Advanced Calculus of a Single Variable [1st ed.] 3319278061, 978-3-319-27806-3, 978-3-319-27807-0, 331927807X

Table of contents :
Front Matter....Pages i-xii
Real Numbers, Sequences, and Limits....Pages 1-59
Limits and Continuity of Functions....Pages 61-108
The Derivative....Pages 109-168
The Riemann Integral....Pages 169-259
Infinite Series....Pages 261-311
Sequences and Series of Functions....Pages 313-379
Back Matter....Pages 381-382

Citation preview

Tunc Geveci

Advanced Calculus of a Single Variable

Advanced Calculus of a Single Variable

Tunc Geveci

Advanced Calculus of a Single Variable

123

Tunc Geveci Department of Mathematics and Statistics San Diego State University San Diego, CA, USA

ISBN 978-3-319-27806-3 DOI 10.1007/978-3-319-27807-0

ISBN 978-3-319-27807-0 (eBook)

Library of Congress Control Number: 2015959713 Springer Cham Heidelberg New York Dordrecht London © Springer International Publishing Switzerland 2016 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper Springer International Publishing AG Switzerland is part of Springer Science+Business Media (www. springer.com)

Dedicated to Simla Muge and Zehra

Preface

This book is based on a one-semester single variable advanced calculus course that I have been teaching at San Diego State University for many years. Mathematics departments in many schools offer such a course. The aim is a rigorous discussion of the concepts and theorems that are dealt with informally in the first two semesters of a beginning calculus course. As such, students are expected to gain a deeper understanding of the fundamental concepts of calculus, such as limits, continuity, the derivative and the Riemann integral. Success in this course is expected to prepare them for more advanced courses in real and complex analysis. The first semester of advanced calculus can be followed by a rigorous course in multivariable calculus and an introductory real analysis course that treats the Lebesgue integral and metric spaces, with special emphasis on Banach and Hilbert spaces. I believe that each course requires a separate text. Chapter 1 begins with a quick review of the properties of the set of real numbers as an ordered field. The concept of the limit of a sequence and the relevant rules are discussed rigorously. The completeness of the field of real numbers is introduced as the existence of the limit of a Cauchy sequence. I believe that this is better than the introduction of the notion of completeness via the existence of the least upper bound of a subset of real numbers that is bounded above. After all, students have been dealing with Cauchy sequences in the form of decimal approximations all along. An added advantage is the fact that the notion of completeness as the existence of the limit of a Cauchy sequence appears time and again within the general framework of a metric space that may not have an order relation, as in the cases of the field of complex numbers, Banach or Hilbert spaces. The least upper bound principle and the special nature of the convergence or divergence of a monotone sequence are also treated in Chapter 1. The notion of an infinite limit is discussed carefully since the convenient symbol 1 can be misunderstood and mistreated by the student. Chapter 2 discusses the continuity and limits of functions. I have chosen to limit the discussion to functions defined on intervals. I believe that the point set topology of more general sets belongs to a more advanced real analysis course. Many students encounter serious difficulties in the transition from informal to rigorous calculus anyway. vii

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I emphasize the "-ı definitions. Such emphasis is essential for the appreciation of the difference between mere pointwise continuity and uniform continuity on a set. One of the highlights of the chapter is the Intermediate Value Theorem that has bearing on the definitions of basic inverse functions that figure prominently in beginning calculus. Chapter 3 takes up the derivative. The emphasis is on the nature of the error in local linear approximations to functions. That renders a rigorous proof of the celebrated chain rule quite straightforward. The student is also prepared for the generalization of the concept of the derivative to functions of more than a single real variable, even to functions between normed vector spaces. I have included a detailed discussion of convexity as a nice application of the Mean Value Theorem. Rigorous proofs of various versions of L’Hôpital’s rule are not neglected. Chapter 4 is on the Riemann integral. Integrability criteria in terms of upper and lower sums and the oscillation of a function are discussed. The two approaches complement each other. I establish the link with the usual introduction of the integral via arbitrary Riemann sums in a beginning calculus course, unlike some popular advanced calculus texts that neglect to mention that connection as if a new type of integral is being discussed. I have included a detailed discussion of improper integrals, including the comparison and Dirichlet tests. Some of the most important improper integrals that are encountered in practice require such tests. I emphasize Cauchy-type criteria for the convergence of an improper integral. Chapter 5 is a review of a series of real numbers. I chose to provide details since this topic challenges students in a beginning calculus course where rigorous proofs are not provided. I emphasize Cauchy-type criteria for the convergence of series. Chapter 6 discusses the convergence of sequences and series of real-valued functions on intervals. The distinction between mere pointwise and uniform convergence is emphasized, with ample examples. The nice behavior of sequences and series of functions with respect to integration and differentiation are not valid unless certain uniform convergence conditions are satisfied. The analyticity of functions defined via power series follows smoothly once the appropriate foundation involving uniform convergence is established. The chapter is concluded with the definition of familiar special functions via power series. This book is an undergraduate text and not a monograph on a special topic. My writing has been inspired and influenced by a variety of authors over many years since my initial encounter with analysis as a student. I have been fortunate to have had teachers such as Stefan Warschawski, Errett Bishop, and William F. Lucas. The following is a short list of books that are relevant to the way I treated the topics that are included in this book (excellent classical texts in this classical subject): 1. Introduction to Calculus and Analysis, Vol. 1, by Richard Courant and Fritz John, Springer, 1998 2. Theory and Application of Infinite Series, by Konrad Knopp, Dover, 1990 3. The Elements of Real Analysis, Second Edition, by Robert G. Bartle, Wiley, 1976 San Diego, CA, USA

Tunc Geveci

Contents

1

Real Numbers, Sequences, and Limits. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 Terminology and Notation .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.1 Set Theoretic Terminology and Notation ... . . . . . . . . . . . . . . . . . . . 1.1.2 Functions .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.1 Rules of Arithmetic .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.2 The Order Axioms .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.3 The Number Line . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.4 The Absolute Value and the Triangle Inequality . . . . . . . . . . . . . . 1.2.5 The Archimedean Property of R . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 The Limit of a Sequence .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3.1 The Definition of the Limit of a Sequence . . . . . . . . . . . . . . . . . . . . 1.3.2 The Limits of Combinations of Sequences . . . . . . . . . . . . . . . . . . . . 1.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.4 The Cauchy Convergence Criterion .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.4.1 Basic Facts about Cauchy Sequences .. . . . .. . . . . . . . . . . . . . . . . . . . 1.4.2 Irrational Numbers Are Uncountable . . . . . .. . . . . . . . . . . . . . . . . . . . 1.4.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5 The Least Upper Bound Principle .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5.1 The Least Upper Bound Principle . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5.2 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.6 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.6.1 The Definition of the Infinite Limit . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.6.2 Propositions for the Evaluation of Infinite Limits .. . . . . . . . . . . . 1.6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

1 1 1 3 7 7 9 11 12 16 17 19 19 24 29 30 30 37 41 42 42 46 52 53 53 55 59

2 Limits and Continuity of Functions . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1.1 The Definition of Continuity . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

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2.1.2 Uniform Continuity .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 67 2.1.3 The Continuity of Basic Functions and Their Combinations . 70 2.1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 73 2.2 The Limit of a Function at a Point . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 75 2.2.1 The Definition of the Limit of a Function .. . . . . . . . . . . . . . . . . . . . 75 2.2.2 Basic Facts about Limits . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 77 2.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 84 2.3 Infinite Limits and Limits at Infinity . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 85 2.3.1 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 85 2.3.2 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 92 2.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 96 2.4 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 97 2.4.1 The Extreme Value Theorem .. . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 97 2.4.2 The Intermediate Value Theorem . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 98 2.4.3 The Existence and Continuity of Inverse Functions.. . . . . . . . . . 100 2.4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 108 3 The Derivative .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1 The Derivative .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.1 The Definition of the Derivative . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.2 The Derivative as a Function .. . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.3 The Leibniz Notation . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.4 Higher-Order Derivatives.. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2 Local Linear Approximations and the Differential .. . . . . . . . . . . . . . . . . . . 3.2.1 Local Linear Approximations .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.2 The Differential .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.3 The Traditional Notation for the Differential .. . . . . . . . . . . . . . . . . 3.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3 Rules of Differentiation .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.1 The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.2 Differentiation is a Linear Operation . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.3 Product and Quotient Rules . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.5 The Derivative of an Inverse Function .. . . .. . . . . . . . . . . . . . . . . . . . 3.3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4.1 The Proof of the Mean Value Theorem .. . .. . . . . . . . . . . . . . . . . . . . 3.4.2 Some Consequences of the Mean Value Theorem . . . . . . . . . . . . 3.4.3 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.5 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

109 109 109 114 116 119 121 122 122 128 131 133 135 135 136 138 141 143 147 148 148 152 153 160 161 168

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4 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 The Riemann Integral .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.1 Definition of the Riemann Integral.. . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.2 Continuous Functions and Monotone Functions are Integrable.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2 Basic Properties of the Riemann Integral . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.1 The Integrals of Combinations of Functions . . . . . . . . . . . . . . . . . . 4.2.2 Additivity of the Integral with Respect to Intervals .. . . . . . . . . . 4.2.3 Mean Value Theorems for Integrals . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 The Fundamental Theorem of Calculus. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3.1 The Fundamental Theorem of Calculus: Part 1 . . . . . . . . . . . . . . . 4.3.2 The Fundamental Theorem: Part 2 . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.4 The Substitution Rule and Integration by Parts . . . .. . . . . . . . . . . . . . . . . . . . 4.4.1 The Substitution Rule for Indefinite Integrals .. . . . . . . . . . . . . . . . 4.4.2 The Substitution Rule for Definite Integrals .. . . . . . . . . . . . . . . . . . 4.4.3 Integration by Parts for Indefinite Integrals . . . . . . . . . . . . . . . . . . . 4.4.4 Integration by Parts for Definite Integrals .. . . . . . . . . . . . . . . . . . . . 4.4.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.5 Improper Integrals: Part 1. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.5.1 Improper Integrals on Unbounded Intervals .. . . . . . . . . . . . . . . . . . 4.5.2 Improper Integrals That Involve Discontinuous Functions . . . 4.5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.6 Improper Integrals: Part 2. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.6.1 Comparison Tests for Improper Integrals on Unbounded Intervals.. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.6.2 Comparison Tests for Improper Integrals That Involve Discontinuous Functions . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1 Infinite Series of Numbers .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.2 Criteria for the Convergence of Infinite Series . . . . . . . . . . . . . . . . 5.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Convergence Tests for Infinite Series: Part 1 . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.1 The Ratio Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.2 The Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.3 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.4 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Convergence Tests for Infinite Series: Part 2 . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3.1 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

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169 169 169 181 184 185 185 191 193 196 197 197 204 209 210 210 213 217 219 221 222 222 231 238 239 239 252 259 261 261 261 266 274 274 275 280 282 289 293 295 295

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5.3.2 Dirichlet’s Test and Abel’s Test . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 300 5.3.3 A Strategy to Test Infinite Series for Convergence or Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 306 5.3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 310 6 Sequences and Series of Functions . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1 Sequences of Functions .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1.1 The Convergence of Sequences of Functions . . . . . . . . . . . . . . . . . 6.1.2 Some Properties of Uniformly Convergent Sequences . . . . . . . 6.1.3 Proof of Remark 2 . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.2 Infinite Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.2.1 The Convergence of Series of Functions . .. . . . . . . . . . . . . . . . . . . . 6.2.2 Tests for the Convergence of Series of Functions . . . . . . . . . . . . . 6.2.3 Continuity, Integrability, and Differentiability of Sums . . . . . . 6.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.3.1 The Convergence of Power Series . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.3.2 Termwise Integration and Differentiation of Power Series . . . 6.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.4 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.4.1 Taylor’s Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.4.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5 Another Look at Special Functions . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5.1 The Natural Exponential Function .. . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5.2 The Natural Logarithm . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5.3 Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

313 313 313 320 327 329 330 330 331 337 337 339 340 345 350 351 351 367 369 370 372 374

Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 381

Chapter 1

Real Numbers, Sequences, and Limits

1.1 Terminology and Notation In this section we will review some notation and terminology that will be used in this book.

1.1.1 Set Theoretic Terminology and Notation We will use standard terminology and notation for sets: A set A is a collection of objects with clearly expressed properties that qualify them for membership in A. We express the fact that x is an element of A by writing “x 2 A” (you can also read this as “x belongs to A”). For example, the set Q of rational numbers is the collection of numbers of the form p=q where p and q are integers and q ¤ 0. Thus, 2=3 2 Q. We can refer to Q as Q D fx W x D p=q where p and q are integers and q ¤ 0g : When we describe a set A we need to be clear about the meaning of the equality of the elements of A. Equality is a relation between the elements of A that satisfies the following conditions: .i/ x D x .ii/ If x D y then y D x (equality is reflexive) .iii/ If x D y and y D z then x D z (equality is transitive).

© Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0_1

1

2

1 Real Numbers, Sequences, and Limits

Thus equality is an equivalence relation. For example, given rational numbers p1 =q1 and p2 =q2 we have p1 p2 D if and only if p1 q2 D p2 q1 : q1 q2 One may also declare that a rational number is an equivalence class Œp=q corresponding to the above equivalence relation. In that case we will need to pick a representative from each equivalence class and define the basic arithmetic operations in terms of the representatives. It is more practical to define equality of rational numbers as above and work with them in the way we have since early school years. Sets A and B are equal if they contain the same elements. In this case we write A D B. A set A is included in the set B (or, “A is contained in B”) if each element of A is also an element of B. If we use the symbol “)” to denote implication this fact can be expressed as follows: x2A)x2B (this can be read as “if x is in A then x is in B”). We will denote the inclusion of the set A in B by writing A  B. This notation will not exclude the possibility that A D B. In some books the notation “A  B” is used. If we wish to indicate that A is contained in B but A is not equal to B we will use the notation “A ¤ B.” We may use the notation “,” to indicate the equivalence of statements. Thus “,” can be read as “if and only if.” For example we have A D B , A  B and B  A (“A is equal to B if and only if A is contained in B and B is contained on A”). We can abbreviate “if and only if” as “iff.” The union of the sets A and B consists of elements that belong to A or B and the notation is “A [ B.” Thus A [ B D fx W x 2 A or x 2 Bg (read “the set of all x such that x belongs to A or x belongs to B”). The “or” in the above statement is “inclusive or.” It does not exclude the possibility that x belongs to both A and B. We will indicate the union of an arbitrary collection of sets as [A2F A: where F denotes that collection. Thus [A2F A D fx W x 2 A for some A 2 Fg

1.1 Terminology and Notation

3

In particular, if F consists of finitely many sets Ak , k D 1; 2; : : : ; n, we denote the union of these sets by A1 [ A2 [    [ An or [nkD1 Ak : If the collection consists of infinitely many sets Ak , k D 1; 2; 3; : : : we denote their union as A1 [ A2 [ : : : Ak [ : : : or [1 kD1 Ak : The intersection of the sets A and B consists of elements that belong to A and B, and the notation is “A \ B.” Thus A \ B D fx W x 2 A and x 2 Bg : Similarly, \A2F A denotes the intersection of the sets in the collection F so that \A2F A D fx W x 2 A for each A 2 F g : In particular, if F consists of finitely many sets Ak , k D 1; 2; : : : ; n, we denote the intersection of these sets by A1 \ A2 \    \ An or \nkD1 Ak : If the collection consists of infinitely many sets Ak , k D 1; 2; 3; : : : we denote their intersection as A1 \ A2 \ : : : \ Ak \ : : : or \1 kD1 Ak : We will mark the end of a proof by the symbol , the end of an example by , and the end of a remark by Þ.

1.1.2 Functions Recall that a function f from a set U to a set V is a rule that assigns to each element of U an element of V. We will denote this by writing f W U ! V. The set U is the domain of f and the set V is the codomain of f . We will denote the element of V that is assigned to u 2 U as f .u/. The set all such elements of V is the range of f . Thus Range of f D fv 2 V W there exists u 2 U with v D f .u/g :

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1 Real Numbers, Sequences, and Limits

In this book we will deal with real-valued functions of a single real variable. We can refer to such a function as f W U  R ! R. Here R denotes the set of real numbers and U denotes the domain of f . Thus the codomain of f is R. We will denote the range of f as f .U/. The graph of f is the set of all points in the Cartesian coordinate plane of the form .x; f .x// where x is in the domain of f : Assume that a function f is defined so that f .x/ is the same expression for each x where it makes sense. In this pcase we refer to all such x as the natural domain of f . For example, if f .x/ D x for each x  0 the natural domain of f is the set of nonnegative real numbers. If g .x/ D 1=x for each x ¤ 0 the natural domain of g consists p of all real numbers that are nonzero. We may take a shortcut and refer to f as “ x” and refer to g as “1=x.” In such a case, it should be understood that the domain of the relevant function is its natural domain. Assume that f and g are both defined on a set U  R. We form the sum f C g and the product fg of f and g by performing these operations pointwise: For each x2U .f C g/ .x/ D f .x/ C g .x/ and .fg/ .x/ D f .x/ g .x/ : The quotient of f and g is also defined pointwise: If x 2 D and g .x/ ¤ 0 we set   f .x/ f .x/ D : g g .x/ If f .x/ is in the domain of g for each x 2 D we define the composition of f and g as the function f ı g (“f composed with g”) such that .f ı g/ .x/ D f .g .x// for each x 2 D: Recall that in general g ı f is different from f ı g (composition is not commutative). For example, if f .x/ D sin .x/ for each x 2 R and g .x/ D 1=x for each x ¤ 0 then     1 1 D sin for each x ¤ 0; .f ı g/ .x/ D f .g .x// D f x x whereas .g ı f / .x/ D g .f .x// D g .sin .x// D

1 sin .x/

for each x 2 R such that sin .x/ ¤ 0, i.e., for each x that is not an odd multiple of ˙=2. Special functions such as sine and cosine should be familiar from beginning calculus. We may refer to some of their properties as needed, even though they may not have been derived rigorously in such a course. We will outline rigorous proofs as appropriate machinery is developed.

1.1 Terminology and Notation

5

The trigonometric functions sine and cosine are periodic with period 2, i.e. sin .x C 2/ D sin .x/ and cos .x C 2/ D cos .x/ for each x 2 R. The number 2 is the fundamental period, i.e., the smallest positive period of these functions. The domain of sine and cosine is the set of all real numbers R. Since 1  sin .x/  1 and 1  cos .x/  1 for each x 2 R the range of both functions is the interval Œ1; 1 D fx 2 R W 1  x  1/g Figure 1.1 displays the graphs of sine and cosine on the interval Œ2; 2. y 1

Fig. 1.1

y = sin(x) 2π

π

−π

2

π

2π

x

1 y 1

y =cos(x) 2π

π

−π

2

π

2π

x

1

The function tangent is defined in terms of sine and cosine: tan .x/ D

sin .x/ if cos .x/ ¤ 0: cos .x/

The fundamental period of tangent is . Since the only points at which cosine vanishes are odd multiples of ˙=2 the domain of tangent consists of all real numbers x such that x ¤ ˙ .2n C 1/

 for any nonnegative integer n: 2

The range of tangent is the set of all real numbers R. Figure 1.2 shows the graph of tangent on the interval .3=2; 3=2/.

6

1 Real Numbers, Sequences, and Limits y 20

Fig. 1.2

10 -p

-3p/2

-p/2

p

p/2

x

3p/2

10 20

The natural exponential function exp is defined for all real numbers and attains all positive numbers (Fig. 1.3). Usually it is practical to use the exponential notation so that exp .x/ D ex > 0 for each x 2 R. Fig. 1.3 The natural exponential function

20

y

10

y = ex

1

2

1

1

2

3

x

The natural logarithm is the inverse of the natural exponential function: y D ln .x/ for x > 0 , x D ey : Thus the domain of the natural logarithm is the set of positive real numbers (Fig. 1.4). Fig. 1.4 The natural logarithm

y 2 1

y = ln(x)

( e , 1) 1

2

e 3

4

5

6

7

x

−2 −3 −4

If a > 0 is an arbitrary base then the exponential function with respect to the base a is defined as ax D ex ln.a/ for each x 2 R.

1.2 Real Numbers

7

The logarithm with respect to the base a is defined so that y D loga .x/ for x > 0 , x D ax : Thus logarithm with respect to the base a is the inverse of the exponential function ln.x/ with respect to the base. As derived in beginning calculusloga .x/ D ln.a/ for each x > 0:

1.2 Real Numbers In this section we will summarize the basic rules of arithmetic and the order properties of numbers.

1.2.1 Rules of Arithmetic The set of positive integers (natural numbers) 1; 2; 3; : : : will be denoted by N and the set of all integers : : : ; 3; 2; 1; 0; 1; 2; 3; : : : will be denoted by Z. Rational numbers are numbers which can be expressed as fractions of the form p=q where p and q are integers and q ¤ 0. The set of rational numbers will be denoted by Q. Even though the set of rational numbers is closed under arithmetic operations (sums, products and quotients of rational numbers are also rational numbers), it is not adequate for the purposes of calculus. Indeed, even simple geometric problems lead to irrational numbers, i.e., numbers which are not fractions of integers, as the ancient Greeks knew. For example, the length ofpthe diagonal of a square whose sides are of unit length is the irrational number 2: The circumference of a circle of unit diameter is the irrational number . We will refer to the set of all rational or irrational numbers as the set of real numbers and denote this set by R. We will take it for granted that the set R exists and that the familiar rules for the arithmetic operations of addition, subtraction, multiplication and division are valid: If x, y, and z are arbitrary real numbers 1. 2. 3. 4. 5.

x C y D y C x (addition is commutative) x C .y C z/ D .x C y/ C z (addition is associative) x C 0 D x (0 is the additive identity) x C .x/ D 0 (x is the additive inverse: we write x  y for x C .y/) xy D yx (multiplication is commutative)

8

1 Real Numbers, Sequences, and Limits

6. x .yz/ D .xy/ z (multiplication is associative) 7. 1x D x (1 is the multiplicative identity) 8. For each x ¤ 0 there exists 1=x such that x .1=x/ D 1: 1=x is the multiplicative inverse of x; we write x=y for x .1=y/ 9. x .y C z/ D xy C xz (distributive law) Thus, the set of real numbers R is a field. Note that the set of rational numbers Q is a subfield of R: When we add, subtract, multiply, or divide fractions of integers we can express the results also as fractions of integers. In the next section we will introduce a crucial property of R that is referred to as completeness that is lacking if we stay within the framework of Q. In the mean time, let us show that the set of rational numbers are inadequate even if we try to compute the square roots of certain integers: Proposition 1. There is no rational number x such that x2 D 2: Proof. Assume that there is such a rational number. We can assume that x > 0 and xD

m n

where m and n are positive integers. We can also assume that not both m and n are even since we can cancel any common factor in the numerator and denominator. Now, x2 D 2 )

m2 D 2 ) m2 D 2n2 : n2

Thus m2 is an even positive integer. If m were odd, we could have written m as 2kC1 for some nonnegative integer k so that   m2 D .2k C 1/2 D 4k2 C 4k C 1 D 2 2k2 C 2k C 1: This would have implied that m2 is odd as well. Therefore m must be even, say m D 2k for some positive integer k. Thus m2 D 2n2 ) .2k/2 D 2n2 ) 4k2 D 2n2 ) 2k2 D n2 : Therefore n2 is even. As in the case of m, this implies that n is also even. Thus we have started with the assumptions that m and n and are not both even, m2 =n2 D 2, and ended up with the conclusion that both are even. This is a contradiction. Therefore there is no rational number x such that x2 D 2. 

1.2 Real Numbers

9

1.2.2 The Order Axioms Real numbers are equipped with an order relationship. We write “a < b” to express that a is less than b and we write “a > b” to express that a is bigger than b. The order relationship has the following properties: 1. Given real numbers a and b exactly one of the relationships a D b; a < b, a > b holds. 2. If a < b then for any c 2 R we have a C c < b C c (we can add the same number to both sides of an inequality without changing the direction of the inequality). 3. If a > 0 and b > 0 then ab > 0 (multiplication of positive numbers yields a positive number). 4. If a > b and b > c then a > c (the inequality relationship is transitive). The above properties lead to the other familiar rules for inequalities. For example, a > b and c > 0 ) ac > bc (when we multiply both sides of an inequality by the same positive number the direction of the inequality is not changed), a > b and c < 0 ) ac < bc (when we multiply both sides of an inequality by the same negative number the direction of the inequality is reversed), 0 b or a D b In some proofs we will make use of the following proposition:

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1 Real Numbers, Sequences, and Limits

Proposition 2. If a < b C " for each " > 0 then a  b. If a > b  " for each " > 0 then a  b. Proof. It is sufficient to prove the first statement. Indeed, if a > b  " for each " > 0 then a < b C " for each " > 0: By the first statement we have a  b. Therefore a  b. Thus, let us assume that a < b C " for each " > 0: We will prove the contrapositive of the statement .a < b C " for each " > 0/ ) a  b: Thus, we will prove that a > b ) (there exists " > 0 such that a  b C ") Indeed, if we assume that a > b then "D

ab > 0: 2

We have bC"DbC

aCb aCa ab D < Da 2 2 2

so that a > b C ".  Corollary 1. If ja  bj < " for each " > 0 then a D b: Proof. We have ja  bj < " for each " > 0 , b  " < a < b C " for each " > 0: By Proposition 2 a  b and a  b: Therefore a D b. 

1.2 Real Numbers

11

1.2.3 The Number Line There is one–one correspondence between the set of real numbers and points on a line. Points on a line are associated with real numbers as follows: A point on the line is selected as the origin. The origin corresponds to 0. A unit length is selected and the point corresponding to 1 is placed at unit distance from the origin. The origin and the point corresponding to 1 determine the positive direction along the line, and the opposite direction is the negative direction. Usually we place the line horizontally and select the positive direction to the right. If x is a positive number the point corresponding to x is placed at a distance x from the origin, in the positive direction. If x is a negative number the corresponding point is at the distance x from the origin, in the negative direction. Thus, we establish a correspondence between the set of real numbers and a line. We will refer to the line as the number line and identify the number x with the point that corresponds to x. For example, we may refer to “the point 2” or “the number 2.” We have a < b if and only if a is to the left of b on the number line (assuming that the positive direction of the line is towards the right). Intervals are subsets of the set of real numbers which occur frequently in calculus. If a < b the open interval .a; b/ with endpoints a and b is the set of all points between a and b: .a; b/ D fx 2 R W a < x < bg : We will usually write .a; b/ D fx W a < x < bg; if it is clear that we are referring to subsets of the set of real numbers R. Note that the open interval .a; b/ does not contain the endpoints a and b. The closed interval Œa; b consists of the points which lie between a and b and the endpoints a and b W Œa; b D fx W a  x  bg : We may also consider half-open intervals of the form Œa; b/ D fx W a  x < bg ; .a; b D fx W a < x  bg : An unbounded interval that consists of all numbers less than a given number b is denoted by .1; b/: .1; b/ D fx W x < bg :

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1 Real Numbers, Sequences, and Limits

There is no need to try to attach a mystical meaning to the symbol 1: Within the context of intervals, the symbol merely indicates that the interval contains negative numbers whose distance from the origin is arbitrarily large. Similarly, .a; C1/ D fx W x > ag ; .1; b D fx W x  bg ; Œa; C1/ D fx W x  ag: If J denotes an arbitrary interval, the interior of J is the interval which is obtained by deleting those endpoints of J which belong to J: For example, the interior of the open interval .a; b/ coincides with itself, and the interior of Œa; b/ is the open interval .a; b/.

1.2.4 The Absolute Value and the Triangle Inequality The absolute value of a number is a measure of the distance of the corresponding point on the number line from the origin: Definition 1. If x is an arbitrary real number the absolute value of x is denoted by jxj and defined as  jxj D

x if x  0; x if x < 0:

For example, j3j D 3; j3j D  .3/ D 3: Given (real) numbers a and b; we have  ja  bj D

a  b if a  b; b  a if a < b:

Geometrically, ja  bj is the distance between the points a and b on the number line. For example, the distance between the points 2 and 4 is j2  4j D j2j D 2; and the distance between the points 1 and 5 is j1  5j D j4j D 4:

1.2 Real Numbers

13

Example 1. Let us express the set A D fx W jx  1j  2g as a union of intervals. The set A consists of all x whose distance from 1 is at least 2. This means that x  1 or x  3. Therefore, A is the union of the intervals .1; 1 and Œ3; C1/, i.e., A D .1; 1 [ Œ3; C1/: We can reach the same conclusion by working with the relevant properties of inequalities: if x  1 then jx  1j D x  1 so that jx  1j  2 ) x  1  2 ) x  3: Thus x 2 Œ3; 1/: If x < 1 then jx  1j D 1  x so that jx  1j  2 ) 1  x  2 ) 1  x: Thus x 2 .1; 1. Therefore A D .1; 1 [ Œ3; C1/.  Proposition 3. Assume that r > 0. Then jxj < r if and only if r < x < r. Similarly, jxj  r if and only if r  x  r. Proof. Since jxj is the distance of x from 0 we should have jxj < r if and only if r < x < r, as in the statement of the proposition. We can reach that conclusion by making use of the relevant properties of inequalities: Assume that jxj < r. If x  0 then x D jxj < r. Since r > 0 we have r < 0  x. Thus r < x < r. If x < 0 then x D jxj < r so that x > r. Since r > 0 we have x < 0 < r. Thus r < x < r. Conversely, assume that r < x < r. If x  0 then jxj D x < r. If x < 0 then jxj D x < r. The proof of the second statement is similar.  We will encounter intervals that are described in the following proposition frequently: Proposition 4. Assume that a 2R and r > 0. We have fx W jx  aj < rg D .a  r; a C r/:

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1 Real Numbers, Sequences, and Limits

Thus, fx W jx  aj < rg is the open interval of length 2r that is centered at a. We have fx W jx  aj  rg D Œa  r; a C r so that fx W jx  aj  rg is the closed interval of length 2r centered at a. Proof. Proposition 4 is an immediate consequence of Proposition 3: We have jx  aj < r , r < x  a < r and r < x  a < r , a  r < x < a C r Thus fx W jx  aj < rg D .a  r; a C r/ : The proof of the second statement is similar.  We will make use of the following fact about the absolute value: Proposition 5. The absolute value of a product is the product of the absolute values: jabj D jajjbj: Proof. We will consider the following cases: 1. 2. 3. 4.

a  0 and b  0; a  0 and b  0; a  0 and b  0 a  0 and b  0. In the first case, jaj D a, jbj D b, and ab  0 so that jabj D ab D jaj jbj : In the second case, jaj D a, jbj D b, and ab  0 so that jabj D  .ab/ D a .b/ D jaj jbj :

In the third case, jaj D a, jbj D b, and ab  0 so that jabj D  .ab/ D .a/ .b/ D jaj jbj :

1.2 Real Numbers

15

In the fourth case, jaj D a, jbj D b, and ab  0 so that jabj D ab D .a/ .b/ D jaj jbj :  We will use the triangle inequality frequently: Theorem 1 (The Triangle Inequality). If a and b are arbitrary real numbers ja C bj  jaj C jbj: Thus, the absolute value of a sum is less than or equal to the sum of the absolute values. Proof. Since a D jaj or a D  jaj, and b D jbj or b D  jbj, we have  jaj  a  jaj ;  jbj  b  jbj : Therefore,  jaj  jbj  a C b  jaj C jbj ; i.e.,  .jaj C jbj/  a C b  jaj C jbj : If a C b  0, ja C bj D a C b  jaj C jbj : If a C b < 0,  .jaj C jbj/  a C b ) jaj C jbj   .a C b/ D ja C bj : Therefore, in all cases, ja C bj  jaj C jbj :  Corollary 2 (Corollary to the Triangle Inequality). If a and b are arbitrary real numbers jjaj  jbjj  ja  bj:

16

1 Real Numbers, Sequences, and Limits

Proof. By the triangle inequality, jaj D ja  b C bj  ja  bj C jbj ; so that jaj  jbj  ja  bj : Similarly, jbj D jb  a C aj  jb  aj C jaj D ja  bj C jaj ; so that jbj  jaj  ja  bj ) jaj  jbj   ja  bj : Thus,  ja  bj  jaj  jbj  ja  bj : As in the proof of the triangle inequality, the above inequality leads to the inequality jjaj  jbjj  ja  bj : 

1.2.5 The Archimedean Property of R We will assume the principle of mathematical induction: Let S be a subset of the set of natural numbers N. Assume that N 2 S and n C 1 2 S if n 2 S . Then S D fn 2 N W n  Ng : In particular, if 1 2 S and n 2 S implies that n C 1 2 S then S D N. Definition 2. A subset S of R is said to be well-ordered (or “has the well-ordering property”) if each nonempty subset of S has a smallest element. Proposition 6. The set N of natural numbers has the well-ordering property. Proof. Assume that S is a nonempty subset of N that does not have a smallest element. Let T denote the complement of S. We will show that T is all of N and contradict the fact that S is nonempty. If 1 2 S then 1 is the smallest element. Therefore 1 does not belong to S. Hence 1 belongs to T. Let T 0 be the subset of

1.2 Real Numbers

17

T consisting of all n 2 T such that 1; 2; : : : ; n all belong to T. We will show that T 0 D N so that T D N as well. We have 1 2 T 0 . Assume that n 2 T 0 . We need to show that n C 1 2 T 0 as well. Assume that this is not the case. Thus n C 1 2 S. Since 1; 2; : : : ; n all belong to T which is the complement of S, the number n C 1 must be the smallest element of S. But we have assumed that S does not have a smallest element. Thus n C 1 2 T 0 . Therefore T 0 D N. so that T D N. Thus S is empty and we have reached a contradiction. Therefore S must have a smallest element as claimed.  Definition 3. Assume that S is a subset of R. We say that S has the Archimedean property if for each x 2 S there exists an integer n such that x < n: Note that the set of rational numbers Q has the Archimedean property. Indeed, if x 2 Q and x  0 we can set n D 1. If x > 0 and we express x as p=q where p and q are natural numbers. Then xD

p  p < p C 1: q

We will assume that the set of all real numbers has the Archimedean property: If x is an arbitrary real number there exists an integer n such that n > x: Proposition 7. If x and y are positive real numbers there exists a positive integer nsuch that nx > y. In particular, for each positive real number x there exists a positive integer m such that 1=m < x. Proof. By the Archimedean property of real numbers there exists a positive integer n such that n>

y x

Thus nx > y: In particular, for any x > 0 there exists a positive integer m such that mx > 1: Thus x > 1=m. 

1.2.6 Problems 1. Let A D fx 2 R W jx  1j < 3g Express A as an interval.

18

1 Real Numbers, Sequences, and Limits

2. Let A D fx 2 R W jx  2j  6g Express A as an interval. 3. Let A D fx 2 R W jx  4j > 2g Express A as a union of intervals. 4. Let A D fx 2 R W x  3 > 4g : Express A as an interval. 5. Let A D x 2 R W x C 3  5: Express A as an interval. 6. Assume that x  7. Show that 1 1  : x5 2 7. Assume that 3 < x < 4. Show that 3x 4. Show that x3  3x2  4 >

3 3 x: 16

9. Assume that x > 3. Show that ˇ ˇ ˇ ˇ 1 2h ˇ ˇ ˇ .x  2/ .x C 3/ ˇ < 3 jhj 10. Assume that x1  2; x2  2. Show that ˇ ˇ ˇ x2  x1 ˇ 1 ˇ ˇ ˇ x2 .x  1/ ˇ  4 jx2  x1 j 1 2

1.3 The Limit of a Sequence

19

1.3 The Limit of a Sequence The concept of the limit is fundamental in calculus. We will begin by discussing the limits of sequences.

1.3.1 The Definition of the Limit of a Sequence Let us begin by recalling the definition of a sequence: Definition 1. A sequence is a function whose domain is a subset of integers of the form fN; N C 1; N C 2; N C 3; : : :g, where N is a given integer. If we refer to this function as f , then f .n/ is usually denoted as an for n D N; N C 1; N C 2; : : :. We may denote a sequence as aN ; aNC1 ; aNC2 ; : : : ; an ; : : :, or fan g1 nDN , or simply as fan g if we don’t feel the need to specify the starting value N of the index n. Thus, the sequence 1 1 1 1 1; ; ; ; : : : ; ; : : : 2 3 4 n can be denoted as  1   1 1 : or n nD1 n The index n is a “dummy index” and can be replaced by any other letter. Thus,  1  1 1 1 and n nD1 k kD1 denote the same sequence. The starting value of the index can be an integer other than 1. For example, if we consider the sequence n 5 6 7 ; ; ;:::; ;:::; 1 2 3 n4 the starting value of the index is 5. We can denote the sequence as n n o1 . n  4 nD5 We may even refer to a sequence simply as “the sequence an .” In this case, it should be understood that the starting value of the index is its smallest value such that the

20

1 Real Numbers, Sequences, and Limits

expression an is defined. For example, if we refer to “the sequence n= .n  4/” , it is understood that the starting value of n is 5. The number an is referred to as the nth term of the sequence a1 ; a2 ; a3 ; : : : ; an ; : : :. Thus, 1=n is the nth term of the sequence f1=ng1 nD1 . In the sequence 5 6 7 n ; ; ;:::; ;::: 1 2 3 n4 the nth term is not n= .n  4/. In such a case we will refer to an as the term corresponding to n. The definitions of the graph and the range of a sequence are consistent with the view of a sequence as a function: Definition 2. The graph of the sequence fan g1 nDN is the set of points of the form .n; an / in the Cartesian coordinate plane, where n D N; N C1; N C2; : : :. The range of the sequence fan g1 nDN is the range of the function f such that f .n/ D an , n  N. Just as in the case of a function that is defined on an interval, the graph of a sequence helps us visualize the sequence. The graph of a sequence consists of isolated points in the plane, unlike the graph of a function that is defined at all points of an interval. We may also visualize a sequence simply by sketching its range on the number line. Example 1. Let an D

n ; n D 5; 6; 7; : : : n4

The graph of the sequence fan g1 nD5 is the set of points in the Cartesian coordinate plane in the form  n;

n ; n4

where n D 5; 6; 7; : : :. Figure 1.5 shows the points in the graph of the sequence corresponding to n D 5; 6; 7; : : : ; 20. Figure 1.6 shows the points in the range of the sequence corresponding to n D 5; 6; : : : ; 12.  y

Fig. 1.5 5 4 3 2 1

5

10

15

20

x

1.3 The Limit of a Sequence

21

Fig. 1.6

1

2

3

4

5

x

Informally, the limit of the sequence fan g1 nD1 exists and is the number L if an is as close to L as desired provided that n is sufficiently large. Here is the precise definition: Definition 3. The limit of the sequence fan g1 nD1 is L if for each " > 0 there exists a positive integer N such that jan  Lj < " if n  N: Example 2. Let an D

n ; n D 5; 6; : : : ; n4

as in Example 1. Show that limn!1 an D 1 (in accordance with the definition of the limit of a sequence). Solution. Let " > 0 be given. If n  N  5 then ˇ ˇˇ n  n C 4 ˇˇ ˇ n ˇ ˇ ˇD 4  4 :  1ˇ D ˇˇ jan  1j D ˇ n4 n4 ˇ n4 N 4 Thus, in order to have jan  1j < " for n  N it is sufficient to choose N so that 4 < ": N 4 This is the case if N4>

4 4 ,N > C4 " "

Such an integer N exists by the Archimedean property of real number. If n  N then jan  1j D

4 < ": N 4

Therefore, n D 1; n!1 n  4

lim an D lim

n!1

as claimed. 

22

1 Real Numbers, Sequences, and Limits

Proposition 1. The limit of a sequence is unique. Proof. Assume that limn!1 an D L1 and limn!1 an D L2 . We will prove that L1 D L2 by showing that jL1  L2 j < " for an arbitrary " > 0. Thus, let " > 0 be given. Since limn!1 an D L1 , there exists N1 2 N such that n  N1 ) jan  L1 j
0 be given. Since limn!1 an D L there exists N 2 N such that

n  N ) jan  Lj < ":1 There exists K 2 N such that nk  N if k  K. Thus k  K ) jank  Lj < ": Therefore limk!1 ank D L, as claimed.  Example 4. Show that the sequence n sin

 n o1 2

nD1

does not have a limit. Solution. Let’s see what the first few terms of the sequence look like: sin

 2

 ; sin ./ ; sin

3 2



 ; sin .2/ ; sin

5 2



 ; sin

6 2



 ; sin

7 2



 ; sin

8 2

i.e. 1; 0; 1; 0; 1; 0; 1; 0; 1; : : : We have  n  n   D sin C 2 D sin : sin .n C 4/ 2 2 2



 ; sin

9 2

 ;:::

24

1 Real Numbers, Sequences, and Limits

Thus, the pattern 1; 0; 1; 0 is repeated. The sequences 1; 1; 1; 1; : : : 0; 0; 0; 0; : : : 1; 1; 1; 1; : : : are subsequences of the given sequence and have the limits 1, 0, and 1, respectively. Since we displayed subsequences with different limits, the sequence does not have a limit. 

1.3.2 The Limits of Combinations of Sequences Now we will establish the rules about the limits of certain combinations of sequences. These rules are intuitively plausible and you must have been using them since beginning Calculus. Here we will provide rigorous proofs. Proposition 3. The limit of a constant sequence c is c. Proof. Assume that an D c for each n 2 N. We need to show that limn!1 an D c. Let " > 0 be given. We have jan  cj D jc  cj D 0 < " for each n  1. Therefore, limn!1 an D c. Proposition 4 (The Constant Multiple Rule for Limits). Assume that c is a constant and limn!1 an exists. Then lim .can / D c lim an :

n!1

n!1

Proof. If c D 0 then can D 0 for each n so that lim .can / D lim .0/ D 0:

n!1

n!1

Thus, let us assume that c ¤ 0 and that limn!1 an D L. Let " > 0 be given. Since limn!1 an D L there exists N 2 N such that n  N ) jan  Lj
0 be given. There exists N1 2 N and N2 2 N such that n  N1 ) jan  Aj
0 be given. Since limn!1 an D L1 and limn!1 bn D L2 , there exists N 2 N such that n  N ) jan  L1 j
0 be arbitrary. Since limn!1 an D L1 and limn!1 bn D L2 , there exists N 2 N such that jan  L1 j
0. Thus L1  L2 < jL1  aN j C jbN  L2 j
0 there exists N 2 N such that n  N ) jan  anCk j < " for k D 1; 2; 3; : : : Intuitively, a sequence satisfies the Cauchy condition if its terms are arbitrarily close to each other provided that they correspond to indices that are sufficiently large, no matter how far apart they may be. Þ Proposition 1. A sequence that converges is a Cauchy sequence. Proof. Assume that limn!1 an D L. For a given " > 0 there exists N 2 N such that jan  Lj
0. Thus we are able to determine arbitrarily small intervals that contain the limit x and approximate x with desired accuracy even though we may not know the exact value of x. Þ Proposition 3. Any real number x can be represented by a decimal. Proof. It is sufficient to assume that x is positive. By the Archimedean property of real numbers there exists a positive integers m > x. By the well-ordering property of positive integers there exists a smallest such integer. Let us label that integer as a0 C 1 so that a0  x < a0 C 1. Let us subdivide the interval Œa0 ; a0 C 1/ into 10 subintervals 

j jC1 where j D 0; 1; 2; : : : ; 9: a0 C ; a0 C 10 10 The number x belongs to exactly one of these disjoint subintervals of Œa0 ; a0 C 1/. Let us label that interval as 

a1 a1 C 1 a0 C ; a0 C 10 10 so that a1 2 f0; 1; 2; : : : 9g. We proceed to produce a sequence of intervals 

an an a1 a1 an C 1 C    C n ; a0 C CC n C Jn D a 0 C 10 10 10 10 10nC1 such that x 2 Jn for each n. The “decimal digits” a1 ; a2 ; : : : ; an ; : : : are integers between 0 and 9. We have shown that the sequence Sn D a0 C

an a1 C    C n ; n D 1; 2; 3; : : : 10 10

is a Cauchy sequence (Proposition 2). Therefore there exists y 2 R such that y D lim Sn n!1

Since Sn  x < SnC1 for each n

1.4 The Cauchy Convergence Criterion

35

we have y D lim Sn  x  lim SnC1 D y: n!1

n!1

Therefore  an a1 a0 C CC n : n!1 10 10

x D y D lim Note that

ˇ  an ˇˇ a1 1 ˇ C C n ˇ < n: ˇx  a0 C 10 10 10 Indeed, ˇ   a1 a1 10 1 an ˇˇ an an C 1 ˇ  nC1 D n : C    C n ˇ D x  a0 C C C n < ˇx  a0 C 10 10 10 10 10nC1 10 10

 Remark 3. The procedure described above produces one of the possible decimal expansions for x. We can obtain another decimal expansion for x by modifying the procedure whereby we consider intervals   j jC1 a0 C ; a0 C where j D 0; 1; 2; : : : ; 9 10 10 that are closed on the right. For example, 1 D 0:125000 : : : 8 and 1 D 0:124999 : : : 8 as well. Þ

p Example 2. We showed p that 2 is irrational (Proposition 1 of Sect. 1.2). Proposition 3 shows that 2 has a decimal expansion. System is p A Computer Algebra p capable of displaying decimal expansions of 2 that approximate 2 with desired accuracy. For example, p 2 Š 1:414213562373095048801688724209698078570 

36

1 Real Numbers, Sequences, and Limits

Definition 2. A sequence fJn g1 nD1 is a nested sequence of intervals if JnC1  Jn for n D 1; 2; 3; : : : The Cauchy convergence criterion implies the following important fact that has nice geometric content: Theorem 1. Assume that fŒan ; bn g1 nD1 is a nested sequence of closed intervals such that lim .bn an / D 0:

n!1

Then there exists a unique real number x such that x2 Œan ; bn  for each n (Fig. 1.7). Thus \1 nD1 Œan ; bn  D fxg: We have lim an D lim bn D x:

n!1

n!1

Fig. 1.7

Proof. Since ŒanCk ; bnCk  is contained in Œan ; bn  for any positive integers n and k we have janCk  an j D anCk  an  bn  an : Since limn!1 .bn  an / D 0, given " > 0 we can choose N so that 0 < bn  an < " if n  N: Therefore janCk  an j  bn  an < " if n  N and k D 1; 2; 3; : : :. Thus fan g1 nD1 is a Cauchy sequence. Therefore there exists x such that limn!1 an D x: Similarly, jbnCk  bn j D bn  bnCk  bn  an so that fbn g1 nD1 is a Cauchy sequence as well. Therefore there exists y such that limn!1 bn D y. We have jy  xj D lim jbn  an j D 0 n!1

so that y D x. Thus limn!1 an D limn!1 bn D x.

1.4 The Cauchy Convergence Criterion

37

Since an  anCk < bnCk  bn for k D 1; 2; 3; : : : , we have an  lim anCk D x D lim bnCk  bn k!1

k!1

Thus x 2 Œan ; bn  for each n. The number x is the unique such number: Assume that y 2 Œan ; bn  for each n. Let " > 0 be arbitrary. Since limn!1 .bn  an / D 0 there exists N 2 N such that bN  aN < ".. Since x and y are both in ŒaN ; bN  we have jx  yj  bN  aN < ": Since " is arbitrary we must have y D x. Thus \1 nD1 D fxg :  The Cauchy convergence principle is one of the ways the completeness of real numbers is expressed. With reference to Theorem 1, you can imagine that the holes on the number line are pugged once we augment the field of rational numbers with irrational numbers.

1.4.2 Irrational Numbers Are Uncountable Definition 3. A set is countable if its elements can be listed as a sequence s1 ; s2 ; s3 ; : : : Proposition 4. The set of rational numbers is countable. Proof. We can list integers as 0; 1; 1; 2; 2; 3; 3; 4; 4; : : : , and fractions with corresponding denominators, expressed in lowest terms : 1 1 1 2 1 2 1 3 1 3 0; 1; 1; 2; 2; ;  ; 3; 3; ; ;  ;  ; 4; 4; ; ;  ;  ; : : : 2 2 3 3 3 3 4 4 4 4  Even though rational numbers are countable, there are infinitely many rational numbers in any interval. The relevant term is described by the following definition:

38

1 Real Numbers, Sequences, and Limits

Definition 4. A subset S of R is dense in R if, given any interval .a; b/, there exists s 2 S such that s 2 .a; b/ : Proposition 5. The set of rational numbers is dense in R. Proof. By the Archimedean property of real numbers there exists a positive integer n such that 1 < b  a: n Again by the Archimedean property of real numbers, and the well-ordering of positive integers, there exists a smallest positive integer k such that k > na so that

k >a n

Thus k k1 a< : n n We have a
an or yn < an ;

1.4 The Cauchy Convergence Criterion

39

(iii) yn  xn
an or yn < an . If xn > an and m  n then

jx  an j D lim jxm  an j D lim .xm  an /  xn  an > 0: m!1

m!1

Therefore x ¤ an . If yn < an and m  n then jx  an j D lim jym  an j D lim .an  ym /  an  yn > 0: m!1

m!1

Therefore x ¤ an . 1 Now let us carry out the construction of the sequences fxn g1 nD1 and fyn gnD1 with properties (i), (ii) and (iii): Since .x0 ; C1/ [ .1; y0 / D R we have a1 > x0 or a1 < y0 . If a1 > x0 set x1 be a rational number such that

x0 < x1 < min .a1 ; y0 / (such a number exists since we showed that the set of rational numbers is dense in R). Let y1 be a real number such that x1 < y1 < min .a1 ; y0 ; x1 C 1/ :

40

1 Real Numbers, Sequences, and Limits

Then x0  x1 < y1 < y0 and y1  x1 < 1: If a1 < y0 let y1 be a rational number with max .a1 ; x0 / < y1 < y0 and let x1 be a rational number with max .a1 ; x0 ; y1  1/ < x1 < y1 : In all cases, we have x0  x1 < y1  y0 ; x1 > a1 or y1 < a1 ; y1  x1 < 1 Now assume that we have constructed x0 ; x1 ; x2 ; : : : ; xn1 and y0 ; y1 ; : : : ; yn1 such that xk1  xk < yk  yk1 ; xk > ak or yk < ak ; yk  xk
xn1 or an < yn1 . If an > xn1 we set xn to be a rational number such that xn1 < xn < min .an ; yn1 / and let yn be a rational number such that   1 : xn < yn < min an ; yn1 ; xn C n If an < yn1 we set yn to be a rational number such that max .an ; xn1 / < yn < yn1

1.4 The Cauchy Convergence Criterion

41

and let xn be a rational number such that   1 max an ; xn1 ; yn  < xn < yn : n In all cases xn1  xn < yn < yn1 ; xn > an or yn < an ; yn  xn
x  ". Since aN 2 S the number x is the least upper bound of S.

1.5 The Least Upper Bound Principle

45

Now let us carry out the construction of the nested intervals with the required properties: Since S is nonempty we can pick a 1 2 S. If x  M for each x 2 S we can pick a real number b1 > M. Thus b1 is an upper bound of S such that b1 > a1 . If the midpoint m of the interval Œa1 ; b1  is an upper bound for S we set b2 D m D a1 C

1 .b1  a1 / and a2 D a1 : 2

Fig. 1.10

If there exists a 2 S such that a is greater than the midpoint of Œa1 ; b1  we set a2 D a and b2 D b1 :

Fig. 1.11

Having constructed a1 ; : : : ; an and b1 ; : : : ; bn with properties (i) and (ii), if the midpoint m of Œan ; bn  is an upper bound for S we set bnC1 D an C

1 .bn  an / and anC1 D an : 2

Fig. 1.12

If there exists a 2 S such that a is greater than the midpoint of Œan ; bn  we set anC1 D a and bnC1 D bn :

Fig. 1.13

Note that 0 < bn  an 

1 .b1  a1 / 2n1

46

1 Real Numbers, Sequences, and Limits

Thus we have constructed a sequence of nested intervals fjan ; bn jg such that lim .bn  an / D 0;

n!1

as desired.  The statement about the existence of a greatest lower bound is established in a similar manner. 

1.5.2 Monotone Sequences Let us begin by introducing some terminology. A sequence fan g1 nD1 is nondecreasing if an  anC1 for each n. We will simply say that such a sequence is increasing. The sequence is strictly increasing if an < anC1 for each n. Similarly, a sequence fan g1 nD1 is said to be nonincreasing if an  anC1 for each n. We will simply say that such a sequence is decreasing. The sequence is strictly decreasing if an > anC1 for each n. In any of these cases we may refer to the sequence as a monotone sequence. The sequence fan g1 nD1 is bounded above if there exists a number M such that an  M for each n 2 N. We refer to any such number as an upper boundfor the sequence. A sequence fan g1 nD1 is bounded below if there exists a number m such that m  an for each n 2 N. We refer to any such number as a lower bound for the sequence. The least upper bound principle leads to the monotone convergence principle: Theorem 2 (The Monotone Convergence Principle). A monotone increasing sequence of real numbers fan g1 nD1 that is bounded above has a limit L and an  L for each n 2N. A monotone decreasing sequence of real numbers fan g1 nD1 that is bounded below has a limit L and an  L for each n 2N. Proof of Theorem 2. Assume that fan g1 nD1 is an increasing sequence that is bounded above. By the least upper bound principle (Theorem 1) the set fan W n D 1; 2; 3; : : :g has a least upper bound L. We claim that L must be the limit of the sequence fan g1 nD1 : Since L is an upper bound for fan W n D 1; 2; 3; : : :g we have an  L for each n: Let " > 0 be given. Since L is the least upper bound of fan W n D 1; 2; 3; : : :g there exists N such that L  " < aN  L: For any n  N L  " < aN  an  L:

1.5 The Least Upper Bound Principle

47

since L is an upper bound for fan W n D 1; 2; 3; : : :g and fan g1 nD1 is increasing (Fig. 1.14). Fig. 1.14

Thus jL  an j < " if n  N: Therefore limn!1 an D L. Now assume that fan g1 nD1 is a decreasing sequence that is bounded below. Then the set fan W n D 1; 2; 3; : : :g has a greatest lower bound l. We claim that l must be the limit of the sequence fan g1 nD1 : Since l is a lower bound for fan W n D 1; 2; 3; : : :g we have an  l for each n: Let " > 0 be given. Since l is the greatest lower bound of fan W n D 1; 2; 3; : : :g there exists N such that l  aN < l C ": For any n  N l  an  aN < l C " since l is a lower bound for fan W n D 1; 2; 3; : : :g and fan g1 nD1 is decreasing (Fig. 1.15). Fig. 1.15

Thus jl  an j < " if n  N: Therefore limn!1 an D l.  Example 1. Set Sn D 2 C

1 1 1 1 1 C C C C    C ; n D 2; 3; 4; : : : 2 6 4Š 5Š nŠ

Make use of the monotone convergence principle to show that the sequence has a limit.

48

1 Real Numbers, Sequences, and Limits

Solution. The sequence is clearly an increasing sequence: SnC1 D 2 C

1 1 1 1 1 1 1 C C C CC C D Sn C > Sn : 2 3Š 4Š 5Š nŠ .n C 1/Š .n C 1/Š

In order to apply the monotone convergence principle we need to show that it is bounded above. If k  2 kŠ D .2/ .3/ .4/ : : : .k  1/ .k/  .2/ .2/ : : : .2/ D 2k1 so that 1 1  k1 : kŠ 2 Therefore 1 1 1 1 1 C C C CC 2 3Š 4Š 5Š nŠ 1 1 1 1 1  2 C C 2 C 3 C 4 C    C n1 2 2 2 2 2   1 1 1 1 1 D 1 C 1 C C 2 C 3 C 4 C    C n1 2 2 2 2 2

Sn D 2 C

1   1 n 2 D 1C D1C2 1 n b. Assume that x < a. Since limn!1 an D a there exists N such that aN > x. Therefore x … ŒaN ; bN . Similarly, if x > b there exists N 0 such that bN 0 < x. Thus x … ŒaN 0 ; bN 00 . Therefore Œa; b D \1 nD Œan ; bn .  Definition 2. A point x is a cluster point (or accumulation point) of the sequence fxn g1 nD1 if given any " > 0 there are infinitely many indices n such that jxn  xj < ": Thus, given any open interval J centered at x and any integer n there exists m > n such that xm 2 J. We can characterize a cluster point of fxn g1 nD1 in terms of its convergent subsequences: Proposition 2. A point x is a cluster point of the sequence fxn g1 nD1 if and only if there exists a subsequence fxnk g1 such that lim x D x: k!1 nk nD1 Proof. Assume that x is a cluster point of the sequence fxn g1 nD1 . There exists xn1 such that jxn1  xj < 1: There exists n2 > n1 such that jxn2  xj
nk such that ˇ ˇx n

kC1

ˇ  xˇ
0 there exists K such that ba < ": 2K

52

1 Real Numbers, Sequences, and Limits

Thus jxnk  xj 

ba ba  K < " for each k  K: 2k 2

Therefore lim xnk D x:

k!1

Thus x is a cluster point of fxn g1 nD1 .  Definition 3. A subset F of R is sequentially compact if, given any sequence 1 fxn g1 nD1 , where each xn 2 F, there is a subsequence fxnk gkD1 and x 2 F such that lim xnk D x:

k!1

Theorem 5. A closed and bounded interval Œa; b is sequentially compact. Proof. Assume that a sequence fxn g1 nD1 is given, where each xn 2 Œa; b. By the Bolzano-Weierstrass Theorem (Theorem 4), there exists a convergent subsequence fxnk g1 kD1 . Assume that limk!1 xnk D x. Since a  xnk  b for each k, we have a  lim xnk  b k!1

i.e., x 2 Œa; b. 

1.5.3 Problems 1. Let an D 1 C

1 1 1 C 3 C    C 3 for n D 1; 2; 3; : : : 23 3 n

Make use of the monotone convergence principle to prove that the sequence fan g has a limit. p 2. Let a1 D 1 and anC1 D 3an for each n 2 N. a) Make use of the monotone convergence principle to prove that the sequence fan g has a limit. b) Determine limn!1 an : 3. Let a1 D 2 and anC1

  1 2 an C for n D 1; 2; 3; : : : D 2 an

1.6 Infinite Limits

53

a) Make use of the monotone convergence principle to prove that the sequence fan g has a limit. b) Determine limn!1 an :

1.6 Infinite Limits In this section we will discuss cases in which the terms of a sequence attain positive or negative values of arbitrarily large magnitude.

1.6.1 The Definition of the Infinite Limit Definition 1. We say that the limit of the sequence fan g1 nD1 is C1 (or an tends to C1) and write limn!1 an D C1 if, given any M > 0, there exists a positive integer N such that an > M if n  N. We say that the limit of the sequence fan g1 nD1 is 1 (or an tends to 1) and write limn!1 an D 1 if, given any M > 0 there exists a positive integer N such that an < M if n  N. Remark 1 (Caution). In any of the cases covered by Definition 1 the limit of the sequence fan g1 nD1 does not exist as a (finite) real number. Indeed, if limn!1 an D L, then the sequence fan g1 nD1 is bounded. In the context of Definition 1 we are using the same word “limit” and the same symbol “lim” to indicate a different behavior of the terms of the sequence. This “doublespeak” is traditional and convenient, and we will use it. The particular context should clarify which usage of the word “limit” we have in mind. Nevertheless, if there is any possibility of confusion, we may stress that we are talking about a “finite limit,” or an “infinite limit” in the sense of Definition 1. Þ Example 1. Show that a) n2 D C1; n!1 n C 1 lim

b) n2 D 1: n!1 1  n lim

Solution. a) We have n2 D nC1

n n2 D  1 1 1C n 1C n n

54

1 Real Numbers, Sequences, and Limits

and 1C

1 2 n

so that 1 1C

1 n



1 : 2

Therefore n2 D nC1

n 1 1C n



N n  2 2

if n  N. Thus, in order to have n2 >M nC1 for a given M > 0 it is sufficient to choose a positive integer N such that N=2 > M. If n  N then n N n2   > M: nC1 2 2 Therefore n2 D C1; n!1 n C 1 lim

as claimed. b) Note that n2 1n is defined if n  2. We need to show that n2 < M 1n for a given M > 0 if n is greater than a sufficiently large N. We have n2 n2 n2 < M ,  >M, > M: 1n 1n n1

1.6 Infinite Limits

55

Now, n2 D n1

n2 n D  : 1 1 1  n 1 n n

We have 0 0 it is sufficient to choose a positive integer N such that N  2 and N  M. 

1.6.2 Propositions for the Evaluation of Infinite Limits When we wish to prove that limn!1 an D 1 it is usually more convenient to show that limn!1 .an / D C1, as in our response to part b) of Example 1. Let us make a note of this: Proposition 1. We have lim an D 1 if and only if lim .an / D C1:

n!1

n!1

Proof. Let us assume that limn!1 .an / D C1 and show that limn!1 an D 1 (the proof of the converse is similar). Let M > 0 be given. Since limn!1 .an / D C1, there exists N 2 N such that an > M if n  N. Then an < M if n  N: Therefore limn!1 an D 1. 

56

1 Real Numbers, Sequences, and Limits

A few such general observations are helpful in dealing with infinite limits. Proposition 2. Assume that an > 0 if n  N0 2 N and limn!1 an D 0. Then lim

n!1

1 D C1: an

Proof. Let M > 0 be given. Since an > 0 if n  N0 2 N and limn!1 an D 0, there exists N  N0 such that 1 1 ) > M if n  N: M an

0 < an < Therefore

lim

n!1

1 D C1 an

as claimed.  Proposition 3. Assume that limn!1 an limn!1 bn D C1. Then

> 0 or limn!1 an

D C1 and

lim an bn D C1:

n!1

Proof. If limn!1 an D L > 0 there exists N1 2 N such that jan  Lj


L L ) an > if n  N1 : 2 2

Let M > 0 be given. Since limn!1 bn D C1 there exists N2 2 N such that bn >

  2 M if n  N2 : L

Therefore, if n  N D max .N1 ; N2 / then an bn >

   2 L M D M: 2 L

Thus, limn!1 an bn D C1, as claimed.

1.6 Infinite Limits

57

If limn!1 an D C1, there exists N1 2 N such that an > 1 if n  N1 : Let M > 0 be given. Since limn!1 bn D C1 there exists N2 2 N such that bn > M if n  N2 : Therefore, if n  N D max .N1 ; N2 / then an bn > M: Thus, limn!1 an bn D C1.  Remark 2 (Caution). Thanks to Proposition 3, we can declare that a .C1/ D C1 for any real number a > 0 and .C1/ .C1/ D C1: On the other hand, the expression .0/ .C1/ is indeterminate, as you have seen in beginning Calculus. If limn!1 an D 0 and limn!1 bn D 1 then limn!1 an bn can be any real number, C1 or 1. Þ Proposition 4. Assume that limn!1 an D L (finite) or limn!1 an D C1 and limn!1 bn D C1. Then lim .an C bn / D C1:

n!1

Proof. Assume that limn!1 an D L and limn!1 bn D C1. Let M > 0 be given. There exists N1 such that if n  N1 then jan  Lj < 1 ) 1 < an  L < 1 ) an > L  1: Since limn!1 bn D C1 there exists N2 such that if n  N2 then bn > M  .L  1/ : Let us set N D max .N1 ; N2 /. If n  N then an C bn > .L  1/ C M  .L  1/ D M: Therefore limn!1 .an C bn / D C1:

58

1 Real Numbers, Sequences, and Limits

If limn!1 an D C1 there exists N1 such that an > 0 if n  N1 . Given M > 0 there exists N2 such that if n  N2 then bn > M: Let us set N D max .N1 ; N2 /. If n  N then an C bn > M: Therefore limn!1 .an C bn / D C1.  Example 2. Show that  lim

n!1

n3 n3  3 n C 1 n C 4n C 1

 D C1:

Solution. We have n3 n3 n2  D lim D lim  D C1 1 n!1 n C 1 n!1 n!1 1 1 C n 1C n n lim

and  lim 

n!1

 n3 D  lim n!1 n3 C 4n C 1

n3  4 n3 1 C 2 C n

1 n3

 D  lim

n!1

1 4 1C 2 C n

1 n3

D  1:

By Proposition 3  lim

n!1

n3 n3  3 n C 1 n C 4n C 1

 D C1:

 Remark 3 (Caution). Thanks to Proposition 4, we can declare that a C 1 D 1 for any real number a and 1 C 1 D 1: On the other hand, an expression such as 1  1 is “indeterminate,” as you have seen in beginning calculus. If limn!1 an D 1 and limn!1 bn D 1 then limn!1 .an  bn / can be any real number, C1 or 1. Þ

1.6 Infinite Limits

59

1.6.3 Problems 1. Assume that limn!1 .an / D C1. Prove that lim an D 1:

n!1

In problems 2–5 prove the assertion in accordance with the relevant precise definition: 2.   lim 2n3  n D C1

n!1

3. n2 D C1 n!1 n C 1 lim

4. n4  3 D C1 n!1 n2  2 lim

5. Prove that 2n3 D 1: n!1 4  n2 lim

You may use the result of problem 1.

Chapter 2

Limits and Continuity of Functions

2.1 Continuity As you have seen in beginning calculus, informally, a function is said to be continuous if its values change by small amounts corresponding to small changes in the value of its independent variable. We will give the precise definition. The discussion will be restricted to functions that are defined on intervals or unions of intervals since our discussion in later chapters will involve functions defined on such sets. We will introduce the important concept of the uniform continuity of a function on an interval. We will also discuss the continuity of some basic functions and their combinations.

2.1.1 The Definition of Continuity Definition 1. Assume that f is a real-valued function that is defined in an open interval that contains the point x0 . We say that f is continuous at x0 if for each " > 0 there exists ı > 0 such that jf .x/  f .x0 /j < " if jx  x0 j < ı: You can think of the number ı as a positive number that is sufficiently small so that the magnitude of the difference between f .x/ and f .x0 / is smaller than a given positive number " that can be arbitrarily small, i.e., as small as desired, provided that the distance between x and x0 is less than ı. You can also think of " > 0 as an “error tolerance” in approximating f .x0 / with the value of f at a nearby point x. Note that jf .x/  f .x0 /j < " , f .x0 /  " < f .x/ < f .x0 / C " © Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0_2

61

62

2 Limits and Continuity of Functions

and jx  x0 j < ı , x0  ı < x < x0 C ı: Thus, f is continuous at x0 if f .x/ is between f .x0 /  " and f .x0 / C " provided that x is between x0  ı and x0 C ı. Remark 1. Note that the “ 0 be given. Let us set  ı D min 1;

 " : 1 C 2 jx0 j

If jx  x0 j < ı then  jf .x/  f .x0 /j < .1 C 2 jx0 j/ jx  x0 j < .1 C 2 jx0 j/

" 1 C 2 jx0 j

 D ":

Therefore, f is continuous at x0 , as claimed. Note that the choice of ı for a given " is not unique. For example, we could have restricted x so that jx  x0 j < 0:1. Then jxj D j.x  x0 / C x0 j  jx  x0 j C jx0 j < 0:1 C jx0 j so that jf .x/  f .x0 /j  .jxj C jx0 j/ jx  x0 j < .0:1 C jx0 j C jx0 j/ jx  x0 j D .0:1 C 2 jx0 j/ jx  x0 j : Then we are led to the choice  ı D min 0:1;

" 0:1 C 2 jx0 j



so that jf .x/  f .x0 /j < " if jx  x0 j < ı:  Remark 2. The definition of continuity can be rephrased as follows by setting x D x0 C h: A function f that is defined in an open interval containing x0 is continuous at x0 if, given any " > 0, there exists ı > 0 such that jf .x0 C h/  f .x0 /j < " provided that jhj < ı: In some cases this expression may be more convenient in confirming the continuity of a function. Þ Example 2. Let f .x/ D x3 for each x 2 R. Show that f is continuous at any x0 2 R.

64

2 Limits and Continuity of Functions

Solution. We have ˇ ˇ ˇ ˇ ˇ ˇ jf .x0 C h/  f .x0 /j D ˇ.x0 C h/3  x30 ˇ D ˇx30 C 3x20 h C 3x0 h2 C h3  x30 ˇ ˇ ˇ D jhj ˇ3x20 C 3x0 h C h2 ˇ    jhj 3x20 C 3 jx0 j jhj C h2 : Let us restrict h so that jhj < 1. Then,   jf .x0 C h/  f .x0 /j < jhj 3x20 C 3 jx0 j C 1 Thus, given " > 0 we can set  ı D min 1;

 " : 3x20 C 3 jx0 j C 1

If jhj < ı then     jf .x0 C h/  f .x0 /j < jhj 3x20 C 3 jx0 j C 1 < ı 3x20 C 3 jx0 j C 1    2  " < 3x0 C 3 jx0 j C 1 D ": 2 3x0 C 3 jx0 j C 1 Therefore, f is continuous at x0 .  In the above example we made a choice for ı that yielded the inequality jf .x0 C h/  f .x0 /j < ": We could have made another choice for ı as min .1; "/. Then jhj < ı implies that     jf .x0 C h/  f .x0 /j < jhj 3x20 C 3 jx0 j C 1 < 3x20 C 3 jx0 j C 1 ": Would that be sufficient to prove the continuity of f at x0 ? Indeed it would: Once we have such an inequality, given " > 0, we can easily amend the choice of ı by setting  ı D min 1;

" 2 3x0 C 3 jx0 j C 1



in order to have jf .x0 C h/  f .x0 /j < " if jhj < ı. There are “one-sided” versions of continuity: Definition 2. Assume that f .x/ is defined on an interval Œx0 ; x0 C ı0 / for some ı0 > 0. The function f is continuous at x0 from the right if for any " > 0 there exists ı > 0 such that jf .x/  f .x0 /j < " if x0  x < x0 C ı. We say that f is

2.1 Continuity

65

continuous at x0 from the left If f .x/ is defined on an interval .x0  ı0 ; x0  for some ı0 > 0 and for any " > 0 there exists ı > 0 such that jf .x/  f .x0 /j < " if x0  ı < x  x0 . Assume that f .x/ is defined for each x in an interval J. We will say that f is continuous on J if f is continuous at any point in the interior of J (i.e., a point of J that is not an endpoint of J) and f is continuous from the right or from the left at an endpoint of J that belongs to J, depending on which concept is applicable. p Example 3. Let f .x/ D x for each x  0. Show that f is continuous on the interval Œ0; C1/: Solution. Assume that x > 0 and that jhj < x. Then p p ! p p p p xChC x f .x C h/  f .x/ D x C h  x D xCh x p p xChC x .x C h/  x h D p p D p p : xChC x xChC x Since p

xChC

p p x> x

we have jf .x C h/  f .x/j D p

jhj jhj p < p x xChC x

if jhj < x. Given " > 0 in order to have jf .x C h/  f .x/j < " it is sufficient to have p jhj p < " , jhj < x": x Thus we will set  p  ı D min x; x" : If jhj < ı then p x" ı jhj p p p D ": < D .x C h/  f .x/j < jf x x x Therefore f is continuous at each x in the interior of the interval Œ0; C1/: Now let us consider the continuity of f at the endpoint 0 of Œ0; C1/. The relevant concept is continuity at 0 from the right. We need to be able to choose ı > 0 so that jf .x/  f .0/j D

p x 0 let us set ı D "2 . If 0  x < "2 then 0 

p x < ":

Thus f is continuous at 0 from the right. Therefore we have shown that f is continuous on the interval Œ0; C1/.  There is a connection between the concepts of continuity and limits of sequences: Theorem 1 (The Sequential Characterization of Continuity). Assume that f is defined in an open interval that contains x0 . The function f is continuous at x0 if and only if lim xn D x0 ) lim f .xn / D f .x0 /

n!1

n!1

Proof. Assume that f is continuous at x0 . Let fxn g1 nD1 be a sequence such that limn!1 xn D x0 . Given " > 0 there exists ı > 0 such that jf .x/  f .x0 /j < " if jx  x0 j < ı. Since limn!1 xn D x0 there exists a positive integer N such that jxn  x0 j < ı if n  N: In that case jf .xn /  f .x0 /j < ". Since we have shown that for a given " > 0 there exists N 2 N such that jf .xn /  f .x0 /j < " we have limn!1 f .xn / D f .x0 / : Conversely, assume that limn!1 f .xn / D f .x0 / for any sequence fxn g1 nD1 such that limn!1 xn D x0 2 D. We would like to prove the continuity of f at x0 . We will prove the contrapositive of the implication. Thus, assume that f is not continuous at x0 . We will show that the statement about sequences that converge to x0 is not true: Since f is not continuous at x0 , there exists "0 > 0 such that for any ı > 0 there exists x where jx  x0 j < ı and jf .x/  f .x0 /j  "0 : Thus, for any n 2 N there exists xn such that jxn  x0 j
0 that works for all points in D: Definition 3. A function f is uniformly continuous on an interval J  R if, given any " > 0 there exists ı > 0 such that jf .x1 /  f .x2 /j < " if x1 and x2 are in J and jx1  x2 j < ı. Example 5. Let f .x/ D 1=x. Show that f is uniformly continuous on Œ1=2; C1/.

68

2 Limits and Continuity of Functions

Solution. Assume that x1  1=2 and x2  1=2. We have ˇ ˇ ˇ ˇ ˇ1 1 ˇ ˇ x1  x2 ˇˇ jx2  x1 j D : jf .x2 /  f .x1 /j D ˇˇ  ˇˇ D ˇˇ ˇ x2 x1 x1 x2 x1 x2 Since x1  1=2 and x2  1=2 we have x1 x2  1=4. Therefore jf .x2 /  f .x1 /j D

jx2  x1 j jx2  x1 j D 4 jx2  x1 j :  x1 x2 1=4

Given " > 0 let us set ı D "=4. If x1  1=2 and x2  1=2 and jx1  x2 j < ı then jf .x2 /  f .x1 /j  4 jx2  x1 j < 4ı D 4

" 4

D ":

Thus f is uniformly continuous on Œ1=2; 1/.  Remark 4. As in the case of continuity at a point, we can rephrase uniform continuity as follows: A function f is uniformly continuous on the interval J if, given any " > 0, there exists ı > 0 such that x 2 J; x C h 2 J and jhj < ı ) jf .x C h/  f .x/j < ": Þ There is a sequential characterization of uniform continuity: Theorem 2. A function f W J !R is uniformly continuous on the interval J if and only if the following condition is satisfied: 1 If fun g1 nD1 and fvn gnD1 are sequences in J and limn!1 .un v n / D 0 then lim .f .un / f .vn // D 0:

n!1

1 Proof. Assume that f is uniformly continuous and fun g1 nD1 and fvn gnD1 are sequences in J and limn!1 .un  vn / D 0. We will show that

lim .f .un /  f .vn // D 0:

n!1

Let " > 0 be given. By the uniform continuity of f on J, there exists ı > 0 such that x1 2 J; x2 2 J and jx1  x2 j < ı ) jf .x1 /  f .x2 /j < ": Since limn!1 .un  vn / D 0, there exists N 2 N such that jun  vn j < ı if n  N: Then jf .un /  f .vn /j < " Therefore limn!1 .f .un /  f .vn // D 0:

2.1 Continuity

69

To prove the converse, we will assume that f is not uniformly continuous on J. Then there exists " > 0 such that for each n 2 N there exists un and vn in J with jun  vn j < 1=n and jf .un /  f .vn /j  ". Thus limn!1 .un  vn / D 0 but it is not true that limn!1 .f .un /  f .vn // D 0.  The sequential characterization of uniform continuity is useful in ruling out uniform continuity, as in the following example: Example 6. Let f .x/ D 1=x for each x ¤ 0. Show that f is not uniformly continuous on .0; 1: Solution. Set un D

1 1 and vn D : n nC1

Then  lim .un  vn / D lim

n!1

n!1

1 1  n nC1



 D lim

n!1

D lim

n!1

nC1n n .n C 1/



1 D 0: n .n C 1/

On the other hand, f .un /  f .vn / D n  .n C 1/ D 1; so that lim .f .un /  f .vn // D lim .1/ D 1 ¤ 0:

n!1

n!1

Therefore, f is not uniformly continuous on .0; 1, even though f is continuous at each point in .0; 1 (confirm).  A continuous function on a closed and bounded interval is guaranteed to be uniformly continuous: Theorem 3. Assume that f is continuous on a closed and bounded interval Œa; b (i.e., f is continuous at each point of J). Then f is uniformly continuous on Œa; b. Proof. Given the interval Œa; b, the contrapositive of the statement f is continuous at each point of Œa; b ) f is uniformly continuous on Œa; b is f is not uniformly continuous on Œa; b ) there exists a point of Œa; b where f is not continuous.

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2 Limits and Continuity of Functions

We will prove the above statement. Thus, assume that f is not uniformly continuous on Œa; b. Then there exists " > 0 such that for each n 2 N there exists points xn and yn in Œa; b where jxn  yn j < 1=n but jf .xn /  f .yn /j  ". Since Œa; b is sequentially compact (Theorem 5 of Sect. 1.5), there exists a subsequence fxnk g1 kD1 of fxn g1 x0 2 Œa; b. Again by the compactness of Œa; b nD1 that converges to n a point o there exists a subsequence ynkj

1

jD1

that converges to y0 2 J. Since

ˇ ˇ 1 ˇ ˇ ˇxnkj  ynkj ˇ < n kj we have jx0  y0 j  lim

j!1

1 D 0: n kj

Thus x0 D y0 . The function f is not continuous at x0 . Indeed, if f were continuous at x0 , we would have   lim f xnkj D f .x0 / and lim f ynkj D f .x0 / j!1

j!1

since limj!1 xnkj D limj!1 ynkj D x0 . Thus ˇ   ˇ ˇ ˇ lim ˇf xnkj  f ynkj ˇ D 0:

j!1

But ˇ   ˇ ˇ ˇ ˇf xnkj  f ynkj ˇ  " for each j 2 N. Thus f is not continuous at x0 2 D. 

2.1.3 The Continuity of Basic Functions and Their Combinations Many functions that are encountered frequently are continuous on their natural domains. Let us begin by noting that a constant function is continuous on the entire number line: Let f .x/ D c for each x 2 R. We have jf .x C h/  f .x/j D jc  cj D 0:

2.1 Continuity

71

Thus jf .x C h/  f .x/j < " for any positive number ". Therefore given " > 0 we have complete freedom in choosing a corresponding ı > 0. For example, ı D 1 will do. Positive integer powers of x define continuous functions: Proposition 1. Assume that n is a positive integer and fn .x/ D xn for each x 2 R. Then fn is continuous on the entire number line. Proof. We have f1 .x/ D x for each x 2 R. Since jf1 .x C h/  f1 .x/j D jhj it is sufficient to set ı D " for a given " > 0. Let x be an arbitrary real number. If n  2 we have fn .x C h/  fn .x/ D .x C h/n  xn   n .n  1/ n2 2 n n1 n x h C    C h  xn D x C nx h C 2 by the Binomial Theorem. Therefore ˇ ˇ ˇ ˇ n .n  1/ n2 x h C    C hn1 ˇˇ jfn .x C h/  fn .x/j D jhj ˇˇnxn1 C 2   n .n  1/ n2  jhj n jxjn1 C jxj jhj C    C jhjn2 ; 2 by the triangle inequality. Let us restrict h so that jhj < 1. Then   n .n  1/ n2 n1 C CC1 : jxj jfn .x C h/  fn .x/j < jhj n jxj 2 If we set C .x/ D n jxjn1 C

n .n  1/ n2 CC1 jxj 2

we have jfn .x C h/  fn .x/j < C .x/ jhj : Thus, given " > 0, it is sufficient to choose ı so that  ı D min 1;

 " : C .x/

Since jfn .x C h/  fn .x/j < " if jhj < ı the function fn is continuous at x.  The following theorem states that arithmetic operations on continuous functions lead to continuous functions:

72

2 Limits and Continuity of Functions

Theorem 4. Assume that the functions f and g are continuous at x0 . Then 1. f C g is continuous at x0 ; 2. fg is continuous at x0 ; 3. f =g is continuous at x0 if g .x0 / ¤ 0. The assertions of Theorem 4 follow easily from the corresponding facts for sequences (exercise). It is also instructive to prove each statement by referring directly to "-ı definition of continuity. Corollary 1. A polynomial is continuous at each x 2R. Proof. A polynomial is defined by an expression of the form p .x/ D a0 C a1 x C a2 x2 C    C an xn ; where the coefficients a0 ; a0 ; : : : ; an are given numbers. We have shown that constants and functions defined by positive integer powers of x are continuous functions on R. Since sums and products of continuous functions are continuous a polynomial defines a continuous function on R (we may simply say that a polynomial is continuous on R).  Proposition 2. A rational function is continuous on its natural domain. Proof. A rational function f is the quotient of polynomials: If P .x/ and Q .x/ are polynomials f .x/ D

P .x/ for each x 2 R such that Q .x/ ¤ 0: Q .x/

Since P .x/ and Q .x/ are continuous at each x 2 R the continuity of f on its natural domain rule follows from Theorem 4.  Remark 5. Thanks to Theorem 3 a polynomial is uniformly continuous on any closed and bounded interval. A rational function g is uniformly continuous on any closed and bounded interval that does not contain a point where the denominator in the expression for g .x/ vanishes. Þ Remark 6. The trigonometric functions sine and cosine are continuous at any x 2R. We are not in a position to provide a rigorous proof for this statement since we have not even provided precise definitions for sine and cosine. In Chap. 5 we will be able to discuss these functions rigorously via power series (Sect. 5.6). In any case, it may be worth mentioning that jsin .x C h/  sin .x/j  jhj and jcos .x C h/  cos .x/j  jhj for each x and h in R. These inequalities lead to the uniform continuity of sine and cosine on the entire number line (we can set ı D " for any " > 0.

2.1 Continuity

73

The trigonometric functions tangent and secant are continuous at any x 2 R that is not an odd integer multiple of =2. Indeed, tan .x/ D

sin .x/ 1 and sec .x/ D ; cos .x/ cos .x/

and a quotient of continuous functions is continuous at any point where the denominator does not vanish (cos .x/ D 0 iff x is an odd integer multiple of =2). Þ Remark 7. We will also take it for granted that exponential functions are continous on the entire number line and logarithms are continuous on .0; C1/. We will provide the justification for these facts in Chaps. 4 and 5. Þ Composition of continuous functions leads to continuous functions: Theorem 5. Assume that f is continuous at x0 and g is continuous at f .x0 /. Then the composite function g ı f is continuous at x0 . Proof. Let " > 0 be given. Since g is continuous at f .x0 /, we can choose ı1 > 0 so that ju  f .x0 /j < ı1 ) jg .u/  g .f .x0 //j < ": Since f is continuous at x0 , we can choose ı > 0 such that jx  x0 j < ı ) jf .x/  f .x0 /j < ı1 : Thus, jx  x0 j < ı ) jg .f .x//  g .f .x0 //j < ": This shows that g ı f is continuous at x0 .    Example 7. Let F .x/ D sin x2 for each x 2 R. Then f is continuous on the entire number line. Indeed, if we set f .x/ D x2 and g .u/ D sin .u/ then F D g ı f is continuous on R since both f and g have that property. 

2.1.4 Problems In problems 1 and 2 prove that f is continuous at x0 in accordance with the "  ı definition of continuity (Suggestion: It may be easier to work with the definition that involves f .x0 C h/).

74

2 Limits and Continuity of Functions

1.

2. f .x/ D

f .x/ D x4 I x0 D 3:

x2

x I x0 D 2: C1

In problems 3 and 4, prove that f is continuous for each x0 2 D in accordance with the "  ı definition of continuity. 3. f .x/ D

1 ; D D fx 2 R W x > 4g x4

Hint: Restrict h so that jhj < .x  4/ =2. 4. f .x/ D x1=3 ; D D Œ0; C1/: Hint: Consider the cases x D 0 and x > 0 separately. If x > 0 multiply and divide f .x C h/  f .x/ by .x C h/2=3 C .x C h/1=3 x1=3 C x2=3 and restrict h so that jhj < x=2: In problems 5 and 6, show that f is uniformly continuous on D. You may find it convenient to make use of the following version of the "  ı definition of uniform continuity: A function f W D ! R is uniformly continuous on D if, given any " > 0 there exists ı > 0 such that x 2 D, x C h 2 D and jhj < ı ) jf .x C h/  f .x/j < ı: 6.

5. f .x/ D x2 C x  2, D D Œ0; 3 :

f .x/ D

1 ; D D Œ1; C1/ x2

In problems 7 and 8 show that f is not uniformly continuous on D by appealing to the sequential characterization of uniform continuity. 7. f .x/ D

1 ; D D .4; 8: x2  16

Hint: Consider sequences that converge to 4.

2.2 The Limit of a Function at a Point

75

8. 1 ; D D .0; 2: x4 Hint: Consider sequences that converge to 0. f .x/ D

2.2 The Limit of a Function at a Point In the previous section we discussed the continuity of a function at a point. Informally f is continuous at a if f .x/ approaches f .a/ as x approaches a. In some cases f .x/ may approach a definite value as x approaches a point a even though f may not be defined at a. Even if f is defined at a that value may not be the same as f .a/. The relevant concept is the limit of a function at a point. That is the topic that we will discuss in this section.

2.2.1 The Definition of the Limit of a Function Definition 1. Assume that f .x/ is defined for each x in an open interval that contains the point x0 , with the possible exception of x0 . The limit of f at x0 is L if, given any " > 0, there exists ı > 0 such that jf .x/  Lj < " if x 2 D; x ¤ x0 and jx  x0 j < ı: If the limit of f at x0 is L we write lim f .x/ D L

x!x0

(read “the limit of f .x/ as xapproaches x0 is L”). Example 1. Let f .x/ D

  4 x2  9 if x ¤ 3: x3

The function is not defined at 3 but we are entitled to investigate the limit of f at 3 since it is defined if x ¤ 3. We have f .x/ D

4 .x C 3/ .x  3/ D 4 .x C 3/ x3

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2 Limits and Continuity of Functions

if x ¤ 3. This expression indicates that f .x/ approaches 24 as x approaches 3. Let us prove that in accordance with Definition 1. We have jf .x/  24j D j4 .x C 3/  24j D j4x  12j D j4 .x  3/j D 4 jx  3j : Given " > 0 we need to choose ı > 0 so that 4 jx  3j < " if jx  3j < ı. It is sufficient to set ı D "=4. If x ¤ 3 and jx  3j < ı then jf .x/  24j D 4 jx  3j < 4ı < ": Therefore limx!3 f .x/  24, as claimed.  Remark 1. Note that if f .x/ is defined in an open interval that contains x0 then f is continuous at x0 if and only if limx!x0 f .x/ D f .x0 /. Þ Remark 2. If we set x D x0 Ch we can rephrase the definition of the limit as follows: The limit of f at x0 is L if, given any " > 0 there exists ı > 0 such that jf .x0 C h/  Lj < " if x0 C h 2 D, h ¤ 0 and jhj < ı: Þ Example 2. Let f .x/ D

x2  4 if x ¤ 2: 3 .x  2/

Show that limx!2 f .x/ D 4=3: Solution. If x ¤ 2 f .x/ D

xC2 .x C 2/ .x  2/ D : 3 .x  2/ 3

This expression for f .x/ indicates that limx!2 f .x/ D 4=3. Let us justify this in accordance with the alternative definition of the limit as in Remark 2: If x D 2 C h where h ¤ 0 then f .2 C h/ D

.2 C h/ C 2 4Ch D : 3 3

Thus ˇ ˇ ˇf .2 C h/  ˇ

ˇ ˇ ˇ ˇ ˇ 4 ˇˇ ˇˇ 4 C h 4 ˇˇ ˇˇ h ˇˇ 1  D D ˇ ˇ D jhj : ˇ ˇ ˇ 3 3 3 3 3

Therefore, for a given " > 0 it is sufficient to have h ¤ 0 and 1 jhj < ": 3

2.2 The Limit of a Function at a Point

77

Thus, we can set ı D 3". If h ¤ 0 and jhj < ı then ˇ ˇ ˇf .2 C h/  ˇ

ˇ 1 1 4 ˇˇ D jhj < .3"/ D ": 3ˇ 3 3

Therefore limx!2 f .x/ D 4=3, as claimed. The choice of ı corresponding to a given " is not unique, of course. For example, the choice ı D " is also sufficient. 

2.2.2 Basic Facts about Limits As in Example 1 and Example 2, in many cases we can show that a given function has a certain limit at a point by identifying a continuous function with values that coincide with the values of the given function near that point: Proposition 1. Assume that f .x/ is defined in an open interval J that contains x0 with the possible exception of x0 . If the function g is continuous at x0 and g .x/ D f .x/ for each x 2 J other than x0 then lim f .x/ D g .x0 / :

x!x0

Proof. Let " > 0 be given. Since g is continuous at x0 , lim g .x/ D g .x0 / :

x!x0

Since f .x/ D g .x/ if x ¤ x0 and x 2 J, we have lim f .x/ D lim g .x/ D g .x0 / :

x!x0

x!x0

 Example 3. Let

p x2 f .x/ D if x > 0 and x ¤ 4: x4

a) Determine limx!4 f .x/ by finding a function g that is continuous at 4 such that g .x/ D f .x/ if x ¤ 4 and x is in some open interval containing 4. b) Justify your assertion that g is continuous at 4 in accordance with the "  ı definition of continuity. Solution. a) We have p p  p  1 x2 x2 xC2 x4 p D p D D f .x/ D p x4 x4 xC2 xC2 .x  4/ x C 2

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2 Limits and Continuity of Functions

if x > 0 and x ¤ 4. Set 1 g .x/ D p : xC2 p Then g is continuous at 4 since x defines a function that is continuous at 4 (Example 3 of Sect. 2.1). We have 1 1 1 D D : g .4/ D p 2C2 4 4C1 Since f .x/ D g .x/ if x > 0 and x ¤ 4; lim f .x/ D lim g .x/ D g .4/ D

x!4

x!4

1 : 4

b) If jhj < 4 and h ¤ 0; g .4 C h/ 

p 1 4 4Ch2 1 1  D p  D p 4 4ChC2 4 4 4ChC2 p 2 4Ch D p  4 4ChC2 ! ! p p 2C 4Ch 2 4Ch p  p D 4 4ChC2 2C 4Ch D

h 4  .4 C h/  p 2 D   p 2 : 4 2C 4Ch 4 2C 4Ch

Therefore, ˇ ˇ ˇg .4 C h/  ˇ

ˇ 1 ˇˇ 1 jhj jhj D  D jhj : p 2 < ˇ 2 4 4 .2 / 16 4 2C 4Ch

Let " > 0 be given. Let ı D min .16"; 4/ : If h ¤ 0 and jhj < ı then ˇ ˇ ˇg .4 C h/  ˇ 

ˇ 1 1 1 ˇˇ .16"/ D ": < jhj < 4 ˇ 16 16

2.2 The Limit of a Function at a Point

79

We can define “one-sided limits,” just as we can refer to one-sided continuity: Definition 2. Assume that there exists ı0 > 0 such that f .x/ is defined in the open interval .x0 ; x0 C ı0 /. The limit of f .x/ as x approaches x0 from the right is LC if given " > 0 there exists ı > 0 such that jf .x/  LC j < " provided that x0 < x < x0 C ı. In this case we write lim f .x/ D LC

x!x0 C

(read “the limit of f .x/ as x approaches x0 from the right is LC ). Similarly, the limit of f .x/ as x approaches x0 from the left is L if given " > 0 there exists ı > 0 such that jf .x/  L j < " if x0  ı < x < x0 . In this case we write lim f .x/ D L

x!x0 

(read “the limit of f .x/ as ax approaches x0 from the left is L ). Remark 3. Clearly, the limit of f .x/ as x approaches x0 exists if and only if the limits of f .x/ as x approaches x0 from the right and from the left exist and have the same value. Also note that f is continuous at x0 from the right if and only if lim f .x/ D f .x0 / :

x!x0 C

Similarly, f is continuous at x0 from the left if and only if lim f .x/ D f .x0 / :

x!x0 

Þ Just as there is a sequential characterization of continuity (Theorem 1 of Sect. 2.1) there is a sequential characterization of a limit: Theorem 1. Assume that f .x/ is defined for each x in an open interval that contains the point x0 , with the possible exception of x0 . Then limx!x0 f .x/ D L if and only if for any sequence fxn g1 nD1 such that each xn ¤x0 , f .xn / is defined at each xn and limn!1 xn Dx0 we have limn!1 f .xn / D L: The proof is similar to the sequential characterization of continuity and is left as an exercise. The rules for the limits of sums, products, and quotients of functions are similar to the corresponding rules for sequences. The following theorem is relevant to the composition of functions: Theorem 2. Assume that f .x/ is defined for each x in an open interval J containing x0 with the possible exception of x0 and that limx!x0 f .x/ D y0 . Assume that g is continuous at y0 . Then the limit of g ı f at x0 exists and

80

2 Limits and Continuity of Functions

lim g .f .x// D g .y0 / ;

x!x0

i.e.,

 lim g .f .x// D g

x!x0

 lim f .x/ :

x!x0

Proof. The proof of Theorem 2 is similar to the proof of the corresponding theorem for the composition of continuous functions (Theorem 5 of Sect. 2.1): Let " > 0 be given. Since g is continuous at y0 , we can choose ı1 > 0 so that ju  y0 j < ı1 ) jg .u/  g .y0 /j < ": Since limx!x0 f .x/ D y0 , we can choose ı > 0 such that x ¤ x0 and jx  x0 j < ı ) jf .x/  y0 j < ı1 : Thus, x ¤ x0 and jx  x0 j < ı ) jg .f .x//  g .y0 /j < ": This shows that limx!x0 g .f .x// D g .y0 /.  Example 4. Evaluate !   x2  1 lim cos x!1 6 .x  1/ Assume that cosine is continuous at any real number. Solution. We have    x2  1 2   .x  1/ .x C 1/  .x C 1/ lim D lim D lim D D : x!1 6 .x  1/ x!1 x!1 6 .x  1/ 6 6 3 Since cosine is continuous at =3, we can apply Theorem 2:  !  !   x2  1  x2  1 1 lim cos D cos lim D cos D : x!1 x!1 6 .x  1/ 6 .x  1/ 3 2  Now we will state and prove two theorems that will be useful when we discuss integrals in Chap. 4, especially in our discussion of improper integrals. The first such theorem is about monotone functions. Such functions may have discontinuities but one-sided limits exist:

2.2 The Limit of a Function at a Point

81

Theorem 3. Assume that f W Œa; b !R is an increasing or decreasing function. Then lim f .x/ and lim f .x/

x!c

x!cC

exist at each c 2 .a; b/. In case f is increasing we have lim f .x/  f .c/  lim f .x/ :

x!c

x!cC

In case f is decreasing lim f .x/  f .c/  lim f .x/

x!c

x!cC

The limits limx!aC f .x/ and limx!b f .x/ exist as well. Proof. We will consider the case of an increasing function and c 2 .a; b/. Let S D ff .x/ W c < x  bg : Since f is an increasing function f .c/ is a lower bound for SC . Therefore L D inf S exists. We claim that lim f .x/ D L:

x!cC

Indeed, let " > 0 be given. By the definition of the greatest lower bound of a set there exists ı > 0 such that c C ı  b and f .c/  f .c C ı/ < L C " Since f is increasing we have f .c/  f .x/  f .c C ı/ < L C " if c < x < c C ı. Since L is a lower bound of the values f .x/ in the interval .c; b we have L  f .x/  f .c C ı/ < L C " if c < x < c C ı. Therefore limx!cC f .x/ exists and lim f .x/ D L:

x!cC

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2 Limits and Continuity of Functions

Since f .c/ < L C " for each " > 0 we have f .c/  L D lim f .x/ : x!cC

The proof of the existence of limx!c f .x/ and the fact that lim f .x/  f .c/

x!c

is along similar lines. We need to consider sup .ff .x/ W a  x < cg/ (confirm as an exercise).  Example 5. Let ( f .x/ D

1 n

if 0 if

1 nC1

< x  1n ; n D 1; 2; 3; : : : : xD0

Then f is monotone increasing on Œ0; 1. Note that f has infinitely many discontinuities,  1 1 [ f0g : n nD1 At each point of discontinuity 1=n we have lim f .x/ D

x!1=n

1 and n

lim f .x/ D

x!1=nC

1 , n D 2; 3; 4; n1

We also have lim f .x/ D 0 and lim f .x/ D 1:

x!0C

x!1

 Another criterion for the existence of the limit of a function is a Cauchy condition that is applicable to functions that are not necessarily monotone increasing.

2.2 The Limit of a Function at a Point

83

Theorem 4 (Cauchy Condition for the Limit of a Function). Assume that f .x/ is defined for each x 2 .c; c C ı 0 / for some ı0 > 0. Also assume that given any " > 0 there exists ı > 0 such that ı  ı 0 and jf .u/ f .v/j < " if c < u < c C ı and c < v < c C ı: Then limx!cC f .x/ exists. The obvious counterpart of the statement is valid for the existence of limx!c f .x/. Proof. By the given condition, if " D 1 there exists positive ı1 < 1 such that jf .u/  f .v/j < 1 if c < u < c C ı1 and c < v < c C ı1 : Select such a point x1 . Thus c < x1 < c C ı1 < c C 1 and jf .u/  f .x1 /j < 1 if c < u < c C ı1 Again by the given condition, if " D 1=2 there exists a positive ı2 < min .1=2; ı1 / such that jf .u/  f .v/j
: : : > xn and positive numbers ı1 > ı2 > : : : > ın such that c < xk < c C ı k < c C

1 k

and jf .u/  f .xk /j
0 be given. Pick N 2 N such that N > 2=". Set ı D ıN . If c < x < c C ı then jf .x/  Lj  jf .x/  f .xn /j C jf .xn /  Lj for any n. If n  N then jf .x/  Lj  jf .x/  f .xn /j C jf .xn /  Lj
0, there exists ı > 0 such that f .x/ > M provided that 0 < jx  aj < ı. We write

86

2 Limits and Continuity of Functions

lim f .x/ D C1

x!a

The limit of f at a is 1 if, given any M > 0, there exists ı > 0 such that f .x/ < M provided that 0 < jx  aj < ı. We write lim f .x/ D 1

x!a

The definition of a one-sided infinite limit such as limx!aC f .x/ D C1 is a modification of the definition of limx!aC f .x/ D C1 by restricting x so that x > a. Note that lim f .x/ D 1 if and only if lim .f .x// D C1:

x!a

x!a

Remark 1 (Caution). In any of the cases covered by Definition 1, the relevant limit of f does not exist as a real number. Indeed, if limx!a f .x/ D L, there exists ı > 0 and M > 0 such that jf .x/j < M if 0 < jx  aj < ı. Here, we have an example of “mathematical doublespeak”: We are using the same word “limit”, and the same symbol “lim” in connection with “finite limits” and “infinite limits” . The doublespeak is traditional and convenient, and we will use it. The particular context should clarify which usage of the word “limit” we have in mind. Nevertheless, if there is any possibility of confusion, we may stress that we are talking about a “finite limit”, or an “infinite limit” in the sense of Definition . Þ Example 1. Let f .x/ D

1 : .x C 3/ .x  2/

Prove that limx!2C f .x/ D C1. and limx!2 f .x/ D 1: Solution. In order to prove that limx!2C f .x/ D C1, we will restrict x so that 2 < x < 3. Thus 5 < x C 3 < 6 so that 1 1 > : xC3 6 Therefore, f .x/ D

1 1 > .x C 3/ .x  2/ 6 .x  2/

Thus, given M > 0, in order to ensure that 1 > M; .x C 3/ .x  2/

2.3 Infinite Limits and Limits at Infinity

87

it is sufficient to have 1 > M: 6 .x  2/ This is the case if 0

1 > 6 .x  2/

1  D M: 1 6 6M 

Therefore limx!2C f .x/ D C1: Now let us prove that limx!2 f .x/ D 1. It may be more convenient to prove the equivalent statement that limx!2 .f .x// D C1: We will restrict x so that 0 < x < 2. Thus 0 < x C 3 < 5 so that 1 1 > : xC3 5 Therefore f .x/ D

1 1 > .x C 3/ .2  x/ 5 .2  x/

Thus, given M > 0, in order to ensure that f .x/ > M it is sufficient to have 1 > M: 5 .2  x/ This is the case if 1 >2x 5M i.e., x>2

1 . 5M

88

2 Limits and Continuity of Functions

We can set   1 ı D min 2; 5M to ensure that x > 0. If 2  ı < x < 2 then f .x/ >

1 > M: 5 .2  x/

Therefore limx!2 .f .x// D C1.  The following proposition is helpful in the determination of infinite limits: Proposition 1. Assume that f .x/ > 0 if x is in an open interval that contains a, x ¤ a and limx!a f .x/ D 0. Then lim

x!a

1 D C1: f .x/

Similarly, if f .x/ < 0 when x is in an open interval that contains a, x ¤ a, and limx!a f .x/ D 0 then lim

x!a

1 D 1: f .x/

Proof. The proof is similar to the proof of Proposition 2 of Sect. 1.6. Let us establish the statement about f that has positive values near a. Let M > 0 be given. Since f .x/ > 0 if x is in an open interval that contains a, x ¤ a and limx!a f .x/ D 0 there exists ı > 0 such that 0 < f .x/
M if x ¤ a and jx  aj < ı. f .x/ Thus lim

x!a

1 D C1: f .x/

The statement about f that has negative values near follows by considering f : Then   1 D C1 lim  x!1 f .x/

2.3 Infinite Limits and Limits at Infinity

89

so that lim

x!a

1 D 1; f .x/

as we noted before.  One-sided versions of Proposition 1are valid. Example 2. Determine limx!=2˙ sec .x/ : Solution. We have sec .x/ D

1 : cos .x/

If 0 < x < =2 then cos .x/ > 0, and lim cos .x/ D lim cos .x/ D cos

x!=2

x!=2

 2

D 0:

Therefore, lim sec .x/ D

x!=2

1 D C1: x!=2 cos .x/ lim

If =2 < x < 3=2 then cos .x/ < 0, and limx!=2 cos .x/ D 0. Therefore, lim sec .x/ D

x!=2C

1 D 1 x!=2C cos .x/ lim

Figure 2.1 shows the graph of secant on the interval Œ; . y

Fig. 2.1

5

-p/2

p/2

x

5

The picture is consistent with the infinite limits that we calculated. The line x D =2 is a vertical asymptote for the graph of secant. Since secant is an even function, the graph is symmetric with respect to the vertical axis, and the line x D =2 is also a vertical asymptote. 

90

2 Limits and Continuity of Functions

Proposition 2. Assume that limx!a f .x/ D L > 0 or limx!a f .x/ D C1 and limx!a g.x/ D C1. Then lim f .x/g .x/ D C1:

x!a

Proof. Assume that limx!a f .x/ D L > 0. Then there exists ı1 > 0 such that jx  aj < ı1 and x ¤ a ) jf .x/  Lj


L L L ) f .x/ > L  D : 2 2 2

Let M > 0 be given. Since limx!1 g .x/ D C1 there exists ı2 > 0 such that g .x/ >

2M if jx  aj < ı2 and x ¤ a: L

Let us set ı D min .ı1 ; ı2 /. If jx  aj < ı and x ¤ a then f .x/ g .x/ >

   2M L D M: 2 L

Therefore limx!a f .x/ g .x/ D C1. The case where limx!1 f .x/ D C1 is handled similarly. Since limx!a f .x/ D C1 there exists ı3 > 0 such that jx  aj < ı3 and x ¤ a ) f .x/ > 1: Given M > 0 there exists ı4 > 0 such that g .x/ > M if jx  aj < ı4 and x ¤ a since limx!1 g .x/ D C1. If we set ı D min .ı3 ; ı4 / we have f .x/ g .x/ > M if jx  aj < ı and x ¤ a. Therefore limx!a f .x/ g .x/ D C1.  One-sided versions of Proposition 2 are valid. Example 3. Determine lim tan .x/ and

x!=2

lim tan .x/ :

x!=2C

2.3 Infinite Limits and Limits at Infinity

91

Proof. We have tan .x/ D

sin .x/ : cos .x/

As in Example 2, lim

x!=2

1 D C1: cos .x/

We also have lim sin .x/ D sin

x!=2

 2

D 1 > 0:

Therefore 

1 lim tan .x/ D lim sin .x/ x!=2 x!=2 cos .x/

 D C1;

by the one-sided version of Proposition 2. As in Example 2 lim

x!=2C

1 D 1: cos .x/

Therefore  lim tan .x/ D

x!=2C

lim sin .x/

x!=2C

1 cos .x/

 D 1;

by Proposition 2. Figure 2.2 is consistent with our assertions.  y

Fig. 2.2

6

-p

-p/2

p/2

−6

p

x

92

2 Limits and Continuity of Functions

Remark 2. A word of caution: Even though lim f .x/ > 0 and lim g .x/ D C1

x!a

x!a

implies that limx!a f .x/ g .x/ D C1, we cannot make a general statement about limx!aC f .x/ g .x/ if limx!aC f .x/ D 0 and limx!aC g .x/ D C1. The expression 0  1 is indeterminate. For example, we have   lim x2  1 D 0; lim

x!1

x!1C

1 D C1 x1

and 

2

lim x  1



x!1C



1 x1

 D lim .x C 1/ D 2: x!1C

In this case the indeterminate expression 1  0 seems to hide the number 2. Similarly, 1 D C1; x!1C x  1

lim .x  1/ D 0; lim

x!1

and 

1 lim .x  1/ x!1C x1

 D lim 1 D 1: x!1C

In this case, 1  0 seems to hide the number 1. Þ Proposition 3. Assume that limx!a g.x/ D C1. Then

limx!a f .x/ D L

or

limx!a f .x/ D C1

and

lim .f .x/ C g .x// D C1:

x!a

The proof is similar to the proof of Proposition 4 of Sect. 1.6 (exercise).

2.3.2 Limits at Infinity The behavior of a function for large positive or negative values of the independent variable are usually of interest. The relevant concepts are limits at C1 or 1. We will provide the precise definitions. The evaluation of such limits is similar to the evaluation of the limits of sequences. Thus our discussion will be brief.

2.3 Infinite Limits and Limits at Infinity

93

Definition 2. The limit of f at C1 is L if, given any " > 0 there exists A > 0 such that jf .x/ Lj < " for each x > A. The limit of f at 1 is L if, given any " > 0 there exists A > 0 such that jf .x/ Lj < " for each x < A. Example 4. Let f .x/ D

2x : xC3

Show that limx!C1 f .x/ D 2. Solution. We have ˇ ˇ ˇ ˇ ˇ ˇ 2x  2 .x C 3/ ˇ ˇ 2x 6 ˇ ˇD ˇ ˇ  2ˇ D ˇ : jf .x/  2j D ˇ ˇ xC3 xC3 jx C 3j Therefore, if x > 3, jf .x/  2j D

6 : xC3

Let " > 0 be given, and assume that x > 3. Then, xC3 1 6 6 6 6=". With reference to Definition 2, we can set A D 6=". By the above calculations, if x > A we have jf .x/  2j < ": Therefore, limx!C1 f .x/ D 2.  We may also speak of infinite limits at infinity: Definition 3. The limit of f at C1 is C1 if, given any M > 0, there exists A > 0 such that f .x/ > M for each x > A. The limit of f at C1 is 1 if, given any M > 0, there exists A > 0 such that f .x/ < M for each x > A. Example 5. Let f .x/ D x2  x. Show that limx!C1 f .x/ D C1. Solution. We have   1 : f .x/ D x2  x D x2 1  x

94

2 Limits and Continuity of Functions

If x > 2, then 1 1 1 1 < ) > : x 2 x 2 Therefore,     x2 1 1 > x2 1  D : f .x/ D x2 1  x 2 2 Let M > 0 be given. By the above inequality, in order to have f .x/ > M it is sufficient to have x > 2 and x2 > M: 2 p This is the case if x > 2 and x > p2M. With reference to Definition 3, we can set A to be the maximum of 2 and 2M. Thus, f .x/ > M if x > A. Therefore, limx!C1 f .x/ D C1.  There is a counterpart of Theorem 3 of Sect. 2.2 for limits at infinity: Theorem 1. Assume that f is monotone increasing on ŒA; C1/. If there exists M > 0 such that f .x/ < M for each x  A then limx!C1 f .x/ exists (as a finite limit). If there is no upper bound on the values of f on ŒA; C1/ then limx!C1 f .x/ D C1. Proof. Assume that there exists a number M such that f .x/  M for each x  a. Then L D sup ff .x/ W x  ag is finite. We claim that limx!C1 f .x/ D L. Indeed, by the definition of the least upper bound, given any " > 0 there exists x such that   L  " < f x  L: Since f is monotone increasing, if x  x   L  " < f x  f .x/  L: Therefore, limx!C1 f .x/ D L, as claimed.

2.3 Infinite Limits and Limits at Infinity

95

On the other hand, if the set ff .x/ W x  agis not bounded above, given any M there exists x  a such that f .x / > M. Since f is increasing, we have   f .x/  f x > M for any x  x . Therefore, limx!C1 f .x/ D C1.  There is also a counterpart of Theorem 4 of Sect. 2.2 for limits at infinity: Theorem 2 (Cauchy Condition for a Limit at Infinity). Assume that f .x/ is defined if x  a. Then limx!C1 f .x/ exists (as a finite number) if and only if given " > 0 there exists A such that c > b  A ) jf .c/ f .b/j < ": Proof. Assume that limx!1 f .x/ D L. Let " > 0 be given. Pick A so that b  A ) jf .b/  Lj
b  n1 ) jf .c/  f .b/j < 1: There exists n2 > n1 such that c > b  n2 ) jf .c/  f .b/j
n2 such that c > b  n3 ) jf .c/  f .b/j
nk1 >    > n1 such that c > b  nj ) jf .c/  f .b/j
n such that c > b  nkC1 ) jf .c/  f .b/j
A and jf .nK /  Lj
n: Since the closed and bounded interval Œa; b is sequentially compact, there exists a 1 convergent subsequence fxnk g1 kD1 of fxn gnD1 and x0 2 Œa; b such that lim xnk D x0 :

k!1

Since f is continuous at x0 , we have lim f .xnk / D f .x0 / :

k!1

98

2 Limits and Continuity of Functions

Since a convergent sequence is bounded, the sequence ff .xnk /g1 kD1 must be bounded. But this contradicts the statement, f .xnk / > nk for each k 2 N. Thus, f is bounded above on Œa; b. By the least upper bound principle, the set ff .x/ W a  x  bg has a least upper bound. Set M D supaxb f .x/. We will show that there exists x0 2 Œa; b such that f .x0 / D M: By the definition of the least upper bound, for each n 2 N there exists xn 2 Œa; b such that 1 M  < f .xn /  M: n Since the closed and bounded interval Œa; b is sequentially compact, there exists a 1 subsequence fxnk g1 kD1 of fxn gnD1 and x0 2 Œa; b such that lim xnk D x0 :

k!1

By the continuity of f at x0 , we have lim f .xnk / D f .x0 / :

k!1

On the other hand, the inequalities M

1 < f .xnk /  M nk

imply that lim f .xnk / D M:

k!1

By the uniqueness of the limit, we must have f .x0 / D M. 

2.4.2 The Intermediate Value Theorem Theorem 2. Assume that f is continuous on Œa; b and f .a/ ¤ f .b/. If c is in the open interval with endpoints f .a/ and f .b/ then there exists x0 2 .a; b/ such that f .x0 / D c:

2.4 The Intermediate Value Theorem

99

Proof. We will assume that f .a/ < c < f .b/ : Set g .x/ D f .x/  c so that g .a/ < 0 and g .b/ > 0: We need to show the existence of x0 2 .a; b/ such that g .x0 / D 0. We will make use of the bisection method. Let a1 D a and b1 D b. Set m1 D

aCb a1 C b1 D ; 2 2

so that m1 is the midpoint of Œa; b. If g .m1 / D 0, we are done: We can set x0 D m. Otherwise, • If g .m1 / < 0 we set a2 D m1 and b2 D b1 D b: • If g .m1 / > 0 we set a2 D a1 D a and b2 D m1 : Thus, Œa2 ; b2   Œa1 ; b1  D Œa; b and g .a2 / < 0, g .b2 / > 0 Having determined Œa2 ; b2 , we set m2 D

a2 C b2 ; 2

so that m2 is the midpoint of Œa1 ; b1 . If g .m2 / D 0, we set x0 D m2 . Otherwise, • If g .m2 / < 0 we set a3 D m2 and b3 D b2 : • If g .m2 / > 0 we set a3 D a2 and b3 D m2 : Note that Œa3 ; b3   Œa2 ; b2  and g .a3 / < 0; g .b3 / > 0 This procedure is terminated at step n if mn D

an C bn ; 2

and g .mn / D 0. In this case, we set x0 D mn . Otherwise, • If g .mn / < 0 we set anC1 D mn and bnC1 D bn : • If g .mn / > 0 we set anC1 D an and bnC1 D mn :

100

2 Limits and Continuity of Functions

Thus, ŒanC1 ; bnC1   Œan ; bn  and g .anC1 / < 0; g .bnC1 / > 0: If the procedure is never terminated, we obtain the nested sequence of intervals Œan ; bn  ; n D 1; 2; 3; : : : ; such that g .an / < 0; g .bn / > 0; and bn  an D

ba : 2n1

By the nested interval property, there exists a unique x0 such that a n  x0  b n for all n, since ba D 0: n!1 2n1

lim .bn  an / D lim

n!1

We have lim an D lim bn D x0 :

n!1

n!1

By the continuity of g, lim g .an / D lim g .bn / D g .x0 / :

n!1

n!1

Since g .an / < 0 for each n, we have g .x0 /  0. Since g .bn / > 0 for each n, we have g .x0 /  0. Therefore, g .x0 / D 0. 

2.4.3 The Existence and Continuity of Inverse Functions Let us begin by recalling the definition of the inverse of a function. Definition 1. Assume that for each x in the range of f there is a unique y in the domain of f such that f .y/ D x. The inverse f 1 of f is defined by the following relationship:

2.4 The Intermediate Value Theorem

101

y D f 1 .x/ , x D f .y/: Thus, the value of f 1 at x is the solution of the equation x D f .y/, provided that the solution exists and is unique. Figure 2.3 illustrates the relationship between f and the inverse f 1 graphically in the yx-plane (the y-axis is horizontal and the x-axis is vertical). x

Fig. 2.3

x = f (y)

y

y = f –1(x)

By the relationship between a function f and its inverse f 1 , the domain of f 1 is the same as the range of f , and the range of f 1 is the same as the domain of f . We must emphasize that the notation f 1 in the present context should not be confused with the reciprocal 1=f of the function f . The meaning of the notation should be clear within a particular context. Assume that the function f has an inverse. Then 

 f 1 ı f .y/ D f 1 .f .y// D y for each y in the domain of f ;     f ı f 1 .x/ D f f 1 .x/ D x for each x in the domain of f 1 . If the scale on the vertical axis is the same as the scale on the horizontal axis, the graph of f 1 appears as the reflection of the graph of f with respect to the diagonal y D x, as illustrated in Fig. 2.4: Fig. 2.4 (y, x)

y

x

f f −1

(x, y)

There is a general fact about the existence and continuity of the inverse of a function: Theorem 3. Assume that f is strictly increasing or decreasing and continuous on the interval J. The range I of f is also an interval. The inverse of f exists and f 1 is continuous on I. The function f 1 is increasing if f is increasing, and decreasing if f is decreasing.

102

2 Limits and Continuity of Functions

Proof. We will assume that f is increasing on the interval J (the case of a decreasing function is similar). Let us first show that the range of f is an interval. Thus, assume that x1 D f .y1 / and x2 D f .y2 / are points in the range of f , and that x1 < x2 . Let x 2 .x1 ; x2 / D .f .y1 / ; f .y2 //. By the Intermediate Value Theorem there exists y 2 J such that f .y / D x . Therefore the range of f is an interval. Let us label it as I. x

Fig. 2.5

x = f (y) x2 x x1

y1

y

y

y2

Let x 2 I. We have shown that there exists y 2 J such that f .y / D x . Since f is strictly increasing if y > y then f .y/ > f .y /, if y < y then f .y/ < f .y /. Therefore y is the only point in J at which f attains the value x . Therefore f has an inverse f 1 W I ! J. The inverse of f is also (strictly) increasing: Assume that x1 and x2 are in I and x1 < x2 . Let y1 D f 1 .x1 / and y2 D f 1 .x2 /. Thus x1 D f .y1 / and x2 D f .y2 /. if y1 > y2 then f .y1 / > f .y2 / so that x1 > x2 . Since that is not the case we must have y1 < y2 . Now let us show that f 1 is continuous. We will consider the case of a point a in the interior of I (the appropriate one-sided continuity is discussed in a similar manner at an endpoint of I which belongs to I). Let c D f 1 .a/, so that a D f .c/. Let " > 0 be a given. With reference to Fig. 2.6, let c  " D f 1 .a1 / and c C " D f 1 .a2 /. Since f is increasing, so is f 1 . Therefore, if a1 < x < a2 , then c  " < f 1 .x/ < c C ". Set ı to be the minimum of ja  a1 j and ja  a2 j. Then, jx  aj < ı ) x 2 .a1 ; a2 / ) c  " < f 1 .x/ < c C " ˇ ˇ ˇ ˇ ) ˇf 1 .x/  cˇ D ˇf 1 .x/  f 1 .a/ˇ < ": This establishes the continuity of f 1 .  y

Fig. 2.6 c c c

y a1

a

a2

f

1

x x

2.4 The Intermediate Value Theorem

103

Example 1. Let f .y/ D y2 . We restrict y so that y  0. The function f is continuous and increasing on Œ0; C1/. The range of f is also Œ0; C1/. We have yD

p x where x  0 , x D f .y/ D y2 and y  0:

p p Therefore, f 1 .x/ D x. Figure 2.7 illustrates the definition of x graphically. The square-root function f 1 is continuous on Œ0; 1/.  x

Fig. 2.7

x

f y

y2

y

f

1

x

x

y

The square-root function illustrates a function defined by x1=n , where n is an even positive integer. We have y D x1=n , x D yn where x  0 and y  0. Therefore, if we set f .y/ D yn , where y  0, then f 1 .x/ D x1=n ; x  0. Example 2. Let f .y/ D y3 , where y is an arbitrary real number. The function is continuous and increasing on R. The range of f is R. We have f 1 .x/ D x1=3 for each x 2 R. Figure 2.8 illustrates the relationship between x D f .y/ D y3 and y D f 1 .x/ D x1=3 .  x

Fig. 2.8 x

f y

y3

y

f

1

x

x1 3

y

Example 2 illustrates x1=n where n is an odd positive integer. If n is an odd positive integer and f .y/ D yn for each y 2 R then f 1 .x/ D x1=n for each x 2 R. Appropriate restrictions of sine, cosine, and tangent have inverses. Let us recall the definitions of these important special functions of mathematics (Fig. 2.9).

104

2 Limits and Continuity of Functions

Let us begin with the sine function. The equation x D sin .y/ has infinitely many solutions for a given x 2 Œ1; 1. Indeed, if y is a solution of the equation x D sin .y/, then y C 2n, n D ˙1; ˙2; : : : are also solutions, since sine is periodic with period 2: sin.y C 2n/ D sin.y/ D x: Thus, the sine function does not have an inverse. x 1

Fig. 2.9

x

2p

-p

sin y

p

y

2p

−1

Let us restrict sine to the interval Œ=2; =2 and call the resulting function f . The function f is continuous and increasing on Œ=2; =2, and the range of f is the interval Œ1; 1. Figure 2.10 shows the graph of f . Fig. 2.10 x D f .y/ D sin.y/ on Œ=2; =2

x 1 x

sin y

π

π

2

2

y

1

By Theorem 3, the inverse of f exists and f 1 is continuous on Œ1; 1. We have y D f 1 .x/ , x D sin .y/ ; where 1  x  1 and =2  y  =2. We will refer to f 1 as arcsine, and abbreviate it as arcsin. Thus, y D arcsin.x/ , x D sin.y/ where 1  x  1and =2  y  =2. Figure 2.11 illustrates the definition of arcsine.

2.4 The Intermediate Value Theorem

105 x 1

Fig. 2.11 x

sin y

π 2

y

arcsin x

π 2

y

1

Figure 2.12 shows the graph of y D arcsin .x/. Fig. 2.12 y D arcsin .x/

π 2

y

π 6 1

0.5

π 6

0.5

1

x

π 2

Another notation for arcsin.x/ is sin1 .x/: We will favor the notation arcsin.x/ in order to avoid any confusion with the reciprocal of sine. Just as in the case of sine, we cannot define the inverse of the periodic function cosine. On the other hand, cosine is continuous and decreasing on Œ0; , so that the restriction of cosine to the interval Œ0;  has an inverse. We will refer to that function as arccosine, and use the abbreviation arccos: y D arccos.x/ , x D cos.y/; where 1  x  1 and 0  y  . Thus, the value of arccosine at x 2 Œ1; 1 is the unique solution y of the equation cos.y/ D x that is in the interval Œ0; . Figure 2.13 illustrates the definition of arccosine. x 1

Fig. 2.13 x

cos y

y

1

arccos x

π

y

106

2 Limits and Continuity of Functions

By Theorem 3, arccosine is continuous on Œ1; 1. Another notation for arccos.x/ is cos1 .x/. We will favor the notation arccos.x/. Figure 2.14 shows the graph of arccosine. y π

Fig. 2.14

π 2

1

1

x

Remark 1. The functions arcsine and arccosine are related to each other in a simple manner. It can be shown that arccos.x/ C arcsin.x/ D

 ; 1  x  1: 2

Þ The function tangent is periodic with period , and its range is the entire number line. Therefore, the equation tan.y/ D x has infinitely many solutions for any real number x, so that the inverse of tangent does not exist. On the other hand, the restriction of tangent to the open interval .=2; =2/ is continuous, increasing, and has range equal to R, so that it has an inverse that is continuous on the entire number line. We will refer to that function as arctangent, and use the abbreviation arctan. Thus, y D arctan.x/ , x D tan.y/, where x is an arbitrary real number and =2 < y < =2. You may think of y as the unique angle between =2 and =2 such that tan .y/ D x. Figure 2.15 illustrates the definition of arctangent. Another notation for arctan.x/ is tan1 .x/. Fig. 2.15

x 10

x

tan(y)

-p / 2

y 10

arctan(x) p / 2

y

2.4 The Intermediate Value Theorem

107

Figure 2.16 shows the graph of arctangent. Fig. 2.16 y D arctan.x/

y p /2

x

10

−10

-p / 2

We have lim arctan .x/ D

x!C1

  and lim arctan .x/ D  : x!1 2 2

These facts are parallel to the facts, lim tan .y/ D C1 and

y! 2 

lim tan .y/ D 1:

y! 2 C

The natural exponential and the natural logarithm are inverses of each other: We have y D ln .x/ , x D ey where x > 0 and y 2 R (Fig. 2.17). x

Fig. 2.17

6

x

e

y

4

2

1

1

y

ln x

2

y

As we will discuss in Chap. 4, both functions are continuous on their respective domains (Fig. 2.18).

108

2 Limits and Continuity of Functions y

Fig. 2.18

2

y

1

ln x

e,1 1

2

e

3

4

5

6

7

x

2 3 4

2.4.4 Problems 1. Let f .x/ D x3 C 2x  7. Show that there exists x0 2 .0; 2/ such that f .x0 / D 0. 2. Let f .x/ D 2 sin .3x/ C 3 cos .2x/ Show that there exists c 2 .0; 2/ such that f .c/ D 2. 3. Let f .x/ D x3 C 2x  7. Show that the inverse function f 1 exists (you may use the derivative test for monotonicity from beginning calculus). What is the domain of f 1 ? 4. Let f .x/ D sin3 .x/ where x 2 Œ0; =2. a) Show that the inverse f 1 exists. What is the domain of f 1 ? What is the range of f 1 ? b) Determine f 1 .x/ explicitly in terms of arcsin.x/.

Chapter 3

The Derivative

3.1 The Derivative In this section we will review the definition of the derivative.

3.1.1 The Definition of the Derivative Assume that f .x/ is defined for each x in some open interval that contains the point a. If h ¤ 0 and jhj is small enough so that f .a C h/ is defined, the slope of the secant line that passes through the points .a; f .a// and .a C h; f .a C h// is f .a C h/  f .a/ : h y

a 7

h, f a

h f a h

a, f a

h a

a h

f a x

Fig. 3.1

Since the secant line that passes through the points .a; f .a// and .a C h; f .a C h// is almost “tangential” to the graph of f at .a; f .a// if jhj is small, it seems reasonable to define the slope of the tangent line to the graph of f at .a; f .a// as © Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0_3

109

110

3 The Derivative

lim

h!0

f .a C h/  f .a/ : h

We will refer to the ratio f .a C h/  f .a/ h as a difference quotient, since f .a C h/  f .a/ is the difference between the values of f at a C h and a, and h is the difference between a C h and a. We may refer to the difference quotient as the average rate of change of f corresponding to the change in the value of the independent variable from a to a C h. It is reasonable to interpret lim

h!0

f .a C h/  f .a/ h

as the rate of change of the function at a. The slope of a tangent line to the graph of a function and its rate of change can be treated within the framework of the concept of the derivative: Definition 1. Assume that f .x/ is defined for each x in some open interval that contains the point a. The derivative of f at a is f .a C h/  f .a/ h!0 h lim

provided that the limit exists. We denote the derivative of f at a as f 0 .a/ (read “f prime at a”), so that f 0 .a/ D lim

h!0

f .a C h/  f .a/ : h

Thus, f 0 .a/ can be interpreted as the slope of the tangent line to the graph of f at .a; f .a// or the rate of change of f at a. Example 1. Assume that f is a linear function, so that f .x/ D mx C b, where m and b are given constants. The graph of f is a line with slope m. Therefore, we should have f 0 .a/ D m at each point a. Indeed, f .a C h/  f .a/ Œm .a C h/ C b  Œma C b ma C mh C b  ma  b D D h h b mh D m: D h Therefore, f .a C h/  f .a/ D lim m D m: h!0 h!0 h

f 0 .a/ D lim 

3.1 The Derivative

111 y

f a h f a h f a f a

h a

x

a h

Fig. 3.2 The derivative of a linear function is the slope of its graph

Let us look at a nonlinear example: Example 2. Let f .x/ D x2  2x C 4. Determine the derivative of f at 3 and the tangent line to the graph of f at .3; f .3//. Solution. The relevant difference quotient is  .3 C h/2  2 .3 C h/ C 4  7 f .3 C h/  f .3/ D h h   2 7 C 4h C h  7 h .4 C h/ D D 4 C h: D h h Therefore, f 0 .3/ D lim

h!0

f .3 C h/  f .3/ D lim .4 C h/ D 4: h!0 h

Thus, the slope of the tangent line to the graph of f at .3; f .3// is 4. The tangent line is the graph of the equation y D f .3/ C f 0 .3/ .x  3/ D 7 C 4 .x  3/ . Figure 3.3 shows the graph of f and the tangent line at .3; f .3//. The picture is consistent with our intuitive notion of a tangent line.  y

20 7

3, f 3 3

Fig. 3.3

6

x

112

3 The Derivative

Remark 1. If we set x D a C h, then x approaches a as h approaches 0. Therefore, f .a C h/  f .a/ f .x/  f .a/ D lim : x!a h!0 h xa

f 0 .a/ D lim

We will favor the expression in terms of h.Þ Definition 2. We say that a function f is differentiable at a point a if the derivative of f at a exists. A function need not be differentiable at a point even if it is continuous at that point, as in the following example. Example 3. Let f be the absolute-value function so that f .x/ D jxj. Show that f is not differentiable at 0. Solution. Note that f is continuous at 0, since limh!0 f .h/ D limh!0 jhj D 0 D f .0/. Figure 3.4 shows the graph of f .

4

y

2

f x

4

2

2

x

4

x

Fig. 3.4

We have f .1=n/  f .0/ 1=n D lim D lim 1 D 1; n!1 n!1 1=n n!1 1=n lim

and f .1=n/  f .0/ 1=n D lim D lim .1/ D 1: n!1 n!1 1=n n!1 1=n lim

Therefore, lim

h!0

f .h/  f .0/ h

does not exist, so that f is not differentiable at 0.  Here is another example of continuity without differentiability: Example 4. Let f .x/ D x2=3 . Show that f is not differentiable at 0.

3.1 The Derivative

113

Solution. The relevant difference quotient is f .h/ h2=3 1 f .h/  f .0/ D D D 1=3 : h h h h Thus f .h/  f .0/ 1 f .h/  f .0/ 1 D lim 1=3 D C1 and lim D lim 1=3 D 1: h!0C h!0C h h!0 h!0 h h h lim

Therefore, f is not differentiable   at 0. Graphically, the secant line that passes through .0; 0/ and .h; f .h// D h; h2=3 becomes steeper and steeper as h approaches 0.  4

y

2

fh 8

h

4

4

8

x

Fig. 3.5

Even though continuity does not imply differentiability, differentiability implies continuity. Proposition 1. Assume that f is differentiable at a. Then f is continuous at a. Proof. We have



 f .a C h/  f .a/ h; h



 f .a C h/  f .a/ h: h

f .a C h/  f .a/ D so that f .a C h/ D f .a/ C Therefore,

  f .a C h/  f .a/ h h!0 h   f .a C h/  f .a/ D f .a/ C lim lim h h!0 h!0 h 

lim f .a C h/ D f .a/ C lim

h!0

D f .a/ C f 0 .a/ .0/ D f .a/: Since limh!0 f .a C h/ D f .a/ the function f is continuous at a. 

114

3 The Derivative

3.1.2 The Derivative as a Function If x denotes the independent variable of f , it is natural to use the same letter to denote the variable basepoint at which the derivative is evaluated. Thus, f 0 .x/ D lim

h!0

f .x C h/  f .x/ : h

We treat x as being fixed in the evaluation of the limit. You can think of h as “the dynamic variable.” Definition 3. The domain of the derivative function corresponding to the function f consists of all x such that f is differentiable at x. The value of the derivative function at such an x is f 0 .x/. We will denote the derivative function corresponding to f as f 0 , so that you may read f 0 .x/ as “f prime of x,” as well as “f prime at x.” Graphically, the value of the derivative function f 0 at x is the slope of the tangent line to the graph of f at .x; f .x//. Usually, we will simply refer to “the derivative of f ,” instead of “the derivative function corresponding to f .” Example 5. Let f be a linear function, so that f .x/ D mx C b, where m and b are constants. In Example 1 we showed that f 0 .a/ D m at each a 2 R. If we replace a by the variable x, we have f 0 .x/ D m for each x 2 R. Thus, the derivative of a linear function is a constant function whose value is the slope of the line that is the graph of the function. As a special case, if f is a constant function, then f 0 .x/ D 0 for each x 2 R:  Example 6. Let f .x/ D x2 . Determine the derivative function f 0 . Solution. If x is an arbitrary point on the number line and h ¤ 0, .x C h/2  x2 x2 C 2xh C h2  x2 f .x C h/  f .x/ D D h h h h .2x C h/ D 2x C h: D h Therefore, f .x C h/  f .x/ D lim .2x C h/ D 2x: h!0 h!0 h

f 0 .x/ D lim

Thus, f 0 .x/ D 2x for each x 2 R. We see that the derivative function that corresponds to the quadratic function f is a linear function. 

3.1 The Derivative

115

Example 7. Let f .x/ D x3 . Determine f 0 . Solution. If x is an arbitrary point on the number line and h ¤ 0, .x C h/3  x3 x3 C 3x2 h C 3xh2 C h3  x3 f .x C h/  f .x/ D D h h h  2  h 3x C 3xh C h2 D h D 3x2 C 3xh C h2 : Therefore, f 0 .x/ D lim

h!0

  f .x C h/  f .x/ D lim 3x2 C 3xh C h2 D 3x2 : h!0 h

Thus, the derivative function that corresponds to f is the quadratic function defined by 3x2 .  The determination of the derivative function f 0 corresponding to f is referred to as the differentiation of f . Thus, differentiation is an operation that assigns a function to a given function, as in the above examples. Example 8. Let f be the absolute-value function, so that f .x/ D jxj for each x 2 R. Determine f 0 (you must specify the domain of f 0 ). Solution. In Example 3 we showed that f is not differentiable at 0. Let x > 0. Then x C h is also positive if jhj is small enough. Therefore, f 0 .x/ D lim

h!0

f .x C h/  f .x/ jx C hj  jxj D lim h!0 h h .x C h/  x h D lim D lim D lim .1/ D 1: h!0 h!0 h!0 h h

If x < 0, we also have x C h < 0 if jhj is small enough. Therefore, f .x C h/  f .x/ jx C hj  jxj D lim h!0 h!0 h h  .x C h/  .x/ h D lim D lim D lim .1/ D  1: h!0 h!0 h h!0 h

f 0 .x/D lim

Thus, 0

f .x/ D



1 if x > 0; 1 if x < 0:

116

3 The Derivative

Figure 3.6 shows the graphs of f and f 0 . The slope of the graph at .x; f .x// is 1 if x > 0, and the slope of the graph of f at .x; f .x// is 1 if x < 0. 

4

y

2

fx

4

2

x

2

4

x

y

1

f' 4

2

2

4

x

1

Fig. 3.6 The absolute-value function and its derivative

3.1.3 The Leibniz Notation Leibniz and Newton are recognized as the cofounders of calculus. Newton used the notation fP for the derivative of f . You may come across Newton’s notation in older books on mechanics. The notation that was devised by Leibniz has been more popular and did not lose its popularity over the centuries, since it is practical to use, as you will see in the following sections. We will continue using “the prime notation” as well. We have f .x C x/  f .x/ : x!0 x

f 0 .x/ D lim

We can replace f .x C x/  f .x/ by f , as illustrated in Fig. 3.7. Thus, f : x!0 x

f 0 .x/ D lim

3.1 The Derivative

117 y

f x x

x

Fig. 3.7

x

x

df f D limx!0 dx x

The Leibniz notation for the derivative of f at x is df .x/ : dx Thus, f df .x/ D lim : x!0 x dx Note that we have replaced  in the expression for the difference quotient by the letter d. We may write simply df : dx The symbol df dx is not a genuine fraction, i.e., it is not the ratio of some quantity df and some quantity dx. We may refer to it as a “symbolic fraction.” As long as we are aware of the fact that we are not dealing with an ordinary fraction, an initial advantage of the Leibniz notation is that it reminds us of the definition of the derivative: The derivative is obtained as the limit of a genuine fraction, namely, f =x, as x approaches 0. We may type the derivative of f in the Leibniz notation as df df .x/ d df .x/ ; ; or f .x/ : dx dx dx dx The Leibniz notation is convenient in expressing differentiation rules. Let us display the results of some of the examples of this section by using the Leibniz notation:

118

3 The Derivative

d .mx C b/ D m, where m and b are constants, dx d  2 x D 2x; dx d  3 x D 3x2 . dx We may refer to a function by the name of the dependent variable. Assume that x is the independent variable and y is the dependent variable of a function. We may speak of “the function y D y.x/.” In such a case, we will denote the difference quotient as y.x C x/  y.x/ y D ; x x so that y denotes the increment of the dependent variable corresponding to the increment x of the independent variable, as illustrated in Fig. 3.8. This leads to the Leibniz notation dy=dx for the derivative of y as a function of x: dy y D lim : x!0 dx x

y

y x x

x

x

x

Fig. 3.8

For example, if y D x2 , dy y .x C x/2  x2 D lim D lim D 2x: x!0 x x!0 dx x If we use the Leibniz notation for the derivative and we wish to indicate that the derivative of f is to be evaluated at a specific point a, we may use the notations ˇ df .x/ ˇˇ df : .a/ or dx dx ˇxDa

3.1 The Derivative

119

For example, if f .x/ D x3 , then df df .x/ D 3x2 ) .2/ D 12: dx dx We may also express this fact as follows:  ˇ d x3 ˇˇ ˇ dx ˇ

ˇ D 3x2 ˇxD2 D 12: xD2

3.1.4 Higher-Order Derivatives Definition 4. The second derivative of a function f is the derivative of f 0 . In the “prime notation,” the second derivative is denoted as f 00 . Thus, f 00 .x/ D .f / .x/ if f 0 is differentiable at x. If we use the Leibniz notation to denote the derivative, we have   d d f 00 .x/ D f .x/ : dx dx 0 0

This suggests the Leibniz notation d2 f dx2 for the second derivative of f . We may type this as d2 f .x/ d2 d2 f .x/ , or f .x/ : dx2 dx2 dx2 Just as in the case of the derivative, the Leibniz notation for the second derivative is convenient to use, as long as you don’t try to attach a meaning other than d dx



d f .x/ dx



to the symbol d2 f : dx2 The above expression does not involve raising a quantity d=dx to the second power. Example 9. Let f .x/ D x3 . Determine the second derivative of f .

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3 The Derivative

Solution. In Example 7 we showed that f 0 .x/ D 3x2 . Therefore, d2 f f 0 .x C h/  f 0 .x/ 3 .x C h/2  3x2 D lim .x/ D lim h!0 h!0 dx2 x x 3x2 C 6xh C 3h2  3x2 h!0 h h .6x C 3h/ D lim h!0 h

D lim

D lim .6x C h/ D 6x: h!0

Thus, f 00 .x/ D 6x for each x 2 R.  The derivative of f 00 is the third derivative f 000 of f :  0 f 000 .x/ D f 00 .x/ : The Leibniz notation for the third derivative is d3 f : dx3 Thus, d3 f d D 3 dx dx



 d2 f .x/ : dx2

This may be typed as d 3 f .x/ d3 or f .x/ : dx3 dx3 Example 10. With reference to the function f of Example 9, we showed that f 00 .x/ D 6x. Thus, f 00 is a linear function. We have f 000 .x/ D

d3 f d D dx3 dx



d2 f dx2

 D

d .6x/ D 6: dx

 The second derivative of f is also referred to as the second-order derivative of f , and the third derivative is the third-order derivative of f . More generally, we obtain the nth order derivative of f by differentiating the derivative of order n  1. As n increases, the prime notation becomes unwieldy. We may denote the nth order derivative of f by f .n/ . There is no difficulty to express higher-order derivatives by the Leibniz notation:

3.1 The Derivative

121

f

.n/

dn d .x/ D n f .x/ D dx dx



 d n1 f .x/ : dxn1

3.1.5 Problems In problems 1–3 determine the derivative function f 0 directly from the definition of f 0 (you need to specify the domain of f 0 ) 1. f .x/ D 6x C x3 ; 2. f .x/ D

1 ; x ¤ 3; x3

3. f .x/ D 2x2  x4 : 4. Let  f .x/ D

x2 if x  0; x if x < 0:

a) Show that f is continuous at 0. b) Is f differentiable at 0? Justify your response. Determine f 0 .0/ if you claim that f is differentiable at 0. 5. Let  f .x/ D

x2 C 1 if x  1; 2x if x > 1:

a) Show that f is continuous at 1. b) Is f differentiable at 1? Justify your response. Determine f 0 .4/ if you claim that f is differentiable at 4.

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3 The Derivative

3.2 Local Linear Approximations and the Differential In this section we will discuss the link between differentiability and local linear approximations. The magnitude of the error in such approximations will be examined in detail.

3.2.1 Local Linear Approximations Given a function f that is differentiable at the point a, the tangent line to the graph of f at .a; f .a// is the graph of the equation y D f .a/ C f 0 .a/ .x  a/ : We will give a name to the underlying linear function: Definition 1. The linear approximation to f based at a is La .x/ D f .a/ C f 0 .a/.x  a/: We refer to a as the basepoint (Fig. 3.9).

y

a, f a La a

x

Fig. 3.9 The graph of La is a tangent line

Example 1. Let f .x/ D x2  2x C 4, as in Example 2 of Sect. 3.1. There we showed that f 0 .3/ D 4 and the tangent line to graph of f at .3; f .3// is the graph of the equation y D f .3/ C f 0 .3/ .x  3/ D 7 C 4 .x  3/ : Thus, the linear approximation to f based at 3 is L3 .x/ D 7 C 4 .x  3/ :

3.2 Local Linear Approximations and the Differential

123

y

20 10 7

3, f 3 3

6

x

9

2.6

3, f 3

3.4

5

7.5

2.8

3, f 3

3.2

6.5

Fig. 3.10

Figure 3.10 illustrates the effect of zooming in towards the point .3; f .3// D .3; 7/. Note that we can hardly distinguish between the graphs of f and L3 in the third frame. This indicates that L3 .x/ approximates f .x/ very well if x is close to the basepoint 3. On the other hand, we do not expect L3 .x/ to approximate f .x/ when x is far from 3. The linear function L3 is a “local approximation” to f . Let us assess the error in the approximation of f .x/ by L3 .x/ algebraically. Since L3 .x/ is expected to be a good approximation to f when x is near 3, it is convenient to set x D 3 C h, so that h D x  3 represents the deviation of x from the basepoint 3. We have L3 .3 C h/ D 7 C 4 .x  3/jxD3Ch D 7 C 4h: Therefore, f .3 C h/  L3 .3 C h/ D .3 C h/2  2 .3 C h/ C 4  .7 C 4h/ D 9 C 6h C h2  6  2h C 4  7  4h D h2 Thus, the absolute error is jf .3 C h/  L3 .3 C h/j D h2 :

124

3 The Derivative

 2 Note that h2 is much smaller than jhj if jhj is small. For example, 102 D 104  3 2 and 10 D 106 . Thus, the absolute error in the approximation of f .x/ by L3 .x/ is much smaller than the distance of x from the basepoint 3 if x is close to 3. This numerical fact is consistent with our graphical observation.  Remark 1. Since the graph of a function f and the tangent line to the graph of f at .a; f .a// are hardly distinguishable from each other near .a; f .a//, we will identify the slope of the graph of f at .a; f .a// with the slope of the tangent line at .a; f .a//, i.e., with f 0 .a/. Þ Example 1 illustrates the following general fact: Theorem 1. Assume that f is differentiable at a, and that La is the linear approximation to f based at a. We have f .a C h/ D La .a C h/ ChQa .h/ ; where lim Qa .h/ D 0:

h!0

Proof. We have ˇ La .a C h/ D f .a/ C f 0 .a/ .x  a/ˇxaDh D f .a/ C f 0 .a/ h: Therefore,   f .a C h/  La .a C h/ D f .a C h/  f .a/ C f 0 .a/ h D .f .a C h/  f .a//  f 0 .a/ h   f .a C h/  f .a/  f 0 .a/ Dh h Let’s set Qa .h/ D

f .a C h/  f .a/  f 0 .a/ ; h

so that hQa .h/ D f .a C h/  La .a C h/ : Therefore, f .a C h/ D La .a C h/ C hQa .h/ :

3.2 Local Linear Approximations and the Differential

125

Thus, the expression hQa .h/ represents the error on the approximation of f .a C h/ by the corresponding value of the linear approximation based at a. Note that Qa .h/ denotes the difference between f 0 .a/ and the relevant difference quotient. Since the difference quotient approaches the derivative as h approaches 0,  lim Qa .h/ D lim

h!0

h!0

 f .a C h/  f .a/  f 0 .a/ D 0: h

 Remark 2. Since jf .a C h/  La .a C h/j D jhQa .h/j D jhj jQa .h/j ; and limh!0 Qa .h/ D 0, the magnitude of the error in the approximation f .x/ by La .x/ is much smaller than the distance of x from the basepoint a. Therefore, we can hardly distinguish between the graph of f and the line that is tangent to the graph of f at .a; f .a// when the viewing window is small. Þ Example 2. Let f .x/ D x3 :

a) Determine f 0 .2/. b) Determine L2 , the linear approximation to f based at 2. c) Determine Q2 .h/, such that f .2 C h/ D L2 .2 C h/ C hQ2 .h/ . Confirm that limh!0 Q2 .h/ D 0: Solution. a) The relevant difference quotient is     23 C 3 22 h C 3 .2/ h2 C h3  23 f .2 C h/  f .2/ .2 C h/3  23 D D h h h 12h C 6h2 C h3 h   h 12 C 6h C h2 D h D

D 12 C 6h C h2 :

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3 The Derivative

Therefore,   .2 C h/3  23 D lim 12 C 6h C h2 D 12: h!0 h!0 h

f 0 .2/ D lim b)

L2 .x/ D f .2/ C f 0 .2/ .x  2/ D 8 C 12 .x  2/ : The graph of L2 is the tangent line to the graph of f at .2; 8/. y

16 8

−2

2, f 2

1

−1

2

3

x

−8

Fig. 3.11

b) As in Theorem 1 f .2 C h/  L2 .2 C h/ D hQ2 .h/ , where Q2 .h/ D

f .2 C h/  f .2/  f 0 .2/ : h

Thus, Q2 .h/ D

  f .2 C h/  f .2/  f 0 .2/ D 12 C 6h C h2  12 h D 6h C h2 :

Therefore   lim Q2 .h/ D lim 6h C h2 D 0:

h!0

h!0

Note that the error is   f .2 C h/  L2 .2 C h/ D hQ2 .h/ D h 6h C h2 D 6h2 C h3 Š 6h2 if jhj is small, since jhj3 is much smaller than h2 . 

3.2 Local Linear Approximations and the Differential

127

We would have discovered the derivative if we had posed a “best approximation” problem: Theorem 2. Assume that f .x/ is defined for each x in some open interval that contains the point a. Let L.x/ D f .a/ C m.x  a/, so that L is a linear function and the graph of L is a line with slope m that passes through .a; f .a//. If f .a C h/ D L .a C h/ ChQa .h/ where limh!0 Qa .h/ D 0, the function f is differentiable at a and f 0 .a/ D m, so that L D La : Proof. We have L .a C h/ D f .a/ C mh: Therefore, f .a C h/  L.a C h/ D f .a C h/  .f .a/ C mh/ D .f .a C h/  f .a//  mh: Thus, .f .a C h/  f .a//  mh D hQa .h/ ; so that f .a C h/  f .a/  m D Q .a; h/ : h Therefore,  lim

h!0

 f .a C h/  f .a/  m D lim Qa .h/ D 0: h!0 h

Thus, f 0 .a/ D lim

h!0

f .a C h/  f .a/ D m; h

as claimed. Therefore, L .x/ D f .a/ C m .x  a/ D f .a/ C f 0 .a/ .x  a/ D La .x/ : 

128

3 The Derivative

3.2.2 The Differential It is useful to consider all the local linear approximations to a given function at once by considering the basepoint to be a variable. In this case it is convenient to work with differences and a change in the notation seems to be in order. We will denote an increment along the x-axis by x. Thus, f 0 .x/ D lim

h!0

f .x C h/  f .x/ f .x C x/  f .x/ D lim : x!0 h x

Therefore, f .x C x/  f .x/ Š f 0 .x/ x if jxj is small, so that f .x C x/  f .x/ Š f 0 .x/ x: Note that f 0 .x/ x is the change corresponding to the increment x along the tangent line to the graph of f at .x; f .x//, as illustrated in Fig. 3.12. The quantity f 0 .x/x depends on two variables, the variable basepoint x and the increment x. We will give this expression a special name: Definition 2. The differential df of the function f depends on the variable basepoint x and the increment x. If we denote the value of the differential of f corresponding to x and x by df .x; x/ we set df .x; x/ D f 0 .x/ x: Thus, f .x C x/ f .x/ Š df .x; x/ if jxj is small. y

x

x, f x

x

f x

x

f x

df x, x x, f x

x

x

Fig. 3.12

x

x

x

3.2 Local Linear Approximations and the Differential

129

Note that the idea behind the differential is the same as the idea of local linear approximations. The differential merely keeps track of local linear approximations to a function as the basepoint varies. The analysis of the error is essentially the same, with a change of notation: Theorem 3. Assume that f is differentiable at x. Then f .x C x/  f .x/ D df .x; x/ C xQx .x/ where lim Qx .x/ D 0:

x!0

Proof. We set Qx .x/ D

f .x C x/  f .x/  f 0 .x/ : x

Thus, xQx .x/ D f .x C x/  f .x/  f 0 .x/ x: Therefore, f .x C x/  f .x/ D f 0 .x/ x C xQx .x/ D df .x; x/ C xQx .x/ We have  lim Qx .x/ D lim

x!0

x!0

 f .x C x/  f .x/ 0  f .x/ D 0; x

since f .x C x/  f .x/ D f .x/ : x!0 x lim

 Example 3. Let f .x/ D

p x:

a) Determine f 0 .x/ p if x > 0. b) Approximate 4:1 via the differential of f . Solution. a) You will recall from elementary calculus that 1 f 0 .x/ D p for any x > 0: 2 x

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3 The Derivative

Let us derive this result from scratch anyway. We have f .xCx/ f .x/ D x

p

p p p ! p p ! xCx x xCx x xCxC x D p p x x x C xC x D D

x

.xCx/ x p p  xCxC x

x

x p p  x C x C x

1 D p p : x C x C x Therefore, 1 f .x C x/  f .x/ 1 D lim p p D p x!0 x!0 x 2 x x C x C x

f 0 .x/ D lim

for any x > 0. b) The differential of f is 1 x df .x; x/ D f 0 .x/ x D p .x/ D p : 2 x 2 x It is natural top set x D 4 and x D 0:1 for the approximation of since f .4/ D 4 D 2. Thus,

p 4:1 D f .4:1/

p 0:1 0:1 D 0:025: 4:1  2 D f .4:1/  f .4/ Š df .4; 0:1/ D p D 4 2 4 Therefore, p

4:1 D 2 C

p 4:1  2 Š 2 C 0:025 D 2:025:

We have p

4:1 Š 2: 024 85;

rounded to 6 significant digits, and ˇ ˇp ˇ ˇ ˇ 4:1  2:025ˇ Š 1:5  10 4 : p Note that the absolute error in the approximation of 4:1 via the differential is much smaller than x D 0:1, as predicted by Theorem 3. 

3.2 Local Linear Approximations and the Differential

131

Remark 3. Since f .x C x/  f .x/ D df .x; x/ C xQx .x/ D f 0 .x/ x C xQx .x/ ; xQx .x/ is the error in the approximation of f .x C x/f .x/ by f 0 .x/ x. Since limx!0 Qx .x/ D 0, we noted that the magnitude of the error is much smaller than jxj if jxj is small. Thus, the approximation, change in f .x/ D f .x C x/  f .x/ Š f 0 .x/ x is very good if jxj is small. This should make the identification of the rate of change with the derivative even more plausible. Þ

3.2.3 The Traditional Notation for the Differential The differential of a function f depends on the variable basepoint x and the increment x: df .x; x/ D f 0 .x/x: Traditionally, the increment x is denoted by dx within the context of differentials. Thus, df .x; dx/ D f 0 .x/dx: If we use the Leibniz notation for f 0 .x/, we have df .x; dx/ D

df .x/ dx: dx

We usually do not bother to indicate that the differential depends on x and dx, and write df D

df dx: dx

This is convenient and traditional notation, but you should keep in mind that the “fraction” df dx is a symbolic fraction, and that the symbol dx that appears as the denominator does not have the same meaning as dx that stands for the increment in the value of the independent variable (Fig. 3.13).

132

3 The Derivative y x

dx, f x

dx

f x

dx

f x

df x, f x

dx

x

x

x

dx

Fig. 3.13

The expression df D

df dx dx

is analogous to the expression f D

f x; x

where x ¤ 0 and f D f .x C x/  f .x/. If we refer to the function as y D y.x/, we can write dy D

dy dx dx

The above expression is analogous to the expression y D

y x; x

where x ¤ 0 and y D y .x C x/  y .x/, and y Š dy if jxj is small. Example 4. Let f .x/ D 1=x. a) Determine f 0 .x/ if x ¤ 0. b) Express the differential of f in the traditional notation. c) Approximate 1 1:9 via the differential of f .

3.2 Local Linear Approximations and the Differential

133

Solution. a) Let x ¤ 0. We have 1 1    1 x  .x C x/ f .x C x/  f .x/ D x C x x D x x x x .x C x/ D

1 : x .x C x/

Therefore,   df 1 1 f .x C x/  f .x/ D lim D lim  D  2: x!0 x!0 dx x x .x C x/ x b)   1 df df D dx D  2 dx: dx x c) Since 1:9 is close to 2, and f .2/ D 1=2, it is natural to set x D 2 so that dx D 0:1. Thus,   ˇ ˇ 1 1 1 1  D f .1:9/  f .2/ Š  2 dxˇˇ .0:1/ D 0:025 D 1:9 2 x 4 xD2;dxD0:1 Therefore, 1 D 1:9



1 1  1:9 2

 C

1 Š 0:025 C 0:5 D 0:525: 2

We have 1 Š 0:526 316; 1:9 rounded to 6 significant digits. The absolute error is ˇ ˇ ˇ 1 ˇ ˇ ˇ Š 1:3  103 :  0:525 ˇ 1:9 ˇ This is much smaller than jdxj D 0:1. 

3.2.4 Problems In problems 1 and 2 a) Determine the linear approximation to f based at a,

134

3 The Derivative

b) Determine Qa such that f .a C h/ D f .a/ C f 0 .a/ h C hQa .h/ ; and confirm that lim Qa .h/ D 0

h!0

(you can make use of the differentiation rules from beginning calculus). 1. f .x/ D 3x2 C 5x; a D 2 2. f .x/ D x4 ; a D 1 In problems 3–5, a) Determine the differential of f , b) Determine Qx such that f .x C x/  f .x/ D df .x; x/ C xQx .x/ ; and confirm that lim Qx .x/ D 0

x!0

for each x in the domain of f 0 (you can make use of the differentiation rules from beginning calculus). 3. f .x/ D 6x C x3 ; 4. f .x/ D

1 ; x ¤ 3; x3

5. f .x/ D 2x2  x4 :

3.3 Rules of Differentiation

135

3.3 Rules of Differentiation In this section we will provide the proofs for some of the rules of differentiation that are familiar from beginning calculus. Let us begin with a special case of the power rule:

3.3.1 The Power Rule Proposition 1 (The Power Rule). If r is an integer, then d r x D rxr1 dx provided that xr and xr1 are defined. Proof. If n D 0, we have xn D x0 D 1 for each x, so that d 0 d x D .1/ D 0: dx dx   Therefore, the power rule is valid in this case as long as we interpret .0/ x1 simply as 0. Now let n be a positive integer. By the Binomial Theorem, .x C h/ D x C nx n

n

n1

! n.n  1/ n2 2 n nk k x h CC hC x h C    C hn : 2 k

For any x 2 R and h ¤ 0, f .x C h/  f .x/ .x C h/n  xn D h h   n.n  1/ n2 2 n n1 n x C nx h C x h C    C h  xn 2 D h   n.n  1/ n2 x h C    C hn1 h nxn1 C 2 D h n.n  1/ xn2 h C    C hn1 : D nxn1 C 2 Therefore,   n.n  1/ n2 n1 n1 f .x/ D lim nx C x hCCh D nxn1 : h!0 2 0

136

3 The Derivative

Now let f .x/ D xn , where n is a positive integer. If x ¤ 0, and jhj is small enough, 

 1 1  .x C h/n xn   1 xn  .x C h/n D h .x C h/n xn    1 .x C h/n  xn : D  h .x C h/n xn

f .x C h/  f .x/ .x C h/n  xn 1 D D h h h

Therefore,    1 .x C h/n  xn f .x C h/  f .x/ D lim  h!0 h!0 h h .x C h/n xn     .x C h/n  xn 1 lim : D  lim n h!0 h!0 .x C h/ xn h

f 0 .x/ D lim

We have d .x C h/n  xn D  xn D nxn1 ; h!0 h dx

 lim

and limh!0 .x C h/n D xn . Therefore,   f .x/ D nxn1 0



1 xn xn

 D

nxn1 D nxn1 : x2n



3.3.2 Differentiation is a Linear Operation Proposition 2 (The Constant Multiple Rule for Differentiation). Assume that f is differentiable at x, and that c is a constant. Then cf is also differentiable at x, and we have .cf /0 .x/ D cf 0 .x/ : In the Leibniz notation, d d .cf .x// D c f .x/ : dx dx

3.3 Rules of Differentiation

137

Proof. The difference quotient corresponding to cf , x, and h ¤ 0 is   cf .x C h/  cf .x/ f .x C h/  f .x/ .cf / .x C h/  .cf / .x/ D Dc : h h h By the constant multiple rule for limits,    f .x C h/  f .x/ f .x C h/  f .x/ D c lim D cf 0 .x/: .cf / .x/ D lim c h!0 h!0 h h 0

 Proposition 3 (The Sum Rule for Differentiation). Assume that f and g are differentiable at x. Then, the sum f C g is also differentiable at x, and we have .f C g/0 .x/ D f 0 .x/ Cg0 .x/ : In the Leibniz notation, d d d .f .x/ Cg .x// D f .x/ C g .x/ : dx dx dx Proof. The difference quotient corresponding to f C g, x, and h ¤ 0 is .f C g/ .x C h/  .f C g/ .x/ f .x C h/ C g.x C h/  .f .x/ C g.x// D h h f .x C h/ C g.x C h/  f .x/  g.x/ D h g.x C h/  g.x/ f .x C h/  f .x/ C : D h h By the sum rule for limits, .f C g/0 .x/ D lim

h!0



g.x C h/  g.x/ f .x C h/  f .x/ C h h



f .x C h/  f .x/ g.x C h/  g.x/ C lim h!0 h!0 h h

D lim

D f 0 .x/ C g0 .x/:  A linear combination of the functions f and g is a function of the form c1 f C c1 g, where c1 and c2 are constants. The derivative of a linear combination of functions is the linear combination of the corresponding derivatives, with the same coefficients (in the language of linear analysis, differentiation is a linear transformation):

138

3 The Derivative

Theorem 1 (Linearity of Differentiation). Assume that f and g are differentiable at x. If c1 and c2 are constants, then the linear combination c1 f C c2 g is also differentiable at x, and we have .c1 f C c2 g/0 .x/ D c1 f 0 .x/ Cc2 g0 .x/ : In the Leibniz notation, d d d .c1 f .x/ Cc2 g .x// D c1 f .x/ Cc2 g .x/ : dx dx dx Proof. We apply the sum rule and the constant multiple rule for differentiation: d d d d d .c1 f .x/ C c2 g .x// D .c1 f .x// C .c2 g .x// D c1 f .x/ C c2 g .x/ : dx dx dx dx dx  As you saw in beginning calculus, the following differentiation formulas for sine and cosine are valid: We have d d sin .x/ D cos .x/ and cos .x/ D  sin .x/ dx dx for each real number x. We will not repeat the plausibility argument that is given in beginning calculus texts. You have used the product rule quite often in beginning calculus:

3.3.3 Product and Quotient Rules Theorem 2 (The Product Rule). Assume that f and g are differentiable at x. The product fg is also differentiable at x and we have .fg/0 .x/ D f 0 .x/ g .x/ Cf .x/ g0 .x/ : In the Leibniz notation,

    df .x/ dg .x/ d .f .x/ g .x// D g .x/ Cf .x/ : dx dx dx

Proof. Let h ¤ 0. We will express the difference quotient for the product fg as follows: f .x C h/ g .x C h/  f .x/ g .x/ h f .x C h/ g .x C h/  f .x/ g .x C h/ C f .x/ g .x C h/  f .x/ g .x/ D h     g .x C h/  g .x/ f .x C h/  f .x/ g .x C h/ C f .x/ : D h h

3.3 Rules of Differentiation

139

Thus, f .x C h/ g .x C h/  f .x/ g .x/ h!0 h     g .x C h/  g .x/ f .x C h/  f .x/ D lim g .x C h/ C f .x/ h!0 h h      f .xCh/ f .x/ g .xCh/ g .x/ D lim lim g .xCh/ Cf .x/ lim h!0 h!0 h!0 h h   D f 0 .x/ lim g .x C h/ C f .x/ g0 .x/ :

.fg/0 .x/ D lim

h!0

Since g is differentiable at x, g is continuous at x. Therefore, lim g .x C h/ D g .x/ :

h!0

Thus,   .fg/0 .x/ D f 0 .x/ lim g .x C h/ C f .x/ g0 .x/ h!0

0

D f .x/ g .x/ C f .x/ g0 .x/ ; as claimed.  You have also differentiated quotients. Let’s begin with a special case: Proposition 4 (The Derivative of a Reciprocal). Assume that g is differentiable at x and g.x/ ¤ 0. Then 1=g is also differentiable at x, and we have  0 1 g0 .x/ : .x/ D  2 g g .x/ In the Leibniz notation, d dx



dg .x/  1 D  2dx : g .x/ g .x/

Proof. The relevant difference quotient is 1 1    1 1 1 g.x C x/ g .x/ D  x x g .x C x/ g .x/   1 g .x/  g .x C x/ D x g .x C x/ g .x/    g .x C x/  g.x/ 1 D  : x g .x C x/ g .x/

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3 The Derivative

Therefore, 1 1  0  1 g.x C x/ g .x/ .x/ D lim x!0 g x    1 g .x C x/  g.x/  D lim x!0 x g .x C x/ g .x/     g .x C x/  g.x/ 1 D lim  lim x!0 x!0 g .x C x/ g .x/ x   D g0 .x/

1 : .limx!0 g.x C x//g.x/

Since g is differentiable at x, g is continuous at x. Therefore, limx!0 g.x C x/ D g .x/. Thus,  0     g0 .x/ 1 1 1 D g0 .x/ D 2 : .x/ D g0 .x/ g .limx!0 g.x C x//g.x/ g.x/g.x/ g .x/  The general case follows from the above special case and the product rule: Theorem 3 (The Quotient Rule). Assume that f and g are differentiable at x, and g2 .x/¤ 0. Then  0 f 0 .x/ g .x/ f .x/ g0 .x/ f : .x/ D g g2 .x/ In the Leibniz notation, 

d dx



   df .x/ dg .x/ g .x/ f .x/ f .x/ dx dx D : g .x/ g2 .x/ 

Proof. We apply the product rule and the rule for the differentiation of reciprocals: d dx



f .x/ g.x/



   d 1 f .x/ dx g .x/      df 1 d 1 D C f .x/ dx g .x/ dx g .x/ D

3.3 Rules of Differentiation

141

D

df dx 

D



1 g .x/



0

dg 1 B C C f .x/ @ 2dx A g .x/

   df dg g .x/  f .x/ dx dx : g2 .x/



3.3.4 The Chain Rule Probably you have not seen a rigorous proof of the chain rule in elementary calculus. Now we can make up for that omission. The following proof is based on the differential approximation: Theorem 4 (The Chain Rule). Assume that g is differentiable at x and f is differentiable at g.x/. Then f ı g is differentiable at x and we have .f ı g/0 .x/ D f 0 .g .x// g0 .x/ : Proof. We set u D g .x/ and u D g.x C x/  g.x/, so that g.x C x/ D u C u. Then, .f ı g/ .x C x/  .f ı g/ .x/ D f .g.x C x//  f .g.x// D f .u C u/  f .u//: Now, f .u C u/  f .u/ D f 0 .u/ u C uQu .u/ ; where limu!0 Qu .u/ D 0 (Theorem 3 of Sect. 3.2). Therefore, f .u C u/  f .u/ f .g .x C x//  f .g .x// D x x 0 f .u/ u C uQu .u/ D x 0 f .g.x// u C uQg.x/ .u/ D x u u C Qg.x/ .u/ ; D f 0 .g.x// x x where x ¤ 0. We have lim

x!0

u g .x C x/  g .x/ D lim D g0 .x/ : x!0 x x

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3 The Derivative

Since g is differentiable at x, it is continuous at x. Thus, lim u D lim .g .x C x/  g .x// D 0:

x!0

x!0

Therefore, lim Qg.x/ .u/ D 0:

x!0

Thus, f .g .x C x//  f .g .x// x   u u 0 D lim f .g.x// C Qg.x/ .u/ x!0 x x    u u C lim lim Qg.x/ .u/ D f 0 .g .x// lim x!0 x x!0 x x!0

.f ı g/0 .x/ D lim

x!0

D f 0 .g.x// g0 .x/ C g0 .x/ .0/ D f 0 .g .x// g0 .x/ :  We can express the chain rule in the Leibniz notation: ! ˇ  dg .x/ d df .u/ ˇˇ f .g .x// D : dx du ˇuDg.x/ dx In the implementation of the chain rule, we usually refer to the functions by the letters that represent the dependent variables and write the chain rule cryptically: Thus, y is a function of u that is a function of x. Then y .u .x// defines a function of x and we write dy du dy D ; dx du dx where it is understood that dy=du will be evaluated at u .x/. The above expression involves an ambiguity of notation in the sense that y stands for two different functions on the right-hand side and on the left-hand side of the formula. Usually such ambiguity does not lead to errors. On the other hand, in the use of the chain rule in deriving other general facts it pays to use precise notation such as ! ˇ  d dg .x/ df .u/ ˇˇ f .g .x// D : dx du ˇuDg.x/ dx Example 1. Let us differentiate f .x/ D sin

  1 , x ¤ 0: x

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143

If we set u .x/ D 1=x we have f .x/ D sin .u .x// : Therefore !    ˇ ˇ d d 1 sin .u/ˇˇ du dx x uD1=x      1 1 cos .1=x/ D cos :  2 D x x x2

df .x/ D dx



3.3.5 The Derivative of an Inverse Function You may have differentiated inverse functions via implicit differentiation in elementary calculus without any justification for implicit differentiation. Here is a rigorous derivation of the formula for the derivative of an inverse that does not rely on implicit differentiation: Theorem 5 (The Derivative of an Inverse Function). Assume that f is increasing or decreasing and differentiable on an open interval I. If y D f 1 .x/ 2 I and f 0 .y/ ¤ 0, then f 0 is differentiable at x and d 1 f .x/ D dx

ˇ df .y/ ˇ dy ˇ

1

:

yDf 1 .x/

In “the prime notation,” we can express the above relationship as  1 0 f .x/ D

1 : f 0 .f 1 .x//

Proof. To begin with, it makes sense to speak of f 1 . Since f is differentiable on the open interval I, f is continuous on I. Since f is also increasing or decreasing on I, the inverse f 1 exists. Assume that f 0 .y/ ¤ 0, where y D f 1 .x/ so that x D f .y/. In order to compute the derivative of the inverse function at x, we must form the difference quotient f 1 .x C x/  f 1 .x/ : x We set y C y D f 1 .x C x/ so that x C x D f .y C y/. Therefore, x D f .y C y/  x D f .y C y/  f .y/. Figure 3.14 illustrates the case of an increasing function.

144

3 The Derivative y

y y

f −1 x

y y y

x

x x

x

x

Fig. 3.14

Thus, we can express the difference quotient for f 1 as .y C y/  y y f 1 .x C x/  f 1 .x/ D D : x x x We can write y 1 D ; x x y if x=y ¤ 0. Since lim

y!0

x f .y C y/  f .y/ D lim D f 0 .y/ ¤ 0; y!0 y y

we do have x=y ¤ 0 if y ¤ 0 and jyj is small enough. Both f and f 1 are continuous. Therefore, x D f .y C y/  f .y/ approaches 0 if and only if y D f 1 .x C x/  f 1 .x/ approaches 0. Thus, f 1 .x C x/  f 1 .x/ y 1 D lim D lim x x!0 x!0 x x!0 x y lim

D

1 limx!0

x y

1

D

limy!0

x y

D

1 ; f 0 .y/

as claimed.  Remark 1. If we refer to f and f 1 in terms of their dependent variables, i.e., we set y D y.x/ D f 1 .x/ , x D x.y/ D f .y/; the relationship

3.3 Rules of Differentiation

145

d 1 1 ˇ f .x/ D ˇ d dx f .y/ˇˇ dy yDf 1 .x/ takes the form 1 dy D dx dx dy where dx=dy is evaluated at y D y .x/. Note that the above expression is “formally correct,” in the sense that it is correct if we treat the symbolic fractions as if they were genuine fractions. Furthermore, the expression has the appealing feature that it can be considered to be the limiting case of the fraction 1 y D x x y as x ! 0 and y ! 0. Þ Now we can derive the power rule for rational exponents rigorously: Theorem 6 (The General Power Rule). If r is a rational number, then d r x D rxr1 dx provided that xr and xr1 are defined. Proof. We have established the power rule if r is an integer. Now we will show that 1 d 1=n x D x1=n1 dx n for any integer n  1, provided that x1=n and x1=n1 are defined. Once this is established, the general rule can be derived with the help of the chain rule (exercise). The expression x1=n defines the inverse of the function defined by yn : We have y D x1=n , x D yn (with appropriate restrictions). Therefore, 1 dy 1 1 1  1=n 1n 1 1 D x D n1 D y1n D D x1=n1 : D dx d dx ny n n n yn dy dy 

146

3 The Derivative

Let us illustrate the use of the formula for the derivative of an inverse function in deriving the expression for the derivative of arcsine: Proposition 5. d 1 if 1 < x < 1: arcsin .x/ D p dx 1  x2 Proof. We have y D arcsin .x/ , x D sin .y/ where 1  x  1 and 

  y : 2 2

We have 1 1 dy 1 D D D dx d dx cos .y/ sin .y/ dy dy if cos .y/ ¤ 0. Now, cos2 .y/ D 1  sin2 .y/ D 1  x2 ; so that p cos .y/ D ˙ 1  x2 : Since 

  y ) cos .y/  0, 2 2

we must disregard the ./ sign. Therefore, dy 1 1 D D p dx cos .y/ 1  x2 if cos .y/ ¤ 0, i.e., if 1  x2 > 0. This is the case if 1 < x < 1. Therefore, d 1 arcsin .x/ D p dx 1  x2 if 1 < x < 1, as claimed. 

3.3 Rules of Differentiation

147

The derivation of the formula for the derivatives of arccosine and arctangent are similar (exercise): 1 d arccos .x/ D  p if  1 < x < 1: dx 1  x2 1 d arctan .x/ D for each x 2 R. dx 1 C x2

3.3.6 Problems In problems 1–4 compute the derivative of the given function. You need to refer to the relevant rules of differentiation that are being used and specify the domain of the derivative function. 1. f .x/ D

4x  3 x2  9

2. f .x/ D

p x2  16

3. 1 f .x/ D p sin .x/ 4. f .x/ D .cos .x//1=3 5. Show that d 1 arccos .x/ D  p if  1 < x < 1 dx 1  x2 6. Show that d 1 arctan .x/ D for each x 2 R. dx 1 C x2

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3 The Derivative

3.4 The Mean Value Theorem In this section we will prove the Mean Value Theorem and review some of its consequences.

3.4.1 The Proof of the Mean Value Theorem Let us begin by recalling the definition of local maxima and minima: Definition 1. A function f has a local maximum at a if there exists an open interval J that contains a such that f .a/  f .x/ for each x 2 J. The function f has a local minimum at a if there exists an open interval J that contains a such that f .a/  f .x/ for each x 2 J. In either case, we say that f has a local extremum at a. The absolute maximum of f on a set D is M if there exists cM 2 D such that f .cM / D M and M  f .x/ for each x 2 D. The absolute minimum of f on a set D is m if there exists cm 2 D such that f .cm / D m and m  f .x/ for each x 2 D. We may refer to absolute maxima and minima as absolute extrema. You have made use of the following fact in elementary calculus in your search for local extrema: Theorem 1 (Fermat’s Theorem). If f has a local maximum or minimum at a and f is differentiable at a we have f 0 .a/ D 0. Proof. Assume that f attains a local maximum at a. Then f .a C h/  f .a/, if jhj is small enough. Therefore, f .a C h/  f .a/  0 if h > 0 and h is sufficiently small. Thus. f .a C h/  f .a/ 0 h under these conditions. Therefore, f 0 .a/ D lim

h!0

f .a C h/  f .a/ f .a C h/  f .a/ D lim  0: h!0C h h

Similarly, if we consider h < 0 such that jhj is sufficiently small, f .a C h/  f .a/ so that f .a C h/  f .a/  0. Since h < 0; we have f .a C h/  f .a/  0; h

3.4 The Mean Value Theorem

149

so that f 0 .a/ D lim

h!0

f .a C h/  f .a/ f .a C h/  f .a/ D lim  0: h!0 h h

Since we deduced that f 0 .a/  0 and f 0 .a/  0; we must have f 0 .a/ D 0; as claimed. The proof of the fact that f 0 .a/ D 0 at a point a at which f attains a local minimum is similar.  Remark 1. In the above proof we have used the fact that limh!0C g .h/  0 if g .h/  0 for all h sufficiently small and positive (and a similar fact for limh!0 g .h/) (confirm). Þ Theorem 2 (Rolle’s Theorem). Assume that f is continuous on Œa; b, differentiable at each point in .a; b/and f .a/ D f .b/. There exists c 2 .a; b/ such that f 0 .c/ D 0: Graphically, Rolle’s Theorem predicts the existence of at least one point c between a and b such that the tangent line to the graph of f at .c; f .c// is horizontal, if f .a/ D f .b/ (there may be more than one such point). This is illustrated in Fig. 3.15. c, f c

a, f a

a

b, f b)

c

b

x

Fig. 3.15

The proof of Rolle’s Theorem. If f is a constant on Œa; b, we have f 0 .x/ D 0 for each x 2 .a; b/. Therefore, we need to consider the case of a function f that is not constant on Œa; b. By the extreme value theorem, f attains its maximum and minimum values on Œa; b. We cannot have both of these values equal to the common value of the function at a and b: This would imply that f is constant on Œa; b. Assume that the maximum value of f on Œa; b is different from f .a/, and therefore different from f .b/ D f .a/. Therefore, if f .c/ is that maximum value, we must have c 2 .a; b/. This implies that f has a local maximum at c, so that we must have f 0 .c/ D 0. Similarly, if the minimum value of f on Œa; b is different from f .a/ and f .b/, and f attains that value at c 2 .a; b/, we must have f 0 .c/ D 0. 

150

3 The Derivative

The Mean Value Theorem follows from Rolle’s Theorem: Theorem 3 (The Mean Value Theorem). Assume that f is continuous on Œa; b and differentiable on .a; b/. There exists c 2 Œa; b such that f .b/ f .a/ D f 0 .c/ .b  a/ : We can express the Mean Value Theorem by writing f 0 .c/ D

f .b/  f .a/ : ba

Since the expression on the right is the slope of the secant line that connects the point .a; f .a// to .b; f .b//, the Mean Value predicts the existence of at least one point in the open interval .a; b/ such that the tangent line at the corresponding point on the graph of f is parallel to that secant line. Figure 3.16 illustrates the graphical meaning of the Mean Value Theorem. y

b, f b

a, f a

c, f c

a

b

x

Fig. 3.16

The Proof of the Mean Value Theorem. Let g be the linear function whose graph is the secant line that passes through the points .a; f .a// and .b; f .b//, as in the above figure. The slope of the secant line is f .b/  f .a/ : ba Therefore,  g.x/ D f .a/ C

 f .b/  f .a/ .x  a/ ba

(the above expression is the point-slope form of the equation of the line that passes through the points .a; f .a// and .b; f .b//, with basepoint a). Set  h .x/ D f .x/  g.x/ D f .x/  f .a/ 

 f .b/  f .a/ .x  a/ : ba

3.4 The Mean Value Theorem

151

We have h .a/ D h .b/ D 0. By Rolle’s Theorem, there exists c 2 .a; b/ such that h0 .c/ D 0. Since h0 .x/ D f 0 .x/ 

f .b/  f .a/ , ba

we have h0 .c/ D 0 , f 0 .c/ D

f .b/  f .a/ ; ba

so that f .b/  f .a/ D f 0 .c/ .b  a/ :  The following generalization of the Mean Value Theorem is useful as well: Theorem 4 (The Generalized Mean Value Theorem). Assume that f and g are continuous on Œa; b and differentiable in .a; b/. Then there exists c 2 .a; b/ such that f 0 .c/ Œg .b/ g .a/ D g0 .c/ Œf .b/ f .a/ : Proof. Set h .x/ D Œf .x/  f .a/ Œg .b/  g .a/  Œf .b/  f .a/ Œg .x/  g .a/ : Then, h .a/ D 0 and h .b/ D 0: By Rolle’s Theorem, there exists c 2 .a; b/ such that h0 .c/ D 0. We have h0 .x/ D f 0 .x/ Œg .b/  g .a/  g0 .x/ Œf .b/  f .a/ : Therefore, h0 .c/ D 0 , f 0 .c/ Œg .b/  g .a/ D g0 .c/ Œf .b/  f .a/ :  Note that the Mean Value Theorem follows from the Generalized Mean Value Theorem if we set g.x/ D x. In this case, g0 .x/ D 1, so that f 0 .c/ Œg .b/  g .a/ D g0 .c/ Œf .b/  f .a/ ) f 0 .c/ .b  a/ D f .b/  f .a/:

152

3 The Derivative

3.4.2 Some Consequences of the Mean Value Theorem The Mean Value Theorem is an “existence theorem” that is used to prove other theorems such as the derivative test for monotonicity: Theorem 5. Assume that f is continuous on the interval J and that f is differentiable at each x in the interior of J. If f 0 .x/ > 0 for each x in the interior of J, then f is increasing on J. If f 0 .x/ < 0 for each x in the interior of J, then f is decreasing on J. Proof. Let x1 and x2 belong to the interval J, and x1 < x2 . Then, the interval Œx1 ; x2  is contained in J, and the open interval .x1 ; x2 / is in the interior of J. Thus, f is continuous on Œx1 ; x2  and differentiable in the interior of Œx1 ; x2 , so that the Mean Value Theorem is applicable to f on the interval Œx1 ; x2 : There exists a point c 2 .x1 ; x2 / such that f .x2 /  f .x1 / D f 0 .c/ .x2  x1 / : If we assume that f 0 .x/ > 0 for each x in the interior of J, we have f 0 .c/ > 0. Therefore, f .x2 /  f .x1 / D f 0 .c/ .x2  x1 / > 0; so that f .x2 / > f .x1 /. This shows that f is increasing on J if f 0 .x/ > 0 for each x in the interior of J: If we assume that f 0 .x/ < 0 for each x in the interior of J, we have f .x2 /  f .x1 / D f 0 .c/ .x2  x1 / < 0; so that f .x2 / < f .x1 /. This shows that f is decreasing on J.  Intuitively, a function whose derivative has the constant value 0 should be a constant. This is indeed the case: Proposition 1. Assume that f 0 .x/ D 0 for each x in an interval J. Then f must be a constant on J. Proof. Let a be a point in J. By the Mean Value Theorem, if x is an arbitrary point in J, and x > a, there exists a point c between a and x such that f .x/  f .a/ D f 0 .c/ .x  a/ Since f 0 .c/ D 0, this implies that f .x/  f .a/ D 0, i.e., f .x/ D f .a/. Similarly, if x < a, there exists a point between x and a such that f .a/  f .x/ D f 0 .c/ .a  x/ ;

3.4 The Mean Value Theorem

153

so that f .a/  f .x/ D 0, i.e., f .x/ D f .a/. Thus, we have shown that f .x/ D f .a/ for any x in J, so that f has the constant value f .a/ on the interval J.  Two functions that have the same derivative differ at most by a constant: Corollary 1. Assume that f 0 .x/ D g0 .x/ for each x in an interval J. Then there exists a constant C such that g.x/ D f .x/ C Cfor each x 2 J. Proof. Set h .x/ D g .x/  f .x/. Then h0 .x/ D g0 .x/  f 0 .x/ D 0 for each x 2 J. By Proposition 1 there exists a constant C such that h .x/ D C, i.e., g .x/  f .x/ D C for each x 2 J. Therefore, g .x/ D f .x/ C C for each x 2 J.  Definition 2. A function ' W D  R ! R is uniformly Lipschitz continuous on D if there exists a constant K such that j' .u/  ' .v/j  K ju  vj for each u and v in D. The constant K is referred to as a Lipschitz constant. Proposition 2. Assume that f is differentiable on the interval J (the appropriate one-sided derivative exists at an endpoint that belongs to J) and there exits K > 0 such that ˇ ˇ ˇf 0 .x/ˇ  K for each x 2 J. Then f is uniformly Lipschitz continuous on J. Proof. Assume that u 2 J and v 2 J and v < u. By the Mean Value Theorem there exists c 2 .u; v/ such that f .u/  f .v/ D f 0 .c/ .u  v/ : Therefore ˇ ˇ jf .u/  f .v/j D ˇf 0 .x/ˇ .u  v/  K .u  v/ D K ju  vj : 

3.4.3 Convexity Assume that x < y. We can express z 2 Œx; y as z D x C .1  / y where 0    1: Indeed, if we set D

yz yx

154

3 The Derivative

then 0    1. We have y  x D y  z so that z D x C .1  / y: Note that z D x if  D 1 and z D y if  D 0, and that 1 D

zx : yx

Definition 3. A function f is convex on the interval J if f .xC .1  / y/  f .x/ C .1  –/ f .y/ for each x and y in J such that x < y and  2 Œ0; 1 : A function is said to be concave on the interval J if f .xC .1  / y/  f .x/ for each x and y in J such that x < y and for each  2 Œ0; 1. The secant line that connects .x; f .x// to .y; f .y// is the graph of the linear function g such that g .x C .1  / y/ D f .x/ C .1  / f .y/ for each  2 Œ0; 1 : Thus f is convex on the interval J if and only if for f .x C .1  / y/  f .x/ C .1  / f .y/ D g .x C .1  / y/ Therefore the graph of f on any interval Œx; y contained in J is below the secant line that connects .x; f .x// to .y; f .y// (Fig. 3.17).

(y,f(y))

(x,f(x))

x

Fig. 3.17 f is convex

y

3.4 The Mean Value Theorem

155

The function f is concave on J if the graph of f on any interval Œx; y contained in J is above the secant line that connects .x; f .x// to .y; f .y// (Fig. 3.18). (x,f (x))

(y,f (y))

x

y

Fig. 3.18 f is concave

Note that f is concave on the interval J if and only if f is convex on J. Therefore any assertion about concave functions can be deduced from a corresponding assertion about convex functions. Texts in beginning calculus usually refer to the graph of a convex function as concave up and to the graph of a concave function as concave down. Proposition 3. A function f is convex on the interval J if and only if for each x and y in Jsuch that x < y and each z 2 .x; y/ we have f .z/ f .x/ f .y/ f .z/  : zx yz Note that f .z/  f .x/ zx is the slope of the secant line that connects .x; f .x// to .z; f .z// and f .y/  f .z/ yz is the slope of the secant line that connects .z; f .z// to .y; f .y//. Figure 3.19 indicates that the slope of the second secant line should be greater than the slope of the first secant line if f is convex.

(y,f(y))

(z,f(z)) (x,f(x)) x

Fig. 3.19

z

y

156

3 The Derivative

Proof of Proposition 3. Assume that f is convex on J and g is the linear function whose graph is the secant line that connects .x; f .x// to .y; f .y// so that g .z/  f .x/ C .1  / f .y/ where z D z D x C .1  / y. We have g .z/ g .x/ g .y/ g .z/ f .z/ f .x/ f .z/ g .x/ D  D zx zx zx yz D

f .y/ f .z/ f .y/ g .z/  yz yz

Thus f .y/  f .z/ f .z/  f .x/  : zx yz Conversely, assume that f .z/  f .x/ f .y/  f .z/  zx yz for each z 2 .x; y/. Since z D x C .1  / y; where 0 <  < 0 where D

zx yz and 1   D : yx yx

we have f .y/  f .z/ f .z/  f .x/  : .1  / .y  x/  .y  x/ Thus f .z/  f .x/ f .y/  f .z/  1 

3.4 The Mean Value Theorem

157

This implies that f .z/  f .x/ C .1  / f .y/ Thus f .x C .1  / y/  f .x/ C .1  / f .y/ for each  2 Œ0; 1 Therefore f is convex on J.  Theorem 6. Assume that f is differentiable on .a; b/. Then f is convex on .a; b/ if and only if f 0 is increasing on .a; b/. Proof. Let us assume that f is convex and differentiable on the interval .a; b/. Let x and y be in .a; b/ and x < y. By Proposition 3 f .y/  f .z/ f .z/  f .x/  for each z 2 .x; y/ : zx yz If we set z D x C " where " > 0 then f .x C "/  f .x/ f .y/  f .x C "/  : " y  .x C "/ Therefore f .y/  f .x/ f .x C "/  f .x/ f .y/  f .x C "/  lim D : "!0C "!0C " y  .x C "/ yx

f 0 .x/ D lim

If we set z D y  " where " > 0 then f .y  "/  f .x/ f .y/  f .y  "/ f .y  "/  f .y/  D .y  "/  x " " so that lim

"!0C

f .y  "/  f .x/ f .y  "/  f .y/  lim : "!0C .y  "/  x "

Therefore f .y/  f .x/  f 0 .y/ : yx

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3 The Derivative

Thus f 0 .x/ 

f .y/  f .x/  f 0 .y/ yx

so that f 0 .x/  f 0 .y/ as claimed. Conversely, assume that f 0 is monotone increasing on .a; b/. We need to show that f is convex on .a; b/. Assume that x and y are in .a; b/ and x < y. By Proposition 3 it is sufficient to show that f .y/  f .z/ f .z/  f .x/  zx yz for any z 2 .x; y/. By the Mean value Theorem there exists  2 .x; z/ such that f .z/  f .x/ D f 0 ./ ; zx and there exists  2 .z; y/ such that f .y/  f .z/ D f 0 ./ : yz Since f 0 is monotone increasing on .a; b/ we have f 0 ./  f 0 ./. Thus f .y/  f .z/ f .z/  f .x/  zx yz Therefore f is convex on .a; b/.  Remark 2. Theorem 6 is also valid on an interval of the form Œa; b if f is differentiable on .a; b/ and continuous on Œa; b. Indeed, if f satisfies these conditions and f 0 is increasing on .a; b/ then f is convex on .a; b/: Assume that a  x < y  b. We need to show that f .z/  f .z/ C .1  / f .z/ if 0    1 for each x  z  y: Assume that zn 2 .a; b/ for each n and limn!1 zn D z. We have f .zn /  f .zn / C .1  / f .zn / for each n. Thus lim f .zn /   lim f .zn / C .1  / lim f .zn / :

n!1

n!1

n!1

3.4 The Mean Value Theorem

159

By the continuity of f on Œa; b : f .z/  f .z/ C .1  / f .z/ : Therefore f is convex on Œa; b. Þ Corollary 2. Assume that f 00 .x/ exists and that f 00 .x/ > 0 for each x 2 .a; b/. Then f is convex on .a; b/ : Proof. Since .f 0 /0 .x/ D f 00 .x/ > 0 the derivative of f is increasing on .a; b/. By Theorem 6 f is convex on .a; b/.  Remark 3. By Remark 2, Corollary 2 is valid on Œa; b as well if f is continuous at the endpoints. Þ Proposition 4. Assume that f is differentiable and convex on .a; b/. The graph of f on .a; b/ lies above the tangent line to the graph of f at any x0 2 .a; b/ :

(z,f(z))

a

z

b

Fig. 3.20

Proof of Proposition 4. Assume that x0 < x < b. By the Mean Value Theorem f .x/  f .x0 / D f 0 ./ .x  x0 / for some  2 .x0 ; x/ : Since f 0 is increasing on .a:b/ f .x/  f .x0 /  f 0 .x0 / .x  x0 / so that f .x/  f .x0 / C f 0 .x0 / .x  x0 / : Now assume that a < x < x0 . By the Mean Value Theorem f .x0 /  f .x/ D f 0 ./ .x0  x/ for some  2 .x; x0 / :

160

3 The Derivative

Since f 0 is increasing on .a; b/ we have f 0 ./  f 0 .x0 /. Therefore f .x0 /  f .x/  f 0 .x0 / .x0  x/ : Thus f .x/  f .x0 /  f 0 .x0 / .x  x0 / so that f .x/  f .x0 / C f 0 .x0 / .x  x0 / as in the previous case. If g .x/ D f .x0 / C f 0 .x0 / .x  x0 / then the graph of g is the tangent line to the graph of f at .x0 ; f .x0 //. Thus the graph of f on .a; b/ lies above the tangent line to the graph of f at x0 2 .a; b/. 

3.4.4 Problems 1. Let 2

f .x/ D ex for each x 2 R: Show that f is uniformly continuous on the entire number line .1; C1/ : Hint: Mean Value Theorem. 2. Let f .x/ D cos

  1 if x ¤ 0: x

Show that f is uniformly continuous on .1=; 2=/ in accordance with the "-ı definition of uniform continuity, Hint: Mean Value Theorem. 3. Assume that f .x/ is continuous in some open interval J that contains the point a, f 0 .x/ exists for each x 2 Jnfag and limx!a f 0 .x/ exists. Prove that f is differentiable at a and f 0 .a/ D lim f 0 .x/ : x!a

Hint: Mean Value Theorem.

3.5 L’Hôpital’s Rule

161

4. Show that the natural exponential function is convex on R. 5. Show that the natural logarithm is concave on .0; C1/. 6. Let f .x/ D 1 C x C

x3 x4 x2 C C : 2 6 24

Show that f is convex on R.

3.5 L’Hôpital’s Rule L’Hôpital’s Rule is discussed in beginning calculus without proof and used in the evaluation of certain limits that involve indeterminate forms. In this section we will prove some versions of the rule. Let us begin by proving the version of L’Hôpital’s rule that is relevant to the indeterminate form 0=0: Theorem 1. Assume that f and g are differentiable at each x in an open interval J that contains the point a, with the possible exception of aitself, and that g0 .x/ ¤ 0 for each x ¤ a and x 2 J. If limx!a f .x/D limx!a g.x/ D 0 and f 0 .x/ x!a g0 .x/ lim

exists then f .x/ f 0 .x/ D lim 0 x!a g .x/ x!a g .x/ lim

Proof. Since limx!a f .x/ D limx!a g .x/ D 0, the functions f and g are continuous on J if we declare that f .a/ D g .a/ D 0. To begin with, let’s show that g .x/ ¤ 0 if x 2 J and x ¤ a. Indeed, if x 2 J, x ¤ a and g .x/ D 0, we must have g0 .c/ D 0 for some c between a and x, by Rolle’s Theorem, since g .x/ D g .a/ D 0. But, g0 .c/ ¤ 0 since c 2 J. This is a contradiction. Let x 2 J and x ¤ a. By the Generalized Mean Value Theorem there exists cx between x and a such that f 0 .cx / Œg .x/  g .a/ D g0 .cx / Œf .x/  f .a/ : Therefore, f 0 .cx / g .x/ D g0 .cx / f .x/ :

162

3 The Derivative

Since g .x/ ¤ 0 and g0 .cx / ¤ 0, we can divide and obtain the equality, f .x/ f 0 .cx / D : 0 g .cx / g .x/ Therefore, f .x/ f 0 .cx / f 0 .x/ D lim 0 D lim 0 ; x!a g .x/ x!a g .cx / x!a g .x/ lim

since cx is between x and a.  The above version of L’Hôpital’s rule is also valid for one-sided limits, as you can confirm easily. Example 1. Determine lim

cos .x  1/  1

x!1

.x  1/2

:

Solution. We have lim .cos .x  1/  1/ D cos .0/  1 D 1  1 D 0 and lim .x  1/2 D 0;

x!1

x!1

so that we are led to the indeterminate form 0=0. By L’Hôpital’s rule,

lim

x!1

cos .x  1/  1 .x  1/2

d .cos .x  1/  1/  sin .x  1/ dx ; D lim D lim d x!1 x!1 2 .x  1/ .x  1/2 dx

provided that the limit on the right-hand side exists (finite or infinite). Since limx!1 sin .x  1/ D sin .0/ D 0 and limx!1 2 .x  1/ D 0, we are still confronted by the indeterminate form 0=0. If we apply L’Hôpital’s rule again, d . sin .x1// cos .0/ 1  sin .x1/  cos .x1/ dx D lim lim D lim D D : d x!1 x!1 x!1 2 .x1/ 2 2 2 .2 .x1// dx Therefore, lim

x!1

cos .x  1/  1 2

.x  1/

1  sin .x  1/ D : x!1 2 .x  1/ 2

D lim

3.5 L’Hôpital’s Rule

163 y

x

1

−0.5

Fig. 3.21

Figure 3.21 shows the graph of cos .x  1/  1 .x  1/2

:

The picture is consistent with the fact that limx!1 f .x/ D 1=2. The discontinuity of f at 1 can be removed by declaring that f .1/ D 1=2.  Here is an analog of Theorem 1 that is applicable to limits at infinity: Theorem 2. Assume that f and g are differentiable on an interval of the form ŒA; C1/ with A > 0 and that g0 .x/ ¤ 0 if x > A. If limx!C1 f .x/ D limx!C1 g.x/ D 0 and f 0 .x/ x!C1 g0 .x/ lim

exists then f .x/ f 0 .x/ D lim 0 : x!C1 g .x/ x!C1 g .x/ lim

Proof. We will make use of Theorem 1 by the following device: Let us set u D 1=x, F .u/ D f .1=u/ and G .u/ D g .1=u/. Then x>A,0 A. Thanks to Theorem 1, f .x/ F .u/ F 0 .u/ f 0 .x/ D lim D lim 0 D lim 0 : x!1 g .x/ x!C1 g .x/ u!0C G .u/ u!0C G .u/ lim

 There is an obvious counterpart of Theorem 2 for limits at 1. There are also versions of L’Hôpital’s rule that are applicable to cases that lead to the indeterminate form 1=1. We will state and prove such a case that involves limits at C1. A similar version is applicable to a limit at a point and its proof is similar to the proof of the following theorem: Theorem 3. Assume that f and g are differentiable on an interval of the form ŒA; C1/with A > 0 and that g0 .x/ ¤ 0 if x > A. If limx!C1 f .x/ D limx!C1 g.x/ D C1 and f 0 .x/ x!C1 g0 .x/ lim

exists then f .x/ f 0 .x/ D lim 0 : x!C1 g .x/ x!C1 g .x/ lim

3.5 L’Hôpital’s Rule

165

Proof. Let f 0 .x/ x!C1 g0 .x/

L D lim

and let " > 0 be given. We can assume that " < 1. Let us pick b > A such that ˇ ˇ 0 ˇ " ˇ f .x/ ˇ ˇ ˇ g0 .x/  Lˇ < 2 if x > b. Let x > b. Since limx!C1 f .x/ D C1 and limx!1 g .x/ D C1 and we are interested in limits at C1, we can assume that f .x/ > f .b/ > 0 and g .x/ > g .b/ > 0. By the generalized mean value theorem, there exists cx such that b < cx < x and f 0 .cx / Œg .x/  g .b/ D g0 .cx / Œf .x/  f .b/ : Therefore f .x/  f .b/ f 0 .cx / D 0 g .cx / g .x/  g .b/ (note that g0 .cx / ¤ 0 and g .x/  g .b/ > 0 so that the above expression is valid). Thus   f .b/ f .x/ 1  f 0 .cx / f .x/   D g .b/ g0 .cx / g .x/ 1  g .x/ so that 0

1 g .b/ 1  f 0 .cx / B f .x/ g .x/ C B C: D 0 @ f .b/ A g .x/ g .cx / 1 f .x/ Note that g .b/ g .x/ lim D1 x!1 f .b/ 1 f .x/ 1

166

3 The Derivative

since limx!C1 f .x/ D limx!C1 g.x/ D C1. Since cx > x and f 0 .x/ DL x!C1 g0 .x/ lim

we also have f 0 .cx / D L: x!C1 g0 .cx / lim

Therefore 0

1 g .b/ 1  f .x/ f 0 .cx / B g .x/ C B C D L: lim D lim 0 f .b/ A x!C1 g .x/ x!C1 g .cx / @ 1 f .x/ If the assertion f 0 .cx / DL x!C1 g0 .cx / lim

does not appear to be convincing, here is a more formal argument ˇ 1 ˇ g .b/ ˇ ˇ g .x/ C C  Lˇ ˇ A f .b/ ˇ ˇ f .x/ ˇ ˇ 0 1 ˇ ˇ g .b/ ˇ 0 ˇ 1  0 0 ˇ f .cx / B ˇ C f f .c / .c / g .x/ C x x ˇ B D ˇˇ 0  C  L ˇ @ A 0 0 g .cx / g .cx / ˇ g .cx / 1  f .b/ ˇ ˇ ˇ f .x/ ˇ ˇ ˇ ˇ g .b/ ˇ 0 ˇ ˇ 0 ˇ ˇ1  ˇ ˇ f .cx / ˇ ˇ ˇ ˇ f .cx / ˇ g .x/ ˇCˇ ˇˇ ˇ  ˇˇ 0  L  1 ˇ ˇ ˇ ˇ ˇ 0 f .b/ g .cx / ˇ g .cx / ˇ ˇ 1  f .x/ ˇ ˇ ˇ ˇ ˇ g .b/ ˇ 0 ˇ ˇ1  ˇ ˇ f .cx / ˇ ˇ ˇ " g .x/ ˇ ˇ ˇ x > b. We also have ˇ 0 ˇ ˇ ˇ ˇ f .cx / ˇ ˇ f 0 .cx / ˇ 1 ˇ ˇˇ ˇ ˇ g0 .c / ˇ ˇ g0 .c /  Lˇ C jLj < 2 C jLj : x x Since g .b/ g .x/ lim D1 x!1 f .b/ 1 f .x/ 1

there exists b > b so that ˇ ˇ ˇ ˇ g .b/ ˇ ˇ1  ˇ ˇ g .x/ ˇ<  " ˇ   1 ˇ ˇ ˇ 2 1 C jLj ˇ 1  f .b/ ˇ ˇ f .x/ 2 if x > b . Therefore if x > b ˇ ˇ ˇ ˇ 0 ˇˇ1  ˇ ˇ ˇ f .cx / ˇ ˇ ˇ f .x/ ˇ ˇ ˇˇ ˇ ˇ g .x/  Lˇ < ˇ g0 .c / ˇ ˇ x ˇ ˇ1  0   B 1 C jLj B < @ 2

ˇ ˇ g .b/ ˇ ˇ " g .x/  1ˇˇ C f .b/ ˇ 2 ˇ f .x/ 1  2

" 1 C jLj 2

C " C A C 2 D ":

Therefore f .x/ f 0 .x/ D L D lim 0 x!1 g .x/ x!C1 g .x/ lim

as claimed.  Example 2. Evaluate x2 : x!1 ex lim

168

3 The Derivative

Solution. We have lim x2 D lim ex D C1:

x!1

x!1

By Theorem 3 x2 2x D lim x x x!1 e x!1 e lim

provided that the latter limit exists. We still have lim 2x D lim ex D C1:

x!1

x!1

Again by Theorem 3 lim

x!1

2x 2 D lim x D 0: x x!1 e e

Therefore x2 2x D lim x D 0: x!1 ex x!1 e lim



3.5.1 Problems In problems 1–6 make use of L’Hôpital’s rule to determine the indicated limit (you need to indicate the version of the rule that you use): 1.

4. sin .x/  x x!0 x3

ex=4 x!C1 x2

lim

2.

lim

5. ex  1  x x!0 x2

ln .x/ x!C1 x1=3

lim

3.

lim

6. arcsin .x/  x x!0 x3 lim

lim

x!0C

p x ln .x/

Chapter 4

The Riemann Integral

4.1 The Riemann Integral In this section we will give a precise definition of the Riemann integral. We will consider functions that are bounded on closed and bounded intervals, even though we may not always state this restriction explicitly. We will prove the integrability of continuous and monotone functions on closed and bounded intervals.

4.1.1 Definition of the Riemann Integral Definition 1. A partition P of the interval Œa; b is a set of points fxk gnkD0 such that a D x0 < x1 < x2 <    < xk1 < xk <    < xn D b: The kth subinterval determined by the partition P is the interval Jk D Œxk1 ; xk . The length of the kth subinterval is jJk j D xk D xk  xk1 : Definition 2. The upper sum for the bounded function f W Œa; b ! R corresponding to the partition P D fxk gnkD0 of Œa; b is defined as U .f ; P/ D

n X

Mk .f / xk

kD1

where Mk .f / D

sup 2Œxk1 ;xk 

© Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0_4

f ./ :

169

170

4 The Riemann Integral

The lower sum corresponding to f and the partition P D fxk gnkD0 of Œa; b is defined as L .f ; P/ D

n X

mk .f / xk

kD1

where mk .f / D

inf

2Œxk1 ;xk 

f ./ :

We will use the notations Mk and mk if there is no ambiguity about the function f in a particular context. If f is positive-valued on the interval Œa; b and we interpret the integral of f on Œa; b as the area between the graph of f and ja; bj, as you did in beginning calculus, the integral should be a number between any upper sum and lower sum for f on Œa; b. In Fig. 4.1 the sum of the areas of the shaded rectangles is a lower sum and the sum of the areas of the taller rectangles is an upper sum. Fig. 4.1

a

b

In order to define the Riemann integral we will need to establish certain facts with regard to upper sums, lower sums, and Riemann sums. Let us begin with a definition: Definition 3. We say that a partition Q of Œa; b is a refinement of the partition P of Œa; b (or Q is finer than P) if Q is obtained by adding a finite number of points to P. Upper sums decrease and lower sums increase as partitions are refined: Lemma 1. Given f W Œa; b !R, if the partition P0 of Œa; b is a refinement of the partition P of Œa; b then   U f ; P0  U .f ; P/

4.1 The Riemann Integral

171

and   L f ; P0  L .f ; P/ : If P and P0 are arbitrary partitions of Œa; b we have   L f ; P0  U .f ; P/ : Proof. In order to confirm the first statement, it is sufficient to consider a “onepoint refinement” of a partition. Thus, let P D fx0 ; x1 ; : : : ; xn g and assume that Q is obtained by adding to P the point z such that xk1 < z < xk . Then L .f ; P/ D

k1 X jD1

D

mj xj

jDkC1

k1 X

 mj xj C

inf

xk1 xxk

jD1 n X

C

n X

mj xj C mk xk C

  f .x/ .z  xk1 / C

inf

xk1 xxk

 f .x/ .xk  z/

mj xj

jDkC1



k1 X

 mj x C

jD1

C

n X

 inf

xk1 xz

f .x/ .z  xk1 / C



 inf f .x/ .xk  z/

zxxk

mj xj

jDkC1

D L .f ; Q/ : Similarly, U .f ; Q/  U .f ; P/ : Now assume that P and P0 are arbitrary partitions of Œa; b. Then P [ P0 is a common refinement of P and P0 . Therefore,       L f ; P0  L f ; P [ P0  U f ; P [ P0  U .f ; P/ ; so that   L f ; P0  U .f ; P/ ; as claimed. 

172

4 The Riemann Integral

Let us set inf U .f ; P/ D inf fU .f ; P/ W P is a partition of Œa; bg P

and sup L .f ; P/ D sup fL .f ; P/ W P is a partition of Œa; bg : P

These quantities exist since f is bounded on Œa; b. We have U .f ; P/  .b  a/ inf f .x/ x2Œa;b

and L .f ; P/  .b  a/ sup f .x/ x2Œa;b

for any partition P of Œa; b. Definition 4. Assume that f W Œa; b ! R is bounded. We define the upper integral of f on Œa; b as Z

b

f .x/ dx D inf U .f ; P/ P

a

The lower integral of f on Œa; b is defined as Z

b

f .x/ dx D sup L .f ; P/ : P

a

Since   L f ; P0  U .f ; P/ for arbitrary partitions P and P0 of Œa; b we have   sup L f ; P0  U .f ; P/ P0

for any partition P of Œa; b. Therefore   sup L f ; P0  inf U .f ; P/ : P0

P

4.1 The Riemann Integral

173

Thus 

0



Z

b

L f;P 

Z

b

f .x/ dx 

a

f .x/ dx  U .f ; P/

a

for an arbitrary partitions P and P0 of Œa; b : Definition 5. The norm of a partition P D fxk gnkD0 is the maximum value of the lengths of the subintervals determined by P and will be denoted by jjPjj: jjPjj D max jJk j D max xk : kD1;:::;n

kD1;:::;n

Definition 6. A bounded function f W Œa; b ! R is Riemann integrable on Œa; b if given any " > 0 there exists ı > 0 such that if jjPjj < ı then U .f ; P/ L .f ; P/ < ": Remark 1. Let P D fxk gnkD0 be a partition of Œa; b. We will define the oscillation of f on Œxk1 ; xk  as ! .f ; Œxk1 ; xk / D sup .jf ./ f ./j where  and  are in Œxk1 ; xk / : Thus ! .f ; Œxk1 ; xk / is indeed a measure of the oscillation of the values of f on Œxk1 ; xk . For future reference, let us note that U .f ; P/  L .f ; P/ D

n X

! .f ; Œxk1 ; xk / xk :

kD1

Thus, the condition of the (Riemann) integrability of f on Œa; b can be rephrased as follows: Given " > 0 there exists ı > 0 such that if P D fxk gnkD0 is a partition of Œa; b and jjPjj < ı then n X

! .f ; Œxk1 ; xk / xk < ":

kD1

Þ Proposition 1. Assume that the bounded function f W Œa; b !R is Riemann integrable on Œa; b. Then Z

Z

b

b

f .x/ dx D a

f .x/ dx: a

174

4 The Riemann Integral

Proof. Let " > 0 be given and assume that U .f ; P/  L .f ; P/ < ": We have Z

Z

b

b

f .x/ dx 

L .f ; P/ 

f .x/ dx  U .f ; P/ :

a

a

Therefore Z 0

Z

b

b

f .x/ dx  a

f .x/ dx  U .f ; P/  L .f ; P/ < ": a

Since " > 0 is arbitrary we must have Z

b

Z

b

f .x/ dx 

a

f .x/ dx D 0

a

so that Z

Z

b

b

f .x/ dx D a

f .x/ dx: a

 Definition 7. Assume that the bounded function f W Œa; b ! R is Riemann integrable on Œa; b. The Riemann integral of f on Œa; b is defined as the common value of the upper and lower integrals of f on Œa; b. We denote the Riemann integral of f on Œa; b by Z b f .x/ dx: a

Thus Z

b

Z

b

f .x/ dx D

a

Z

b

f .x/ dx D

a

f .x/ dx

a

Note that Z

b

L .f ; P/ 

f .x/ dx  U .f ; P/

a

for arbitrary partitions P and P0 of Œa; b :

4.1 The Riemann Integral

175

Definition 8. Let f W Œa; b ! R be a bounded function and let P D fxk gnkD0 be a partition of Œa; b. Assume that k 2 Œxk1 ; xk  for k D 1; 2; : : : ; n. The Riemann sum for f corresponding to the partition P D fxk gnkD0 and the set of intermediate points fk gnkD1 is n X

f .k / xk :

kD1

Theorem 1. Assume that f is Riemann integrable on Œa; b. Given " > 0 there exists ı > 0 such that ˇ n ˇ Z b ˇX ˇ ˇ ˇ f .k / xk  f .x/ dxˇ < " ˇ ˇ ˇ a kD1

where n X

f .k / xk

kD1

is a Riemann sum for corresponding to a partition P D fxk gnkD1 of Œa; b with jjPjj < ı. Proof. Assume that jjPjj < ı so that U .f ; P/  L .f ; P/ < ": Since mk  f .k /  Mk for k D 1; 2; : : : n We have n X

mk xk 

kD1

n X

f .k / xk 

kD1

n X

Mk xk :

kD1

Thus L .f ; P/ 

n X

f .k / xk  U .f ; P/ :

kD1

We also have Z

b

L .f ; P/  a

f .x/ dx  U .f ; P/

176

4 The Riemann Integral

Therefore ˇ n ˇ Z b ˇX ˇ ˇ ˇ f .k / xk  f .x/ dxˇ  U .f ; P/  L .f ; P/ < " ˇ ˇ ˇ a kD1

 Remark 2. Assume that Pn is the partition of Œa; b that leads to n subintervals of equal length: 



ba Pn D a C k n



 W k D 0; 1; 2; : : : n; :

Let us denote a specific choice of a point in the kth subinterval determined by Pn by .n/ .n/ k . For example, we can set k to be the midpoint 

1 aC kC 2



 ba : n

Since lim jjPn jj D lim

n!1

n!1

ba D0 n

the sequence of the corresponding Riemann sums Sn converges to the Riemann integral of f on Œa; b W Z

b

lim Sn D

n!1

f .x/ dx a

In beginning calculus texts this is the usual “working definition” of the Riemann integral. The above expression is usually written as lim

maxk xk !0

N X

Z

b

f .k / xk D

kD1

f .x/ dx:

a

The above symbolism can be misleading: Riemann sums need not correspond to partitions of the interval Œa; b that lead to subintervals of the same length. Even for the same partition there are many choices for the intermediate points. Þ There is a converse to the statement of Theorem 1:

4.1 The Riemann Integral

177

Theorem 2. Assume that f W Œa; b !R is bounded and there exists a number I with the property that given any " > 0 there exists ı > 0 such that ˇ n ˇ ˇX ˇ ˇ ˇ f .k / xk I ˇ < " ˇ ˇ ˇ kD1

Pn where kD1 f .k / xk is a Riemann sum for f corresponding to the partition P D fxk gnkD0 with jjPjj < ı. Then f is Riemann integrable on Œa; b and Z

b

f .x/ dx D I: a

Proof. Let " > 0 be given. Choose ı > 0 such that if P D fxk gnkD0 is a partition of Œa; b with jjPjj < ı then ˇ n ˇ ˇX ˇ " ˇ ˇ f .k / xk  I ˇ < ˇ ˇ ˇ 4 kD1 for any set of intermediate points fk gnkD1 . For each k choose k 2 Œxk1 ; xk  so that sup 2Œxk1 ;xk 

f ./ 

" < f .k /  sup f ./ 4n .xk / 2Œxk1 ;xk 

(this is possible by the definition of the least upper bound). Thus Mk 

" < f .k /  Mk : 4n .xk /

Therefore 0  Mk  f .k /
0 be given. Let us enclose these points in closed subintervals I1 ; I2 ; : : : ; IN of Œa; b such that N X ˇ ˇ ˇIj ˇ < ": jD1

180

4 The Riemann Integral

There exists ı > 0 such that ˇ ˇ n Z b ˇ ˇX ˇ ˇ f .k / xk  f .x/ dxˇ < " ˇ ˇ ˇ a kD1

and ˇ ˇ n Z b ˇ ˇX ˇ ˇ g .k / xk  g .x/ dxˇ < " ˇ ˇ ˇ a kD1

if P D fxk gnkD1 is a partition of Œa; b such that jjPjj < ı and fk gnkD1 are corresponding intermediate points. We can assume that the subintervals I1 ; I2 ; : : : ; IN are among the subintervals Jk D Œxk1 ; xk , k D 1; 2; : : : ; n. Let us set n X

‚…„ƒ X P f .k / xk D b f .k / jJk j C f .k / jJk j ;

kD1 n X

‚…„ƒ X P b g .k / xk D g .k / jJk j C g .k / jJk j ;

kD1

where the first sum is over those subintervals other than I1 ; I2 ; : : : ; IN and the second sum is over the subintervals I1 ; I2 ; : : : ; IN . Note that P b f .k / jJk j D P b g .k / jJk j since f .x/ D g .x/ if x does not belong to any of the intervals I1 ; I2 ; : : : ; IN . If jf .x/j  M and jg .x/j  M for each x 2 Œa; b then ˇ ˇ ˇ‚…„ƒ ˇ n ‚…„ƒ n ˇ ˇ ˇ X ˇX X X ˇ ˇ ˇ ˇ f .k / jJk j  f .k / xk  g .k / xk ˇ D ˇ g .k / jJk jˇ ˇ ˇ ˇ ˇ ˇ kD1 kD1 ˇ‚…„ƒ ˇ ˇ‚…„ƒ ˇ ˇ X ˇ ˇ X ˇ ˇ ˇ ˇ ˇ ˇ f .k / jJk jˇ C ˇ g .k / jJk jˇ ˇ ˇ ˇ ˇ  M" C M" D 2M":

4.1 The Riemann Integral

181

We have ˇZ ˇ ˇ ˇ

Z

b

b

f .x/ dx  a

a

ˇ ˇ g .x/ dxˇˇ 

ˇ ˇZ n ˇ ˇ b X ˇ ˇ f .x/ dx  f .k / xk ˇ ˇ ˇ ˇ a kD1 ˇ ˇ n n ˇ ˇX X ˇ ˇ f .k / xk  g .k / xk ˇ Cˇ ˇ ˇ kD1 kD1 ˇ n ˇ Z b ˇX ˇ ˇ ˇ Cˇ g .k / xk  g .x/ dxˇ ˇ ˇ a kD1

< " C 2M" C " D .2M C 2/ ": Since " > 0 is arbitrary we must have Z

b a

Z

b

f .x/ dx D

g .x/ dx

a

as claimed.  Continuous functions are Riemann integrable:

4.1.2 Continuous Functions and Monotone Functions are Integrable Theorem 3. A function f W Œa; b !R that is continuous on a closed and bounded interval Œa; b is Riemann integrable. Proof. Since f is continuous on the closed and bounded interval Œa; b it is bounded and uniformly continuous. Therefore, given " > 0 there exists ı > 0 such that ;  2 Œa; b and j  j < ı ) jf ./  f ./j
0 be given. Choose ı1 so that ı1
f .a/. Given " > 0 let us set " : ıD f .b/  f .a/

184

4 The Riemann Integral

Let P D fxk gnkD1 be a partition of Œa; b with jjPjj < ı. We have Mk  f .xk / and mk  f .xk1 / since f is nondecreasing. Therefore U .f ; P/  L .f ; P/ D

n X

.Mk  mk / xk 

kD1

n X

.f .xk /  f .xk1 // xk

kD1


0 be given. Choose ı > 0 such that ˇ ˇ n Z b ˇ ˇX " ˇ ˇ f .k / xk  f .x/ dxˇ < ˇ ˇ jcj ˇ a kD1

4.2 Basic Properties of the Riemann Integral

187

if jjPjj < ı. We have n X

.cf / .k / xk D

kD1

n X

cf .k / xk D c

kD1

n X

f .k / xk :

kD1

Therefore, ˇ ˇ n ˇ ˇ n Z b Z b ˇ ˇ X ˇ ˇX ˇ ˇ ˇ ˇ .cf / .k / xk  c f .x/ dxˇ D ˇc f .k / xk  c f .x/ dxˇ ˇ ˇ ˇ ˇ ˇ a a kD1 kD1 ˇ n ˇ Z b ˇX ˇ ˇ ˇ D jcj ˇ f .k / x  f .x/ dxˇ ˇ ˇ a kD1   " < jcj D" jcj if jjPjj < ı. Thus, cf is integrable and Z

b a

Z

b

cf .x/ dx D c

f .x/ dx:

a

 Let us recall a definition: Definition 1. A function ' W D R!R is uniformly Lipschitz continuous if there exists a constant K such that j' .u/ ' .v/j  K ju  vj for each u and v in D. The constant K is referred to as a Lipschitz constant. Note that any continuously differentiable function is uniformly Lipschitz continuous on any closed and bounded interval (see Proposition 2 of Sect. 3.4). Proposition 1. Assume that f W Œa; b !R is Riemann integrable and ' is a Lipschitz continuous function on its domain that contains the range of f . Then ' ı f is Riemann integrable on Œa; b. Proof. Assume that j' .u/  ' .v/j  K ju  vj for each u and v in the domain of '. We can assume that K > 0. Let " > 0 be given. Since f is Riemann integrable on Œa; b there exists ı > 0 such that if P D fxk gnkD0 is a partition of Œa; b with jjPjj < ı then U .f ; P/  L .f ; P/
0 be given. Let us choose ı > 0 so that if P is a partition of Œa; b with jjPjj < ı then U .f ; P/  L .f ; P/
0 we can choose ı > 0 such that if P D fxk gnkD0 is a partition of Œa; b with jjPjj < ı then U .g; P/  L .g; P/ < m2 ": By the previous inequality  U

    1 1 1 1  ;P  L ; P < 2 .U .g; P/  L .g; P// < 2 m2 " D ": g g m m

Therefore 1=g is Riemann integrable on Œa; b. 

4.2 Basic Properties of the Riemann Integral

191

Remark 3. The statements of Proposition 2 follow from the continuity of fg and f =g in the case of continuous functions f and g (we need to have minx2Œa;b jg .x/j > 0 in the case of f =g). Þ

4.2.2 Additivity of the Integral with Respect to Intervals Proposition 3. Assume that f W Œa; b !R is Riemann integrable. Then f is Riemann integrable on any subinterval of Œa; b. Proof. Let Œc; d be a subinterval of Œa; b and let " > 0 be given. Since f is integrable on Œa; b there exists ı > 0 such that if P is a partition of Œa; b with jjPjj < ı then U .f ; P/  L .f ; P/ < ": Assume that Q is a partition of Œc; d with jjQjj < ı. We can add points to Q and form a partition P of Œa; b with jjPjj < ı. Then U .f ; Q/  L .f ; Q/  U .f ; P/  L .f ; P/ < " Therefore f is Riemann integrable on Œc; d.  Theorem 2 (Additivity of the Integral with Respect to Intervals). If f is Riemann integrable on Œa; b and Œb; c then f is integrable on Œa; c and Z

c

Z

b

f .x/ dx D

a

a

Z

c

f .x/ dxC

f .x/ dx:

b

Proof. Let " > 0 be given. Let us pick ı > 0 such that if P1 and P2 be partitions of Œa; b and Œb; c, respectively, with jjP1 jj < ı and jjP2 jj < ı then U .f ; P1 /  L .f ; P1 /
0 be given. Since F 0 is Riemann integrable on Œa; b there exists ı > 0 such that if P D fxk gnkD0 is a partition of Œa; b with jjPjj < ı and fk gnkD1 is a corresponding set of intermediate points then ˇ ˇ n Z b ˇ ˇX ˇ ˇ F 0 .k / xk  F 0 .x/ dxˇ < ": ˇ ˇ ˇ a kD1

Let’s pick such a partition P D fxk gnkD0 . We can express the change in the value of F over the interval Œa; b as the sum of the changes in the value of F over the subintervals determined by P. Since F .b/ D F .xn / and F .a/ D F .x0 / we have

198

4 The Riemann Integral

F.b/  F.a/ D ŒF .xn /  F .xn1 / C ŒF .xn1 /  F .xn2 / C    C ŒF.x2 /  F .x1 / C ŒF .x1 /  F .x0 / D

n X

ŒF .xk /  F .xk1 / :

kD1

By the Mean Value Theorem there exists k 2 .xk1 ; xk / such that F .xk /  F .xk1 / D F 0 .k / .xk  xk1 / D F 0 .k / xk : Therefore, F.b/  F.a/ D

n X

F 0 .k / xk :

kD1

Thus, ˇZ ˇ ˇ ˇ

a

b

ˇ ˇ ˇˇZ b n ˇ X ˇ ˇ ˇ 0 0 F .x/ dx  .F .b/  F .a//ˇˇ D ˇ F .x/ dx  F .k / xk ˇ < ": ˇ a ˇ 0

kD1

Since

ˇZ ˇ ˇ ˇ

b a

ˇ ˇ F .x/ dx  .F .b/  F .a//ˇˇ < " 0

and " is an arbitrary positive number, we must have Z b F 0 .x/ dx D F .b/  F .a/ a

 We may refer to the corollary to the Fundamental Theorem of Calculus simply as “the Fundamental Theorem of Calculus.” Remark 1. We may have Z

b

f .x/ dx  0 if f .x/  0 for each x 2 Œa; b a

and Z

b

f .x/ dx  0 if f .x/  0 for each x 2 Œa; b : a

Rb Rb In general a f .x/ dx may be positive, negative, or 0. We may refer to a f .x/ dx as the signed area between the graph of f and the interval Œa; b. Þ

4.3 The Fundamental Theorem of Calculus

199

Example 1. Let F .x/ D

2 3=2 x : 3

By the power rule, d F .x/ D dx 0



   p 2 3=2 2 d 3=2 2 3 1=2 x x D x D D x 3 3 dx 3 2

0 .0/). Thus, F 0 if x  0 (we have to interpret F 0 .0/ as the one-sided derivative FC is continuous, hence Riemann integrable on Œ0; 1. The Fundamental Theorem of Calculus is applicable on Œ0; 1: Z 1 Z 1 p 2 xdx D F 0 .x/ dx D F .1/  F .0/ D : 3 0 0

Thus, the area of the region between the graph of y D 2=3. The region is illustrated in Fig. 4.2. 

p x and the interval Œ0; 1 is

Fig. 4.2

Example 2. Evaluate Z

p 0

=4

  d cos x2 dx dx

  Solution. If we set F .x/ D cos x2 , we have Z

p 0

=4

  d cos x2 dx D dx

by Theorem 1 (Fig. 4.2).

Z

p =4 0

r    F .0/ 4 p  2 D cos  cos .0/ D  1; 4 2

d F .x/ dx D F dx

200

4 The Riemann Integral

If we set f .x/ D then f .x/  0 if 0  x  p

Z 0

=4

    d cos x2 D 2x sin x2 dx

p =4. We have

p p 2 2 1 a, then F.x/ is the signed area of the region between the graph of f and the interval Œa; x. If x < a, we have Z

a

f .x/dx;

F.x/ D  x

so that F.x/ is .1/ times the signed area of the region between the graph of f and the interval Œa; x. Also note that Z

a

F.a/ D

f .t/dt D 0:

a

We are able to differentiate F .x/: Theorem 2 (The Fundamental Theorem of Calculus, Part 2). Assume that f is integrable on the interval J, and a is a point in J. If Z

x

F .x/ D

f .t/ dt for each x 2 J

a

and f is continuous at x0 2 J then F 0 .x0 / D f .x0 / (the derivative is the appropriate one-sided derivative if x0 is an endpoint of J that belongs to J). The derivative should be interpreted as the appropriate one-sided derivative at an endpoint of J. Remark 2. The second part of the Fundamental Theorem of Calculus asserts that d dx

Z

x

f .t/ dt D f .x/ a

for each x 2 J if f is continuous on J. Therefore Z

x

F .x/ D

f .t/ dt a

defines an antiderivative of f on J: Þ

4.3 The Fundamental Theorem of Calculus

205

A plausibility argument for Theorem 2: We have Z

x0 Cx

F.x0 C x/  F.x0 / D

Z

x0

f .t/dt 

a

Z

x0 Cx

f .t/dt D

a

f .t/dt:

x0

Fig. 4.4 R x0 Cx f .t/ dt Š f .x0 / x x0 f(x0)

a

x0

x x0

x

t

With reference to Fig. 4.4, if x > 0 and small, this quantity is approximately the area of the rectangle that has as its base the interval Œx0 ; x0 C x and has height f .x0 /. Therefore F.x0 C x/  F.x0 / Š f .x0 /x; so that F.x0 C x/  F.x0 / Š f .x0 / x if x is small. Thus, it is plausible that F 0 .x0 / D lim

x!0

F.x0 C x/  F.x0 / D f .x0 /: x

The Proof of Theorem 2. We will show that F 0 .x0 / D f .x0 / at a point x0 in the interior of J. If x0 is an endpoint of J, the equality of the appropriate one-sided derivative of F and f .x0 / is established in a similar manner. Let x > 0. As in our plausibility argument, Z x0 Cx F.x0 C x/  F.x0 / D f .t/dt: x0

Therefore, F.x0 C x/  F.x0 / 1 D x x

Z

x0 Cx x0

f .t/dt:

206

4 The Riemann Integral

Note that  f .x0 / D f .x0 /

1 x

Z

x0 Cx x0

 Z x0 Cx 1 1dt D f .x0 /dt: x x0

Therefore ˇ ˇ ˇ F.x0 C x/  F.x0 / ˇ ˇ ˇ  f .x / 0 ˇ D ˇ x

ˇ ˇ Z x0 Cx Z x0 Cx ˇ ˇ 1 1 ˇ ˇ f .t/dt  f .x /dt 0 ˇ ˇ x x x0 x0 ˇ ˇ Z x0 Cx ˇ ˇ 1 .f .t/  f .x0 // dtˇˇ D ˇˇ x x0 ! 1  sup jf .t/  f .x0 /j x x x0 tx0 Cx

D

sup

x0 tx0 Cx

jf .t/  f .x0 /j :

Since f is continuous at x0 we have lim

x!0

sup

!

x0 tx0 Cx

jf .t/  f .x0 /j D 0:

Therefore lim

x!0C

F.x0 C x/  F.x0 / D f .x0 / : x

If x < 0 ˇ ˇ ˇ ˇ Z x0 Cx ˇ ˇ 1 ˇ ˇ F.x0 C x/  F.x0 / ˇ ˇ ˇ  f .x0 /ˇ D ˇ .f .t/  f .x0 // dtˇˇ : ˇ x x x0 as before. We have ˇ ˇZ ˇ ˇ Z x0 Cx ˇ 1 ˇ ˇ 1 ˇˇ x0 Cx ˇ ˇ ˇ .f .t/  f .x // dt .f .t/  f .x // dt D 0 0 ˇ x ˇ ˇ ˇ .x/ x0 x0 ˇ Z x0 Cx ˇ ˇ 1 ˇˇ ˇ  .f .t/  f .x // dt D 0 ˇ ˇ .x/ x0 ˇZ ˇ ˇ 1 ˇˇ x0 ˇ D .f .t/  f .x // dt 0 ˇ ˇ .x/ x0 Cx 1  .x/ D

sup

sup

x0 Cxtx0

x0 Cxtx0

jf .t/  f .x0 /j .x/

jf .t/  f .x0 /j :

!

4.3 The Fundamental Theorem of Calculus

207

Therefore F.x0 C x/  F.x0 / D f .x0 / x!0 x lim

as well. Thus, F 0 .x0 / as D f .x0 / as claimed.  Example 5. The sine integral function Si is defined by the expression Z x sin .t/ Si .x/ D dt t 0 Determine Si0 .x/. Solution. Since sin .t/ D 1; t!0 t lim

if we set

8 sin .t/ ˆ ˆ < t f .t/ D ˆ ˆ : 1

if

t ¤ 0;

if

t D 0;

then f is continuous on the entire number line. We can interpret the integral Z x sin .t/ dt t 0 as

Z

x 0

f .t/ dt:

Thus, the second part of the Fundamental Theorem of Calculus is applicable: 8 sin .x/ ˆ ˆ Z x if x D 0; < d d x Si .x/ D f .t/ dt D ˆ dx dx 0 ˆ : 1 if x D 0: Figure 4.5 shows the graph of f and Fig. 4.6 shows the graph of the sine integral function Si. 

208 Fig. 4.5 y D

4 The Riemann Integral y

sin.t/ t

1

0.5

−4p

2p

−2p

4p

t

y

Fig. 4.6 The sine integral function

1.5

4π

2π

2π

4π

x

−1.5

Example 6. In beginning calculus you may have seen the definition of the natural logarithm as an integral: Z ln .x/ D

x 1

1 dt for each x > 0: t

Thus d 1 ln .x/ D if x > 0 dt x so that the natural logarithm is an antiderivative of 1=x on .0; C1/. All the basic properties of the natural logarithm can be deduced from the above definition. Then the natural exponential function can be defined as the inverse of the natural logarithm. That program is carried out in many calculus texts. In Sect. 6.6 we will discuss an alternative approach and define the natural exponential function in terms of a power series. Then the natural logarithm can be defined as the inverse of the natural exponential function. 

4.3 The Fundamental Theorem of Calculus

209

Now that we have established the second part of the Fundamental Theorem of Calculus, let us display both parts of the Theorem in a symmetric fashion (assuming that f is continuous and f 0 is integrable on an interval that contains the relevant points): The Fundamental Theorem of Calculus 1. Z

x a

df .t/ dt D f .x/  f .a/: dt

2. d dx

Z

x

f .t/dt D f .x/: a

It is worthwhile to repeat the meaning of the Fundamental Theorem: The first part of the theorem states that the integral of the derivative of a function on an interval is the difference between the values of the function at the endpoints. The second part of the theorem states that the derivative of the function Z x f .t/dt a

is the value of the integrand at the upper limit. We can say that differentiation and integration are inverse operations in the precise sense of the Fundamental Theorem. We may refer to either part of the Fundamental Theorem of Calculus simply as “the Fundamental Theorem of Calculus.”

4.3.3 Problems 1. Assume that f is continuous on Œa; b, g is differentiable on Œc; d, g .Œc; d/  Œa; b and Z g.x/ f .t/ dt F .x/ D a

for each x 2 Œc; d. Prove that F 0 .x/ D f .g .x// g0 .x/ for each x 2 .c; d/. Hint: The Fundamental Theorem of Calculus and the chain rule.

210

4 The Riemann Integral

2. Determine the derivative: a) d dx

Z

x 0

2 2 p et dt 

b) d dx

Z 1p 4 C t2 dt x

c) d dx

Z

x2 0

q

1  dt  .1  t2 / 1  14 t2

d) d dx

Z

x3 p x2

t2 C 1dt

4.4 The Substitution Rule and Integration by Parts In this section we will prove the substitution rule and integration by parts that you are familiar with from beginning calculus. For simplicity we will assume that the relevant functions are continuous on the relevant intervals.

4.4.1 The Substitution Rule for Indefinite Integrals The substitution rule is the counterpart of the chain rule for differentiation: Theorem 1 (The Substitution Rule for Indefinite Integrals). Assume that f is continuous on the interval I, u is a differentiable function on the interval J and u.x/ 2 I if x 2 J. Then Z f .u .x// where x 2 J.

du dx D dx

Z f .u/ du

4.4 The Substitution Rule and Integration by Parts

211

The expression Z f .u/du denotes a function of u. It should be understood that the above equality is valid, provided that u is replaced by its expression in terms of x. Since the equality involves indefinite integrals, we are entitled to add arbitrary constants to either side. The Proof of Theorem 1. By the second part of the Fundamental Theorem of Calculus, a continuous function f has an antiderivative F on the interval I. Thus, d F .u/ D f .u/ , F .u/ D du

Z f .u/ du

on I. Let us consider the composite function F ı u on J. By the chain rule, ! ˇ  ˇ d d du ˇ F.u/ˇ u.x/ D f .u.x// du dx dx uDu.x/

d d .F ı u/ .x/ D F.u.x// D dx dx for each x 2 J. Therefore Z f .u.x//

du dx D F .u .x// dx

on J. Since Z F .u/ D

f .u/ du

we have Z f .u.x//

du dx D F .u .x// D dx

Z

ˇ ˇ f .u/ duˇˇ

uDu.x/

on J. Therefore, Z f .u .x//

du dx D dx

Z f .u/du;

with the understanding that the right-hand side is evaluated at u .x/.  It is easy to remember the substitution rule: In the expression Z f .u .x//

du dx; dx

212

4 The Riemann Integral

we can treat du dx as a “symbolic fraction,” carry out “symbolic cancellation” and write Z f .u .x//

du dx D dx

Z f .u/du:

Thus, we can set du D

du dx dx

when we implement the substitution rule. There is no need to try to attach a mystical meaning to the symbolic manipulation, though: We are merely describing a practical way to remember the substitution rule. Within the present context, the symbol du dx dx does not express the value of the differential du .x; dx/ that we discussed in Sect. 3.1, even though the notation is the same. Nevertheless, we will establish a connection between the two usages of the same notation at the end of this section. Example 1. Determine Z

p x 4  x2 dx:

Solution. Let us set u D 4  x2 . Then  d  du D 4  x2 D 2x: dx dx Therefore, du D

du dx D 2xdx: dx

By the substitution rule, Z

Z Z Z p 1 p 1 p du 1 2 2 x 4  x dx D  4  x .2x/ dx D  u dx D  u1=2 du: 2 2 dx 2

4.4 The Substitution Rule and Integration by Parts

213

By the reverse power rule, 

1 2

Z

u1=2 du D 

1 2



u3=2 3=2



1 C C D  u3=2 C C; 3

where C is an arbitrary constant. Therefore, Z

ˇ p ˇ 3=2 1 1 D  4  x2 C C: x 4  x2 dx D  u3=2 C Cˇˇ 3 3 uD4x2



4.4.2 The Substitution Rule for Definite Integrals An indefinite integral that is determined with the help of the substitution rule can be used to evaluate a definite integral, as in the above examples. There is also a version of the substitution rule which applies directly to definite integrals: Theorem 2 (The Substitution Rule for Definite Integrals). Assume that f is continuous on the interval determined by u.a/ and u .b/,and that du=dx is continuous on the interval Œa; b.Then Z

b a

du f .u .x// dx D dx

Z

u.b/

f .u/ du:

u.a/

Proof. The substitution rule for definite integrals is derived in a way that is similar to the derivation of the substitution rule for indefinite integrals. Let F be an antiderivative of f in the interval determined by u.a/ and u.b/. Thus, d F.u/ D f .u/ du if u between u.a/ and u.b/. By the chain rule, ! ˇ dF ˇˇ du du D f .u .x// du ˇuDu.x/ dx dx

d F .u .x// D dx

if x 2 Œa; b. The first part of the Fundamental Theorem of Calculus implies that Z F .u.b//  F .u .a// D a

b

d F .u .x// dx D dx

Z a

b

f .u .x//

du dx: dx

214

4 The Riemann Integral

The first part of the Fundamental Theorem of Calculus also implies that Z

u.b/

F .u.b//  F .u .a// D u.a/

Therefore, we must have Z b

f .u .x//

a

dF.u/ du D du Z

du.x/ dx D dx

u.b/

Z

u.b/

f .u/du: u.a/

f .u/du;

u.a/

as claimed.  Example 2. Evaluate Z

=2 0

cos2=3 .x/ sin .x/ dx

by using the substitution rule for definite integrals. Solution. We set u D cos .x/ so that   du d dx D cos .x/ dx D  sin .x/ dx: du D dx dx Therefore, Z

=2

2=3

cos 0

Z

uDcos.=2/

.x/ sin .x/ dx D

u

2=3

uDcos.0/

Z

0

D

u2=3 du D

1

  du dx  dx Z

1 0

ˇ 3 5=3 ˇˇ1 3 D u ˇ D : 5 5 0

ˇ1 ˇ ˇ u ˇ 2=3 u du D ˇ 2 ˇ C 1ˇ 3 0 2 3 C1

 The definite integral version of the substitution rule does not offer an advantage over the indefinite integral version of the rule if Z f .u .x//

du dx D dx

Z

and Z f .u/ du

f .u/ du

4.4 The Substitution Rule and Integration by Parts

215

can be expressed in terms of familiar functions. On the other hand, the substitution rule for definite integrals leads to useful facts about integrals, as in the following proposition: Proposition 1. a) If f is even and continuous on Œa; a, then Z

Z

a a

f .x/dx D 2

a 0

f .x/dx:

b) If f is odd and continuous on Œa; a, then Z

a a

f .x/ dx D 0:

Both parts of Proposition 1 are plausible. If f is even, the graph of f is symmetric with respect to the vertical axis. With reference to Fig. 4.7, the area of GL is the same as the area of GR . Fig. 4.7

Thus, Z

a a

Z f .x/ dx D

0 a

Z f .x/ dx C

a 0

f .x/ dx D .area of the GL / C .area of GR / Z D 2  .area of GR / D 2

0

a

f .x/dx:

If f is odd, the graph of f is symmetric with respect to the origin. With reference to Fig. 4.8, the signed area of G is .1/  the area of GC .

216

4 The Riemann Integral

Fig. 4.8

Thus, Z

a

a

Z f .x/ dx D

0

a

Z f .x/dx C

a 0

f .x/ dx D .the signed area of G / C .the area of GC / D 0:

The Proof of Proposition 1. We will prove part a) and leave the similar proof of part b) as an exercise. Thus, assume that f is even. By the additivity of integrals with respect to intervals, Z

Z

a

a

f .x/dx D

Z

0 a

f .x/dx C

a 0

f .x/dx:

Since f is even, we have f .x/ D f .x/. Therefore, Z

Z

0 a

f .x/dx D

0 a

f .x/ dx:

Let us apply the substitution rule to this integral by setting u D x. Then, du=dx D 1, so that Z

0 a

Z f .x/dx D 

0

a

Z

Z

u.0/

D

0

du dx dx a Z 0 Z a f .u/du D  f .u/du D f .u/du:

f .u/.1/dx D 

u.a/

f .u/

0

a

Thus, Z

0 a

Z f .x/dx D

0

a

Z f .u/du D

a 0

f .x/dx

4.4 The Substitution Rule and Integration by Parts

217

(the variable of integration is a dummy variable). Therefore, Z

Z

a a

f .x/dx D

0 a

Z f .x/dx C

Z

a 0

f .x/dx D

Z

a

f .x/dx C

0

Z

D2

a 0

f .x/dx

a

f .x/dx;

0

as claimed. 

4.4.3 Integration by Parts for Indefinite Integrals Integration by parts is the counterpart of the product rule for differentiation: Theorem 3 (Integration by Parts for Indefinite Integrals). Assume that f and g are differentiable in the interval J. Then, Z

dg f .x/ dx D f .x/g.x/ dx

Z g.x/

df dx dx

for each x 2 J. We can use the ‘prime notation,” of course: Z

f .x/ g0 .x/ dx D f .x/ g.x/ 

Z

g .x/ f 0 .x/ dx

Proof. By the product rule, d df dg .f .x/ g .x// D g .x/ C f .x/ dx dx dx for each x 2 J. This is equivalent to the statement that f 0 g C fg0 is an antiderivative of fg. Thus, Z  f .x/g.x/ D

 df dg g.x/ C f .x/ dx dx dx

for each x 2 J. By the linearity of indefinite integrals, Z f .x/g.x/ D

df g.x/dx C dx

Z f .x/

dg dx dx

218

4 The Riemann Integral

Therefore, Z

dg f .x/ dx D f .x/g.x/  dx

Z g.x/

df dx dx

for each x 2 J.  The symbolic expression du D

du dx dx

is helpful in the implementation of the substitution rule. This symbolism is also helpful in the implementation of integration by parts. In the expression Z f .x/

dg dx D f .x/g.x/  dx

Z g.x/

df dx; dx

let us replace f .x/ by u and g .x/ by v. Thus, Z

dv u dx D uv  dx

Z v

du dx: dx

Let us also replace du dx dx by du, and dv dx dx by dv. Therefore, we can express the formula for integration by parts as follows: Z

Z udv D uv 

vdu:

Note that Z vD

dv dx D dx

Z

Example 3. Determine Z x sin .4x/ dx:

dv:

4.4 The Substitution Rule and Integration by Parts

219

Solution. We will apply the formula for integration by parts in the form Z

Z udv D uv 

vdu;

with u D x and dv D sin .4x/ dx. Therefore, du D

du dx D dx; dx

and Z vD

Z dv D

1 sin.4x/dx D  cos .4x/ : 4

Therefore, Z

Z x sin .4x/ dx D

udv Z

D uv 

vdu

  Z   1 1 D x  cos .4x/   cos .4x/ dx 4 4 Z 1 1 D  x cos .4x/ C cos .4x/ dx 4 4   1 1 1 D  x cos .4x/ C sin .4x/ C C 4 4 4 1 1 D  x cos .4x/ C sin .4x/ C C; 4 16 where C is an arbitrary constant. The above expression is valid on the entire number line. 

4.4.4 Integration by Parts for Definite Integrals Theorem 4 (The Definite Integral Version of Integration by Parts). Assume that f 0 and g0 are continuous on Œa; b. Then, Z

b

f .x/ a

dg dx D Œf .b/g.b/  f .a/g.a/  dx

Z

b

g.x/ a

df dx dx

220

4 The Riemann Integral

Proof. As in the proof of the indefinite integral version of integration by parts, the starting point is the product rule for differentiation: d df dg .f .x/ g .x// D g .x/ C f .x/ dx dx dx Since f 0 and g0 are continuous on Œa; b, fg, f 0 g and fg0 are all continuous, hence integrable, on Œa; b. By the Fundamental Theorem of Calculus, Z

b a

d .f .x/ g .x// dx D f .b/ g .b/  f .a/ g .a/ : dx

Therefore, Z

b

 dg df g .x/ C f .x/ dx dx dx Z b dg df g .x/ dx C f .x/ dx: dx dx a



f .b/ g .b/  f .a/ g .a/ D Z

a b

D a

Thus, Z

b a

dg f .x/ dx D Œf .b/ g .b/  f .a/ g .a/  dx

Z

b

g .x/ a

df dx dx

 As in the case of indefinite integrals, we can express the definite integral version of integration by parts by using the symbolism du D

du dx: dx

Indeed, if we replace f .x/ by u and g .x/ by v, we have Z

Z

b

b

udv D Œu .b/ v .b/  u .a/ v .a/  a

Z D uvjba 

vdu a

b

vdu:

a

Example 4. Assume that f 00 and g00 are continuous on the interval Œa; b and that f .a/ D f .b/ D g .a/ D g .b/ D 0. Show that Z

b a

f 00 .x/g.x/dx D

Z

b a

f .x/g00 .x/dx:

4.4 The Substitution Rule and Integration by Parts

221

Solution. We apply integration by parts: Z

b

f 00 .x/g.x/dx D

Z

a

 0 0 f .x/g.x/dx

b a

  D f 0 .b/g.b/  f 0 .a/g.a/ 

Z

b

f 0 .x/g0 .x/dx

a

Z

b

D

f 0 .x/g0 .x/dx;

a

since g.a/ D g.b/ D 0. We apply integration by parts again, by setting u D g0 .x/ and dv D f 0 .x/ dx. Therefore, du D g00 .x/ dx and v D f .x/ : Thus, Z

b

g0 .x/f 0 .x/dx D

a

Z

b

udv a

Z

b

D u.b/v.b/  u.a/v.a/  a 0

0

vdu

Z

b

D g .b/ f .b/  g .a/ f .a/ 

f .x/ g00 .x/ dx

a

Z

b

D

f .x/ g00 .x/ dx;

a

since f .a/ D f .b/ D 0. Therefore, Z

b

f 00 .x/g.x/dx D 

a

Z

b

 Z f 0 .x/g0 .x/dx D  

a

b

 Z f .x/ g00 .x/ dx D

a

a

as claimed. 

4.4.5 Problems 1. The function erf is defined by the expression 2 erf .x/ D p  for each x 2 R.

Z 0

x

b

2

et dt

f .x/ g00 .x/ dx;

222

4 The Riemann Integral 2

a) Determine an antiderivative of ex in terms of erf : b) Make use of the result of part a) to evaluate Z

4

2

ex dx

2

2. Assume that a > 0 and f is an odd function that is continuous on the interval Œa; a. Prove that Z

a

f .x/ dx D 0: a

3. Assume that f is continuous on R and periodic with period p. Show that Z

Z

aCp

f .x/ dx D a

p 0

f .x/ dx

for any a 2 R. 4. Assume that f , f 0 and f 00 are continuous on Œa; b and f .a/ D f .b/ D 0. Then Z

b

Z

00

b

f .x/ f .x/ dx D 

a

 0 2 f .x/ dx:

a

4.5 Improper Integrals: Part 1 In this section we will expand the scope of the integral to cover some cases that involve unbounded intervals or functions with discontinuities.

4.5.1 Improper Integrals on Unbounded Intervals Definition 1. Assume that f is (Riemann) integrable on R 1any interval of the form Œa; b where b > a. We say that the improper integral a f .x/ dx converges if Z lim

b!C1 a

b

f .x/ dx

exists (as a finite limit). In this case we define the value of the improper integral as the above limit, and denote it by the same symbol. Thus,

4.5 Improper Integrals: Part 1

Z

223 1

Z f .x/ dx D lim

b

b!C1 a

a

We say that the improper integral

R1 a

f .x/ dx diverges if Z

lim

f .x/ dx:

b!C1 a

b

f .x/ dx

does not exist. In the case of convergence we may interpret the value of an improper integral as the signed area of the region between the graph of f and the interval Œa; C1/, just as in the case of a bounded interval. Example 1. Let f .x/ D 1=x2 . We have ˇ Z b Z b 1 1 ˇˇb 1 f .x/ dx D dx D  ˇ D  b C 1: 2 x x 1 1 1 Therefore, Z lim

b!C1 1

b

  1  C 1 D 1: b!C1 b

f .x/ dx D lim

Thus the improper integral of f on the interval Œ1; C1/ converges and has the value 1: Z 1 Z b 1 1 dx D lim dx D 1: 2 b!C1 1 x2 x 1 Since we can interpret the integral of f from 1 to b as the area between the graph of f and the interval Œ1; b, it is reasonable to say that the area of the region G between the graph of f and the interval Œ1; C1/ is 1. The region G is illustrated in Fig. 4.9.  Fig. 4.9

224

4 The Riemann Integral

Example 2. Determine whether the improper integral Z

1 1

1 dx x

converges or diverges, and its value if it converges. Solution. For any b > 1, Z

b

1

1 dx D ln .b/ x

Therefore, Z

b

lim

b!C1 1

1 dx D lim ln .b/ D C1: b!C1 x

This is merely shorthand for the statement that ln .b/ is arbitrarily large if b is large enough. Thus, the improper integral Z

1 1

1 dx x

diverges. 

R1 Remark 1. Note that 1 1=x dx diverges even though limx!1 1=x D 0. Thus the fact thatRlimx!1 f .x/ D 0 is not sufficient for the convergence of the improper 1 integral a f .x/ dx: Þ R1 Examples 1 and 2 involve improper integrals of the type a 1=xp dx. Let us record the general case for future reference: Proposition 1. Assume R 1 that p is an arbitrary real number and that a > 0. The improper integral a 1=xp dx converges if p > 1 and diverges if p  1: Proof. Let p D 1. We have Z

b a

1 dx D ln .b/  ln .a/ , x

so that Z lim

b!C1 a

b

1 dx D lim .ln .b/  ln .a// D C1: b!C1 x

Therefore the improper integral

R1 a

1=x dx diverges.

4.5 Improper Integrals: Part 1

225

Now assume that p ¤ 1. We have Z a

b

1 dx D xp

Z

b a

ˇb apC1 xpC1 ˇˇ bpC1  : x dx D D p C 1 ˇa p C 1 p C 1 p

If p > 1, Z lim

b

b!C1 a

  pC1 apC1 b 1  dx D lim b!C1 p C 1 xp p C 1   1 1 D lim  C b!C1 .p  1/bp1 .p  1/ ap1 1 ; .p  1/ ap1

D

since limb!C1 1=bp1 D 0. Therefore, the improper integral and Z 1 1 1 dx D : p x .p  1/ ap1 a

R1 a

1=xp dx converges

Now assume that p < 1. We have Z

b a

apC1 b1p a1p 1 bpC1  D  : dx D xp p C 1 p C 1 1p 1p

Since 1  p > 0, we have limb!C1 b1p D C1. Therefore, Z

b

lim

b!C1 a

so that the improper integral

R1 a

1 dx D C1; xp

1=xp dx diverges. 

Example 3. Determine whether the improper integral Z

1 0

sin .x/ dx

converges or diverges. Compute its value if it converges. Solution. Since Z sin .x/ dx D  cos .x/ ;

226

4 The Riemann Integral

we have Z 0

b

sin .x/ dx D  cos .x/jb0 D  cos .b/ C 1

for any b. The limit of  cos .b/ C 1 as x tends to C1 does not exists. Indeed,   C1D1  cos .2n C 1/ 2 and  cos .2n/ C 1 D 0 for n D 1; 2; 3; : : :. Therefore, the improper integral Z

1

sin .x/ dx

0

diverges. Note that Z 0

b

sin .x/ dx  2

0

for each b > 0, since 0   cos .b/ C 1  2.  Remark 2. Example 3 shows that an improper integral Z

1

f .x/ dx

a

may diverge, even though Z

b

f .x/ dx a

does not diverge to infinity as b tends to infinity. Þ We can also consider improper integrals of the form

Ra 1

f .x/ dx:

Definition 2. Assume that f is integrable Ron any interval of the form Œb; a where a b < a. We say that the improper integral 1 f .x/ dx converges if Z lim

b!1 b

a

f .x/ dx

4.5 Improper Integrals: Part 1

227

exists (as a finite limit). In this case we define the value of the improper integral as the limit and denote it by the same symbol. Thus, Z a Z a f .x/ dx D lim f .x/ dx: 1

b!1 b

If the above limit does not exist, we say that the improper integral diverges.

Ra 1

f .x/ dx

Example 4. Determine whether the improper integral Z 1 1 dx 2C1 x 1 converges or diverges, and its value if it converges. Solution. We have Z 1 1  dx D arctan .x/j1  arctan .b/ : b D arctan .1/  arctan .b/ D  2 x C1 4 b Therefore, Z lim

b!1 b

1

   1  dx D lim  arctan .b/ D   lim arctan .b/ 2 b!1 x C1 4 4 b!1     D   D : 4 2 4

Thus, the given improper integral converges, and we have Z

1 1

1 dx D lim b!1 x2 C 1

Z

1 b

 1 dx D : x2 C 1 4

  Figure 4.10 shows the graph of f , where f .x/ D 1= x2 C 1 . The area between the graph f and the interval .1; 1 is =4.  Fig. 4.10

-1

We may also consider “two-sided” improper integrals:

x

228

4 The Riemann Integral

Definition 3. Assume that f is integrable on any closed and bounded interval. We R C1 if there exists an a 2 R say that the improper integralR 1 f .x/ dx converges R1 a such that the improper integrals 1 f .x/ dx and a f .x/ dx converge. If this is the case, we define the value of the improper integral to be the sum of the values of the “one-sided” improper integrals and set Z

C1 1

The improper integral diverges.

Z f .x/ dx D

R C1 1

a 1

Z f .x/ dxC

1

f .x/ dx:

a

f .x/ dx is said to diverge if

Ra 1

f .x/ dx or

R1 a

f .x/ dx

R C1 Remark 3. The convergence or divergence of the improper integral 1 f .x/ dx, and its value in the case of convergence, do not depend on “the intermediate point” a. This assertion should be plausible due to the geometric interpretation of the integral (Fig. 4.11). Þ

Fig. 4.11 The area under the graph of f does not depend on the choice of a or b as the intermediate point

Remark 4 (Caution). We have not defined the improper integral Z

b

lim

b!C1 b

Let’s consider the improper integral Z

b

lim

b!C1 0

R C1 1

f .x/ dx:

xdx. Since

b2 D C1, b!C1 2

xdx D lim

this improper integral diverges. On the other hand, we have Z

b

lim

b!C1

b

 xdx D lim .0/ D 0: b!C1

R C1 1

f .x/ dx as

4.5 Improper Integrals: Part 1

229

Note that f .x/ D x defines an odd function, so that the graph of f is symmetric with Rb respect to the origin (Fig. 4.12). Therefore, it is not surprising that b f .x/ dx D 0. Fig. 4.12

Our definition of the convergence of the improper integral to the requirement that Z

R C1 1

f .x/ dx amounts

c

f .x/ dx b

approaches a definite number as b ! 1 and c ! C1, with no prior restriction such as b D c. Þ Example 5. Determine whether the improper integral Z

1 1

x dx x2 C 1

converges or diverges, and its value if it converges. Solution. We will select 0 as the intermediate point. Both improper integrals Z

0 1

x dx and 2 x C1

Z

1

x2

0

x dx C1

must converge for the convergence of the given “two-sided” improper integral. We set u D x2 C 1 so that du=dx D 2. By the definite integral version of the substitution rule, Z

b 0

1 x dx D 2 x C1 2

Z 0

b

1 x du dx D 2 x C 1 dx 2

Z

b2 C1 1

 1  1 du D ln b2 C 1 : u 2

230

4 The Riemann Integral

Therefore, Z lim

b!C1 0

b

   1  2 x dx D lim ln b C 1 D C1; b!C1 2 x2 C 1

Thus, the improper integral Z

1

x2

0

x dx C1

diverges. This is sufficient to conclude that the improper integral Z

1 1

x dx x2 C 1

diverges.   Note that f .x/ D x= x2 C 1 defines an odd function, so that the graph of f is symmetric with respect to the origin, as shown in Fig. 4.13. Fig. 4.13

0.4

4

−4

−0.4

We have Z

b b

Z f .x/ dx D

b b

x2

x dx D 0 C1

for each b 2 R, so that Z lim

b

b!C1 b

Nevertheless, the improper integral

x dx D 0: x2 C 1

R C1 1

  x= x2 C 1 dx does not converge. 

x

4.5 Improper Integrals: Part 1

231

4.5.2 Improper Integrals That Involve Discontinuous Functions A function that is bounded on an interval Œa; b and continuous in the interior .a; b/ of Œa; b is Riemann integrable on that interval. We will extend the definition of the integral to certain cases that involve functions with unbounded discontinuities. Definition 4. Assume that f is integrable on any interval of the form Œa C "; b that is contained in Œa; b and that limx!aC f .x/ does not exist (as a finite number). The Rb improper integral a f .x/ dx is said to converge if Z

b

f .x/ dx

lim

"!0C

aC"

exists. In this case we define the value of the improper integral as the above limit and denote it by the same symbol: Z

Z

b

b

f .x/ dx D lim

"!0C aC"

a

f .x/ dx:

If Z

b

f .x/ dx

lim

"!0C

aC"

does not exist, we say that the improper integral

Rb a

f .x/ dx diverges.

With reference to Definition 4, we can set c D a C ", so that c approaches a from the right as " approaches 0 through positive values. Therefore, we have Z

Z

b

Z

b

f .x/ dx D lim a

"!0C aC"

f .x/ dx D lim

c!aC c

The improper integral diverges if Z lim

c!aC c

b

f .x/ dx

does not exist. Example 6. Let 1 f .x/ D p ; x > 0: x

b

f .x/ dx:

232

4 The Riemann Integral

We have limx!0C f .x/ D C1. Since f is continuous on any interval of the form Œ"; 1 where 0 < " < 1, the integral of f on such an interval exists, and corresponds to the area between the graph of f and the interval Œ"; 1, as indicated in Fig. 4.14. Fig. 4.14

We have Z

1 "

Z f .x/ dx D

1

x

1=2

"

ˇ1 p x1=2 ˇˇ dx D D 2  2 ": ˇ 1=2 "

Therefore the improper integral of f on Œ0; 1 exists and we have Z 1 Z 1  p  1 p dx D lim f .x/ dx D lim 2  2 " D 2: "!0C "!0C x 0 " We can interpret the result as the area between the graph of f and the interval Œ0; 1.  Example 7. Consider Z

1 0

ln .x/ dx

a) Why is the integral an improper integral? b) Determine whether the improper integral converges or diverges, and its value if it converges. Solution. a) The given integral is improper since lim ln .x/ D 1:

x!0C

b) Let 0 < " < 1. We have Z 1 ln .x/ dx D x ln .x/  xj1" D 1  ." ln ."/  "/ D 1  " ln ."/ C ": "

4.5 Improper Integrals: Part 1

233

Recall that lim"!0C " ln ."/ D 0 (you may appeal to L’Hôpital’s rule). Therefore, Z 1 ln .x/ dx D lim .1  " ln ."/ C "/ D 1: lim "!0C

"!0C

"

Thus, the given improper integral converges, and we have Z 1 Z 1 ln .x/ dx D lim ln .x/ dx D 1: "!0C

0

"

We can interpret the value of the improper integral as the signed area of the region G between the graph of the natural logarithm and the interval Œ0; 1. G is illustrated in Fig. 4.15.  Fig. 4.15

R1 In Example 6 we showed that 0 1=x1=2 dx converges. Let us state the following generalization for future reference: Proposition 2. Assume that p > 0 and a > 0. The improper integral Z

a 0

1 dx xp

converges if p < 1 and diverges if p  1: Proof. Let’s begin with the case p D 1. Let 0 < " < a. We have Z

a

lim

"!0C "

1 dx D lim .ln .a/  ln ."// D C1; "!0C x

since lim"!0C ln ."/ D 1. Therefore, the improper integral Assume that p ¤ 1 and 0 < " < a. We have Z "

a

1 dx D xp

Z

a "

xp dx D

Ra 0

1=x dx diverges.

ˇa "pC1 xpC1 ˇˇ apC1  : D ˇ p C 1 " p C 1 p C 1

234

4 The Riemann Integral

If 0 < p < 1 then 1  p > 0, so that lim"!0C "1p D 0. Therefore, Z

a

lim

"!0C "

  1p "1p a1p 1 a  D : dx D lim "!0C 1  p xp 1p 1p

Ra Thus, the improper integral 0 1=xp dx converges (and has the value a1p = .1  p/). If p > 1 then 1  p < 0 so that lim"!0C "1p D C1. Therefore, Z lim

"!0C "

a

  1p "1p a 1  D C1: dx D lim "!0C 1  p xp 1p

Thus, the improper integral

Ra 0

1=xp dx diverges. 

If f is integrable on any interval of the form Œa; b  " that is contained in Œa; b and Rb limx!b f .x/ does not exist, the integral a f .x/ dx is improper: Definition 5. Assume that f is integrable on any interval of the form Œa; b  " that is contained in Œa; b and that limx!b f .x/ does not exist (as a finite number). The Rb improper integral a f .x/ dx is said to converge if Z

b"

f .x/ dx

lim

"!0C

a

exists. In this case the value of the improper integral is Z b Z b" f .x/ dx D lim f .x/ dx: "!0C a

a

If Z

b"

f .x/ dx

lim

"!0C

a

does not exist, we say that the improper integral

Rb a

f .x/ dx diverges.

With reference to Definition 5, we can set c D b  ", so that c approaches b from the left as " approaches 0 through positive values. Therefore, we have Z

Z

b a

Z

b"

f .x/ dx D lim

f .x/ dx D lim

"!0C a

c!b a

The improper integral diverges if Z lim

c!b a

does not exist.

c

f .x/ dx

c

f .x/ dx:

4.5 Improper Integrals: Part 1

235

Example 8. Consider Z

2

1 .x  2/1=3

1

dx

a) Why is the integral an improper integral? b) Determine whether the improper integral converges or diverges, and its value if it converges. Solution. a) The integral is improper since lim

1

x!2

.x  2/1=3

D 1:

b) If 0 < " < 1, we have Z

2" 1

Z

1 .x  2/1=3

dx D

" 1

Z

1 u1=3

du D

"

u

1=3

1

ˇ" 3u2=3 ˇˇ 3"2=3 3  : du D D 2 ˇ1 2 2

Therefore Z

2"

lim

"!0C

1



1 .x  2/

1=3

dx D lim

"!0C

3"2=3 3  2 2



3 D : 2

Thus, the given improper integral converges, and we have Z 1

2

1 1=3

.x  2/

3 dx D  : 2

We can interpret the result as the signed area of the region between the graph of y D .x  2/1=3 and the interval Œ1; 2, as indicated in Fig. 4.16.  Fig. 4.16

236

4 The Riemann Integral

A function may have a discontinuity at an interior point of the given interval: Definition 6. Assume that a < c < b and f is integrable on any closed interval that is contained in Œa; c/ or .c; b. Also assume that at least one of the limits, f .x/ or limx!cC f .x/, fails to exists. In this case, the improper integral lim R b x!c a f .x/ dx converges if and only if Z

Z

c

b

f .x/ dx and

f .x/ dx

a

c

exist as ordinary integrals or converge as improper integrals. If the improper integral converges, we define its value as Z

Z

b

Z

c

f .x/ dx D

b

f .x/ dx C

a

a

f .x/ dx: c

Rc Rb Rb The improper integral a f .x/ dx diverges if a f .x/ dx or c f .x/ dx diverges. The extension of Definition 6 to more general cases is obvious: We must take into account the discontinuities of the integrand and examine the improper integrals on the intervals that are separated from each other by the points of discontinuity of the function separately. Rb Remark 5. If a f .x/ dx is an improper integral that converges we may simply say that f is integrable on Œa; b unless we need to emphasize that the integral is an improper integral. Þ Example 9. Consider the improper integral Z

2 1

1 dx x2=3

a) Why is the integral an improper integral? b) Determine whether the improper integral converges or diverges, and its value if it converges. Solution. a) The integral is improper since lim

1

x!0 x2=3

D C1;

and 0 2 Œ1; 2. b) We must examine the improper integrals Z

0

1

1

x

dx and 2=3

Z

2 0

1 x2=3

dx

4.5 Improper Integrals: Part 1

237

separately. We have Z

0

Z

1

x

1

dx D lim 2=3

"

"!0C 1

x2=3 dx D lim







"!0C

ˇ" 3x1=3 ˇ1

D lim 3"1=3 C 3 D 3; "!0C

and Z

2

0

Z

1

x

dx D lim 2=3

"!0C

2

x2=3 D lim

"!0C

"



ˇ2 3x1=3 ˇ"

      D lim 3 21=3  3"1=3 D 3 21=3 : "!0

Therefore, the given improper integral converges, and we have Z

2

1

Z

1

x

dx D 2=3

0

1

1

x

dx C 2=3

Z

2 0

  1 dx D 3 C 3 21=3 : x2=3

We can interpret the result as the area of the region between the graph of y D x2=3 and the interval Œ1; 2, as illustrated in Fig. 4.17.  Fig. 4.17

Remark 6 (Caution). The convergence of an improper integral on Œa; b requires the convergence of the improper integrals on Œa; c and Œc; b. We did not Rb define a f .x/ dx as Z

c"

lim

"!0C

a

Z f .x/ dx C

b

 f .x/ dx :

cC"

Our definition of the convergence of the improper integral the limits Z c Z c" f .x/ dx D f .x/ dx lim "!0C

a

a

Rb a

f .x/ dx requires that

238

4 The Riemann Integral

and Z

Z

b

ı!0C

b

f .x/ dx D

lim

f .x/ dx

cCı

c

exist independently, without any relationship between " and ı (such as " D ı). For example, the improper integral Z

1 1

1 dx x

does not converge since the improper integral Z

1 0

1 dx x

diverges. On the other hand, 1=x defines an odd function, and we have Z

" 1

1 dx C x

Z

1 "

1 dx D 0 x

for each " > 0, so that Z

"

lim

"!0C

1

1 dx C x

Z

1 "

 1 dx D 0: x

Þ

4.5.3 Problems In problems 1–5, determine whether the given improper integral converges or diverges, and the value of the improper integral in case of convergence. 3.

1. Z

1

x .x2 C 4/

0

Z dx 2

1

x2 ex dx

0

4.

2. Z

1 1

1 dx xC4

Z

1 e2

ln .x/ dx x2

4.6 Improper Integrals: Part 2

239

Hint:

5. Z

Z

0 1

ex sin .x/ dx

1 1 ex sin .x/ dx D  ex cos .x/ C ex sin .x/ 2 2

In problems 6–9, a) Explain why the given integral is an improper integral, b) Determine whether the given improper integral converges or diverges, and the value of the improper integral in case of convergence. 6.

8. Z

2

0

p

1 2x

Z dx

2

1

1

.x  1/4=5

dx

9.

7. Z

1 0

Z

x dx 1  x2

3 1

1 .x  2/4=3

dx

4.6 Improper Integrals: Part 2 Many important improper integrals cannot be evaluated exactly with the help of familiar antiderivatives. In such a case we can compare the given improper integral with an improper integral that can be computed easily and conclude whether the improper integral of interest converges or diverges.

4.6.1 Comparison Tests for Improper Integrals on Unbounded Intervals Let us begin by stating and proving the Cauchy condition for improper Integrals: Proposition 1 (Cauchy Condition for an Improper Integral). Assume that f is integrable on any interval of the form Œa; b. The improper integral Z

1 a

f .x/ dx

240

4 The Riemann Integral

converges if and only if given " > 0 there exists A such that ˇZ c ˇ ˇ ˇ c > b  A ) ˇˇ f .x/ dxˇˇ < ": b

Proof. Assume that

R1 a

f .x/ dx converges. Set Z F .x/ D

x

f .t/ dt.

a

Then limx!1 F .x/ exists. By the Cauchy condition for the existence of a limit at infinity (Theorem 2 of Sect. 2.3), given " > 0 there exists A such that c > b  A ) jF .c/  F .b/j < ": Thus, ˇZ c Z ˇ f .t/ dt  c > b  A ) ˇˇ a

b

a

ˇZ c ˇ ˇ ˇ ˇ ˇ f .t/ dtˇˇ < " ) ˇˇ f .t/ dtˇˇ < ": b

Conversely, assume the given condition. Again, this means that c > b  A ) jF .c/  F .b/j < ": Thus, limx!1 F .x/ exists. This means that Z

x

f .t/ dt

lim

x!1 a

exists, so that

R1 a

f .x/ dx converges. 

Definition 1. We say that the improper integral R1 if a jf .x/j dx converges.

R1 a

f .x/ dx converges absolutely

Absolute convergence implies convergence: Proposition 2. Assume that fRis integrable on any interval of the form Œa; b and R1 1 .x/j dx converges. Then a f .x/ dx converges. jf a Proof. Note that jf j is integrable on any interval of the form Œa; b (Corollary 1 of R1 Sect. 4.2). Assume that a jf .x/j dx converges. By the Cauchy condition for an improper integral, given " > 0 there exists A such that Z

c

c>bA)

jf .x/j dx < ": b

4.6 Improper Integrals: Part 2

241

By the triangle inequality for integrals ˇZ c ˇ Z c ˇ ˇ ˇ f .x/ dxˇˇ  jf .x/j dx < ": ˇ b

b

R1

Therefore a f .x/ dx converges, thanks to the Cauchy condition for improper integrals.  R1 Definition R2. An improper integral a f .x/ dx converges conditionally if it con1 verges but a jf .x/j dx diverges. Example 1. Let us consider the Dirichlet integral Z 1 sin .x/ dx: x 0 The integral is improper only because the interval is unbounded. There is no problem at 0 since sin .x/ D 1: x!0 x lim

We will show that the Dirichlet integral converges when we state and prove the Dirichlet’s test (naturally!) later in this section. The integral converges only conditionally since it does not converge absolutely. Indeed, if n  2 we have ˇ ˇ ˇ ˇ Z n ˇ Z 2 ˇ Z 3 ˇ Z n ˇ ˇ sin .x/ ˇ ˇ sin .x/ ˇ ˇ sin .x/ ˇ ˇ sin .x/ ˇ ˇ dx  ˇ dx C ˇ dx C    C ˇ ˇ ˇ ˇ ˇ ˇ x ˇ ˇ x ˇ ˇ x ˇ ˇ x ˇ dx .n1/ 0  2 Since Z

.kC1/

jsin .x/j dx D 2

k

and 1 1  if x 2 Œk; .k C 1/  x .k C 1/  we have Z 0

n

ˇ ˇ Z 2 Z 3 Z n ˇ sin .x/ ˇ 1 1 ˇ ˇ dx  1 .x/j .x/j jsin dx C jsin dx C    C jsin .x/j dx ˇ x ˇ 2  3 2 n .n1/

 2 1 1 1 1 D C C C C  2 3 4 n

 2 1 1 1 1 2 D 1C C C C C   2 3 4 n 

242

4 The Riemann Integral

Note that 1C

1 1 1 1 C C CC > 2 3 4 n

Z

nC1 1

1 dx: x

Indeed, Z

nC1 1

1 dx D x

Z

2 1

Z

1 dx C x

3 2

1 dx C x

Z

4 3

1 dx C    C x

Z

nC1 n

1 dx x

1 1 1 < 1C C CC : 2 3 n Thus Z

n 0

ˇ ˇ

 ˇ sin .x/ ˇ ˇ dx > 2 1 C 1 C 1 C 1 C    C 1  2 ˇ ˇ x ˇ  2 3 4 n  Z nC1 2 1 2 dx  >  1 x  D

2 2 ln .n C 1/   

Since Z

n 0

ˇ ˇ ˇ sin .x/ ˇ 2 ˇ ˇ ˇ x ˇ dx >  .ln .n C 1/  1/ for any n  2

and lim

n!1

2 .ln .n C 1/  1/ D C1 

we have Z

n

lim

n!1 0

ˇ ˇ ˇ sin .x/ ˇ ˇ ˇ ˇ x ˇ dx D C1

as well. Therefore Z

1 0

ˇ ˇ ˇ sin .x/ ˇ ˇ ˇ ˇ x ˇ dx

diverges.  In the case of nonnegative functions the convergence or divergence of an improper integral can be characterized as follows:

4.6 Improper Integrals: Part 2

243

Proposition 3. Assume that f is integrable on any interval of the form Œa; b and f .x/  0 for each x  a. If there exists M > 0 such that Z

b

f .x/ dx  M a

for each b  a the improper integral Z

1

R1 a

f .x/dx converges, and Z

b

f .x/ dx D sup ba

a

f .x/ dx: a

Otherwise, Z

b

f .x/ dx D C1:

lim

b!C1 a

so that

R1 a

f .x/dx diverges.

Proof. Set Z

x

F.x/ D

f .t/ dt:

a

The function F is monotone increasing on Œa; C1/. Indeed, if x2 > x1 > a, Z F.x2 / D

x2

Z

x1

f .t/ dt D

a

Z f .t/ dt C

a

x2

f .t/ dt;

x1

and Z

x2

f .t/ dt  0

x1

since f .t/  0 for each t  a. Therefore, Z

x1

F.x2 / 

f .t/ dt D F .x1 / :

a

Thus Theorem 1 of Sect. 2.3 is applicable to the function F: If there exists M such that Z

b

f .x/ dx  M

F.b/ D a

244

4 The Riemann Integral

for each b  a, then limx!C1 F.x/ exists and limx!C1 F.x/ D sup fF .x/ W x  ag for each b  a. This means that Z b lim f .x/ dx b!C1 a

exists, so that the improper integral Z

1

R1 a

f .x/dx converges and Z

b

f .x/ dx D sup ba

a

f .x/ dx:

a

On the other hand, if there does not exist M > 0 such that Z b f .x/ dx  M F.b/ D a

for each b  a, we have Z

b

lim F.b/ D lim

b!C1

Thus, the improper integral

b!C1 a

R1 a

f .x/dx D C1:

f .x/ dx diverges. 

The above criteria lead to a useful “comparison theorem: Theorem 1 (The Basic Comparison Test). Assume that f and g are integrable on any interval of the form Œa; b R1 a) (The convergence clause) If jf .x/j  g.x/ for each x  a and a g .x/ dx R1 converges then the improper integral a f .x/ dx converges absolutely, and we have ˇZ 1 ˇ Z 1 Z 1 ˇ ˇ ˇ ˇ f .x/ dx g .x/ dx: .x/j dx  jf ˇ ˇ a

a

a

R1 b) (The divergence clause) If fR.x/  g.x/  0 for each x  a and a g.x/dx diverges 1 then the improper integral a f .x/ dx diverges as well. R1 Proof. a) Let " > 0 be given. Since a g.x/dx converges, there exists A such that Z

c

c>bA)

g .x/ dx < "; b

by the Cauchy condition for improper integrals. Since jf .x/j  g.x/ Z

c

c>bA) b

Z

c

jf .x/j dx  b

g.x/dx < ":

4.6 Improper Integrals: Part 2

Therefore

R1 a

245

f .x/ dx converges absolutely. We have ˇZ ˇ ˇ ˇ

b a

ˇ Z ˇ f .x/ dxˇˇ 

Z

b

b

g .x/ dx;

jf .x/j dx  a

a

so that ˇZ ˇ lim ˇ b!1 ˇ

b a

ˇ Z ˇ ˇ f .x/ dxˇ  lim b!1

b

Z jf .x/j dx  lim

b!1 a

a

b

g .x/ dx:

Thus, ˇZ ˇ ˇ ˇ

1 a

ˇ Z ˇ f .x/ dxˇˇ 

Z

1

1

jf .x/j dx 

a

g .x/ dx;

a

as claimed. R1 b) Since g.x/  0 for each x  a and the improper integral a g.x/ diverges, we have Z

b

g.x/dx D C1:

lim

b!C1 a

Since f .x/  g.x/ for each x  a, Z

Z

b

b

f .x/dx 

g.x/dx

a

a

for each b  a. Therefore, Z

b

lim

b!C1 a

as well. Thus, the improper integral

f .x/dx D C1

Rb a

f .x/dx diverges. 

Example 2. Show that the improper integral Z

1

2

ex dx

1

converges. Solution. If x  1 then x2  x, so that x2  x. Since the natural exponential function is increasing on the entire number line, we have 2

ex  ex if x  1:

246

4 The Riemann Integral

The improper integral Z 1

b

R1 1

ex dx converges by direct calculation. Indeed,

ex dx D ex jb1 D eb C e1 D

so that Z

1



x

e dx D lim

b!C1

1

1 1  e eb

 D

1 1  ; e eb

1 : e

By the convergence clause of the comparison test (Theorem 1, part a), the improper R1 2 integral 1 ex dx converges as well (Fig. 4.18).  y

Fig. 4.18

1

1

y y 1

e

e

x

x2

x

Example 3. Show that Z

1 1

2

ex cos .3x/ dx

converges. Solution. We have

ˇ ˇ 2 2 2 ˇ ˇ x ˇe cos .3x/ˇ D ex jcos .3x/j  ex

since jcos .u/j  1 for each u 2 R. In Example 2 we showed that Z

1

2

ex dx

1

ˇ R 1 ˇˇ 2 ˇ converges. Therefore, the improper integral 0 ˇex cos .3x/ˇ dx converges. Thus, R1 2 the integral 1 ex cos .3x/ dx converges absolutely.  In some cases the limit comparison test is more convenient to apply than the basic comparison test:

4.6 Improper Integrals: Part 2

247

Theorem 2 (The Limit Comparison Test). Assume that f and g are integrable on any interval of the form Œa; b and g .x/ > 0 for each x 2 Œa; C1/: R1 a) (The convergence clause) If a g .x/ dx converges, and lim

x!C1

jf .x/j g .x/

R1 exists (as a finite limit) then the improper integral a f .x/ dx converges absolutely. R1 b) (The divergence clause) If f .x/ > 0; g .x/ > 0, a g .x/ dx diverges and f .x/ f .x/ > 0 or lim D C1 x!C1 g .x/ g .x/ R1 then the improper integral a f .x/ dx diverges as well. lim

x!C1

Proof. a) Let lim

x!C1

jf .x/j D L: g .x/

Then there exists B > a such that ˇ ˇ ˇ jf .x/j ˇ ˇ ˇ ˇ g .x/  Lˇ < 1 if x  B. In particular, 0

jf .x/j < LC1 g .x/

if x  B. Thus jf .x/j < .L C 1/ g .x/ if x  B. Since Z

1

Z

1

.L C 1/ g .x/ dx D .L C 1/

B

B

converges, Z

1 B

jf .x/j dx

g .x/ dx

248

4 The Riemann Integral

converges as well, thanks to the basic comparison test. Therefore Z

1

Z

B

jf .x/j dx D

a

Z

1

jf .x/j dx C

a

jf .x/j dx

B

R1 converges as well. Thus, a f .x/ dx converges absolutely. b) If f .x/ > 0, g .x/ > 0 and lim

x!C1

f .x/ > 0 or g .x/

lim

x!C1

f .x/ D C1; g .x/

there exists B > a and C > 0 such that f .x/ C g .x/ if x  B. Therefore, f .x/  Cg .x/ if x  B. Since Z

b

g .x/ D C1;

lim

b!C1 B

and Z

Z

b

Z

b

f .x/ dx  B

B

g .x/ dx; B

where C > 0, we have Z

b

f .x/ D C1

lim

b!C1 B

as well. Thus, Z

1

f .x/ dx

B

diverges, and so does Z

b

f .x/ dx: a



b

Cg .x/ dx D C

4.6 Improper Integrals: Part 2

249

Example 4. Discuss the convergence or divergence of the following improper integrals: a) Z

1

p

xex dx

1

b) Z

1

1 dx ln .x/

x1=3

2

Solution. a) We have p x xe x5=2 lim D lim D0 1 x!C1 x!C1 ex x2 (for example, by L’Hôpital’s rule). Since Z

1

1

1 dx x2

converges so does Z

1

p x xe dx:

1

b) We have

lim

x!C1

1 x2=3 x1=3 ln .x/ D C1 D lim 1 x!C1 ln .x/ x

(for example, by L’Hôpital’s rule). Since Z

1 2

1 dx x

diverges so does Z 2



1

x1=3

1 dx: ln .x/

250

4 The Riemann Integral

We have the obvious counterparts of the above definitions and propositions for improper integrals of the form Z a Z 1 f .x/ dx and f .x/ dx: 1

1

The following test is useful in cases where an improper integral does not converge absolutely but converges conditionally: Theorem 3 (Dirichlet’s Test). Assume that f is continuous and g is continuously differentiable on Œa; C1/. Also assume (i) There exists M > 0 such that ˇ ˇZ b ˇ ˇ ˇ  M for each b > a; ˇ f .x/ dx ˇ ˇ a

(ii) g .x/ > 0 if x  a; g is decreasing on Œa; 1/ and limx!1 g .x/ D 0. Then the improper integral Z 1 f .x/ g .x/ dx a

converges. Proof. Set Z

x

F .x/ D

f .t/ dt a

so that F 0 .x/ D f .x/ for each x  a. We have ˇ ˇZ x ˇ ˇ ˇ f .t/ dtˇˇ  M for each x  a. jF .x/j D ˇ a

Let c > b  a. By the integration by parts formula Z c Z c Z c f .x/ g .x/ dx D F 0 .x/ g .x/ dx D ŒF .c/ g .c/  F .b/ g .b/  F .x/ g0 .x/ dx: b

b

b

Therefore ˇZ c ˇZ c ˇ ˇ ˇ ˇ ˇ ˇ 0 ˇ ˇ ˇ f .x/ g .x/ dxˇ  jF .c/j jg .c/j C jF .b/j jg .b/j C ˇ F .x/ g .x/ dxˇˇ : ˇ b

Z

b

c

 jF .c/j jg .c/j C jF .b/j jg .b/j C b

ˇ ˇ jF .x/j ˇg0 .x/ˇ dx

4.6 Improper Integrals: Part 2

251

Since jF .x/j  M for each x  a ˇZ c ˇ Z c ˇ ˇ ˇ 0 ˇ ˇ ˇ  M .jg .c/j C jg .b/j/ C M ˇg .x/ˇ dx: f .x/ g .x/ dx ˇ ˇ b

b

Since g .x/ > 0 if x  a and g is decreasing on Œa; C1/ ˇZ c ˇ Z c ˇ ˇ  0  ˇ ˇ  Mg .c/ C Mg .b/ C M g .x/ dx f .x/ g .x/ dx ˇ ˇ b

b

D Mg .c/ C Mg .b/ C M .g .b/  g .c// D 2Mg .b/ : Since limb!1 g .b/ D 0, given " > 0, there exists A > a such that 0 < g .b/ < " if b  A: Therefore ˇZ c ˇ ˇ ˇ ˇ ˇ  2Mg .b/ < 2M" if c > b  A: f .x/ g .x/ dx ˇ ˇ b

Thus the Cauchy condition is satisfied and the improper integral Z

1

f .x/ g .x/ dx

a

converges.  Example 5. In Example 1 we showed that the Dirichlet integral Z

1 0

sin .x/ dx x

does not converge absolutely. Show that the integral converges. Solution. Since Z



sin .x/ dx x

1

sin .x/ dx x

0

exists it is sufficient to show that Z 

252

4 The Riemann Integral

converges. We set f .x/ D sin .x/ and g .x/ D 1=x. Thus g .x/ > 0 if x  ; g is decreasing on Œ; 1/ and lim g .x/ D lim

x!1

If b > 

ˇZ ˇ ˇ ˇ

b 

x!1

1 D 0: x

ˇ ˇ sin .x/ dxˇˇ D j cos .b/ C 1j  1 C 1 D 2:

Therefore the Dirichlet’s test implies that Z 1 sin .x/ dx x  converges. It is known that Z

1 0

 sin .x/ dx D : x 2



4.6.2 Comparison Tests for Improper Integrals That Involve Discontinuous Functions We have similar comparison tests for improper integrals that involve functions with discontinuities on bounded intervals. We will state almost all propositions and theorems for integrals that are improper due to a discontinuity at the right endpoint of an interval. The generalization to other cases is straightforward. We may also refer to a function f as being integrable on an interval Œa; b even though the integral may be an improper integral (Remark 5 of Sect. 4.5). Let us begin with a Cauchy-type condition for the convergence of an improper integral on a bounded interval: Proposition 4. Assume that f is integrable on any interval of the form Œa; b  ı that is contained in Œa; b/. The improper integral Z

b

f .x/ dx a

converges if and only if given " > 0 there exists ı > 0 such that ˇZ ˇ b  ı < c < d < b ) ˇˇ

d c

ˇ ˇ f .x/ dxˇˇ < ":

4.6 Improper Integrals: Part 2

Proof. Assume that

Rb a

253

f .x/ dx converges. Set Z F .x/ D

x

f .t/ dt.

a

Then limx!b F .x/ exists. By the Cauchy condition of the existence of the limit (Theorem 4 of Sect. 2.2), given " > 0, there exists ı > 0 such that b  ı < c < d < b ) jF .d/  F .c/j < ": But Z

d

F .d/  F .c/ D

f .x/ dx: c

Therefore ˇZ ˇ b  ı < c < d < b ) ˇˇ

d c

ˇ ˇ f .x/ dxˇˇ < ":

Conversely, assume the given condition. This means that b  ı < c < d < b ) jF .d/  F .c/j < ": Thus, limx!b F .x/ exists by the Cauchy condition of the existence of the limit (Theorem 4 of Sect. 2.2). This means that Z lim

x!b a

exists so that

Rb a

x

f .t/ dt

f .x/ dx converges. 

Proposition 5. Assume that f is integrable on any interval of the form Œa; b  ı that is contained in Œa; b/, f .x/  0 if a  x < b; and there exists M such that Z

d

f .x/ dx  M for each d 2 Œa; b/: a

Then the improper integral

Rb a

Z

f .x/ dx converges and Z

b

f .x/ dx D sup a

ad 0 there exists N 2N such that jcnC1 CcnC2 C    C cnCk j < " if n N and k is an arbitrary positive integer. We can paraphrase the Cauchy condition for the convergence of a series by setting n C k D m: Given " > 0 there exists N 2 N such that jcnC1 CcnC2 C    C cm j < " if m > n  N. Proof of Theorem 2. Assume that the series

P

cn converges. If

S n D c1 C c2 C    C cn is the nth partial sum of the series, the sequence fSn g1 nD1 is a Cauchy sequence. Therefore, given " > 0 there exits N 2 N such that n  N ) jSnCk  Sn j < " for k D 1; 2; 3; : : : :

270

5 Infinite Series

But jSnCk  Sn j D jcnC1 C cnC2 C    C cnCk j : Therefore, n  N ) jcnC1 C cnC2 C    C cnCk j < " for k D 1; 2; 3; : : : : Conversely, assume that the given condition is satisfied. Just as in the first part of the proof, this implies that the sequence of partial sums fSn g1 nD1 is a Cauchy sequence. By the completeness of the set of real numbers, lim Sn exists. This means that n!1 P the series cn converges.  If the series that is formed by the absolute values of the terms of a series converges, so does the series itself: P P Theorem 3. Assume that the series jcn j converges. Then cn converges. Proof. By the triangle inequality, jcnC1 C cnC2 C    C cnCk j  jcnC1 j C jcnC2 j C    C jcnCk j : P

Since jcn j converges it satisfies the Cauchy condition for series. Thus, given " > 0 there exists N 2 N such that n  N ) jcnC1 j C jcnC2 j C    C jcnCk j < " for k D 1; 2; 3; : : : : By the triangle inequality, jcnC1 C cnC2 C    C cnCk j  jcnC1 j C jcnC2 j C    C jcnCk j < " P if n  N and Pk D 1; 2; 3; : : : :. Thus, jcn j satisfies the Cauchy condition for series. Therefore, cn converges.  P P Definition 5. The series cn converges absolutely if jcn j converges. P P Since we showed that cn converges if jcn j does, a series that converges absolutely is convergent (thus, the terminology makes sense). On the other hand, a series may converge even though it is not absolutely convergent: Example 4. Let us consider the series 1 X nD1

.1/n1

1 1 1 1 1 1 D 1 C  C  C n 2 3 4 5 6

We have 1 ˇ X ˇ ˇ.1/n1 ˇ nD1

ˇ 1 1 ˇˇ X 1 : D n ˇ nD1 n

5.1 Infinite Series of Numbers

271

The harmonic series diverges (Remark 1) so that the given series does not converge absolutely. Nevertheless, the series converges: If Sn D

n X kD1

.1/k1

1 k

is the nth partial sum of the series then 1 1 1 1 C .1/nC1 C .1/nC2 C    C .1/nCk1 nC1 nC2 nC3 nCk 

1 1 1 1 1 n k1 D .1/  C  C    C .1/ : nC1 nC2 nC3 nC4 nCk

SnCk  Sn D .1/n

We claim that 0
1 and diverges if p  1: Proof. Let us first eliminate the cases where p  0. In such a case the series diverges, since lim

n!1

1 ¤ 0: np

If p > 0, set f .x/ D

1 : xp

Then, f is continuous, positive, and decreasing on the interval Œ1; C1/, so that the integral test is applicable to the series 1 1 X X 1 D f .n/ : np nD1 nD1

We know that the improper integral Z

1 1

Z f .x/ dx D

1 1

1 dx xp

5.2 Convergence Tests for Infinite Series: Part 1

289

converges if p > 1 and diverges if p  1. Therefore, the integral test implies that the infinite series 1 X 1 np nD1

converges if p > 1 and diverges if 0 < p  1. 

5.2.4 Comparison Tests We may encounter some cases where it is not easy to apply any of the previous tests. It may be convenient to compare a given series with a series that is known to be convergent or divergent. Theorem 4 (The Basic Comparison Test). n D N; N C 1; N C 2; : : : 1. If there exists a positive P integer N such that jcn j  dn forP 1 and the series 1 d converges, then the series n nD1 nD1 cn converges absolutely. 2. If there exists a positive for n D N; N C 1; N C 2; P integer N such that cn  dn  0 P 1 : : : and the series 1 d diverges, then the series n nD1 nD1 cn diverges as well. Proof. 1. If we assume the first set of conditions, we have jcnC1 j C jcnC2 j C jcnC3 j C    jcnCk j  dnC1 C dnC2 C dnC3 C    C dnCk P for any n  N and k D 1; 2; 3; : : :. Since dn converges, the Cauchy condition is satisfied. Thus, given " > 0 there exists N2  N such that 0  dnC1 C dnC2 C dnC3 C    C dnCk < " if n  N2 and k D 1; 2; 3; : : :.Therefore, jcnC1 j C jcnC2 j C jcnC3 j C    jcnCk j  dnC1 C dnC2 C dnC3 C    C dnCk < " under the same conditions.P Thus, the Cauchy condition is fulfilled by the series P jcn j as well. Therefore, cn converges absolutely. 2. Now we assume the second set of conditions. Then, cN C cNC1 C cNC2 C    C cNCk  dN C dNC1 C dNC2 C    C dNCk P for k D 0; 1; 2; : : :. Since 1 nD1 dn diverges the sequence fdN C dNC1 C dNC2 C 1    C dNCk gkD0 tends to 1. By the above inequality, the sequence fcN CP cNC1 C 1 cNC2 C    C cNCk g1 tends to 1 as well. Therefore the infinite series kD0 nD1 cn diverges. 

290

5 Infinite Series

Example 5. Show that the series 1 X cos .n/ nD1

n2

converges absolutely. Solution. We have ˇ ˇ ˇ cos .n/ ˇ 1 jcos .n/j ˇ ˇ  2 ; n D 1; 2; 3; : : : ; ˇ n2 ˇ D n2 n since jcos .x/j  1 for any real number x. We know that the series 1 X 1 n2 nD1

converges. By the comparison test, the given series converges absolutely.  Example 6. Determine whether the series 1 X ln .n/ p n2  1 nD2

converges or diverges. Solution. Since ln .n/ > 1 for n  3 > e, we have ln .n/ 1 p >p ; n D 3; 4; 5; : : : : 2 2 n 1 n 1 We also have p p 1 1 > : n2  1 < n2 D n ) p 2 n n 1 Therefore, ln .n/ 1 > if n  3: p 2 n n 1 P The harmonic series 1 nD1 1=n diverges. Therefore, the divergence clause of the comparison test is applicable, and we conclude that the given series diverges.  Sometimes it is more convenient to apply the limit-comparison test, rather than the comparison test:

5.2 Convergence Tests for Infinite Series: Part 1

291

Theorem 5 (The Limit-Comparison Test). 1. Assume that jcn j n!1 dn lim

exists (as a finite Plimit), dn > 0; n D 1; 2; 3; : : :, and that Then, the series 1 nD1 cn converges absolutely. P 2. If dn > 0; n D 1; 2; 3; : : :, 1 nD1 dn diverges, and lim

n!1

then the series

P1

P1 nD1

dn converges.

cn cn D L exists and L > 0 or lim D C1 n!1 dn dn

nD1 cn

diverges.

Proof. 1. If we set jcn j ; n!1 dn

L D lim there exists an integer N such that

jcn j  LC1 dn for all n  N. Therefore, jcn j  .L C 1/dn ; n  N: P1 P1 P1 Since n . Therefore, nDN dn converges, so does nDN .L C 1/d nDN jcn j P 1 converges by the comparison test. This implies that converges, i.e., jc j n nD1 P1 nD1 cn converges absolutely. 2. Now let us assume the conditions of the divergence clause of the limitcomparison test. If cn cn D L > 0 or lim D1 n!1 dn n!1 dn lim

there exists a > 0 and an integer N such that cn a dn for all n  N. Therefore, cn  adn ; n D N; N C 1; N C 2; : : : :

292

5 Infinite Series

Therefore, cN C cNC1 C    C cNCk  a .dN C dNC1 C    C dNCk / ; for k D 0; 1; 2; : : :. Since the series

P1

nD1 dn

diverges and a > 0 we have

lim a .dN C dNC1 C    C dNCk / D C1:

n!1

By the above inequality, we also have lim .cN C cNC1 C    C cNCk / D C1;

k!1

Therefore,

P1

nD1 cn

diverges. 

Remark 7. In the divergence clause of the limit-comparison test it is essential that the limit in question is positive (or C1). For example, we have 1=n2 1 D lim D 0; n!1 1=n n!1 n lim

and the series

P

1=n diverges, but the series

P

1=n2 converges. Þ

Example 7. Determine whether the series 1 X ln .n/ nD2

n3=2

converges or diverges. Solution. We will compare the given series with the convergent series 1 X 1 : 5=4 n nD2

We have ln .n/ 3=2 ln .n/ lim n D lim 1=4 D 0: 1 n!1 n!1 n n5=4 Therefore, the given series converges. 

5.2 Convergence Tests for Infinite Series: Part 1

293

Example 8. Determine whether the series 1 X nD3

1 p 2 n 4

converges or diverges. Solution. Since p

1

1 1 Šp D n n2  4 n2

if n is large, the harmonic series is a good candidate for the series which will be chosen for comparison. We have 1 p 2 n 1 n  4 D lim p n lim D lim r D lim r D 1 > 0: 1 n!1 n!1 n!1 n!1 4 4 n2  4 n 1 2 1 2 n n n The divergencePclause of the limit-comparison test implies that the given series diverges, since 1 nD3 1=n diverges. 

5.2.5 Problems In problems 1–4 use the ratio test in order to determine whether the given series converges absolutely or whether it diverges, provided that the test is applicable. 3.

1. X1 3n nD0 nŠ 2.

X1 nD1

.1/n

4n n2

4. X1 nŠ nD0 3n

1 X nD1

.1/n1

1 : n2n

In problems 5–8 use the root test in order to determine whether the given series converges absolutely or whether it diverges, provided that the test is applicable.

294

5 Infinite Series

5.

7. X1

X1 10n nD1 nn

nD1

n4 en

8.

6.

1 X

X1 n2 nD1 2n

.1/n1

nD1

.1:2/n : n

In problems 9 and 10 apply the integral test to determine whether the given series converges absolutely or whether it diverges. You need to verify that the conditions for the applicability of the integral test are met: 10.

9.

1 X

X1

1 nD1 n2 C 1

nD2

1 n ln3 .n/

In problems 11–13 use the comparison test to establish the absolute convergence or the divergence of the series. 11.

12. 1 X nD1

1 X

1 p : 2 n C n

  en=4 sin n2 :

nD1

13. 1 X cos .4n/ 3

nD1

n2

:

In problems 14 and 15 use the limit-comparison test to show that the given series converges absolutely or that the series diverges. 15.

14. 1 X nD2

1 p : n n

1 X nD2

1 : p n2  1

5.3 Convergence Tests for Infinite Series: Part 2

295

5.3 Convergence Tests for Infinite Series: Part 2 In Sect. 5.2 we discussed tests for infinite series that predicted the absolute convergence of a series when certain conditions are fulfilled. In this section we will discuss tests that enable us to demonstrate the convergence of a series even though the series may not converge absolutely. Thus, these tests are able to demonstrate conditional convergence.

5.3.1 Alternating Series Definition 1. A series in the form 1 X

.1/

n1

an D a1  a2 C a3  a4 C    or

1 X

nD1

.1/n an D a1 C a2  a3 C   

nD1

where an  0 for each n is called an alternating series. Since 1 X

.1/n an D 

nD1

1 X

.1/n1 an ;

nD1

and the multiplication of a series by 1 does not alter the convergence or divergence of a series, state the general results of this section in terms of series in the P we will n1 form 1 .1/ an . nD1 The following theorem predicts the convergence of an alternating series under certain conditions: Theorem 1 (The Theorem on Alternating Series). Assume that an  0 and that the sequence fan g1 nD1 is decreasing, i.e., an  anC1 for each n. If limn!1 an D 0 the alternating series 1 X

.1/n1 an D a1 a2 Ca3 a4 C   

nD1

converges. If S is the sum of the series and Sn D

n X kD1

.1/k1 ak

296

5 Infinite Series

is the nth partial sum of the series, we have jS  Sn j  anC1 for n D 1; 2; 3; : : : Thus, the magnitude of the error in the approximation of the sum of the series by the nth partial sum is at most equal to the magnitude of the first term that is left out. Note theorem on alternating series predicts the convergence of the P that the n1 series 1 an if the necessary condition for the convergence of the series, nD1 .1/ limn!1 an D 0, is met, provided that the magnitude of the nth term decreases towards 0 monotonically. If an does not approach 0 as n ! 1, the series diverges anyway. The Proof of Theorem 1. Consider the partial sums S2k ; k D 1; 2; 3; : : :, so that each to the addition of an even number of the terms of the series P1 S2k corresponds n1 .1/ a . We claim that the sequence S2k ; k D 1; 2; 3; : : : is increasing: n nD1 S2  S4  S6      S2k  S2kC2     : Indeed, S2kC2 D a1  a2 C a3  a4 C    C a2k1  a2k C a2kC1  a2kC2 D S2k C .a2kC1  a2kC2 /  S2k ; since a2kC1  a2kC2 : On the other hand, the sequence S2kC1 ; k D 0; 1; 2; : : :, corresponding to the addition of odd numbers of terms, is decreasing: S1  S3  S5      S2kC1  S2kC3     : Indeed, S2kC3 D a1  a2 C    C a2kC1  a2kC2 C a2kC3 D S2kC1  .a2kC2  a2kC3 /  S2kC1 ; since a2kC2  a2kC3 : The following is also true: Any partial sum which corresponds to the addition of an even number of terms does not exceed any partial sum which is obtained by adding an odd number of terms: S2k  S2mC1 ; where k is an arbitrary positive integer, and m is an arbitrary nonnegative integer. Let us first assume that k  m. Then, S2mC1 D a1  a2 C a3  a4 C     a2m C a2mC1 D S2m C a2mC1  S2m  S2k :

5.3 Convergence Tests for Infinite Series: Part 2

297

If k  m C 1, we have 2k  1  2m C 1, so that S2k D S2k1  a2k  S2k1  S2mC1 : Thus, the sequence of partial sum of the series are lined up on the number line as shown in Fig. 5.5. Fig. 5.5

S2

S4

S6

S

S5

S3

S1

Since the sequence fS2k g1 kD1 is an increasing sequence and S2k  S1 D a1 for each k, L1 D limk!1 S2k exists and S2k  L1 for each k, by the monotone convergence principle. Since fS2kC1 g1 kD0 is a decreasing sequence and S2  S2mC1 for each m, L2 D limm!1 S2mC1 exists and L2  S2mC1 for each m, again by the monotone convergence principle. Since S2k  S2mC1 for each k and m, we have L1 D lim S2k  S2mC1 k!1

for each m. Therefore, L1  lim S2mC1 D L2 : m!1

We claim that L1 D L2 . Indeed, S2k  L1  L2  S2kC1 for each k. Therefore, 0  L2  L1  S2kC1  S2k D a2kC1 for each k, Since limk!1 a2kC1 D 0, L2  L1 is a nonnegative real number that is arbitrarily small. This is the case if and only if that number is 0. Thus, L2  L1 D 0, i.e., L2 D L1 D P S. Therefore, the sequence of partial sums that corresponds to P the infinite series .1/n1 an has to converge to S, i.e., the series .1/n1 an converges and has the sum S. We have S2k  L1 D S D L2  S2kC1 ; so that S2k  S  S2kC1 for each k, as claimed.

298

5 Infinite Series

Now let us establish the estimate of the error in the approximation of the sum S by a partial sum. Since S2k  S  S2kC1 ; we have 0  S  S2k  S2kC1  S2k D a2kC1 : Since S2kC2  S  S2kC1 also, we have 0  S2kC1  S  S2kC1  S2kC2 D a2kC2 : Therefore, ˇ ˇ n ˇ ˇ X ˇ ˇ .1/k1 ak ˇ  anC1 jS  Sn j D ˇS  ˇ ˇ kD1

for n D 1; 2; 3; : : :.  Example 1. Consider the series, 1 X

1 1 1 1 .1/n1 p D 1  p C p  p C    : n 2 3 4 nD1

Does the series converge absolutely or conditionally? Solution. The series does not converge absolutely, since ˇ 1 ˇ 1 1 X X ˇ ˇ X 1 1 ˇ.1/n1 p1 ˇ D D p ˇ ˇ 1=2 n n n nD1 nD1 nD1 is a p-series with p D 1=2 < 1. On the other hand, the theorem on the convergence of an alternating series is applicable, since 1 1 1 p < p and lim p D 0: n!1 n n nC1 Therefore, the series converges. The series converges conditionally since it does not converge absolutely. 

5.3 Convergence Tests for Infinite Series: Part 2

299

You should not get the impression that a series is conditionally convergent whenever the theorem on alternating series is applicable to the series: Example 2. Consider the series 1 X

.1/n1

nD1

n : 2n

a) Show that the series converges absolutely. b) Apply the theorem on alternating series in order to determine n so that jS  Sn j < 103 , where S is the sum of the series and Sn is its nth partial sum, and an interval of length less than 103 that contains S. Solution. a) We will apply the ratio test: ˇ ˇ ˇ ˇ ˇ.1/n n C 1 ˇ    n     ˇ 2 nC1 1 nC1 2nC1 ˇ ˇ D lim lim ˇ D lim nˇ n!1 ˇ n!1 n!1 2 2nC1 n n ˇ.1/n1 n ˇ 2 1 nC1 D lim 2 n!1 n 1 1 D .1/ D < 1: 2 2 Therefore, the series converges absolutely. b) We can also apply the theorem on alternating series to the given series. Indeed, .1/n1

n D .1/n1 f .n/ ; 2n

where f .x/ D

x : 2x

The function f is decreasing function on Œ2; C1/ (confirm with the help of the derivative test for monotonicity), and lim f .x/ D 0

x!C1

(with or without L’Hôpital’s rule). By Theorem 1 jS  Sn j  f .n C 1/ : We have f .13/ Š 1:558691  103 and f .14/ Š 8:54492  104 < 103 :

300

5 Infinite Series

Therefore, it is sufficient to approximate S by S13 . We have jS  S13 j  f .14/ < 103 : By the proof of Theorem 1 S14  S  S13 ; and 0  S13  S14 D f .14/ < 103 : Thus, the interval ŒS14 ; S13  contains S and has length less than 103 . We have S13 Š 0:222778 and S14 Š 0:221924: Therefore, 0:221924  S  0:222778 (The actual value of S is 2=9 Š 0:222 : : :). 

5.3.2 Dirichlet’s Test and Abel’s Test In this subsection we will discuss two tests that may be used to establish the convergence of some series for which the previous tests are not adequate. The following “summation by parts” formula will be essential: Proposition 1 (Abel’s Partial Summation Formulas). Assume that fbn g1 nD1 is a sequence of real numbers and An D a1 Ca2 C    C an is the nth partial sum of the series kDm X

P1 nD1

an . Then kDm X

ak bk D .Am bmC1 An bnC1 / 

kDnC1

Ak .bkC1 bk /

kDnC1

and kDm X

ak bk D .Am An / bnC1 C

kDnC1

for integers m > n  1.

kDm X kDnC1

.Am Ak / .bkC1 bk /

5.3 Convergence Tests for Infinite Series: Part 2

301

Proof. We have ak bk D .Ak  Ak1 / bk D Ak bk  Ak1 bk D Ak .bk  bkC1 / C Ak bkC1  Ak1 bk Therefore kDm X

ak bk D

kDnC1

kDm X

kDm X

Ak .bk  bkC1 / C

kDnC1

.Ak bkC1  Ak1 bk /

kDnC1

The second term on the right-hand side is a telescoping sum: kDm X

.Ak bkC1  Ak1 bk / D .AnC1 bnC2  An bnC1 / C .AnC2 bnC3  AnC1 bnC2 / C   

kDnC1

C .Am1 bm  Am2 bm1 / C .Am bmC1  Am1 bm / D An bnC1 C Am bmC1 Thus kDm X

ak bk D .Am bmC1  An bnC1 / C

kDnC1

kDm X

Ak .bk  bkC1 /

kDnC1

D .Am bmC1  An bnC1 / 

kDm X

Ak .bkC1  bk /

kDnC1

We have kDm X

.bkC1  bk / D .bnC2  bnC1 / C .bnC3  bnC2 / C    C .bm1  bm2 /

kDnC1

C .bmC1  bm / D bmC1  bnC1 : Therefore bmC1 D

kDm X kDnC1

.bkC1  bk / C bnC1

302

5 Infinite Series

Thus kDm X

kDm X

ak bk D .Am bmC1  An bnC1 / 

kDnC1

Ak .bkC1  bk /

kDnC1

D Am

kDm X

!

.bkC1  bk / C bnC1  An bnC1 

kDnC1

kDm X

Ak .bkC1  bk /

kDnC1

D .Am  An / bnC1 C

m X

.Am  Ak / .bkC1  bk / :

kDnC1



P1 Theorem 2 (Abel’s Test). Assume that the series nD1 an converges and the 1 sequence is bounded and monotone (decreasing or increasing). Then the fb g n nD1 P series 1 nD1 an bn converges. Proof. AsP in Proposition 1, let An D a1 C a2 C    C an be the nth partial sum of the series 1 nD1 an so that kDm X

ak bk D .Am  An / bnC1 C

kDnC1

m X

.Am  Ak / .bkC1  bk /

kDnC1

for integers m > n  1. P Let " > 0 be given. Since an converges there exists N 2 N such that jAk  An j < " if k > n  N: Since the sequence fbn g1 nD1 is bounded there exists M > 0 such that jbn j  M for each n 2 N. Thus, for each n  N ˇ ˇ kDm m ˇ ˇ X X ˇ ˇ ak bk ˇ  jAm  An j jbnC1 j C jAm  Ak j jbkC1  bk j ˇ ˇ ˇ kDnC1

kDnC1

< M" C "

m X

jbkC1  bk j

kDnC1

Since the sequence fbn g1 nD1 is monotone the sum on the right is a telescoping sum so that m X kDnC1

jbkC1  bk j D jbmC1  bnC1 j  2M:

5.3 Convergence Tests for Infinite Series: Part 2

303

Therefore ˇ kDm ˇ m ˇ X ˇ X ˇ ˇ ak bk ˇ < M" C " jbkC1  bk j  3M": ˇ ˇ ˇ kDnC1

Thus, the series

P

kDnC1

an bn satisfies the Cauchy condition for convergence. 

Example 3. Let us consider the series  1  X 1 n 1 1C : n n2 nD1 The sequence  1  1 n 1C n nD1 is monotone increasing (confirm) and we have   1 n 1C D e: n!1 n lim

P The series 1=n2 converges. Thus, Abel’s test is applicable and the given series converges.  P The next test relaxes the hypothesis on the series an but tightens the hypothesis on the sequence fbn g: P Theorem 3 (Dirichlet’s Test). Assume that the partial sums of the series 1 nD1 an are bounded andPthe sequence fbn g1 is a monotone sequence that converges to 0. nD1 Then the series 1 a b converges. n n nD1 Proof. We will use the same notation as in the statement and proof of Abel’s partial summation formula (Proposition 1): An D a 1 C a 2 C    C a n is the nth partial sum of the series kDm X kDnC1

P1 nD1

an and we have

ak bk D .Am bmC1  An bnC1 / 

kDm X kDnC1

Ak .bkC1  bk /

304

5 Infinite Series

for each m > n  1: Thus ˇ ˇ kDm kDm ˇ ˇ X X ˇ ˇ ak bk ˇ  jAm bmC1  An bnC1 j C jAk j jbkC1  bk j ˇ ˇ ˇ kDnC1

kDnC1

 jAm j jbmC1 j C jAn j jbnC1 j C

kDm X

jAk j jbkC1  bk j :

kDnC1

Since we are given that the partial sums of the series exists M > 0 such that jAk j  M for each k 2 N. Thus

P1

nD1 an

are bounded there

ˇ ˇ kDm kDm ˇ ˇ X X ˇ ˇ ak bk ˇ  M jbmC1 j C M jbnC1 j C M jbkC1  bk j : ˇ ˇ ˇ kDnC1

kDnC1

Since the sequence fbn g1 nD1 is monotone the sum on the right is a telescoping sum so that kDm X

jbkC1  bk j D jbmC1  bnC1 j  jbmC1 j C jbnC1 j :

kDnC1

Thus ˇ ˇ ˇ kDm ˇ ˇ X ˇ ˇ ˇ a b k k ˇ  M jbmC1 j C M jbnC1 j C M .jbmC1 j C jbnC1 j/ D 2M jbmC1 j C 2M jbnC1 j : ˇ ˇkDnC1 ˇ

Let " > 0 be given. Since limn!1 bn D 0 there exists N 2 N such that jbk j < " if k  N: Therefore ˇ ˇ kDm ˇ ˇ X ˇ ˇ ak bk ˇ  2M jbmC1 j C 2M jbnC1 j < 4M" if m > n  N: ˇ ˇ ˇ kDnC1

Thus the series

P

an bn satisfies the Cauchy condition, hence converges. 

There are nice applications of Dirichlet’s test to series that involve cos .nx/ and sin.nx/. In such cases the following identities about sums of cosines and sines are useful:

5.3 Convergence Tests for Infinite Series: Part 2

305

Proposition 2. If x is not an integer multiple of 2 then sin cos .x/ C cos .2x/ C    C cos .nx/ D

   x 1 x  sin nC 2 2 x 2 sin 2

and    1 cos  cos nC x 2 2 x sin .x/ C sin .2x/ C    C sin .nx/ D 2 sin 2 x

for each n 2 N. Proof. Assume that x is an arbitrary real number and k is an integer. By the addition formula for sine       x 1 1 D sin kC x  sin k x : 2 cos .kx/ sin 2 2 2 Therefore    x 1 2 sin Œcos .x/ C cos .2x/ C    C cos .nx/ D sin nC x  sin : 2 2 2 x

Thus, if x is not an integer multiple of 2 then    x 1 x  sin sin nC 2 2 x cos .x/ C cos .2x/ C    C cos .nx/ D : 2 sin 2 Similarly, by the addition formula for cosine 2 sin .kx/ sin

x 2

 D cos

     1 1 k x  sin kC x : 2 2

Therefore    1 x : 2 sin Œsin .x/ C sin .2x/ C    C sin .nx/ D cos  cos nC 2 2 2 x

x

306

5 Infinite Series

Thus, if x is not an integer multiple of 2 then cos

x 2

sin .x/ C sin .2x/ C    C sin .nx/ D

 cos 2 sin

   1 nC x 2 x : 2

Example 4. Show that the series 1 X 1 nD1

n

cos .nx/ and

1 X 1 nD1

n

sin .nx/

converge if x is not an integer multiple of 2. Solution. Assume that x is not an integer multiple of 2. We have   ˇ   x ˇˇ 1 ˇ ˇ ˇ sin n C x  sin ˇ 1 2 2 ˇˇ x jcos .x/ C cos .2x/ C    C cos .nx/j D ˇˇ ˇ  ˇˇ  x ˇˇ ˇ ˇsin ˇ 2 sin ˇ ˇ ˇ 2 2 P for each n. Therefore the partial sums of the series 1 nD1 cos .nx/ are bounded. The sequence is decreasing and lim 1=n D 0. By Dirichlet’s test the series f1=ng n!1 P1 1 cos .nx/ converges. nD1 n Similarly,   ˇ ˇ x 1 ˇ ˇ ˇ cos  cos n C x ˇˇ ˇ 2 2 ˇ  ˇ 1 ˇ x jsin .x/ C sin .2x/ C    C sin .nx/j D ˇˇ ˇ ˇ x ˇ ˇ ˇsin ˇ 2 sin ˇ ˇ ˇ 2 2 if an integer multiple of 2 and n is a positive integer. Therefore the series Px1is not 1 sin .nx/ converges by the Dirichlet’s test.  nD1 n

5.3.3 A Strategy to Test Infinite Series for Convergence or Divergence Let us map a strategy to test infinite Pseries for convergence or divergence based on all the tests that we discussed. Let an be a given series: P • If limn!1 an ¤ 0, the series an diverges, since a necessary condition for the convergence of an infinite series is that the nth term should converge to 0 as n tends to infinity. There is nothing more to be done.

5.3 Convergence Tests for Infinite Series: Part 2

307

P • If limn!1 an D 0, we can begin by testing the series an for absolute convergence. If we conclude that the series converges absolutely, we are done, since absolute convergence implies convergence. Usually, the ratio test is the easiest test to apply, provided that it is conclusive. In some cases, the root test may be more convenient. If these tests are not conclusive, we may try the integral test. We may also try the comparison test or the limit comparison test, if we spot a “comparison series” without too much difficulty. Usually, the limit comparison test is easier to apply than the “basic” comparison test. • If we conclude that the given series does not converge absolutely, we may still investigate conditional convergence. We can appeal to the theorem on alternating series, Abel’s test or the Dirichlet’s test. Let us illustrate the implementation of the suggested strategy by a few examples. Example 5. Determine whether the series 1 X

.1/n1

nD1

10n .nŠ/2

converges absolutely, converges conditionally, or diverges. Solution. Whenever we see the factorial sign in the expression for the terms of a series, it is a good idea to try the ratio test: ˇ ˇ nC1 ˇ ˇ ˇ.1/n 10 ˇ 2 !  nC1   ˇ 2ˇ nŠ 10 ..n C 1/Š/ ˇ D lim lim ˇ n ˇ ˇ n!1 n!1 10n .n C 1/Š ˇ.1/n1 10 ˇ ˇ 2ˇ .nŠ/ 2 !  1 D lim 10 n!1 nC1 2  1 D 10 lim D 10 .0/ D 0 < 1: n!1 n C 1 Therefore, the series converges absolutely.  Example 6. Determine whether the series 1 X nD2

.1/

1 n ln2 .n/

converges absolutely, converges conditionally, or diverges.

308

5 Infinite Series

Solution. The ratio test and the root test are inconclusive, since the required limit in either case is 1 (confirm). Let’s try the integral test for absolute convergence. Thus, we set f .x/ D

1 ; x ln2 .x/

so that ˇ ˇ ˇ 1 1 ˇˇ ˇ.1/ D D f .n/ : ˇ ˇ 2 n ln .n/ n ln2 .n/ The function f is continuous, positive-valued, and decreasing on Œ2; C1/. Therefore, the integral test is applicable. If we set u D ln .x/, Z

1

1 dx D 2 x .ln .x//

Z

1 du dx D u2 dx

Z

1 du D u2

Z

u2 du D u1 D 

1 1 D : u ln .x/

Therefore, Z

Z

b 2

f .x/ dx D

ˇ 1 1 1 1 ˇˇb C : D dx D  ˇ 2 ln .x/ ln .b/ ln .2/ x ln .x/ 2

b 2

Thus, Z lim

b!1 2

b

  1 1 1 f .x/ dx D lim  C D : b!1 ln .b/ ln .2/ ln .2/

Therefore, the improper integral Z 2

Z

b

f .x/ dx D

2

1

1 dx x ln2 .x/

converges. Therefore the series 1 X nD2

.1/

1 n ln2 .n/

converges absolutely. Note that the theorem on alternating series is applicable to the given series, but does not lead to the fact that the series converges absolutely. 

5.3 Convergence Tests for Infinite Series: Part 2

309

Example 7. Determine whether the series 1 X

1 .1/n1 p 2 n C 4n C 1 nD1

converges absolutely, converges conditionally, or diverges. Solution. The ratio test and the root test are inconclusive, since the required limits are equal to 1 (confirm). The integral test will lead to the conclusion that the series does not converge absolutely after some hard work involving antidifferentiation. We will choose to apply the limit-comparison test. Since 1 1 p Š s  2 n C 4n C 1 4 n2 1 C C n

1 1 Šp 2 D n n 1

n2

for large n, the harmonic series appears to be a good choice as a “comparison series.” Let’s evaluate the limit that is required by the limit-comparison test: 1 p n n2 C 4n C 1 D lim p lim 2 1 n!1 n!1 n C 4n C 1 n n D lim s  n!1 4 n2 1 C C n D lim

n!1

n

r

n 1C

4 C n

1 n2



D lim r 1 n2

n!1

Since the harmonic series diverges, so does the series 1 X nD1

p

1 n2

C 4n C 1

Therefore, the series 1 X

1 .1/n1 p 2 n C 4n C 1 nD1

does not converge absolutely.

1C

1 4 C n

D 1 ¤ 0: 1 n2

310

5 Infinite Series

The theorem on alternating series is applicable to the given alternating series. Clearly, the sequence 1  1 p n2 C 4n C 1 nD1 is decreasing, and we have 1 D 0: lim p 2 n C 4n C 1

n!1

Therefore, the series converges. Since the series does not converge absolutely, the series converges conditionally. 

5.3.4 Problems In problems 1 and 2 use the theorem on alternating series to show that the given series converges. You need to verify that the conditions of the theorem are met: 2.

1. 1 X nD1

.1/

n1

1 X

1 2n  1

nD1

.1/n1

1 .n C 1/ ln .n C 1/

In problems 3 and 4 make use of Abel’s test to show that the given series converges. You need to verify that the conditions of the theorem are met: 3.  1  X 1 1 1C n 2n nD1 4. 1 X nD1

n1=n

1 n2 C 1

5. Make use of Dirichlet’s test to show that 1 X 1 p cos .n/ n nD1

converges (you need to verify that the conditions of the theorem are met).

5.3 Convergence Tests for Infinite Series: Part 2

311

In problems 6–12 determine whether the given series diverges, or whether it converges absolutely or conditionally. Use any means at your disposal. 6.

10. 1 X

1 X

1 .1/n1 p : n nD1

.1/n1

nD1

1 : n4n

11.

7. 1 X

.1/n

nD0

1 X

10n nŠ

8.

.1/n

nD2

1 : ln .n/ 2

12. 1 X

.1/k1

kD0

3k : k

9. 1 X nD2

.1/n1

1 : n ln .n/

1 X nD1

2

en cos .10n/

Chapter 6

Sequences and Series of Functions

6.1 Sequences of Functions In this section we will discuss the pointwise and uniform convergence of sequences of real-valued functions on subsets of the number line. The distinction between uniform versus pointwise convergence is important with regard to the preservation of properties such as continuity, differentiability, and integrability.

6.1.1 The Convergence of Sequences of Functions Definition 1. Assume that fn is a real-valued function that is defined on the set D  R for each n 2 N. The sequence of functions ffn g1 nD1 converges to the function f pointwise on the set D if lim f n .x/ D f .x/ for each x 2 D:

n!1

Thus, given a point x 2 D and " > 0 there exists a positive integer N .x; "/ such that jfn .x/  f .x/j < " if n  N .x; "/ : Convergence is said to be uniform if we can choose N independently of the particular x 2 D: Definition 2. The sequence of functions ffn g1 nD1 converges to the function f uniformly on the set D if, given any " > 0, there exists a positive integer N ."/ such that jfn .x/  f .x/j < " for each x 2 D if n  N ."/ : © Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0_6

313

314

6 Sequences and Series of Functions

Remark 1. It is worthwhile to express the negation of uniform convergence: The sequence ffn g1 nD1 does not converge to f uniformly on D if there exists " > 0 such that for each N 2 N there exists n  N and x 2 D such that jfn .x/  f .x/j  ": Example 1. Let fn .x/ D xn , n D 1; 2; 3; : : : a) Determine the pointwise limit f of the sequence ffn g1 nD1 on Œ0; 1. b) Show that ffn g1 nD1 converges to f uniformly on Œ0; 1  ı for any ı such that 0 < ı < 1: c) Show that ffn g1 nD1 does not converges to f uniformly on Œ0; 1. Solution. a) We have fn .0/ D 0 for each n so that limn!1 fn .0/ D 0. We also have fn .1/ D 1 for each n so that limn!1 fn .1/ D 1. If 0 < x < 1 then lim fn .x/ D lim xn D 0:

n!1

n!1

Therefore, the pointwise limit of the sequence ffn g is  f .x/ D

0 if 0  x < 1; 1 if x D 1:

Figure 6.1 shows f4 ; f8 ; f16 , and f32 . Fig. 6.1

1 0.8 0.6 0.4

f4

0.2

f8 0.2

0.4

0.6

f16 0.8

f32 1

b) Since xn is increasing on Œ0; 1, we have 0  fn .x/ D xn  .1  ı/n for each x 2 Œ0; 1  ı . Since limn!1 .1  ı/n D 0, given " > 0 there exists a positive integer N such that n  N ) 0  fn .x/  0 < " for each x 2 Œ0; 1  ı :

6.1 Sequences of Functions

315

Therefore, ffn g1 nD1 converges to f uniformly on Œ0; 1  ı. With reference to Fig. 6.1, if you imagine a narrow band centered around the graph of the limit function 0 on an interval of the form Œ0; 1  ı, the graphs of fn appear to be in such a band if n is sufficiently large. This is typical for the graphical implication of uniform convergence. c) Let " D 1=2. For each n lim fn .x/ D lim xn D 1:

x!1

n!1

Therefore, there exists x 2 .0; 1/ such that fn .x/  f .x/ D xn  0 >

1 : 2

Thus, ffn g1 nD1 does not converge to f uniformly on Œ0; 1. Figure 6.1 indicates that the graph of fn on the interval Œ0; 1 does not lie in a band centered around the graph of the limit function 0 on the interval Œ0; 1, no matter how large n may be. That is a graphical indication of nonuniform convergence.  We can rephrase the uniform convergence of a sequence of functions as follows: Proposition 1. The sequence of functions ffn g1 nD1 converges to f uniformly on a set D R if and only if lim

n!1

  sup jfn .x/ f .x/j D 0: x2D

Proof. Assume that ffn g1 nD1 converges to f uniformly on D. Given " > 0 there exists N 2 N such that jfn .x/  f .x/j < " for each x 2 D if n  N: Therefore sup jfn .x/  f .x/j  " if n  N: x2D

Thus lim

n!1

  sup jfn .x/  f .x/j D 0: x2D

Conversely, assume the above condition. Let " > 0 be given. Choose N such that sup jfn .x/  f .x/j  " if n  N: x2D

316

6 Sequences and Series of Functions

Then jfn .x/  f .x/j  " for each x 2 D if n  N Therefore ffn g1 nD1 converges to f uniformly on D.  Here is a useful sufficient condition for uniform convergence: Corollary 1. Assume that jfn .x/ f .x/j  Bn for each n 2N and x 2 D and limn!1 Bn D 0. Then the sequence ffn g1 nD1 converges to f uniformly on D. Proof. Let " > 0 be given. Since limn!1 Bn D 0 there exists N 2 N such that Bn < " if n  N. Thus jfn .x/  f .x/j  Bn < " for each n 2 N Therefore ffn g1 nD1 converges to f uniformly on D.  Example 2. Let fn .x/ D enx : Prove that ffn g1 nD1 converges to 0 uniformly on Œa; C1/ for any a > 0: Solution. Let a > 0. Each fn is monotone decreasing on the entire number line. Therefore we have 0 < fn .x/ D enx  ena : if x  a: Since limn!1 ena D 0 the sequence ffn g1 nD1 converges to 0 uniformly on Œa; C1/. Figure 6.2 shows the graphs of f1 ; f2 and f3 on Œ0; 4. The picture indicates that the graph of fn is an arbitrarily narrow band around the graph of the limit function 0 on an interval Œ0; A of arbitrarily large length when n is sufficiently large. That graphical observation is consistent with uniform convergence on Œ0; 1/:  Fig. 6.2

1

0.5

f1 f2 f3 1

2

3

4

6.1 Sequences of Functions

317

Example 3. Let fn .x/ D

xCn : 1 C nx

a) Determine the pointwise limit f of the sequence ffn g1 nD1 on .0; C1/. b) Show that ffn g1 converges to f uniformly on Œ1; C1/. nD1 c) Show that ffn g1 nD1 does not converge to f uniformly on .0; 1. Solution. a) For fixed x > 0, we have C 1 xCn  D lim lim fn .x/ D lim D lim  n n!1 n!1 1 C nx n!1 n!1 1 Cx n n n

x

x C1 1 n D : 1 x Cx n

Therefore, if f .x/ D 1=x, ffn g converges to pointwise to f on .0; C1/. Figure 6.3 shows the graphs of f4 , f8 and f . Fig. 6.3 2

1

f4 f8 f 1

2

3

b) We have ˇ ˇ 2 ˇ ˇ ˇ ˇ ˇx  1 ˇ ˇ xCn 1 ˇˇ ˇˇ x2 C nx  1  nx ˇˇ ˇ  D jfn .x/  f .x/j D ˇ ˇ D x .1 C nx/ 1 C nx x ˇ ˇ x .1 C nx/ Set gn .x/ D

ˇ ˇ 2 ˇx  1 ˇ x .1 C nx/

4

318

6 Sequences and Series of Functions

Figure 6.4 shows the graph of g4 on Œ0; 7. Fig. 6.4 0.25

1

2

3

4

5

6

7

If x  1 then gn .x/ D

x2  1 x .1 C nx/

so that g0n .x/

D

  2x2 .1 C nx/  x2  1 .1 C 2nx/ x2 .1 C nx/2

D

x2 C 2nx C 1 x2 .1 C nx/2

>0

Therefore, gn is increasing on Œ1; C1/. We have   1 1 x 1 2 1 2 x2  1 x x D 1:  D lim  D lim lim gn .x/ D lim 1 x!1 x!1 x .1 C nx/ x!1 x!1 1 n Cn Cn x2 x x 2

Therefore, jfn .x/  f .x/j D gn .x/
0 be given. Choose a positive integer N such that n  N and m  N ) jfn .x/  fm .x/j
0 such that jfN .x/  fN .x0 /j
0 be given. Since ffn g1 nD1 converges to f uniformly on Œa; b there exists N 2 N such that jfN .x/  f .x/j < " if x 2 Œa; b . We will make use of the characterization of integrability in terms of the oscillations of a function on intervals (Remark 1 of Sect. 4.1): Since fN is integrable on Œa; b there exists ı > 0 such that m X

! .fN ; Œxk1 ; xk / xk < "

kD1

if P D fx0 ; x1 ; x2 ; : : : :xm g is a partition of Œa; b with jjPjj < ı and ! .fN ; Œxk1 ; xk / D sup f.jfN ./  fN ./j where  and  are in Œxk1 ; xk /g : If  and  are in Œxk1 ; xk  jf ./  f ./j  jf ./  fN ./j C jfN ./  fN ./j C jfN ./  f ./j < jfN ./  fN ./j C 2":

6.1 Sequences of Functions

323

Therefore ! .f ; Œxk1 ; xk / xk  ! .fN ; Œxk1 ; xk / xk C 2"xk : Thus m X

! .f ; Œxk1 ; xk / xk 

kD1

m X

! .fN ; Œxk1 ; xk / xk C 2" .b  a/ < " C 2" .b  a/ :

kD1

if jjPjj < ı. Therefore f is integrable on Œa; b. Now let us show that Z

Z

b

b

fn .x/ dx D

lim

n!1 a

f .x/ dx: a

Let " > 0 be given. Since ffn g1 nD1 converges to f uniformly on Œa; b there exists a positive integer N such that jfn .x/  f .x/j < ": for each x 2 Œa; b if n  N. If n  N, by the triangle inequality for integrals, ˇZ ˇ ˇ ˇ

Z

b

b

fn .x/ dx  a

a

ˇ ˇZ ˇ ˇ f .x/ dxˇˇ D ˇˇ

b a

ˇ Z b ˇ .fn .x/  f .x// dxˇˇ  jfn .x/  f .x/j dx a Z b "dx D " .b  a/  a

Thus, Z

Z

b

b

fn .x/ dx D

lim

n!1 a

f .x/ dx; a

as claimed.  Corollary 2. If each fn is continuous on Œa; b and ffn g1 nD1 converges to f uniformly on Œa; b then f is integrable on Œa; b and Z lim

n!1 a

b

Z fn .x/ dx D a

b

f .x/ dx:

324

6 Sequences and Series of Functions

Proof. Since ffn g1 nD1 converges to f uniformly on Œa; b the limit f is also continuous (Theorem 2). Thus f is certainly integrable and Z lim

n!1 a

Z

b

b

fn .x/ dx D

f .x/ dx: a

 We cannot claim that

Z

lim

n!1 a

b

Z

b

fn .x/ dx D a



lim fn .x/ dx

n!1

if the sequence ffn g1 nD1 does not converge uniformly on Œa; b: Example 5. Let fn .x/ D

2n2 x .n2 x2 C 1/2

:

Figure 6.6 displays the graphs of f4 and f16 : Fig. 6.6

10 5

f16 f4 0.5

−0.5 −5

−10

For each x 2 R lim fn .x/ D lim

n!1

n!1

D lim

n!1

2n2 x .n2 x2 C 1/2

2n2 x   n!1 1 2 n 4 x2 C 2 n

D lim

2x   D 0: 1 2 2 2 n x C 2 n

Thus, the pointwise limit of the sequence ffn g on R is the constant function 0.

6.1 Sequences of Functions

325

We have Z

1 0

Z fn .x/ D

1

0

2n2 x .n2 x2 C 1/2

dx:

If we set u D n2 x2 C 1, then du D 2n2 x so that Z

1 0

Z

2n2 x

dx D .n2 x2 C 1/2

n2 C1 1

1 du D u2

Z

n2 C1

u2 du

0

ˇ2 n2 1 ˇˇn C1 1 C1D 2 D  ˇ D 2 u 1 n C1 n C1

Therefore, Z

1

lim

n!1 0

n2 D 1; n!1 n2 C 1

fn .x/ dx D lim

even though Z

1

Z lim fn .x/ dx D

0 n!1

1 0

.0/ dx D 0:

Note that the sequence ffn g1 nD1 does not converge to f uniformly on Œ0; 1. Indeed, fn

  n 1 D : n 2

Thus, given any n, we can find x 2 Œ0; 1 such that jf .x/  0j 

1 : 2

This rules out the uniform convergence of the sequence ffn g1 nD1 to 0 on Œ0; 1.  The issue of the convergence of the derivatives of a sequence of functions is more delicate than the convergence of the integrals: Theorem 4. Assume that ˚ 0 1each fn has a continuous derivative on Œa; b, the sequences ffn g1 nD1 and fn nD1 converge uniformly on Œa; b to f and g, respectively. Then f is differentiable on Œa; b and we have g .x/ D f 0 .x/ for each x 2 Œa; b. Thus, lim fn0 .x/ D

n!1

 lim fn

n!1

0

.x/ for each x 2 Œa; b :

326

6 Sequences and Series of Functions

Proof. By Theorem 2 f and g are continuous on Œa; b. By the Fundamental Theorem of Calculus, Z x fn .x/ D fn .a/ C fn0 .t/ dt a

for each x 2 Œa:b. By Theorem 3 Z f .x/ D lim fn .x/ D lim fn .a/ C lim n!1

n!1

D f .a/ C

Z x

fn0 .t/ dx

lim fn0 .t/ dx

n!1

a

Z

n!1 a

x

x

D f .a/ C

g .t/ dt: a

Again, by the Fundamental Theorem of Calculus, f 0 .x/ D g .x/ for each x 2 Œa; b. Thus f 0 .x/ D

 lim fn

0

n!1

.x/ for each x 2 Œa; b :

 Remark 2. The hypotheses of Theorem 4 can be weakened: Assume that each fn is ˚ 1 differentiable on Œa; b, fn0 nD1 convergence uniformly to g on Œa; b, x0 2 Œa; b and 1 the sequence of numbers ffn .x0 /g1 nD1 converges. Then ffn gnD1 converges uniformly 0 0 to a function f on Œa; b and f .x/ D limn!1 fn .x/ D g .x/ for each x 2 Œa; b. The proof is somewhat more delicate than the proof of Theorem 4 and can be found at the end of this section. For our purposes Theorem 4 will be adequate. Þ ˚ 1 If we do not assume the uniform convergence of fn0 nD1 we cannot conclude that lim fn0 .x/ D

n!1

 lim fn

n!1

0

.x/ ;

as shown by the following example: Example 6. Let fn .x/ D

1 arctan .nx/ : n

6.1 Sequences of Functions

327

Since jfn .x/j D

1  for each x 2 R; jarctan .nx/j < n 2n

the sequence ffn g1 nD1 converges to the constant function 0 uniformly on R. We have fn0 .x/ D

1 n



 1 1 .n/ D 2 2 1Cn x 1 C n 2 x2

for each x 2 R. In Example 4 we showed that lim f 0 n!1 n

1 .x/ D lim D n!1 1 C n2 x2



0 1

if if

x ¤ 0; x D 0:

Thus, lim f 0 n!1 n

.0/ D 1:

˚ 1 But .limn!1 fn /0 .0/ D 0. In Example 4 we showed that the sequence fn0 nD1 does not converge uniformly on any interval that contains 0. 

6.1.3 Proof of Remark 2 To begin with, we will show that the sequence ffn g1 nD1 converges uniformly on Œa; b. Given m and n in N and x 2 Œa; b, we will apply the Mean Value Theorem to the interval determined by x and x0 : There exists a point c between x and x0 such that   .fm .x/  fn .x//  .fm .x0 /  fn .x0 // D fm0 .c/  fn0 .c/ .x  x0 / : Therefore ˇ ˇ   j.fm .x/  fn .x//j D ˇ.fm .x0 /  fn .x0 // C fm0 .c/  fn0 .c/ .x  x0 /ˇ ˇ ˇ  jfm .x0 /  fn .x0 /j C jx  x0 j ˇfm0 .c/  fn0 .c/ˇ ˇ ˇ  jfm .x0 /  fn .x0 /j C .b  a/ sup ˇfm0 .y/  fn0 .y/ˇ y2Œa;b

˚ 1 for each x 2 Œa; b. Since limn!1 fn .x0 / exists and fn0 nD1 converges uniformly on Œa; b, the above inequality shows that the sequence ffn g1 nD1 satisfies the Cauchy criterion for uniform convergence. Therefore there exists a function f such that ffn g1 nD1 converges to f uniformly on Œa; b. Since each fn is differentiable on Œa; b each fn is continuous on Œa; b. By the uniform convergence of ffn g1 nD1 to f the function f is also continuous on Œa; b.

328

6 Sequences and Series of Functions

Now we will show that f is differentiable at each point x 2 Œa; b and f 0 .x/ D g .x/. We will consider x 2 .a; b/. The statement about the relevant one-sided derivative at the endpoints a and b requires a trivial modification. Let " > 0 be given. Let us apply the Mean Value Theorem to fm  fn on the interval determined by x and x C h, assuming that jhj ¤ 0 is small enough: There exists c in the interval determined by x and x C h such that

 Œfm .x C h/  fn .x C h/  Œfm .x/  fn .x/ D fm0 .c/  fn0 .c/ h: Therefore ˇ ˇ ˇ ˇ ˇ fm .x C h/  fm .x/ fn .x C h/  fn .x/ ˇ ˇ  sup ˇf 0 .y/  f 0 .y/ˇ ˇ  m n ˇ ˇ h h y2Œa;b ˚ 1 Since fn0 nD1 converges uniformly on Œa; b there exists N1 2 N such that ˇ ˇ ˇ fm .x C h/  fm .x/ fn .x C h/  fn .x/ ˇ ˇ ˇ < " if m  N1 and n  N1 :  ˇ ˇ h h Since limm!1 fm .x/ D f .x/ ˇ ˇ ˇ f .x C h/  f .x/ fn .x C h/  fn .x/ ˇ ˇ  " if n  N1 : ˇ  ˇ ˇ h h ˇ ˇ Since limn!1 fn0 .x/ D g .x/ there exists N  N1 such that ˇfN0 .x/  g .x/ˇ < ". Since fN is differentiable at x there exists ı > 0 such that ˇ ˇ ˇ fN .x C h/  fN .x/ ˇ 0 ˇ  fN .x/ˇˇ < " ˇ h if 0 < jhj < ı. In this case ˇ ˇ ˇ ˇ ˇ ˇ f .x C h/  f .x/ fN .x C h/  fN .x/ ˇ ˇ f .x C h/  f .x/ ˇ ˇ ˇ ˇ  g .x/ˇ  ˇ  ˇ ˇ h h h ˇ ˇ ˇ fN .x C h/  fN .x/ ˇ ˇ ˇ C ˇˇ  fN0 .x/ˇˇ C ˇfN0 .x/  g .x/ˇ h < 3": Therefore f 0 .x/ exists and we have f 0 .x/ D g .x/. 

6.1 Sequences of Functions

329

6.1.4 Problems 1. Let fk .x/ D

sin .kx/ ; k D 1; 2; 3; : : : : k

a) Determine the pointwise limit f of the sequence ffk g1 kD1 on R: b) Show that the sequence ffk g1 converges to f uniformly to f on R: kD1 2. Let fk .x/ D x1=k , k D 1; 2; 3; : : :. a) Determine the pointwise limit f of the sequence ffk g1 kD1 on the interval Œ0; 1. b) Show that the sequence ffk g1 does not converge to f uniformly to f on Œ0; 1. kD1 c) Show that the sequence ffk g1 converges to f uniformly to f on Œı; 1, where kD1 0 < ı < 1. 3. Let 2

fk .x/ D

ex =k ; k D 1; 2; 3; : : : : k

a) Determine the pointwise limit f of the sequence ffk g1 kD1 on R: b) Show that the sequence ffk g1 converges to f uniformly to f on R: kD1 4. Let fk .x/ D kxekx ; k D 1; 2; 3; : : : : a) Determine the pointwise limit f of the sequence ffk g1 kD1 on Œ0; C1/: b) Show that the sequence ffk g1 does not converges to f uniformly to f on kD1 Œ0; C1/: c) Show that the sequence ffk g1 kD1 converges to f uniformly to f on Œı; 1/, where ı > 0. 5. Let fk .x/ D

xk ; k D 1; 2; 3; : : : : 1 C x2k

a) Determine the pointwise limit f of the sequence ffk g1 kD1 on Œ0; C1/: b) Show that the sequence ffk g1 does not converges to f uniformly to f on kD1 Œ0; C1/: c) Show that the sequence ffk g1 kD1 converges to f uniformly to f on Œ1 C ı; 1/, where ı > 0.

330

6 Sequences and Series of Functions

6.2 Infinite Series of Functions In this section we will discuss the convergence of series of functions. Uniform convergence on a subset of the number line will be important with regard to the continuity, differentiability, and the integrability of their sums on that set. We will state and prove useful tests that ensure uniform convergence.

6.2.1 The Convergence of Series of Functions The convergence of a series of functions is defined in terms of the convergence of the sequence of its partial sums: Definition 1. Assume that each fk is defined on D  R and sn .x/ D

n X

fk .x/ for each x 2 D:

kD1

P The series 1 kD1 fk converges pointwise onPD if the sequence of partial sums converges pointwise on D. The series 1 on D if fsn g1 nD1 kD1 fk converges uniformlyP 1 the sequence of partial sums fsn g1 converges uniformly on D. The series nD1 kD1 fk P1 converges absolutely and uniformly on D if kD1 jfk j converges uniformly on D. Just as in the case of sequences of functions, the uniform Cauchy condition for series of functions is essential: Theorem P 1 (Uniform Cauchy Criterion for Series of Functions). The series of functions fn converges uniformly on D R if and only if given " > 0 there exists N 2N such that ˇ ˇ n  N ) ˇfnC1 .x/ Cf nC2 .x/ C    C f nCk .x/ˇ < " for each x 2 D and k D 1; 2; 3; : : : Proof. Let sn .x/ D

n X

fk .x/

kD1

be the nth partial sum of the series

P

fk . We have

snCk .x/  sn .x/ D f1 .x/ C f2 .x/ C    C fn .x/ C fnC1 .x/ C fnC2 .x/ C    C fnCk .x/  .f1 .x/ C f2 .x/ C    C fn .x/ C fnC1 .x// D fnC1 .x/ C fnC2 .x/ C    C fnCk .x/ :

6.2 Infinite Series of Functions

331

Therefore jsnCk .x/  sn .x/j D jfnC1 .x/ C fnC2 .x/ C    C fnCk .x/j : Given " > 0 there exists N such that jsnCk .x/ sn .x/j D jfnC1 .x/ CfnC2 .x/ C    CfnCk .x/j 0 there exists N 2 N such that jfnC1 .x/j C jfnC2 .x/j C    C jfnCk .x/j < " for each x 2 D, n  N and k D 1; 2; 3; : : : Then jfnC1 .x/ CfnC2 .x/ C    CfnCk .x/j  jfnC1 .x/j C jfnC2 .x/j C    C jfnCk .x/j 0 be given. Since Mk converges it satisfies the Cauchy condition. Thus there exists a positive integer N such that if n  N then 0  MnC1 C MnC2 C    C MnCk < " for k D 1; 2; 3; : : : Therefore, if x 2 D and n  N the jfnC1 .x/j C jfnC2 .x/j C    C jfnCk .x/j  MnC1 C MnC2 C    C MnCk < " for k D 1; 2; 3; : : :

Thus the series gence. 

P

jfn j satisfies the uniform Cauchy condition on D, hence conver-

332

6 Sequences and Series of Functions

Example 1. Show that 1 X .1/k

kŠ

kD0

xk

converges uniformly on the interval ŒA; A where A is an arbitrary positive integer. Solution. If x 2 ŒA; A then ˇ ˇ ˇ .1/k ˇ 1 1 ˇ ˇ xk ˇ D jxjk  Ak : ˇ ˇ ˇ kŠ kŠ kŠ The series 1 X 1 k A kŠ kD0

converges, as you can confirm via the ratio test (as a matter of fact, the sum of the series is eA , as we will discuss in Sect. 6.4). Therefore the Weierstrass M-test is applicable and the given series of functions converges absolutely and uniformly on ŒA; A.  Example 2. Show that 1 X

2

ek t sin .kx/

kD1

converges uniformly on R for each t > 0: Solution. We have ˇ 2 ˇ 2 2 ˇ k t ˇ sin .kx/ˇ D ek t jsin .kx/j  ek t for each x 2 R: ˇe By the Weierstrass M-test, it is sufficient to show that 1 X

ek

2t

kD1

converges for each t > 0. Indeed,  2 1=k lim ek t D lim ekt D 0:

k!1

Therefore, the series

P1 kD1

k!1

2

ek t converges by the root test. 

6.2 Infinite Series of Functions

333

In Chap. 5 we discussed Abel’s test and Dirichlet’s test (Theorem 2 and Theorem 3 of Sect. 5.3, respectively). Those tests were useful in establishing the convergence of series of numbers even in cases where convergence was conditional (i.e., absolute convergence was lacking). The counterparts of those tests for series of functions are useful in establishing the uniform convergence of a series of functions in cases where the series converges uniformly on a set but the series of absolute values does not. Let us begin with Abel’s test for series of functions. The proof is a modification of the proof of the test for series of real numbers. Theorem 4 (Abel’s Test for Series of Functions). Assume that each fn is a realP valued function defined on the set D R and the series 1 f nD1 n converges uniformly on D. Let fgn g1 be a uniformly bounded monotone (increasing nD1 P1 or decreasing) sequence of real-valued functions on D. Then the series nD1 fn gn converges uniformly on D: Proof. The proof will be based on the uniform Cauchy condition for uniform convergence. We will make use of Abel’s partial summation formula (Proposition 1 of Sect. 5.3): If Fn .x/ D f1 .x/ C f2 .x/ C    C fn .x/ P is the nth partial sum of the series fn .x/ we have kDm X

fk .x/ gk .x/ D .Fm .x/  Fn .x// gnC1 .x/

kDnC1 m X

C

.Fm .x/  Fk .x// .gkC1 .x/  gk .x//

kDnC1

for integers m > n  1 and x 2P D. Let " > 0 be given. Since fn converges uniformly on D there exists N 2 N such that jFk .x/  Fn .x/j < " if k > n  N and x 2 D: Since the sequence fgn g1 nD1 is uniformly bounded on D there exists M > 0 such that jgn .x/j  M for each n 2 N and x 2 D. Thus, for each n  N and x 2 D ˇ kDm ˇ ˇ X ˇ ˇ ˇ fk .x/ gk .x/ˇ  jFm .x/  Fn .x/j jgnC1 .x/j ˇ ˇ ˇ kDnC1

C

m X

jFm .x/  Fk .x/j jgkC1 .x/  gk .x/j

kDnC1

< M" C "

m X kDnC1

jgkC1 .x/  gk .x/j :

334

6 Sequences and Series of Functions

Since the sequence fgn .x/g1 nD1 is monotone the sum on the right is a telescoping Pm sum so that kDnC1 jgkC1 .x/  gk .x/j D jgmC1 .x/  gnC1 .x/j  2M: Therefore ˇ kDm ˇ m ˇ X ˇ X ˇ ˇ fk .x/ gk .x/ˇ < M" C " jgkC1 .x/  gk .x/j  3M": ˇ ˇ ˇ kDnC1

kDnC1

P for each m > n  N and x 2 D. Thus, the series fn gn converges uniformly on D since it satisfies the uniform Cauchy condition on D.  Example 3. Show that the series 1 X

1 .1/n enx n nD1

converges uniformly on Œ0; C1/: Solution. Note that the series does not converge absolutely at x D 0. Thus the Weierstrass M-Test is not applicable to the given series on Œ0; C1/. We will apply Abel’s test: The alternating series 1 X nD1

.1/n

1 n

converges since the sequence f1=ng1 nD1 is decreasing and limn!1 1=n D 0. Since the series has constant terms convergence is uniform, of course. If we set gn .x/ D enx then the sequence fgn .x/g1 nD1 is decreasing and jgn .x/j  1 for each x  0. Therefore the given series converges uniformly, thanks to Abel’s test.  Here is the counterpart of Dirichlet’s test for series of functions: Theorem 5 (Dirichlet’s Test for Series of Functions). Assume that each fn is a real-valued function defined on the set D R and the partial sums of the series P 1 1 of nD1 fn are uniformly bounded on D. Let fgn gnD1 be a monotone sequence P real-valued functions that converge uniformly to 0 on D.Then the series 1 f g nD1 n n converges uniformly on D. Proof. We will make use of Abel’s partial summation formula (Proposition 1 of Sect. 5.3): If Fn .x/ D f1 .x/ C f2 .x/ C    C fn .x/

6.2 Infinite Series of Functions

335

is the nth partial sum of the series kDm X

P

fn .x/ we have

fk .x/ gk .x/ D Fm .x/ gmC1 .x/  Fn .x/ gnC1 .x/

kDnC1

C

m X

Fk .x/ .gkC1 .x/  gk .x//

kDnC1

for integers m > n  1 and x 2 D: Thus ˇ ˇ kDm ˇ ˇ X ˇ ˇ fk .x/ gk .x/ˇ  jFm .x/ gmC1 .x/  Fn .x/ gnC1 .x/j ˇ ˇ ˇ kDnC1

C

m X

jFk .x/j jgkC1 .x/  gk .x/j

kDnC1

 jFm .x/j jgmC1 .x/j C jFn .x/j jgnC1 .x/j C

m X

jFk .x/j jgkC1 .x/  gk .x/j :

kDnC1

P Since we are given that the partial sums of the series 1 nD1 fn are uniformly bounded on D there exists M > 0 such that jFk .x/j  M for each k 2 N and x 2 D. Thus ˇ ˇ kDm m ˇ ˇ X X ˇ ˇ fk .x/ gk .x/ˇ  M jgmC1 .x/j C M jgnC1 .x/j C M jgkC1 .x/  gk .x/j : ˇ ˇ ˇ kDnC1

kDnC1

Since the sequence fgn .x/g1 nD1 is monotone the sum on the right is a telescoping sum so that m X

jgkC1 .x/  gk .x/j D jgmC1 .x/  gnC1 .x/j  jgmC1 .x/j C jgnC1 .x/j

kDnC1

Thus ˇ kDm ˇ ˇ X ˇ ˇ ˇ fk .x/ gk .x/ˇ  M jgmC1 .x/j C M jgnC1 .x/j C M .jgmC1 .x/j C jgnC1 .x/j/ ˇ ˇ ˇ kDnC1

D 2M jgmC1 .x/j C 2M jgnC1 .x/j :

336

6 Sequences and Series of Functions

Let " > 0 be given. Since fgn g converges to 0 uniformly on D there exists N 2 N such that jgk .x/j < " if k  N and x 2 D: Therefore ˇ ˇ kDm ˇ ˇ X ˇ ˇ fk .x/ gk .x/ˇ  2M jgmC1 .x/j C 2M jgnC1 .x/j < 4M" ˇ ˇ ˇ kDnC1

P if m > n  N and x 2 D. Thus fn gn satisfies the uniform Cauchy P the series condition on that set. Therefore fn gn converges uniformly on D.  Example 4. In Example 3 of Sect. 5.3 we showed that the series 1 X 1 nD1

n

cos .nx/

converges if x is not an integer multiple of 2 by applying Dirichlet’s test. Let us show that the given series of functions converge uniformly on an interval of the form Œı; 2  ı where 0 < ı < 2: We showed that    x 1 sin nC x  sin 2 2 x cos .x/ C cos .2x/ C    C cos .nx/ D 2 sin 2 if x is not an integer multiple of 2 (Proposition 2 of Sect. 5.3). Therefore jcos .x/ C cos .2x/ C    C cos .nx/j 

2 1 ˇ  x ˇ D ˇ  x ˇ : ˇ ˇ ˇ ˇ 2 ˇsin ˇsin ˇ ˇ 2 2

If x 2 Œı; 2  ı and 0 < ı < 2 there exists m > 0 such that ˇ  x ˇ ˇ ˇ ˇm ˇsin 2 P (confirm). Thus, the partial sums of the series cos .nx/ are uniformly bounded on Œı; 2  ı. The monotone sequence f1=ng1 nD1 converges to 0. Therefore Theorem 5 is applicable and confirms that the given series converges uniformly on Œı; 2  ı. 

6.2 Infinite Series of Functions

337

6.2.3 Continuity, Integrability, and Differentiability of Sums The following facts follow from the corresponding facts about sequences of functions. The proofs are straightforward exercises. Theorem 6. Assume that each fk is continuous on Œa; b, f .x/ D

1 X

fk .x/ for each x 2 Œa; b ;

kD1

and the convergence of the series of functions is uniform on Œa; b. Then f is continuous on Œa; b. Theorem 7 (Term-by-Term Integration). We have Z

b

Z f .x/ dx D

a

1 bX

a

! fk .x/ dx D

kD1

1 Z X kD1

b

fk .x/ dx:

a

Theorem 8 (Term-by-Term Differentiation). Assume that each fk is continuously differentiable on Œa; b g .x/ D

1 X

fk0 .x/ for each x 2 Œa; b

kD1

and the convergence of

P

0

fk is uniform on Œa; b. Then d f .x/ D g .x/ ; dx

i.e., 1 1 X d X d fk .x/ fk .x/ D dx kD1 dx kD1

for each x 2 Œa; b.

6.2.4 Problems In problems 1–3 show that the given series of functions converges uniformly on the given set D:

338

6 Sequences and Series of Functions

1. 

1 1 .1/n1 xn D x  x2 C x3  x4 C    ; D D  ; 2 2 nD1

1 X

(do not rely on a general fact about power series). 2. 1 X sin .nx/

nŠ

nD1

; DDR

3. 1 X

enx D ex C e2x C e3x C    ; D D .1; 1

nD1

4. Make use of Abel’s Theorem on uniform convergence show that the series 1 X

1 .1/n p 2nx n nD1

converges uniformly on Œ0 C 1/: 5. Make use of Dirichlet’s test to show that 1 X 1 p sin .nx/ n nD1

converges uniformly on Œı; 2  ı if 0 < ı < 2. 6. Provide the proof: Assume that each fk is continuous on Œa; b, f .x/ D

1 X

fk .x/ for each x 2 Œa; b ;

kD1

and the convergence of the series of functions is uniform on Œa; b. Then f is continuous on Œa; b. 7. Provide the proof: Assume that each fk is continuous on Œa; b, f .x/ D

1 X kD1

fk .x/ for each x 2 Œa; b ;

6.3 Power Series

339

and the convergence of the series Z

b

Z f .x/ dx D

a

P1

kD1 fk

1 bX

a

is uniform on Œa; b. Then

! fk .x/ dx D

kD1

1 Z X kD1

b

fk .x/ dx:

a

8. Provide the proof: Assume that each fk is continuous on Œa; b, f .x/ D

1 X

fk .x/ for each x 2 Œa; b ;

kD1

P and the convergence of 1 kD1 fk functions is uniform on Œa; b. Also assume P 0 fk that each fk is continuously differentiable on Œa; b and the convergence of is uniform on Œa; b. Then 1

1

kD1

kD1

X d d X d f .x/ D fk .x/ fk .x/ D dx dx dx for each x 2 Œa; b. 9. Let Z f .x/ D

x 0

1 du for each x 2 .1; 1/ : 1  u2

Determine a power series for f that converges for each x 2 .1; 1/. 10. Let f .x/ D

1 X

en sin .nx/ :

nD1

a) Show that the series converges uniformly and absolutely on R . b) Determine f 0 .x/ for each x 2 R by differentiating the series termwise. Justify the fact that termwise differentiation is valid for each x 2 R.

6.3 Power Series Many special functions are expressed as series in powers of .x  x0 / for some x0 2 R. In this section we will discuss the basic facts about such power series and functions that are defined as their sums.

340

6 Sequences and Series of Functions

6.3.1 The Convergence of Power Series Definition 1. Given x0 2 R, a power series in powers of x  x0 is of the form 1 X

ak .x  x0 /k D a0 Ca1 .x  x0 / Ca2 .x  x0 /2 Ca3 .x  x0 /3 C    ;

kD0

where the coefficients a0 ; a1 ; a2 ; a3 ; : : : are given numbers. Example 1. The power series 1 X

xk D 1 C x C x2 C x3 C   

kD0

is the geometric series corresponding to x.  Example 2. 1 X

.1/k1

kD1

1 1 1 1 .x  1/k D .x  1/  .x  1/2 C .x  1/3  .x  1/4 C    k 2 3 4

is a power series in powers of x  1.  The following proposition signals the nice convergence behavior of power series: P k Proposition 1. Assume that the power series 1 kD0 ak .x  x0 / converges at x1 ¤ x0 . If 0 < r < jx1 x0 j the series converges absolutely and uniformly in the interval Œx0 r; x0 Cr. Proof. If x 2 Œx0  r; x0 C r then jx  x0 j  r < jx1  x0 j so that r jx  x0 j  < 1: jx1  x0 j jx1  x0 j Thus, if we set qD

r jx1  x0 j

we have 0  q < 1 and jx  x0 j  q for each x 2 Œx0  r; x0 C r : jx1  x0 j Since

P1

kD0 ak

.x1  x0 /k converges we have lim jak j jx1  x0 jk D 0:

k!1

6.3 Power Series

341

Therefore there exists a positive integer N such that k  N ) jak j jx1  x0 jk  1: For any k  N and x 2 Œx0  r; x0 C r, jak j jx  x0 j D jak j k

jx  x0 jk jx1  x0 jk

! jx1  x0 jk D jak j qk jx1  x0 jk D jak j jx1  x0 jk qk  qk :

P k Since 0  q 0 such that the series converges absolutely and uniformly on any interval of the form Œx0  r; x0 C r where 0 < r < R, and the series diverges if x < x0  R or x > x0 C R: Proof. a) The series may converge only at x D x0 . For example 1 X

kŠxk

kD0

converges if and only if x D 0 (confirm by the ratio test). b) Assume that r is an arbitrary positive number. Since the series converges at any real number, it converges at x D x0 Cr C1. By Proposition 1 the series converges absolutely and uniformly on Œx0  r; x0 C r. c) Set n o X SD r2RW ak .x  x0 /k converges if jx  x0 j < r and let R D sup S. We claim that 0 < R < C1. Indeed, assume that the series converges at x1 ¤ x0 . By Proposition 1 the series converges if jx  x0 j < jx1  x0 j. This shows that R > 0. We are also given that there exists x2 2 R

342

6 Sequences and Series of Functions

P such that ak .x  x0 /k diverges at x D x2 . Then R  jx2  x0 j. Indeed, if jx2  x0 j < R, the number jx2  x0 j is not an upper bound for the set S, since R is the least upper bound of S. Therefore, there exists r 2 S such that jx2  x0 j < r < R. P By the definition of S, ak .x  x0 / converges if jx  x0 j < r. P In particular, P ak .x  x0 / converges if x D x2 , contradicting the fact that ak .x  x0 / diverges at x2 . Thus, 0 < R < 1: If 0 < r < R there exists 2 S such P that r < < R since R is the least upper bound of S. By the definition of S, ak .x  x0 /k converges if jx  x0 j < . By Proposition 1 the series converges absolutely and uniformly on Œx0  r; x0 C r.  Definition 2. With reference to Theorem 1, if there are points other than x0 at which the series converges and points at which the series diverges,.the radius of P k convergence of the power series 1 a kD0 k .x  x0 / is n o X R D sup r 2 R W ak .x  x0 /k converges if jx  x0 j < r : The open interval .x0  R; x0 C R/ is the open interval of convergence of the series. If the series converges only at x D x0 then the radius of convergence of the series is declared to be 0, and the open interval of convergence is empty. If the series converges at each x 2 R, the radius of convergence of the series is declared to be C1 and the open interval of convergence is the entire number line. By Theorem P 1, if the radius of convergence is a finite positive number R the power series an .x  x0 /n converges absolutely and uniformly on any closed and bounded interval that is contained in the open interval .x0  R; x0 C R/. If the radius of convergence is C1 the power series converges absolutely and uniformly on any closed and bounded interval. If the radius of convergence is 0 the power series does not have an open interval of convergence. We can implement the ratio test or the root test in order to determine the radius of convergence and the open interval of convergence. We need to use other tests if we are interested in the convergence or divergence of the series at the endpoints of the open interval of convergence. Example 3. Show that the power series 1 X 1 k 1 1 x D 1 C x C x2 C x3 C    kŠ 2 3Š kD0

converges absolutely and uniformly on any bounded interval.

6.3 Power Series

343

Solution. Let x be an arbitrary real number. We have ˇ ˇ ˇ ˇ 1 kC1 ˇ ! ˇ   x ˇ .k C 1/Š ˇ kŠ 1 jxjkC1 ˇ ˇ D lim D lim lim jxj ˇ1 ˇ k!1 k!1 .k C 1/Š jxjk k!1 k C 1 ˇ xk ˇ ˇ kŠ ˇ D jxj lim

k!1

1 D 0 < 1: kC1

By the ratio test, the power series converges. By Theorem 1 the power series converges absolutely and uniformly on any interval of the form Œx0  r; x0 C r, where r is an arbitrary positive number. Therefore the same is true on any bounded interval (in Sect. 6.4 we will show that the sum of the series is ex for each x 2 R).  Example 4. Consider the power series 1 X nD0

1 1 1 1 xn D 1 C x C x2 C x3 C    : 2n C 1 3 5 7

a) Show that the series converges uniformly on any closed interval Œr; r, where 0 < r < 1, and that the series diverges if jxj > 1: b) Determine whether the power series converges absolutely or conditionally at the endpoints of its open interval of convergence, or whether it diverges at any of these points. Solution. a) We will apply the ratio test: ˇ ˇ ˇ 1 ˇ nC1 ˇ ˇ   x ˇ 2n C 3 ˇ 2n C 1 2n C 1 ˇ ˇ lim D jxj : D lim jxj D jxj lim ˇ n!1 ˇ n!1 2n C 3 n!1 2n C 3 1 nˇ ˇ x ˇ 2n C 1 ˇ Therefore, the radius of convergence of the series is 1, and the open interval of the series is .1; 1/. Thus, the series converges absolutely and uniformly on any closed interval Œr; r, where 0 < r < 1, and diverges if jxj > 1: b) If x D 1, 1 X nD0

1

X 1 1 1 1 1 xn D .1/n D 1  C  C    : 2n C 1 2n C 1 3 5 7 nD0

This series does not converge absolutely, since 1 X nD0

1 2n C 1

344

6 Sequences and Series of Functions

diverges: We can apply the limit comparison test: Since 1 1 n lim 2n C 1 D lim D > 0; 1 n!1 n!1 2n C 1 2 n P P and the harmonic series 1=n diverges, 1=.2n C 1/ diverges also. On the other hand, the series converges conditionally. Indeed, the sequence 1=.2n C 1/ is a decreasing sequence, and lim

n!1

1 D 0; 2n C 1

so that the series converges by the theorem on alternating series. If we set x D 1, we obtain the series 1 X nD0

1 1 1 1 D 1C C C C ; 2n C 1 3 5 7

which has been shown to be divergent.  Example 5. Consider the power series 1 X 1 .x  2/n : 2 n nD1

a) Show that the series converges uniformly on any closed interval Œ2  r; 2 C r, where 0 < r < 1, and that the series diverges if jx  2j > 1: b) Determine whether the power series converges absolutely or conditionally at the endpoints of its open interval of convergence, or whether it diverges at any of these points. Solution. a) We will apply the root test: ˇ ˇ1=n ˇ1 ˇ 1 1 nˇ ˇ .x  2/ ˇ D lim 2=n jx  2j D jx  2j lim  lim 2 D jx  2j : n!1 ˇ n2 n!1 n n!1 n1=n Therefore, radius of convergence of the series is 1. The series converges uniformly on any closed interval Œ2  r; 2 C r, where 0 < r < 1, and the series diverges if jx  2j > 1. The open interval of convergence of the power series is fx W jx  2j < 1g D .1; 3/ :

6.3 Power Series

345

b) If we set x D 1, 1 1 X X 1 1 1 1 n .x  2/ D .1/n 2 D 1 C 2  2 C    : 2 n n 2 3 nD1 nD1

The series converges absolutely since 1 X 1 2 n nD1

converges. If x D 3, 1 1 X X 1 1 n .x  2/ D ; 2 2 n n nD1 nD1

so that the series converges absolutely. 

6.3.2 Termwise Integration and Differentiation of Power Series Power series can be integrated term by term: P k Theorem 2. Assume that the power series 1 kD0 ak .x  x0 / converges in the open interval J that contains x0 and x 2 J. Then Z xX 1 x0 kD0

ak .t  x0 /k dt D

1 X ak .x  x0 /k : k C 1 kD0

P k Proof. Since 1 kD0 ak .t  x0 / converges uniformly for t in the interval determined by x0 and x, thanks to Theorem 7 of Sect. 6.2, we have ˇtDx ! Z xX 1 1 Z x 1 X X ˇ 1 k k kC1 ˇ .t  x0 / ˇ ak .t  x0 / dt D ak .t  x0 / dt D ak kC1 x0 kD0 tDx0 kD0 x0 kD0 D

1 X ak .x  x0 /kC1 k C 1 kD0

 Example 6. Determine a power series whose sum is arctan .x/ if 1 < x < 1 by making use of the fact that d 1 arctan .t/ D : dt 1 C t2

346

6 Sequences and Series of Functions

Solution. By the Fundamental Theorem of Calculus, Z arctan .x/ D arctan .x/  arctan .0/ D

x

0

d arctan .t/ dt D dt

Z

x 0

1 dt; x 2 R: 1 C t2

Now, 1 1 D 2 1Ct 1  .t/2    2  3  n D 1 C t2 C t2 C t2 C    C t2 C    D 1  t2 C t4  t6 C    C .1/n t2n C    D

1 X

.1/n t2n

nD0

ˇ ˇ if ˇt2 ˇ D t2 < 1, i.e., if 1 < t < 1, since the above series is a geometric series in the variable t2 . Therefore, if 1 < x < 1; Z x   arctan .x/ D 1  t2 C t4  t6 C    C .1/n t2n C    dt 0

1 1 1 x2nC1 D x  x3 C x5  x7 C    C .1/n C 3 5 7 2n C 1 D

1 X nD0

.1/n

x2nC1 ; 2n C 1

thanks to Theorem 2.  Power series can be differentiated term by term: P k Theorem 3. Assume that the power series 1 kD0 ak .x  c/ has the nonempty open interval of convergence J and we set f .x/ D a0 Ca1 .x  c/ Ca2 .x  c/2 Ca3 .x  c/3 Ca4 .x  c/4 C    C an .x  c/n C    for each x 2 J. Then f has derivatives of all orders in the interval J and its derivatives can be computed by differentiating the power series termwise: f 0 .x/ D a1 C 2a2 .x  c/ C 3a3 .x  c/2 C 4a4 .x  c/3 C    C nan .x  c/n1 C    ;

6.3 Power Series

347

f 00 .x/ D 2a2 C .2/.3/a3 .x  c/ C .3/ .4/ a4 .x  c/2 C    C .n  1/nan .x  c/n2 C    ; f .3/ .x/ D .2/.3/a3 C .2/ .3/ .4/ a4 .x  c/ C    C .n  2/ .n  1/ nan .x  c/n3 C    ; :: :: The power series for f .n/ has the same radius of convergence as the power series that defines f . Proof. It is sufficient to prove the statement for f 0 . We can assume that c D 0, since we can set X D x  c and obtain the result easily. We will show that the series that is obtained by termwise differentiation converges uniformly in any closed and bounded interval that is contained in J. Then the theorem on the termwise differentiation of a series of functions is applicable: 1

1

1

1

X d X   X d X .an xn / D a n xn D an nxn1 D nan xn1 dx nD0 dx nD0 nD1 nD1 for each x 2 J. Thus, let Œr; r be a closed and bounded interval that is contained in the open interval of convergence of thePpower series. Choose > r so that the interval Œ ;  is still contained in J. Since an n converges, there exists M > 0 such that jan n j D jan j n  M for each n 2 N: Then n ˇ n1 ˇ ˇan ˇ D jan j n1  jan j  M for each n 2 N:



Since 0 < r < , we have r D q , where 0 < q < 1. Thus, if jxj  r we have ˇ ˇ ˇnan xn1 ˇ D n jan j jxjn1  n jan j rn1  n jan j q



n1 n1

  D jan j n1 nqn1 

for n D 1; 2; 3; : : :. Since 0 < q < 1, the series  1  X M nD1



nqn1 D

1 M X n1 nq

nD1



M

 nqn1

348

6 Sequences and Series of Functions

converges. Indeed, .n C 1/ qn nC1 D q < 1: D q lim n1 n!1 n!1 nq n lim

Since ˇ ˇ ˇnan xn1 ˇ 



 M nqn1

for each x 2 Œr; r, the series 1 X

nan xn1

nD1

converges absolutely and uniformly on Œr; r by the Weierstrass M-test.  The series that is obtained by the termwise differentiation of a power series has the same radius of convergence: P n1 is the same as the Proposition 2. The radius of 1 nD1 nan x P1of convergence n radius of convergence of nD0 an x . P1 n1 Proof. By Theorem converges if x is in P the open interval nD1 nan x P31the series n n1 of convergence of nD0 an x . Thus,P the radius of convergence of 1 is nD1 nan x 1 n C1 if the radius of convergence of nD0 an x is C1. P n Now assume that radius of convergence of 1 nD0 an x is R and the radius Pthe 1 n1 0 of convergence of nD1 nan x is R . Theorem 3 implies that R0  R. We will establish equality by showing that R0  R. Let " > 0 be given. By the of Pdefinitionn1 R0 as a least upper bound, there exists such that R0 -" <  R0 and 1 na x n nD1 converges absolutely and uniformly on Œ ; . By Theorem 1 we can integrate termwise, so that for any x 2 Œ ;  we have Z xX 1 0 nD1

Thus,

P1 nD0

nan tn1 dt D

1 X

Z nan

nD1

x

tn1 dt D 0

1 X nD1

 nan

 X 1 1 n x D a n xn : n nD1

an xn converges for any x 2 Π; . This implies that R0  " <  R:

Since R0 < R C " for any " > 0 we have R0  R. 

6.3 Power Series

349

Example 7. We know that f .x/ D

1 D 1 C x C x2 C x3 C    C xn C    if  1 < x < 1 1x

(geometric series). a) Find a representation of f 0 .x/ D

1 .1  x/2

as a power series in powers of x by differentiating the power series for f termwise. Confirm that the power series has the same open interval as the geometric series, i.e., the interval .1; 1/: b) Find a representation of d2 f .x/ D 2 dx 00



1 1x

 D

2 .1  x/3

as a power series in powers of x by differentiating the power series for f 0 termwise. Confirm that the power series has the same open interval as the geometric series, i.e., the interval .1; 1/. Solution. a) By Theorem 3, d f .x/ D dx 0



1 1x

 D

1 .1  x/

2

D

 d  1 C x C x2 C x3 C    C xn C    dx

D 1 C 2x C 3x2 C    C nxn1 C    : The open interval of convergence of the above series is the same as the open interval of convergence of the original series, i.e., .1; 1/. Let us confirm this by the ratio test:   nC1 nC1 j.n C 1/ xn j D jxj .1/ D jxj : lim D lim jxj D jxj lim n1 n!1 n!1 n!1 n n jnx j Therefore, the series converges absolutely if jxj < 1 and diverges if jxj > 1.

350

6 Sequences and Series of Functions

b) Again, by Theorem 3, 

d2 dx2

f 00 .x/ D

1 1x







1

D

d dx

D

 d  1 C 2x C 3x2 C    C nxn1 C    dx

.1  x/2

D 2C.2/.3/xC    C .n1/ .n/ xn2 C    ; x2 .1; 1/ : Let us apply the ratio test the above power series: ˇ ˇ n .n C 1/ ˇxnC1 ˇ nC1 D jxj : lim D jxj lim n!1 .n  1/ n jxn j n!1 n  1 Therefore, the open interval of convergence of the series is .1; 1/, as predicted by Theorem 3. 

6.3.3 Problems In problems 1–4, determine the radius of convergence and the open interval of convergence of the given power series (you need not investigate convergence at the endpoints of the interval). 1.

3. 1 X .x C 2/n nD0

1 X x2n .2n/Š nD0

n2 C 1

2.

4. 1 X n3 n1

2n

1 X

.x  4/n

nD0

.1/n

2n 2nC1 x n2

5. Determine a power series in powers of x whose sum is ln .1 C x/ if 1 < x < 1. Make use of the fact that d 1 ln .1 C t/ D dt 1Ct if t > 1.

6.4 Taylor Series

351

6.4 Taylor Series In section 6.3 we discussed the basic facts about power series and functions that are defined as their sums. In this section we will discuss the representation of a given function via a power series. Let us begin by establishing a crucial link between the derivatives of a function that is defined via a power series and the coefficients of that series:

6.4.1 Taylor’s Formula Proposition 1. If f .x/ D

1 X

an .x  x0 /n

nD0

for each x in an open interval J that contains x0 then a0 D f .x0 / and an D

1 .n/ f .x0 / ; n D 1; 2; 3; : : : nŠ

Proof. We showed that f is infinitely differentiable in J and we can compute its derivatives by termwise differentiation (Theorem 3 of Sect. 6.3). Thus, f .x/ D a0 C a1 .x  x0 / C a2 .x  x0 /2 C a3 .x  x0 /3 C a4 .x  x0 /4 C    C an .x  x0 /n C    f 0 .x/ D a1 C 2a2 .x  x0 / C 3a3 .x  x0 /2 C 4a4 .x  x0 /3 C    C nan .x  x0 /n1 C    ; f 00 .x/ D 2a2 C .2/.3/a3 .x  x0 / C .3/ .4/ a4 .x  x0 /2 C    C .n  1/nan .x  x0 /n2 C    ; f .3/ .x/ D .2/.3/a3 C .2/ .3/ .4/ a4 .x  x0 / C    C .n  2/ .n  1/ nan .x  x0 /n3 C    ; :: :: Therefore, f .x0 / D a0 ; f 0 .x0 / D a1 ;

352

6 Sequences and Series of Functions

1 00 f .x0 / ; 2 1 f .3/ .x0 / D .2/ .3/ a3 ) a3 D f .3/ .x0 / ; 3Š :: : f 00 .x0 / D 2a2 ) a2 D

You can confirm by induction that a0 D f .x0 / and an D

1 .n/ f .x0 / for n D 1; 2; 3; 4; : : : nŠ

 Definition 1. Assume that f is infinitely differentiable at x0 . The Taylor series for f based at x0 is the power series 1 X 1 .n/ f .x0 / .x  x0 /n : nŠ nD0

Remark 1 (Uniqueness of power series representation). By Proposition 1, if f .x/ D

1 X

an .x  x0 /n

nD0

P n for each x in an open interval that contains x0 then 1 nD0 an .x  x0 / is the Taylor series for f based at x0 . Thus, the power series that defines a function is the Taylor series of that function.Þ Definition 2. The Maclaurin series for f is the Taylor series of f based at 0, i.e., 1 X 1 .n/ f .0/ xn : nŠ nD0

Example 1. If f .x/ D

1 1x

the Maclaurin series for f is 1 X nD0

xn :

6.4 Taylor Series

353

P n Indeed, f .x/ is the sum of the geometric series 1 nD0 x in the interval .1; 1/. By the uniqueness of the Maclaurin series, the geometric series is the Maclaurin series for f .  Example 2. If f .x/ D arctan .x/ the Maclaurin series for f is 1 X

.1/n

nD0

x2nC1 : 2n C 1

Indeed, arctan .x/ D

1 X

.1/n

nD0

x2nC1 2n C 1

if x 2 .1; 1/, as we showed in Example 6 of Sect. 6.3. By the uniqueness of the Maclaurin series, the above power series is the Maclaurin series for arctangent.  Example 3. If f .x/ D ex the Maclaurin series for f is 1 X 1 n x : nŠ nD0

Indeed, f .n/ .x/ D ex so that f .n/ .0/ D e0 D 1 for n D 0; 1; 2; 3; : : : Therefore, Maclaurin series for the natural exponential function is 1 1 X X 1 .n/ 1 n n f .0/ x D x : nŠ nŠ nD0 nD0



354

6 Sequences and Series of Functions

The series in Example 3 converges for each x 2 R, as you can confirm by the ratio test. But how do we know that the sum is ex for each x 2 R? We will be able to answer this question in the affirmative with the help of the following theorem: Theorem 1 (Taylor’s Formula for the Remainder). Assume that f has continuous derivatives up to order n C 1 in an open interval J containing the point c. If x 2 J there exists a point cn .x/ between x and c such that f .x/ D Pc;n .x/ CRc;n .x/ D f .c/ Cf 0 .c/ .x  c/ Cf 00 .c/ .x  c/2 C    C

1 .n/ f .c/ .x  c/n CRc;n .x/ ; nŠ

where Z

x

Rc;n .x/ D c

1 .nC1/ 1 f f .nC1/ .cn .x// .x  c/nC1 : .t/ .x  t/n dt D nŠ .n C 1/ Š

The expression Pc;n .x/ D f .c/ Cf 0 .c/ .x  c/ Cf 00 .c/ .x  c/2 C    C

1 .n/ f .c/ .x  c/n nŠ

is referred to as the Taylor polynomial for f based at c. Thus, the Taylor polynomial for f based at c is the .n C 1/st partial sum of the Taylor series for f based at c. The expression Rc;n .x/ D f .x/  Pc;n .x/ represents the error in the approximation of f .x/ by Pc;n .x/, and it is referred to as the remainder in such an approximation. In almost all the examples, we will be interested in Maclaurin series, so that c D 0. In such a case we will use the notations, Pn .x/ D f .0/ C f 0 .0/ x C

1 00 1 f .0/ x2 C    C f .n/ .0/ xn ; 2Š nŠ

Rn .x/ D f .x/  Pn .x/ ; The polynomial Pn .x/ is the Maclaurin polynomial for f . The proof of Theorem 1. The starting point is the Fundamental Theorem of Calculus: Z x f 0 .t/dt: f .x/  f .c/ D c

6.4 Taylor Series

355

Let us apply integration by parts to the integral by setting u D f 0 .t/ and dv D dt. Ordinarily, we would set v D t. In the present case, we will be somewhat devious, and set v D t  x. We still have   d dv dt D .t  x/ dt D dt: dv D dt dt Thus, Z

x

f .x/  f .c/ D

f 0 .t/dt D

Z

x

udv

c

c

D



 uvjxc

Z

x



vdu c

  D f 0 .x/.x  x/  f 0 .c/.c  x/  Z

0

D f .c/ .x  c/ C

Z

x

.t  x/ f 00 .t/ dt

c x

f 00 .t/ .x  t/ dt:

c

Therefore, Z

0

x

f .x/ D f .c/ C f .c/.x  c/ C

00

Z

x

f .t/.x  t/dt D Pc;1 .x/ C c

f 00 .t/.x  t/dt:

c

Let us focus our attention on the integral, and apply integration by parts, setting u D f 00 .t/ and dv D .x  t/ dt. Then, Z vD

Z dv D

1 .x  t/ dt D  .x  t/2 : 2

Thus, Z x Z x f 00 .t/.x  t/dt D udv c

c

D



 uvjxc

Z

x



vdu c

ˇx  Z x ˇ 1 1  f 00 .t/ .x  t/2 ˇˇ   .x  t/2 f .3/ .t/dt 2 2 c c   Z x 1 1 1 D  f 00 .x/ .x  x/2 C f 00 .c/ .x  c/2 C .x  t/2 f .3/ .t/dt 2 2 2 c Z x 1 00 1 D f .c/ .x  c/2 C .x  t/2 f .3/ .t/dt: 2 c 2 

D

356

6 Sequences and Series of Functions

Therefore, Z

0

x

f .x/ D f .c/ C f .c/.x  c/ C

f 00 .t/.x  t/dt

c

1 D f .c/ C f 0 .c/.x  c/ C f 00 .c/ .x  c/2 C 2 Z x 1 .x  t/2 f .3/ .t/dt: D Pc;2 .x/ C c 2

Z

x c

1 .x  t/2 f .3/ .t/dt 2

Let us integrate by parts again, setting u D f .3/ .t/ and dv D

1 .x  t/2 dt: 2

Thus, Z vD

1 1 1 .x  t/2 dt D  .x  t/3 D  .x  t/3 : 2 .2/.3/ 3Š

Therefore, Z x Z x 1 f .3/ .t/ .x  t/2 dt D udv 2 c c Z x  x D uvjc  vdu c

ˇx  Z x ˇ 1 .3/ 1 f .t/ .x  t/3 ˇˇ   .x  t/3 f .4/ .t/dt 3Š 3Š c c Z x 1 .3/ 1 .x  t/3 f .4/ .t/dt: D f .c/ .x  c/3 C 3Š c 3Š 

D



Thus, 1 f .x/ D f .c/ C f 0 .c/.x  c/ C f 00 .c/ .x  c/2 C 2

Z

x c

1 .x  t/2 f .3/ .t/dt 2

1 1 D f .c/ C f 0 .c/.x  c/ C f 00 .c/ .x  c/2 C f .3/ .c/ .x  c/3 2 3Š Z x 1 .x  t/3 f .4/ .t/dt C c 3Š Z x 1 .x  t/3 f .4/ .t/dt: D Pc;3 .x/ C c 3Š

6.4 Taylor Series

357

The general pattern is emerging: f .x/ D f .c/ C f 0 .c/.x  c/ C    C

1 .n/ f .c/ .x  c/n C Rc;n .x/ nŠ

D Pc;n .x/ C Rc;n .x/; where Z

x

Rc;n .x/ D c

1 .nC1/ f .t/ .x  t/n dt: nŠ

It is not difficult to supply an inductive proof. Let us assume that the statement is valid for n. We set u D f .nC1/ .t/ and dv D

1 .x  t/n dt: nŠ

Then, Z

1 1 1 .x  t/n dt D  .x  t/nC1 D  .x  t/nC1 ; nŠ nŠ .n C 1/ .n C 1/Š

vD so that Z c

x

f .nC1/ .t/

1 .x  t/n dt D nŠ

Z

x

udv c

  D uvjxc   D

 Z

 c

x

vdu c

ˇx  ˇ 1 f .nC1/ .t/ .x  t/nC1 ˇˇ .n C 1/Š c

x



Z

1 .x  t/nC1 f .nC2/ .t/dt .n C 1/Š

1 f .nC1/ .c/ .x  c/nC1 .n C 1/Š Z x 1 .x  t/nC1 f .nC2/ .t/dt: C c .n C 1/Š D

Therefore, 1 f .x/ D f .c/ C f 0 .c/.x  c/ C    C f .n/ .c/ .x  c/n nŠ Z x 1 C f .nC1/ .t/ .x  t/n dt nŠ c

358

6 Sequences and Series of Functions

1 .n/ f .c/ .x  c/n nŠ Z x 1 1 nC1 .nC1/ f .x  t/nC1 f .nC2/ .t/dt C .c/ .x  c/ C .n C 1/Š .n C 1/Š c D f .c/ C f 0 .c/.x  c/ C    C

D Pc;nC1 .x/ C Rc;nC1 .x/ ; where Z

x

Rc;nC1 .x/ D c

1 .x  t/nC1 f .nC2/ .t/dt: .n C 1/Š

This completes the induction. We have shown that f .x/ D Pc;n .x/ C Rc;n .x/ ; where Z

x

Rc;n .x/ D c

1 .nC1/ f .t/ .x  t/n dt: nŠ

The above expression for the remainder in the approximation of f .x/ by the Taylor polynomial of order n based at c for f is referred to as the integral form of the remainder. We can obtain the other form of the remainder by appealing to the generalized mean value theorem for integrals (Theorem 3 of Sect. 4.2). If x > c, .x  t/n  0 for t 2 Œc; x . Therefore, there exists cx 2 Œc; x such that Z

x c

1 .nC1/ 1 f .t/ .x  t/n dt D f .nC1/ .cx / nŠ nŠ

Z

x

.x  t/n dt

c

D

  1 1 .nC1/ f .x  c/nC1 .cx / nŠ nC1

D

1 f .nC1/ .cx / .x  c/nC1 : .n C 1/Š

Similarly, if x < c, .x  t/n D .1/n .t  x/n does not change sign on the interval Œx; c, so that

6.4 Taylor Series

Z

x c

359

1 .nC1/ f .t/ .x  t/n dt D  nŠ

Z

1 .nC1/ f .t/ .x  t/n dt nŠ x Z c 1 .nC1/ D f .cx / .x  t/n dt nŠ x   1 .nC1/ 1 nC1 .x  c/ D f .cx /  nŠ nC1 D

c

1 f .nC1/ .cx / .x  c/nC1 .n C 1/Š

for some cx in Œx; c.  Remark 2. Since f .x/ D Pc;n .x/ C Rc;n .x/ We have f .x/ D

1 X

an .x  x0 /n

nD0

if and only if lim Rc;n .x/ D 0:

n!1

Þ The following fact will enable us to assess the error in the approximation of the natural exponential function, sine and cosine with the corresponding Maclaurin polynomials: Proposition 2. We have xn D0 n!1 nŠ lim

for each x 2R. Proof. Since xn jxjn D 0 , lim D 0; n!1 nŠ n!1 nŠ lim

360

6 Sequences and Series of Functions

and the limit of the constant sequence 0 is 0, it is enough to show that rn D0 n!1 nŠ lim

for any r > 0. Given r > 0, there exists a positive integer N such that N > 2r, no matter how large r may be. If n  N then n  2r, so that 1 r  : n 2 Therefore, if n  N,     rn rN rnN rN r r r D D  nŠ NŠ.N C 1/.N C 2/    .n/ NŠ N C 1 NC2 n     N  N    N  r 1 1 1 1 1 .2r/ r   D : D nN NŠ 2 2 2 NŠ 2 NŠ 2n ƒ‚ … „ .nN/ factors

Thus, 0

rn .2r/N  nŠ NŠ



1 2n



if n  N. We keep N fixed, once it is chosen so that N  2r, and let n ! 1. Since lim

n!1

1 D 0; 2n

the above inequality shows that rn D0 n!1 nŠ lim

also.  Now we can establish that the sum of the Maclaurin series for ex is ex : Example 4. We have   1 2 1 3 1 n lim 1 C x C x C x C    C x D ex n!1 2Š 3Š nŠ for each real number x.

6.4 Taylor Series

361

Solution. For each x 2 R there exists cn .x/ between 0 and x such that Rn .x/ D

1 ecn .x/ xnC1 : .n C 1/Š

If x > 0, we have 0  cn .x/  x. Since the natural exponential function is an increasing function on the entire number line, we have 0 < Rn .x/ 

ex xnC1 : .n C 1/Š

If x < 0, we have x  cn .x/  0, so that exp .cn .x// nC1 exp .0/ 1  jxj jxjnC1  jxjnC1 : .n C 1/Š .n C 1/Š .n C 1/Š

jRn .x/j D Thus,

8 ˆ
0 2 3Š nŠ if z  0. Fix z > 0 and set f .x/ D

exCz for each x 2 R: ez

We expect that f .x/ D ex . By the chain rule f 0 .x/ D

exCz D f .x/ for each x 2 R: ez

We also have f .0/ D

ez D 1: ez

Therefore f .x/ D ex . Thus exCz D ex ez so that exCz D ex ez for each x 2 R and z  0: If x and y are arbitrary real numbers we can choose z > 0 so that z C y > 0. Then exCy ez D exCyCz D ex eyCz D ex ey ez : Thus exCy D ex ey :

372

6 Sequences and Series of Functions

We can conclude that ex > 0 for each x 2 R: If x  0 then ex  1. If x < 0 we can choose z > 0 such that x C z > 0. We have exCz > 1 and ez > 0. Therefore 1 xCz e >0 ez

ex D Thus ex > 0 for each x 2 R.

6.5.2 The Natural Logarithm Assuming that our starting point is the natural exponential function we can define the natural logarithm as the inverse of the natural exponential function. We should have eln.x/ D x for each x > 0. By the chain rule 1D

d d d d .x/ D eln.x/ D eln.x/ ln .x/ D x ln .x/ dx dx dx dx

so that we should have d 1 ln .x/ D for each x > 0: dx x Therefore we set Z ln .x/ D

x 1

1 dt for each x > 0: t

By the Fundamental Theorem of Calculus d 1 ln .x/ D for each x > 0: dx x We have Z ln .1/ D

1 1

1 dt D 0: t

We can show that ln .xy/ D ln .x/ C ln .y/ for each x > 0 and y > 0: Let us fix y > 0 and set g .x/ D ln .xy/  ln .y/ :

6.5 Another Look at Special Functions

373

Then 1 and g .1/ D 0: x

g0 .x/ D Therefore

Z g .x/ D

x 1

1 dt D ln .x/ : t

Thus ln .xy/ D ln .x/ C ln .y/ for each x > 0 and y > 0: We need to confirm that ln is the inverse of exp. We have d ln .ex / D dx



1 ex



  .ex / D 1 for each x 2 R and ln e0 D ln .1/ D 0:

Therefore ln .ex / D x for each x 2 R: Now let x > 0 and set y D eln.x/ . We need to show that y D x. We have  ln .y/ D ln eln.x/ D ln .x/ so that Z

y

0 D ln .y/  ln .x/ D x

1 dt: t

If y > x then Z

y

x

1 dt  t

Z

y x

yx 1 dt D > 0: x x

But Z

y

0 D ln .y/  ln .x/ D x

1 dt: t

Therefore 0 < y  x. If we assume that x > y then Z

x y

xy 1 dt  > 0: t y

374

6 Sequences and Series of Functions

But Z

x

0 D ln .x/  ln .y/ D y

1 dt: t

Therefore we must have x  y as well. Therefore x D y so that eln.x/ D x for each x > 0: Since ln .ex / D x for each x 2 R and eln.x/ D x for each x > 0 the functions ln and exp are inverses of each other. Once we have defined the logarithm with base e we can define ax and lna .x/ for any positive base: ax D ex ln.a/ for any x 2 R. The inverse of the above function is loga .x/. As you have seen in beginning calculus loga .x/ D

ln .x/ for each x > 0: ln .a/

6.5.3 Sine and Cosine We have seen that sin .x/ D

1 X kD0

1

X .1/k .1/k 2kC1 x x2k for each x 2 R: and cos .x/ D .2k C 1/Š .2k/Š kD0

These functions are usually introduced informally via the unit circle picture. We can introduce them rigorously in terms of the above power series. Since the series converge for all x 2 R they define analytic functions on the entire number line. It is easily verified that d d sin .x/ D cos .x/ and cos .x/ D  sin .x/ for each x 2 R: dx dx

6.5 Another Look at Special Functions

375

We can confirm the addition formulas for sine and cosine by performing the relevant operations on the power series. Let us consider an alternative route to show that sin .x C y/ D sin .x/ cos .y/ C cos .x/ sin .y/ ; cos .x C y/ D cos .x/ cos .y/  sin .x/ sin .y/ for each x and y in R: Let us fix y 2 R and set f .x/ D sin .x C y/ and g .x/ D sin .x/ cos .y/ C cos .x/ sin .y/ : We will show that f .x/ D g .x/ for each x 2 R by showing that f and g have the same Taylor series in powers of x. This will be done by showing that f .n/ .0/ D g.n/ .0/ for n D 0; 1; 2; 3; : : : We have f .x/ D sin .x C y/ ; f 0 .x/ D cos .x C y/ ; f .2/ .x/ D  sin .x C y/ ; f .3/ .x/ D  cos .x C y/ ; f .4/ .x/ D sin .x C y/ ; : : : Therefore f .0/ D sin .y/ ; f 0 .0/ D cos .y/ ; f .2/ .0/ D  sin .y/ ; f .3/ .0/ D  cos .y/ ; f .4/ .x/ D sin .y/ ; : : : We also have g .x/ D sin .x/ cos .y/ C cos .x/ sin .y/ ; g0 .x/ D cos .x/ cos .y/  sin .x/ sin .y/ ; g

.2/

.x/ D  sin .x/ cos .y/  cos .x/ sin .y/ ; g.3/ .x/ D  cos .x/ cos .y/ C sin .x/ sin .y/ ;

g

.4/

.x/ D sin .x/ cos .y/ C cos .x/ sin .y/ ; : : :

Thus g .0/ D sin .y/ ; g0 .0/ D cos .y/ ; g.2/ .0/ D  sin .y/ ; g.3/ .0/ D  cos .y/ ; g.4/ .0/ D sin .y/ ; : : : We see that f .2k/ .0/ D .1/k sin .y/ and f .2kC1/ .0/ D .1/k cos .y/ for k D 0; 1; 2; : : :

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6 Sequences and Series of Functions

We also see that g.2k/ .0/ D .1/k sin .y/ and g.2kC1/ .0/ D .1/k cos .y/ for k D 0; 1; 2; : : : Thus f .n/ .0/ D g.n/ .0/ for n D 0; 1; 2; 3; : : : so that f .x/ D g .x/ for each x 2 R, as claimed. We conclude that sin .x C y/ D sin .x/ cos .y/ C cos .x/ sin .y/ for each x and y in R. We differentiate both sides with respect to x:  cos .x C y/ D  cos .x/ cos .y/  sin .x/ sin .y/ so that cos .x C y/ D cos .x/ cos .y/  sin .x/ sin .y/ ; as claimed. When we set y D x in the above identity we have 1 D cos .x/ cos .x/  sin .x/ sin .x/ D cos .x/ cos .x/  sin .x/ . sin .x// D cos2 .x/ C sin2 .x/ : The identity cos2 .x/ C sin2 .x/ D 1 for each x 2 R places the point .cos .x/ ; sin .x// on the unit circle u2 C v 2 D 1. This is the usual starting point in the informal geometric definition of sine and cosine. Note that the above identity shows that jcos .x/j  1 and jsin .x/j  1 for each x 2 R: We can define  as twice the smallest positive zero of cosine. We need to show that such a number exists. Assume that cos .x/ ¤ 0 for each x > 0 and there exists x > 0 such that cos .x/ < 0. Since cos .0/ D 1 the intermediate value theorem implies that there exists  > 0 such that cos ./ D 0 contradicting the assumption that cos .x/ ¤ 0 for each x > 0. Therefore the assumption that cos .x/ ¤ 0 for each x > 0 leads to the conclusion that cos .x/ > 0 for each x > 0. Since d sin .x/ D cos .x/ dx

6.5 Another Look at Special Functions

377

we need to have sine to be increasing on Œ0; 1/. Therefore 0 D sin .0/ < sin .1/  sin .x/ if x  1: By the Mean Value Theorem there exists  2 .1; x/ such that cos .1/  cos .x/ D . sin .// .1  x/ Therefore cos .x/ D cos .1/ C sin ./ .x  1/ > cos .1/ C sin .1/ .x  1/ : Since sin .1/ > 0 we conclude that lim cos .x/ D C1:

x!1

This contradicts the fact that jcos .x/j  1 for each x 2 R: Since cos .0/ D 1 and cosine is continuous at 0 there exists ı > 0 such that cos .x/  1=2 if 0  x < ı: Since we showed that there exists x > 0 such that cos .x/ D 0 the set S D fx 2 R W x  ı and cos .x/ D 0g is not empty. Therefore S has a greatest lower bound   ı > 0. Since cosine is continuous on S we have cos ./ D 0. Now we can assert that  is the smallest zero of cosine. We set  D 2. Thus cos

 2

D0

and cos.x/ > 0 if 0  x < =2. Since d sin .x/ D cos .x/ dx sine is increasing on Œ0; =2. Since sin .0/ D 0 we must have sin .x/  0 on Œ0; =2. In particular, sin .=2/  0. Since sin2 sin.=2/ D 1:

 2

D 1  cos2

 2

D1

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6 Sequences and Series of Functions

By the addition formulae for sine and cosine,     D sin .x/ cos C cos .x/ sin D cos .x/ ; sin x C 2 2 2 

    cos .x/ C cos sin  D cos .x/ ;  x D sin 2 2 2 2     cos x C D cos .x/ cos  sin .x/ sin D  sin .x/ 2 2 2 sin

for each x 2 R. Therefore      sin .x C / D sin x C C D cos x C D  sin .x/ : 2 2 2 Thus sin .x C 2/ D sin .x C  C / D  sin .x C / D sin .x/ for each x 2 R: We have established the periodicity of sine with period 2: We also have     cos x C D cos .x/ cos  sin .x/ sin D  sin .x/ ; 2 2 2   D  cos .x/ ; cos .x C / D  sin x C 2 and cos .x C 2/ D  cos .x C / D cos .x/ for each x 2 R. Thus we have established the periodicity of sine and cosine with period 2: Note that    cos ./ D cos C D  sin D 1 2 2 2 and sin ./ D sin

 2

C

  D cos D 0: 2 2

6.5 Another Look at Special Functions

379

We can show that sine does not have a period smaller than 2: Let us begin by showing that cosine is decreasing on Œ0;  and increasing on Œ; 2. We have d cos .x/ D  sin .x/ : dx If 0 < x < =2 we have sin .x/ > 0 so that cosine is decreasing on Œ0; =2. If =2 < x <  then      D cos x  > 0: sin .x/ D sin x  C 2 2 2 Therefore cosine is decreasing on Œ=2;  as well. Thus cosine is decreasing on Œ0; . If  < x < 2 then sin .x/ D sin .x   C / D  sin .x  / < 0: Therefore d cos .x/ D  sin .x/ > 0 dx if  < x < 2. Therefore cosine is increasing on Œ; 2 Now let us assume that ˛ > 0 and sin.x C ˛/ D sin .x/ for each x 2 R. Then 1 D sin

 2

D sin



  C ˛ D sin cos .˛/ C cos sin .˛/ D cos .˛/ 2 2 2

Thus cos .˛/ D 1. Since cos .2/ D cos .0/ D 1 and cosine is decreasing on Œ0;  and increasing on Œ; 2 we must have ˛  2: Similarly, 2 is the smallest period of cosine.

Index

A Absolute Value, 12 Accumulation Point, 49 Archimedean Property, 16

B Bolzano-Weierstrass Theorem, 50

C Cauchy Condition for a sequence of numbers, 30 Cluster Point, 49 Compactness, 52 Completeness of Real Numbers, 37 Continuity, 61 continuity from the left, 65 continuity from the right, 64 Lipschitz continuity, 153 sequential characterization, 66 uniform continuity, 67 Convexity, 153

D Decimal Expansions, 34 Derivative, 109 chain rule, 141 derivative of an inverse function, 143 Differential, 128 Generalized Mean Value Theorem, 151 Leibniz notation, 116 linearity of differentiation, 137 Local Linear Approximation, 122

Mean Value Theorem, 148 power rule, 135 produce rule, 138 quotient rule, 140 Rolle’s Theorem, 149

E Equality, 1 Exponential Functions the natural exponential function, 5, 370 Extreme Value Theorem, 97

F Field Axioms, 8 Function, 3 codomain of a function, 3 composition of functions, 4 domain of s function, 3 natural domain, 4 product of functions, 4 quotient of functions, 4 range of a function, 3 sum of functions, 4 Fundamental Theorem of Calculus, 197

G Greatest Lower Bound, 42

I Indeterminate Form, 57 Infimum, 42

© Springer International Publishing Switzerland 2016 T. Geveci, Advanced Calculus of a Single Variable, DOI 10.1007/978-3-319-27807-0

381

382 Infinite Series, 261 Abel’s test, 302 absolute convergence, 270 alternating series, 295 Cauchy condition for series, 269 comparison tests, 289 conditional convergence, 272 Dirichlet’s test, 303 geometric series, 263 harmonic series, 268 integral test, 282 limit comparison test, 290 ratio test, 275 root test, 280 Integral, 169 antiderivative, 200 Dirichlet Test, 250 improper integrals, 222, 239 comparison tests, 244 indefinite integral, 202 integration by parts, 217 lower sum, 170 mean value theorem, 196 properties of the integral, 185 Riemann sum, 175 substitution rule, 210 triangle inequality for integrals, 194 upper sum, 169 Intermediate Value Theorem, 98 Inverse Function, 100 inverse trigonometric functions, 103

L L’Hôpital’s Rule, 161 Least Upper Bound, 42 least upper bound principle, 43 Limit infinite limit of a function, 85 infinite limit of a sequence, 53 limit at infinity, 85 limit from the left, 79 limit from the right, 79 limit of a function, 75 of a sequence, 21 Logarithmic Functions the natural logarithm, 6, 372

Index M Monotone Sequences, 46 decreasing sequences, 46 increasing sequences, 46 Monotone sequences monotone convergence principle, 46 N Nested Sequences, 35, 48 O Order Axioms, 9 P Power Series, 339 open interval of convergence, 342 radius of convergence, 342 Taylor series, 351 Taylor’s formula for the remainder, 354 S Sequence of Functions uniform Cauchy condition, 319 Sequences of Functions, 313 pointwise convergence, 313 uniform convergence, 313 Series of Functions, 330 Abel’s Test, 333 Dirichlet’s Test, 334 pointwise convergence, 330 uniform convergence, 330 Weierstrass M-Test, 331 Sets, 1 intersection of sets, 3 Union of sets, 2 Supremum, 42 T Triangle Inequality, 15 Trigonometric Functions cosine, 5, 374 periodicity, 5 sine, 5, 374 tangent, 5