yn = 0 if xn is even and yn = 1 otherwise. The new sequence y1, y2, . . . , yn, . . . is formed by the residues of the numbers xn modulo 2. If x1, x2, . . . , xn, . . . is periodic, then so is y1, y2, . . . , yn, . . . . We shall prove that y1, y2, . . . , yn, . . . is not periodic, which implies that the answer to the question is negative. Let us consider the sequence y1, y3, y5, . . . , y2n+1, . . . . Its term y2n+1 can be obtained as follows. Write down √ 2 in base-2, multiply by 2n (this gives ( √ 2)2n+1), and discard the fractional part of the result to get ( √ 2)2n+1 . Then take the last (binary) digit of this integer; it is y2n+1. But multiplying by 2n in base 2 amounts simply to shifting the binary point n positions to the right. This implies that y2n+1 is in fact the nth digit of √ 2 after the binary point. Since √ 2 is irrational, we conclude that the sequence y1, y3, . . . , y2n+1, . . . is not periodic. It is easy to infer from here that y1, y2, . . . , yn, . . . is not periodic too, and we are done. 39. [Erd¨ os-Suranyi] Prove that every integer n can be represented in infinitely many ways as n = ±12 ± 22 ± · · · ± k2 for a convenient k and a suitable choice of the signs + and −. www.riazisara.ir 164 104 Number Theory Problems Proof: It suffices to prove the statement for nonnegative n’s, because for negative n’s we can simply change all the signs. The proof goes by induction of step 4; that is, establishing the statement for n = k + 4 based on the induction hypothesis for n = k. We first show that the statement holds for n = 0, 1, 2, and 3. We consider representations 0, 1, 2 and 3: 0 = 12 + 22 −32 + 42 −52 −62 + 72, 1 = 12, 2 = −12 −22 −32 + 42, 3 = −12 + 22. If n is representable in the desired form then so is n + 4, because 4 can be written as 4 = (k + 1)2 −(k + 2)2 −(k + 3)2 + (k + 4)2 (∗) for any k. It follows inductively that a representation of the desired form can be written for any nonnegative integer n. Note: From (∗) it also follows that (k + 1)2 −(k + 2)2 −(k + 3) + (k + 4)2 −(k + 5)2 + (k + 6)2 + (k + 7)2 −(k + 8)2 = 0 for every integer k; hence it can be easily inferred that the number of repre-sentations of an integer in the desired form is infinite. 40. [China 2004] Let n be a given integer with n ≥4. For a positive integer m, let Sm denote the set {m, m + 1, . . . , m + n −1}. Determine the minimum value of f (n) such that every f (n)-element subset of Sm (for every m) contains at least three pairwise relatively prime elements. First Proof: The answer is f (n) = n + 1 2 + n + 1 3 − n + 1 6 + 1. (∗) Let us call a set T good if T contains three (distinct) elements that are relatively prime. In the first step, we establish two simple claims: (a) f (n) exists and f (n) ≤n; (b) f (n + 1) ≤f (n) + 1. www.riazisara.ir 5. Solutions to Advanced Problems 165 Since n ≥4, m, m+1, m+2, m+3 are distinct element in Sm. If m is even, then the set {m + 1, m + 2, m + 3} is good; if m is odd, {m, m + 1, m + 2} is good. Hence the n-element set Sm is good for all m, and so f (n) ≤n, establishing (a). Claim (b) follows directly from the relation {m, m + 1, . . . , m + n} = {m, m + 1, . . . , m + n −1} ∪{m + n}. Next we give a lower bound for f (n). Consider S2 = {2, 3, . . . , n + 1} and its subset T2 that contains those elements in S2 that are multiples of either 2 or 3 or both. By the pigeonhole principle, any three elements in T must share a common factor (of either 2 or 3). Hence T2 is not good. But by the inclusion and exclusion principle, |T2| = n + 1 2 + n + 1 3 − n + 1 6 , and so f (n) ≥ n + 1 2 + n + 1 3 − n + 1 6 + 1, (∗∗) where ⌊x⌋is the greatest integer less than or equal to x. We claim that this lower bound is in fact the exact value of f (n). Since n ≥4, we know that m +1, m +2, m +3, m +4 are distinct elements in Sm. If m is even, then {m + 1, m + 2, m + 3} is good; if m is odd, then {m + 2, m + 3, m + 4) is good. Hence the n-element set Sm is good for all m. Using this fact with (∗∗) gives us f (4) = 4 and f (5) = 5. By simple computation, the last inequality gives f (4) ≥4, f (5) ≥5, f (6) ≥5, f (7) ≥6, f (8) ≥7, and f (9) ≥8. Since f (n) ≤n, we conclude that f (4) = 4 and f (5) = 5. We claim that f (6) = 5. Then by claim (b), we have f (7) = 6, f (8) = 7, and f (9) = 8. We now show that f (6) = 5; that is, any 5-element subset T of a set of 6 consecutive numbers is good. Among these 6 numbers, 3 are odd consecutive numbers (which is a good triple) and 3 are even consecutive numbers. If all three odd numbers are in T , then T is good and we are done. Otherwise, T must contain all the even numbers, and two of the three odd numbers. If the two odd numbers in T are consecutive (of the form 2x + 1 and 2x + 3), then T is good since (2x + 1, 2x + 2, 2x + 3) is in T ; otherwise, the two odd numbers in T are of the form 2x + 1 and 2x + 5, and T is good since T contains both (2x + 1, 2x + 2, 2x + 5) and (2x + 1, 2x + 4, 2x + 5), and at least one of these two triples is good (since at least one of 2x + 1 and 2x + 5 is not divisible by 3). www.riazisara.ir 166 104 Number Theory Problems Since it is clear that f (n + 1) ≤f (n) + 1, this combined with (∗∗) gives f (7) = 6, f (8) = 7, and f (9) = 8. Finally, we prove the result (∗) by induction on n. The above arguments show that the base cases for n ≤9 are true. Assume that (∗) is true for some n = k, where k is an integer greater than or equal to 9. For n = k +1, note that Sm = {m, m + 1, . . . , m + k} = {m, m + 1, . . . , m + k −6} ∪{m + k −5, . . . , m + k}. Hence, by the pigeonhole principle, f (k + 1) ≤f (k −5) + f (6) −1. Applying the induction hypothesis to f (k −5), and using f (6) = 5, we have f (k + 1) ≤ k −4 2 + k −4 3 − k −4 6 + 5 = k + 2 2 + k + 2 3 − k + 2 6 + 1. This combined with (∗∗) establishes (∗) for n = k + 1, and our induction is complete. Second Proof: (By Kevin Modzelewski) We maintain the same notation as in the first solution. As we have shown in the first proof, all 5-element subsets of a set of 6 consecutive integers are good. Now we consider some cases. (i) In this case we assume that n ≡0 (mod 6). We write n = 6k. We can partition the set Sm into k subsets of 6 consecutive integers. If 4k + 1 numbers are chosen, by the pigeonhole principle one these subsets contains 5 of the chosen numbers, and hence is good. On the other hand, each subset contains 4 numbers that are either divisible by 2 or 3 (those numbers that are congruent to 0, 2, 3, 4 modulo 6). The 4k-element subset consisting of these numbers is not good. Hence f (n) = 4k + 1 = 4 ! n 6 " + 1. (ii) In this case we assume that n ≡1 (mod 6). We write n = 6k +1. By (i) and observation (b) in the first solution, we have f (n) = 4k + 1 or f (n) = 4k + 2. On the other hand, there are 4k + 1 elements in S1 = {2, 3, . . . , n + 1} = {2, 3, . . . , 6k + 2} that are divisible by 2 or 3. Hence f (n) = 4k + 2 = 4 ! n 6 " + 2. (iii) In this case we assume that n ≡2 (mod 6). We write n = 6k +2. By (ii) and observation (b), we have f (n) = 4k +2 or f (n) = 4k +3. On www.riazisara.ir 5. Solutions to Advanced Problems 167 the other hand, there are 4k + 2 elements in S1 = {2, 3, . . . , 6k + 3} that are divisible by 2 or 3. Hence f (n) = 4k + 3 = 4 ! n 6 " + 3. (iv) In this case we assume that n ≡3 (mod 6). We write n = 6k + 3. Again, we have f (n) = 4k + 2 or f (n) = 4k + 3. On the other hand, there are 4k + 2 elements in S1 = {2, 3, . . . , 6k + 4} that are divisible by 2 or 3. Hence f (n) = 4k + 4 = 4 ! n 6 " + 4. (v) In this case we assume that n ≡4 (mod 6). We write n = 6k + 4. We can partition set Sm into {6k + 1, 6k + 2, 6k + 3, 6k + 4} and k subsets of 6 consecutive integers. Let T be a subset of Sm that is not good. Each of the 6-element subsets can have 4 elements in T . Also note that 6k + 1 and 6k + 3 cannot be both in T . Hence T can have at most 4k + 3 elements. Hence f (n) ≤4k + 4. By (iv) and observation (b), we conclude that f (n) = 4k + 4 = 4 ! n 6 " + 4. (vi) In this case we assume that n ≡5 (mod 6). We write n = 6k + 5. Again, we have f (n) = 4k + 4 or f (n) = 4k + 5. On the other hand, there are 4k + 4 elements in S1 = {2, 3, . . . , 6k + 6} that are divisible by 2 or 3. Hence f (n) = 4k + 5 = 4 ! n 6 " + 5. Combining the above, we conclude that f (n) = 4 · n 6 + ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 n ≡0 (mod 6), 2 n ≡1 (mod 6), 3 n ≡2 (mod 6), 4 n ≡3 (mod 6), 4 n ≡4 (mod 6), 5 n ≡5 (mod 6). It is then not difficult to check that f (n) = n + 1 2 + n + 1 3 − n + 1 6 + 1. Note: Note also that f (n) can be expressed as f (n) = n − n 6 − n + 2 6 + 1. Indeed, the above expression might be more convenient in the second solu-tion. The equivalence of the two expressions can be established by repeated www.riazisara.ir 168 104 Number Theory Problems applying the Hermite identity (Proposition 1.48) as follows: n = 2 · n 2 = n 2 + n 2 + 1 2 = n 2 + n + 1 2 , n 2 = 3 · n 6 = n 6 + n + 2 6 + n + 4 6 , n + 1 3 = 2 · n + 1 6 = n + 1 6 + n + 4 6 . 41. [China 1999] Find the least positive integer r such that for any positive integers a, b, c, d, ((abcd)!)r is divisible by the product of (a!)bcd+1, (b!)acd+1, (c!)abd+1, (d!)abc+1, ((ab)!)cd+1, ((bc)!)ad+1, ((cd)!)ab+1, ((ac)!)bd+1, ((bd)!)ac+1, ((ad)!)bc+1, ((abc)!)d+1, ((abd)!)c+1, ((acd)!)b+1, ((bcd)!)a+1. Solution: Let p denote the product of the 14 numbers. Setting b = c = d = 1, then p = (a!)2+3·2+3·2 = (a!)14, implying that r ≥14. We claim that r = 14. It suffices to show that p divides ((abcd)!)14. We pair numbers (a!)bcd+1 and ((bcd)!)a+1. Indeed, we have (a!)bcd+1 · ((bcd)!)a+1 = 6 (a!)bcd · (bcd)! 7 8 ((bcd)!)a · (a)! 9 and its cyclic analogous forms. Likewise, we have ((ab!)cd+1 · ((cd)!)ab+1 = 6 ((ab)!)cd · (cd)! 7 6 ((cd)!)ab · (ab)! 7 and its cyclic analogous forms. It is then not difficult to see that our claim follows Example 1.74 (1). 42. Two classics on L.C.M. (1) Let a0 < a1 < a2 < · · · < an be positive integers. Prove that 1 lcm(a0, a1) + 1 lcm(a1, a2) + · · · + 1 lcm(an−1, an) ≤1 −1 2n . (2) Several positive integers are given not exceeding a fixed integer con-stant m. Prove that if every positive integer less than or equal to m is not divisible by any pair of the given numbers, then the sum of the reciprocals of these numbers is less than 3 2. www.riazisara.ir 5. Solutions to Advanced Problems 169 Proof: While it is clear that (1) is a property of L.C.M., it is not obvious that (2) is also related to L.C.M. (1) We induct on n. The base case n = 1 is trivial, since lcm(a0, a1) ≥ lcm(1, 2) = 2. We assume that the statement is true for n = k; that is, if a0 < a1 < a2 < · · · < ak are positive integers, then 1 lcm(a0, a1) + 1 lcm(a1, a2) + · · · + 1 lcm(ak−1, ak) ≤1 −1 2k . Now we consider the case n = k + 1. Let a0 < a1 < a2 < · · · < ak < ak+1 be positive integers. We consider two cases. • In the first case, we assume that ak+1 ≥2k+1. Then we have lcm(ak, ak+1) ≥ak+1 ≥2k+1. It follows by the induction hy-pothesis that 1 lcm(a0, a1) + · · · + 1 lcm(ak−1, ak) + 1 lcm(ak, ak+1) ≤1 −1 2k + 1 2k+1 = 1 − 1 2k+1 , establishing the inductive step. • In the second case, we assume that ak < 2k+1. We have 1 lcm(ai−1, ai) = gcd(ai−1, ai) ai−1ai ≤a1 −ai−1 ai−1a1 = 1 ai−1 −1 ai . Adding the above inequalities for i from 1 through k + 1 gives 1 lcm(a0, a1) + · · · + 1 lcm(ak−1, ak) + 1 lcm(ak, ak+1) ≤1 a0 − 1 ak+1 ≤1 − 1 2k+1 , again establishing the inductive step. (2) The key is to interpret the sentence “every positive integer less than or equal to m is not divisible by any pair of the given numbers.” Indeed, this implies that the least common multiple of every two of the given numbers is greater than m. Given n numbers, denote them by x1, x2, . . . , xn. For a given i, there are m xi multiples of xi among 1, 2, . . . , m. None of them is a mul-tiple of x j for j ̸= i, since the least common multiple of xi and x j is greater than m. Hence there are m x1 + m x2 + · · · + m xn www.riazisara.ir 170 104 Number Theory Problems distinct elements in the set {1, 2, . . . , m} that are divisible by one of the numbers x1, x2, . . . , xn. None of these elements can be 1 (unless n = 1, in which case the claim is obvious). Hence m x1 + m x2 + · · · + m xn ≤m −1. Taking into account that m xi < m xi + 1 for each i, we obtain m  1 x1 + 1 x2 + · · · + 1 xn < m + n −1. We now claim that n ≤m+1 2 , which will imply 1 x1 + 1 x2 + · · · + 1 xn < 1 + n −1 m < 3 2. Indeed, note that the greatest odd divisors of x1, x2, . . . , xn are all distinct. Otherwise, if some two of the given numbers shared the same greatest odd divisor, one of them would be a multiple of the other, contradicting the hypothesis. Hence n does not exceed the number of odd integers among 1, 2, . . . , m, and our claim n ≤m+1 2 follows. 43. For a positive integer n, let r(n) denote the sum of the remainders of n divided by 1, 2, . . . , n. Prove that there are infinitely many n such that r(n) = r(n −1). Solution: By Proposition 1.46 (a), the remainder when n is divided by k is equal to # n k $ · k = n − ! n k " · k. Hence we have r(n) =  k=1 n  n − n k · k  . Thus, the condition r(n) = r(n −1) is equivalent to the equation  k=1 n  n − n k · k  =  k=1 n −1  n −1 − n −1 k · k , or 2n −1 = n + n−1  k=1 [n −(n −1)] = n  k=1 n k · k − n−1  k=1 n −1 k · k. (∗) www.riazisara.ir 5. Solutions to Advanced Problems 171 If k does not divide n, then ! n k " = n−1 k , and so ! n k " · k = n−1 k · k; if k divides n, then ! n k " = n−1 k + 1, and so ! n k " · k = n−1 k · k + k. We conclude that the equation (∗) is equivalent to 2n −1 =  k|n k. But the last equation can easily be satisfied by setting n = 2m, where m is a nonnegative integer. Indeed, 2n −1 = 2m+1 −1 = 1 + 2 + 22 + · · · + 2m. Therefore, if n is a perfect power of 2, then r(n) = r(n −1). 44. Two related IMO problems. (1) [IMO 1994 Short List] A wobbly number is a positive integer whose digits are alternately nonzero and zero with the units digit being non-zero. Determine all positive integers that do not divide any wobbly numbers. (2) [IMO 2004] A positive integer is called alternating if among any two consecutive digits in its decimal representation, one is even and the other is odd. Find all positive integers n such that n has a multiple that is alternating. Solution: This is a continuation of introductory problem 52. (1) If n is a multiple of 10, then the last digit of any of its multiples is 0, and so n does not divide any wobbly numbers. If n is a multiple of 25, then the last two digits of any of its multiples are 25, or 50, or 75, or 00, and so n does not divide any wobbly numbers. We now prove that these are the only numbers not dividing any wobbly numbers. First, we consider odd numbers m not divisible by 5. Then gcd(m, 10) = 1 and gcd((10k −1)m, 10) = 1 for every positive integer k. By Euler’s theorem, there exists an integer ℓsuch that 10ℓ≡1 (mod (10k −1)m), implying that 10kℓ≡1 (mod (10k −1)m). www.riazisara.ir 172 104 Number Theory Problems Since 10kℓ−1 =  10k −1   10k(ℓ−1) + 10k(ℓ−2) + · · · + 10k + 1  , we conclude that wk = 101010 . . . 1   2ℓ−1 digits = 102(ℓ−1) + 102(ℓ−2) + · · · + 102 + 1 is divisible by m. In particular, w2 is a wobbly number (with digits 0 and 1) divisible by m. Second, we consider odd numbers m′ that are divisible by 5. Since the number is not a multiple of 25, we can write m′ = 5m. Then 5w2 is a wobbly number (with digits 0 and 5) divisible by m′. Next, we consider perfect powers of 2. It suffices to show that 22t+1 (for every nonnegative integer t) divides a (2t −1)-digit wobbly num-ber. We induct on t. The base case t = 1 is trivial by considering wobbly numbers v1 = a1 = 8. For t = 2, we consider numbers in the form v2 = a208 = 100a2 + 8 = 4(25a2 + 2). We need to find a nonzero digit a2 such that 25a2 + 2 ≡0 (mod 8). It is easy to check that a2 = 6 satisfies the condition, and so 608 is a wobbly multiple of 25. In general, assume that 22t+1 divides wobbly number vt = at0at−1 . . . 0a1. We write vt = 22t+1ut. Consider the numbers in the form at+10at0at−1 . . . 0a1 = at+1 · 102t + 22t+1ut = 22t 52tat+1 + 2ut  . We need to find a digit at+1 such that 52tat+1 + 2ut ≡8. Since S = {0, 1, 2, 3, 4, 5, 6, 7, 8} forms a complete set of residue classes modulo 8, there is an element at+1 in S such that 52tat+1 + 2ut ≡8, and for this at+1, the (2t + 1)-digit wobbly number vt+1 = at+10at0at−1 . . . 0a1 is divisible by 22t+3, completing the induction. Finally, we consider the number of the form 2tm, where t ≥1 and gcd(m, 10) = 1. It suffices to show that 22t+1m divides a wobbly number. We claim that the concatenation of ℓ−1 vt0 = vt · 10’s and a vt will do the job. Indeed, vt0vt0 . . . vt   ℓvt’s = vt · w2t, because 22t+1 divides vt and m divides w2t. www.riazisara.ir 5. Solutions to Advanced Problems 173 (2) The answers are those positive integers that are not divisible by 20. We call an integer n an alternator if it has a multiple that is alternating. Because any multiple of 20 ends with an even digit followed by 0, multiples of 20 are not alternating. Hence multiples of 20 are not alternators. We show that all other numbers are alternators. Let n be a positive integer not a multiple of 20. Note that all divisors of an alternator are alternators. We may assume that n is a even number. We first establish the following key fact: If n = 2ℓor 2 · 5ℓ, for some positive integer ℓ, then there exists a multiple X(n) of n such that X(n) is alternating and X(n) has n digits. Indeed, we can set m = 10n+1 −10 99 = 101010 . . . 10   n digits . For every integer k = 0, 1, . . . , n −1, there exists a sequence e0, e1, . . . , ek ∈{0, 2, 4, 6, 8} such that M + k  i=0 ei · 10i is divisible by 2k+2 if n is of the form 2ℓ, or by 2 · 5k+1 if n = 2 · 5ℓ. This is straightforwardly proved by induction on k (as we did in the proof of part (1) or Example 1.53). In particular, there exist e0, . . . , en−1 ∈{0, 2, 4, 6, 8} such that X(n) = m + n−1  i=0 ei · 10i is divisible by n. This X(n) is alternating and has n digits, establishing this fact. Now we prove our main result. Because n is even and not divisible by 20, we write n in the form n′m, where n′ = 2ℓor 2 · 5ℓand gcd(m, 10) = 1. (Clearly, n′ ≥ℓ.) Let c ≥n′ be an integer such that 10c ≡1 (mod m). (Such a c exists because 10ϕ(m) ≡1 (mod m), by Euler’s theorem.) Let M be the concatenation of 1010 . . . and X′(n). More precisely, we set M = 102mc+1 −10 99 · 10n′ + X(n′) = 101010 . . . 10   2mc digits X(n′). www.riazisara.ir 174 104 Number Theory Problems Since X(n′) is an alternating number with exactly n′ digits, M is clearly an alternating number. Because n′ ≥ℓ, M is divisible by n′. Since gcd(2, m) = 1, there exists k ∈{0, 1, 2, . . . , m −1} such that M ≡−2k (mod m). We consider the number X(n) = M + k  i=1 2 · 10ci. Note that 10c > X(n′), because c ≥n′ and X(n′) has exactly n′ digits (by the key fact we established earlier). It is not difficult to show that X(n) is also alternating. It is clear that X(n) ≡m+2k ≡0 (mod m), that is, X(n) is divisible by m. This X(n) is also divisible by n′ (since n′ divides 10n′ , which divides 10c) and is alternating. Thus X(n) is an alternating number divisible by n; that is, n is an alternator. Note: There are different approaches to both parts. Nevertheless, all these methods work on powers of 2 and 5 first, and applying certain concatena-tions of wobbly/alternating numbers. These particular methods have been chosen because they are independent of the Chinese Remainder Theorem. 45. [USAMO 1995] Let p be an odd prime. The sequence (an)n≥0 is defined as follows: a0 = 0, a1 = 1, . . . , ap−2 = p −2, and for all n ≥p −1, an is the least positive integer that does not form an arithmetic sequence of length p with any of the preceding terms. Prove that for all n, an is the number obtained by writing n in base-(p −1) and reading the result in base-p. Proof: We say that a subset of positive integers is p-progression-free if it does not contain an arithmetic progression of length p. Denote by bn the number obtained by writing n in base-(p −1) and reading it in base-p. One can easily prove that an = bn for all n = 0, 1, 2, . . . by induction, using the following properties of the set B = {b0, b1, . . . , bn, . . . }: (a) B is p-progression-free; (b) If bn−1 < a < bn for some n ≥1, then the set {b0, b1, . . . , bn−1, a} is not p-progression-free. Indeed, assume that (a) and (b) hold. By the definitions of ak and bk, we have ak = bk for k = 0, 1, . . . , p −2. Let ak = bk for all k ≤n −1, where n ≥p −1. By (a), the set {a0, a1, . . . , an−1, bn} = {b0, b1, . . . , bn−1, bn} is p-progression-free, so an ≤bn. Also, the inequality an < bn is impossi-ble in view of (b). Hence an = bn and we are done. www.riazisara.ir 5. Solutions to Advanced Problems 175 It remains to establish properties (a) and (b). Let us note first that B consists of all numbers whose base-p representation does not contain the digit p−1. Hence (a) follows from the fact that if a, a + d, . . . , a + (p −1)d is any arithmetic progression of length p, then all base-p digits occur in the base-p representation of its terms. To see this, represent d in the form d = pmk, where gcd(k, p) = 1. Then d ends in m zeros, and the digit δ preceding them is nonzero. It is easy to see that if α is the (m + 1)st digit of a (from right to left), then the corresponding digits of a, a + d, . . . , a + (p −1)d are the remainders of α, α + δ, . . . , α + (p −1)δ modulo p, respectively. It remains to note that α, α +δ, . . . , α +(p −1)δ is a complete set of residues modulo p, because δ is relatively prime to p. This finishes the proof of (a). We start proving (b) by the remark that bn−1 < a < bn implies that a is not in B. Since B consists precisely of the numbers whose base-p representa-tions do not contain the digit p −1, this very digit must occur in the base-p representation of a. Let d be the number obtained from a by replacing each of its digits by 0 if the digit is not p −1, and by 1 if it is p −1. Consider the progression a −(p −1)d, a −(p −2)d, . . . , a −d, a. As the definition of d implies, the first p −1 terms do not contain p −1 in their base-p representation. Hence, being less than a, they must belong to the set {b0, b1, . . . , bn−1}. Therefore the set {b0, b1, . . . , bn−1, a} is not p-progression-free, and the proof is complete. 46. [IMO 2000] Determine whether there exists a positive integer n such that n is divisible by exactly 2000 different prime numbers, and 2n +1 is divisible by n. Solution: The answer is positive. We claim the following key fact: For any integer a > 2 there exists a prime p such that p divides (a3 + 1) but p does not divide (a + 1). Indeed, since a3 + 1 = (a + 1)(a2 −a + 1), we need to show that there exists a prime p such that p | (a2 −a + 1) but p ∤(a + 1). Since a2 −a + 1 = (a + 1)(a −2) + 3, it follows that gcd(a2 −a + 1, a + 1) = 1 or gcd(a2 −a + 1, a + 1) = 3. In the first case, our claim is clearly true. In the second case, we note that 3 www.riazisara.ir 176 104 Number Theory Problems divides both a +1 and a −2, and so 3 fully divides a2 −a +1. Since a > 2, a2 −a + 1 > 3, and so there is a prime p ̸= 3 that divides a2 −a + 1, and this prime p satisfies the conditions of the claim. By our claim, there exist (odd) distinct primes p1, p2, p3, . . . , p2000 such that p1 = 3, p2 ̸= 3, p2 | (232 + 1), and pi+1 | (23i+1 + 1), pi+1 ∤(23i + 1), for every 2 ≤i ≤1999. It is not difficult to see that n = p2000 1 · p2 · · · p2000 = 32000 p2 · p2000 satisfies the conditions of the problem. Indeed, for every 2 ≤i ≤2000, 3i | 32000, and so pi | 23i + 1 | 232000 + 1. By a simple induction, we also note that 3k+1 fully divides 23k +1 for every positive integer k, because 3 fully divides a2 −a + 1 for a = 23k (as we have shown in the proof of our claim). Therefore, n | 232000 + 1 | 2n + 1, since n is a odd multiple of 32000. 47. Two cyclic symmetric divisibility relations. (1) [Russia 2000] Determine whether there exist pairwise relatively prime integers a, b, and c with a, b, c > 1 such that b | 2a + 1, c | 2b + 1, a | 2c + 1. (2) [TST 2003, by Reid Barton] Find all ordered triples of primes (p, q,r) such that p | qr + 1, q | r p + 1, r | pq + 1. Solution: Order is the key word to this problem. (1) The answer is negative. We claim that no such integers exist. Assume for the sake of the contradiction that we did have pairwise relatively prime integers a, b, c > 1 such that b divides 2a + 1, c divides 2b + 1, and a divides 2c + 1. Then a, b, and c are all odd. www.riazisara.ir 5. Solutions to Advanced Problems 177 To make our life a bit easier, we first assume that all a, b, c are primes. By cyclic conditions given in the problem, we may assume that a < b and a < c. By Fermat’s little theorem and Proposition 1.30, orda(2) | gcd(2c, a −1) = 2, by noting that c is a prime greater than a. Since a is an odd prime, orda(2) must then be 2, implying that a = 3, and so b | 2a + 1 = 9, which is a contradiction. What if a, b, c are not all primes? We try to generalize our previ-ous method. Let π(n) denote the smallest prime factor of a positive integer n. We make the following claim: If p is a prime such that p | (2y + 1) and p < π(y), then p = 3. The proof of our claim is similar to our previous discussion for the case that all a, b, c are primes. Then ordp(2) | gcd(2y, p −1) = 2. Again, we have ordp(2) = 2 and p = 3, establishing our claim. Now we solve our main problem. Since a, b, c are pairwise relatively prime, π(a), π(b), and π(c) are distinct. Without loss of generality, assume that π(a) < π(b), π(c). Applying the claim with (p, y) = (π(a), c), we find that π(a) = 3. Write a = 3a0. We claim that 3 fully divides a0. Otherwise, 9 would divide 2c + 1 and hence 22c −1. Because 2n ≡1 (mod 9) only if 6 | n, we must have 6 | 2c. Then 3 | c, contradicting the assumption that a and c are relatively prime. Thus, 3 does not divide a0, b, or c. Let q = π(a0bc), so that π(q) = q ≤min{π(b), π(c)}. Suppose, for the sake of contradiction, that q divides a. Because a and c are relatively prime, q cannot divide c, implying that π(q) = q is not equal to π(c). Because π(q) ≤π(c), we must have π(q) < π(c). Furthermore, q must divide 2c +1 because it divides a factor of 2c +1 (namely, a). Applying our claim with (p, y) = (q, c), we find that q = 3, a contradiction. Hence, our assumption was wrong, and q does not divide a. Similarly, q does not divide c. It follows that q must divide b. Now, let e be the order of 2 modulo q. Then e ≤q −1, so e has no prime factors less than q. Also, q divides b and hence 2a + 1 and 22a −1, implying that e | 2a. The only prime factors of 2a less than q are 2 and 3, so e | 6. Thus, q | (26 −1), and q = 7. However, 23 ≡1 (mod 7), so 2a + 1 ≡(23)a0 + 1 ≡1a0 + 1 ≡2 (mod 7). Hence, q does not divide 2a + 1, contradicting the assumption that q divides b, which divides 2a + 1. www.riazisara.ir 178 104 Number Theory Problems (2) The answers are (2, 5, 3) and cyclic permutations. We check that this is a solution: 2 | 126 = 53 + 1, 5 | 10 = 32 + 1, 3 | 33 = 25 + 1. Now let p, q, r be three primes satisfying the given divisibility rela-tions. Since q does not divide qr + 1, p ̸= q, and similarly q ̸= r, r ̸= p, so p, q, and r are all distinct. We apply introductory problem 1.49 (1) in this solution. We first consider the case that p, q, and r are all odd. Since p | qr +1, by introductory problem 49 (1), either 2r | p −1 or p | q2 −1. But 2r | p −1 is impossible because 2r | p −1 leads to p ≡1 (mod r), or 0 ≡pq + 1 ≡2 (mod r), which contradicts the fact that r > 2. Thus we must have p | q2 −1 = (q −1)(q + 1). Since p is an odd prime and q −1, q + 1 are both even, we must have p | q−1 2 or p | q+1 2 ; either way, p ≤q+1 2 < q. But then by a similar argument we may conclude that q < r, r < p, a contradiction. Thus, at least one of p, q, r must equal 2. By a cyclic permutation we may assume that q = 2. Now p | 2r + 1, so by introductory problem 49 (1) again, either 2r | p −1 or p | 22 −1 = 3. But 2r | p −1 is impossible as before, because r divides pq+1 = p2+1 = (p2−1)+2 and r > 2. Hence, we must have p = 3, and r | pq +1 = 32+1 = 10. Because r ̸= q, we must have r = 5. Hence (2, 5, 3) and its cyclic permutations are the only solutions. 48. [IMO 2002 Short List] Let n be a positive integer, and let p1, p2, . . . , pn be distinct primes greater than 3. Prove that 2p1 p2···pn + 1 has at least 4n divisors. First Proof: We induct on n. For n = 1, we consider the number a1 = 2p1+1. Since p1 is odd, 2p1+1 ≡ −1 + 1 ≡0 (mod 3). Hence a1 has distinct divisors 1, 3, and a1 itself. Since p1 > 3, it follows that a1 > 9, and so a1 3 is another divisor of a1, implying that a1 has at least 4 distinct divisors 1, 3, a1 3 , and a1, establishing the base case. Assume that the statement is true for n = k for some positive integer k; that is, ak = 2p1 p2···pk + 1 has at least 4k divisors. We consider the case n = k +1. Since p1, p2, . . . , pk+1 are odd, 3 divides both ak and 2pk+1 +1. Also, by introductory problem 38 (3), gcd(ak, 2pk+1 + 1) = gcd(2p1 p2···pk + 1, 2pk+1 + 1) = 3, www.riazisara.ir 5. Solutions to Advanced Problems 179 or gcd  ak, 2pk+1 + 1 3 = 1. (∗) We note that both ak and 2pk+1 +1 divide ak+1, because p1 · · · pk and pk+1 are odd. We conclude that ak+1 = ak · 2pk+1 + 1 3 · bk (∗∗) for some integer bk. By the induction hypothesis, and by (∗), we conclude that the product ak · 2pk+1 + 1 3 has has at least 4k · 2 divisors, namely, those 4k divisors d1, d2, . . . , d4k of ak and another 4k divisors di · 2pk+1 + 1 3 for every 1 ≤i ≤4k. We arrange these 2·4k divisors in increasing order as d1 < d2 < · · · < d2·4k. By (∗∗), these numbers are also divisors of ak+1. We now consider numbers d1bk, d2bk, . . . , d2·4kbk. They are also divisors of ak+1. We claim that d1, d2, . . . , d2·4k, d1bk, d2bk, . . . , d2·4kbk are distinct divisors of ak+1, from which our inductive step follows, since we find 4k+1 divisors of ak+1. To establish our claim, it suffices to show that d1bk ≥d2·4k. Since d1 ≥1 and d2·4k ≤ak · 2pk+1+1 3 , it suffices to show that bk ≥ak · 2pk+1 + 1 3 , or  ak · 2pk+1 + 1 3 2 ≤ak+1, www.riazisara.ir 180 104 Number Theory Problems by (∗∗). The last inequality is equivalent to (2p1 p2···pk + 1)2(2pk+1 + 1)2 ≤9(2p1 p2···pk+1 + 1), which follows from the inequality (2u + 1)2(2v + 1)2 ≤9(2uv + 1) for integers u and v with u and v greater than or equal to 5. Indeed, we have (2u + 1)2(2v + 1)2 ≤(22u + 2 · 2u + 1)(22v + 2 · 2v + 1) < (3 · 22u + 1)(3 · 22v + 1) < 9(22u + 1)(22v + 1) = 9(22u+2v + 22u + 22v + 1) < 9(22u+2v+2 + 1) < 9(2uv + 1), since uv −2u −2v −2 = (u −2)(v −2) −6 > 3. Second Proof: (Based on work by Hyun Soo Kim) Call an integer “tene-brous” if it is odd, square-free, not divisible by 3, and at least 5. For any integer m, let ψ(m) denote the number of distinct prime factors of m, and let d(m) denote the number of factors of m. We wish to prove that d(2a + 1) ≥4τ(a) for all tenebrous integers a. Induct on τ(a). For the base case τ(a) = 1, 2a + 1 is divisible by 3 exactly once and is greater than 3, so τ(2a + 1) ≥2 and d(2a + 1) ≥4. Now let a, b be relatively prime tenebrous integers such that the claim holds for both a and b. Clearly 2ab + 1 is divisible by both 2a + 1 and 2b + 1, so we can write 2ab + 1 = C · lcm[2a + 1, 2b + 1]. Because ab −2a −2b −4 = (a −2)(b −2) −8 > 0, 2ab + 1 > 22a+2b+4 > (2a + 1)2(2b + 1)2 > lcm[2a + 1, 2b + 1]2, so C ≥lcm[2a + 1, 2b + 1]. From the comment, we have gcd(2a + 1, 2b + 1) = 3, so as 3 divides each of 2a + 1 and 2b + 1 exactly once, d(lcm[2a + 1, 2b + 1]) = d(2a + 1)d(2b + 1) 2 ≥22τ(a)+2τ(b)−1. For every divisor m of lcm[2a + 1, 2b + 1], both m and Cm are divisors of 2ab + 1. Since C > lcm[2a + 1, 2b + 1], d(2ab + 1) ≥2 · d(lcm[2a + 1, 2b + 1]) ≥4τ(a)+τ(b), www.riazisara.ir 5. Solutions to Advanced Problems 181 completing the induction. Third Proof: (Based on work by Eric Price) Following the notation of the second solution, a stronger claim is that for any tenebrous integer a, τ(2a + 1) ≥2τ(a). We proceed by induction on τ(a). The base case is the same as in the first solution. Now let a, b be coprime tenebrous integers. We claim that τ(2ab + 1) ≥ τ(2a + 1) + τ(2b + 1). Note that 2ab + 1 2a + 1 = b  i=1 b i (−2a −1)i−1 ≡b − b 2 (2a + 1) (mod (2a + 1)2), so if a prime p divides 2a + 1 exactly k ≥1 times, then p divides 2ab + 1 either k times (if p doesn’t divide b) or k + 1 times (if p divides b). In any case p divides 2ab + 1 at most twice as many times as p divides 2a + 1. The same is true for prime factors of 2b + 1. As in the first solution, 2ab + 1 > (2a + 1)2(2b + 1)2, so in light of the above, 2ab + 1 must have a prime factor dividing neither 2a + 1 nor 2b + 1. Clearly 2ab + 1 is divisible by lcm[2a + 1, 2b + 1]. Because 2ab + 1 has a prime factor not dividing lcm[2a + 1, 2b + 1], we have τ(2ab + 1) ≥τ(lcm[2a + 1, 2b + 1]) + 1 = τ(2a + 1) + τ(2b + 1) −τ(gcd(2a + 1, 2b + 1)) + 1 = τ(2a + 1) + τ(2b + 1) −τ(3) + 1 = τ(2a + 1) + τ(2b + 1), completing the induction. 49. [Zhenfu Cao] Let p be a prime, and let {ak}∞ k=0 be a sequence of integers such that a0 = 0, a1 = 1, and ak+2 = 2ak+1 −pak for k = 0, 1, 2, . . . . Suppose that −1 appears in the sequence. Find all possible values of p. www.riazisara.ir 182 104 Number Theory Problems Solution: The answer is p = 5. It is not difficult to see that it is a solution. For p = 5, a3 = −1. Now we prove that it is the only solution. Assume that am = −1 for some nonnegative integer m. Clearly, p ̸= 2, because otherwise ak+2 = 2ak+1 −2ak is even and −1 will not appear in the sequence. Thus, we can assume that gcd(2, p) = 1. We consider the recursive relation ak+2 = 2ak+1 −pak modulo p, and then modulo p −1. First, we obtain ak+2 ≡2ak+1 (mod p), implying that ak+1 ≡2ka1 mod p. In particular, we have −1 ≡am ≡2m−1a1 ≡2m−1 (mod p). (∗) Second, we obtain ak+2 ≡2ak+1 −ak (mod p −1), or ak+2 −ak+1 ≡ak+1 −ak (mod p −1); that is, the sequence is arithmetic modulo p −1. Hence ak+1 ≡(k + 1)(a1 −a0) + a0 ≡k + 1 (mod p −1). In particular, we have −1 ≡am ≡m (mod p −1), or m + 1 ≡0 (mod p −1). Since gcd(2, p) = 1, by Fermat’s little theorem, we have 2p−1 ≡1 (mod p). Combining the last two congruence relations and (∗), we have 1 ≡2m+1 ≡4 · 2m−1 ≡−4 (mod p), implying that 5 ≡0 (mod p); that is, p = 5 is the only possible value. www.riazisara.ir 5. Solutions to Advanced Problems 183 50. [Qinsan Zhu] Let F be a set of subsets of the set {1, 2, . . . , n} such that (a) if A is an element of F, then A contains exactly three elements; (b) if A and B are two distinct elements in F, A and B share at most one common element; Let f (n) denote the maximum number of elements in F. Prove that (n −1)(n −2) 6 ≤f (n) ≤(n −1)n 6 . Proof: We will begin with the upper-bound inequality, since it is easier to prove. For such a set F, let us count the number of distinct doubletons {x, y} ⊂{1, 2, . . . , n} that are subsets of some element of F. Since any set A ∈F contains 3 such distinct doubletons, and no two elements of F can share a common doubleton, it means that 3 f (n) ≤ n 2 = n(n −1) 2 , so the right inequality is proved. Now we prove the lower-bound inequality. The set S = {1, 2, . . . , n} has n 3  = n(n−1)(n−2) 6 3-element subsets. Let T denote the set of all these 3-element subsets. We consider the subsets Ti = {{a, b, c} | {a, b, c} ∈T , a + b + c ≡i (mod n)} , for i = 0, 1, . . . , n −1. It is clear that these subsets are nonintersecting and their union is T ; that is, they form a partition of T . Since T has n 3  = n(n−1)(n−2) 6 elements, we may conclude by the pigeonhole principle that at least one of these n subsets has at least n(n−1)(n−2) 6n = (n−1)(n−2) 6 elements. Say T j is such a subset. We claim that T j satisfies both conditions (a) and (b). It is clear T j satisfies condition (a). For (b), assume (for contradiction) that there are two distinct elements A and B in T j that share at least two elements. Assume that A = {x, y, z1} and B = {x, y, z2}. Since A and B are elements of T j, we have x + y + z1 ≡x + y + z2 ≡j (mod n), implying that z1 ≡z2 (mod n). But recall that 1 ≤z1, z2 ≤n. So we must have z1 = z2, and thus A = B, a contradiction. It follows that we can set F = T j, and so f (n) is at least the number of elements in T j; that is, f (n) ≥(n −1)(n −2) 6 . www.riazisara.ir 184 104 Number Theory Problems Note: Under the same conditions, the last problem in the 6th Balkan Math-ematics Olympiad (1989) was asking for n(n −4) 6 ≤f (n) ≤(n −1)n 6 . Qinsan Zhu improved this result when he encountered this problem during his preparation for the IMO 2004. 51. [IMO 1998] Determine all positive integers k such that τ(n2) τ(n) = k, for some n. Note: Let n = pa1 1 pa2 2 · · · par r be a prime decomposition of n. Then τ(n) = (a1 + 1)(a2 + 1) · · · (ar + 1) and τ(n2) = (2a1 + 1)(2a2 + 1) · · · (2ar + 1). It follows that τ(n2) is always odd, so if k is an integer, then it must be odd. We now prove that the converse is also true; that is, if k is an odd positive integer, then k = τ(n2) τ(n) = (2a1 + 1)(2a2 + 1) · · · (2ar + 1) (a1 + 1)(a2 + 1) · · · (ar + 1) (∗) for some nonnegative integers a1, a2, . . . , ar. (Since there are infinitely many primes, we can always set n = pa1 1 pa2 2 · · · par r .) We call a positive integer acceptable if can be written in the above form. First Solution: A natural approach is strong induction on k. The result is trivial for k = 1 by setting n = 1, r = 1, and a1 = 0. For any odd integer k > 1, if it is of the form 4m + 1, then k = 4m + 1 2m + 1 · 2m + 1. Since 2m + 1 < k, it is acceptable by the induction hypothesis. Hence k is also acceptable. www.riazisara.ir 5. Solutions to Advanced Problems 185 However, if k is of the form 4m + 3, then we further assume that it is of the form 8m + 3. Then we have k = 24m + 9 12m + 5 · 12m + 5 6m + 3 · (2m + 1), and so k is acceptable by applying the induction hypothesis to 2m + 1 < k. Our proof remains open for the case k = 8m + 7. We have to split into two more cases again. To terminate this process, we reformulate the above idea as follows. Since every odd positive integer k can be written in the form 2sx −1 for some positive integer x, it suffices to show that if x is acceptable, then so is 2sx −1 for every s ≥1. Let ℓbe such that τ(ℓ2) τ(ℓ) = x. If s = 1, then k = 2sx −1 = 2x −1 = 2x −1 x · x shows that k = 2x −1 is acceptable. For s > 1, then 2sx −1 = 2s · 3x −3 2s−1 · 3x −1 · 2s−132x −3 2s−232x −1 · 2s−233x −3 2s−333x −1 · · · 223k−2x −3 2 · 3k−2x −1 · 2 · 3k−1x −3 3k−1x · x shows that k = 2sx −1 is acceptable. Our induction is thus complete. Second Proof: The proof is again by strong induction. Clearly the asser-tion is true for k = 1. Next assume that k > 1 is an odd positive integer and that the assertion is true for all positive odd integers less than k. As in the first solution, write k = 2sx −1, where x is an odd integer less than k. By the induction hypothesis, k0 is acceptable. It suffices to find a1, a2, . . . , ar such that k = x · 2a1 + 1 a1 + 1 · 2a2 + 1 a2 + 1 · · · 2at + 1 at + 1 . (∗∗) Note that if we set a2 = 2a1, a3 = 2a2, and so on, the equation (∗∗) can be simplified to 2sx −1 = k = x · 2ta1 + 1 a1 + 1 , www.riazisara.ir 186 104 Number Theory Problems or 1 = 2sx −2ta1 + 1 a1 + 1 · x = 2sa1 + 2s −2ta1 −1 a1 + 1 · x. It is convenient to set t = s, and further reduce the above equation to 1 = 2s −1 a1 + 1 · x, or a1 + 1 = (2s −1)x. Combining the above, we conclude that equation (∗∗) can be satisfied by setting t = s, a1 = (2s −1)x −1, a2 = 2a1, a3 = 2a2, . . . , at = 2at−1. Our induction is complete. 52. [China 2005] Let n be a positive integer greater than two. Prove that the Fermat number fn has a prime divisor greater than 2n+2(n + 1). Proof: For 1 ≤n ≤4, we know that fn are primes, and the conclusion is trivial. Now we assume that n ≥5. By introductory problem 49, we may assume that fn = pk1 1 pk2 2 · · · pkm m , (∗) where k is some positive integer, p1, . . . , pk are distinct primes, and k1, . . . , km are positive integers with pi = 2n+1xi + 1 for some positive integer xi, for every 1 ≤i ≤m. It suffices to show that xi ≥2(n + 1) (∗∗) for some 1 ≤i ≤m. First, we give an upper bound for the sum k1 + k2 + · · · + km. Note that for every i, pi ≥2n+1 + 1. It follows from (∗) and the binomial theorem that 22n + 1 = fn ≥(2n+1 + 1)k1+k2+···+km ≥2(n+1)(k1+k2+···+km) + 1, implying that k1 + k2 + · · · + km ≤ 2n n + 1. (†) Second, we give a lower bound for the sum x1k1 + x2k2 + · · · + xmkm. By the binomial theorem again, we have pki i ≡(2n+1xi + 1)ki ≡2n+1xiki + 1 (mod 22n+2). www.riazisara.ir 5. Solutions to Advanced Problems 187 Since 2n > 2n + 2 for n ≥5, we have fn ≡1 (mod 22n+2). Taking the equation (∗) modulo 22n+2 gives 1 ≡(2n+1x1k1 + 1)(2n+1x2k2 + 1) · · · (2n+1xmkm + 1) ≡1 + 2n+1x1k1 + 2n+1x2k2 + · · · + 2n+1x2k2 (mod 22n+2), or 0 ≡2n+1(x1k1 + x2k2 + · · · + xmkm) (mod 22n+2). It follows that 0 ≡x1k1 + x2k2 + · · · + xmkm (mod 2n+1). Since the xi’s and ki’s are nonnegative, we conclude that x1k1 + x2k2 + · · · + xmkm ≥2n+1. (‡) Let xi = max{x1, x2, . . . , xm}. Then inequality (‡) implies that xi(k1 + k2 + · · · + km) ≥2n+1. By inequality (†), we conclude that xi ≥ 2n+1 k1 + k2 + · · · + km ≥2n+1 2n n+1 = 2(n + 1), establishing the desired inequality (∗∗). www.riazisara.ir Glossary Arithmetic function A function defined on the positive integers that is complex valued. Arithmetic-Geometric Means Inequality If n is a positive integer and a1, a2, . . . , an are nonnegative real numbers, then 1 n n  i=1 ai ≥(a1a2 · · · an)1/n, with equality if and only if a1 = a2 = · · · = an. This inequality is a special case of the power mean inequality. Base-b representation Let b be an integer greater than 1. For any integer n ≥1 there is a unique system (k, a0, a1, . . . , ak) of integers such that 0 ≤ai ≤b −1, i = 0, 1, . . . , k, ak ̸= 0 and n = akbk + ak−1bk−1 + · · · + a1b + a0. Beatty’s theorem Let α and β be two positive irrational real numbers such that 1 α + 1 β = 1. The sets {⌊α⌋, ⌊2α⌋, ⌊3α⌋, . . . }, {⌊β⌋, ⌊2β⌋, ⌊3β⌋, . . . } form a partition of the set of positive integers. www.riazisara.ir 190 104 Number Theory Problems Bernoulli’s inequality For x > −1 and a > 1, (1 + x)a ≥1 + ax, with equality when x = 0. B´ ezout’s identity For positive integers m and n, there exist integers x and y such that mx + by = gcd(m, n). Binomial coefficient n k = n! k!(n −k)!, the coefficient of xk in the expansion of (x + 1)n. Binomial theorem The expansion (x + y)n = n 0 xn + n 1 xn−1y + n 2 xn−2y +· · ·+  n n −1 xyn−1 + n n yn. Canonical factorization Any integer n > 1 can be written uniquely in the form n = pα1 1 · · · pαk k , where p1, . . . , pk are distinct primes and α1, . . . , αk are positive integers. Carmichael numbers The composite integers n satisfying an ≡a (mod n) for every integer a. Complete set of residue classes modulo n A set S of integers such that for each 0 ≤i ≤n −1 there is an element s ∈S with i ≡s (mod n). www.riazisara.ir Glossary 191 Congruence relation Let a, b, and m be integers, with m ̸= 0. We say that a and b are congruent modulo m if m | (a −b). We denote this by a ≡b (mod m). The relation “≡” on the set Z of integers is called the congruence relation. Division algorithm For any positive integers a and b there exists a unique pair (q,r) of nonnegative integers such that b = aq + r and r < a. Euclidean algorithm Repeated application of the division algorithm: m = nq1 + r1, 1 ≤r1 < n, n = r1q2 + r2, 1 ≤r2 < r1, . . . rk−2 = rk−1qk + rk, 1 ≤rk < rk−1, rk−1 = rkqk+1 + rk+1, rk+1 = 0 This chain of equalities is finite because n > r1 > r2 > · · · > rk. Euler’s theorem Let a and m be relatively prime positive integers. Then aϕ(m) ≡1 (mod m). Euler’s totient function The function ϕ(m) is defined to be the number of integers between 1 and n that are relatively prime to m. Factorial base expansion Every positive integer k has a unique expansion k = 1! · f1 + 2! · f2 + 3! · f3 + · · · + m! · fm, where each fi is an integer, 0 ≤fi ≤i, and fm > 0. www.riazisara.ir 192 104 Number Theory Problems Fermat’s little theorem Let a be a positive integer and let p be a prime. Then a p ≡a (mod p). Fermat numbers The integers fn = 22n + 1, n ≥0. Fibonacci sequence The sequence defined by F0 = 1, F1 = 1, and Fn+1 = Fn + Fn−1 for every positive integer n. Floor function For a real number x there is a unique integer n such that n ≤x < n + 1. We say that n is the greatest integer less than or equal to x or the floor of x and we write n = ⌊x⌋. Fractional part The difference x −⌊x⌋is called the fractional part of x and is denoted by {x}. Fundamental theorem of arithmetic Any integer n greater than 1 has a unique representation (up to a permutation) as a product of primes. Hermite’s identity For any real number x and for any positive integer n, ⌊x⌋+ +1 n + +2 n + · · · + +n −1 n = ⌊nx⌋. Legendre’s formula For any prime p and any positive integer n, ep(n) =  i≥1 n pi . www.riazisara.ir Glossary 193 Legendre’s function Let p be a prime. For any positive integer n, let ep(n) be the exponent of p in the prime factorization of n!. Linear Diophantine equation An equation of the form a1x1 + · · · + anxn = b, where a1, a2, . . . , an, b are fixed integers. Mersenne numbers The integers Mn = 2n −1, n ≥1. M¨ obius function The arithmetic function µ defined by µ(n) = ⎧ ⎨ ⎩ 1 if n = 1, 0 if p2 | n for some prime p > 1, (−1)k if n = p1 · · · pk, where p1, . . . , pk are distinct primes. M¨ obius inversion formula Let f be an arithmetic function and let F be its summation function. Then f (n) =  d|n µ(d)F n d  . Multiplicative function An arithmetic function f ̸= 0 with the property that for any relatively prime positive integers m and n, f (mn) = f (m) f (n). Number of divisors For a positive integer n denote by τ(n) the number of its divisors. It is clear that τ(n) =  d|n 1. www.riazisara.ir 194 104 Number Theory Problems Order modulo m We say that a has order d modulo m, denoted by ordm(a) = d, if d is the smallest positive integer such that ad ≡1 (mod m). Perfect number An integer n ≥2 with the property that the sum of its divisors is equal to 2n. Pigeonhole Principle If n objects are distributed among k < n boxes, some box contains at least two objects. Prime number theorem The relation lim n→∞ π(n) n/ log n = 1, where π(n) denotes the number of primes less than or equal to n. Prime number theorem for arithmetic progressions For relatively prime integers a and r, let πa,d(n) denote the number of primes in the arithmetic progression a, a +d, a +2d, a +3d, . . . that are less than or equal to n. Then lim n→∞ πa,d(n) n/ log n = 1 ϕ(d). This result was conjectured by Legendre and Dirichlet and proved by Charles De la Vall´ ee Poussin. Sum of divisors For a positive integer n denote by σ(n) the sum of its positive divisors including 1 and n itself. It is clear that σ(n) =  d|n d. www.riazisara.ir Glossary 195 Summation function For an arithmetic function f the function F defined by F(n) =  d|n f (d). Wilson’s theorem For any prime p, (p −1)! ≡−1 (mod p). Zeckendorf representation Each nonnegative integer n can be written uniquely in the form n = ∞  k=0 αk Fk, where αk ∈{0, 1} and (αk, αk+1) ̸= (1, 1) for each k. www.riazisara.ir Further Reading 1. Andreescu, T.; Feng, Z., 101 Problems in Algebra from the Training of the USA IMO Team, Australian Mathematics Trust, 2001. 2. Andreescu, T.; Feng, Z., 102 Combinatorial Problems from the Training of the USA IMO Team, Birkh¨ auser, 2002. 3. Andreescu, T.; Feng, Z., 103 Trigonometry Problems from the Training of the USA IMO Team, Birkh¨ auser, 2004. 4. Andreescu, T.; Feng, Z., A Path to Combinatorics for Undergraduate Stu-dents: Counting Strategies, Birkh¨ auser, 2003. 5. Feng, Z.; Rousseau, C.; Wood, M., USA and International Mathematical Olympiads 2005, Mathematical Association of America, 2006. 6. Andreescu, T.; Feng, Z.; Loh, P., USA and International Mathematical Olympiads 2004, Mathematical Association of America, 2005. 7. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2003, Mathematical Association of America, 2004. 8. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2002, Mathematical Association of America, 2003. 9. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2001, Mathematical Association of America, 2002. 10. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2000, Mathematical Association of America, 2001. 11. Andreescu, T.; Feng, Z.; Lee, G.; Loh, P., Mathematical Olympiads: Prob-lems and Solutions from Around the World, 2001–2002, Mathematical As-sociation of America, 2004. www.riazisara.ir 198 104 Number Theory Problems 12. Andreescu, T.; Feng, Z.; Lee, G., Mathematical Olympiads: Problems and Solutions from Around the World, 2000–2001, Mathematical Association of America, 2003. 13. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutions from Around the World, 1999–2000, Mathematical Association of America, 2002. 14. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutions from Around the World, 1998–1999, Mathematical Association of America, 2000. 15. Andreescu, T.; Kedlaya, K., Mathematical Contests 1997–1998: Olympiad Problems from Around the World, with Solutions, American Mathematics Competitions, 1999. 16. Andreescu, T.; Kedlaya, K., Mathematical Contests 1996–1997: Olympiad Problems from Around the World, with Solutions, American Mathematics Competitions, 1998. 17. Andreescu, T.; Kedlaya, K.; Zeitz, P., Mathematical Contests 1995–1996: Olympiad Problems from Around the World, with Solutions, American Mathematics Competitions, 1997. 18. Andreescu, T.; Enescu, B., Mathematical Olympiad Treasures, Birkh¨ auser, 2003. 19. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkh¨ auser, 2000. 20. Andreescu, T., Andrica, D., An Introduction to Diophantine Equations, GIL Publishing House, 2002. 21. Andreescu, T.; Andrica, D., 360 Problems for Mathematical Contests, GIL Publishing House, 2003. 22. Andreescu, T.; Andrica, D., Complex Numbers from A to Z, Birkh¨ auser, 2004. 23. Beckenbach, E. F.; Bellman, R., An Introduction to Inequalities, New Math-ematical Library, Vol. 3, Mathematical Association of America, 1961. 24. Coxeter, H. S. M.; Greitzer, S. L., Geometry Revisited, New Mathematical Library, Vol. 19, Mathematical Association of America, 1967. 25. Coxeter, H. S. M., Non-Euclidean Geometry, The Mathematical Associa-tion of America, 1998. www.riazisara.ir Further Reading 199 26. Doob, M., The Canadian Mathematical Olympiad 1969–1993, University of Toronto Press, 1993. 27. Engel, A., Problem-Solving Strategies, Problem Books in Mathematics, Springer, 1998. 28. Fomin, D.; Kirichenko, A., Leningrad Mathematical Olympiads 1987– 1991, MathPro Press, 1994. 29. Fomin, D.; Genkin, S.; Itenberg, I., Mathematical Circles, American Math-ematical Society, 1996. 30. Graham, R.L.; Knuth, D.E.; Patashnik, O., Concrete Mathematics, Addison-Wesley, 1989. 31. Gillman, R., A Friendly Mathematics Competition, The Mathematical As-sociation of America, 2003. 32. Greitzer, S.L., International Mathematical Olympiads, 1959–1977, New Mathematical Library, Vol. 27, Mathematical Association of America, 1978. 33. Holton, D., Let’s Solve Some Math Problems, A Canadian Mathematics Competition Publication, 1993. 34. Kazarinoff, N.D., Geometric Inequalities, New Mathematical Library, Vol. 4, Random House, 1961. 35. Kedlaya, K; Poonen, B.; Vakil, R., The William Lowell Putnam Mathemat-ical Competition 1985–2000, The Mathematical Association of America, 2002. 36. Klamkin, M., International Mathematical Olympiads, 1978–1985, New Mathematical Library, Vol. 31, Mathematical Association of America, 1986. 37. Klamkin, M., USA Mathematical Olympiads, 1972–1986, New Mathemat-ical Library, Vol. 33, Mathematical Association of America, 1988. 38. K¨ ursch´ ak, J., Hungarian Problem Book, volumes I & II, New Mathematical Library, Vols. 11 & 12, Mathematical Association of America, 1967. 39. Kuczma, M., 144 Problems of the Austrian–Polish Mathematics Competi-tion 1978–1993, The Academic Distribution Center, 1994. 40. Kuczma, M., International Mathematical Olympiads 1986–1999, Mathe-matical Association of America, 2003. www.riazisara.ir 200 104 Number Theory Problems 41. Larson, L.C., Problem-Solving Through Problems, Springer-Verlag, 1983. 42. Lausch, H. The Asian Pacific Mathematics Olympiad 1989–1993, Aus-tralian Mathematics Trust, 1994. 43. Liu, A., Chinese Mathematics Competitions and Olympiads 1981–1993, Australian Mathematics Trust, 1998. 44. Liu, A., Hungarian Problem Book III, New Mathematical Library, Vol. 42, Mathematical Association of America, 2001. 45. Lozansky, E.; Rousseau, C. Winning Solutions, Springer, 1996. 46. Mitrinovic, D.S.; Pecaric, J.E.; Volonec, V. Recent Advances in Geometric Inequalities, Kluwer Academic Publisher, 1989. 47. Mordell, L.J., Diophantine Equations, Academic Press, London and New York, 1969. 48. Niven, I., Zuckerman, H.S., Montgomery, H.L., An Introduction to the The-ory of Numbers, Fifth Edition, John Wiley & Sons, Inc., New York, Chich-ester, Brisbane, Toronto, Singapore, 1991. 49. Savchev, S.; Andreescu, T. Mathematical Miniatures, Anneli Lax New Mathematical Library, Vol. 43, Mathematical Association of America, 2002. 50. Sharygin, I.F., Problems in Plane Geometry, Mir, Moscow, 1988. 51. Sharygin, I.F., Problems in Solid Geometry, Mir, Moscow, 1986. 52. Shklarsky, D.O; Chentzov, N.N; Yaglom, I.M., The USSR Olympiad Prob-lem Book, Freeman, 1962. 53. Slinko, A., USSR Mathematical Olympiads 1989–1992, Australian Mathe-matics Trust, 1997. 54. Szekely, G.J., Contests in Higher Mathematics, Springer-Verlag, 1996. 55. Tattersall, J.J., Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999. 56. Taylor, P.J., Tournament of Towns 1980–1984, Australian Mathematics Trust, 1993. 57. Taylor, P.J., Tournament of Towns 1984–1989, Australian Mathematics Trust, 1992. www.riazisara.ir Further Reading 201 58. Taylor, P.J., Tournament of Towns 1989–1993, Australian Mathematics Trust, 1994. 59. Taylor, P.J.; Storozhev, A., Tournament of Towns 1993–1997, Australian Mathematics Trust, 1998. 60. Yaglom, I.M., Geometric Transformations, New Mathematical Library, Vol. 8, Random House, 1962. 61. Yaglom, I.M., Geometric Transformations II, New Mathematical Library, Vol. 21, Random House, 1968. 62. Yaglom, I.M., Geometric Transformations III, New Mathematical Library, Vol. 24, Random House, 1973. www.riazisara.ir Index arithmetic functions, 36 base-b representation, 41 Beatty’s theorem, 60 Bernoulli’s inequality, 145 B´ ezout’s identity, 13 binomial theorem, 5 canonical factorization, 8 Carmichael numbers, 32 ceiling, 52 Chinese remainder theorem, 22 complete set of residue classes, 24 composite, 5 congruence relation, 19 coprime, 11 decimal representation, 41 Diophantine equations, 14 division algorithm, 4 Euclidean algorithm, 12 Euler’s theorem, 28 Euler’s totient function, 27 factorial base expansion, 45 Fermat numbers, 22, 70 Fermat’s little theorem, 28 Fibonacci numbers, 45 sequence, 45 fractional part, 52 fully divides, 9 fundamental theorem of arithmetic, 7 geometric progression, 9 greatest common divisor, 11 Hermite identity, 63 inverse of a modulo m, 26 least common multiple, 16 of a1, a2, . . . , an, 16 Legendre function, 65 linear combinations, 14 linear congruence equation, 22 linear congruence system, 22 linear Diophantine equation, 38 Mersenne numbers, 71 multiplicative arithmetic functions, 18 M¨ obius function, 36 M¨ obius inversion formula, 37 number of divisors, 17 order d modulo m, 32 perfect cube, 2 perfect numbers, 72 perfect power, 2 perfect square, 2 pigeonhole principle, 93 prime, 5 203 www.riazisara.ir 204 Index prime number, 5 quotient, 4 reduced complete set of residue classes, 28 relatively prime, 11 remainder, 4 square free, 2 sum of positive divisors, 18 summation function, 36 twin primes, 6 Wilson’s theorem, 26 Wolstenholme’s theorem, 115 Zeckendorf representation, 45 www.riazisara.ir ﺩﺭﺳﻨﺎﻣﻪ ﻫﺎ ﻭ ﺟﺰﻭﻩ ﻫﺎﻱ ﺩﺭﻭﺱ ﺭﻳﺎﺿﻴﺎﺕ ﺩﺍﻧﻠﻮﺩ ﻧﻤﻮﻧﻪ ﺳﻮﺍﻻﺕ ﺍﻣﺘﺤﺎﻧﺎﺕ ﺭﻳﺎﺿﻲ ﻧﻤﻮﻧﻪ ﺳﻮﺍﻻﺕ ﻭ ﭘﺎﺳﺨﻨﺎﻣﻪ ﻛﻨﻜﻮﺭ ﺩﺍﻧﻠﻮﺩ ﻧﺮﻡ ﺍﻓﺰﺍﺭﻫﺎﻱ ﺭﻳﺎﺿﻴﺎﺕ ﻭ... ﺳﺎﻳﺖ ﻭﻳﮋﻩ ﺭﻳﺎﺿﻴﺎﺕ www.riazisara.ir
187865
https://www.youtube.com/watch?v=24CLbZYKCvQ
cm2 to m2, m2 to cm2 physics manibalan 34600 subscribers 1845 likes Description 171377 views Posted: 16 Sep 2020 physicsmanibalan Transcript: the student today we are going to learn one centimeter square is equal to dash meter square okay if you like my channel please subscribe okay so 1 meter is 100 centimeter so 1 meter square will take it as a meter square it's a hundred square centimeter square under square is a 10 power 10 like a 4 0 will get 10 power 4 centimeter square and meter square if you want what is centimeter square you can bring 10 as a denominator so 1 divided by 10 power 4 meter square and is equal to centimeter square from denominator if you are taken 10 power 4 as a numerator you will get a certain power minus 4 so we'll get centimeter square is equal to 10 power minus 4 meter square so 1 centimeter square is a 10 power minus 4 meter square okay one meter square is equal to 10 power plus 4 centimeter square okay thank you for watching my channel please subscribe
187866
https://scholars.unf.edu/en/datasets/the-iupac-gold-book-a-compendium-of-chemical-terminology
UNF Scholar Research Profiles - Thomas G. Carpenter Library Home Help & FAQ English Español The IUPAC Gold Book: A Compendium of Chemical Terminology Stuart J. Chalk (Creator) Leah McEwen (Cornell University) (Creator) University of North Florida Chemistry and Biochemistry Dataset Description The IUPAC Compendium of Chemical Terminology (the "Gold Book") is an aggregation of the formally defined terms in the International Union of Pure and Applied Chemistry's Recommendations published in the journal of Pure and Applied Chemistry. The talk highlights the status of the Gold Book and the current major update, machine access to the definitions via an API, and the inclusion of these terms in domain ontologies. | | | --- | | Date made available | Nov 29 2023 | | Publisher | ZENODO | DOI 10.5281/zenodo.10219434 Access Dataset open Cite this DataSetCite Chalk, S. (Creator), McEwen, L. (Creator) (Nov 29 2023). The IUPAC Gold Book: A Compendium of Chemical Terminology. ZENODO. 10.5281/zenodo.10219434
187867
https://encyclopediaofmath.org/wiki/Parameter-dependent_integral
Log in www.springer.com The European Mathematical Society Navigation Main page Pages A-Z StatProb Collection Recent changes Current events Random page Help Project talk Request account Tools What links here Related changes Special pages Printable version Permanent link Page information Namespaces Page Discussion Variants Views View View source History Actions Parameter-dependent integral From Encyclopedia of Mathematics Jump to: navigation, search An integral of the type J(y)=∫f(x,y)dx, J(y)=∫f(x,y)dx, in which the point x=(x1,…,xn) ranges over the space Rn (if the point ranges only over a certain domain D in Rn, the function f(x,y) may be assumed to vanish for x∈Rn∖D), while the point y=(y1,…,ym), representing a set of parameters y1,…,ym, varies within some domain G of the space Rm. The main concern of the theory of such integrals is to determine conditions for the continuity and differentiability of J(y) with respect to the parameters y1,…,ym. If J(y) is interpreted as a Lebesgue integral, one obtains less restrictive conditions for its continuity and differentiability. The following propositions are valid. 1) If f(x,y) is continuous in y in the domain G⊂Rm for almost-all x∈Rn and if there exists an integrable function g on Rn such that |f(x,y)|≤g(x) for every y∈G and almost-all x∈Rn, then J(y) is continuous in G. 2) Let f(x,t) be a function defined for x∈Rn, t∈(a,b). Assume that the derivative ∂f(x,t)/∂t exists for almost-all x∈Rn and every t∈(a,b) and that is a continuous function of t on (a,b) for almost-all x∈Rn. Assume, moreover, that there exists an integrable function g on Rn such that |∂f(x,t)/∂t|≤g(x) for every t∈(a,b) and almost-all x∈Rn. Finally, assume that for some t0∈(a,b) the integral ∫f(x,t0)dx exists. Then the function J(t)=∫f(x,t)dx is differentiable with respect to t on (a,b), and its derivative J′(t) may be evaluated by differentiating under the integral sign: J′(t)=∫∂f∂t(x,t)dx. These two propositions imply a series of simpler propositions about the continuity and differentiability of integrals with parameters, relating to the interpretation of the integral as a Riemann integral and to more specific cases (see –). Parameter-dependent improper integrals. For the simplest improper integral of the first kind, J(t)=∞∫af(x,t)dx, one introduces the notion of uniform convergence with respect to the parameter t in a closed interval c≤t≤d. This integral is said to be uniformly convergent in t on [c,d] if, for each ϵ>0, there exists an A(ϵ)>0 such that |∞∫Rf(x,t)dx|<ϵ for all R≥A(ϵ). The following propositions are valid. a) If f(x,t) is continuous in a half-strip [a≤x<∞,c<t≤d] and if the integral () is uniformly convergent in t on [c,d], then J(t) is continuous in c<t≤d. b) If f(x,t) and the derivative ∂f(x,t)/∂t are continuous in a half-strip [a≤x<∞,c≤t≤d], if the integral () is convergent for some t∈[c,d] and if the integral ∞∫a∂f∂t(x,t)dx is uniformly convergent in t on [c,d], then the function J(t) is differentiable on [c,d] and its derivative may be evaluated by the formula J′(t)=∞∫a∂f∂t(x,t)dx. Analogous propositions hold for improper integrals of the second kind. References | | | --- | | | V.S. Vladimirov, "Equations of mathematical physics" , MIR (1984) (Translated from Russian) | | | V.A. Il'in, E.G. Poznyak, "Fundamentals of mathematical analysis" , 2 , MIR (1982) (Translated from Russian) | | | L.D. Kudryavtsev, "Mathematical analysis" , 2 , Moscow (1970) (In Russian) | | | S.M. Nikol'skii, "A course of mathematical analysis" , 2 , MIR (1977) (Translated from Russian) | | | A.N. [A.N. Tikhonov] Tichonoff, A.A. Samarskii, "Differentialgleichungen der mathematischen Physik" , Deutsch. Verlag Wissenschaft. (1959) (Translated from Russian) | Comments The propositions stated are simple consequences of Lebesgue's dominated convergence principle (see Lebesgue theorem 2)). How to Cite This Entry: Parameter-dependent integral. Encyclopedia of Mathematics. URL: This article was adapted from an original article by V.A. Il'in (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article Retrieved from " Category: TeX done
187868
https://en.wikipedia.org/wiki/Bond_energy
Jump to content Search Contents (Top) 1 Bond energy versus bond-dissociation energy 2 Predicting the bond strength by radius 3 Factors affecting ionic bond energy 4 See also 5 References Bond energy العربية Bosanski Català Čeština Eesti Ελληνικά Español فارسی Français Gaeilge 한국어 हिन्दी Bahasa Indonesia עברית کٲشُر پنجابی Polski Português Simple English Srpskohrvatski / српскохрватски தமிழ் Türkçe اردو Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Strength of a chemical bond In chemistry, bond energy (BE) is one measure of the strength of a chemical bond. It is sometimes called the mean bond, bond enthalpy, average bond enthalpy, or bond strength. IUPAC defines bond energy as the average value of the gas-phase bond-dissociation energy (usually at a temperature of 298.15 K) for all bonds of the same type within the same chemical species. The bond dissociation energy (enthalpy) is also referred to as bond disruption energy, bond energy, bond strength, or binding energy (abbreviation: BDE, BE, or D). It is defined as the standard enthalpy change of the following fission: R—X → R + X. The BDE, denoted by Dº(R—X), is usually derived by the thermochemical equation, This equation tells us that the BDE for a given bond is equal to the energy of the individual components that make up the bond when they are free and unbonded minus the energy of the components when they are bonded together. These energies are given by the enthalpy of formation ΔHfº of the components in each state. The enthalpy of formation of a large number of atoms, free radicals, ions, clusters and compounds is available from the websites of NIST, NASA, CODATA, and IUPAC. Most authors use the BDE values at 298.15 K. For example, the carbon–hydrogen bond energy in methane BE(C–H) is the enthalpy change (∆H) of breaking one molecule of methane into a carbon atom and four hydrogen radicals, divided by four. The exact value for a certain pair of bonded elements varies somewhat depending on the specific molecule, so tabulated bond energies are generally averages from a number of selected typical chemical species containing that type of bond. Bond energy versus bond-dissociation energy [edit] Bond energy (BE) is the average of all bond-dissociation energies of a single type of bond in a given molecule. The bond-dissociation energies of several different bonds of the same type can vary even within a single molecule. For example, a water molecule is composed of two O–H bonds bonded as H–O–H. The bond energy for H2O is the average energy required to break each of the two O–H bonds in sequence: Although the two bonds are the equivalent in the original symmetric molecule, the bond-dissociation energy of an oxygen–hydrogen bond varies slightly depending on whether or not there is another hydrogen atom bonded to the oxygen atom. Thus, the bond energy of a molecule of water is 461.5 kJ/mol (110.3 kcal/mol). When the bond is broken, the bonding electron pair will split equally to the products. This process is called homolytic bond cleavage (homolytic cleavage; homolysis) and results in the formation of radicals. Predicting the bond strength by radius [edit] The strength of a bond can be estimated by comparing the atomic radii of the atoms that form the bond to the length of bond itself. For example, the atomic radius of boron is estimated at 85 pm, while the length of the B–B bond in B2Cl4 is 175 pm. Dividing the length of this bond by the sum of each boron atom's radius gives a ratio of . This ratio is slightly larger than 1, indicating that the bond itself is slightly longer than the expected minimum overlap between the two boron atoms' valence electron clouds. Thus, we can conclude that this bond is a rather weak single bond. In another example, the atomic radius of rhenium is 135 pm, with a Re–Re bond length of 224 pm in the compound [Re2Cl8]−2. Taking the same steps as above gives a ratio of . This ratio is notably lower than 1, indicating that there is a large amount of overlap between the valence electron clouds of the two rhenium atoms. From this data, we can conclude that this is a very strong bond. Experimentally, the Re-Re bond in [Re2Cl8]−2 was found to be a quadruple bond. This method of determination is most useful for covalently bonded compounds. Factors affecting ionic bond energy [edit] In ionic compounds, the electronegativity of the two atoms bonding together has a major effect on their bond energy. The extent of this effect is described by the compound's lattice energy, where a more negative lattice energy corresponds to a stronger force of attraction between the ions. Generally, greater differences in electronegativity correspond to stronger ionic bonds. For example, the compound sodium chloride (NaCl) has a lattice energy of −786 kJ/mol with an electronegativity difference of 2.23 between sodium and chlorine. Meanwhile, the compound sodium iodide (NaI) has a lower lattice energy of −704 kJ/mol with a similarly lower electronegativity difference of 1.73 between sodium and iodine. See also [edit] Bond-dissociation energy Binding energy Ionization energy Isodesmic reaction Lattice energy References [edit] ^ a b Treptow, Richard S. (1995). "Bond Energies and Enthalpies: An Often Neglected Difference". Journal of Chemical Education. 72 (6): 497. Bibcode:1995JChEd..72..497T. doi:10.1021/ed072p497. ^ Christian, Jerry D. (1973-03-01). "Strength of Chemical Bonds". Journal of Chemical Education. 50 (3): 176. Bibcode:1973JChEd..50..176C. doi:10.1021/ed050p176. hdl:2060/19980004003. ISSN 0021-9584. ^ March, Jerry (1985). Advanced Organic Chemistry: Reactions, Mechanisms, and Structure (3rd ed.). New York: Wiley. ISBN 9780471854722. OCLC 642506595. ^ Haynes, William (2016–2017). CRC Handbook of Chemistry and Physics, 97th Edition (CRC Handbook of Chemistry & Physics) 97th Edition (97th ed.). CRC Press; 97 edition. ISBN 978-1498754286. ^ Luo, Yu-Ran and Jin-Pei Cheng "Bond Dissociation Energies". In Lide, David R. (ed) 2017, CRC Handbook of Chemistry and Physics, 97th edition (2016–2017). Boca Raton: Taylor & Francis Group. 9-73. ^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "Bond energy (mean bond energy)". doi:10.1351/goldbook.B00701 ^ Madhusha (2017), Difference Between Bond Energy and Bond Dissociation Energy, Pediaa, Difference Between Bond Energy and Bond Dissociation Energy ^ Lehninger, Albert L.; Nelson, David L.; Cox, Michael M. (2005). Lehninger principles of biochemistry (4th ed.). New York: W.H. Freeman. ISBN 978-0-7167-4339-2. ^ "Illustrated Glossary of Organic Chemistry - Homolytic cleavage (homolysis)". www.chem.ucla.edu. Retrieved 2019-11-27. ^ a b Slater, J. C. (1964-11-15). "Atomic Radii in Crystals". The Journal of Chemical Physics. 41 (10): 3199–3204. Bibcode:1964JChPh..41.3199S. doi:10.1063/1.1725697. ISSN 0021-9606. ^ Atoji, Masao; Wheatley, Peter J.; Lipscomb, William N. (1957-07-01). "Crystal and Molecular Structure of Diboron Tetrachloride, B2Cl4". The Journal of Chemical Physics. 27 (1): 196–199. doi:10.1063/1.1743668. ISSN 0021-9606. ^ Cotton, F. A.; Curtis, N. F.; Harris, C. B.; Johnson, B. F. G.; Lippard, S. J.; Mague, J. T.; Robinson, W. R.; Wood, J. S. (1964-09-18). "Mononuclear and Polynuclear Chemistry of Rhenium (III): Its Pronounced Homophilicity". Science. 145 (3638): 1305–1307. Bibcode:1964Sci...145.1305C. doi:10.1126/science.145.3638.1305. ISSN 0036-8075. PMID 17802015. ^ Alcock, N. W. (1990). Bonding and Structure: Structural Principles in Inorganic and Organic Chemistry. New York: Ellis Horwood. pp. 40–42. ISBN 9780134652535. ^ Handbook of Chemistry & Physics (65th ed.). CRC Press. 1984-06-27. ISBN 0-8493-0465-2. ^ Atkins; et al. (2009). Shriver and Atkins' Inorganic Chemistry (Fifth ed.). New York: W. H. Freeman and Company. ISBN 978-1-4292-1820-7. ^ Huheey, James E.; Keiter, Ellen A.; Keiter, Richard L. (2009). Inorganic chemistry: principles of structure and reactivity (4th ed.). Cambridge: Harper. ISBN 978-0-06-042995-9. | Authority control databases | | International | | | Other | Yale LUX | Retrieved from " Categories: Chemical bond properties Binding energy Hidden categories: Articles with short description Short description matches Wikidata Bond energy Add topic
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Page Toolbox Search Casework In combinatorics, casework is a counting method that involves splitting a problem into several parts, counting these parts individually, then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. Contents Examples Here are some examples that demonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an intermediate step across all of mathematics, not just in competitions. While there are problems where casework produces the most elegant solution, in those where a shorter answer exists, casework may be considered brute force. This is especially true if that alternative solution uses complementary counting. Example 1 How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters. Solution: We divide the problem into cases, based on how long the word is. Case 1: The word is one letter long. Clearly, there are of these words. Case 2: The word is two letters long. Constructing the set of these words, there are options for the first letter and options for the second letter, so there are of these words. Case 3: The word is three letters long. By similar logic as above, we have options for the first letter, options for the second, and options for the third. Then there are of these letters. Adding all our cases up, there are words that are less than four letters long and contain only the letters A, B, C, D, and E. Example 2 How many positive integers satisfy the equation ? Solution: We use casework, based on the value of x. Case 1: . Then this problem becomes , so there are possible values for , and thus solutions when is one. Case 2: . Then this problem becomes , so there are possible values for , and thus solutions when is two. If , the problem becomes . But the question mandates that is positive, so there are no solutions when . Thus, there are positive integer solutions to the equation. Example 3 2004 AIME II Problem 4: How many positive integers less than 10,000 have at most two different digits? Solution: Let and be the two digits of the number. Use casework, based on how many digits the number has. Case 1: The number is one digit. All numbers in this category satisfy the given condition, so there are of these. Case 2: The number is two-digit. Again, all numbers in this category have two different digits, so there are of these. Case 3: The number is three-digit. The possible cases in this category are and Using a constructive approach, there are digits for what the first number can be, zero excluded to keep the number three-digit. The second digit can be digits, with zero included and the first digit's number removed. Then there are of these numbers. Case 4: The number is four-digit. The possible cases in this category are and Using the same logic as before, the first digit has options, as does the second. Then there are of these numbers. Thus, there are integers less than that have at most two different digits. More examples Resources See also Something appears to not have loaded correctly. Click to refresh.
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https://files.jofph.com/files/article/20250912-454/pdf/OFPH20250416001.pdf
This is an open access article under the CC BY 4.0 license ( J Oral Facial Pain Headache 2025 vol.39(3), 191-199 ©2025 The Author(s). Published by MRE Press. www.jofph.com Submitted: 16 April, 2025 Accepted: 04 June, 2025 Published: 12 September, 2025 DOI:10.22514/jofph.2025.061 O R I G I N A L R E S E A R C H Cervicogenic headache in forward head posture: frequency and associated factors in a cross-sectional study Ahmet Usen1, , Merve Demiroz Gunduz1 1Department of Physical Medicine and Rehabilitation, Faculty of Medicine, Istanbul Medipol University, 34214 Istanbul, Türkiye Correspondence ahmet.usen@medipol.edu.tr (Ahmet Usen) Abstract Background: This study aimed to determine the frequency and associated factors of cervicogenic headache (CGH) in individuals with forward head posture (FHP). Additionally, craniovertebral angle (CVA)-related factors were examined in patients diagnosed with CGH. Methods: This cross-sectional study included 117 patients aged 18–45 years who presented with neck pain and were identified with FHP. CGH diagnosis was based on the International Classification of Headache Disorders (ICHD-3) criteria. CVA was measured using posture analysis software, and assessments included the Neck Disability Index (NDI), Henry Ford Headache Disability Inventory (HDI-T), Headache Impact Scale (HIT-6), Pittsburgh Sleep Quality Index (PSQI), Beck Depression Inventory (BDI), and Visual Analog Scale (VAS) for pain. Statistical analyses included independent t-tests, chi-square tests and logistic regression models. Results: The frequency of CGH in patients with FHP was 53.8%. Compared to the non-CGH group, those with CGH had significantly lower CVA (p = 0.030) and higher PSQI (p = 0.001) and BDI scores (p < 0.001). Logistic regression analysis identified low CVA (Odds Ratio (OR): 0.878, p = 0.014) and poor sleep quality (OR: 1.140, p = 0.025) as independent predictors of CGH. Additionally, Body Mass Index and VAS scores were negatively correlated with CVA (p < 0.05). Conclusions: FHP may be associated with CGH, possibly through increased biomechanical load and neuromechanical sensitivity. Interventions such as corrective exercises, weight management, and improving sleep quality may be considered as supportive strategies in CGH management; however, causal relationships cannot be inferred from this study. Further studies are needed to explore the long-term effects of postural interventions on CGH. Keywords Cervicogenic headache; Head; Posture; Cervical vertebrae 1. Introduction Cervicogenic headache (CGH) is a type of secondary headache that originates in the cervical spine and is felt in the head . It is usually defined as a unilateral headache triggered by neck movements and neck-related symptoms. Its basic mechanism is based on the neural convergence between the trigeminal nerve and the upper three cervical nerves (C1, C2 and C3) . This process is also defined as trigeminocervical convergence at the trigeminal nucleus caudalis level . This leads to the perception of pain signals from the neck as headaches. The source of pain is often thought to be structures such as upper cervical joints, muscles, ligaments, intervertebral discs, and veins. However, current perspectives suggest that muscular sources may be more characteristic of tension-type headache, while upper cervical facet joints are increasingly emphasized in CGH . Dysfunction in these anatomical structures can generate pains that are reflected in the head through the com-plex nerve network of the cervical spine . Understanding the mechanism of CGH is of great importance for both diagnosis and treatment. While the one-year prevalence in the adult population ranges from 0.2% to 2.2% [5, 6], this rate increases to 10.4% in the young population and has been associated with sustained neck loading during prolonged use of digital devices such as smartphones and computers . Forward head posture (FHP) may predispose individuals to the development of CGH by increasing the biomechanical load on the cervical spine . The increasing use of information and communication technologies such as mobile phones and computers causes individuals to keep their necks tilted forward for long periods of time and contributes to the formation of FHP . This posture, which is characterized by a decrease in the craniovertebral angle (CVA) and the forward placement of the head in the sagittal plane relative to the shoulders, leads 192 to a deterioration of the balance between the cervical muscles and an increase in mechanosensitivity . In the literature, it has been shown that FHP is associated with restriction of cervical flexion movement, increased neck pain severity, and postural balance disorders . Additionally, FHP in-volves upper cervical extension, leading to shortening and increased activation of the rectus capitis posterior major and minor muscles. These changes may contribute to nociceptive input from the upper cervical joints, a mechanism implicated in cervicogenic headaches . Furthermore, this posture is reported to trigger chronicity of CGH through shortening of the suboccipital muscles, tension in the trapezius muscle, and structural changes . In addition, it has been stated that the negative effects of FHP on neck pain and loss of functionality increase pain severity, especially in adults, and adversely affect musculoskeletal health . Such findings suggest the need for extensive research to understand the relationship between FHP and CGH better and to develop effective approaches to the management of neck pain. Although previous studies have investigated the relationship between FHP and CGH using postural or clinical measures, their findings were limited by methodological inconsistencies and a lack of integration between postural, clinical, and psychosocial variables [15, 16]. The aim of this study was to determine the frequency and associated factors of CGH in patients who presented to our outpatient clinic with neck pain and were identified as hav-ing FHP. In particular, the study explored the prevalence of CGH in this population, the demographic and clinical features associated with its occurrence, and whether specific CVA-related characteristics could differentiate patients with CGH from those without. The findings are expected to contribute new insights into the mechanisms underlying CGH in the context of postural disorders and improve the clinical man-agement of patients presenting with FHP-related neck pain. We hypothesized that CVA-related postural parameters, sleep quality, and BMI would be significantly associated with the presence of CGH in this population. 2. Materials and methods 2.1 Study design and participants This was a cross-sectional study conducted in the Department of Physical Medicine and Rehabilitation, Medipol Mega Hos-pitals Complex. Data collection was carried out between July 2024 and January 2025. The sample size was determined via power analysis with 80% power, 5% significance level, and an effect size of 0.30. As a result of the analysis, it was calculated that at least 105 participants should be included in order to provide reliable and valid results. Considering a dropout rate of 10%, a total of 117 participants were required. All participants presented directly to the physical medicine and rehabilitation outpatient clinic without prior referral by a neurologist or general practitioner. Patients who applied to the outpatient clinic with neck pain and had a CVA measurement of ≤50◦were included in this study. Patients with spinal deformi-ties such as cervical vertebral fractures, scoliosis, or kyphosis, patients with vertigo, osteoporosis, tumors or inflammatory diseases affecting the cervical region, and patients with known symptoms of vertebrobasilar insufficiency or cervical spine instability were excluded from the study. 2.2 Data collection The following assessment tools were applied to each partici-pant. 2.2.1 Neck disability index (NDI) The NDI is a scale that evaluates the impact of neck pain on daily life. The Turkish validity and reliability study was conducted by Aslan et al. (Intraclass Correlation Coef-ficient (ICC) = 0.979). It consists of a total of 10 questions and is scored between 0–100, with higher scores meaning more disability. 2.2.2 Henry Ford headache disability inventory (HDI-T) HDI-T, a 21-item scale with a two-dimensional structure, emo-tional and functional, is used to measure headache disability. It is scored on a scale of 0–100, with high scores indicating more disability. The Turkish validity and reliability study was conducted by Kılınç et al. (ICC = 0.901). Unlike its original form, it contains 21 items, not 25. 2.2.3 Headache impact scale (HIT-6) This is a 6-item scale that evaluates the impact of headache on social and physical functioning. The Turkish validity and reli-ability study was conducted by Dikmen et al. (Cronbach’s α = 0.753–0.864; r = 0.437). Each question is graded between 6 and 13 points, and the total score ranges from 36 to 78. As the score increases, the severity of the headache’s impact on an individual’s life also increases. 2.2.4 Pittsburgh sleep quality index (PSQI) This is a self-report scale that evaluates sleep quality and dis-orders over the past month. The Turkish validity and reliability study was conducted by Ağargün et al. (Cronbach’s α = 0.80; r = 0.98). It consists of a total of 24 questions; 19 are self-reported, and five are answered by a spouse or roommate. The PSQI total score ranges from 0 to 21, and sleep quality is considered to worsen as the score increases. 2.2.5 Beck depression inventory (BDI) This scale evaluates the emotional, cognitive, somatic, and motivational components of depression. The Turkish validity and reliability study was conducted by Hisli (Cronbach’s α = 0.80). The scale consists of 21 items and is scored on a scale of 0–63. High scores indicate an increase in the severity of depression. Demographic (age, sex, height, weight, smoking, alcohol consumption) and clinical information including headache du-ration (per year), frequency (how many days in the past week and how many hours in total), and severity (evaluated with the Visual Analog Scale (VAS)) were also collected from participants. 2.3 Evaluation and measurements 193 2.3.1 Craniovertebral angle (CVA) CVA was measured by evaluating photographs taken with a reflex camera (Nikon Model D5300 SLR, Tokyo, Japan) using posture analysis software . The camera was fixed at a distance of 3 meters from the standing participant, and the participant was asked to take a comfortable posture. The cran-iovertebral angle (α) is defined as the angle formed between a horizontal line drawn parallel to the ground at the level of the spinous process of C7 and an oblique line extending from the tragus of the ear to the spinous process of C7 (Fig. 1). Measurements were reported to provide high reliability with an ICC value of 0.98 and good validity compared to radiography (r = 0.89) . 2.3.2 Cervical joint range of motion (ROM) Flexion, extension, right lateral flexion, left lateral flexion, right rotation and left rotation movements were measured using a goniometer. 2.3.3 Diagnosis of cervicogenic headache Participants’ headaches were evaluated according to the crite-ria of the International Classification of Headache Disorders (ICHD-3) . CGH was diagnosed based on the following criteria: (1) temporal onset of headache in relation to a cervical disorder or trauma, (2) headache triggered by cervical move-ment or external pressure, (3) ipsilateral neck pain accompa-nying the headache, and (4) cervical pathology confirmed by imaging. All assessments were conducted by a physiatrist with clinical experience in headache management. It is important to note that patients with overlapping headache types (e.g., coex-isting migraine or tension-type headache) were not excluded, as CGH may coexist with other headache disorders in real-world clinical settings. 2.4 Statistical analysis The data obtained in this study were analyzed using IBM SPSS Statistics software (Version 26.0, IBM Corp., Armonk, NY, USA). The normality distribution of the data was evaluated by the Kolmogorov-Smirnov test. Continuous variables are expressed as mean ± standard deviation because they meet the assumption of normality. Independent sample t-test was used to compare the groups. Categorical variables were summarized as frequency (n) and percentage (%). The Chi-square test or, where appropriate, Fisher’s Exact Test was used to compare categorical variables. The relationships between clinical and functional parame-ters (e.g., VAS, headache frequency and duration, CVA, ROM, F IG URE 1. The measurement of craniovertebral angle (α) defined as the angle formed between a horizontal line drawn parallel to the ground at the level of the spinous process of the seventh cervical vertebra (C7) and an oblique line extending from the tragus of the ear to the spinous process of C7. 194 PSQI, BDI, NDI, HDI-T, HIT-6) were analyzed using the Pearson correlation coefficient. Logistic regression analysis was performed to evaluate the effects of independent variables (CVA, PSQI, BDI, BMI and age) on the dependent variable (presence of CGH). The goodness-of-fit of each logistic regres-sion model was evaluated using the Hosmer-Lemeshow test. For the univariable models constructed with age (73.5% ac-curacy, 77.8% sensitivity, 68.5% specificity; χ2(7) = 36.124, p < 0.001), CVA (65.8% accuracy, 81.0% sensitivity, 48.1% specificity; χ2(7) = 40.893, p < 0.001), and BDI (75.2% accu-racy, 81.0% sensitivity, 68.5% specificity; χ2(8) = 32.983, p < 0.001), the Hosmer-Lemeshow test indicated poor model fit due to statistically significant results. In contrast, the model based on PSQI (64.1% accuracy, 69.8% sensitivity, 57.4% specificity; χ2(8) = 12.842, p = 0.117) and the multivariable model including age, CVA and PSQI (65.0% accuracy, 81.0% sensitivity, 46.3% specificity; χ2(8) = 7.802, p = 0.453) did not show significant lack of fit, indicating an acceptable alignment between observed and predicted values. The significance level was accepted as p < 0.05 with Odds Ratio (OR) and 95% Confidence Intervals (CI). 3. Results Table 1 compares the demographic and sociocultural data of the CGH and non-CGH groups. In the analysis, there was no statistically significant difference between the two groups in terms of BMI, sex, smoking status and alcohol consumption (all p > 0.05). However, the mean age was higher in the CGH group than in the non-CGH group (33.25 ± 5.56 years vs. 28.54 ± 8.35 years, p < 0.001). Table 2 compares the clinical and functional data of CGH and non-CGH groups. CVA value was significantly lower in the CGH group compared to the non-CGH group (38.95 ± 4.65 vs. 41.11 ± 5.97, p = 0.030). PSQI (6.96 ± 4.18 vs. 9.48 ± 3.92, p = 0.001) and BDI scores (9.74 ± 7.37 vs. 14.79 ± 6.48, p < 0.001) were significantly higher in the CGH group than in the non-CGH group. Notably, no significant difference was observed between the two groups in terms of cervical joint range of motion (flexion, extension, lateral and rotation movements) and NDI scores (p > 0.05). According to the results of the univariable logistic regression analysis, age (OR = 1.10, 95% CI: 1.04–1.16, p = 0.001), CVA (OR = 0.926, 95% CI: 0.863–0.994, p = 0.032), PSQI (OR = 1.167, 95% CI: 1.059–1.287, p = 0.002) and BDI (OR = 1.115, 95% CI: 1.049–1.184, p < 0.001) showed significant relationships with CGH (Table 3). In multivariable logistic regression analysis, CVA (OR = 0.892, 95% CI: 0.812–0.979, p = 0.017) was determined as an independent predictor for CGH (Table 3). When PSQI was added to the multivariable model, it was determined that PSQI (OR = 1.140, 95% CI: 1.017–1.279, p = 0.025) was an independent predictor for CGH in addition to CVA (OR = 0.878, 95% CI: 0.792–0.974, p = 0.014) (Table 3). According to the results of the analysis in Table 4, there was a significant negative correlation between CVA and BMI (r = −0.449; p < 0.001). While there was no significant relationship between CVA and age (r = −0.121; p = 0.344), a statistically significant negative correlation was found be-tween CVA and VAS (r = −0.275; p = 0.029). There was no significant relationship between CVA and headache duration (years) and frequency (number of days seen in the last week and total duration) (p > 0.05). There was also no statistically significant difference between cervical range of motion and CVA (p > 0.05). In addition, a significant positive correlation was found between CVA and PSQI (r = 0.323; p = 0.010). Although the correlations between CVA and variables such as VAS (r = −0.275) and PSQI (r = 0.323) were statistically significant, their explanatory power was modest, accounting for approximately 7% and 10% of the variance, respectively. However, there was no statistically significant relationship between CVA and HDI-T, NDI, HIT and BDI (p > 0.05). According to the results of the multiple linear regression analysis in Table 5, it is seen that the model is significant (F = 8.130; p < 0.001). In this model, increased BMI (B = −0.645; p = 0.001) and increased VAS (B = −0.571; p = 0.026) were found to be independent predictors of low CVA. While BMI and VAS were significantly negatively correlated with CVA, PSQI did not have a significant effect on the model (p = 0.504). TAB L E 1. Distribution of demographic data by CGH and non-CGH groups. Variables Groups non-CGH Group CGH Group p Sex, n (%) Female 24 (40.7) 35 (59.3) 0.231 Male 30 (51.7) 28 (48.3) Smoking, n (%) Yes 15 (51.7) 14 (48.3) 0.488 No 39 (44.3) 49 (55.7) Alcohol, n (%) Yes 8 (53.3) 7 (46.7) 0.550 No 46 (45.1) 56 (54.9) Age (yr) (Mean ± SD) 28.54 ± 8.35 33.25 ± 5.56 <0.001 BMI (Mean ± SD) 24.26 ± 4.28 23.68 ± 3.09 0.899 CGH: Cervicogenic Headache; BMI: Body Mass Index (kg/m2); SD: Standard Deviation; p: p-value. : p < 0.05 refers to the level of statistical significance. 195 TAB L E 2. Comparison of clinical and functional data by CGH and non-CGH groups. Variables non-CGH Group CGH Group t p CVA 41.11 ± 5.97 38.95 ± 4.65 2.196 0.030 FLEX 57.98 ± 3.40 57.19 ± 4.46 1.064 0.289 EXT 65.39 ± 4.55 65.51 ± 5.43 −0.127 0.899 R LAT 37.72 ± 3.59 37.02 ± 2.77 1.200 0.232 L LAT 38.78 ± 3.20 38.81 ± 3.12 −0.054 0.957 R ROT 70.11 ± 5.45 68.30 ± 6.23 1.658 0.100 L ROT 69.52 ± 5.52 67.73 ± 6.26 1.625 0.107 NDI 32.63 ± 10.94 35.49 ± 9.55 −1.511 0.134 PSQI 6.96 ± 4.18 9.48 ± 3.92 −3.352 0.001 BDI 9.74 ± 7.37 14.79 ± 6.48 −3.945 <0.001 VAS 6.30 ± 2.17 −21.351 N/A Headache Duration (yr) 7.70 ± 4.55 −12.417 N/A Headache Days (Last wk) 2.46 ± 1.53 −11.788 N/A Headache Hours (Last wk) 17.57 ± 19.86 −6.496 N/A HDI-T 35.60 ± 16.03 −16.311 N/A HIT 62.29 ± 5.09 −37.942 N/A CGH: Cervicogenic Headache; CVA: Craniovertebral Angle; FLEX: Cervical Flexion (degrees); EXT: Cervical Extension (degrees); R LAT: Right Lateral Flexion (degrees); L LAT: Left Lateral Flexion (degrees); R ROT: Right Rotation (degrees); L ROT: Left Rotation (degrees); NDI: Neck Disability Index; PSQI: Pittsburgh Sleep Quality Index; BDI: Beck Depression Inventory; VAS: Visual Analog Scale; HDI-T: Henry Ford Headache Disability Inventory; HIT: Headache Impact Test; N/A: Not applicable. : Refers to the statistical significance level as p < 0.05. TAB L E 3. Factors associated with cervicogenic headache: univariable and multivariable logistic regression analysis results. Variables Univariable analysis Multivariable Model 1 Multivariable Model 2 OR 95% CI p OR 95% CI p OR 95% CI p Age (yr) 1.101 1.041–1.165 0.001 1.062 0.995–1.134 0.072 1.052 0.982–1.128 0.151 CVA 0.926 0.863–0.994 0.032 0.892 0.812–0.979 0.017 0.878 0.792–0.974 0.014 PSQI 1.167 1.059–1.287 0.002 1.140 1.017–1.279 0.025 BDI 1.115 1.049–1.184 <0.001 OR: Odds Ratio; CI: Confidence Interval; p: p-value; CVA: Craniovertebral Angle; PSQI: Pittsburgh Sleep Quality Index; BDI: Beck Depression Inventory. : p < 0.05 refers to the statistical significance level. Model 1: Multivariable logistic regression analysis including Age and CVA. Model 2: Multivariable logistic regression analysis including Age, CVA and PSQI. 4. Discussion In this study, we examined the frequency and associated factors of CGH in patients who presented to our outpatient clinic with neck pain and were identified as having FHP. According to our findings, increased age, decreased CVA, and poor sleep quality were significantly associated with CGH. CVA and PSQI were found to be independently associated factors with CGH. However, although both were identified as independent predictors of CGH, the modest odds ratios suggest that other unmeasured variables may also contribute to CGH risk, in-dicating a multifactorial etiology. In addition, CVA showed a significant negative independent association with BMI and VAS, suggesting the potential impact of these parameters on CGH-related mechanisms. The prevalence of CGH has been reported with wide vari-ation in previous studies, largely depending on study pop-ulations and diagnostic methods. The prevalence of CGH ranges from 0.2% to 64.1%, depending on the type of study population [7, 24, 25]. For example, Kanniappan et al. reported that the prevalence of CGH in the young population was 10.4%. On the other hand, Xu et al. reported that the prevalence of CGH in individuals over 50 years of age can be up to 42% and suggested that this increase may be due to biomechanical and postural changes depending on age groups. Similarly, in a study conducted among college students, concomitant neck pain was detected in 64.10% of 196 TAB L E 4. Correlation of craniovertebral angle (CVA) with clinical and functional parameters in patients with cervicogenic headache (CGH). Variables r p Age (yr) −0.121 0.344 BMI −0.449 <0.001 Headache Duration (yr) 0.055 0.669 Headache Days (Last wk) 0.017 0.897 Headache Hours (Last wk) 0.123 0.339 VAS −0.275 0.029 FLEX 0.165 0.195 EXT 0.200 0.117 R LAT 0.099 0.440 L LAT 0.026 0.839 R ROT 0.171 0.181 L ROT 0.199 0.118 HDI-T 0.035 0.785 PSQI 0.323 0.010 NDI 0.205 0.106 HIT −0.173 0.176 BDI 0.133 0.299 CVA: Craniovertebral Angle; BMI: Body Mass Index; VAS: Visual Analog Scale; FLEX: Flexion; EXT: Extension; R LAT: Right Lateral Flexion; L LAT: Left Lateral Flexion; R ROT: Right Rotation; L ROT: Left Rotation; HDI-T: Henry Ford Headache Disability Inventory; PSQI: Pittsburgh Sleep Quality Index; NDI: Neck Disability Index; HIT: Headache Impact Test; BDI: Beck Depression Inventory. : p < 0.05. TAB L E 5. Results of multiple linear regression analysis of craniovertebral angle (CVA) with clinical parameters. Variables B SE t p F BMI −0.645 0.180 −3.583 0.001 F = 8.130, p < 0.001 VAS −0.571 0.250 −2.281 0.026 PSQI 0.100 0.149 0.672 0.504 R2 (Coefficient of Determination): 0.292. BMI: Body Mass Index (kg/m2); VAS: Visual Analog Scale; PSQI: Pittsburgh Sleep Quality Index; B: Unstandardized Regression Coefficient; SE: Standard Error; t: t-statistic; p: p-value; F: F-statistic; : Refers to the level of statistical significance as p < 0.05. individuals with headaches . Although this may initially suggest a higher prevalence of CGH in the younger population, it should be noted that neck pain is also frequently reported in primary headache disorders such as migraine and tension-type headache due to referred pain mechanisms, which complicates differential diagnosis . In line with the studies mentioned above but with even higher rates, in this study, the frequency of CGH was found to be 53.8% among patients who visited our outpatient clinic for neck pain and were diagnosed with FHP. One of the main reasons for the high prevalence of CGH in our study is that all individuals participating in the study presented with complaints of neck pain. In addition, the evaluation of only patients with FHP may have contributed to this high rate. Beyond prevalence, the relationship between CGH and cer-vical alignment, including changes in cervical curvature and postural deviations, has also been widely studied. In the study of Farmer et al. , which evaluated the relationship between CGH and cervical posture by radiographic analysis, it was stated that an increase in general cervical lordosis increased the likelihood of CGH. In our study, it was found that the CVA value was significantly lower in the CGH group than in the non-CGH group. FHP is known to be associated with decreased lordosis, especially in the lower cervical region (C2– C7). This postural change can increase the biomechanical load on the cervical spine and is associated with greater neck muscle tension and reduced muscle function. Furthermore, FHP can trigger pain mechanisms by causing an imbalance in the cervical musculoskeletal system. Similarly, in the study of Delen and İlter , it was reported that the duration of headache was longer in patients with loss of cervical lordosis, but it was not directly related to other pain parameters such as severity and frequency. These findings in the literature 197 support the potential role of changes in cervical lordosis in the development of CGH but do not reveal a definitive causality. In particular, the fact that FHP causes muscle imbalance in the upper cervical segments and increased tension in the suboc-cipital muscles may be effective in the emergence of CGH by increasing neuromechanical sensitivity. The biomechanical and physiological mechanisms behind these postural changes are also worth considering. Mechanical stress, postural disorders, and neurophysiological effects play an essential role in the etiology of CGH. In the study conducted by Çoban et al. , a significant inverse relationship was found between CVA and VAS, and it was shown that CVA is an important parameter in the evaluation of CGH symp-toms. Similarly, in our study, a negative correlation was found between low CVA and increased VAS, and these findings are consistent with the results reported in the literature. In this context, Martinez-Merinero et al. , in their study of 102 patients, reported that the decrease in CVA may increase the mechanosensitivity of tissues through mechanical stress, which may contribute to pain mechanisms in CGH patients. In support of this mechanism, Patwardhan et al. suggested that FHP may increase neuromechanical sensitivity in the C2 nerve root by creating shortening of the suboccipital muscles. Furthermore, Chua et al. reported that forward head posture exerts continuous pressure on the joint surfaces of the upper cervical vertebrae, which may trigger both peripheral and central sensitization processes. These literature findings support the negative association between low CVA and in-creased VAS in our study. In addition, Martinez-Merinero et al. emphasized that mechanosensitivity is more pro-nounced in certain regions, especially at the C2 level, reinforc-ing the importance of CVA in terms of postural and mechanical effects. Moreover, although the CVA difference between CGH and non-CGH groups in our study was approximately 2◦, it is notable that Heydari et al. reported a minimal clinically important difference (MCID) of 1.40◦for CVA, suggesting that this observed change may also have clinical relevance in addition to statistical significance. In addition to structural and mechanical aspects, psychoso-cial components such as sleep and depression should also be addressed. In our study, sleep quality was found to be significantly lower, and depression levels were significantly higher in the CGH group. Although impaired sleep quality has also been reported in patients with migraine and tension-type headache , observational studies suggest that CGH may involve more specific associations with postural impairments and central sensitization. It has been observed that sleep qual-ity is lower in CGH patients, and the significant relationship between CVA and PSQI suggests a link between postural align-ment and sleep quality. Considering the multifactorial nature of CGH, which involves structural, postural, and psychosocial factors, it is also possible that sleep disturbances may emerge as a consequence of persistent cervical discomfort and pain, rather than serving as an initial trigger. Similarly, in the study of Mingels et al. , which examined the relationships between pain processing, lifestyle, and psychosocial factors in CGH patients, it was shown that sleep quality was significantly worse in the CGH group than in the control group, and this was associated with central sensitivity symptoms. In another study by Mingels et al. , the relationship between individual differences in mechanical pain sensation and biopsychosocial lifestyle factors in CGH patients was investigated. It was emphasized that modifiable factors like poor sleep quality and depression are associated with more frequent headaches and a higher likelihood of chronic CGH. These findings suggest that CGH should be considered within the framework of a biopsychosocial model. Although poor sleep and depression are common features of chronic pain in general, addressing these factors in CGH may still help reduce the risk of chroni-fication. Another factor influencing postural alignment is BMI, which has biomechanical implications. In this study, a significant negative correlation was found between BMI and CVA, and it was determined that an increase in BMI is an independent predictor of CVA decrease. In the literature, it has been reported that an increase in BMI alters cervical sagittal alignment, increasing the sagittal vertical axis and causing a forward shift in cervical posture . This change may increase the mechanical load on the cervical spine, leading to a reduction in lordosis. Indeed, patients with a higher BMI were found to have reduced C2–C7 cervical lordosis, and this difference became more pronounced during follow-up . Furthermore, high BMI was shown to increase the risk of adjacent segment degeneration, which may contribute to cervical postural instability and increased musculoskeletal load, ultimately leading to postural dysfunction . Therefore, weight management in CGH patients can be considered together with approaches to correct postural dysfunction. Finally, the therapeutic implications of postural correction are worth emphasizing. It has been shown in the literature that exercise programs focusing on the correction of postural disorders and maintaining muscle balance are effective in the treatment of CGH. Nobari M et al. reported that correc-tive exercises in individuals with FHP significantly reduced headache severity, duration, and frequency, while also im-proving the neck disability index. Similarly, in another study, it was stated that postural arrangements and exercises posi-tively affected the mechanical properties of the upper cervical muscles and contributed to the relief of headache symptoms, with improvements in the craniovertebral angle . These findings in the literature align with our results regarding the effects of FHP on CGH symptoms. For this reason, it is recommended to apply specific exercise programs in order to correct postural disorders and maintain muscle balance in the treatment of CGH. This study has some limitations worth acknowledging. First, the absence of a control group limits the ability to compare differences between individuals with CGH and healthy indi-viduals. Therefore, controlled studies are needed to understand better the relationship between CGH and clinical and environ-mental factors. Second, the observational and retrospective design of our study limits the ability to draw definitive causal inferences, which would ideally require an experimental ap-proach such as a randomized controlled trial. As this study has a cross-sectional design, no causal relationship can be established between forward head posture and cervicogenic headache. In particular, prospective research is needed to 198 understand better the changes in the effects of parameters such as BMI, CVA, and sleep quality on CGH over time. Third, the generalizability of the findings is limited since the participants were selected only from patients who applied to our outpatient clinic with complaints of neck pain. Finally, the risk of type I error due to multiple comparisons should be acknowledged, and future studies using correction methods are recommended to confirm these findings. 5. Conclusions In this study, important findings suggesting that postural changes may be related to CGH have been obtained. In particular, it was determined that decreased CVA was an independent predictor of CGH, while BMI and pain severity were associated with postural changes. It is thought that FHP may be associated with CGH, possibly through increased biomechanical load on the cervical spine. In this context, clinical implementation of posture-corrective exercises and structured weight management programs may serve as supportive strategies to improve biomechanical balance and reduce CGH symptoms in daily practice. In addition, poor sleep quality was found to be an independent predictor of CGH. These results suggest that CGH may be shaped not only by biomechanical factors but also by biopsychosocial factors. Therefore, multidisciplinary management approaches that integrate physiotherapy, psychological support, and lifestyle interventions could be more effective in the clinical care of CGH. Furthermore, future prospective and randomized controlled trials are needed to evaluate the long-term effects of postural rehabilitation, sleep quality enhancement, and weight control interventions on CGH incidence and symptom severity. AVAILABILITY OF DATA AND MATERIALS The data that support the findings of this study are available on request from the corresponding author. The data are not publicly available due to privacy or ethical restrictions. AUTHOR CONTRIBUTIONS AU—conceptualization; methodology. MDG—data curation. AU and MDG—formal analysis; software; validation; investi-gation; writing-original draft; writing-review & editing. ETHICS APPROVAL AND CONSENT TO PARTICIPATE The study protocol was approved by the Ethics Committee of Istanbul Medipol University on 25 June 2024 (reference number: E-10840098-202.3.02-3742). The research was con-ducted in accordance with the Declaration of Helsinki, and informed consent was obtained from all participants. ACKNOWLEDGMENT Not applicable. FUNDING This research received no external funding. CONFLICT OF INTEREST The authors declare no conflict of interest. REFERENCES Headache Classification Committee of the International Headache Society (IHS) The international classification of headache disorders, 3rd edition. Cephalalgia. 2018; 38: 1–211. Bogduk N. Cervicogenic headache: anatomic basis and pathophysiologic mechanisms. Current Pain and Headache Reports. 2001; 5: 382–386. Piovesan EJ, Utiumi MAT, Grossi DB. Cervicogenic headache—how to recognize and treat. Best Practice & Research Clinical Rheumatology. 2024; 38: 101931. Verma S, Tripathi M, Chandra PS. Cervicogenic headache: current perspectives. Neurology India. 2021; 69: S194–S198. Sjaastad O, Bakketeig LS. Prevalence of cervicogenic headache: vågå study of headache epidemiology. Acta Neurologica Scandinavica. 2008; 117: 173–180. Knackstedt H, Bansevicius D, Aaseth K, Grande RB, Lundqvist C, Russell MB. Cervicogenic headache in the general population: the Akershus study of chronic headache. Cephalalgia. 2010; 30: 1468–1476. Kanniappan V, Abraham SS, Veeragoudhaman TS. Prevalence of cervicogenic headache among young population. International Journal of Research-GRANTHAALAYAH. 2022; 10: 14–26. Szeto GPY, Straker LM, O’Sullivan PB. A comparison of symptomatic and asymptomatic office workers performing monotonous keyboard work—1: neck and shoulder muscle recruitment patterns. Manual Therapy. 2005; 10: 270–280. Gustafsson E, Thomée S, Grimby-Ekman A, Hagberg M. Texting on mobile phones and musculoskeletal disorders in young adults: a five-year cohort study. Applied Ergonomics. 2017; 58: 208–214. Kendall FP, McCreary EK, Provance PG, Rodgers MM, Romani WA. Muscles testing and function with posture and pain. In Kendall FP, McCreary EK, Provance PG, Rodgers MM, Romani WA (eds.) Kendall, muscles (pp. 49–118). 5th edn. Lippincott Williams & Wilkins: Philadelphia, PA. 2005. Lee J. Effects of forward head posture on static and dynamic balance control. Journal of Physical Therapy Science. 2016; 28: 274–277. Hallgren RC, Pierce SJ, Sharma DB, Rowan JJ. Forward head posture and activation of rectus capitis posterior muscles. Journal of Osteopathic Medicine. 2017; 117: 24–31. Patwardhan AG, Khayatzadeh S, Havey RM, Voronov LI, Smith ZA, Kalmanson O, et al. Cervical sagittal balance: a biomechanical perspective can help clinical practice. European Spine Journal. 2018; 27: 25–38. Mahmoud NF, Hassan KA, Abdelmajeed SF, Moustafa IM, Silva AG. The relationship between forward head posture and neck pain: a systematic review and meta-analysis. Current Reviews in Musculoskeletal Medicine. 2019; 12: 562–577. Farmer PK, Snodgrass SJ, Buxton AJ, Rivett DA. An investigation of cervical spinal posture in cervicogenic headache. Physical Therapy. 2015; 95: 212–222. Zito G, Jull G, Story I. Clinical tests of musculoskeletal dysfunction in the diagnosis of cervicogenic headache. Manual Therapy. 2006; 11: 118– 129. Aslan E, Karaduman A, Yakut Y, Aras B, Simsek IE, Yaglý N. The cultural adaptation, reliability and validity of neck disability index in patients with neck pain: a Turkish version study. Spine. 2008; 33: E362– E365. Kılınç HE, Öz M, Berberoğlu U, Özel Aslıyüce Y, Onan D, Fanuscu A, et al. Cross-cultural adaptation, reliability and validity of the Turkish version of the henry ford hospital headache disability inventory (HDI/T) 199 in patients with cervicogenic headache. Disability and Rehabilitation. 2024; 46: 820–827. Dikmen PY, Bozdağ M, Güneş M, Koşak S, Taşdelen B, Uluduz D, et al. Reliability and validity of Turkish version of headache impact test (HIT-6) in patients with migraine. Archives of Neuropsychiatry. 2020; 58: 300– 307. Ağargün MY, Kara H, Anlar O. The validity and reliability of the Pittsburgh Sleep Quality Index. Turkish Journal of Psychiatry. 1996; 7: 107–115. Hisli Şahin N. Beck a study on the validity of the Beck Depression Inventory. Turkish Journal of Psychology. 1988; 6: 118–126. Ruivo RM, Pezarat-Correia P, Carita AI. Intrarater and interrater reliability of photographic measurement of upper-body standing posture of adolescents. 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Medical Science Monitor. 2023; 29: e939427. Çoban G, Çöven İ, Çifçi BE, Yıldırım E, Yazıcı AC, Horasanlı B. The importance of craniovertebral and cervicomedullary angles in cervicogenic headache. Diagnostic and Interventional Radiology. 2014; 20: 172–177. Martínez-Merinero P, Aneiros Tarancón F, Montañez-Aguilera J, Nuñez-Nagy S, Pecos-Martín D, Fernández-Matías R, et al. Interaction between pain, disability, mechanosensitivity and cranio-cervical angle in subjects with cervicogenic headache: a cross-sectional study. Journal of Clinical Medicine. 2021; 10: 159. Chua NHL, van Suijlekom HA, Vissers KC, Arendt-Nielsen L, Wilder-Smith OH. Differences in sensory processing between chronic cervi-cal zygapophysial joint pain patients with and without cervicogenic headache. Cephalalgia. 2011; 31: 953–963. Heydari Z, Sheikhhoseini R, Shahrbanian S, Piri H. Establishing minimal clinically important difference for effectiveness of corrective exercises on craniovertebral and shoulder angles among students with forward head posture: a clinical trial study. BMC Pediatrics. 2022; 22: 230. Rains JC, Davis RE, Smitherman TA. Tension-type headache and sleep. Current Neurology and Neuroscience Reports. 2015; 15: 520. Mingels S, Dankaerts W, van Etten L, Bruckers L, Granitzer M. Exploring multidimensional characteristics in cervicogenic headache: relations between pain processing, lifestyle, and psychosocial factors. Brain and Behavior. 2021; 11: e2339. Mingels S, Dankaerts W, Bruckers L, Granitzer M. Inter-individual variability in mechanical pain sensation in patients with cervicogenic headache: an explorative study. Scientific Reports. 2022; 12: 20635. Basques BA, Khan JM, Louie PK, Mormol J, Heidt S, Varthi A, et al. Obesity does not impact clinical outcome but affects cervical sagittal alignment and adjacent segment degeneration in short term follow-up after an anterior cervical decompression and fusion. The Spine Journal. 2019; 19: 1146–1153. Nobari M, Arslan SA, Hadian MR, Ganji B. Effect of corrective exercises on cervicogenic headache in office workers with forward head posture. Journal of Modern Rehabilitation. 2017; 11: 201–208. Rani M, Kaur J. Effectiveness of spinal mobilization and postural correction exercises in the management of cervicogenic headache: a randomized controlled trial. Physiotherapy Theory and Practice. 2023; 39: 1391–1405. How to cite this article: Ahmet Usen, Merve Demiroz Gunduz. Cervicogenic headache in forward head posture: frequency and associated factors in a cross-sectional study. Journal of Oral & Facial Pain and Headache. 2025; 39(3): 191-199. doi: 10.22514/jofph.2025.061.
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Skip to content Watch Live King Charles III: A guide to his Accession Council and proclamation Jon Kelly BBC News The new king has been proclaimed by an Accession Council. What exactly does that mean? Charles is already king - under the terms of the Act of Settlement 1701, he automatically became so on the death of his mother. So the purpose of the Accession Council is basically a ceremonial one - it officially announces the name of the new monarch. Normally, this happens within 24 hours of the death of the sovereign. But on this occasion a little more time will have passed between the death of Queen Elizabeth II and the ceremony at St James's Palace in London. In another break with tradition, King Charles III has decided that for the first time the Accession Council will be televised. Who attended, and what took place? The council is attended by members of the Privy Council (a group of senior politicians that formally advise the monarch), the Lord Mayor of the City of London, and senior judges and officials. The Accession Council is divided into two parts, and Charles will only be present for the second. In the first part, the Lord President - the MP Penny Mordaunt, appointed by Liz Truss on 6 September - announced the monarch's death. She then asked the clerk of the council, currently retired civil servant Richard Tilbrook, to read out the Accession Proclamation, which confirms the name of the new monarch. The proclamation was then signed by the members of the Royal Family, the prime minister, the Archbishop of Canterbury, the Lord Chancellor and the Earl Marshall - the Duke of Norfolk, who is responsible for organising state ceremonies. After the signing, Ms Mordaunt called for silence then read out the outstanding items of business, which include disseminating the proclamation and directing that artillery guns will be fired at Hyde Park in central London and the Tower of London. The proclamation was then read out from a balcony above the Friary Court at St James's Palace by the Garter King of Arms, England's senior herald. He was joined by the Earl Marshall and other officials wearing traditional heraldic garments. The proclamation was also read in Belfast, Cardiff and Edinburgh, and other locations around the country. What is Charles' role? Part two of the Accession Council was the first meeting of the Privy Council held by the new monarch, and Charles began with a personal declaration. According to Robert Hazell, professor of government and the constitution at University College London, this usually has three elements. The king commemorates the Queen, affirms his fidelity to the constitution and expresses "hope for the support of the nation in his heavy responsibilities". Under the terms of the Act of Union, new monarchs are also required to make an oath to maintain and preserve the Church of Scotland. Usually, this takes place at part two of the Accession Council and the new monarch signs two copies of the oath. Charles can say if he objects to any of the oaths he is asked to make, although the last time a new monarch did so was in 1910 when King George V rejected the anti-Catholic wording of the declaration oath. The government of the time evidently agreed with him as they changed the wording in the Accession Declaration Act 1910. The Accession Council has now concluded its business, but it may be some time before Charles's coronation takes place. Some 16 months passed between the death of Queen Elizabeth's father, King George VI, in February 1952 and her coronation in June 1953. Coronation of King Charles III
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http://buzzard.ups.edu/courses/2016spring/projects/woody-resultants-ups-434-2016.pdf
Polynomial Resultants Henry Woody Department of Mathematics and Computer Science University of Puget Sound Copyright ©2016 Henry Woody. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.3 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license can be found at Abstract This paper covers various aspects of the resultant of polynomials. Starting with a definition, we move to a practical method of calculating the resultant, specifically through the use of the Sylvester matrix, whose entries are the coefficients of the two polynomials, and whose determinant gives the resultant of two polynomials. We focus on whether or not two univariate polynomials have a common factor, and finding solutions to homogeneous systems of multivariate polynomials. We cover, as well, the discriminant of polynomials, which is the resultant of a polynomial and its derivative. 1 Introduction Polynomials are covered in elementary algebra, so many people are familiar with their general properties and principles, but, as is the case with almost all topics covered in elementary algebra, they have much more going on than is shown in high school mathematics. The study of polynomials often centers around the search for roots, or zeros, of those polynomials. An integral part of the study of abstract algebra, polynomials possess interesting qualities and give rise to important topics, such as field extensions. Throughout this paper we let F be a field. We start with a simple definition. Definition 1.1 (Polynomial). A polynomial f(x) ∈F[x] is defined as f(x) = n Y i=0 (x −αi) = anxn + an−1xn−1 + ... + a1x + a0, an ̸= 0 where each ai ∈F is a coefficient, each αi is a root of f(x), and the degree of f(x), deg(f(x)), is n. For more information about polynomials, their properties, and polynomial rings, we refer the reader to . 1 2 RESULTANTS 2 Resultants We are often concerned with knowing whether or not two polynomials share a root or common factor. One way to determine this is to factor both polynomials completely and compare the sets of roots. While perhaps the most obvious method, it is not the most efficient since factoring polynomials entirely, especially high degree or multivariate polynomials, can be difficult if not impossible through algebraic methods. A better way to accomplish this is to calculate the greatest common divisor of the two polynomials using the Euclidean algorithm, but this requires that the polynomials be in a Euclidean domain. This is the case for polynomials over fields, however not all polynomials are defined over fields and not all polynomial rings are Euclidean domains. Thus we would like a method of determining if two polynomials share a common factor that will work efficiently for any polynomial. Resultants satisfy these criteria. Definition 2.1 (Resultant). Given two polynomials f(x) = anxn+...+a1x+a0, g(x) = bmxm + ... + b1x + b0 ∈F[x], their resultant relative to the variable x is a polynomial over the field of coefficients of f(x) and g(x), and is defined as Res(f, g, x) = am n bn m Y i,j (αi −βj), where f(αi) = 0 for 1 ≤i ≤n, and g(βj) = 0 for 1 ≤j ≤m. There are a few things to take away from this definition. First, Res(f, g, x) is an element of F. Second, Res(f, g, x) = am n Q i g(αi) = (−1)nmbn m Q j f(βj), which is to say the resultant is the product of either polynomial evaluated at each root, including multiplicity, of the other polynomial. Finally, the reader should note what happens in the case that the two polynomials share a root, as outlined in the following lemma. Lemma 2.2. The resultant of f(x) and g(x) is equal to zero if and only if the two polynomials have a root in common. Proof. To see this, suppose γ is a root shared by both polynomials, then one term of the product is (γ −γ) = 0, hence the whole product vanishes. Conversely, suppose Res(f, g, x) = 0, then, since F is a field and therefore has no zero divisors, at least one term of the product zero. Suppose that term is (αk −βl), this implies αk = βj, so this is a root shared by f(x) and g(x). It is clear that the resultant of two polynomials will tell us whether or not two polynomials share a root. However, by this definition, calculating the resultant requires us to factor each polynomial completely in order to find roots, and doing so would render the resultant redundant. We will now work toward a more efficient way of determining the resultant. Lemma 2.3. Let f(x), g(x) ∈F[x] have degrees n and m, both greater than zero, respectively. Then f(x) and g(x) have a non-constant common factor if and only if there exist nonzero polynomials A(x), B(x) ∈F[x] such that deg(A(x)) ≤m −1, deg(B(x)) ≤n −1 and A(x)f(x) + B(x)g(x) = 0. Proof. Assume f(x) and g(x) have a common, non-constant, factor h(x) ∈F[x]. Then f(x) = h(x)f1(x) and g(x) = h(x)g1(x), for some f1(x), g1(x) ∈F[x]. 2 Henry Woody 2 RESULTANTS Consider, g1(x)f(x) + (−f1(x))g(x) = g1(x)(h(x)f1(x)) −f1(x)(h(x)g1(x)) = 0. Notice deg(g1(x)) ≤m −1 and deg(f1(x)) ≤n −1 since h(x) has at least degree 1, therefore these are the polynomials we seek. Note A(x) is equal to the product of factors of g(x) that are not shared by f(x) and B(x) is the product of factors of f(x) that g(x) does not share (multiplied by -1). We will prove the left implication by contradiction. First we assume the poly-nomials A(x) and B(x) that satisfy the criteria above exist. Now suppose, to the contrary of the lemma, that f(x) and g(x) have no non-constant common factors, then gcd(f(x), g(x)) = 1. Hence there exist polynomials r(x) and s(x) in F[x] such that r(x)f(x)+s(x)g(x) = 1. We also know that A(x)f(x)+B(x)g(x) = 0 = ⇒A(x)f(x) = −B(x)g(x). Then A(x) = 1A(x) = (r(x)f(x) + s(x)g(x))A(x) = r(x)f(x)A(x) + s(x)g(x)A(x) = r(x)(−B(x)g(x)) + s(x)g(x)A(x) = (s(x)A(x) −r(x)B(x))g(x) Since A(x) ̸= 0, we know (s(x)A(x)−r(x)B(x))g(x) ̸= 0. We also know deg(g(x)) = m, which means the degree of A(x) is at least m, but this is a contradiction since A(x) was defined to have degree strictly less than m. Lemma 2.3 can be translated from polynomials to the integers. If we consider the irreducible polynomial factors of polynomials as prime factors of integers, then if we have two integers a and b that share at least one prime factor, we can find two other integers c and d such that ac+bd = 0. Then c is the product of prime factors of a that are not shared with b, and d is the product of prime factors that b has but a lacks, and one will be negative if necessary. We now look more in depth at the equation A(x)f(x) + B(x)g(x) = 0 from Lemma 2.3, and let f(x) = anxn + ... + a1x + a0, an ̸= 0 g(x) = bmxm + ... + b1x + b0, bm ̸= 0 A(x) = cm−1xm−1 + ... + c1x + c0, B(x) = dn−1xn−1 + ... + d1x + d0. By the definition of polynomial equality, the equation A(x)f(x) + B(x)g(x) = 0 is a homogeneous system of n + m equations in n + m variables, which are the ci and dj. This system is shown below. ancm−1 + bmdn−1 = 0 (coefficients of xn+m−1) ancm−2 + an−1cm−1 + bmdn−2 + bm−1dn−1 = 0 (coefficients of xn+m−2) ancm−3 + an−1cm−2 + an−2cm−1 + bmdn−3 + bm−1dn−2 + bm−2dn−1 = 0 (coefficients of xn+m−3) . . . a0c0 + b0d0 = 0 (coefficients of x0) 3 Henry Woody 2 RESULTANTS The (n + m) × (n + m) coefficient matrix of this system is called the Sylvester matrix, and is shown below. Syl(f, g, x) =                    an bm an−1 an bm−1 bm an−2 an−1 ... bm−2 bm−1 ... . . . . . . ... an . . . . . . ... bm . . . . . . an−1 . . . . . . bm−1 a0 a1 b0 b1 a0 ... . . . b0 ... . . . ... a1 ... b1 a0 b0                    Zeros fill the blank spaces. The first m columns of Syl(f, g, x) are populated by the coefficients of f(x), and the last n columns are filled with the coefficients of g(x). Though it is not clear from the figure above, the entries along the diagonal are an for the first m slots, and b0 for the last n. It is clear that an is along the diagonal for the first m columns, but it may be less obvious for b0. To see this, consider the m + 1 column, which has b0 in the m + 1 row, since there are m terms above it, therefore b0 is along the diagonal. The Sylvester matrix is sometimes written as the transpose of the matrix given above, but the distinction is not important, as the next theorem indicates. Theorem 2.4. The determinant of the Sylvester matrix Syl(f, g, x) is a polynomial in the coefficients ai,bj of the polynomials f(x) and g(x). Further, det(Syl(f, g, x)) = Res(f, g, x). Proof. Proof can be found in , though Van Der Waerden, and many other authors, defines the resultant by Theorem 2.4 and gets to our definition through theorems. Corollary 2.5. det(Syl(f, g, x)) = 0 if and only if f(x) and g(x) have a common factor. Corollary 2.6. For f(x), g(x) ∈F[x], there exist polynomials A(x), B(x) ∈F[x] so that A(x)f(x) + B(x)g(x) = Res(f, g, x). Proof. If Res(f, g, x) = 0 then the statement is trivially true by A(x) = B(x) = 0, since it puts no restrictions on the degrees of these polynomials. So we consider Res(f, g, x) ̸= 0, which implies f(x) and g(x) are relatively prime. So we have gcd(f(x), g(x)) = r(x)f(x) + s(x)g(x) = 1, for some r(x), s(x) ∈F[x]. This equation describes a system of equations with coefficient matrix Syl(f, g, x) and a vector of constants with n+m−1 zeros and a one in the bottom entry. Since Res(f, g, x) ̸= 0, we know det(Syl(f, g, x)) ̸= 0, therefore the coefficient matrix is nonsingular and there is a unique solution to this system. In particular the polynomials A(x) and B(x) of Corollary 2.6 are simply A(x) = Res(f, g, x)r(x) and B(x) = Res(f, g, x)s(x). So these polynomials are just multiples of the polynomials from the extended greatest common divisor of f(x) and g(x). 4 Henry Woody 3 APPLICATIONS Theorem 2.4 gives us a way to determine whether or not two polynomials have a common factor and all we need to know is the coefficients of those polynomials. This is a powerful result because of how simple it is to set up. So now we can easily tell if two polynomials share a factor, but the information conveyed by the resultant is binary, meaning it will tell us only if a factor is shared or not, but no more information than this. The following theorem gives us a way to find more information regarding the common factor of two polynomials, still without using the Euclidean algorithm. Theorem 2.7. If f(x) is the characteristic polynomial of a square matrix M, and g(x) is any polynomial, then the degree of the common factor of f(x) and g(x) is the nullity of the matrix g(M). Proof. Proof of this theorem can be found in . This theorem may seem as though it is only applicable in a narrow range of cases, but it is more general than it appears. For any monic polynomial f(x), we can construct a matrix M so that f(x) is the characteristic polynomial. Note also that if we start with two polynomials that are both not monic, we can simply factor out the leading coefficient from one of the polynomials, which will not affect the degree of the common factor. Of course this requires division in the ring of the coefficients, so this theorem will only hold for polynomials over a division ring, which is still less restrictive than a Euclidean domain, or polynomials over any ring if at least one of those polynomials is monic. Let f(x) = xn + an−1xn−1... + a1x + a0, then the square matrix M is given below. M =        −an−1 −an−2 · · · −a1 −a0 1 0 · · · 0 0 0 1 · · · 0 0 . . . . . . ... . . . . . . 0 0 · · · 1 0        This method is easily implemented and will tell us the degree of the polynomial factor shared by two polynomials. It does not, however, actually give us the common factor, but if we are working in a Euclidean domain, the Euclidean algorithm can be used to find the greatest common factor. 3 Applications We have already seen the most straightforward application of the resultant, namely that it indicates whether or not two polynomials share any non-constant factors, but we will now introduce two more. Not surprisingly, both of these focus on roots of polynomials. 3.1 Discriminant The first application involves the discriminant of a polynomial and the resultant’s connection to it. The discriminant, D, of a polynomial gives some insight into the nature of the polynomial’s roots. For example the discriminant of a quadratic of the 5 Henry Woody 3.1 Discriminant 3 APPLICATIONS form f(x) = ax2 + bx + c ∈R[x] is D = b2 −4ac. In this case, if D is positive, then f(x) has two distinct real roots. If D = 0, then f(x) has one real root with multiplicity 2. If D is negative, then f(x) has no real roots, but has two complex roots that are conjugate. For higher degree polynomials, we can tell if a polynomial has a multiple root, meaning a root with multiplicity greater than 1, if the discriminant vanishes. A polynomial of degree n is separable if it has n distinct roots in its splitting field. It is also the case that a polynomial is separable if the polynomial and its derivative are relatively prime. We can related these ideas to the discriminant of a polynomial. Lemma 3.1. A polynomial f(x) ∈F[x] is separable if and only if its discriminant is nonzero. Since the discriminant is zero only if a polynomial has a multiple root, and, equiv-alently if the polynomial and its derivative share a common factor, there is evidence to suggest the discriminant and the resultant are in some way related. Lemma 3.2. A polynomial f(x) ∈F[x] has a zero discriminant if and only if Res(f, f′, x) = 0, where f′(x) is the formal derivative of f(x). The previous lemma indicates that the resultant and the discriminant are closely connected, in fact the discriminant is a multiple of a specific kind of resultant, specifi-cally that of a polynomial and its derivative, as shown in the following definition. Definition 3.3. For a polynomial f(x) ∈F[x], where f(x) = anxn + ... + a1x + a0, the discriminant is given by D = (−1)n(n−1)/2 an Res(f, f′, x), where f′(x) is the derivative of f(x). At this point, it should not come as much of a surprise that the determinant is defined in terms of the resultant. The determinant is zero if f(x) and f′(x) share a root, which is only possible if f(x) has a multiple root. In fact, the term determinant was coined by James Joseph Sylvester, for whom the Sylvester matrix is named. Example 3.4. Let f(x) = ax2+bx+c, then f′(x) = 2ax+b, thus we have the equation for the determinant as follows D = (−1)2(2−1)/2 a a 2a 0 b b 2a c 0 b = −1 a (a(b2) −b(2ab) + c(4a2)) = −1 a (ab2 −2ab2 + 4a2c) = −1 a (−ab2 + 4a2c) = b2 −4ac, which is the explicit formula for the determinant of a quadratic. Finally we conclude this subsection with a corollary that connects determinants, resultants, and the study of fields. Corollary 3.5. A polynomial f(x) ∈F[x] is separable if and only if Res(f, f′, x) ̸= 0. Equivalently, f(x) is separable if and only if Syl(f, f′, x) is nonsingular. 6 Henry Woody 3.2 Elimination 3 APPLICATIONS 3.2 Elimination So far we have been working exclusively with polynomials in one variable, in other words those in F[x], but multivariate polynomials deserve consideration as well, and they are often more difficult to analyze than their univariate cousins. One important application of the resultant is in elimination theory, which focuses on eliminating variables from systems of multivariate polynomials. As we have seen in previous examples, and from the definition, the resultant takes two polynomials in the variable x and gives one polynomial with no variables. In multiple variables this outcome is generalized to remove one variable. First we look at an example. Example 3.6. Let f(x, y) = x2y +x2 +3x−1, g(x, y) = xy2 +y −5 ∈F[x, y]. We will examine the resultant of these two polynomials by considering them as polynomials in x with coefficients that are polynomials in y. We compute the resultant relative to x through the Sylvester matrix. Res(f, g, x) = y + 1 y2 0 3 y −5 y2 −1 0 y −5 = (y + 1)(y −5)2 −3y2(y −5) + (−1)y4 = −y4 −2y3 + 6y2 + 15y + 25 The resultant of these two bivariate polynomials is a single univariate polynomial, so the variable x has been eliminated. This resulting polynomial shares properties with its parents, f(x, y) and g(x, y), but is easier to analyze than its parents. The example above illustrates an interesting property of the resultant of two multi-variate polynomials. Whereas in the single variable case the resultant is a polynomial in the coefficients of the two starting polynomials, in the two variable case, the resultant relative to one variable is a polynomial in the other variable. This follows from the fact that F[x, y] = F[y][x], which is to say the polynomial ring F[x, y] of polynomials with coefficients from F and with variables x and y can be thought of as the polynomial ring F[y][x], which has polynomials with coefficients from F[y], meaning polynomials in y, in the variable x. Hence the placement of the x in Res(f, g, x) to indicate the variable to be eliminated. Since two polynomials share a root if and only if their resultant is zero, we take the resultant polynomial and set it equal to zero. By finding roots of the resultant relative to x, we are finding values of y that will make the resultant relative to x zero, therefore making the two polynomials share a non-constant factor in the variable x. After taking the resultant of two polynomials f(x, y) and g(x, y) relative to x in F[x, y], and solving for roots of the resulting polynomial, we can take the resultant of these polynomials again, but this time relative to y and solve for this resultant’s roots. Now we have two sets of partial solutions, and we can take combinations of solutions to Res(f, g, x)(y) = 0 and Res(f, g, y)(x) = 0 and test them in f(x, y) and g(x, y). Alternatively, we can evaluate f(x, y) and g(x, y) at each of the partial solutions given by the first resultant. This will eliminate the variable y, and we can then attempt to find roots of each of the simplified polynomials in F[x]. Neither of these is an elegant solution, but they are solutions nonetheless. 7 Henry Woody 3.2 Elimination 3 APPLICATIONS Example 3.7. Suppose f(x, y) = x2y2 −25x2 + 9 and g(x, y) = 4x + y are two polynomials in F[x, y]. We now employ the resultant in order to find a common root of f(x, y) and g(x, y). Res(f, g, x) = y2 −25 4 0 0 y 4 9 0 y = y4 −25y2 + 144 Res(f, g, y) = x2 1 0 0 4x 1 −25x2 + 9 0 4x = 16x4 −25x2 + 9 The four roots of Res(f, g, x) are y = ±3, ±4, and those of Res(f, g, y) are x = ± 3 4, ±1. By testing each of the 16 possible combinations of partial solutions, we find that the solutions to the homogeneous system f(x, y) = 0 g(x, y) = 0 are (x, y) = (1, −4), (−1, 4), ( 3 4, −3), (−3 4, 3). So we can use resultants to simplify polynomial systems in multiple variables, but how many polynomials are required to actually solve a system in n variables? In order to answer this question, we must first consider the manner in which homogeneous polynomial systems can be solved using the resultant for polynomials in more than two variables. Variable elimination can be generalized further from polynomials in two variables to polynomials in n variables. If we consider the resultant as a function explicitly, then Resi : F[x1, ..., xn] × F[x1, ..., xn] →F[x1, ..., xi−1, xi+1, ..., xn], where Resi is the resul-tant relative to the variable xi. In other words, the resultant takes two polynomials in n variables and returns one polynomial in n −1 variables. If we have a sufficient number of polynomials to start with, we can continue to eliminate variables until there is only one remaining, where it is significantly easier to solve for roots. In order to solve a homogeneous polynomial system in n variables, n polynomials are required. Any fewer than n polynomials will cause the resultants to become zero before the univariate stage. Solving a homogeneous system of n polynomials in n vari-ables in a manner similar to Example 3.7 is possible, however it can become extremely computationally expensive, especially in more than three variables. Since many de-terminants will be taken, the powers on the other variables can get quite large, thus we can end up with a univariate polynomial of high degree (5 or greater) which may not be solvable by radicals. In this case, an approximate computational method must be employed. For example, for a polynomial in four variables, where each variable has a power of at most two, the single variable polynomial obtained through multiple resultants can easily have degree greater than 100. For polynomials in more than three variables, it is more practical to eliminate variables using the multivariate resultant, which takes n polynomials in n variables as an argument, rather than just two univariate polynomials. Note that by univariate here, we mean the single variable resultant only considers one variable, so a polynomial can be in multiple variables, but will be considered as a polynomial in one variable with 8 Henry Woody 4 CONCLUSION coefficients that are themselves polynomials. In combination with Gr¨ obner bases, the multivariate resultant is one of the main tools utilized in elimination theory. For more on the multivariate resultant we refer the reader to . Returning to the single variable resultant, assuming we have n polynomials in n variables, we now have a new question. Can we even be sure that we will find a common root of these polynomials? The next theorem answers this question in the affirmative. Theorem 3.8. If (α1, ..., αi−1, αi+1, ..., αn) is a solution to a homogeneous system of polynomials in F[x1, ..., xi−1, xi+1, ..., xn] obtained by taking resultants of polynomials in F[x1, ..., xn] with respect to xi, then there exists αi ∈E, where E is the field in which all polynomials in the system split, such that (α1, ..., αi, ..., αn) is a solution to the system in F[x1, ..., xn]. Proof. Proof can be found in . We can take Theorem 3.8 and essentially work from the bottom up. So if we have the appropriate number of polynomials in F[x1, ..., xn] then we can eliminate all variables except x1 and find roots of the polynomial f1(x1) ∈F[x1], which was obtained through repeated resultants. Then let α1 be a root of this polynomial and consider the partially eliminated system with only x1 and x2, i.e. polynomials in F[x1, x2], where we know there exists an α2 such that (α1, α2) is a root shared by all polynomials in the system in F[x1, x2]. We then continue this process until we have a full solution. Example 3.7 illustrates this result as there is one full solution for each of the partial solutions given by the each of the resultant polynomials. 4 Conclusion What started as a simple question - do two polynomials share a root? - has led to some interesting results. The resultant has several different forms, each of which enhance our understanding of its properties and how it can be used. Considering the resultant as the determinant of the Sylvester matrix gives us a method for computing it. This method, along with Theorem 2.7, gives us some information about shared factors of two polynomials, without having to compute the greatest common factor using the Euclidean algorithm. This is useful if we are only interested in some properties of the shared factors of two polynomials, but do not actually need to know the greatest common factor explicitly. The form of the resultant given in Definition 2.1, or the equivalent definitions given below it, lends itself to a more insightful conceptual under-standing of what the resultant is, and why it is equal to zero if the two polynomials have a common factor. The applications of the resultant are both interesting and far-reaching. The re-sultant of a polynomial and its derivative is a multiple of the determinant of that polynomial. The determinant gives information about the nature of one polynomial’s roots rather than about the roots of two polynomials together. The second applica-tion covered in this paper, variable elimination, shows the power of the resultant. The resultant can be employed to reshape complicated polynomial systems in multiple vari-ables into univariate polynomials, which are significantly easier to analyze and solve for roots. In this application, the resultant represents a common theme in mathemat-ics - taking complicated objects and ideas and translating them into versions that are simpler and better understood. 9 Henry Woody REFERENCES REFERENCES References Cox, David, John Little, and Donal O’Shea. Ideals, Varieties, and Algorithms. 3rd ed. New York: Springer, 2007. Print. ISBN-10: 0-387-35650-9 Judson, Thomas. Abstract Algebra: Theory and Applications. 2015 ed. Ann Arbor, MI: Orthogonal Publishing, 2015. Print. ISBN 978-0-9898975-9-4 Kalorkoti, K. ”On Macaulay’s Form of the Resultant,” School of Informatics, Uni-versity of Edinburgh. Web. 25 April 2016. Parker, W. V. ”The Degree of the Highest Common Factor of Two Polynomials,” The American Mathematical Monthly 42(3) (1935): 164-166. Web. 12 April 2016. Sylvester, J.J. ”On a remarkable discovery in the Theory of Canonical Forms and of Hyperdeterminants,” The London, Edinburgh and Dublin Philosophical Magazine and Journal of Science 4 (1851): 391-410. Print. Van Der Waerden, B. L. Modern Algebra. Vol. I. Trans. Fred Blum. New York: Frederick Ungar Publishing Co., 1953. Print. Van Der Waerden, B. L. Modern Algebra. Vol. II. Trans. Theodore J. Benac. New York: Frederick Ungar Publishing Co., 1950. Print. 10 Henry Woody
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AP® Calculus AB AP® Calculus BC Free-Response Questions and Solutions 1989 – 1997 Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. 1989 BC1 Let f be a function such that ′ ′ f (x) = 6x + 8. (a) Find f(x) if the graph of f is tangent to the line 3 . 2 at the point (0, 2) x y − = − (b) Find the average value of f(x) on the closed interval [ 1 . ,1] − Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC1 Solution (a) ( ) ( ) ( ) ( ) 2 3 2 3 2 3 8 0 3 3 4 3 2 4 3 f x x x C f C 2 f x x x x d f x x x x ′ = + + ′ = = = + + + = − = + + − d (b) ( ) ( ) 1 3 2 1 1 4 3 2 1 1 4 3 2 1 1 1 1 4 3 2 2 4 3 2 1 1 4 3 1 4 3 2 2 2 4 3 2 4 3 2 2 3 x x x dx x x x x − − + + − −−   = + + −           = + + − − − + +             = − ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC2 Let R be the region enclosed by the graph of y = x 2 x 2 +1, the line x = 1, and the x-axis. (a) Find the area of R . (b) Find the volume of the solid generated when R is rotated about the y-axis. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC2 Solution (a) 1 2 2 0 1 2 0 1 0 Area 1 1 1 1 arctan 1 4 x dx x dx x x x π = + = − + = − = − ⌠  ⌡ ⌠  ⌡ (b) ( ) 1 2 2 0 1 2 0 1 2 2 0 Volume 2 1 2 1 1 2 ln 2 2 1 ln 2 x x dx x x x dx x x x π π π π   =   +   = − +   = − +     = − ⌠  ⌡ ⌠  ⌡ 1 or ( ) ( ) 1/ 2 0 1/ 2 0 Volume 1 1 2 ln 1 1 ln 2 y dy y y y π π π   = −   −   = + − = − ⌠  ⌡ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC3 Consider the function f defined by f(x) = e x cos x with domain [0 . ,2 ] π (a) Find the absolute maximum and minimum values of f(x). (b) Find the intervals on which f is increasing. (c) Find the x-coordinate of each point of inflection of the graph of f . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC3 Solution (a) ( ) [ ] ( ) sin cos cos sin 5 0 when sin cos , , 4 4 x x x f x e x e x e x x f x x x x π π ′ = − + = − ′ = = = 0 4 5 4 2 x π π π ( ) / 4 5 /4 2 1 2 2 2 2 f x e e e π π π − Max: 2 5 2 ; Min: 2 π π − / 4 e e (b) 4 π 0 2π 5 4 π + − + ( ) f x ′ Increasing on 5 0, , ,2 4 4 π π π             ] (c) ( ) [ ] [ ( ) sin cos cos sin 2 sin 0 when 0, , 2 x x x f x e x x e x x e x f x x π π ′′ = − − + − = − ′′ = = Point of inflection at x π = Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC4 Consider the curve given by the parametric equations 3 2 3 2 3 and 12 x t t y t = − = − t (a) In terms of t , find dy dx . (b) Write an equation for the line tangent to the curve at the point where t = −1. (c) Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC4 Solution (a) ( )( ( ) ) 2 2 2 2 2 2 3 12 6 6 2 2 3 12 4 6 6 2 2 2 1 dy t dt dx t t dt t t dy t t dx t t t t t t = − = − + − − − = = = − − − (b) x y ( ) 5, 11 3 4 3 11 5 4 or 3 29 4 4 4 3 29 dy dx y x y x y x = − = = − − = − + = − + + = (c) ( ) ( ) ( ) ( ) ( ) , type 2 28,16 horizontal 0 0,0 vertical 1 1, 11 vertica 2 4, 16 horizontal t x y − − − − − l Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC5 At any time t ≥0 , the velocity of a particle traveling along the x-axis is given by the differential equation dx dt −10x = 60e 4 t . (a) Find the general solution x(t) for the position of the particle. (b) If the position of the particle at time t = 0 is x = −8, find the particular solution x(t) for the position of the particle. (c) Use the particular solution from part (b) to find the time at which the particle is at rest. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC5 Solution (a) Integrating Factor: 10 10 dt t e e − − ∫ = ( ) ( ) 10 4 10 10 6 4 1 60 10 10 t t t t t t t d xe e e dt 0 xe e C x t e C − − − − = = − + = − + e or ( ) ( ) 10 4 4 4 10 4 4 10 60 10 10 t h t p t t t t x t Ce x Ae Ae Ae e A x t Ce e = = − = = − = − 4t 2 (b) ( ) 10 4 8 10; 2 10 t t C C x t e e − = = − −= (c) 10 4 10 4 20 40 20 40 0 1 ln 2 6 t t t t dx e e dt e e t = − − = = or ( ) 4 10 4 10 4 10 10 2 60 0 100 20 60 1 ln 2 6 t t t t t dx e e e dt e e e t − − + = + − = = 4t Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC6 Let f be a function that is everywhere differentiable and that has the following properties. (i) f(x + h) = f (x) + f (h) f (−x) + f (−h) for all real numbers h and x . (ii) f(x) > 0 for all real numbers x . (iii) ′ f (0) = −1. (a) Find the value of f(0) . (b) Show that f(−x) = 1 f (x) for all real numbers x . (c) Using part (b), show that f(x + h) = f (x) f (h) for all real numbers h and x . (d) Use the definition of the derivative to find ( ) in terms of ( ) f x f ′ x . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC6 Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) Let 0 0 0 0 0 0 0 0 x h f f f f f f = = + = + = = + 1 (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Let 0 0 0 0 1 Use 0 1 and solve for h f x f f x f x f x f f f x f x = + + = = − + − = = − or Note that ( ) (0 ( 0) ( ) (0) ) f x f f x f x f − + −+ = + is the reciprocal of f(x). (c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 f x f h f x h f x f h f x f h f x f h f h f x f x f h + + = + + = + = (d) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 lim lim 1 lim 0 h h h f x h f x f x h f x f h f x h f h f x h f x f f x → → → + − ′ = − = − = ′ = = − Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1990 BC1 A particle starts at time t = 0 and moves along the x-axis so that its position at any time t ≥0 is given by x(t) 1) 3(2t −3). = (t − (a) Find the velocity of the particle at any time t ≥0. (b) For what values of t is the velocity of the particle less than zero? (c) Find the value of t when the particle is moving and the acceleration is zero. 1990 BC1 Solution (a) v t ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 3 1 2 3 2 1 1 8 11 x t t t t t t ′ = = − − + − = − − (b) ( ) ( ) ( ) 2 0 when 1 8 11 0 Therefore 8 11 0 and 1 11 or and 1 8 11 Since 0, answer is 0 , except 1 8 t t t t t t t t < − − < − < ≠ < ≠ ≥ ≤< v t t = (c) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 2 2 1 8 11 8 1 6 1 4 5 5 0 when 1, 4 5 but particle not moving at 1 so 4 a t v t t t t t t a t t t t t ′ = = − − + − = − − = = = = = 1990 BC2 Let R be the region in the xy-plane between the graphs of y = e x and y = e −x from . 0 to 2 x x = = (a) Find the volume of the solid generated when R is revolved about the x-axis. (b) Find the volume of the solid generated when R is revolved about the y-axis. 1990 BC2 Solution (a) ( ) 2 2 2 0 2 2 2 0 4 4 4 4 1 1 2 2 1 1 1 1 2 2 2 2 2 2 x x x x e dx e e e e e e π π π π − − − − = −   = +         = + − +           = + −   ∫ V e (b) V x ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 0 2 0 2 2 2 2 2 2 2 2 2 2 2 0 1 1 2 3 x x x x x x x x x x e e dx x e e e e dx x e e e e e e e e e e π π π π π − − − − − − − −   = −     = + − +       = + − −     = + − − − − −         = +   ∫ ∫ 1990 BC3 Let f(x) =12 −x 2 for x ≥0 and f (x) ≥0. (a) The line tangent to the graph of f at the point ( , intercepts the x-axis at ( )) k f k . What is the value of k ? x = 4 (b) An isosceles triangle whose base is the interval from (0 to has its vertex on ,0) ( ,0) c the graph of f . For what value of c does the triangle have maximum area? Justify your answer. 1990 BC3 Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 12 ; 2 slope of tangent line at , 2 line through 4,0 & , has slope 0 12 4 4 12 so 2 8 12 0 4 2 or 6 but 6 24 so 6 is not in the domain. 2 f x x f x x k f k k k f k f k k k k k k k k k k k f k ′ = − = − = − − − = − − − − = ⇒ − + = − = = = − = (b) 2 3 2 2 1 1 12 2 2 2 4 6 on 0,4 3 8 3 3 6 ; 6 0 when c 8 8 c c A c f c c c dA c c dc     = ⋅ = −           = −   = − − = = 4. Candidate test First derivative 0 0 4 16 4 3 0 c A Max − A′ + 0 4 4 3 second derivative 2 2 4 3 0 so 4 gives a relative max. c d A c dc = = −< = c = 4 is the only critical value in the domain interval, therefore maximum 1990 BC4 Let R be the region inside the graph of the polar curve r = 2 and outside the graph of the polar curve r = 2(1−sinθ). (a) Sketch the two polar curves in the xy-plane provided below and shade the region R . −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 y x (b) Find the area of R . 1990 BC4 Solution (a) −5 5 −5 5 y x R O 5 − 5 − 5 5 y x ( ) ( ) ( ) ( ) ( ) ( ) [ ] 2 2 0 2 0 0 0 0 0 1 2 2 1 sin 2 2 2sin sin 4 sin 1 cos2 1 4cos sin 2 2 4 1 4 1 0 8 A d d d d π π π π π π θ θ θ θ θ θ θ θ θ θ θ θ π π   = − −   = − = − −   = − − −     = − − + − −     = − ⌠ ⌡ ∫ ∫ ∫ (b) 1990 BC5 Let f be the function defined by f(x) = 1 x −1 . (a) Write the first four terms and the general term of the Taylor series expansion of f(x) about . x = 2 (b) Use the result from part (a) to find the first four terms and the general term of the series expansion about for ln x = 2 x −1 . (c) Use the series in part (b) to compute a number that differs from ln 3 2 by less than 0.05. Justify your answer. 1990 BC5 Solution (a) Taylor approach Geometric Approach ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 2 1 2 2 1 1 2 2 2 2 1 2; 1 2! 2 2 6 2 1 6; 3! f f f f f f − − − = ′ = − − = − ′′ ′′ = − = = ′′′ ′′′ = − − = − = −1 ( ) ( ) 2 3 1 1 1 1 2 1 1 where 2 n n x x u u u u u x = − + − = − + − + + − + = − " " ( ) ( ) ( ) ( ) ( ) 2 3 1 Therefore 1 2 2 2 1 2 1 n n x x x x x = − − + − − − + + − − + − " " (b) Antidifferentiates series in (a): ( ) ( ) ( ) ( ) ( ) 1 2 3 4 1 2 1 1 1 ln 1 2 2 2 2 3 4 1 0 ln 2 1 2 n n x x C x x x x n C + − − − = + − − + − − − + + + + = − ⇒ = − " " Note: If , “first 4 terms” need not include 0 C ≠ ( ) 4 1 2 4 x − − (c) 2 3 3 5 1 1 1 1 1 ln ln 1 2 2 2 2 2 3 2 1 1 1 2 8 24 1 1 1 1 since , 0.375 is sufficient. 24 20 2 8     = − = − + −         = − + − < − = " " Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. ( ) 1 1 1 1 1 1 , where 2 1 2 2 1 1 1 1 2 1 when 2 20 n n n n R C n C n n + + +   = <   +   − < + < ≥ 5 < Alternate Justification: 1990 BC6 Let f and g be continuous functions with the following properties. (i) ( ) ( ) where is a constant g x A f x A = − (ii) 2 3 1 2 ( ) ( ) f x dx g x dx = ∫ ∫ (iii) 3 2 ( ) 3 f x dx A = − ∫ (a) Find 3 1 ( ) f x dx ∫ in terms of A . (b) Find the average value of g(x) in terms of A , over the interval [1 . ,3] (c) Find the value of k if 1 0 ( 1) f x dx k + = A ∫ . 1990 BC6 Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 1 1 2 3 3 2 2 3 3 2 2 3 3 2 2 f x dx f x dx f x dx g x dx f x dx A f x dx f x dx A f x dx f x dx A = + = + = − + = − + = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ (b) ( ) ( ) ( ) ( ) [ ] 3 3 1 1 3 1 1 1 verage value 2 2 1 2 2 1 1 2 2 2 g x dx A f x dx A f x dx A A A = = −   = −     = − = ∫ ∫ ∫ A (c) ( ) ( ) ( ) 1 2 0 1 3 2 1 3 4 Therefore 4 kA f x dx f x dx g x dx A A k = + = = = + = = ∫ ∫ ∫ A 1990 BC5 Let f be the function defined by f(x) = 1 x −1 . (a) Write the first four terms and the general term of the Taylor series expansion of f(x) about . x = 2 (b) Use the result from part (a) to find the first four terms and the general term of the series expansion about for ln x = 2 x −1 . (c) Use the series in part (b) to compute a number that differs from ln 3 2 by less than 0.05. Justify your answer. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1990 BC5 Solution (a) Taylor approach Geometric Approach ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 2 1 2 2 1 1 2 2 2 2 1 2; 1 2! 2 2 6 2 1 6; 3! f f f f f f − − − = ′ = − − = − ′′ ′′ = − = = ′′′ ′′′ = − − = − = −1 ( ) ( ) 2 3 1 1 1 1 2 1 1 where 2 n n x x u u u u u x = − + − = − + − + + − + = − " " ( ) ( ) ( ) ( ) ( ) 2 3 1 Therefore 1 2 2 2 1 2 1 n n x x x x x = − − + − − − + + − − + − " " (b) Antidifferentiates series in (a): ( ) ( ) ( ) ( ) ( ) 1 2 3 4 1 2 1 1 1 ln 1 2 2 2 2 3 4 1 0 ln 2 1 2 n n x x C x x x x n C + − − − = + − − + − − − + + + + = − ⇒ = − " " Note: If , “first 4 terms” need not include 0 C ≠ ( ) 4 1 2 4 x − − (c) 2 3 3 5 1 1 1 1 1 ln ln 1 2 2 2 2 2 3 2 1 1 1 2 8 24 1 1 1 1 since , 0.375 is sufficient. 24 20 2 8     = − = − + −         = − + − < − = " " trance Examination Board. All rights reserved. ( ) 1 1 1 1 1 1 , where 2 1 2 2 1 1 1 1 2 1 when 2 20 n n n n R C n C n n + + +   = <   +   − < + < ≥ Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. 5 < Alternate Justification: Copyright © 2003 by College En Available at apcentral.collegeboard.com 1990 BC6 Let f and g be continuous functions with the following properties. (i) ( ) ( ) where is a constant g x A f x A = − (ii) 2 3 1 2 ( ) ( ) f x dx g x dx = ∫ ∫ (iii) 3 2 ( ) 3 f x dx A = − ∫ (a) Find 3 1 ( ) f x dx ∫ in terms of A . (b) Find the average value of g(x) in terms of A , over the interval [1 . ,3] (c) Find the value of k if 1 0 ( 1) f x dx k + = A ∫ . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1990 BC6 Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 1 1 2 3 3 2 2 3 3 2 2 3 3 2 2 f x dx f x dx f x dx g x dx f x dx A f x dx f x dx A f x dx f x dx A = + = + = − + = − + = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ (b) ( ) ( ) ( ) ( ) [ ] 3 3 1 1 3 1 1 1 Average value 2 2 1 2 2 1 1 2 2 2 g x dx A f x dx A f x dx A A A = = −   = −     = − = ∫ ∫ ∫ (c) k ( ) ( ) ( ) 1 2 0 1 3 2 1 3 4 Therefore 4 A f x dx f x dx g x dx A A k = + = = = + = = ∫ ∫ ∫ A Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC1 A particle moves on the x-axis so that its velocity at any time is given by . At , the particle is at the origin. t ≥0 v(t) = 12t 2 −36t +15 t =1 (a) Find the position x(t) of the particle at any time . t ≥0 (b) Find all values of t for which the particle is at rest. (c) Find the maximum velocity of the particle for 0 . ≤t ≤2 (d) Find the total distance traveled by the particle from t = 0 to t = 2. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC1 Solution (a) ( ) ( ) ( ) 3 2 3 2 4 18 15 0 1 4 18 15 Therefore 1 4 18 15 1 x t t t t C x C C x t t t t = − + + = = − + + = − = − + − (b) ( ) ( )( ) 2 2 36 3 2 1 2 5 0 1 5 , 2 2 v t t t t t t = = − + − − = = 0 1 15 (c) dv ( ) ( ) 24 36 3 0 when 2 0 15 3 12 2 2 9 Maximum velocity is 15 t dt dv t dt v v v = − = = =  = −     = − (d) Total distance ( ) ( ) ( ) ( ) ( ) 1/2 2 0 1/ 2 1 1 0 2 2 2 5 5 1 11 17 2 2 v t dt v t dt x x x x −        − − −                 −− −− − =        = = = ∫ ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC2 Let f be the function defined by for all real numbers x . f(x) = xe 1−x (a) Find each interval on which f is increasing. (b) Find the range of f . (c) Find the x-coordinate of each point of inflection of the graph of f . (d) Using the results found in parts (a), (b), and (c), sketch the graph of f in the xy-plane provided below. (Indicate all intercepts.) −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 y x Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC2 Solution (a) ( ) ( ) ( ) ( ] 1 1 1 1 increases on ,1 x x 1 x f x xe e x e f − − ′ = − + = − −∞ − −∞ (b) ( ) ( ) ( ] 1 1; lim Range: ,1 x f f x →−∞ = = −∞ (c) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 2 x x x f x e x e x e − − − ′′ = − + − − = − Point of inflection at x = 2. (d) −1 1 2 3 −2 −1 1 x y Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC3 Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y = sin x and y = cos x, as shown in the figure above. (a) Find the area of R . (b) Find the volume of the solid generated when R is revolved about the x-axis. (c) Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC3 Solution (a) ( ) ( ) /4 0 / 4 0 Area cos sin sin cos 2 2 0 1 2 2 2 1 x x dx x x π π = − = +   = + − +       = − ∫ (b) ( ) / 4 2 2 0 / 4 0 /4 0 cos sin cos2 sin 2 2 1 0 2 2 x dx x π π π π π π π π = − = = = − = ∫ ∫ V x xdx ( ) ( ) ( ) / 4 2 0 / 4 0 /4 2 0 (c) cos sin 1 2sin cos sin 1 0 0 4 2 1 4 2 V x x x x dx x x π π π π π = − = − = − = − − − = − ∫ ∫ dx Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC4 Let F(x) = t 2 + t dt 1 2x . (a) Find F ′ (x) . (b) Find the domain of F . (c) Find lim x→1 2 F(x) . (d) Find the length of the curve y = F(x) for 1 ≤x ≤2 . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC4 Solution (a) ( ) 2 2 4 2 F x x x ′ = + (b) t t ( ) ( 2 0; therefore 0 or 1 Since 1 0, want 2 0 Therefore 0 t t x x + ≥ ≥ ≤− ≥ ≥ ≥ ) (c) ( ) 1 2 1 m 0 2 x F x F →   = =   li   (d) ( ) ( ) 2 2 1 2 2 1 2 1 2 2 1 1 1 16 8 4 1 2 7 L F x dx x x dx x dx x x ′ = + = + + = + = + = ⌠ ⌡ ∫ ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC5 Let f be the function given by f(t) = 4 1 + t2 and G be the function given by G(x) = f (t) dt 0 x . (a) Find the first four nonzero terms and the general term for the power series expansion of f(t) about t = 0. (b) Find the first four nonzero terms and the general term for the power series expansion of G(x) about x = 0. (c) Find the interval of convergence of the power series in part (b). (Your solution must include an analysis that justifies your answer.) Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC5 Solution (a) ( ) ( ) ( ) 2 2 2 4 6 2 4 , geometric with 4, 1 4 4 4 4 1 4 n n f t a t f t t t t t = = + = − + − + + − + " " r t = − (b) ( ) ( ) ( ) ( ) 2 4 6 2 0 0 2 1 3 5 7 0 2 1 3 5 7 4 4 4 4 4 1 1 4 4 4 4 4 3 5 7 2 1 1 4 4 4 4 4 3 5 7 2 1 x x x n n n n dt t t t dt t t t t t t n x x x x x n + + = = − + − + +   − = − + − + + +     +   − = − + − + + + + ⌠  ⌡ ∫ " " " " " G x (c) By Ratio Test, ( ) ( ) ( ) ( ) 1 2 3 2 2 1 2 2 2 1 4 2 1 2 1 2 3 2 3 1 4 2 1 lim ; 1 for 1 1 2 3 4 4 4 Check endpoints: 1 4 Converges by Alternating Series Test 3 5 7 4 4 1 4 Converges by Al 3 5 n n n n n x n n x n n x n x x x x n G G + + + →∞ − + + ⋅ = + + − + = < −< < + = − + − + − = −+ − + " " ternating Series Test Converges for 1 1 x −≤ ≤ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC6 A certain rumor spreads through a community at the rate dy dt = 2y(1−y) , where y is the proportion of the population that has heard the rumor at time t . (a) What proportion of the population has heard the rumor when it is spreading the fastest? (b) If at time t = 0 ten percent of the people have heard the rumor, find y as a function of t . (c) At what time t is the rumor spreading the fastest? Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1991 BC6 Solution (a) ( ) 2 2 1 2 2 is largest when 2 4 0 1 so proportion is 2 y y y y y y − = − − = = (b) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 1 2 1 1 1 2 1 ln ln 1 2 ln 2 1 1 1 0 0.1 9 9 t t t dy dt y y dy dt y y dy dt y y y y t y t C y y ke y y k e y e = − = − + = − − − = + = + − = − = ⇒ = = + ⌠  ⌡ ⌠  ⌡ ∫ ∫ C (c) 2 2 1 1 2 1 9 1 2 1 1 9 1 ln9 ln3 2 t t e e t = − = = = Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 AB4/BC1 Consider the curve defined by the equation y + cosy = x +1 for 0 ≤y ≤2π . (a) Find dy dx in terms of y . (b) Write an equation for each vertical tangent to the curve. (c) Find d 2y dx2 in terms of y . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 AB4/BC1 Solution (a) dy − = ( ) sin 1 1 sin 1 1 1 sin dy y dx dx dy y dx dy dx y − = = − (b) unde dy fined when sin 1 2 0 1 2 1 2 y dx y x x π π π = = + = + = − (c) d y ( ) ( ) ( ) 2 2 2 2 3 1 1 sin cos 1 sin 1 cos 1 sin 1 sin cos 1 sin d y dx dx dy y dx y y y y y y     −   =   −−     = −     −   = − = − Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 AB5/BC2 Let f be the function given by f(x) = e −x , and let g be the function given by g(x) = kx , where k is the nonzero constant such that the graph of f is tangent to the graph of g . (a) Find the x-coordinate of the point of tangency and the value of k . (b) Let R be the region enclosed by the y-axis and the graphs of f and g . Using the results found in part (a), determine the area of R . (c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated by revolving the region R , given in part (b), about the x-axis. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 AB5/BC2 Solution ( ) x f x e− = ( ) g x kx = x y (a) ( ) ( ) ; x x x f x e g x e k e kx − − − ′ ′ = − = − = = 1 and x k e = − = − k (b) ( ) ( ) ( ) ( ) 0 0 1 1 0 2 1 2 1 0 2 1 2 x x x e ex dx e ex dx ex e e e e − − − − − − −− = +   = − +       = −+ −−+     = − ∫ ∫ (c) ( π ⌠  ⌡ ( ) ( ) ) 0 2 2 1 x e ex − − −− dx Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC3 At time t , 0 ≤t ≤2π , the position of a particle moving along a path in the xy-plane is given by the parametric equations x = e t sint and y = e t cost. (a) Find the slope of the path of the particle at time 2 t π = . (b) Find the speed of the particle when t =1. (c) Find the distance traveled by the particle along the path from t = 0 to t =1. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC3 Solution (a) ( ) ( ) ( ) ( ) / 2 /2 sin cos cos sin cos sin / / sin cos 0 1 at , 1 2 1 0 t t t t t t e t e t dt dy e t e t dt e t t dy dy dt dx dx dt e t t e dy t dx e π π π = − − = = + − = = = + dx = + − ) (b) ( ) ( ( ) ( ) 2 2 2 2 d sin cos cos sin when 1 speed is sin1 cos1 cos1 sin1 2 t t t t e t e t e t e t t e e e e e + − = + + − = spee = + (c) ( ) ( ) ( ) ( ) 1 2 2 0 1 1 2 2 2 0 0 1 0 ance is sin cos cos sin 2 sin cos 2 2 2 1 t t t t t t t e t e t e t e t d e t t dt e dt e e + + − = + = = = − ⌠  ⌡ ∫ ∫ dist t Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC4 Let f be a function defined by f(x) = 2x −x 2 for x ≤1, x 2 + kx + p for x >1. (a) For what values of k and p will f be continuous and differentiable at x =1 ? (b) For the values of k and p found in part (a), on what interval or intervals is f increasing? (c) Using the values of k and p found in part (a), find all points of inflection of the graph of f . Support your conclusion. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC4 Solution (a) For continuity at x = 1, ( ) ( ) ( ) 2 2 1 1 lim 2 1 lim Therefore 1 1 x x x x f x kx k p − + → → − = = + + = + + p Since f is continuous at x = 1 and is piecewise polynomial, left and right derivatives exist. ( ) ( ) 1 0 and 1 2 f f k − + ′ ′ = = + 1 For differentiability at x = 1, 0 . 2 k = + Therefore 2, 2 k p = − = (b) ( ) ( ) 2 2 , 1 2 2 0 1 2 2, 2 2 0 1 f x x x x x f x x x x x ′ = − ≤ − > < ′ = − > − > > Since f increases on each of ( and and is continuous at x = 1, ) ) ,1 −∞ (1,∞ f is increasing on ( ). , −∞∞ (c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2, 1 2, 1 Since 0 on ,1 and 0 on 1, and 1 is defined, 1, 1 1,1 is a point of inflection. f x x f x x f x f x f f ′′ = − < ′′ = > ′′ < −∞ ′′ > ∞ = Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC5 The length of a solid cylindrical cord of elastic material is 32 inches. A circular cross section of the cord has radius 1 2 inch. (a) What is the volume, in cubic inches, of the cord? (b) The cord is stretched lengthwise at a constant rate of 18 inches per minute. Assuming that the cord maintains a cylindrical shape and a constant volume, at what rate is the radius of the cord changing one minute after the stretching begins? Indicate units of measure. (c) A force of 2x pounds is required to stretch the cord x inches beyond its natural length of 32 inches. How much work is done during the first minute of stretching described in part (b)? Indicate units of measure. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC5 Solution (a) 2 2 1 32 8 2 h π π   = = ⋅ =   V r π   ( ) ( ) 2 2 2 0 2 ; at 1, 50 and so 8 50, 2 so 5 2 2 Therefore 0 2 50 18 5 5 72 40 25 9 in/min 125 dV dr dh rh r dt dt dt t h r r dr dt dr dt dr dt π π π π π π π = = + = = = ⋅ =     = +           = +     = − (b) or ( ) 2 1 2 2 1 2 8 8 , so 1 8 8 Therefore 2 at 1, 50 so 1 8 8 18 2 50 2500 9 in/min 125 V r h r h dr dh dt h h dt t h dr dt π π − − = = = −      = ⋅ ⋅           = = −     = ⋅         = − (c) 18 18 2 2 0 0 Work 2 18 324 in-pounds 27 foot-pounds xdx x = = = = = ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC6 Consider the series 1 n p ln(n) n= 2 ∞ , where p ≥0. (a) Show that the series converges for p > 1. (b) Determine whether the series converges or diverges for p = 1. Show your analysis. (c) Show that the series diverges for 0 ≤p < 1. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1992 BC6 Solution (a) ( ) ( ) 1 1 for ln 1, for 3 ln p p n n n n n < < > 0 ≥ by p-series test, 1 p n ∑ converges if p > 1 and by direct comparison, ( ) 2 1 ln p n n n ∞ = ∑ converges. (b) ( ) ( ) 2 2 2 1 et , so series is ln 1 limln ln lim[ln(ln( )) ln(ln 2)] ln n b b b f x f n x x dx x b x x ∞ = ∞ →∞ →∞ = = = − ⌠  ⌡ ∑ L = ∞ Since f (x) monotonically decreases to 0, the integral test shows 2 1 ln n n n ∞ = ∑ diverges. (c) 1 1 0 for 1 ln ln p p n n n n > > < , so by direct comparison, 2 1 ln p n n n ∞ = ∑ diverges for 0 1 p ≤ < Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 AB3/BC1 Consider the curve and chord AB joining the points on the curve. y 2 = 4 + x ( 4,0) and (0,2) A B − (a) Find the x- and y-coordinates of the point on the curve where the tangent line is parallel to chord AB . (b) Find the area of the region R enclosed by the curve and the chord AB . (c) Find the volume of the solid generated when the region R , defined in part (b), is revolved about the x-axis. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 AB3/BC1 Solution (a) 2 1 ope of 4 2 AB = = sl ( ) B 0, 2 ( ) A 4, 0 − Method 1: 1 1 4 ; ; 2 2 4 2 4 so 3, 1 dy y x dx x x x y = + = = + + = − = 1 Method 2: 1 2 1; so 2 1 and 1, 3 2 dy y y y x dx   = = =   = −   (b) Method 1: ( ) ( ) ( ) 0 0 3/ 2 2 4 4 3/ 2 1 2 1 4 2 4 2 3 4 2 16 4 4 4 8 4 3 3 3 2 x x dx x x x − −     + − + = + − −         = −−+ = − = ⌠  ⌡ Method 2: ( ) ( ) 2 3 2 2 2 0 0 2 4 4 3 8 4 4 3 3 y y y dy y   − − − = −   = − = ∫ Method 3: 0 4 16 4 ; Area of triangle 4 3 16 4 Area of region 4 3 3 x dx − + = = = − = ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 AB3/BC1 Solution, continued (c) Method 1: ( ) 0 2 2 4 0 2 4 1 4 2 2 1 4 2 4 16 8 8 8.378 3 3 4 x x dx x x x d π π π π − −     + − +               = + − + +           = − = ≈     ⌠   ⌡ ⌠  ⌡ x Method 2: ( ) ( ) 2 2 0 8 2 2 4 4 3 y y y dy π π   − − − =   ∫ Method 3: ( ) ( ) ( ) ( ) 0 2 0 2 4 4 2 4 4 2 0 16 8 8 1 1 Volume of cone 2 4 3 3 16 8 Volume 8 3 3 x x dx x π π π π π π π π π − −   + = +     = − − + = = = = − = ⌠ ⌡ 6 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 AB4/BC3 Let f be the function defined by f(x) = ln(2 + sin x) for π ≤x ≤2π. (a) Find the absolute maximum value and the absolute minimum value of f . Show the analysis that leads to your conclusion. (b) Find the x-coordinate of each inflection point on the graph of f . Justify your answer. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 AB4/BC3 Solution (a) ( ) [ ] 1 cos ; 2 sin 3 In , 2 , cos 0 when ; 2 f x x x x x π π π ′ = + = = ( ) ( ) ( ) ln 2 0.693 ln 2 ln 1 0 = = 2 3 2 π π π ( ) f x x absolute maximum value is ln 2 absolute minimum value is 0 (b) ( ) ( )( ) ( ) ( ) ( ) 2 2 sin 2 sin cos cos 2 sin 2sin 1 ; 2 sin 1 0 when sin 2 7 11 , 6 6 x x x f x x x x f x x x π π − + − ′′ = + − − = + ′′ = = − = x up sign of f ′′ concavity − − + π down down 7 6 π 2 11 6 π π 7 1 and 6 6 x π = 1π since concavity changes as indicated at these points Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC2 The position of a particle at any time is given by t ≥0 x(t) = t 2 −3 and y(t) = 2 3 t 3. (a) Find the magnitude of the velocity vector at . t = 5 (b) Find the total distance traveled by the particle from t = 0 to t = 5 . (c) Find dy dx as a function of x . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC 2 Solution (a) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 5 10 5 50 5 10 50 2600 10 26 50.990 x t t y t t x y v ′ ′ = = ′ ′ = = = + = = ≈ (b) ( ) ( ) 2 4 0 5 2 0 5 3/ 2 2 0 3/ 2 2 1 2 1 3 2 26 1 87.716 3 t t dt t t dt t + = + = + = − ≈ ∫ 5 4 4 ∫ (c) dy = ( ) ( ) 2 2 2 2 2 3; 3 3 3 y t t t dx x t t x t t x t x dy x dx ′ = = ′ = − = + = + = + or ( ) 2 3/ 2 3 3; 3 2 2 ; 3 3 3 3 x t t x y t y x dy x dx = − = + = = + = + Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC4 Consider the polar curve r = 2sin(3θ) for 0 ≤θ ≤π . (a) In the xy-plane provided below, sketch the curve. O −2 −1 1 2 −2 −1 1 2 y x (b) Find the area of the region inside the curve. (c) Find the slope of the curve at the point where θ = π 4 . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC4 Solution 003 by College Entrance Examination Board. All rights re apcentral.collegeboard.com ( ) ( ) 2 2 2 2 3 2 2 2 2 6 2 2 2 2 6 12 6 At , 1 and 1 so 4 4 2 2 6 12 6 1 8 8 12 2 x y x y y dy dy dy x y x y x xy y dx dx dx x y dy dy dy dx dx dx dy dy dx dx π θ + = −   + + = + −     = = =   + = + −     + = ⇒ = served. Available at (a) (b) ( ) 2 0 0 0 /3 2 0 /6 2 0 1 1 4sin 3 1 cos6 sin 6 2 6 3 or 4sin 3 2 6 or 4sin 3 2 A d d d d π π π π π θ θ θ θ θ θ π θ θ π θ θ π   = = − = −     = = = = ∫ ∫ ∫ ∫ " " = −2 −1 1 2 −2 −1 1 2 y x O (c) = 2sin3 cos 2sin3 sin 2sin3 sin 6cos3 cos 2sin3 cos 6cos3 sin At , 2 and 4, so 4 2 1 4 2 x y dx d dy d dy dx d d dy θ θ θ θ θ θ θ θ θ θ θ θ θ θ π θ θ θ = = − + = + = = − = − − = = − dx or Copyright © 2 1993 BC5 Let f be the function given by f(x) = e x 2 . (a) Write the first four nonzero terms and the general term for the Taylor series expansion of f(x) about x = 0. (b) Use the result from part (a) to write the first three nonzero terms and the general term of the series expansion about x = 0 for g(x) = e x 2 −1 x . (c) For the function g in part (b), find and use it to show that ′ g (2) n 4(n +1)! = 1 4 n=1 ∞ . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC 5 Solution (a) ( ) ( ) ( ) 2 3 2 3 / 2 2 3 2 3 1 2! 3! ! / 2 / 2 / 2 1 2 2! 3! ! 1 2 2 2! 2 3! 2 ! n x n x n n x x x n x x x x e n x x x x n = + + + + + + = + + + + + + = + + + + + + " " " " " " e x (b) e 2 3 / 2 2 3 2 1 2 3 1 2 2 2! 2 3! 2 ! 1 2 2 2! 2 3! 2 ! n x n n n x x x x n x x x x x n − + + + + + −= = + + + + + " " " " (c) g x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 3 2 2 2 3 1 / 2 / 2 /2 2 1 1 2 2 2! 2 3! 2 ! 1 1 8 24 2 ! 1 2 1 2 2 2 2 2! 2 3! 2 ! 1 1 1 8 12 4 ! 4 1 ! 1 Also, 1 1 1 2 1 2 1 1 2 2 4 4 n n n n n n n x x x n x x n n x x n n g n n n n n e g x x x e e g x x e e g − − − ∞ = − ′ = + + + + − = + + + + − ⋅ ′ = + + + + − = + + + + = + − = ⋅ − − ′ = ⋅ − − ′ = = ∑ " " " " " " " " Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC6 Let f be a function that is differentiable throughout its domain and that has the following properties. (i) f(x + y) = f (x) + f (y) 1−f (x) f (y) for all real numbers , , and x y x + y in the domain of f (ii) lim h→0 f (h) = 0 (iii) lim h→0 f (h) h = 1 (a) Show that f(0) . = 0 (b) Use the definition of the derivative to show that [ 2 ( ) 1 ( )] f x f x ′ = + . Indicate clearly where properties (i), (ii), and (iii) are used. (c) Find f(x) by solving the differential equation in part (b). Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC6 Solution (a) Method 1: f is continuous at 0, so ( ) ( ) 0 0 lim x f f x → = = 0 or Method 2: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 0 0 0 0 0 1 0 0 0 1 0 2 0 0 1 0 0 0 0 f f f f f f f f f f f f + = + = − − =     −− =     = (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 2 0 2 2 lim 1 lim [By (i)] 1 lim 1 1 1 [By (iii) & (ii)] 1 0 1 h h h f x h f x f x h f x f h f x f x f h h f x f h h f x f h f x f x f x → → → + − ′ = + − − =   +       = ⋅−     +     = ⋅ − ⋅ = +     (c) Method 1: Le ( ) ( ) ( ) [ ] ( ) ( ) ( ) 2 2 1 t ; 1 1 tan tan 0 0 0 or , tan or tan dy y f x y dx dy dx y y x C y x C f C C n n f x x f x x n π π − = = + = + = + = + = ⇒ = = ∈ = =     Z + Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1993 BC6 Solution, continued or ( ) ( ) ( ) ( ) ( ) 2 2 2 that tan 1 1 tan sec 0 tan 0 0 f x x f x x x f = ′ + = + = =     = = f x Method 2: Guess Since the solution to the D.E. is unique ( ) tan f x = x is the solution. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 AB 2-BC 1 Let R be the region enclosed by the graphs of , and the lines x = 0 and x = 4. , x y e y x = = (a) Find the area of R . (b) Find the volume of the solid generated when R is revolved about the x-axis. (c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 AB 2- BC 1 4 R x y (a) ( ) 4 0 4 2 0 4 0 4 Area 2 16 0 2 9 45.598 x x e x dx x e e e e = − = −   = − − −     = − ≈ ∫ OR Using geometry (area of triangle) 4 0 1 4 4 2 x e dx − ⋅⋅ ∫ or Using y-axis 4 1 4 0 1 4 ln 4 ln e y dy y y dy y dy + − + − ∫ ∫ ∫ ( ) ( ) 4 2 2 0 4 2 2 0 4 3 2 0 8 0 8 (b) 1 2 3 1 64 1 0 2 3 2 1 131 2 6 1468.646 4613.886 x x x x V e x dx e x dx x e e e e π π π π π π = − = −   = −          = − − −             = −     ≈ ≈ ∫ ∫    Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 AB 2- BC 1 (continued) [or] Using geometry (Volume of the cone) ( ) 4 2 2 0 4 2 0 8 1 4 4 3 1 64 2 3 1 1 64 2 2 3 x x e dx e e π π π π π π − ⋅ ⋅   ⋅       = − −   = −   ∫ Using y- axis ( ) ( ) 4 1 4 0 1 4 2 ln 4 e y y dy y y y dy y y dy π   ⋅ + − + −     ∫ ∫ ∫ ln dx ( ) 4 0 (c) 2 x y V x e x π = − ∫ or ( ) ( ) 4 1 4 2 2 2 2 0 1 4 ln 16 ln e y V y dy y y dy y dy π   = + − + −     ∫ ∫ ∫ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 AB 5-BC 2 A circle is inscribed in a square as shown in the figure above. The circumference of the circle is increasing at a constant rate of 6 inches per second. As the circle expands, the square expands to maintain the condition of tangency. (Note: A circle with radius r has circumference and area ) 2 C = πr 2 A r = π (a) Find the rate at which the perimeter of the square is increasing. Indicate units of measure. (b) At the instant when the area of the circle is square inches, find the rate of increase in the 25π area enclosed between the circle and the square. Indicate units of measure. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 AB 5-BC 2 (a) 8 8 6 2 3 24 ; inches/second 7.639 inches/second dP dR dt dt dC dR dt dt dR dP dt dt π π π = = = = = = ≈ P R R (b) Ar ( ) ( ) ( ) ( ) 2 2 2 2 2 2 ea 4 Area 8 2 4 2 Area of circle 25 5 Area 120 30 inches /second 30 4 inches /second 8.197 inches /second R R d dR dR R R dt dt dt dR R dt R R d dt π π π π π π π π = − = − = − = = = = − = − ≈ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 3 A particle moves along the graph of y = cos x so that the x-coordinate of acceleration is always 2. At time t = 0, the particle is at the point π ,−1 ( and the velocity vector of the particle is ( ). ) 0,0 (a) Find the x- and y-coordinates of the position of the particle in terms of t. (b) Find the speed of the particle when its position is ( ). 4,cos4 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 3 (a) x t ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 0 0 0; 2 , 0 cos x t t C x C x t t x t t k x k x t t y t t π π π ′′ ′ = ⇒ = + ′ ′ = ⇒ = = = + = = = + = + (b) dy = − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 sin 2 2 sin 4 4 sin t t dt dx dy s t dt dt t t t t t t π π π +     = +         = + − + = + + ( ) ( ) 2 2 2 when 4, 4; 4 4 4 4 4 sin 4 2.324 x t t s π π π π = + = = − = − + − ≈ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 4 Let . For 0 f(x) = 6 −x 2 < w < 6 , let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point w,6 ( −w 2). (a) Find A(1) . (b) For what value of w is A(w) a minimum? Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 4 (a) ( ) ( ) ( ) ( ) ( ) 2 6 ; 2 1 2 5 2( 1) or 2 7 int : int :7 2 1 7 49 1 7 2 2 4 f x x f x x f y x y x x y A ′ = − = − ′ = − − = − − = − +   = =   7   (b) ( ) ( ) 2 2 2 ( ) 2 ; 6 2 6 int: int: 6 2 f w w y w w x w x y w w = − − − = − − + + w ′ ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 6 4 4 2 6 2 4 6 16 0 when 6 3 6 0 2 w w w w w w A w w A w w w w + = + − + ′ = ′ = + − = = A w A′ w 6 2 0 − + Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 5 Let f be the function given by . f(x) = e −2x 2 (a) Find the first four nonzero terms and the general term of the power series for f(x) about x = 0. (b) Find the interval of convergence of the power series for f(x) about x = 0. Show the analysis that leads to your conclusion. (c) Let g be the function given by the sum of the first four nonzero terms of the power series for f(x) about x = 0. Show that f (x) −g(x) < 0.02 for −0.6 ≤x ≤0.6 . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 5 (a) ( ) 2 2 3 2 4 6 2 2 1 2 3! ! 1 2 4 8 1 2 2 3! ! n u n n n x u u u e u n x x x e x n − = + + + + + + − = − + − + + + " " " " (b) The series for converges for u e u −∞< < ∞ So the series for converges for −∞ 2 2x e− 2 2x < − < ∞ And, thus, for x −∞< < ∞ Or ( ) ( ) ( ) ( ) 1 2 1 1 1 2 2 1 2 ! lim lim 1 ! 1 2 2 lim 0 1 1 n n n n n n n n n n n x a n a n x x n + + + + →∞ →∞ →∞ − = ⋅ + − = = < + So the series for converges for 2 2x e− x −∞ < < ∞ (c) ( ) ( ) 8 16 16 32 4! 5! x x f x g x − = − +" This is an alternative series for each x, since the powers of x are even. Also, 2 1 2 1 for 0.6 0.6 1 n n a x a n + = < − ≤ ≤ + x so terms are decreasing in absolute value. Thus f x ( ) ( ) ( ) 8 8 16 0.6 16 4! 4! 0.011 0.02 x g x − ≤ ≤ = < " Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 6 Let f and g be functions that are differentiable for all real numbers x and that have the following properties. (i) ′ f (x) = f (x) −g(x) (ii) ′ g (x) = g(x) −f (x) (iii) f(0) = 5 (iv) g(0) =1 (a) Prove that f(x) + g(x) = 6 for all x . (b) Find f(x) and g(x). Show your work. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1994 BC 6 (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 so is constant. 0 0 6, so 6 f x g x f x g x g x f x f g f g f x g x ′ ′ + = − + − = + + = + = (b) f x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 6 so 6 2 2 6; 2 6 1 ln 2 6 2 ln 2 6 2 2 6 2 6 0 1 so 4 2 4 6 3 2 6 3 2 x K x x x x g x g x g x g x g x dy dy y dx dx y y x C y x K y e y Ae x y A y e y e g x f x g x e + = − ′ = − + = − = − = − − = + − = + − = − = = ⇒ = − = = − + = − = = − = + 6 Or Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 AB4/BC2 Note: Figure not drawn to scale. The shaded regions 1 2 and R R shown above are enclosed by the graphs of 2 ( ) f x x and ( ) 2x g x . (a) Find the x- and y-coordinates of the three points of intersection of the graphs of f and g . (b) Without using absolute value, set up an expression involving one or more integrals that gives the total area enclosed by the graphs of f and g . Do not evaluate. (c) Without using absolute value, set up an expression involving one or more integrals that gives the volume of the solid generated by revolving the region 1 R about the line 5 y . Do not evaluate. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 AB4/BC2 Solution (a) 2,4 4,16 0.767,0.588  or (–0.766, 0.588) (b) 2 2 2 4 0.767 2 2 2 x x x dx x dx     ³ ³ or 0.588 4 16 0 0.588 4 ln ln ln 2 ln 2 2 y y y dy y dy y dy     ³ ³ ³ (c) 2 2 2 2 0.767 5 5 2x x dx S     ³ or 0.588 4 0 0.588 ln ln 2 2 5 2 2 5 y y y y dy y dy S S § ·  ¨ ¸ © ¹    ³ ³ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 AB5/BC3 As shown in the figure above, water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area 400S square feet. The depth h , in feet, of the water in the conical tank is changing at the rate of ( 12) h  feet per minute. (The volume V of a cone with radius r and height h is 2 1 3 V r h S .) (a) Write an expression for the volume of water in the conical tank as a function of h . (b) At what rate is the volume of water in the conical tank changing when 3 h ? Indicate units of measure. (c) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when 3 h ? Indicate units of measure. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 AB5/BC3 Solution (a) 2 3 4 1 1 12 3 3 1 1 3 3 27 r r h h h V h h S S § · ¨ ¸ © ¹ (b) 2 2 9 12 9 9 dV h dh dt dt h h S S S   V is decreasing at 3 9 ft / min S (c) Let W = volume of water in cylindrical tank 400 ; 400 400 9 dW dy W y dt dt dy dt S S S S y is increasing at 9 ft/min 400 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC1 Two particles move in the xy-plane. For time 0 t t , the position of particle A is given by 2 x t  and 2 ( 2) y t  , and the position of particle B is given by 3 4 2 t x  and 3 2 2 t y  . (a) Find the velocity vector for each particle at time 3 t . (b) Set up an integral expression that gives the distance traveled by particle A from t = 0 to t = 3. Do not evaluate. (c) Determine the exact time at which the particles collide; that is, when the particles are at the same point at the same time. Justify your answer. (d) In the viewing window provided below, sketch the paths of particles A and B from 0 t until they collide. Indicate the direction of each particle along its path. Viewing Window 5 5  7 7  [ 7,7] [ 5,5]  u  Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC1 Solution (a) 1,2 4 ; 3 1,2 3 3 3 3 , ; 3 , 2 2 2 2 A A B B V t V V V  § · § · ¨ ¸ ¨ ¸ © ¹ © ¹ (b) 3 2 2 0 distance 1 2 4 t dt   ³ (c) Set 3 2 4; 4 2 t t t   When 4 t , the y-coordinates for A and B are also equal. Particles collide at (2,4) when 4 t . (d) 7 5 5  7  Viewing Window [ 7,7] [ 5,5]  u  Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC4 Let f be a function that has derivatives of all orders for all real numbers. Assume (1) 3, (1) 2, (1) 2 f f f c cc  , and (1) 4 f ccc . (a) Write the second-degree Taylor polynomial for f about 1 x and use it to approximate (0.7) f . (b) Write the third-degree Taylor polynomial for f about 1 x and use it to approximate (1.2) f . (c) Write the second-degree Taylor polynomial for f c , the derivative of f , about 1 x and use it to approximate (1.2) f c . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC4 Solution (a) 2 2 2 3 2 1 1 2 0.7 3 0.6 0.09 3.69 T x x x f      |   (b) 2 3 3 4 3 2 1 1 1 6 2 1.2 3 0.4 0.04 0.008 2.645 3 T x x x x f       |    (c) 2 3 2 2 1 2 1 1.2 2 0.4 0.08 1.52 T x x x f c      c |     Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC5 Let 2 2 ( ) , ( ) cos , and ( ) cos f x x g x x h x x x  . From the graphs of f and g shown above in Figure 1 and Figure 2, one might think the graph of h should look like the graph in Figure 3. (a) Sketch the actual graph of h in the viewing window provided below. 40 6  6 6  Viewing Window [ 6, 6] x [ 6, 40] (b) Use ( ) h x cc to explain why the graph of h does not look like the graph in Figure 3. (c) Prove that the graph of 2 cos( ) y x kx  has either no points of inflection or infinitely many points of inflection, depending on the value of the constant k . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC5 Solution (a) 40 6  6  6 (b) 2 sin ; 2 cos h x x x h x x c cc   2 cos 0 x  ! for all x , so graph must be concave up everywhere (c) 2 2 cos y x k kx cc  If 2 2, 0 k ycc d t for all x , so no inflection points. If 2 2, k ycc ! changes sign and is periodic, so changes sign infinitely many times. Hence there are infinitely many inflection points. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC6 Graph of f Let f be a function whose domain is the closed interval > @ 0,5 . The graph of f is shown above. Let 3 2 0 ( ) ( ) x h x f t dt  ³ . (a) Find the domain of h . (b) Find (2) hc . (c) At what x is ( ) h x a minimum? Show the analysis that leads to your conclusion. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1995 BC6 Solution (a) 0 3 5 2 6 4 x x d  d  d d (b) 1 3 2 2 1 3 2 4 2 2 x h x f h f § · c  ˜ ¨ ¸ © ¹ c ˜  (c) hc is positive, then negative, so minimum is at an endpoint 0 0 5 0 6 0 4 0 h f t dt h f t dt  ³ ³ since the area below the axis is greater than the area above the axis therefore minimum at 4 x Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 AB3/BC3 The rate of consumption of cola in the United States is given by ( ) kt S t Ce , where S is measured in billions of gallons per year and t is measured in years from the beginning of 1980. (a) The consumption rate doubles every 5 years and the consumption rate at the beginning of 1980 was 6 billion gallons per year. Find C and k . (b) Find the average rate of consumption of cola over the 10-year time period beginning January 1, 1983. Indicate units of measure. (c) Use the trapezoidal rule with four equal subdivisions to estimate 7 5 ( ) S t dt ³ . (d) Using correct units, explain the meaning of 7 5 ( ) S t dt ³ in terms of cola consumption. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 AB3/BC3 Solution (a) 5 5 0 6 6 5 12 12 6 2 ln 2 0.138 or 0.139 5 kt k k S t Ce S C S e e k Ÿ Ÿ (b) ln 2 13 5 3 2.6ln2 0.6ln 2 1 13 3 3 ln 2 ( ) Average rate 6 billion gal/yr (19.680 billion gal/yr) t e dt e e  ª º  ¬ ¼ ³ (c) 7 5 1 5 2 5.5 2 6 2 6.5 7 4 S t dt S S S S S     ª º ¬ ¼ ³  (d) This gives the total consumption, in billions of gallons, during the years 1985 and 1986. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 AB4/BC4 This problem deals with functions defined by ( ) sin f x x b x  , where b is a positive constant and 2 2 x  S d d S . (a) Sketch the graphs of two of these functions, sin y x x  and 3sin y x x  . y x 6  6  6 6 y x 6  6  6 6 (b) Find the x-coordinates of all points, 2 2 x  S d d S , where the line y x b  is tangent to the graph of ( ) sin f x x b x  . (c) Are the points of tangency described in part (b) relative maximum points of f ? Why? (d) For all values of 0 b ! , show that all inflection points of the graph of f lie on the line y x . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 AB4/BC4 Solution (a)         6 6 6 6 6  6  6  6  y y x x (b) 1 1 cos cos 0 cos 0 y b x b x x c  sin sin 1 sin y x b x b x b b x x   3 or 2 2 x S S  (c) No, because 1 or 0 f x f x c c z at x-coordinates of points of tangency (d) sin 0 sin 0 0 f x b x x f x x b x cc   ˜ at x-coordinates of any inflection points Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC1 Consider the graph of the function h given by 2 ( ) x h x e for 0 x d f . (a) Let R be the unbounded region in the first quadrant below the graph of h . Find the volume of the solid generated when R is revolved about the y-axis. (b) Let ( ) A w be the area of the shaded rectangle shown in the figure above. Show that ( ) A w has its maximum value when w is the x-coordinate of the point of inflection of the graph of h . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC1 Solution (a) 2 0 Volume 2 x xe dx S  f ³ 2 2 2 0 0 0 2 lim 1 1 2 lim 2 lim 2 2 1 2 2 b x b b x b b b xe dx 1 2 e e e S S S S S  of   of of ª º § ·    ¨ ¸ « » ¬ ¼ © ¹ § · ¨ ¸ © ¹ ³ or 2 1 0 1 0 Volume ln lim ln a a y dy y dy S S S  o   ³ ³ (b) Maximum: 2 2 2 2 2 2 , 2 1 2 . w w w w A w we A w e w e e w     c   1 2 1 2 1 2 0 when , 0 when , 0 when . A w w A w w A w w c ! c c ! Therefore, max occurs when 1 2 w Inflection: 2 2 2 2 2 2 , 2 , 2 2 2 2 1 2 . x x x x x h x e h x xe h x e x x e e x      c  cc      1 2 1 2 1 2 0 when 0 when 0 when . , , h x x h x x h x x cc cc cc ! ! Therefore, inflection point when 1 2 . x Therefore, the maximum value of ( ) A w and the inflection point of ( ) h x occur when x and w are 1 2 . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC2 The Maclaurin series for ( ) f x is given by 2 3 2! 3! 4! ( 1)! n x x x x n 1        " " (a) Find (17) (0) and (0) f f . c (b) For what values of x does the given series converge? Show your reasoning. (c) Let ( ) ( ) g x x f x . Write the Maclaurin series for ( ) g x , showing the first three nonzero terms and the general term. (d) Write ( ) g x in terms of a familiar function without using series. Then, write ( ) f x in terms of the same familiar function. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC2 Solution (a) 1 17 17 0 1 ! 1 ! 1 0 2 1 1 0 17! 17! 18! 18 n n f a n n f a f a  c § · ¨ ¸ © ¹ (b) 1 2 ! lim lim 0 1 2 1 ! n n n n x n x n x n  of of    Converges for all x , by ratio test (c) 2 3 1 2! 3! 1 ! n g x xf x x x x x n        " " (d) 2 1 2! ! 1 1 if 0 1 if 0 n x x x x x e x n e g x xf x e x f x x x       ­  z ° ® ° ¯ " " Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC5 An oil storage tank has the shape as shown above, obtained by revolving the curve 4 9 625 y x from 0 to 5 x x about the y-axis, where x and y are measured in feet. Oil weighing 50 pounds per cubic foot flowed into an initially empty tank at a constant rate of 8 cubic feet per minute. When the depth of oil reached 6 feet, the flow stopped. (a) Let h be the depth, in feet, of oil in the tank. How fast was the depth of oil in the tank increasing when 4 h ? Indicate units of measure. (b) Find, to the nearest foot-pound, the amount of work required to empty the tank by pumping all of the oil back to the top of the tank. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC5 Solution (a) 0 25 3 25 3 25 at 4, 8 4 3 12 ft/min 25 h V y dy dV dh h dt dt dh h dt dh dt S S S S ´ µ ¶ (b) 6 0 6 1 3 2 2 0 6 3 5 2 2 0 25 50 9 3 25 50 9 3 25 2 2 50 9 3 3 5 69,257.691 ft-lbs W y y dy W y y dy W y W S S S § ·  ¨ ¸ © ¹ y § · § ·  ¨ ¸ ¨ ¸ © ¹ © ¹ § · § · ˜  ¨ ¸ ¨ ¸ © ¹© ¹ ´ µ ¶ ´ µ ¶ to the nearest foot-pound 69,258 ft-lbs Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC6 The figure above shows a spotlight shining on point ( , ) P x y on the shoreline of Crescent Island. The spotlight is located at the origin and is rotating. The portion of the shoreline on which the spotlight shines is in the shape of the parabola 2 y x from the point 1,1 to the point 5,25 . Let T be the angle between the beam of light and the positive x-axis. (a) For what values of T between 0 and 2S does the spotlight shine on the shoreline? (b) Find the x- and y-coordinates of point P in terms of tanT . (c) If the spotlight is rotating at the rate of one revolution per minute, how fast is the point P traveling along the shoreline at the instant it is at the point 3,9 ? Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1996 BC6 Solution (a) 1 1 1 2 2 1 1 tan or 0.785 1 4 25 tan tan 5 or 1.373 5 Therefore, tan 5 4 S T T T T S T   Ÿ Ÿ d d (b) 2 2 2 tan Therefore, tan tan y x x x x x y x T T T (c) 2 2 2 sec ; 2tan sec At 3,9 : 10 2 20 2 3 10 2 120 d dt dx d dy d dt dt dt dt dx dt dy dt T S T T T T S S S S ˜ ˜ ˜ ˜ T 2 2 Speed 20 120 20 37 or 382.191 S S S  Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 AB5/BC5 The graph of the function f consists of a semicircle and two line segments as shown above. Let g be the function given by 0 ( ) ( ) x g x f t dt ³ . (a) Find (3) g . (b) Find all the values of x on the open interval 2,5  at which g has a relative maximum. Justify your answer. (c) Write an equation for the line tangent to the graph of g at 3 x . (d) Find the x-coordinate of each point of inflection of the graph of g on the open interval 2,5  . Justify your answer. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 AB5/BC5 Solution (a) 3 0 2 3 ( ) 1 1 1 2 4 2 2 g f t dt S S ˜ ˜   ³ (b) ( ) has a relative maximum at 2 because ( ) changes from the positive to negative at 2 g x x g' x f x x (c) 1 (3) 2 3 (3) 1 1 1 3 2 g g' f y x S S   § ·     ¨ ¸ © ¹ (d) graph of g has points of inflection with x-coordinates x = 0 and x = 3 because ( ) changes from the positive to negative at 0 and from negative to positive at 3 or because changes from increasing to decreasing at 0 and from decreasing to increasing at 3 g x f x x x g x f x x x cc c c Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 AB6/BC6 Let ( ) v t be the velocity, in feet per second, of a skydiver at time t seconds, 0 t t . After her parachute opens, her velocity satisfies the differential equation 2 32 dv v dt   , with initial condition (0) 50 v  . (a) Use separation of variables to find an expression for v in terms of t , where t is measured in seconds. (b) Terminal velocity is defined as lim ( ) t v t of . Find the terminal velocity of the skydiver to the nearest foot per second. (c) It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed? Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 AB6/BC6 Solution (a) 2 32 2 16 dv v v dt     2 2 2 0 2 2 16 ln 16 2 16 16 50 16 ; 34 34 16 t A A t t t dv dt v v t A v e e e v Ce Ce C v e                  (b) 2 lim lim 34 16 16 t t t v t e of of    (c) 2 2 34 16 20 2 1 2 ; ln 1.070 17 2 17 t t v t e e t      § ·  ¨ ¸ © ¹ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC1 During the time period from 0 to 6 t t seconds, a particle moves along the path given by ( ) 3 cos( ) and ( ) 5 sin( ) x t t y t t S S . (a) Find the position of the particle when 2.5 t . (b) On the axes provided below, sketch the graph of the path of the particle from t = 0 to t = 6 . Indicate the direction of the particle along its path. (c) How many times does the particle pass through the point found in part (a)? (d) Find the velocity vector for the particle at any time t . (e) Write and evaluate an integral expression, in terms of sine and cosine, that gives the distance the particle travels from t = 1.25 to t = 1.75 . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC1 Solution (a) 2.5 3cos 2.5 0 2.5 5sin 2.5 5 x y S S (b)                     y x (c) 3 (d) ( ) 3 sin ( ) 5 cos 3 sin ,5 cos x t t y t t v t t t S S S S S S S S c c   J G (e) distance 1.75 2 2 2 2 1.25 9 sin 25 cos 5.392 t t dt S S S S ³  Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC2 Let 2 3 4 ( ) 7 3( 4) 5( 4) 2( 4) 6( 4) P x x x x x         be the fourth-degree Taylor polynomial for the function f about 4. Assume f has derivatives of all orders for all real numbers. (a) Find (4) and (4) f f ccc . (b) Write the second-degree Taylor polynomial for f c about 4 and use it to approximate (4.3) f c . (c) Write the fourth-degree Taylor polynomial for 4 ( ) ( ) x g x f t dt ³ about 4. (d) Can (3) f be determined from the information given? Justify your answer. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC2 Solution (a) 4 4 7 4 2, 4 12 3! f P f f ccc ccc   (b) 2 3 3 2 3 2 7 3 4 5 4 2 4 3 10 4 6 4 4.3 3 10 0.3 6 0.3 0.54 P x x x x P x x x f       c      c |     (c) 4 3 4 2 3 4 2 3 4 , ( ) 7 3 4 5 4 2 4 3 5 1 7 4 4 4 4 2 3 2 x x P g x P t dt t t t dt x x x x ª º       ¬ ¼        ³ ³ (d) No. The information given provides values for 4 4 , 4 , 4 , 4 and 4 f f only. f f f c cc ccc Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC3 Let R be the region enclosed by the graphs of 2 ln 1 y x  and cos y x . (a) Find the area of R . (b) Write an expression involving one or more integrals that gives the length of the boundary of the region R . Do not evaluate. (c) The base of a solid is the region R . Each cross section of the solid perpendicular to the x-axis is an equilateral triangle. Write an expression involving one or more integrals that gives the volume of the solid. Do not evaluate. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC3 Solution (a)       x y R 2 ln 1 cos 0.91586 let 0.91586 x x x B  r 2 area cos ln 1 1.168 B B x x  ª º   ¬ ¼ ³ dx (b) 2 2 2 2 1 1 1 B B B B x sin L dx x dx x   § ·     ¨ ¸  © ¹ ³ ³ (c) 3 base 2 base area of cross section 2 2 1 3 cos ln 1 cos ln 1 2 2 x x x x ª º ª º   u   « » ¬ ¼ ¬ ¼ volume 2 2 3 cos ln 1 4 B B x x dx  ª º   ¬ ¼ ´ µ ¶ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC4 Let 2 2 x ky  , where 0 k ! . (a) Show that for all 0 k ! , the point 2 4, k is on the graph of 2 2 x ky  . (b) Show that for all 0 k ! , the tangent line to the graph of 2 2 x ky  at the point 2 4, k passes through the origin. (c) Let R be the region in the first quadrant bounded by the x-axis, the graph of 2 2 x ky  , and the line 4 x . Write an integral expression for the area of the region R and show that this area decreases as k increases. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1997 BC4 Solution (a) 4 2 4 4 (b 2/ 1 2 2 y k dy dx k the tangent line is 2 1 4 2 2 1 which contains (0,0) 2 2 y x k k y x k ª « « «   « « « « ¬ or 2/ 1 slope of the line through 0,0 and 4, 2/ is 4 2 2 which is the same as the slope of the tangent line k k k ª « « « ¬ (c) 4 thus the area decreases as k increases 2 k k § · ¨ ¸ © ¹  2 ) 2 1 2 x ky dy ky dx  2/ 2 0 4 2 0.5 1.5 2 or 1 2 4 2 3 2 2 0 for all 0 3 k A ky dy A x dx k A k dA k k dk       ! ³ ³ R 2 4, k § · ¨ ¸ ¨ ¸ © ¹ 4 2 x y Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com
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https://mathematica.stackexchange.com/questions/278412/summation-over-a-restricted-domain
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Summation over a restricted domain Ask Question Asked Modified 2 years, 8 months ago Viewed 219 times 3 $\begingroup$ i would like to sum n=0 to n=(some value) over all integers except those that are generated by some function with integer inputs and outputs, like f[n]=(n^2+3n)/2 so it would sum when n=0, n=1, skip n=2, n=3, n=4, skip n=5 ... the best I've been able to do so far is make a range for both values, then delete my special cases from the first list, but I feel there is a better, more mathematical way to do this so i can start playing with limits and other fun stuff. summation Share Improve this question edited Jan 11, 2023 at 6:48 user64494 1 asked Jan 10, 2023 at 17:47 WomblesWombles 80244 silver badges88 bronze badges $\endgroup$ 1 1 $\begingroup$ Is the resulting list such that you can write a function that gives the $i$-th value? $\endgroup$ JimB – JimB 2023-01-10 18:40:56 +00:00 Commented Jan 10, 2023 at 18:40 Add a comment | 3 Answers 3 Reset to default 2 $\begingroup$ The comment that you want to sum a convergent function over this domain suggests a different approach. First, an example. Say you want to sum g[n_] := 1/n^2; over the domain of f[n] for all n. This can be done Sum[g[f[n]], {n, 1, Infinity}] If you want to sum over the complement of the f[n], then: f[n_] := (n^2 + 3 n)/2; g[n_] := 1/n^2; Sum[g[n], {n, 1, Infinity}] - Sum[g[f[n]], {n, 1, Infinity}] Your suggested convergent function 1/10^n can be summed over the complementary domain like this: f1[n_] := n^2/2; f2[n_] := 3/2 n; g[n_] := 1/10^n; Sum[g[n], {n, 1, Infinity}] - Sum[g[f1[n]], {n, 1, Infinity}] - Sum[g[f2[n]], {n, 1, Infinity}] Your answer is in terms of the EllipticTheta function. Share Improve this answer answered Jan 13, 2023 at 21:56 bill sbill s 70.1k44 gold badges105105 silver badges200200 bronze badges $\endgroup$ 1 $\begingroup$ Thank you, this is exactly what i was looking for! $\endgroup$ Wombles – Wombles 2023-01-17 17:53:28 +00:00 Commented Jan 17, 2023 at 17:53 Add a comment | 3 $\begingroup$ One approach is to define your function f f[n_] := (n^2 + 3 n)/2; Then n = 10; Complement[Range[n], f /@ Range[n]] gives the set of elements that you want to sum over. Hence, you can sum them using: Sum[i, {i, Complement[Range[n], f /@ Range[n]]}] Share Improve this answer answered Jan 10, 2023 at 20:38 bill sbill s 70.1k44 gold badges105105 silver badges200200 bronze badges $\endgroup$ 3 $\begingroup$ Is it possible to take the limit as this goes to infinity? $\endgroup$ Wombles – Wombles 2023-01-12 17:45:27 +00:00 Commented Jan 12, 2023 at 17:45 $\begingroup$ The limit of the sum would be infinite, no? $\endgroup$ bill s – bill s 2023-01-12 23:15:00 +00:00 Commented Jan 12, 2023 at 23:15 $\begingroup$ well, i want to sum a convergent series over this domain, something like 1/10^(n) $\endgroup$ Wombles – Wombles 2023-01-13 18:01:18 +00:00 Commented Jan 13, 2023 at 18:01 Add a comment | 3 $\begingroup$ Clear[cSum, k, n] f[n_] := (n^2 + 3 n)/2; cSum[k_] := Plus @@ Extract[Range[k], Position[(Solve[f[n] == #, n, PositiveIntegers] & /@ Range[k] ), {}]] cSum /@ Range $${1,1,4,8,8,14,21,29,29,39,50,62,75,75,90,106,123,141,160,160,181,203,226,250,275,301,301,329,358,388,419,451,484,518,518,554,591,629,668,708,749,791,834,834,879,925,972,1020,1069,1119,1170,1222,1275,1275,1330,1386,1443,1501,1560,1620,1681,1743,1806,1870,1870,1936,2003,2071,2140,2210,2281,2353,2426,2500,2575,2651,2651,2729,2808,2888,2969,3051,3134,3218,3303,3389,3476,3564,3653,3653,3744,3836,3929,4023,4118,4214,4311,4409,4508,4608}$$ Share Improve this answer answered Jan 10, 2023 at 21:46 SyedSyed 62.9k55 gold badges4747 silver badges103103 bronze badges $\endgroup$ Add a comment | Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions summation See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related Problems with Symbolic summation over unknown values Summation over integers satisfying some conditions 2 A restricted summation over a function of integer n-tuple 1 Perform an explicit summation, if possible, over a restricted range of two indices both varying from 0 to infinity Summation over an index and a set 2 Summation over specific tuples Double summation 1 Summation over elements 0 Mathematica ruins domain after explicit summation? 3 Double Summation over Lattice Hot Network Questions Origin of Australian slang exclamation "struth" meaning greatly surprised What can be said? How to use \zcref to get black text Equation? 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https://www.teacherspayteachers.com/browse/free?search=money%20menu%20worksheet
Money Menu Worksheet | TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Money Menu Worksheet 47+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Free Format All filters (1) Filters Free Clear all Grade Elementary 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Not grade specific Subject English language arts Grammar Informational text Reading Writing Health Math Algebra Applied math Arithmetic Basic operations Decimals Measurement Mental math Money math Numbers Other (math) Science Family consumer sciences Social emotional Character education Social emotional learning More Social studies U.S. history For all subjects Price Free Under $5 $5-10 $10 and up Format Google Apps Microsoft Microsoft Word PDF More Resource type Teacher tools Lessons Homeschool curricula Printables Hands-on activities Activities Centers Projects Bell ringers Instruction Handouts Student practice Independent work packet Worksheets Graphic organizers Homework More Standard Theme Seasonal Back to school Winter Holiday Thanksgiving More Audience Homeschool More Supports Special education Life skills Specialty Occupational therapy Vocational education More Adding money worksheet with a realistic food menu Created by Smart Fox Homeschool This is a worksheet that helps students add up money in a real life situation. For this worksheet, we have included a menu on one side, and questions on the other. This is a great worksheet to help students understand the importance of adding money! Ideal for 1st-2nd grade students. 1 st - 2 nd Arithmetic, Basic Operations, Math FREE Rated 5 out of 5, based on 2 reviews 5.0(2) Log in to Download Wish List Menu Math Real World Money Pet Store Addition Subtraction Multiplication FREEBIE Created by Heather Johnson 33 This bundle includes parts of ALL THREE of my addition, subtraction, & multiplication Pet Store math bundles. This freebie is part of a complete Pet Store Math Bundle. Click HERE to see the complete Pet Store Math Add, Subtract, Multiply: Money: Real-World Word Problem Solving Bundle. This is part of a great bundle that I use in my classroom (see thumbnail photos to see what is included in the FREEBIE)! The complete version includes a fun and colorful "Price Sheet" with a variety of differen 2 nd - 5 th Basic Operations, Decimals, Other (Math) FREE Rated 4.75 out of 5, based on 323 reviews 4.8(323) Log in to Download Wish List Get more with resources under $5 See all Interactive Grocery Store Shopping Ad Money Addition Menu Worksheets Meaningful Sped Teaching $4.00 Original Price $4.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Menu Money Math Worksheets – Identifying Prices & Counting One Dollar Bills LifeSkillsClassroom $3.00 Original Price $3.00 Addition money Worksheets. Coffee shop menu math. ABA talks store $3.00 Original Price $3.00 Rated 5 out of 5, based on 2 reviews 5.0(2) St. Patrick's Day Donut Shop Menu Math Money Addition Worksheets Meaningful Sped Teaching $4.00 Original Price $4.00 Free Menu Math Special Education Life Skills Money Math Worksheet Created by Adulting Life Skills Resources This Free SPED Money Math Life Skills Worksheet for Reading Restaurant Menus is designed for middle and high school students. A no-prep life skills worksheet that teaches students how to determine the cost of eating out at a restaurant calculate the tip, and apply sales tax. This money math worksheet will engage your students with product illustrations, relatable dialog, and real-world scenarios. What's includedHere is how you can calculate the total cost of breakfast.How to include sales tax, 6 th - 12 th, Adult Education, Higher Education Also included in:High School Special Education Basic Life Skills Bundle for Teens and Adults FREE Rated 5 out of 5, based on 4 reviews 5.0(4) Log in to Download Wish List FREE Fast Food Money Menu Math Multiplication Special Ed Functional Life Skills Created by Adapted4Adulthood Have students who need to work on their money math skills and reading a menu? Practice menu math multiplication with your students! Write down the prices and multiply them by the quantity needed to find the total! This is great for high school or transition special education classrooms needing extra practice, progress monitoring, or individual work time. What's Included?5 worksheets 2 menu levels: Level 1- Whole Dollar Level 2- Dollar and change Interested in more money math? These monthly them 9 th - 12 th, Adult Education Math Also included in:Life Skills Money Math Functional Year Long Curriculum Bundle Special Ed FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Free Spring Stay Within Budget Life Skills Functional Shopping Money Math Created by Adapted4Adulthood Practice budgeting, money math, and shopping skills with your students with this FREE set of stay within budget worksheets! This free packet includes: Level 1 menu (whole dollar) Level 2 menu (dollar and change) 4 worksheets These engaging worksheets are perfect for high school and transition age special education students needing support and instruction in the following concepts: Community Based Instruction (CBI)Functional Life SkillsConsumer MathMoney MathYour opinion is important to me! P 9 th - 12 th, Adult Education Math Also included in:Life Skills Money Math Functional Year Long Curriculum Bundle Special Ed FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List Café Menu Maths - Practical Arithmetic and Money Practice UK - Free Sample Created by DoingMathsUK This free sample set contains the beginner menu and questions set from the Cafe Menu Maths pack. Pupils use the menu to answer the questions involving finding total cost. The sample set includes prices rounded to the nearest 50p and simple questions about the total cost. The full pack contains two further menus (one rounded to the nearest 25p, one to the nearest 5p), two further sets of questions of increasing difficulty and a bonus table order sheet for pupils to add together the prices on the 2 nd - 5 th Applied Math, Arithmetic, Math CCSS 2.MD.C.8 , 2.NBT.B.5 , 2.NBT.B.6 +4 FREE Rated 4.83 out of 5, based on 6 reviews 4.8(6) Log in to Download Wish List Menu Math-Life Skills Math-Money Math-Real World Math Created by Life Skills with Logan This product is perfect for introducing adding prices together with simple numbers. This simple worksheet allows students to begin adding numbers together for items they will see at a local coffee shop. 2 nd - 5 th Math, Numbers, Other (Math) FREE Log in to Download Wish List Back to School Addition FREEBIE Functional Life Skills Money Math Special Ed Created by Adapted4Adulthood Use these FREE Back to School themed Addition Worksheets to practice important money math and shopping skills with your students! This resource is great for a high school special education or transition classroom! This resource includes:Store Menu 3 two item addition worksheets These engaging worksheets are perfect for high school and transition age special education students needing support and instruction in the following concepts:Community Based Instruction (CBI)Functional Life SkillsConsu 9 th - 12 th, Adult Education Math Also included in:Back to School Money Math Worksheet & Task Card BUNDLE Life Skills Special Ed FREE Log in to Download Wish List Sleepover Math Menu Money Project Based Learning Created by Mad's Monkeys Using a flier or your local store's online website, students are asked to plan a meal and snacks for their friends at a sleepover. After planning items and recording their choices, students will write down the price and add the amounts together. They are then asked to demonstrate what coins and bills they could use to purchase the dinner and snack products. 1 st - 4 th Math CCSS 2.MD.C.8 , MP4 FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Next Dollar Up and Subtract For Change A Menu Math Freebie Created by Carol Bell - Saved By A Bell A REAL WORLD MATH FREEBIE! These print and go worksheets simulate a real life restaurant shopping experience for kids who are working on functional math skills! Kids decide how much money they'll need to pay for their food, figure out how much change they'll get back, and Voila! they just went shopping! TARGET SKILLS:• Round Up to the Next Dollar to Pay • Subtract to Find Change This modified resource is great for older kids with special needs and younger kids who are working on mastering functi 2 nd - 3 rd Math Also included in:Money Worksheets and Word Problems Bundle Menu Math Add Subtract & Dollar Up FREE Rated 4.86 out of 5, based on 72 reviews 4.9(72) Log in to Download Wish List Next Dollar Up and Subtract For Change A Menu Math BBQ Freebie Created by Carol Bell - Saved By A Bell A REAL WORLD MATH FREEBIE! These print and go worksheets simulate a real life restaurant shopping experience for kids who are working on functional math skills! Kids decide how much money they'll need to pay for their food, figure out how much change they'll get back, and Voila! they just went out to eat! TARGET SKILLS:• Next Dollar Up to Pay • Add to Find Total Amount Spent • Subtract to Find Change This modified resource is great for older kids with special needs and younger kids who are wo 2 nd - 3 rd Math Also included in:Financial Literacy Bundle Menu Math BBQ Add Subtract & Dollar Up FREE Rated 4.42 out of 5, based on 12 reviews 4.4(12) Log in to Download Wish List Menu Math Special Education Life Skills Shopping Activity FREE Created by A Little Help Please No Prep! Print and Go! This FREE resource includes 5 worksheets to practice basic budgeting skills. Students are given a small budget and the price of an item and then asked to determine if they have enough money to make the purchase. All my materials are appropriate for older students and adults with intellectual and developmental disabilities: age-appropriate graphics, larger fonts, clean design, and room to write. If you're looking for more consumer math activities that will keep your student Not Grade Specific Math FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List Daily Marketplace Skills: Create Your Own Budget - WORKSHEET Created by Classroom Complete Press THIS IS AN INSTANT DOWNLOADThis FREE worksheet includes 1 page from our Daily Marketplace Skills Gr. 6-12 title.This worksheet challenges students to create their own monthly budget. This worksheet can be used on its own, or paired with the individual resource. And the best part is, it's FREE. About the full resource, Daily Health & Hygiene Skills:Get a sense on how to best spend your hard-earned money with our engagin 6 th - 12 th, Adult Education Character Education, Family Consumer Sciences FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Dollar Up - Chipotle Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori Perfect to incorporate into math lessons for Cinco De Mayo, Spanish/Hispanic/Mexican Heritage Month, Día de los Muertos, Three Kings Day and Día de Reyes. A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Educati 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 2 reviews 5.0(2) Log in to Download Wish List Life Skills: Reading a Menu Created by Adult Speech Therapy I'm a huge proponent of functional therapy tools when working with the adult population to improve participation and help relate tasks to personal experiences. Restaurant menus are full of language and can be adapted to target a variety of cognitive-linguistic goals: 1. Categorization: naming food groups each item would be placed in, generative naming of additional items in category (i.e. naming vegetables, soup ingredients, dairy) 2. Money Concepts: give your patient a budget and choose items Adult Education FREE Rated 4.67 out of 5, based on 6 reviews 4.7(6) Log in to Download Wish List Menu Math Created by Travel Time This NO PREP printable worksheet is designed to calculate snack prices and determine whether or not you have enough money. Requires students to apply real-world mathematical principles including decimal addition and subtraction of money, while incorporating concepts such as more than, less than, and budgeting. MORE ABOUT THIS PRODUCT:Did you know that theme parks can be full of incredible learning opportunities? Planned to be used at the Florida theme park, but versatile enough to be used in 2 nd - 5 th Math FREE Rated 4.5 out of 5, based on 6 reviews 4.5(6) Log in to Download Wish List Dollar Up - Crumbl Cookie Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Dollar Up - Papa John's Pizza Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Dollar Up - Five Guys Inspired Menu Math | Rounding & Bar Graphs, Burgers Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Dollar Up - Jersey Mike's Subs Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Prince's Place Menu Created by Jessie Prince This is a fake menu to be used in many different lessons. It can be used as a non-fiction reading resource or in many different math lessons. It is a companion to the Percent and Money worksheet posted. 5 th - 8 th Algebra, Informational Text FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List Dollar Up - Wingstop Inspired Menu Math | Rounding & Bar Graphs, Chicken Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Dollar Up - Domino's Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Dollar Up - Raising Cane's Inspired Menu Math | Rounding & Bar Graphs Created by Spirit-Filled Montessori A super fun and practical way to teach rounding up, money, and plotting on a bar graph. This worksheet helps with reading and interpreting graphs and representing data using a bar graph. Students can also solve for the medium, maximum, minimum, and mode while using their bar graphs. This is also a great activity for High School Special Education classes and for students who are on the spectrum. These worksheets will assist students with special needs in determining the amount of money required t 1 st - 3 rd, Adult Education Math Also included in:Dollar Up - Fast Food Restaurant Menu Math | Rounding & Bar Graphs FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List 1 2 Showing 1-24 of 47+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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https://web2.qatar.cmu.edu/cs/15317/lectures/09-simplified.pdf
Restricted Sequent Calculus Constructive Logic (15-317) Instructor: Giselle Reis Lecture 09 1 Motivation This is the sequent calculus we have worked with so far: Γ ⇒A Γ ⇒B Γ ⇒A ∧B ∧R Γ, A ∧B, A ⇒C Γ, A ∧B ⇒C ∧L1 Γ, A ∧B, B ⇒C Γ, A ∧B ⇒C ∧L2 Γ ⇒A Γ ⇒A ∨B ∨R1 Γ ⇒B Γ ⇒A ∨B ∨R2 Γ, A ∨B, A ⇒C Γ, A ∨B, B ⇒C Γ, A ∨B ⇒C ∨L Γ, A ⇒B Γ ⇒A ⊃B ⊃R Γ, A ⊃B ⇒A Γ, A ⊃B, B ⇒C Γ, A ⊃B ⇒C ⊃L Γ ⇒⊤⊤R Γ, ⊥⇒C ⊥L Γ, P ⇒P init(for atomic P) It looks a bit bulky and, after proving cut-elimination, you might have noticed that the repetition of the main formula in the premises can become quite a burden. Indeed, if we are automating proof search, keeping these formulas around is not a very good thing to do. The computer can choose to work on the same formula over and over again, like a dog chasing after its tail. Of course we can come up with smart heuristics in an implementation, such as mark-ing how many times a formula was used and keeping track of the sequents that have appeared in our proof, but as we discussed before, it is hard to guarantee soundness and completeness in an implementation level. The smarter the heuristic, the harder to check if it is not missing anything. We will thus follow the same stragtegy when coming up with the verification calculus: change the calculus and enforce syntactically the fact that we should not use a formula again and again. 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 2 Γ, A ∧B, A ⇒C Γ, A ∧B ⇒C ∧L1 Γ, A ∧B, A ⇒C Γ, A ∧B, A ∧B ⇒C ∧L1∗ Γ, A ∧B ⇒C cont Figure 1: Implicit and explicit use of contraction: on the derivation on the right, ∧L1∗is the rule that does not implicitly copies the main formula. This duplication of the main formula is sometimes called auto- or implicit contraction: it is equivalent to having a system with the contraction rule (on left-side formulas) and conservatively duplicate a formula before decomposing it, just in case (see Figure 1). Our goal is to have a contraction-free calculus: no contraction rule is allowed and no auto-contractions occur in the other rules. 2 A Leaner Sequent Calculus To distinguish the sequents in our new simplified calculus, we will use the leaner arrow →. Since the right rules have no auto-contraction, they remain the same. We restrict our analysis thus to the left rules. Conjunction Removing the auto-contracted formulas from the conjunction rule gives us the following: Γ, A →C Γ, A ∧B →C ∧L1 Γ, B →C Γ, A ∧B →C ∧L2 Are these rules ok? Actually no. They both result on information loss. You start from the fact that you know two things, but continue with one (and only one!) of them. If the conjunction formula is no longer there, the other conjunct is lost forever. It might as well just be the case that we need both of them in a proof. This happens, for instance, in the proof of A∧(A ⊃B) ⇒B. Using the proposed rules above, the sequent A∧(A ⊃B) →B is not derivable. Luckily, this can be easily fixed by simply keeping both conjuncts in the premise, and getting rid of the main formula at the same time: Γ, A, B →C Γ, A ∧B →C ∧L 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 3 Disjunction The simplified disjunction rule will be: Γ, A →C Γ, B →C Γ, A ∨B →C ∨L In this case, there is no loss of information. Keeping the disjunction around and using it again will give us nothing new. So we keep this rule as is. Implication Removing the duplication of the main formula on the ⊃L rule results in: Γ →A Γ, B →C Γ, A ⊃B →C ⊃L On the right premise, having B certainly supersedes the fact A ⊃B. The latter requires an A to use B whereas the former is simply B itself. On the left premise, can a proof of A make use of A ⊃B? Definitely. Maybe it is the case that A involves a choice, and we can choose one option at the first application of ⊃L and another option at the second application. If we were to throw this implication away, we would not be able to have both. This is what happens, for instance, in a proof of →¬¬(a ∨¬a). Thus we have the following rule: Γ, A ⊃B →A Γ, B →C Γ, A ⊃B →C ⊃L Unfortunately we cannot get rid of the auto-contraction in this rule just yet. True There is no left rule for ⊤, therefore we do not need to analyse it. False The left rule for ⊥does not copy formulas to premises because it does not have premises, so it remains unchanged. After analysing all the rules, our new proposed calculus looks like this: Γ →A Γ →B Γ →A ∧B ∧R Γ, A, B →C Γ, A ∧B →C ∧L Γ →A Γ →A ∨B ∨R1 Γ →B Γ →A ∨B ∨R2 Γ, A →C Γ, B →C Γ, A ∨B →C ∨L 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 4 Γ, A →B Γ →A ⊃B ⊃R Γ, A ⊃B →A Γ, B →C Γ, A ⊃B →C ⊃L Γ →⊤⊤R Γ, ⊥→C ⊥L Γ, P →P init(for atomic P) 3 Soundness and Completeness We formally show that the →calculus is correct by showing soundness and completeness w.r.t. the ⇒calculus. The soundness proof is quite straightforward, using heavily the fact that weakening is admissible. Theorem 1. (Soundness) If Γ →C then Γ ⇒C. Proof. As usual, we prove by structural induction on the proof tree. Base case: Γ →C init ⇝ Γ ⇒C init Induction hypothesis: If there are derivations D and E of sequents Γ1 →C and Γ2 →D, respectively, then there are derivations D′ and E′ of sequents Γ1 ⇒C and Γ2 ⇒D, respectively. Inductive cases: We will not show the cases for the right-side rules or ⊥L, as these are exactly the same for both calculi. 1. Case ∧L D Γ, A, B →C Γ, A ∧B →C ∧L By the inductive hypothesis, we have a derivation D′ of Γ, A, B ⇒C. By admissibility of weakening, this can be transformed into a derivation D′′ of Γ, A∧B, A, B ⇒C which, in turn, can be used to get a derivation of Γ, A∧B ⇒ C: D′′ Γ, A ∧B, A, B ⇒C Γ, A ∧B, A ⇒C ∧L2 Γ, A ∧B ⇒C ∧L1 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 5 2. Case ∨L Γ, A →C Γ, B →C Γ, A ∨B →C ∨L By the inductive hypothesis, we have derivations D′ and E′ of Γ, A ⇒C and Γ, B ⇒C, respectively. By admissibility of weakening, these can be trans-formed into derivations D′′ and E′′ of Γ, A ∨B, A ⇒C and Γ, A ∨B, B ⇒C. Using those, we get: D′′ Γ, A ∨B, A ⇒C E′′ Γ, A ∨B, B ⇒C Γ, A ∨B ⇒C ∨L 3. Case ⊃L Γ, A ⊃B →A Γ, B →C Γ, A ⊃B →C ⊃L By IH we have derivations D′ and E′ of Γ, A ⊃B ⇒A and Γ, B ⇒C, respec-tively. By admissibility of weakening, we can transform E′ into a derivation E′′ of Γ, A ⊃B, B ⇒C. We can thus apply the ⊃L rule from the first calculus and obtain: D′ Γ, A ⊃B ⇒A E′′ Γ, A ⊃B, B ⇒C Γ, A ⊃B ⇒C ⊃L The completeness proof is a little more involved. Theorem 2. (Completeness) If Γ ⇒C then Γ →C. Proof. Again we proceed by structural induction on the proof tree (i.e. on the derivation). Base case: Γ, C ⇒C init ⇝ Γ, C →C init Induction hypothesis: If there are derivations D and E of sequents Γ1 ⇒C and Γ2 ⇒D, respectively, then there are derivations D′ and E′ of sequents Γ1 →C and Γ2 →D, respectively. 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 6 Inductive cases: We will not show the cases for the right-side rules or ⊥L, as these are exactly the same for both calculi. 1. Case ∧L1 D Γ, A ∧B, A ⇒C Γ, A ∧B ⇒C ∧L1 By IH we have a derivation D′ of Γ, A ∧B, A →C. This can be used to derive Γ, A ∧B →C in the following way: Γ, A, B →A id Γ, A ∧B →A ∧L D′ Γ, A ∧B, A →C Γ, A ∧B →C cut The ∧L2 case is analogous. 2. Case ∨L D Γ, A ∨B, A ⇒C E Γ, A ∨B, B ⇒C Γ, A ∨B ⇒C ∨L By IH we have derivations D′ and E′ of Γ, A ∨B, A →C and Γ, A ∨B, B →C, respectively. We can use those and cut to construct the following derivation: Γ, A →A id Γ, A →A ∨B ∨R1 D′ Γ, A ∨B, A →C Γ, A →C cut Γ, B →B id Γ, B →A ∨B ∨R2 E′ Γ, A ∨B, B →C Γ, B →C cut Γ, A ∨B →C ∨L 3. Case ⊃L (see homework) D Γ, A ⊃B ⇒A E Γ, A ⊃B, B ⇒C Γ, A ⊃B ⇒C ⊃L 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 7 In order to make the transformations local and clearer for the completeness proofs, we have used the cut and id rules. To be precise, we need to show that they are admissible for this calculus. Admissibility of id is straightforward. The proof of cut-elimination follows the same lines as the one before: induction on the cut formula and structure of the trees above it; except that, in the case of this calculus we would need to use some invertibility lemmas. We will see what those are next class. Using contraction There is also another strategy to prove this result without using cut (directly at least). We can in fact get the transformations by using structural rules of contraction and weakening. The case for ∧L would thus be: D′ Γ, A ∧B, A →C Γ, A ∧B, A, B →C weak Γ, A ∧B, A ∧B →C ∧L Γ, A ∧B →C cont Afterwards, of course, we need to show the admissibility of those rules in the simpli-fied calculus. Weakening is simple, following the same lines as before. But contraction is more tricky. We only outline the proof here. Theorem 3. If D :: Γ, A, A →C then D′ :: Γ, A →C. Proof sketch. The proof is a nested induction on the structure of A first, and then D. • BASE CASE 1: A is an atom. We need to show that D :: Γ, A, A →C implies D′ :: Γ, A →C for atomic A. The proof proceeds by induction on D. – BASE CASE 1.1: D is init. There are two subcases, one where A is the main formula of init and another where it is not. Both cases can be proved easily. – INDUCTIVE CASES 1.1: each case corresponds to the lowermost rule in D. IH: The property holds for proof structurally smaller than D. These cases follow the usual transformation where we apply the inductive hy-pothesis to the premises of a rule, and then use the rule itself to get the conclu-sion we want. 15-317 LECTURE NOTES LEC 09 Restricted Sequent Calculus 8 • INDUCTIVE CASES 1: We case on the structure of the contracted formula. IH: The property holds for formulas A and B structurally smaller than A ⋆B, for ⋆∈{∧, ∨, ⊃}. At each case, we need an inner induction on D. We show here only the case for ⊃. – Case A ⊃B: We proceed by structural induction on D. If the lowermost rule of D cannot operate on A ⊃B, then this case is analogous to the atomic one. The only really interesting cases are the ones where the lowermost rule of D can operate on the contracted formula. For the case of A ⊃B, this would be: E Γ, A ⊃B, A ⊃B →A F Γ, A ⊃B, B →C Γ, A ⊃B, A ⊃B →C ⊃L Can we transform this into a proof of Γ, A ⊃B →C? Using cut we can, and then we are back at the problem of having to show cut-admissibility for this calculus. Using the wrong rules Interestingly, the wrong rules for ⊃L (without copying the ⊃formula to the left premise) and ∧L (keeping only one of the two conjuncts) seem to make it through the proof of completeness of the →-calculus w.r.t. ⇒(at least the version using contraction). That may give us the impression that the rules are just fine. But we cannot forget that we used weakening, contraction, cut, etc. for these transformations, and we still need to show that these rules are admissible in the calculus. Eventually, one of these long proofs will fail (in maybe one or two out of dozens of cases), so we must always be careful about these things! 15-317 LECTURE NOTES LEC 09
187877
https://physics.stackexchange.com/questions/523344/control-volume-and-momentum-theorem
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Control volume and momentum theorem Ask Question Asked Modified 2 months ago Viewed 538 times 0 $\begingroup$ I'm studying fluid and propulsion mechanics by myself. I stumbled upon this website from MIT: It states that "Newton€™s second law for a control volume of fixed mass" is $ \sum \vec{F}=\int_V\rho \frac{D}{dt}(\vec{u}_0+\vec{u}) dV $ but it's said that this is valid for a fixed mass control volume. $\vec{u}_0$ is the velocity of a reference frame attached to the control volume and $\vec{u}$ is the velocity of fluid relative to this moving frame. The notes then goes on to derive this formula: $\sum {F}_x-{F}_{0_x}=\int_V \left[\frac{\partial}{\partial t}(\rho{u}_x)\right]dV+\int_S u_x(\rho \vec{u}) \cdot \vec{n} dA$ where ${F}_{0_x}$ is basically $m\vec{a}$. So far so good. However, I still don't understand why this equation is only valid for a control volume with fixed mass. Moreover, we're allowing the control volume to change its mass with by having the boundary term. This is even stressed in the quizz accompanying these notes: where the solution starts by remarking the validity of this equation depending on this assumption. This seems to contradict books on Fluid Mechanics, where the mass can vary and they reach this similar equation (or maybe it's not the same equation?). For example Frank White's book equation 3.35: $\sum \vec{F}=\frac{d}{d t}\int_V \left[(\rho \vec{V})\right]dV+\int_S \vec{V}(\rho \vec{V_r}) \cdot \vec{n} dA$ I can see that White's equation is not exactly the same but I'm trying to prove they are by expanding V and V_r (V relative to Earth, inertial frame and V_r is a relative velocity). I think I'm missing something here. Is the requirement of fixed mass even right? Considering that there is a boundary term. fluid-dynamics momentum flow Share Improve this question edited Jan 6, 2020 at 3:57 Qmechanic♦ 222k5252 gold badges636636 silver badges2.6k2.6k bronze badges asked Jan 6, 2020 at 2:05 Manuel JesusManuel Jesus 3333 bronze badges $\endgroup$ Add a comment | 1 Answer 1 Reset to default 0 $\begingroup$ In the term $$\int_V \frac{\partial}{\partial t} (\rho u_x) dV,$$ the time derivative is calculated using the body frame, i.e. the reference frame attached to the control volume. Imagine that you are sitting in the fueltank of a moving vehicle which consumes the liquid fuel, observing the level of the fuel contained in it. The rate of change of the fuel level and consequently, the rate of change of the fuel mass is going to be negligible. If this mass rate of change was not expected to be negligible, the accurate term which would replace this term would be $$\frac{\partial}{\partial t} \int_V \rho u_x dV.$$ Share Improve this answer answered Feb 24, 2020 at 22:46 kbakshi314kbakshi314 2,46022 gold badges1212 silver badges3838 bronze badges $\endgroup$ Add a comment | Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions fluid-dynamics momentum flow See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related Euler equation of fluid dynamics Why is momentum conserved in a control volumen? 1 External Forces from Fluid Motion 1 Unconventional thrust generation? 0 Reynolds transport theorem and steady flow 3 How can mass diffuse out of a lagrangian control volume when it always tracks the same mass? 1 Interchange of Lagrangian/material derivative and volume integral 0 How does the rate of change of momentum due to unsteady flow inside a control volume not always equal 0 due to momentum conservation? Hot Network Questions Storing a session token in localstorage Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Matthew 24:5 Many will come in my name! 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187878
https://www.desmos.com/calculator/mlhbhtb26g
r=2acos(theta) plot | Desmos Loading... r=2acos(theta) plot Save Copy Log In Sign Up Expression 1: "r" equals 2 "a" cosine theta r=2 a c o s θ 0 0 domain \theta Minimum: less than or equal to theta less than or equal to≤θ≤ domain \theta Maximum: 2 pi 2 π 1 Expression 2: "r" equals 2 "a" cosine theta left brace, 0 less than or equal to theta less than or equal to "b" , right brace r=2 a c o s θ 0≤θ≤b 0 0 domain \theta Minimum: less than or equal to theta less than or equal to≤θ≤ domain \theta Maximum: 2 pi 2 π 2 Expression 3: "b" equals 0 b=0 0 0 StartFraction, pi Over 2 , EndFraction π 2​ 3 Expression 4: "a" equals 1 a=1 0 0 10 1 0 4 Expression 5: "r" equals 2 "a" cosine theta left brace, StartFraction, 3 pi Over 2 , EndFraction less than or equal to theta less than or equal to "c" , right brace r=2 a c o s θ 3 π 2​≤θ≤c 3 pi / 2 3 π/2 domain \theta Minimum: less than or equal to theta less than or equal to≤θ≤ domain \theta Maximum: 2 pi 2 π 5 Expression 6: "c" equals 4.7 1 2 3 8 8 9 8 0 3 8 4 6 9 c=4.7 1 2 3 8 8 9 8 0 3 8 4 6 9 StartFraction, 3 pi Over 2 , EndFraction 3 π 2​ 2 pi 2 π 6 7 powered by powered by "x"x "y"y "a" squared a 2 "a" Superscript, "b" , Baseline a b 7 7 8 8 9 9 divided by÷ functions (( )) less than< greater than> 4 4 5 5 6 6 times× | "a" ||a| ,, less than or equal to≤ greater than or equal to≥ 1 1 2 2 3 3 negative− A B C StartRoot, , EndRoot pi π 0 0 .. equals= positive+
187879
https://www.hufsd.edu/assets/pdfs/academics/algebra_text/chapter14.pdf
CHAPTER 1 4 539 CHAPTER TABLE OF CONTENTS 14-1 The Meaning of an Algebraic Fraction 14-2 Reducing Fractions to Lowest T erms 14-3 Multiplying Fractions 14-4 Dividing Fractions 14-5 Adding or Subtracting Algebraic Fractions 14-6 Solving Equations with Fractional Coefficients 14-7 Solving Inequalities with Fractional Coefficients 14-8 Solving Fractional Equations Chapter Summary Vocabulary Review Exercises Cumulative Review ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS Although people today are making greater use of decimal fractions as they work with calculators, computers, and the metric system, common fractions still surround us. We use common fractions in everyday measures: -inch nail, -yard gain in football, pint of cream, cups of flour. We buy dozen eggs, not 0.5 dozen eggs. We describe 15 minutes as hour,not 0.25 hour.Items are sold at a third off, or at a fraction of the original price. Fractions are also used when sharing.For example, Andrea designed some beautiful Ukrainian eggs this year.She gave one-fifth of the eggs to her grandparents.Then she gave one-fourth of the eggs she had left to her parents.Next,she presented her aunt with one-third of the eggs that remained. Finally, she gave one-half of the eggs she had left to her brother, and she kept six eggs. Can you use some problem-solving skills to discover how many Ukrainian eggs Andrea designed? In this chapter, you will learn operations with algebraic fractions and methods to solve equations and inequalities that involve fractions. A 1 3 B 1 4 1 2 11 3 1 2 21 2 1 4 A fraction is a quotient of any number divided by any nonzero number. For example, the arithmetic fraction indicates the quotient of 3 divided by 4. An algebraic fraction is a quotient of two algebraic expressions. An alge-braic fraction that is the quotient of two polynomials is called a fractional expression or a rational expression. Here are some examples of algebraic frac-tions that are rational expressions: The fraction means that the number represented by a, the numerator, is to be divided by the number represented by b, the denominator. Since division by 0 is not possible, the value of the denominator, b, cannot be 0.An algebraic frac-tion is defined or has meaning only for values of the variables for which the denominator is not 0. EXAMPLE 1 Find the value of x for which is not defined. Solution The fraction is not defined when the denominator, x 9, is equal to 0. x 9 0 x 9 Answer Writing About Mathematics 1. Since any number divided by itself equals 1, the solution set for 1 is the set of all real numbers. Do you agree with this statement? Explain why or why not. 2. Aaron multiplied by (equal to 1) to obtain the fraction . Is the fraction equal to the fraction for all values of b? Explain your answer. Developing Skills In 3–12, find, in each case, the value of the variable for which the fraction is not defined. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 3 x 2 2 5x 2 14 1 x 2 2 4 2y 1 3 4y 1 2 10 2x 2 1 y 1 5 y 1 2 7 2 2 x 1 x 2 5 12 y2 25 6x 2 x b2 2 b b 1 1 b 1 1  1 b b2 2 b b 1 1 b b b 1 1  1 b x x EXERCISES 12 x 2 9 12 x 2 9 a b x2 1 4x 1 3 x 1 1 x 1 5 x 2 2 4c 3d 2 x x 2 3 4 14-1 THE MEANING OF AN ALGEBRAIC FRACTION 540 Algebraic Fractions, and Equations and Inequalities Involving Fractions Applying Skills In 13–17, represent the answer to each problem as a fraction. 13. What is the cost of one piece of candy if five pieces cost c cents? 14. What is the cost of 1 meter of lumber if p meters cost 980 cents? 15. If a piece of lumber 10x  20 centimeters in length is cut into y pieces of equal length, what is the length of each of the pieces? 16. What fractional part of an hour is m minutes? 17. If the perimeter of a square is 3x  2y inches, what is the length of each side of the square? A fraction is said to be reduced to lowest terms or is a lowest terms fraction when its numerator and denominator have no common factor other than 1 or 1. Each of the fractions and can be expressed in lowest terms as . The arithmetic fraction is reduced to lowest terms when both its numer-ator and denominator are divided by 5: The algebraic fraction is reduced to lowest terms when both its numera-tor and denominator are divided by a, where a  0: Fractions that are equal in value are called equivalent fractions.Thus, and are equivalent fractions, and both are equivalent to , when a  0. The examples shown above illustrate the division property of a fraction: if the numerator and the denominator of a fraction are divided by the same nonzero number, the resulting fraction is equal to the original fraction. In general, for any numbers a, b, and x, where b  0 and x  0: When a fraction is reduced to lowest terms, we list the values of the vari-ables that must be excluded so that the original fraction is equivalent to the reduced form and also has meaning. For example: (where x  0) (where y  0, d  0) cy dy 5 cy 4 y dy 4 y 5 c d 4x 5x 5 4x 4 x 5x 4 x 5 4 5 ax bx 5 ax 4 x bx 4 x 5 a b a 2a 1 2 5 10 a 2a 5 a 4 a 2a 4 a 5 1 2 a 2a 5 10 5 5 4 5 10 4 5 5 1 2 5 10 1 2 a 2a 5 10 14-2 REDUCING FRACTIONS TO LOWEST TERMS Reducing Fractions to Lowest T erms 541 When reducing a fraction, the division of the numerator and the denomina-tor by a common factor may be indicated by a cancellation. Here, we use cancellation to Here, we use cancellation to divide the numerator and the divide the numerator and the denominator by 3: denominator by : (where a  3) By re-examining one of the examples just seen, we can show that the multi-plication property of one is used whenever a fraction is reduced: However, when the multiplication property of one is applied to fractions, it is referred to as the multiplication property of a fraction. In general, for any numbers a, b, and x, where b  0 and x  0: a b 5 a b ? x x 5 a b ? 1 5 a b 3(x 1 5) 18 5 3 ? (x 1 5) 3 ? 6 5 3 3 ? (x 1 5) 6 5 1 ? (x 1 5) 6 5 x 1 5 6 a2 2 9 3a 2 9 5 (a 2 3)(a 1 3) 3(a 2 3) 5 a 1 3 3 3(x 1 5) 18 5 3(x 1 5) 18 5 x 1 5 6 (a 2 3) 542 Algebraic Fractions, and Equations and Inequalities Involving Fractions 1 1 1 6 Procedure T o reduce a fraction to lowest terms: METHOD 1 1. Factor completely both the numerator and the denominator. 2. Determine the greatest common factor of the numerator and the denominator. 3. Express the given fraction as the product of two fractions, one of which has as its numerator and its denominator the greatest common factor deter-mined in step 2. 4. Use the multiplication property of a fraction. METHOD 2 1. Factor both the numerator and the denominator. 2. Divide both the numerator and the denominator by their greatest common factor. EXAMPLE 1 Reduce to lowest terms. Solution METHOD 1 METHOD 2 Answer (x  0) EXAMPLE 2 Express as a lowest terms fraction. Solution METHOD 1 METHOD 2 Answer (x  0) EXAMPLE 3 Reduce each fraction to lowest terms. a. b. Solution a. Use Method 1: b. Use Method 2: Answers a. (x  1, x  4) b. (x  2) 21 4 x 1 4 x 2 1 5 x 1 4 x 2 1 5 x 1 4 x 2 1 ? 1 5 x 1 4 x 2 1 ? x 2 4 x 2 4 2 2 x 4x 2 8 5 21(x 2 2) 4(x 2 2) x2 2 16 x2 2 5x 1 4 5 (x 1 4)(x 2 4) (x 2 1)(x 2 4) 2 2 x 4x 2 8 x2 2 16 x2 2 5x 1 4 x 2 3 5 5 x 2 3 5 5 1 ? (x 2 3) 5 5 2x 2x ? (x 2 3) 5 2x2 2 6x 10x 5 2x(x 2 3) 10x 2x2 2 6x 10x 5 2x(x 2 3) 2x ? 5 2x2 2 6x 10x 3 7x2 5 3 7x2 5 3 7x2 ? 1 15x2 35x4 5 3 ? 5x2 7x2 ? 5x2 15x2 35x4 5 3 7x2 ? 5x2 5x2 15x2 35x4 Reducing Fractions to Lowest T erms 543 1 5 3 ? 5x2 7x2 ? 5x2 5 3 7x2 1 1 5 2x(x 2 3) 10x 5 x 2 3 5 5 1 5 21 4 5 21(x 2 2) 4(x 2 2) 1 Writing About Mathematics 1. Kevin used cancellation to reduce to lowest terms as shown below. What is the error in Kevin’s work? 2. Kevin let a 4 to prove that when reduced to lowest terms, . Explain to Kevin why his reasoning is incorrect. Developing Skills In 3–54, reduce each fraction to lowest terms. In each case, list the values of the variables for which the fractions are not defined. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. x2 2 7xy 1 12y2 x2 1 xy 2 20y2 2x2 2 7x 1 3 (x 2 3)2 48 1 8x 2 x2 x2 1 x 2 12 r2 2 4r 2 5 r2 2 2r 2 15 2x2 2 50 x2 1 8x 1 15 a2 2 6a a2 2 7a 1 6 a2 2 a 2 6 a2 2 9 x2 2 25 x2 2 2x 2 15 x2 2 3x x2 2 4x 1 3 3y 2 3 y2 2 2y 1 1 x2 1 x 2 2 x2 1 4x 1 4 x2 1 7x 1 12 x2 2 16 r2 2 r 2 6 3r 2 9 2b(3 2 b) b2 2 9 x2 2 y2 3y 2 3x 16 2 a2 2a 2 8 2s 2 2r s2 2 r2 3 2 b b2 2 9 1 2 x x 2 1 x2 2 1 5x 2 5 x2 2 9 3x 1 9 12a 1 12b 3a 1 3b 14 7r 2 21s 2a2 6a2 2 2ab 5y 5y 1 5x 7d 7d 1 14 4x 4x 1 8 18b2 1 30b 9b3 6x2y 1 9xy2 12xy 12ab 2 3b2 3ab 5a2 2 10a 5a2 2ax 1 2bx 6x2 8m2 1 40m 8m 5x 2 35 5x 8y 2 12 6 3x 1 6 4 5xy 45x2y2 232a3b3 148a3b3 220x2y2 290xy2 112a2b 28ac 8xy2 24x2y 27a 36a2 5x2 25x4 15x2 5x 2abc 4abc 5xy 9xy 3ay2 6by2 ab cb 9r 10r 24c 36d 27y2 36y 4x 12x a 1 4 a 1 8 5 2 3 a 1 4 a 1 8 5 a 1 4 a 1 8 5 1 1 1 1 1 2 5 2 3 a 1 4 a 1 8 EXERCISES 544 Algebraic Fractions, and Equations and Inequalities Involving Fractions 1 1 1 2 55. a. Use substitution to find the numerical value of , then reduce each numerical frac-tion to lowest terms when: (1) x 7 (2) x 10 (3) x 20 (4) x 2 (5) x 4 (6) x 10 b. What pattern, if any, do you observe for the answers to part a? c. Can substitution be used to evaluate when x 5? Explain your answer. d. Reduce the algebraic fraction to lowest terms. e. Using the answer to part d, find the value of , reduced to lowest terms, when x 38,756. f. If the fraction is multiplied by to become , will it be equivalent to ? Explain your answer. The product of two fractions is a fraction with the following properties: 1. The numerator is the product of the numerators of the given fractions. 2. The denominator is the product of the denominators of the given fractions. In general, for any numbers a, b, x, and y, when b  0 and y  0: We can find the product of and in lowest terms by using either of two methods. METHOD 1. METHOD 2. Notice that Method 2 requires less computation than Method 1 since the reduced form of the product was obtained by dividing the numerator and the denominator by a common factor before the product was found. This method may be called the cancellation method. The properties that apply to the multiplication of arithmetic fractions also apply to the multiplication of algebraic fractions. 7 27 3 9 4 5 7 27 3 3 9 1 4 5 7 12 7 27 3 9 4 5 7 3 9 27 3 4 5 63 108 5 7 3 9 12 3 9 5 7 12 3 9 9 5 7 12 3 1 5 7 12 9 4 7 27 a b ? x y 5 ax by 14-3 MULTIPLYING FRACTIONS x2 2 5x x 2 5 x(x2 2 5x) x(x 2 5) x x x2 2 5x x 2 5 x2 2 5x x 2 5 x2 2 5x x 2 5 x2 2 5x x 2 5 x2 2 5x x 2 5 Multiplying Fractions 545 To multiply by and express the product in lowest terms, we may use either of the two methods. In this example, x  0 and y  0. METHOD 1. METHOD 2. (the cancellation method) While Method 1 is longer, it has the advantage of displaying each step as a property of fractions. This can be helpful for checking work. EXAMPLE 1 Multiply and express the product in reduced form: Solution METHOD 1 (1) Multiply the numerators and denominators of the given fractions: (2) Reduce the resulting fraction to lowest terms: 5 10a 3 ? 3a2bx 3a2bx 5 10a 3 ? 1 5 10a 3 5a3 9bx ? 6bx a2 5 5a3 ? 6bx 9bx ? a2 5 30a3bx 9a2bx 5a3 9bx ? 6bx a2 5x2 7x ? 14y2 15x3 5 5x2 1 7y 1 ? 14y2 2y 15x3 3x 5 2y 3x 5x2 7y ? 14y2 15x3 5 5x2 ? 14y2 7y ? 15x3 5 70x2y2 105x3y 5 2y 3x ? 35x2y 35x2y 5 2y 3x ? 1 5 2y 3x 14y2 15x3 5x2 7y 546 Algebraic Fractions, and Equations and Inequalities Involving Fractions Procedure T o multiply fractions: METHOD 1 1. Multiply the numerators of the given fractions. 2. Multiply the denominators of the given fractions. 3. Reduce the resulting fraction, if possible, to lowest terms. METHOD 2 1. Factor any polynomial that is not a monomial. 2. Use cancellation to divide a numerator and a denominator by each common factor. 3. Multiply the resulting numerators and the resulting denominators to write the product in lowest terms. METHOD 2 (1) Divide the numerators and denominators by the common factors 3bx and a2: (2) Multiply the resulting numerators and the resulting denominators: Answer (a  0, b  0, x  0) EXAMPLE 2 Multiply and express the product in simplest form: Solution Think of 12a as . Answer (a  0) EXAMPLE 3 Multiply and simplify the product: Solution Answer (x  0, 3) Writing About Mathematics 1. When reduced to lowest terms, a fraction whose numerator is x2 3x  2 equals 1. What is the denominator of the fraction? Explain your answer. 2. Does for all values of x and z? Explain your answer. x2 xz 1 z2 ? x2 2 z2 x2 2 xz 5 x z EXERCISES x 2 2 6x x2 2 5x 1 6 3x ? 2 4x 2 12 5 (x23) 1 (x 2 2) 3x ? 2 1 4 2(x 2 3) 1 5 x 2 2 6x x2 2 5x 1 6 3x ? 2 4x 2 12 9 2 5 4a 4a ? 9 2 5 1 ? 9 2 5 9 2 12a ? 3 8a 5 12a 1 ? 3 8a 5 36a 8a 12a 1 12a ? 3 8a 10a 3 5 10a 3 5a3 9bx ? 6bx a2 5 5a3 a 9bx 3 ? 6bx 2 a2 1 Multiplying Fractions 547 Developing Skills In 3–41, find each product in lowest terms. In each case, list any values of the variable for which the fractions are not defined. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. What is the value of when x 65,908? We know that the operation of division may be defined in terms of the multi-plicative inverse, the reciprocal. A quotient can be expressed as the dividend times the reciprocal of the divisor. Thus: and We use the same rule to divide algebraic fractions. In general, for any num-bers a, b, c, and d, when b  0, c  0, and d  0: a b 4 c d 5 a b ? d c 5 ad bc 8 7 4 5 3 5 8 7 3 3 5 5 8 3 3 7 3 5 5 24 35 8 4 5 5 8 1 3 1 5 5 8 3 1 1 3 5 5 8 5 14-4 DIVIDING FRACTIONS x2 2 4 6x 1 12 ? 4x 2 12 x2 2 5x 1 6 a2 1 12a 1 36 a2 2 36 ? 36 2 a2 36 1 a2 d2 2 25 4 2 d2 ? 5d2 2 20 d 1 5 b2 1 81 b 2 2 81 ? 81 2 b2 81 1 b 2 x2 2 3x 1 2 2x2 2 2 ? 2x x 2 2 2 2 x 2x ? 3x 3x 2 6 8x 2x 2 2 8 ? 8x 1 16 32x 2 y2 2 81 (y 1 9)2 ? 10y 1 90 5y 2 45 4x 1 8 6x 1 18 ? 5x 1 15 x2 2 4 x2 2 25 4x2 2 9 ? 2x 1 3 x 2 5 4a 2 6 4a 1 8 ? 6a 1 12 5a 2 15 y2 2 2y 2 3 2c3 ? 4c2 2y 1 2 x2 1 6x 1 5 9y2 ? 3y x 1 1 a2 2 7a 2 8 2a 1 2 ? 5 a 2 8 (a 2 2)2 4b ? 16b3 4 2 a2 a(a 2 b)2 4b ? 4b a(a2 2 b2) x2 2 x 2 2 3 ? 21 x2 2 4 a2 2 9 3 ? 12 2a 2 6 1 x2 2 1 ? 2x 1 2 6 8x 2x 1 6 ? x 1 3 x2 7s s 1 2 ? 2s 1 4 21 2r r 2 1 ? r 2 1 10 x2 2 1 x2 ? 3x2 2 3x 15 ab 2 a b2 ? b3 2 b2 a 12a 2 4 b ? b3 12 5x 2 5y x2y ? xy2 25 3a 1 9 15a ? a3 18 7 8 ? 2x 1 4 21 24a3b2 7c3 ? 21c2 12ab 30m2 18n ? 6n 5m 6r2 5s2 ? 10rs 6r3 m2 8 ? 32 3m 12x 5y ? 15y2 36x2 24x 35y ? 14y 8x mn ? 8 m2n2 x2 36 ? 20 5 d ? d2 1 2 ? 20x 36 ? 5y 9y 8 12 ? 30a 36 548 Algebraic Fractions, and Equations and Inequalities Involving Fractions Procedure T o divide by a fraction, multiply the dividend by the reciprocal of the divisor. EXAMPLE 1 Divide: Solution How to Proceed (1) Multiply the dividend by the reciprocal of the divisor: (2) Divide the numerators and denominators by the common factors: (3) Multiply the resulting numerators and the resulting denominators: Answer (c  0, d  0) EXAMPLE 2 Divide: Solution How to Proceed (1) Multiply the dividend by the reciprocal of the divisor: (2) Factor the numerators and denominators, and divide by the common factors: (3) Multiply the resulting numerators and the resulting denominators: Answer (x  0, 5, 5, 3) Note: If x 5, 5, or 3, the dividend and the divisor will not be defined. If x 0, the reciprocal of the divisor will not be defined. 2(x 1 3)2 x(x 2 5) 5 2(x 1 3)2 x(x 2 5) 5 8 2 (x 1 3) (x 1 5) 1 (x 2 5) ? (x 1 5) 1 (x 1 3) 4 1x 8x 1 24 x2 2 25 4 4x x2 1 8x 1 15 5 8x 1 24 x2 2 25 ? x2 1 8x 1 15 4x 8x 1 24 x2 2 25 4 4x x2 1 8x 1 15 4d 9c 5 4d 9c 5 16c3 2 21d 2 3 ? 14d3 2d 24c4 3c 16c3 21d 2 4 24c4 14d 3 5 16c3 21d 2 ? 14d3 24c 4 16c3 21d 2 4 24c4 14d 3 Dividing Fractions 549 Writing About Mathematics 1. Explain why the quotient is undefined for x 2 and for x 3. 2. To find the quotient , Ruth canceled (x 4) in the numerator and denomi-nator and wrote . Is Ruth’s answer correct? Explain why or why not. Developing Skills In 3–27, find each quotient in lowest terms. In each case, list any values of the variables for which the quotient is not defined. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. For what value(s) of a is undefined? 29. Find the value of when y 70. 30. If x  y a and , what is the value of x  z? We know that the sum (or difference) of two arithmetic fractions that have the same denominator is another fraction whose numerator is the sum (or differ-ence) of the numerators and whose denominator is the common denominator of the given fractions.We use the same rule to add algebraic fractions that have the same nonzero denominator. Thus: 14-5 ADDING OR SUBTRACTING ALGEBRAIC FRACTIONS y 4 z 5 1 a y2 2 6y 1 9 y2 2 9 4 10y 2 30 y2 1 3y a2 2 2a 1 1 a2 4 a2 2 1 a (a 1 b)2 a2 2 b2 4 a 1 b b2 2 a2 ? a 2 b (a 2 b)2 2a 1 6 a2 2 9 4 3 1 a 3 2 a ? a 1 3 4 x 1 y x2 1 y2 ? x x 2 y 4 (x 1 y)2 x4 2 y4 x 2 1 x 1 1 ? 2x 1 2 x 1 2 4 4x 2 4 x 1 2 (9 2 y2) 4 y2 1 8y 1 15 2y 1 10 x2 2 4x 1 4 3x 2 6 4 (2 2 x) x2 2 2xy 2 8y2 x2 2 16y2 4 5x 1 10y 3x 1 12y (x 2 2)2 4x2 2 16 4 21x 3x 1 6 12y 2 6 8 4 (2y2 2 3y 1 1) a2 2 ab 4a 4 (a2 2 b2) b2 2 b 2 6 2b 4 b2 2 4 b2 4a2 2 9 10 4 10a 1 15 25 x2 2 5x 1 4 2x 4 2x 2 2 8x2 x2 2 1 5 4 x 2 1 10 a3 2 a b 4 a3 4b3 3y2 1 9y 18 4 5y2 27 4x 1 4 9 4 3 8x 6a2b2 8c 4 3ab xy2 x2y 4 x y3 7ab2 10cd 4 14b3 5c2d2 3x 5y 4 21x 2y x 9 4 x 3 8 4 x 2y 12 35 4 4b 7 7a 10 4 21 5 3 2 4 1 5 5 3 2 ? 5 1 5 15 2 3 2(x 2 4) 4 x 2 4 5 2 x 2 2 4 x 2 3 5 EXERCISES 550 Algebraic Fractions, and Equations and Inequalities Involving Fractions Arithmetic fractions Algebraic fractions EXAMPLE 1 Add and reduce the answer to lowest terms: Solution Answer (x  0) EXAMPLE 2 Subtract: Solution Answer (x  0) Note: In Example 2, since the fraction bar is a symbol of grouping, we enclose numerators in parentheses when the difference is written as a single fraction. In this way, we can see all the signs that need to be changed for the subtraction. In arithmetic, in order to add (or subtract) fractions that have different denominators, we change these fractions to equivalent fractions that have the same denominator, called the common denominator.Then we add (or subtract) the equivalent fractions. For example, to add and , we use any common denominator that has 4 and 6 as factors. METHOD 1. Use the product of the denominators as the common denomina-tor. Here, a common denominator is 4  6, or 24. Answer 3 4 1 1 6 5 3 4 3 6 6 1 1 6 3 4 4 5 18 24 1 4 24 5 22 24 5 11 12 1 6 3 4 2x 1 11 6x 4x 1 7 6x 2 2x 2 4 6x 5 (4x 1 7) 2 (2x 2 4) 6x 5 4x 1 7 2 2x 1 4 6x 5 2x 1 11 6x 4x 1 7 6x 2 2x 2 4 6x 7 2x 5 4x 1 9 4x 5 5 1 9 4x 5 14 4x 5 7 2x 5 4x 1 9 4x a x 2 b x 5 a 2 b x 5 7 2 1 7 5 5 2 1 7 5 4 7 a x 1 b x 5 a 1 b x 5 7 1 1 7 5 5 1 1 7 5 6 7 Adding or Subtracting Algebraic Fractions 551 Procedure T o add (or subtract) fractions that have the same denominator: 1. Write a fraction whose numerator is the sum (or difference) of the numer-ators and whose denominator is the common denominator of the given fractions. 2. Reduce the resulting fraction to lowest terms. METHOD 2. To simplify our work, we use the least common denominator (LCD), that is, the least common multiple of the given denomi-nators. The LCD of is 12. Answer To find the least common denominator of two fractions, we factor the denominators of the fractions completely. The LCD is the product of all of the factors of the first denominator times the factors of the second denominator that are not factors of the first. Then, to change each fraction to an equivalent form that has the LCD as the denominator, we multiply by , where x is the number by which the original denominator must be multiplied to obtain the LCD. Note that the LCD is the smallest possible common denominator. Algebraic fractions are added in the same manner as arithmetic fractions, as shown in the examples that follow. 1 6 A 2 2 B 5 2 12 3 4 A 3 3 B 5 9 12 1 6 A x x B 5 12 3 4 A x x B 5 12 x x 4 2 2 6 2 3 LCD 2 2 3 3 4 1 1 6 5 3 4 3 3 3 1 1 6 3 2 2 5 9 12 1 2 12 5 11 12 3 4 and 1 6 552 Algebraic Fractions, and Equations and Inequalities Involving Fractions Procedure T o add (or subtract) fractions that have different denominators: 1. Choose a common denominator for the fractions. 2. Change each fraction to an equivalent fraction with the chosen common denominator. 3. Write a fraction whose numerator is the sum (or difference) of the numerators of the new fractions and whose denominator is the common denominator. 4. Reduce the resulting fraction to lowest terms. EXAMPLE 3 Add: Solution How to Proceed (1) Find the LCD of the fractions: (2) Change each fraction to an equivalent fraction with the least common denominator, a2b2: (3) Write a fraction whose numerator is the sum of the numerators of the new fractions and whose denominator is the common denominator: Answer (a  0, b  0) EXAMPLE 4 Subtract: Solution LCD 3 4 12 Answer EXAMPLE 5 Express as a fraction in simplest form: Solution LCD y – 1 Answer (y  1) y2 2 2 y 2 1 y 1 1 2 1 y 2 1 5 y 1 1 1 ? A y 2 1 y 2 1 B 2 1 y 2 1 5 y2 2 1 y 2 1 2 1 y 2 1 5 y2 2 1 2 1 y 2 1 5 y2 2 2 y 2 1 y 1 1 2 1 y 2 1 5 5x 1 26 12 5 8x 1 20 2 3x 1 6 12 5 (8x 1 20) 2 (3x 2 6) 12 5 8x 1 20 12 2 3x 2 6 12 2x 1 5 3 2 x 2 2 4 5 4 4 ? 2x 1 5 3 2 3 3 ? x 2 2 4 2x 1 5 3 2 x 2 2 4 5b 1 2a a2b2 5 5b 1 2a a2b2 5 5b a2b2 1 2a a2b2 5 a2b 1 2 ab2 5 5 a2b ? b b 1 2 ab2 ? a a a2b a a b ab2 a b b LCD a a b b a2b2 5 a 2b 1 2 ab 2 Adding or Subtracting Algebraic Fractions 553 EXAMPLE 6 Subtract: Solution x2 4 (x 2)(x  2) x 2 (x 2) LCD (x 2)(x  2) Answer (x  2, 2) Writing About Mathematics 1. In Example 2, the answer is . Can we divide 2x and 6x by 2 to write the answer in lowest terms as ? Explain why or why not. 2. Joey said that if a  x. Do you agree with Joey? Explain why or why not. Developing Skills In 3–43, add or subtract the fractions as indicated. Reduce each answer to lowest terms. In each case, list the values of the variables for which the fractions are not defined. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 9a 8b 2 3a 4b 1 2x 2 1 x 1 1 8x 9 4x 1 3 2x a 7 1 b 14 5a 6 2 3a 4 8x 5 2 3x 4 1 7x 10 ab 5 1 ab 4 y 6 1 y 5 2 y 2 5x 6 2 2x 3 x 3 1 x 2 r2 1 4r r2 2 r 2 6 1 8 2 r2 r2 2 r 2 6 6x 2 5 x2 2 1 2 5x 2 6 x2 2 1 9d 1 6 2d 1 1 2 7d 1 5 2d 1 1 6y 2 4 4y 1 3 1 7 2 2y 4y 1 3 x x 1 1 1 1 x 1 1 6 10c 1 9 10c 2 3 10c 5r t 2 2s t 11 4c 1 5 4c 2 6 4c 2 2 x 2 a a 2 x 5 3 x 1 11 3x 2x 1 11 6x EXERCISES 3 x 1 2 5 3 x 1 2 5 3(x 2 2) (x 2 2)(x 1 2) 5 3x 2 6 (x 2 2)(x 1 2) 5 6x 2 3x 2 6 (x 2 2)(x 1 2) 5 6x 2 3(x 1 2) (x 2 2)(x 1 2) 6x x2 2 4 2 3 x 2 2 5 6x (x 2 2)(x 1 2) 2 3(x 1 2) (x 2 2)(x 1 2) 6x x2 2 4 2 3 x 2 2 554 Algebraic Fractions, and Equations and Inequalities Involving Fractions 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. Applying Skills In 44–46, represent the perimeter of each polygon in simplest form. 44. The lengths of the sides of a triangle are represented by , , and . 45. The length of a rectangle is represented by , and its width is represented by . 46. Each leg of an isosceles triangle is represented by , and its base is represented by . In 47 and 48, find, in each case, the simplest form of the indicated length. 47. The perimeter of a triangle is , and the lengths of two of the sides are and . Find the length of the third side. 48. The perimeter of a rectangle is , and the measure of each length is . Find the mea-sure of each width. 49. The time t needed to travel a distance d at a rate of speed r can be found by using the for-mula . a. For the first part of a trip, a car travels x miles at 45 miles per hour. Represent the time that the car traveled at that speed in terms of x. b. For the remainder of the trip, the car travels 2x  20 miles at 60 miles per hour. Represent the time that the car traveled at that speed in terms of x. c. Express, in terms of x, the total time for the two parts of the trip. 50. Ernesto walked 2 miles at a miles per hour and then walked 3 miles at (a 1) miles per hour. Represent, in terms of a, the total time that he walked. 51. Fran rode her bicycle for x miles at 10 miles per hour and then rode (x  3) miles farther at 8 miles per hour. Represent, in terms of x, the total time that she rode. t 5 d r x 1 2 3 14x 15 2x 2 5 12 3x 8 17x 24 6x 2 18 21 2x 2 3 7 x 2 4 3 x 1 3 4 7x 10 3x 5 x 2 2a 1 7 a2 2 2a 2 15 2 3a 2 4 a2 2 7a 1 10 7a (a 2 1)(a 1 3) 1 2a 2 5 (a 1 3)(a 1 2) x 1 2y 3x 1 12y 2 6x 2 y x2 1 3xy 2 4y2 2x 2 1 x 1 2 1 2x 2 3 x 2 5 2 x x 1 3 a 1 1 1 1 a 1 1 1 y 2 3 1 2 y 1 4 1 2 3 x x2 2 36 2 4 3x 1 18 6 y2 2 16 2 5 y 1 4 5 y2 2 9 2 3 y 2 3 1x 8x 2 8 2 3x 4x 2 4 10 3x 2 6 1 3 2x 2 4 2 3a 2 1 1 7 15a 2 5 9 y 1 1 2 3 4y 1 4 5 x 2 3 1 7 2x 2 6 5 2 2x x 1 y 3 1 5 x 1 1 3c 2 7 2c 2 3c 2 3 6c2 y 2 4 4y2 1 3y 2 5 3y b 2 3 5b 2 b 1 2 10b 3y 2 4 5 2 y 2 2 4 a 2 3 3 1 a 1 1 6 d 1 7 5d Adding or Subtracting Algebraic Fractions 555 The following equations contain fractional coefficients: Each of these equations can be solved by finding an equivalent equation that does not contain fractional coefficients. This can be done by multiplying both sides of the equation by a common denominator for all the fractions present in the equation. We usually multiply by the least common denominator, the LCD. Note that the equation 0.5x 10 can also be written as , since a dec-imal fraction can be replaced by a common fraction. EXAMPLE 1 Solve and check: Solution How to Proceed Check (1) Write the equation: (2) Find the LCD: LCD 3 5 15 (3) Multiply both sides of the equation by the LCD: (4) Use the distributive property: (5) Simplify: 5x  3x 120 (6) Solve for x: 8x 120 x 15 Answer x 15 15A x 3 B 1 15A x 5 B 5 15(8) 15A x 3 1 x 5 B 5 15(8) x 3 1 x 5 5 8 x 3 1 x 5 5 8 1 2x 5 10 x 3 1 60 5 5x 6 1 3x 1 60 5 5 6x x 2 5 10 1 2x 5 10 14-6 SOLVING EQUATIONS WITH FRACTIONAL COEFFICIENTS 556 Algebraic Fractions, and Equations and Inequalities Involving Fractions Procedure T o solve an equation that contains fractional coefficients: 1. Find the LCD of all coefficients. 2. Multiply both sides of the equation by the LCD. 3. Solve the resulting equation using the usual methods. 4. Check in the original equation. 5  3 8 8 ✔ 5 ? 8 15 3 1 15 5 5 ? 8 x 3 1 x 5 5 8 EXAMPLE 2 Solve: a. b. Solution a. b. LCD 4 LCD 30 5(2x  7) 3(2x 9) 90 10x  35 6x + 27 90 Answer 4x  62 90 4x 28 x 7 Answer In Example 2, the check is left to you. EXAMPLE 3 A woman purchased stock in the PAX Company over 3 months. In the first month, she purchased one-half of her present number of shares. In the second month, she bought two-fifths of her present number of shares. In the third month, she purchased 14 shares. How many shares of PAX stock did the woman purchase? Solution Let x total number of shares of stock purchased. Then number of shares purchased in month 1, number of shares purchased in month 2, 14 number of shares purchased in month 3. The sum of the shares purchased over 3 months is the total number of shares.   14 x 10   10(x) 5x  4x  140 10x 9x  140 10x 140 x 14B 2 5x A 1 2x 2 5x 1 2x month 1 1 month 2 1 month 3 5 total 2 5x 1 2x x 5 40 2x 5 80 3x 5 80 1 x 30A 2x 1 7 6 B 2 30A 2x 2 9 10 B 5 30(3) 4A 3x 4 B 5 4(20) 1 4A x 4 B 30A 2x 1 7 6 2 2x 2 9 10 B 5 30(3) 4A 3x 4 B 5 4A20 1 x 4 B 2x 1 7 6 2 2x 2 9 10 5 3 3x 4 5 20 1 x 4 2x 1 7 6 2 2x 2 9 10 5 3 3x 4 5 20 1 x 4 Solving Equations with Fractional Coefficients 557 Check month 1 70, month 2 56, month 3 14 70  56  14 140 ✔ Answer 140 shares EXAMPLE 4 In a child’s coin bank, there is a collection of nickels, dimes, and quarters that amounts to $3.20. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there? Solution Let x the number of nickels. Then 3x the number of quarters, and x  5 the number of dimes. Also, 0.05x the value of the nickels, 0.25(3x) the value of the quarters, and 0.10(x  5) the value of the dimes. Write the equation for the value of the coins. To simplify the equation, which contains coefficients that are decimal fractions with denominators of 100, multiply each side of the equation by 100. The total value of the coins is $3.20. 0.05x  0.25(3x)  0.10(x  5) 3.20 100[0.05x  0.25(3x)  0.10(x  5)] 100(3.20) 5x  25(3x)  10(x  5) 320 5x  75x  10x  50 320 90x  50 320 90x 270 x 3 Check There are 3 nickels, 3(3) 9 quarters, and 3  5 8 dimes. The value of 3 nickels is $0.05(3) $0.15 The value of 9 quarters is $0.25(9) $2.25 The value of 8 dimes is $0.10(8) $0.80 $3.20 ✔ Answer There are 3 nickels, 9 quarters, and 8 dimes. 2 5(140) 1 2(140) 558 Algebraic Fractions, and Equations and Inequalities Involving Fractions Note: In a problem such as this, a chart such as the one shown below can be used to organize the information: Writing About Mathematics 1. Abby solved the equation 0.2x 0.84 3x as follows: 8.4 x Is Abby’s solution correct? Explain why or why not. 2. In order to write the equation 0.2x 0.84 3x as an equivalent equation with integral coefficients, Heidi multiplied both sides of the equation by 10. Will Heidi’s method lead to a correct solution? Explain why or why not. Compare Heidi’s method with multiplying by 100 or multiplying by 1,000. Developing Skills In 3–37, solve each equation and check. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 0.03y 1.2 8.7 24. 0.4x  0.08 4.24 25. 2c  0.5c 50 26. 0.08y 0.9 0.02y 27. 1.7x 30  0.2x 28. 0.02(x  5) 8 29. 0.05(x 8) 0.07x 3m 1 1 4 5 2 2 3 2 2m 6 t 2 3 6 2 t 2 25 5 5 4 y 1 2 4 2 y 2 3 3 5 1 2 7y 12 2 1 4 5 2y 2 5 3 a 2 1 a 3 1 a 4 5 26 3t 4 2 6 5 t 12 r 3 2 r 6 5 2 10 5 x 3 1 x 7 x 5 1 x 3 5 8 15 3y 1 1 4 5 44 2 y 5 2x 1 1 3 5 6x 2 9 5 m 2 5 35 5 5 7 5x 2 5 15 4 5y 2 30 7 5 0 2r 1 6 5 5 24 m 2 2 9 5 3 x 1 8 4 5 6 3x 5 5 15 1 6t 5 18 x 7 5 3 0.2x 0.84 3x 0.2x 0.2x 0.84 0.1x EXERCISES Solving Equations with Fractional Coefficients 559 Coins Number of Coins Value of One Coin T otal Value Nickels x 0.05 0.05x Quarters 3x 0.25 0.25(3x) Dimes x  5 0.10 0.10(x  5) 30. 0.4(x 9) 0.3(x  4) 31. 0.06(x 5) 0.04(x  8) 32. 0.04x  0.03(2,000 x) 75 33. 0.02x  0.04(1,500 x) 48 34. 0.05x  10 0.06(x  50) 35. 0.08x 0.03(x  200) 4 36. 37. 38. The sum of one-half of a number and one-third of that number is 25. Find the number. 39. The difference between one-fifth of a positive number and one-tenth of that number is 10. Find the number. 40. If one-half of a number is increased by 20, the result is 35. Find the number. 41. If two-thirds of a number is decreased by 30, the result is 10. Find the number. 42. If the sum of two consecutive integers is divided by 3, the quotient is 9. Find the integers. 43. If the sum of two consecutive odd integers is divided by 4, the quotient is 10. Find the integers. 44. In an isosceles triangle, each of the congruent sides is two-thirds of the base. The perimeter of the triangle is 42. Find the length of each side of the triangle. 45. The larger of two numbers is 12 less than 5 times the smaller. If the smaller number is equal to one-third of the larger number, find the numbers. 46. The larger of two numbers exceeds the smaller by 14. If the smaller number is equal to three-fifths of the larger, find the numbers. 47. Separate 90 into two parts such that one part is one-half of the other part. 48. Separate 150 into two parts such that one part is two-thirds of the other part. Applying Skills 49. Four vegetable plots of unequal lengths and of equal widths are arranged as shown. The length of the third plot is one-fourth the length of the second plot. The length of the fourth plot is one-half the length of the second plot. The length of the first plot is 10 feet more than the length of the fourth plot. If the total length of the four plots is 100 feet, find the length of each plot. 50. Sam is now one-sixth as old as his father. In 4 years, Sam will be one-fourth as old as his father will be then. Find the ages of Sam and his father now. 51. Robert is one-half as old as his father. Twelve years ago, he was one-third as old as his father was then. Find their present ages. 0.1a 6 2 0.3a 4 5 3 0.4a 3 1 0.2a 4 5 2 560 Algebraic Fractions, and Equations and Inequalities Involving Fractions 1 2 3 4 52. A coach finds that, of the students who try out for track, 65% qualify for the team and 90% of those who qualify remain on the team throughout the season. What is the smallest num-ber of students who must try out for track in order to have 30 on the team at the end of the season? 53. A bus that runs once daily between the villages of Alpaca and Down makes only two stops in between, at Billow and at Comfort. Today, the bus left Alpaca with some passengers. At Billow, one-half of the passengers got off, and six new ones got on. At Comfort, again one-half of the passengers got off, and, this time, five new ones got on. At Down, the last 13 pas-sengers on the bus got off. How many passengers were aboard when the bus left Alpaca? 54. Sally spent half of her money on a present for her mother. Then she spent one-quarter of the cost of the present for her mother on a treat for herself. If Sally had $6.00 left after she bought her treat, how much money did she have originally? 55. Bob planted some lettuce seedlings in his garden. After a few days, one-tenth of these seedlings had been eaten by rabbits. A week later, one-fifth of the remaining seedlings had been eaten, leaving 36 seedlings unharmed. How many lettuce seedlings had Bob planted originally? 56. May has 3 times as many dimes as nickels. In all, she has $1.40. How many coins of each type does she have? 57. Mr. Jantzen bought some cans of soup at $0.39 per can, and some packages of frozen veg-etables at $0.59 per package. He bought twice as many packages of vegetables as cans of soup. If the total bill was $9.42, how many cans of soup did he buy? 58. Roger has $2.30 in dimes and nickels. There are 5 more dimes than nickels. Find the number of each kind of coin that he has. 59. Bess has $2.80 in quarters and dimes. The number of dimes is 7 less than the number of quarters. Find the number of each kind of coin that she has. 60. A movie theater sold student tickets for $5.00 and full-price tickets for $7.00. On Saturday, the theater sold 16 more full-price tickets than student tickets. If the total sales on Saturday were $1,072, how many of each kind of ticket were sold? 61. Is it possible to have $4.50 in dimes and quarters, and have twice as many quarters as dimes? Explain. 62. Is it possible to have $6.00 in nickels, dimes, and quarters, and have the same number of each kind of coin? Explain. 63. Mr. Symms invested a sum of money in 7% bonds. He invested $400 more than this sum in 8% bonds. If the total annual interest from these two investments is $257, how much did he invest at each rate? 64. Mr. Charles borrowed a sum of money at 10% interest. He borrowed a second sum, which was $1,500 less than the first sum, at 11% interest. If the annual interest on these two loans is $202.50, how much did he borrow at each rate? Solving Equations with Fractional Coefficients 561 In our modern world, many problems involve inequalities.A potential buyer may offer at most one amount for a house, while the seller will accept no less than another amount. Inequalities that contain fractional coefficients are handled in much the same way as equations that contain fractional coef-ficients. The chart on the right helps us to translate words into algebraic symbols. EXAMPLE 1 Solve the inequality, and graph the solution set on a number line: a. b. Solution a. Since no domain was given, use the domain of real numbers. Answer: x  12 0 2 4 6 8 10 12 14 16 x . 12 2x 2 x . 12 6A x 3 B 2 6A x 6 B . 12 6A x 3 2 x 6 B . 6(2) x 3 2 x 6 . 2 3y 2 1 8 2 4y 7 # 3 x 3 2 x 6 . 2 14-7 SOLVING INEQUALITIES WITH FRACTIONAL COEFFICIENTS 562 Algebraic Fractions, and Equations and Inequalities Involving Fractions Words Symbols a is greater than b a  b a is less than b a b a is at least b a b a is no less than b a is at most b a b a is no greater than b Procedure T o solve an inequality that contains fractional coefficients: 1. Find the LCD, a positive number. 2. Multiply both sides of the inequality by the LCD. 3. Solve the resulting inequality using the usual methods. b. Since no domain was given, use the domain of real numbers. Answer: y 2 –1 0 1 2 3 y # 2 13y # 26 21y 1 16 2 8y # 42 14A 3y 2 B 1 14A 8 2 4y 7 B # 42 14A 3y 2 1 8 2 4y 7 B # 14(3) 3y 2 1 8 2 4y 7 # 3 EXAMPLE 2 Two boys want to pool their money to buy a comic book. The younger of the boys has one-third as much money as the older. Together they have more than $2.00. Find the smallest possible amount of money each can have. Solution Let x the number of cents that the older boy has. Then the number of cents that the younger boy has. The sum of their money in cents is greater than 200. x   200 3x + x  600 4x  600 x  150  50 The number of cents that the younger boy has must be an integer greater than 50.The number of cents that the older boy has must be a multiple of 3 that is greater than 150. The younger boy has at least 51 cents. The older boy has at least 153 cents. The sum of 51 and 153 is greater than 200. Answer The younger boy has at least $0.51 and the older boy has at least $1.53. Writing About Mathematics 1. Explain the error in the following solution of an inequality. 2. In Example 2, what is the domain for the variable? Developing Skills In 3–23, solve each inequality, and graph the solution set on a number line. 3. 4. 5. 6. 7. 8. 9. 10. 11. 2.5x 1.7x  4 1 1 2x 3 $ x 2 t 10 # 4 1 t 5 y 9 2 y 4 . 5 36 y 6 $ y 12 1 1 x 4 2 x 8 # 5 8 5 6c . 1 3c 1 3 y 2 2 3y , 5 1 4x 2 1 5x . 9 20 x . 6 23A x 23 B . 23(22) x 23 . 2 2 EXERCISES 1 3x 3Ax 1 1 3xB . 3(200) 1 3x 1 3x Solving Inequalities with Fractional Coefficients 563 12. 2y  3 0.2y 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. If one-third of an integer is increased by 7, the result is at most 13. Find the largest possible integer. 25. If two-fifths of an integer is decreased by 11, the result is at least 4. Find the smallest possi-ble integer. 26. The sum of one-fifth of an integer and one-tenth of that integer is less than 40. Find the greatest possible integer. 27. The difference between three-fourths of a positive integer and one-half of that integer is greater than 28. Find the smallest possible integer. 28. The smaller of two integers is two-fifths of the larger, and their sum is less than 40. Find the largest possible integers. 29. The smaller of two positive integers is five-sixths of the larger, and their difference is greater than 3. Find the smallest possible integers. Applying Skills 30. Talk and Tell Answering Service offers customers two monthly options. Measured Service Unmeasured Service base rate is $15 base rate is $20 each call costs $0.10 no additional charge per call Find the least number of calls for which unmeasured service is cheaper than measured service. 31. Paul earned some money mowing lawns. He spent one-half of this money for a book, and then one-third for a CD. If he had less than $3 left, how much money did he earn? 32. Mary bought some cans of vegetables at $0.89 per can, and some cans of soup at $0.99 per can. If she bought twice as many cans of vegetables as cans of soup, and paid at least $10, what is the least number of cans of vegetables she could have bought? 33. A coin bank contains nickels, dimes, and quarters. The number of dimes is 7 more than the number of nickels, and the number of quarters is twice the number of dimes. If the total value of the coins is no greater than $7.20, what is the greatest possible number of nickels in the bank? 34. Rhoda is two-thirds as old as her sister Alice. Five years from now, the sum of their ages will be less than 60. What is the largest possible integral value for each sister’s present age? OPTION 2 OPTION 1 2 2 a 2 # a 1 2 5 2 2a 1 3 6 3t 2 4 3 $ 2t 1 4 6 1 5t 2 1 9 2r 2 3 5 2 r 2 3 3 # 2 2y 2 3 3 1 y 1 1 2 , 10 6x 2 3 2 . 37 10 1 x 1 2 5 3x 2 30 6 , x 3 2 2 2m 3 $ 7 2 m 4 1 1 4c 3 2 7 9 $ c 2 1 7 6 2d 1 1 4 , 7d 12 1 5 3 5y 2 30 7 # 0 3x 2 1 7 . 5 564 Algebraic Fractions, and Equations and Inequalities Involving Fractions 35. Four years ago, Bill was times as old as his cousin Mary. The difference between their present ages is at least 3. What is the smallest possible integral value for each cousin’s present age? 36. Mr. Drew invested a sum of money at interest. He invested a second sum, which was $200 less than the first, at 7% interest. If the total annual interest from these two invest-ments is at least $160, what is the smallest amount he could have invested at ? 37. Mr. Lehtimaki wanted to sell his house. He advertised an asking price, but knew that he would accept, as a minimum, nine-tenths of the asking price. Mrs. Patel offered to buy the house, but her maximum offer was seven-eighths of the asking price. If the difference between the seller’s lowest acceptance price and the buyer’s maximum offer was at least $3,000, find: a. the minimum asking price for the house; b. the minimum amount Mr. Lehtimaki, the seller, would accept; c. the maximum amount offered by Mrs. Patel, the buyer. 38. When packing his books to move, Philip put the same number of books in each of 12 boxes. Once packed, the boxes were too heavy to lift so Philip removed one-fifth of the books from each box. If at least 100 books in total remain in the boxes, what is the minimum number of books that Philip originally packed in each box? An equation is called an algebraic equation when a variable appears in at least one of its sides. An algebraic equation is a fractional equation when a variable appears in the denominator of one, or more than one, of its terms. For example, are all fractional equations.To simplify such an equation, clear it of fractions by multiplying both sides by the least common denominator of all fractions in the equation. Then, solve the simpler equation. As is true of all algebraic fractions, a fractional equation has meaning only when values of the variable do not lead to a denominator of 0. KEEP IN MIND When both sides of an equation are multiplied by a variable expression that may represent 0, the resulting equation may not be equivalent to the given equation. Such equations will yield extraneous solutions, which are solutions that satisfy the derived equation but not the given equation. Each solu-tion, therefore, must be checked in the original equation. 1 y2 1 2y 2 3 1 1 y 2 2 5 3 a2 1 1 a 2 1 1 a 2 5 a 1 2 2 3d 1 1 3 5 11 6d 2 1 4 1 3 1 1 x 5 1 2 14-8 SOLVING FRACTIONAL EQUATIONS 71 2% 71 2% 11 4 Solving Fractional Equations 565 EXAMPLE 1 Solve and check: Solution Multiply both sides of the equation Check by the least common denominator, 6x. Answer x 6 EXAMPLE 2 Solve and check: Solution Multiply both sides of the equation by Check the least common denominator, x  2. The only possible value of x is a value for which the equation has no meaning because it leads to a denominator of 0. Therefore, there is no solution for this equation. Answer The solution set is the empty set, or { }. EXAMPLE 3 Solve and check: 2 x 5 6 2 x 4 x 5 22 22x 5 4 5x 1 10 5 7x 1 14 (x 1 2) A 5x 1 10 x 1 2 B 5 (x 1 2)(7) 5x 1 10 x 1 2 5 7 5x 1 10 x 1 2 5 7 6 5 x 2x 1 6 5 3x 6xA 1 3 B 1 6xA 1 x B 5 6xA 1 2 B 6xA 1 3 1 1 x B 5 6xA 1 2 B 1 3 1 1 x 5 1 2 1 3 1 1 x 5 1 2 566 Algebraic Fractions, and Equations and Inequalities Involving Fractions ✔ 1 2 5 1 2 3 6 5 ? 1 2 2 6 1 1 6 5 ? 1 2 1 3 1 1 6 5 ? 1 2 1 3 1 1 x 5 1 2 ✘ 0 0 5 7 210 1 10 0 5 ? 7 5(22) 1 10 22 1 2 5 ? 7 5x 1 10 x 1 2 5 7 Solution METHOD 1 Multiply both sides of the equation by the LCD, 4x: x 2 0 x 4 0 x 2 x 4 Check x 2 x 4 ✔ ✔ Answer x 2 or x 4 EXAMPLE 4 Solve and check: Solution Multiply both sides of the equation by the LCD, : Answer x 0 x 5 0 3x 5 0 4x 1 2 5 x 1 2 2(x 2 1) 1 2(x 1 2) 5 x 1 2 2(x 1 2)(x 2 1) x 1 2 1 2(x 1 2)(x 2 1) x 2 1 5 2(x 1 2)(x 2 1) 2(x 2 1) 2(x 1 2)(x 2 1) x 1 2 1 2(x 1 2)(x 2 1) x 2 1 5 2(x 1 2)(x 2 1) 2(x 2 1) 2(x 1 2)(x 2 1) A 1 x 1 2 1 1 x 2 1 B 5 A 1 2(x 2 1) B2(x 1 2)(x 2 1) 2(x 1 2)(x 2 1) 1 x 1 2 1 1 x 2 1 5 1 2(x 2 1) 1 2 5 1 2 1 5 1 2 4 5 6 2 4 4 2 2 5 6 2 2 4 2 x 5 6 2 x 4 2 x 5 6 2 x 4 (x 2 2)(x 2 4) 5 0 x2 2 6x 1 8 5 0 8 5 6x 2 x2 8 5 x(6 2 x) 4xA 2 x B 5 4xA 6 2 x 4 B Solving Fractional Equations 567 METHOD 2 Use the rule for proportion: the product of the means equals the product of the extremes. x 2 0 x 4 0 x 2 x 4 (x 2 2)(x 2 4) 5 0 x2 2 6x 1 8 5 0 2x2 1 6x 2 8 5 0 6x 2 x2 5 8 x(6 2 x) 5 8 2 x 5 6 2 x 4 Check ✔ 21 2 5 21 2 1 2 1 1 21 5 ? 1 2(21) 1 0 1 2 1 1 0 2 1 5 ? 1 2(0 2 1) Writing About Mathematics 1. Nathan said that the solution set of is the set of all real numbers. Do you agree with Nathan? Explain why or why not. 2. Pam multiplied each side of the equation by (y + 5)(y 5) to obtain the equation y  5 3y 15, which has as its solution y 10. Pru said that the equation is a proportion and can be solved by writing the product of the means equal to the product of the extremes. She obtained the equation 3(y2 25) (y  5)2, which has as its solution 10 and 5. Both girls used a correct method of solution. Explain the differ-ence in their answers. Developing Skills In 3–6, explain why each fractional equation has no solution. 3. 4. 5. 6. In 7–45, solve each equation, and check. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 1 x 1 2 1 1 2x 1 4 5 23 2x 1 x 2 1 9 5 x 2 5 18 1 b 2 1 1 1 6 5 b 1 2 12 3 2b 1 1 5 b 2 1 b 5 b 2 1 2 a 2 1 6 5 1 a 2 x 5 x 2 3 2 x x2 2 9 5 1 x 1 3 y y 1 1 2 1 y 5 1 12y 8y 1 5 5 2 3 12 2 2 x 5 15 7 1 x 2 a 2 4 5 5 a 2 1 2 m 5 5 3m 2 1 5 a 5 7 a 2 4 3 y 5 2 5 2 y 1 2 r 1 1 r 5 2 3 4z 7 1 5z 5 1 3 3 5 2 3a 5 1 2 5x x 1 1 5 4 2 3x 2 4 5 1 4 6 3x 2 1 5 3 4 x 1 1 2x 1 2x 1 1 3x 5 1 a a 1 2 2 a 2 2 a 5 1 a 2 1 x 6x 5 3 5x 1 1 30 5 1 x 2x 2 1 5 x 1 1 x 7 8 2 x 2 3 x 5 1 4 b 2 6 b 2 1 6 5 4 b 1 3a 1 5 12 5 2 a y 1 9 2y 1 3 5 15 y x 2 5 x 1 3 x 5 2 3 y 2 2 2y 5 3 8 30 x 5 7 1 18 2x 9 2x 5 7 2x 1 2 15 y 2 3 y 5 4 10 x 1 8 x 5 9 15 4x 5 1 8 3 2x 5 1 2 15 y 5 3 10 x 5 5 x x 2 1 1 2 5 1 x 2 1 2 x 5 4 1 2 x 4a 1 4 a 1 1 5 5 6x x 5 3 y 1 5 y2 2 25 5 3 y 1 5 y 1 5 y2 2 25 5 3 y 1 5 2 r 2 5 5 10 5r 2 25 EXERCISES 568 Algebraic Fractions, and Equations and Inequalities Involving Fractions In 46–49, solve each equation for x in terms of the other variables. 46. 47. 48. 49. 50. If , , a  0, and c  0, is it possible to know the numerical value of x without knowing numerical values of a, b, c, and y? Explain your answer Applying Skills 51. If 24 is divided by a number, the result is 6. Find the number. 52. If 10 is divided by a number, the result is 30. Find the number. 53. The sum of 20 divided by a number, and 7 divided by the same number, is 9. Find the num-ber. 54. When the reciprocal of a number is decreased by 2, the result is 5. Find the number. 55. The numerator of a fraction is 8 less than the denominator of the fraction. The value of the fraction is . Find the fraction. 56. The numerator and denominator of a fraction are in the ratio 3 : 4. When the numerator is decreased by 4 and the denominator is increased by 2, the value of the new fraction, in sim-plest form, is . Find the original fraction. 57. The ratio of boys to girls in the chess club is 4 to 5. After 2 boys leave the club and 2 girls join, the ratio is 1 to 2. How many members are in the club? 58. The length of Emily’s rectangular garden is 4 feet greater than its width. The width of Sarah’s rectangular garden is equal to the length of Emily’s and its length is 18 feet. The two gardens are similar rectangles, that is, the ratio of the length to the width of Emily’s garden equals the ratio of the length to the width of Sarah’s garden. Find the possible dimensions of each garden. (Two answers are possible.) CHAPTER SUMMARY An algebraic fraction is the quotient of two algebraic expressions. If the algebraic expressions are polynomials, the fraction is called a rational expres-sion or a fractional expression. An algebraic fraction is defined only if values of the variables do not result in a denominator of 0. Fractions that are equal in value are called equivalent fractions. A fraction is reduced to lowest terms when an equivalent fraction is found such that its numerator and denominator have no common factor other than 1 or 1. This fraction is considered a lowest terms fraction. 1 2 3 5 b 5 a c x 5 by c , y 5 c2 a d x 5 d 2 1 x 2 e a 1 b x 5 c t x 2 k 5 5k t x 2 k 5 0 Chapter Summary 569 570 Algebraic Fractions, and Equations and Inequalities Involving Fractions Operations with algebraic fractions follow the same rules as operations with arithmetic fractions: Multiplication Division (x  0, y  0) (x  0, y  0, b  0) Addition/subtraction with the same denominator (c  0), (c  0) Addition/subtraction with different denominators (first, obtain the common denominator): (b  0, d  0) A fractional equation is an equation in which a variable appears in the denominator of one or more than one of its terms. To simplify a fractional equa-tion,or any equation or inequality containing fractional coefficients,multiply both sides by the least common denominator (LCD) to eliminate the fractions. Then solve the simpler equation or inequality and check for extraneous solutions. VOCABULARY 14-1 Algebraic fraction • Fractional expression • Rational expression 14-2 Reduced to lowest terms • Lowest terms fraction • Equivalent fractions • Division property of a fraction • Cancellation • Multiplication property of a fraction 14-3 Cancellation method 14-5 Common denominator • Least common denominator 14-8 Algebraic equation • Fractional equation • Extraneous solution REVIEW EXERCISES 1. Explain the difference between an algebraic fraction and a fractional expression. 2. What fractional part of 1 centimeter is x millimeters? 3. For what value of y is the fraction undefined? 4. Factor completely: 12x3 27x y 2 1 y 2 4 a b 1 c d 5 a b ? d d 1 c d ? b b 5 ad bd 1 bc bd 5 ad 1 bc bd a c 2 b c 5 a 2 b c a c 1 b c 5 a 1 b c a x 4 b y 5 a x ? y b 5 ay bx a x ? b y 5 ab xy In 5–8, reduce each fraction to lowest terms. 5. 6. 7. 8. In 9–23, in each case, perform the indicated operation and express the answer in lowest terms. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. If the sides of a triangle are represented by , , and , express the perimeter of the triangle in simplest form. 25. If a 2, b 3, and c 4, what is the sum of ? In 26–31, solve each equation and check. 26. 27. 28. 29. 30. 31. In 32–34, solve each equation for r in terms of the other variables. 32. 33. 34. 35. Mr. Vroman deposited a sum of money in the bank. After a few years, he found that the interest equaled one-fourth of his original deposit and he had a total sum, deposit plus interest, of $2,400 in the bank. What was the original deposit? 36. One-third of the result obtained by adding 5 to a certain number is equal to one-half of the result obtained when 5 is subtracted from the number. Find the number. 37. Of the total number of points scored by the winning team in a basketball game, one-fifth was scored in the first quarter, one-sixth was scored in the second quarter, one-third was scored in the third quarter, and 27 was scored in the fourth quarter. How many points did the winning team score? 38. Ross drove 300 miles at r miles per hour and 360 miles at r  10 miles per hour. If the time needed to drive 300 miles was equal to the time needed to drive 360 miles, find the rates at which Ross drove. (Express the time needed for each part of the trip as .) t 5 d r a r 2 n 5 0 c 2r 5 p S h 5 2pr 1 a 2 1 2 1 a 5 1 20 2t 5 2 t 2 2 10 5 2 6 m 5 20 m 2 2 y 2 2 y 6 5 4 x 2 3 10 5 4 5 k 20 5 3 4 b a 1 a c 2b 3 5b 6 b 2 3a 2 9a2 a 4 (1 2 9a2) c 2 3 12 1 c 1 3 8 x2 2 25 12 4 x2 2 10x 1 25 4 y 1 7 5 2 y 1 3 4 2a a 1 b 1 2b a 1 b x2 2 5x x2 ? x 2x 2 10 4x 1 5 3x 1 2x 1 1 2x 5 xy 2 2 yz ax 3 1 ax 4 9 k 2 3 k 1 4 k 5m 6 2 m 6 3a 7b 4 18a 35 6c2 4 c 2 2x 2 2 3y ? 3xy 2x 3x2 4 ? 8 9x 8y2 2 12y 8y 5x2 2 60 5 14d 7d2 8bg 12bg Review Exercises 571 39. The total cost, T, of n items that cost a dollars each is given by the equa-tion T na. a. Solve the equation T na for n in terms of T and a. b. Use your answer to a to express n1, the number of cans of soda that cost $12.00 if each can of soda costs a dollars. c. Use your answer to a to express n2, the number of cans of soda that cost $15.00 if each can of soda costs a dollars. d. If the number of cans of soda purchased for $12.00 is 4 less than the number purchased for $15.00, find the cost of a can of soda and the number of cans of soda purchased. 40. The cost of two cups of coffee and a bagel is $1.75. The cost of four cups of coffee and three bagels is $4.25. What is the cost of a cup of coffee and the cost of a bagel? 41. A piggybank contains nickels, dimes, and quarters. The number of nickels is 4 more than the number of dimes, and the number of quarters is 3 times the number of nickels. If the total value of the coins is no greater than $8.60, what is the greatest possible number of dimes in the bank? Exploration Some rational numbers can be written as terminating decimals and others as infinitely repeating decimals. (1) Write each of the following fractions as a decimal: (2) What do you observe about the decimals written in (1)? (3) Write each denominator in factored form. (4) What do you observe about the factors of the denominators? (5) Write each of the following fractions as a decimal: What do you observe about the decimals written in (5)? (7) Write each denominator in factored form. (8) What do you observe about the factors of the denominators? (9) Write a statement about terminating and infinitely repeating decimals based on your observations. 1 3, 1 6, 1 9, 1 11, 1 12, 1 15, 1 18, 1 22, 1 24, 1 30 1 2, 1 4, 1 5, 1 8, 1 10, 1 16, 1 20, 1 25, 1 50, 1 100 572 Algebraic Fractions, and Equations and Inequalities Involving Fractions CUMULATIVE REVIEW CHAPTERS 1–14 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The product of 3a2 and 5a5 is (1) 15a10 (2) 15a7 (3) 8a10 (4) 8a7 2. In the coordinate plane, the point whose coordinates are (2, 1) is in quadrant (1) I (2) II (3) III (4) IV 3. In decimal notation, 3.75  10–2 is (1) 0.0375 (2) 0.00375 (3) 37.5 (4) 375 4. The slope of the line whose equation is 3x y 5 is (1) 5 (2) 5 (3) 3 (4) 3 5. Which of the following is an irrational number? (1) 1.3 (2) (3) (4) 6. The factors of x2 7x 18 are (1) 9 and 2 (3) (x 9) and (x  2) (2) 9 and 2 (4) (x  9) and (x 2) 7. The dimensions of a rectangular box are 8 by 5 by 9. The surface area is (1) 360 cubic units (3) 157 square units (2) 360 square units (4) 314 square units 8. The length of one leg of a right triangle is 8 and the length of the hypotenuse is 12. The length of the other leg is (1) 4 (2) (3) (4) 80 9. The solution set of is (1) {1, 2} (2) {2} (3) {0, 1} (4) {1, 2} 10. In the last n times a baseball player was up to bat, he got 3 hits and struck out the rest of the times. The ratio of hits to strike-outs is (1) (2) (3) (4) Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitu-tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numer-ical answer with no work shown will receive only 1 credit. n 2 3 3 3 n 2 3 n 2 3 n 3 n a2 a 1 1 2 1 5 1 a 1 1 4"13 4"5 "5 "9 2 3 Cumulative Review 573 11. Mrs. Kniger bought some stock on May 1 for $3,500. By June 1, the value of the stock was $3,640. What was the percent of increase of the cost of the stock? 12. A furlong is one-eighth of a mile. A horse ran 10 furlongs in 2.5 minutes. What was the speed of the horse in feet per second? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitu-tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numer-ical answer with no work shown will receive only 1 credit. 13. Two cans of soda and an order of fries cost $2.60. One can of soda and two orders of fries cost $2.80. What is the cost of a can of soda and of an order of fries? 14. a. Draw the graph of y x2 2x . b. From the graph, determine the solution set of the equation x2 2x 3. Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitu-tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numer-ical answer with no work shown will receive only 1 credit. 15. If the measure of the smallest angle of a right triangle is 32° and the length of the shortest side is 36.5 centimeters, find the length of the hypotenuse of the triangle to the nearest tenth of a centimeter. 16. The area of a garden is 120 square feet. The length of the garden is 1 foot less than twice the width. What are the dimensions of the garden? 574 Algebraic Fractions, and Equations and Inequalities Involving Fractions
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Skip to lesson content High school geometry Course: High school geometry>Unit 6 Lesson 5: Equations of parallel & perpendicular lines Parallel lines from equation Parallel lines from equation (example 2) Parallel lines from equation (example 3) Perpendicular lines from equation Parallel & perpendicular lines from equation Writing equations of perpendicular lines Writing equations of perpendicular lines (example 2) Write equations of parallel & perpendicular lines Proof: parallel lines have the same slope Proof: perpendicular lines have opposite reciprocal slopes Analytic geometry FAQ Math> High school geometry> Analytic geometry> Equations of parallel & perpendicular lines © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Parallel & perpendicular lines from equation GA.Math: A.GSR.3.1 Google Classroom Microsoft Teams Problem What do the following two equations represent? y=4 3 x+5 3‍ y−8=4 3(x+6)‍ Choose 1 answer: Choose 1 answer: (Choice A) The same line A The same line (Choice B) Distinct parallel lines B Distinct parallel lines (Choice C) Perpendicular lines C Perpendicular lines (Choice D) Intersecting, but not perpendicular lines D Intersecting, but not perpendicular lines Related content Video 3 minutes 48 seconds 3:48 Parallel lines from equation Video 3 minutes 3 seconds 3:03 Parallel lines from equation (example 2) Video 2 minutes 42 seconds 2:42 Parallel lines from equation (example 3) Video 3 minutes 29 seconds 3:29 Perpendicular lines from equation Report a problem Do 4 problems Skip Check Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. 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Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: oHA-fraction of species in the form HA oA- action of species in the form A oxHA The fraction denoted here as aA is the same thing we called the fraction of dissociation (a) previously 0.9 0.8 E 0.7 0,5 0.4 0.3 0.2 pKa 0.0 FIGURE 9-3 Fractional composition diagram of a monoprotic system with pKa 5.00. Below pH 5, HA is the dominant form, whereas, above pH 5, Using the fractional composition equations for diprotic acids from class (also on p. 198 of textbook) and the Kas from the textbook, create a distribution diagram for the forms of adipic acid (also known as 1,6-hexanedioic acid). Your y axis should be alpha (i.e. fraction in a form). Plot values for each pH (integer values) in the range pH 1-14 This AI-generated tip is based on Chegg's full solution. Sign up to see more! Start by using the fractional composition equations for a diprotic acid, such as , , and , then calculate these values for integer pH values from 1 to 14. The adipic acid (diprotic acid) fract… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
187882
https://www.sciencedirect.com/science/article/pii/0304392477900144
Scenic assessment: An overview - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract References (61) Cited by (93) Landscape Planning Volume 4, 1977, Pages 109-129 Scenic assessment: An overview Author links open overlay panel Louise M.Arthur∗, Terry C.Daniel∗∗, Ron S.Boster∗∗∗ Show more Add to Mendeley Share Cite rights and content Abstract Arthur, L.M., Daniel, T.C. and Boster, R.S., 1977. Scenic assessment: An overview. Landscape Plann., 4: 109–129. The authors present a synthesis and overview of techniques developed for evaluating the scenic beauty of natural resources. Literature is grouped into three categories: descriptive inventories, public evaluations, and economic analyses. Both quantitative and non-quantitative methods within each category are discussed, strengths and weaknesses of the general approaches noted, and, occasionally, alternatives suggested. Discussions are focused on methodological soundness and on utility of the evaluative systems for management of scenic resources. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Recommended articles References (61) C.A. Acking et al. How do we verbalize what we see? Landscape Archit. (1973) L.M. Arthur Predicting scenic beauty of forest environments For. Sci. (1977) L.M. Arthur et al. Measuring Scenic Beauty: A Selected Annotated Bibliography Battelle Columbus Laboratories Development of the Arizona Environmental and Economic Trade-Off Model M.A. Boster Colorado River Trips within the Grand Canyon National Park and Monument: A Socio-economic Analysis J.S. Calvin et al. An attempt at assessing preferences for natural landscapes Environ. Behav. (1972) G.J. Cherem Looking through the eyes of the public or public images as social indications of aesthetic opportunity R.N. Clark et al. Values, behavior and conflict in modern camping culture J. Leisure Res. (1971) R.N. Clark et al. An Introduction to CODINVOLVE: A System for Analyzing, Storing, and Retrieving Public Input to Resource Decisions W.L. Cook An evaluation of the aesthetic quality of forest trees J. Leisure Res. (1972) N.H. Coomber et al. Evaluation of Environmental Intangibles (1973) R.E. Coughlin et al. The Extent of Agreement among Observers on Environmental Attractiveness Regional Science Research Institute Discussion Paper No 37 (1970) F.D. Craighead et al. River systems: recreational classification, inventory and evaluation Naturalist, J. Nat. Hist. Soc. Minn. (1962) K.H. Craik A System of Landscape Dimensions: Appraisal of its Objectivity and Illustration of its Scientific Application K.H. Craik Psychological factors in landscape appraisal Environ. Behav. (1972) T.C. Daniel et al. Measuring Landscape Esthetics: The Scenic Beauty Estimation Method T.C. Daniel et al. Quantitative evaluation of landscapes: an application of signal detection analysis to forest management alternatives Man-Environ. Syst. (1973) P. Davidson An exploratory study to identify and measure the benefits derived from scenic enhancement of federal aid highways Highw. Res. Rec. (1967) J.A. Dearinger Aesthetic and Recreational Potential of Small Naturalistic Streams near Urban Areas D.A. Dillman et al. Increasing mail questionnaire response: a four state comparison Am. Sociol. Rev. (1974) J.G. Fabos An analysis of environmental quality ranking systems K.D. Fines Landscape evaluations: a research project in East Sussex Reg. Stud. (1968) M.A. Gratzer et al. Adaptation of an Eye Movement Recorder to Esthetic Environmental Mensuration H.H. Hancock Recreation preference: its relation to user behavior J. For. (1973) G. Hardin The tragedy of the commons Science (1968) W.J. Hart et al. How to rate and rank landscapes Landscape Archit. (1967) T.A. Heberlein Social psychological assumptions of user attitude surveys: the case of the wildernism scale J. Leisure Res. (1973) J.C. Hendee et al. Wilderness users — what do they think? Am. For. (1968) J.C. Hendee et al. Wilderness users — what do they think? Am. For. (1968) View more references Cited by (93) Interactions of ecosystem properties, ecosystem integrity and ecosystem service indicators: A theoretical matrix exercise 2013, Ecological Indicators Citation Excerpt : Norton et al. (2012) have stated that there are only few cultural service indicators and that they are mostly based on the collection of qualitative data from local people and/or visitors of a specific area. Arthur et al. (1977) provide methods for evaluating scenic beauty of natural resources and Ode et al. (2008) describe in detail visual indicators to capture landscape perception. However, there might be huge differences in valuation of cultural services in different cultural areas. Show abstract The ecosystem service concept is becoming more and more acknowledged in science and decision-making, resulting in several applications in different case studies and in environmental management, but still it is developing in terms of definitions, typologies and understanding its complexity. By examining the interrelations between ecosystem properties, ecosystem integrity, biodiversity, ecosystem services and human well-being qualitatively, the mutual influences on each constituent of the ‘ecosystem service cascade’ are illuminated, giving an impulse for further discussions and improvements for a better understanding of the complexity of human–environmental systems. Results of the theoretical interactions are among others the assumption that provisioning services exclude or compete with each other, while the role of biodiversity was found to be supporting for regulating services and cultural services. Ecosystem services meet the criteria of being adequate human–environmental system indicators and therefore, they are an appropriate instrument for decision-making and management. ### Assessing the visual quality of rural landscapes 2004, Landscape and Urban Planning Citation Excerpt : Moreover, this method does not capture any interactive effects of the individual components (Dunn, 1976). Likewise, Crofts (1975) describes two types of technique for landscape evaluation: preference and surrogate component techniques, whereas, Arthur et al. (1977) used the terminology of public preference models and descriptive inventories methods. These classifications are similar to those of direct and indirect methods, respectively. Show abstract This paper presents a methodology for assessing the visual quality of agricultural landscapes through direct and indirect techniques of landscape valuation. The first technique enables us to rank agricultural landscapes on the basis of a survey of public preferences. The latter weighs the contribution of the elements and attributes contained in the picture to its overall scenic beauty via regression analysis. An application based on two Mediterranean rural areas in Andalusia in Southern Spain is presented. The photos used in the survey included man-made elements, positive and negative, agricultural fields, mainly of cereals and olive trees, and a natural park. There were 10 panels, each containing 16 photos, and 226 participants ranked the best four and worst four pictures of each panel. Each participant ranked an average of 7.3 panels. The results show that perceived visual quality increases, in decreasing order of importance, with the degree of wilderness of the landscape, the presence of well-preserved man-made elements, the percentage of plant cover, the amount of water, the presence of mountains and the colour contrast. ### Whither scenic beauty? Visual landscape quality assessment in the 21st century 2001, Landscape and Urban Planning Show abstract The history of landscape quality assessment has featured a contest between expert and perception-based approaches, paralleling a long-standing debate in the philosophy of aesthetics. The expert approach has dominated in environmental management practice and the perception-based approach has dominated in research. Both approaches generally accept that landscape quality derives from an interaction between biophysical features of the landscape and perceptual/judgmental processes of the human viewer. The approaches differ in the conceptualizations of and the relative importance of the landscape and human viewer components. At the close of the 20th century landscape quality assessment practice evolved toward a shaky marriage whereby both expert and perceptual approaches are applied in parallel and then, in some as yet unspecified way, merged in the final environmental management decision making process. The 21st century will feature continued momentum toward ecosystem management where the effects of changing spatial and temporal patterns of landscape features, at multiple scales and resolutions, will be more important than any given set of features at any one place at any one time. Valid representation of the visual implications of complex geo-temporal dynamics central to ecosystem management will present major challenges to landscape quality assessment. Technological developments in geographic information systems, simulation modeling and environmental data visualization will continue to help meet those challenges. At a more fundamental level traditional landscape assessment approaches will be challenged by the deep ecology and green philosophy movements which advocate a strongly bio-centric approach to landscape quality assessment where neither expert design principles nor human perceptions and preferences are deemed relevant. On the opposite side of the landscape–human interaction, social/cultural construction models that construe the landscape as the product of socially instructed human interpretation leave little or no role for biophysical landscape features and processes. A psychophysical approach is advocated to provide a more appropriate balance between biophysical and human perception/judgement components of an operationally delimited landscape quality assessment system. ### Landscape and the philosophy of aesthetics: Is landscape quality inherent in the landscape or in the eye of the beholder? 1999, Landscape and Urban Planning Show abstract The paper proposes that landscape quality assessment may be approached on the basis of two contrasting paradigms, one which regards quality as inherent in the physical landscape, and the other which regards quality as a product of the mind – eye of the beholder. These are termed, respectively, the objectivist and subjectivist paradigms. These paradigms underlie the surveys of the physical landscape and studies of observer preferences. Examination of these paradigms through the approaches taken by philosophers from Plato to modern times demonstrates the ubiquity of the paradigms in underlying human perception of landscape. Until recent centuries, the objectivist paradigm provided philosophers with the basis for understanding beauty, including landscape beauty. However, the philosophers Locke, Hume, Burke and particularly Immanuel Kant identified beauty as lying in the eyes of the beholder rather than in the object. The parallels between Kant's aesthetic philosophy and contemporary theories of landscape quality based on an evolutionary perspective are examined. Most philosophers over recent centuries have adopted the subjectivist view of aesthetics. The paper concludes by proposing that only the subjectivist model should be used in research of landscape quality. ### Landscape perception: Research, application and theory 1982, Landscape Planning Show abstract Landscape perception research during the past two decades has responded to legislative mandates and landscape management, planning and design issues in a number of countries. It has also engaged the interests of individuals from a variety of disciplines and professions. This paper presents an analysis of the paradigms that have been followed in assessing perceived landscape values, and identifies the theoretical or conceptual bases which underlie these approaches. Four paradigms are identified from review of over 160 articles published in 20 journals during the period 1965–1980. Publications in each paradigm (expert, psychophysical, cognitive and experiential) are reviewed with reference to contributions to pragmatic landscape planning and management issues and to the evolution of a general theory of landscape perception. Trends in publications within the paradigms are indicated over time and by professional-disciplinary orientation. Overall, the absence of an explicit theoretical foundation is noted. Arguments in support of the development of a theoretical framework for landscape perception research are advanced and a proposed framework based on an interactive perception process is presented. ### Why do preferences differ between scene types? 2001, Environment and Behavior View all citing articles on Scopus View full text Copyright © 1977 Published by Elsevier B.V. Recommended articles pySAPC, a python package for sparse affinity propagation clustering: Application to odontogenesis whole genome time series gene-expression data Biochimica et Biophysica Acta (BBA) - General Subjects, Volume 1860, Issue 11, Part B, 2016, pp. 2613-2618 Huojun Cao, Brad A.Amendt ### Restorative urban forests: Exploring the relationships between forest stand structure, perceived restorativeness and benefits gained by visitors to coastal Pinus pinea forests Ecological Indicators, Volume 90, 2018, pp. 594-605 Antonio Tomao, …, Mariagrazia Agrimi ### Single-cell RNA-seq analysis reveals the multi-step process of cellular senescence Biochemistry and Biophysics Reports, Volume 42, 2025, Article 102042 Minseo Ahn, …, Jae Ho Seo ### Establishment of an Iron Deficiency Model by Iron Depletion in Pregnant Rats Biomedical and Environmental Sciences, Volume 37, Issue 2, 2024, pp. 210-215 Xi CHEN, …, Jun Sheng HUO ### Single-cell transcriptomic insights into endosulfan-induced liver injury: Key pathways and inflammatory responses Liver Research, Volume 9, Issue 2, 2025, pp. 144-156 Pan Huang, …, Jigang Wang ### Participation of tumor suppressors long non-coding RNA MEG3, microRNA-377 and PTEN in glioma cell invasion and migration Pathology - Research and Practice, Volume 215, Issue 10, 2019, Article 152558 Dong Wang, …, De-Qing Fan Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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187883
https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/a0889c5ca8a479c3e56c544d646fb770_MIT8_06S18ch1.pdf
Chapter 1 Perturbation theory c ⃝B. Zwiebach It is often the case that the Hamiltonian of a system differs slightly from a Hamiltonian that is well studied and completely understood. This is a situation where perturbation the-ory can be useful. Perturbation theory allows us to make statements about the Hamiltonian of the system using what we know about the well studied Hamiltonian. The well studied Hamiltonian could be the that of the simple harmonic oscillator in one, two, or three dimensions. In a diatomic molecule, for example, the potential that controls the vibrations is not exactly quadratic; it has extra terms that make the vibrations slightly anharmonic. In that situation the extra terms in the potential represent perturbations of the Hamiltonian. The hydrogen atom Hamiltonian is also a well understood system. If we place the atom inside a weak external magnetic field or electric field, the situation is described by adding some small terms to the hydrogen Hamiltonian. Similarly, the interaction between the magnetic moments of the proton and the electron can be incorporated by modifying the original hydrogen atom Hamiltonian. The interaction between two neutral hydrogen atoms at a distance, leading to the van der Waals force can be studied in perturbation theory by thinking of the two atoms as electric dipoles. The Hamiltonian of interest is written as the understood, original Hamiltonian H(0), plus a perturbation δH: H(0) + δH . (1.0.1) Since H(0) is Hermitian and the sum must be a Hermitian Hamiltonian, the perturbation operator δH must also be Hermitian. It is convenient to introduce a unit-free constant λ ∈[0, 1] and to consider, instead, a λ-dependent Hamiltonian H(λ) that takes the form H(λ) = H(0) + λ δH . (1.0.2) When λ = 1 we have the Hamiltonian of interest, but λ allows us to consider a family of Hamiltonians that interpolate from H(0), when λ is equal to zero, to the Hamiltonian of interest for λ equal to one. In many cases perturbations can be turned on and off; think, 1 2 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY for example, of an atom in an external magnetic field that can be varied continuously. In that case we can view λ as the parameter that allows us to turn on the perturbation by letting λ ̸= 0. The parameter λ is also useful in organizing the perturbation analysis, as we will see below. We spoke of a Hamiltonian that differs slightly from H(0). In order to use perturbation theory we need λδH to be a ‘small’ perturbation of the Hamiltonian H(0). We will have to deal with the meaning of small. At first sight we may imagine that small means that, viewed as matrices, the largest entries in λδH are smaller than the largest entries in H(0). While this is necessary, more is needed, as we will see in our analysis. An additional advantage of using λ is that by taking it to be sufficiently small we can surely make λδH small. We assume that the Hamiltonian H(0) is understood, namely, we know the eigenstates and eigenvalues of H(0). We want to know the eigenstates and eigenvalues of H(λ). One may be able to calculate those exactly, but this is seldom a realistic possibility. Diagonalizing δH is seldom useful, since δH and H do not generally commute and therefore δH eigenstates are not eigenstates of H(λ). In perturbation theory the key assumption is that the eigenvalues and eigenvectors of H(λ) can be found as series expansions in powers of λ. We hope, of course, that there are some values of λ for which the series converges, or at least gives useful information. Figure 1.1: The energy eigenvalues of H(λ) as λ varies from zero to one. On the λ = 0 vertical axis the H(0) eigenstates are represented by heavy dots. By the time λ = 1 the dots have shifted. In Figure 1.1 we illustrate some of the phenomena that we may see in the spectrum of a system with Hamiltonian H(λ). We show how the energies of the various states may change as the parameter λ is increased from zero. The two lowest energy eigenstates are non-degenerate and their energies can go up and down as λ varies. Next up in energy we have two degenerate states of H(0) (the Hamiltonian as λ = 0). The perturbation splits the 1.1. NONDEGENERATE PERTURBATION THEORY 3 two levels, and that happens generically. In the figure, the perturbation splits the levels to first order in λ, as shown by the different slopes of the two curves that meet at λ = 0. In other words, viewed as power series in λ the energies of the two states have different linear terms in λ. The last level shown corresponds to four degenerate states. The perturbation to first order in λ splits the states into a group of three states and a fourth. To second order in λ the three states split further. A single Hamiltonian can exhibit behavior like this, with many possible variations. To analyze the evolution of states and energies as functions of λ we have two possible cases: (i) we are following a non-degenerate state or, (ii) we are following a collection of degenerate states. The challenges are quite different and therefore we must analyze them separately. Clearly both situations can occur for a single Hamiltonian, depending on the spectrum of H(0). To follow a non-degenerate state we use non-degenerate perturbation theory. To follow a set of degenerate states we use degenerate perturbation theory. Since Hamiltonians H(0) generally have both non-degenerate and degenerate states we need to consider both types of perturbation theory. We begin with non-degenerate perturbation theory. 1.1 Nondegenerate perturbation theory We begin by describing the original Hamiltonian H(0). We assume this Hamiltonian has a discrete spectrum with an orthonormal basis |k(0)⟩of energy eigenstates, where k ∈Z is a label that ranges over a possibly infinite set of values: H(0) |k(0)⟩= E(0) k |k(0)⟩, ⟨k(0)|l(0)⟩= δkl . (1.1.1) We will let k = 0 denote the ground state and we order the states so that the energies generally increase as the value of the label increases, so that E(0) 0 ≤E(0) 1 ≤E(0) 2 ≤E(0) 3 ≤. . . (1.1.2) The equal signs are needed because some states may be degenerate. In this section we focus on a non-degenerate state |n(0)⟩with fixed n. This means that |n(0)⟩is a single state that is separated by some finite energy from all the states with more energy and from all the states with less energy. In other words the following must be part of the sequence of inequalities in (1.1.2) . . . ≤E(0) n−1 < E(0) n < E(0) n+1 ≤. . . (1.1.3) If the chosen state is the ground state, we have instead E(0) 0 < E(0) 1 . As the perturbation is turned on by making λ different from zero, the energy eigenstate |n(0)⟩of H(0) will evolve into some energy eigenstate |n⟩λ of H(λ) with energy En(λ): H(λ)|n⟩λ = En(λ) |n⟩λ , (1.1.4) 4 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY where |n⟩λ=0 = |n(0)⟩, and En(λ = 0) = E(0) n . (1.1.5) As we said, the solution is assumed to take the form of a regular power series expansion in λ. To make this clear consider a function f(λ) such that its derivatives to all orders exist for λ = 0. In that case we have a Taylor expansion f(λ) = ∞ X n=0 1 n!f (n)(0) λn = f(0) + f ′(0) λ + 1 2f ′′(0) λ2 + 1 3!f ′′′(0) λ3 + · · · (1.1.6) The expansion is a power series in λ, with coefficients f(0), f ′(0), etc, that are λ independent and reflect the value of the function and its derivatives at λ = 0. For our problem we note the values of |n⟩λ and En(λ) for λ = 0 (1.1.5) and write: |n⟩λ = |n(0)⟩+ λ|n(1)⟩+ λ2|n(2)⟩+ λ3|n(3)⟩+ . . . , En(λ) = E(0) n + λ E(1) n + λ2 E(2) n + λ3 E(3) n + . . . (1.1.7) The superscripts on the states and energies denote the power of λ that accompanies them in the above expressions. The above equations are a natural assumption; they state that the perturbed states and energies, being functions of λ, admit a Taylor expansion around λ = 0. Our aim is to calculate the states |n(1)⟩, |n(2)⟩, |n(3)⟩, . . . (1.1.8) and the energies E(1) n , E(2) n , E(3) n , . . . (1.1.9) Note that all these states and energies are, by definition, λ independent. Here |n(1)⟩is the leading correction to the state |n(0)⟩as we turn on λ, and E(1) n is the leading correction to the energy as we turn on λ. We will not impose the requirement that |n⟩λ is normalized. It suffices that |n⟩λ is normalizable, which it will be for sufficiently small perturbations. For λ = 1 we would find the solution for H(1) = H(0) + δH in the form |n⟩≡|n⟩1 = |n(0)⟩+ |n(1)⟩+ |n(2)⟩+ |n(3)⟩+ . . . , En ≡En(1) = E(0) n + E(1) n + E(2) n + E(3) n + . . . (1.1.10) Substituting the ansatz (1.1.7) into the Schr¨ odinger equation (1.1.4) we will find the conditions for such solution to exist: H(0) + λδH −En(λ)  |n⟩λ = 0 , (1.1.11) which more explicitly takes the form  (H(0) −E(0) n ) −λ(E(1) n −δH) −λ2E(2) n −λ3E(3) n −. . . −λk E(k) n + . . .   |n(0)⟩+ λ|n(1)⟩+ λ2|n(2)⟩+ λ3|n(3)⟩+ . . . + λk|n(k)⟩+ . . .  = 0 . (1.1.12) 1.1. NONDEGENERATE PERTURBATION THEORY 5 Multiplying out we get a series in λ with coefficients λ-independent vectors in the state space of the theory. If this is to vanish for all values of λ those coefficients must be zero. Collecting the coefficients for each power of λ we λ0 : (H(0) −E(0) n ) |n(0)⟩= 0 , λ1 : (H(0) −E(0) n ) |n(1)⟩= (E(1) n −δH)|n(0)⟩, λ2 : (H(0) −E(0) n ) |n(2)⟩= (E(1) n −δH)|n(1)⟩+ E(2) n |n(0)⟩, λ3 : (H(0) −E(0) n ) |n(3)⟩= (E(1) n −δH)|n(2)⟩+ E(2) n |n(1)⟩+ E(3) n |n(0)⟩, . . . . . . . . . λk : (H(0) −E(0) n ) |n(k)⟩= (E(1) n −δH)|n(k−1)⟩+ E(2) n |n(k−2)⟩+ . . . + E(k) n |n(0)⟩. (1.1.13) Each equation is the condition that the coefficient multiplying the power of λ indicated to the left vanishes. That power is reflected as the sum of superscripts on each term, counting δH as having superscript one. This gives a simple consistency check on our equations. These are equations for the kets |n(1)⟩, |n(2)⟩, . . . as well as the energy corrections E(1) n , E(2) n , . . .. Note again that λ does not enter into the equations, and thus the kets and energy corrections are λ independent. The first equation, corresponding to λ0, is satisfied by construction. The second equa-tion, corresponding to λ1, should allow us to solve for the first correction |n(1)⟩to the state and the first correction E(1) n to the energy. Once these are known, the equation corre-sponding to λ2 involves only the unknowns |n(2)⟩and E(2) n , and should determine them. At each stage each equation has only two unknowns: a state correction |n(k)⟩and an energy correction E(k) n . A useful choice. We now claim that without loss of generality we can assume that all the state corrections |nk⟩, with k ≥1 contain no vector along |n(0)⟩. Explicitly: 0 = ⟨n(0)|n(1)⟩= ⟨n(0)|n(2)⟩= ⟨n(0)|n(3)⟩= . . . . (1.1.14) To show this we explain how we can manipulate a solution that does not have this property into one that does. Suppose you have solution in which the state corrections |n(k)⟩ have components along |n(0)⟩: |n(k)⟩= |n(k)⟩′ −ak|n(0)⟩, k ≥1 , (1.1.15) with some constants ak and with |n(k)⟩′ orthogonal to |n(0)⟩. Then the solution for the full 6 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY corrected state is |n⟩λ = |n(0)⟩+ λ |n(1)⟩′ −a1|n(0)⟩  + λ2 |n(2)⟩′ −a2|n(0)⟩  + . . . = 1 −a1λ −a2λ2 −. . .  |n(0)⟩+ λ|n(1)⟩′ + λ2|n(2)⟩′ + . . . (1.1.16) Since this is an eigenstate of the Hamiltonian H(λ), it will still be an eigenstate if we change its normalization by dividing it by any function of λ. Dividing by the coefficient of |n(0)⟩ we have the physically identical solution |n⟩′ λ given by |n⟩′ λ = |n(0)⟩+ 1 1 −a1λ −a2λ2 −. . .  λ|n(1)⟩′ + λ2|n(2)⟩′ + . . . (1.1.17) We can expand the denominator so that we get |n⟩′ λ = |n(0)⟩+ λ|n(1)⟩′ + λ2|n(2)⟩′ + a1|n(1)⟩′ + . . . (1.1.18) The explicit expressions do not matter, the key point, actually visible in (1.1.17), is that we have a physically identical solution of the same equation in which the state corrections are all orthogonal to |n(0)⟩. This shows that we can impose the conditions (1.1.14) without loss of generality. Solving the equations. Let us finally begin solving equations (1.1.13). For this we note that the Schrodinger equation for the ket |n(0)⟩implies that for the bra we have ⟨n(0)|(H(0) −E(0) n ) = 0 . (1.1.19) This means that acting with ⟨n(0)| on the left-hand side of any of the equations in (1.1.13) will give zero. Consistency requires that acting with ⟨n(0)| on the right-hand side of any of the equations in (1.1.13) also give zero, and presumably some interesting information. For the λ-equation this gives: 0 = ⟨n(0)|(E(1) n −δH)|n(0)⟩. (1.1.20) Since |n(0)⟩is normalized and E(1) n is a number, this means that E(1) n = ⟨n(0)|δH|n(0)⟩. (1.1.21) This is the most famous result in perturbation theory: the first correction to the energy of an energy eigenstate is simply the expectation value of the correction to the Hamiltonian in the uncorrected state. You need not know the correction to the state to determine the first correction to the energy! Note that the hermicity of δH implies the required reality of the energy correction. We can actually find some interesting formulae (but not yet fully explicit!) for the higher energy corrections. For the λ2 equation, acting with ⟨n(0)| on the right-hand side gives 0 = ⟨n(0)|  (E(1) n −δH)|n(1)⟩+ E(2) n |n(0)⟩  . (1.1.22) 1.1. NONDEGENERATE PERTURBATION THEORY 7 Recalling our orthogonality assumption, we have ⟨n(0)|n(1)⟩= 0 and the term with E(1) n drops out. We get E(2) n = ⟨n(0)|δH|n(1)⟩, (1.1.23) which states that the second correction to the energy is determined if we have the first correction |n(1)⟩to the state. Note that this expression is not explicit enough to make it manifest that E(2) n is real. This and the earlier result for the first correction to the energy have a simple generalization. Acting with ⟨n(0)| on the last equation of (1.1.13) we get 0 = ⟨n(0)|  (E(1) n −δH)|n(k−1)⟩+ E(2) n |n(k−2)⟩+ . . . + E(k) n |n(0)⟩  . (1.1.24) Using the orthogonality of |n(0)⟩and all the state corrections, we have 0 = −⟨n(0)|δH)|nk−1⟩+ E(k) n , (1.1.25) and therefore we have E(k) n = ⟨n(0)| δH |n(k−1)⟩. (1.1.26) At any stage of the recursive solution, the energy at a fixed order is known if the state correction is known to previous order. So it is time to calculate the corrections to the states! Let us solve for the first correction |n(1)⟩to the state. This state must be some particular superposition of the original energy eigenstates |k(0)⟩. For this we look at the equation (H(0) −E(0) n ) |n(1)⟩= (E(1) n −δH)|n(0)⟩. (1.1.27) This is a vector equation: the left-hand side vector set equal to the right-hand side vector. As in any vector equation, we can check it using a basis set of vectors. Forming the inner product of each and every basis vector with both the left-hand side and the right-hand side, we must get equal numbers. We already acted on the above equation with ⟨n(0)| to figure out E(1) n . The remaining information in this equation can be obtained by acting with all the states ⟨k(0)| with k ̸= n: ⟨k(0)|(H(0) −E(0) n ) |n(1)⟩= ⟨k(0)|(E(1) n −δH)|n(0)⟩. (1.1.28) On the left-hand side we can let H(0) act on the bra. On the right-hand side we note that with k ̸= n the term with E(1) n vanishes (E(0) k −E(0) n ) ⟨k(0)| n(1)⟩= −⟨k(0)|δH|n(0)⟩. (1.1.29) To simplify notation we define the matrix elements of δH in the original basis δHmn ≡⟨m(0)|δH|n(0)⟩. (1.1.30) 8 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY Note that the Hermiticity of δH implies that δHnm = (δHmn)∗. (1.1.31) With this notation, equation (1.1.29) gives ⟨k(0)| n(1)⟩= − δHkn E(0) k −E(0) n , k ̸= n . (1.1.32) Since we now know the overlap of |n(1)⟩with all basis states, this means that the state has been determined. Indeed we can use the completeness of the basis to write |n(1)⟩= X k |k(0)⟩⟨k(0)|n(1)⟩= X k̸=n |k(0)⟩⟨k(0)|n(1)⟩, (1.1.33) since the term with k = n does not contribute because of the orthogonality assumption. Using the overlaps (1.1.32) we now get |n(1)⟩= − X k̸=n |k(0)⟩δHkn E(0) k −E(0) n . (1.1.34) This shows that the first correction |n(1)⟩can have components along all basis states, except |n(0)⟩. The component along a state |k(0)⟩vanishes if the perturbation δH does not couple |n(0)⟩to |k(0)⟩, namely, if δHkn vanishes. Note that the assumption of non-degeneracy is needed here. We are summing over all states |k(0)⟩that are not |n(0)⟩and if any such state has the same H(0) energy as |n(0)⟩the energy denominator will vanish causing trouble! Now that we have the first order correction to the states we can compute the second order correction to the energy. Using (1.1.23) we have E(2) n = ⟨n(0)|δH|n(1)⟩= − X k̸=n ⟨n(0)|δH|k(0)⟩δHkn E(0) k −E(0) n . (1.1.35) In the last numerator we have ⟨n(0)|δH|k(0)⟩= δHnk = (δHkn)∗and therefore E(2) n = − X k̸=n |δHkn|2 E(0) k −E(0) n . (1.1.36) This is the second-order energy correction. This explicit formula makes the reality of E(2) n manifest. In summary, going back to (1.1.7), we have that the states and energies for H(λ) = H(0) + λδH are, to this order, |n⟩λ = |n(0)⟩−λ X k̸=n δHkn E(0) k −E(0) n |k(0)⟩+ O(λ2) , En(λ) = E(0) n + λ δHnn −λ2 X k̸=n |δHkn|2 E(0) k −E(0) n + O(λ3) , (1.1.37) 1.1. NONDEGENERATE PERTURBATION THEORY 9 Remarks: 1. The first order corrected energy of the (non-degenerate) ground state overstates the true exact ground state energy. To see this consider the first order corrected ground state energy E(0) 0 + λE(1) 0 . Writing this in terms of expectation values, with |0(0)⟩ denoting the unperturbed ground state, we have E(0) 0 + λE(1) 0 = ⟨0(0)|H(0)|0(0)⟩+ λ⟨0(0)|δH|0(0)⟩ = ⟨0(0)|(H(0) + λδH)|0(0)⟩ = ⟨0(0)|H(λ)|0(0)⟩. (1.1.38) By the variational principle, the expectation value of the Hamiltonian on an arbitrary (normalized) state is larger than the ground state energy E0(λ), therefore E(0) 0 + λE(1) 0 = ⟨0(0)|H(λ)|0(0)⟩≥E0(λ) , (1.1.39) which is what we wanted to prove. Given this overestimate at first order, the second order correction to the ground state energy is always negative. Indeed, −λ2 X k̸=0 |δHk0|2 E(0) k −E(0) 0 , (1.1.40) and each term is negative because the unperturbed excited state energies E(0) k (k ̸= 0) exceed the unperturbed ground state energy E(0) 0 . 2. The second order correction to the energy of the |n(0)⟩eigenstate exhibits level repul-sion: the levels with k > n push the state down and the levels with k < n push the state up. Indeed, −λ2 X k̸=n |δHkn|2 E(0) k −E(0) n = −λ2 X k>n |δHkn|2 E(0) k −E(0) n + λ2 X k<n |δHkn|2 E(0) n −E(0) k . (1.1.41) The first term gives the negative contribution from the higher energy states and the second term gives the contribution from the lower energy states (see Figure 1.2). The systematics of solving the equations is now apparent. For each equation we take inner products with all states in the state space. That gives the full content of the equation. We first take the inner product with ⟨n(0)|, as this makes the left-hand side equal to zero and is thus simpler. Then we take the inner product with ⟨k(0)| with all k ̸= n and that gives the remainder of the information that is contained in the equation. Exercise 1. Calculate |n(2)⟩and E(3) n . 10 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY Figure 1.2: The second order corrections to the energy of the state |n(0)⟩receives negative contri-butions from the higher energy states and positive contributions from the lower energy states. We have, effectively, a repulsion preventing the state |n(0)⟩from approaching the neighboring states. Exercise 2. The state |n⟩λ is not normalized. Use (1.1.37) to calculate to order λ2 the quantity Zn(λ) defined by 1 Zn(λ) ≡ λ⟨n|n⟩λ . (1.1.42) What is the probability that the state |n⟩λ will be observed to be along its unperturbed version |n(0)⟩? 1.1.1 Validity of the perturbation expansion We now return to a question we did not address: What do we mean when we say that λδH is small? We have said that λδH must be small compared to the original Hamiltonian H(0), but it is not clear what this means, as both expressions are operators. For some insight into this matter consider an example where H(0) is a two-by-two diagonal matrix with non-degenerate eigenvalues H(0) = E(0) 1 0 0 E(0) 2 ! . (1.1.43) The perturbation, called λ ˆ V , only has off-diagonal elements so that H(λ) = H(0) + λ ˆ V ≡ E(0) 1 λV λV ∗ E(0) 2 ! . (1.1.44) 1.1. NONDEGENERATE PERTURBATION THEORY 11 In this simple example there is no need to use perturbation theory since the eigenvalues, E+ and E−, can be calculated exactly as functions of λ E±(λ) = 1 2(E(0) 1 + E(0) 2 ) ± 1 2(E(0) 1 −E(0) 2 ) v u u t1 + " λ|V | 1 2(E(0) 1 −E(0) 2 ) #2 . (1.1.45) The perturbative expansion of the energies is obtained by Taylor expansion of the square Figure 1.3: The Taylor expansion of the function f(z) = √ 1 + z2 about z = 0 has a radius of convergence equal to one. root in powers of λ. To perform this expansion we need the result f(z) ≡ p 1 + z2 = 1 + z2 2 −z4 8 + z6 6 + 5 128z8 + O(z10) . (1.1.46) The function f(z) exhibits branch cuts at z = ±i (see Figure 1.3), thus the expansion of f(z) around z = 0 has radius of convergence equal to one: the series converges for |z| < 1 and diverges for |z| > 1. Table 1 shows f(z) evaluated for z = 0.9, 1.2, and 1.5. The various approximations to the full series are shown. For our expansion of (1.1.45), convergence for |z| < 1 implies convergence when |λ||V | 1 2|E(0) 1 −E(0) 2 | < 1 = ⇒|λV | < 1 2|E(0) 1 −E(0) 2 | . (1.1.47) For |λV | > 1 2|E(0) 1 −E(0) 2 | the perturbation series does not converge. We learn that for convergence the perturbation must be small compared with energy differences in H(0). It is not sufficient that the magnitude of the matrix elements of λδH be small compared to those in H(0), energy differences matter. This result leads us to expect complications when energy differences go to zero and H(0) has degeneracies. 12 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY z 0.9 1.2 1.5 f(z) 1.34536 1.56205 1.80278 f8(z) 1.33939 1.47946 1.20297 f14(z) 1.33939 1.67280 4.82288 f20(z) 1.34490 1.36568 -18.4895 f30(z) 1.34545 2.23047 641.772 Table 1.1: f(z) ≡ √ 1 + z2 = P i cizi and fn(z) = Pn i=0 cizi. 1.1.2 Example: Anharmonic oscillator Consider the simple harmonic oscillator H(0) = ˆ p2 2m + 1 2mω2ˆ x2 . (1.1.48) We want to explore the effect of a perturbation proportional to ∼ˆ x4. This has the effect of changing the original quadratic potential for a more complicated potential that includes a quartic term. To do analysis in a clear way, we must consider units. Using the constants ℏ, m, ω of the harmonic oscillator a length scale d can be uniquely build: d2 = ℏ mω . (1.1.49) The unit-free coordinate ˆ x/d then has a simple expression in terms of creation and annihi-lation operators: ˆ x = r ℏ 2mω (ˆ a + ˆ a†) = d 1 √ 2(ˆ a + ˆ a†) → ˆ x d = 1 √ 2(ˆ a + ˆ a†) . (1.1.50) It follows that an ˆ x4 perturbation with units of energy takes the form δH = ℏω ˆ x4 d4 = m2ω3 ℏ ˆ x4 = 1 4 ℏω(ˆ a + ˆ a†)4 . (1.1.51) Using the unit-free parameter λ, the perturbed Hamiltonian will therefore be H(λ) = H(0) + λ m2ω3 ℏ ˆ x4 = H(0) + λ 1 4 ℏω(ˆ a + ˆ a†)4 . (1.1.52) We will identify the states |k(0)⟩of H(0) with the number eigenstates |k⟩, k = 0, 1, . . . , of the harmonic oscillator. Recall that E(0) k = ℏω(k + 1 2) , |k⟩= (ˆ a†)k √ k! |0⟩. (1.1.53) 1.1. NONDEGENERATE PERTURBATION THEORY 13 The Hamiltonian H(λ) defines an anharmonic oscillator. In a classical anharmonic oscil-lator the frequency of oscillation depends on the amplitude of oscillation. In the quantum harmonic oscillator all levels are equally spaced. The frequencies associated with transi-tions between various levels are therefore integer multiples (i.e. harmonics) of the basic frequency associated to a transition between the first excited state and the ground state. In the quantum anharmonic oscillator the spacing between the energy levels is not uniform. First the simplest question: What is the first-order correction E(1) 0 to the energy of the ground state? For this, following (1.1.21) we simply calculate the expectation value of the perturbation on the ground state: E(1) 0 = ⟨0|1 4 ℏω(ˆ a + ˆ a†)4|0⟩= 1 4 ℏω⟨0|(ˆ a + ˆ a†)4|0⟩= 3 4 ℏω , (1.1.54) where we used ⟨0|(ˆ a + ˆ a†)4|0⟩= 3 , (1.1.55) as you should verify. It follows that the corrected energy is E0(λ) = E(0) 0 + λE(1) 0 + O(λ2) = 1 2ℏω + λ 3 4ℏω + O(λ2) = 1 2ℏω 1 + 3 2λ + O(λ2)  (1.1.56) We note that the energy of the ground state increases with λ > 0. This is reasonable as the quartic term in the modified potential squeezes the ground state. How about second order correction to the ground state energy? For this we use (1.1.36) taking n = 0: E(2) 0 = − X k̸=0 |δHk0|2 E(0) k −E(0) 0 . (1.1.57) The sum is over all k ≥1 such that δHk0 is non-vanishing. Here δHk0 = 1 4ℏω ⟨k|(ˆ a + ˆ a†)4|0⟩. (1.1.58) We consider (ˆ a + ˆ a†)4|0⟩which corresponds, up to constants to acting on the ground state wavefunction ϕ0 with x4. This should give an even wavefunction. So (ˆ a + ˆ a†)4|0⟩must be a superposition of |0⟩, |2⟩, and |4⟩. We cannot get states with higher number because there are at most four creation operators acting on the vacuum. A short calculation (do it!) confirms that (ˆ a + ˆ a†)4|0⟩= 3|0⟩+ 6 √ 2 |2⟩+ √ 4! |4⟩. (1.1.59) This immediately gives δH00 = 3 4 ℏω , δH20 = 3 √ 2 2 ℏω , δH40 = √ 6 2 ℏω , (1.1.60) the first of which we had already determined and is not needed for the second order com-putation. Back in (1.1.57) we have E(2) 0 = −|δH20|2 2ℏω −|δH40|2 4ℏω = −(ℏω)2 2ℏω 9 2 −(ℏω)2 4ℏω 3 2 = − 9 4 + 3 8  ℏω = −21 8 ℏω . (1.1.61) 14 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY Therefor the corrected ground state energy to quadratic order is E(0) 0 + λE(1) 0 + λ2E(2) 0 = 1 2ℏω 1 + 3 2λ −21 4 λ2 . (1.1.62) The computation can be carried to higher order, as first done by Bender and Wu (Phys. Rev.184 (1969)1231). They find that1 E0(λ) = 1 2ℏω 1 + 3 2λ −21 4 λ2 + 333 8 λ3 −30885 64 λ4 + 916731 128 λ5 −65518401 512 λ6 + O(λ7)  (1.1.63) As it turns out the coefficients keep growing and the series does not converge for any nonzero λ; the radius of convergence is actually zero! This does not mean the series is not useful. It is an asymptotic expansion. This means that for a given small value of λ the magnitude of successive terms generally decrease until, at some point, they start growing again. A good approximation to the desired answer is obtained by including only the part of the sum where the terms are decreasing. Exercise 3. Calculate the first order correction E(1) n to the energy for the state |n⟩of number n. Exhibit the anharmonicity of the oscillator by using this result to find, to first order in λ, the energy separation ∆En(λ) = En(λ) −En−1(λ) between levels. Let us now find the first order correction to the ground-state wavefunction. Using (1.1.34) with n = 0 we have |0(1)⟩= − X k̸=0 δHk0 E(0) k −E(0) 0 |k⟩. (1.1.64) We then find |0(1)⟩= −δH20 2ℏω |2⟩−δH40 4ℏω |4⟩= −3 4 √ 2|2⟩−1 16 √ 4!|4⟩ = −3 4 ˆ a†ˆ a†|0⟩−1 16 ˆ a†ˆ a†ˆ a†ˆ a†|0⟩. (1.1.65) This means that to first order the ground state of the perturbed oscillator is |0⟩λ = |0⟩−λ  3 4 ˆ a†ˆ a†|0⟩+ 1 16 ˆ a†ˆ a†ˆ a†ˆ a†|0⟩  + O(λ2) . (1.1.66) 1.2 Degenerate perturbation theory If the spectrum of H(0) has degenerate states, as shown in Figure 1.1, tracking the evolution of those states as λ becomes nonzero presents new challenges. We first show that naive extrapolation of our results for a non-degenerate state do not work. We will also be able to appreciate the basic difficulty. 1Bender and Wu’s results, in eqns. (2.12) of their paper must all be multiplied by a factor of 2, as they take A0 = 1/2. 1.2. DEGENERATE PERTURBATION THEORY 15 1.2.1 Degenerate toy model Consider an example with two-by-two matrices. The unperturbed matrix H(0) will be set equal to the identity matrix: H(0) = 1 0 0 1  . (1.2.1) We have a degeneracy here as the two eigenvalues are identical (and equal to one). The perturbation matrix δH is chosen to be offdiagonal: δH = 0 1 1 0  . (1.2.2) We then have H(λ) = H(0) + λδH = 1 λ λ 1  . (1.2.3) Using labels n = 1, 2 the unperturbed eigenstates can be taken to be |1(0)⟩= 1 0  , |2(0)⟩= 0 1  , E(0) 1 = E(0) 2 = 1 , (1.2.4) with the corresponding eigenvalues indicated as well. To first order in λ, the eigenvalues predicted from non-degenerate perturbation theory (1.1.37) are En(λ) = E(0) n + λδHnn. This gives E1(λ) = E(0) 1 + λδH11 = 1 + λ · 0 = 1 ? E1(λ) = E(0) 2 + λδH22 = 1 + λ · 0 = 1 ? (1.2.5) The eigenvalues are unperturbed to first order in λ since the matrix δH is off-diagonal. These answers, however, are wrong. We can compute the exact eigenvalues of H(λ) and they are 1 ± λ. There is also a problem with the state corrections. Equation (1.1.37) states that |n⟩λ = |n(0)⟩−λ X k̸=n δHkn E(0) k −E(0) n |k(0)⟩. (1.2.6) but this time, with E(0) 1 = E(0) 2 the denominator is zero, and the δH matrix element is also zero, giving us an ambiguous result. So what can we do? A direct calculation shows that H(λ) has eigenvectors 1 √ 2 1 1  with eigenvalue = 1 + λ , (1.2.7) and 1 √ 2  1 −1  with eigenvalue = 1 −λ . (1.2.8) You may think the eigenvectors jump from those of H(0) indicated in (1.2.4) to those of H(λ) as soon as λ becomes nonzero. Such discontinuity is totally against the spirit of 16 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY perturbation theory. Happily, this is not really true. The eigenvectors of H(0) are in fact ambiguous, precisely due to the degeneracy. The eigenvectors of H(0) are actually the span of the two vectors listed in (1.2.4). The perturbation selected a particular combination of these eigenvectors. This particular combination is the one that we should use even for λ = 0. The lesson is that to get states that vary continuously as λ is turned on we must choose the basis in the degenerate subspace of H(0) carefully. We will call that carefully selected basis the “good” basis. 1.2.2 Systematic Analysis Again, we are looking at the perturbed Hamiltonian H(λ) = H(0) + λδH , (1.2.9) where H(0) has known eigenvectors and eigenvalues. We will focus this time on a degenerate subspace of eigenvectors of dimension N > 1, that is, a space with N linearly independent eigenstates of the same energy. In the basis of eigenstates, H(0) is a diagonal matrix that contains a string of N > 1 identical entries: H(0) = diag { E(0) 1 , E(0) 2 , . . . , E(0) n , . . . , E(0) n | {z } N , . . . } . (1.2.10) In the degenerate subspace we choose a collection of N orthonormal eigenstates |n(0); 1⟩, |n(0); 2⟩, . . . , |n(0); N⟩. (1.2.11) Accordingly, we have ⟨n(0); p |n(0); l⟩= δp,l , (1.2.12) H(0)|n(0); k⟩= E(0) n |n(0); k⟩. (1.2.13) This set of vectors span a degenerate subspace of dimension N that we will call VN VN ≡span{|n(0); k⟩, k = 1, . . . N} . (1.2.14) The total state space of the theory, denoted by H is written as a direct sum: H = VN ⊕ˆ V , (1.2.15) where ˆ V is spanned by those eigenstates of H(0) that are not in VN. We denote by |p(0)⟩ with p ∈Z a basis for ˆ V . That basis may include both degenerate and non degenerate states. Together with the states in VN we have an orthonormal basis for the whole state space: ⟨p(0)|q(0)⟩= δpq , ⟨p(0)|n(0); k⟩= 0 . (1.2.16) 1.2. DEGENERATE PERTURBATION THEORY 17 Our notation distinguishes the states in VN from those in ˆ V because the former have two labels and the latter have only one. We now consider the evolution of the degenerate eigenstates as we turn on the pertur-bation. Again we assume that the states vary continuously in λ and thus write: |n(0); k⟩→ |n; k⟩λ = |n(0); k⟩+ λ|n(1); k⟩+ λ2|n(2); k⟩+ O(λ3) , E(0) n → En,k(λ) = E(0) n + λ E(1) n,k + λ2 E(2) n,k + O(λ3) . (1.2.17) These equations hold for k = 1, . . . , N. Note that for each value of k the energy corrections might be different and that’s why the energy corrections carry the label k. Our goal is to find the state corrections |n(p); k⟩and the energy corrections E(p) n,k for p ≥1 and for each k. As before we demand that |n(p); k⟩for p ≥1 has no component along |n(0); k⟩, i.e. ⟨n(0); k|n(p); k⟩= 0 for p ≥1 (1.2.18) Note, however, that |n(p); k⟩may have components along |n(0); ℓ⟩with ℓ̸= k. So |n(0); k⟩ may and in fact will have a component in VN. The perturbed eigenstates must satisfy H(λ) |n; k⟩λ = En,k(λ) |n; k⟩λ , (1.2.19) and substituting the perturbative expansions above we obtain equations completely analo-gous to the ones in Eq.(1.1.13) λ0 : (H0 −E(0) n ) |n(0); k⟩= 0 , (1.2.20) λ1 : (H0 −E(0) n ) |n(1); k⟩= (E(1) n,k −δH)|n(0); k⟩, (1.2.21) λ2 : (H0 −E(0) n ) |n(2); k⟩= (E(1) n,k −δH)|n(1); k⟩+ E(2) n,k |n(0); k⟩, (1.2.22) . . . . . . . . . In the following we will discuss a solution to first order for the case in which the de-generacy in VN is completely broken to first order in perturbation theory; that is, the first order corrections to the energies split the N states completely. Our solution will proceed in three steps: 1. Hit the O(λ) equation with ⟨n(0); ℓ| to learn that δH must be diagonal in the chosen basis for VN and to determine the first-order energy shifts. 2. Use the O(λ) equation to calculate the components of |n(1); k⟩in ˆ V . 3. Hit the O(λ2) equation with ⟨n(0); ℓ| to determine the second order energy correction E(2) n,k and the component of |n(1); k⟩in VN. 18 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY Step 1. Recalling that ⟨n(0); ℓ|(H(0) −E(0) n ) = 0, as we hit the O(λ) equation with ⟨n(0); ℓ| the left-hand side vanishes and we find ⟨n(0); ℓ|(E(1) n,k −δH)|n(0); k⟩= 0 . (1.2.23) Since the basis states in VN are orthonormal, this implies that ⟨n(0); ℓ| δH |n(0); k⟩= E(1) n,k δℓ,k . (1.2.24) This equation holds for all k, ℓ= 1, . . . , N. Remarkably, this equation is telling us that the basis |n(0); k⟩must be chosen to make the matrix δH diagonal in the subspace VN! This is required in order to get the perturbation theory going. Setting ℓequal to k we read the values of the first order energy shifts E(1) n,k = ⟨n(0); k|δH|n(0); k⟩= δHnk,nk , (1.2.25) where the last equality is a definition. The energies to first order are then En,k(λ) = E(0) n + λ δHnk,nk . (1.2.26) A few remarks: 1. The above result for the first order energy shifts is true always, even if the degeneracy is not lifted. The degeneracy is lifted when E(1) n,k ̸= E(1) n,ℓ, whenever k ̸= ℓ, (1.2.27) for all values of k, ℓ= 1, . . . , N. This assumption will be used in the later steps. If the degeneracy is lifted, the basis states |n(0); k⟩that make δH diagonal in VN are called “good states” or a “good basis”. This means that they are the basis states in VN that get deformed continuously as λ becomes non-zero. If the degeneracy is not lifted to first order the determination of the good basis has to be attempted to second order. 2. The perturbation δH is diagonalized in the subspace VN. The perturbation δH is not diagonal on the whole space H, only within the block representing VN is δH a diagonal matrix. Alternatively we can see this via the action of δH on the basis states. Introducing a resolution of the identity, we have δH|n(0); ℓ⟩= X q |n(0); q⟩⟨n(0); q|δH|n(0); ℓ⟩+ X p |p(0)⟩⟨p(0)|δH|n(0); ℓ⟩ = X q E(1) n,ℓδℓ,q|n(0); q⟩+ X p |p(0)⟩⟨p(0)|δH|n(0); ℓ⟩ = E(1) n,ℓ|n(0); ℓ⟩+ X p |p(0)⟩⟨p(0)|δH|n(0); ℓ⟩. (1.2.28) This shows that the states |n(0); ℓ⟩are almost δH eigenstates with eigenvalues equal to the first order energy corrections. The failure is an extra state along ˆ V . 1.2. DEGENERATE PERTURBATION THEORY 19 3. We can sometimes assess without computation that a certain basis in VN makes δH diagonal. Here is a rule: the matrix δH is diagonal for a choice of basis in VN if there is a Hermitian operator K that commutes with δH for which the chosen basis vectors are K eigenstates with different eigenvalues. This is quickly established. Consider two different basis states in VN: |n(0); p⟩and |n(0); q⟩, with p ̸= q. Assume these have K eigenvalues λp and λq, respectively. Since [δH, K] = 0: 0 = ⟨n(0); p|[δH, K]|n(0); q⟩= (λq −λp)⟨n(0); p| δH |n(0); q⟩. (1.2.29) Since the eigenvalues λp and λq are presumed to be different, the non-diagonal matrix elements of δH vanish. Step 2. The O(λ) equation cannot determine the component of |n(1); k⟩along VN. As we will see later, such piece is required by consistency and gets determined from the O(λ2) equation. We now determine the piece of |n(1); k⟩along ˆ V . For this we hit the O(λ) equation with ⟨p(0)| to find ⟨p(0)|(H(0) −E(0) n )|n(1); k⟩= ⟨p(0)| ( |p(0)⟩⊥VN E(1) n,k −δH ) |n(0); k⟩ (1.2.30) and we get (E(0) p −E(0) n )⟨p(0)|n(1); k⟩= −δHp,nk , (1.2.31) where we introduced the matrix element δHp,nk ≡⟨p(0)| δH |n(0); k⟩. Our equation above means that the piece of |n(1); k⟩in ˆ V is now determined: |n(1); k⟩= − X p δHp,nk E(0) p −E(0) n |p(0)⟩+ |n(1); k⟩ VN , (1.2.32) where we included explicitly the still undetermined piece of |n(1); k⟩along VN. Step 3. We now hit we hit the O(λ2) equation with ⟨n(0); ℓ|. The left-hand side vanishes and using the above expression for |n(1); k⟩we find 0 = −⟨n(0); ℓ|(E(1) n,k −δH) X p |p(0)⟩ δHp,nk E(0) p −E(0) n + ⟨n(0); ℓ|(E(1) n,k −δH)|n(1); k⟩ VN +E(2) n,kδk,ℓ. (1.2.33) In the first term on the right-hand side, the part proportional to E(1) n,k vanishes by orthonor-mality. On the second line, the term including δH can be simplified because δH is diagonal within VN. Recalling (1.2.28) we have ⟨n(0); ℓ|δH = E(1) n,ℓ⟨n(0); ℓ| + X p ⟨n(0); ℓ|δH|p(0)⟩⟨p(0)| . (1.2.34) 20 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY The piece in ˆ V vector drops out for our case of interest: ⟨n(0); ℓ|δH|n(1); k⟩ VN = E(1) n,ℓ⟨n(0); ℓ|n(1); k⟩ VN . (1.2.35) Back into equation (1.2.33) we now get X p δHnℓ,pδHp,nk E(0) p −E(0) n +  E(1) n,k −E(1) n,ℓ  ⟨n(0); ℓ|n(1); k⟩ VN + E(2) n,kδk,ℓ= 0 . (1.2.36) Setting ℓ= k we can determine the second correction to the energies: E(2) n,k = − X p |δHp,nk|2 E(0) p −E(0) n . (1.2.37) For k ̸= ℓwe get X p δHnℓ,pδHp,nk E(0) p −E(0) n +  E(1) n,k −E(1) n,ℓ  ⟨n(0); ℓ|n(1); k⟩ VN = 0 . (1.2.38) Had we not included the piece of |n(1); k⟩along the degenerate subspace we would have had an inconsistency, since there is no reason why the first term on the left-hand side must be zero. Now the above equation just fixes the components of |n(1); k⟩in the degenerate subspace as long as E(1) n,k ̸= E(1) n,ℓ: ⟨n(0); ℓ|n(1); k⟩ VN = − 1 E(1) n,k −E(1) n,ℓ X p δHnℓ,pδHp,nk E(0) p −E(0) n , k ̸= ℓ. (1.2.39) We thus have |n(1); k⟩ VN = − X ℓ̸=k |n(0); ℓ⟩ 1 E(1) n,k −E(1) n,ℓ X p δHnℓ,pδHp,nk E(0) p −E(0) n . (1.2.40) It may seem that this extra piece, found by using the O(λ2) equation, is higher order than it should in the perturbation: its numerator contains two powers of δH. But this expression also has a curious energy denominator, E(1) n,k −E(1) n,ℓ, in which each term has a power of δH. All in all, the correction to the state is properly first order in δH. Summarizing our result we have Degenerate perturbation theory with degeneracies lifted at O(λ): |n; k⟩λ = |n(0); k⟩−λ  X p δHp,nk E(0) p −E(0) n |p(0)⟩+ X ℓ̸=k |n(0); ℓ⟩ E(1) n,k −E(1) n,ℓ X p δHnℓ,pδHp,nk E(0) p −E(0) n  + O(λ2) En,k(λ) = E(0) n + λ δHnk,nk −λ2 X p δHnk,p δHp,nk E(0) p −E(0) n + O(λ3) , E(1) n,k = δHnk,nk . (1.2.41) 1.2. DEGENERATE PERTURBATION THEORY 21 1.2.3 Degeneracy lifted at second order We now investigate the case when the degeneracy is completely unbroken to first order. The situation and the setup is similar to the one we just considered: we have a degenerate subspace VN of dimension N and the rest of the space is called ˆ V . This time, however, we will assume that the degeneracy of H(0) is not broken to first order in the perturbation δH. Concretely, this means that on the VN basis |n(0); k⟩with k = 1, . . . , N, we have ⟨n(0); ℓ| δH |n(0); k⟩= E(1) n δℓ,k . (1.2.42) The first order energy correction is the same, and equal to E(1) n , for all basis states in VN. You should compare with (1.2.24), where the energy had an extra subscript to distinguish its various possible values. Because the degeneracy is not broken to first order we do not know at this point what is the good basis in VN. We will consider here the case when the degeneracy is completely lifted to second order. We express our ignorance about good basis vectors by stating that we are searching for the right linear combinations: |ψ(0)⟩= N X k=1 |n(0); k⟩a(0) k . (1.2.43) For some values of the constants a(0) k with k = 1, . . . , N the state |ψ(0)⟩will be good. We can think of a(0) as the column vector representation of |ψ(0)⟩in VN. We have written just one state, |ψ(0)⟩, even though we are expecting to find N good states to span the degenerate subspace. We therefore adjust the notation to reflect this. We introduce a new index I = 1, . . . , N and write |ψ(0) I ⟩= N X k=1 |n(0); k⟩a(0) Ik , I = 1, . . . , N . (1.2.44) The index I now labels the different good states and their different vector representations a(0) I . Our most immediate goal is to find those vectors a(0) I and thus the good basis. To do so we will have to consider second order energy corrections. The states |ψ(0) I ⟩form an orthonormal basis in VN if the coefficients a(0) I satisfy ⟨ψ(0) J |ψ(0) I ⟩= δIJ → X k (a(0) Jk)∗a(0) Ik = δIJ . (1.2.45) We set up the perturbation theory as usual |ψI⟩λ = |ψ(0) I ⟩+ λ|ψ(1) I ⟩+ λ2|ψ(2) I ⟩+ . . . , EnI(λ) = E(0) n + λ E(1) n + λ2 E(2) nI + λ3 E(3) nI + . . . . (1.2.46) 22 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY Note that in the energy expansion we have accounted for the degeneracy to zeroth and first order: the index I first appears in the second-order corrections to the energy. Using the Schr¨ odinger equation H(λ)|ψI⟩λ = EnI(λ)|ψI⟩λ , (1.2.47) gives the by now familiar equations, of which we list the first four: λ0 : (H(0) −E(0) n ) |ψ(0) I ⟩= 0 , λ1 : (H(0) −E(0) n ) |ψ(1) I ⟩= (E(1) n −δH)|ψ(0) I ⟩, λ2 : (H(0) −E(0) n ) |ψ(2) I ⟩= (E(1) n −δH)|ψ(1) I ⟩+ E(2) nI |ψ(0) I ⟩, λ3 : (H(0) −E(0) n ) |ψ(3) I ⟩= (E(1) n −δH)|ψ(2) I ⟩+ E(2) nI |ψ(1) I ⟩+ E(3) nI |ψ(0) I ⟩. (1.2.48) The zero-th order equation is trivially satisfied. For the order λ equation the overlap with ⟨n(0); ℓ| works out automatically, without giving any new information. Indeed, the left-hand side vanishes and we thus get 0 = ⟨n(0); ℓ|(E(1) n −δH)|ψ(0) I ⟩. (1.2.49) Since ⟨n(0); ℓ| is a δH eigenstate with eigenvalue E(1) n , up to a vector in ˆ V (see (1.2.34)), the above right-hand side vanishes. Acting on the order λ equation with ⟨p(0)| gives useful information: (E(0) p −E(0) n )⟨p(0)|ψ(1) I ⟩= ⟨p(0)|(E(1) n −δH)|ψ(0) I ⟩= −⟨p(0)|δH|ψ(0) I ⟩, (1.2.50) using the orthogonality of ˆ V and VN. Letting δHpI ≡⟨p(0)|δH|ψ(0) I ⟩, (1.2.51) we then have ⟨p(0)|ψ(1) I ⟩= − δHpI E(0) p −E(0) n . (1.2.52) Since the ket |ψ(0) I ⟩is still undetermined, it makes sense to write this information about |ψ(1) I ⟩in terms of the unknown a(0) I coefficients. We have δHpI ≡ N X k=1 ⟨p(0)|δH|n(0); k⟩a(0) Ik = N X k=1 δHp,nk a(0) Ik , (1.2.53) Back into (1.2.52) we get ⟨p(0)|ψ(1) I ⟩= − 1 E(0) p −E(0) n N X k=1 δHp,nk a(0) Ik . (1.2.54) 1.2. DEGENERATE PERTURBATION THEORY 23 This gives the piece of |ψ(1) I ⟩in ˆ V in terms of the unknown zeroth order eigenstates. We have now extracted all the information from the order λ equation. We look now at the order λ2 equation, which contains the second order corrections to the energy and therefore should help us determine the zeroth order good states. We hit that equation with ⟨n(0); ℓ| and we get 0 = ⟨n(0); ℓ|(E(1) n −δH)|ψ(1) I ⟩ ˆ V +⟨n(0); ℓ|(E(1) n −δH)|ψ(1) I ⟩ VN + E(2) nI a(0) Iℓ. (1.2.55) Happily, the second term, involving the components of |ψ(1) I ⟩along VN, vanishes because of the by now familiar property (1.2.28) adapted to this case. The piece with E(1) n on the first term also vanishes. We are thus left with 0 = −⟨n(0); ℓ|δH|ψ(1) I ⟩ ˆ V + E(2) nI a(0) Iℓ. (1.2.56) Introducing a resolution of the identity to the immediate right of δH, only the basis states in ˆ V contribute and we get 0 = − X p ⟨n(0); ℓ|δH|p(0)⟩⟨p(0)|ψ(1) I ⟩+ E(2) nI a(0) Iℓ, (1.2.57) where there is no need to copy the | ˆ V anymore. Using the result in (1.2.54) we now get 0 = X p δHnl,p 1 E(0) p −E(0) n N X k=1 δHp,nk a(0) Ik + N X k=1 E(2) nI δℓka(0) Ik (1.2.58) Reordering sums and multiplying by minus one we get N X k=1 − X p δHnl,pδHp,nk E(0) p −E(0) n −E(2) nI δℓk ! a(0) Ik = 0 . (1.2.59) To understand better this equation we define the N × N Hermitian matrix M(2) M(2) ℓ,k ≡− X p δHnℓ,p δHp,nk E(0) p −E(0) n . (1.2.60) The equation then becomes N X k=1  M(2) ℓ,k −E(2) nI δℓk  a(0) Ik = 0 . (1.2.61) Recalling that the Kronecker delta is the matrix representation of the identity, we have  M(2) −E(2) nI 1  a(0) I = 0 . (1.2.62) 24 CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY This is an eigenvalue equation that tells us that the energy corrections E(2) nI are the eigen-values of M(2) and the vectors a(0) I are the associated normalized eigenvectors. These determine, via (1.2.44), our orthonormal basis of good zeroth order states. If δH is known, the matrix M(2) is computable and Hermitian and can therefore be diagonalized. We will leave the computation of the component of |ψ(1) I ⟩on the degenerate subspace. That can be done if the degeneracy is completely broken to quadratic order (the eigenvalues of M(2) are all different). Still, it takes some effort and one must use the order λ3 equation. Our results so far are |ψI⟩λ = |ψ(0) I ⟩+ λ  X p |p(0)⟩ δHpI E(0) n −E(0) p + X J̸=I |ψ(0) J ⟩a(1) I,J  + O(λ2) , EIn(λ) = E(0) n + λ E(1) n + λ2 E(2) In + λ3 E(3) In + . . . O(λ3) . (1.2.63) Here the a(1) I,J are still unknown coefficients that determine the component of the first cor-rection to the states along the degenerate subspace. If you followed the discussion, all other symbols in the above equations have been defined and are computable given δH. The answer for the coefficients a(1) I,J turns out to be a(1) I,J = 1 E(2) nI −E(2) nJ "X p,q δHJp δHpq δHqI (E(0) p −E(0) n )(E(0) q −E(0) n ) −E(1) n X p δHJp δHpI E(0) p −E(0) n 2 # . (1.2.64) The third order corrections to the energy are E(3) In = X p,q δHIp δHpq δHqI (E(0) p −E(0) n )(E(0) q −E(0) n ) −E(1) n X p |δHpI|2 E(0) p −E(0) n 2 . (1.2.65) MIT OpenCourseWare 8.06 Quantum Physics III Spring 2018 For information about citing these materials or our Terms of Use, visit:
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Root Test - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Aptitude Engineering Mathematics Discrete Mathematics Operating System DBMS Computer Networks Digital Logic and Design C Programming Data Structures Algorithms Theory of Computation Compiler Design Computer Org and Architecture Sign In ▲ Open In App Root Test Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report The Root Test is a method used in the calculus to the determine the convergence or divergence of the infinite series. It is particularly useful for the series where the terms involve exponential functions or factorials. The test provides the criterion based on the nth root of the terms in the series to the assess whether the series converges absolutely. Table of Content What is Root Test? Proof of Root Test How the Root Test Works Applications of the Root Test Solved Examples on Root Test What is Root Test? The Root Test also known as the nth Root Test is a convergence test for the infinite series. It is used to the determine whether a series converges or diverges based on the nth root of the absolute value of its terms. The test is particularly useful for the series with the terms that involve exponentials or factorials. By applying the Root Test, we can simplify the process of the evaluating the convergence of the complex series. Statement of Root Test To prove the Root Test, consider a series ∑a n. The Root Test examines the limit of nth root of the absolute value of the terms: L=lim⁡n→∞∣a n∣n L = \lim_{n \to \infty} \sqrt[n]{|a_n|} L=lim n→∞​n∣a n​∣​ Then, If L < 1, then ∑a n converges absolutely. If L > 1 or L = ∞ then ∑a n diverges. If L = 1 the test is inconclusive, and the series may converge or diverge. Proof of Root Test Let's discuss each case in detail. Convergence Case (L < 1) Assume L < 1. This means there exists a constant c such that L=lim⁡n→∞∣a n∣n<c<1 L = \lim_{n \to \infty} \sqrt[n]{|a_n|} < c < 1 L=lim n→∞​n∣a n​∣​<c<1. Since L < c < 1 there exists an integer N such that for the all n > N ∣a n∣n<c\sqrt[n]{|a_n|} < c n∣a n​∣​<c. Raising both sides to the power of the n we get ∣a n∣<c n.|a_n| < c^n.∣a n​∣<c n. Since c < 1, c n converges to the 0 as n→∞n \to \infty n→∞. Therefore, is eventually smaller than a term of the convergent geometric series with the ratio c implying that ∑a n\sum a_n ∑a n​converges absolutely. Divergence Case (L > 1) Assume L > 1. This means there exists a constant d such that L=lim⁡n→∞∣a n∣n>d>1 L = \lim_{n \to \infty} \sqrt[n]{|a_n|} > d > 1 L=lim n→∞​n∣a n​∣​>d>1. Since L > d > 1 there exists an integer N such that for the all n > N ∣a n∣n>d\sqrt[n]{|a_n|} > d n∣a n​∣​>d. Raising both sides to the power of the n we get ∣a n∣>d n|a_n| > d^n∣a n​∣>d n. Since d > 1, d n diverges to infinity as n→∞.n \to \infty.n→∞. Therefore, ∣a n∣|a_n|∣a n​∣ is eventually larger than a term of the divergent geometric series implying that ∑a n\sum a_n ∑a n​diverges. Inconclusive Case (L = 1) When L = 1 the test does not provide the enough information about the behavior of the ∑a n\sum a_n∑a n​. The series could converge or diverge and additional tests or methods are needed to the determine its nature. How the Root Test Works? For a given infinite series ∑a n\sum a_n∑a n​ the Root Test involves evaluating the following limit: L=lim⁡n→∞∣a n∣n L = \lim_{n \to \infty} \sqrt[n]{|a_n|} L=lim n→∞​n∣a n​∣​ Where ∣a n∣n\sqrt[n]{|a_n|}n∣a n​∣​ denotes the n-th root of the absolute value of the n-th term a n a_n a n​ of the series. Conditions for Root Test If L<1: The series converges absolutely. If L>1 or L=∞: The series diverges. If L=1: The test is inconclusive and series may either converge or diverge. Other tests should be used. Applications of the Root Test The Root Test is useful in the various scenarios including: Series withExponential Terms: When dealing with the series involving e n or similar terms. Factorials in Series: For series involving the factorial terms such as the ∑n=1∞n n!\sum_{n=1}^{\infty} \frac{n}{n!}∑n=1∞​n!n​. Power Series: To determine the radius of the convergence of power series. Solved Examples on Root Test Example 1: Series:∑n 2 n\sum \frac{n}{2^n}∑2 n n​ Solution: a n=n 2 n a_n = \frac{n}{2^n} a n​=2 n n​ ∣a n∣n=n 2 n n=n n 2\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n}{2^n}} = \frac{\sqrt[n]{n}}{2} n∣a n​∣​=n 2 n n​​=2 n n​​ L=lim⁡n→∞n n 2=1 2 L = \lim_{n \to \infty} \frac{\sqrt[n]{n}}{2} = \frac{1}{2} L=lim n→∞​2 n n​​=2 1​ Since L < 1 the series converges. Example 2: Series:∑2 n n 2\sum \frac{2^n}{n^2}∑n 2 2 n​ Solution: a n=2 n n 2 a_n = \frac{2^n}{n^2} a n​=n 2 2 n​ ∣a n∣n=2 n n 2 n=2 n 2 n=2 n 2/n\sqrt[n]{|a_n|} = \sqrt[n]{\frac{2^n}{n^2}} = \frac{2}{\sqrt[n]{n^2}} = \frac{2}{n^{2/n}} n∣a n​∣​=n n 2 2 n​​=n n 2​2​=n 2/n 2​ L=lim⁡n→∞2 n 2/n=2 L = \lim_{n \to \infty} \frac{2}{n^{2/n}} = 2 L=lim n→∞​n 2/n 2​=2 Since L > 1 the series diverges. Example 3: Series:∑n 2 3 n\sum \frac{n^2}{3^n}∑3 n n 2​ Solution: a n=n 2 3 n a_n = \frac{n^2}{3^n} a n​=3 n n 2​ ∣a n∣n=n 2 3 n n=n 2 n 3=n 2/n 3\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n^2}{3^n}} = \frac{\sqrt[n]{n^2}}{3} = \frac{n^{2/n}}{3} n∣a n​∣​=n 3 n n 2​​=3 n n 2​​=3 n 2/n​ L=lim⁡n→∞n 2/n 3=1 3 L = \lim_{n \to \infty} \frac{n^{2/n}}{3} = \frac{1}{3} L=lim n→∞​3 n 2/n​=3 1​ Since L < 1 the series converges. Example 4: Series:∑5 n n!\sum \frac{5^n}{n!}∑n!5 n​ Solution: a n=5 n n!a_n = \frac{5^n}{n!} a n​=n!5 n​ ∣a n∣n=5 n n!n=5 n!n\sqrt[n]{|a_n|} = \sqrt[n]{\frac{5^n}{n!}} = \frac{5}{\sqrt[n]{n!}} n∣a n​∣​=n n!5 n​​=n n!​5​ Using Stirling's approximation: n!≈2 π n(n e)n n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n n!≈2 πn​(e n​)n, n!n≈n e\sqrt[n]{n!} \approx \frac{n}{e} n n!​≈e n​ L=lim⁡n→∞5 n e=0 L = \lim_{n \to \infty} \frac{5}{\frac{n}{e}} = 0 L=lim n→∞​e n​5​=0 Since L < 1 the series converges. Example 5: Series:∑n!4 n\sum \frac{n!}{4^n}∑4 n n!​ Solution: a n=n!4 n a_n = \frac{n!}{4^n} a n​=4 n n!​ ∣a n∣n=n!4 n n=n!n 4\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n!}{4^n}} = \frac{\sqrt[n]{n!}}{4} n∣a n​∣​=n 4 n n!​​=4 n n!​​ Using Stirling's approximation n!n≈n e\sqrt[n]{n!} \approx \frac{n}{e}n n!​≈e n​ L=lim⁡n→∞n e 4=∞L = \lim_{n \to \infty} \frac{\frac{n}{e}}{4} = \infty L=lim n→∞​4 e n​​=∞ Since L > 1 the series diverges. Practice Questions on Root Test Q1. Determine whether the series ∑3 n n 3\sum \frac{3^n}{n^3}∑n 3 3 n​ converges or diverges using the Root Test. Q2.Analyze the series ∑(n+1)!2 n\sum \frac{(n+1)!}{2^n}∑2 n(n+1)!​ to check its convergence or divergence. Q3.Apply the Root Test to the series ∑2 n n 2 n!\sum \frac{2^n}{n^2 n!}∑n 2 n!2 n​ and determine the behavior. Q4.Use the Root Test to evaluate the series ∑n n 1 0 n\sum \frac{n^n}{10^n}∑1 0 n n n​. Q5.Check the convergence of the series ∑n 2 5 n\sum \frac{n^2}{5^n}∑5 n n 2​ using the Root Test. Q6.Determine the behavior of the series ∑4 n n 4\sum \frac{4^n}{n^4}∑n 4 4 n​ using the Root Test. Q7.Analyze whether the series ∑5 n n 2\sum \frac{5^n}{n^2}∑n 2 5 n​ converges or diverges. Q8. Apply the Root Test to the series ∑n 3 3 n\sum \frac{n^3}{3^n}∑3 n n 3​ and state the result. Q9. Evaluate the series ∑n!n 2\sum \frac{n!}{n^2}∑n 2 n!​ to determine if it converges or diverges using the Root Test. Q10.Check the convergence of ∑6 n n!\sum \frac{6^n}{n!}∑n!6 n​ using the Root Test. Conclusion The Root Test is a powerful tool for the determining the convergence or divergence of the infinite series particularly useful for the series involving the exponential functions and factorials. By evaluating the n th root of the terms it provides the clear criterion for the series behavior. When the limit L is greater than 1 the series diverges when L is less than 1 the series converges. However, if L = 1 the test is inconclusive and other methods should be employed. Overall, mastering the Root Test enhances one's ability to the analyze series efficiently and effectively. 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187885
https://www.lutheranpioneers.org/courses/model-rocketry/lessons/model-rocketry-lesson-5/topic/model-rocketry-lesson-5/
Model Rocketry Lesson 5 Topic Progress: ← Back to Lesson For those of who are good at math I have the calculations listed below that are supplied by NASA to calculate the thrust of a model rocket engine. I do understand that there will be few who will be inspired by what is listed here, but I found it interesting and hopefully someone else will also. Thrust is the force which moves a rocket through the air. Thrust is generated by the rocket engine through the reaction of accelerating a mass of gas. The gas is accelerated to the rear and the rocket is accelerated in the opposite direction. To accelerate the gas, we need some kind of propulsion system. We will discuss the details of the propulsion system on some other pages. For right now, let us just think of the propulsion system as some machine which accelerates a gas. From Newton’s second law of motion, we can define a force to be the change in momentum of an object with a change in time. Momentum is the object’s mass times the velocity. When dealing with a gas, the basic thrust equation is given as: F = mdot e Ve – mdot 0 V0 + (pe – p0) Ae Thrust F is equal to the exit mass flow rate mdot e times the exit velocity Ve minus the free stream mass flow rate mdot 0 times the free stream velocity V0 plus the pressure difference across the engine pe – p0 times the engine area Ae. For liquid or solid rocket engines, the propellants, fuel and oxidizer, are carried on board. There is no free stream air brought into the propulsion system, so the thrust equation simplifies to: F = mdot Ve + (pe – p0) Ae where we have dropped the exit designation on the mass flow rate. Using algebra, let us divide by mdot: F / modt = Ve + (pe – p0) Ae / mdot We define a new velocity called the equivalent velocity Veq to be the velocity on the right hand side of the above equation: Veq = Ve + (pe – p0) Ae / mdot Then the rocket thrust equation becomes: F = mdot Veq The total impulse (I) of a rocket is defined as the average thrust times the total time of firing. On the slide we show the total time as “delta t”. (delta is the Greek symbol that looks like a triangle): I = F delta t Since the thrust may change with time, we can also define an integral equation for the total impulse. Using the symbol (Sdt) for the integral, we have: I = S F dt Substituting the equation for thrust given above: I = S (mdot Veq) dt Remember that mdot is the mass flow rate; it is the amount of exhaust mass per time that comes out of the rocket. Assuming the equivalent velocity remains constant with time, we can integrate the equation to get: I = m Veq where m is the total mass of the propellant. We can divide this equation by the weight of the propellants to define the specific impulse. The word “specific” just means “divided by weight”. The specific impulse Isp is given by: Isp = Veq / g0 where g0 is the gravitational acceleration constant (32.2 ft/sec^2 in English units, 9.8 m/sec^2 in metric units). Now, if we substitute for the equivalent velocity in terms of the thrust: Isp = F / (mdot g0) Mathematically, the Isp is a ratio of the thrust produced to the weight flow of the propellants. A quick check of the units for Isp shows that: Isp = m/sec / m/sec^2 = sec Why are we interested in specific impulse? First, it gives us a quick way to determine the thrust of a rocket, if we know the weight flow rate through the nozzle. Second, it is an indication of engine efficiency. Two different rocket engines have different values of specific impulse. The engine with the higher value of specific impulse is more efficient because it produces more thrust for the same amount of propellant. Third, it simplifies our mathematical analysis of rocket thermodynamics. The units of specific impulse are the same whether we use English units or metric units. Fourth, it gives us an easy way to “size” an engine during preliminary analysis. The result of our thermodynamic analysis is a certain value of specific impulse. The rocket weight will define the required value of thrust. Dividing the thrust required by the specific impulse will tell us how much weight flow of propellants our engine must produce. This information determines the physical size of the engine. The forces on a rocket change dramatically during a typical flight. This figure shows a derivation of the change in velocity during powered flight while accounting for the changing mass of the rocket. During powered flight the propellants of the propulsion system are constantly being exhausted from the nozzle. As a result, the weight of the rocket is constantly changing. In this derivation, we are going to neglect the effects of aerodynamic lift and drag. We can add these effects to the final answer. Let us begin with Newton’s second law of motion, shown in blue on the figure: d (M u) / dt = F net where M is the mass of the rocket, u is the velocity of the rocket, F net is the net external force on the rocket and the symbol d / dt denotes that this is a differential equation in time t. The only external force which we will consider is the thrust from the propulsion system. On the web page describing the specific impulse, the thrust equation is given by: F = mdot Veq where mdot is the mass flow rate, and Veq is the equivalent exit velocity of the nozzle which is defined to be: Veq = V exit + (p exit – p0) Aexit / mdot where V exit is the exit velocity, p exit is the exit pressure, p0 is the free stream pressure, and A exit is the exit area of the nozzle. Veq is also related to the specific impulse Isp: Veq = Isp g0 where g0 is the gravitational constant. m dot is mass flow rate and is equal to the change in the mass of the propellants mp on board the rocket: mdot = d mp / dt Substituting the expression for the thrust into the motion equation gives: d (M u) / dt = V eq d mp / dt d (M u) = Veq d mp Expanding the left side of the equation: M du + u dM = Veq d mp Assume we are moving with the rocket, then the value of u is zero: M du = Veq d mp Now, if we consider the instantaneous mass of the rocket M, the mass is composed of two main parts, the empty mass me and the propellant mass mp. The empty mass does not change with time, but the mass of propellants on board the rocket does change with time: M(t) = me + mp (t) Initially, the full mass of the rocket mf contains the empty mass and all of the propellant at lift off. At the end of the burn, the mass of the rocket contains only the empty mass: M initial = mf = me + mp M final = me The change on the mass of the rocket is equal to the change in mass of the propellant, which is negative, since propellant mass is constantly being ejected out of the nozzle: dM = – d mp If we substitute this relation into the motion equation: M du = – Veq dM du = – Veq dM / M We can now integrate this equation: delta u = – Veq ln (M) where delta represents the change in velocity, and ln is the symbol for the natural logarithmic function. The limits of integration are from the initial mass of the rocket to the final mass of the rocket. Substituting for these values we obtain: delta u = Veq ln (mf / me) This equation is called the ideal rocket equation. There are several additional forms of this equation which we list here: Using the definition of the propellant mass ratio MR MR = mf / me delta u = Veq ln (MR) or in terms of the specific impulse of the engine: delta u = Isp g0 ln (MR) If we have a desired delta u for a maneuver, we can invert this equation to determine the amount of propellant required: MR = exp (delta u / (Isp g0) ) where exp is the exponential function. If you include the effects of gravity, the rocket equation becomes: delta u = Veq ln (MR) – g0 tb where tb is the time for the burn. Posted in .
187886
https://www.thecreativeeducator.com/2017/lessons/simple-surveys-and-great-graphs
Math - Grades K-2 Simple Surveys and Great Graphs Students create and complete surveys, graph the data, and share the results with an audience outside the classroom. Task In this age of overwhelming amounts of information, data is playing a larger and larger role in our culture. While young students may not be having sophisticated conversations about data yet, they are hearing statistics and numbers on television, radio, and adult conversations. In this lesson, students work in small teams to choose a question and survey their classmates to collect data. Then, the teams create bar graphs of the data, analyze the results, and share their findings with an audience that can use the information. Engage Young students are just beginning to learn to read, and may not be ready to read to learn. That doesn’t mean that they don’t need to think about information or can’t work with it. A bar graph is a simple way to present data to someone in a way that makes it easy to understand. Teaching young students how to create and read a graph gets them thinking about information now and builds a foundation for future understanding of more complex data. Show your students examples of graphs that you have found in the world outside of your classroom. For example, USA Today regularly shares data in visual form through their Snapshots. Explore an example of a bar graph like the one above. See if students can already “read” it and answer questions like: What is this graph about? Which is the most popular option? Which is the least popular option? Who would care about this information? The last question is more difficult because the answer is not in the graph. If students can see how the owner of a bagel shop could use this information to order the right type of bagels in the right amounts for their customers, you help them make connections between what they are learning in the classroom and the world beyond it. Next, introduce your students to the concept of a survey. Read a story like Tally O’Malley by Stuart J. Murphy to get students thinking about using tally marks to collect data. Create a two- or three-option survey and have students come up individually to make their tally mark. Younger students can make a simple tally mark, but older students should make a diagonal tally mark to show groups of five. You can easily create this survey using a digital tool like Wixie. Then you can open it on an interactive white board and when students come into class have them make a tally mark as a bell ringer activity. Take the information from the survey you created and show students how to create a bar chart to show the information. You can use chart paper with grid lines or create one from a template in Wixie. As you look at the bar chart and discuss the results of your survey, connect the data in it to an audience that would benefit from the information. For example, a bar chart created from the survey above would quickly show a Kindergarten teacher which book would be a good choice to buy next for their classroom library. Create Let students know that they will be working in small teams to create their own survey and collect data they will display in a bar graph. To make the project more engaging, let the students collect data about something they want to know. For example: Which flavor of ice cream do students like the most - vanilla, chocolate, or strawberry? or What kind of class pet should your teacher get for your room? You might want to share some ideas with your students. Depending on the age of your students you can suggest ideas with two, three (CCSS 1st grade), or four (CCSS 2nd Grade) options, such as: Do you want to visit the moon? (yes or no) Who will win the game this weekend? (names of two sports teams) What is your favorite school subject? (reading, math, or science) What type of pet do you have? (dog, cat, fish, other) Ask students to come up with some of their own ideas and create a list everyone can see. Choose students to be your team leaders and ask them to be the first to select a topic they want to survey. Then, have other students form teams of 3-4 students around each topic chosen. If you want to make real world connections, you might focus their questions around food. This way the data can be used to help the cafeteria make decisions or a parent choose the right food for a birthday party. For example: What is your favorite fruit? What is your favorite lunch meal? Do you prefer ice cream or a cookie for dessert? Do you drink milk, water, juice with your lunch? After teams have chosen a topic or question they want to ask, have them create a survey they can use to collect data in the form of tally marks. Next, have them collect data from each student in the class. While students should create and collect survey data as a team, have individual team members each create a bar graph to display the data. This will make it easier to assess individual understanding. Share Once individual graphs are complete, have each student print their bar graph and share it with the rest of their group. Have students work together to explore each of their graphs and ask and answer questions about the data similar to those you asked in the initial graph discussion. Have teams present what they think is their best graph to an outside audience, sharing their findings and the recommendations they have for actions they should take based on the information displayed in the bar graph. For example, if a team surveyed most popular fruits, they could share their bar graph with cafeteria staff to help them make decisions about which fruits are likely to be most popular in school lunches. If a team surveyed students about favorite playground activities, they could share the data with the principal and parent association as they consider new playground equipment to purchase, update, or replace. Assessment In this project, you are evaluating student understanding of why to use a survey, how to create one, and how to collect survey data. You are also evaluating student’s ability to create a bar graph that represents the data they collect and interpret the data on the graph. Creating a tally mark survey is perhaps the easiest. Even in Kindergarten, students understand one-to-one correspondence, so creating a mark for each piece of data should be easy for most students. Attention to detail and clear headings that organize the data will more likely be places you need to evaluate student comprehension. Depending on the age and ability of your students, you may have created many of the necessary features of their graphs for them. Even if you have added in the title, labels for the axes, and images to represent the options surveyed or used grid lines or graph paper to create a consistent scale, be sure to discuss the need for each of these elements to help the viewer more clearly understand the information. You may want to create a _______ (blank) space in the position of the title and axis labels so that students have to do more work and think more deeply about what the graphs is supposed to be about and what information is being shown. As students are sharing their graphs with a partner, listen to their discussions to get a sense of their understanding. You might want to provide them with prompts like: ”Which option had the most responses?” but they will internalize more deeply if they have to work to come up with their own questions. If students have worked in small teams to create a graph together, take time to prepare the questions as a group to challenge another group. Then, pair with that group for questions and answers. Resources Stuart J. Murphy. Tally O’Malley.. ISBN: 0060531649 BrainPOP: Tally Charts and Bar Graphs Standards Common Core State Standards for Math: CCSS.MATH.CONTENT.1.MD.C.4 Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. CCSS.MATH.CONTENT.K.G.A.2. Correctly name shapes regardless of their orientations or overall size. CCSS.MATH.CONTENT.2.MD.D.10 Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems1using information presented in a bar graph. ISTE NETS for Students 2016: 3. Knowledge Constructor Students critically curate a variety of resources using digital tools to construct knowledge, produce creative artifacts and make meaningful learning experiences for themselves and others. Students: d. build knowledge by actively exploring real-world issues and problems, developing ideas and theories and pursuing answers and solutions. 6. Creative Communicator Students communicate clearly and express themselves creatively for a variety of purposes using the platforms, tools, styles, formats and digital media appropriate to their goals. Students: d. publish or present content that customizes the message and medium for their intended audiences. Lessons by Subject Language Arts Math Science Social Studies Lessons by Grade K 1 2 3 4 5 MS CE Newsletter Get creative classroom ideas delivered straight to your inbox once a month. Follow us on Instagram for daily inspiration### Create a Graphic Organizer Create a thought web, cluster, flowchart, or other graphic organizer for a lesson Most Read Classroom constitutionLesson Plan Hero's JourneyLesson Plan Set SMART GoalsLesson Plan Character Coat of ArmsLesson Plan InfographicsLesson Plan Embrace action research Support learners throughout all 5 stages of language acquisition Design a book coverLesson Plan Create a vision boardLesson Plan Creative ways to use technology with students Sponsored by Wixie Creating timelines with Wixie's Mind Map tool Creative, digital book reviews Fun and powerful ideas with animated characters More from Creative Educator Language Arts Lesson: My Favorite Relative Mathematics Lesson: Tangram Tales Science Lesson: Visit Our Solar System Social Studies Lesson: Create a Vision Board Science Lesson: Fantastic Field Guides Language Arts Lesson: Commemorative Ceremonies More sites to help you find success in your classroom Wixie Share your ideas, imagination, and understanding through writing, art, voice, and video. 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187887
https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_1e_(OpenStax)/12%3A_Introduction_to_Calculus/12.02%3A_Finding_Limits_-_Properties_of_Limits
Skip to main content 12.2: Finding Limits - Properties of Limits Last updated : Sep 26, 2025 Save as PDF 12.1: Finding Limits - Numerical and Graphical Approaches 12.3: Continuity Page ID : 1404 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Consider the rational function The function can be factored as follows: which gives us Does this mean the function is the same as the function The answer is no. Function does not have in its domain , but does. Graphically, we observe there is a hole in the graph of at , as shown in Figure and no such hole in the graph of , as shown in Figure. (left) The graph of function contains a break at and is therefore not continuous at . (Right)The graph of function is continuous. So, do these two different functions also have different limits as approaches 7? Not necessarily. Remember, in determining a limit of a function as approaches , what matters is whether the output approaches a real number as we get close to . The existence of a limit does not depend on what happens when equals . Look again at Figure and Figure. Notice that in both graphs, as approaches 7, the output values approach 8. This means Remember that when determining a limit, the concern is what occurs near , not at . In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition. Finding the Limit of a Sum, a Difference, and a Product Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits. Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions. properties of limits Let and represent real numbers, and and be functions, such that and For limits that exist and are finite, the properties of limits are summarized in Table | Constant, k | | | Constant times a function | | | Sum of functions | | | Difference of functions | | | Product of functions | | | Quotient of functions | | | Function raised to an exponent | , where is a positive integer | | nth root of a function , where n is a positive integer | | | Polynomial function | | Example : Evaluating the Limit of a Function Algebraically Evaluate Solution Exercise : Evaluate the following limit: Solution 26 Finding the Limit of a Polynomial Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function , we can find the limits of the individual terms of the function , and then add them together. Also, the limit of a polynomial function as approaches is equivalent to simply evaluating the function for . how to: Given a function containing a polynomial, find its limit Use the properties of limits to break up the polynomial into individual terms. Find the limits of the individual terms. Add the limits together. Alternatively, evaluate the function for . Example : Evaluating the Limit of a Function Algebraically Evaluate Solution Exercise : Evaluate Solution 59 Example : Evaluating the Limit of a Polynomial Algebraically Evaluate Solution Exercise : Evaluate the following limit: Solution 10 Finding the Limit of a Power or a Root When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function ; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function ; the same holds true for higher roots. Example : Evaluating a Limit of a Power Evaluate Solution We will take the limit of the function as approaches 2 and raise the result to the 5th power. Exercise : Evaluate the following limit: Solution −64 Q & A: If we can’t directly apply the properties of a limit, for example in , can we still determine the limit of the function as approaches ? Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function . Finding the Limit of a Quotient Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify. Example : Evaluating the Limit of a Quotient by Factoring Evaluate Solution Factor where possible, and simplify. Analysis When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated. Notice, the function is equivalent to the function Notice that the limit exists even though the function is not defined at . Exercise Evaluate the following limit: Solution Example : Evaluating the Limit of a Quotient by Finding the LCD Evaluate Solution Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator. Analysis When determining the limit of a rational function that has terms added or subtracted in either the numerator or denominator, the first step is to find the common denominator of the added or subtracted terms; then, convert both terms to have that denominator, or simplify the rational function by multiplying numerator and denominator by the least common denominator. Then check to see if the resulting numerator and denominator have any common factors. Exercise : Evaluate Solution how to: Given a limit of a function containing a root, use a conjugate to evaluate If the quotient as given is not in indeterminate form, evaluate directly. Otherwise, rewrite the sum (or difference) of two quotients as a single quotient, using the least common denominator (LCD). If the numerator includes a root, rationalize the numerator; multiply the numerator and denominator by the conjugate of the numerator. Recall that are conjugates. Simplify. Evaluate the resulting limit. Example : Evaluating a Limit Containing a Root Using a Conjugate Evaluate Solution Analysis When determining a limit of a function with a root as one of two terms where we cannot evaluate directly, think about multiplying the numerator and denominator by the conjugate of the terms. Exercise Evaluate the following limit: . Solution Example : Evaluating the Limit of a Quotient of a Function by Factoring Evaluate Solution Analysis Multiplying by a conjugate would expand the numerator; look instead for factors in the numerator. Four is a perfect square so that the numerator is in the form and may be factored as Exercise Evaluate the following limit: Solution how to: Given a quotient with absolute values, evaluate its limit Try factoring or finding the LCD. If the limit cannot be found, choose several values close to and on either side of the where the function is undefined. 3. Use the numeric evidence to estimate the limits on both sides. Example : Evaluating the Limit of a Quotient with Absolute Values Evaluate Solution The function is undefined at , so we will try values close to 7 from the left and the right. Left-hand limit: Right-hand limit: Since the left- and right-hand limits are not equal, there is no limit. Exercise Evaluate Solution Key Concepts The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves. See Example. The limit of a polynomial function can be found by finding the sum of the limits of the individual terms. See Example and Example. The limit of a function that has been raised to a power equals the same power of the limit of the function . Another method is direct substitution. See Example. The limit of the root of a function equals the corresponding root of the limit of the function . One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify. See Example. Another method of finding the limit of a complex fraction is to find the LCD. See Example. A limit containing a function containing a root may be evaluated using a conjugate. See Example. The limits of some functions expressed as quotients can be found by factoring. See Example. One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful. See Example. Glossary properties of limits : a collection of theorems for finding limits of functions by performing mathematical operations on the limits 12.1: Finding Limits - Numerical and Graphical Approaches 12.3: Continuity
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle
Skip to main content Le Chatelier's Principle Last updated : Jan 30, 2023 Save as PDF Reversible vs. Irreversible Reactions Case Study: The Manufacture of Ethanol from Ethene Page ID : 1345 ( \newcommand{\kernel}{\mathrm{null}\,}) Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position. Case Study: The Manufacture of Ethanol from Ethene : This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction. Effect of Temperature on Equilibrium : A temperature change occurs when temperature is increased or decreased by the flow of heat. This shifts chemical equilibria toward the products or reactants, which can be determined by studying the reaction and deciding whether it is endothermic or exothermic. : + Exothermic vs. Endothermic and K ICE Tables : An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation). Le Chatelier's Principle and Dynamic Equilbria : This page looks at Le Chtelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium. Le Chatelier's Principle Fundamentals : Le Chtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. The Contact Process : The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. The Effect of Changing Conditions : This page looks at the relationship between equilibrium constants and Le Chtelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same. The Haber Process : This page describes the Haber Process for the manufacture of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. Template:HideTOC Reversible vs. Irreversible Reactions Case Study: The Manufacture of Ethanol from Ethene
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https://math.stackexchange.com/questions/2272225/solving-int-01-sqrtx21-with-euler-substitution
analysis - Solving $\int_0^1 \sqrt{x^2+1}$ with Euler substitution - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Solving \int_0^1 \sqrt{x^2+1} with Euler substitution Ask Question Asked 8 years, 4 months ago Modified8 years, 4 months ago Viewed 233 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. \begingroup So I was trying to solve this integral \int_0^1 \sqrt{x^2+1} \, dx in two different ways with Euler substitution. So: (x^2+1)=x+t, so that x=(1-t^2)/(2t), (x^2+1)=xt+1, x=2t/(1-t^2). But I am not sure how to proceed, and how do the integral and the limits change... Any tips? Thanks :) analysis definite-integrals Share Cite Follow Follow this question to receive notifications edited May 8, 2017 at 22:09 TMM 10.1k 3 3 gold badges 38 38 silver badges 54 54 bronze badges asked May 8, 2017 at 22:04 Simon JachsonSimon Jachson 406 2 2 silver badges 14 14 bronze badges \endgroup 5 1 \begingroup is it x^2-1 or 1-x^2\endgroup hamam_Abdallah –hamam_Abdallah 2017-05-08 22:06:00 +00:00 Commented May 8, 2017 at 22:06 \begingroup it's + , sorry for the mistake\endgroup Simon Jachson –Simon Jachson 2017-05-08 22:07:04 +00:00 Commented May 8, 2017 at 22:07 \begingroup Please use MathJax.\endgroup Em. –Em. 2017-05-08 22:09:04 +00:00 Commented May 8, 2017 at 22:09 \begingroup It does not look like those changes are going to work. You could try x=\tan t but there is still some work ahead.\endgroup Miguel –Miguel 2017-05-08 22:18:21 +00:00 Commented May 8, 2017 at 22:18 \begingroup@MiguelAtencia Why wouldn't it work? It is much more work than the conventional x=tant sub, but Euler sub can work:en.wikipedia.org/wiki/Euler_substitution\endgroup imranfat –imranfat 2017-05-08 22:25:25 +00:00 Commented May 8, 2017 at 22:25 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. \begingroup Put x=\sinh (t)=\frac {e^t-e^{-t}}{2} then x=0 gives t=0 x=1 gives t such that e^t-e^{-t}=2 that is (e^t)^2-2e^t-1=0 t=\ln (1+\sqrt {2}) thus the integral becomes \int_0^{\ln (1+\sqrt {2})}\cosh^2 (t)dt =\int_0^{\ln (1+\sqrt {2})}\frac {1+\cosh (2t)}{2}dt =\frac {1}{2}\ln (1+\sqrt {2})+\frac {1}{8}((1+\sqrt {2})^2-(1-\sqrt {2})^2) i am sure you can finish. Share Cite Follow Follow this answer to receive notifications edited May 8, 2017 at 22:17 answered May 8, 2017 at 22:14 hamam_Abdallahhamam_Abdallah 63.8k 4 4 gold badges 29 29 silver badges 50 50 bronze badges \endgroup 2 1 \begingroup i already did this. My task is to do it with Euler substitution...\endgroup Simon Jachson –Simon Jachson 2017-05-08 22:17:42 +00:00 Commented May 8, 2017 at 22:17 \begingroup@SimonJachson What do you mean by Euler substitution. e^{it}... \endgroup hamam_Abdallah –hamam_Abdallah 2017-05-08 22:19:40 +00:00 Commented May 8, 2017 at 22:19 Add a comment| This answer is useful 0 Save this answer. Show activity on this post. \begingroup @Salahamam With Euler substitution it is meant a particular kind of substitution discovered by Euler for integrals containing a squareroot and a quadratic polynomial inside. See my link in a comment. It is not about e^{it} here. The substitution in question is \sqrt{x^2+1}=x+t. Now solving this for x to get: x=\frac{1-t^2}{2t} and so dx=\frac{-2t^2-2}{4t^2}dt. The given integral is of the form \int\frac{1-t^2+2t^2}{2t}\frac{-2t^2-2}{4t^2}dt. Enfin, this is now all polynomial work where the denominator is a monomial and so after the numerator is worked out, one can split fractions and integrate. Please give it a try from here. This is just (annoying) algebra... Share Cite Follow Follow this answer to receive notifications answered May 8, 2017 at 22:34 imranfatimranfat 10.2k 4 4 gold badges 23 23 silver badges 35 35 bronze badges \endgroup 4 \begingroup Thanks but it is much easier with hyperbolic sinus.\endgroup hamam_Abdallah –hamam_Abdallah 2017-05-08 22:46:49 +00:00 Commented May 8, 2017 at 22:46 \begingroup Thanks!! and how does the limits changes here , 1/4 lower and 1 upper ?\endgroup Simon Jachson –Simon Jachson 2017-05-08 22:48:11 +00:00 Commented May 8, 2017 at 22:48 \begingroup@SimonJachson i think the lower is \sqrt {2}-1\endgroup hamam_Abdallah –hamam_Abdallah 2017-05-08 22:57:10 +00:00 Commented May 8, 2017 at 22:57 \begingroup@Salahamam_Fatima I agree that the Euler substitution is a bit cumbersome and that other methods like the hyperbolic sine or tangent substitutions are easier. I wouldn't choose the Euler sub either for this case, but that is what the OP wanted. And limit changes, that shouldn't be to difficult.You can find that from the given sub\endgroup imranfat –imranfat 2017-05-08 23:03:32 +00:00 Commented May 8, 2017 at 23:03 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions analysis definite-integrals See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 3Integral \int_0^1\sqrt[2\,n\,]{\frac x{1-x}}\,\mathrm dx 8Prove that \int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0 (without trigonometric substitution) 7To Find \int_0^{\pi} \frac{\sin(x)}{1+\sin(x)} why the substitution of \sin(x)=t Gives wrong answer? 2Does this question require double u-substitution? 4Ahmed's Integral Variation \int_0^1 \frac{\tan^{-1}(\sqrt{2x^2+1})}{(1+3x^2)\sqrt{2x^2+1}}dx 2Tangent half-angle substitution for \int_0^{2 \pi} \frac{1- \cos x}{3 + \cos x} 5Find \int_0^\infty \frac{\sqrt x e^{-x}}{b^2 +x^2} dx 11Integral \int_0^\infty \frac{\phi^2 x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}\mathrm dx Hot Network Questions Why are LDS temple garments secret? Can you formalize the definition of infinitely divisible in FOL? 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187890
https://par.nsf.gov/servlets/purl/10173221
Effect of quantum mechanical global phase factor on error versus sensitivity limitation in quantum routing E. Jonckheere, S. Schirmer, and F. Langbein Abstract — In this paper, we explore the effect of the purely quantum mechanical global phase factor on the problem of controlling a ring-shaped quantum router to transfer its excitation from an initial spin to a specified target spin. “Quantum routing” on coherent spin networks is achieved by shaping the energy landscape with static bias control fields, which already results in the nonclassical feature of purely oscillatory closed-loop poles. However, more to the point, it is shown that the global phase factor requires a projective re-interpretation of the traditional tracking error where the wave function state is considered modulo its global phase factor. This results in a time-domain relaxation of the conflict between small tracking error and small sensitivity of the tracking error to structured uncertainties. While fundamentally quantum routing is achieved at a specific final time and hence calls for time-domain techniques, we also develop a projective s-domain limitation. I. I NTRODUCTION We consider a spintronic network of N XX-coupled spins in its single excitation subspace. The latter means that one spin and one spin only is excited, “up,” while all others are “down.” In this subspace, we choose a basis such that the wave function |Ψ〉 = en, where {en}Nn=1 is the natural basis of CN over C, denotes the quantum state where the sole excitation is on spin # n. In the chosen basis, for XX-couplings, the Hamiltonian H is the adjacency matrix of the graph of the spin couplings, weighted by the coupling strengths, with zeros on the diagonal. A simple example is given by the XX-ring structure, where the Hamiltonian has tridiagonal-like structure, H = ⎛⎜⎜⎜⎜⎜⎝ 0 J1,2 . . . 0 J1,N J2,1 0 . . . 0 0 ... ... . . . ... ... 0 0 . . . 0 JN −1,N JN, 1 0 . . . JN,N −1 0 ⎞⎟⎟⎟⎟⎟⎠ . (1) In the above, Jm,n = Jn,m to make the Hamiltonian Hermitian. We operate in a system of units where ¯h = 1 and the network has uniform couplings with strengths Jmn , m = n, normalized to 1. The “open-loop” Schr¨ odinger equation reads ∣∣∣ ˙Ψ( t) 〉 = −jH |Ψ( t)〉, subject to some initial condition |Ψ(0) 〉 = |IN 〉, where |IN 〉 = ei denotes the quantum state with the E. Jonckheere is with the Dept. of Elec. and Comp. Eng., Univ. of Southern California, jonckhee@usc.edu. E. J. was supported by NSF grant IRES 1829078. S. Schirmer is with the Physics Dept., Singleton Campus, Swansea University, UK, lw1660@gmail.com F. Langbein is with the School of Computer Science and Informatics, Cardiff University, UK, langbeinfc@cardiff.ac.uk excitation on some “input” spin i. The control objective is to transfer the |IN 〉 state to some |OUT 〉 = eo state where the excitation is on some “output” spin o. This is to be accomplished in a short amount of time tf and with maximum fidelity 1 , F(tf ) := |〈 OUT |Ψ( tf )〉| . This is achieved by i-o selectively modifying the energy landscape with static bias fields {Dn}Nn=1 applied to the respective spins, resulting in the Hamiltonian HD = H + D, where D = diag {Dn}Nn=1 . The controlled Schr¨ odinger equation becomes ∣∣∣ ˙Ψ( t) 〉 = −j(H + D) |Ψ( t)〉 , |Ψ(0) 〉 = |IN 〉 , = −jH |Ψ( t)〉 + u(t), u(t) = −jD |Ψ( t)〉 . (2) It is observed that the right-hand side is split, somewhat artificially, into an open-loop term −jH |Ψ( t)〉 and a “con-trol” term u(t). Despite the appearance of this control as a classical measurement-mediated feedback, it does not need measurement of the state (and does not create back-action of the measurements); indeed, the feedback is field-mediated by the physical interaction between the spins and the bias fields. Nevertheless, u(t) has the mathematical structure of a classical feedback and as such the question is whether it is subject to some of the classical error-versus-sensitivity limitations. Classically, such limitations refer to a tracking error |OUT 〉 − | Ψ( t)〉 and its sensitivity to uncertainties, but in the quantum context, the error is made smaller by considering the wave function modulo its phase factor. This paper investigates the impact of such global phase factor on the log-sensitivity of the error and points to a relaxation of the traditional conflict. This paper follows in the footsteps of , where classical limitations are shown to be defeated in the time-domain, whereas here we also investigate the frequency-domain limi-tations. All data is from the database . Proofs are omitted, but available in . A. Notation Throughout the paper, we consider three feedback con-figurations: the CLASSICAL configuration of Fig. 1, the QUAN T UM configuration of Fig. 2 with the global phase factor shown in the shaded areas, and the semi-classical configuration of Fig. 2 but with the global phase factors removed. The relevant quantities are as follows: • L(s),̂ S(s), T(s) := I −̂ S(s): classical (Fig. 1) loop ma-trix, sensitivity and complementary sensitivity matrices, resp. 1 Sometimes the fidelity is defined as |〈 OUT |Ψ( tf )〉| 2. 2019 IEEE 58th Conference on Decision and Control (CDC) Palais des Congrès et des Expositions Nice Acropolis Nice, France, December 11-13, 2019 978-1-7281-1398-2/19/$31.00 ©2019 IEEE 1339 Authorized licensed use limited to: University of Southern California. Downloaded on July 20,2020 at 04:17:51 UTC from IEEE Xplore. Restrictions apply. • L(s),̂ S(s), T (s) = I − S (s): projective loop ma-trix, sensitivity and complementary sensitivity matrices, resp., with global phase factor (shaded boxes in Fig 2). ˆS(s) is defined analytic in s > 0. • S(t): inverse Laplace transform of ̂ S(s), vanishing for t < 0. • L(s),̂ S(s), T (s) = I − ˆS(s): loop matrix, sensitivity and complementary sensitivity matrices, resp., without global phase factor (after removal of shaded boxes in Fig 2). ̂ S(s) is analytic in s > 0. • S(t): inverse transform of ˆS(s) with S(t < 0) = 0 .Throughout the paper we use the Dirac bra and ket notations. II. C LASSICAL VERSUS QUANTUM ARCHITECTURE Laplace domain technique are of limited use in quantum control as most of the fidelity specification are rather in the time domain. Nevertheless, as shown in , Laplace techniques are still useful to study steady-state ( s ≈ 0)behavior. Besides, a quick review of the Laplace domain limitations are necessary to explore the classical-quantum discrepancies. A. Classical The fundamental limitation looked at in the present paper is the quantum mechanical equivalent, if any, of ˆS(s) + T(s) = I, where ˆS(s) = ( I+L(s)) −1 is the sensitivity matrix, L(s) is the loop matrix, and T(s) is the complementary sensitivity L(s)( I + L(s)) −1 of the classical loop shown in Fig. 1. Note that the disturbance ˆr(s) could be anything and does not support the notion of selectivity , that is, when ˆr(s) is restricted to be a terminal target as shown in Fig. 2, nor does Fig. 1 support the initial conditon |IN 〉 of Fig. 2. Given the classical tracking error e(t) = r(t) − y(t), we have ˆe(s) = ˆ S(s)ˆ r(s) indicating that ˆS(s) is the transmission from the disturbance ˆr(s) to the error ˆe(s). T(s) on the other hand is related to the log-sensitivity of ˆS(s) to errors dL(s) in the loop matrix. Precisely, ˆS−1(s)dˆS(s) = −(( dL(s)) L−1(s)) T(s). ˆS(s) + T(s) = I hence quantifies the well known con-flict between achieving simultaneously small tracking error and small log-sensitivity of tracking error to uncertainties, disregarding selectivity. If dL is structured to represent an uncertainty on a parameter, say J, then the above is rewritten as d ln ˆ S(s) d ln J = − d ln L(s) d ln J T(s), (3) where d ln ˆ S = ˆ S−1(dˆS) and d ln L = ( dL)L−1. In either case, it is observed that T(s) is related to the log-sensitivity of ˆS(s). B. Phase factor and complex projective space CP N −1 In the quantum control problem of moving the system from one quantum state to another one, there is no tracking error to be minimized, but a fidelity F(tf ) = |〈 OUT |Ψ( tf )〉| to be maximized relative to D. However, the maximization ( ) ( ) ( ) LPKsss )(sr )(track se 1ˆ ( ) (( )) sIsSL +െ ( ) y s ( ) y s 6 Fig. 1: Classical single-degree-of-freedom loop of the fidelity can be related to minimization of a tracking error understood in some projective sense. Theorem 1: The optimal controller achieving the maxi-mum fidelity max D |〈 OUT |Ψ( tf )〉| is the same as the controller achieving min D ( min φ ‖ | OUT 〉 − ejφ |Ψ( tf )〉 ‖ ) , (4) where φ is a global phase factor with the minimum achieved for φ∗(tf ) = −∠〈OUT |Ψ( tf )〉, e jφ ∗(tf ) = 〈OUT |Ψ( tf )〉† |〈 OUT |Ψ( tf )〉| . (5) The preceding theorem states that controllers can as well be optimized (although in a somewhat computationally clumsy way) on the basis of the projective tracking error eproj (t) = |OUT 〉 − ejφ ∗(t) |Ψ( t)〉 (6) with the already perceived reward that the above connects with classical concepts. More formally speaking, since ‖ | Ψ〉 ‖ CN = 1 and since a phase factor exp( −jφ ) does not fundamentally change the quantum state, |Ψ〉 lives in S2N −1/S1 = CP N −1,the complex projective space. Observe that the fidelity |〈 OUT |Ψ( tf )〉| is the cosine of the Fubini-Study metric on CP N −1. More closely related to (6), observe the following: Corollary 1: δ(|OUT 〉 , |Ψ〉) := min φ ‖ | OUT 〉 − e−jφ |Ψ〉 ‖ CN is a metric on CP N −1. Remark 1: The global phase φ∗(t) could be viewed as an ad hoc trick to think maximum fidelity in terms of δ-minimum tracking error. However, for it to have its classical quantum mechanical interpretation, it needs to be constant, which could be accomplished by limiting it to φ∗(tf ).However, a time-varying global phase φ∗(t) could have the quantum mechanical interpretation of change of the zero energy level. Indeed, a shift of energy level HD → HD + cI yields a phase factor exp( −jct ). From (5), under near perfect 1340 Authorized licensed use limited to: University of Southern California. Downloaded on July 20,2020 at 04:17:51 UTC from IEEE Xplore. Restrictions apply. state transfer, it follows that this specific global phase factor could be associated with a shift c = 〈OUT |HD |IN 〉. C. Projective sensitivity Observing from (2) that |Ψ( t)〉 = e−jH D t |IN 〉 and defin-ing the output-input swapping operator P = |IN 〉 〈 OUT | , the projective tracking error leads to the concept of projective sensitivity function S(t), eproj (t) = ( I − ejφ (t)e−jH D tP )︸ ︷︷ ︸ S(t) |OUT 〉 . (7) The connection with the classical relationship ˆe(s) = ˆS(s)ˆ r(s) is obvious, but note the selectivity feature of the above that the disturbance |OUT 〉 is selectively restricted to be a natural basis of CN . In fact, the selectivity is 2-fold, as contrary to a classical controller, D is not universal, as it is selectively optimized for |OUT 〉. To connect the above with the fidelity, observe that 〈OUT |S (t)|OUT 〉 = 〈OUT |eproj (t)〉 = 1 − (〈OUT |e−jH D t |IN 〉) ejφ ∗(t) = 1 − F (t), (8) where the third equality is seen by observing that φ∗(t) is chosen so as to make (〈OUT |e−jH D t |IN 〉) ejφ ∗(t) real and positive. Here we are at the crucial point. Even though quantum transport is usually formulated in terms of fidelity, Eq. (8) reveals that we could equally argue in terms of the projective time-sensitivity function S(t).Fidelity is usually formulated as above in the time-domain; however, Laplace domain techniques have also been used but in the very specific context of steady-state behavior (s ≈ 0). Nevertheless, to better connect with the classical concepts, usually formulated in the Laplace domain, we define ˆS(s) via ˆeproj (s) = ( I/s −̂ ejφ (t)  (sI + jH D )−1P )︸ ︷︷ ︸ ˆS(s) |OUT 〉 , (9) where the widehat notation denotes the unilateral Laplace transform and  denotes the Laplace domain convolution ( ˆX  ˆY )( s) = 12πj ∫ c+j∞ c−j∞ ˆX(s − z) ˆY (z)dz, (10) where the integration path is a vertical line in the common z-domain of convergence of ˆX(s − z) and ˆY (z), assuming such a nonempty intersection exists. Relevant results are summarized in Appendix A. The problem is that ˆS(s) does not naturally lend itself to a representation of the form (I + L(s)) −1 with the idea that L(s) factors as P(s)K(s), where P(s) is some plant and K(s) some controller. At best, ˆS(s) can be related to the architecture shown in Fig. 2, which is certainly not of the single degree of freedom configuration, but could be interpreted as a 3-degree of freedom one, notwithstanding the feedbacks involved in the phase function. Following the classical path of ideas, we define a fictitious loop matrix L to reproduce the classical relation ˆS(s) = ( I + L)−1(s), that is, L = ˆS−1 − I; explicitly, L(s) = ( I/s −̂ ejφ (t)  (sI + jH D )−1P )−1 − I. D. Classical oscillatory systems Schr¨ odinger’s equation (2) is, after all, an Ordinary Differ-ential Equation (ODE) over Cn and should the eigenvalues of jH D come in complex conjugate pairs, it could be interpreted as a lossless spring mechanical system or a LC oscillatory circuit. Moreover, “energy landscape” techniques have been popular in robotics and electromechanical systems , , where the energy is shaped so as to put its minimum at the target by local feedbacks bearing similarity with uk = −jD kΨk. Such classical systems follow the architecture of Fig. 2—without the global phase factors in the shaded areas—with relevant tracking error defined as, reverting to classical notation, e(t) = (I − e−jH D tP )︸ ︷︷ ︸ S(t) |OUT 〉 , (11) or taking the unilateral Laplace transform ˆe(s) = (I/s − (sI + jH D )−1P )︸ ︷︷ ︸̂ S(s) |OUT 〉 . (12) The importance of this case-study is that comparison between the two sensitivity matrices ̂ S(s) and ˆS(s) would highlight the quantum mechanical effects. This is essentially what is addressed in Sec. V-B. III. S ENSITIVITY —L APLACE DOMAIN A. Selective sensitivity Taking the Laplace transform of (8) and using (9) yields 〈OUT | ˆS(s)|OUT 〉 = 1 /s − ˆF(s). Taking the log-differential, while remembering that nomi-nally Jmn = 1 , yields d〈OUT | ˆS(s)|OUT 〉 dJ mn 1 〈OUT | ˆS(s)|OUT 〉 (13) = − 〈 OUT ∣∣∣∣∣ d ˆF(s) dJ mn ∣∣∣∣∣ OUT 〉 11/s − ˆF(s) , (14) where the expressions for ˆS(s) and d ˆS(s) are summarized in Appendix A. Such quantities are numerically explored in Sec. V-B. 1341 Authorized licensed use limited to: University of Southern California. Downloaded on July 20,2020 at 04:17:51 UTC from IEEE Xplore. Restrictions apply. jD 1 sI jH  P  OUT w < u IN je I < proj e j e I < je I 666 1 Fig. 2: The projective error eproj embedded in a semi-classical 3-degree-of-freedom loop. The top paths ( P and 1) are indeed two additional degrees of freedom relative to the single degree of freedom configuration. The shaded areas refer to the “global phase factor.” Note that the e±jφ ∗ operations have to be interpreted in the time-domain. B. Motivation for Laplace techniques: Asymptotic results Here we provide motivation for the sensitivity analysis of ˆS(s). We proceed from the explicit expressions for ˆS(s) and ˆdS(s) of Appendix A and use a generalized Laplace final value theorem to derive some asymptotic behavior of S(t), dS(t) as t → ∞ . Moreover, in the quest for a quantum enhancement, we contrast those results with the limiting behavior of S(t), dS (t) when they do not include the global phase factor (no shaded boxes in Fig. 2). Since our systems are not closed-loop stable in the classical sense, we need a generalized Laplace final value theorem 2 : Theorem 2: Nonclassical Laplace final value theorem [3, Th. 2]. Let ˆf (s) be the Laplace transform of f (t). If lim s→0 ∫ ∞ s ( ˆf (ξ)/ξ )dξ = ∞, then lim t→∞ 1 t ∫ t 0 f (τ )dτ = lim s→0 s ˆf (s). In the following, we highlight the difference between the two cases: with and without global phase factor (with and without shaded boxes in Fig. 2) as a way to gauge quantum effects. Theorem 3: Regarding the average steady-state error in the sense of Th. 3, we have 1) With global phase factor: lim s↓0 〈OUT |s ˆS(s)|OUT 〉 = 1 − ∑ k |〈 OUT |Πk|IN 〉| 2. 2) Without global phase factor: lim s↓0 〈OUT |s ˆS(s)|OUT 〉 = 1 − 〈 OUT |IN 〉, = { 1 for transfer, 0 for localization. 2Ph. Anderson in his famous localization paper was aware of and utilized this result, but not with the level of rigor as in . Theorem 4: For the differential 〈OUT |sd ˆS(s)|OUT 〉, we have 1) With global phase factor: lim s↓0 〈OUT |sd ˆS(s)|OUT 〉 = { ∞ for transfer, 0 for localization. 2) Without global phase factor and with λk(HD ) = 0 : lim s↓0 〈OUT |sd ˆS(s)|OUT 〉 = 0 . C. Comparison with classical, nonselective sensitivity Because the relationship between L and ˆS is the same as that between ˆL(s) and S(s), the log-sensitivity of ˆS(s) with respect to coupling parameters in L is structurally the same as (3), d ln ˆS(s) d ln Jmn = − d ln L(s) d ln Jmn T (s), (15) where T (s) = L(s)( I + L(s)) −1. The above comes together with the obvious relationship ˆS(s) + T (s) = I. (16) The above might be called the s-domain quantum mechanical error versus sensitivity limitation. IV. S ENSITIVITY —T IME DOMAIN The starting point of the time-domain analysis is the sensitivity of the matrix exponential to variation in the matrix exponent, as given by the Zassenhaus formula : exp( −j(HD +dH D )t) = exp( −jH D t) exp( −jdH D t) × Π∞ p=2 exp( Zp(HD , dH D )( −jt )p), where Zp(·, ·) is a homogeneous Lie polynomial of degree p, and the decomposition is unique. Note that Zp(HD , dH D ) 1342 Authorized licensed use limited to: University of Southern California. Downloaded on July 20,2020 at 04:17:51 UTC from IEEE Xplore. Restrictions apply. contains a linear term in dH D , which should be taken into consideration when computing sensitivity. Explicitly, e−j(HD +dH D )t =e−jH D te−jdH D te 12 [HD ,dH D ]t2 × e jt 36 [[ HD ,dH D ],H D ] × e− t424 [[[ HD ,dH D ],H D ],H D ]... Setting dH D = dJ mn Smn , where dJ mn is the variation of the parameter Jmn and Smn the associated structure and utilizing the above formula with its expansion restricted to include polynomials up to Z2 yields de −jH D t dJ mn ≈ e−jH D t ( −jS mn t + 12 [HD , S mn ]t2 ) . (17) While approximate, this formula has the merit that it reveals the role of the commutator [HD , S mn ].From (8), the time-domain log-sensitivity is set up as d(1 − F ) dJ mn 11 − F = − 〈 OUT ∣∣∣ dS(t) dJ mn ∣∣∣ OUT 〉 〈OUT |S (t)|OUT 〉 , (18) where dS(t)/dJ mn is computed from Eqs. (7), (17), and ejφ ∗(t) is evaluated as given by Th. 1. Note that for numerical computations, Eq. (17) might not be accurate enough, in which case we have to revert to [11, Eq. 32]. The details are left out. V. N UMERICAL RESULTS A. Time-domain Fig. 3 shows a N = 10 , |OUT 〉 = |2〉, instantaneous readout (as opposed to windowed readout as in ) case study with J45 uncertainty, with an error 1 − F (as opposed to 1 − F 2 as in ). It confirms the anti-classical trend of concordance between error and log-sensitivity especially from controller 1 to 200. Such a trend was already observed in , but here it is in a context that relates better to the “tracking error,” modified with a phase factor to make it relevant to quantum systems. controller index 0500 1000 1500 -40 -35 -30 -25 -20 -15 -10 -5 0510 N=10, |OUT>=|2>, t, J 54 uncertainty, JT=31.5480 log(sensitivity) log(logarithmic sensitivity) log(error) Fig. 3: Case N = 10 , |OUT 〉 = |2〉, J45 uncertain, with S(t) defined by (7) 050 100 150 200 250 300 controller index -35 -30 -25 -20 -15 -10 -5 05 N=10, |OUT>=|2>, t, J 54 uncertainty, JT=0.4551, no factor log(sensitivity) log(logarithmic sensitivity) log(error) Fig. 4: Case N = 10 , |OUT 〉 = |2〉, J45 uncertain, with S(t) defined by (11) We now suppress the phase factor ejφ ∗ (remove the shaded boxes in Fig. 2) and obtain Fig. 4. Comparing Figs. 3 and 4, it is noted that, not surprisingly, the latter error has significantly increased, because of the removal of ejφ ∗ in (6). Surprisingly, the log-sensitivity has also increased in the 1:300 range of controllers. More impor-tantly, the latter log-sensitivity does not show an increasing trend with the error, confirmed by the Jonckheere-Terpstra (JT) test that accepts the H0 hypothesis of no trend. “No trend” in the log-sensitivity while the error increases is rather classical. B. s-domain The Laplace domain approach is useful to investigate asymptotic behavior, as made precise by Theorem 2. More-over, it especially makes sense in the localization case (|OUT 〉 = |IN 〉). By symmetry, we set |IN 〉 = |1〉. Nu-merical exploration reveals two cases: 1) The case where the spin to hold the excitation, |IN 〉 = |1〉, has an uncertain coupling strength with its neigh-bor; by symmetry, the uncertainty is on J12 . Represen-tatives of such case are Figs. 5-6. 2) The case where the uncertain strength Jmn is between spins not holding the excitation; by symmetry, m, n =1.Common to Cases 1) and 2) is an error/log-sensitivity trend reversal associated with the removal of the global phase factor, obvious from comparing Figs. 5 and 6. Specifically in Case 1), with phase factor, the log-sensitivity is nearly “flat” at 100%, but the error is very small; without the phase factor, the trend is completely reversed; the error is “flat” and the sensitivity is significantly reduced. However, a close inspection of Fig. 5 around controllers #1900-2000 reveals an abrupt increase of the sensitivity together with a decrease of the log-sensitivity, a manifestation of the projective limitation (16). In Case 2), the trend reversal is the same, but not as “brutal” as in Case 1). Nevertheless, with removal of the phase factor the error increases while the log-sensitivity decreases. 1343 Authorized licensed use limited to: University of Southern California. Downloaded on July 20,2020 at 04:17:51 UTC from IEEE Xplore. Restrictions apply. 0200 400 600 800 1000 1200 1400 1600 1800 2000 controller index -12 -10 -8 -6 -4 -2 02 N=11, localization, dt, J 12 uncertainty, JT=45.7682, s=0.1 log(||) log(|/|) log(||) Fig. 5: Case N = 11 , |OUT 〉 = |1〉, J12 uncertainty, ˆS(s) defined as in Eq. (9), with phase factor 0200 400 600 800 1000 1200 1400 1600 1800 2000 controller index -35 -30 -25 -20 -15 -10 -5 05 N=11, localization, dt, J 12 uncertainty, JT=24.6124, s=0.1, no  log(||) log(|/|) log(||) Fig. 6: Case N = 11 , |OUT 〉 = |1〉, J12 uncertainty, ˆS(s) defined as in Eq. (12), without phase factor VI. C ONCLUSION In this paper, we have studied robustness of energy land-scape control for excitation transport in ring shaped quantum routers. The fundamental stumbling block in comparing classical versus quantum robustness is that energy landscape control does not fit in the paradigm of Fig. 1, which has been the basic architecture upon which classical error versus log-sensitivity limitations, e.g., ˆS(s) + T(s) = I, were built. The closest-to-classical feedback structure to model landscape control is the one of Fig. 2, where a projective tracking error has been substituted for the classical tracking error to accommodate the quantum mechanical global phase factor shown in the shaded boxes. In this architecture, a quantum limitation ˆS(s) + T (s) = I holds, and manifested itself at the extreme right of Fig. 5. The real question that is answered here in the affirmative here is whether the anti-classical behavior observed in the time-domain is quantum mechanical. The only way to answer such a question is to remove the quantum mechanical global phase factor from Fig. 2, which results in a complete reversal of the trends. As our major result, this demonstrates the quantum mechanical origin of the anti-classical behavior. Other ways to gauge the quantum effect, like those sug-gested in Sec. III-C, will be explored in a further paper. A PPENDIX A. Explicit expressions for projective ̂ S and d̂ S The details of the convolutions (9)-(10) are left out, but available in . The starting point is the eigendecomposition HD = ∑ k Πkωk, where Πk is the projection on the eigenspace of the eigenvalue ωk. The relevant results, under the hypothesis that |〈 OUT |Ψ( t)〉| ≈ 1, are the following: ̂ S = ( I/s − ∑ k (e† i Πkeo) ΠPs + j(ω − ωk) ) (19) and upon differentiating ̂ S as defined by (9), d̂ S = − j ∑ km (e† i ΠkdH D Πeo)Π mP (s + j(ωm − ωk))( s + j(ωm − ω)) + j ∑ km ΠkdH D ΠP (e† i Πmeo)(s + j(ωk − ωm))( s + j(ω − ωm)) . (20) R EFERENCES P. W. Anderson. Absence of diffusion in certain random lattices. Physical Review , 109(5):1492–1505, March 1958. F. Casas, A. Murua, and Mladen Nadinic. Efficient computation of the Zassenhaus formula. Computer Physics Communications ,183(11):2386–2391, November 2012. Available at arXiv:1204.0389v2 [math-ph] 15 Jun 2012. E. Gluskin and S. Miller. On the recovery of the time average of contionuous and discrete time functions from their Laplace and z-transform. International Journal of Circuit Theory and Applica-tions , 41(9):988–997, 2013. doi: 10.1002/cta.1877, arXiv:1109.3356v4 [math-ph]. A. Jonckheere. A distribution-free k-sample test against ordered alternatives. Biometrika , 41:133145, 1954. doi:10.2307/233301. E. Jonckheere, S. Schirmer, and F. Langbein. Structured singular value analysis for spintronics network information transfer control. IEEE Transactions on Automatic Control , 62(12):6568–6574, December 2017. Available at arXiv:1706.03247v1 [quant-ph] 10 Jun 2017. E. Jonckheere, S. Schirmer, and F. Langbein. Jonckheere-Terpstra test for nonclassical error versus log-sensitivity relationship of quan-tum spin network controllers. International Journal of Robust and Nonlinear Control , 28(6):2383–2403, April 2018. Available at arXiv:1612.02784 [math.OC]. E. Jonckheere, S. Schirmer, and F. Langbein. Effect of quantum mechanical global phase factor on error versus sensitivity limitation in quantum routing. Full paper, available at jonckhee, 2019. E. A. Jonckheere. Lagrangian theory of large scale systems. In European Conference on Circuit Theory and Design , pages 626–629, The Hague, the Netherlands, August 25-28 1981. invited paper. Frank C. Langbein, Sophie G. Schirmer, and Edmond Jonckheere. Static bias controllers for XX spin-1/2 rings. Data set, figshare ,DOI:10.6084/m9.figshare.3485240.v1, July 3 2016. R. Ortega, A. Loria, Per J. Nicklasson, and H. Sira-Ramirez. Passivity-based control of Euler-Lagrange systems: Mechanical, electrical and electromechanical applications . Springer, London, Berlin, Heidelberg, ..., 1998. S. Schirmer, E. Jonckheere, and F. Langbein. Design of feed-back control laws for spintronics networks. IEEE Transactions on Automatic Control , 63(8):2523–2536, August 2018. Available at arXiv:1607.05294. 1344
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Symmetry-Breaking Constraints for Grid-Based Multi-Agent Path Finding ∗ Jiaoyang Li1, Daniel Harabor2, Peter J. Stuckey2, Hang Ma1, Sven Koenig1 1University of Southern California 2Monash University jiaoyanl@usc.edu, {daniel.harabor,peter.stuckey}@monash.edu, {hangma,skoenig}@usc.edu Abstract We describe a new way of reasoning about symmetric colli-sions for Multi-Agent Path Finding (MAPF) on 4-neighbor grids. We also introduce a symmetry-breaking constraint to resolve these conflicts. This specialized technique allows us to identify and eliminate, in a single step, all permutations of two currently assigned but incompatible paths. Each such per-mutation has exactly the same cost as a current path, and each one results in a new collision between the same two agents. We show that the addition of symmetry-breaking techniques can lead to an exponential reduction in the size of the search space of CBS, a popular framework for MAPF, and report significant improvements in both runtime and success rate versus CBSH and EPEA – two recent and state-of-the-art MAPF algorithms. 1 Introduction Multi-Agent Path Finding (MAPF) is the planning problem of finding a set of paths for a team of agents. Each agent is required to move from an initial start location to a specified goal location, while avoiding conflicts with other agents. A conflict (i.e., collision) happens when two agents stay at the same vertex or traverse the same edge at the same time. Such problems appear in a range of application areas, includ-ing warehouse logistics (Wurman, D’Andrea, and Mountz 2008), office robots (Veloso et al. 2015), aircraft-towing ve-hicles (Morris et al. 2016) and computer games (Silver 2005; Ma et al. 2017). MAPF is known to be NP-hard on general graphs (Yu and LaValle 2013b; Ma et al. 2016b), planar graphs (Yu 2016) and grids (Banfi, Basilico, and Amigoni 2017). De-spite these intractability results and due to the substantial in-terest in applications, numerous optimal MAPF algorithms have been proposed in recent years. Approaches include re-ducing MAPF to instances of other well known problems ∗The research at the University of Southern California was sup-ported by the National Science Foundation (NSF) under grant num-bers 1409987, 1724392, 1817189 and 1837779 as well as a gift from Amazon. The views and conclusions contained in this docu-ment are those of the authors and should not be interpreted as rep-resenting the official policies, either expressed or implied, of the sponsoring organizations, agencies or the U.S. government. Copyright c ⃝2019, Association for the Advancement of Artificial Intelligence (www.aaai.org). All rights reserved. (a) (b) Figure 1: Two situations involving symmetric conflicts be-tween two agents. (a) highlights the problem in general: ev-ery shortest path for one agent conflicts with every shortest path for the other agent somewhere in the yellow rectangular area. (b) shows a cardinal conflict, a related class of conflicts which requires that all shortest paths for each agent must pass through a common location at the same timestep: here location (3, 3) at timestep 3. (e.g., multi-commodity flow (Yu and LaValle 2013a), sat-isfiability (Surynek et al. 2016) and Answer Set Program-ming (Erdem et al. 2013)); solving MAPF with a single inte-grated A-search (Standley 2010; Wagner and Choset 2011; Goldenberg et al. 2014); and solving MAPF with a two-level search (Sharon et al. 2013; 2015; Boyarski et al. 2015; Felner et al. 2018), which constructs a plan by keeping track of constraints between agents at a high level and computing paths consistent with those constraints at a low level, one agent at a time. More detailed surveys are given in (Ma et al. 2016a; Felner et al. 2017). In this paper, we introduce a new way of reasoning about symmetric conflicts between two agents for MAPF on 4-neighbor grids (which are arguably the most common way of representing the environment for MAPF). Our approach exploits grid symmetries: equivalences between sets of paths or path segments which have the same start and goal loca-tions, the same cost, and which differ only in the order in which grid actions (up, down, left, right, or wait) appear on them. Figure 1 shows two examples. All shortest paths for the two agents conflict somewhere inside the yellow rect-angular area. The optimal strategy here is for one agent to wait for the other. We refer to such cases as cardinal rect-angle conflicts. In this paper, we propose several efficient algorithms to detect cardinal rectangle conflicts as well as two other types of conflicts, semi-cardinal rectangle con-flicts and non-cardinal rectangle conflicts. We also introduce barrier constraints that are able to resolve these rectangle conflicts in a single step and demonstrate, in principle and in practice, that the addition of barrier constraints can achieve an exponential reduction in the number of nodes expanded by Conflict-Based Search (CBS), a popular state-of-the-art MAPF framework. 2 Preliminaries A MAPF problem is defined by a graph G = (V, E) and a set of m agents {a1, . . . , am}. Each agent ai has a start ver-tex si ∈V and a goal vertex gi ∈V . Time is discretized into timesteps. At each timestep, every agent can either move to an adjacent vertex or wait at its current vertex. Both move and wait actions have unit cost unless the agent terminally waits at its goal vertex, which has zero cost. We call the tuple ⟨ai, aj, v, t⟩a vertex conflict iff agents ai and aj oc-cupy the same vertex v ∈V at the same timestep t, and ⟨ai, aj, u, v, t⟩an edge conflict iff agents ai and aj traverse the same edge (u, v) ∈E in opposite directions at the same timestep t. Our task is to find a set of conflict-free paths which move all agents from their start vertices to their goal vertices while minimizing the sum of their individual path costs (SIC). In this paper, graph G is always a 4-neighbor grid whose vertices are unblocked cells and whose edges connect vertices corresponding to adjacent unblocked cells in the four main compass directions. 3 Conflict-Based Search Conflict-Based Search (CBS) (Sharon et al. 2015) is a two-level search algorithm for MAPF. At the low level, CBS in-vokes a space-time A search to find a shortest path for each agent that satisfies some spatio-temporal constraints added by the high level. It break ties by preferring the path that has the fewest conflicts with the paths of other agents. At the high level, CBS performs a best-first search on a bi-nary constraint tree (CT). Each CT node contains a set of current paths, one for each agent, and also a set of spatio-temporal constraints that are used to coordinate agents and avoid conflicts. The cost of a CT node is the SIC of its cur-rent paths. CBS proceeds from one CT node to the next, checking for conflicts and calling its low-level search to re-plan paths one at a time. CBS succeeds when the current CT node is conflict-free, which corresponds to an optimal solu-tion. Constraints: A constraint is a spatio-temporal restriction introduced by CBS to resolve situations where the paths of two agents are in conflict. Specifically, a vertex constraint ⟨ai, v, t⟩means that agent ai is prohibited from occupy-ing vertex v at timestep t. Similarly, an edge constraint ⟨ai, u, v, t⟩means that agent ai is prohibited from travers-ing edge (u, v) at timestep t. Splits: When CBS expands a CT node N, it checks for pairwise conflicts among the current paths. If there are none, then N is a goal CT node and CBS terminates. Otherwise, CBS chooses one of the conflicts (by default, randomly) and resolves it by splitting N into two child CT nodes. In each child CT node, one agent from the conflict is forbidden to use the contested vertex or edge by way of an additional constraint. The path of this agent becomes invalidated and must be replanned by a low-level search. All other paths re-main unchanged. With two child CT nodes per conflict, CBS guarantees optimality, exploring both ways of resolving each conflict. Cardinal, semi-cardinal and non-cardinal conflicts: In (Boyarski et al. 2015), conflicts are categorized into three different types, and it is shown that prioritizing among them improves performance. The highest priority is given to car-dinal conflicts, which Boyarski et al. (2015) define as fol-lows: [A conflict] C = ⟨ai, aj, v, t⟩is cardinal if all the con-sistent optimal paths for both [agents] ai and aj include vertex v at timestep t. An example of such a conflict is shown in Figure 1(b). Every possible way of resolving the cardinal conflict ⟨a1, a2, (3, 3), 3⟩requires one of the agents to wait for the other or take a detour. That means, when CBS splits on a car-dinal conflict, it produces two child CT nodes whose costs are both strictly higher than the current CT node. In this work, we show that there exist other types of conflicts which have the same result when splitting on them but which can-not be detected using the present definition, such as shown in the example in Figure 1(a). We therefore introduce a revised and more general definition: Definition 1. A conflict C is cardinal iff replanning for any agent involved in the conflict increases the SIC. Once all cardinal conflicts are processed, the next highest priority is given to semi-cardinal conflicts, which Boyarski et al. (2015) define as: [A conflict] C = ⟨ai, aj, v, t⟩is semi-cardinal if all the consistent optimal paths of one agent include vertex v at timestep t, but the other agent has such a path that does not include v at timestep t. Similarly, we give a revised and more general definition: Definition 2. A conflict C is semi-cardinal iff replanning for one agent involved in the conflict always increases the SIC while replanning for the other agent does not. Any conflict which is not cardinal or semi-cardinal is said to be non-cardinal. These can be processed in any order after the other conflicts, though a popular strategy involves choosing the earliest non-cardinal conflict first. Admissible heuristics: The high-level of CBS consists of a best-first search that prioritizes for expansion CT nodes having the smallest SIC. Felner et al. (2018) show that the efficiency of the high-level search can be improved through the addition of admissible heuristics. The suggested algo-rithm, CBSH, proceeds by building a conflict graph, whose vertices represent agents and edges represent cardinal con-flicts of the current paths. It can be shown that the value of the minimum vertex cover of the conflict graph is an ad-missible and consistent lower bound on the cost-to-go. The addition of heuristics to the high-level search often produces smaller CTs and decreases the runtime of CBS by a large factor. Figure 2: The CT of CBS and CBSH (without the 2 blue CT nodes for CBSH) when resolving a 1×3 cardinal rectangle conflict. Table 1: Number of CT nodes expanded by CBSH on MAPF instances where 2 agents are involved in one cardinal rect-angle conflict. The first column and first row are the width and length of the rectangular area. 1 2 3 4 5 6 7 8 9 1 1 1 2 3 4 5 6 7 8 2 3 7 14 26 46 79 133 221 3 22 53 116 239 472 904 1,692 4 142 392 1,016 2,651 6,828 17,747 5 1,015 2,971 8,525 23,733 65,236 6 7,447 24,275 78,002 254,173 7 62,429 222,524 795,197 8 573,004 >1,518,151 4 Inefficiency of CBS and CBSH when Resolving Cardinal Rectangle Conflicts In this section, we demonstrate how CBS and CBSH resolve cardinal rectangle conflicts, and illustrate the large number of CT nodes resulting from it. Figure 2 illustrates the issue. All shortest paths of agents a1 and a2 cross the 1×3 yellow rectangular area. a1 has 1 shortest path with cost 4 while a2 has 6 shortest paths with cost 4. Thus, the cost of the root CT node of CBS is 8. How-ever, each of the 6 combinations of these paths has a vertex conflict in one of the yellow cells. Consequently, this is a cardinal rectangle conflict, and the optimal solution cost is 9. The CT of CBS consists of 3 non-goal CT nodes with cost 8 and 4 goal CT nodes with cost 9. The CT of CBSH only saves the last 2 goal CT nodes (in blue). When the rectangular area is larger, CBS performs worse. Sharon et al. (2015) show that, to resolve the 2×2 cardinal rectangle conflict in Figure 1(a), CBS generates 5 non-goal CT nodes and 6 goal CT nodes (and, CBSH generates 5 CT non-goal nodes and 2 CT goal nodes). To illustrate this is-sue further, we ran CBSH on MAPF instances where two agents are involved in a cardinal rectangle conflict of differ-ent sizes. Surprisingly, the number of expanded CT nodes, as shown in Table 1, is exponential in the length and width of the rectangular area. For a small 8×9 rectangular area, CBSH expands already more than 1 million CT nodes and fails to solve the MAPF instance within 5 minutes. 5 Cardinal Rectangle Reasoning for Entire Paths In this section, we present a simple algorithm for identify-ing cardinal rectangle conflicts and introduce a new type of constraints, called barrier constraints, to resolve such con-flicts efficiently. We refer to a node S as a three-element tu-ple (S.x, S.y, S.t) corresponding to an agent staying in lo-cation (S.x, S.y) at timestep S.t. We refer to a valid path (or path for short) of an agent as a path (i.e., sequences of nodes whose locations can repeat and whose timesteps are 0, 1, 2, . . . ) from its start location to its goal location that satisfies its constraints in the CT node but ignores paths of other agents and an optimal path of an agent as its shortest valid path. 5.1 Identify Cardinal Rectangle Conflicts Assume that two agents ai and aj have a vertex conflict ⟨ai, aj, v, t⟩. Let nodes Si, Sj, Gi and Gj be the corre-sponding start and goal nodes (Figure 3(a)). We define the rectangular area (or rectangle for short) as the intersec-tion of the Si-Gi rectangle and the Sj-Gj rectangle, where Sk-Gk rectangle (k = i, j) represents the rectangle whose diagonal corners are in location (Sk.x, Sk.y) and location (Gk.x, Gk.y), respectively. The first two requirements for a cardinal rectangle conflict are intuitive: (1) both agents fol-low their Manhattan-optimal paths, i.e., the cost of each path equals the Manhattan distance from its start node to its goal node, and (2) the distances from each location inside the rectangle to the locations of the two start nodes are equal, which can be simplified to the requirement that both agents move in the same direction in both dimensions (because we already know that the distances from location v to the loca-tion of node Si and the location of node Sj are equal): |Si.x −Gi.x| + |Si.y −Gi.y| = Gi.t −Si.t > 0 (1) |Sj.x −Gj.x| + |Sj.y −Gj.y| = Gj.t −Sj.t > 0 (2) (Si.x −Gi.x)(Sj.x −Gj.x) ≥0 (3) (Si.y −Gi.y)(Sj.y −Gj.y) ≥0. (4) However, these two requirements do not guarantee that all combinations of optimal paths conflict. Figures 3(b) and 3(c) are two counterexamples where at least one agent has a by-pass through which the agent can reach its goal node without entering the rectangle, and thus does not conflict with the other agent. The difference between these two conflicts and the cardinal rectangle conflict in Figure 3(a) is that their goal nodes are located differently compared to their start nodes. Therefore, the third requirement is that the start and goal nodes have opposite relative locations in both dimensions: (Si.x −Sj.x)(Gi.x −Gj.x) ≤0 (5) (Si.y −Sj.y)(Gi.y −Gj.y) ≤0. (6) To sum up, if agents ai and aj have a vertex conflict and their corresponding start and goal nodes satisfy Equations (1) to (6), then agents ai and aj are involved in a cardinal rect-angle conflict. 5.2 Calculate Corner Nodes of the Rectangle We refer to the four corner nodes of the rectangle as Rs, Rg, Ri and Rj, where Rs and Rg are the corner nodes closest to the start and goal nodes, respectively, and Ri and Rj are (a) Cardinal conflict (b) Semi-cardinal conflict (c) Non-cardinal conflict (d) Cardinal conflict (e) Semi-cardinal conflict (f) No rectangle conflict Figure 3: Some examples of rectangle conflicts. The locations of the start and goal nodes are shown in the figures. Gk.t = Sk.t + |Gk.x −Sk.x| + |Gk.y −Sk.y|, k = i, j. In (a), (b) and (c), Si.t = Sj.t; in (d) and (e), Si.t = Sj.t −1; and, in (e), Si.t = Sj.t −2. the other corner nodes on the opposite borders of Si and Sj, respectively (Figure 3(a)). The timestep of each node is defined as the timestep when an optimal path of agent ai or aj reaches the location of the node. We analyze all combinations of relative locations of start and goal nodes and come up with the following way to calculate them: For the locations of Rs and Rg: Rs.x = ( Si.x, Si.x = Gi.x max{Si.x, Sj.x}, Si.x < Gi.x min{Si.x, Sj.x}, Si.x > Gi.x (7) Rg.x = ( Gi.x, Si.x = Gi.x min{Gi.x, Gj.x}, Si.x < Gi.x max{Gi.x, Gj.x}, Si.x > Gi.x. (8) We can calculate Rs.y and Rg.y by replacing all x by y in Equations (7) and (8). Next, for the locations of Ri and Rj, if (Si.x−Sj.x)(Sj.x−Rg.x) ≥0, then Ri.x = Rg.x, Ri.y = Si.y, Rj.x = Sj.x and Rj.y = Rg.y; else, Ri.x = Si.x, Ri.y = Rg.y, Rj.x = Rg.x and Rj.y = Sj.y. Finally, for the timesteps of all corner nodes Rk (k = i, j, s, g), Rk.t = Si.t + |Si.x −Rk.x| + |Si.y −Rk.y|. 5.3 Add Barrier Constraints Since all combinations of the optimal paths of the agents conflict, we resolve the cardinal rectangle conflict by giv-ing one agent priority within the rectangle and forcing the other agent to leave it later or take a detour. To integrate this idea into CBS, we introduce the barrier constraint, B(ak, Rk, Rg) (k = i, j), which is a set of vertex con-straints that prohibits agent ak from occupying all loca-tions along the border of the rectangle that is opposite of its start node (i.e., from Rk to Rg) at the timestep when ak would optimally reach the location. For example, in Figure 3(a), two barrier constraints are B(ai, Ri, Rg) = {⟨ai, (2 + n, 4), 3 + n⟩|n = 0, 1} and B(aj, Rj, Rg) = {⟨aj, (3, 2 + n), 2 + n⟩|n = 0, 1, 2}. B(ak, Rk, Rg) blocks all possible paths for ak that reach its goal node Gk via the rectangle, and thus forces ak to wait or take a detour. When resolving a cardinal rectangle conflict, we generate two child CT nodes and add B(ai, Ri, Rg) to one of them and B(aj, Rj, Rg) to the other one. We now present two ob-vious properties of barrier constraints. Property 1. For all combinations of paths of agents ai and aj with a cardinal rectangle conflict, if one path violates B(ai, Ri, Rg) and the other path violates B(aj, Rj, Rg), then the two paths have one or more vertex conflicts within the rectangle. Proof. We assume that the vertex conflict between agents ai and aj that underlies the cardinal rectangle conflict is ⟨ai, aj, (C.x, C.y), C.t⟩. We then assume Si.x ≤C.x and Si.y ≤C.y without loss of generality (because the prob-lem is invariant under rotations of axes). According to Equa-tions (1) to (4), max{Si.x, Sj.x} ≤C.x ≤min{Gi.x, Gj.x} (9) max{Si.y, Sj.y} ≤C.y ≤min{Gi.y, Gj.y} (10) (C.x −Si.x) + (C.y −Si.y) = (C.x −Sj.x) + (C.y −Sj.y). (11) From Equation (11), we know Si.x + Si.y = Sj.x + Sj.y. (12) We can assume that Si.x ≥Sj.x without loss of generality (because the problem is invariant under swaps of the indexes of agents), which implies Si.y ≤Sj.y. From Equations (9) and (10) and the method for calculating rectangle corner nodes in Section 5.2, we have Rg.x = min{Gi.x, Gj.x} ≥ Si.x, Rg.y = min{Gi.y, Gj.y} ≥Sj.y, Ri.x = Si.x, Ri.y = Rg.y, Rj.x = Rg.x and Rj.y = Sj.y. Thus, Sj.x ≤Si.x = Ri.x ≤Rg.x = Rj.x (13) Si.y ≤Sj.y = Rj.y ≤Rg.y = Ri.y. (14) Consequently, the relative locations of the start, goal and rectangle corner nodes are exactly the same as given in Figure 3(a). For every node Ni on the border Ri-Rg (i.e., Ri.x ≤Ni.x ≤Rg.x, Ni.y = Rg.y, Ni.t = Ri.t + Ni.x − Ri.x) and every node Nj on the border Rj-Rg (i.e., Nj.x = Rg.x, Rj.y ≤Nj.y ≤Rg.y, Nj.t = Rj.t + Nj.y −Rj.y), we need to prove that the path from Si to Ni and the path from Sj to Nj have at least one node in common within the rectangle. Since Sj.x ≤Si.x ≤Ni.x ≤Nj.x and Si.y ≤Sj.y ≤Nj.y ≤Ni.y, the Si-Ni rectangle and the Sj-Nj rectangle consist of a cross shape, which implies that the path from Si to Ni and the path from Sj to Nj have at least one location in common within the intersection of the Si-Ni rectangle and the Sj-Nj rectangle, i.e., this location is within the rectangle. By the definition of rectangle conflicts, the two paths traverse this location at the same timestep, i.e., they have at least one node in common within the rectan-gle. Property 2. If agents ai and aj have a cardinal rectangle conflict, then the cost of any path of agent ak (k = i, j) that satisfies B(ak, Rk, Rg) is larger than the cost of an optimal path of agent ak. Proof. We use the same assumptions as in the proof for Property 1. Then, Equations (13) and (14) also hold here. According to Equation (5), Si.x = Ri.x ≤Rg.x = Gi.x. Any optimal path that connects locations (Si.x, Si.y) and (Gi.x, Gi.y) has at least one of the nodes {(Ri.x + n, Ri.y, Ri.t+n)|n = 0, . . . , Rg.x−Ri.x}. But all of these nodes are constrained by B(ai, Ri, Rg). Therefore, the cost of any path of agent ai that satisfies B(ai, Ri, Rg) is larger than the cost of an optimal path of agent ai. The proof for k = j can be derived analogously using Equation (6) instead of Equation (5). Property 1 is important because CBS requires the con-straints added to child CT nodes to not block any conflict-free paths, which is why we add constraints that force an agent to leave the rectangle later rather than enter the it later. 5.4 CBSH-CR We now present our first algorithm, CBSH with cardinal rectangle reasoning (CBSH-CR). It is identical to CBSH ex-cept for the following four modifications. Perform splits: When the chosen conflict is a cardi-nal rectangle conflict, CBSH-CR adds B(ai, Ri, Rg) to one child CT node and B(aj, Rj, Rg) to the other child CT node. Then, in both child CT nodes, the rectangle conflict is re-solved by one of the agents increasing its cost. Classify conflicts: CBSH-CR first classifies vertex/edge conflicts into cardinal, semi-cardinal and non-cardinal con-flicts. It then finds cardinal rectangle conflicts among all semi- and non-cardinal vertex conflicts. Prioritize conflicts: It follows from Definition 1 and Properties 1 and 2 that both cardinal rectangle conflicts and cardinal vertex/edge conflicts are cardinal conflicts. There-fore, CBSH-CR chooses cardinal conflicts first, then semi-cardinal conflicts and last non-cardinal conflicts. It breaks ties by preferring the earliest conflict, where we define Rs.t as the timestep of a cardinal rectangle conflict. Calculate heuristics: It uses all cardinal conflicts (includ-ing cardinal rectangle conflicts) to compute the heuristics for the high-level search. Now we show that CBSH-CR is complete and optimal. Lemma 1. For every cost c, there is a finite number of CT nodes with cost c. Proof. The number of conflicts within c timesteps is finite, and, once a conflict is chosen at a CT node N, it never ap-pears again in the subtree of N. Therefore, the number of CT nodes is also finite. Theorem 2. CBSH-CR is complete and optimal. Proof. The proof is similar to the proof for the optimality and completeness of CBS (Sharon et al. 2015). The low-level search always returns an optimal path, the high-level search always chooses a CT node with minimum f-value to expand, and the expansion does not lose any conflict-free paths (Property 1). Therefore, the first chosen CT node whose paths are conflict-free has a set of conflict-free paths with minimum SIC (i.e., CBSH-CR is optimal). Besides, the f-value of CT nodes are non-decreasing in expansion order. It follows from Lemma 1 that, if there exist solutions, a so-lution must be found after expanding a finite number of CT nodes whose costs are no more than the optimal cost (i.e., CBSH-CR is complete). 6 Rectangle Reasoning for Entire Paths Reasoning about cardinal rectangle conflicts does not elimi-nate all symmetric conflicts on grids for CBS. For instance, the conflict in Figure 3(b) is not a cardinal rectangle conflict because agent aj has an optimal bypass outside of the rect-angle. However, if location (2, 5) at timestep 4 and location (3, 5) at timestep 5 are occupied by other agents, whenever the low-level search of CBS replans agent aj’s path, it al-ways returns a path that conflicts with agent ai’s path, be-cause the low-level search uses the number of conflicts with other agents as the tie-breaking rule. Therefore, CBS again generates many CT nodes before finally finding conflict-free paths. We refer to such cases as semi-cardinal rectangle con-flicts. Similarly, we refer to cases with symmetric conflicts where both agents have bypasses as non-cardinal rectangle conflicts, like the case in Figure 3(c). Together with cardinal rectangle conflicts, we refer to these three types of conflicts as rectangle conflicts. We now show how to identify and classify rectangle con-flicts. If agents ai and aj have a vertex conflict, then they are involved in a rectangle conflict iff their start and goal nodes satisfy Equations (1) to (4). Moreover, if they also satisfy Equations (5) and (6), it is cardinal; if they also satisfy only one of these equations, it is semi-cardinal; and if they sat-isfy neither equation, it is non-cardinal. Property 1 holds for all types of rectangle conflicts. It follows from the proof for Property 2 that, when resolving a semi-cardinal rectangle conflict by barrier constraints, at least one of the child CT nodes has to increase its SIC. But for a non-cardinal rectan-gle conflict, both child CT nodes may not change their SICs. 6.1 CBSH-R We now introduce the second algorithm, CBSH with rectan-gle reasoning (CBSH-R). It is identical to CBSH-CR except for the following three modifications. Perform splits: CBSH-R uses barrier constraints to re-solve all rectangle conflicts (not only cardinal ones). Classify conflicts: After classifying all vertex/edge con-flicts, CBSH-R checks all semi- and non-cardinal vertex conflicts to identify and classify rectangle conflicts. If a semi-/non-cardinal rectangle conflict has been resolved in one of the ancestors of the current CT node, it ignores this rectangle conflict, otherwise it could always choose to re-solve the same rectangle conflict and thus be in a cycle for-ever. A semi-/non-cardinal rectangle conflict can be found multiple times in a CT branch because its barrier constraint does not disallow all optimal paths that traverse locations in-side the rectangle. For example, in Figure 3(c), both agents ai and aj have optimal paths that contains node Rs but do not contain nodes that are constrained by B(ai, Ri, Rg) and B(aj, Rj, Rg), respectively. So, after adding barrier con-straints, a vertex conflict could still happen between two op-timal paths within the rectangle and then it is identified as a semi-/non-cardinal rectangle conflict again. (a) (b) Figure 4: Rectangle conflicts between path segments. In Fig-ure (b), a2 follows the red solid arrow but waits at (1, 4) or (2, 4) for one timestep because of constraints. Prioritize conflicts: CBSH-R uses the same conflict pri-oritization as CBSH-CR, except that it adds a tie-breaking rule for semi-/non-cardinal rectangle conflicts. Since our reasoning method ignores obstacles and constraints inside the rectangle, it is possible that both child CT nodes increase their costs when CBSH-R resolves a semi-/non-cardinal rectangle conflict using barrier constraints. Therefore, for all semi-cardinal conflicts, it prefers semi-cardinal rectangle conflicts to semi-cardinal vertex/edge conflicts. Similarly, for all non-cardinal conflicts, it prefers non-cardinal rectan-gle conflicts to non-cardinal vertex/edge conflicts. The sec-ondary tie-breaking rule is still to prefer the earliest conflict, where we define Rs.t as the timestep of a rectangle conflict. Theorem 3. CBSH-R is complete and optimal. Proof. Since all chosen rectangle conflicts are different in any CT branch, Lemma 1 still holds. Property 1 also holds for semi-/non-cardinal rectangle conflicts. Therefore, we can directly use the proof for Theorem 2 without changes. 7 Rectangle Reasoning for Path Segments Our rectangle reasoning methods so far ignore obstacles and constraints, so they can reason only about the rectangle con-flicts for entire paths. In some cases, however, rectangle con-flicts exist for path segments but not entire paths, such as the cardinal rectangle conflict in Figure 4(a). Since the paths are not Manhattan-optimal, our rectangle reasoning methods so far fail to identify the rectangle conflict. Therefore, in this section, we discuss a rectangle reasoning method for path segments using MDDs. 7.1 Identify Rectangle Conflicts using MDDs A Multi-Valued Decision Diagram (MDD) (Sharon et al. 2013) MDDi for agent ai is a directed acyclic graph that consists of all optimal paths of agent ai. The nodes at depth t in MDDi correspond to all possible locations at timestep t in these paths. If MDDi has only one node (x, y, t) at depth t, we call this node a singleton, and all optimal paths of agent ai traverse location (x, y) at timestep t. CBSH uses singletons in MDDs to classify cardinal, semi-cardinal and non-cardinal vertex/edge conflicts. MDDs offer information about the impact of obstacles in the grid and constraints imposed on an agent, and thus help us to reason about its path segments. We extend rectan-gle reasoning to reasoning about rectangle conflicts between two path segments, each of which starts at a singleton (called Algorithm 1: Identify rectangle conflicts for path segments. Input: A semi/non-cardinal vertex conflict ⟨ai, aj, v, t⟩. // Collect start and goal node candidates. 1 N S i ←singletons in MDDi no later than timestep t; 2 N G i ←singletons in MDDi no earlier than timestep t; 3 N S j ←singletons in MDDj no later than timestep t; 4 N G j ←singletons in MDDj no earlier than timestep t; 5 type′ ←Not-Rectangle; area′ ←0; // Try all combinations. 6 foreach Si ∈N S i , Sj ∈N S j , Gi ∈N G i , Gj ∈N G j do 7 if isRectangle(Si, Sj, Gi, Gj) then 8 {Ri, Rj, Rs, Rg} ←getVertices(Si, Sj, Gi, Gj); 9 type ←classifyRect(Ri, Rj, Rg, Si, Sj, Gi, Gj); 10 area ←|Ri.x −Rj.x| × |Ri.y −Rj.y|; 11 if type′ = Not-Rectangle or type is better than type′ or (type = type′ and area > area′) then 12 type′ ←type; area′ ←area; 13 {R′ i, R′ j, R′ s, R′ g} ←{Ri, Rj, Rs, Rg}; 14 if type′ ̸= Not-Rectangle and no ancestor CT node has chosen this rectangle conflict {R′ i, R′ j, R′ s, R′ g} before then 15 return type′ and {R′ i, R′ j, R′ s, R′ g}; 16 return Not-Rectangle; its start node) and ends at another singleton (called its goal node). If we find a rectangle conflict for a combination of start and goal nodes, we can impose barrier constraints. Algorithm 1 shows the pseudo-code. It first treats all sin-gletons as start and goal node candidates (Lines 1-4) and then tries all combinations to find rectangle conflicts. If mul-tiple rectangle conflicts are identified, it prefers one of the highest priority type and breaks ties by preferring a con-flict with the largest rectangle area (Line 11). Line 14 pro-hibits choosing the same rectangle conflict more than once in any CT branch. We discuss details of the three functions on Lines 7, 8 and 9 in Section 7.3. 7.2 Add Modified Barrier Constraints When reasoning about entire paths, all paths of agent ai al-ways traverse its start node Si. However, path segments do not necessarily traverse its start node Si. In this case, barrier constraints may disallow pairs of conflict-free paths and thus lose the completeness and optimality guarantees. Figure 4(b) provides a counterexample where a CT node N has the set of constraints listed in the figure. The con-straints force agent a2 to wait for at least one timestep before reaching its goal location. It can either wait before entering the rectangle, which leads to a conflict with agent a1, or en-ter the rectangle without waiting and wait later, which might avoid conflicts with agent a1. However, all optimal paths of agent a2 in N (whose costs are 6) have to wait for one timestep before entering the rectangle (see MDD2 shown in the figure). Therefore, node S2 = (2, 4, 2) is a single-ton, and agents a1 and a2 have a cardinal rectangle conflict. If this conflict is resolved using barrier constraints, the CT subtree of N disallows the pair of conflict-free paths where agent a1 directly follows the blue arrow (which traverses node (3, 5, 4) constrained by B(a1, R1, Rg)) and agent a2 follows the dotted red arrow but waits at location (4, 4) for 2 timesteps (which traverses node (4, 4, 4) constrained by B(a2, R2, Rg)). Barrier constraints fail here because the constrained node (4, 4, 4) is not in MDD2 and thus agent a2 could have a path with a larger cost that does not traverse node S2 but traverses node (4, 4, 4). Therefore, we add a barrier constraint only for nodes that are in the current MDD of the agent. We call this a modi-fied barrier constraint B′(ak, Rk, Rg) = {⟨ak, (x, y), t⟩∈ B(ak, Rk, Rg)|(x, y, t) ∈MDDk} (k = i, j), and its prop-erties are discussed in Section 7.3. 7.3 CBSH-RM Our last algorithm is CBSH with rectangle reasoning by MDDs (CBSH-RM), which reasons about rectangle con-flicts between path segments. It uses Algorithm 1 to iden-tify and classify rectangle conflicts and uses modified barrier constraints to resolve them. Previously, all start nodes were at timestep 0, and thus their distances to the rectangle were equal. However, now we allow start nodes to be at different timesteps, e.g., S1 = (5, 2, 1) and S2 = (5, 3, 2) in Figure 4(a). We thus need to modify how to identify rectangle conflicts, calculate rect-angle corner nodes and classify rectangle conflicts, corre-sponding to the three functions on Lines 7, 8 and 9 of Algo-rithm 1, respectively. Identify rectangle conflicts: The start and goal nodes of a rectangle conflict have to satisfy not only Equations (1) to (4) but also (Si.x −Sj.x)(Si.y −Sj.y)(Si.x −Gi.x)(Si.y −Gi.y) ≤0. (15) This guarantees that the start nodes are on different borders of the rectangle since, otherwise, adding modified barrier constraints might lose a pair of paths that allow both agents to reach the constrained border without waiting, such as in the example of Figure 3(f). We also require that Si ̸= Sj, otherwise the two agents have a cardinal vertex conflict at node Si and CBS constraints can resolve it in a single step. Calculate rectangle corner nodes: The method in Sec-tion 5.2 can miscalculate Ri and Rj when Si.x = Sj.x, such as in Figures 3(d) and 3(e). Instead, we calculate Ri and Rj with the following method when Si.x = Sj.x: If (Si.y −Sj.y)(Sj.y −Rg) ≤0, then Ri.x = Rg.x, Ri.y = Si.y, Rj.x = Sj.x and Rj.y = Rg.y; otherwise, Ri.x = Si.x, Ri.y = Rg.y, Rj.x = Rg.x and Rj.y = Sj.y. Classify rectangle conflicts: Similarly, Equations (5) and (6) misclassify rectangle conflicts when Si.x = Sj.x or Si.y = Sj.y. Instead, we classify rectangle conflicts us-ing the corner nodes of their rectangles. Since we always add modified barrier constraints along two adjacent borders of the rectangle, we only need to compare the length and width of the rectangle with those of the Si-Gi and Sj-Gj rectangles. Consider the two equations: Rk.x −Rg.x = Sk.x −Gk.x (16) Rk.y −Rg.y = Sk.y −Gk.y. (17) If one holds for k = i and the other one holds for k = j, the rectangle conflict is cardinal; if only one of them holds for k = i or k = j, it is semi-cardinal; otherwise, it is non-cardinal. Lemma 4. If agents ai and aj have a rectangle conflict, any path of agent ak (k = i, j) that traverses a node constrained by B′(ak, Rk, Rg) also traverses its start node Sk. Proof. Let Nk be a node constrained by B′(ak, Rk, Rg). Thus node Nk is in MDDk. Then, any node before timestep Nk.t on any path of agent ak that traverses node Nk is also in MDDk. Since node Sk is a singleton of MDDk, any path of agent ak that traverses node Nk also traverses its start node Sk. Property 3. For all combinations of paths of agents ai and aj with a rectangle conflict, if one path violates B′(ai, Ri, Rg) and the other path violates B′(aj, Rj, Rg), then the two paths have one or more vertex conflicts within the rectangle. Proof. By Lemma 4, we need to prove that any path of agent ai from its start node Si to one of the nodes constrained by B′(ai, Ri, Rg) and any path of agent aj from its start node Sj to one of the nodes constrained by B′(aj, Rj, Rg) have at least one node in common within the rectangle. This holds by applying the proof for Property 1 after replacing Equations (11) and (12) by Equation (15) and replacing the method for calculating rectangle corner nodes in Section 5.2 by the method in this section. Property 4. If agents ai and aj have a rectangle conflict and one of the Equations (16) and (17) holds for k (k = i, j), the cost of any path of agent ak that satisfies B′(ak, Rk, Rg) is larger than the cost of an optimal path of agent ak. Proof. Since nodes Sk and Gk are singletons, all optimal paths contain these two nodes. If one of Equations (16) and (17) holds, any path from node Sk to node Gk traverses at least one node constrained by B′(ak, Rk, Rg). So, all op-timal paths violate B′(ak, Rk, Rg). Therefore, the cost of any path of agent ak that satisfies B′(ak, Rk, Rg) is larger than the cost of an optimal path of agent ak. Theorem 5. CBSH-RM is complete and optimal. Proof. The proof for Theorem 3 applies after replacing Properties 1 and 2 by Properties 3 and 4, respectively. 8 Experimental Results In this section, we compare CBSH-CR, CBSH-R and CBSH-RM with CBSH on grids with randomly blocked cells and benchmark grids. Previous research found that A-based solvers usually run faster than CBS-based solvers on sparse grids, where many rectangle conflicts exist (Sharon et al. 2015). Therefore, we compare our algorithms also with EPEA (Goldenberg et al. 2014), a state-of-the-art A-based solver. Following Boyarski et al. (2015), we enhance EPEA with Independence Detection (ID) (Standley 2010), which identifies independent groups of agents and runs the solver for each group. We ran experiments on a 2.80 GHz Intel Table 2: Results on 20×20 grids. The first “Ins” column shows the number of instances solved by both CBSH and CBSH-RM, and the following columns show results on these instances. Similarly, the second “Ins” column shows the number of instances solved by CBSH-CR, CBSH-R and CBSH-RM, and the following columns show results on these instances. m Ins Runtime (s) #CT Nodes Ins Runtime (s) #CT Nodes CBSH RM CBSH RM CR R RM CR R RM 0% 30 46 6.2 0.02 29,506 87 49 0.06 0.03 0.02 222 89 82 40 40 2.1 0.02 10,889 105 50 2.2 2.1 2.1 11,282 11,029 10,140 50 27 14.8 1.4 92,627 5,925 39 0.6 1.6 1.1 3,454 6,770 4,327 60 9 28.4 2.9 169,916 16,194 26 11.0 9.6 7.5 54,300 45,691 37,210 10% 20 50 2.1 0.002 9,567 8 50 0.002 0.002 0.002 10 9 8 30 50 4.5 2.2 19,322 8,702 50 1.8 2.0 2.2 7,250 7,962 8,702 40 43 17.7 4.4 96,121 21,384 46 8.2 6.0 4.1 37,425 28,686 20,232 50 16 19.0 16.4 97,553 79,975 20 14.4 11.1 14.8 70,517 56,762 73,624 Core i7-7700 laptop with 8 GB RAM with a runtime limit of 5 minutes. For every grid and every number of agents, we average over 50 instances with random start and goal loca-tions. 8.1 Results on Small Grids Figure 5 presents the success rates and runtimes of all al-gorithms on a 20×20 empty grid and a 20×20 grid with 10% randomly blocked cells. On both grids, many opti-mal paths are Manhattan-optimal. As expected, EPEA runs faster than CBSH on sparse grids (with no blocked cells and few agents). The success rates of EPEA drop dramatically as grids get denser. The success rates of CBSH, however, has higher success rates on the non-empty grid than the empty grid when the number of agents is at most 40, indicating that rectangle conflicts significantly slow down CBSH on sparse grids. The three new algorithms run significantly faster than CBSH and EPEA on both grids. In particular, CBSH-R and CBSH-RM perform similarly, and both of them run faster than CBSH-CR, especially on the empty grid with many agents. This observation implies that these instances have many semi- or non-cardinal rectangle conflicts. Table 2 provides additional details. It first compares CBSH-RM with CBSH by showing their runtimes and num-bers of expanded CT nodes on instances solved by both algorithms, i.e., instances that are relatively easy to solve. CBSH-RM wins on both metrics in all cases, by factors of up to three orders of magnitude, and its overhead due to rea-soning with rectangle conflicts appears negligible. CBSH-RM improves CBSH by two techniques, using barrier con-straints to resolve rectangle conflicts and using cardinal rect-angle conflicts to calculate heuristics. In order to see the improvements of the two techniques independently, we also compare CBSH and CBSH-RM with two modified versions of CBSH-RM where one of the two techniques is turned off. The results show that using cardinal rectangle conflicts to calculate heuristics speeds up CBSH, and using barrier constraints to resolve rectangle conflicts speeds up CBSH more. The combination of the two techniques, i.e., CBSH-RM, runs faster than all of them. Table 2 also compares CBSH-CR, CBSH-R and CBSH-RM on both metrics on instances solved by all three algo-rithms. All of them perform similarly. In a few instances, CBSH-CR even expands fewer CT nodes than CBSH-R and CBSH-RM, because our reasoning methods ignore blocked cells and constraints inside the rectangles. So, sometimes rectangle conflicts do not have many symmetries and are faster to solve with CBS constraints than with barrier con-straints. 8.2 Results on Large Grids We also compare the algorithms on two standard benchmark game grids, den520d and lak503d, from (Sturtevant 2012). Figures 6(a) and 6(b) present the success rates and run-times on map den520d, a 257×256 grid with 28,178 empty cells and 37,614 blocked cells. This grid has a large open space and many large obstacles around the open space. Thus, many optimal paths are not Manhattan-optimal. Therefore, although CBSH-CR and CBSH-R run faster than CBSH, EPEA runs faster than all of them. However, CBSH-RM, which reasons about rectangle conflicts between path seg-ments, runs faster than CBSH, CBSH-CR and CBSH-R as well as, in most cases, EPEA. Figures 6(c) and 6(d) present the success rates and run-times on map lak503d, a 192×192 grid with 17,953 empty cells and 18,911 blocked cells. This grid also has large open spaces. But it has many narrow corridors as well, which A-based solvers cannot handle efficiently. EPEA, CBSH, CBSH-CR and CBSH-R perform similarly, while CBSH-RM runs faster than all of them. 9 Conclusions and Future Work In this paper, we introduced a new way of reasoning about a special class of symmetric conflicts, called rectangle con-flicts, between two agents in grid-based MAPF problems. We demonstrated the poor performance of CBS and CBSH when resolving them. We then proposed three methods, CBSH-CR, CBSH-R and CBSH-RM, for identifying such conflicts and resolving them efficiently. Experimental re-sults showed that all three proposed algorithms improve sig-nificantly on CBSH and, among them, CBSH-RM runs the fastest and also runs faster than the A-based MAPF solver EPEA. We suggest the following future research directions: (1) Generalize the symmetry reasoning methods to gen-eral graphs; (2) study symmetric conflicts among multiple agents; and (3) apply symmetry reasoning methods to sub-optimal MAPF solvers. References Banfi, J.; Basilico, N.; and Amigoni, F. 2017. Intractability of time-optimal multirobot path planning on 2D grid graphs with holes. IEEE Robotics and Automation Letters 2(4):1941–1947. Boyarski, E.; Felner, A.; Stern, R.; Sharon, G.; Tolpin, D.; Betza-lel, O.; and Shimony, S. E. 2015. ICBS: Improved conflict-based search algorithm for multi-agent pathfinding. In IJCAI, 740–746. Erdem, E.; Kisa, D. G.; Oztok, U.; and Schueller, P. 2013. A general formal framework for pathfinding problems with multiple agents. In AAAI, 290–296. Felner, A.; Stern, R.; Shimony, S. E.; Boyarski, E.; Goldenberg, M.; Sharon, G.; Sturtevant, N. R.; Wagner, G.; and Surynek, P. 2017. (a) Success rate on 0%-blocked. (b) Runtime on 0%-blocked. (c) Success rate on 10%-blocked. (d) Runtime on 10%-blocked. Figure 5: Results on 20×20 grids with 0% and 10% blocked cells. (a) and (c) plot the success rates within 5 minutes. (b) and (d) plot the runtimes, where the runtime limit of 5 minutes is included in the average for unsolved instances. Many parts of the blue lines in (a) and (b) are hidden by the yellow lines. (a) Success rate on den520d. (b) Runtime on den520d. (c) Success rate on lak503d. (d) Runtime on lak503d. Figure 6: Results on the game grids den520d and lak503d. (a) and (c) plot the success rates within 5 minutes. (b) and (d) plot the runtimes, where the runtime limit of 5 minutes is included in the average for unsolved instances. Search-based optimal solvers for the multi-agent pathfinding prob-lem: Summary and challenges. In SoCS, 29–37. Felner, A.; Li, J.; Boyarski, E.; Ma, H.; Cohen, L.; Kumar, T. K. S.; and Koenig, S. 2018. Adding heuristics to conflict-based search for multi-agent path finding. In ICAPS, 83–87. Goldenberg, M.; Felner, A.; Stern, R.; Sharon, G.; Sturtevant, N. R.; Holte, R. C.; and Schaeffer, J. 2014. Enhanced partial expan-sion A. Journal of Artificial Intelligence Research 50:141–187. Ma, H.; Koenig, S.; Ayanian, N.; Cohen, L.; H¨ onig, W.; Kumar, T. K. S.; Uras, T.; Xu, H.; Tovey, C.; and Sharon, G. 2016a. Overview: Generalizations of multi-agent path finding to real-world scenarios. In IJCAI-16 Workshop on Multi-Agent Path Find-ing. Ma, H.; Tovey, C.; Sharon, G.; Kumar, T. K. S.; and Koenig, S. 2016b. Multi-agent path finding with payload transfers and the package-exchange robot-routing problem. In AAAI, 3166–3173. Ma, H.; Yang, J.; Cohen, L.; Kumar, T. K. S.; and Koenig, S. 2017. Feasibility study: Moving non-homogeneous teams in congested video game environments. In AIIDE, 270–272. Morris, R.; Pasareanu, C.; Luckow, K.; Malik, W.; Ma, H.; Kumar, S.; and Koenig, S. 2016. Planning, scheduling and monitoring for airport surface operations. In AAAI-16 Workshop on Planning for Hybrid Systems. Sharon, G.; Stern, R.; Goldenberg, M.; and Felner, A. 2013. The increasing cost tree search for optimal multi-agent pathfinding. Ar-tificial Intelligence 195:470–495. Sharon, G.; Stern, R.; Felner, A.; and Sturtevant, N. R. 2015. Conflict-based search for optimal multi-agent pathfinding. Arti-ficial Intelligence 219:40–66. Silver, D. 2005. Cooperative pathfinding. In AIIDE, 117–122. Standley, T. S. 2010. Finding optimal solutions to cooperative pathfinding problems. In AAAI, 173–178. Sturtevant, N. 2012. Benchmarks for grid-based pathfinding. Transactions on Computational Intelligence and AI in Games 4(2):144 – 148. Surynek, P.; Felner, A.; Stern, R.; and Boyarski, E. 2016. Efficient SAT approach to multi-agent path finding under the sum of costs objective. In ECAI, 810–818. Veloso, M. M.; Biswas, J.; Coltin, B.; and Rosenthal, S. 2015. Cobots: Robust symbiotic autonomous mobile service robots. In IJCAI, 4423. Wagner, G., and Choset, H. 2011. M: A complete multirobot path planning algorithm with performance bounds. In IROS, 3260– 3267. Wurman, P. R.; D’Andrea, R.; and Mountz, M. 2008. Coordinating hundreds of cooperative, autonomous vehicles in warehouses. AI Magazine 29(1):9–20. Yu, J., and LaValle, S. M. 2013a. Planning optimal paths for mul-tiple robots on graphs. In ICRA, 3612–3617. Yu, J., and LaValle, S. M. 2013b. Structure and intractability of optimal multi-robot path planning on graphs. In AAAI, 1444–1449. Yu, J. 2016. Intractability of optimal multirobot path planning on planar graphs. IEEE Robotics and Automation Letters 1(1):33–40.
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https://math.stackexchange.com/questions/2023284/is-it-enough-to-show-that-lim-x-rightarrow-0-cos1-x-doesnt-exist-to-show
calculus - Is it enough to show that $\lim_{x\rightarrow 0}\cos(1/x)$ doesn't exist to show that $\lim_{x \rightarrow0}(2x\sin(1/x)-\cos(1/x))$ doesn't exist? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Is it enough to show that lim x→0 cos(1/x)lim x→0 cos⁡(1/x) doesn't exist to show that lim x→0(2 x sin(1/x)−cos(1/x))lim x→0(2 x sin⁡(1/x)−cos⁡(1/x)) doesn't exist? Ask Question Asked 8 years, 10 months ago Modified8 years, 10 months ago Viewed 779 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. My initial thought would have been yes, but my professor's solution to proving that lim x→0(2 x sin(1/x)−cos(1/x))lim x→0(2 x sin⁡(1/x)−cos⁡(1/x)) doesn't exist has me thinking otherwise. Professor's solution: Assume for a contradiction that there exists lim x→0(2 x sin(1/x)−cos(1/x))=l lim x→0(2 x sin⁡(1/x)−cos⁡(1/x))=l for some l∈R.l∈R. Notice that cos(1/x)=(cos(1/x)−2 x sin(1/x))+2 x sin(1/x).cos⁡(1/x)=(cos⁡(1/x)−2 x sin⁡(1/x))+2 x sin⁡(1/x). Then by our assumption, the fact that lim x→0 2 x sin(1/x)=0 lim x→0 2 x sin⁡(1/x)=0 (proof by Sandwich Theorem omitted) and using the Algebra of Limits we have that lim x→0(cos(1/x)−2 x sin(1/x))+2 x sin(1/x)=lim x→0(cos(1/x)−2 x sin(1/x))+lim x→0 2 x sin(1/x)lim x→0(cos⁡(1/x)−2 x sin⁡(1/x))+2 x sin⁡(1/x)=lim x→0(cos⁡(1/x)−2 x sin⁡(1/x))+lim x→0 2 x sin⁡(1/x) But observe that lim x→0 cos(1/x)lim x→0 cos⁡(1/x) does not exist (proof using sequences omitted). Hence we have a contradiction. Ok so I have to admit that I don't even see where the contradiction is nor why we had to go through such a long process, just to end up with needing to show that the limit doesn't exist because lim x→0 cos(1/x)lim x→0 cos⁡(1/x) doesn't exist. Couldn't we just have done that from the start? calculus limits Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 20, 2016 at 21:29 Alex.FAlex.F asked Nov 20, 2016 at 20:55 Alex.FAlex.F 359 3 3 silver badges 12 12 bronze badges 13 1 "Couldn't we just have done that from the start?" Okay, and how would you do that?user223391 –user223391 2016-11-20 20:56:39 +00:00 Commented Nov 20, 2016 at 20:56 Well, I agree with you. lim x→0 cos(1/x)lim x→0 cos⁡(1/x) doesn't exist, so the whole limit doesn't exist. Perhaps you are missing part of the original question?Tim Thayer –Tim Thayer 2016-11-20 21:00:15 +00:00 Commented Nov 20, 2016 at 21:00 2 The limit as x x tends to 0 0 of 1=(x+1)/x−1/x 1=(x+1)/x−1/x exists and yet...Guest –Guest 2016-11-20 21:00:49 +00:00 Commented Nov 20, 2016 at 21:00 2 @Alex.F That is lim x→0(x+1 x−1 x)lim x→0(x+1 x−1 x). This is lim x→0 1 lim x→0 1 which exists and equals 1 1. Yet, the second term does not tend to a limit as x→0 x→0(lim x→0 1 x does not exist)(lim x→0 1 x does not exist). You seem to think that this last fact only would imply that the first limit doesn't exist, which is false in general.Guest –Guest 2016-11-20 21:14:18 +00:00 Commented Nov 20, 2016 at 21:14 1 If the limit of the summands exists, then the limit of the sum exists. The converse is false.MathematicsStudent1122 –MathematicsStudent1122 2016-11-20 21:16:32 +00:00 Commented Nov 20, 2016 at 21:16 |Show 8 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. The contradiction comes from the following theorem: If lim x→a f(x)lim x→a f(x) and lim x→a g(x)lim x→a g(x) both exist, then lim x→a(f(x)+g(x))lim x→a(f(x)+g(x)) exists (and is equal to lim x→a f(x)+lim x→a g(x)lim x→a f(x)+lim x→a g(x)). This theorem is being applied with f(x)=cos(1/x)−2 x sin(1/x)f(x)=cos⁡(1/x)−2 x sin⁡(1/x) and g(x)=2 x sin(1/x)g(x)=2 x sin⁡(1/x). You have assumed that lim x→0 f(x)lim x→0 f(x) exists, and you have proved that lim x→0 g(x)lim x→0 g(x) exists. The theorem then tells you that lim x→0(f(x)+g(x))=lim x→0 cos(1/x)lim x→0(f(x)+g(x))=lim x→0 cos⁡(1/x) exists. Since it doesn't exist, this is a contradiction. Note that it is absolutely essential to prove that lim x→0 g(x)lim x→0 g(x) exists here. In particular, the following statement which it seems you intend to use (with h(x)=cos(1/x)h(x)=cos⁡(1/x) and g(x)=2 x sin(1/x)g(x)=2 x sin⁡(1/x)) is not true in general: (FALSE) If lim x→a h(x)lim x→a h(x) does not exist, then lim x→a(h(x)−g(x))lim x→a(h(x)−g(x)) does not exist. For instance, lim x→0 1 x lim x→0 1 x does not exist, but lim x→0(1 x−1 x)lim x→0(1 x−1 x) does exist since 1 x−1 x=0 1 x−1 x=0 for all x≠0 x≠0. The FALSE statement above, however, is true if lim x→a g(x)lim x→a g(x) exists. The proof is exactly the argument given above: define f(x)=h(x)−g(x)f(x)=h(x)−g(x), and suppose for a contradiction that lim x→a f(x)lim x→a f(x) does exist. Then lim x→a(f(x)+g(x))lim x→a(f(x)+g(x)) would exist, but this is lim x→a h(x)lim x→a h(x) which we know does not exist. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 20, 2016 at 22:42 answered Nov 20, 2016 at 21:34 Eric WofseyEric Wofsey 343k 28 28 gold badges 485 485 silver badges 701 701 bronze badges 3 Many thanks for the fantastic explanation. I hope it's okay if I state my understanding just to check I'm now on the right lines. Define cos(1/x)=(cos(1/x)−2 x sin(1/x))+2 x sin(1/x)cos⁡(1/x)=(cos⁡(1/x)−2 x sin⁡(1/x))+2 x sin⁡(1/x). Assume for a contradiction that lim x→0(cos(1/x)−2 x sin(1/x))lim x→0(cos⁡(1/x)−2 x sin⁡(1/x)) exists and we certainly know that lim x→0 2 x sin(1/x)lim x→0 2 x sin⁡(1/x) exists. So lim x→0(cos(1/x)−2 x sin(1/x)+2 x sin(1/x))lim x→0(cos⁡(1/x)−2 x sin⁡(1/x)+2 x sin⁡(1/x)) should exist and be equal to lim x→0 cos(1/x)lim x→0 cos⁡(1/x) but this last limit doesn’t exist, hence contradiction?Alex.F –Alex.F 2016-11-20 22:02:27 +00:00 Commented Nov 20, 2016 at 22:02 Yes, that's right. (Though it doesn't make sense to say "Define" the way you did.)Eric Wofsey –Eric Wofsey 2016-11-20 22:03:55 +00:00 Commented Nov 20, 2016 at 22:03 Nailed it. I think the misconception is that "does not exist" is some magic quantity that annihilates everything else. As in, "does not exist plus anything equals does not exist."Matthew Leingang –Matthew Leingang 2016-11-20 22:36:37 +00:00 Commented Nov 20, 2016 at 22:36 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus limits See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 3Properly arguing for basic limit laws: We have to do it backwards? I mean, we're not sure the limits exist... Related 2show that lim x→0 sin 1 x lim x→0 sin⁡1 x doesn't exist by using the Squeeze Theorem 6Why doesn't lim(x,y)→(0,0)x y sin(1 x y)lim(x,y)→(0,0)x y sin⁡(1 x y) exist? 2Using sequential definition of functional limits, show that lim x→0 1/x lim x→0 1/x does not exist 3Prove that lim x→0 s g n sin(1 x)lim x→0 s g n sin⁡(1 x) does not exist. 1Why does the limit lim x→0 1−cos 3 x x sin 2 x lim x→0 1−cos 3⁡x x sin⁡2 x exist? 4Does lim x→0 ln(x)cot(x)lim x→0 ln⁡(x)cot⁡(x) exist or not? 2lim(x,y)→0 x ln(1+x 3)y(x 2+y 2)lim(x,y)→0 x ln⁡(1+x 3)y(x 2+y 2) doesn't exist (?) 3Double limit lim x→0 lim y→x y 2 f(x)−x 2 f(y)(1−cos x)sin(x−y)lim x→0 lim y→x y 2 f(x)−x 2 f(y)(1−cos⁡x)sin⁡(x−y) 1Proof verification lim x→0 sin(π x)lim x→0 sin⁡(π x) does not exist. 1Finding limit lim x→0,y→0 cos x y−2 x y 2 x 2 y 2 lim x→0,y→0 cos⁡x y−2 x y 2 x 2 y 2 Hot Network Questions The rule of necessitation seems utterly unreasonable Is it ok to place components "inside" the PCB Does the Mishna or Gemara ever explicitly mention the second day of Shavuot? 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187893
https://www.columbia.edu/cu/physics/pdf-files/BubbleChamberPhysics.pdf
1 Bubble Chamber Physics at Columbia Introduction. The desire to identify the basic building blocks of matter and to identify the forces amongst them has long motivated people. In the 1930’s this effort seemed to have succeeded. The proton (p), the electron (e), the neutrino (ν), and the Yukawa particle (needed to understand the nuclear force and originally identified as the particle we now call the muon µ) had been identifies and were thought to be sufficient to explain the structure of matter. By the late 1940’s it became obvious that this was not the case. First the muon was shown not to be the carrier of the nuclear force. The muon was shown to be the product of the Yukawa particle, now called the pion (π). (This prompted Professor Rabi’s famous remark about the muon, “Who ordered that?") The discovery of the pion was followed soon afterwards by the discovery of “V” and “K” particles in events initiated by cosmic rays. These experiments used emulsions and cloud chambers for their detectors. Looking at the results today one has to admire the skill and tenacity that the experimenters brought to their work. The results have stood the test of time, though of course they have been superseded by later experiments. The properties of these new particles were strange, copious production yet long life times. Obviously they required much more investigation and the new particle accelerators then under construction, the Cosmotron at Brookhaven and the Bevatron at Berkley, seemed capable of producing large numbers of these particles. The problem was that the available detectors, cloud chambers and nuclear emulsions, while useful, would not be adequate for the investigations that people wanted perform to study the newly discovered particles. Fortunately, in 1953, the bubble chamber had been invented by Donald Glaser. The bubble chamber had several advantages over both the cloud chamber and nuclear emulsions for the study of the new particles. The material in the bubble chamber acted as the target for the incoming beam, which meant that the point where the strange particles were produced could be observed. Most of the materials used in the bubble chamber had a long radiation length so that the secondary particles could be tracked through the bubble chamber and their momentum measured using the magnetic field in the chamber. The identity of the secondary particles could often be determined by the visual density of the tracks in the bubble chamber. The use of the bubble chamber yielded many of the discoveries from the 1950’s until the 1970's that contributed to our current understanding of elementary particles. The activities of the bubble chamber program at Columbia were centered at the Nevis Laboratories of the Columbia Physics Department though most of the early pictures analyzed were taken at BNL, first at the Cosmotron and then at the AGS. In its final experiments film from SLAC and FNAL was analyzed. The program was initiated by Prof. Jack Steinberger in 1956. Unlike many other bubble chamber groups which used the film as a survey tool to search for new effects, the group under Prof. Steinberger focused its efforts in answering specific questions and the discoveries resulting from the group's work provided the answers to these questions. 2 Following his work on the 2 experiment and the discovery of CP violation the problems that interested Prof. Steinberger could no longer be answered using the bubble chamber. He moved on to using other techniques, and eventually left Columbia for CERN, and the responsibility for leadership of the bubble chamber group was assumed by Prof. Charles Baltay. Prf. Baltay led the group until the effort was ended in the 1980's. During the forty years of research using the bubble chamber many physicists and staff members contributed to the research. I will try to identify as many of them as I can at the conclusion of this essay. The fruits of their efforts resulted in many discoveries and other contributions to our understanding of particle physics. I have also collected the papers published by the group in Physical Review and Physical Review Letters. If I left some out, please let me know and I will add them to the list. As I discovered early in my days with the bubble chamber group Prof. Steinberger preferred to be called Jack, and I will often refer to him by his preferred name in what follows. The Early Years. After the invention of the bubble chamber by Donald Glaser in 1953, Prof. Jack Steinberger recognized very quickly its potential use in studying strange particles.1 With his graduate students, Jack Leitner2, Nick Samios3 and Melvyn Schwartz4 This required that they first learn how to make a bubble chamber able to do an experiment at an accelerator, for no such no such device then existed. One of the challenges was the recompression of the chamber, after the expansion that sensitized it to particles, that was fast enough so that the chamber could operate with the accelerator cycle of a few seconds. This was resolved when the Columbia group discovered the Barkesdale valve, which made it possible to recompress the chamber in a fraction of a second. , he began in 1954 the design of an experiment using a bubble chamber to study strange particle production at the Cosmotron. They then constructed a 6” chamber filled with propane. In this effort they collaborated with Glaser (who was then at BNL) and his graduate student David Rahm. This early collaborative effort was to characterize much of the work of the bubble chamber group over the next two decades; the experiments were done in collaboration with physicists from other institutions. 1 Much of the material in this section comes from Jack’s book “Learning About Particles – 50 Privileged Years.” (Steinberger, 2005) 2 Jack Leitner joined the faculty at Syracuse University after receiving his degree. He started his own group there. His promising career was cut short by his death at an early age. 3 Nick Samios left Nevis and joined the Shutt group at BNL. He led the effort that led to the discovery of the Ω- baryon which completed the lowest lying decuplet in SU3. He later became director of BNL. 4 Mel Schwartz eventually joined the faculty at Columbia where he continued working with data from bubble chambers. His interest soon turned to ν physics. His insights and calculations led to the first accelerator neutrino beam and the discovery that there were (at least) two neutrinos. For this work he, along with Jack Steinberger and Leon Lederman were awarded the Nobel Prize. 3 The 6” chamber had operational difficulties and lacked a magnetic field but it permitted the measurement of the cross sections and angular distributions of the associated production reactions 𝜋−𝑝→Σ−+ 𝐾+ and 𝜋−𝑝→Λ0 + 𝐾0. This experiment in 1956 with the 6" chamber was the first bubble chamber experiment ever! The experiment following the suggestion of T.D, Lee,, searched for parity violation in lambda decay in the angular correlation of the decay and production angle. , A 2 ½ σ parity violating effect was observed , some 6 months or so before parity violation was discovered by C.S. Wu et al.5 The early successes with the bubble chamber, despite its obvious deficiencies, led the group at Columbia to construct a new 12” Propane Given the importance of the physics and the substantial uncertainty, no claim the discovery was made. 6 A 12" Hydrogen chamber was also built and shared the magnet and optics with the propane chamber. Data from both chambers were used in the determinations of the spin and lifetimes of the Λ and Σ hyperons. Parity violation in the decay of the  was first observed by the Columbia group. There were in addition numerous systematic measurements of cross section and decay modes of , and K particles. Bubble Chamber with a 1.3T magnetic field which meant that the momentum of charged particles could be measured. This chamber also the first to use three rather than two cameras (which permitted simpler reconstruction of the tracks) to observe the particles. This innovation at Columbia was used in all future bubble chambers. The first instance of a Σ0 hyperon was seen in this chamber. The γ from the decay Σ0Λ0+γ converted in the propane into an e+e- pair. Other neutral decays of hyperons and K0 particles were also observed in the propane chamber along with rough estimates of the branching fractions. Among the other results of the group's efforts were discovery of the reduced rate in the leptonic - decay which contributed to the ideas that led to the Cabibbo angle. This chamber was also used on a series of experiments using beam from the Nevis cyclotron. In 1959 there was a run of the 12-inch H2 chamber at Nevis which yielded half a million pictures in one week! This rate was at least an order of magnitude greater than any chamber had taken previously. The experiments included measurements on the decay of the µ, which yielded one of the early reliable and accurate measurements of the Michel rho value, and a definitive study of Dalitz pairs from the decay of the 0 . The 0decays with two Dalitz pairs yielded a definitive determination of 0 parity. Measurements at Nevis on the low energy scattering of of π+p and π-p were also performed using the chamber. The discoveries and results from these early experiments using bubble chambers, both at Nevis7 5 Wu, C.S., Ambler, E., Hayward, R.W., D.D. Hoppes, and Hudson, R.P., Experimental test of parity conservation in beta decay, and elsewhere, led to a desire to have larger bubble chambers to study more and rarer Physical Review, 105(4), 1957 6 The methodology for use of liquid hydrogen in a bubble chamber was not fully developed at that time. Propane, while not without its dangers, could be used without cryogenic equipment, had a greater hydrogen density then did liquid hydrogen and its shorter radiation length made the observation of γ rays in the bubble chamber more likely. 7 The experiments using beams from the Nevis Synchrocyclotron involved a determination of the parity of the π0 meson and a search for the  decay of the π. 4 interactions as well as improving, with increased statistics, on the results already obtained. A bigger bubble chamber would produce more interactions for a given number of particles incident in the chamber, utilizing the accelerator that produced the particles initiating the reactions, more efficiently, allow more track length for more precise measurement of the momenta of the secondary particles, and convert more γ rays in the chamber. The higher energy beams that would soon become available at the Brookhaven AGS seconded the need for larger chambers so that there would be more path length to measure the secondary particles with the necessary precision and to observe hyperon decays. With this in mind work started at Nevis on the construction of two 30” bubble chambers, one filled with propane (which began operation in 1961)8, the second with Hydrogen (which proably began operation in 1962). Physicists at BNL worked on the design and construction of these chambers. Among the people who contributed to the effort to design and build the chambers were Dr. Al Prodell (BNL) 9and Dr. Derek Colley (Nevis)10 . A Technology Aside. The increased number of events anticipated from the larger chambers also necessitated a change in the way the pictures were studied and analyzed. Whereas with the earlier chambers much of the scanning and measuring of the pictures was done by physicists now that work was to be done by trained technicians. Our scanners (technicians trained to scan the pictures and recognize event typologies of interest to the physicists doing the experiments) and measurers were excellent and very dedicated workers who were an essential part of the bubble chamber program.11 Initially the curvature of tracks was measured using templates. The development of computers and advances in electronics led to major changes in the way the events were analyzed. To measure the coordinates of the tracks in the pictures, film projectors with moveable stages were constructed. The motion of the stage was digitized and coordinates of the measured points were punched on to IBM cards. The IBM cards with the data were taken to an IBM 65012 computer at Nevis and the spatial coordinates of the measured coordinates were reconstructed. Using these coordinates and measurements of the magnetic field in the chamber, the computer calculated the momenta of the measured tracks.13 8 There was an accidental spill of several hundred liters of propane from this chamber. Fortunately it evaporated into the air and did not cause an major incident. 9 Al Prodell received his graduate education at Columbia and was a student of Prof. Polykarp Kusch. 10 Derek Colley was a research associate, having received his degree from the University of Birmingham (UK). He was to return to Birmingham where he started his own successful bubble chamber group. 11 Several of the scanners and measurers at Nevis were people who had escaped from the USSR. 12 It is amusing to recall the specifications of the IBM 650. It had 4K words of 10 digits each. It could do a fixed point add in 0.4ms (that is 10-3s). 13 The primary reconstruction program, NP54, was developed by Richard J. Plano (later a professor at Rutgers) and Don Burd of the Nevis computing department. Other members of the department who contributed to 5 With the use of higher energy beams and the wider range of reactions being studied the topology of an event could not always be relied upon to identify the reaction. For example a simple two prong event in a π-p of the following could correspond to any of the following final states; the work of the Bubble Chamber Group were Fred Wuensch and Dot Palmer. There must have been others but their names escape me. My apologies to them. 6 π-p π-p π0 π- π+n π-K+Λ0 etc. To help correctly identify the correct reaction topology computer programs were written first at LBL14 The first event fitting program that we used at Nevis was the program GUTS developed at LBL. It was installed at the IBM 7090 of the Institute for Space Studies and then at CERN to apply the constraints of energy and momentum conservation to the measurements assuming the identity, i.e. mass, of each track in the event. For each mass assignment hypothesis the program minimized a test function computed from the changes in the measured momenta that needed to be made to satisfy the energy and momentum constraints. A comparison of the values of the test function for each of the hypotheses meant that a assignment of the event to a particular reaction could often be made unambiguously. When an unambiguous identification was not possible visual estimates of the ionization of a track could be used to reduce the ambiguity. In most cases a unique identification of the particles in the final state, and hence the reaction, could be made. 15 Fortunately CERN had developed a multi-vertex fitting program, GRIND. Rudy Bock of CERN came to Nevis and helped us install it on the computers used by the Bubble Chamber Group. We also had access to the source code so that we modify it to work with the Nevis programs. by Sandy Wolf. The GUTS program could only fit one vertex at a time, a serious limitation in studying events with multiple vertices, such as were found in many events with strange particles. In addition we did not have the source code for GUTS which made its use awkward since it could not be easily adapted to the Nevis analysis system. The Middle Years The 30” Propane Chamber. The 30” propane chamber was finished before the 30” Hydrogen Chamber and used in an experiment using a 2GeV/c π- beam at the BNL Cosmotron. The objective was to study the properties on the newly discovered Y(1385)16 14 The Lawrence Berkeley Laboratory. resonance that had been discovered by the group 15 The Institute of Space Studies, a NASA facility, was at that time located in the Interchurch Center. The computer room was located on the top floor of the center with a great view of the Hudson River. Time on the Institutes 7090 was provided to us by its then director Robert Jastrow. The staff of the computing center there were very helpful to us. The Bubble Chamber Group continued to use the facilities of the ISS until Columbia opened its computing center with its own IBM 7090. Prior to sing the computers at the ISS members of the group could frequently be found after midnight, usually on a Friday , at the IBM computing center in the basement of the Time-Life building in Rockefeller Center where we had free use of the 7090 there. 16 I am using the nomenclature used at the time. This resonance is now known as the Σ(1385) and is part of the lowest mass SU3 decuplet. 7 at Wisconsin using the beam at the Berkeley Bevatron. Despite the small size of the exposure, as measured by later experiments, the experiment was a success. The first discovery from the exposure was the first observation of the β decay of a Σ- hyperon The event was analyzed and submitted for publication in less than a week. Additional results would take longer to obtain. The Y(1385) was observed decaying into a Λ + π. When the Λ decay was analyzed it was discovered that it was polarized which could only occur if the Y(1385) was also polarized in the production process. The size of the Λ polarization meant that the Y must have spin 3/2 and not 1/2. This was the first determination of the spin of a hyperon resonance. The exposure was also used to look for polarization of 0 in the productions plane. If the 0 were polarized it would indicate that parity was not conserved in the production process. No polarization was observed. Experiments in the BNL 20" Chamber While work progressed on the 30” Hydrogen chamber and the development of a low energy separated beam at the AGS there was a need for new pictures for the enlarged bubble chamber group to analyze. Therefore we obtained an exposure of the BNL 20” hydrogen BC to a separated π+ beam at the AGS. Although we took pictures at three values of beam momentum most of the data were obtained at an incident momentum of 2.9 BeV/c. The primary result from these exposures was the first observation of simultaneous production of a boson resonance (in this case we observed principally the ρ, and the ω but also the ) along with the N(1238) baryon resonance. We also found evidence, based on studying the decay of the ρ, that the reaction π+ + pρ0 + N++ proceeded primarily via one pion exchange. The predictions of the model were not in complete agreement with the theoretical model which lead to the one pion exchange model with absorption. The reaction π+ + p  ω + N++ was hypothesized to proceed via ρ exchange though the angular distributions required significant absorption in calculations in order to obtain the observed decay distributions. Another experiment in the 20" chamber, this time filled with D2 measured the isotopic spin of the f(1250) meson to be 0. The 30" Hydrogen Chamber While rectangular chambers like the 20" at BNL or the 72" chamber at LRL were well suited to studies of high energy interactions where the produced particles come out close to the direction of the incoming beam, the circular 30" chamber had distinct advantages in the study of the interactions from stopped or low energy beams. A low energy/stopping beam was designed at Nevis and installed on the inside the ring of the AGS at BNL. With the chamber in place pictures were taken with stopped K- and p-bar beams. (Later other groups obtained exposures using the beam and the chamber filled with D2.) 8 The exposure of the 30 H2 chamber to a stopped K- beam was performed, among other things, to measure the - relative parity. The experiment was a success and the relative parity was found to be even. The stopped p-bar exposure turned out to be very productive. The first measurements of the widths of the  and  mesons were obtained in this exposure. In addition the data were used to test charge conjugation invariance in the annihilation reaction. None was found. These experiments ended Jack's active participation in bubble chamber experiments as he redirected his efforts towards understanding more about the recently discovered CP violation. The mantle of the leadership of the Columbia bubble chamber group devolved on to Charles Baltay. The Latter Years. Under Prof. Baltay's leadership the group continued to flourish. Work continued on the existing exposures while new exposures were obtained. The stopping K- film was used to determine the branching ratio of the 0. Approximately 50 examples of the decay e in the exposure were used to measure the branching rates and more significantly a matrix element of a strangeness conserving weak current between baryon states. A precise measurement of the lifetime and decay branching ration of the 0 was made using this exposure. Multi-pion final states were studied in the film with the stopped p-bar. The frequency for annihilations into different final states, both resonant and non-resonant, were measured. The capture rates from the different p-pbar states were determined for several final state configurations. Charge conjugation invariance in the annihilation was tested and no violation was observed. The B meson at a mass of 1200Mev/c2 was observed decaying into 0π+π-. The data from both stopping beams was combined in an unsuccessful attempt to find tachyons. Measurement were made on the lifetimes and branching ratios of the  and the Ks. New Exposures Over the years additional film was obtained from Brookhaven, SLAC and Fermilab. One of the first of these new experiments was an exposure of the 30" D2 chamber to a π+ beam at a momentum of 820MeV/c. In a study of the 0 they observed and measured the rate of decay of the 0 into neutral particles and in to the final state π+π-. Charge conservation in the decay of the 0 was also tested in this exposure. The 31" chamber at BNL (an enlarged version of the 20" chamber) was exposed to a 1.7GeV/c K- beam and the properties and rare decay modes of the Ξ- and Ξ0 were studied. This exposure also yielded a measurement of the decay of the Ks into 2π. The ' was also observed and an inconclusive attempt was made to determine its spin and parity. 9 In an exposure of the BNL 80" bubble chamber to π+ beams at 6.95 and 8.5 GeV/c multi-pion resonances were observed at masses of 1630 and 1720Mev/c2. In another exposure of the 80" chamber, this time to a proton beam at 24.8GeV/c, resonant pπ+π- states were observed at 1423 and 1688 GeV/c2. The 82" hydrogen bubble chamber (the old LBL 72" chamber) at SLAC was illuminated by at 15GeV/c π+ beam and an unsuccessful search for charmed particles decaying into a strange particle + pions. The exposure did reveal a new meson resonance at 2340Mev/c2 which decayed primarily into  with iso-spin of 1 or 2. The production of the A1, A3 and A4 resonances was analyzed and these states were found to have properties usually associated with resonant states. A detailed study of the '(1675) was performed and compared with the A2 produced where both where produced in the reaction π+p++π+π-π0. The experiment, in the spirit of the times, also studied pion production in terms of the Feynman variable x and pperp. The interests of the Bubble Chamber group shifted after this experiment to studies of  interactions. Interactions were observed at both BNL, in the 7' bubble chamber, and at FNAL in the 15' Bubble Chamber using both narrow band and broad band beams. The chambers were filled with a mixture of Ne and H2. An early experiment in the FNAL 15' chamber with a broad band  beam failed to find evidence for a heavy lepton (such as the ) but did observe production of the D0 charmed meson. The cross section for the reaction µe-µe- was measured and found to be in agreement with the Weinberg-Salam model with sin2 𝜃𝑤= 0.2. Cross sections and production distributions were also measured for strange particles in charged current interactions. The experiment confirmed the existence of the c ++ and the c + charmed baryons. Observation of opposite sign dilepton production was consistent with expectations of charmed particle production in the  interactions. An excess of 0 events in the dilepton sample was evidence of charmed baryon production. Like sign dilepton events could all be assigned to the background of conventional sources. Evidence was also sought for  oscillations but none was found and upper limits were set on the mixing angles and mass differences. The 7' chamber at BNL was used to search for prompt neutrinos or other penetrating neutral particles, such as an axion, produced in a beam dump at BNL. None were found. the same chamber was exposed to a narrow band  beam and both neutral and charged current -Nucleon interactions were studied. With increasing intensity at the Fermilab accelerator and improved beams it was possible to use the 15' chamber in a narrow band neutrino beam. A series of successful measurements of cross sections and particle distributions was carried out. No evidence for neutrino oscillations was found in the data. The data also allowed the observation of neutral current single π0 production. Closing Comments The experiments done with bubble chambers at Columbia spans 40 years and resulted in many significant contributions to both to our understanding of elementary particle physics and to the 10 list of accomplishments of the Columbia Physics Department. Many of the individuals who worked on these experiments either as student or as young researchers went on to illustrious careers at other institutions. They brought great credit to the Physics Department at Columbia. As we learn and as technology advances, old techniques and questions are abandoned to be replaced by new ones. Such has been the case with bubble chambers but those of us who used them can be proud of what we accomplished. If anyone reads this and wishes to add or modify anything here please contact me. Norman Gelfand nmg38@columbia.edu Appendix A Institutions Which Have Collaborated With the Columbia Bubble Chamber Group. Bloomsburg State University at Bloomsburg, Pennsylvania Brandeis University, Waltham, Massachusetts Brookhaven National Laboratory, Upton, New York Case Western Reserve University, Cleveland, Ohio Institute for Advanced Study, Princeton, New Jersey Institute for High Energy Physics, University of Heidelberg, Heidelberg, Germany Institute for Space Studies, Goddard Space Flight Center, National Aeronautics and Space Administration, New Istituto di Fisica, Bologna, Italy Istituto di Fisica, Pisa, Italy Louisiana State University, Baton Rouge, Louisiana Max-Planck-Institut für Physik, D-8000 Munich 40, Federal Republic of Germany Oak Ridge National Laboratory, Oak Ridge, Tennessee Physics Department, State University of New York at Binghamton, Binghamton, New York Princeton University, Princeton, New Jersey Princeton-Pennsylvania Accelerator, Princeton University, Princeton, New Jersey Rutgers, The State University, New Brunswick, New Jersey State University of New York at Binghamton, Binghamton, New York Stevens Institute of Technology, Hoboken, New Jersey The City College of New York, New York, New York The State University of New York at Stony Brook, Stony Brook, New York The University of Tennessee, Knoxville, Tennessee University of Bologna, Bologna, Italy University of California, San Diego, La Jolla, California University of Kentucky, Lexington, Kentucky University of Maryland, College Park, Maryland 20742; and University of Rochester, Rochester, New York Yale University, New Haven, Connecticut 11 Appendix B- Members of the Columbia Bubble Chamber Group Cynthia Alff-Steinberger Charles Baltay Naomi Barash P. Bassi Dave Berley Enid Bierman T. Böhringer V. Borelli M. Bregman A. Bridgewater G.L. Brown E. B. Brucker D. Caroumbalis C. V. Cautis L. D. Chen Max Chretien D. Cohen Derek C. Colley M. Conversi W. A. Cooper S. Csorna Robert Ehrlich Fred Eisler Gary Feinberg J. Feinman G.E. Fisher Paolo Franzini H. French Jean Marc Gaillard Norman M. Gelfand L. K. Gershwin D. A. Glaser C. Graves M. Habibi K. Han S. W. Herb M. Hibbs R. Hylton P. Igo-Kemenesa G. Impeduglia 12 P. F. Jacques E. W. Jenkins M. Kalelkar Joe Keren Jewan Kim Lawrence Kirsch E. L. Koller H. H. Kung T. D. Lee Juliet Lee Jack Leitner J. T. Liu Gunther Lütjens G. Mageras I. Mannelli Dave Miller Uriel Nauenberg R. Newman H. Norton Mike Nussbaum J. Okamitsu J. Okamitsuc W. Orance G. Ormazabal D. Peterson D. Pisello Richard J. Plano Al Prodell G. Puppi Joseph T. Ratau E. Rice Nick P. Samios R. Santangelo A. C. Schaffer R. D. Schaffer A. C. Schaffere Peter Schmidt "Jay" Schmidt Jonas Schultz Mel Schwartz J. C. (Hans) 13 Severiens K. Shastri V. Silvestrini J. Spitzer P. E. Stamer Jack Steinberger Tai Ho Tan H. Tanaka Dan Tycko A. Vogel P. Waloscher Sandy Wolf N. Yeh J. K. Yoh Dino Zanello V. Zomboli Appendix C- List of Publications form The Columbia Bubble Chamber Group in the Physical Review, Physical Review Letters and Il Nuovo Cimento. 1956 Properties of Heavy Unstable Particles Produced by 1.3-Bev π- Mesons R. Budde, M. Chretien, J. Leitner, N. P. Samios, M. Schwartz, and J. Steinberger Phys. Rev. 103, 1827 (1956) – Published September 15, 1956 A propane bubble chamber has been exposed to a π- beam of 1.3-Bev kinetic energy. The reactions π-+p→Σ-+K+, π-+p→Λ0+θ0, π-+p→Σ0+θ0, can be experimentally distinguished from carbon events. Results based on the first 55 such events are presented. The center-of-mass production distribution of the Σ- is peaked forward, that of the Λ0 backward. No large anisotropies in the angular correlation of production and decay were found, so that we have no evidence for spin in excess of ½ for any of the three particles: Σ-, Λ0, or θ0. A study of the relative abundance of single and double V production indicates that both Λ0 and θ0 have either long-lived "states" or neutral decay modes. A statistical analysis gives α̅ Λ 0=0.3-0.12 +0.15, α̅ θ 0=0.3-0.12 +0.19, for the normal charged decay probabilities (Λ0→π-+p; θ0→π++π-) of the Λ0 and θ0, respectively. One event was analyzed to obtain the energy released in Σ- decay. Σ-→π-+n+Q; Q=118±2.6 Mev. The Σ- lifetime on the basis of 16 decays is (1.4-0.5 +1.6)×10-10 sec. 1957 Possible Detection of Parity Nonconservation in Hyperon Decay 14 T. D. Lee, J. Steinberger, G. Feinberg, P. K. Kabir, and C. N. Yang Phys. Rev. 106, 1367 (1957) – Published June 15, 1957 Cited 52 times No abstract available. Experimental Situation on Parity Doubling F. Eisler, R. Plano, N. Samios, M. Schwartz, and J. Steinberger Phys. Rev. 107, 324 (1957) – Published July 1, 1957 No abstract available. π--p Elastic Scattering at 1.44 Bev M. Chretien, J. Leitner, N. P. Samios, M. Schwartz, and J. Steinberger Phys. Rev. 108, 383 (1957) – Published October 15, 1957 An investigation of π-+p elastic scattering, made in a liquid propane bubble chamber, is reported. Identification of events is made on the basis of kinematics. The problem of contamination by pion scattering from protons bound in carbon is considered in some detail; it is shown that the latter requires a correction of only 4±2.5% of the total number of events. The angular distribution is presented. It shows a large diffraction peak at small angles and an approximately isotropic plateau over the backward hemisphere. The forward peak is fitted to a black-sphere diffraction pattern with a radius of (1.08±0.06)×10-13 cm. The total elastic cross section is found to be σe=10.1±0.80 mb. Demonstration of Parity Nonconservation in Hyperon Decay F. Eisler, R. Plano, A. Prodell, N. Samios, M. Schwartz, J. Steinberger, P. Bassi, V. Borelli, G. Puppi, G. Tanaka, P. Woloschek, V. Zoboli, M. Conversi, P. Franzini, I. Mannelli, R. Santangelo, V. Silvestrini, D. A. Glaser, C. Graves, and M. L. Perl Phys. Rev. 108, 1353 (1957) – Published December 1, 1957 No abstract available. Demonstration of the Existence of the Σo Hyperon and a Measurement of its Mass R Plano, N Samios, M Schwartz and J Steinberger Il Nuovo Cimento (1955-1965), 1957, Volume 5, Number 1, Pages 216-219 Abstract Three events, demonstrating the existence of the Σ0 hyperon, have been found in a propane bubble chamber. The Q-value for the decay Σ0 → Λ0+γ+Q has been measured to be (73.0 ± 3.5) MeV. 15 Systematics of Λ0 and θ0 decay • F. Eisler, R. Plano, N. Samios, M. Schwartz and J. Steinberger Il Nuovo Cimento (1955-1965), 1957, Volume 5, Number 6, Pages 1700-1715 Systematic observations on 528 Λ0-θ0 production events in propane have yielded the following information: 1) There exists a neutral Λ0 decay; Λ0→π0 + n and the fraction of Λ0’s decaying in this mode is .32±.05. 2) The θ1 0 component has a neutral decay mode, very probably θ1 0 →π0+π0. The θ0 therefore very probably has even spin. The fraction P(θ1 0 → π0 + π0) /P(θ1 0 → 2π) is .14±.06. The branching ratio P(Λ0 → π0 + n) /P(Λ0 → π− + p) is in good agreement with the ∆I = 1/2 selection rule; the branching ratio P(θ0 → 2π0) /P(θ0 → π+ + π-) seems to be in disagreement. 3) (51±7.5)%, or just one half of the θ0 mesons escape the chamber and are to be identified with the θ2 0 -meson proposed by GELL-MANN and PAIS (4). The question of the existence of additional (parity doublet) decay modes is discussed. 4) The decays Λ0 → µ− + p + ν / e− + p + ν are not observed and the fraction of such decays is less than 2%. 5) The 3 particle decay modes of the θ1 0 have not been observed and are therefore very unlikely to exceed 2% of all θ1 0 decays. 6) Preliminary measurements yield lifetimes τθ1 0 =(.95 ± .08) · 10−=( s and τΛ 0=(2.8 ± .2)·10−(2 s. 7) For the θ2 0 we obtain a lower limit of the lifetime, which together with the results of LANDE et al. (5) brackets the θ2 0 lifetime (3 < θ2 0 < 10)·10−< s. 1958 Associated Production of Σ0 and θ2 0; Mass of the Σ0 F. Eisler, R. Plano, N. Samios, J. Steinberger, and M. Schwartz Phys. Rev. 110, 226 (1958) – Published April 1, 1958 An event in which θ2 0 and Σ0 production and decay are both observed, is described. This event yields a mass value for the Σ0 and demonstrates the associated production of the θ2 0. Additional events yielding mass values of the Σ0 are reported. β Decay of the Pion G. Impeduglia, R. Plano, A. Prodell, N. Samios, M. Schwartz, and J. Steinberger Phys. Rev. Lett. 1, 249 (1958) – Published October 1, 1958 No abstract available. Leptonic Decay Modes of the Hyperons F. Eisler, R. Plano, A. Prodell, N. Samios, M. Schwartz, J. Steinberger, M. Conversi, P. Franzini, I. Manelli, R. Santangelo, and V. Silvestrini Phys. Rev. 112, 979 (1958) – Published November 1, 1958 We have searched for the leptonic decay of the Λ0 and Σ-. The sensitivity of the experiment was such that 5-6 events 16 should have been found according to the predictions of the "universal" V-A model of β decay. No examples of leptonic decay were observed. Experimental determinations of the Λ0 and Σ- spins F. Eisler, R. Plano, A. Prodell, N. Samios and M. Schwartz, et al. Il Nuovo Cimento (1955-1965), 1958, Volume 7, Number 2, Pages 222-230. Abstract We discuss the applicability of the argument of Adair (1) to the determination of the hyperon spins on the basis of the observed distribution in the production angle for the process π+N→Y + θ. Because of the pronounced backward and forward peaking of these distributions it is found possible to use a large fraction of the events without jeopardizing the validity of the argument. We find from measurement of the distribution in the correlation angles between incident and outgoing pions that the spins of both the Λ0 and Σ hyperons are one half. The only assumptions necessary to this result are 1) that the θ spin is zero and 2) that the interaction radius for strange particle production is not pathologically large. Bubble chamber study of unstable particle production in π−-p collisions at 910, 960, 1200 and 1300 MeV F. Eisler, R. Plano, A. Prodell, N. Samios and M. Schwartz, et al. Il Nuovo Cimento (1955-1965), 1958, Volume 10, Number 3, Pages 468-489 Abstract Results are reported on the total and differential cross-sections for the reactions π− + P → Λ 0 + θ0 ; Σ0 + θ0  Σ- + K+ at 910, 960, 1200 and 1300 MeV using hydrogen and propane bubble chambers in a magnetic field of 13.4 kG. The chambers and their operation, as well as the method of analysis are described in detail. Lifetime of Λ0 ϑ0, and Σ− F. Eisler, R. Plano, A. Prodell, N. Samios and M. Schwartz, et al. Il Nuovo Cimento (1955-1965), 1958, Volume 10, Number 1, Pages 150-154 The lifetimes of the Λ0 ϑ0, and Σ− observed in bubble chamber photographs have been calculated. The following values are obtained: τΛ°=(2.29 -.13 +.15 )·10-10 s on the basis of 454 events; τθ°=(1.06 -.06 +0.8 ) ·10-10s on the basis of 259 events and τΣ-=(1.89 -.25 +.33 )·10-10 s on the basis of 107 events. A comparison with previous values of the mean lives is presented. 17 1959 Search for Rare Decay Modes of the μ+ Meson Juliet Lee and N. P. Samios Phys. Rev. Lett. 3, 55 (1959) – Published July 1, 1959 No abstract available. Scattering of 3.7-25 Mev Positive Pions by Hydrogen G. E. Fischer and E. W. Jenkins Phys. Rev. 116, 749 (1959) – Published November 1, 1959 The Columbia University hydrogen bubble chamber was used to investigate the π+-p scattering cross section in a laboratory energy range from 3.7 to 25 Mev. A total of 950 events were measured, of which 338 were caused by incident pions that would have come to rest in the chamber. Treating the small p-wave and large Coulomb contributions as known, the s-wave phase shift is found to deviate from a linear dependence on momentum only by one and a half standard deviations. Parity of the Neutral Pion R. Plano, A. Prodell, N. Samios, M. Schwartz, and J. Steinberger Phys. Rev. Lett. 3, 525 (1959) – Published December 1, 1959 No abstract available. 1960 Panofsky Ratio N. P. Samios Phys. Rev. Lett. 4, 470 (1960) – Published May 1, 1960 No abstract available. 18 Richard J. Plano Momentum and Asymmetry Spectrum of μ-Meson Decay Phys. Rev. 119, 1400 (1960) – Published August 15, 1960 The beta decay of the positive μ meson was studied using a liquid hydrogen bubble chamber in a magnetic field of 8800 gauss. An analysis of 9213 events used in the momentum spectrum yielded ρ=0.780±0.025. This number includes the internal radiative corrections and is to be compared directly with the 0.75 predicted by two component theory. The analysis of 8354 events used in the asymmetry spectrum gave for the magnitude of the asymmetry |ξ|=0.94±0.07 and for the shape parameter δ=0.78±0.05. π0-π− mass difference as determined from double Dalitz pairs N. P. Samios Il Nuovo Cimento (1955-1965), 1960, Volume 18, Supplement 1, Pages 154-159 Abstract The π0, π− mass difference was determined by measuring the momenta of the two internally converted electron positron pairs from π0-decay. A total of 119 events gave a value π--π0=(4.69±0.07) MeV 1961 Dynamics of Internally Converted Electron-Positron Pairs N. P. Samios Phys. Rev. 121, 275 (1961) – Published January 1, 1961 The reactions studied were π-+p→n+π0 (π0→γ+e++e-) and π-+p→n+e++e-. From a sample of ∼15 000 internally converted electron-positron pairs, 7000 were measured, of which 4200 were used in the detailed analysis. The differential distributions in y (the energy partition) and x2 (the virtual mass of the photon) agree with the theoretical quantum electrodynamic calculations. A measure of the form factor for the π0→2γ reaction gave a value Γ(x/μ)=1-(0.24±0.16)x2/μ2, where μ is the mass of the pion. It was further demonstrated that the number of events necessary 19 to determine the contribution of the longitudinally polarized virtual γ rays in the second reaction is of the order of 50 times that in the present experiment. Leptonic Decay of a Σ- Hyperon Paolo Franzini and Jack Steinberger Phys. Rev. Lett. 6, 281 (1961) – Published March 15, 1961 No abstract available. Example of the Decay Λ0→p+μ-+ν̅ F. Eisler, J. M. Gaillard, J. Keren, M. Schwartz, and S. Wolf Phys. Rev. Lett. 7, 136 (1961) – Published August 15, 1961 No abstract available. 1962 Parity of the Neutral Pion and the Decay π0→2e++2e- N. P. Samios, R. Plano, A. Prodell, M. Schwartz, and J. Steinberger Phys. Rev. 126, 1844 (1962) – Published June 1, 1962 Two hundred and six electronic decays of the π0, π0→e++e-+e++e-, have been observed in a hydrogen bubble chamber. The decay distributions of the electron pairs and the total rate for this process are shown to be in good agreement with theory. An examination of correlations of the e+e- pair decay planes on the basis of electrodynamic predictions is in agreement with the hypothesis that the π0 is pseudoscalar, but disagrees for scalar pions by 3.6 standard deviations. Scattering of 6-24 MeV Negative Pions by Hydrogen Enid Bierman Phys. Rev. 127, 599 (1962) – Published July 15, 1962 The scattering of negative pions at laboratory energies between 6 and 24 MeV has been observed in a liquid hydrogen bubble chamber. The energy of each scattering was deduced from the ranges and angles of the scattered pion and recoil proton. Flux was measured by direct count of tracks of stopped pions. A maximum likelihood analysis was performed treating the strength of the small P-wave contribution as known. Based on 246 scattering events at a median energy of 13 MeV, the S-wave scattering length a=(2α1+α3)/3η is found to be 0.090±0.005 when the large Coulomb contribution is assumed to be the nonrelativistic amplitude for pure Coulomb 20 scattering. Results are also presented with further Coulomb corrections. Values of α1 are computed from the combination of this experimental result with experiments on π+-p scattering. Evidence for π+π- Resonances at 395- and 520-MeV Effective Mass N. P. Samios, A. H. Bachman, R. M. Lea, T. E. Kalogeropoulos, and W. D. Shephard Phys. Rev. Lett. 9, 139 (1962) – Published August 1, 1962 Production of Pion Resonances in π+ p Interactions C. Alff, D. Berley, D. Colley, N. Gelfand, U. Nauenberg, D. Miller, J. Schultz, J. Steinberger, T. H. Tan, H. Brugger, P. Kramer, and R. Plano Phys. Rev. Lett. 9, 322 (1962) – Published October 1, 1962 No abstract available. Decays of the ω and η Mesons C. Alff, D. Berley, D. Colley, N. Gelfand, U. Nauenberg, D. Miller, J. Schultz, J. Steinberger, T. H. Tan, H. Brugger, P. Kramer, and R. Plano Phys. Rev. Lett. 9, 325 (1962) – Published October 1, 1962 No abstract available. Resonances in Strange-Particle Production D. Colley, N. Gelfand, U. Nauenberg, J. Steinberger, S. Wolf, H. R. Brugger, P. R. Kramer, and R. J. Plano Phys. Rev. 128, 1930 (1962) – Published November 15, 1962 In an exposure of propane to 2.0-BeV/c π- mesons at the Cosmotron in the Columbia 30-in. chamber, reactions have been analyzed for resonances between the particles present in the final state. The reactions studied were sufficiently overdetermined to permit a separation of hydrogen events from carbon. We find definite evidence for resonances in the Λπ system (Y1 ) with M0=1392±7 MeV and Γ/2=40±10 MeV; in the Kπ system (K) with M0=897±10 MeV and Γ/2=30±10 MeV. We also seem to see the 1404- and 1525-MeV Σπ resonances. The data indicate that the Y1 has spin 3/2 and its parity is even. 1963 Experimental Study of Double-Charged Pion Production in (π-,p) Collisions at 900 MeV Derek C. Colley and Joseph T. Ratau 21 Phys. Rev. 130, 357 (1963) – Published April 1, 1963 Eight hundred and forty events of the kind π-+p→π-+p+π-+π+ produced in the 20-in. BNL hydrogen bubble chamber by 900-MeV pions have been unambiguously identified using spatial reconstruction and kinematic fitting programs as well as ionization density estimates. The π+,π- and π-,π- combined mass distributions can be fitted by smooth curves, with no deviation beyond statistical fluctuations; no indication has been found of any prominent pion-pion resonance in this interaction, which covers a mass range up to 610 MeV. The π+,p combined mass distribution differs markedly from the four-body phase-space curve, but can be well fitted by weighting the π+,p total cross-section curve at each point according to the amount of phase space available for production of an isobar of corresponding mass. Assuming that the interaction proceeds exclusively via formation of the (π+,p) isobar, one can get a good fit to the π-,π- mass spectrum. This isobar model is also consistent with all of the observed angle and momentum distributions for both pions and protons. The momentum distributions show no indication of any pion-pion-proton resonance in the range up to 1550 MeV, or of any three-pion resonance in the range up to 750 MeV. The cross section for the events studied was measured and found to be (0.33±0.04) mb. Lifetime of the ω Meson N. Gelfand, D. Miller, M. Nussbaum, J. Ratau, J. Schultz, J. Steinberger, T. -H. Tan, L. Kirsch, and R. Plano Phys. Rev. Lett. 11, 436 (1963) – Published November 1, 1963 No abstract available. Width of the φ Meson N. Gelfand, D. Miller, M. Nussbaum, J. Ratau, J. Schultz, J. Steinberger, T. H. Tan, L. Kirsch, and R. Plano Phys. Rev. Lett. 11, 438 (1963) – Published November 1, 1963 No abstract available. 1964 Experimental Study of Parity Conservation in Λ∘ Production in Carbon Nuclei Using Incident π- of 2.0 BeV/c Momentum. R. Ehrlich and J. K. Kim Phys. Rev. 133, B132 (1964) – Published January 13, 1964 Parity conservation has been tested for the Λ∘ production processes: π-+N→Λ∘+anything, where the nucleon N is in a carbon nucleus. Approximately 120 000 pictures were taken of the Columbia-BNL 30-in. propane bubble chamber exposed to a π- beam of 2.0 BeV/c momentum at the Cosmotron. On the basis of 486 events, of which all but 23 were identified without ambiguity as Λ∘, no indication of parity nonconservation was found. This result rests on the fact that the 22 average Λ∘ polarization P ̅ in the production plane was found to be consistent with the value zero within the statistical error. The values found for two components of P ̅ in the production plane are: -0.08±0.12 and 0.00±0.12, using the value +0.63 for α, the asymmetry parameter. Associated Production of ΛK at the ΣK Threshold Joseph Keren Phys. Rev. 133, B457 (1964) – Published January 27, 1964 The reaction π-+p→Λ+K has been studied in the liquid-hydrogen bubble chamber at the threshold energy for the reaction π-+p→Σ+K. The differential cross section for Λ production has been found to be dσ/dΩ=50-10 cosθ-25 cos2θ-51 cos3θ+56 cos4θ μb/sr with a total cross section of 0.67±0.04 mb. The Λ's produced are nearly completely polarized normal to the production plane, and their decay is characterized by |αP ̅ |=0.60±0.05. Two leptonic Λ decays have been identified giving a rate for the leptonic decay consistent with one in a thousand. An unsuccessful attempt has been made to detect cusp-like effects. This attempt has failed because of the presence of high angular momentum states in the ΛK production process. Compilation of Results on the Two-Pion Decay of the ω G. Lütjens and J. Steinberger Phys. Rev. Lett. 12, 517 (1964) – Published May 4, 1964 No abstract available. Neutral Decay and Isotopic Spin of the f0 N. Gelfand, G. Lütjens, M. Nussbaum, J. Steinberger, H. O. Cohn, W. M. Bugg, and G. T. Condo Phys. Rev. Lett. 12, 567 (1964) – Published May 18, 1964 No abstract available. β Decay of the Σ+ and Σ- Hyperons and the Validity of the ΔS=ΔQ Law U. Nauenberg, P. Schmidt, J. Steinberger, S. Marateck, R. J. Plano, Henry Blumenfeld, and Leo Seidlitz Phys. Rev. Lett. 12, 679 (1964) – Published June 15, 1964 No abstract available. Complication of Results on the Two-pion Decay of the ω G. Lütjens and J. Steinberger Phys. Rev. Lett. 12, 717 (1964) – Published June 22, 1964 No abstract available. Test of the Validity of ΔS=ΔQ Rule in K0 Decay 23 L. Kirsch, R. J. Plano, J. Steinberger, and P. Franzini Phys. Rev. Lett. 13, 35 (1964) – Published July 6, 1964 No abstract available. 1965 Determination of Σ-Λ Relative Parity Cynthia Alff, N. Gelfand, U. Nauenberg, M. Nussbaum, J. Schultz, J. Steinberger, H. Brugger, L. Kirsch, R. Plano, D. Berley, and A. Prodell Phys. Rev. 137, B1105 (1965) – Published February 22, 1965 An experiment has been performed to determine the Σ-Λ relative parity, through a study of the decay mode Σ0→Λ0+e++e-. The Σ0 were produced by stopping K- mesons in the Brookhaven National Laboratory-Columbia 30-in. hydrogen chamber, in the reaction K-+p→Σ0+π0, and 314 events were identified. The experimental distribution of the combined mass of the electron-positron pair was compared to that predicted by Feinberg, by Feldman and Fulton, and by Evans. If it is assumed that the dependence of the form factors on the combined mass of the electron-positron pair can be neglected, and that the ratio of the electric form factor F1 to the magnetic form factor F2 is less than 6, then the data show that the Σ-Λ relative parity is even. Strange-Particle Production in π+-p Collisions David Berley and Norman Gelfand Phys. Rev. 139, B1097 (1965) – Published August 23, 1965 Strange particles produced in interactions of positive pions with protons have been studied with the Brookhaven 20-in. bubble chamber, which was exposed to π+ beams of 2.35, 2.62, and 2.90 BeV/c. Cross sections are presented and the production of resonances is discussed. The outstanding feature of the multi-particle final states is that they are dominated by K-π, K-K ̅ , and Y-π resonances. The isotopic spin of the ϕ is confirmed to be zero and no evidence is found for a ϕ decay into three pions. Antiproton Annihilation in Hydrogen at Rest. I. Reaction p ̅ +p→K+K ̅ +π N. Barash, P. Franzini, L. Kirsch, D. Miller, J. Steinberger, T. H. Tan, R. Plano, and P. Yaeger Phys. Rev. 139, B1659 (1965) – Published September 20, 1965 In a study of 735 000 antiproton annihilations at rest in the hydrogen bubble chamber, 182 examples of the reaction K1K1π0 and 851 examples of the reaction K1K±π∓ were recorded. The distributions in the internal variables of these reactions are presented. A substantial fraction of the latter reaction proceeds through an intermediate K state; p ̅ +p→K+K. The theory of the interference effects in this reaction is presented and compared with the experimental result. It is concluded that the KK annihilation proceeds dominantly from the 3S, I=1 state of the N ̅ N 24 system. The fraction of p ̅ p annihilations into KK is given as fKK =(2.1±0.3)×10-3. Annihilations of Antiprotons in Hydrogen at Rest Into Two Mesons C. Baltay, N. Barash, P. Franzini, N. Gelfand, L. Kirsch, G. Lütjens, D. Miller, J. C. Severiens, J. Steinberger, T. H. Tan, D. Tycko, D. Zanello, R. Goldberg, and R. J. Plano Phys. Rev. Lett. 15, 532 (1965) – Published September 20, 1965 No abstract available. Annihilations of Antiprotons in Hydrogen at Rest into Two Mesons C. Baltay, N. Barash, P. Franzini, N. Gelfand, L. Kirsch, G. Lütjens, D. Miller, J. C. Severiens, J. Steinberger, T. H. Tan, D. Tycko, D. Zanello, R. Goldberg, and R. J. Plano Phys. Rev. Lett. 15, 597 (1965) – Published October 4, 1965 No abstract available. Test of Charge-Conjugation Invariance in p ̅ -p Annihilations at Rest C. Baltay, N. Barash, P. Franzini, N. Gelfand, L. Kirsch, G. Lütjens, J. C. Severiens, J. Steinberger, D. Tycko, and D. Zanello Phys. Rev. Lett. 15, 591 (1965) – Published October 4, 1965 No abstract available. Some Features of K0 Decay: The ΔS=ΔQ Rule, and the |ΔI|=1/2 Rules for Leptonic and Three-Pion Decays P. Franzini, L. Kirsch, P. Schmidt, J. Steinberger, and R. J. Plano Phys. Rev. 140, B127 (1965) – Published October 11, 1965 In a sample of ∼36 000 K0 and K ̅ 0 mesons produced in antiproton annihilations in a liquid-hydrogen chamber, the K0 leptonic decay rate and the time distribution of K0 leptonic decays as well as the K0→π++π-+π0 decay rates were studied. These results permit tests of the ΔS=ΔQ rule and CP violation in leptonic decays, as well as tests of the ΔI=1/2 rule for both leptonic and nonleptonic decays when compared with published results on other charge channels. The data, within their experimental error, are found to be consistent with all three rules. Annihilation of Antiprotons in Hydrogen at Rest. II. Analysis of the Annihilation into Three Pions C. Baltay, P. Franzini, N. Gelfand, G. Lütjens, J. C. Severiens, J. Steinberger, D. Tycko, and D. Zanello Phys. Rev. 140, B1039 (1965) – Published November 22, 1965 Analysis of 823 events attributed to the reaction p ̅ +p(at rest )→π++π-+π0 yields the following results: (a) The channel accounts for 7.8% of the annihilations; (b) 0.55±0.05 of the channel proceeds via ρ production, and the capture to ρπ is from the 3S1 state; and (c) 0.45±0.05 25 of the channel is nonresonant and this nonresonant production is from the 1S0 state. Annihilation of Antiprotons in Hydrogen at Rest. III. The Reactions p ̅ +p→ω0+π++π- and p ̅ +p→ω0+ρ0 C. Baltay, P. Franzini, G. Lütjens, J. C. Severiens, J. Steinberger, D. Tycko, and D. Zanello Phys. Rev. 140, B1042 (1965) – Published November 22, 1965 The reactions (a) p ̅ +p→ω0+π++π-(π+π- nonresonating), and (b) p ̅ +p→ω0+ρ0 have been studied for antiprotons at rest. It is found that reaction (a) proceeds from the 3S p ̅ p state, whereas reaction (b) is allowed only for the 1S state. Reaction (a) accounts for 0.039±0.005 of all annihilations, and reaction (b) for 0.007±0.003 of all annihilations. . Measurement of the K0 Mass and the K0-K- Mass Difference J. K. Kim, L. Kirsch, and D. Miller Phys. Rev. 140, B1334 (1965) – Published December 6, 1965 From the decay of the K10 in a hydrogen bubble chamber, we have measured the mass of the K0 to be 497.44±0.33 MeV. From the reaction {K-p→K0,n} { ↘} { π +π -} we have measured the K0-K- mass difference to be 3.71±0.35 MeV. . Σ Radiative Decay and the Angular Momentum of Σ Pionic Decay M. Bazin, H. Blumenfeld, U. Nauenberg, L. Seidlitz, R. J. Plano, S. Marateck, and P. Schmidt Phys. Rev. 140, B1358 (1965) – Published December 6, 1965 We have studied the pion spectrum in the Σ±→n+π±+γ decay in order to determine the angular-momentum channel of the Σ pionic decay. We discuss the results from measurements of a sample of 14 800 Σ+→π++n decays and 25 000 Σ-→π-+n decays. After subtraction of the background, we find 26 Σ+ radiative decays and 28 Σ- radiative decays with Pc.m.<166 MeV/c. The combination Σ+→π++n decays via P wave and Σ-→π-+n decays via S wave is 45 times more likely than the combination Σ+→π++n decays via S wave and Σ-→π-+n decays via P wave. This means that our result is 2.7 standard deviations in favor of the first combination. Search for a Neutral Scalar Meson H. O. Cohn, W. M. Bugg, G. T. Condo, R. D. McCulloch, G. Lütjens, and N. Gelfand 26 Phys. Rev. Lett. 15, 906 (1965) – Published December 6, 1965 No abstract available. The one-pion exchange model and an attempt to measure the spin of the f0 N. Gelfand and G. Lütjens Il Nuovo Cimento A (1965-1970), 1965, Volume 40, Number 4, Pages 1034-1040 Abstract We have compiled data on π−π scattering distributions from the reaction π±+Ν→π++π−+Ν. For each 50 MeV interval in ππ mass the ππ angular distribution is fitted to a polynomial in cosθππ. The simple one-pion exchange model fails to explain the π−π angular distribution in the ρ0 mass region. Using the calculations of Gottfired and Jackson a fair agreement is obtained. Above 1150 MeV in ππ mass cos4θππ terms are needed. The coefficient of cos θππ  hows no evidence of a resonant shape in the f0 region; it does not decrease until well past the f0 mass. 1966 Nonstrange-Resonance Production in π+p Collisions at 2.35, 2.62, and 2.90 BeV/c C. Alff-Steinberger, D. Berley, D. Colley, N. Gelfand, D. Miller, U. Nauenberg, J. Schultz, T. H. Tan, H. Brugger, P. Kramer, and R. Plano Phys. Rev. 145, 1072 (1966) – Published May 27, 1966 In an exposure of the Brookhaven National Laboratory 20-in. hydrogen bubble chamber to a separated π+ beam at π+ momenta of 2.35 BeV/c (center-of-mass energy E=2.30 BeV), 2.62 BeV/c (E=2.41 BeV), and 2.90 BeV/c (E=2.52 BeV), we have observed production of the ω0, ρ0, and η0 mesons. The production of the ω0, ρ0, and η0 is often accompanied by simultaneous production of the N++. The momentum transfer in ω0 and ρ0 production is characteristic of peripheral collisions and suggests a single-particle exchange for the production mechanism. The decay distributions for the ω0, ρ0, and the ρ+ demonstrate the importance of modifying the single-particle-exchange model to include absorptive effects. An upper limit on the two-π decay of the ω0 is set at 2%. The width of the η0 is found to be less than 10 MeV. Elastic-scattering distributions are presented. Annihilations of Antiprotons at Rest in Hydrogen. IV. p̅ p→K̅ Kππ N. Barash, L. Kirsch, D. Miller, and T. H. Tan Phys. Rev. 145, 1095 (1966) – Published May 27, 1966 27 In a study of 735 000 antiproton annihilations at rest in the hydrogen bubble chamber, 3424 events of the reaction p̅ +p→KK̅ ππ were observed. We present here the invariant-mass distributions and scatterplots for this reaction, separated according to the three channels K1K1π+π-, K1(K0)π+π-, and K1K±π∓π0. Also presented are branching ratios into the various channels. K production is found to dominate in all cases. The fraction of p̅ p annihilations into KK̅ is (4.5±0.9)×10-3, and into KKπ is (7.7±1.7)×10-3. Annihilations of Antiprotons at Rest in Hydrogen. V. Multipion Annihilations C. Baltay, P. Franzini, G. Lütjens, J. C. Severiens, D. Tycko, and D. Zanello Phys. Rev. 145, 1103 (1966) – Published May 27, 1966 Approximately 45 000 pionic annihilations of stopped antiprotons in hydrogen have been measured and analyzed. The relative annihilation rates into the various multipion final states are presented. The experimental distributions of the invariant masses of all the possible pion combinations have been obtained. Strong production of the ρ, ω, and η meson has been observed, and their production rates have been established. The rates for simultaneous production of ρ0ρ0, ρ0ω0, and ρ0η0 are discussed. Experimental Evidence Concerning Charge-Conjugation Noninvariance in the Decay of the η Meson Charles Baltay, Paolo Franzini, Jewan Kim, Lawrence Kirsch, Dino Zanello, Juliet Lee-Franzini, Richard Loveless, John McFadyen, and Harold Yarger Phys. Rev. Lett. 16, 1224 (1966) – Published June 27, 1966 No abstract available. A Measurement of the K1 0 Lifetime L. Kirsch and P. Schmidt Phys. Rev. 147, 939 (1966) – Published July 29, 1966 On the basis of about 5000K0 decays into two charged pions in a hydrogen bubble chamber, we have measured the K1 0 lifetime to be τ1=(0.843±0.013)×10-10 sec. Experimental Test of Time-Reversal Invariance in Σ0→Λ0+e++e- R. G. Glasser, B. Kehoe, P. Engelmann, H. Schneider, and L. E. Kirsch Phys. Rev. Lett. 17, 603 (1966) – Published September 12, 1966 28 No abstract available. Σ± Lifetime and the Branching Ratio BΣ +≡(Σ+→π++n)/(Σ+→All) C. Y. Chang Phys. Rev. 151, 1081 (1966) – Published November 25, 1966 Using 11 000 charged Σ's produced from K-+p→Σ±+π∓ interactions taking place in the 30-in. Columbia-BNL hydrogen bubble chamber, we have determined the Σ± lifetimes and the branching ratio BΣ +≡(Σ+→π++n)/(Σ+→all). By means of a least-squares fit to the differential decay distributions for the in-flight sigma decays, the following values were obtained: τΣ +=(0.830±0.018)×10-10 sec, τΣ -=(1.666±0.026)×10-10 sec, and BΣ +=0.46±0.02. 1967 Observation of the B Meson in the Reaction p ̅ +p→ω0+π++π- C. Baltay, J. C. Severiens, N. Yeh, and D. Zanello Phys. Rev. Lett. 18, 93 (1967) – Published January 16, 1967 The B meson, decaying into ω0+π±, has been observed in p ̅ p annihilations at rest in the reaction p ̅ +p→ω0+π++π-. The mass and width of the B meson as observed in this reaction are M=1200±20 MeV and Γ=100±30 MeV. Annihilations of Antiprotons at Rest in Hydrogen. VI. Kaonic Final States N. Barash, L. Kirsch, D. Miller, and T. H. Tan Phys. Rev. 156, 1399 (1967) – Published April 25, 1967 We present here experimental results on the annihilation of stopped antiprotons into KK̅ +3π, KK̅ η, and KK̅ ω. We find the branching ratios for p̅ +p→K1K1η and p̅ +p→K1K1ω to be (0.25±0.04) × 10-3 and (1.08±0.16) × 10-3, respectively. From events of the latter reaction we find the width of the ω meson to be 12.3±2.0 MeV. The E0 meson is observed and evidence is seen for the assignment of even charge conjugation and even G parity to this resonance. In addition, we have used events from the reaction p̅ +p→K1K±π± to measure the mass difference between the charged and neutral K. We find MK 0,K― 0-MK ±=6.3±4.1 MeV. Neutral Decay Branching Ratios of the η0 Meson 29 C. Baltay, P. Franzini, J. Kim, R. Newman, N. Yeh, and L. Kirsch Phys. Rev. Lett. 19, 1495 (1967) – Published December 25, 1967 The branching ratios for the decay of the η0 meson into 3π0, π0γγ, and γγ have been measured. Under the assumption that no other neutral decays are significant, the results are (η0→3π0)/(η0→γγ)=0.88±0.16 and (η0→π0γγ)/(η0→γγ)<~0.28 (95% confidence-level upper limit). Partial Decay Rates of the η0 Meson C. Baltay, P. Franzini, J. Kim, L. Kirsch, R. Newman, N. Yeh, J. A. Cole, J. Lee-Franzini, and H. Yarger Phys. Rev. Lett. 19, 1498 (1967) – Published December 25, 1967 We have measured the ratios (η0→neutral decay modes )/(η0→charged decay modes ) and (η0→π+π-γ)/(η→π+π-π0) to be, respectively, 2.64 ± 0.23 and 0.28 ± 0.04. By combining these ratios with the results of the preceding Letter for the neutral modes, we obtain partial decay rates for the various η0 decays. In particular, we obtain (η0→3π0)/(η0→π+π-π0)=1.58±0.25, in good agreement with the assumption that the three-pion final state has I=1. 1968 Observation of Multipion Resonances at 1630 and 1720 MeV in High-Energy π+p Collisions C. Baltay, H. H. Kung, N. Yeh, T. Ferbel, P. F. Slattery, M. Rabin, and H. L. Kraybill Phys. Rev. Lett. 20, 887 (1968) – Published April 15, 1968 No abstract available. pπ+π- Enhancements in the Reaction pp→ppπ+π- at 24.8 GeV/c R. Ehrlich, R. Nieporent, R. J. Plano, J. B. Whittaker, C. Baltay, J. Feinman, P. Franzini, R. Newman, and N. Yeh Phys. Rev. Lett. 21, 1839 (1968) – Published December 30, 1968 A study of the reaction pp→ppπ+π- at 24.8 GeV/c, based on 3250 events, gives strong support for the production of resonant pπ+π- states at 1.423±0.027 and 1.688±0.023 GeV. 1969 THE DECAY Σ±→Λe±ν C. Baltay, P. Franzini, R. Newman, H. Norton, N. Yeh, J. Cole, J. Lee-Franzini, R. Loveless, and J. McFadyen Phys. Rev. Lett. 22, 615 (1969) – Published March 24, 1969 We have observed 46 examples of the decay Σ-→Λe-ν and six of the decay Σ+→Λe+ν. The branching ratios are, respectively, (0.52±0.09)×10-4 and (0.16±0.07)×10-4. A study of the 30 internal variables distribution yields GV/GA=-0.7±0.4 for the combined sample of Σ±→Λe±ν. 1970 C. Baltay, G. Feinberg, N. Yeh, and R. Linsker Phys. Rev. D 1, 759 (1970) – Published February 1, 1970 An experiment has been carried out to search for uncharged particles with spacelike four-momentum which presumably travel faster than light. No evidence for such particles has been found. The results can be expressed as upper limits on the production rates for such particles by stopped K- and p ̅ compared to production rates of pions in similar reactions: (K-+p→Λ0+t0)/(K-+p→Λ0+π0)≤2×10-3, (K-+p→Λ0+t0+t̅ 0)/(K-+p→Λ0+π0)≤2.5×10-3, (p ̅ +p→π++π-+t0)/(p ̅ +p→3π)≤2×10-3, (p ̅ +p→π++π-+t0+t̅ 0)/(p ̅ +p→4π)≤1×10-3. Other sources of information placing limits on the interactions of tachyons are discussed. 1971 J. Canter, J. Cole, J. Lee-Franzini, R. J. Loveless, and P. Franzini β Decay of the Λ Hyperon Phys. Rev. Lett. 26, 868 (1971) – Published April 5, 1971 We have obtained a sample of 141 kinematically determined Λ→peν events. The internal-variable distribution for these events has been used to determine the magnitude of the ratio of the coupling constants, |gA/gV|=0.75-0.15 +0.18. The branching ratio was also obtained as (0.78±0.09)×10-3. Particle Momentum Distributions for 8.5-GeV/c π+p Interactions J. Cole, J. Lee-Franzini, R. J. Loveless, P. Franzini, and H. H. Kung Phys. Rev. D 4, 627 (1971) – Published August 1, 1971 We have measured two-, four-, six-, and eight-prong events from 8.5-GeV/c π+p interactions and have prepared laboratory distributions of particle momenta both for the individual classes of events and for the combined sample. Strangeness-Changing Leptonic Decay Rates for the Σ± Hyperons J. Cole, J. Lee-Franzini, R. J. Loveless, C. Baltay, P. Franzini, R. Newman, H. Norton, and N. Yeh Phys. Rev. D 4, 631 (1971) – Published August 1, 1971 The strangeness-changing leptonic branching ratios of the Σ± hyperon were determined from a stopping K- exposure in the BNL-Columbia 30-in. hydrogen bubble chamber. The ratio (Σ-→neν)/(all Σ- decays) is (0.97 ± 0.15) × 10-3, and the ratio (Σ-→nμν)/(all Σ- decays) is (0.38 ± 0.11) × 10-3. No evidence is found for ΔS=ΔQ violations, and an upper limit for the ratio of 31 decay rates [(Σ+→neν)/(Σ-→neν)] is determined to be <12.6% with 95% confidence. Precision Measurement of the Lifetime and Decay Branching Ratio of the Λ0 Hyperon C. Baltay, A. Bridgewater, W. A. Cooper, M. Habibi, and N. Yeh Phys. Rev. D 4, 670 (1971) – Published August 1, 1971 A precision measurement of the Λ0 lifetime τΛ and the branching ratio R=(Λ→pπ-)/(Λ→all) has been carried out. The results are τΛ=(2.54±0.04)×10-10 sec and R=0.646±0.008. The errors include both the statistical errors and our estimate of the systematic errors. Measurement of the KS 0→2π Decay Branching Ratio C. Baltay, A. Bridgewater, W. A. Cooper, L. K. Gershwin, M. Habibi, and N. Yeh Phys. Rev. Lett. 27, 1678 (1971) – Published December 13, 1971 We have measured the branching ratio (KS 0→π+π-)/(KS 0→π0π0) to be 2.22 ± 0.095, using a total of ∼ 32 000 K ̅ 0's produced by K- charge exchange. 1972 Measurement of the Ratio of the Axial-Vector to the Vector Coupling in the Decay Σ-→ne-ν̅ C. Baltay, J. Feinman, P. Franzini, R. Newman, N. Yeh, J. Canter, J. Cole, J. Lee-Franzini, and R. J. Loveless Phys. Rev. D 5, 1569 (1972) – Published April 1, 1972 From a sample of 393 Σ-β decays, we have selected 63 events in which a proton recoil from a neutron interaction in the chamber is observed. From the measured values of the electron-neutrino angle we conclude that for the Σ-→ne-ν̅ , |gA/gV|=0.29-0.29 +0.28. This result is obtained from a maximum-likelihood calculation which includes the effect of a well-understood background of 27±6 events contained in our sample. Decay Σ±→Λe±ν P. Franzini, R. Newman, F. Eisele, R. Engelmann, H. Filthuth, F. Föhlisch, V. Hepp, E. Leitner, V. Linke, W. Presser, H. Schneider, M. L. Stevenson, G. Zech, N. Barash, T. B. Day, R. G. Glasser, B. Kehoe, B. Knopp, B. Sechi-Zorn, G. A. Snow, J. A. Cole, J. Lee-Franzini, R. Loveless, and J. McFadyen Phys. Rev. D 6, 2417 (1972) – Published November 1, 1972 The ratio of the vector to axial-vector coupling constant for Σ±→Λe±ν decays using 186 events is determined to be -0.37±0.20. The branching ratio for Σ-→Λe-ν is (0.62±0.07) ×10-4 and for Σ+→Λe+ν is (0.21±0.05)×10-4. An upper limit on the magnitude of the ratio of the axial-32 magnetism to axial-vector coupling constants is 3.2. 1974 Properties of Ξ- and Ξ0 hyperons C. Baltay, A. Bridgewater, W. A. Cooper, L. K. Gershwin, M. Habibi, M. Kalelkar, N. Yeh, and A. Gaigalas Phys. Rev. D 9, 49 (1974) – Published January 1, 1974 We report a measurement of the Ξ- and Ξ0 weak-decay parameters, mean lifetimes, and spins, based on 4303 Ξ- and 652 Ξ0 decays. We find for the Ξ-, α=-0.376±0.038, φ=11∘±9∘, τ=(1.63±0.03)×10-10 sec, and spin equal to ½ with higher spins excluded by more than 7 standard deviations; for the Ξ0, α=-0.54±0.10, φ=16∘±17∘, τ=(2.88-0.19 +0.21)×10-10 sec, and spin equal to ½ with higher spins excluded by more than 4 standard deviations. The results are consistent with the requirements of T invariance, and they are in fair agreement with the ΔI=1/2 rule. Study of the decay distributions of the η′ meson C. Baltay, D. Cohen, S. Csorna, M. Habibi, M. Kalelkar, W. D. Smith, and N. Yeh Phys. Rev. D 9, 2999 (1974) – Published June 1, 1974 We have carried out a study of the decay distributions of η′(958) mesons produced in the reaction K-p→Λη′ at 1.75 GeV/c, utilizing both the ηπ+π- and π+π-γ decay modes of the η′. A Dalitz-plot analysis of the ηπ+π- decay channel rules out all spin-parity assignments except 0- and 2-, but is unable to distinguish between them. We find no evidence for the existence of anisotropies in the η′ decay angular distributions, and thus our data do not support the recent conjecture, based on the observation of such anisotropies, that the η′ has spin 2. Observation of rare decay modes of the Ξ hyperons N. Yeh, A. Gaigalas, W. D. Smith, H. Zendle, C. Baltay, A. Bridgewater, S. Csorna, W. A. Cooper, L. K. Gershwin, M. Habibi, and M. Kalelkar Phys. Rev. D 10, 3545 (1974) – Published December 1, 1974 In an experiment based on the production of 8150 Ξ-1 and 2975 Ξ0 hyperons, we have detected one example each of the decays Ξ-→Λμ-ν̅ and Ξ0→Λγ. The branching ratios of these hitherto unobserved decay modes are Γ(Ξ-→Λμ-ν̅ )/Γ(Ξ-→Λπ-)=(3±3)×10-4 and Γ(Ξ0→Λγ)/Γ(Ξ0→Λπ0)=(5±5)×10-3. One event of the decay Ξ-→Λe-ν̅ has also been observed. There is no evidence for ΔS=2 decays; upper limits on these and other rare decay modes are presented. 1975 Search for Charmed-Particle Production in 15-BeV/c π+p Interactions C. Baltay, C. V. Cautis, D. Cohen, S. Csorna, M. Kalelkar, D. Pisello, E. Schmidt, W. D. 33 Smith, and N. Yeh Phys. Rev. Lett. 34, 1118 (1975) – Published April 28, 1975 A search for the production of charmed particles in 15-BeV/c π+p interactions has been carried out. The search was sensitive to charmed particles in the 1.5 to 4.0 BeV mass range, with lifetimes ≲10-11 sec, decaying into a strange particle with up to eight additional pions. No evidence for the production of such particles was found. Search for Charmed-Particle Production in 15-BeV/c π+p Interactions. C. Baltay, C. V. Cautis, D. Cohen, S. Csorna, M. Kalelkar, D. Pisello, E. Schmidt, W. D. Smith, and N. Yeh Phys. Rev. Lett. 34, 1205 (1975) – Published May 5, 1975 Evidence for a New Meson Resonance at 2340 MeV C. Baltay, C. V. Cautis, D. Cohen, M. Kalelkar, D. Pisello, W. D. Smith, and N. Yeh Phys. Rev. Lett. 35, 891 (1975) – Published October 6, 1975 Evidence is presented for a new meson resonance at 2340±20 MeV, with a width of 180±60 MeV, decaying primarily into ρρπ. The resonance, which is observed in 15-GeV/c π+p interactions, has isotopic spin 1 or 2 and odd G parity. The cross section for production of the ρρπ state is 7.3±1.7 μb. Branching ratios into ρρπ and other decay modes are given. 1976 Distribution of charge in π+p interactions at 15 GeV/c C. Baltay, C. V. Cautis, D. Cohen, M. Kalelkar, D. Pisello, W. D. Smith, and N. Yeh Phys. Rev. D 14, 55 (1976) – Published July 1, 1976 Inclusive and semi-inclusive distributions of charge from a 15-GeV/c π+p experiment are presented in terms of the Feynman variable x and the transverse momentum p⊥. The charge distributions are found to have different nch and p⊥ dependences in different kinematic regions. Our distributions are compared to similar ones calculated from published single-particle distributions in π-p and pp experiments at several energies. 1977 Dilepton Production by Neutrinos in Neon C. Baltay, M. Hibbs, R. Hylton, M. Kalelkar, W. Orance, E. Schmidt, A. -M. Cnops, P. L. Connolly, S. A. Kahn, M. J. Murtagh, R. B. Palmer, N. P. Samios, and T. T. Tso Phys. Rev. Lett. 39, 62 (1977) – Published July 11, 1977 In an exposure of the Fermilab 15-ft bubble chamber filled with a heavy neon-hydrogen mixture to a broadband neutrino beam, we have observed 81 dilepton events of the type νμ+Ne→μ-+e++…. This corresponds to (0.5±0.15)% of the total charged-current neutrino 34 interactions. A total of fifteen neutral strange-particle decays (Ks 0→π+π-, Λ0→pπ-) were found in these dilepton events. When corrected for detection efficiencies and unobservable strange particles, this is consistent with the production of approximately one strange particle per event. Diffractive and Nondiffractive A1, A3, and A4 Production in π+p Interactions at 15 GeV/c C. Baltay, C. V. Cautis, and M. Kalelkar Phys. Rev. Lett. 39, 591 (1977) – Published September 5, 1977 We have studied the spin-parity structure of the 3π system produced opposite a proton or Δ++ in π+p interactions at 15 GeV/c. Our results suggest that the broad enhancement at 1.1 GeV, traditionally associated with the A1, does not have the properties usually associated with a resonant state. We obtain similar results for the A3 and A4 enhancements. 1978 Meson-resonance production in π+p interactions at 15 GeV/c C. Baltay, C. V. Cautis, D. Cohen, S. Csorna, M. Kalelkar, D. Pisello, W. D. Smith, and N. Yeh Phys. Rev. D 17, 62 (1978) – Published January 1, 1978 We report on a study of 15-GeV/c π+p interactions of all topologies using the SLAC 82-in. hydrogen bubble chamber. A description is given of the automatic pattern-recognition techniques used to measure the events. Cross sections are given for meson-resonance production in all topologies. Evidence is presented for a new resonance decaying to five pions. A measurement is made of the branching ratios of the g meson. A study is made of low-mass enhancements in the diffractively produced ρπ, fπ, and gπ channels, and a search is made for nondiffractive production of the A1 meson. Production of ω′(1675) in the Reaction π+p→Δ++π+π-π0 at 15 GeV/c C. Baltay, C. V. Cautis, and M. Kalelkar Phys. Rev. Lett. 40, 87 (1978) – Published January 9, 1978 We present the results of a detailed study of ω′(1675) production in the reaction π+p→Δ++π+π-π0 from a high-statistics bubble-chamber experiment at 15 GeV/c. We have measured the mass, width, and cross section as well as differential cross section and spin density matrix elements and compare then to A2 0 production in the same reaction. We show clear evidence for the resonant phase increase of the 3- (ρπ)fI=0 amplitude with ω′(1675) production. Experimental Limits on Heavy Lepton Production by Neutrinos A. M. Cnops, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, T. T. Tso, C. Baltay, D. Caroumbalis, H. French, M. Hibbs, R. Hylton, M. Kalelkar, W. Orance, and E. Schmidt Phys. Rev. Lett. 40, 144 (1978) – Published January 16, 1978 35 We present upper limits on the production of heavy leptons (L±) by neutrinos via the process νμ+Ne→L±+⋯, L±→e±+ν+ν̅ . These limits imply that the L- and L+, if they couple in full strength to νμ, are heavier than 7.5 and 9 GeV, respectively. They also imply that the coupling strength νμ to the recently discovered 1.9-GeV heavy lepton τ is less than 0.025 of the normal νμ-μ coupling. Charmed-D-Meson Production by Neutrinos C. Baltay, D. Caroumbalis, H. French, M. Hibbs, R. Hylton, M. Kalelkar, W. Orance, E. Schmidt, A. M. Cnops, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, and M. Tanaka Phys. Rev. Lett. 41, 73 (1978) – Published July 10, 1978 We have observed the production of the D0 meson by neutrinos followed by the decay D0→KS 0+π++π-. Correcting for detection efficiencies and K0 decay branching ratios, we find that the production of the D0 followed by decay into K0π+π- corresponds to (0.7 ± 0.2)% of all charged-current neutrino interactions. Measurement of the Cross Section for the Process νμ+e-→νμ+e- at High Energies A. M. Cnops, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, C. Baltay, D. Caroumbalis, H. French, M. Hibbs, R. Hylton, M. Kalelkar, and K. Shastri Phys. Rev. Lett. 41, 357 (1978) – Published August 7, 1978 We have observed eleven events of the reaction νμe-→νμe- in a sample of 106 000 charged-current neutrino interactions in a heavy neon-hydrogen mixture in the 15-ft. bubble chamber at Fermilab. We obtain a cross section for this process of [(1.8±0.8)×10-42 cm2/GeV]Eν. This result is in good agreement with the prediction of the Weinberg-Salam model with sin2θW=0.2. 1979 Confirmation of the Existence of the Σc ++ and Λc + Charmed Baryons and Observation of the Decay Λc +→Λπ+ and K ̅ 0p C. Baltay, D. Caroumbalis, H. French, M. Hibbs, R. Hylton, M. Kalelkar, K. Shastri, A. Vogel, A. M. Cnops, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, and M. Tanaka Phys. Rev. Lett. 42, 1721 (1979) – Published June 25, 1979 In a broadband neutrino exposure of the Fermilab 15-ft bubble chamber, we observe the production of the Σc ++(2426) charmed baryon followed by its decay to Λc +(2260) and π+. We find the mass of the Λc + to be 2257±10 MeV and the m(Σc ++)-m(Λc +) mass difference to be 168±3 MeV. Previously unseen two-body decay modes of the Λc +(2260) are observed. 1980 36 Search for prompt neutrinos and penetrating neutral particles in a beam-dump experiment at Brookhaven National Laboratory P. F. Jacques, M. Kalelkar, P. A. Miller, R. J. Plano, P. Stamer, E. B. Brucker, E. L. Koller, S. Taylor, C. Baltay, H. French, M. Hibbs, R. Hylton, K. Shastri, and A. Vogel Phys. Rev. D 21, 1206 (1980) – Published March 1, 1980 Results are presented from a beam-dump experiment at Brookhaven National Laboratory using the 7-ft bubble chamber to search for prompt sources of neutrinos and the interactions or decays into γγ or e+e- of light, penetrating neutral particles (such as the axion). The observed events are all consistent with representing neutrinos from ordinary sources. We place upper limits on axion production that are below theoretical estimates. We also set upper limits on associated production of the charmed D meson. Cross Sections and Scaling-Variable Distributions of Neutral- and Charged- Current Neutrino-Nucleon Interactions from a Low-Energy Narrow Band Beam C. Baltay, H. French, M. Hibbs, R. Hylton, K. Shastri, A. Vogel, P. F. Jacques, M. Kalelkar, P. A. Miller, R. J. Plano, P. E. Stamer, E. B. Brucker, E. L. Koller, and S. Taylor Phys. Rev. Lett. 44, 916 (1980) – Published April 7, 1980 This Letter compares neutral-current and charged-current scaling-variable distributions in neutrino-nucleon interactions induced by a narrow-band beam at Brookhaven National Laboratory; the x distribution of neutral-current events has been reported previously. The first measurement of flux-normalized neutrino cross sections from a narrow-band beam in the energy range Eν=3-9 GeV is also presented. 1981 Cross-section ratio σ(νn)/σ(νp) for charged-current and neutral-current interactions below 10 GeV P. F. Jacques, M. Kalelkar, P. A. Miller, R. J. Plano, P. Stamer, E. B. Brucker, E. L. Koller, S. Taylor, C. Baltay, H. French, M. Hibbs, R. Hylton, K. Shastri, and A. Vogel Phys. Rev. D 24, 1067 (1981) – Published September 1, 1981 We have measured the cross-section ratio σ(νn)/σ(νp) for both charged-current and neutral-current interactions at low energy. The experiment used the wide-band neutrino beam at Brookhaven National Laboratory. The detector was the 7-foot bubble chamber filled with a 62% neon-hydrogen mixture. For charged-current events we find that the ratio reaches an asymptotic value of 1.80±0.19 for neutrino energies above 1 GeV. For neutral-current events we measure the ratio to be 1.07±0.24. Both of these results are in agreement with the quark model. Experimental Limits on Neutrino Oscillations N. J. Baker, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, C. Baltay, M. Bregman, D. Caroumbalis, L. D. Chen, H. French, M. Hibbs, 37 R. Hylton, J. T. Liu, J. Okamitsu, G. Ormazabal, R. D. Schaffer, K. Shastri, and J. Spitzer Phys. Rev. Lett. 47, 1576 (1981) – Published November 30, 1981 A search for neutrino oscillations in a wide-band neutrino beam at Fermilab with use of the 15-ft bubble chamber is reported. No evidence is found for neutrino oscillations and upper limits are set on the mixing angles and neutrino mass differences in the transitions νμ→νe, νμ→ντ, and νe→ν∼e, where ∼e denotes "not e." 1983 Measurement of the νμ Charged-Current Cross Section N. J. Baker et al. Phys. Rev. Lett. 51, 735 (1983) – Published August 29, 1983 The Fermilab 15-ft bubble chamber, filled with a heavy neon-hydrogen mix, was exposed to a narrow-band νμ beam. Based on the observation of 830 charged-current νμ interactions, the cross section was found consistent with a linear rise with the neutrino energy in the interval 10 GeV<~Eν≲240 GeV. The average slope was determined to be σν/Eν=(0.62±0.05)×10-38 cm2 GeV-1. 1984 Measurement of the Bjorken x and y Distributions in Neutral- and Charged-Current νμ Interactions C. Baltay et al. Phys. Rev. Lett. 52, 1948 (1984) – Published May 28, 1984 Distributions of the Bjorken scaling variables x and y, and the structure function F+(x), are presented both for neutral-current and for charged-current νμ interactions. The data were obtained by use of the Fermilab 15-ft neon bubble chamber exposed to a narrow-band νμ beam. Results are based on 151 neutral-current and 683 charged-current events. An important feature of the neutral-current analysis is the event-by-event reconstruction of the outgoing neutrino. 1985 Opposite-sign dilepton production in νμ interactions N. J. Baker, A.-M. Cnops, P. L. Connolly, S. A. Kahn, H. G. Kirk, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, C. Baltay, M. Bregman, D. Caroumbalis, L. D. Chen, H. French, M. Hibbs, R. Hylton, M. Kalelkar, J. T. Liu, J. Okamitsu, G. Ormazabal, A. C. Schaffer, K. Shastri, and J. Spitzer Phys. Rev. D 32, 531 (1985) – Published August 1, 1985 We report on a high-statistics bubble-chamber experiment using the Fermilab 15-ft 38 bubble chamber filled with a heavy neon/hydrogen mixture exposed to a wide-band neutrino beam. In a sample of 106 000 νμ &>0.3 GeV/c) and a negative muon were observed. After corrections, the rate for opposite-sign dilepton production for νμ charged-current events is (0.52±0.09)%. The Eν dependence of this rate from threshold to ≊200 GeV is presented. The kinematic distributions and strange-particle content of the dilepton events are consistent with those expected from charm-particle production in neutrino interactions. A total of 58 neutral strange particles (Λ→pπ-, KS 0→π+π-) are observed in these events, where less than 16 are expected from conventional charged-current interactions. The presence of a significant excess of Λ’s is evidence for substantial charmed-baryon production. Limits on Like-Sign Dilepton Production in νμ Interactions C. Baltay, M. Bregman, M. Hibbs, J. T. Liu, J. Okamitsu, A. C. Schaffer, K. Shastri, N. J. Baker, P. L. Connolly, S. A. Kahn, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, E. B. Brucker, P. F. Jacques, M. Kalelkar, E. L. Koller, R. J. Plano, and P. E. Stamer Phys. Rev. Lett. 55, 2543 (1985) – Published December 2, 1985 We have searched for the production of like-sign dilepton events (νμ+Ne→μ-+e-+…) in a wide-band neutrino beam at Fermilab using the 15-ft bubble chamber. We observe no signal above the background arising from conventional sources. We set 90%-confidence-level upper limits for the production rates of (νμ+Ne→μ-+e-+…)/(νμ+Ne→μ-+…)<~0.76×10-4 and (νμ+Ne→μ-+e-+…)/(νμ+Ne→μ-+e++…)<~5.3×10-2. 1986 Evidence for a new state produced in antiproton annihilations at rest in liquid deuterium D. Bridges, H. Brown, I. Daftari, R. Debbe, A. deGuzman, W. Fickinger, L. Gray, T. Kalogeropoulos, R. Marino, D. Peaslee, Ch. Petridou, D. K. Robinson, G. Tzanakos, and R. Venugopal Phys. Rev. Lett. 56, 211 (1986) – Published January 20, 1986 Inclusive charged-pion spectra from p¯ d annihilations at rest have been measured in a high-statistics experiment in search of broad states. Analysis of these spectra reveals an enhancement of the π- spectrum at 315 MeV/c. This may be interpreted as a production of a new state of mass 1485 MeV/c2 and width at most 200 MeV/c2 recoiling against the π-. This quasi two-body final state accounts for a large fraction of the p¯ n annihilations. Strange-particle production in neutrino-neon charged-current interactions 39 N. J. Baker, P. L. Connolly, S. A. Kahn, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, C. Baltay, M. Bregman, L. Chen, M. Hibbs, R. Hylton, J. Okamitsu, A. C. Schaffer, and M. Kalelkar Phys. Rev. D 34, 1251 (1986) – Published September 1, 1986 We report a study of strange-particle production from 61?0 charged-current νμNe interactions in the Fermilab 15-ft bubble chamber. The observed sample consists of 2279 K0, 1843 Λ (including 94 Σ0), 93 Λ¯ , and 4 Ξ-. We give inclusive production rates for each of these, as well as the rates for specific single-, double-, and triple-vee channels. From these we derive the rates for associated production of ss¯ quark pairs, and for single-s-quark production via charm decay. The dependence of K0 and Λ production on Eν, Q2, W2, xB, and yB are given. Normalized distributions of Feynman x, rapidity, fragmentation variable z, and transverse momentum squared are obtained for K0 and Λ, and compared with those for charged pions. QCD predictions on the behavior of 〈PT 2〉 are tested. Limits on neutrino oscillations in the Fermilab narrow-band beam E. B. Brucker, P. F. Jacques, M. Kalelkar, E. L. Koller, R. J. Plano, P. E. Stamer, N. J. Baker, P. L. Connolly, S. A. Kahn, M. J. Murtagh, M. Tanaka, C. Baltay, M. Bregman, D. Caroumbalis, L. D. Chen, M. Hibbs, P. Igo-Kemenes, J. T. Liu, J. Okamitsu, G. Ormazabal, A. C. Schaffer, K. Shastri, and J. Spitzer Phys. Rev. D 34, 2183 (1986) – Published October 1, 1986 A search for neutrino oscillations was made using the Fermilab narrow-band neutrino beam and the 15-ft bubble chamber. No positive signal for neutrino oscillations was observed. The 90%-C.L. upper limits for νμ→νe, νμ→ντ, and νμ↛νe were found to be Rμ→e<0.56×103, Rμ→τ<4.4×10-2, and Re↛e<0.27. Evidence for Coherent Neutral-Pion Production by High-Energy Neutrinos C. Baltay, M. Bregman, D. Caroumbalis, L. D. Chen, M. Hibbs, J. T. Liu, J. Okamitsu, G. Ormazabal, A. C. Schaffer, K. Shastri, N. J. Baker, P. L. Connolly, S. A. Kahn, M. J. Murtagh, N. P. Samios, and M. Tanaka Phys. Rev. Lett. 57, 2629 (1986) – Published November 24, 1986 We have observed a signal of 55 isolated γ-conversion pairs produced in a wide-band neutrino beam using the Fermilab 15-ft bubble chamber filled with a heavy Ne-H2 mixture. The signal is consistent with coherent neutral-current single-π0 production followed by decay of the π0 with one of the decay γ's being lost. This signal corresponds to a rate of (2.0 ± 0.4) × 10-4 of the total νμ charged-current cross section at our average Eν of 20 GeV. From this result, we obtain a value of β=0.98±0.24 for the weak-neutral-current isovector axial-vector coupling. 1989 40 Measurement of muon-neutrino—electron elastic scattering in the Fermilab 15-foot bubble chamber N. J. Baker, P. L. Connolly, S. A. Kahn, M. J. Murtagh, R. B. Palmer, N. P. Samios, M. Tanaka, C. Baltay, M. Bregman, M. Hibbs, M. Kalelkar, J. Okamitsu, and A. C. Schaffer Phys. Rev. D 40, 2753 (1989) – Published November 1, 1989 A total of 22 muon-neutrino-electron elastic-scattering events (νμeνμe) have been observed in an exposure of the Fermilab 15-foot bubble chamber filled with a heavy neon-hydrogen mixture to a wide-band neutrino beam. The elastic-scattering cross section is measured to be 1.67±0.44×10-42Eν cm2 GeV-1. The value of the weak mixing angle (sin2θW) determined from this cross section, which is consistent with other measurements of this angle, is 0.20-0.05 +0.06. Appendix D Early Articles Published in Il Nuovo Cimento Experimental determinations of the 0 and - spins F. Eisler, R. Plano, A. Prodell, N. Samios and M. Schwartz, et al. Il Nuovo Cimento (1955-1965), 1958, Volume 7, Number 2, Pages 222-230 We discuss the applicability of the argument of Adair (1) to the determination of the hyperon spins on the basis of the observed distribution in the production angle for the process π+N→Y + θ. Because of the pronounced backward and forward peaking of these distributions it is found possible to use a large fraction of the events without jeopardizing the validity of the argument. We find from measurement of the distribution in the correlation angles between incident and outgoing pions that the spins of both the 0 and - hyperons are one half. The only assumptions necessary to this result are 1) that the θ spin is zero and 2) that the interaction radius for strange particle production is not pathologically large. Bubble chamber study of unstable particle production in π−-p collisions at 910, 960, 1200 and 1300 MeV F. Eisler, R. Plano, A. Prodell, N. Samios, M. Schwartz, J. Steinberger, P. Bassi, V. Borelli, G. Puppi and H. Tanaka, et al. Results are reported on the total and differential cross-sections for the reactions π− + P →  0+ 0; 0+0 ; - + K+ 'at 910, 960, 1200 and 1300 MeV using hydrogen and propane bubble chambers in a magnetic field of 13.4 kG. The chambers and their operation, as well as the method of analysis are described in detail. Lifetime of Λ0 ϑ0, and Σ− F. Eisler, R. Plano, A. Prodell, N. Samios and M. Schwartz, et al. Il Nuovo Cimento (1955-1965), 1958, Volume 10, Number 1, Pages 150-154 The lifetimes of the Λ0, ϑ0, and Σ− observed in bubble chamber photographs have been calculated. The following values are obtained: τΛ°=(2.29 -.13 +.15 )·10-10 s on the 41 basis of 454 events; τθ°=(1.06 -.06 +0.8 ) ·10-10s on the basis of 259 events and τΣ°=(1.89 -.25 +.33 )·10-10 s on the basis of 107 events. A comparison with previous values of the mean lives is presented. π0-π− mass difference as determined from double Dalitz pairs N. P. Samios Il Nuovo Cimento (1955-1965), 1960, Volume 18, Supplement 1, Pages 154-159 Abstract The π−, π0 mass difference was determined by measuring the momenta of the two internally converted electron positron pairs from π0-decay. A total of 119 events gave a value m π −−π0 =(4.69±0.07) MeV Demonstration of the existence of the Σ0 hyperon and a measurement of its mass R. Plano, N. Samios, M. Schwartz and J. Steinberger Il Nuovo Cimento (1955-1965), 1957, Volume 5, Number 1, Pages 216-219 Three events, demonstrating the existence of the Σ0 hyperon, have been found in a propane bubble chamber. The Q-value for the decay Σ0 → Λ0+γ+Q has been measured to be (73.0 ± 3.5) MeV. Il Nuovo Cimento (1955-1965) Volume 5, Number 6, 1700-1715, DOI: 10.1007/BF02856062 Systematics of Λ0 and θ0 decay F. Eisler, R. Plano, N. Samios, M. Schwartz and J. Steinberger Systematic observations on 528 Λ0-θ0 production events in propane have yielded the following information: 1) There exists a neutral Λ0 decay; Λ0→π0 + n and the fraction of Λ0’s decaying in this mode is .32±.05. 2) The θ 1 0 component has a neutral decay mode, very probably θ 1 0 → π0+π0. The θ0 therefore very probably has even spin. The fraction P(θ 1 0 → π0 + π0) /P(θ 1 0 → 2π) is .14±.06. The branching ratio P(Λ0 → π0 + n) /P(Λ0 → π− + n) is in good agreement with the I = 1/2 selection rule; the branching ratio P(θ0 → 2π0) /P(θ0 → π+ + π−) seems to be in disagreement. 3) (51±7.5)%, or just one half of the θ0 mesons escape the chamber and are to be identified with the θ 2 0 -meson proposed by GELL-MANN and PAIS (4). The question of the existence of additional (parity doublet) decay modes is discussed. 4) The decays Λ0 → μ− + p + ν / e− + p + ν are not observed and the fraction of such decays is less than 2%. 5) The 3 particle decay modes of the θ 1 0 have not been observed and are therefore very unlikely to exceed 2% of all θ 1 0 decays. 6) Preliminary measurements yield lifetimes τ θ 1 0 =(.95 ± .08) · 10−10 s and τ  0\=(2.8 ± .2)·10−10 s. 7) For the θ 2 0 we obtain a lower limit of the lifetime, which together with the results of LANDE et al. (5) brackets the θ 2 0 lifetime (3 < θ 2 0 < 10)·10−8 s. Appendix E- Scanners and Measurers. 42 The work of the bubble chamber group was dependant on the work of our talented and dedicated measurers and scanners. I will list the names of those I can recall and encourage others to send me the names that I have forgotten. Alex Rytov Margaret-Anne Lewis Ann Therrien Cecile Margolis Irene Yozdez Fran Joyce Alex Wosney Alice Sttewart Gerry Mauk
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https://sites.millersville.edu/bikenaga/calculus2/alternating-series/alternating-series.html
Alternating Series If a series has only positive terms, the partial sums get larger and larger. If they get large too rapidly, the series will diverge. However, if some of the terms are negative, the negative terms may cancel with the positive terms and prevent the partial sums from "blowing up". Here's an example. This is the harmonic series: We've seen that it diverges. This is the alternating harmonic series The alternating harmonic series converges. I computed the , , and partial sums of the harmonic series and the alternating harmonic series. Here's what I got: Of course, this is just numerical evidence, not a proof. However, you can see that while the partial sums of the harmonic series are getting steadily larger, the partial sums for the alternating harmonic series seem to be converging. The pictures below show the first 20 and the first 100 partial sums of the alternating harmonic series. Notice how the partial sums appear to converge by oscillation to a value around 0.69. A series alternates if the signs of the terms alternate in sign. The Alternating Series Test provides a way of testing an alternating series for convergence. Theorem. ( Alternating Series Test) Suppose is an alternating series (so the 's are positive). Suppose in addition that: (a) The 's decrease. (b) . Then the series converges. It is usually easy to see by inspection that a series alternates. To check that the 's decrease, look at the corresponding function . Compute the derivative , and use the fact that a function decreases if its derivative is negative. is the same limit that appears in the Zero Limit Test. By itself, the condition that is not enough to make a series converge. (If the limit isn't 0, the Zero Limit Test says the series diverges.) The Alternating Series Rule augments the computation of this limit with other conditions, and all together these conditions are enough to ensure convergence. Proof. Consider the alternating series The partial sum is The odd partial sums , , decrease: The reason is that, since the terms of the series decrease, for all k, so for all k. Thus So at each step I'm subtracting a sequence of positive numbers from . Moreover, the odd partial sums are all greater than 0: The terms in parentheses are all positive, because for all k. Thus, the odd partial sums , , , ... form a decreasing sequence that is bounded below. Therefore, they have a limit: In similar fashion, the even partial sums , , , ... form an increasing sequence bounded above by . Hence, they have a limit: But Then This means that Their common value is . Since the partial sums approach a limit, the series converges. Remark. A common mistake is to try to apply the conditions of the Alternating Series Rule, and then, upon discovering that some condition doesn't hold, conclude that the series diverges. The rule only says that if the conditions are true, then the series converges; it does not say what happens if the conditions are not true. On the other hand, if , you can conclude that the series diverges, by the Zero Limit Test. Example. Apply the Alternating Series Test to the alternating harmonic series . The ensures that the terms alternate. Let . Then , which is always negative. Since f is always decreasing, the terms of the series decrease in magnitude. I could also see this by graphing . Note that in considering whether the terms decrease, I ignore the part --- the . Finally, The conditions of the Alternating Series Test hold. Therefore, the series converges. Example. Does converge or diverge? The series alternates, and Set . Then Since for and the other factors in the last expression are positive, I have and the terms of the series decrease in absolute value. By the Alternating Series Test, the series converges. Example. Does converge or diverge? This is an alternating series, though it's disguised by the absence of the usual . But note that In fact, , so the series can be rewritten as . The series alternates. If , then Hence, the terms decrease in magnitude. Finally, The conditions of the Alternating Series Test are satisfied. Hence, the series converges. When a series converges, you can approximate the sum to an arbitrary degree of accuracy by adding up sufficiently many terms. How many terms do you need to add up in order to approximate the sum to within a given tolerance? When the series is an alternating series, there's an easy way to find out. As the picture shows, the partial sums of an alternating series converge by oscillation to the actual sum. Alternate partial sums are greater than the sum, less than the sum, greater than the sum, and so on. If is the n-th partial sum (the sum of the terms through the n-th), the error is smaller than the next "jump". The size of the jump from to is . (a) The n-th partial sum of a convergent alternating series is in error by no more than (the absolute value of) the next term in the series. (b) The actual sum of an alternating series lies between any two consecutive partial sums. Example. Approximate using the and partial sums. I showed in the last example that this alternating series converges. Here's the sum of the first 10 terms. The next term in the series is . Ignoring the sign, this is approximately 0.11715. The estimate 1.15141 is in error by no more than 0.11715, around . In fact, more is true. The last term was subtracted, so the estimate is too small. The next partial sum is too large: The actual sum is between 1.15141 and 1.26856. Example. (a) Estimate the error if the first 100 terms of the sum are used to approximate the sum. (b) How many terms would be required to estimate the sum to within 0.001? (a) The error in using the partial sum is less than the absolute value of the term. Hence, the error is less than (b) The error in using the partial sum is less than the absolute value of the term, which is . So I want Now . I want the first integer larger than this, and that is . Example. Consider the convergent alternating series . Estimate the smallest value of n so that the partial sum approximates the actual value of the sum with an error of no more than 0.001. The partial sum is in error by no more than the (absolute value of the) next term, which is I want this to be no more than 0.001, so I want the smallest n such that I can't solve this inequality algebraically, so I will use trial and error. The first value for which the inequality holds is . Contact information Bruce Ikenaga's Home Page Copyright 2019 by Bruce Ikenaga
187895
https://www.fortbendisd.com/cms/lib09/TX01917858/Centricity/Domain/11981/Geometry%20Practice%207.2%20Angle%20Relationships.pdf
Name Date Class Angle Relationships Practice Identify each pair of angles as complementary, vertical or adjacent. 1. 2. 3. 4. 5. 6. Find the measure of angle b. 7. 8. 9. Find the value of x. 10. 11. 12. 13. Use the figure to the right to answer the following questions. a. Name a pair of vertical angles. b. Name a pair of adjacent angles c. Find the sum of <a and <b? d. Find the sum of <e, <f, <g and <h? 14. Name two pair of complementary and supplementary angles in the figure. a. <_ and < _ are complementary. b. <_ and < _ are complementary. c. <_ and < _ are supplementary. d. <_ and < _ are supplementary. 1. An electrician charges $75 for the first hour of labor and $60 for each additional hour of labor. A customer calculates that the electrician charges $225 for a total of 3½ hours of labor. Is the customer correct? A Yes, because (75 × − 3.5) 60 = 225 B No, because 60 + × (75 3.5) = 322.50 C Yes, because 75 + × (60 2.5) = 225 D No, because (75 × − 2.5) 60 = 127.50 2. Regina owns a drum that has a diameter of 14 inches and a height of 5.5 inches, as shown below. She wants to design a new drum by dilating the dimensions of the original drum by a scale factor of 1.4. What will be the diameter, d, and the height, h, of the new drum? A d = 19.6 in. and h = 7 7 in. B d = 15.4 in. and h = 6 9 in. C d = 12.5 in. and h = 4 in. D d = 22.5 in. and h =14 in. Maintain Your Skills ANSWERS 1. adjacent 2. adjacent 3. complementary 4. vertical 5. adjacent 6. vertical 7. b = 145 8. b = 61 9. b = 23 10. x = 20 11. x = 23 12. x = 59 13. a. <c & <d; <a & <b; <e & < h; <g & < g b. <a & <b; <b & <d; <d & < c; <c & <a <e & <f; <f & <h; <h & <g; <g & < e c. 180o d. 360o 14. a. <AFB & <BFC are complementary b. <EFD & <BFC are complementary c. <AFB & <BFD are supplementary d. <BFC & <CFE are supplementary Maintain Your Skills 1. C 2. A
187896
https://chem.libretexts.org/Courses/College_of_the_Canyons/Chem_151%3A_Preparatory_General_Chemistry_OER/09%3A_Gases/9.09%3A_Mixtures_of_Gases_-_Dalton's_Law_of_Partial_Pressures
9.9: Mixtures of Gases - Dalton's Law of Partial Pressures - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 9: Gases Chem 151: Preparatory General Chemistry OER { } { "9.01:Kinetic_Molecular_Theory-_A_Model_for_Gases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.02:_Pressure-The_Result_of_Constant_Molecular_Collisions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.03:_Boyles_Law-Pressure_and_Volume" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.04:_Charless_Law-_Volume_and_Temperature" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.05:_The_Combined_Gas_Law-_Pressure_Volume_and_Temperature" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.06:_Avogadros_Law-_Volume_and_Moles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.07:_The_Ideal_Gas_Law-_Pressure_Volume_Temperature_and_Moles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.08:_Gay-Lussac\'s_Law-_Temperature_and_Pressure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.09:_Mixtures_of_Gases-_Dalton\'s_Law_of_Partial_Pressures" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9.10:_Gases_in_Chemical_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Chemistry_and_Measurements" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Units_and_Dimensional_Analysis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Matter_and_Energy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Atoms_and_Elements" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Molecules_and_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Chemical_Composition" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Chemical_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Quantities_in_Chemical_Reactions(Stoichiometry)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:Gases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Solutions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Acids_and_Bases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Electrons_in_Atoms_and_the_Periodic_Table" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Chemical_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Liquids_Solids_and_Intermolecular_Forces" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Appendix-_Math_Fundementals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Fri, 01 Aug 2025 00:49:30 GMT 9.9: Mixtures of Gases - Dalton's Law of Partial Pressures 371661 371661 Joshua Halpern { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "source-chem-47541" ] [ "article:topic", "showtoc:no", "source-chem-47541" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. College of the Canyons 4. Chem 151: Preparatory General Chemistry OER 5. 9: Gases 6. 9.9: Mixtures of Gases - Dalton's Law of Partial Pressures Expand/collapse global location Chem 151: Preparatory General Chemistry OER Front Matter 1: Chemistry and Measurements 2: Units and Dimensional Analysis 3: Matter and Energy 4: Atoms and Elements 5: Molecules and Compounds 6: Chemical Composition 7: Chemical Reactions 8: Quantities in Chemical Reactions (Stoichiometry) 9: Gases 10: Solutions 11: Acids and Bases 12: Electrons in Atoms and the Periodic Table 13: Chemical Bonding 14: Liquids, Solids, and Intermolecular Forces Appendix - Math Fundementals Back Matter 9.9: Mixtures of Gases - Dalton's Law of Partial Pressures Last updated Aug 1, 2025 Save as PDF 9.8: Gay-Lussac's Law- Temperature and Pressure 9.10: Gases in Chemical Reactions Page ID 371661 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Dalton's Law of Partial Pressures 3. Collecting Gases Over Water 4. Gas Collection by Water Displacement 1. Example 14.14.1 Summary Contributions & Attributions Learning Objectives Explain Dalton's Law of Partial Pressures. The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are 96.5% carbon dioxide and 3% nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over 2700 mm Hg. And there is no oxygen present, so we couldn't breathe there. Not that we would want to go to Venus, as the surface temperature is usually over 460 o⁢C. Dalton's Law of Partial Pressures Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm. The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by P N 2. Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton's Law can be expressed with the following equation: (9.9.1)P total=P 1+P 2+P 3+⋯ The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, P 1 and P 2, reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If P 1=300 mm Hg and P 2=500 mm Hg, then P total=800 mm Hg. Figure 9.9.1: Dalton's Law states that the pressure of a gas mixture is equal to the partial pressures of the combining gases. Collecting Gases Over Water You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don't. All you need is the atmospheric pressure in the room. As the gas pushes out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside. Gas Collection by Water Displacement Gases that are produced in laboratory experiments are often collected by a technique called water displacement (Figure 9.9.2). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing, which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid. Figure 9.9.2: A gas produced in a chemical reaction can be collected by water displacement. Because the gas is collected over water, it is not pure, but is mixed with vapor from the evaporation of the water. Dalton's Law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor. (9.9.2)P Total=P g+P H 2⁡O P g is the pressure of the desired gas P g=P T⁢o⁢t⁢a⁢l−P H 2⁡O In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see table below). The sample problem illustrates the use of Dalton's Law when a gas is collected over water. Table 9.9.1: Vapor Pressure of Water (mm Hg) at Selected Temperatures (o⁢C)051015202530354045505560 4.58 6.54 9.21 12.79 17.54 23.76 31.82 42.18 55.32 71.88 92.51 118.04 149.38 Example 14.14.1 A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20 o⁢C and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP. Solution Step 1: List the known quantities and plan the problem. Known V Total=2.58 L T=20 o⁢C=293 K P Total=98.60 kPa=739.7 mm Hg Unknown V H 2 at STP =?L The atmospheric pressure is converted from kPa to mm Hg in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law. Step 2: Solve. P H 2=P Total−P H 2⁡O=739,7 mm Hg−17.54 mm Hg=722.2 mm Hg Now the combined gas law is used, solving for V 2, the volume of hydrogen at STP. V 2=P 1×V 1×T 2 P 2×T 1=722.2 mm Hg×2.58 L×273 K 760 mm Hg×293 K=2.28 L⁢H⁡A 2 Step 3: Think about your result. If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes. Summary Dalton's Law of Partial Pressures states that the total pressure in a system is equal to the sum of the partial pressures of the gases present. The vapor pressure due to water in a sample can be corrected for, in order to get the true value for the pressure of the gas. Contributions & Attributions This page was constructed from content via the following contributor(s)and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: Marisa Alviar-Agnew(Sacramento City College) Henry Agnew(UC Davis) 9.9: Mixtures of Gases - Dalton's Law of Partial Pressures is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 9.8: Gay-Lussac's Law- Temperature and Pressure 9.10: Gases in Chemical Reactions Was this article helpful? Yes No Recommended articles 9: GasesGases have no definite shape or volume; they tend to fill whatever container they are in. They can compress and expand, sometimes to a great extent. G... Article typeSection or PageShow Page TOCno on page Tags source-chem-47541 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 9.8: Gay-Lussac's Law- Temperature and Pressure 9.10: Gases in Chemical Reactions
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https://www.youtube.com/watch?v=4u5k7BhbD7w
verify (1 + sin(x))/cos(x) + cos(x)/(1 + sin(x)) = 2sec(x) MSolved Tutoring 66100 subscribers 61 likes Description 4765 views Posted: 2 Apr 2020 verify (1 + sin(x))/cos(x) + cos(x)/(1 + sin(x)) = 2sec(x) 1 comments Transcript: all right so let's go ahead and verify the following so what we can do here hmm well we get a common denominator so we'll start off with on the denominator cosine theta and then one plus sine of theta so we get one plus two sine theta plus sine squared theta plus cosine squared theta this becomes 1 so we end up with two plus two sine theta over cosine theta all right yes that makes sense then that we can factor out it 2 1 plus sine theta divided by cosine theta which is the same thing as just saying to seek secant theta and that's it that proves re that identity okay I hope it's helped you out appreciate you watching have yourself a great day
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https://digestiblenotes.com/physics/motion/equations.php
Equations for Uniform Acceleration Subjects | Physics Notes | A-Level Physics Summary ⇒ Velocity-time graphs can be used to derive various different formulas ⇒ This velocity-time graph shows a vehicle increasing its velocity with a consent acceleration, a, from an initial velocity, u, to a final velocity, v, over a time, t The first equation The second equation ⇒ Displacement = average velocity x time ⇒ The average velocity is half way between the initial and final velocities The third equation ⇒ The displacement can also be calculated from the area under the velocity-time graph The fourth equation ⇒ This links velocity, acceleration and displacement... ⇒ There is no need to be able to derive these equations as they will be given in the exam. However, the derivation of the fourth equation is provided below: Example Example 1 Example 2 Example 3 Acceleration due to gravity, g ⇒ Galileo Galilei demonstrated that gravity accelerates all masses at the same rate, provided air resistance is very small ⇒ Knowing this, you could calculate the acceleration due to gravity using the following: ⇒ For example, you could imagine Galileo dropping a ball from the Leaning Tower of Pisa (55m high) and taking 3.25s to fall: Terminal Speed ⇒ Terminal speed is the speed reached when the weight of an object in free fall is balanced by the drag forces (e.g. wind resistance) acting upwards ⇒ Drag is the name given to resistive forces experienced by an object moving through a fluid such as air or water ⇒ The size of the drag on a falling object increases with: Speed Surface area ⇒ This image shows how drag affects two similar balls - they are the same size/shape, but the blue ball weighs 10N and the red ball weighs 1N ⇒ Both balls fall at the same speed and both have an upward drag of 1N, but the blue ball continues to accelerate as it has a resultant downward force whereas the red ball moves at a constant speed because the force of gravity is balanced by the drag In other words, the red ball has reached its terminal speed Extra ⇒ Also see our notes on: Motion in a Straight Line Projectile Motion About Us Digestible Notes was created with a simple objective: to make learning simple and accessible. We believe that human potential is limitless if you're willing to put in the work. © 2025 Digestible Notes All Rights Reserved. Our Socials Say Hello
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https://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html
Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Liquids - Latent Heat of Evaporation Latent heat of vaporization for fluids like alcohol, ether, nitrogen, water and more. The input energy required to change the state from liquid to vapor at constant temperature is called the latent heat of vaporization. When a liquid vaporize at the normal boiling point the temperature of the liquid will not rise beyond the temperature of the boiling point. The latent heat of vaporization is the amount of "heat required to convert a unit mass of a liquid into vapor without a change in temperature". Liquids - Latent Heat of Evaporation | Product | Latent Heat of Evaporation) - he - | | (kJ/kg) | (Btu/lb) | | Acetic acid | 402 | 173 | | Acetone | 518 | 223 | | Alcohol | 896 | 385 | | Alcohol, ethyl (ethanol) | 846 | 364 | | Alcohol, methyl (methanol alcohol, wood alcohol, wood naphtha or wood spirits) | 1100 | 473 | | Alcohol, propyl | 779 | 335 | | Ammonia | 1369 | 589 | | Aniline | 450 | 193 | | Benzene | 390 | 168 | | Bromine | 193 | 83 | | Carbon bisulphide | 160 | | Carbon dioxide | 574 | 247 | | Carbon disulphide | 351 | 151 | | Carbon tetrachloride | 194 | 83 | | Chlorine | 293 | | Chloroform | 247 | 106 | | Decane | 263 | 113 | | Dodecane | 256 | 110 | | Ether | 377 | 162 | | Ethylene glycol | 800 | 344 | | Trichlorofluoromethane refrigerant R-11 | 180 | 77 | | Dichlorodifluoromethane refrigerant R-12 | 165 | 71 | | Chlorodifluoromethane refrigerant R-22 | 232 | 100 | | Glycerine | 974 | 419 | | Helium | 21 | 9 | | Heptane | 318 | 137 | | Hexane | 365 | 157 | | Hydrogen | 461 | 198 | | Iodine | 164 | 71 | | Kerosene | 251 | 108 | | Mercury | 295 | 127 | | Methyl chloride | 406 | | Nitrogen | 199 | 86 | | Octane | 298 | 128 | | Oxygen | 214 | 92 | | Propane | 428 | 184 | | Propylene | 342 | 147 | | Propylene glycol | 914 | 393 | | Sulphur | 1510 | 650 | | Sulfur dioxide | 164 | | Toluene | 351 | 151 | | Turpentine | 293 | 126 | | Water | 2256 | 970.4 | 1 kJ/kg = 0.43 Btu/lbm = 0.24 kcal/kg ) The latent heats of evaporation are based on fluid boiling point temperatures at atmospheric pressure. Evaporation Heat The heat required to evaporate a fluid can be calculated as: q = he m (1) q = evaporation heat (kJ, Btu) he = evaporation heat (kJ/kg, Btu/lb) m = mass of liquid (kg, lb) Example - Calculate heat required to evaporate 10 kg of water The latent heat of evaporation for water is 2256 kJ/kg at atmospheric pressure and 100 oC. The heat required to evaporate 10 kg can be calculated as q = (2256 kJ/kg) (10 kg) = 22560 kJ Unit Converter . Make Shortcut to Home Screen? Cookie Settings