Calculus of One Variable [2nd ed. 2021] 3030886360, 9783030886363

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M. Thamban Nair

Calculus of One Variable Second Edition

Calculus of One Variable

M. Thamban Nair

Calculus of One Variable Second Edition

M. Thamban Nair Department of Mathematics Indian Institute of Technology Madras Chennai, Tamil Nadu, India

ISBN 978-3-030-88636-3 ISBN 978-3-030-88637-0 (eBook) https://doi.org/10.1007/978-3-030-88637-0 Jointly published with ANE Books Pvt. Ltd. In addition to this printed edition, there is a local printed edition of this work available via Ane Books in South Asia (India, Pakistan, Sri Lanka, Bangladesh, Nepal and Bhutan) and Africa (all countries in the African subcontinent). ISBN of the ANE Books Pvt. Ltd. edition: 9788194891840 1st edition: © Author 2015 2nd edition: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of 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 publishers, 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 publishers 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. The publishers remain neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Dedicated to the fond memories of my Father P. Chandu Nair (Late), Mother Meloth Parvathi Amma (Late), Father-in-law N. Gopalan Nair (Late) and Mother-in-law Santha G Nair (Late)

Preface to the Second Edition

Since its first publication, more than five years ago, the book has been used by many students all over India for their undergraduate courses in mathematics, science and engineering, either as a text or as a reference book. The present version is a rewritten version of the original one with many additions here and there making it better readable and also giving better motivations and clarifications for many of the results. Some of the subsections are renamed, and a few subsections are added. I am thankful to my Ph.D. student Subhankar Mondal for critically reading the revised text and suggesting many improvements. Unlike in the first edition, where results (theorems and corollaries), examples, exercises and remarks are numbered as p.q, where p and q denote the chapter number and occurrence, respectively; in this edition, we number them as p.q.r, where p, q, r denote the chapter number, section number and occurrence, respectively. Chennai, India October 2020

M. Thamban Nair

vii

Preface to the First Edition

This book grew out of my class notes for the first semester course in mathematics for B.Tech. students at IIT Madras, which I have been teaching several times since 1996. The notes have been in circulation among the students of IIT Madras, and they have been using them as supplementary material along with the prescribed texts. Students have found the notes to be very useful, particularly the manner in which new concepts are introduced, the style of presentation, the way examples have been worked out and the interspersed remarks. The book can be used as a first course in calculus for science and engineering students. It can also be used as a supplementary text for B.Sc. mathematics course.

Why another book? There are many books on calculus meant for students of engineering and science. So, why another book? To my assessment, most of the books meant for engineers are designed only for a single course or a combination of courses under a particular university with a few exceptions. They mostly consist of mathematical techniques for solving problems without much elaboration on the underlying mathematical principles. On the other hand, books meant for students pursuing bachelors of science contain elaborate description of underlying mathematical principles which may not be suitable for the bachelor of engineering courses. In this book, I have tried to strike a balance between the above two approaches.

Key features. The book provides clear understanding of the basic concepts of differential and integral calculus starting with the concepts of sequences and series of numbers and also ix

x

Preface to the First Edition

introduces slightly advanced topics such as sequences and series of functions, power series and Fourier series which would be of use for other courses in mathematics for science and engineering programs. Here are some of the salient features of the book: • Precise definitions of basic concepts are given. • Several motivating examples are provided for understanding the concepts and also for illustrating the results. • Proofs of theorems are given with sufficient motivation—not just for the sake of proving them alone. • Remarks in the text supply additional information on the topics under discussion. • Exercises are interspersed within the text for making the students attempt them while the lectures are in progress. • A large number of problems at the end of each chapter are meant as home assignments. The student-friendly approach of the exposition of the book would definitely be of great use not only for students, but also for the teachers. IIT Madras June 2014

M. Thamban Nair [email protected]

Acknowledgments

While preparing the notes, I have benefited from suggestions and comments from many of my colleagues, especially professors S. H. Kulkarni, A. Singh and P. Veeramani. Also, Research Scholars Viswanathan and Ajoy Jana red the manuscript thoroughly and pointed out many typos and made some suggestions for improvements. I am grateful to all of them. Also, curtesy to Google/Wikipedia for all the figures in the text. I gratefully acknowledge the support received from the Centre for Continuing Education (CCE), IIT Madras, under its book writing scheme. Also, I thank my wife Dr. Sunita Nair and daughters Priya and Sneha for their constant support in all my academic endeavors and my mother Meloth Parvathi Amma and mother-in-law Shantha G. Nair for their blessings. Comments and suggestions from the readers are most welcome.

xi

Note to the Reader

Set-Theoretic Notations. Throughout the book, we shall use standard set-theoretic notations such as ∪,

∩,

⊆,

⊂,

∈

to denote “union,” “intersection,” “subset of,” “proper subset of,” “belong(s) to,” respectively. For sets S1 and S2 , the set {x ∈ S1 : x S2 } is denoted by S1 \S2 . If f is a function with domain S1 and codomain S2 , then we use the notation f : S1 → S2 , and it is also called a “map” from S1 to S2 . Also, we use the standard notations such as N Z Q R C := ∀ ∃ ⇒ ⇔ →

: : : : : : : : : :

:

set of all positive integers set of all integers set of all rotational numbers set of all real numbers set of all complex numbers is defined by for all there exists or there exist implies or imply if and only if maps to

For “if and only if,” sometimes we shall also use the symbol “iff”. To mark the end of a proof of a Theorem or Corollary, we use the symbol ❚, and to mark the end of an Example or a Definition or a Remark, the symbol ♦ is used. The symbol  is used to signal the end of an exercise. Bold face is often used for defining a term, and italics is used when a new term is used in a sentence which may be defined subsequently.

xiii

xiv

Note to the Reader

About Numbering. Four sequences of numbers have been used for specifying mathematical results (Theorems and Corollaries), examples, exercises and remarks, respectively.

Contents

1 Sequence and Series of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Sequence of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Convergence and Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Some Tests for Convergence and Divergence . . . . . . . . . . . . . 1.1.3 Monotonic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.4 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.5 Further Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.6 Cauchy Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Series of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Convergence and Divergence of Series . . . . . . . . . . . . . . . . . . 1.2.2 Some Tests for Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.4 Madhava-Nilakantha Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.5 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 15 24 32 36 40 43 44 48 57 60 62 65 65 67

2 Limit, Continuity and Differentiability of Functions . . . . . . . . . . . . . . . 71 2.1 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.1.1 Limit Point of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.1.2 Limit of a Function at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . 73 2.1.3 Limit of a Function in Terms of Sequences . . . . . . . . . . . . . . 79 2.1.4 Some Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 2.1.5 Left Limit and Right Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 2.1.6 Limit at ±∞ and Limit ±∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 2.2 Continuity of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 2.2.1 Definition and Some Basic Results . . . . . . . . . . . . . . . . . . . . . 95 2.2.2 Some More Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 2.2.3 Some Properties of Continuous Functions . . . . . . . . . . . . . . . 104 2.2.4 Exponential and Logarithm Functions . . . . . . . . . . . . . . . . . . . 112

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Contents

2.3 Differentiability of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Left and Right Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Some Properties of Differentiable Functions . . . . . . . . . . . . . 2.3.4 Local Maxima and Local Minima . . . . . . . . . . . . . . . . . . . . . . 2.3.5 Rolle’s Theorem and Mean Value Theorems . . . . . . . . . . . . . 2.3.6 A Sufficient Condition for a Local Extremum Point . . . . . . . 2.3.7 L’Hospital’s Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.8 Higher Derivatives and Taylor’s Formula . . . . . . . . . . . . . . . . 2.3.9 Determination of Shapes of a Curves . . . . . . . . . . . . . . . . . . . . 2.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

119 119 125 129 135 138 143 144 153 163 167 167 168 169

3 Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Integrability and Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Lower and Upper Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 The Integral and Its Characterizations . . . . . . . . . . . . . . . . . . . 3.1.4 Some Basic Properties of Integral . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Integral of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 3.2 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fundamental Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Some Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Computing Area Under the Graph of a Function . . . . . . . . . . 3.5.2 Computing the Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Computing Volume of a Solid . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 Computing the Volume of Solid of Revolution . . . . . . . . . . . . 3.5.5 Computing the Area of Surface of Revolution . . . . . . . . . . . . 3.5.6 Centre of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

173 173 173 175 177 183 192 198 199 206 214 215 220 225 227 230 233 240 243

4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Integrals over Unbounded Intervals . . . . . . . . . . . . . . . . . . . . . 4.1.2 Improper Integrals over Bounded Intervals . . . . . . . . . . . . . . . 4.1.3 Typical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Tests for Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Integrability by Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Integral Test for Series of Numbers . . . . . . . . . . . . . . . . . . . . . 4.2.3 Integrability Using Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

251 251 251 254 258 264 264 267 269 271 271

Contents

xvii

4.3.2 Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 4.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 5 Sequence and Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Sequence of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Pointwise Convergence and Uniform Convergence . . . . . . . . 5.1.2 Uniform Convergence and Continuity . . . . . . . . . . . . . . . . . . . 5.1.3 Uniform Convergence and Integration . . . . . . . . . . . . . . . . . . . 5.1.4 Uniform Convergence and Differentiation . . . . . . . . . . . . . . . 5.2 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Dominated Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Convergence and Absolute Convergence . . . . . . . . . . . . . . . . 5.3.2 Term by Term Differentiation and Integration . . . . . . . . . . . . 5.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

277 277 277 285 286 289 292 294 298 298 303 309

6 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Fourier Series of 2π -Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . 6.2 Best Approximation Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Fourier Series for Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . 6.4 Sine and Cosine Series Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Fourier Series of 2-Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Fourier Series on Arbitrary Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

313 313 319 320 324 328 329 331

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

About the Author

M. Thamban Nair is a Professor of Mathematics at the Indian Institute of Technology Madras, India. Prof. Nair has been a research mathematician and teacher for over 30 years for postgraduate- and undergraduate-level mathematics courses. He won many awards including the C. L. Chandana Award for distinguished and outstanding contributions in mathematics research and teaching in India for the year 2003. He was also a Post Doctoral Fellow at the University of Grenoble (France) and Visiting Professor at Australian National University, Australia, University of Kaiserslautern, Germany, Sun Yat-sen University, Guangzhou, China, and University of Saint Etienne, France. He gave several invited talks at various conferences in India and abroad. Professor Nair has got over 80 journal publications and 5 books to his credit. The broad area of Prof. Nair’s research is in applicable functional analysis, more specifically, spectral approximation, operator equations, and inverse and ill-posed problems.

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Chapter 1

Sequence and Series of Real Numbers

The concept of convergence of sequences and series is fundamental in mathematics. They appear naturally while studying other concepts such as continuity, differentiation and integration of functions, and also while approximating certain unknown quantities using known quantities obtained via numerical or experimental methods. We may recall that, at the school level, one comes across numbers and mathematical constants that do not have finite decimal places. For instance, 1/3 cannot be expressed in terms of finite number of decimal places. In fact, we were taught to write 1/3 in decimal expansion, using repeated division, as 1 = 0.333 · · · . 3 If we take only a finite number of decimal places, how much close that number would be to the number 1/3? By taking more and more decimal places, is it possible to reach closer and closer to 1/3? We expect so. How do we justify? In this chapter, we shall deal with such issues in detail.

1.1 Sequence of Real Numbers Suppose that, for each positive integer n, we are given a real number an . Then, the list of numbers a1 , a2 , . . . , an , . . . is called a sequence. A more precise definition of a sequence is the following: Definition 1.1.1 A sequence of real numbers is a function from the set N of natural numbers to the set R of real numbers. ♦ Notation 1.1.1 If f : N → R is a sequence, and if an = f (n) for n ∈ N, then we write the sequence f as (a1 , a2 , . . .) or (an ) or {an }, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_1

1

2

1 Sequence and Series of Real Numbers

and the term an is called the n th term of the sequence (an ). In specific cases, where one knows an expression for an , one may write (a1 , a2 , . . . , an , . . .) instead of

(a1 , a2 , . . .).

♦

Remark 1.1.1 It is to be borne in mind that a sequence (a1 , a2 , . . .) is different from the set {an : n ∈ N}. For instance, a number may be repeated in a sequence (an ), but it need not be written repeatedly in the set {an : n ∈ N}. As an example, (1, 21 , 1, 13 , . . . , 1, n1 , . . .) is a sequence (an ) with a2n−1 = 1 and a2n = 1/(n + 1) for ♦ each n ∈ N, whereas the set {an : n ∈ N} is same as the set {1/n : n ∈ N}. Remark 1.1.2 We can also talk about a sequence of elements from any non-empty set S, such as sequence of sets, sequence of functions and so on. Thus, given a non-empty set S, a sequence in S is a function f : N → S. In this chapter, we shall consider only sequence of real numbers. In some of the later chapters we shall consider sequences of functions as well. ♦ Example 1.1.1 Let us consider a few examples of sequences (Fig. 1.1): (i) (an ) with an = 1 for all n ∈ N. (ii) (an ) with an = n for all n ∈ N. (iii) (an ) with an = 1/n for all n ∈ N. (iv) (an ) with an = n/(n + 1) for all n ∈ N. (v) (an ) with an = (−1)n for all n ∈ N. This sequence takes values 1 and −1 alternately. ♦

Fig. 1.1 (an ) converges to a

1.1 Sequence of Real Numbers

3

1.1.1 Convergence and Divergence In certain sequences the n th term comes closer and closer to a particular number as n becomes larger and larger. For example, in the sequence ( n1 ), the n th term comes n ), the n th term comes closer and closer to 1 as closer and closer to 0, whereas in ( n+1 n becomes larger and larger. If you look at the sequence ((−1)n ), the terms oscillate between −1 and 1 as n varies, whereas in (n 2 ) the terms become larger and larger. Now, we make precise the the statement “an comes closer and closer to a number a as n becomes larger and larger”, that is, “an can be made arbitrarily close to a by taking n large enough”, by defining the notion of convergence of a sequence. Definition 1.1.2 A sequence (an ) of real numbers is said to converge to a real number a if for every ε > 0, there exists a positive integer N , that may depend on ε, such that ∀ n ≥ N. |an − a| < ε A sequence that converges is called a convergent sequence, and a sequence that does not converge is called a divergent sequence. ♦ Notation 1.1.2 (i) If (an ) converges to a, then we write an → a as n → ∞ that we may read as “an tends to a as n tends to infinity”, that we also write in short as an → a. ♦ (ii) If (an ) does not converge to a, then we write an → a. Remark 1.1.3 We must keep in mind that the symbol ∞ is not a number; it is only a notation used in the context of describing some properties of real numbers, such as in Definition 1.1.2. ♦ Remark 1.1.4 In Definition 1.1.2, the expression |an − a| < ε can be replaced by |an − a| ≤ ε or by |an − a| < c0 ε for some c0 > 0. In other words, the following statements are equivalent. (i) For every ε > 0, there exists N ∈ N such that |an − a| < ε for all n ≥ N . (ii) For every ε > 0, there exists N ∈ N such that |an − a| ≤ ε for all n ≥ N . (iii) For every ε > 0, there exists N ∈ N such that |an − a| ≤ c0 ε for all n ≥ N for some c0 > 0. Clearly, (i) implies (ii). To see (ii) implies (i), assume (ii) and let ε > 0 be given. Then, by (ii), with ε/2 in place of ε, there exists N ∈ N such that |an − a| ≤ ε/2 for all n ≥ N . In particular, (i) holds. Now, (iii) follows from (i) by taking c0 ε in place ♦ of ε, and (i) follows from (iii) by taking ε/c0 in place of ε. Before further discussion on convergence of sequences, let us observe an important property of convergent sequences.

4

1 Sequence and Series of Real Numbers

Theorem 1.1.1 If (an ) converges to a, then (an ) cannot converge to b with b = a. Proof Suppose an → a and an → b as n → ∞. Let ε > 0 be given and let N1 and N2 in N be such that |an − a| < ε ∀ n ≥ N1 and |an − b| < ε ∀ n ≥ N2 . Then, we have |a − b| = |(a − an ) + (an − b)| ≤ |a − an | + |an − b| < 2ε for any n ≥ N := max{N1 , N2 }. Thus, we have proved that |a − b| < 2ε for every ε > 0. This is possible only when a = b. Remark 1.1.5 In the proof of Theorem 1.1.1, we used an important property of real numbers: For a ∈ R, if 0 ≤ a < r for every r > 0, then a = 0.

This follows from the fact that, if a > 0, then 0 ≤ a < r is not satisfied for r = a/2. Similarly, it can be shown that a, b ∈ R, a < b + r ∀ r > 0

⇒

a ≤ b.

♦

If (an ) converges, then there exists a unique a ∈ R such that an → a Thus, we can define the concept of the limit of a convergent sequence. Definition 1.1.3 If (an ) converges to a, then a is called the limit of (an ), and we write ♦ lim an = a. n→∞

It can be easily seen that an → a ⇐⇒ |an − a| → 0 ⇒ |an | → |a|. Exercise 1.1.1 Verify the above.



Suppose (an ) converges to a. Then, by Definition 1.1.2, taking ε = 10−k for some k ∈ N, there exists some Nk ∈ N such that |an − a| < 10−k ∀ n ≥ Nk .

1.1 Sequence of Real Numbers

5

In other words: If n ≥ Nk and if an and a are expressed in decimal expansion, then the first k decimal places of an and a are the same Note that, in the definition of convergence, different ε can result in different N , i.e., the number N may vary as ε varies. This fact will be revealed in the following examples. Example 1.1.2 Consider the sequence (an ) with an = 1/n for n ∈ N. We expect that an → 0 as n → ∞. Let us prove this. For this, let ε > 0 be given. Then 1 1 < ε ⇐⇒ n > . n ε Thus, taking a positive integer N > 1/ε, we obtain |an − 0| < ε for all n ≥ N . Similarly, for any ε > 0,  1  (−1)n   − 0 = < ε ∀ n ≥ N ,  n n where N is a positive integer such that N > 1/ε. Thus we have shown that 1 (−1)n → 0 and → 0. n n Note that the positive integer N in both the examples depend on the ε chosen. Using the above arguments, it can also be verified that for any real number x, x/n → 0 as n → ∞. ♦ Example 1.1.3 Consider the sequence (an ) with an = n/(n + 1) for n ∈ N. We claim that an → 1 as n → ∞. Note that |an − 1| =

1 n+1

so that, for ε > 0, |an − 1| < ε ⇐⇒ n + 1 >

1 . ε

Thus, taking a positive integer N with N ≥ 1/ε, we obtain |an − 1| < ε for all n ≥ N . Note that, 1 , n ≥ 100 =⇒ |an − 1| < 100 n ≥ 1000

=⇒

|an − 1|
1/ε. Then, we have |an − 1| < ε for all n ≥ N . ♦ Therefore, an → 1 as n → ∞. Example 1.1.5 For n ∈ N, let an =

3 3 3 + + ··· + n, 10 102 10

i.e., an = 0.33 · · · 3 with 3 repeated n times. Note that

1.1 Sequence of Real Numbers

7

Fig. 1.2 Sequence (1/2n )

3 3 3 3 + + ··· + n = 10 102 10 10



1− 1

1 10n 1 − 10



  1 1 1− n . = 3 10

Thus, |an − 1/3| = 1/(3 × 10n ) so that, for any given ε > 0, we can choose a positive integer N (depending on ε) such |an − 1/3| < ε for all n ≥ N . Therefore, an → 1/3 as n → ∞. ♦ In the examples above we have used an important property of real numbers, known as the Archimedean Property : Archimedean Property: Given any x ∈ R, there exists n ∈ N such that n > x. We shall use this property throughout the text without mentioning it explicitly. Let us show the convergence of a few more sequences which we shall come across very often (Fig. 1.2). Example 1.1.6 Consider the sequence (1/n 2 ). Note that, for any ε > 0, 1 1 1 ⇐⇒ n ≥ √ . < ε ⇐⇒ n 2 > 2 n ε ε √ Hence, if we take N ≥ 1/ ε, we obtain 1 < ε ∀ n ≥ N. n2 Thus, 1/n 2 → 0. More generally, for any k ∈ N, we see that 1 < ε ∀ n ≥ Nk , nk where Nk ∈ N is such that N ≥ 1/ε1/k , so that 1/n k → 0. Example 1.1.7 Consider the sequence (1/2n ). Note that, for any ε > 0, 1/2n < ε ⇐⇒ 2n > 1/ε.

♦

8

1 Sequence and Series of Real Numbers

Since 2n > n, if we take N ∈ N such that N > 1/ε, we obtain n≥N

⇒

2n ≥ 2 N > N >

1 ε

⇒

1 < ε. 2n

Thus, 1/2n → 0. We shall show that, 1/a n → 0 for any a > 1.

♦

It can happen that for a convergent sequence (an ), same N works for different ε’s. To illustrate this point, look at the following examples. Example 1.1.8 Let an = 1 for all n ∈ N. We can easily guess that (an ) converges to 1. Note that for any ε > 0, |an − 1| < ε ∀ n ≥ 1. Thus, N = 1 works for any ε > 0.

♦

Example 1.1.9 For n ∈ N, let an =

n for 1 ≤ n ≤ 99, 1 for n ≥ 100.

You must have guessed that, (an ) converges to 1. Yes; it is true: For any given ε > 0, |an − 1| < ε ∀ n ≥ 100. Thus, N = 100 works for any ε > 0.

♦

In Example 1.1.8, all the terms are the same, whereas in Example 1.1.9, the terms are the same after some stage. This observation prompts us to make the following definition. Definition 1.1.4 A sequence (an ) is said to be a constant sequence if an = a1 for all n ∈ N; and it is called an eventually constant sequence if there exists k ∈ N such ♦ that an = ak for all n ≥ k. Clearly, every constant sequence is eventually constant, and the converse does not hold. Further, every eventually constant sequence converges. Remark 1.1.6 Throughout the book, by an interval we mean a subset I of R with the property that for every x, y ∈ I with x < y, z ∈ R with x < z < y

⇒

z ∈ I.

1.1 Sequence of Real Numbers

9

Thus, for a, b ∈ R with a < b, the sets (a, b) := {x ∈ R : a < x < b}; [a, b) := {x ∈ R : a ≤ x < b}; (a, b] := {x ∈ R : a < x ≤ b}; [a, b] := {x ∈ R : a ≤ x ≤ b} are intervals of finite lengths, and the sets (a, ∞) := {x ∈ R : a < x < ∞}; [a, ∞) := {x ∈ R : a ≤ x < ∞}; (−∞, b) := {x ∈ R : −∞ < x < b}; (−∞, b] := {x ∈ R : −∞ < x ≤ b} for any a, b ∈ R and (−∞, ∞) := R are intervals of infinite length. Among them, (a, b), (a, ∞) and (−∞, b) are called open intervals, and [a, b] is called a closed interval. ♦ Remark 1.1.7 We may observe that, for ε > 0, |an − a| < ε ⇐⇒ an ∈ (a − ε, a + ε). Thus, (an ) converges to a ∈ R if and only if for every ε > 0, there exists a positive integer N such that an ∈ (a − ε, a + ε) ∀ n ≥ N . In other words, an → a as n → ∞ if and only if for every ε > 0, an belongs to the open interval (a − ε, a + ε) for all n after some finite stage, and this finite stage may vary according as ε varies. ♦ In view of Remark 1.1.7, the following theorem can be proved easily (see also Figures 1.3 and 1.4). Theorem 1.1.2 For a given sequence (an ) and a ∈ R, we have the following. (i) an → a if and only if for every open interval I with a ∈ I , there exists N ∈ N (depending on a and I ) such that an ∈ I for all n ≥ N . (ii) an → a if and only if there exists an open interval Ia with a ∈ I such that an ∈ / Ia for infinitely many n. Exercise 1.1.2 Supply details of the proof of the above theorem.



A convergent sequence remains convergent to the same point even if we replace or delete a finite number of terms. Equivalently,

10

1 Sequence and Series of Real Numbers

Fig. 1.3 an ∈ (L − ε, L + ε) for all n ≥ N

Fig. 1.4 There exists ε > 0 such that an ∈ / (L − ε, L + ε) for infinitely many n

an → a if and only if ak+n → a as n → ∞ for any k ∈ N.

Exercise 1.1.3 Verify the above statement.



Now, let us give a few examples of divergent sequences. Example 1.1.10 Let an = (−1)n+1 for n ∈ N. The terms of the sequence change alternately between 1 and −1. So, one may guess that it cannot converge to any number. To see this, we observe that, for any a ∈ R, if we take 0 < ε < 1/2, either 1 or −1 will be outside the interval (a − ε, a + ε); in other words, infinitely many ♦ of the terms of the sequence (an ) lie outside (a − ε, a + ε). Example 1.1.11 Let (an ) be defined by a2n−1 = 1/n and a2n = 1. If we take ε, a positive number less than 1, then, for any a ∈ R, infinitely many of the terms of the sequence (an ) lie outside (a − ε, a + ε). For instance, let us take ε = 1/2. If a ≥ 1/2, then infinitely many of a2n−1 lie outside (a − ε, a + ε), and if a < 1/2, then every ♦ a2n lie outside (a − ε, a + ε) for all n ∈ N. Thus, an → a for any a ∈ R.

1.1 Sequence of Real Numbers

11

Example 1.1.12 Let an = n for n ∈ N. In this case, for any given ε > 0 and for any a ∈ R, infinitely many of the terms of the sequence (an ) lie outside (a − ε, a + ε). Similar is the case if an = n 2 or an = n 2 /(n + 1) (verify). Thus, an → a for any a ∈ R. ♦ Definition 1.1.5 A sequence (an ) is said to be an alternating sequence if an changes ♦ sign alternately, that is, an an+1 < 0 for every n ∈ N. For example, ((−1)n ) and ((−1)n /n) are alternating sequences. An alternating sequence may converge or diverge. We have seen that the sequence ((−1)n ) diverges, whereas ((−1)n /n) converges to 0. A divergent sequence may have some specific properties such as the ones defined below. Definition 1.1.6 Consider a sequence (an ). (1) We say that (an ) diverges to infinity if for every M > 0, there exists N ∈ N such that ∀ n ≥ N, an > M and in that case, we write an → ∞. (2) We say that (an ) diverges to minus infinity if for every M > 0, there exists N ∈ N such that an < −M ∀ n ≥ N , and in that case we write an → −∞.

♦

It can be easily seen that an → ∞ ⇐⇒ −an → −∞. Example 1.1.13 Let us look at two sequences which converge to ∞. (i) The sequence (an ) with an = n 2 /(n + 1) diverges to infinity: Note that n n2 ≥ ∀ n ∈ N, n+1 2 so that for any M > 0, an > M whenever n > 2M. Hence, taking a positive integer N such that N > 2M, we have the relation an > M for all n ≥ N . Thus, n2 → ∞ as n → ∞. n+1 (ii) The sequence (an ) with an = 2n diverges to infinity: Note that 2n ≥ n for all n ∈ N so that for any M > 0, 2n > M whenever n > M. Hence, taking a positive integer N > M, we have an > M for all n ≥ N . Thus,

12

1 Sequence and Series of Real Numbers

2n → ∞ as n → ∞. Similarly, for any k ∈ N, (k + 1)n → ∞ as n → ∞.

♦

Observe the following facts: 1. If an → ∞ or an → −∞ as n → ∞, then (an ) is a divergent sequence. 2. If (an ) and (bn ) are such that an ≤ bn for all n ∈ N, then (a) an → ∞ as n → ∞ implies bn → ∞ as n → ∞, (b) bn → −∞ as n → ∞ implies an → −∞ as n → ∞. Exercise 1.1.4 Prove the above three statements.



The following theorem is useful for showing convergence of certain sequences using the convergence of some other sequences. Theorem 1.1.3 Suppose an → a and bn → b as n → ∞. Then the following results hold. (i) (ii) (iii) (iv)

an + bn → a + b as n → ∞. c an → c a as n → ∞ for any real number c. If an ≤ bn for all n ∈ N, then a ≤ b. (Sandwich theorem) If an ≤ cn ≤ bn for all n ∈ N, and if a = b, then cn → a as n → ∞.

Proof Let ε > 0 be given. (i) Note that, for every n ∈ N, |(an + bn ) − (a + b)| = |(an − a) + (bn − b)| ≤ |an − a| + |bn − b|. Since an → a and bn → b, the above inequality suggests that we may take ε1 = ε/2, and consider N1 , N2 ∈ N such that |an − a| < ε1 ∀ n ≥ N1 and |bn − b| < ε1 ∀ n ≥ N2 so that |(an + bn ) − (a + b)| ≤ |an − a| + |bn − b| < 2ε1 = ε for all n ≥ N := max{N1 , N2 }. (ii) Note that |can − ca| = |c| |an − a| ∀ n ∈ N. If c = 0, then (can ) is a constant sequence with every term 0, and hence can → 0. Next, suppose c = 0. Then we have

1.1 Sequence of Real Numbers

13

|can − ca| < ε ⇐⇒ |an − a|
1, then a n → ∞. (ii) If 0 < a < 1, then a n → 0.

14

1 Sequence and Series of Real Numbers

Proof (i) Suppose a > 1. Writing a = 1 + r with r > 0, we have a n = (1 + r )n = 1 + nr +

n(n − 1) 2 r + · · · + rn 2!

for all n ∈ N. Hence, a n ≥ r n for all n ∈ N. Since r n → ∞, we have a n → ∞. (ii) Let 0 < a < 1 and ε > 0 be given. Then b = 1/a satisfies b > 1 so that bn → ∞. Hence there exists N ∈ N such that bn > 1/ε for all n ≥ N . But, bn > 1/ε if and only if a n < ε. Hence, 0 < an < ε ∀ n ≥ N . Thus, a n → 0. Example 1.1.14 For 0 < r < 1 and n ∈ N, let an = 1 + r + r 2 + · · · + r n . Note that 1 − r n+1 ∀ n ∈ N. an = 1−r By Theorem 1.1.4, r n → 0. Hence, by Theorem 1.1.3 (i)-(ii), an →

1 as n → ∞. 1−r

♦

Exercise 1.1.5 Show that |a| < 1 implies a n → 0.



Exercise 1.1.6 For a ≥ 0, prove the following. (i) a n → 0 ⇐⇒ a < 1. (ii) a n → ∞ ⇐⇒ a > 1.



Exercise 1.1.7 If an → a and a = 0, then show that there exists k ∈ N such that  an = 0 for all n ≥ k. 1/n  Exercise 1.1.8 Consider the sequence (an ) with an = 1 + n1 , n ∈ N. Then show that lim an = 1. n→∞

[Hint: Observe that 1 ≤ an ≤ (1 + 1/n) for all n ∈ N.] Exercise 1.1.9 Consider the sequence (an ) with an = any given k ∈ N, lim an = 0 (Fig. 1.5). n→∞

[Hint: Observe that 1 ≤ an ≤ 1/n for all n ∈ N.]

1 , nk



n ∈ N. Then show that for 

1.1 Sequence of Real Numbers

15

Fig. 1.5 2n → ∞ as n→∞

1.1.2 Some Tests for Convergence and Divergence Theorem 1.1.5 (Ratio test) Suppose an > 0 for all n ∈ N such that lim for some  ≥ 0. Then the following hold.

n→∞

an+1 = an

(i) If  < 1, then an → 0. (ii) If  > 1, then an → ∞. Proof (i) Suppose  < 1. Let q be such that  < q < 1. Then, taking for example the open interval I containing  as I = ( − 1, q), there exists N ∈ N such that an+1 ≤ q ∀ n ≥ N. an Hence, 0 < an ≤ q n−N a N ∀ n ≥ N . By Theorem 1.1.4, q n−N → 0 as n → ∞. Hence, by Sandwich theorem, an → 0 as n → ∞. (ii) Suppose  > 1. Let q be such that 1 < q < . Then, taking for example the open interval I containing  as I = (q,  + 1), there exists N ∈ N such that

16

1 Sequence and Series of Real Numbers

an+1 > q ∀ n ≥ N. an Hence, an ≥ q n−N a N ∀ n ≥ N . By Theorem 1.1.4, q n−N → ∞ as n → ∞. Hence, an → ∞. Example 1.1.15 Let 0 < a < 1. Then na n → 0 as n → ∞. To see this, let an := na n for n ∈ N. Then we have an+1 (n + 1)a n+1 (n + 1)a = = an na n n

∀ n ∈ N.

an+1 = a < 1. Thus, by Theorem 1.1.5, na n → 0. an Similarly, it can be shown that, for any k ∈ N, n k a n → 0 as n → ∞.

Hence, lim

n→∞

♦

Remark 1.1.9 The converse of the results (i) and (ii) in Theorem 1.1.5 does not hold. To see this, consider the following examples: (i) Consider (an ) with an = 1/n. Then an → 0, but an+1 /an → 1. (ii) Consider (an ) with an = n. Then an → ∞, but an+1 /an → 1. In fact, we shall see in the next section that the condition  < 1 in Theorem 1.1.5 is too strong, in the sense that, not only we have the convergence of (an ) to 0, but also we can show the convergence of the sequence (sn ), where sn = a1 + a2 + · · · + an . Note that, for any sequence (an ), if sn = a1 + a2 + · · · + an , then the convergence ♦ of (sn ) implies an → 0, but an → 0 does not imply the convergence of (sn ). Exercise 1.1.10 Establish the statement in the last paragraph of the above remark.  The following theorem gives a sufficient condition for certain number to be a limit of a given sequence. Theorem 1.1.6 Let (an ) be a sequence such that |an+1 − a| ≤ r |an − a| ∀ n ∈ N for some a ∈ R and for some r with 0 < r < 1. Then an → a. Proof For each n ∈ N, we have |an+1 − a| ≤ r |an − a| ≤ · · · ≤ r n |a1 − a|. Thus, 0 ≤ |an+1 − a| ≤ r n |a1 − a| ∀ n ∈ N. Since 0 < r < 1, by Theorem 1.1.4, r n → 0. Hence, by Sandwich theorem, an → a.

1.1 Sequence of Real Numbers

17

Example 1.1.16 Let a sequence (an ) be defined iteratively as follows : a1 = 1, an+1 =

2 + an , n = 1, 2, . . . . 1 + an

Let us assume for a moment that (an ) converges to a ∈ R. Then we obtain a=

2+a , 1+a

2 i.e., a(1 √ the limit must be √ + a) = 2 + a, i.e., a = 2. Thus, if (an ) converges, then a = 2. Now, we prove that (an ) actually converges to a := 2. Note that

2 + an 2+a − 1 + an 1+a a − an . = (1 + an )(1 + a)

an+1 − a =

Note that an ≥ 1 for all n ∈ N and a ≥ 1. Hence, we obtain |an+1 − a| ≤

|an − a| |a − an | ≤ ∀ n ∈ N. (1 + an )(1 + a) 4

Therefore, by Theorem 1.1.6, an → a =

√

2.

♦

Remark 1.1.10 In view of Theorem 1.1.6, one may ask the following question: If (an ) is such that |an+1 − a| < |an − a| for all n ∈ N for some a ∈ R, then does (an ) converge to a?

Not necessarily! To see this, consider the sequence (an ) with an = Since

n+1 , n ∈ N. n

n+2 n+1 < ∀ n ∈ N, n+1 n

taking a = 0, we have |an+1 − a| < |an − a| for all n ∈ N. But (an ) does not con♦ verge to 0. In fact, an → 1. Remark 1.1.11 In Theorem 1.1.6, the sufficient condition given for inferring the convergence of a sequence (an ) involves its limit. It would be better if we have a condition on (an )independent of its limit. In this connection, we shall show at the end of Sect. 1.1.4 that |an+2 − an+1 | ≤ r |an+1 − an | ∀ n ∈ N

18

1 Sequence and Series of Real Numbers

is one such condition, where 0 < r < 1.

♦

Remark 1.1.12 In Example 1.1.16, the terms of the sequence are defined iteratively, in the sense that the (n + 1)-th term of the sequence is defined in terms of the n th term of the sequence. More generally, a sequence (an ) defined iteratively can be of the form (1.1) an+1 = ϕ(an ), n ∈ N, where ϕ is some function defined on R. Suppose, ϕ has the property that an → a

⇒

ϕ(an ) → ϕ(a).

Thus, if we assume that an → a, then (1.1) implies that a = ϕ(a).

(1.2)

Thus, following the procedure as in Example 1.1.16, convergence of a sequence (an ) which satisfies (1.1) can be established by looking for a number a satisfying (1.2), and then proving that (an ) actually converges to a. Note that, by (1.1) and (1.2), an+1 − a = ϕ(an ) − ϕ(a). Thus, an → a if and only if ϕ(an ) → ϕ(a). In case, there is a positive number r < 1 such that |ϕ(an ) − ϕ(a)| ≤ r |an − a| for all n ∈ N, then Theorem 1.1.6 leads to the convergence of (an ) to a. The above procedure also can be used to show the divergence of certain sequences, by showing that there is no a which satisfies (1.2). For example, suppose (an ) is defined by a1 = 1 and

an+1 =

1 + an2 , n ∈ N.

√ Since there is no a satisfying a = 1 + a 2 , we can assert that the given sequence does not converge. Existence of a satisfying (1.2) alone does not imply that the sequence converges. For example, consider the sequence (an ) defined by a1 = 1 and an+1 = −an , n ∈ N. We know that a = 0 satisfies the equation a = −a, but an → 0. In fact, in this case, ♦ the sequence is the divergent sequence ((−1)n+1 ). Let us consider another example to illustrate the procedure described in Remark 1.1.12.

1.1 Sequence of Real Numbers

19

Example 1.1.17 Consider the sequence (an ) defined by an+1 = 1 +

1 , n ∈ N, an

with a1 = 1. Here ϕ(x) := 1 + x1 , x = 0. Observe that if an → a for some number a, then a = ϕ(a), and √ 1+ 5 . a = ϕ(a) ⇐⇒ a = 2 Now, let us try to find a number r with 0 < r < 1 such that |ϕ(an ) − ϕ(a)| ≤ r |an − a| for all n ∈ N. Note that 1 1   |ϕ(an ) − ϕ(a)| ≤ r |an − a| ⇐⇒  −  ≤ r |an − a| an a a − a  n  ⇐⇒   ≤ r |an − a| an a 1 ⇐⇒ an a ≥ . r Thus, it is enough to find r such that 0 < r < 1 and an a ≥ 1/r for all n ∈ N. Since an ≥ 1 and a ≥ 3/2, we have an a ≥ a ≥ 3/2 so that we may take any r satisfying 2/3 ≤ r < 1. Thus, √ 1+ 5 . ♦ an → a := 2 Remark 1.1.13 (Hemachandra–Fibonacci sequence) We observe that the terms of the sequence (an ) in Example 1.1.17 are 1 , 1

2 , 1

3 , 2

5 , 3

8 , 5

13 , .... 8

These terms are ratios of consecutive terms of the sequence (bn ), given by 1, 1, 2, 3, 5, 8, 13, . . . . Note that the terms of the sequence (bn ) are iteratively defined by b1 = 1, b2 = 1, and for n = 3, 4, . . ., bn = bn−1 + bn−2 . The number

√ bn+1 1+ 5 , := lim n→∞ bn 2

20

1 Sequence and Series of Real Numbers

the limit of the sequence (an ), is called the golden ratio. The golden ratio is approx√ imately equal to 1.618033, correct to (1 + 5)/2 up to six decimal places, i.e.,   1 + √5 1   − 1.618033 < 6 .  2 10

√ In fact, the number (1 + 5)/2 is the ratio α/β of the sides α and β of a rectangle with β < α such that when the square of side β is cut out of the rectangle, the remaining rectangle has the same ratio as its sides, i.e., α β = . β α−β Indeed, if γ = α/β, then the above relation gives γ = 1/(γ − 1), i.e., γ = 1 + 1/γ, equivalently, γ(γ − 1) = 1. √ so that γ = (1 + 5)/2. Such a rectangle is called a golden rectangle. The sequence (bn ) defined above, that is the sequence, 1, 1, 2, 3, 5, 8, 13, . . . is known as the Fibonacci sequence, after the Italian mathematician Leonardo of Pisa, also known as Fibonacci (1170 - 1250). However, this sequence of numbers appeared in Indian mathematics as early as 200 BC in the work of Pingala on enumerating possible patterns of Sanskrit poetry formed from syllables of two lengths, called laghu (short) and guru (long), and extensively used by ¯ ar Ach ¯ ya H emachandra (1088–1173) about fifty years prior to Fibonacci (cf. Perkins [10]). Because of this, Fibonacci sequence is also known as Hemachandra– Fibonacci sequence. ♦ Next we prove a necessary condition for the convergence of a sequence. Theorem 1.1.7 (Boundedness test) If (an ) converges, then there exists M > 0 such that

1.1 Sequence of Real Numbers

21

Fig. 1.6 (an ) is a bounded sequence

|an | ≤ M ∀ n ∈ N. (Fig. 1.6) Proof Suppose an → a. Note that, for all n ∈ N, |an | = |(an − a) + a| ≤ |an − a| + |a|. Since |an − a| → 0, (by taking ε = 1) there exists N ∈ N such that |an − a| ≤ 1 for all n ≥ N . Then we have |an | ≤ 1 + |a|

∀ n ≥ N.

Thus, |an | ≤ M := max{1 + |a|, |a1 |, |a2 |, . . . , |a N |} for all n ∈ N. The converse of Theorem 1.1.7 is not true. Note that the sequence (an ) with an = (−1)n satisfies |an | = 1 for all n ∈ N, but it is not convergent. For later use we introduce the following definition. Definition 1.1.7 A sequence (an ) is said to be (1) bounded above if there exists a real number M such that an ≤ M for all n ∈ N; (2) bounded below if there exists a real number M such that an ≥ M for all n ∈ N; (3) bounded if it is bounded above and bounded below. A sequence that is not bounded is called an unbounded sequence. Thus, according to Theorem 1.1.7: Every convergent sequence is bounded The following statements can be verified easily:

♦

22

1 Sequence and Series of Real Numbers

1. A sequence (an ) is bounded if and only if there exists M > 0 such that |an | ≤ M for all n ∈ N. 2. A sequence (an ) is unbounded if and only if it is either not bounded above or not bounded below. 3. If (an ) diverges to ∞, then it is not bounded above, and if it diverges to −∞, then it is not bounded below. 4. A sequence (an ) is not bounded above if and only if for every k ∈ N, there exists n k ∈ N such that an k > k. 5. A sequence (an ) is not bounded below if and only if for every k ∈ N, there exists n k ∈ N such that an k < −k. 

Exercise 1.1.11 Verify the above statements.

Example 1.1.18 Let us look at a few examples of bounded and unbounded sequences (verify). (i) (an ) with an = (−1)n is a bounded sequence. (ii) (an ) with an = (−1)n n is neither bounded above nor bounded below, and it neither diverges to +∞ nor diverges to −∞. (iii) (an ) with an = −n is bounded above, but not bounded below, and it diverges to −∞. (iv) (an ) with an = n is bounded below, but not bounded above, and it diverges to +∞. ♦ Theorem 1.1.7 can be used to show that certain sequence is not convergent, as in the following example. Example 1.1.19 For n ∈ N, let an = 1 +

1 1 1 + + ... + . 2 3 n

Then (an ) diverges: To see this, observe that 1 1 1 + + ... + n 2 3 2 1 1 1 1 1 1 1 + + + + + + = 1+ + 2 3 4 5 6 7 8  1 1 . . . + n−1 + ... + n 2 +1 2 n ≥ 1+ . 2

a2 n = 1 +

Note that 1 1 1 + ≥ , 3 4 2

1 1 1 1 1 + + + ≥ , ··· , 5 6 7 8 2

1 1 1 + ... + n ≥ . 2n−1 + 1 2 2

Hence, an ≥ 1 + n2 . Thus, (an ) is not bounded, so that it diverges.

♦

1.1 Sequence of Real Numbers

23

Theorem 1.1.7 will be used for proving the following. Theorem 1.1.8 Suppose an → a and bn → b. Then we have the following. (i) an bn → ab as n → ∞. (ii) If bn = 0 for all n ∈ N and b = 0, then an 1 1 a and → → . bn b bn b Proof (i) Note that, for every n ∈ N, |an bn − ab| = |an (bn − b) + (an − a)b| ≤ |an | |bn − b| + |an − a| |b|. By Theorem 1.1.7, (an ) is bounded, say |an | ≤ M for all n ∈ N. Hence, we have 0 ≤ |an bn − ab| ≤ M |bn − b| + |b||an − a|. Since an → a and bn → b, by Theorem 1.1.3, we obtain an bn → ab as n → ∞. (ii) Suppose bn = 0 for all n ∈ N and b = 0. Note that, for every n ∈ N,    1  − 1  = |bn − b| . b b |bn | |b| n Since bn → b and b = 0, we have |bn | → |b| so that, taking ε = |b|/2, there exists N ∈ N such that |bn | ≥ |b|/2 ∀ n ≥ N . Hence,

   1  − 1  ≤ |bn − b| = 2 |bn − b| ∀ n ≥ N . b b  (|b|/2) |b| |b|2 n

Hence, by Theorem 1.1.3, we obtain 1/bn → 1/b as n → ∞. Now, using (i), we also obtain an /bn → a/b as n → ∞. Exercise 1.1.12 Prove the following. (i) If (an ) is not bounded above, then there exists a sequence (kn ) of natural numbers such that kn ≤ kn+1 for all n ∈ N and akn → +∞ as n → ∞. (ii) If (an ) is not bounded below, then there exists sequence (kn ) of natural numbers  such that kn ≤ kn+1 for all n ∈ N and akn → −∞ as n → ∞. If (an ) converges to a and a = 0, then show that there Exercise 1.1.13 Let (an ) and (bn ) are such that an ≤ bn for all n ∈ N. Prove the following:

24

1 Sequence and Series of Real Numbers

(i) If (bn ) is bounded above, then (an ) bounded above. (ii) If (an ) is bounded below, then (bn ) bounded below.



1.1.3 Monotonic Sequences We have seen that a bounded sequence need not converge. However, we shall show that, if the terms an of a bounded sequence (an ) either increase or decrease as n increases, then the sequence does converge. Definition 1.1.8 A sequence (an ) is said to be a (1) monotonically increasing sequence if an ≤ an+1 for all n ∈ N; (2) monotonically decreasing sequence if an ≥ an+1 for all n ∈ N; (3) monotonic sequence if it is either monotonically increasing or monotonically decreasing. If strict inequality occur in (1) (resp. (2)), then we say that the sequence is strictly increasing (resp. strictly decreasing). A monotonically increasing (respectively, a monotonically decreasing) sequence is also called an increasing (respectively, a decreasing) sequence. ♦ We may observe that: 1. A sequence (an ) is monotonically increasing and bounded above if and only if (−an ) is monotonically decreasing and bounded below. 2. Every monotonically increasing sequence is bounded below, and every monotonically decreasing sequence is bounded above.

Fig. 1.7 Strictly decreasing

1.1 Sequence of Real Numbers

25

Fig. 1.8 Neither increasing nor decreasing

Example 1.1.20 The following statements can be easily verified (verify) (Figs. 1.7 and 1.8): (i) (an ) with an = n/(n + 1) is monotonically increasing. (ii) (an ) with an = (n + 1)/n is monotonically decreasing. (iii) (an ) with an = (−1)n n/(n + 1) is neither monotonically increasing nor monotonically decreasing. ♦ Exercise 1.1.14 Let (an ) be a monotonic sequence. Prove the following. (i) Suppose (an ) converges. Then it is bounded above by the limit if it is increasing, and bounded below by the limit if it is decreasing. (ii) Suppose (an ) diverges. Then it diverges to either infinity or minus infinity depending on whether it is increasing or decreasing.  Note that a convergent sequence need not be monotonically increasing or monotonically decreasing. For example, the sequence ((−1)n /n) is convergent, but it is neither monotonically increasing nor monotonically decreasing. However, we have the following theorem. It helps us to show the convergence of many of the standard sequences. Theorem 1.1.9 (Monotone convergence theorem) Every bounded monotonic sequence is convergent. For the proof of the above theorem we shall make use of an important property of the set of real numbers, that is called the least upper bound property. For its statement, we need to introduce the notion of the least upper bound. We have already defined the concepts of bounded above, bounded below and bounded corresponding to a sequence. Now, we extend these notions to an arbitrary subset of the set of real numbers. Definition 1.1.9 Let S be a subset of R. Then S is said to be (1) bounded above if there exists b ∈ R such that x ≤ b for all x ∈ S, and in that case b is called an upper bound of S;

26

1 Sequence and Series of Real Numbers

(2) bounded below if there exists a ∈ R such that a ≤ x for all x ∈ S, and in that case a is called a lower bound of S; (3) bounded if it is bounded above and bounded below; (4) unbounded if it is not bounded. ♦ For a set S ⊆ R, we observe that, 1. S is bounded iff there exists M > 0 such that |x| ≤ M for all x ∈ S; 2. S is unbounded iff either it is not bounded above or it is not bounded below; 3. b is an upper bound of S iff −b is a lower bound of {−x : x ∈ S}. Exercise 1.1.15 Verify the above statements.



Before going further, let us illustrate the above notions by a few simple examples. Example 1.1.21 We may observe the following: (i) Intervals (a, b), [a, b), (a, b], [a, b] with a, b ∈ R and a < b are bounded sets. (ii) For a, b ∈ R, the intervals (−∞, b), (−∞, b] are bounded above, but not bounded below, and the intervals (a, ∞), [a, ∞) are bounded below but not bounded above. (iii) The set N is bounded below, but not bounded above. (iv) The sets Z of integers are neither bounded above nor bounded below. ♦ (v) The set {(−1)n n : n ∈ N} is neither bounded above nor bounded below. We may also observe: 1. If S ⊆ R is bounded above (respectively, bounded below), then any subset of S is bounded above (respectively, bounded below). 2. If S ⊆ R is not bounded above (respectively, not bounded below), then any super set of S in R is not bounded above (respectively, not bounded below). Exercise 1.1.16 Verify the above statements.



Let S ⊆ R. If S is bounded above and if b ∈ R is an upper bound of S, then any number greater than b is also an upper bound. So, one may seek a least of the upper bounds. Definition 1.1.10 Let S be a subset of R. If S is bounded above, and if b0 is an upper bound of S such that b0 ≤ b for every upper bound b of S, then b0 is called the least upper bound or supremum for S, and it is denoted by lub(S) or sup(S). ♦ Given any set of numbers having a certain property, it is not obvious whether there is a least number having the same property. For instance, if we consider the set {x ∈ Q : 0 < x < 1} does not have any least number. However, the set of all upper bounds for a given set S ⊆ R, does have a least of the upper bound, provided S is non-empty. This is, in fact, one of the important properties of the set of all real numbers, called the least upper bound property (cf. Pugh [12]):

1.1 Sequence of Real Numbers

27

Least upper bound property: Every non-empty subset of R which is bounded above has the least upper bound.

Analogously, if S is bounded below with a lower bound a ∈ R, then any number less than a is also a lower bound. In this case, we define the greatest lower bound as follows. Definition 1.1.11 Let S be a subset of R. If S is bounded below, and if a0 is a lower bound of S such that a0 ≥ a for every lower bound a of S, then a0 is called the greatest lower bound or infimum for S, and it is denoted by glb(S) or inf(S). ♦ We may observe that, for a non-empty set S ⊆ R, b0 is a least upper bound for S if and only if −b0 is a greatest lower bound for S˜ := {−x : x ∈ S}.

Thus, we also have the greatest lower bound property of R: Greatest lower bound property: Every non-empty subset of R which is bounded below has the greatest lower bound.

For a non-empty set S ⊆ R, the following can be verified easily. 1. If S is bounded above, then S has only one least upper bound. 2. If S is bounded below, then S has only one greatest lower bound. Exercise 1.1.17 Show that, every subset of R that is bounded above has the least upper bound if and only if every subset of R that is bounded below has the greatest lower bound.  As per the Exercise 1.1.17, least upper bound property is equivalent to greatest lower bound property. 1. If S is not bounded above, then S does not have the supremum, and in that case we write sup(S) = ∞. 2. If S is not bounded below, then S does not have the infimum, and in that case we write inf(S) = −∞. Example 1.1.22 The following can be verified easily: (i) If S is any of the intervals (0, 1), [0, 1), (0, 1], [0, 1], then 1 is the supremum for S and 0 is the infimum of S. (ii) If S = {1/n : n ∈ N}, then 1 is the supremum of S and 0 is the infimum for S. (iii) For k ∈ N, if Sk = {n ∈ N : n ≥ k}, then k is the infimum of Sk , and Sk has no supremum. (iv) For k ∈ N, if Sk = {n ∈ Z : n ≤ k}, then k is the supremum of Sk , and Sk has no infimum. ♦

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1 Sequence and Series of Real Numbers

The above examples would have convinced you the following: The supremum (respectively, infimum) of a setS, if exists, need not belong toS

Exercise 1.1.18 Let S ⊆ R. Prove the following. (i) If b0 is the supremum of S, then there exists a sequence (xn ) in S which converges to b0 . (ii) If a0 is the infimum of S, then there exists a sequence (xn ) in S which converges  to a0 . Now, we prove Theorem 1.1.9. Proof of Theorem 1.1.9 Suppose (an ) is a bounded monotonic sequence. Then either it is monotonically increasing and bounded above or it is monotonically decreasing and bounded below. Assume that (an ) is monotonically increasing and bounded above. Then the set S := {an : n ∈ N} is bounded above. Hence, by the least upper bound property of R, S has the least upper bound (the supremum). Let b0 := sup(S), and let ε > 0 be given. Then, b0 − ε cannot be an upper bound of S. Hence, there exists k ∈ N such that b0 − ε < ak . Since (an ) is monotonically increasing, we have b0 − ε < ak ≤ an ∀ n ≥ k. But, an ≤ b0 for all n ∈ N. Hence, we obtain b0 − ε < ak ≤ an < b0 + ε ∀ n ≥ k. Thus we have proved that an → b0 as n → ∞. Next, suppose that (an ) is monotonically decreasing and bounded below. Then the sequence (−an ) is monotonically increasing and bounded above. Therefore, (−an ) converges to b := sup{−an : n ∈ N}. Consequently, an → a := −b = inf{an : n ∈ N}. In fact, we have proved the following: (i) If (an ) is monotonically increasing and bounded above, then (an ) converges to sup{an : n ∈ N} (ii) If (an ) is monotonically decreasing and bounded below, then (an ) converges to inf{an : n ∈ N}

A monotonic sequence that is not bounded cannot converge. Can we say something more? Yes:

1.1 Sequence of Real Numbers

29

(i) A monotonically increasing sequence either converges or diverges to ∞ (ii) A monotonically decreasing sequence either converges or diverges to − ∞ Exercise 1.1.19 Prove the statements (i) and (ii) above.



In the following examples, Theorem 1.1.9 has been used for showing convergence of certain sequences. Example 1.1.23 We have already seen that if 0 < a < 1, then a n → 0 as n → ∞ (See Theorem 1.1.4). This can also be seen by making use of Theorem 1.1.9, as follows: Let xn = a n . Then 0 ≤ xn+1 = a n+1 = axn ≤ xn ∀ n ∈ N. Thus, (xn ) is monotonically decreasing and bounded below. Hence (xn ) converges ♦ to some x ∈ R. Then xn+1 = axn → ax. Therefore, x = ax so that x = 0. Example 1.1.24 For n ∈ N, let an = 1 +

1 1 1 + 2 + ··· + 2. 2 2 3 n

Clearly, (an ) is monotonically increasing. Now, we show that it is bounded above, so that by Theorem 1.1.9, it is convergent: We note that 1 1 2 1 + 2 ≤ 2 = , 2 2 3 2 2 1 1 1 1 4 1 + 2+ 2+ 2 ≤ 2 = , 42 5 6 7 4 4 and more generally, 1 (2n−1 )2

+

1 1 2n−1 1 + · · · + ≤ = n−1 . (2n−1 + 1)2 (2n − 1)2 (2n−1 )2 2

Hence, 1 1 1 + 2 + ··· + n 2 22 3 (2 ) 1 1 1 1 ≤ 1 + + 2 + · · · + n−1 + 2n 2 2 2 2   1 1 = 2 1 − n + 2n 2 2 ≤ 2.

a2 n = 1 +

30

1 Sequence and Series of Real Numbers

Since an ≤ a2n for all n ∈ N, we obtain that (an ) is monotonically increasing and also bounded above, and hence, by Theorem 1.1.9, it converges. We shall see in the final chapter of this book that the number to which this (an ) converges is π 2 /6. ♦ Example 1.1.25 Given a sequence (an ) with an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, consider the sequence (bn ) with bn =

a1 an a2 + ··· + n + 10 102 10

for n ∈ N. Does (bn ) converge? By our knowledge in decimal expansion of a number, we can believe that (bn ) would converge to a number in the interval [0, 1]. Let us prove this: Note that bn ≤ bn+1 for all n ∈ N. Therefore, to assert the convergence of (bn ), by Theorem 1.1.9, it is enough to show that (bn ) is bounded above. For this end, we observe that a1 a2 + 2 + ··· + 10 10 9 9 + 2 + ··· + ≤ 10  10  9 1 − 101n = 1 10 1 − 10

bn =

= 1−

an 10n 9 10n

1 10n

for all n ∈ N. Thus, (bn ) is bounded above by 1. Therefore, (bn ) converges to a number in the interval [0, 1]. More generally, for a given positive integer k ≥ 2, let (an ) be a sequence with an ∈ {0, 1, . . . , k − 1}, and consider (bn ) with bn =

a2 a1 an + 2 + ··· + n k k k

for n ∈ N. Then, as in the case of k = 10, (bn ) is monotonically increasing and bounded above. In fact, a2 a1 an + 2 + ··· + n k k  k 1 k − 1 1 − kn ≤ k 1 − k1

bn =

= 1−

1 kn

for all n ∈ N, so that (bn ) is bounded above by 1, and hence it converges to a number b ∈ [0, 1]. ♦

1.1 Sequence of Real Numbers

31

Example 1.1.26 (The number e) Consider the sequences (an ) and (bn ) defined by  1 n 1 1 1 1 , bn = 1 + + + + · · · + . an = 1 + n 1! 2! 3! n! We show that both (an ) and (bn ) are monotonically increasing and bounded above. Hence, by Theorem 1.1.9, they converge. Further, we shall show that they have the same limit. For this we show that (i) bn ≤ bn+1 ≤ 3 for all n ∈ N, (ii) an ≤ bn for all n ∈ N, (iii) an ≤ an+1 ≤ 3 for all n ∈ N. Clearly, bn ≤ bn+1 for all n ∈ N. Also, bn ≤ 1 + 1 +

1 1 1 + 2 + · · · + n−1 < 3. 2 2 2

Thus, (i) is proved. Next, 1 n(n − 1) 1 n(n − 1) . . . 2.1 + ··· + 2 2! n n! nn 1 1 1 ≤ 1 + 1 + + + ··· + 2! 3! n! = bn

an = 1 + 1 +

and  1 n an = 1 + n 1 n(n − 1) 1 n(n − 1) . . . 2.1 1 = 1 + n. + + ··· + n 2! n2 n! nn 1 1  1  2 1 1− + 1− 1− = 1+1+ 2! n 3! n n     1 1 n−1 +··· + 1− ··· 1 − n! n n ≤ an+1 . Thus, the proofs of (i)–(iii) are over. From, (i)–(iii), we see that both (an ) and (bn ) are monotonically increasing bounded above. Hence, by Theorem 1.1.9, both (an ) and (bn ) converge. Let a and b be their limits. We show that a = b. We have already observed that an ≤ bn . Hence, taking limits, we obtain a ≤ b. Notice that 1 1  1  2 1 1− + 1− 1− 2! n 3! n n 1  n − 1 1 1− ... 1 − . +... + n! n n

an = 1 + 1 +

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Hence, for m, n with m ≤ n, we have 1 1  1  2 1 1− + 1− 1− 2! n 3! n n 1 1  m − 1 +··· + 1− ... 1 − . m! n n

an ≥ 1 + 1 +

Taking limit as n → ∞, we get (cf. Theorem 1.1.3 (c)) a ≥1+

1 1 1 1 + + + ··· + = bm . 1! 2! 3! m!

Now, taking limit as m → ∞, we get a ≥ b. Thus we have proved a = b. The common limit of the two sequence (an ) and (bn ) above is denoted by the ♦ letter e, after the great mathematician Euler.1   1 1 1 1 n 1 . = lim 1 + + + + · · · + e := lim 1 + n→∞ n→∞ n 1! 2! 3! n! Further examples are given in Sect. 1.1.5.

1.1.4 Subsequences There are divergent sequences which contain terms which form convergent sequences. For example, the sequence (an ) with an = (−1)n is divergent, but the sequences (a2n−1 ) and (a2n ) are convergent. Those sequences extracted from a given sequence, retaining the order in which the terms occur, are called subsequences. More precisely, we have the following definition. Definition 1.1.12 A subsequence of a sequence (an ) is a sequence of the form (akn ), ♦ where (kn ) is a strictly increasing sequence of positive integers. A sequence (bn ) is a subsequence of a sequence (an ) if and only if there is a strictly increasing sequence (kn ) of positive integers such that bn = akn for all n ∈ N.

For example, given a sequence (an ), the sequences (a2n ), (a2n+1 ), (an 2 ), (a2n ) are some of the subsequences of (an ). As concrete examples, 1

Leonhard Euler (15 April 1707–18 September 1783) was a Swiss mathematician and physicist. He made important discoveries in various fields in mathematics. He introduced many modern mathematical terminology and notation, including the notion of a mathematical function (Courtsey to Wikipedia).

1.1 Sequence of Real Numbers

33

1 1 (1) ( 2n ), ( 2n+1 ), ( n12 ) and ( 21n ) are subsequences of ( n1 ); n ) are subsequences of (1, 21 , 21 , 23 , 13 , 43 , . . .). (2) ( n1 ) and ( n+1 (3) The sequence (bn ) with bn = 1 for all n ∈ N and the sequence (cn ) with cn = −1 for all n ∈ N are subsequences of the sequence ((−1)n ). (4) (2n), (2n + 1), (n 2 ) and (2n ) are subsequences of the sequence (an ) with an = n for all n ∈ N.

In (1), the given sequence and all the subsequences listed are convergent; in (2), the subsequences are convergent, but the original sequence is not convergent; same is the case with the case in (3); in (4), the original sequence and all the subsequences listed are divergent. In fact, in (4), every subsequence of the given sequence has to diverge to infinity (verify). However, we have the following result. Theorem 1.1.10 If a sequence (an ) converges to a, then all its subsequences converge to the same limit a. Proof Suppose an → a. Consider a subsequence (akn ) of (an ). Let ε > 0 be given. Since an → a, there exists N ∈ N such that |an − a| < ε ∀ n ≥ N . In particular, since k N ≥ N , |an − a| < ε ∀ n ∈ {k N , k N +1 , . . .}. Thus, |akn − a| < ε ∀n ≥ N . Hence, akn → a. What about the converse of the above theorem? Obviously, if all subsequences of (an ) converge, then (an ) also has to converge, since (an ) is a subsequence of itself. Thus, we have proved: (an ) converges to a iff every subsequence of (an ) converges to a Theorem 1.1.10 can be used to assert the divergence of certain sequences. Let us consider the following example. Example 1.1.27 Let sn = 1 + 21 + · · · + n1 for n ∈ N. We know that the sequence (sn ) is not convergent (cf. Example 1.1.19). Now, for every n ∈ N, s2n − sn =

1 1 1 1 + + ··· + ≥ . n+1 n+2 2n 2

Hence, by Theorem 1.1.10, (sn ) cannot converge to any point.

♦

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We know that convergence of some subsequences, even if the limits are same, does not imply the convergence of the original sequence. However, in certain cases, convergence of a few of subsequences does imply the convergence of the original sequence. Theorem 1.1.11 Let (an ) be such that a2n → a and a2n+1 → a for some a ∈ R. Then an → a. Proof Let ε > 0 be given. Since a2n → a and a2n+1 → a for some a ∈ R, there exists N ∈ N such that |a2n − a| < ε, |a2n+1 − a| < ε ∀ n ≥ N . Then we have |an − a| < ε ∀ n ≥ 2N + 1. Thus, an → a. Remark 1.1.14 In the above theorem, to obtain convergence of (an ) it is important that both the sequences (a2n ) and (a2n+1 ) converge to the same point. For instance if an = (−1)n , then we have the convergence of (a2n ) and (a2n+1 ), but (an ) does not ♦ converge. In this case, the limits of (a2n ) and (a2n+1 ) are different. Exercise 1.1.20 Let (an ) be such that, for some k ∈ N, the sequences (akn+1 ),  (akn+2 ), . . . , (akn+k ) converge to the same limit a. Show that an → a. The following result on non-convergence of a sequence to a particular a ∈ R will be of use in later chapters. Theorem 1.1.12 Let (an ) be a sequence of real numbers and a ∈ R. If an → a, then there exists ε > 0 and a subsequence (akn ) such that |akn − a| ≥ ε for all n ∈ N. Proof Suppose an → a. Then there exists ε > 0 such that |an − a| ≥ ε for infinitely many n’s in N. Thus, we can find a sequence (kn ) in N such that kn < kn+1 for all n ∈ N and |akn − a| ≥ ε for all n ∈ N . In fact, if S = {n ∈ N : |an − a| ≥ ε}, then we may take k1 = inf(S), k2 = inf(S \ {k1 }), k3 = inf(S \ {k1 , k2 }), and having chosen k1 , k2 , . . . , kn , take kn+1 = inf(S \ {k1 , . . . , kn }). Clearly, (kn ) is a strictly increasing sequence of positive integers such that |akn − a| ≥ ε for all n ∈ N .

1.1 Sequence of Real Numbers

35

We know that a divergent sequence can have convergent subsequences. But, this cannot happen for monotonic sequences. If (an ) is a monotonic sequence having at least one convergent subsequence, say with limit α, then (an ) itself converge to a Exercise 1.1.21 Prove the above statement.



We have seen in Theorem 1.1.7 that every convergent sequence is bounded, but a bounded sequence need not be convergent. However, we have the following theorem, which is called the Bolzano–Weierstrass theorem. Theorem 1.1.13 (Bolzano-Weierstrass theorem2 ). Every bounded sequence of real numbers has at least one convergent subsequence. The idea involved in the proof of the above theorem is to extract a bounded monotone subsequence and then apply Monotone convergence theorem (Theorem 1.1.9). For the details of the proof, the reader may find its proof in Binmore [3]. We use Theorem 1.1.13 to prove the following theorem, which gives another sufficient condition for the convergence of a sequence, as promised in Remark 1.1.15. Theorem 1.1.14 Let (an ) be a sequence such that |an+2 − an+1 | ≤ r |an+1 − an | ∀ n ∈ N for some r with 0 < r < 1. Then (an ) converges, and if a = limn→∞ an , then |an − a| ≤

r n−1 |a2 − a1 | ∀ n ∈ N. 1−r

Proof From the given property of (an ) we obtain |an+1 − an | ≤ r n−1 |a2 − a1 | ∀ n ∈ N. Hence, for any n, m with n < m, we have |am − an | ≤

2

r n−1 |a2 − a1 |. 1−r

(∗)

Bernard Bolzano (October 5, 1781–December 18, 1848), was a Bohemian mathematician, logician, philosopher, theologian and Catholic priest of Italian Origins, and Karl Weierstrass (October 31, 1815–February 19, 1897) was a German mathematician who is often cited as the “father of modern analysis”.

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1 Sequence and Series of Real Numbers

Indeed, |am − an | ≤ |am − am−1 | + · · · + |an+1 − an | ≤ (r m−2 + r m−3 + · · · + r n−1 )|a2 − a1 | r n−1 |a2 − a1 |. ≤ 1−r From (∗), taking n = 1, we have |am − a1 | ≤ |a2 − a1 |/(1 − r ) for all m ∈ N so that (an ) is a bounded sequence. Hence, by Theorem 1.1.13, it has a convergent subsequence, say akn → a for some a ∈ R. Taking m = kn in (∗), we have |akn − an | ≤

r n−1 |a2 − a1 | ∀ n ∈ N. 1−r

In particular, |akn − an | → 0 as n → ∞. Now, since |a − an | ≤ |a − akn | + |akn − an |, it follows that |a − an | → 0. Remark 1.1.15 A convergent sequence (an ) need not satisfy the condition in Theorem 1.1.14 for any r with 0 < r < 1. To see this one may consider the sequence (an ) with an = 1/n for n ∈ N. In this case, we see that |an+2 − an+1 | ≤ r |an+1 − an | ⇐⇒

n ≤r n+2

which is satisfied for all n ∈ N if and only if r ≥ 1.

♦

1.1.5 Further Examples Example 1.1.28 Let a sequence (an ) be defined iteratively as follows: a1 = 1 and an+1 =

2an + 3 4

n ∈ N. We show that (an ) is monotonically increasing and bounded above. Note that an+1 =

an 3 2an + 3 = + ≥ an 4 2 4

⇐⇒

Thus it is enough to show that an ≤ 3/2 for all n ∈ N.

an ≤

3 . 2

1.1 Sequence of Real Numbers

37

Clearly, a1 ≤ 3/2. If an ≤ 3/2, then an+1 = an /2 + 3/4 < 3/4 + 3/4 = 3/2. Thus, we have proved that an ≤ 3/2 for all n ∈ N. Hence, by Theorem 1.1.9, (an ) converges. Let its limit be a. Then taking limit on both sides of an+1 = 2an4+3 we have 2a + 3 3 a= i.e., 4a = 2a + 3 so that a = . 4 2 Another solution: Since a := 3/2 satisfies a= we obtain an+1 − a =

2a + 3 , 4

1 2an + 3 2a + 3 − = (an − a). 4 4 2

Thus, by Theorem 1.1.6, an → a = 3/2.

♦

Example 1.1.29 Let a sequence (an ) be defined as follows : a1 = 2 and an+1 =

1 2 an + 2 an

for n ∈ N. Note that, if the sequence converges, then its limit a ≥ 0, and then a=

2 1 a+ 2 a

√ so that a = 2. Since a1 = 2, one may try to show that (an ) is monotonically decreasing and bounded below. Note that an+1 :=

1 2 an + ≤ an 2 an

⇐⇒

an2 ≥ 2, i.e., an ≥

√

2,

and an+1 :=

√ 1 2 √ an + ≥ 2 ⇐⇒ an2 − 2 2an + 2 ≥ 0 2 an √ ⇐⇒ (an − 2)2 ≥ 0.

√ Hence, an+1 ≥ 2 for all n ∈ N so that an+1 ≤ an for all n ≥ 2. Hence, (an ) is monotonically decreasing and √ bounded below, so that by Theorem 1.1.9, (an ) converges ♦ and hence its limit is 2. Example 1.1.30 The sequence (n 1/n ) converges and the limit is 1:

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1 Sequence and Series of Real Numbers

For each n ∈ N, since n 1/n ≥ 1, there exists rn ≥ 0 such that n 1/n = 1 + rn . Then we have n = (1 + rn )n ≥ so that rn2 ≤

n(n − 1) 2 rn , 2

2 ∀ n ≥ 2. n−1

Since 2/(n − 1) → 0, by Theorem 1.1.3(c), rn → 0. Hence n 1/n → 1.

♦

Example 1.1.31 For any a > 0, (a 1/n ) converges to 1: If a > 1, then we can write a 1/n = 1 + rn for some sequence (rn ) of positive reals. Then we have a = (1 + rn )n ≥ nrn so that rn ≤ a/n. Since a/n → 0, by Theorem 1.1.3(c), rn → 0 and a 1/n = 1 + rn → 1. In case 0 < a < 1, then 1/a > 1. Hence, by the first part, 1/a 1/n = (1/a)1/n → 1, so that by Theorem 1.1.8 (ii), a n → 1.

♦

Example 1.1.32 Let (an ) be a bounded sequence of non-negative real numbers. Then (1 + an )1/n → 1 as n → ∞: Let M > 0 be such that 0 ≤ an ≤ M for all n ∈ N. Then, 1 ≤ (1 + an )1/n ≤ (1 + M)1/n

∀ n ∈ N.

Using the result in Example 1.1.31, (1 + M)1/n → 1. Hence, by the Sandwich the♦ orem (Theorem 1.1.3 (iv)), (1 + an )1/n → 1. Example 1.1.33 Let (an ) be a sequence of non-negative terms such that 1 ≤ an ≤ n for all n ∈ N. Then (1 + an )1/n → 1: Note that 1 ≤ (1 + an )1/n ≤ (1 + n)1/n ≤ (2n)1/n = 21/n n 1/n ∀ n ∈ N. By the results in Examples 1.1.30 and 1.1.31, 21/n → 1 and n 1/n → 1. Hence, by the Sandwich theorem (Theorem 1.1.3 (iv)), we have the convergence ♦ (1 + an )1/n → 1. 2

Example 1.1.34 Let an = (n!)1/n , n ∈ N. Then an → 1 as n → ∞. We give two proofs for this.

1.1 Sequence of Real Numbers

39

(i) Note that, for every n ∈ N, 2

2

1 ≤ (n!)1/n ≤ (n n )1/n = n 1/n . Since n 1/n → 1, by the Sandwich 2 we have (n!)1/n → 1. (ii) By GM-AM inequality, for n ∈ N, (n!)1/n = (1.2. . . . .n)1/n ≤

theorem

(Theorem

1.1.3

(iv)),

n+1 1 + 2 + ... + n = ≤ n. n 2

2

Thus, 1 ≤ (n!)1/n ≤ n 1/n . Since n 1/n → 1, by the Sandwich theorem (Theorem 2 ♦ 1.1.3 (iv)), we have (n!)1/n → 1. What about the convergence of the sequence (an ) with an = (n!)1/n ? Example 1.1.35 Let an = (n!)1/n , n ∈ N. We show that an → 1. In fact, we show that (an ) is unbounded. Note that, for any k, n ∈ N with n ≥ k, (n!)1/n ≥ (k!)1/n (k n−k )1/n = (k!)1/n k 1−k/n = k

 k! 1/n kk

.

 1/n → k as n → ∞, we can conclude that (n!)1/n → 1. Since for any fixed k, k kk!k Now, for k ∈ N, let n k ∈ N be such that n k ≥ k and  k! 1/n kk Thus, (n!)1/n ≥

≥

1 ∀ n ≥ nk . 2

k 2

∀ n ≥ nk .

 Therefore, the sequence (n!)1/n is unbounded. In fact (n!)1/n → ∞ as n → ∞. ♦ Remark 1.1.16 Suppose for each k ∈ N, an(k) → 0, bn(k) → 1 as n → ∞, and also an(n) → 0, bn(n) → 1 as n → ∞. In view of Theorems 1.1.3 and 1.1.8, one may think that an(1) + an(2) + · · · + an(n) → 0 as n → ∞ and

bn(1) bn(2) · · · bn(n) → 1 as n → ∞.

Unfortunately, that is not the case. To see this consider an(k) =

k , n2

bn(k) = k 1/n for k, n ∈ N.

40

1 Sequence and Series of Real Numbers

Then, for each k ∈ N, we have an(k) → 0, bn(k) → 1 as n → ∞. Also

an(n) → 0, bn(n) → 1 as n → ∞.

But, an(1) + an(2) + · · · + an(n) =

1 1 2 n n+1 → as n → ∞ + 2 + ··· + 2 = 2 n n n 2n 2

and from Example 1.1.35, bn(1) bn(2) · · · bn(n) = 11/n 21/n · · · n 1/n = (n!)1/n → 1 as n → ∞.

♦

1.1.6 Cauchy Criterion In Theorem 1.1.14, we have given a sufficient condition for the convergence of a sequence (an ), namely, that if (an ) satisfies |an+2 − an+1 | ≤ r |an+1 − an | ∀ n ∈ N for some r with 0 < r < 1, then it converges. In the proof we have observed that, (an ) satisfies r m−1 |a2 − a1 | |an − am | ≤ 1−r for all n, m ∈ N with n > m. Thus, |an − am | can be made arbitrarily small for all large enough n, m ∈ N. Now, we show that any sequence (an ) such that |an − am | can be made arbitrarily small for all large enough n, m ∈ N actually converges. First, let us formally define the requirement on the sequence. Definition 1.1.13 A a sequence (an ) is said to be a Cauchy sequence3 if for every ε > 0, there exists N ∈ N such that |an − am | < ε ∀ n, m ≥ N .

♦

We have already observed in Remark 1.1.15 that if (an ) converges, then it need not satisfy the assumption in Theorem 1.1.14. However, we have the following theorem. Theorem 1.1.15 Every convergent sequence is a Cauchy sequence. 3

Augustin-Louis Cauchy (21 August 1789 – 23 May 1857) was a French mathematician who made many contributions to calculus, specifically in terms of its rigorous foundation.

1.1 Sequence of Real Numbers

41

Proof Suppose (an ) converges to a. Let ε > 0 be given. Then we know that there exists N ∈ N such that |an − a| < ε/2 for all n ≥ N . Hence, we have |an − am | ≤ |an − a| + |a − am | < ε ∀ n, m ≥ N . This completes the proof. Now, we show that the converse of Theorem 1.1.15 is also true. The idea of the proof is akin to the idea used in the proof of Theorem 1.1.14, namely, we first show that (an ) is a bounded sequence, so that by Bolzano–Weierstrass theorem (Theorem 1.1.13), (an ) has a subsequence which converges to some a, and then show that (an ) itself converges to a. Theorem 1.1.16 (Cauchy criterion) Every Cauchy sequence of real numbers converges. Proof Let (an ) be a Cauchy sequence. Taking ε = 1, there exists N ∈ N such that |an − am | < 1 ∀ n, m ≥ N . In particular, for all n ≥ N , |an | ≤ |(an − a N ) + a N | ≤ |an − a N | + |a N | ≤ 1 + |a N |. Therefore, |an | ≤ max{|a1 |, |a2 |, . . . , |a N |, 1 + |a N |}. Thus, (an ) is a bounded sequence. As already mentioned, by Bolzano–Weierstrass theorem (Theorem 1.1.13), (an ) has a subsequence (akn ) which converges to some a. Now, let ε > 0 be given. Then there exist positive integers N1 , N2 such that |akn − a| < ε/2 ∀ n ≥ N1 , |an − akn | < ε/2 ∀ n ≥ N2 . Therefore, |an − a| ≤ |an − akn | + |akn − a| < ε/2 + ε/2 = ε for all n ≥ N3 := max{N1 , N2 }. Thus, an → a. This completes the proof. An alternative proof of Theorem 1.1.16. Let (an ) be a Cauchy sequence and ε > 0 be given. Let n ε ∈ N be such that |an − am | < ε for all n, m ≥ n ε . In particular, |an − an ε | < ε for all n ≥ n ε . Hence, for any ε1 , ε2 > 0, a n ε1 − ε 1 < a n < a n ε2 + ε 2

∀ n ≥ max{n ε1 , n ε2 }.

(1.1)

From this, we see that the set {an ε − ε : ε > 0} is bounded above. Let a := sup{an ε − ε : ε > 0}. Then, by (1.1), we have

42

1 Sequence and Series of Real Numbers

an ε − ε ≤ a ≤ an ε + ε

∀ ε > 0.

(1.2)

From (1.1) and (1.2), we obtain |an − a| ≤ ε for all n ≥ n ε . Thus, in view of Remark 1.1.4, (an ) converges to a. Remark 1.1.17 The above alternative proof of Theorem 1.1.16, which does not directly make use of Bolzano-Weierstrass theorem, was suggested by my colleague Professor P. Veeramani. ♦ Example 1.1.36 Let sn = 1 + 21 + . . . + n1 for n ∈ N. We have seen in Example 1.1.27 that 1 1 1 1 s2n − sn = + + ··· + ≥ n+1 n+2 2n 2 for every n ∈ N. This also shows that (sn ) is not a Cauchy sequence, and hence it diverges. ♦ Exercise 1.1.22 Suppose f is a function defined on an interval J such that | f (x) − f (y)| ≤ r |x − y| ∀ x, y ∈ J, for some r with 0 < r < 1. For a given a0 ∈ J , let (an ) be defined by a1 = f (a0 ), an+1 := f (an ) ∀ n ∈ N. Show that (an ) is a Cauchy sequence. Show also that the limit of the sequence (an )  is independent of the choice of a0 . Remark 1.1.18 We know that every Cauchy sequence of real numbers converges to a real number. However, we cannot expect a Cauchy sequence of rational numbers to have a rational limit. We have seen many such examples. If S is a subset of R such that limit of every convergent sequence in S belongs to S, then we say that S is a closed subset of R. Thus, the set of rational numbers is not a closed set. Similarly, intervals (a, b], [a, b), (a, b) for a, b ∈ R with a < b and intervals (a, ∞), (−∞, b) for any a, b ∈ R are not closed subsets of R, whereas for any a, b ∈ R with a < b, the interval [a, b] is a closed set. To see that [a, b] is a closed set let (u n ) be a sequence in [a, b] that converges to some u ∈ R. If u ∈ / [a, b], then we can find an ε > 0 such that (u − ε, u + ε) ∩ [a, b] = ∅. As u n → u, there exists N ∈ N such that u n ∈ (u − ε, u + ε) for all n ≥ N . This is not possible as (u n ) is a sequence in [a, b]. It can be shown that if S ⊆ R, then 1. S is a closed set if and only if S contains all those points x ∈ R such that (x − ε, x + ε) ∩ S = ∅ for every ε > 0; 2. S is not a closed set if and only if there exists x ∈ R \ S such that (x − ε, x + ε) ∩ S = ∅ for every ε > 0.

1.1 Sequence of Real Numbers

43

Exercise 1.1.23 Justify the above two statements.



A subset S of R is called an open set if for every x ∈ S, there exists ε > 0 such that (x − ε, x + ε) ⊆ S. It can be shown that if S ⊆ R, then S is open if and only if its complement R \ S is closed. Exercise 1.1.24 Justify the above statement.



In mathematics and its applications, the notion of convergence, closed sets and open sets are defined not only for numbers but also for more general objects such as vectors and functions (see, e.g., Nair [8]). We shall also encounter some such situations in the later chapters of this book. ♦

1.2 Series of Real Numbers In the last section we have come across sequences whose terms involve some other sequences. For example, we had sequences such as (an ) with 3 (i) an = 10 + 1032 + · · · + 103n , (ii) an = 1 + 21 + · · · + n1 , (iii) an = 1 + 212 + · · · + n12 .

Recall that, the sequence in (i) and (iii) converge whereas the sequence in (ii) diverge. In (i), we have also seen that the sequence converges to 13 . Because of this we may represent the number 13 as 3 3 1 = + 2 + ··· . 3 10 10 More generally, we may have a sequence (an ) of real numbers, and we may form a new sequence (sn ) by defining its n th term as sum of the first n terms of (an ), that is, sn = a 1 + · · · + a n . Then we may enquire whether this new sequence converges or not. In case this sequence (sn ) converge, then we may write its limit as a1 + a2 + · · · . This expression has a special name!

44

1 Sequence and Series of Real Numbers

Definition 1.2.1 A series of real numbers is an expression of the form a1 + a2 + a3 + . . . , or more compactly,

∞

an ,

n=1

where (an ) is a sequence of real numbers. The number an is called the n th term of the series and the sum of the first n terms of (an ), that is, sn := a1 + · · · + an , is called the n th partial sum of the series.

♦

Remark 1.2.1 Some authors denote a series as the sequence (an , sn ), where sn is ♦ n th partial sum of the sequence (an ). Example 1.2.1 Following are some examples of series: 3 (i) 10 + 1032 + · · · , (ii) 1 + 21 + 13 + · · · , (iii) 1 + 212 + 312 + · · ·

Note that in (i), (ii), (iii) above the partial sums are n n n

3 1 1 , , , 10k k k2 k=1 k=1 k=1

♦

respectively.

1.2.1 Convergence and Divergence of Series The following definition is on expected lines (Fig. 1.9):  Definition 1.2.2 A series ∞ n=1 an is said to be a convergent series if the corre) of partial sums converges. If sn → s, then we say that the sponding sequence (s n  a converges to s, and s is called the sum of the series, and we write series ∞ n n=1 this fact as ∞

s= an . n=1

1.2 Series of Real Numbers

45

Fig. 1.9 sn =  2  n 1 + 23 + 23 + · · · + 23

A series which does not converge is called a divergent series.

♦

Observe the following: Suppose an ≥ 0 for all n ∈ N. Then the sequence (sn ) of partial sums of the series ∞ n=1 an is monotonically increasing. Hence, in this case, either (sn ) converges or sn → ∞. Example 1.2.2 Consider the three series given in Example 1.2.1, i.e., ∞ ∞ ∞

3 1 1 , , . 10n n n2 n=1 n=1 n=1

Note that the partial sums of these series, say (sn(1) ), (sn(2) ), (sn(3) ) with sn(1) :=

n n n

3 1 1 (2) (3) , s , s := := , n n k 10 k k2 k=1 k=1 k=1

are the sequences considered in Examples 1.1.5, 1.1.19, 1.1.24, respectively, and we 3 have seen that (sn(1) ) and (sn(3) ) converge, whereas (sn(2) ) diverges. Thus, ∞ n=1 10n ∞ 1 ∞ 1 and n=1 n 2 are convergent series, whereas n=1 n is a divergent series. ♦ Example 1.2.3 Consider the geometric series 1 + q + q2 + · · · for q ∈ R. We show that this series converges if and only if |q| < 1: Note that n if q = 1, sn = 1 + q + · · · + q n−1 = (1 − q n )/(1 − q) if q = 1.

46 Fig. 1.10 sn = 1 +

1 Sequence and Series of Real Numbers

1 2

+

1 22

+ ··· +

1 2n

Thus, if q = 1, then (sn ) is not bounded; hence not convergent (Fig. 1.10). If q = −1, then we have 1 if n odd, sn = 0 if n even. Thus, (sn ) diverges for q = −1 as well. Now, let |q| = 1. Then sn =

1 qn 1 − qn = − . 1−q 1−q 1−q

Recall that (q n ) converges if and only if |q| < 1, and in that case q n → 0. Hence, (sn ) converges if and only if |q| < 1, and in that case lim sn =

n→∞

1 . 1−q

 1 n−1 In particular, if |q| < 1, then ∞ converges with its sum 1−q , and if |q| ≥ 1, n=1 q ∞ n−1 then n=1 q diverges. Further, we have the error estimate corresponding to the partial sums as  |q|n 1   . ♦ sn − = 1−q |1 − q| Here is a necessary condition for the convergence of a series, which can help to infer the divergence of certain series.  Theorem 1.2.1 (A necessary condition) If ∞ n=1 an converges, then an → 0 as n → ∞. The converse is not true.  Proof Suppose the series ∞ n=1 an converges to s, that is, sn → s, where sn is the n th partial sum of the series. Since an = sn − sn−1 for n = 2, 3, . . ., we have an = sn − sn−1 → 0 as n → ∞. Thus, an → 0.

1.2 Series of Real Numbers

47

To see that the converse does not hold, consider the series an = 1/n → 0, but the series diverges.

∞ n=1

1/n. In this case,

Exercise 1.2.1 Prove the following.  n (i) The series ∞ n=1 n+1 diverges.  (ii) If an+1 > an > 0 for all n ∈ N, then ∞ n=1 an diverges.



The following theorem shows that if we remove the first few terms from a convergent (respectively, divergent) series, then the resulting series remain convergent (respectively, divergent). ∞ Theorem ∞1.2.2 For each k ∈ N, the series n=1 an converges if and only if the series n=1 ak+n converges, and if they converge, then ∞

an = (a1 + · · · + ak ) +

n=1

Proof Let k ∈ N. Let sn =

∞

ak+n .

n=1

n i=1

ai and sn =

n i=1

ak+i . Then we have

sn = sk+n − (a1 + · · · + ak ) ∀ n ∈ N. Recall that a sequence (bn ) converges to b if and only if, for any k ∈ N, the sequence (bk+n ) converges to b. Hence, from the above relation between (sn ) and (sn ) we obtain sn → s ⇐⇒ sn → s − (a1 + · · · + ak ). Thus, we have proved the theorem. Modifying the arguments ∞in the proof of the above theorem, it follows that, if ∞ b is obtained from n n=1 n=1 an by omitting or adding a finite number of terms, then ∞ ∞

an converges ⇐⇒ bn converges. n=1

n=1

The following theorem is an immediate corollary to Theorem 1.2.2. Theorem 1.2.3 Suppose (a that for some k ∈ N, n ) and (bn ) are sequences such  ∞ a converges if and only if an = bn for all n ≥ k. Then ∞ n n=1 n=1 bn converges. ∞  Theorem 1.2.4 (Sum-rule) Suppose n=1 an converges to s and ∞ n=1 bn converges to s . Then for any α, β ∈ R, ∞

(α an + β bn ) converges to α s + β s . n=1

48

1 Sequence and Series of Real Numbers

In particular, any c ∈ R.

∞

n=1 (an

+ bn ) converges to s + s and

∞ n=1

c an converges to cs for

Proof Let sn(1) , sn(2) and sn be the n th partial sums of the series ∞ and n=1 (α an + β bn ) respectively. Then we obtain sn =

∞ n=1

an ,

∞ n=1

bn

n n n

(α ak + β bk ) = α ak + β β bk = αsn(1) + βsn(2) ∀ n ∈ N. k=1

k=1

k=1

Since sn(1) → s and sn(2) → s , we obtain sn = αsn(1) + βsn(2) → αs + βs . Exercise 1.2.2 As proof of Theorem 1.2.4, suppose we write ∞

(α an + β bn ) = α

∞

n=1

an + β

n=1

∞

bn = αs + βs .

n=1

What is wrong with it?

n



Theorem 1.2.5 Let (an ) be such that an → 0 and let sn := k=1 ak for n ∈ N. Then for x ∈ R, sn → x ⇐⇒ s2n → x ⇐⇒ s2n−1 → x. Proof Clearly, if sn → x then s2n → x and s2n−1 → x, as (s2n ) and (s2n−1 ) are subsequences of (sn ). Next, we note that s2n = s2n−1 + a2n and s2n+1 = s2n + a2n+1 ∀ n ∈ N. Hence, s2n → x if and only if s2n−1 → x, and in that case, by Theorem 1.1.11 , sn → x.

1.2.2 Some Tests for Convergence Theorem 1.2.6 (Comparison test) Let 0 ≤ an ≤ bn for all n ∈ N. Then ∞

n=1

bn converges

⇒

∞

an converges .

n=1

 ∞ Proof Suppose sn and sn are the n th partial sums of the series ∞ n=1 an and n=1 bn , respectively. By the assumption, we have 0 ≤ sn ≤ sn for all n ∈ N, and both (sn ) monotonically increasing. and (sn ) are

Suppose ∞ n=1 bn converges, that is, (sn ) converges. Then, (sn ) is bounded. Hence,

by the relation 0 ≤ sn ≤ sn for all n ∈ N, (sn ) is bounded as well as monotonically increasing. Therefore, by Theorem 1.1.9, (sn ) converges. The following corollary is immediate from the above theorem.

1.2 Series of Real Numbers

49

Corollary 1.2.7 ( Comparison test) Let 0 ≤ an ≤ bn for all n ∈ N. Then ∞

an diverges

n=1

⇒

∞

bn diverges.

n=1

Corollary 1.2.8 Suppose (an ) and (bn ) are sequences of positive terms. an (i) Suppose  := lim exists. n→∞ bn  ∞ (a) If  > 0, then ∞ n=1 bn converges ⇐⇒ n=1 an converges.  ∞ b converges ⇒ (b) If  = 0, then ∞ n=1 n n=1 an converges.  ∞ an (ii) Suppose lim = ∞. Then ∞ n=1 an converges ⇒ n=1 bn converges. n→∞ bn an Proof (i) Assume that  := lim exists, i.e., 0 ≤  < ∞. n→∞ bn (a) Suppose  > 0. Then for any 0 < ε <  there exists n ∈ N such that 0≤−ε
0 be given. Then, there exists n ∈ N such that −ε < abnn < ε for all n ≥ N . In particular, an < εbn ∀ n ≥ N . ∞ Again, by comparison ∞ test (Theorem 1.2.6), convergence of n=1 bn implies the convergence of n=1 an . (ii) Assume that abnn → ∞. Then there exists N ∈ N such that abnn ≥ 1 for all n ≥ N , i.e., bn ≤ an ∀ n ≥ N . Hence, comparison test can be applied in this case as well to obtain the required result.  Example 1.2.4 We have seen that the sequence (sn ) with sn = nk=1 k!1 converges. This also follows from comparison test, since 1 1 ≤ n−1 ∀ n ∈ N n! 2 and

∞

1 n=1 2n−1

converges.

♦

with an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, The, using Example 1.2.5 Let (an ) be   n ∈ aN. n the sequence (sn ) with sn = nk=1 ak /10k , we see that the series ∞ n=1 10n converges. This also follows by comparison test, since

50

1 Sequence and Series of Real Numbers

an 9 ≤ n ∀n ∈ N n 10 10 ∞ 1  an and n=1 10n converges. Note that if s = ∞ n=1 10n , then the series represents the decimal expansion of s. By similar arguments, we can assert the following (Verify): For any k ∈ N, an ∈ {0, 1, . . . , k − 1}, n ∈ N

⇒

∞

an n=1

kn

converges.

 an Suppose s = ∞ n=1 k n . If k = 2, then the series represents the binary expansion of s and if k = 3, then the series represents the ternary expansion of s. ♦  1 Example 1.2.6 (i) Since √1n ≥ n1 for all n ∈ N, and since the series ∞ n=1 n diverges, ∞ 1 by comparison test, the series n=1 √n also diverges. More generally, let p ≤ 1. Since 1 1 ≥ ∀ n ∈ N, np n by comparison test,

∞

1 diverges for p ≤ 1. p n n=1

 1 1 (ii) Let p ≥ 2. We know that ∞ k=1 n 2 converges (Example 1.1.24). Since n p ≤ for all n ∈ N, by comparison test, the series ∞

1 converges for p ≥ 2. np n=1

1 n2

♦

Now, we use comparison test to show the convergence of certain sequences with positive and negative terms. Example 1.2.7 We show that the series 1−

1 1 1 1 1 + − + ··· + − + ··· 2 3 4 2n − 1 2n

converges. For this let sn be the n-th partial sum of this series. Then we see that s2n+1 = s2n + where

1 ∀ n ∈ N, 2n − 1

1.2 Series of Real Numbers

51

 1  1 1 1 1 + − + ··· + − s2n = 1 − 2 3 4 2n − 1 2n 1 1 1 = + + ··· + 1×2 3×4 (2n − 1) × 2n 1 1 1 ≤ 2 + 2 + ··· + 1 3 (2n − 1)2 ≤ σ2n−1 ,  2 where σn is the n-th partial sum of the convergent series ∞ n=1 (1/n ) (cf. Example 1.1.24). Hence, by the comparison test (Theorem 1.2.6) (s2n ) converges, and hence by Theorem 1.2.5, (sn ) also converges. Thus we have proved that the given series converges. ♦ Example 1.2.8 The series 1 1 1 (−1)n+1 + − + ··· + + ··· , 3 5 7 2n − 1

(1.1)

(−1)n+1 1 1 1 1 − + − + ··· + + ··· , 2 4 6 8 2n

(1.2)

1− and

converge: We observe that, if sn is the n-th partial sum of the series in (1.1), then (−1)2n+1 1 1 1 s2n = 1 − + − + · · · + 3 5 7 4n − 1  1  1  1 1 1 + − + ··· + − = 1− 3 5 7 4n − 3 4n − 1 2 2 2 = + + ··· + 1×3 5×7 (4n − 3) × (4n − 1) 2 2 2 ≤ 2 + 2 + ··· + ≤ σ4n−3 , 1 5 (4n − 3)2  2 where σn is the n-th partial sum of the convergence series ∞ n=1 (1/n ) (cf. Example 1.1.24). Hence, by the comparison test (Theorem 1.2.6) (s2n ) converges, and hence by Theorem 1.2.5, (sn ) also converges. Thus we have proved that the given series in (1.1) converges. To see the convergence of the series in (2), we first observe that this series is same as ∞ 1 (−1)n+1 . 2 n=1 n We already know that the series 1 ∞ (−1)n+1 also converges. n=1 2 n

∞ n=1

(−1)n+1 n

converges (cf. Example 1.2.7). Hence ♦

52

1 Sequence and Series of Real Numbers

 1 In the following example we show the convergence of the series ∞ n=1 n p for any p > 1. We shall use the familiarity of the integral to the extent that the reader knows the result:  k+1 b b x bk+1 − a k+1 k for k = −1. x dx = = k+1 a k+1 a

 1 Example 1.2.9 Consider the series ∞ n=1 n p for p > 1. To discuss this general case, consider the function f (x) := 1/x p , x ≥ 1. Note that, for each k ∈ N, we have k−1≤ x ≤k

⇒

1 1 ≤ p p k x

⇒

k

1 ≤ kp

dx . xp

k−1

Hence,

 n n 

1 1 dx dx n 1− p − 1 ≤ . ≤ = = p p p k x x 1− p p−1 k=2 k=2 n

k

k−1

Thus, sn :=

1

n

p 1 1 +1= . ≤ p k p−1 p−1 k=1

Hence, (sn ) is monotonically increasing and bounded above, and hence (sn ) converges. Thus, ∞

1 converges for p > 1. np n=1 A more general result on convergence of series in terms of integrals, known as integral test, will be considered in Chap. 4. ♦ Example 1.2.10 In this example, we make use of some simple properties of the logarithm function log x, x > 0, namely, the relation log x k = k log x for x ≥ 1, for showing the divergence of the series ∞

1 . (n + 1) log(n + 1) n=1

1.2 Series of Real Numbers

53

1 Let an = n log for n = 2, 3, . . ., and sn = n an+1 for all n ∈ N, we have

n k=2

ak . Then, using the fact that an ≥

s2 n = a 2 + a 3 + · · · + a 2 n = a2 + (a3 + a4 ) + (a5 + a6 + a7 + a8 ) + · · · + (a2n−1 +1 + · · · a2n ) ≥ a2 + 2a4 + 22 a8 + · · · + 2n−1 a2n . Thus, using the relation a2 k = we have s2 n ≥

2k

n

1 1 , = k k log 2 2 k log 2

k=2

 Since nk=2 diverges.

1 k

1 1 . 2 log 2 k=2 k n

2k−1 a2k ≥

→ ∞, by comparison test, s2n → ∞. Hence, the sequence (sn ) also ♦

Another consequence of the comparison test is the following. Theorem 1.2.9 Let (an ) be a sequence of positive numbers. If there exists r with 0 < r < 1 and a positive integer N such that at least one of the following conditions is satisfied: (i)

an+1 an 1/n an

≤ r ∀ n ≥ N,

≤ r ∀ n ≥ N.  Then the series ∞ n=1 an converges.

(ii)

Proof Let 0 < r < 1. First, let N ∈ N be such that an+1 ≤ r ∀ n ≥ N. an Then an+1 ≤ ran for all n ≥ N so that an+1 ≤ r n−N +1 a N =

 a  N rn ∀ n ≥ N. N r −1

 ∞ n Since ∞ n=1 r converges, the comparison test shows that, the series n=1 an converges. Next, let N ∈ N be such that an1/n ≤ r ∀ n ≥ N . Then an ≤ r n for all n ≥ N . Again, since  ∞ n=1 an converges.

∞

n=1 r

n

converges, by comparison test,

54

1 Sequence and Series of Real Numbers

a

n+1



1/n

or (an ) rather than an finding an upper bound. In such cases, the following two tests due to d’Alembert4 and Cauchy, respectively, are useful to decide the convergence of a series. Sometimes it may be easier to find the limit of either

Theorem 1.2.10 (D’Alembert’s ratio test) Suppose (an ) is a sequence of positive an+1 =  exists. terms such that lim n→∞ an  (i) If  < 1, then the series ∞ n=1 an converges.  (ii) If  > 1, then an → 0; in particular, ∞ n=1 an diverges. Proof (i) Suppose  < 1. Let r be such that  < r < 1. Then there exists N ∈ N such that an+1 ≤ r ∀ n ≥ N. an  Hence, by Theorem 1.2.9, the series ∞ n=1 an converges. (ii) Let  > 1. Then there exists N ∈ N such that an+1 > 1 ∀ n ≥ N. an Hence, an+1 > an for all n ≥ N . Therefore, an → 0. Theorem 1.2.11 (Cauchy’s root test) Suppose (an ) is a sequence of positive terms such that lim an 1/n =  exists. n→∞  (i) If  < 1, then the series ∞ n=1 an converges.  (ii) If  > 1, then an → 0; in particular, ∞ n=1 an diverges. Proof (i) Suppose  < 1. Let r be such that  < r < 1. Then there exists N ∈ N such that an 1/n ≤ r ∀ n ≥ N .  Hence, by Theorem 1.2.9, the series ∞ n=1 an converges. (ii) Let  > 1. Then there exists N ∈ N such that an 1/n > 1 ∀ n ≥ N . Hence, an ≥ 1 for all n ≥ N . Therefore, an → 0. Remark 1.2.2 We remark that both d’Alembert’s ratio test and Cauchy’s root test are silent for the case  = 1. In this case, we cannot assert either way. For example, 4

Jean-Baptiste le Rond d’Alembert (16 November 1717 – 29 October 1783) was a French mathematician, physicist and philosopher. A particular method of solution of wave equation is named after him - courtsey Wikipedia.

1.2 Series of Real Numbers

we know that the series the cases, we have

55

∞

1 n=1 n

diverges whereas

∞

1 n=1 n 2

converges, and in both

an+1 → 1, an1/n → 1. an

However, for such cases, we may be able to infer the convergence or divergence by some other means. ♦ Example 1.2.11 Let us test the convergence of the series ∞

n2 n=1

(i)

∞

n2 n=1

2n

: In this case an =

n2 2n

2n

and

∞

n! . 2n n=1

so that

(n + 1)2 /2n+1 1  n + 1 2 1 an+1 = = → . 2 n an n /2 2 n 2 ∞

n! Hence, by d’Alembert’s ratio test, the series converges. (ii) : In this case n 2 n=1

an =

n! 2n

so that an+1 (n + 1)!/2n+1 n+1 → ∞. = = an n!/2n 2 ♦

Hence, by d’Alembert’s ratio test, the series diverges. Example 1.2.12 For x ∈ R, the series

∞

|x|n n=1

n!

converges:

Clearly the series converges if x = 0. For x = 0, let an =

|x|n . n!

Then we have

an+1 |x| → 0. = an n+1 Hence, by d’Alembert’s ratio test, the series converges. We shall also show that the ∞

xn series converges for any x ∈ R. ♦ n! n=1 Example 1.2.13 The series

∞ 

n=1

n n converges: In this case, we have 2n + 1

an 1/n =

1 n → < 1. 2n + 1 2

56

1 Sequence and Series of Real Numbers

Hence, by Cauchy’s root test, the series converges. The convergence of the above series can also be proved using comparison test, since n  n n  1 1 = ≤ n an = 2n + 1 2 + 1/n 2 ∞

1 for all n ∈ N and converges. n 2 n=1

♦

Example 1.2.14 Consider the series

∞

n=1

lim

n→∞

1 . In this case, we have n(n + 1)

an+1 = 1 = lim an 1/n . n→∞ an

Hence, we are not in a position to apply ratio test and root test. However, sn :=

n

k=1

1 1 1 1  =1− = − → 1. k(k + 1) k k+1 n+1 k=1 n

♦

Thus, the series converges to 1. Exercise 1.2.3 Assert the convergence of the series and Example 1.2.13 by showing that

∞ n=1

an in Example 1.2.11(i)

an+1 8 1 ∀ n ≥ 3 and an1/n ≤ ∀ n ∈ N, ≤ an 9 2 

respectively.

 Exercise 1.2.4 Assert the divergence of the series ∞ n=1 an in Example 1.2.11(ii) by showing either an+1 ≥ 1 ∀ n ∈ N or an → ∞.  an Exercise 1.2.5 Show the convergence of the series ∞ 1 n=1 n(n+1) .  Example 1.2.15 Consider the series ∞ n=1 an with a2n−1 = for n ∈ N. Note that

1 6n−1

, a2n =

a2n 1 = , a2n−1 3

∞

1 n=1 n 2

1 3 × 6n−1

a2n+1 1 = a2n 2

by comparing it with 

1.2 Series of Real Numbers

57

an+1 does not exist. However, an  that by Theorem 1.2.9, the series ∞ n=1 an converges.  Example 1.2.16 Consider the series ∞ n=1 an with

for all n ∈ N so that lim

n→∞

an =

≤ 1/2 for all n ∈ N so ♦

⎧ n n ⎨ ( 2n+1 ) , n odd, ⎩

Note that 1/(2n+1) lim a n→∞ 2n+1

an+1 an

= lim

n→∞

1 , 3n

n even.

1 2n + 1 = , 4n + 3 2 1/n

so that lim an1/n does not exist. However, an n→∞  orem 1.2.9, the series ∞ n=1 an converges.

1/2n lim a n→∞ 2n

=

1 3

≤ 1/2 for all n ∈ N. Hence, by The♦

We close this subsection with another application of the comparison test (Theorem 1.2.6). Theorem 1.2.12 Let (an ) be a sequence of non-negative numbers, and let (bn ) be a sequence obtained from (an ) by a rearrangement of the terms of (an ). Then ∞

an converges ⇐⇒

n=1

∞

bn converges,

n=1

and in that case, both the series have the same limit.

 ∞ Proof Suppose sn and tn are the n-th partial sums of ∞ n=1 an and n=1 bn , respectively, that is, n n

ai , tn = bi sn = i=1

i=1

for all n ∈ N. Suppose sn → s. Since (bn ) is a rearrangement of the terms of (an ), for each n ∈ N, there exists m ≥ n such that tn ≤ sm . Then we have tn ≤ s. This shows that (tn ) converges, say to t, and we obtain t ≤ s. Similarly, changing the roles of (sn ) and (tn ), we see that convergence of (tn ) to some t implies (sn ) converges to some s and s ≤ t. Thus, we obtain s = t, and this completes the proof.

1.2.3 Alternating Series In the last subsection we have described some tests for asserting the convergence or divergence of series of non-negative terms. In this subsection we provide a sufficient condition for convergence of series with alternatively positive and negative terms.

58

1 Sequence and Series of Real Numbers

 n+1 Definition 1.2.3 A series of the form ∞ u n , where (u n ) is a sequence of n=1 (−1) positive terms, is called an alternating series. ♦ We have seen in Example 1.2.7 that the alternating series 1−

(−1)n+1 1 1 1 1 + − + + ··· + + ··· 2 3 4 5 n

(1.1)

is convergent. Note that the series 1+

1 1 1 1 1 1 1 − − + ··· + + − − + ··· 2 3 4 4n − 3 4n − 2 4n − 1 4n

(1.2)

is not an alternating series. As you can see, in the latter case, though the series in not an alternating series, it is of the form ∞

[(−1)n+1 u n + (−1)n+1 vn ] n=1

with un =

1 1 , vn = . 2n − 1 2n

 n+1 We know results in Example 1.2.8 that the alternating series ∞ un n=1 (−1) ∞ from the n+1 and n=1 (−1) vn are convergent. Hence, we can assert the convergence of the n+1 u n + (−1)n+1 vn ]. original series ∞ n=1 [(−1) The next theorem, due to Leibnitz,5 provides such a sufficient condition for the convergence of alternating series. Theorem 1.2.13 (Leibnitz’s theorem) Suppose (u n ) is a sequence of positive terms for all n ∈ N and u n → 0 as n → ∞. Then the alternating such that  u n ≥ u n+1 n+1 u n converges, and in that case, series ∞ n=1 (−1) |s − sn | ≤ u n+1 where sn =

∀ n ∈ N,

n ∞

(−1) j+1 u j and s = (−1)n+1 u n , j=1

n=1

Proof We observe that s2n+1 = s2n + u 2n+1 ∀ n ∈ N. 5

Gottfried Wilhelm von Leibnitz (July 1, 1646 – November 14, 1716) was a German mathematician and philosopher. He developed the infinitesimal calculus independently of Isaac Newton - courtsey Wikipedia.

1.2 Series of Real Numbers

59

Since u n → 0 as n → ∞, it is enough to show that (s2n ) converges, and in that case, both (s2n ) and (s2n+1 ) converge to the same limit. Note that, for every n ∈ N, s2n+2 = s2n + (u 2n+1 − u 2n+2 ) ≥ s2n and s2n = u 1 − (u 2 − u 3 ) − . . . (u 2n−2 − u 2n−1 ) − u 2n ≤ u 1 . increasing and bounded above. Therefore (s2n ) conHence, (s2n ) is monotonically  n+1 (−1) u n converges. verges. Thus, the series ∞ n=1 To obtain the remaining part, note also that, for n ∈ N, s2n+1 = s2n−1 − (u 2n − u 2n+1 ) ≤ s2n−1 , so that {s2n−1 } is a monotonically decreasing sequence. Thus, s2n−1 = s2n + u 2n ≤ s + u 2n ,

s ≤ s2n+1 = s2n + u 2n+1 ,

s ≤ s2n−1 , s2n ≤ s. Hence, s2n−1 − s ≤ u 2n ,

s − s2n ≤ u 2n+1 ,

Consequently, |s − sn | ≤ u n+1

∀ n ∈ N.

This completes the proof. Corollary 1.2.14 Suppose (an ) is a sequence of positive terms such that an ≥ an+1 for all n ∈ N and an → 0 as n → ∞. Then the series a1 + a2 − a3 − a4 + · · · + a4n−3 + a4n−2 − a4n−1 − a4n + · · · converges. Proof The given series can be written as ∞

[(−1)n+1 u n + (−1)n+1 vn ], n=1

where (u n ) and (vn ) are decreasing sequences of positive terms such that u n → 0 and vn → 0. Hence, by Theorem 1.2.13 can be applied to obtain the conclusion. Remark 1.2.3 The relation |s − sn | ≤ u n+1 in Theorem 1.2.13 shows the rate of convergence of the partial sums to the sum of the series. In particular, for a given

60

1 Sequence and Series of Real Numbers

k ∈ N, if n is large enough such that u n < 1/10k , then first k decimal places of sn and s are the same. ♦ The following example shows that the condition u n ≥ u n+1 for all n ∈ N in Theorem 1.2.13 is not redundant, that is, the conclusion in the theorem need not hold if the above condition is dropped. Example 1.2.17 Consider the series 1 1 1 1 1 1 −√ + ··· −√ +√ −√ + ··· + √ √ n−1 n+1 3−1 3+1 4−1 4+1  n+1 which is of the form ∞ u n with n=1 (−1) u 2n−1 = √

1 n+2−1

, u 2n = √

1 n+2+1

.

Note that, in this case, we have u n ≥ 0 for all n ∈ N and u n → 0. However, the series diverges. To see this, we observe that s2n =

n 

k=1

√

1 n+2−1

−√

1 n+2+1



=

n

k=1

2 → ∞. n+1

Thus, (sn ) has a subsequence that diverges. Therefore, (sn ) diverges. Note that (u n ) is not a decreasing sequence. ♦

1.2.4 Madhava-Nilakantha Series The convergence of the series 1−

(−1)n 1 1 1 + − + ··· + + ··· , 3 5 7 2n + 1

which is generally known as Leibnitz-Gregory series, was known to Indian mathematicians as early as in 15-th century, and the value of the above series was proved to be π4 . The above series appeared in the work of a Kerala mathematician Madhava around 1425 that was presented later in the year around 1550 by another Kerala mathematician Nilakantha (cf. [7]). The discovery of the above series is normally attributed to Leibnitz and James Gregory after nearly 300 years of its discovery. Respecting the chronology of its discovery, we shall refer this series as MadhavaNilakantha series.

1.2 Series of Real Numbers

61

Let us give a simple proof for the equality π 1 1 1 (−1)n+1 = 1 − + − + ··· + + ··· 4 3 5 7 2n − 1 using some elementary rules of integration that one studies in school, which we shall study in detail in Chap. 6. We know that (1 − r )(1 + r + + · · · + r n ) = (1 − r n+1 ) so that for r = 1, 1 r n+1 = 1 + r + · · · + rn + . 1−r 1−r Now, taking r = −x 2 we have 2n+2 1 2 4 6 n 2n n+1 x = 1 − x + x − x + · · · + (−1) x + (−1) . 1 + x2 1 + x2

On integration 

2n+1 x5 x7 dx x3 n x + − + · · · + (−1) + = x − 1 + x2 3 5 7 2n + 1

Now, recalling

1 0



(−1)n+1 x 2n+2 d x. 1 + x2

dx π = tan−1 (1) = , 1 + x2 4

we have 1 1 1 1 π = 1 − + − + · · · + (−1)n + 4 3 5 7 2n + 1

1 0

(−1)n+1 x 2n+1 d x. 1 + x2

Now, observe that 1  1 x 2n+2  1 2n+2 1   n+1 x . dx ≤ d x ≤ x 2n+2 d x =  (−1) 2 2 1+x 1+x 2n + 3 0

0

0

Thus, π  1 1 1 1 1   →0 ≤  − 1 − + − + · · · + (−1)n 4 3 5 7 2n + 1 2n + 3 Thus, we have proved that

62

1 Sequence and Series of Real Numbers ∞

π (−1)n =1+ . 4 2n + 1 n=1

Using the procedure used above and the fact that log 2 =

∞

(−1)n+1

n

n=1

1

dx 0 1+x

= log 2, we see that

.

Exercise 1.2.6 Derive the above series representation for log 2.



1.2.5 Absolute Convergence  ∞ We know that for a sequence (an ), the series ∞ n=1 an may converge, but n=1 |an | ∞ (−1)n+1  1 can diverge. For example we have seen that n=1 n converges whereas ∞ n=1 n diverges.  Definition 1.2.4 Let (an ) be a sequence of real numbers. Then the series ∞ n=1 an is said to be  (1) absolutely convergent, if ∞ n=1 |an | is convergent, (2) conditionally convergent, if it converges, but not absolutely. ♦ Example 1.2.18 (i) the series ∞

(−1)n n=1

n2

,

∞

(−1)n+1

n!

n=1

,

∞

sin n n=1

n2

are absolutely convergent, so also the series ∞

an n=1

n!

for any a ∈ R. (ii) We already observed in Sect. 1.2.4 that the series ∞

(−1)n+1 n=1

n

and

∞

(−1)n 2n − 1 n=1

are convergent series, which also from Leibnitz theorem, but they are not absolutely convergent. Thus, these series are conditionally convergent. ♦ From the above observations we conclude:

1.2 Series of Real Numbers

63

A convergent series need not converge absolutely.

However, we have the following theorem. Theorem 1.2.15 Every absolutely convergent series is convergent.  Proof Suppose ∞ is an absolutely convergent series. Let sn and sn be the n th n=1 an  ∞ ∞ partial sums of the series n=1 an and n=1 |an | respectively. Then, for n > m, we have n n  

  an  ≤ |an | = |sn − sm |. |sn − sm | =  j=m+1

j=m+1

Since, (sn ) converges, it is a Cauchy sequence. Hence, from the above relation it follows that (sn ) is also a Cauchy sequence. Therefore, by the Cauchy criterion, it converges.  Another proof without using Cauchy criterion.Suppose ∞ n=1 an is an absolutely convergent series. For each n ∈ N, let us write an as difference of two positive numbers, namely, an = (an + |an |) − |an |.  Since ∞ n=1 |an | converges, by the sum rule (Theorem 1.2.4), it is enough to show that ∞ n=1 (an + |an |) converges. This is true, since 0 ≤ an + |an | ≤ 2|an |  for all n ∈ N and ∞ n=1 2|an | converges, by applying the comparison test for series of non-negative terms (Theorem 1.2.6). In view of Theorem 1.2.15, the series ∞

xn converges for each x ∈ R. n! n=1

Remark 1.2.4 We shall prove in the next chapter that the value of the above sum is same as  x n , lim 1 + n→∞ n ♦

and it is generally denoted by exp(x) or e x . In view of Theorem 1.2.15 we can assert the following. Let (an ) be a sequence of nonzero numbers ∞ a 

 n+1  (i) If lim  an x n converges absolutely  =  exists and  < 1, then n→∞ an n=1 ∞

1/n (ii) If lim |an | =  exists and  < 1, then an x n converges absolutely. n→∞

n=1

64

1 Sequence and Series of Real Numbers

 ∞ Definition 1.2.5 By a rearrangement of a series ∞ n=1 an we mean a series n=1 bn obtained by rearranging the terms of ∞ a , in the sense that b = a , n ∈ N, n ϕ(n) n=1 n where ϕ is a bijection on N. ♦  Recall from Theorem 1.2.12 that if ∞ n=1 an is a series of non-negative terms, then this series converges if and only if any rearrangement of it converges and all rearranged series have the same sum. However, rearrangements of a conditionally convergent series need not converge to the same limit, as the following example shows.  Example 1.2.19 Consider the series ∞ n=1 an with an :=

(−1)n+1 , n ∈ N. n

We know that this series converges to s := log 2 (Sect. 1.2.4). Let us see if a rearrangement of it converges to a different number. For this, let us consider the following rearrangement: 1 1 1 1 1 1 1 1 − + − − + ··· + − − + ··· . 2 4 3 6 8 2n − 1 4n − 2 4n  Thus, the rearranged series is written as ∞ n=1 bn , with 1−

b3n−2 =

1 , 2n − 1

b3n−1 = −

1 , 4n − 2

b3n = −

for n ∈ N. Let sn and sn be the n th partial sums of the series respectively. Then we have

∞ n=1

1 4n an and

∞ n=1

1 1 1 1 1 1 1 1

− − = 1 − − + − − + ··· + s3n 2 4 3 6 8 2n − 1 4n − 2 4n  1  1 1 1 1 1 1 1 + − − + ··· + − − = 1− − 2 4 3 6 8 2n − 1 4n − 2 4n  1 1 1 1 1 1 − + − + ··· + − = 2 4 6 8 4n − 2 4n    1 1 1 1 1 1  1− + − + ··· + − = 2 2 3 4 2n − 1 2n   1 1 1 1 1 1 1 − + − + ··· + − = 2 2 3 4 2n − 1 2n 1 = s2n . 2

bn

1.2 Series of Real Numbers

65

Also, we have

= s3n + s3n+1

1 , 2n + 1

s3n+2 = s3n +

1 1 − . 2n + 1 4n + 2

Since, an → 0 as n → ∞, we obtain lim s n→∞ 3n

=

s , 2

lim s n→∞ 3n+1

=

s , 2

lim s n→∞ 3n+2

=

s . 2

Hence, we can infer that sn → s/2 as n → ∞.

♦

Not only that. Look at the following astonishing result!  Theorem 1.2.16 Suppose ∞ convergent series. Then, for n=1 an is a conditionally  a such that the rearranged series each s ∈ R, there exists a rearrangement of ∞ n n=1 converges to s. The proof of the above two theorem (Theorems 1.2.16 is quite involved, and hence we omit its proof. Interested readers may refer Delninger [4] for the proof.

1.3 Additional Exercises 1.3.1 Sequences 1. Let (an ) be a sequence of real numbers which converges to a, i.e., an → a as n → ∞. Prove: (a) |an − a| → 0 and |an | → |a| as n → ∞. (b) There exists k ∈ N such that |an | > |a|/2 for all n ≥ k. (c) If a = 0, then an = 0 for all large enough n and 1/an → 1/a. 2. Prove that (a) an → a if and only if for every open interval I containing a, there exists a positive integer N (depending on I ) such that an ∈ I for all n ≥ N . (b) an → a if and only if there exists an open interval I containing a such that infinitely may an ’s are not in I . 3. In each of the following, establish the convergence or divergence of the sequence (an ), where an is: (i)

(−1)n n+1

(ii)

2n , 2 3n + 1

(iii)

2n 2 + 3 . 3n 2 + 1

4. Suppose an → a and an ≥ 0 for all n ∈ N. Show that a ≥ 0 and

√ √ an → a.

66

1 Sequence and Series of Real Numbers

5. Let 0 < a < 1. If bn > 0 for all n ∈ N and bbn+1 →  with 0 ≤  < 1/a, then n show that bn a n → 0. 6. Let (an ) be a sequence defined recursively by an+2 = an+1 + an for n ∈ N with a1 = a2 = 1. Show that √ an → ∞. √ √ 7. For n ∈ N, let an = n + 1 − n. Show that (an ) and ( nan ) are convergent sequences. Find theirlimits. 1 . Show that (xn ) is convergent. 8. For n ∈ N, let xn = nk=1 n+k 9. Prove the following: (a) If (an ) is increasing and unbounded, then an → +∞. (b) If (an ) is decreasing and unbounded, then an → −∞. 10. If every subsequence of (an ) has at least one subsequence which converges to x, then (an ) also converges to x. 11. Suppose (an ) is an increasing sequence. Prove the following. (a) If (an ) has a bounded subsequence, then (an ) is convergent. (b) If (an ) does not diverge to +∞, then (an ) has a subsequence which is bounded above. (c) If (an ) is divergent, then an → +∞. 12. If (an ) is a sequence with positive terms, prove that an → 0 ⇐⇒

an → 0. 1 + an

√ 13. Let a1 = 1 and an+1 = 2 + an for all n ∈ N. Show that (an ) converges. Also, find its limit. 14. Let a1 = 1 and an+1 = 14 (2an + 3) for all n ∈ N. Show that (an ) is monotonically increasing and bounded above. Find its limit. an for all n ∈ N. Show that (an ) converges. Find its 15. Let a1 = 1 and an+1 = 1 + an limit. 16. Suppose (an ) is a sequence such that the subsequences (a2n−1 ) and (a2n ) converge to the same limit, say a. Show that (an ) also converges to a. 17. Let (an ) be a monotonically increasing sequence such that (a3n ) is bounded. Is (an ) convergent? Why? 18. Let (an ) be defined by 1 a1 = 0, an+1 = an + (1 − 2an2 ), n = 1, 2, . . . . 4 √ Show that (an ) converges to 1/ 2. 19. Give an example in support of the statement: If (an ) is a sequence such that an+1 − an → 0 as n → ∞, then (an ) need not converge. 20. For a > 0, let an − 1 , n ∈ N. bn = n a +1

1.3 Additional Exercises

67

Find values of a for which (bn ) converges, and also find the limits for such values. 21. For 0 < a < b, let an = (a n + bn )1/n , n ∈ N. Show that (an ) converges to b. [Hint: Note that (a n + bn )1/n = b(1 + 22. For 0 < a < b, let

 n a b

)1/n .]

an+1 = (an bn )1/2 and bn+1 =

an + bn , n∈N 2

with a1 = a, b1 = b. Show that (an ) and (bn ) converge to the same limit. [Hint: First observe that an ≤ an+1 ≤ bn+1 ≤ bn for all n ∈ N.] 23. Let a1 = 1/2 and b = 1, and for n ∈ N, let an+1 = (an bn )1/2 , bn+1 =

2an bn . an + bn

Show that (an ) and (bn ) converge to the same limit. [Hint: First observe that bn ≤ bn+1 ≤ an+1 ≤ an for all n ∈ N.] 24. Let D ⊆ S ⊆ R, and let a ∈ S. Show that the following statements are equivalent: (a) Every open interval containing a contains at least one point from D. (b) There exists a sequence (an ) in D such that an → a. [If every a ∈ D has the above listed property, then D is said to be dense in S.] 25. Show that given any a ∈ R, there exists a sequence (an ) of rational numbers such that an → a; in other words, the set of rational numbers is dense in R. [Hint: Use Archimedean Property.]

1.3.2 Series 1. Suppose (an ) and (bn ) are sequences of positive ∞ numbers. Prove that, if (an ) is b is convergent, then bounded and ∞ n=1 n n=1 an2bn convergent. ∞ 2 are sequences such that ∞ 2. Let (an ) and (bn )  n=1 an and n=1 bn are convergent |a b | converges. series. Show that ∞ n=1 n n [Hint: Observe 2ab ≤ a 2 + b2 .] 3. In each of the following check whether the statement or its converse is true. Justify the answers.   2 (a) Convergence of  ∞ convergence of ∞ n=1 an imply n=1 an . the  ∞ ∞ (b) Convergence of n=1 an and n=1 bn imply the convergence of ∞ n=1 an bn .

68

1 Sequence and Series of Real Numbers

(c) Divergence of

∞ n=1

an and

∞ n=1

bn imply the divergence of

∞ n=1

4. Test the following series for convergence (a) (b) (c) (d) (e) (f) (g)

1 2 3 n + + 3 + ··· + n + ··· 2 22 2 2 1 1 1 1 + ··· √ + √ + √ + ··· + √ 10n 10 20 30 1 1 1 1 + ··· +√ +√ + ··· + √ √ 3 3 3 3 n+6 7 8 9 1 2 3 n + + + ··· + 2 + ··· 2 5 10 n +1 1 2 3 n + + 3 + ··· + n + ··· 2 22 2 2 1 1 1 1 + ··· √ + √ + √ + ··· + √ 10n 10 20 30 1 1 1 1 + ··· +√ +√ + ··· + √ √ 3 3 3 3 n+6 7 8 9

5. Test the following series for convergence: (a) (c) (e)

∞

(n!)2 (2n)! n=1 ∞

(−2)n n=1 ∞

(b) (d)

n2

n=1

(−1)(n−1) √ n

n=1 ∞

(k)

n=1 ∞ 

3

n3 + 1 − n

n=1



n!

∞

1 √ (n + 1) n n=1 ∞

√  1 √ (h) n+1− n n n=1  2 ∞ 

1 n 1+ (j) n n=1 ∞ √

√  (l) (−1)n+1 n+1− n (f)

(−1)n−1 √ n(n + 1)(n + 2) n=1 ∞   

n4 + 1 − n4 − 1 (i)

(g)

∞

(n!)2 n 5 (2n)! n=1 √ ∞

2n!

n=1

6. Find the sum of the series: 1 1 1 + + ··· + + ··· 1.2.3 2.3.4 n.(n + 1).(n + 2)

an bn .

1.3 Additional Exercises

69

7. Find out whether the following series converge absolutely or conditionally: (a)

∞

(−1)n+1 n=1

1 log(n + 1)

∞

(−1)n+1 n , (c) 3n−1 n=1 ∞

(−1)n log n (e) , n log log n n=1

∞

(−1)n+1 , (2n − 1)2 n=1 ∞

(−1)n (d) , n(log n)2 n=1 ∞

1 (−1)n+1 n (f) n2

(b)

n=1

8. Let (an ) be a sequence of non-negative numbers and (akn ) be a subsequence (an ). ∞  a converges, then a Show that, if ∞ n=1 n n=1 kn also converges. Is the converse true? Why? a   n+1  9. Let (an ) be a sequence of nonzero numbers. If lim   =  exists, then show n→∞ an that  n (a) ∞ n=1 an x converges absolutely if |x| < 1/, ∞ n (b) n=1 an x diverges if |x| > 1/. 10. Find the set of all points x such that the following series convergent/absolutely convergent: ∞ ∞ ∞

xn xn xn (i) , (ii) , (iii) 2 n n n! n=1 n=1 n=1

Chapter 2

Limit, Continuity and Differentiability of Functions

In this chapter, we deal with the concepts of limit, continuity and differentiability of functions. Apart from establishing some of the basic properties related to these concepts, we shall also consider many of its applications to the study of qualitative nature of functions such as increasing or decreasing, and determination of points at which they attain maxima and minima. Also, the issue of approximating a function by a polynomial is considered.

2.1 Limit of a Function Suppose f is a real valued function defined on a subset D of R. We are going to define the limit of f (x) as x ∈ D approaches a point a which is not necessarily in D. In the last chapter we have considered the notion of convergence of a sequence of numbers to a particular number so that the terms of the sequence approaches that number. Now, the question: What do we mean by saying that points of a set D ⊆ R approaches a particular point a ∈ R?

2.1.1 Limit Point of a Set By saying the points of a set D ⊆ R approaches a particular point a ∈ R, we shall mean that a is a limit point of the set D, in the following sense. Definition 2.1.1 Let D ⊆ R and a ∈ R. A point a ∈ R is said to be a limit point of D if every open interval containing the point a contains at least one point from D other than a. ♦ Thus, a ∈ R is a limit point of D if and only if for any δ > 0, D ∩ {x ∈ R : 0 < |x − a| < δ} = ∅.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_2

71

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2 Limit, Continuity and Differentiability of Functions

Remark 2.1.1 Very often, instead of saying “x is an element of D”, we may say that “x is a point in D”, with the geometrical connotation associated with it, as we identify R by, the so called, real line. ♦ Example 2.1.1 The reader is urged to verify the following: (i) For a, b ∈ R with a < b, the interval [a, b] is the set of all limit points of each of the intervals (a, b), (a, b], [a, b) and [a, b]. (ii) For a ∈ R, the interval [a, ∞) is the set of all limit points of each of the intervals (a, ∞) and [a, ∞). (iii) For b ∈ R, the interval (−∞, b] is the set of all limit points of each of the intervals (−∞, b) and (−∞, b]. (iv) The set of limit points of R is itself. (v) The set of all limit points of the set D = (0, 1) ∪ {2} is the closed interval [0, 1]. (vi) If D = { n1 : n ∈ N}, then 0 is the only limit point of D. n : n ∈ N}, then 1 is the only limit point of D. (vii) If D = { n+1 (viii) A finite subset of R does not have any limit points. (ix) The set N and Z have no limit points. (x) The set R is the set of all limit points of the set Q and R \ Q. ♦ Definition 2.1.2 Let a ∈ R. (1) Any open interval containing a point a is called a neighbourhood of a or an open neighbourhood of a. (2) If the point a is deleted from a neighbourhood of a, then the remaining part of that neighbourhood is called a deleted neighbourhood of a. ♦ Thus, for a ∈ R and δ > 0, the open interval (a − δ, a + δ) is a neighbourhood of a, which is also called a δ-neighbourhood of a, and the set (a − δ, a + δ) \ {a} is a deleted neighbourhood of a, which is also called a deleted δ-neighbourhood of a. With the above terminologies, we can state the following: A point a ∈ R is a limit point of D ⊆ R if and only if every deleted neighbourhood of a contains at least one point of D

In particular, if D contains either a deleted neighbourhood of a or if D contains an open interval with a as one of its end points, then a is a limit point of D. Notation 2.1.1 In the sequel, for a ∈ R, we shall use the notation Ia for an open ♦ neighbourhood of a, and Iˆa for a deleted neighbourhood of a. Now we give a characterization of limit points in terms of convergence of sequences. Theorem 2.1.1 Let D ⊆ R. A point a ∈ R is a limit point of D if and only if there exists a sequence (an ) in D \ {a} such that an → a as n → ∞.

2.1 Limit of a Function

73

Proof Suppose a ∈ R is a limit point of D. Then for each n ∈ N, there exists an ∈ D \ {a} such that an ∈ (a − 1/n, a + 1/n), i.e., |an − a| < 1/n. Hence, an → a. Conversely, suppose that there exists a sequence (an ) in D \ {a} such that an → a. Hence, for every δ > 0, there exists N ∈ N such that an ∈ (a − δ, a + δ) for all n ≥ N . In particular, every deleted δ-neighbourhood of a contains an for every n ≥ N . Hence, a is a limit point of D. Exercise 2.1.1 Let D ⊆ R. Prove that a point a ∈ R is a limit point of D if and only if there exists a sequence (an ) in D such that (an ) is not eventually constant and

an → a as n → ∞.

2.1.2 Limit of a Function at a Point Definition 2.1.3 Let f be a real valued function defined on a set D ⊆ R, and let a ∈ R be a limit point of D. We say that b ∈ R is a limit of f (x) as x approaches a or limit of f at a, if for every ε > 0, there exists δ > 0 such that | f (x) − b| < ε whenever x ∈ D, 0 < |x − a| < δ.

♦

Let us observe the following important property. Theorem 2.1.2 A function cannot have more than one limit at a given point. Proof Suppose b1 and b2 are limits of a function f : D → R at a given point a, where a is a limit point of D. Let ε > 0 be given. By the definition of the limit, there exist δ1 > 0 and δ2 > 0 such that x ∈ D, 0 < |x − a| < δ1

⇒

| f (x) − b1 | < ε,

x ∈ D, 0 < |x − a| < δ2

⇒

| f (x) − b2 | < ε.

Thus, |b1 − b2 | = |(b1 − f (x)) + ( f (x) − b2 )| ≤ |b1 − f (x)| + | f (x) − b2 | < 2ε whenever, x ∈ D, and 0 < |x − a| < δ := min{δ1 , δ2 }. Notation 2.1.2 If b is the limit of f (x) as x approaches a, the we denote this fact as lim f (x) = b

x→a

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2 Limit, Continuity and Differentiability of Functions

Fig. 2.1 lim x→0 |x| = 0

or (Fig. 2.1) f (x) → b as x → a.

♦

Observe that (verify): lim x→a f (x) = b if and only if for every open interval Ib containing b there exists an open interval Ia containing a such that x ∈ Iˆa ∩ D ⇒ f (x) ∈ Ib . Convention: In the following, whenever we talk about limit of a function f as x approaches a ∈ R, we assume that f is defined on a set D f ⊆ R and a is a limit point of D f . Theorem 2.1.3 Suppose lim x→a f (x) = b. If b = 0, then there exists a deleted neighbourhood Iˆa of a such that f (x) = 0 for every x ∈ Iˆa ∩ D f . In fact, for every r with 0 < r < |b|, there exists a deleted neighbourhood Iˆa of a such that | f (x)| ≥ r ∀ x ∈ Iˆa ∩ D f . Proof Suppose lim x→a f (x) = b with b = 0. Let 0 < r < |b|. Now, for every x ∈ Df, | f (x)| = |b − (b − ( f (x))| ≥ |b| − | f (x) − b|. Hence, | f (x)| ≥ r whenever |b| − | f (x) − b| ≥ r. But, |b| − | f (x) − b| ≥ r ⇐⇒ | f (x) − b| ≤ |b| − r. Now, let 0 < ε ≤ |b| − r . Then there exists δ > 0 such that | f (x) − b| < ε whenever x ∈ D f , 0 < |x − a| < δ. Thus, for x ∈ D with 0 < |x − a| < δ, we have | f (x) − b| < ε ≤ |b| − r so that | f (x)| ≥ r .

2.1 Limit of a Function

75

If lim x→a f (x) = b = 0, then there exists a deleted neighbourhood Iˆa of a such that f (x) = 0 for all x ∈ Iˆa ∩ D f

Example 2.1.2 In (i) and (ii) below, let I be an interval and a is either in I or a is an end point of I . (i) Let f (x) = x. We show that lim x→a f (x) = a: Since | f (x) − a| = |x − a| ∀ x ∈ I, it is clear that, for any ε > 0, taking δ = ε, x ∈ I, 0 < |x − a| < δ

⇒

| f (x) − a| < ε.

Hence, lim x→a f (x) = a. (ii) Let f (x) = x 2 . We show that lim x→a f (x) = a 2 : We have | f (x) − a 2 | ≤ |x + a| |x − a| ≤ (|x − a| + 2|a|)|x − a|. Now, let ε > 0 be given. Then, (|x − a| + 2|a|)|x − a| < ε implies | f (x) − a 2 | < ε. Note that |x − a| ≤ 1

⇒

(|x − a| + 2|a|)|x − a| < (1 + 2|a|)|x − a|

Thus, |x − a| ≤ 1 and (1 + 2|a|)|x − a| < ε implies | f (x) − a 2 | < ε. Hence, taking δ = min{1, ε/(1 + 2|a|)}, we have | f (x) − a 2 | < ε whenever x ∈ I, |x − a| < δ. Hence, lim x→a f (x) = a 2 . (iii) Let f (x) = 1/x for x ∈ R \ {0}. Let a ∈ R \ {0}. We show that lim x→a Let x ∈ R \ {0}. W observe that   1 1  |x − a|  | f (x) − f (a)| =  −  = . x a |ax| Note that |ax| = |a||a + (x − a)| ≥ |a|(|a| − |x − a|).

1 1 = : x a

76

2 Limit, Continuity and Differentiability of Functions

Hence, |a|2 2 |x − a| 2|x − a| ⇒ | f (x) − f (a)| = . ≤ |ax| |a|2

|x − a| ≤ |a|/2 ⇒ |ax| ≥ |a|(|a| − |x − a|) ≥

Thus, for any ε > 0, |x − a| ≤ |a|/2 and

2|x − a| 0, taking δ = min{|a|/2, ε|a|2 /2}, we have 0 < |x − a| < δ showing that lim x→a

1 x

=

⇒

| f (x) − f (a)| =

|x − a| < ε, |ax| ♦

1 a

More examples will be considered in Sect. 2.1.4 after proving some properties of the limit. Before that let us look at what the statement “lim x→a f (x) does not exist” means. From the definition of the limit, we can state the the following. Theorem 2.1.4 Suppose f is a real valued function defined on a set D and a ∈ R is a limit point of D. Then lim x→a f (x) does not exist if and only if for any b ∈ R, there exists an ε0 > 0 with the property that for any δ > 0, there is at least one / (b − ε0 , b + ε0 ). xδ ∈ (a − δ, a + δ) ∩ D with xδ = a, but f (xδ ) ∈ Exercise 2.1.2 Write the details of the proof of the above theorem.



We illustrate Theorem 2.1.4 by two simple examples. Example 2.1.3 Let f : [−1, 1] → R be defined by  f (x) =

0, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1.

We show that lim x→0 f (x) does not exist. For this let b ∈ R. Let us consider the following cases: Case (i) : b = 0. In this case, if 0 < ε < 1, then (b − ε, b + ε) does not contain 1 so that f (x) ∈ / (b − ε, b + ε) for any x > 0. Case (ii):b = 1. In this case, if 0 < ε < 1, then (b − ε, b + ε) does not contain 0 so that f (x) ∈ / (b − ε, b + ε) for any x < 0. Case (iii) : b = 0, b = 1. In this case, if 0 < ε < min{|b|, |b − 1|}, then (b − ε, b + ε) does not contain 0 and 1 so that f (x) ∈ / (b − ε, b + ε) for any x = 0. Thus, b is not a limit of f (x) as x approaches 0. ♦

2.1 Limit of a Function

77

Fig. 2.2 lim x→0 1/x does not exist

Example 2.1.4 Let f : R \ {0} → R is defined by f (x) =

1 , x ∈ R \ {0}. x

We show that lim x→0 f (x) does not exist: Let b ∈ R. Note that | f (x) − b| ≥ | f (x)| − |b| > 1 whenever | f (x)| > 1 + |b|. But, (Fig. 2.2) | f (x)| > 1 + |b| ⇐⇒ |x|
1 whenever |x|
0 such that | f (x) − b| < 1 for all x with |x| < δ. This is true for any b ∈ R. Hence, lim x→0 f (x) does not exist. By similar argument, we see that, for any k ∈ N,

78

2 Limit, Continuity and Differentiability of Functions

lim

x→0

1 does not exist. xk

♦

In Example 2.1.4, we see that the function takes arbitrary large values in any deleted neighbourhood of 0. In the following theorem, we show that, if a function has such property in a deleted neighbourhood of a point, then the function cannot have a limit at that point. Theorem 2.1.5 (Boundedness test) If lim x→a f (x) = b, then there exists a deleted neighbourhood Iˆa of a and M > 0 such that | f (x)| ≤ M for all x ∈ Iˆa ∩ D. Proof Suppose lim x→a f (x) = b. Then there exists a deleted neighbourhood Iˆa of a such that | f (x) − b| < 1 for all x ∈ Iˆa ∩ D. Hence, | f (x)| ≤ | f (x) − b| + |b| < 1 + |b| ∀ x ∈ Iˆa ∩ D. Thus, | f (x)| ≤ M = 1 + |b| for all x ∈ Iˆa ∩ D. The necessary condition given in Theorem 2.1.5 may be called as boundedness of f in a deleted neighbourhood of a point. More generally, we have the following definition. Definition 2.1.4 Let f : D → R and S ⊆ D. (1) The function f is said to be bounded on a subset S ⊆ D if the set { f (x) : x ∈ S} is a bounded subset of R. If f is bounded on D, then we say that f is a bounded function. (2) The function f is said to be unbounded on a subset S ⊆ D if it is not bounded on S. ♦ Thus, f : D → R is bounded on S ⊆ D if and only if there exists M > 0 (in general, depends on S) such that | f (x)| ≤ M ∀ x ∈ S, and f : D → R is unbounded on S if and only if there exists a sequence (xn ) in S such that | f (xn )| → ∞ as n → ∞. In view of the Definition 2.1.4, Theorem 2.1.5 can be restated as follows: If lim x→a f (x) exists, then f is bounded on some deleted neighbourhood of a

2.1 Limit of a Function

79

2.1.3 Limit of a Function in Terms of Sequences Let a be a limit point of D ⊆ R and f : D → R. Suppose lim x→a f (x) = b. Since a is a limit point of D, we know by Theorem 2.1.1 that there exists a sequence (xn ) in D \ {a} such that xn → a. Does f (xn ) → b? The answer is in the affirmative. Theorem 2.1.6 If lim x→a f (x) = b, then for every sequence (xn ) in D with xn → a, we have f (xn ) → b. Proof Suppose lim x→a f (x) = b. Let (xn ) be a sequence in D such that xn → a. Let ε > 0 be given. We have to show that there exists N ∈ N such that | f (xn ) − b| < ε for all n ≥ N . Since lim x→a f (x) = b, we know that there exists δ > 0 such that x ∈ D, 0 < |x − a| < δ

⇒

| f (x) − b| < ε.

Also, since xn → a, corresponding to the above δ, there exists N ∈ N such that |xn − a| < δ for all n ≥ N . Hence, we have | f (xn ) − b| < ε for all n ≥ N . The converse of the above theorem is also true. Theorem 2.1.7 Suppose that, for every sequence (xn ) in D for which xn → a, we have f (xn ) → b. Then lim x→a f (x) = b. Proof Assume for a moment that f does not have the limit b as x approaches a. Then, by Theorem 2.1.4, there exists ε0 > 0 such that for every δ > 0, there exists at least one xδ ∈ D such that 0 < |xδ − a| < δ and | f (xδ ) − b| ≥ ε0 . In particular, for every n ∈ N, there exists xn ∈ D such that 0 < |xn − a|
0. Let xn = 2/nπ, n ∈ N. Then we have xn → 0, and f (xn ) = sin(nπ/2) ∀ n ∈ N.

2.1 Limit of a Function

81

Note that f (x2n−1 ) = (−1)n−1 and f (x2n ) = 0 for all n ∈ N. Hence, the sequence ( f (xn )) does not converge. Therefore, by Theorem 2.1.6, lim x→0 f (x) does not exist. ♦

2.1.4 Some Properties For considering certain properties of the limit, and also for later use, we recall some definitions from set theory: Suppose f and g are (real valued) functions with domains D f and Dg , respectively, and let α ∈ R. Suppose D f ∩ Dg = ∅. Then, we define functions f + g, f g and α f as ( f + g)(x) = f (x) + g(x), x ∈ D f ∩ Dg , ( f g)(x) = f (x)g(x), x ∈ D f ∩ Dg , (α f )(x) = α f (x), x ∈ D f . The function f + g is called the sum of f and g, and f g is called the product of f and g. If f is a nonzero function, that is, f (x) = 0 for some x ∈ D f , then we define the function 1/ f by 1 1 (x) = , x ∈ D f , f f (x) where D f := {x ∈ D f : f (x) = 0}. Thus, we can also define the function f /g by f 1 = f g g on the set D f /g := {x ∈ D f ∩ Dg : g(x) = 0}. If D := {x ∈ D f : f (x) ∈ Dg } = ∅, then we define the composition of g and f , denoted by g ◦ f , by (g ◦ f )(x) = g( f (x)), x ∈ D. Thus, domain of g ◦ f is the set Dg◦ f := {x ∈ D f : f (x) ∈ Dg }. In the following, when we talk about the functions f + g, f g, f /g and g ◦ f, we mean their definitions as above with appropriate domains of definitions; we may not write the domains explicitly. Functions are assumed to be defined on subsets of the set R of real numbers, and are real valued.

The following three theorems can be proved using Theorems 2.1.6 and 2.1.7, and the results on convergence of sequences of real numbers (supply details). However, to get accustomed with the ε-δ arguments, we provide the detailed proof using the definition itself.

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2 Limit, Continuity and Differentiability of Functions

Theorem 2.1.8 Suppose lim x→a f (x) = b and lim x→a g(x) = c. Then we have the following. (i) lim x→a [ f (x) + g(x)] = b + c, (ii) lim x→a f (x)g(x) = bc. (iii) If c = 0, then g(x) = 0 in a deleted neighbourhood of a and lim

x→a

b f (x) = . g(x) c

Proof (i) Note that for x ∈ D f ∩ Dg , |( f (x) + g(x)) − (b + c)| = |[ f (x) − b] + [g(x) − c]| ≤ | f (x) − b| + |g(x) − c|.

(2.1) Since lim x→a f (x) = b and lim x→a g(x) = c, there exist δ1 > 0, δ2 > 0 such that | f (x) − b| < ε/2 whenever x ∈ D f , 0 < |x − a| < δ1 ,

(2.2)

|g(x) − c| < ε/2 whenever x ∈ Dg , 0 < |x − a| < δ2 .

(2.3)

From (2.1), (2.2), (2.3), we obtain |( f (x) + g(x)) − (b + c)| < ε whenever x ∈ D f ∩ Dg with 0 < |x − a| < δ := min{δ1 , δ2 }. Thus, lim x→a [ f (x) + g(x)] = b + c. (ii) Note that for x ∈ D f ∩ Dg , | f (x)g(x) − bc| = | f (x)[g(x) − c] + [ f (x) − b]c|. Since lim x→a f (x) = b, there exists M > 0 and δ0 > 0 such that | f (x)| ≤ M whenever x ∈ D f , 0 < |x − a| < δ0 . Hence, | f (x)g(x) − bc| ≤ M|g(x) − c| + | f (x) − b||c|

(2.1)

whenever x ∈ D f ∩ Dg with 0 < |x − a| < δ0 . Now, since lim x→a f (x) = b and lim x→a g(x) = c, there exist δ1 > 0, δ2 > 0 such that M|g(x) − c| < ε/2 whenever x ∈ Dg , 0 < |x − a| < δ1 ,

(2.2)

|c| | f (x) − b| < ε/2 whenever x ∈ D f , 0 < |x − a| < δ2 .

(2.3)

2.1 Limit of a Function

83

From (2.1), (2.2), (2.3), we obtain | f (x)g(x) − bc| < ε whenever x ∈ D f ∩ Dg with 0 < |x − a| < δ := min{δ0 , δ1 , δ2 }. Thus, lim x→a f (x) g(x) = bc. (iii) Since lim x→a g(x) = c = 0, by Theorem 2.1.3, there exists δ0 > 0 such that |g(x)| ≥ |c|/2 for all x ∈ Dg with 0 < |x − a| < δ0 . Hence for x ∈ D f ∩ Dg with 0 < |x − a| < δ0 , we have  f (x) b  |c f (x) − bg(x)|   − =  g(x) c |cg(x)| |c[ f (x) − b] + [c − g(x)]b| ≤ |cg(x)| |c| | f (x) − b| + |c − g(x)| |b| ≤ (|c|2 /2) Now, since lim x→a f (x) = b and lim x→a g(x) = c, from the above we can conclude that there exists δ > 0 such that  f (x) b    −  < ε whenever x ∈ D f ∩ Dg , 0 < |x − a| < δ.  g(x) c Thus, lim x→a [ f (x)/g(x)] = b/c. Proof Alternative proof using sequences Let (xn ) be any sequence in D f ∩ Dg such that xn → a. Hence, by Theorem 2.1.6, we have limn→∞ f (xn ) = b and limn→∞ g(xn ) = c. Hence, by Theorem 1.1.3 (i) and Theorem 1.1.8 (i), we have limn→∞ [ f (xn ) + g(xn )] = b + c, limn→∞ f (xn )g(xn ) = bc. Also, since c = 0 and limn→∞ g(xn ) = c, there exists N ∈ N such that g(xn ) = 0 for all n ≥ N . Therefore, by Theorem 1.1.8 (ii), we have lim

n→∞

b f (xn ) = . g(xn ) c

Therefore, by Theorem 2.1.7, we have lim [ f (x) + g(x)] = b + c,

x→a

This completes the proof.

lim f (x)g(x) = bc,

x→a

lim f (x)/g(x) = b/c.

x→a

84

2 Limit, Continuity and Differentiability of Functions

Theorem 2.1.9 (Sandwich theorem) If f and g have the same limit b as x approaches a, and if h is a function such that f (x) ≤ h(x) ≤ g(x) for all x in a deleted neighbourhood of a, then lim x→a h(x) = b. Proof Suppose lim x→a f (x) = b = lim x→a g(x). Let ε > 0 be given. Then there exists δ0 > 0 such that if x ∈ D f ∩ Dg and 0 < |x − a| < δ0 , then f (x) ∈ (b − ε, b + ε) and g(x) ∈ (b − ε, b + ε). Since f (x) ≤ h(x) ≤ g(x) for all x ∈ D f ∩ Dg ∩ Dh with 0 < |x − a| < δ1 for some δ1 , we obtain that h(x) ∈ (b − ε, b + ε) whenever x ∈ D f ∩ Dg ∩ Dh with 0 < |x − a| < δ := min{δ0 , δ1 }. Thus, lim x→a h(x) = b. Proof Alternative proof using sequences Let (xn ) be any sequence in D f ∩ Dg ∩ Dh such that xn → a. By assumption lim x→a f (x) = b = lim x→a g(x). Hence, by Theorem 2.1.6, we have limn→∞ f (xn ) = b = limn→∞ g(xn ). Again by assumption, we have f (xn ) ≤ h(xn ) ≤ g(xn ) for all n ∈ N. Therefore, by the Sandwich theorem Theorem, 1.1.3 (iv), for sequences, we have lim x→a h(x) = b. Hence, by Theorem 2.1.7, lim x→a h(x) = b. Theorem 2.1.10 Suppose lim x→a f (x) = b, lim x→a g(x) = c and f (x) ≥ g(x) for all x in a deleted neighbourhood of a. Then b ≥ c. Proof Suppose b < c. Since lim x→a f (x) = b and lim x→a g(x) = c, for ε > 0 small enough, say 0 < ε < (c − b)/2, there exists δ > 0 such that f (x) ∈ (b − ε, b + ε) and g(x) ∈ (c − ε, c + ε) for all x ∈ D f ∩ Dg with 0 < |x − a| < δ0 . Hence, f (x) < g(x) for all x ∈ D f ∩ Dg with 0 < |x − a| < δ0 , which is a contradiction to the assumption in the theorem. Hence, b ≥ c. Exercise 2.1.3 Provide an alternative proof for the above theorem using sequences.

Theorem 2.1.11 Suppose lim x→a f (x) = b and lim y→b g(y) = c. Further, assume that f (x) ∈ Dg \ {b} for every x ∈ D f \ {a}. Then lim (g ◦ f )(x) = c.

x→a

Proof Let ε > 0 be given. Then there exists δ1 > 0 such that 0 < |y − b| < δ1

⇒

|g(y) − c| < ε.

⇒

| f (x) − b| < δ1 .

Also, let δ2 > 0 be such that 0 < |x − a| < δ2

2.1 Limit of a Function

85

Hence, along with the given condition that f (x) ∈ D2 \ {b} for every x ∈ D1 \ {a}, 0 < |x − a| < δ2

⇒

0 < | f (x) − b| < δ1

⇒

|g( f (x)) − c| < ε.

This completes the proof. Proof Alternative proof using sequences By Theorem 2.1.7, it is enough to prove that for any sequence (xn ) in Dg \ {a} which converges to a, the sequence (g( f (xn ))) converges to c. So, let (xn ) be in D f \ {0} such that xn → a. Since lim x→a f (x) = b, by Theorem 2.1.6, f (xn ) → b. Let yn = f (xn ), n ∈ N. By the assumption, yn ∈ Dg \ {b} for all n ∈ N. Since lim y→b g(y) = c and yn → b, again by Theorem 2.1.6, g(yn ) → c. Thus we obtained g( f (xn )) → c, completing the proof. Example 2.1.8 If f (x) is a polynomial, say f (x) = a0 + a1 x + . . . + ak x k , then for any a ∈ R, lim f (x) = f (a). x→a

This follows from the results in Theorem 2.1.8 by making use of the fact that lim x→a x = a. ♦ Example 2.1.9 Let f (x) =

x 2 −4 x−2

for x ∈ D = R \ {2}. Then lim f (x) = 4.

x→2

Note that, for x = 2, f (x) =

(x + 2)(x − 2) = (x + 2). x −2

Hence, lim x→2 f (x) = lim x→2 (x + 2) = 4.

♦

Example 2.1.10 We show that lim sin(x) = 0 and

x→0

lim cos(x) = 1.

x→0

From the graph of the function sin x, it is clear that 0 < |x|
0 such that | f (x) − b| < ε ∀ x ∈ D ∩ (a − δ, a), ♦ and in that case we write lim x→a − f (x) = b. (2) Suppose D ∩ (a, ∞) = ∅ and a is a limit point of D ∩ (a, ∞). Then we say that f (x) has the right limit b ∈ R as x approaches a ∈ R from right if for every ε > 0, there exists δ > 0 such that | f (x) − b| < ε ∀ x ∈ D ∩ (a, a + δ), and in that case we write lim x→a + f (x) = b. Notation 2.1.3 We shall use the notations: f (a−) := lim− f (x), x→a

whenever the above limits exist.

f (a+) := lim+ f (x) x→a

♦

2.1 Limit of a Function

89

We have the following characterizations in terms of sequences (Verify): 1. lim x→a− f (x) = b if and only if for every sequence (xn ) in D \ {a}, xn < a ∀ n ∈ N, xn → a

⇒

f (xn ) → b.

2. lim x→a+ f (x) = b if and only if for every sequence (xn ) in D \ {a}, xn > a ∀ n ∈ N, xn → a

⇒

f (xn ) → b.

The proof of the following theorem is left as an exercise. Theorem 2.1.12 Let f be a real valued function defined on a set D ⊆ R, and let a ∈ R be a limit point of D. Then lim x→a f (x) exists if and only if lim x→a − f (x) and lim x→a + f (x) exist and lim x→a − f (x) = lim x→a + f (x), and in that case lim f (x) = lim− f (x) = lim+ f (x).

x→a

x→a

x→a

lim x→a f (x) does not exist in the following cases: (i) lim x→a − f (x) does not exist (ii) lim x→a + f (x) does not exist, (iii) lim x→a − f (x) and lim x→a + f (x) exist, but they are not equal

We may observe that 1. lim x→a − f (x) = b ⇐⇒ limh→0+ f (a − h) = b, 2. lim x→a + f (x) = b ⇐⇒ limh→0+ f (a + h) = b. Exercise 2.1.6 Verify the above two results. Example 2.1.12 We consider a few examples to illustrate Theorem 2.1.12. (i) Let f : R → R be defined by  f (x) =

1/x, x > 0, 1, x ≤ 0.

Then, lim x→0− f (x) = 1, but lim x→0+ f (x) does not exist. (ii) Let f : R → R be defined by  f (x) =

1/x, x < 0, 1, x ≥ 0.

Then, lim x→0+ f (x) = 1, but lim x→0− f (x) does not exist.



90

2 Limit, Continuity and Differentiability of Functions

(iii) Let f : R → R be defined by  f (x) =

1/x, x = 0, 1, x = 0.

Then, lim x→0+ f (x) and lim x→0− f (x) do not exist. (iv) Let f be as in Example 2.1.3, that is, f : [−1, 1] → R defined by  f (x) =

0, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1.

Then, lim x→0− f (x) and lim x→0+ f (x) exist, but lim x→0 f (x) does not exist. ♦

2.1.6 Limit at ±∞ and Limit ±∞ Definition 2.1.6 Let f be a real valued function defined on D f ⊆ R. (1) If D f contains (a, ∞) for some a ∈ R, then f is said to have the limit b as x → ∞, if for every ε > 0, there exists M > a such that | f (x) − b| < ε whenever x > M, and in that case we write lim x→∞ f (x) = b. (2) If D f contains (−∞, a) for some a ∈ R, then f is said to have the limit b as x → −∞, if for every ε > 0, there exits M < a such that | f (x) − b| < ε whenever x < M, and in that case we write lim x→−∞ f (x) = b.

♦

Notation 2.1.4 When we write lim x→∞ f (x) = b we mean that D f contains an interval of the form (a, ∞) for some a ∈ R and the limit is in the sense of Definition 2.1.6(1). Similarly, when we write we lim x→−∞ f (x) = b, it is assumed that D f contains an interval of the form (−∞, a) for some a ∈ R, and the limit is in the sense of Definition 2.1.6(2). Also, for a sequence (xn ) in R, if we write f (xn ) → b for some b ∈ R, we assume that xn belongs to D f , and the limit of ( f (xn )) is b. Now, we give the sequential characterizations of limits as given in Definition 2.1.6. ♦ Theorem 2.1.13 The following hold. (i) lim x→∞ f (x) = b if and only if for every sequence (xn ), xn → ∞ implies f (xn ) → b.

2.1 Limit of a Function

91

(ii) lim x→−∞ f (x) = b if and only if for every sequence (xn ), xn → −∞ implies f (xn ) → b. Proof Suppose lim x→∞ f (x) = b, and let (xn ) be such that xn → ∞. Let ε > 0 be given. To show that there exists N ∈ N such that | f (xn ) − b| < ε for all n ≥ N . Since lim x→∞ f (x) = b, there exists M > 0 such that x ∈ Df, x > M

⇒

| f (x) − b| < ε.

(2.1)

Since xn → ∞, there exists n 0 ∈ N such that n ≥ n0

⇒

xn > M.

(2.2)

From (2.1) and (2.2) above we have n ≥ n0

⇒

| f (xn ) − b| < ε.

Conversely, suppose that for every (xn ) with xn → ∞, we have f (xn ) → b. Assume for a moment that lim x→∞ f (x) = b. Then there exists ε0 > 0 such that for every n ∈ N, there exists xn > n and | f (xn ) − b| ≥ ε0 . Thus, we obtained a sequence (xn ) with xn → ∞, but f (xn ) → b. This is a contradiction to the hypothesis. Thus, we have proved (i). Proof of (ii) follows by similar arguments. The following results can be verified as in the case of limits at points in R. 1. If lim x→∞ f (x) = b and lim x→∞ g(x) = c, then lim [ f (x) + g(x)] = b + c,

x→∞

lim f (x)g(x) = bc.

x→∞

2. If lim x→∞ f (x) = b, lim x→∞ g(x) = c and c = 0, then there exists M0 > 0 such that g(x) = 0 for all x > M0 and lim

x→∞

b f (x) = . g(x) c

Exercise 2.1.7 Verify the above results. Example 2.1.13 (i) For any k ∈ N, lim

x→∞

1 = 0, xk

lim

x→−∞

1 = 0. xk

From the results on sequences, for any sequence (xn ) of nonzero numbers, we have 1 1 xn → ∞ ⇐⇒ k → 0 and xn → −∞ ⇐⇒ k → 0. xn xn

92

2 Limit, Continuity and Differentiability of Functions

Hence, we obtain the required limits, by using Theorem 2.1.13. 1+x 1+x = 0: Let f (x) = for x ∈ R. By (i), (ii) lim x→∞ 1 + x2 1 + x2 f (x) =

(iii) lim x→∞

1+x 1+x = −1: Let f (x) = for x = 1. By (i), 1−x 1−x f (x) =

(iv) lim x→∞

0 1+x 1/x 2 + 1/x → = 0. = 1 + x2 1/x 2 + 1 1

1/x + 1 1 1+x = → = −1. 1−x 1/x − 1 −1

1 + 2x 2 1 + 2x = : Let f (x) = for x = −1/3. By (i), 1 + 3x 3 1 + 3x f (x) =

1 + 2x 1/x + 2 2 = → . 1 + 3x 1/x + 3 3

♦

Exercise 2.1.8 Prove the limits in Example 2.1.13(i) by using ε, M arguments, without using Theorem 2.1.13.

Definition 2.1.7 We define the following: (1) lim x→a f (x) = ∞ if for every M > 0, there exists δ > 0 such that 0 < |x − a| < δ

⇒

f (x) > M.

(2) lim x→a f (x) = −∞ if for every M > 0, there exists δ > 0 such that 0 < |x − a| < δ

⇒

f (x) < −M.

(3) lim x→+∞ f (x) = ∞ if for every M > 0, there exists α > 0 such that x >α

⇒

f (x) > M.

(4) lim x→+∞ f (x) = −∞ if for every M > 0, there exists α > 0 such that x >α

⇒

f (x) < −M.

(5) lim x→−∞ f (x) = ∞ if for every M > 0, there exists α > 0 such that x < −α

⇒

f (x) > M.

(6) lim x→−∞ f (x) = −∞ if for every M > 0, there exists α > 0 such that

2.1 Limit of a Function

93

x < −α

⇒

f (x) < −M.

♦

It can be easily shown that 1. lim x→a f (x) = ∞ ⇐⇒ lim x→a [− f (x)] = −∞, 2. lim x→+∞ f (x) = ∞ ⇐⇒ lim x→+∞ [− f (x)] = −∞, 3. lim x→−∞ f (x) = ∞ ⇐⇒ lim x→−∞ [− f (x)] = −∞. Exercise 2.1.9 Verify the above statements.



1 1 Example 2.1.14 f(i) lim x→0 2 = ∞: Taking f (x) = 2 for x = 0 and M > 0, x x we observe that f (x) > M ⇐⇒

1 1 > M ⇐⇒ |x| < √ . 2 x M

√ Hence, for 0 < δ < 1/ M, |x| < δ

1 |x| < √ M

⇒

⇒

f (x) > M.

1 = ∞. x 2 1 + x    1 + x   (ii) lim x→1   = ∞: Let f (x) =   for x = 1. Then for M > 0, 1−x 1−x

Thus, lim x→0

1 + x  |1 + x|   f (x) =  .  > M ⇐⇒ |1 − x| < 1−x M Note that

|1 + x| |2 − (1 − x)| 2 − |1 − x| = ≥ . M M M

Hence, |x − 1|
M.

1 + x    Thus, lim x→1   = ∞. 1−x (iii) Let f (x) = x 2 , x ∈ R. For M > 0, f (x) = x 2 > M ⇐⇒ |x| > Thus,

√

M.

94

2 Limit, Continuity and Differentiability of Functions

x> and

√

M

√ x M ⇒ f (x) > M.

Thus, lim x→∞ f (x) = ∞ and lim x→−∞ f (x) = ∞.

♦

Example 2.1.15 (The number e) We have seen in Example 1.1.26 that limn→∞  1 n 1+ exists, and we denoted its value by e. Now we show that n  1 x = e. lim x→∞ 1 + x Let ε > 0 be given. We have to find an M > 0 such that  1 x < e + ε whenever x > M. e−ε < 1+ x Note that, for every n ∈ N, if x ∈ R is such that n ≤ x ≤ n + 1, then 1+ so that

i.e.,

 1+

1 1 1 ≤1+ ≤1+ n+1 x n

1 n  1 x  1 n+1 ≤ 1+ ≤ 1+ , n+1 x n  1 x ≤ βn , αn ≤ 1 + x

where  1 n+1 1 n  1 −1  1+ αn = 1 + = 1+ , n+1 n+1 n+1  1 1 n+1  1 n  βn = 1 + . 1+ = 1+ n n n −1   1 Since (1 + n+1 → 1 and 1 + n1 → 1, we have αn → e and βn → e. Therefore, there exists N ∈ N such that e − ε < αn < e + ε,

e − ε < βn < ε + ε

2.1 Limit of a Function

95

for all n ≥ N . Now, for x > N , let n ≥ N be such that n ≤ x ≤ n + 1. Then we have  1 x ≤ βn < e + ε. e − ε < αn ≤ 1 + x Thus,  1 x e−ε < 1+ < e + ε whenever x > N . x  1 x = e. Thus, we have proved lim x→∞ 1 + x

♦

  1 x 1 n lim x→∞ 1 + = e = limn→∞ 1 + . x n Exercise 2.1.10 Suppose (αn ) and (βn ) are sequences of positive real numbers and f is a (real valued) function defined on (0, ∞) having the following property: For n ∈ N, x ∈ R, n < x < n + 1 ⇒ αn ≤ f (x) ≤ βn . If (αn ) and (βn ) converge to the same limit, say b, then lim x→∞ f (x) = b. (Hint: Use the arguments used in the Example 2.1.15.)

2.2 Continuity of a Function Suppose f is defined on an interval I and x0 ∈ I . If lim x→x0 f (x) exists, then a natural question would be whether the limit is equal to f (x0 ).

2.2.1 Definition and Some Basic Results Definition 2.2.1 Let f be a real valued function defined on a set D ⊆ R. Then f is said to be continuous at a point x0 ∈ D if for every ε > 0, there exists a δ > 0 such that | f (x) − f (x0 )| < ε whenever x ∈ D, |x − x0 | < δ. The function f is said to be continuous on D if it is continuous at every point in D. ♦ Note that, in the above definition, we did not assume that x0 is a limit point of D. However, if we assume that x0 is a limit point of D, then we have the following characterization of continuity.

96

2 Limit, Continuity and Differentiability of Functions

Fig. 2.9 Limit does not exist at 0

Theorem 2.2.1 Let x0 ∈ D be a limit point of D. Then, for a function f : D → R, the following are equivalent. (i) f is continuous at x0 . (ii) lim x→x0 f (x) exists and it is equal to f (x0 ). (iii) For every sequence (xn ) in D, xn → x0 implies f (xn ) → f (x0 ). Recall that a point x0 in an interval I is a limit point of I if and only if either x0 ∈ I or if x0 is an endpoint of I . In this book, we shall consider continuity of functions which are defined on intervals. Convention: When we say that f is continuous at a point x0 ∈ R, we mean that f is defined on an interval containing x0 and f is continuous at x0 . Example 2.2.1 Let f (x) = a0 + a1 x + · · · ak x k for some a0 , a1 , . . . , ak in R and k ∈ N. We know (cf. Example 2.1.8) that, for any x0 ∈ R, lim f (x) = f (x0 ).

x→x0

Hence, by Theorem 2.2.1, f is continuous at every x0 ∈ R (Fig. 2.9).

♦

Example 2.2.2 Let f : [−1, 1] → R be as in Example 2.1.3, i.e.,  f (x) =

0, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1.

We have seen that lim x→0 f (x) does not exist. Hence, by Theorem 2.2.1, f is not continuous at 0. ♦ Example 2.2.3 We have seen in Examples 2.1.10 and 2.1.11 that lim sin x = 0,

x→0

lim cos x = 1,

x→0

sin x = 1. x→0 x lim

2.2 Continuity of a Function Fig. 2.10 lim x→0

97

sin x =1 x

Hence, by Theorem 2.2.1, the functions f (x) := sin x, g(x) := cos x, h(x) :=

 sin x x

1,

, x = 0, x =0 ♦

are continuous at 0 (Fig. 2.10). The following theorem is a consequence of Theorems 2.1.8 and 2.2.1.

Theorem 2.2.2 Suppose f and g are defined on an interval I and both f and g are continuous at x0 ∈ I . Then f + g and f g are continuous at x0 . The following Theorem is analogous to Theorem 2.1.11. Theorem 2.2.3 Suppose f : I → R is continuous at a point x0 ∈ I and g : J → R is continuous at the point y0 := f (x0 ), where J is an interval such that f (I ) ⊆ J . Then g ◦ f : I → R is continuous at x0 . Proof Let (xn ) be any sequence in I such that xn → x0 . Since f is continuous at x0 , we have f (xn ) → f (x0 ). Let yn = f (xn ), n ∈ N. Since g is continuous at y0 := f (x0 ), g(yn ) → g(y0 ). Thus, we have proved that for every sequence (xn ) in I with xn → x0 , (g ◦ f )(xn ) → (g ◦ f )(x0 ). Hence, by Theorem 2.2.1, g ◦ f is continuous at x0 . Here is another characterization of continuity at a point. Theorem 2.2.4 A function f : I → R is continuous at a point x0 ∈ I if and only if for every open interval J containing f (x0 ), there exists an open interval I0 containing x0 such that f (x) ∈ J whenever x ∈ I0 ∩ I. Proof Suppose f is continuous at x0 and J be such that f (x0 ) ∈ J . For ε > 0, let δ > 0 be such that x ∈ I, |x − x0 | < δ

⇒

| f (x) − f (x0 )| < ε,

98

2 Limit, Continuity and Differentiability of Functions

i.e., taking I0 = (x0 − δ, x0 + δ), x ∈ I0 ∩ I

⇒

f (x) ∈ ( f (x0 ) − ε, f (x0 ) + ε).

Choosing ε > 0 small enough such that ( f (x0 ) − ε, f (x0 ) + ε) ⊆ J , we obtain x ∈ I0 ∩ I

⇒

f (x) ∈ J.

Conversely, suppose that for every open interval J containing f (x0 ), there exists an open interval I0 containing x0 such that x ∈ I0 ∩ I implies f (x) ∈ J . So, given ε > 0, we may take J = ( f (x0 ) − ε, f (x0 ) + ε). Then, by taking δ > 0 such that (x0 − δ, x0 + δ) ⊆ I0 , we obtain x ∈ I, |x − x0 | < δ

⇒

| f (x) − f (x0 )| < ε.

Thus, f is continuous at x0 . Theorem 2.2.5 Suppose f is a continuous function defined on an interval I and x0 ∈ I . Suppose f (x0 ) = 0. Then there exists an open interval I0 containing x0 such that f (x) = 0 for every x ∈ I0 ∩ I . Further, the function g : I0 ∩ I → R defined by g(x) = 1/ f (x) is continuous at x0 . Proof Suppose f (x0 ) = 0. Let J = (a, b) be an open interval containing f (x0 ) such that 0 ∈ / J . Then by Theorem 2.2.4, there exists an open interval I0 containing x0 such that f (x) ∈ J whenever x ∈ I0 ∩ I . In particular, f (x) = 0 for all x ∈ I0 ∩ I and g(x) = 1/ f (x) is defined on I0 ∩ I . Next, we observe that for every x ∈ I0 ∩ I , 1 1 f (x0 ) − f (x) − = . f (x) f (x0 ) f (x) f (x0 ) Since f (x) = 0 for all x ∈ I0 ∩ I we have | f (x)| > c := min{|a|, |b|} for all x ∈ I0 ∩ I . Therefore,    1 1  | f (x0 ) − f (x)| | f (x0 ) − f (x)|   f (x) − f (x )  = | f (x) f (x )| ≤ c2 0 0 for all x ∈ I0 ∩ I . Now, by continuity of f at x0 , for every ε > 0, there exists δ > 0 such that | f (x0 ) − f (x)| < c2 ε whenever x ∈ I0 ∩ I, |x − x0 | < δ. Hence,    1 1   < ε whenever x ∈ I0 ∩ I, |x − x0 | < δ. −  f (x) f (x0 ) 

2.2 Continuity of a Function

99

Thus, 1/ f is continuous at x0 . Theorems 2.2.2 and 2.2.5 imply the following theorem. Theorem 2.2.6 Suppose f : I → R and g : I → R are continuous at a point x0 ∈ I and g(x0 ) = 0. Then there exists an open interval I0 containing x0 such that f /g is well defined on I0 ∩ I and f /g is continuous at x0 . Exercise 2.2.1 Suppose f is a continuous function defined on an interval I and x0 ∈ I . Prove the following. 1. If α ≥ 0 is such that | f (x0 )| > α, then there exists a subinterval I0 of I containing x0 such that | f (x)| > α for all x ∈ I0 . 2. If f (x0 ) > 0, then there exists a subinterval I0 of I containing x0 such that f (x) ≥ f (x0 )/2 for all x ∈ I0 . 3. If f (x0 ) < 0, then there exists a subinterval I0 of I containing x0 such that

f (x) ≤ f (x0 )/2 for all x ∈ I0 .

2.2.2 Some More Examples In the following examples a particular procedure is adopted to show continuity or discontinuity of a function. The reader may adopt any other alternate procedure, for instance, any one of the characterizations in Theorem 2.2.1. Example 2.2.4 For given x0 ∈ R, let f (x) = |x − x0 |, x ∈ R. Then f is continuous on R. To see this, note that, for a ∈ R, | f (x) − f (a)| = ||x − x0 | − |a − x0 || ≤ |(x − x0 ) − (a − x0 )| = |x − a|. Hence, for every ε > 0, we have |x − a| < ε ⇒ | f (x) − f (a)| < ε. Example 2.2.5 Recall from Example 2.2.9 that lim

x→2

Hence, f defined by f (x) :=

x2 − 4 = 4. x −2  x 2 −4 x−2

4,

, x = 2, x =2

♦

100

2 Limit, Continuity and Differentiability of Functions

is continuous at 2. However, the function g(x) :=

 x 2 −4 x−2

α,

, x = 2, x =2

is not continuous at 2 for any α = 4. Also, if a = 2, then x − 2 is nonzero in a neighbourhood of a, and the functions x − 2 and x 2 − 4 are continuous on R. Hence, by Theorem 2.2.6, f is continuous at every a = 2 as well. By similar arguments, g is continuous at every point a = 2. ♦ Example 2.2.6 We already observed in Example 2.2.3 that the functions f, g, h defined by f (x) = sin x, g(x) = cos x, h(x) =

 sin x x

1,

, x = 0, x =0

are continuous at 0. Now, we show that they are continuous at every point in R. Note that for x, y ∈ R, sin x − sin y = 2 sin

x + y x − y cos 2 2

so that | sin x − sin y| ≤ |x − y|

∀x, y ∈ R.

Hence, for every ε > 0 and for every x0 ∈ R, x ∈ R, |x − x0 | < ε

⇒

| sin x − sin x0 | < ε.

Thus, f is continuous at every point in R. Since cos x = 1 − 2 sin2 (x/2), x ∈ R, it also follows that g is continuous at every point in R. Now, let x0 = 0. Then the continuity of h at x0 follows from Theorem 2.2.6, since h = f / f 0 where f 0 (x) = x ♦ in nonzero in a neighbourhood of x0 . Exercise 2.2.2 Prove continuity of the function f (x) := point in R by using ε − δ arguments.

sin x , x

x = 0 at every nonzero

Example 2.2.7 Let f be defined by f (x) = 1/x on (0, 1]. Then there does not exist a continuous function g on [0, 1] such that g(x) = f (x) for all x ∈ (0, 1]: Suppose there is a function g defined on [0, 1] such that g(x) = f (x) for all x ∈ (0, 1]. Then we have 1/n → 0 but g(1/n) = f (1/n) = n → ∞. Thus, g(1/n) → g(0). ♦ Exercise 2.2.3 Show by ε − δ arguments that f defined by f (x) = 1/x, x = 0, is continuous at every x0 = 0.

2.2 Continuity of a Function

101

√ Example 2.2.8 The function f defined by f (x) = x, x ≥ 0 is continuous at every x0 ≥ 0: Let ε > 0 be given. First consider the point x0 = 0. Then we have | f (x) − f (x0 )| =

√

x < ε whenever |x| < ε2 .

√ Thus, √ f is continuous at x0 = 0. Next assume that x0 > 0. Since |x − x0 | = ( x + √ √ x0 )| x − x0 |, we have √ √ |x − x0 | |x − x0 | . | x − x0 | = √ √ ≤ √ x0 x + x0 Thus,

√ √ √ | x − x0 | < ε whenever |x − x0 | < δ := ε x0 . ♦

More generally, we have the following example.

Example 2.2.9 Let k ∈ N. Then the function f defined by f (x) = x 1/k , x ≥ 0 is continuous at every x0 ≥ 0: Let ε > 0 be given. First consider the point x0 = 0. Then we have | f (x) − f (x0 )| = x 1/k < ε whenever |x| < εk . Thus, f is continuous at x0 = 0. Next assume that x0 > 0. Let y = x 1/k and y0 = 1/k x0 . Since y k − y0k = (y − y0 )(y k−1 + y k−2 y0 + · · · + yy0k−2 + y0k−1 ), so that 1/k

x − x0 = (x 1/k − x0 )(y k−1 + y k−2 y0 + · · · + yy0k−2 + y0k−1 ). Hence, |x − x0 |

1/k

|x 1/k − x0 | =

y k−1 + y k−2 y0 + . . . + yy0k−2 + y0k−1

≤

|x − x0 | y0k−1

.

Thus, 1/k

1−1/k

|x 1/k − x0 | < ε whenever |x − x0 | < δ := εy0k−1 = εx0 Thus, f is continuous at every x0 > 0.

. ♦

Remark 2.2.1 In the above two examples we assumed the knowledge of the squareroot and the k th -root of a positive number for k ∈ N. In school, for a > 0 and k ∈ N, the number a 1/k is defined as a number whose k th power is a, that is, (a 1/k )k = a.

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2 Limit, Continuity and Differentiability of Functions

However, existence of such a number is not established. In the next section, we shall prove that the function f k : (0, ∞) → (0, ∞) defined by f k (x) = x k , x ∈ (0, ∞) is bijective so that for every a > 0, there exists a unique number b > 0 such that ♦ bk = a. Such a number b is denoted by a 1/k , and called the k th -root of a. Example 2.2.10 For a rational number r , let f (x) = x r for x > 0. Then using Example 2.2.9 together with Theorem 2.2.3, we see that f is continuous at every ♦ x0 > 0. Remark 2.2.2 We know that for any given r ∈ R, there exists a sequence (rn ) of rational numbers such that rn → r . For n ∈ N, let f n (x) = x rn , x > 0. Since each f n is continuous on [0, ∞) one may ask whether lim f n (x)

n→∞

exists for each x ∈ [0, ∞), and if this limit exists, then one may define x r as x r = lim f n (x), x ∈ [0, ∞). n→∞

Also, one may want to know whether the function x → x r continuous. We shall discuss this issue in a latter section, where we shall introduce two important classes of functions, namely exponential and logarithm functions. In fact, our discussion will also include, as special cases, the Examples 2.2.8–2.2.10. ♦ In all the examples given above, either the function is continuous everywhere on the domain of the function or discontinuous at a one point in the domain of definition. Functions with only a finite number of discontinuities can be easily constructed. For instance, the function ⎧ 4x, 0 ≤ x < 1/4 ⎪ ⎪ ⎨ 4x − 1, 1/4 ≤ x < 1/2, f (x) = 4x − 2, 1/2 ≤ x < 3/4, ⎪ ⎪ ⎩ 4x − 3, 3/4 ≤ x ≤ 1 defined on the interval [0, 1] has three points of discontinuity. The discontinuities are at 1/4, 1/2 and 3/4. What about functions with infinite number of discontinuities?

2.2 Continuity of a Function

103

Look at the following examples: Example 2.2.11 Let f : [0, ∞) → R be defined by f (x) = x − n whenever n ≤ x < n + 1. Then f has discontinuities at every positive integer.

♦

Example 2.2.12 Let J = {1, 1/2, . . .} and f : [0, 1] → R be defined by  f (x) =

0, x ∈ / J, x, x ∈ J,

that is, f (1/n) = 1/n for all n ∈ N and f (x) = 0 for x ∈ / {1, 1/2, . . .}. Clearly, this function is not continuous at 1/n for any n ∈ N, and it is continuous at every x ∈ (0, 1] \ J . Also, it is continuous at 0. To see the last statement, consider a sequence (xn ) in [0, 1] such that xn → 0. Now, for a given ε > 0, let k ∈ N be such that 1/k < ε, and let Nk ∈ N be such that 0 ≤ xn < 1/k for all n ≥ Nk . Then we have | f (xn )| ≤ 1/k < ε ∀ n ≥ Nk . Thus, f (xn ) → 0 as n → ∞.

♦

Example 2.2.13 Let I be an interval, S := {an : n ∈ N} ⊆ I and let f : I → R be / S. Clearly, this function defined by f (an ) = 1/n for all n ∈ N and f (x) = 0 for x ∈ is not continuous at every x ∈ S. We show that f is continuous at every x ∈ I \ S. Let x0 ∈ I \ S. Then f (x0 ) = 0. For ε > 0, we have to find a δ > 0 such that |x − x0 | < δ implies | f (x)| < ε. For δ > 0, let Jδ := (x0 − δ, x0 + δ) ∩ I . For ε > 0, let k ∈ N be such that 1/k < ε. Choose δ > 0 such that / Jδ . a1 , a2 , . . . , ak ∈ For instance, we may choose 0 < δ < min{|x0 − ai | : i = 1, . . . , k}. Then we have Jδ ∩ {a1 , a2 , . . .} = {ak+1 , ak+2 , . . .}. Hence, for x ∈ Jδ , we have either f (x) = 0 or f (x) = 1/n for some n > k. Thus, | f (x)| ≤

1 < ε. k

Thus we have proved that f is continuous at x0 . If we take S to be the set of all rational numbers in I , then the corresponding function is continuous at every irrational number in I and discontinuous at every rational number in I . ♦

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2 Limit, Continuity and Differentiability of Functions

2.2.3 Some Properties of Continuous Functions Recall that a subset S of R is said to be bounded if there exists M > 0 such that |s| ≤ M for all s ∈ S, and a set which is not bounded is called an unbounded set. Recall also that if S is a bounded subset of R, then S has the infimum and the supremum, not necessarily in S. For S ⊆ R, we have the following: 1. Suppose S is bounded, and say α := inf S and β := sup S. Then there exist sequences (sn ) and (tn ) in S such that sn → α and tn → β. 2. S is unbounded if and only if there exists a sequence (sn ) in S which is unbounded. 3. S is unbounded if and only if there exists a sequence (sn ) in S such that |sn | → ∞ as n → ∞. 4. If (sn ) is a sequence in S which is unbounded, then there exists a subsequence (skn ) of (sn ) such that |skn | → ∞ as n → ∞. 5. If (sn ) is a sequence in S such that |sn | → ∞ as n → ∞, and if (skn ) is a subsequence of (sn ), then |skn | → ∞ as n → ∞. Exercise 2.2.4 Prove the above statements.



Theorem 2.2.7 Suppose f is a real valued continuous function defined on a closed and bounded interval [a, b]. Then f is a bounded function. Proof Assume for a moment that f is not a bounded function. Then, there exists a sequence (xn ) in [a, b] such that | f (xn )| → ∞. Since (xn ) is a bounded sequence, by Bolzano–Weierstrass theorem (Theorem 1.1.13), there exists a subsequence (xkn ) of (xn ) such that xkn → x for some x ∈ [a, b]. Therefore, by the continuity of f , f (xkn ) → f (x). In particular, ( f (xkn )) is a bounded sequence. This is a contradiction to the fact that | f (xn )| → ∞. Thus, we have proved that f cannot be unbounded. Remark 2.2.3 The conditions in Theorem 2.2.7 are only sufficient conditions; they are not necessary conditions. To see this consider the function  f (x) =

1, 0 < x ≤ 1, 2, 1 < x < ∞.

Then f defined on I = (0, ∞) is not continuous and I is neither closed nor bounded, but f is a bounded function. It is also true that, if we drop any of the conditions in the theorem, then the conclusion need not be true. To see this, consider the unbounded functions in the following examples: 

1. Let f (x) =

1/x, x ∈ (0, 1], 1, x = 0.

2.2 Continuity of a Function

105

In this case f is not continuous, though it is defined on a closed and bounded interval [0, 1]. 2. Let f (x) = 1/x, x ∈ (0, 1]. In this case f is continuous, but its domain (0, 1] is not a closed set. 3. Let f (x) = x, x ∈ [0, ∞). In this case f is continuous, but its domain [0, ∞) is not bounded. ♦ Maximum and Minimum Suppose f is a bounded function defined on an interval I . Then, we know that inf f (x) := inf{ f (x) : x ∈ I } and sup f (x) := sup{ f (x) : x ∈ I } x∈I

x∈I

exist. Definition 2.2.2 A function f defined on an interval I is said to attain (1) maximum at a point x1 ∈ I if f (x1 ) = supx∈I f (x), (2) minimum at a point x2 ∈ I if f (x2 ) = inf x∈I f (x). The function f is said to attain extremum at a point x0 ∈ I if f attains either ♦ maximum or minimum at x0 . If f attains maximum (respectively, minimum) at a point in I ,then we write maxx∈I f (x) for supx∈I f (x) (respectively, min x∈I f (x) for inf x∈I f (x)). A natural question would be the following: Under what conditions on f and I does the function attain maximum and minimum at some points in I?

In this regard, we have the following theorem. Theorem 2.2.8 Suppose f is a continuous function defined on a closed and bounded interval [a, b]. Then there exists x0 , y0 in [a, b] such that f (x0 ) = inf f (x) and f (y0 ) = sup f (x). a≤x≤b

a≤x≤b

Proof Let α = inf a≤x≤b f (x). By the definition of the infimum of a set, there exists a sequence (xn ) in [a, b] such that f (xn ) → α. Since (xn ) is a bounded sequence, by Bolzano–Weierstrass theorem, there exists a subsequence (xkn ) such that xkn → x0 for some x0 ∈ [a, b]. By continuity of f , f (xkn ) → f (x0 ). But, we already have f (xkn ) → α. Hence, α = f (x0 ), that is, f (x0 ) = inf a≤x≤b f (x). Similarly, using the definition of supremum, it can be shown that there exists y0 ∈ [a, b] such that f (y0 ) = supa≤x≤b f (x). Every continuous function f : [a, b] → R attain maximum and Minimum at some points in [a, b]

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2 Limit, Continuity and Differentiability of Functions

Remark 2.2.4 The conclusion in Theorem 2.2.8 need not hold if the domain of the function is not of the form [a, b] or if f is not continuous. For example, f : (0, 1] → R defined by f (x) = 1/x for x ∈ (0, 1] is continuous, but does not attain supremum. Same is the case if g : [0, 1] → R is defined by  g(x) =

1/x, x ∈ (0, 1], 1, x = 0.

Thus, neither continuity nor the fact that the domain is a closed and bounded interval can be dropped. This does not mean that the conclusion in the theorem does not hold for all those functions which are either discontinuous or their domains are not closed intervals. For example, consider f : [0, 1) → R defined by  f (x) =

0, x ∈ [0, 1/2), 1, x ∈ [1/2, 1).

Then we see that neither f is continuous, nor its domain is of the form [a, b]. But, f attains both its maximum and minimum. ♦ Intermediate Value Theorem Suppose f is a continuous real valued function defined on an interval. We ask the following question: Given any number γ lying between any two values α and β of the function f, does f attain the value γ at some point in the domain of definition of f?

The answer is in the affirmative. This result is known as the Intermediate value theorem. Theorem 2.2.9 (Intermediate value theorem (IVT)) Suppose f is a continuous function defined on a closed and bounded interval [a, b]. Let γ be a number between f (a) and f (b). Then there exists c ∈ [a, b] such that f (c) = γ (Fig. 2.11). Before giving its proof, let us look at the interpretations of the theorem geometrically and algebraically. Geometric interpretation: Consider the curve C represented by the equation y = f (x), where f is a continuous function on [a, b]. If γ lies between the values f (a) and f (b), then the curve C intersects with the straight line represented by the equation y = γ

Algebraic interpretation:

2.2 Continuity of a Function

107

Fig. 2.11 Intermediate Value Property

If f is a continuous function defined on [a, b] and γ lies between the values f (a) and f (b), then the equation f (x) = γ has at least one solution in [a, b]

Theorem 2.2.9 ensures that if γ lies between f (a) and f (b), then there exists at least one point c between a and b such that f (c) = γ . There can be more than one such points. Obviously, if f is a constant function, then every point in [a, b] is having this property. Even if the function is not constant in any subinterval, then also there can be more than one such points. In this regard, we may look at the Fig. 2.12. Also, for the function f given by f (x) = x 2 , x ∈ [−1, 1], we have f (−1/2) = 1/4 = f (1/2). Also, a function can take all intermediate values without being continuous. For example, the function f : [−1, 1] → R defined by  f (x) =

x, x ∈ [−1, 0], 1 − x, x ∈ (0, 1]

is not continuous at 0, but it takes all intermediate values between any two values. Now, we give the proof of Theorem 2.2.9 adapted from Ghorpade and Limaye [6]. Proof of Theorem 2.2.9 Without loss of generality, assume that f (a) < f (b). If γ is either f (a) or f (b), then there is nothing to prove. So, let f (a) < γ < f (b). Let

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2 Limit, Continuity and Differentiability of Functions

Fig. 2.12 Intermediate value property

S = {x ∈ [a, b] : f (x) < γ }. Note that S is non-empty (since a ∈ S) and bounded above (since x ≤ b for all x ∈ S). Let u := sup S. Then u ∈ [a, b] and there exists a sequence (u n ) in S such that u n → u. Hence, by continuity of f , f (u n ) → f (u). Since f (u n ) < γ for all n ∈ N, we have f (u) ≤ γ . Note that u = b, since f (u) ≤ γ < f (b). Now, let (vn ) be a sequence in (u, b) such that vn → u. Then, again by continuity of / S and hence f (vn ) ≥ γ . Therefore, f (u) ≥ γ . f , f (vn ) → f (u). Since vn > u, vn ∈ Thus, a ≤ u ≤ b and f (u) = γ . Example 2.2.14 Consider the function f (x) = x 5 + 2x 3 + 1, x ∈ R. Does there exists x0 ∈ R such that f (x0 ) = 0? The answer is in the affirmative by IVT, since f (−1) = −2 and f (1) = 4. Not only that, since f (x) → −∞ as x → −∞ and f (x) → ∞ as x → ∞, f attain every real value, that is, for every y ∈ R, there exists x ∈ R such that f (x) = y. In other words, f is an onto function. ♦ Example 2.2.15 Consider the function f (x) = x 3 + sin x + 1, x ∈ R. Does there exists x0 ∈ R such that f (x0 ) = 0? Note that f (0) = 1 and f (−2) = −8 + sin(−2) + 1 < 0.

2.2 Continuity of a Function

109

Hence, by IVT, there exists x0 ∈ R such that f (x0 ) = 0. As in the case of the last example, we see that f attain every real value. ♦ The following two corollaries are immediate consequences of IVT. Corollary 2.2.10 Let f be a continuous function defined on an interval. Then range of f is an interval. Corollary 2.2.11 Suppose f is a continuous function defined on an interval I . If a, b ∈ I are such that f (a) and f (b) have opposite signs, then there exists x0 ∈ I such that f (x0 ) = 0. Now, we derive another important property of continuous functions. Theorem 2.2.12 Suppose f is a continuous function defined on a closed and bounded interval I . Then its range is a closed and bounded interval. Proof We know, by Theorem 2.2.7 Corollary 2.2.10, that range of f is a bounded interval, say J . Hence, it is enough to show that J contains its endpoints. For this, let c be an end point of J . Let (yn ) in J such that yn → c. Let xn ∈ I be such that f (xn ) = yn , n ∈ N. Since I is closed and bounded, (xn ) has a subsequence (xkn ) which converges to some point x0 ∈ I . By continuity of f , ykn = f (xkn ) → f (x0 ). Thus, we obtain c = f (x0 ) ∈ J . This completes the proof. Continuity of the Inverse of a Function Suppose f is defined on a set D ⊆ R. We may recall the following from elementary set theory: If f is injective, i.e., one-one, then we know that a function g can be defined on the range E := f (D) of f by g(y) = x for y ∈ E, where x ∈ D is the unique element in D such that f (x) = y. The above function g is called the inverse of f . Note that the domain of the inverse of f is the range of f . By Corollary 2.2.10, we know that range of a continuous function defined on an interval I is also an interval. Suppose f is also injective. Then a natural question one would like to ask is whether its inverse is also continuous. First we answer this question affirmatively by assuming that the domain of the function is a closed and bounded interval. Theorem 2.2.13 (Inverse function theorem (IFT)) Let f be a continuous injective function defined on a closed and bounded interval I . Then its inverse from its range is continuous. Proof Suppose J = f (I ), the range of f . Let y0 ∈ J and (yn ) be an arbitrary sequence in J which converges to y0 . Let xn = f −1 (yn ), n ∈ N and x0 = f −1 (y0 ). We have to show that xn → x0 . Suppose, on the contrary, that xn → x0 . Then there exists ε0 > 0 and a subse/ (x0 − ε0 , x0 + ε0 ) for all n ∈ N. Since I is a quence (u n ) of (xn ) such that u n ∈ bounded interval, (u n ) is a bounded sequence. Hence, (u n ) has a subsequence (vn )

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2 Limit, Continuity and Differentiability of Functions

which converges to some v ∈ R. Since I is a closed interval, v ∈ I . Now, continuity of f implies that f (vn ) → f (v). But, since ( f (vn )) is a subsequence of (yn ), and since yn → y0 , we have f (v) = y0 = f (x0 ). Now, since f is injective, v = x0 . Thus we have proved that vn → x0 . This is a contradiction to the fact that / (x0 − ε0 , x0 + ε0 ) for all n ∈ N. vn ∈ Next we shall prove the conclusion in the last theorem by dropping the condition that I is closed and bounded, but assuming an additional condition on f , namely that it is strictly monotonic. Definition 2.2.3 Let f be defined on an interval I . Then (1) f is said to be monotonically increasing on I if x, y ∈ I, x < y

⇒

f (x) ≤ f (y),

(2) f is said to be strictly monotonically increasing on I if x, y ∈ I, x < y

⇒

f (x) < f (y),

(3) f is said to be monotonically decreasing on I if x, y ∈ I, x < y

⇒

f (x) ≥ f (y).

(4) f is said to be strictly monotonically decreasing on I if x, y ∈ I, x < y

⇒

f (x) > f (y).

If f is either monotonically increasing (respectively, strictly monotonically increasing) or monotonically decreasing (respectively, strictly monotonically decreasing) on I , then it is called a monotonic (respectively, strictly monotonic) function. ♦ Observe: f is strictly monotonic on I

⇒

f is injective on I .

The converse of the above statement is not true. For example, the function  f (x) =

x, −1 ≤ x ≤ 0, 1 − x, 0 < x ≤ 1,

is injective but not strictly monotonic on [−1, 1]. Sometimes, the terminology increasing, decreasing, strictly increasing, strictly decreasing, are used in place of monotonically increasing, monotonically decreasing, strictly monotonically increasing, and strictly monotonically decreasing, respectively. Example 2.2.16 We observe the following.

2.2 Continuity of a Function

111

(i) The function f (x) = x is strictly increasing on R. (ii) The function f (x) = −x is strictly decreasing on R. (iii) The function f (x) = x 2 is strictly increasing for x ≥ 0 and strictly decreasing for x ≤ 0. (iv) The function f (x) = x 3 is strictly increasing on R. (v) For k ∈ N, f (x) = x k is strictly increasing on [0, ∞). (vi) The function f (x) = sin x is strictly increasing on [0, π/2] and strictly decreasing on [π/2, π ]. (vii) The function f (x) = cos x is strictly decreasing on [0, π ]. ♦ Now, we prove a companion result to Theorem 2.2.13 without assuming that the domain of the function is closed and bounded, but assuming that the function is strictly monotonic. Theorem 2.2.14 (Inverse function theorem (IFT)) Let f be a continuous function defined on an interval I . Suppose f is strictly monotonic on I . Then f is injective and its inverse from its range is continuous. Proof We assume that f is strictly monotonically increasing. The case when f is strictly monotonically decreasing will follow by similar arguments. Since f is continuous, its range is also an interval, say J . By the assumption, for x1 , x2 ∈ J , x1 < x2 ⇒ f (x1 ) < f (x2 ). Hence, f is injective. Let g be its inverse from the range J . Let y0 ∈ J and (yn ) in J be such that yn → y0 . Let xn = g(yn ), n ∈ N and x0 = g(y0 ) for that yn = f (xn ) and y0 = f (x0 ) for all n ∈ N. We have to show that xn → x0 . Suppose xn → x0 . Then there exists ε > 0 and a subsequence (xkn ) of (xn ) such that / (x0 − ε, x0 + ε) ∀ n ∈ N. x kn ∈ Hence, for each n ∈ N, either xkn < x0 − ε or xkn > x0 + ε. Hence, using the fact that f is strictly monotonically increasing, we have the following, for each n ∈ N: x kn < x 0 − ε

⇒

x0 − ε ∈ I and f (xkn ) < f (x0 − ε) < f (x0 ),

x kn > x 0 + ε

⇒

x0 + ε ∈ I and f (xkn ) > f (x0 + ε) > f (x0 ).

/ Thus, ( f (x0 − ε), f (x0 + ε)) is an open interval containing f (x0 ) such that f (xkn ) ∈ ( f (x0 − ε), f (x0 + ε)) for every n ∈ N. Hence, we can conclude that ykn → y0 , which is a contradiction. Thus, we have proved that g is continuous. Theorem 2.2.15 (Root function) For every y ∈ [0, ∞) and k ∈ N, there exists a unique x ∈ [0, ∞) such that x k = y. Proof We have observed in Example 2.2.16 that for k ∈ N, the function f (x) = x k , x ≥ 0,

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2 Limit, Continuity and Differentiability of Functions

is strictly monotonically increasing. Also, f (0) = 0 and f (x) → ∞ as x → ∞. Hence, by IVT and IFT (Theorem 2.2.14), f : [0, ∞) → [0, ∞) is bijective and the inverse of f is continuous. In particular, for every y ∈ [0, ∞), there exists a unique x ∈ [0, ∞) such that x k = y. Definition 2.2.4 For k ∈ N and y > 0, the unique number x > 0 such that x k = y ♦ is called the k th -root of y, and it is denoted by y 1/k . Remark 2.2.5 (i) In the proof of Theorem 2.2.14, the continuity of f is used only to assert that its range J is an interval so that its inverse f −1 is defined on an interval. (ii) We know that strict monotonicity of a function implies that it is injective, but injectivity does not implies strict monotonicity. So, one may ask whether strict monotonicity assumption in Theorem 2.2.14 can be replaced by injectivity. The answer is in the affirmative as the following Exercise shows. ♦ Exercise 2.2.5 Let f be an injective function defined on an interval I . Show that if f is continuous, then it is strictly monotonic on I . [Hint: IVT].

2.2.4 Exponential and Logarithm Functions We have already come across the expression a b for a > 0 and b ∈ R, though we have not defined it explicitly. However, from elementary arithmetic we know the definition of a n for n ∈ N, and the relation, a m+n = a m a n for m, n ∈ N. Also, defining a −n = a1n for n ∈ N and using the convention a 0 = 1, we have a m+n = a m a n ∀ m, n ∈ Z. Also, we have defined a 1/n , the n th -root of a for any a > 0 and n ∈ N (cf. Theorem 2.2.15 and Definition 2.2.4). Thus, for any positive rational number r = m/n, we can define 1 a r := (a m )1/n , a −r := r . a Thus, we have defined a r for any a > 0 and r ∈ Q. Now, the question is, can we have a meaningful definition of a x for any x ∈ R? To this end, we shall first define e x , where e is the Euler constant defined by

 1 n e := limn→∞ 1 + n

or e :=

∞ 1 . n! n=0

1 We define e x for x ∈ R in such a way that e1 = ∞ n=0 n! . More specifically, we define the concept of an exponential function exp(x), x ∈ R, as

2.2 Continuity of a Function

113

exp(x) =

∞ xn , x ∈ R. n! n=0

Observe that the above series converges absolutely for every x ∈ R. This is easily seen by using the ratio test. This series plays very significant roles in mathematics. Definition 2.2.5 For x ∈ R, the function exp(x) :=

∞ xn , x ∈ R, n! n=0

♦

is called the exponential function. Clearly, exp(0) = 1, exp(1) = e. Our first attempt is to show that exp(r ) = er

for every rational number. In order to do that we have to derive some of the important properties of the function exp(x). For that purpose, we shall make use of the following result on convergence of series.

∞

Theorem 2.2.16 Suppose that ∞ n=0 an and n=0 bn are absolutely convergent series, and n ak bn−k , n ∈ N ∪ {0}. cn = k=0

Then, the series

∞

n=0 cn

converges absolutely and ∞ 

an

n=0

∞ 

∞  bn = cn .

n=0

n=0

∞

∞

Proof Since n=0 an and n=0 bn are absolutely convergent, they are convergent. Let their sums be A and B, respectively. Let An =

n i=0

ai ,

Bn =

n i=0

bi , Cn =

n

ci .

i=0

Then An → A, Bn → B and An Bn → AB. We have to prove that Cn → AB. First let us assume that the terms of the series are non-negative. Note that, if αi j = ai b j , i, j = 0, 1, . . . , n,

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2 Limit, Continuity and Differentiability of Functions

then An Bn is the sum of all entries of the matrix (αi j ) and Cn is the sum of the entries of the left upper triangular part of the matrix (αi j ), i.e., An Bn =

n n

αi j , Cn =

i=0 j=0

n n−i

αi j .

i=0 j=0

Hence, it follows that Cn ≤ An Bn ≤ C2n

(∗)

for all n ∈ N. Since (An Bn ) converges to AB and (Cn ) is an increasing sequence of nonnegative terms, the relation (∗) implies that (Cn ) is bounded, and hence it converges. Let Cn → C. Since (C2n ) is a subsequence of (Cn ), we also have the convergence C2n → C. Hence, by Sandwich theorem, Cn → AB. This proves the case when the series are with nonnegative terms. Next let us consider the general case. By what we have already proved, we have ∞ 

|ai |

i=0

Let Aˆ n =

n

∞ 



|bi | =

i=0

Bˆ n =

|ai |,

i=0

∞  k k=0

n

 |ai | |bk−i | .

i=0

|bi |, Cˆ n =

i=0

n  k k=0

 |ai | |bk−i | .

i=0

As in the last paragraph, we obtain Cˆ n ≤ Aˆ n Bˆ n ≤ Cˆ 2n . Note that |An Bn − Cn | ≤

n n

|αi j | = Aˆ n Bˆ n − Cˆ n ≤ Cˆ 2n − Cˆ n .

i=1 j=n−i+1

Since (Cˆ n ) converges, we obtain Cˆ 2n − Cˆ n → 0, and since An Bn → AB, we have the convergence Cn → AB.

n

∞ Definition 2.2.6

∞ n=0 cn with cn = k=0 ak bn−k is called the Cauchy

∞ The series ♦ product of n=0 an and n=0 bn . It is to be observed that the Cauchy product of two (non-absolutely)

convergent ∞ series need not be convergent. To see this one may consider the series n=1 an and √

∞ n+1 b with a = b = (−1) / n + 1 for n ∈ N. Since n n n n=1 cn =

n k=0

ak bn−k = (−1)n

n k=0

√

1 √ k+1 n−k+1

√

and (k + 1)(n − k + 1) ≤ n + 1 we have |cn | ≥ 1 for all n ∈ N. Hence, ∞ n=1 cn is not convergent.

2.2 Continuity of a Function

115

Properties of Exponential and Logarithm Functions Recall from Definition 2.2.5 that for x ∈ R, exp(x) :=

∞ xn , x ∈ R. n! n=0

Theorem 2.2.17 Let exp(·) be the function as in Definition 2.2.5. Then the following results hold. (i) exp(x + y) = exp(x) exp(y) ∀ x, y ∈ R (ii) exp(x) = 0 ∀ x ∈ R. 1 (iii) exp(−x) = ∀ x ∈ R. exp(x) (iv) exp(x) > 0 ∀ x ∈ R. (v) exp(kx) = [exp(x)]k ∀ x ∈ R, k ∈ Z. In particular, (a) exp(k) = ek , ∀ k ∈ Z, (b) [exp(1/k)]k = e ∀ k ∈ Z. (c) exp(m/n) = [exp(1/n)]m ∀ m, n ∈ Z with n = 0. (vi) (vii) (viii) (ix)

exp(x) > 1 ⇐⇒ x > 0 and exp(x) = 1 ⇐⇒ x = 0. x > y ⇐⇒ exp(x) > exp(y). exp(x) → ∞ as x → ∞. exp(x) → 0 as x → −∞.

Proof Note that, for x, y ∈ R, x k y n−k (x + y)n 1 n! = x k y n−k = . n! n! k=0 k!(n − k)! k! (n − k)! k=0 n

n

Hence, by Theorem 2.2.16 by taking an = x n /n! and bn = y n /n!, we have ∞ ∞ ∞  (x + y)n x n  y n  = . n! n! n! n=0 n=0 n=0

This proves (i). The results in (ii), (iii) and (v) follow from (i), and the result in (iv) follows from (iii), since exp(x) > 0 for x ≥ 0. To see (vi), observe that x > 0 implies exp(x) > 1. Next, suppose x ≤ 0. If x = 0, then exp(x) = exp(0) = 1. If x < 0, then taking y = −x, we have y > 0, and hence exp(y) > 1. Thus, 1/ exp(x) = exp(−x) > 1 so that exp(x) < 1. Hence, exp(x) > 1 ⇐⇒ x > 0. From the above arguments, we also obtain exp(x) = 1 ⇐⇒ x = 0. The result in (vii) follows from the facts that x > y ⇐⇒ x − y > 0 ⇐⇒ exp(x − y) > 1

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2 Limit, Continuity and Differentiability of Functions

and the relation exp(x − y) = exp(x)/ exp(y), which is a consequence of (i) and (iii). The result in (viii) follows from the relation exp(x) = 1 + x +

∞ xn ≥ 1 + x ∀ x > 0, n! n=2

and (ix) is a consequence of (iii) and (viii). In view of (v)(b) above, we may define e1/k := exp(1/k) ∀ k ∈ N, and hence by (v)(c), em/n := [e1/n ]m ∀ m, n ∈ N. Thus, for every rational number r , we can define er := exp(r ) which satisfies the usual index laws. We know that every real number is a limit of a sequence of rational numbers. Thus, if x ∈ R, there exists a sequence (xn ) of rational numbers that xn → x. So, we may define e x = lim e xn n→∞

provided the above limit exists. Thus, our next attempt is to show that the function exp(x), x ∈ R, is continuous. Theorem 2.2.18 The function exp(·) is continuous on R Proof For brevity of expression, let us use the notation e x for exp(x). Let x, x0 ∈ R. Then, by Theorem 2.2.17(i), we have e x − e x0 = e x0 (e x−x0 − 1) ∞ (x − x0 )n = e x0 n! n=1 ∞ (x − x0 )n−1 . = e (x − x0 ) n! n=1 x0

Thus, if |x − x0 | ≤ 1, then |e x − e x0 | ≤ e x0 |x − x0 |

∞ 1 = e x0 (e − 1)|x − x0 |. n! n=1

2.2 Continuity of a Function

117

From this, we obtain that exp(·) is a continuous function on R. Notation 2.2.1 We know that for every x ∈ R, there exists a sequence (xn ) of rational numbers such that xn → x. In view of Theorem 2.2.18, e xn = exp(xn ) → exp(x). Hence, we shall use the notation e x for exp(x) for every x ∈ R, i.e., e x :=

∞ xn , x ∈ R. n! n=0

♦

With the above notation we have the identity: e x+y = e x e y ∀ x, y ∈ R. Theorem 2.2.19 The function exp(·) is bijective from R to (0, ∞). Proof Using the notation e x := exp(x), x ∈ R, first we observe that e x > 0 for every x ∈ R, and for x1 , x2 in R e x2 − e x1 = e x1 [e x2 −x1 − 1]. Thus,

e x2 = e x1 ⇐⇒ e x2 −x1 = 1 ⇐⇒ x1 = x2 ,

showing that the function x → e x is one-one. Next, we show that the function is onto. Let y ∈ (0, ∞). Recall that (cf. Theorem 2.2.17) e x → ∞ as x → ∞. e x → 0 as x → −∞, Hence, there exists M1 > 0 such that e x > y for all x > M1 , and there exists M2 > 0 such that e x < y for all x < −M2 . Now, taking x1 > M1 and x2 < −M2 , we obtain e x2 < y < e x1 . Hence, by the intermediate value property, there exists x ∈ (x2 , x1 ) such that e x = y. Definition 2.2.7 For b > 0, the unique a ∈ R such that ea = b is called the natural logarithm of b, and it is denoted by ln b. The function ln x, x > 0,

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2 Limit, Continuity and Differentiability of Functions

♦

is called the natural logarithm function. Definition 2.2.8 For a > 0 and x ∈ R, we define a x := e x ln a .

♦

Remark 2.2.6 We note that ln e = 1 so that if a = e, then the Definition 2.2.8 matches with Definition 2.2.5. ♦ Theorem 2.2.20 Let a > 0. Then the function x → a x is continuous and bijective from R to (0, ∞). Proof By Definition 2.2.8, a x := e x ln a for x ∈ R. Hence, the result is a consequence of Theorems 2.2.18 and 2.2.19, and using the fact that composition of two continuous functions is continuous. The following corollary is immediate from the above theorem. Corollary 2.2.21 Let a > 0 be given. Then for every x > 0, there exists a unique b ∈ R such that a b = x. Definition 2.2.9 Let a > 0. For x > 0, the unique b ∈ R such that a b = x is called the logarithm of x to the base a, and it is denoted by loga x. The function loga x,

x > 0, ♦

is called the logarithm function. Note that, for x > 0,

loge x = ln x.

We observe that following. 1. For y ∈ R, y = ln x ⇐⇒ e y = x. 2. For a > 0 and y ∈ R, y = loga x ⇐⇒ a y = x. ln x . 3. For a > 0 and x > 0, loga x = ln a 4. For a > 0, b > 0, show that (logb a)(loga b) = 1. Exercise 2.2.6 Prove the above results.



Theorem 2.2.22 For a > 0, the function loga x, x > 0, is continuous on (0, ∞). Proof First we show that ln x, x > 0 is continuous. Then the continuity of loga x, x > 0 will follow from the relation loga x = ln x/ ln a. Let x, x0 belong to the interval (0, ∞), and let y = ln x and y0 = ln x0 . Then we have e y = x and e y0 = x0 . Assume, without loss of generality that x > x0 . Since ea > 1 if and only if a > 0, we have y > y0 , and hence

2.2 Continuity of a Function

119

x − x0 = e y − e y0 = e y0 (e y−y0 − 1) ∞ (y − y0 )n ≥ e y0 (y − y0 ). = e y0 n! n=1 Hence, |y − y0 | ≤ e−y0 |x − x0 |, that is, | ln x − ln x0 | ≤ e−y0 |x − x0 |, showing the continuity of ln x, x > 0. Example 2.2.17 Using continuity of the function x → log x, x > 0, we have lim

n→∞

log(1 + n) = 0. n

Indeed, since (1 + n)1/n → 1, we have log(1 + n) = log(1 + n)1/n → log 1 = 0. n

♦

Theorem 2.2.23 For r ∈ R, the function f : (0, ∞) → R be defined by f (x) = x r ,

x ∈ (0, ∞)

is continuous. Proof For r ∈ R and x > 0, we have x r = er ln x . Hence, the result follows from Theorem 2.2.22 and Theorem 2.2.3. Notation 2.2.2 Often, the notation log x is used for the natural logarithm function in place ln x. ♦

2.3 Differentiability of a Function 2.3.1 Definition and Examples Let f be a continuous function defined on an open interval I and x0 ∈ I . In the Cartesian plane, consider the curve represented by the equation y = f (x), x ∈ I.

120

2 Limit, Continuity and Differentiability of Functions

Fig. 2.13 Derivative

Then, for x in a neigbourhood of x0 , the quotient f (x) − f (x0 ) x − x0 is the slop of the line segment joining the points X 0 = (x0 , f (x0 )) and X = (x, f (x)). As x → x0 , the point X approaches X 0 , so that, if the limit lim

x→x0

f (x) − f (x0 ) x − x0

exists, then it can be thought of as the slope of the tangent to the curve at the point X 0 . Definition 2.3.1 Suppose f is a (real valued) function defined on an open interval I and x0 ∈ I . Then f is said to be differentiable at x0 if lim

x→x0

f (x) − f (x0 ) x − x0

exists, and in that case the value of the limit is called the derivative of f at x0 , and ♦ this value is denoted by f  (x0 ) (Fig. 2.13). Notation 2.3.1 The number f  (x0 ), namely, the derivative of f at x0 is also denoted by d df (x0 ) or f (x)|x=x0 . dx dx

2.3 Differentiability of a Function

121

df , introduced by Leibnitz, is useful in realizing that the expression The notation dx d as an operator which associates each differentiable function f to the function d x f . Writing the function f as an equation y = f (x), x ∈ I, and denoting x a small number such that x + x ∈ I whenever x ∈ I , the notation y is used for the difference f (x + x) − f (x) so that f (x + x) − f (x) y = . x x The above expression is sometimes called as the differential quotient of f at x. In view of this, the notation ddyx is used for f  (x), that is, dy y = f  (x) = lim x→0 . dx x

♦

A physical interpretation of the derivative: Suppose a particle is moving along a straight line. Suppose the particle is at points A and B at time t0 and t, which are at distances f (t0 ) and f (t), respectively, from a fixed point O on the line. Then the ratio f (t) − f (t0 ) t − t0 can be thought of as the average velocity of the particle while moving from A to B during the time interval [t0 , t], and the limit lim

t→t0

f (t) − f (t0 ) t − t0

can be thought of as the instantaneous velocity at the point A. Let f be a real valued function defined on an open interval I containing x0 . We observe the equivalence of the following statements. 1. f is differentiable at x0 ∈ I . f (x0 + h) − f (x0 ) exists. 2. The limit limh→0 h 3. For every sequence (xn ) in I \ {x0 } such that xn → x0 , the limit f (xn ) − f (x0 ) exists, and in that case limn→∞ xn − x0 f  (x0 ) = lim

n→∞

f (xn ) − f (x0 ) . xn − x0

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2 Limit, Continuity and Differentiability of Functions

Convention: Whenever we say that “a function f is differentiable at a point x0 ”, we mean that f is a real valued function defined on an open interval I containing x0 and f : I → R is differentiable at x0 . Example 2.3.1 Let us look at the following simple examples. (i) Derivative of a constant function is 0 at every point. Indeed, for c ∈ R, if f (x) = c for all x ∈ R, then we have f (x) − f (x0 ) =0 x − x0 for every x, x0 ∈ R with x = x0 . hence, f  (x0 ) = 0. (ii) Let f (x) = x, x ∈ R. Then for any x0 ∈ R, f (x) − f (x0 ) = 1 ∀ x = x0 . x − x0 Hence f  (x0 ) = 1. More generally, let f be a polynomial function, that is, f (x) = a0 + a1 x + · · · + an x n , x ∈ R for some n ∈ N. Then, for any x0 ∈ R, we have f (x) − f (x0 ) = a1 (x − x0 ) + · · · + an (x n − x0n ). We know that, for any k ∈ N, x k − x0k = (x − x0 )

k

x k−i x0i−1 .

i=1

Thus, f (x) − f (x0 ) =

n

ak (x k − x0k ) = (x − x0 )g(x),

k=1

where g(x) :=

n k=1

Thus,

ak

k

x k−i x0i−1 .

i=1

f (x) − f (x0 ) = lim x→x0 g(x) = g(x0 ) = kak x0k−1 , x − x0 k=1 n

lim

x→x0

2.3 Differentiability of a Function

123

and therefore, f  (x0 ) =

n

kak x0k−1 .

k=1

In particular, for n ≥ 2, f  (x0 ) = a1 + 2a2 x0 + · · · + nan x0n−1 . (iii) Let f (x) = sin x, x ∈ R. Then for any x, x0 ∈ R with x = x0 ,  0  0 2 cos x+x sin x−x f (x) − f (x0 ) 2 2 = x − x0 x − x0  0  x + x0  sin x−x 2 = cos . x−x0 2 2 Then, using the composition rule for limits and the facts that lim cos y = cos y0 and lim y→0

y→y0

we obtain f  (x0 ) = lim

x→x0

sin y = 1, y

f (x) − f (x0 ) = cos x0 . x − x0

(iv) The function e x is differentiable at every x ∈ R and (e x ) = e x

∀ x ∈ R.

To see this, first we note that for h = 0, ∞ ∞ e x+h − e x h n−2 ex ex h n − e x = (eh − 1 − h) = = ex h . h h h n=2 n! n! n=2

Hence, |h| ≤ 1 implies  x+h  ∞ e  − ex 1 x x  ≤ e − e = e x |h|(e − 2). |h|   h n! n=2

Thus, we have proved that the function e x is differentiable at x and its deriva♦ tive is e x . Remark 2.3.1 In deriving the result in Example 2.3.1(iv), we used the following

∞ fact: If (an ) is a sequence such that ∞ n=1 an converges absolutely, then n=1 an converges and

124

2 Limit, Continuity and Differentiability of Functions ∞ ∞     an  ≤ |an |.  n=1

♦

n=1

Many of the functions that occur in mathematics can be constructed with the help of the functions considered in the Example 2.3.1 using some properties of differentiation. In Example 2.3.1(ii), we have seen that if f is a polynomial, then we can write f (x) − f (x0 ) = (x − x0 )g(x), where g is a polynomial, and therefore, f  (x0 ) = g(x0 ). Do we have similar situation for a general differentiable function? In this regard, we have the following theorem. Theorem 2.3.1 (Caratheodory criterion1 ) Suppose f is defined in an open interval I . Then f is differentiable at x0 ∈ I if and only if there exists a function g : I → R which is continuous at x0 and satisfies f (x) = f (x0 ) + g(x)(x − x0 ), and in that case f  (x0 ) = g(x0 ). Proof Suppose f is differentiable at x0 ∈ I . Then the function g defined on I by  g(x) =

f (x)− f (x0 ) , x−x0 f  (x0 ),

x = x0 , x = x0 ,

satisfies the requirements. Conversely, suppose that there exists a function g : I → R which is continuous at x0 and satisfies f (x) = f (x0 ) + g(x)(x − x0 ). Then we have lim

x→x0

f (x) − f (x0 ) = lim g(x) = g(x0 ) x→x0 x − x0

so that f is differentiable at x0 ∈ I and f  (x0 ) = g(x0 ). Let us illustrate the above theorem by some examples. √ Example 2.3.2 Let f (x) = x, x > 0. For x0 , x > 0, we have f (x) − f (x0 ) =

√

x−

√

x0 = √

x − x0 √ = (x − x0 )g(x), x + x0

√ √ where g(x) = 1/( x + x 0 ). Since g√ is continuous at x0 , we can infer from The♦ orem 2.3.1 that f  (x0 ) = g(x0 ) = 1/(2 x 0 ). 1

Constantin Carathéodory (September 13, 1873 February 2,1950) was a Greek mathematician who made significant contributions to the theory of functions of a real variable, the calculus of variations, and measure theory.

2.3 Differentiability of a Function

125

Example 2.3.3 Let f (x) = e x :=

∞ xn , x ∈ R. Then we have n! n=0

f (x) − f (x0 ) =

∞ x n − x0n n! n=1

But, x − n

x0n

= (x − x0 )

n

x n−i x0i−1 .

i=1

Thus, formally, f (x) − f (x0 ) = (x − x0 ) with g(x) =

∞ n 1  n−i i−1  = (x − x0 )g(x), x x0 n! i=1 n=1

∞ n 1  n−i i−1  x x0 . n! i=1 n=1

If we can show that the series defining g(x) actually converges and the limit is a continuous function, then we have g(x0 ) = e x0 . For showing the convergence of  ∞ 1  n x n−i x0i−1 to a continuous function, we need to use the concept n=1 n! i=1 of uniform convergence of sequence of functions which we shall do in a later chapter of this book. ♦

2.3.2 Left and Right Derivatives Recall that in the definition of continuity of a function we considered the domain of the function to be an interval, not necessarily an open interval, whereas in the definition of differentiability we took the interval to be an open interval. Even in the definition of differentiability we could have taken an arbitrary interval I and x0 to be an end point of I if it belongs to that interval. In such case, we have the so called right differentiability or left differentiability at x0 depending on whether x0 is a right end point or left end point of I . In fact right differentiability and left differentiability can be defined at an interior point as well. By an interior point of an interval I we mean those points in I which are not the end points. More generally: Definition 2.3.2 A a point a ∈ R is said to be an interior point of a set D ⊆ R if D contains an open neighbourhood of a. ♦

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2 Limit, Continuity and Differentiability of Functions

Definition 2.3.3 Let f be a real valued function defined on an interval I and x0 ∈ I . (1) If x0 is such that (x0 − δ0 , x0 ] ⊆ I for some δ0 > 0, then f is said to be left differentiable at x0 if f − (x0 ) := lim x→x0 −

f (x) − f (x0 ) exists, x − x0

and in that case, it is called the left derivative of f at x0 . (2) If x0 is such that [x0 , x0 + δ0 ) ⊆ I for some δ0 > 0, then f is said to be right differentiable at x0 if f + (x0 ) := lim

x→x0 +

f (x) − f (x0 ) exists, x − x0 ♦

and in that case, it is called the right derivative of f at x0 . 

Remark 2.3.2 In some of the books in calculus, one may find the notations f (x0 −) and f  (x0 +) for left derivative and right derivative, respectively, at x0 . We preferred to use the notations f − (x0 ) and f + (x0 ) as the notations f  (x0 −) and f  (x0 +) can be confused with the left and right limits of the function f  at the point x0 . Left derivative and right derivative at a point can be characterized as follows: Let f be a real valued function defined on an interval I and x0 ∈ I . (i) If (x0 − δ0 , x0 ] ⊆ I for some δ0 > 0, then f − (x0 ) exists if and only if for every f (x0 ) sequence (xn ) in (x0 − δ0 , x0 ) with xn → x0 , we have limn→∞ f (xxnn)− exists, and −x0 in that case f (xn ) − f (x0 ) . f − (x0 ) = lim n→∞ xn − x0 (ii) If [x0 , x0 + δ0 ) ⊆ I for some δ0 > 0, then f + (x0 ) exists if and only if for every f (x0 ) sequence (xn ) in (x0 , x0 + δ0 ) with xn → x0 , we have limn→∞ f (xxnn)− exists, and −x0 in that case f (xn ) − f (x0 ) . ♦ f + (x0 ) = lim n→∞ xn − x0 In view of the above discussion, we have the following: If x0 is an interior point of I , then f  (x0 ) exists if and only if f + (x0 ) and f − (x0 ) exists and f  (x0 ) = f + (x0 ) = f − (x0 ).

Example 2.3.4 Let

 f (x) =

Then we have following.

0, x ∈ [−1, 0), 1, x ∈ [0, 1].

2.3 Differentiability of a Function

1. 2. 3. 4.

f f f f

127

is differentiable at every x0 ∈ (−1, 0) ∪ (0, 1), and f  (x0 ) = 0, is right differentiable at −1 and f + (−1) = 0, is right differentiable at 0 and f + (0) = 0, left derivative at 0 does not exist. is left differentiable at 1, and f − (1) = 0.

Here are some details of the above statements: (1) If x0 ∈ (−1, 0) ∪ (0, 1), then lim

x→x0

f (x) − f (x0 ) 0 = lim = 0. x→x0 x − x 0 x − x0

(2) If x0 = −1, then lim+

f (x) − f (x0 ) 0−0 = lim x→x0+ = 0. x − x0 x − x0

lim+

f (x) − f (x0 ) 1−1 = lim x→x0+ =0 x − x0 x − x0

x→x0

(3) If x0 = 0, then

x→x0

and lim

x→x0−

f (x) − f (x0 ) 0−1 does not exist. = lim− x − x0 x x→x0

(4) If x0 = 1, then lim

x→x0−

f (x) − f (x0 ) 1−1 = lim− = 0. x − x0 x→x0 x − x 0

♦

Example 2.3.5 Consider the signum function, f (x) = sgn(x), x ∈ R, that is, f : R → R is defined by  x/|x| if x = 0, f (x) = 0 if x = 0. Note that f (x) = 1 for x > 0, f (x) = −1 for x < 0. Hence, we obtain f  (x) = 0 for every x = 0. Note that f (x) − f (0) = x



1/x, x > 0 −1/x, x < 0.

Hence, neither f + (0) nor f − (0) exists. However, lim f  (x) = 0 = lim x→0+ f  (x).

x→0−

♦

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2 Limit, Continuity and Differentiability of Functions

Example 2.3.6 Let f : R → R be defined by  f (x) =

1 − |x| if x ∈ [−1, 1], 0 if x ∈ / [−1, 1].

⎧ ⎨ 1 + x if x ∈ [−1, 0), f (x) = 1 − x if x ∈ [0, 1], ⎩ 0 if x ∈ / [−1, 1].

That is,

/ {−1, 0, 1}. Let us consider the situations at Clearly, f is differentiable at every x0 ∈ the points −1, 0, 1. (i) Let x0 = −1. Then f (x) − f (x0 ) 0−0 = lim = 0, x→x0 − x − (−1) x − x0

lim

x→x0 −

lim

x→x0 +

f (x) − f (x0 ) (1 + x) − 0 = 1. = lim x→x0 + x − (−1) x − x0

Hence, f − (−1) = 0 and f + (−1) = 1. (ii) Let x0 = 0. Then lim

x→x0 −

lim

x→x0 +

f (x) − f (x0 ) (1 + x) − 1 = 1, = lim x→x0 − x − x0 x −0 f (x) − f (x0 ) (1 − x) − 1 = −1. = lim x→x0 + x − x0 x −0

Hence, f − (0) = 1 and f + (0) = −1. (iii) Let x0 = 1. Then lim

x→x0 −

lim

f (x) − f (x0 ) (1 − x) − 0 = −1, = lim x→x − 0 x − x0 x −1

x→x0 +

f (x) − f (x0 ) 0−0 = 0. = lim x→x0 + x − x0 x −1

Hence, f − (1) = −1 and f + (0) = 0. Thus left and right derivatives of f at the points −1, 0, 1 exist, but they are not the same, and hence f is not differentiable at these points. ♦

2.3 Differentiability of a Function

129

2.3.3 Some Properties of Differentiable Functions The proof of the following theorem is easy, and hence it is left as an exercise. Theorem 2.3.2 Suppose f and g are differentiable at a point x0 and α ∈ R. Then f + g and α f are differentiable at x0 , and ( f + g) (x0 ) = f  (x0 ) + g  (x0 ),

(α f ) (x0 ) = α f  (x0 ).

Here is a necessary condition for differentiability. Theorem 2.3.3 (Differentiability implies continuity) Suppose f is differentiable at a point x0 . Then f is continuous at x0 . Proof Let x0 ∈ I . Then for every x ∈ I with x = x0 , we have  f (x) − f (x0 ) =

 f (x) − f (x0 ) (x − x0 ). x − x0

Hence, lim x→x0 ( f (x) − f (x0 )) = f  (x0 ).0 = 0, showing that f is continuous at x0 . Exercise 2.3.1 Prove that if f is left (respectively, right) differentiable at x0 , then it

is left (respectively, right) continuous at x0 . For the following theorem, we may recall that if a function g is continuous at a point x0 and g(x0 ) = 0, then there exists an open interval I0 containing x0 such that g(x) = 0 for all x ∈ I0 (cf. Theorem 2.2.5). Theorem 2.3.4 (Product and quotient rules) Suppose f and g are defined on an open interval I and differentiable at x0 ∈ I . Then f g is differentiable at x0 , and ( f g) (x0 ) = f  (x0 )g(x0 ) + f (x0 )g  (x0 ). If g is nonzero in a neighbourhood of x0 , then f /g is differentiable at x0 , and  f  g(x0 ) f  (x0 ) − f (x0 )g  (x0 ) (x0 ) = . g [g(x0 )]2 Proof Let ϕ := f g and ψ := f /g. For any x ∈ I , we have ϕ(x) − ϕ(x0 ) = f (x)g(x) − f (x0 )g(x0 ) = [ f (x) − f (x0 )]g(x) + f (x0 )[g(x) − g(x0 )] so that, using the facts that f  (x0 ) and g  (x0 ) exist and g is continuous at x0 , we obtain

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ϕ(x) − ϕ(x0 ) f (x) − f (x0 ) g(x) − g(x0 ) = g(x) + f (x0 ) x − x0 x − x0 x − x0 → f  (x0 )g(x0 ) + f (x0 )g  (x0 ) as x → x0 . Hence, ϕ is differentiable at x0 , and ϕ  (x0 ) = f  (x0 )g(x0 ) + f (x0 )g  (x0 ). Also, for x ∈ I with x = x0 , we have f (x)g(x0 ) − f (x0 )g(x) g(x)g(x0 ) [ f (x) − f (x0 )]g(x0 ) − f (x0 )[g(x) − g(x0 )] = , g(x)g(x0 )

ψ(x) − ψ(x0 ) =

so that, using differentiability of f and g at x0 and continuity of g at x0 , we have   f (x) − f (x0 ) 1 g(x) − g(x0 ) ψ(x) − ψ(x0 ) = g(x0 ) − f (x0 ) x − x0 g(x)g(x0 ) x − x0 x − x0   f (x0 )g(x0 ) − f (x0 )g (x0 ) → [g(x0 )]2 as x → x0 . Thus, ψ is differentiable at x0 , and ψ  (x0 ) =

g(x0 ) f  (x0 ) − f (x0 )g  (x0 ) . [g(x0 )]2

This completes the proof. Theorem 2.3.5 (Composition rule) Suppose f is differentiable at x0 and g is differentiable at y0 := f (x0 ). Then g ◦ f is differentiable at x0 and (g ◦ f ) (x0 ) = g  (y0 ) f  (x0 ). Proof Let (xn ) be a sequence in a deleted neighbourhood of x0 which converges to (g ◦ f )(xn ) − (g ◦ f )(x0 ) exists and the limit is x0 . We have to prove that limn→∞ xn − x0   g (y0 ) f (x0 ). For this, let yn := f (xn ) for n ∈ N and y0 = f (x0 ). Let us look at the formal expression (g ◦ f )(xn ) − (g ◦ f )(x0 ) g(yn ) − g(y0 ) = xn − x0 xn − x0 g(yn ) − g(y0 ) f (xn ) − f (x0 ) = × . yn − y0 xn − x0

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131

f (xn ) − f (x0 ) = f  (x0 ). However, we will not be able xn − x0 g(yn ) − g(y0 ) to write limn→∞ = g  (x0 ), because (yn ) may not be in a deleted neighyn − y0 bourhood of y0 , although yn → y0 , by continuity of f at x0 . To take care of this situation, for each n ∈ N, we define Since f  (x0 ) exists, limn→∞

αn =

 g(yn )−g(y0 ) yn −y0

g  (y0 )

if yn = y0 , if yn = y0 .

Then αn → g  (y0 ). Hence, (g ◦ f )(xn ) − (g ◦ f )(x0 ) f (xn ) − f (x0 ) = αn × → g  (y0 ) f  (x0 ) xn − x0 xn − x0 showing that (g ◦ f ) (x0 ) = g  (y0 ) f  (x0 ). In view of the formula in Theorem 2.3.5, the following result is not surprising. Theorem 2.3.6 Suppose g ◦ f is differentiable at x0 , g is differentiable at y0 := f (x0 ) with g  (y0 ) = 0, and f is continuous at x0 . Then f is differentiable at x0 and f  (x0 ) =

(g ◦ f ) (x0 ) . g  (y0 )

Proof Let (xn ) be a sequence in a deleted neighbourhood of x0 which converges to x0 , yn := f (xn ) for n ∈ N and y0 = f (x0 ). Let (αn ) be as in the proof of Theorem 2.3.5, that is,  g(yn )−g(y0 ) if yn = y0 , yn −y0 αn = if yn = y0 . g  (y0 ) Since f is continuous at x0 , yn → y0 . Hence, αn → g  (y0 ) = 0 and αn = 0 for all large enough n. Then, for all such n, we have (g ◦ f )(xn ) − (g ◦ f )(x0 ) f (xn ) − f (x0 ) g(yn ) − g(y0 ) = = αn , xn − x0 xn − x0 xn − x0 so that f (xn ) − f (x0 ) 1 (g ◦ f )(xn ) − (g ◦ f )(x0 ) = × xn − x0 αn xn − x0 (g ◦ f ) (x0 ) as n → ∞. → g  (y0 ) Thus f  (x0 ) exists and f  (x0 ) =

(g ◦ f ) (x0 ) g  (y0 )

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As a corollary to the above theorem we have a formula for the derivative of the inverse of a function. Theorem 2.3.7 Suppose f : I → J is a bijective function between open intervals I and J . Suppose f is differentiable at a point x0 ∈ I and f  (x0 ) = 0 and f −1 is continuous at y0 := f (x0 ). Then f −1 is differentiable at y0 , and ( f −1 ) (y0 ) =

1 . f  (x0 )

Proof Note that ( f ◦ f −1 )(y) = y for every y ∈ J . Hence by Theorem 2.3.6, f −1 is differentiable at y0 and ( f −1 ) (y0 ) = 1/ f  (x0 ). Remark 2.3.3 Recall that in Theorem 2.3.6 and Theorem 2.3.7 we assumed g  (y0 ) = 0 and f  (x0 ) = 0, respectively. Can we obtain at least differentiability without the above assumptions? Theorem 2.3.5 shows that the condition f  (x0 ) = 0 is necessary in Theorem 2.3.7 for the differentiability of f −1 at x0 . What about the case of Theorem 2.3.6? That is, can we say that f is differentiable at x0 and f  (x0 )g  (y0 ) = (g ◦ f ) (x0 )? The answer is not in the affirmative. For example, consider f (x) = |x|, g(x) = x 2 , x ∈ R. Then (g ◦ f )(x) = x 2 . In this case, g ◦ f and g are differentiable at every point in ♦ R, but f is not differentiable at x0 = 0. Note that g  (y0 ) = 0. Example 2.3.7 In the following we consider some applications of Theorems 2.3.2– 2.3.7. (i) For n ∈ N, let f (x) = x n , x ∈ R. We have already seen that f  (x) = nx n−1 for x ∈ R. This can also be seen by the product formula: Let f k (x) := x k for k ∈ N. Then we have f k (x) = x f k−1 (x) for all k ∈ N. Now, f 1 (x) = 1. Assuming that f k (x) = kx k−1 for some k ∈ N, we have  (x) = f k+1

d [x f k (x)] = x f k (x) + f k (x) = x(kx k−1 ) + x k = (k + 1)x k . dx

Thus, by induction, ddx x n = nx n−1 for all x ∈ R. (ii) Let f (x) = cos x. Since cos x = 1 − 2 sin2 (x/2), x ∈ R, applying Theorems 2.3.2 and 2.3.5, d d d [1 − 2 sin2 (x/2)] = [−2 sin2 (x/2)] = −2 sin2 (x/2), dx dx dx d d sin2 (x/2) = 2 sin(x/2) sin(x/2) = 2 sin(x/2)[(1/2) cos(x/2)]. dx dx

2.3 Differentiability of a Function

133

Hence, f  (x) = −2

d sin2 (x/2) = −2 sin(x/2) cos(x/2) = − sin x. dx

(iii) Let f (x) = tan x for x ∈ D := {x ∈ R : cos x = 0}. Then f  (x) =

cos x cos x − sin x(− sin x) 1 d sin x = = = sec2 x. 2 d x cos x cos x cos x

(iv) By Theorem 2.3.7, if y = tan x for −π/2 < x < π/2, then d 1 1 1 1 . tan−1 y = = = =  2 2 dy tan x sec x 1 + tan x 1 + y2 Thus,

d tan−1 y = 1/(1 + y 2 ) for all y ∈ R. dy

♦

Example 2.3.8 Let f : R → R be defined by  f (x) =

x sin(1/x) if x = 0, 0 if x = 0.

From the composition and product rules, it can be seen that f is differentiable at every x0 = 0. Now, let us check the differentiability at x0 = 0. For x in a deleted neighbourhood of 0, we have f (x) − f (0) x sin(1/x) = = sin(1/x). x x Hence f  (0) does not exist (cf. Example 2.1.7).

♦

Example 2.3.9 Let f : R → R be defined by  f (x) =

x 2 sin(1/x) if x = 0, 0 if x = 0.

In this case also, f is differentiable at every x0 = 0 which follows from the composition and product rules. Now, let x0 = 0 and x be in a deleted neighbourhood of 0. Then x 2 sin(1/x) f (x) − f (0) = = x sin(1/x). x x Since 0 ≤ |x sin(1/x)| ≤ |x|, lim x→0 x sin(1/x) exists and it is equal to 0. Hence ♦ f  (0) exists and f  (0) = 0.

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Example 2.3.10 Let f (x) = x|x|, x ∈ R, that is,  f (x) =

x 2 if x ≥ 0, −x 2 if x < 0.

Note that, for x = 0, f is differentiable at x and f  (x) = 2|x|. Now, let us check the differentiability at 0. We have lim

x→0−

lim

x→0+

f (x) − f (0) x2 = lim x→0− = 0, x x f (x) − f (0) −x 2 = lim x→0+ = 0. x x

Thus, f − (0) = 0 = f + (0) so that f is differentiable at 0 and f  (0) = 0. Hence, ♦ f  (x) = 2|x| for every x ∈ R. Example 2.3.11 For a > 0, the function a x is differentiable at every x ∈ R and (a x ) = a x ln a

∀ x ∈ R.

Indeed, by the composition rule (Theorem 2.3.5), (a x ) = (e x ln a ) = e x ln a ln a = a x ln a.

♦

Example 2.3.12 The function ln x is differentiable for every x > 0, and (ln x) =

1 , x

x > 0.

To see this, let f (x) = ln x and g(x) = e x . Then we have g( f (x)) = x for every x > 0. Since g ◦ f is differentiable, and g  (y) = e y = 0 for every y ∈ R, Theorem 2.3.6 implies that f is differentiable at every x > 0 and g  ( f (x)) f  (x) = 1. Thus, 1 = eln x (ln x) = x(ln x) so that (ln x) = 1/x.

♦

Example 2.3.13 For a > 0, the function loga x is differentiable for every x > 0, and 1 , x > 0. (loga x) = x ln a To see this, recall that loga x =

ln x 1 . Hence, (loga x) = for every x > 0. ♦ ln a x ln a

2.3 Differentiability of a Function

135

Example 2.3.14 For r ∈ R, let f (x) = x r for x > 0. Then f is differentiable for every x > 0 and x > 0. f  (x) = r x r −1 , By the composition rule in Theorem 2.3.5, f  (x) = (er ln x ) = er ln x

xr r r = = r x r −1 . x x

♦

Exercise 2.3.2 Prove the following. (i) The function ln |x| is differentiable for every x ∈ R with x = 0, and (ln |x|) =

1 , x

x = 0.

(ii) For a > 0, the function loga |x| is differentiable for every x ∈ R with x = 0, and (loga |x|) =

1 , x ln a

x = 0.



2.3.4 Local Maxima and Local Minima Recall from Theorem 2.2.8 that if f : [a, b] → R is a continuous function, then there exists x0 , y0 in [a, b] such that f (x0 ) ≤ f (x) ≤ f (y0 )

∀x ∈ [a, b].

In Remark 2.2.4 we have seen that a function f defined on an interval I need not attain maximum or minimum if either I is not closed and bounded or if f is not continuous. However, maximum or minimum can attain if we restrict the function to a subinterval. To take care of these cases, we introduce the following definition. Definition 2.3.4 Let f be a (real valued) function defined on an interval I . Then f is said to attain a (1) local maximum at a point x1 ∈ I if f (x) ≤ f (x1 ) for all x in a deleted neighbourhood of x1 , (2) local minimum at a point x2 ∈ I if f (x) ≥ f (x2 ) for all x in a deleted neighbourhood of x2 , (3) strict local maximum and strict local minimum at x1 and x2 , respectively, if strict inequality holds in (a) and (b), respectively. The function f is said to attain a

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2 Limit, Continuity and Differentiability of Functions

(4) local extremum at a point x0 ∈ I if f attains either a local maximum or a local minimum at x0 , (5) strict local extremum at a point x0 ∈ I if f attains either a strict local maximum ♦ or a strict local minimum at x0 . Remark 2.3.4 Sometimes the adjective local in local maximum, local minimum and local extremum are omitted, if there is no confusion with the maximum and minimum on the whole interval. In order to avoid any possible confusion, the maximum and minimum on the whole interval are also called global maximum and global minimum. ♦ Theorem 2.3.8 (A necessary condition) Suppose f is a continuous function defined on an interval I having a local extremum at a point x0 ∈ I . If x0 is an interior point of I and f is differentiable at x0 , then f  (x0 ) = 0. Proof Suppose f attains local maximum at x0 which is an interior point of I . Then there exists δ > 0 such that (x0 − δ, x0 + δ) ⊆ I and f (x0 ) ≥ f (x0 + h) for all h with |h| < δ. Hence, for all h with |h| < δ, f (x0 + h) − f (x0 ) ≥ 0 if h

h < 0,

f (x0 + h) − f (x0 ) ≤ 0 if h > 0. h Letting h → 0, we get f  (x0 ) ≥ 0 and f  (x0 ) ≤ 0. Thus, f  (x0 ) = 0. By similar arguments, it can be shown that if f attains local minimum at a point y0 ∈ (a, b), then f  (y0 ) = 0. If a differentiable function f has a local extremum at a point x0 , then f  (x0 ) = 0

Definition 2.3.5 Suppose f is defined on an interval I and x0 is an interior point of I . If f  (x0 ) exists and f  (x0 ) = 0 or if f  (x0 ) does not exist, then x0 is called a critical point of f . ♦ Points at which the derivative of a function vanish are also called stationary points of the function. Remark 2.3.5 A function can have more than one local maximum and local minimum. For example, consider f (x) = sin(4x),

[0, π ].

We see that f has a local maximum value 1 at π/8 and 5π/8, and has a local minimum value −1 at 3π/8 and 7π/8. ♦

2.3 Differentiability of a Function

137

As a consequence of Theorem 2.3.8, we have the following result on intermediate value property of derivatives. Theorem 2.3.9 (Darboux’s theorem) Suppose f is differentiable in an open interval I containing [a, b] and λ lies between f  (a) and f  (b). Then there exists α ∈ [a, b] such that f  (α) = λ. Proof Assume, without loss of generality, that f  (a) < λ < f  (b). Let g(x) = f (x) − λx, x ∈ [a, b]. Since g is differentiable at every point in [a, b] and g  (x) = f  (x) − λ ∀x ∈ [a, b], we have

g  (a) = f  (a) − λ < 0, g  (b) = f  (b) − λ > 0.

Since g is continuous, by Theorem 2.2.8, it has maximum and minimum at some points in [a, b]. Since lim

x→a

g(x) − g(a) = g  (a) < 0, x −a

lim

y→b

g(y) − g(b) = g  (b) > 0, y−b

there exists δ > 0 such that g(x) < g(a) ∀ x ∈ [a, a + δ] and g(y) < g(b) ∀ y ∈ [b − δ, b]. Hence the point at which g has a local minimum can neither be at a nor at b. Therefore, g has a local minimum at an interior point, say at x = α. Therefore, by Theorem 2.3.8, g  (α) = 0, that is, f  (α) = λ. Darboux’s theorem (Theorem 2.3.9) can be rephrased as follows: The derivative of a differentiable function has intermediate value property Here is an illustration of the Theorem 2.3.9 . Example 2.3.15 Let

 f (x) =

−1, x = 0, 1, x = 0.

Then there is no function g which is differentiable on (−1, 1) such that g  (x) = f (x) for all x ∈ (−1, 1). This follows from Darboux’s theorem (Theorem 2.3.9), since f does not take values between 0 and 1. ♦

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Remark 2.3.6 (a) In view of Theorem 2.3.8, if a function f is differentiable at an interior point x0 of an interval I and f  (x0 ) = 0, then f cannot have local maximum or local minimum at x0 . (b) In order to have a local maximum or local minimum at a point, the function need not be differentiable at that point. For example f (x) = 1 − |x|,

|x| ≤ 1,

has a local maximum at 0 and g(x) = |x|,

|x| ≤ 1,

has a local minimum at 0. Both f and g are not differentiable at 0. (c) Also, if a function is differentiable at a point x0 and f  (x0 ) = 0, then it is not necessary that it has local maximum or local minimum at x0 . For example, consider |x| < 1. f (x) = x 3 , In this example, we have f  (0) = 0. Note that f has neither local maximum nor local minimum at 0. ♦ In Sects. 2.3.6 and 2.3.8, we shall give some sufficient conditions for existence of local extrema of functions. Now, let us derive some important consequences of Theorem 2.3.8.

2.3.5 Rolle’s Theorem and Mean Value Theorems Suppose we have a curve y = f (x) defined on a closed interval [a, b]. If the values of f at both the end points a and b are the same, and if the curve has unique tangents at every point x with a < x < b, can we say that there is a point at which the tangent is parallel to the x-axis? The answer is in the affirmative. That is the Rolle’s theorem (Fig. 2.14). Theorem 2.3.10 (Rolle’s theorem) Suppose f is a continuous function defined on a closed and bounded interval [a, b] such that it is differentiable at every x ∈ (a, b). If f (a) = f (b), then there exists c ∈ (a, b) such that f  (c) = 0. Proof Let g(x) = f (x) − f (a). Then we have g(a) = 0 = g(b) and g  (x) = f  (x) ∀ x ∈ (a, b).

(∗)

Since g is continuous on [a, b], it attains the (global) maximum and (global) minimum at some points x1 and x2 , respectively, in [a, b], i.e., there exists x1 , x2 in [a, b] such that

2.3 Differentiability of a Function

139

Fig. 2.14 Illustration of rolle’s theorem

g(x2 ) ≤ g(x) ≤ g(x1 )

∀ x ∈ [a, b].

If g(x1 ) = g(x2 ), then g is a constant function and hence g  (x) = 0 for all x ∈ [a, b]. Hence, assume that g(x2 ) < g(x1 ). Then, either g(x1 ) = 0 or g(x2 ) = 0. If g(x1 ) = / {a, b}, i.e., x1 ∈ (a, b) so that by Theorem 2.3.8, g  (x1 ) = 0 and 0, then by (∗), x1 ∈  hence, f (x1 ) = 0. Similarly, if g(x2 ) = 0, then we shall arrive at f  (x2 ) = 0. Remark 2.3.7 Rolle’s theorem guarantees the existence of at least one point c such that f  (c) = 0. There can be more than one point as illustrated in the following example. ♦ Example 2.3.16 Let f (x) = x 3 − x + 1. Note that f (−1) = 1 and f (1) = 1. So, Rolle’s theorem ensures that there exists a point c ∈ (−1, 1) such that f  (c) = 0. In fact, √ f  (x) = 0 ⇐⇒ 3x 2 − 1 = 0 ⇐⇒ x = ±1/ 3. √ √ Thus, c = 1/ 3 or c = −1/ 3.

♦

Remark 2.3.8 In Rolle’s theorem, the stated conditions on f are not necessary conditions. That is, in the absence of some of the conditions also, there can be a point in (a, b) at which f is differentiable and its derivative vanishes. For example, consider the function f (x) = x 3 for x ∈ [−1, 1]. In this case, f (x) = f (y) for every x = y. But, f  (0) = 0. Also, if we drop any of the conditions in Rolle’s theorem, then the conclusion need not hold. To see this, consider f (x) = |x| for x ∈ [−1, 1]. Note that, f is continuous

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2 Limit, Continuity and Differentiability of Functions

Fig. 2.15 Illustration of Lagrange’s mean value theorem

and f (−1) = f (1) = 1, but there is no point c ∈ (−1, 1) such that f  (c) = 0. In this case, the continuous function f is not differentiable at 0. Thus, in Rolle’s theorem, the assumption on differentiability of f in the whole open interval cannot be dropped.♦ In Rolle’s theorem, we imposed the condition that f (a) = f (b) and obtained a point in the open interval (a, b) at which f  vanishes. What can we say if f (a) = f (b)? The geometry suggests that there must be a point in the open interval (a, b) at which the tangent to the curve y = f (x) is parallel to line joining the points A(a, f (a)) and B(b, f (b)) (Fig. 2.15). Theorem 2.3.11 (Lagrange’s mean value theorem (MVT)) Let f be a function which is continuous on [a, b] and differentiable on (a, b). Then there exists c ∈ (a, b) such that f (b) − f (a) f  (c) = . b−a Proof Let ϕ(x) := f (x) −

f (b) − f (a) (x − a), b−a

x ∈ [a, b].

Then ϕ is continuous on [a, b], differentiable in (a, b), ϕ(a) = ϕ(b), and ϕ  (x) := f  (x) −

f (b) − f (a) , b−a

x ∈ (a, b).

By Rolle’s theorem (Theorem 2.3.10), there exists c ∈ (a, b) such that ϕ  (c) = 0. Thus, f (b) − f (a) = f  (c)(b − a). Remark 2.3.9 Rolle’s theorem is a special case of MVT, in the sense that, if we ♦ have f (a) = f (b), then we have f  (c) = 0. Example 2.3.17 Let f (x) = 2x 3 − 3x + 1. Note that

2.3 Differentiability of a Function

f (0) = 1,

141

f (1) = 0 and

f (1) − f (0) = −1. 1−0

So, MVT ensures that there exists a point c ∈ (0, 1) such that f  (c) = −1. In fact, √ f  (x) = −1 ⇐⇒ 6x 2 − 3 = −1 ⇐⇒ 3x 2 − 1 = 0 ⇐⇒ x = ±1/ 3. √ Thus, in this case, c = 1/ 3.

♦

Example 2.3.18 Let f be continuous on [a, b] and differentiable at every point in (a, b). Suppose there exists c ∈ R such that f  (x) = c

x ∈ (a, b).

Then there exists b ∈ R such that f (x) = c x + b

∀ x ∈ [a, b].

In particular, if f  (x) = 0 for all x ∈ (a, b), then f is a constant function. To see this consider x0 ∈ (a, b). Then, for any x ∈ [a, b], there exists ξx between x0 and x such that f (x) − f (x0 ) = f  (ξx )(x − x0 ) = c(x − x0 ). Hence, f (x) = c x + b with b = f (x0 ) − c x0 .

♦

Convention: If f is defined in a closed interval [a, b], then by lim x→a f (x) and lim x→b f (x) we mean lim x→a+ f (x) and lim x→b− f (x), respectively. Theorem 2.3.12 Let f be continuous on [a, b] and differentiable on (a, b). Then the following hold. (i) (ii) (iii) (iv)

f f f f

is increasing iff f  (x) ≥ 0 for all x ∈ (a, b). is decreasing iff f  (x) ≤ 0 for all x ∈ (a, b). is strictly increasing if f  (x) > 0 for all x ∈ (a, b). is strictly decreasing if f  (x) < 0 for all x ∈ (a, b).

Proof (i) Suppose f is increasing and x ∈ (a, b). Then f (x + h) − f (x) ≥0 h for all h such that x + h ∈ (a, b). Hence f  (x) ≥ 0. Conversely, suppose f  (x) ≥ 0 for all x ∈ (a, b). Let x1 , x2 ∈ [a, b] with x1 < x2 . Then, by the Lagrange’s MVT, there exists ξ ∈ (x1 , x2 ) such that f (x2 ) − f (x1 ) = f  (ξ )(x2 − x1 ).

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Since f  (ξ ) ≥ 0, the above equation shows that f (x1 ) ≤ f (x2 ). (ii) As in the proof of (i) by reversing the inequalities. (iii) Follows from the converse part of the proof of (i) using f  (ξ ) > 0. (iv) As in the converse part of the proof of (i) by using f  (ξ ) < 0. Example 2.3.19 Consider the function f (x) = x 4 for x ∈ R. Then we have f  (x) = 4x 3 for all x ∈ R. Note that f  (x) > 0 ∀ x > 0 and f  (x) < 0 ∀ x < 0. Hence, f is strictly increasing on (0, ∞), and f is strictly decreasing on (−∞, 0).♦ Suppose f and g are continuous functions on [a, b] which are differentiable on (a, b). Suppose further that g  (x) = 0 for all x ∈ (a, b). Then, by the mean value theorem, there exist c1 , c2 in (a, b) such that f (b) − f (a) f  (c1 ) =  . g(b) − g(a) g (c2 ) One may ask whether there exists a single point c ∈ (a, b) such that f  (c) f (b) − f (a) =  . g(b) − g(a) g (c) Answer is in the affirmative as the following theorem shows. Theorem 2.3.13 (Cauchy’s generalized mean value theorem) Suppose f and g are continuous functions on [a, b] which are differentiable at every point in (a, b). Suppose further that g  (x) = 0 for all x ∈ (a, b). Then, g(a) = g(b) and there exists c ∈ (a, b) such that f  (c) f (b) − f (a) =  . g(b) − g(a) g (c) Proof First note that, since g  (x) = 0 for all x ∈ (a, b), by Lagrange’s MVT, g(b) = g(a). Now, let ϕ(x) := f (x) −

f (b) − f (a) [g(x) − g(a)], g(b) − g(a)

x ∈ [a, b].

Note that ϕ is continuous on [a, b], differentiable in (a, b), ϕ(a) = ϕ(b), and ϕ  (x) := f  (x) −

f (b) − f (a)  g (x), g(b) − g(a)

x ∈ (a, b).

2.3 Differentiability of a Function

143

By Rolle’s theorem (Theorem 2.3.10), there exists c ∈ (a, b) such that ϕ  (c) = 0. This completes the proof. Exercise 2.3.3 Let 0 < a < b. Using Cauchy’s generalized mean value theorem, show that n[bn+1 − a n+1 ] < b ∀ n ∈ N. a< (n + 1)[bn − a n ] [Hint: take f (x) = x n+1 and g(x) = x n .]



2.3.6 A Sufficient Condition for a Local Extremum Point Theorem 2.3.14 Suppose f is continuous on an interval I and x0 is an interior point of I . Further suppose that f is differentiable in a deleted neighbourhood of x0 . (i) If there exists an open interval I0 ⊆ I containing x0 such that f  (x) > 0 ∀ x ∈ I0 , x < x0 and f  (x) < 0

∀ x ∈ I0 , x > x 0 ,

then f has a local maximum at x0 . (ii) If there exists an open interval I0 ⊆ I containing x0 such that f  (x) < 0 ∀ x ∈ I0 , x < x0 and f  (x) > 0 ∀ x ∈ I0 , x > x0 , then f has a local minimum at x0 . Proof (i) Let x ∈ I0 . Then, by mean value theorem, there exists ξx between x0 and x such that f (x) − f (x0 ) = f  (ξx )(x − x0 ). By assumption, x < x0 ⇒ f  (ξx ) > 0 and x > x0 ⇒ f  (ξx ) < 0. Hence, in both the cases, we have f (x) < f (x0 ) so that f has local maximum at x0 . Thus, (i) is proved. Similar arguments will lead to the proof of (ii). Example 2.3.20 Consider f (x) = x 4 ,

g(x) = 1 − x 4 , |x| < 1.

Then f  (x) = 4x 3 is negative for x < 0 and positive for x > 0. Hence, by Theorem 2.3.14, f has local minimum at 0. Also, g  (x) = −4x 3 is positive for x < 0 and negative for x > 0. Hence, by Theorem 2.3.14, g has local maximum at 0. ♦

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2 Limit, Continuity and Differentiability of Functions

Remark 2.3.10 The conditions given in Theorem 2.3.14 cannot be dropped. For example, consider f (x) = x 3 , x ∈ R. Then f  (x) = 3x 2 > 0 for all x = 0. Note that f does not have extremum at 0. ♦

2.3.7 L’Hospital’s Rules Recall from Theorem 2.1.8 that if lim x→a f (x) = b and lim x→a g(x) = c, and if c = 0, then g(x) = 0 in a deleted neighbourhood of a and lim x→a

b f (x) = . g(x) c

Clearly, if c = 0, then the above result is meaningless. However, the limit lim x→a f (x)/ g(x) can exist whenever g(x) = 0 for x in a deleted neighbourhood of x0 . For example, we have seen that lim

x→a

x 2 − a2 = 2a and x −a

sin x = 1. x→0 x lim

Note that, in these examples, the numerator of the quotient f (x)/g(x) also vanish at the point where the limit is taken. So, one may tend to hypothesize that lim x→a f (x)/g(x) may exist if both f (x) and g(x) vanish at a. Again, not always. Look at the quotient x 2 − a2 (x − a)2 for x in a deleted neighbourhood of a. In this case, taking xn = a + 1/n, we have xn → a, but xn2 − a 2 (a + 1/n)2 − a 2 = = n2 (xn − a)2 1/n 2



2a 1 + 2 n n

 = 2an + 1 → ∞.

So, under what additional condition, can one assert the existence of lim x→a f (x)/ g(x) ? Some answers are provided in the following theorems, which are known as L’Hospital’s rules.2 The first one is the simplest among them. Theorem 2.3.15 (L’Hospital’s rule-I) Suppose f and g are differentiable at a point x0 and f (x0 ) = 0 = g(x0 ). If g  (x0 ) = 0, then lim x→x0 f (x)/g(x) exists and

2

L’Hospital (pronounced as Lopital) rule is named after the 17th-century French mathematician Guillaume de l’Hospital, who published the rule in his book Analyse des Infiniment Petits pour l’Intelligence des Lignes Courbes (i.e., Analysis of the Infinitely Small to Understand Curved Lines) (1696), the first textbook on differential calculus - curtsey Wikipedia.

2.3 Differentiability of a Function

145

lim

x→x0

f (x) f  (x0 ) =  . g(x) g (x0 )

Proof We note that, for x = x0 , f (x) − f (x0 ) ( f (x) − f (x0 ))/(x − x0 ) f (x) = = . g(x) g(x) − g(x0 ) (g(x) − g(x0 ))/(x − x0 ) Hence, if g  (x0 ) = 0, then using the differentiability of f and g at x0 , we obtain f (x) f  (x0 ) →  as x → x0 . g(x) g (x0 ) This completes the proof. Example 2.3.21 We have already recalled that lim

x→a

x 2 − a2 sin x = 2a and lim x→0 = 1. x −a x

These results follow from Theorem 2.3.15: (i) Taking f (x) = x 2 − a 2 and g(x) = x − a we have f  (a) = 2a and g  (a) = 1 so that, by Theorem 2.3.15, lim

x→a

f  (a) x 2 − a2 =  = 2a. x −a g (a)

(ii) Taking f (x) = sin x and g(x) = x we have f  (0) = 1 and g  (0) = 1 so that, by Theorem 2.3.15, f  (0) sin x =  = 1. ♦ lim x→0 x g (0) Example 2.3.22 For a ∈ R with a = 0, let us find lim

x→0

log(1 + ax) x

by using Theorem 2.3.15. Note that log(1+ax) is of the form f (x)/g(x) with f (x) = x log(1 + ax) and g(x) = x. Clearly, f is well defined for x in a neighbourhood of 0, say for |x| < 1/2|a| and g is defined for all x ∈ R. Further, f (0) = 0 = g(0). Also, we have a 1 , g  (x) = 1 for |x| < . f  (x) = 1 + ax 2|a| In particular,

f  (0) = a, g  (0) = 1.

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2 Limit, Continuity and Differentiability of Functions

Hence, by Theorem 2.3.15, we obtain log(1 + ax) f  (x) a = lim x→0  = lim = a. x→0 x→0 1 + ax x g (x) lim

♦

Example 2.3.23 Using Theorem 2.3.15, we show that for a ∈ R,  a n lim 1 + = exp(a). n→∞ n This is true for a = 0. So, assume that a = 0. We have shown in Example 2.3.28 = a. Since the function x → exp(x) is continuous on R, that for lim x→0 log(1+ax) x we have lim exp

x→0

 log(1 + ax)  x

 log(1 + ax)  = exp lim x→0 = exp(a). x

From this, since 1/n → 0 as n → ∞, we have lim exp

 log(1 + a/n)  1/n

n→∞

= exp(a).

But, for all n ∈ N such that 1 + a/n > 0, exp

 log(1 + a/n)  1/n

   a n = exp log(1 + a/n)n = 1 + . n

Thus, we have shown that  a n lim 1 + = exp(a). n→∞ n

♦

sin x − cos x x−cos x is of the . Note that sinx−π/4 x − π/4 form f (x)/g(x) with f (x) = √ sin x − cos x and g(x) = x − π/4, and f (π/4) = 0 = g(π/4). Since f  (π/4) = 2 and g  (π/4) = 1, by Theorem 2.3.15, we obtain Example 2.3.24 Let us find lim x→π/4

lim

x→π/4

√ sin x − cos x = 2. x − π/4

♦

sin x . Applying Theorem 2.3.15, where x2 − x 2 f (x) = sin x, g(x) = x − x which satisfy f (0) = 0 = g(0) and f  (0) = 1, g  (0) = −1, we obtain 1 sin x cos x = lim x→0 = = −1. ♦ lim 2 x→0 x − x 2x − 1 −1

Example 2.3.25 Let us find lim x→0

2.3 Differentiability of a Function

147

While applying Theorem 2.3.15, we have to ensure that the conditions in the theorem are satisfied. For example, lim

x→2

1 x −2 1 = lim x→2 = x2 − 2 2x 4

is not correct. The correct answer is lim

x→2

x −2 0 lim x→2 (x − 2) = = 0. = 2 2 x −2 2 lim (x − 2) x→2

Note that, writing (x − 2)/(x 2 − 2) as f (x)/g(x) with f (x) = x − 2 and g(x) = x 2 − 2, we have f (2) = 0, but g(2) = 0. We know that Theorem 2.3.15 cannot be applied if g  (x0 ) = 0. However, the condition g  (x0 ) = 0 is not a necessary condition to obtain lim x→x0 f (x)/g(x). For example, consider question of finding sin x 2 . x→0 x 2 lim

Note that sinx 2x is of the form f (x)/g(x) with g  (0) = 0. So, we cannot apply Theorem 2.3.15. However, the limit does exist. This can be seen as follows: Let ϕ(x) = x 2 sin x . Then and ψ(x) = x 2

sin x 2 = ψ(ϕ(x)) with x2

lim ϕ(x) = 0,

x→0

lim ψ(y) = 1.

y→0

Hence, by composition rule for limits (Theorem 2.1.11), sin x 2 = lim x→0 ψ(ϕ(x)) = 1. x→0 x 2 lim

So, if g  (x0 ) = 0, then under what additional condition, can we assert the existence f (x) ? Here is a another L’Hospital’s rule to deal with such situations. of lim x→a g(x) Theorem 2.3.16 (L’Hospital’s rule-II) Suppose f and g are differentiable in a deleted neighbourhood of x0 . Suppose f (x0 ) = 0, g(x0 ) = 0 and Then lim x→x0 f (x)/g(x) exists and

lim

x→x0

f  (x) exists. g  (x)

148

2 Limit, Continuity and Differentiability of Functions

lim

x→x0

f (x) f  (x) = lim x→x0  . g(x) g (x)

Proof Since lim x→x0 f  (x)/g  (x) exists, there exists a deleted neighbourhood D0 of x0 in the domain of definition of g such that g  (x) = 0 for all x ∈ D0 . By Cauchy’s generalized mean value theorem (Theorem 2.3.13), for every x ∈ D0 , there exists ξx between x and x0 such that f (x) − f (x0 ) f  (ξx ) f (x) = =  . g(x) g(x) − g(x0 ) g (ξx ) Since |ξx − x0 | < |x − x0 | and lim x→x0 f  (x)/g  (x) exists, by using the limits of composition of functions, lim x→x0 f  (ξx )/g  (ξx ) exists and it is equal to lim x→x0 f  (x)/ g  (x). Thus, lim x→x0 f (x)/g(x) exists and lim

x→x0

f (x) f  (x) = lim  . x→x0 g (x) g(x)

This completes the proof. Example 2.3.26 By Theorem 2.3.16, sin x 2 2x cos x 2 = lim x→0 cos x 2 = 1. = lim x→0 x→0 x 2 2x lim

♦

Example 2.3.27 By Theorem 2.3.16, 1 1 1 − cos x sin x sin x = lim x→0 = . = lim x→0 2 x→0 x 2x 2 x 2 lim

♦

We know that if a function f is defined in a neighbourhood of a point x0 and if lim x→x0 f (x) exists, then the above limit need not be f (x0 ). For example, if  f (x) =

0, x = 0, 1, x = 0,

then lim x→0 f (x) = 0, but f (0) = 1. This kind of situation will not occur for the derivative of a function, as the following theorem shows. Theorem 2.3.17 Suppose f is differentiable in a deleted neighbourhood of x0 and lim x→x0 f  (x) exists. Then f  (x0 ) = lim f  (x). x→x0

2.3 Differentiability of a Function

149

Proof Applying L’Hospital’s rule-II (Theorem 2.3.16) to the quotient g(x)/ h(x), where g(x) = f (x) − f (x0 ) and h(x) = x − x0 , we have lim

x→x0

f (x) − f (x0 ) g  (x) f  (x) = lim  = lim = lim f  (x), x→x0 h (x) x→x0 x→x0 x − x0 1

so that f  (x0 ) exists and f  (x0 ) = lim x→x0 f  (x). In the following L’Hospital’s rule, we further relax the conditions on f and g, by replacing the condition that f and g vanish at x0 by requiring the limits of f and g to be 0 as x → x0 . Theorem 2.3.18 (L’Hospital’s rule-III) Suppose f and g are differentiable in a deleted neighbourhood of x0 . Suppose lim f (x) = 0,

x→x0

lim g(x) = 0 and

x→x0

lim

x→x0

f  (x) exists. g  (x)

Then lim x→x0 f (x)/g(x) exists and lim

x→x0

f (x) f  (x) = lim x→x0  . g(x) g (x)

Proof Let f˜(x) =



f (x) if x = x0 , g(x) ˜ = 0 if x = x0



g(x) if x = x0 0 if x = x0 .

Then, the result is obtained from Theorem 2.3.16 by taking f˜ and g˜ in place of f and g, respectively. Next we discus case of finding the limit of f (x)/g(x) as x → ∞. Theorem 2.3.19 (L’Hospital’s rule-IV) Suppose f and g are differentiable at every point in (a, ∞) for some a > 0. Suppose lim f (x) = 0,

x→∞

lim g(x) = 0 and

x→∞

lim

x→∞

Then lim x→∞ f (x)/g(x) exists and lim

x→∞

f (x) f  (x) = lim  . x→∞ g (x) g(x)

f  (x) exists. g  (x)

150

2 Limit, Continuity and Differentiability of Functions

Proof Let f˜(y) = f (1/y) and g(y) ˜ = g(1/y) for 0 < y < 1/a. We note that ˜ lim f (x) = 0 = lim x→∞ g(x) ⇐⇒ lim y→0 f˜(y) = 0 = lim y→0 g(y).

x→∞

Also, since f˜ (y) = [ f (1/y)] = f  (1/y)(−1/y 2 ), we have lim

x→∞

g˜  (y) = [g(1/y)] = g  (1/y)(−1/y 2 ),

f  (x) f˜ (y) exists ⇐⇒ lim exists. y→0 g  (x) g˜  (y)

Hence, applying Theorem 2.3.16 to f˜, g˜ instead of f, g, we obtain the result. Theorem 2.3.20 (L’Hospital’s rule-V) Let x0 ∈ [a, b]. Suppose f and g are differentiable in a deleted neighbourhood of a point x0 . Suppose lim f (x) = ∞, lim x→x0 g(x) = ∞ and

x→x0

lim

x→x0

f  (x) exists. g  (x)

Then lim x→x0 f (x)/g(x) exists and lim

x→x0

Proof Let β := lim x→x0 since

f (x) f  (x) = lim x→x0  . g(x) g (x)

f  (x) . First we consider the case of β = 0. In this case, g  (x)

lim f (x) = ∞ = lim x→∞ g(x) ⇐⇒ lim x→x0 (1/ f (x)) = 0 = lim x→x0 (1/g(x)),

x→x0

the result follows from Theorem 2.3.18 by interchanging the roles of the functions f and g. To consider the general case when β is not necessarily non-zero, let x, y be distinct points in a deleted neighbourhood of x0 . Since g  (x) = 0 for x sufficiently close to x0 , in view of Lagrange’s MVT, we can assume that g(x) = g(y). Note that, 

f (x) 1 − f (x) − f (y)  = g(x) − g(y) g(x) 1 −

f (y) f (x) g(y) g(x)

 

(2.1)

Since f (x) → ∞ and g(x) → ∞ as x → x0 , the above expression is meaningful for each fixed y and x close enough to x0 , and

2.3 Differentiability of a Function

151

  f (y)  g(y)  = 1 = lim x→x0 1 − . lim x→x0 1 − f (x) g(x)

(2.2)

Also, by the Cauchy’s generalized MVT, there exists ξx,y lying between x and y such that f  (ξx,y ) f (x) − f (y) . (2.3) =  g(x) − g(y) g (ξx,y) From (2.1) and (2.3) above we have 

f (ξx,y ) 1 − f (x)  =  g(x) g (ξx,y) 1 −

g(y) g(x)



f (y) f (x)

 .

(2.4)

We observe that |ξx,y − x0 | ≤ |ξx,y − y| + |y − x0 | ≤ |x − y| + |y − x0 |. Hence, ξx,y → x0 as x → x0 and y → y0 . Hence, by using the limits of composition of functions, we obtain lim y→x0

f  (ξx,y ) f  (x) . = lim x→x 0 g  (ξx,y) g  (x)

Therefore, (2.2), (2.4), (2.5) imply that lim x→x0

(2.5)

f (x) exists and g(x)

 f (x) f  (ξx,α ) 1 −  lim = lim y→x0  x→x0 g(x) g (ξx,α) 1 −

f (y) f (x) g(y) g(x)

  = lim

x→x0

f  (x) . g  (x)

This completes the proof. Remark 2.3.11 The cases lim f (x) = 0 = lim g(x),

x→−∞

x→−∞

lim f (x) = −∞ = lim g(x)

x→x0

x→x0

can be treated analogously to the cases already discussed in the above theorems. ♦ We shall make use of Theorem 2.3.20 to prove the following theorem. Theorem 2.3.21 (L’Hospital’s rule-VI) Suppose f and g are differentiable on (0, ∞). Suppose lim f (x) = ∞ = lim x→∞ g(x) and

x→∞

lim

x→∞

f  (x) exists. g  (x)

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2 Limit, Continuity and Differentiability of Functions

Then lim f (x)/g(x) exists and x→∞

lim

x→∞

f (x) f  (x) = lim  . x→∞ g (x) g(x)

Proof We observe that lim x→∞

f (x) f (1/x) exists ⇐⇒ lim x→0+ exists , g(x) g(1/x)

(2.1)

and in that case they are same, and lim x→∞

f  (x) f  (1/x) + exists ⇐⇒ lim exists , x→0 g  (x) g  (1/x)

(2.2)

and in that case they are same. Therefore, it is enough to show that f (1/x) f  (1/x) = lim x→0+  . g(1/x) g (1/x)

lim x→0+

(2.3)

To show this, let ϕ and ψ be defined by ϕ(x) = f (1/x) and ψ(x) = g(1/x) for x > 0. Then we have ϕ  (x) = − so that

f  (1/x) g  (1/x)  , ψ (x) = − x2 x2

ϕ(x) f (1/x) ϕ  (x) f  (1/x) = and = ψ(x) g(1/x) ψ  (x) g  (1/x)

(2.4)



(x) for all x > 0. By assumption, lim x→∞ gf  (x) exists. Hence, by (2.2) and (2.4), 

ϕ(x) exists. Therefore, by Theorem 2.3.20, lim x→∞ ψ(x) exists and are equal, lim x→∞ ψϕ (x) (x) that is ϕ(x) ϕ  (x) = lim  lim x→∞ ψ(x) x→∞ ψ (x)

In view of (2.4), the above equality is same as (2.3), completing the proof. Example 2.3.28 We have seen in Example 2.2.17 that limn→∞ log(1+n) = 0. Now, n using Theorem 2.3.21, we show lim

n→∞

log(1 + n) =0 np

2.3 Differentiability of a Function

153

for any p > 0. In fact, we show lim

x→∞

log(1 + x) =0 xp

for any p > 0. For this, let p > 0 and let f (x) = log(1 + x), g(x) = x p for x > 0. Then we have lim x→∞ f (x) = ∞ = lim x→∞ g(x). Note that, for x > 0, 1/(1 + x)  x  1 f  (x) = = .  g (x) px p−1 1 + x px p Hence, by Theorem 2.3.21,  x  1 log(1 + x) f (x) f  (x) = lim = lim = lim = 0. x→∞ x→∞ g(x) x→∞ g  (x) x→∞ 1 + x px p xp lim

From this we have limn→∞ n1p log(1 + n) = 0.

♦

2.3.8 Higher Derivatives and Taylor’s Formula Suppose f is defined on an open interval I and x0 ∈ I . If f is differentiable in a neighbourhood of x0 , then the derivative f  is a function defined in that neighbourhood, and in that case we can talk about the existence of derivative of f  at x0 , and similarly, we can talk about higher derivatives of f at x0 . Definition 2.3.6 Suppose f is differentiable in a neighbourhood of x0 . If lim

x→x0

f  (x) − f  (x0 ) x − x0

exists, then f is said to be twice differentiable at x0 , and in that case the limit is called the second derivative of f at x0 and it is denoted by f  (x0 ).

♦

Definition 2.3.7 Let f be defined in an open interval I . Then, for k ∈ N with k ≥ 2, f is said to be k times differentiable at x0 ∈ I if f (1) (x), f (2) (x), . . . , f (k−1) (x) are defined iteratively as f ( j) (x) := [ f ( j−1) ] (x),

j = 1, . . . , k − 1

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2 Limit, Continuity and Differentiability of Functions

for x in a neighbourhood of x0 with f (0) (x) = f (x) and f (k) (x0 ) := [ f (k−1) ] (x0 ) exists. Then f (k) (x0 ) is called the k th -derivative of f at x0 .

♦

Note that f (2) (x0 ) is the second derivative of f at x0 . The k th -derivative of f is also denoted by dk f . dxk It is also customary to use the notation D f for f  and D k f for f (k) , so that it is D k f = D(D k−1 f ). Definition 2.3.8 The function f is said to be infinitely differentiable at a point x0 ∈ I if f has k th -derivative at x0 for every k ∈ N. ♦ We may observe the following: If f is infinitely differentiable at a point x0 ∈ I , then for every k ∈ N, f has k th -derivative not only at x0 but also at every point in some neighbourhood of x0 .

Example 2.3.29 For n ∈ N, let f (x) = x n , x ∈ R. Then we know that f (1) (x) = f  (x) = nx n−1 . Hence, for k ≤ n, we have f (k) (x) = n(n − 1) · · · (n − k + 1)x n−k and f (k) (x) = 0 for k > n. Thus, f is infinitely differentiable in R. More generally, if f is a polynomial, then f is infinitely differentiable in R. ♦ Example 2.3.30 Let f (x) = sin x, x ∈ R. Then we have f (1) (x) = cos x,

f (2) (x) = − sin x,

f (3) (x) = − cos x,

f (4) (x) = sin x,

and more generally for any k ∈ N, f (2k−1) (x) = (−1)k+1 cos x, Thus, f is infinitely differentiable in R.

f (2k) (x) = (−1)k sin x. ♦

Example 2.3.31 Let f (x) = e x , x ∈ R. We know that f  (x) = e x . Hence, it follows that f (k) (x) = e x for every k ∈ N so that f is infinitely differentiable in R. ♦ Example 2.3.32 Let f (x) = x|x|, x ∈ R. We have seen in Example 2.3.10 that f is differentiable at every point in R and f  (x) = 2|x|. Thus, f is infinitely differentiable at every x = 0, but differentiable only once at 0. For k ∈ N, if f (x) = x k |x|, x ∈ R, then it can be verified that f is infinitely differentiable at every x = 0 and f (k) (0) exists, but f (k+1) (0) does not exist. ♦

2.3 Differentiability of a Function

155

L’Hospital’s Rule Revisited We know, by L’Hospital’s rule (Theorem 2.3.18) using first derivatives, that if lim f (x) = 0,

x→x0

lim g(x) = 0 and

x→x0

f  (x) exists, g  (x)

lim

x→x0

f (x) f  (x) exists and it is equal to lim x→x0  . g(x) g (x) Suppose it happens that

then lim x→x0

lim f  (x) = 0,

x→x0

lim g  (x) = 0 but

x→x0

lim

x→x0

f  (x) exists. g  (x) 

(x) Then, Theorem 2.3.18 can be applied to f  , g  instead of f, g so that lim x→x0 gf  (x) 



(x) (x) exists and lim x→x0 gf  (x) = lim x→x0 gf  (x) . Hence, applying again Theorem 2.3.18 to f, g, we have f (x) f  (x) f  (x) = lim  = lim  . lim x→x0 g(x) x→x0 g (x) x→x0 g (x)

More generally we have the following result. Theorem 2.3.22 (L’Hospital’s rule) Suppose the functions f and g are k times differentiable in a deleted neighbourhood of x0 , lim f ( j−1) (x) = 0,

x→x0

f (k) (x) (k) x→x0 g (x)

for j = 1, . . . , k and lim

lim

x→x0

lim g ( j−1) (x) = 0

x→x0

exists. Then lim x→x0 f (x)/g(x) exists and f (x) f (k) (x) = lim (k) . x→x0 g (x) g(x)

Remark 2.3.12 Other forms of L’Hospital’s rules in the general setting as in Theorem 2.3.22 are also valid. ♦ Exercise 2.3.4 Supply details of the proof of Theorem 2.3.22. 3



Example 2.3.33 Let us check whether the lim x→∞ xex exists. Let f (x) = x 3 and g(x) = e x . Then, we have the following: lim f (x) = lim x 3 = ∞,

x→∞

x→∞

lim f  (x) = lim 3x 2 = ∞,

x→∞

x→∞

lim g(x) = lim e x = ∞,

x→∞

x→∞

lim g  (x) = lim e x = ∞,

x→∞

x→∞

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2 Limit, Continuity and Differentiability of Functions

lim f  (x) = lim 6x = ∞,

x→∞

x→∞

lim f  (x) = 6,

x→∞

lim g  (x) = lim e x = ∞,

x→∞

x→∞

lim g  (x) = lim e x = ∞.

x→∞

x→∞

x3 6 = lim x→x0 x = 0. More generally, for any k ∈ N, it can be x e e k ♦ shown (verify) that lim x→∞ xex = 0.

Hence, lim x→x0

Taylor’s Formula Recall from Lagrange’s MVT that if f is differentiable in an open interval I and x0 ∈ I , then for every x ∈ I , there exists ξx between x0 and x such that f (x) = f (x0 ) + f  (ξx )(x − x0 ). Can we say something more if f has higher-order derivatives in a neighbourhood of x0 ? Suppose f (x) is a polynomial of degree n ∈ N and x0 ∈ R. Since f (x) − f (x0 ) vanishes at x = x0 , we can write f (x) = f (x0 ) + (x − x0 ) f 1 (x), where f 1 (x) is a polynomial of degree n − 1. By the same argument, if n > 1, then f 1 can be written as f 1 (x) = f 1 (x0 ) + (x − x0 ) f 2 (x), where f 2 (x) is a polynomial of degree n − 2. Thus, f (x) = f (x0 ) + f 1 (x0 )(x − x0 ) + (x − x0 )2 f 2 (x). Continuing this, there are polynomials f 1 (x), f 2 (x), . . . , f n−2 (x), f n−1 (x), f n (x) of degree n − 1, n − 2, . . . , 2, 1, 0, respectively, such that f (x) = f (x0 ) + f 1 (x0 )(x − x0 ) + f 2 (x0 )(x − x0 )2 + · · · + f n (x0 )(x − x0 )n . Note that f (1) (x0 ) = f 1 (x0 ),

f (2) (x0 ) = 2! f 2 (x0 ), . . . , f (n) (x0 ) = n! f n (x0 ),

so that f (x) = f (x0 ) +

f (1) (x0 ) f (2) (x0 ) f (n) (x0 ) (x − x0 ) + (x − x0 )2 + · · · + (x − x0 )n . 1! 2! n!

2.3 Differentiability of a Function

157

Now, suppose that f is a function which is n + 1 times differentiable in a neighbourhood of x0 for some n ∈ N. If we write, P(x) = f (x0 ) +

n f (k) (x0 ) (x − x0 )k , k! k=1

then we can write f (x) = P(x) + R(x) where R(x) := f (x) − P(x) is n + 1 times differentiable and R(x0 ) = 0. We may also observe that R (k) (x0 ) = 0 for k = 1, . . . , n. Taylor’s formula in the following theorem gives a specific expression for R(x) in terms of the (n + 1)th derivative of f at a point ξ lying between x0 and x. Theorem 2.3.23 (Taylor’s formula) Suppose f is defined and has derivatives f (1) (x), f (2) (x), . . . , f (n+1) (x) for x in an open interval I and x0 ∈ I . Then, for every x ∈ I , there exists ξx between x and x0 such that f (x) = f (x0 ) +

n f ( j) (x0 ) f (n+1) (ξx ) (x − x0 ) j + (x − x0 )n+1 . j! (n + 1)! j=1

Proof Let x ∈ I with x = x0 , and let Pn (y) = f (x0 ) +

n f ( j) (x0 ) (y − x0 ) j , j! j=1

y ∈ I.

Then Pn (y) is a polynomial of degree n, Pn (x0 ) = f (x0 ) and Pn( j) (x0 ) = f ( j) (x0 ),

j = 1, . . . , n.

Now, let g(y) = f (y) − Pn (y) − ϕ(x)(y − x0 )n+1 , where ϕ(x) :=

y ∈ I,

f (x) − Pn (x) . (x − x0 )n+1

Note that, by this choice of ϕ(x), we have g(x0 ) = 0 and g(x) = 0. Also, we have g (1) (x0 ) = 0, g (2) (x0 ) = 0, . . . , g (n) (x0 ) = 0.

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2 Limit, Continuity and Differentiability of Functions

Since g(x0 ) = 0 = g(x), by Rolle’s theorem, there exists x1 between x0 and x such that g  (x1 ) = 0. Since g  (x0 ) = 0 = g  (x1 ), again by Rolle’s theorem, there exists x2 between x0 and x1 such that g  (x2 ) = 0. Continuing this, there exists ξx := xn+1 between x0 and xn such that g (n+1) (ξx ) = 0. But, g (n+1) (y) = f (n+1) (y) − Pn(n+1) (y) − ϕ(x)(n + 1)! = f (n+1) (y) − ϕ(x)(n + 1)!. Thus, using the fact that g (n+1) (ξx ) = 0, we have ϕ(x) = Thus, f (x) = Pn (x) +

f (n+1) (ξx ) . (n + 1)! f (n+1) (ξx ) (x − x0 )n+1 , (n + 1)!

and the proof is complete. Definition 2.3.9 With the notations as in Theorem 2.3.23, the polynomial Pn (x) = f (x0 ) +

n f ( j) (x0 ) (x − x0 ) j j! j=1

is called the Taylor’s polynomial of f of degree n around x0 , and the term Rn (x) :=

f (n+1) (ξx ) (x − x0 )n+1 (n + 1)!

is called the remainder term in the formula corresponding to the n-th degree Taylor’s polynomial. ♦ Remark 2.3.13 The Taylor’s formula given in Theorem 2.3.23 is usually known as Lagrange form of the Taylor’s formula. The proof given above is adapted from [6]. In the next chapter, we shall give another form of the Taylor’s formula, called, the Cauchy form of the Taylor’s formula, and derive the formula given in Theorem 2.3.23. ♦ We observe that if f is infinitely differentiable and if |Rn (x)| → 0 as n → ∞ for every x ∈ I , then f (x) = f (x0 ) +

∞ f (n) (x0 ) (x − x0 )n , n! n=1

x ∈ I.

2.3 Differentiability of a Function

159

Definition 2.3.10 If f is infinitely differentiable in a neighbourhood of x0 and if it can be represented as a series f (x) = f (x0 ) +

∞ f (n) (x0 ) (x − x0 )n , n! n=1

x∈I

for all x in a neighbourhood of x0 , then such a series is called the Taylor series of f around the point x0 . The Taylor series with x0 = 0 is called the Maclaurin series of f . ♦ Observe that if f (n+1) is bounded in a neighbourhood of x0 , that is, there exists Mn > 0 such that | f (n+1) (x)| ≤ Mn for all x in that neighbourhood, then | f (x) − Pn (x)| ≤

Mn |x − x0 |n+1 . (n + 1)!

In particular, if f is infinitely differentiable, and if there exists M > 0, independent of n such that | f (n+1) (x)| ≤ M for all x in a neighbourhood I0 of x0 , then | f (x) − Pn (x)| ≤

M|x − x0 |n+1 → 0 as n → ∞. (n + 1)!

Thus, we have proved the following theorem. Theorem 2.3.24 Suppose f is infinitely differentiable in an open interval I and x0 ∈ I . Further, suppose that there exists M > 0 such that | f (k) (x)| ≤ M

∀x ∈ I, ∀k ∈ N ∪ {0}.

Then f has the Taylor series expansion f (x) = f (x0 ) +

∞ f (n) (x0 ) (x − x0 )n n! n=1

for all x ∈ I0 . Remark 2.3.14 A natural question that one may ask is: Does every infinitely differentiable function in a neighbourhood of x0 has a Taylor’s series expansion?

Unfortunately, the answer is in the negative. For example, if we define  f (x) =

e−1/x , x = 0, 0, x = 0, 2

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2 Limit, Continuity and Differentiability of Functions

then it can be seen that f (0) = 0 and f (k) (0) = 0 for all k ∈ N. Thus, f does not have the Taylor’s series expansion around the point 0. ♦ Example 2.3.34 Let f (x) = e x for x ∈ R. Then f (k) (x) = e x so that for any x0 , x ∈ R, f (n+1) (ξx ) e ξx (x − x0 )n+1 = (x − x0 )n+1 . Rn (x) := (n + 1)! (n + 1)! Since

(x − x0 )n+1 → 0 as n → ∞, we have (n + 1)! Rn (x) → 0 as n → ∞.

Hence, f has the Taylor series expansion  e =e x

x0

 ∞ (x − x0 )n 1+ . n! n=1

for every x, x0 ∈ R. Observe that the function which represents the series within the is nothing but e x−x0 . In particular, taking x0 = 0, we have e x = 1 +

∞bracket xn ♦ n=1 n! . Example 2.3.35 Using the Taylor’s formula, we shall show that sin x =

∞ (−1)n x 2n+1 n=0

(2n + 1)!

∀ x ∈ R.

For this, let f (x) = sin x and x0 = 0. Since f is infinitely differentiable, and f (2 j) (0) = 0,

f (2 j−1) (0) = (−1) j

∀ j ∈ N,

we have f (x) = f (x0 ) +

2n+1 j=1

f ( j) (0) j f (2n+2) (ξx ) 2n+2 x + x j! (2n + 2)!

n f (2 j+1) (0) 2 j+1 f (2n+2) (ξx ) 2n+2 x x = f (x0 ) + + (2 j + 1)! (2n + 2)! j=0

= f (x0 ) +

n (−1) j 2 j+1 f (2n+2) (ξx ) 2n+2 x x + (2 j + 1)! (2n + 2)! j=0

Also, since | sin x| ≤ 1, we have  f (2n+2) (ξ )x 2n+2  |x|2n+2 x   → 0 as n → ∞.  ≤ (2n + 2)! (2n + 2)!

2.3 Differentiability of a Function

161

Therefore, n   (−1) j 2 j+1   x  → 0 as n → ∞  f (x) − f (x0 ) + (2 j + 1)! j=0

and hence, sin x =

∞ (−1)n x 2n+1

(2n + 1)!

n=0

∀ x ∈ R.

♦

Exercise 2.3.5 Using the Taylor’s formula, prove the following: (i) cos x = (ii)

∞ (−1)n x 2n n=0 ∞

1 = 1−x

for all x ∈ R.

(2n)!

x n for all x with |x| < 1.

n=0 ∞

(−1)n x 2n+1 for all x ∈ R. 2n + 1 n=0 ∞ π (−1)n (iv) From (iii), deduce = , the Madhava-Nilakantha series for π/4.

4 2n + 1 n=0

(iii) tan

−1

x=

Taylor’s formula will help us to find approximate values for a complicated function by the evaluation of a polynomial. Let us look at a few examples. Example 2.3.36 Let f (x) = sin x, x ∈ R. Let us find approximate value for sin 1: By the Taylor’s formula, we have sin x = Pn (x) + Rn (x), where Pn (x) :=

n (−1)k x 2k+1 k=0

(2k + 1)!

,

Rn (x) =

f (2n+2) (ξx )x 2n+2 . (2n + 2)!

Thus, we obtain  f (2n+2) (ξ )x 2n+2  |x|2n+2 x   = εn (x). |Rn (x)| =  ≤ (2n + 2)! (2n + 2)! Here are the values f n (1) and the corresponding error εn (1) for some values of n. n

0

Pn (1)

1

εn (1)

0.5

1

2

3

1 1− 3! 1  0.4166 4!

1 1 1− + 3! 5! 1  0.001388 6!

1 1 1 1− + − 3! 5! 7! 1  0.0001736 8!

In particular, sin 1  1 − 0.16666 + 0.00833 − 0.0001984  0.3415316

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2 Limit, Continuity and Differentiability of Functions

and | sin 1 − 0.3415316|  0.0001736.

♦

Sufficient Condition for Extremum Points Revisited Using the second derivative of a function and the Taylor’s formula, now we give another sufficient condition for a critical point to be an extreme point. Theorem 2.3.25 Suppose f is defined on an interval I and x0 is an interior point of I . Suppose that x0 is a critical point of f and f has continuous second derivative in a neighbourhood of x0 . Then the following hold. (i) If f  (x0 ) < 0, then f has local maximum at x0 . (ii) If f  (x0 ) > 0, then f has local minimum at x0 . Proof By Taylor’s theorem, there exists an open interval I0 containing x0 such that for every x ∈ I0 , there exists ξx between x0 and x such that f (x) − f (x0 ) = f  (x0 )(x − x0 ) +

f  (ξx ) (x − x0 )2 . 2

Since x0 is a critical point of f and f is differentiable at x0 , f  (x0 ) = 0. Hence, f (x) − f (x0 ) =

f  (ξx ) (x − x0 )2 . 2

(∗)

(i) Suppose f  (x0 ) < 0. Since f  is continuous in a neighbourhood of x0 , there exists an open interval I1 containing x0 such that for all x ∈ I1 , f  (x) ≤

f  (x0 ) . 2

In particular, from (∗), we obtain f (x) − f (x0 ) =

f  (ξx ) (x − x0 )2 < 0 2

∀x ∈ I1 .

Thus, f has a local maximum at x0 . (ii) Suppose f  (x0 ) > 0. In this case, we obtain reverse of the inequalities in the proof of (i), and arrive at the conclusion that f has a local minimum at x0 . Remark 2.3.15 The conditions given in Theorem 2.3.25 are only sufficient conditions. There are functions f for which none of the conditions (i) and (ii) of Theorem 2.3.25 are satisfied at a point x0 , still f can have local extremum at x0 . For example, consider g(x) = 1 − x 4 , |x| < 1. f (x) = x 4 , Then f  (0) = 0 = g  (0), f has local minimum at 0 and g has local maximum at 0. But, f  (0) = 0 = g  (0). ♦

2.3 Differentiability of a Function

163

Remark 2.3.16 How to identify critical points and extreme points of a function? 1. Suppose f is defined on an open interval I . (a) Find those points at which either f is not differentiable or f  vanish. These points are the critical points of f . (b) Suppose f is differentiable in a neighbourhood of x0 and f  (x0 ) = 0. If f  (x) has the same sign for x on both side of x0 , then f does not have an extremum at x0 ; otherwise, use the test for maximum or minimum given in Theorem 2.3.14. 2. Suppose f is continuous on [a, b] and differentiable on (a, b). (a) f can have maximum or minimum only at the end points of [a, b] or at those points in (a, b) at which f  vanishes. (b) Use the tests as in Theorem 2.3.14 or Theorem 2.3.25. ♦

2.3.9 Determination of Shapes of a Curves We shall use conditions on derivatives of a function to find out certain nature of the curve determined by a function. First we spell out what is meant by a curve determined by a function. Definition 2.3.11 Let f be a continuous function defined on an interval I . Then the graph of f , i.e., G f := {(x, f (x)) : x ∈ I }, ♦

is called the curve determined by f . A curve determined by a function f : I → R is often written as an equation y = f (x), x ∈ I.

Definition 2.3.12 Let f be a continuous function defined on an interval I . Then the curve determined by f is said to be (1) convex upwards or concave downwards if f is points of I and the tangent line at each point x ∈ I (2) convex downwards or concave upwards if f is points of I and the tangent line at each point x ∈ I

differentiable at all interior lies above the curve, differentiable at all interior lies below the curve. ♦

Thus, if f is defined on an interval I and differentiable at all interior points of I , then the curve determined by f is

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2 Limit, Continuity and Differentiability of Functions

Fig. 2.16 Concave down and concave up on subintervals

(1) convex upwards if and only if for any interior point x0 of I , x ∈ I \ {x0 }, y = f (x0 ) + f  (x0 )(x − x0 )

⇒

f (x) < y,

(2) convex downwards if and only if for any interior point x0 of I , x ∈ I \ {x0 }, y = f (x0 ) + f  (x0 )(x − x0 )

⇒

f (x) > y.

It is also conventional to define a function to be convex or concave in the following sense. Definition 2.3.13 Let I be an interval. Then a function f : I → R is said to be a convex function if for every x, y ∈ I and 0 < λ < 1, f ((1 − λ)x + λy) ≤ (1 − λ) f (x) + λ f (y). The function f : I → R is said to be a concave function if for every x, y ∈ I and 0 < λ < 1, f ((1 − λ)x + λy) ≥ (1 − λ) f (x) + λ f (y). ♦ Thus, if f : I → R is differentiable at every interior points of I , then (1) f is convex if and only if f is convex downwards if and only if f is concave upwards, and (2) f is concave if and only if f is concave downwards if and only if f is convex upwards. ♦ Exercise 2.3.6 Give an example of a function f defined on some interval I which is (i) convex, but not concave downwards, (ii) concave, but not convex upwards.



Theorem 2.3.26 Let f be a continuous function defined on an interval I . Suppose f has second derivative at all interior points of I . Then the curve determined by f is (i) convex upwards if f  (x) < 0 for all interior points x in I ; (ii) convex downwards if f  (x) > 0 for all interior points x in I .

2.3 Differentiability of a Function

165

Fig. 2.17 Concave down and concave up on subintervals

Proof Suppose f  (x) < 0 for all interior points x in I . Let x0 be any point in the interior of I . We have to show that x ∈ I, y = f (x0 ) + f  (x0 )(x − x0 )

⇒

f (x) < y.

So let x ∈ I and y = f (x0 ) + f  (x0 )(x − x0 ). By Taylor’s theorem, there exists ξx between x and x0 such that f (x) = f (x0 ) + f  (x0 )(x − x0 ) +

f  (ξx ) (x − x0 )2 2

so that, using the fact that f  (ξx ) < 0, f (x) = y +

f  (ξx ) (x − x0 )2 < y. 2

Hence, G f is convex upwards, proving (i). Proof of (ii) follows analogously (Fig. 2.17). Exercise 2.3.7 Prove that the converse statements in (i) and (ii) in Theorem 2.3.26 are also true.

Example 2.3.37 Let us consider a few examples. (i) Let f (x) = x 2 and g(x) = 1 − x 2 for x ∈ R. Then G f is convex downwards and G g is convex upwards. (ii) Let f (x) = e x , x ∈ R. Note that f  (x) > 0 for all x ∈ R. Hence, by the Theorem 2.3.26, y = e x is convex downwards on R. (iii) Let f (x) = x 3 , x ∈ R. Note that f  (x) = 6x so that, by the Theorem 2.3.26, the curve y = x 3 is convex upwards for x < 0 and convex downwards for x > 0.♦

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2 Limit, Continuity and Differentiability of Functions

Fig. 2.18 Point of inflexion

Definition 2.3.14 A point (x0 , y0 ) on the curve determined by a function f is said to be a point of inflexion of the curve if in a neighbourhood of x0 , the curve is convex upwards on one side of x0 and convex downwards on other side of x0 (Fig. 2.18). ♦ Example 2.3.38 In view of the conclusions in Example 2.3.37(iii), the point (0, 0) ♦ on the curve y = x 3 is a point of inflexion. Theorem 2.3.27 Suppose f has second derivative in a deleted neighbourhood of a point x0 . Then the point (x0 , f (x0 )) is a point of inflexion of the curve G f if f  has constant but different signs on each side of x0 , and at the point x0 , either f  (x0 ) does not exist or f  (x0 ) = 0. Proof This is a consequence of Theorem 2.3.26. Theorem 2.3.28 Suppose f has second derivative in a neighbourhood I0 of a point x0 . If (x0 , f (x0 )) is a point of inflexion of the curve G f and if f  is continuous at x0 , then f  (x0 ) = 0. Proof Suppose (x0 , f (x0 )) is a point of inflexion of the curve G f and f  is continuous at x0 . Without loss of generality, we may assume that G f is convex upwards for x ∈ I0 , x < x0 and it is convex downwards for x ∈ I0 , x > x0 . Thus, x ∈ I0 , x < x 0

⇒

f (x) < f (x0 ) + f  (x0 )(x − x0 ),

(2.1)

x ∈ I0 , x > x 0

⇒

f (x) > f (x0 ) + f  (x0 )(x − x0 ).

(2.2)

So, let x ∈ I0 . By Taylor’s theorem, there exists ξx between x and x0 such that f (x) = f (x0 ) + f  (x0 )(x − x0 ) +

f  (ξx ) (x − x0 )2 . 2

2.3 Differentiability of a Function

167

Now, (2.1) implies that f  (ξx ) < 0 so that by letting x → x0 , we have f  (x0 ) ≤ 0. Also, (2.2) implies that f  (ξx ) > 0 so that by letting x → x0 , we have f  (x0 ) ≥ 0. Thus, f  (x0 ) = 0.

2.4 Additional Exercises 2.4.1 Limit x = 1. 1. Using the definition of limit, show that lim x→3 4x  −9 x, if x < 1, 2. Show that the function f defined by f (x) = does not have 1 + x, if x ≥ 1 the limit as x → 1. ⎧ ⎨ 3 − x, if x > 1, if x = 1, Find lim x→1 f (x). Is this limit 3. Let f be defined by f (x) = 1, ⎩ 2x, if x < 1. equal to f (1)? 4. In the following cases, find the left limit and right limit of f at 1 and check whether theyare equal and equal to f (1):  x, if x ≤ 1, x, if x ≤ 1, (i) f (x) = (ii) f (x) = 2, if x > 1. 1, if x > 1. 5. Let f be defined on a deleted neighbourhood D0 of a point x0 and lim x→x0 f (x) = b. If b = 0, then show that there exists δ > 0 such that f (x) = 0 for every x ∈ (x0 − δ, x0 + δ) ∩ D0 .  1, if x ∈ Q, 6. Let f be defined by f (x) = Show that 0, if x ∈ / Q. (a) lim x→a f (x) does not exist for any a ∈ R, and (b) lim x→a (x − a) f (x) = 0 for every a ∈ R. 7. Suppose f, g : R → R and b ∈ R. Show that, if lim x→∞ f (x) = ∞ and lim x→∞ g(x) = b, then lim x→∞ (g ◦ f )(x) = b. 8. Give an example in which lim x→a f (x) = b and lim y→b g(y) = c, but lim x→a (g ◦ f )(x) = c. 9. Let f : (0, ∞) → R. Show that lim x→0 f (x) = b if and only if lim x→∞ f (1/x) = b. 10. Suppose lim x→∞ f (x) = b and lim x→∞ g(x) = c. Verify the following. (a) lim x→∞ [ f (x) + g(x)] = b + c and lim x→∞ f (x)g(x) = bc. (b) If c = 0, then there exists M0 > 0 such that g(x) = 0 for all x > M0 and b f (x) = . lim x→∞ g(x) c 11. State and prove sequential characterization for the following:

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2 Limit, Continuity and Differentiability of Functions

lim x→a f (x) = ∞, lim x→a f (x) = −∞, lim x→+∞ f (x) = ∞, lim x→+∞ f (x) = −∞, lim x→−∞ f (x) = ∞, lim x→−∞ f (x) = −∞. 12. Let f : R → R be such that f (x + y) = f (x) + f (y). Suppose lim x→0 f (x) exists. Prove that lim x→0 f (x) = 0 and lim x→c f (x) = f (c) for any c ∈ R.

2.4.2 Continuity 1. Suppose f : [a, b] → R is continuous. If c ∈ (a, b) is such that f (c) > 0, and if 0 < β < f (c), then show that there exists δ > 0 such that f (x) > β for all x ∈ (c − δ, c + δ) ∩ [a, b]. 2. Let f : R → R satisfy the relation f (x + y) = f (x) + f (y) for every x, y ∈ R. If f is continuous at 0, then show that f is continuous at every x ∈ R, and in that case f (x) = x f (1) for every x ∈ R. 3. There does not exist a continuous function f from [0, 1] onto R. Why? 4. Find a continuous function f from (0, 1) onto R. 5. Prove that if f : [a, b] → R is continuous on the intervals [a, c] and [c, b] for some c ∈ (a, b), then f is continuous on [a, b]. 6. Give an example of a discontinuous function f : [0, 1] → R such that f is continuous on the intervals [0, 1/2] and (1/2, 1]. 7. Suppose f : [a, b] → [a, b] is continuous. Show that there exists c ∈ [a, b] such that f (c) = c. 8. Suppose f : [a, b] → [a, b] is such that there exists r satisfying 0 < r < 1 and | f (x) − f (y)| ≤ r |x − y| for all x, y ∈ [a, b]. Let x1 ∈ [a, b] and for n ∈ N, let xn+1 := f (xn ). Prove that xn → c for some c and f (c) = c. 9. There exists x ∈ R such that 17x 19 − 19x 17 − 1 = 0. Why? 10. Let f : R → R be defined by f (x) = 1 + x + x 2 . Without solving a quadratic equation, can you assert that there is some x0 such that f (x0 ) = 2? 11. If p(x) is a polynomial of odd degree, then there exists at least one x0 ∈ R such that p(x0 ) = 0. Why? 12. Suppose f : R → R is continuous such that f (x) → 0 as |x| → ∞. Prove that f attains either a maximum or a minimum. 13. Suppose f : [a, b] → R is continuous such that for every x ∈ [a, b], there exists a y ∈ [a, b] such that | f (y)| ≤ | f (x)|/2. Show that there exists x0 ∈ [a, b] such that f (x0 ) = 0. 14. Let f : [a, b] → [a, b] be such that | f (x) − f (y)| ≤ |x − y|/2 for all x, y ∈ [a, b]. Show that there exists x0 ∈ [a, b] such that f (x0 ) = x0 . 15. Write details of the proof of Corollary 2.2.10 16. Prove the following. (a) Let f : (a, b) → R be a continuous function. If lim x→a f (x) = c and lim x→b f (x) = d, where c < d, then for every y ∈ (c, d), there exists x ∈ (a, b) such that f (x) = y.

2.4 Additional Exercises

169

(b) Let f : R → R be a continuous function. If lim x→−∞ f (x) = c and lim x→∞ f (x) = d, where c < d, then for every y ∈ (c, d), there exists x ∈ R such that f (x) = y. (c) Let f : R → R be a continuous function. If lim x→−∞ f (x) → c and lim x→∞ f (x) = ∞, where c < d, then for every y ∈ (c, ∞), there exists x ∈ R such that f (x) = y. (d) From (c) above, deduce that for every y ∈ (0, ∞), there exists x ∈ R such that e x = y. 17. Prove that if f is strictly monotonic on an interval I , then f is injective on I . 18. Let n ∈ N. Prove that for every y ≥ 0, there exists a unique x ≥ 0 such that x n = y. 19. Suppose f is a continuous function defined on an interval I and x0 is an interior point of I . Prove the following. (a) If f is increasing on (x0 − h, x0 ) and decreasing on (x0 , x0 + h) for some h > 0, then f attains local maximum at x0 . (b) If “increasing” and “decreasing” in (a) above are interchanged, then in the conclusion “maximum” can be replaced by “minimum”. (c) If “increasing” and “decreasing” in (a) are replaced by “strictly increasing” and “strictly decreasing”, respectively, then we obtain “strict local maximum”. 20. Let f be a continuous function defined on an interval I . Show that if f is injective, then it is strictly monotonic on I [Hint: Use Intermediate Value Theorem].

2.4.3 Differentiation 1. Prove that the function f (x) = |x|, x ∈ R is not differentiable at 0. 2. Let x0 be an interior point of an interval I . Prove that f : I → R is differentiable at x0 if and only if f + (x0 ) and f − (x0 ) exist and f − (x0 ) = f + (x0 ), and in that case f  (x0 ) = f − (x0 ) = f + (x0 ). 3. Consider a polynomial p(x) = a0 + a1 x 2 + . . . + an x n with real coefficients a2 an a1 + + ... + = 0. Show that there a0 , a1 , . . . , an such that a0 + 2 3 n+1 exists x0 ∈ R such that p(x0 ) = 0. [Note that the conclusion need not hold if the condition imposed on the coefficients is dropped. To see this, consider p(x) = 1 + x 2 .] 4. Let I and J be open intervals and f : I → J be bijective and differentiable at every x0 ∈ I . If f  (x0 ) = 0, then show that the inverse function f −1 : J → I is also differentiable at x0 and ( f −1 ) (x0 ) = 1/ f  (x0 ). 5. Let f : R → R be defined by

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2 Limit, Continuity and Differentiability of Functions

 f (x) =

x 2 , x rational, 0, x irrational.

Show that f is differentiable at 0 and f  (0) = 0. 6. Let f : R → R be defined by  f (x) =

x 2 sin(1/x), x = 0, 0, x = 0.

Show that f is differentiable at every x ∈ R. Is f  continuous? 7. Let f : R → R be defined by  f (x) =

x 2 , x ≥ 0, 0, x < 0.

Show that f is differentiable at every x ∈ R. Is f  continuous? 8. The function f : R → R defined by f (x) = 1 + x + x 3 has neither maximum nor minimum at any point. Why? 9. Find the points at which the function f : R → R defined by f (x) = 1 = x − x 3 attain local maxima and local minima. 10. Using Taylor’s theorem, show that (1 + x)n = 1 + nx +

n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + . . . + xn. 2! 3!

11. Show that there does not exist a function f : [0, 1] → R which is differentiable on (0, 1) such that  0, if 0 < x < 1/2,  f (x) = 1, if 1/2 ≤ x < 1. [Hint: Use Example 2.3.18 in the interval [0, 1/2] and [1/2, 1] taking x0 = 1/2, and show that the resulting function f is not differentiable at x0 = 1/2.] 12. Suppose f is differentiable on (0, ∞) and lim x→∞ f  (x) = 0. Prove that lim x→∞ [ f (x + 1) − f (x)] = 0. 13. Let f : R → R be defined by f (x) = 1 + x + x 3 . Prove that f is a bijection. [Hint: Use Intermediate value theorem and mean value theorem.] 14. Let f : [a, b] → R be a continuous function which is differentiable on (a, b). Prove the following. (a) If f  (x) ≥ 0 (respectively, f  (x) > 0) for all x ∈ (a, b), then f is monotonically increasing (respectively, strictly increasing) on [a, b]. (b) If lim x→a f  (x) exists, then f  (a) exists and f  (a) = lim x→a f  (x). [Hint: Use mean value theorem.]

2.4 Additional Exercises

171

15. Let f : [a, b] → R be a continuous function which is differentiable on (a, b). Prove that if f  is strictly positive or strictly negative on (a, b), then it is a one-one function. 16. Let f : [a, b] → R be a continuous function which is differentiable on (a, b). Prove that if f  (x) = 0 for all x ∈ (a, b), then f is either strictly increasing or strictly decreasing on [a, b]. 17. Let f : [0, 1] → R be a continuous function which is differentiable on (0, 1). Prove that, if f (0) = 0 and f (1) = 1, then there exists c ∈ (0, 1) such that f  (c) = 1. 18. Using MVT, prove the following: (a) e x ≥ 1 + x for all x ∈ R. (b) | sin x − sin y| ≤ |x − y| for all x ∈ R. x −1 ≤ ln x ≤ x − 1 for all x ∈ R. (c) x 19. Let f (x) = x 2 + 1 and g(x) = x + 2 for all x ∈ R. What is wrong with the following statement? Since g  (0) = 0, lim x→0

f  (0) f (x) =  = 0. g(x) g (0)

20. Prove that lim x→∞ x ln(1 + 1/x) exists and it is equal to 1. [Hint: L’Hospital’s rule] 21. Find the following limits. cos x x − π/2 1 (b) lim x→0 x(ln x)2 ln x (c) lim x→∞ 2 x ln x (d) lim x→∞ √ x (a) lim x→π/2

22. Let f be a function defined on an interval I . Prove the following. (a) f is convex (i.e., convex downwards) if and only if the line segment joining any tow points on the graph of f lies in the region {(x, y) : f (x) ≤ y, x ∈ I }. (b) f is concave (i.e., concave upwards) if and only if the line segment joining any tow points on the graph of f lies in the region {(x, y) : y ≤ f (x), x ∈ I }. 23. Suppose f is convex on an open interval I . Prove the following. (a) f  (x−) and f  (x+) exist for every x ∈ I . (b) f is continuous on I .

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2 Limit, Continuity and Differentiability of Functions

24. Give an example to show that a convex function defined on a closed interval need not be continuous. 25. Find approximate values of sin x, cos x, tan−1 x at the point x = 1/2 using Taylor’s formula and find the estimates for the errors for different values of n.

Chapter 3

Definite Integral

Integration is as ubiquitous in calculus as differentiation is. As the word suggests, it represents, in some sense, aggregate or sums of certain quantities associated with the values of certain functions. The integration that we deal with is the definite integral which is different from the so called indefinite integral or antiderivative of a function. Definite integral also can be defined using indefinite integral. But, the class of functions that has indefinite integral is much smaller than the class of functions which can be integrated using the concept of definite integral. Moreover, the indefinite integral becomes a special case of definite integral, thanks to an important theorem, called the fundamental theorem of integration.

3.1 Integrability and Integral 3.1.1 Introduction In school one comes across the definition of the integral of a real valued function defined on a closed and bounded interval [a, b] between the limits a and b, that is, b f (x)dx a

as the number g(b) − g(a), where g is a function whose derivative is f at all points in the open interval (a, b). One immediate question that one would like to ask is the following: Given any function f : [a, b] → R, is it possible to find a a function g : [a, b] → R which is differentiable on (a, b) and g  (x) = f (x) for all x ∈ (a, b)?

The answer is in the negative, as following example shows. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_3

173

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3 Definite Integral

Fig. 3.1 Area under the curve y = f (x)

Example 3.1.1 Consider the function  f (x) =

−1, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1.

In this case, there does not exist a function g : [−1, 1] → R which is differentiable ♦ on (−1, 1) and g  (x) = f (x) for all x ∈ (−1, 1) (cf. Theorem 2.3.9). Another point one recalls from school mathematics is that if g is a differentiable function defined on an open interval, then g  (x) has a geometric meaning, namely, it represents the slope of the tangent to the graph of g at the point x. Do we have a geometric meaning to the integral

b a

f (x)dx?

We answer to the above question in the affirmative for certain class of functions as we shall see. Suppose f : [a, b] → R is a bounded function. Our attempt is to associate a number γ to such a function such that, in case f (x) ≥ 0 for x ∈ [a, b], then γ is the area of the region under the graph of f , i.e., the region R bounded by the graph of f , x-axis, and the ordinates at a and b. We may not succeed to do this for all bounded functions f (Fig. 3.1). Suppose, for a moment, that f (x) ≥ 0 for all x ∈ [a, b]. Let us agree that we have some idea about the area A of the region R under the curve y = f (x). For example, we require that the area A lies between two quantities A1 and A2 , where A1 is the area of the rectangle that is inscribed within the region R and A2 is the area of the region which subscribe R, with their sides parallel to the coordinate axes. In particular, A1 = m(b − a), A2 = M(b − a) and m(b − a) ≤ A ≤ M(b − a), where m = inf

x∈[a,b]

f (x) and M = sup f (x). x∈[a,b]

(3.1)

3.1 Integrability and Integral

175

Thus, we get an upper and lower bound for A. To get better estimates, let us consider a point c such that a < c < b. Then we have m 1 ≤ f (x) ≤ M1 ∀x ∈ [a, c]; m 2 ≤ f (x) ≤ M2 ∀x ∈ [c, b], where m 1 = inf f (x),

m 2 = inf f (x),

M1 = sup f (x),

M2 = sup f (x).

x∈[a,c]

x∈[c,b]

x∈[a,c]

x∈[c,b]

Then, we must have m 1 (c − a) + m 2 (b − c) ≤ A ≤ M1 (c − a) + M2 (b − c).

(3.2)

Since m ≤ min{m 1 , m 2 } and max{M1 , M2 } ≤ M, we have m(b − a) = m(c − a) + m(b − c) ≤ m 1 (c − a) + m 2 (b − c), M(b − a) = M(c − a) + M(b − c) ≥ M1 (c − a) + M2 (b − c). Thus, we can infer that the estimates in (3.2) are better than those in (3.1). We may be able to improve these bounds by taking more and more points in [a, b]. This is the basic idea of Riemann integration that we describe below.

3.1.2 Lower and Upper Sums Let f : [a, b] → R be a bounded function and let P be a partition of [a, b], i.e., a finite set P = {x0 , x1 , x2 , . . . , xk } of points in [a, b] such that a = x0 < x1 < x2 < . . . < xk = b. k . Corresponding to the partition We shall denote such partition also by P = {xi }i=0 k P = {xi }i=0 and the function f , we associate two numbers:

L(P, f ) :=

k 

m i (xi − xi−1 ), U (P, f ) :=

k 

i=1

Mi (xi − xi−1 ),

i=1

where mi = for i = 1, . . . , k.

inf

x∈[xi−1 ,xi ]

f (x),

Mi =

sup

x∈[xi−1 ,xi ]

f (x)

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3 Definite Integral

Note that for the definition of L(P, f ) and U (P, f ) above, we used the fact that f is a bounded function (How?). Definition 3.1.1 The quantities L(P, f ) and U (P, f ) are called the lower sum and the upper sum, respectively, of the function f associated with the partition P. ♦ Note that if f (x) ≥ 0 for all x ∈ [a, b], then L(P, f ) is the total area of the rectangles with lengths m i and widths xi − xi−1 , and U (P, f ) is the total area of the rectangle with lengths Mi and widths xi − xi−1 for i = 1, . . . , k. Thus, it is intuitively clear that the required area A must satisfy the relation L(P, f ) ≤ A ≤ U (P, f ) for all partitions P of [a, b] (Figs. 3.2 and 3.3).

Fig. 3.2 Lower sum

Fig. 3.3 Upper sum

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177

Remark 3.1.1 If f is continuous, then L(P, f ) and U (P, f ) can be represented as L(P, f ) :=

k 

f (ci )(xi − xi−1 ), U (P, f ) :=

i=1

k 

f (di )(xi − xi−1 ),

i=1

respectively, for some ci , di , i ∈ {1, . . . , k} (Why?).

♦

3.1.3 The Integral and Its Characterizations Throughout this chapter, functions defined on [a, b] are considered to be bounded functions, and we use the notation P to denote the set of all partitions of [a, b]. Thus we have L(P, f ) ≤ U (P, f ) ∀ P ∈ P. Definition 3.1.2 Let f : [a, b] → R be a bounded function. Then f is said to be Riemann integrable on [a, b], if there exists a unique γ such that for every P ∈ P, L(P, f ) ≤ γ ≤ U (P, f ), and in that case, γ is called the Riemann integral1 of f , and it is denoted by  b f (x)d x

♦

a

CONVENTION: In the due course, Riemann integral will be simply referred to as the integral. Remark 3.1.2 In higher mathematics one will come across another form of integral called Lebesgue integral, which is more general than Riemann integral. It is dealt in a course on Measure and Integration (see, e.g., Nair [9]). ♦ For a bounded function f : [a, b] → R, though L(P, f ) ≤ U (P, f ) for all P ∈ P, it is not obvious that sup L(P, f ) ≤ inf U (P, f ). P∈P

P∈P

In fact, it is true. Before showing this, let us introduce a definition.

1

Named after the German mathematician Georg Friedrich Bernhard Riemann (September 17, 1826 July 20, 1866) who placed the concept of integration in a rigorous footing.

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3 Definite Integral

Definition 3.1.3 Let P and Q be a partitions of [a, b]. Then Q is called a refinement of P if P ⊆ Q. ♦ If Q is a refinement of P, then it can be shown easily that L(P, f ) ≤ L(Q, f ) and U (Q, f ) ≤ U (P, f ). Theorem 3.1.1 Let f : [a, b] → R be a bounded function. Then (i) L(P, f ) ≤ U (Q, f ) ∀ P, Q ∈ P, (ii) sup L(P, f ) ≤ inf U (P, f ). P∈P

P∈P

Proof Let P and Q be partitions of [a, b], and let P˜ be the partition obtained by taking points in P and Q without repetitions. Then we have ˜ f ) ≤ U ( P, ˜ f ) ≤ U (Q, f ). L(P, f ) ≤ L( P, Thus, (i) is proved. By (i), sup L(P, f ) ≤ U (Q, f ) ∀ Q ∈ P. P∈P

Hence, sup L(P, f ) ≤ inf U (Q, f ). Q∈P

P∈P



Thus, (ii) is also proved.

Notation 3.1.1 Let P and Q be partitions of [a, b]. Then we denote by P ∪ Q, the partition obtained by taking points in P and Q without repetitions. ♦ We have already observed in Theorem 3.1.1 that, if P and Q are partitions of [a, b], then P ∪ Q is a refinement of both P and Q. Suppose f is Riemann integrable over [a, b]. Then we know that b L(P, f ) ≤

f (x) dx ≤ U (P, f ) ∀ P ∈ P.

(1)

a

Hence,

b sup L(P, f ) ≤ P∈P

f (x) dx ≤ inf U (P, f ). P∈P

a

3.1 Integrability and Integral

Since γ :=

b a

179

f (x) dx is the only number satisfying (1), we can assert that

sup L(P, f ) = inf U (P, f ). P∈P

(2)

P∈P

Conversely, suppose (2) holds, and let γ := sup L(P, f ) = inf U (P, f ). P∈P

P∈P

Then it is clear that L(P, f ) ≤ γ ≤ U (P, f ) for all P ∈ P. If γ˜ ∈ R is such that L(P, f ) ≤ γ˜ ≤ U (P, f ) for all P ∈ P, then we obtain sup L(P, f ) ≤ γ˜ ≤ inf U (P, f ) P

P

so that by (2), γ˜ = γ . Thus, we have proved the following theorem. Theorem 3.1.2 A bounded function f : [a, b] → R is Riemann integrable if and only if L( f ) := sup L(P, f ) and U ( f ) := inf U (P, f ) P∈P

P∈P

are equal, and in that case, the common value is the integral of f . Notation 3.1.2 In view of Theorem 3.1.2, the quantities L( f ) := sup L(P, f ) and U ( f ) := inf U (P, f ) P∈P

P∈P

are known as lower integral and upper integral, respectively, and they are also denoted by  b b f (x)dx and f (x)dx, a

a

♦

respectively. Definition 3.1.4 Suppose f : [a, b] → R is integrable. Then we define a

b f (x) dx = −

and for any τ ∈ [a, b], we define

f (x) dx, a

b

τ τ

f (x) dx = 0.

♦

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3 Definite Integral

Example 3.1.2 Let f (x) = c for all x ∈ [a, b], for some c ∈ R. Then we have L(P, f ) = c(b − a), U (P, f ) = c(b − a) Hence c(b − a) = L(P, f ) ≤ L( f ) ≤ U ( f ) ≤ U (P, f ) = c(b − a). Thus, f is integrable and

b

f (x)dx = c(b − a).

♦

a

Example 3.1.3 Let f (x) = x for all x ∈ [a, b]. Let us consider an arbitrary partition P : a = x0 < x1 < · · · < xk = b. Then L(P, f ) =

k 

xi−1 (xi − xi−1 ), U (P, f ) =

k 

i=1

xi (xi − xi−1 ).

i=1

Since xi−1 (xi − xi−1 ) ≤

1 (xi + xi−1 )(xi − xi−1 ) ≤ xi (xi − xi−1 ) 2

we have

1 2 2 (x − xi−1 ) ≤ U (P, f ). 2 i=1 i k

L(P, f ) ≤ Thus,

L(P, f ) ≤

b2 − a 2 ≤ U (P, f ) 2

for all partitions P. Also, we have U (P, f ) − L(P, f ) =

k 

(xi − xi−1 )2 .

i=1

Thus, taking xi = a + i (b−a) for i = 0, 1, . . . , k, we have k U (P, f ) − L(P, f ) = k

 b − a 2 k

=

(b − a)2 . k

Since L(P, f ) ≤ L( f ) ≤ U ( f ) ≤ U (P, f ), we obtain, 0 ≤ U ( f ) − L( f ) ≤

(b − a)2 ∀ k ∈ N. k

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181

Thus, L( f ) = U ( f ) so that f is integrable, and the value of

b a

(b2 − a 2 )/2.

f (x)dx is ♦

Remark 3.1.3 Not all functions are integrable! For example, consider f : [a, b] → R defined by  0, x ∈ Q, f (x) = 1, x ∈ / Q. For this function we have L(P, f ) = 0 and U (P, f ) = b − a for any partition P of [a, b]. Thus, in this case L( f ) = 0, U ( f ) = b − a, and hence, f is not integrable. The above function f is called the Dirichlet function. ♦ Another sum corresponding to a partition P of [a, b] is the so-called Riemann sum. Definition 3.1.5 Let P : a = x0 < x1 < · · · < xk = b of [a, b]. Then any set T := {ξ1 , . . . , . . . ξk } of points in [a, b] with ξi ∈ [xi−1 , xi ] for i = 1, . . . , k is called a set of tags on P, and the sum S(P, f, T ) :=

k 

f (ξi )(xi − xi−1 )

i=1

is called the Riemann sum of f corresponding to (P, T ).

♦

We may observe that for any partition P of [a, b], L(P, f ) ≤ S(P, f, T ) ≤ U (P, f ) for every tag T on P. Theorem 3.1.3 A bounded function f : [a, b] → R is Riemann integrable if and only if for every ε > 0, there exists a partition P of [a, b] such that U (P, f ) − L(P, f ) < ε. Proof Suppose f is integrable and γ is its integral. Then γ is the unique number such that for every partition P of [a, b], L(P, f ) ≤ γ ≤ U (P, f ).

(∗)

Hence, if there exists an ε0 > 0 satisfying U (P, f ) − L(P, f ) ≥ ε0 for all partitions P of [a, b], then we arrive at a contradiction to the fact that γ is the unique number satisfying (∗). This proves the necessary part. Conversely, suppose that for every ε > 0, there exists a partition Pε of [a, b] such that U (Pε , f ) − L(Pε , f ) < ε. Since

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3 Definite Integral

L(Pε , f ) ≤ L( f ) ≤ U ( f ) ≤ U (Pε , f ) we obtain that U ( f ) − L( f ) < ε for every ε > 0. This shows that L( f ) = U ( f ), so that by Theorem 3.1.2, f is integrable.  The proof of the following corollary is immediate from Theorem 3.1.3. Corollary 3.1.4 (Archimedes–Riemann theorem)2 A bounded function f : [a, b] → R is Riemann integrable if and only if there exists a sequence (Pn ) of partitions of [a, b] such that U (Pn , f ) − L(Pn , f ) → 0 as n → ∞ and in that case b L(Pn , f ) →

b f (x) dx,

U (Pn , f ) →

a

f (x) dx a

and

b S(Pn , f, Tn ) →

f (x) dx, a

where Tn is any set of tags on Pn , n ∈ N. as n → ∞. Exercise 3.1.1 Supply details of the proof of Corollary 3.1.4.



Here is another application of Theorem 3.1.3. Theorem 3.1.5 Every monotonic function f : [a, b] → R is Riemann integrable. Proof Assume first that f is monotonically increasing, that is, for every x, y ∈ [a, b], x ≤ y implies f (x) ≤ f (y). This assumption, in particular implies that f is a bounded function, as f (a) ≤ f (x) ≤ f (b) for every x ∈ [a, b]. We obtain the proof by using Theorem 3.1.3. For this, let ε > 0 be given. Let P be a partition of [a, b] with equidistant points, that is, xi = a +

i(b − a) , i = 0, 1, . . . , n. n

Then, we have U (P, f ) − L(P, f ) =

n 

(Mi − m i )(xi − xi−1 ),

i=1

2

Attributed to the Greek mathematician of the yore, Archimedes, as he devised procedures to compute areas of non-polygonal regions using the idea of approximation using polygonal regions, more than 2000 years before Riemann considered the procedure in a general framework in the year 1845.

3.1 Integrability and Integral

183

where Mi = f (xi ), m i = f (xi−1 ), xi − xi−1 =

b−a . n

Thus, U (P, f ) − L(P, f ) =

n b−a  b−a [ f (b) − f (a)] [ f (xi ) − f (xi−1 )] = n i=1 n

b−a [ f (b) − f (a)] < ε, we obtain U (P, f ) − n L(P, f ) < ε so that, by Theorem 3.1.3, f is integrable. 

Taking n large enough such that

In the appendix (Sect. 3.6), we shall also give a characterization of Riemann integrability in terms of Riemann sums (see Theorem 3.6.1). Remark 3.1.4 Suppose for each n = 2, 3, . . ., we have a partition Pn of kn subintervals. It is to be observed that, as n varies, not only that kn may vary, but also the partition points in Pn also may vary. Thus, we may write Pn as = b. Pn := x0(n) < x1(n) < · · · < xk(n) n For example, let Pn be the partition obtained by dividing [a, b] into equidistant points with n partition points. Then we have kn = n and xi(n) = a + i

b−a , i = 0, 1, . . . , n. n

Note that Pn+1 is not a refinement of Pn . However, if we take Q n to be the partition with 2n equidistant points, that is, yi(n) = a + i

b−a , i = 0, 1, . . . , 2n , 2n

then Q n+1 is a refinement of Q n . In this case kn = 2n .

♦

3.1.4 Some Basic Properties of Integral Theorem 3.1.6 Suppose f is integrable on [a, b], and m, M are such that m ≤ f (x) ≤ M for all x ∈ [a, b]. Then b f (x) dx ≤ M(b − a).

m(b − a) ≤ a

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3 Definite Integral

Proof We know that for any partition P on [a, b], b m(b − a) ≤ L(P, f ) ≤

f (x) dx ≤ U (P, f ) ≤ M(b − a). a



Hence the result. Observation: For a bounded function ϕ defined on a closed interval I , let m ϕ := inf ϕ(x),

Mϕ := sup ϕ(x).

x∈I

x∈I

Then, we see that, for any two bounded functions f and g on I and for every y ∈ I , inf f (x) + inf g(x) ≤ f (y) + g(y) ≤ sup f (x) + sup g(x). x∈I

x∈I

x∈I

x∈I

Hence, we obtain m f + m g ≤ m f +g ≤ M f +g ≤ M f + Mg . Theorem 3.1.7 Let f and g be integrable over [a, b]. Then f + g is integrable and b

b [ f (x) + g(x)]dx =

a

b f (x) dx +

a

g(x) dx, a

Proof From the observation that we made before the statement of the theorem, we obtain L(P, f ) + L(P, g) ≤ L(P, f + g) ≤ U (P, f + g) ≤ U (P, f ) + U (P, g) for any partition P of [a, b] (Verify). Now, let P1 and P2 be any two partitions of [a, b] and P = P1 ∪ P2 . Since L(P1 , f ) + L(P2 , g) ≤ L(P, f ) + L(P, g) ≤ L(P, f + g), U (P, f + g) ≤ U (P, f ) + U (P, g) ≤ U (P1 , f ) + U (P2 , g), and since L(P, f + g) ≤ L( f + g) ≤ U ( f + g) ≤ U (P, f + g), we obtain

3.1 Integrability and Integral

185

L(P1 , f ) + L(P2 , g) ≤ L( f + g) ≤ U ( f + g) ≤ U (P1 , f ) + U (P2 , g). This is true for any two arbitrary partitions P1 , P2 of [a, b]. Therefore (How?), L( f ) + L(g) ≤ L( f + g) ≤ U ( f + g) ≤ U ( f ) + U (g). But,

b L( f ) + L(g) =

b f (x)dx +

a

g(x)dx = U ( f ) + U (g). a

Hence, L( f + g) = U ( f + g) and it is equal to

b

b f (x)dx + g(x)dx so that f + g

a

a

is integrable and b

b [ f (x) + g(x)]dx =

a

b f (x)dx +

a

g(x)dx. a

Thus, the proof is complete.



Exercise 3.1.2 Prove Theorem 3.1.7 using Corollary 3.1.4.



Observation: For the next theorem, we shall be using the following fact: For any bounded set S ⊆ R, inf{−s : s ∈ S} = − sup{s : s ∈ S}, sup{−s : s ∈ S} = − inf{s : s ∈ S}. Theorem 3.1.8 If f is integrable on [a, b] and c ∈ R, then c f is integrable on [a, b], and b b c f (x)dx = c f (x)dx. a

a

Proof Let c ≥ 0. Then for any given partition P, we have L(P, c f ) = cL(P, f ), U (P, c f ) = cU (P, f ) so that

b sup L(P, c f ) = c P

b f (x)dx, inf U (P, c f ) = c

f (x)dx.

P

a

a

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3 Definite Integral

Hence,

b sup L(P, c f ) = c P

f (x)dx = inf U (P, c f ), P

a

b b showing that a c f (x)dx = c a f (x)dx. Next, let us assume that c < 0. Then we have b

b c f (x)dx =

a

b (−c)(− f )(x)dx = (−c)

a

(− f )(x)dx. a

Thus, once we prove b

b [− f (x)]dx = −

a

f (x)dx,

(1)

a

we obtain b

b

b

c f (x)dx = (−c)(−1) a

f (x)dx = c a

f (x)dx.

(2)

a

To prove (1), let P be any partition of [a, b]. Now, using the observation made before the statement of this theorem, we obtain (Verify) L(P, − f ) = −U (P, f ), −L(P, f ) = U (P, − f ) and sup L(P, − f ) = − inf U (P, f ), inf U (P, − f ) = − sup L(P, f ). Thus,

b f (x)dx = inf U (P, − f )

sup L(P, − f ) = − a

Thus, we have proved (1), and hence (2).



Theorem 3.1.9 The following results hold. (i) If f and g are integrable on [a, b], then f g is integrable on [a, b]. (ii) If f is integrable on [a, b], f (x) = 0 for all x ∈ [a, b] and 1/ f is bounded, then 1/ f is integrable on [a, b].

3.1 Integrability and Integral

187

Proof First we show that if f is integrable on [a, b], then f 2 is integrable on [a, b]. Since f 2 (x) − f 2 (y) = (| f (x)| + | f (y)|)(| f (x)| − | f (y)|) ≤ 2M(| f (x)| − | f (y)|) for all x, y ∈ [a, b], where M > 0 is a constant such that | f (x)| ≤ M for all x ∈ [a, b]. From this, it can be shown that, for any partition P of [a, b], U (P, f 2 ) − L(P, f 2 ) ≤ 2M[U (P, f ) − L(P, f )]. Hence, by Theorem 3.1.3, f 2 is integrable. (i) Let f and g be integrable on [a, b]. We observe that fg =

1 [( f + g)2 − ( f − g)2 ]. 4

Hence, the proof of (ii) follows by applying (i) above, Theorem 3.1.7 and Theorem 3.1.8. (ii) Suppose f (x) = 0 for all x ∈ [a, b] and 1/ f is bounded. Let K > 0 be such that 1/| f (x)| ≤ K for all x ∈ [a, b]. Then, for every x, y ∈ [a, b], we have 1 1 f (y) − f (x) − = ≤ K 2 [ f (y) − f (x)]. f (x) f (y) f (x) f (y) Hence, we obtain U (P, 1/ f ) − L(P, 1/ f ) ≤ K 2 [U (P, f ) − L(P, f )], so that by Theorem 3.1.3, 1/ f is integrable.



Example 3.1.4 For a0 , a1 , . . . , an ∈ R, the function f : R → R defined by f (x) = a0 + a1 x + · + an x n , x ∈ R, is integrable on every closed interval [a, b]. Indeed, we have seen in Example 3.1.3 that the function g(x) = x is integrable on [a, b], so that by Theorems 3.1.7, 3.1.8, 3.1.9, we obtain that f is integrable on every closed interval [a, b]. ♦ In the above example, we saw that every polynomial function f defined by f (x) = a0 + a1 x + · · · + an x n is integrable on every closed and bounded interval [a, b]. b However, we did not obtain the value of the integral a f (x)dx. Theorems 3.1.7 and 3.1.8 do give the relation

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3 Definite Integral

b

 f (x) dx = a0

b

 dx + a1

a

a

b

 x dx + · · · + an

a

b

x n dx. a

b b So, we get the value of a f (x) dx, once we know the value of a x k dx for each k ∈ N. This can also be computed using the procedure adopted in Example 3.1.3. However, there are easier ways. In fact, we have already indicated in the beginning of this chapter that if there is a differentiable function g such that g  (x) = f (x) on (a, b), then b f (x) dx = g(b) − g(a) a

so that

b

b x dx = k

a

d dx



a

x k+1 k+1

dx =

bk+1 − a k+1 . k+1

We shall prove this result soon. Theorem 3.1.10 Suppose f is integrable on [a, c] and [c, b]. Then f is integrable on [a, b], and b c b f (x) dx = f (x) dx + f (x) dx. a

a

c

Proof Let f 1 = f |[a,c] , f 2 = f |[c,b] . Let ε > 0 be given. Since f 1 and f 2 are integrable, there exist partitions P1 and P2 of [a, c] and [c, b] respectively such that U (P1 , f 1 ) − L(P1 , f 1 )
0. Hence the final result.



We shall make use of the following lemma. Lemma 3.1.11 Let f, g, h be bounded functions on [a, b] such that g(x) ≤ f (x) ≤ h(x) ∀ x ∈ [a, b]. Then b

b g(x) dx ≤

a

b f (x) dx ≤

a

b f (x)dx ≤

a

(∗)

h(x)dx. a

Proof Let P and Q be any two partitions of [a, b]. Then, from the assumption that g(x) ≤ f (x) ≤ h(x) for all x ∈ [a, b], we have L(P, g) ≤ L(P, f ) ≤ U (Q, f ) ≤ U (Q, h). From this we have (how?) sup L(P, g) ≤ sup L(P, f ) ≤ inf U (Q, f ) ≤ inf U (Q, h). P∈P

P∈P

P∈P

P∈P

This is exactly the relations in (∗).



The following theorem is a simple consequence of the above lemma. Theorem 3.1.12 Let f be integrable on [a, b]. Then b f (x) ≥ 0 ∀x ∈ [a, b]

⇒

f (x) dx ≥ 0. a

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3 Definite Integral

More generally, if f and g are integrable on [a, b], then b f (x) ≤ g(x) ∀x ∈ [a, b]

b f (x) dx ≤

⇒ a

g(x) dx. a

Proof Suppose f (x) ≥ 0 forall x ∈ [a, b]. Taking g = 0 and h = f in Lemma b 3.1.11, we obtain a f (x) dx ≥ 0. Next, let f and g be integrable on [a, b] such that f (x) ≤ g(x) forall x ∈ [a, b]. Applying the above result for the function g − f , b b we obtain a f (x) dx ≤ a g(x) dx.  Corollary 3.1.13 Let f be integrable on [a, b] such that f (x) ≥ 0 for all x ∈ [a, b]. Then d b [c, d] ⊆ [a, b] ⇒ f (x) dx ≤ f (x) dx. c

a

Proof Let [c, d] ⊆ [a, b]. By Theorem 3.1.10, we have b

c

d

f (x) dx = a

f (x) dx + a

b f (x) dx +

c

f (x) dx d

and by Theorem 3.1.12, w have c

b f (x) dx +

a

f (x) ≥ 0. d



Hence, we obtain the required result.

Theorem 3.1.14 Let f be a bounded function on [a, b], and for n ∈ N, let gn and h n be integrable functions on [a, b] such that gn (x) ≤ f (x) ≤ h n (x) ∀ x ∈ [a, b], ∀ n ∈ N, b [h n (x) − gn (x)]dx → 0. a

Then f is integrable and b

b gn (x) dx =

lim

n→∞ a

b f (x) dx = lim

h n (x)dx.

n→∞

a

a

3.1 Integrability and Integral

191

Proof Since gn (x) ≤ f (x) ≤ h n (x) for all x ∈ [a, b] and gn and h n are integrable for all n ∈ N, it follows that f is a bounded function and by Lemma 3.1.11, we have, b

b gn (x) dx ≤

a



b f (x) dx ≤

a

b

f (x)dx ≤

h n (x)dx

(∗)

a

a

b b for each n ∈ N. From this, since a [h n (x) − gn (x)]dx → 0, we have a f (x) dx = b b f (x)dx, so that f is integrable, and hence by (*), we have limn→∞ a gn (x) dx = a b b  a f (x) dx = lim n→∞ a h n (x)dx. Theorem 3.1.15 Suppose f is integrable on [a, b]. Then | f | is integrable and

b

b





f (x)dx ≤ | f (x)|dx.





a

a

Proof First we show that | f | is integrable. For this, let ε > 0 be given and let n be a partition of [a, b] such that P := {xi }i=0 U (P, f ) − L(P, f ) < ε. For i = 1, 2, . . . , n, let Mi =

sup

x∈[xi−1 ,xi ]

f (x), m i =

inf

x∈[xi−1 ,xi ]

f (x).

Then, for i ∈ {1, . . . , n} and for any x, y ∈ [xi−1 , xi ], we have | f (x)| − | f (y)| ≤ | f (x) − f (y)| = max{ f (x) − f (y), f (y) − f (x)} ≤ Mi − m i . From this, it follows that U (P, | f |) − L(P, | f |) ≤ U (P, f ) − L(P, f ) < ε. Hence, by Theorem 3.1.3, | f | is integrable. Next, since f (x) ≤ | f (x)| and − f (x) ≤ | f (x)| for all x ∈ [a, b], by Theorem 3.1.12, we have b

b f (x)dx ≤

a

b | f (x)|dx and

a

−

b f (x)dx ≤

a

| f (x)|dx. a

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3 Definite Integral

Hence, ⎧ b

⎫

b

 b ⎨ ⎬ b



f (x)dx = max f (x)dx, − f (x)dx ≤ | f (x)|dx.



⎩ ⎭



a

a

a

a



This completes the proof.

3.1.5 Integral of Continuous Functions We have shown that every monotone function on [a, b] is Riemann integrable (Theorem 3.1.5). Our aim in this section is to show that every continuous function defined on [a, b] is Riemann integrable. We shall also deduce some important properties of integrals of continuous functions. Definition 3.1.6 Given a partition P = {xi : i = 0, 1, . . . , k} of [a, b], the quantity μ(P) := max{xi − xi−1 : i = 1, . . . , k} ♦

is called the mesh or norm of the partition P.

Theorem 3.1.16 Suppose f : [a, b] → R is a continuous function. Then f is integrable. Further, if (Pn ) is a sequence of partitions of [a, b] such that μ(Pn ) → 0 as n → ∞, and for each n ∈ N, if Tn is a set of tags on Pn , then the sequences {U (Pn , f )}, b {L(Pn , f )} {S(Pn , f, Tn )} converge to the same limit a f (x) dx. For its proof, we shall make use of a property of continuous functions defined on a closed and bounded intervals. Theorem 3.1.17 Let f be a real valued continuous function defined on a closed and bounded interval [a, b]. Then for every ε > 0, there exists δ > 0 such that x, y ∈ [a, b], |x − y| < δ

⇒

| f (x) − f (y)| < ε.

Proof Suppose f : [a, b] → R is continuous, and let ε > 0 be given. Suppose that the conclusion in the theorem is not true. Then there exists ε0 > 0 such that for every n ∈ N, there exist xn , yn ∈ [a, b] such that |xn − yn | < 1/n but | f (xn ) − f (yn )| ≥ ε0 .

(∗)

Since (xn ) is a bounded sequence, it has a convergent subsequence, say xkn → c for some c ∈ R. Since |xkn − ykn | → 0, we also have the convergence ykn → c. Now, by the continuity of f , we have f (xkn ) → f (c) and f (ykn ) → f (c). Thus, | f (xkn ) − f (ykn )| → 0. This is a contradiction to (∗) above. 

3.1 Integrability and Integral

193

The property described in the conclusion in Theorem 3.1.17 is called uniform continuity. Definition 3.1.7 A real valued function f defined on an interval I is said to be uniformly continuous on I if for every ε > 0, there exists δ > 0 (depending on ε) such that x, y ∈ I, |x − y| < δ ⇒ | f (x) − f (y)| < . ♦ Clearly, every uniformly continuous function is continuous. But, the converse is not true, as the following examples show. Example 3.1.5 Consider the function f (x) =

1 , x

0 < x ≤ 1.

Note that f is continuous on I := (0, 1]. But, it is not uniformly continuous. To see this, consider xn = 1/n and yn = 1/(n + 1). Then we know that |xn − yn | → 0 as n → ∞ and | f (xn ) − f (yn )| = 1 for all n ∈ N. Hence, the condition in the definition of uniform continuity is not satisfied if we take ε < 1. ♦ Example 3.1.6 Consider the function f (x) = sin(1/x),

0 < x ≤ 1.

Here, f is a bounded continuous on I := (0, 1]. But, it is not uniformly continuous. To see this, consider xn = 2/[(2n + 1)π ] for n ∈ N. Then taking yn = xn+1 , we have |xn − yn | → 0 as n → ∞, but π π | f (xn ) − f (xn+1 | = | sin(2n + 1) ) − sin((2n + 5) )| = 2 ∀ n ∈ N. 2 2 Thus, the condition in the definition of uniform continuity is not satisfied if we take ε < 2. ♦ Remark 3.1.5 The above two examples show that each of those functions cannot be extended to a continuous function on [0, 1] by assigning any value at 0. ♦ Exercise 3.1.3 Suppose f is a real valued function defined on an interval I such that there exists K > 0 satisfying | f (x) − f (y)| ≤ K |x − y| ∀ x, y ∈ I. Show that f is uniformly continuous. (Functions satisfying the condition above are called Lipschitz continuous functions, and the constant K is called a Lipschitz constant for f .)

194

3 Definite Integral

Proof (Proofs of Theorem 3.1.16) Let f : [a, b] → R be a continuous function. First we prove that for every ε > 0 there exists a partition P of [a, b] such that U (P, f ) − L(P, f ) < ε so that by Theorem 3.1.3, f is integrable. Let P : a = x0 < x1 < x2 . . . < xk = b be any partition of [a, b]. Then U (P, f ) − L(P, f ) =

k 

(Mi − m i )(xi − xi−1 ).

i=1

Since f is continuous on each closed interval [xi−1 , xi ], there exists ξi , ηi in [xi−1 , xi ] such that Mi = f (ξi ), m i = f (ηi ) for i = 1, . . . , k. Hence, U (P, f ) − L(P, f ) =

k  [ f (ξi ) − f (ηi )](xi − xi−1 ). i=1

Again, since f is uniformly continuous on [a, b], there exists δ > 0 such that | f (t) − f (s)| < ε/(b − a)

whenever |t − s| < δ.

(1)

Hence, if we take P such that μ(P) < δ, then we have U (P, f ) − L(P, f ) =

k  [ f (ξi ) − f (ηi )](xi − xi−1 ) < ε.

(2)

i=1

Thus, we have proved that f is integrable. Now, let (Pn ) be a sequence of partitions such that μ(Pn ) → 0 as n → ∞. Let δ > 0 be as in (1), and let N ∈ N be such that μ(Pn ) < δ for all n ≥ N . Then, taking Pn in place of P in (2), we obtain U (Pn , f ) − L(Pn , f ) < ε ∀ n ≥ N . Thus, U (Pn , f ) − L(Pn , f ) → 0 as n → ∞. Now, the conclusions follow from the observations b L(Pn , f ) ≤ f (x)dx ≤ U (Pn , f ), a

L(Pn , f ) ≤ S(Pn , f, Tn ) ≤ U (Pn , f ) for all n ∈ N.



3.1 Integrability and Integral

195

Example 3.1.7 Let f (x) = e x for all x ∈ [a, b]. Then, f is continuous. Let x0 = a, hn =

(b − a) , xi = a + i h n , ti = xi−1 , i = 1, . . . , n. n

n n and T = {ti }i=1 , we have μ(Pn ) → 0, and Then with Pn = {xi }i=1

S(Pn , f, Tn ) = h n

n 

ea+(i−1)h n = h n ea

i=1

n 

αn(i−1) = h n ea

i=1

αnn − 1 , αn − 1

where αn = eh n . Since αnn = eb−a , we have S(Pn , f, Tn ) = h n ea

αnn − 1 hn hn = ea [eb−a − 1] h = [eb − ea ] h . αn − 1 e n −1 e n −1

eh n − 1 = 1, we have n→∞ hn

Since lim

S(Pn , f, Tn ) → eb − ea . b e x dx = eb = ea .

Hence, by Theorem 3.1.16,

♦

a

Remark 3.1.6 It can be shown that if a bounded function f : [a, b] → R is piecewise continuous,, i.e., there are at most a finite number of points in [a, b] at which f is discontinuous, then f is integrable. For a proof of this, see Theorem 3.6.4 in the appendix (Sect. 3.6). ♦ In view of the above remark, we can assert the following: Every bounded function f : [a, b] → R which is discontinuous at most at a finite number of points in [a, b] is integrable. Example 3.1.8 For example, the function f : [−1, 1] → R defined by  f (x) =

−1, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1

is bounded, and it is discontinuous only at the point 0. Hence, f is integrable, but the function g : [−1, 1] → R defined by  g(x) =

1/x, x = 0, 1, x = 0

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3 Definite Integral

is not integrable, although g is discontinuous only at the point 0, because g is not a bounded function. ♦ Theorem 3.1.18 Suppose f : [a, b] → R and g : [c, d] → R are continuous functions such that f (x) ∈ [c, d] for all x ∈ [a, b]. Then g ◦ f : [a, b] → R is integrable. Proof The proof follows using the fact that composition of two continuous functions is continuous.  As a particular case of Theorem 3.1.18, it follows that, if f : [a, b] → R is a continuous function, then for any n ∈ N, the function f n (x) := [ f (x)]n , x ∈ [a, b], is integrable, so also the functions sin f (x), cos f (x), exp[ f (x)], x ∈ [a, b]. Remark 3.1.7 Theorem 3.1.18 is true without using the continuity of f , but using only its integrability. Interested reader can find the proof in [6] or [4]. Thus, if f is integrable on [a, b], then | f | is integrable. However, integrability of | f | does not imply integrability of f . For example, consider the function f : [0, 1] → R defined by  −1, x rational, f (x) = 1, x irrational. Note that L(P, f ) = −1 and U (P, f ) = 1 for all partitions P of [a, b], so that f is not integrable. However, | f | is integrable on [0, 1]. ♦ Theorem 3.1.19 Suppose f : [a, b] → R is continuous, f (x) ≥ 0 for all x ∈ [a, b]  b and f (x)dx = 0. Then f (x) = 0 for all x ∈ [a, b]. a

Proof Suppose the conclusion is not true. Then there exists x0 ∈ [a, b] such that f (x0 ) = 0. Without loss of generality, assume that f (x0 ) > 0. Then, by continuity of f at x0 , given any α with 0 < α < 1, there exists a closed interval [c, d] ⊆ [a, b] containing x0 such that f (x) ≥ α f (x0 ) for all x ∈ [c, d]. Therefore, b 0=

d f (x)dx ≥

a

b f (x)dx ≥ α f (x0 )

c

dx = α f (x0 )(d − c) = 0, a

which is a contradiction. We close this section by a test for convergence of series using integrals.



3.1 Integrability and Integral

197

Theorem 3.1.20 (I NTEGRAL TEST) Suppose f (x) is a continuous, non-negative n and decreasing function for x ∈ [1, ∞). For each n ∈ N, let an := f (x)dx. Then 1 ∞ 

f (n) converges ⇐⇒ (an ) converges.

n=1

n  k Proof First we observe that an = 1 f (x) dx = nk=2 k−1 f (x) dx. Now, since f (x) is a decreasing function for x ∈ [1, ∞), we have for each k ∈ N, k − 1 ≤ x ≤ k ⇒ f (k) ≤ f (x) ≤ f (k − 1). Hence, for k = 2, 3, . . ., k f (x) dx ≤ f (k − 1)

f (k) ≤ k−1

so that n 

f (k) ≤

k=2

n  

n 

f (k) ≤

n 

f (k − 1).

k=2

n

k=2 n 

f (x) dx ≤

k=2 k−1

Thus,

Now, let sn :=

k

f (x) dx ≤

n−1 

f (k).

k=1

1

f (k) for n ∈ N. Then from the above inequalities, together with

k=1

the fact that (sn ) is a monotonically increasing sequence, it follows that (an ) converges  if and only if (sn ) converges.  n Remark 3.1.8 Suppose f : [a, ∞) → R is a continuous function. If lim f (x)dx n→∞ a  ∞ f (x)dx. Thus, by Theorem 3.1.20, if f : [1, ∞) → exists, then it is denoted by a

R is continuous and decreasing, then ∞

n=1

f (n) converges if and only if

∞ 1

f (x)dx exists.

More on this type of integrals will be discussed in the next chapter.

♦

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3 Definite Integral

3.2 Mean Value Theorems Theorem 3.2.1 (Mean value theorem) Suppose f is a continuous function on [a, b]. Then there exists c ∈ [a, b] such that 1 b−a



b

f (x) dx = f (c).

a

Proof Since f is continuous, we know that there exist u, v ∈ [a, b] such that f (u) = m := min f (x) and f (v) = M := max f (x). Hence, by Theorem 3.1.6, 1 f (u) ≤ b−a

b f (x) dx ≤ f (v). a

Hence, by intermediate value theorem, there exists c ∈ [a, b] such that 1 b−a

b f (x) dx = f (c). a



Hence the result.

Theorem 3.2.2 (Generalized mean value theorem) Suppose f and g are continuous on [a, b] with g(x) ≥ 0 for all x ∈ [a, b]. Then there exists c ∈ [a, b] such that b b f (x)g(x) dx = f (c) g(x)dx. a

a

Proof Let m := inf f (x) and M = sup f (x). Then, since g(x) ≥ 0 for every a≤x≤b

a≤x≤b

x ∈ [a, b], we have b

b g(x)dx ≤

m a

b f (x)g(x) dx ≤ M

a

g(x)dx. a

If g(x) = 0 for all x ∈ [a, b], then the conclusion in the theorem holds trivially for any c ∈ [a, b]. So, assume that g(x0 ) = 0 for some x0 ∈ [a, b]. Now, since g(x) ≥ 0 b for all x ∈ [a, b], it follows (how?) that g(x0 ) > 0 and hence a g(x)dx > 0. Hence,

3.2 Mean Value Theorems

199

b m≤

a

f (x)g(x) dx ≤ M. b a g(x)dx

Therefore, by the intermediate value theorem, there exists c ∈ [a, b] such that b f (c) =

Thus,

b a

f (x)g(x) dx = f (c)

b a

a

f (x)g(x) dx . b a g(x)dx 

g(x)dx.

Exercise 3.2.1 The conclusion of Theorem 3.2.2 holds if g(x) ≥ 0 is replaced by g(x) ≤ 0. How?

3.3 Fundamental Theorems The first theorem that we prove in this section justifies what we do in school for the calculation of integrals. Theorem 3.3.1 (Fundamental theorem-I) Let f be a Riemann integrable function on [a, b]. Suppose there exists a continuous function g on [a, b] such that it is differentiable in (a, b) and g  (x) = f (x) for all x ∈ (a, b). Then b f (x)dx = g(b) − g(a). a

Proof Let P : a = x0 < x1 < . . . < xn = b be any partition of [a, b]. Then by Lagrange’s mean value theorem, there exists ci ∈ (xi−1 , xi ) such that g(xi ) − g(xi−1 ) = g  (ci )(xi − xi−1 ) = f (ci )(xi − xi−1 ). Hence, g(b) − g(a) =

n n   [g(xi ) − g(xi−1 )] = f (ci )(xi − xi−1 ). i=1

i=1

Thus, we have L(P, f ) ≤ g(b) − g(a) ≤ U (P, f )

200

3 Definite Integral

for all partition P of [a, b]. Since

b

f (x)dx is the unique number which lies between

a

L(P, f ) and U (P, f ) for all partition P of [a, b], we obtain b f (x)dx = g(b) − g(a). a



This completes the proof. The concluding formula in Theorem 3.3.1, namely, 

b

g  (x)dx = g(b) − g(a)

a

is known as Newton–Leibnitz formula. The difference g(b) − g(a) is usually  b written as g(x) a , i.e.,  b g(x) a := g(b) − g(a). Definition 3.3.1 Let f : [a, b] → R be a Riemann integrable function. Then a function g : [a, b] → R is called an antiderivative or indefinite integral or primitive ♦ of f if g is differentiable in (a, b) and g  (x) = f (x) for all x ∈ (a, b). Difference of any two antiderivatives of a given function is a constant. Exercise 3.3.1 Justify the above statement.



Notation 3.3.1 Antiderivative of an integrable function f , if exists, is usually denoted by  f (x)dx. ♦ A function can be integrable, but it need not have an antiderivative. To see this, recall the function f in Example 3.1.1, that is,  f (x) =

−1, −1 ≤ x ≤ 0, 1, 0 < x ≤ 1.

We know that f is integrable on [−1, 1], but there does not exist a function g : [−1, 1] → R which is differentiable on (−1, 1) such that g  (x) = f (x) for all x ∈ (−1, 1) (see Example 2.3.15). However, we shall soon show that every continuous function has an antiderivative.

3.3 Fundamental Theorems

201

Remark 3.3.1 See carefully the statement of Theorem 3.3.1. It is different from the following statement: If g : [a, b] → R is such that it is differentiable on (a, b), then

b a

g  (x)dx = g(b) − g(a).

Because, there are functions which are derivatives of some functions, but they need not be integrable. For example, consider the function  g(x) =

sin(1/x), 0 < x ≤ 1, 0, x = 0.

Then g is differentiable at every point in (0, 1), and its derivative is g  (x) = −(1/x 2 ) cos(1/x), x ∈ (0, 1). 1 But, we cannot write 0 g  (x)dx = g(1) − g(0). The reason is that there is no inte♦ grable function f on [0, 1] such that f (x) = g  (x) for every x ∈ (0, 1). We shall use the concept of extension of a function. Definition 3.3.2 Let f be a function defined on an interval I , and let J be another interval such that I ⊆ J . Then a function ϕ : J → R is said to be an extension of f to J if ϕ(x) = f (x) for all x ∈ I . ♦ Indeed, every function f defined on an interval I can be extended to a bigger interval J ⊇ I by defining ϕ : J → R by  ϕ(x) =

f (x), x ∈ I, c, x∈J\I

for any specified c ∈ R. However, in applications we may require the extended function to have certain properties. Let us illustrate these by a few examples. Example 3.3.1 Let f : (0, 1] → R be defined by f (x) =

sin(x) , 0 < x ≤ 1. x

We know that f is continuous. Then, for c ∈ R, the extended function ϕ : [0, 1] → R defined by  f (x), 0 < x ≤ 1, ϕ(x) = c, x =0 is continuous if and only if c = 1. Example 3.3.2 Let f : (0, 1] → R be defined by

♦

202

3 Definite Integral

f (x) = sin(1/x), 0 < x ≤ 1. Then the function f˜ : [0, 1] → R defined by  ϕ(x) =

sin(1/x), 0 < x ≤ 1, 0, x = 0.

is an extension of f to [0, 1]. Note that f is continuous, whereas ϕ is not continuous. In fact, f does not have any continuous extension to all of [0, 1]. Also, f does not have any integrable extension to [0, 1] (Why?) ♦ Example 3.3.3 Consider the function f : [0, 1] → R defined by  f (x) =

x 2 sin(1/x), 0 < x ≤ 1, 0, x = 0.

It is clear that f  (x) exists for every x ∈ (0, 1) and for x = 0, f  (x) = − cos(1/x) + 2x sin(1/x). Since lim x→0 f  (x) does not exist, f  does not have a continuous extension to [0, 1]. However, since f  is continuous and | f  (x)| ≤ 3 on (0, 1), f  has integrable extension to [0, 1] (Exercise). ♦ Recall that a bounded function defined on [a, b] which is continuous except possibly at a finite number of points in [a, b] is integrable. Therefore, as a special case of Theorem 3.3.1 we have the following: Let f : [a, b] → R be differentiable in (a, b) and f  is continuous and bounded  b f  (x)dx = f (b) − f (a). on (a, b). Then a

The following examples have been worked out by knowing the antiderivatives of the functions involved. d k+1 x = (k + 1)x k so that by Theorem Example 3.3.4 Recall that for k ≥ 0, dx 3.3.1,  k+1 b b x bk+1 − a k+1 . ♦ x k dx = = k+1 a k+1 a

Example 3.3.5 Recall that for any c = 0,

d cx (e ) = cecx so that by Theorem 3.3.1, dx

3.3 Fundamental Theorems

203



b ecx dx =

ecx c

b = a

a

ecb − eca . c

♦

Example 3.3.6 Recall that for any c = 0, d d sin(cx) = c cos(cx) and cos(cx) = −c sin(cx), dx dx so that by Theorem 3.3.1,  b  sin cx b (i) cos cx dx = = a c a  b  cos cx b sin cx dx = − (ii) a c a

1 (sin cb − sin ca). c 1 = (cos ca − cos cb). c

♦

d 1 Example 3.3.7 Recall that log x = for x > 0. Hence, by Theorem 3.3.1, we dx x have x 1 dx = log x − log 1 = log x. ♦ x 1

x Remark 3.3.2 In some books, log x is defined as the integral 1 x1 dx, whereas we defined log x in the last chapter as the inverse of the exponential function. ♦ Theorem 3.1.16 together with Theorem 3.3.1 can be used to compute the limit of certain sequences. For example, see the following. Example 3.3.8 For p ≥ 0, we show that3 lim

n→∞

n 

1 n p+1

kp =

k=0

1 . p+1

We note that 1 n p+1

n  k=0

k = p

n  (k/n) p k=0

n

=

n 

(n) f (ξk(n) )(xk(n) − xk−1 ),

k=1

It is to be remarked that, for p ∈ N, this result was first found in the Sanskrit text Yuktibhasha of Jyeshthadeva (AD:1500-1575) of Kerala School of Mathematics, more than hundred years before it was considered by European mathematicians Fermat, Pascal and others in the seventeenth century (cf. [7], Chap. 10).

3

204

3 Definite Integral

where f (x) = x p , 0 ≤ x ≤ 1, 1 n p for k = 1, . . . , n. Thus, n p+1 xk(n) := nk for k=0, 1, . . . , n and ξk(n) := k−1 k=0 k n is a Riemann sum of the continuous function f corresponding to the partition Pn : 0 = x0(n) < x1(n) · · · < xn(n) = 1 with μ(Pn ) → 0 as n → ∞. Hence, by Theorem 3.1.16 and Example 3.3.4, n 

1 n p+1 Thus, lim

k=0

n

1

n→∞ n

k

p

k=0

p+1

n 

f (ξk(n) )(xk(n)

−

(n) xk−1 )

1 →

k=1

x p dx = 0

1 . p+1

k p = 1/( p + 1).

♦

Example 3.3.9 We show that lim

n→∞

n  k=0

n2 3 = . (n + k)3 8

Note that n  k=0

1  1 n2 (n) = = f (ξk(n) )(xk(n) − xk−1 ), 3 3 (n + k) n (1 + k/n) k=0 k=1 n

n

where f (x) =

1 , 0 ≤ x ≤ 1, (1 + x)3

 n2 for k = 1, . . . , n. Thus, nk=0 (n+k) xk(n) := nk for k = 0, 1, . . . , n and ξk(n) := k−1 3 n is a Riemann sum of the continuous function f corresponding to the partition Pn : 0 = x0(n) < x1(n) · · · < xn(n) = 1 with μ(Pn ) → 0 as n → ∞. Hence, by Theorem 3.1.16, n  k=0

We know that

d dx

 n2 (n) = f (ξk(n) )(xk(n) − xk−1 )→ (n + k)3 k=1 n



1 (1+x)2



1 0

=

−2 . (1+x)3

1 0

dx . (1 + x)3

Hence, by Theorem 3.3.1,

1  1 1 dx 3 = − = . 3 2 (1 + x) 2 (1 + x) 0 8

3.3 Fundamental Theorems

Thus, lim

n→∞

n  k=0

205

n2 = 3/8. (n + k)3

♦

Now, we address the question: Does every integrable function have an antiderivative? As stated earlier, we answer this question affirmatively when the function is continuous. Theorem 3.3.2 (Fundamental theorem-II) Suppose f is Riemann integrable on [a, b], and x g(x) = f (t)dt, x ∈ [a, b]. a

Then g is continuous on [a, b]. If, in addition, f is continuous on [a, b], then g differentiable and g  (x) = f (x) ∀x ∈ (a, b). Proof Let x ∈ [a, b] and h ∈ R be such that x + h ∈ [a, b]. Then we have x+h

x f (t)dt −

g(x + h) − g(x) = a

x+h f (t)dt =

a

f (t)dt. x

Thus, since | f | is integrable (Remark 3.1.7), we have x+h |g(x + h) − g(x)| ≤ | f (t)|dt ≤ M|h|, x

where M > 0 is such that | f (x)| ≤ M for all x ∈ [a, b]. Recall that such M exists, as Riemann integrability is defined only for bounded functions. Thus, |g(x + h) − g(x)| → 0 as h → 0, showing that g is continuous at x. Next assume that f is continuous on [a, b]. Then, by mean value theorem, there exists ξh between x and x + h such that 1 h

x+h f (t)dt = f (ξh ). x

206

3 Definite Integral

Since f is continuous at x, we have f (ξh ) → f (x) as h → 0. Hence lim

h→0

g(x + h) − g(x) = lim f (ξh ) = f (x). h→0 h

Thus, g  (x) exists and g  (x) = f (x).



In view of Theorem 3.3.2, Every continuous function has an antiderivative.

Proof (An alternative proof for Theorem 3.3.1 when f continuous) x Let ϕ be the indefinite integral of f , i.e., ϕ(x) = a f (t)dt, x ∈ [a, b]. Then    we have g (x) = f (x) = ϕ (x) for all x ∈ [a, b], i.e., g (x) − ϕ  (x) = 0 for all x ∈ [a, b]. Hence, g − ϕ is a constant function. Hence, in view of the Theorem 3.3.2, we have b f (t)dt. g(b) − g(a) = ϕ(b) − ϕ(a) = a



This completes the proof.

Theorem 3.3.2 leads to the following existence theorem in differential equation: Theorem 3.3.3 Given a continuous function f on [a, b], there exists a function g which is continuous on [a, b] and differentiable on (a, b) such that dg = f (x) ∀ x ∈ (a, b) and g(a) = 0. dx

3.4 Some Consequences In this section we derive some results as consequences of mean value theorems and fundamental theorems. Theorem 3.4.1 (Integral as Riemann sum) Suppose f is a continuous function on [a, b]. Then for every partition P of [a, b], there exists a set T of tags on P such that b f (x) dx.

S(P, f, T ) = a

3.4 Some Consequences

207

Proof Let P = {xi : i = 0, 1, . . . , k} be a partition of [a, b]. Since f is continuous, by the mean value theorem (Theorem 3.2.1), there exists ξi ∈ [xi−1 , xi ] such that xi f (x) dx = f (ξi )(xi − xi−1 ), i = 1, . . . , k. xi−1

Hence, taking T = {ξi : i = 1, . . . , k}, S(P, f, T ) =

k  

b

xi

k 

f (ξi )(xi − xi−1 ) =

i=1

i=1 x

f (x) dx =

i−1

f (x) dx. a



This completes the proof.

Next theorem is a consequence of Theorem 3.3.2 and the formula for the derivative of composition of two differentiable functions. Theorem 3.4.2 Let f : [a, b] → R be a continuous function. Let ϕ(x) F(x) :=

f (t)dt, x ∈ [α, β], a

where ϕ : [α, β] → R is a continuous function which is differentiable on (α, β) such that ϕ(x) ∈ (a, b) for all x ∈ (α, β). Then F  (x) = f (ϕ(x))ϕ  (x) for all x ∈ (α, β). Proof Note that, F(x) = g(ϕ(x)) for x ∈ [α, β], where

y f (t)dt,

g(y) =

y ∈ [a, b].

a

Since f is continuous on [a, b], by Theorem 3.3.2, g  (y) = f (y) for all y ∈ (a, b). Thus,

F  (x) = g  (ϕ(x))ϕ  (x) = f (ϕ(x))ϕ  (x) for all x ∈ (α, β).

This complete the proof.



208

3 Definite Integral

Example 3.4.1 Let us find an expression for the function 3x−2 



d dx

1 + t 2 dt.

0

Let ϕ(x) := 3x − 2 for x ∈ [α, β] and let [a, b] be such that 3x − 2 ∈ (a, b) for all x ∈ (α, β). Then, by Theorem 3.4.2, we have d dx

3x−2 



  1 + t 2 dt = [ 1 + (3x − 2)2 ]3 = 3 9x 2 − 12x + 5.

♦

0

We have seen in Theorem 3.1.9 that the product of two integrable functions is integrable. However, we did not give any formula for the computation of the integral of the product. Now, under some additional condition on one of the functions, we give a formula which would facilitate the computation of the integral of the product. Theorem 3.4.3 (Product formula) Suppose f and g are continuous functions defined on [a, b]. If f is differentiable in (a, b) and its derivative f  has integrable extension to [a, b], then b

b f (x)g(x) dx = [

f (x)G(x)]ab

−

a

f  (x)G(x) dx,

a

where G is an antiderivative of g on (a, b). Proof Since g is continuous on [a, b], it has an antiderivative, say G, i.e., G is differentiable and G  = g on (a, b). Then we have ( f G) = f g + f  G on (a, b). Hence, using fundamental theorem (Theorem 3.3.1), b [ f (x)G(x)]ab =

[ f (x)G(x)] dx

a

b =

b f (x)g(x) dx +

a

a

f  (x)G(x) dx.

3.4 Some Consequences

209

b

b f (x)g(x) dx = [

Thus,

f (x)G(x)]ab

−

a

f  (x)G(x) dx.



a

Another form of the product formula is the following. Theorem 3.4.4 (Product formula) Suppose f and g are continuous functions on [a, b], which are differentiable in (a, b). Also, assume that both f  and g  have Riemann integrable extensions to [a, b]. Then b

b



f (x)g (x) dx = [

f (x)g(x)]ab

−

a

f  (x)g(x) dx.

a

Proof Proof follows as in the proof of Theorem 3.4.4, by replacing g there by g  .  If we use new variables u and v which relate to x by u = f (x), v = g(x), a ≤ x ≤ b, then the product formula in Theorem 3.4.4 can be written, in short, as b



udv = uv

b a

b −

a

vdu. a

Thus, under appropriate conditions, denoting the antiderivative of g by G, we have the following formulae: 

b

 f (x)g(x) dx = [

a



b



f (x)G(x)]ab

f (x)g (x) dx = [

a

b

− 

f (x)g(x)]ab  a

b

f  (x)G(x) dx

a

−

b

f  (x)g(x) dx

a



udv = [uv]ab −

Example 3.4.2 Let us find the value of the integral

 π/2 0

b

vdu.

a

sinn xdx for n ∈ N. Clearly,

π/2 π/2 sin x dx = [− cos x]0 = −[0 − 1] = 1, 0

210

3 Definite Integral

and for n = 2, π/2 π/2 π/2 1 1 π 2 sin x dx = (1 − cos 2x) dx = dx = . 2 2 4 0

For n ≥ 3, we write mula, we obtain

0

 π/2 0

0

sinn x dx =

 π/2 0

sinn−1 x sin x dx, so that by product for-

π/2 π/2 sinn−1 x sin x dx = [sinn−1 x(− cos x)]0 0

π/2 − (n − 1) sinn−2 x cos x(− cos x) dx 0

π/2 =

(n − 1) sinn−2 x cos2 x dx 0

π/2 = (n − 1) sinn−2 x(1 − sin2 x) dx. 0

Hence, if we denote αn :=

 π/2 0

sinn x dx for n ∈ N, then we obtain

αn = (n − 1)αn−2 − (n − 1)αn , i.e., αn = In particular, α3 =

n−1 αn−2 . n

2 2 3 3π α1 = , α4 = α2 = . 3 3 4 44

More generally, for k ∈ N, α2k = α2k+1 =

2k − 1 π(2k)! π 1 3 . . .··· . = 2k+1 , 2 2 4 2k 2 (k!)2

2k 22k (k!)2 π 24 ··· = = . 35 2k + 1 (2k + 1)! 2(2k + 1)α2k

♦

3.4 Some Consequences

211

Example 3.4.3 Let us evaluate the integral

b a

e x sin xdx. By product formula,

b

b e sin xdx = [e x

x

(− cos x)]ab

+

a

e x cos xdx, a

b

b e cos xdx = [e x

x

sin x]ab

a

−

e x sin xdx. a

Hence, b e x sin xdx =

 1 x [e (− cos x)]ab + [e x sin x]ab 2

a

=

1 x [e (sin x − cos x)]ab . 2

In particular, π/2 1 1 π/2 e x sin xdx = [e x (sin x − cos x)]0 = [eπ/2 + 1]. 2 2

♦

0

Theorem 3.4.5 (Change of variable formula) Let f : [a, b] → R be a continuous function. Let ψ : [α, β] → R be a continuous function which is differentiable on (α, β) such that ψ(x) ∈ [a, b] for every x ∈ [α, β] with ψ(α) = a and ψ(β) = b. Then, b β f (x) dx = f (ψ(t))ψ  (t)dt. α

a

Proof Let F be an antiderivative of f , i.e., F  (x) = f (x). Then taking G(t) = F(ψ(t)) for t ∈ [α, β], we have G  (t) = F  (ψ(t))ψ  (t) = f (ψ(t))ψ  (t), t ∈ [α, β]. Hence, by fundamental theorem, β



β

f (ψ(t))ψ (t)dt = α

α

G  (t)dt = G(β) − G(α) = F(ψ(β)) − F(ψ(α)).

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3 Definite Integral

Hence,

β

b



f (ψ(t))ψ (t)dt = F(b) − F(a) = α

f (x) dx. a



This completes the proof. The formula in Theorem 3.4.5 can be written as: 

β

α

f (ψ(t))ψ  (t)dt =



ψ(β) ψ(α)

f (x) dx.

The formula in Theorem 3.4.5 can be used to find the integral of certain complicated functions by expressing it in the form β

f (ψ(t))ψ  (t)dt

α

by using appropriate functions f and ψ.  π/2 Example 3.4.4 Let us find 0 sin x cos2 dx. Writing sin x cos2 = f (ψ(x))ψ  (x), with ψ(x) = − cos x and f (y) = y 2 , Theorem 3.4.5 gives ψ(π/2) π/2 π/2  2  sin x cos xdx = f (ψ(x))ψ (x)dx = f (y)dy. 0

But,

ψ(0)

0

ψ(π/2) 

ψ(π/2) 

f (y)dy = ψ(0)

Thus,

 π/2 0

y 2 dy =

 y 3 0

ψ(0)

3

−1

=

1 . 3

sin x cos2 dx = 1/3.

The method of expressing the integral described as follows:

♦ β α

f (ψ(t))ψ  (t)dt as

b

dx = ψ  (t). dt 2. Write the above relation formally as dx = ψ  (t)dt.

1. Consider the change of variable x = ψ(t) so that

a

f (x)dx can be

3.4 Some Consequences

213



β

3. Replace the expression integral α





f (ψ(t))ψ (t)dt by

ψ(β)

ψ(α)

f (x)dx.

In view of the above remarks, let us look at another example. 1 dx Example 3.4.5 We would like to find the value of −1 (1+x 2 )2 . Write x = tan t so 2 that dx = sec tdt and dx sec2 tdt sec2 tdt 1 = = = cos2 tdt = (cos 2t + 1)dt. (1 + x 2 )2 (1 + tan2 t)2 (sec2 t)2 2 Since tan(π/4) = 1 and tan(−π/4) = −1, we obtain 1 −1

But,

π/4 −π/4

Thus,

dx = (1 + x 2 )2

sec2 tdt = (1 + tan2 t)2

1 −1

π/4 −π/4

sec2 tdt . (1 + tan2 t)2

π/4

π/4 cos tdt = 2

−π/4

−π/4

1 (cos 2t + 1)dt. 2

π/4 dx 1  sin 2t 1 π + t = = + . −π/4 (1 + x 2 )2 2 2 2 2

♦

Theorem 3.4.6 (Taylor’s formula - Cauchy integral form) Let I be an open interval, x0 ∈ I and f has n + 1 continuous derivatives in I . Then for every x ∈ I ,  n  f (k) (x0 ) 1 k (x − x0 ) + f (x) = f (n+1) (t)(x − t)n dt. k! n! k=0 x

x0

Proof Let x ∈ I and x = x0 . By Fundamental theorem-I (Theorem 3.3.1), we have x f (x) = f (x0 ) + x0

Hence, by integration by parts, we have

f  (t)dt.

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3 Definite Integral

x

x



f (t)dt = − x0

f  (t)

d (x − t)dt dt

x0





= − f (t)(x − t)

x x0

x +

f  (t)(x − t)dt

x0

= f  (x0 )(x − x0 ) +

x

f  (t)(x − t)dt.

x0

Thus, theorem holds for n = 1. Now, let m < n and assume that theorem holds for n = m − 1, i.e., f (x) =

m−1  k=0

f (k) (x0 ) 1 (x − x0 )k + k! (m − 1)!

x

f (m) (t)(x − t)m−1 dt.

x0

Then we have x f

(m)

(t)(x − t)

m−1

1 dt = − m

x0

x

f (m) (t)

d (x − t)m dt dt

x0

1 (m) f (x0 )(x − x0 )m = m x 1 + f (m+1) (t)(x − t)m dt. m x0

Thus, the theorem holds for n = m as well.



Exercise 3.4.1 Derive the Taylor’s formula in Theorem 2.3.23 from Theorem 3.4.6 by applying the generalized mean value theorem (Theorem 3.2.2).

3.5 Some Applications In this section we shall make use of the properties of integrals to derive formulae for the area of certain regions in the plane, arc length of curves, volume of certain type of solid domains in the three-dimensional space, area of certain surfaces in the space, and also to obtain formulae for certain physical quantities such as centre of gravity of a material line, centre of gravity of certain planar region, moment of inertia of a material line and moment of inertia of certain planar region. First let us define the notion of a curve in the plane.

3.5 Some Applications

215

Definition 3.5.1 By a curve in the plane we mean a function t → γ (t) := (ϕ(t), ψ(t)) defined on a closed and bounded interval [a, b] taking values in R2 , where ϕ : [a, b] → R and ψ : [a, b] → R are continuous functions. ♦ Given a curve t → γ (t) := (ϕ(t), ψ(t)) defined on [a, b], we may also say that the set C := {γ (t) : a ≤ t ≤ b} is a curve, and it is given by the parametric form x = ϕ(t),

y = ψ(t) for t ∈ [a, b].

Note that, as t varies from a to b, the point γ (t) moves from γ (a) to γ (b) along C. The point γ (a) is called the initial point of C and γ (b) is called the terminal point of C. We may observe that, if f : [a, b] → R is a continuous function, then it defines a curve, namely, t → (t, f (t)). We say that this curve is given by the equation y = f (x), x ∈ [a, b].

3.5.1 Computing Area Under the Graph of a Function Using Cartesian coordinates Motivated from the geometric interpretation of the Riemann integral of a positive function, we define area of certain regions in the plane. Definition 3.5.2 Suppose a curve is given by an equation y = f (x), a ≤ x ≤ b, where f : [a, b] → R is a continuous function with f (x) ≥ 0 for all x ∈ [a, b]. Then the area of the region bounded by the graph of f , namely, {(x, f (x)) : a ≤ x ≤ b}, the x-axis, and the ordinates at x = a and x = b is defined as the integral of f over [a, b], that is, b f (x) dx, a

which is also written as

b a

y dx.

♦

216

3 Definite Integral

Sometimes, the computation of the integral becomes easier if we can express the curve in parametric form. For example, suppose the curve is given in parametric form as x = ϕ(t), y = ψ(t), α ≤ t ≤ β, such that a = ϕ(α), b = ψ(β) and ϕ is differentiable on [α, β] with ϕ  continuous on [α, β]. Then the area under the curve takes the form β

ψ(t)ϕ  (t)dt.

α

If f takes both positive and negative values, but changes sign only at a finite number of points, then the area bounded by the curve, the x-axis, and the ordinates at x = a and x = b, is given by b | f (x)| dx. a

Generalizing Definition 3.5.2, we have the following definition. Definition 3.5.3 Suppose f : [a, b] → R and g : [a, b] → R are continuous functions such that f (x) ≤ g(x) for all x ∈ [a, b]. Then the area of the region bounded by the graphs of f and g, and the ordinates at x = a and x = b is given by b A :=

[g(x) − f (x)] dx.

♦

a

A slight variant of the Definition 3.5.2 is the following. Definition 3.5.4 Suppose f and g are continuous (real valued) functions defined on an interval I such that f (x) ≤ g(x) for all x ∈ I . Suppose that the curves y = f (x),

y = g(x) for x ∈ I

(∗)

intersect only at two points, say f (a) = g(a) and f (b) = g(b) with a < b for some a, b ∈ I . Then the area of the region bounded by the curves in (∗) is defined by the integral b [g(x) − f (x)] dx. ♦ a

3.5 Some Applications

217

Example 3.5.1 We find the area of the region bounded by the curves defined by y=

√

x,

y = x 2, x ≥ 0 :

Note the points of intersection of the curves are at x = 0 and x = 1. Also, √ that x ≥ x 2 for 0 ≤ x ≤ 1. Hence, the required area is 1

√

0

1  3/2  x3 x 1 − x − x 2 dx = = . 3/2 3 0 3

♦

Example 3.5.2 Let us find the area of the region bounded by the straight line y = x and the parabola y = ax 2 , a > 0. Note that the required region is in the first quadrant of the plane, and the limits of integration are obtained by finding the intersection of the curves y = x and y = ax 2 , that is by solving x = ax 2 . Thus, x = 0 and x = 1/a are the limits of integration. Thus, the required area is given by 1/a  x2 x 3 1/a 1 −a [x − ax 2 ]dx = = 2. 2 3 0 6a

♦

0

Example 3.5.3 We find the area of the region bounded by the ellipse x2 y2 + = 1. a2 b2 Since the region is symmetric with the coordinate axes, the required area is 4 times the area under the curve in the first quadrant. We shall compute this area in two different ways: (i) The equation of the curve in the first quadrant is given by  y=b

1−

x2 , 0 ≤ x ≤ a. a2

Hence, the required area is a A := 4 0

4b y dx = a

a  a 2 − x 2 dx. 0

By the change of variable x = a sin t, we obtain

218

3 Definite Integral

4b a

π/2 π/2 1 + cos 2t 2 2 dt = πab. a cos t dt = 4ab 2 0

0

(ii) Note that the ellipse can be represented in parametric form as x(t) = a cos t,

y(t) = b sin t, 0 ≤ t ≤ 2π.

Thus, using this parametrization, the required area is given by π/2 4

π/2 y(t)x (t)dt = 4 (b sin t)(−asint) dt 

0

0

π/2 = 2ab (1 − cos 2t) dt 0

= πab. Thus, the area of the region is πab.

♦

Example 3.5.4 We find the area bounded by one arch of the cycloid x = a(t − sin t),

y = a(1 − cos t).

One arch of the cycloid is obtained by varying t over the interval [0, 2π ]. Thus, the required area is 2πa

2π y dx =

0



2π

y(t)x (t) dt = 0

a 2 (1 − cos t)2 dt = 3πa 2 .

♦

0

Using polar coordinates We may recall that a point in the plane R2 , represented in Cartesian coordinates as (x, y) can also be represented in polar coordinates as (ρ, θ ) such that x = ρ cos θ,

y = ρ sin θ for ρ ≥ 0, 0 ≤ θ < 2π.

Suppose a curve is given in polar coordinates as ρ = ϕ(θ ), α ≤ θ ≤ β,

3.5 Some Applications

219

where ϕ : [α, β] → R is a continuous function. We would like to find an expression for the area of the region bounded by the graph of ϕ and the rays θ = α and θ = β. For this, first consider a partition of [α, β], say P : α = θ0 < θ1 < · · · < θk = β and a set {ξi } of tags on P. Then the area of the region bounded by the graph of ϕ and the rays θ = θi−1 and θ = θi would be close to the quantity 1 [ϕ(ξi )(θi − θi−1 )]ϕ(ξi ). 2 Thus, the required area must be close to the k  1 i=1

2

[ϕ(ξi )]2 (θi − θi−1 ).

Note that lim

μ(P)→0

k  1 i=1

1 [ϕ(ξi )] (θi − θi−1 ) = 2 2

β ρ 2 dθ.

2

α

In view of this we have the following definition. Definition 3.5.5 The area of the region bounded by the graph of ϕ and the rays θ = α and θ = β is defined as β 1 ρ 2 dθ. ♦ 2 α

As in the case of Cartesian coordinates, the above definition can be used to find the area of the certain regions in the plane using polar coordinates. Let us consider a few examples. Example 3.5.5 We find the area bounded by a circle of radius a. Without loss of generality, we may assume that the centre of the circle is the origin. Then, the circle can be represented in polar coordinates as ρ = a, 0 ≤ θ ≤ 2π. Hence the required area is 1 A := 2

2π ρ 2 dθ = πa 2 . 0

♦

220

3 Definite Integral

Example 3.5.6 We find the area bounded by the lemniscate √ ρ = a cos 2θ . Note that the lemniscate has two identical loops, and the curve traces one complete loop when θ varies from −π/4 to π/4. Thus, the required area is ⎡ ⎢1 2⎣ 2

π/4

⎤ ⎥ ρ dθ ⎦ = a 2

π/4

2

−π/4

cos 2θ dθ = a 2 .

♦

−π/4

3.5.2 Computing the Arc Length Suppose a curve C : γ (t) := (x(t), y(t)), a ≤ t ≤ b, is given. In order to compute the length of C, we first consider a polygonal approximation of the curve. What we mean by that is the following: Corresponding to a partition P : a = t0 < t1 < · · · < tk = b of [a, b], consider the length of the polygonal line obtained by joining the points γ (t0 ), γ (t1 ), . . . , γ (tk ), i.e., the quantity  P (C) :=

k   (x(ti ) − x(ti−1 ))2 + (y(ti ) − y(ti−1 ))2 . i=1

Using the above quantity, we define the length of C as follows: Definition 3.5.6 The length of the curve C is defined by (C) := sup  P (C), P

where the supremum is taken over all partitions P of [a, b].

♦

In order to compute the quantity (C), we assume that the curve C : γ (t) := (x(t), y(t)), t ∈ [a, b], is smooth in the sense that the functions t → x(t) and t → y(t) are differentiable in the open interval (a, b) and their derivatives t → x  (t) and t → y  (t) are continuous. Then, for a given a partition P : a = t0 < t1 < · · · < tk = b of [a, b], by mean value theorem, there exist ξi , ηi ∈ (ti−1 , ti ) such that x(ti ) − x(ti−1 ) = x  (ξi )(ti − ti−1 ),

3.5 Some Applications

221

y(ti ) − y(ti−1 ) = y  (ηi )(ti − ti−1 ) for i = 1, 2, . . . , k. Thus, the quantity  P (C) takes the form  P (C) =

k  

x  (ξi )2 + y  (ηi )2 (ti − ti−1 ).

i=1

By our assumptions on the functions x(t) and y(t), the function f (t) :=



x  (t)2 + y  (t)2 , a ≤ t ≤ b,

is integrable over [a, b]. Hence, if we take a sequence (Pn ) of partitions on [a, b], say =b Pn : a = t0(n) < t1(n) < · · · < tk(n) n such that μ(Pn ) → 0 as n → ∞, then S(Pn , f, Tn ) :=

kn " 

(n) x  (ci(n) )2 + y  (ci(n) )2 (ti(n) − ti−1 )

i=1

→

b 

x  (t)2 + y  (t)2 dt

a kn as n → ∞, where Tn := {ci(n) }i=1 is any set of tags on Pn . Corresponding to the partition Pn , we have

 Pn (C) =

kn " 

(n) x  (ξi(n) )2 + y  (ηi(n) )2 (ti(n) − ti−1 )

i=1 (n) for some ξi(n) , ηi(n) ∈ (ti−1 , ti(n) ) for i = 1, . . . , kn . Therefore, by the continuity of the   functions x (t) and y (t), we obtain (Exercise)

|S(Pn , f, Tn ) −  Pn (C)| → 0 as n → ∞. Hence, we can conclude that (C) = sup  P (C) = lim  Pn (C) = lim S(Pn , f, Tn ). P

n→∞

n→∞

222

3 Definite Integral

Thus,  (C) = a

b



dx 2  dy 2 + dt. dt dt

Suppose s(τ ) is the length of the arc which is part of the curve C from the point γ (a) to the point γ (τ ). Then from the above formula, we have s(τ ) =

τ 

x  (t)2 + y  (t)2 dt, a < τ ≤ b.

a

Example 3.5.7 Let us compute the length of the arc of the circle x = r cos θ,

y = r sin θ, 0 ≤ θ ≤ 2π

when θ varies from θ = α to θ = β. Using the formula given above, the required length is

(C) =

β 

x  (θ )2 + y  (θ )2 dθ

α

β  r 2 sin2 θ + r 2 cos2 θ dθ = α

= r (β − α). In particular, the circumference of the circle is r (2π − 0) = 2πr . Example 3.5.8 Let us find the length of the ellipse x = a cos θ,

y = b sin θ, 0 ≤ θ ≤ 2π,

where a > b. The required length is

♦

3.5 Some Applications

223

# π/2  (C) := 4

dx dθ

2

 +

dy dθ

2 dθ

0

π/2 = 4 a 2 sin2 θ + b2 cos2 θ dθ 0

π/2 = 4 a 2 (1 − cos2 θ ) + b2 cos2 θ dθ 0

π/2 = 4 a 2 − (a 2 − b2 ) cos2 θ dθ. 0

Thus, (C) = 4a

π/2

1−

β2

cos2

√ a 2 − b2 θ dθ, β := . a

0

The above integral cannot be computed using standard methods, unless β = 0, i.e., b = a in which case the ellipse is the circle. But, the integral can be approximately computed numerically. ♦ Example 3.5.9 We find the length of the astroid: x = a cos3 t,

y = a sin3 t.

We observe that the astroid consists of four loops, each of length π/2

x  (t)2

+

y  (t)2

dt =

0

π/2

9a 2 cos4 t sin2 t + 9a 2 sin4 t cos2 t dt

0

π/2 = 3a cos2 t sin2 t dt 0

3a . = 2 Hence, the length of the astroid is 6a. Using Cartesian coordinates If the curve C is given by an equation

♦

224

3 Definite Integral

y = f (x), a ≤ x ≤ b, where f is a continuous function on [a, b], then we may write C : γ (t) := (t, f (t)), a ≤ t ≤ b. In this case, the length of the curve C is given by

(C) =

b 

[1 + f  (t)2 ]dt,

a

i.e.,  (C) =

b

 1+

 dy 2 dx

a

dx.

Example 3.5.10 Let us find the circumference of a circle of radius r using the above formula in Cartesian coordinates: Without loss of generality assume that the centre of the circle is the origin, i.e., the circle is given by x 2 + y 2 = r 2 . The required length is #  2 r  dy 1+ dx, y = r 2 − x 2 . (C) := 4 dx 0

Thus,

r (C) := 4r 0

dx = 2πr. √ 2 r − x2

♦

Remark 3.5.1 Curves in parametric form can be assumed to be piecewise smooth, i.e., having unique tangents except possibly at a finite number of points. Note that if a curve is given in parametric form as x = x(t),

y = y(t), α ≤ t ≤ β,

then it has unique tangent at (x(t0 ), y(t0 )), if x  (t0 ), y  (t0 ) exist and |x  (t0 )|2 + ♦ |y  (t0 )|2 = 0. Using polar coordinates Suppose a curve is given in polar coordinates as

3.5 Some Applications

225

ρ = ϕ(θ ), α ≤ θ ≤ β, where ϕ : [α, β] → R is a continuous function. Since x = ρ cos θ, we have

y = ρ sin θ, α ≤ θ ≤ β,

β

#

(C) = α

Note that

dx dθ

2

dx = ρ  cos θ + ρ(− sin θ ), dθ

 +

dy dθ

2 dθ.

dy = ρ  sin θ + ρ cos θ. dθ

Hence, the length of the curve C is given by  (C) =

β α

#



ρ2 +

dρ dθ

2 dθ.

Example 3.5.11 We find the length of the cardioid ρ = a(1 + cos θ ). The required length (C) is given by 2π  ρ 2 + ρ 2 dθ. (C) = 0

Since

ρ 2 = a 2 (1 + cos θ )2 , ρ 2 = a 2 sin2 θ,

we have (C) =

√

2π 2a

√

2π

θ



1 + cos θ dθ = 2a cos dθ = 8a. 2

0

♦

0

3.5.3 Computing Volume of a Solid Suppose that a three-dimensional object, a solid, lies between two parallel planes x = a and x = b with a < b. Let α(x) be the area of the cross section of the solid

226

3 Definite Integral

at the point x, with cross section being parallel to the yz-plane. We assume that the function α(x), x ∈ [a, b], is continuous. = b of the interval [a, b]. Let Consider a partition P : a = x0 < x1 < . . . < xk  k ξi ∈ [xi−1 , xi ] for i = 1, . . . , k. Then the quantity i=1 α(ξi )(xi − xi−1 ) can be thought of as the volume of a slice of the solid with width xi . Note that

lim

μ(P)→0

k 

b α(ξi )(xi − xi−1 ) =

i=1

α(x) dx. a

In view of this relation we have the following definition. Definition 3.5.7 The volume of the solid with cross-sectional area α(x) at the point x with x varies over the interval [a, b] is defined by the integral b V :=

α(x) dx.

♦

a

Example 3.5.12 Let us compute the volume of the solid enclosed by the ellipsoid x2 y2 z2 + + = 1. a2 b2 c2 For a fixed x ∈ [−a, a], the boundary of the cross section at x is given by the equation z2 x2 y2 + 2 = 1− 2. 2 b c a This equation can be written as y2 z2 + = 1, φ(x)2 ψ(x)2 where



 x2 x2 φ(x) = b 1 − 2 , ψ(x) = c 1 − 2 , a a

so that it represents an ellipse. Hence, the cross-sectional area α(x) at x is given by  x2  α(x) = π φ(x)ψ(x) = π bc 1 − 2 . a Therefore, the required volume V is given by

3.5 Some Applications

227

a V = −a

a  x2  4 1 − 2 dx = πabc. α(x) dx = π bc a 3 −a

In particular, the volume of the solid bounded by the sphere of radius a is 43 πa 3 . ♦

3.5.4 Computing the Volume of Solid of Revolution Suppose a solid is obtained by revolving a curve y = f (x), x ∈ [a, b] with x-axis as the axis of revolution. We would like to find the volume of the solid. In this case the area of the cross section at x is given by α(x) = π y 2 = π [ f (x)]2 , a ≤ x ≤ b. Thus, in view of Definition 3.5.7, we have the following definition. Definition 3.5.8 The volume of the solid of revolution obtained by revolving the curve y = f (x), a ≤ x ≤ b about the x-axis is given by b V := π

[ f (x)]2 dx.

♦

a

Analogously, we have the following definition. Definition 3.5.9 The volume of the solid of revolution obtained by revolving the curve x = g(y), c ≤ y ≤ d, about the y-axis is given by d V := π

[g(y)]2 dy.

♦

c

What about volume of the solid of revolution of a curve with a specific straight line, say L, as the axis of revolution?

This can also be found, provided the curve is such that, the perpendicular at each point on the line intersects the curve at most at one point.

228

3 Definite Integral

Let ϕ(x, y) be the perpendicular distance from a point (x, y) on the curve C to the line L. Then, at the point (x, y), the cross-sectional area of the solid of revolution of C about L is given by π [ϕ(x, y)]2 . Now, suppose that the curve C and the line L are given in parametric form as C : x = x(t), y = y(t) L : x = 1 (t), y = 2 (t) for t ∈ [a, b]. Note that 1 (t) and 2 (t) are linear in t, and an elementary segment on L is of the form " 1 (t)2 + 2 (t)2 t. Thus, we have the following definition. Definition 3.5.10 If the curve C and the line L have parametric representations C : x = x(t),

y = y(t),

L : x = 1 (t),

y = 2 (t)

for t ∈ [a, b]. Then the volume of the solid of revolution of C about the line L is given by the integral b V := π

" [ϕ(x(t), y(t))]2 1 (t)2 + 2 (t)2 dt,

a

where ϕ(α, β) is the perpendicular distance from the point (α, β) on the curve C to the line L. ♦ From the above formula in the above definition, we can deduce the expressions in Definition 3.5.8 and Definition 3.5.9 as special cases: (i) Suppose C is given by the equation y = f (x), a ≤ x ≤ b, and L is the x-axis. Then we may take C : x = t,

y = f (t) and L : x = t,

y=0

for t ∈ [a, b]. Then ϕ(x(t), y(t)) = ϕ(t, f (t)) = | f (t)|, 1 (t)2 + 2 (t)2 = 1 so that

3.5 Some Applications

229

b V := π

[ϕ(x(t), y(t))]

2

"

1 (t)2

+

2 (t)2 dt

b =π

a

[ f (t)]2 dt. a

(ii) Suppose C is given by the equation x = g(y), c ≤ y ≤ d, and L is the y-axis. Then we may take C : x = g(t),

y = t and L : x = 0,

y=t

for t ∈ [c, d]. Then ϕ(x(t), y(t)) = ϕ(g(t), t) = |g(t)|, 1 (t)2 + 2 (t)2 = 1 so that b V := π

d " [ϕ(x(t), y(t))]2 1 (t)2 + 2 (t)2 dt = π [g(t)]2 dt.

a

c

Example 3.5.13 Let us compute the volume of the solid of revolution of the curve y = x 2 about x-axis for −a ≤ x ≤ a. The required volume is a V := π

a y dx = π

x 4 dx =

2

−a

−a

2 5 a . 5

♦

Example 3.5.14 We compute the volume of the solid of revolution of the catenary y=

 a  x/a e + e−x/a 2

about x-axis for 0 ≤ x ≤ b. Hence, b

b y dx = 2

0

2 a 2  x/a e + e−x/a dx 4

0

a2 = 4

b

 2x/a  e + e−2x/a + 2

0

 πa 2 b a  2b/a e . = − e−2b/a + 8 2 3

230

3 Definite Integral

The required volume is V =

 πa 2 b πa 3  2b/a e − e−2b/a + . 8 2

♦

Example 3.5.15 Let us compute the volume of the solid of revolution of line segment C : 0 ≤ x ≤ 1 about the line L : y = x. Consider the parametrization of C and L as L : 1 (t) =

C : x(t) = t, y(t) = 0;

t t , 2 (t) = , 0 ≤ t ≤ 1. 2 2

The perpendicular distance ϕ(x, y) from the point (x, y) on the curve to the line L is given by x ϕ(x, y) = √ . 2 Thus, 1 [ϕ(x(t), y(t))]

π

2

"

1 (t)2

+

2 (t)2 dt

1 =π

0

0

t2 2



π 1 1 + dt = √ . 4 4 6 2

√ Thus, the volume is π/6 2.

♦

3.5.5 Computing the Area of Surface of Revolution Suppose a solid is obtained by revolving a curve C : y = f (x), a ≤ x ≤ b, with x-axis as axis of revolution. We would like to find the area of the surface of the solid. We assume that f (x) ≥ 0 for every x ∈ [a, b]. Let us consider a partition P : a = x0 < x1 < . . . < xk = b of the interval [a, b]. For each i = 1, . . . , k, let ξi ∈ [xi−1 , xi ] and let si be the length of the portion of the curve C when x varies over [xi−1 , xi ]. Then the quantity 2π f (ξi )si would be approximately equal to the lateral surface area of the cylindrical piece with base as the cross section at ξi and height si . Hence the area of surface of revolution of the curve C is approximately equal to k  i=1

2π f (ξi )si .

3.5 Some Applications

Since si ≈



231

1 + [ f  (ξi )]2 xi , where xi := xi − xi−1 for i = 1, . . . , n, we have k 

2π f (ξi )si ≈

i=1

k 

 2π f (ξi ) 1 + [ f  (ξi )]2 xi .

i=1

Note that b   dy 2  2π f (ξi ) 1 + [ f  (ξi )]2 xi = 2π y 1 + dx. μ(P)→0 dx i=1 lim

k 

a

Here, y is the perpendicular distance of the point (x, y) on the curve from the x-axis, and the expression   dy 2 1+ dx dx corresponds to the elementary arc length of the curve. In view of the above observation, we have the following definition. Definition 3.5.11 The area of the surface of revolution S of the curve C : y = f (x), a ≤ x ≤ b with x-axis as the axis of revolution is defined as b S := 2π

# y 1+



dy dx

2 dx.

♦

a

Suppose we are interested in finding the area of the surface of revolution of a curve C with a specific straight line L as the axis of revolution. Let C : γ (t) := (x(t), y(t)), a ≤ t ≤ b, be the parametric representation of C. In this case, the expression corresponding to the elementary arc length of the curve is given by 

x  (t)2 + y  (t)2 dt.

Let p(t) be the perpendicular distance from a point (x(t), y(t)) on the curve C to the line L. Then, we obtain the required area as

232

3 Definite Integral

 S = 2π

b

 p(t) x  (t)2 + y  (t)2 dt.

a

The following are the two special cases: (i) Suppose C is given by the equation y = f (x), a ≤ x ≤ b, and L is the x-axis. Then, C : γ (x) := (x, f (x)), a ≤ x ≤ b, is a parametrization of C and f (x) is the perpendicular distance from the point (x, f (x)) on the curve C to the x-axis. Thus, we obtain b S = 2π

 f (x) 1 + f  (x)2 dx.

a

Note that this is the same as in Definition 3.5.11, as it should be. (ii) Suppose C is given by the equation x = g(y), c ≤ y ≤ d, and L is the y-axis. In this case, we can consider the parametrization of the curve as C : γ (y) := (g(y), y), c ≤ y ≤ d. Then the perpendicular distance from the point (g(y), y) on the curve C to the y-axis is g(y), so that d  S = 2π g(y) 1 + g  (y)2 dy. c

Example 3.5.16 We find the area of the surface of revolution of the parabola y 2 = 2 px, 0 ≤ x ≤ a for p > 0 with x-axis as the axis of revolution. The required area S is given by the formula a   dy 2 S = 2π y 1 + dx, dx 0

3.5 Some Applications

where y =

√

233

2 px. Thus,  a  p S = 2π dx 2 px 1 + 2x 0

= 2π

√

p

a 

p + 2x dx

0

 a √ 2 3/2 1 = 2π p (2x + p) 3 2 0 √  2π p  (2a + p)3/2 − p 3/2 . = 3

♦

3.5.6 Centre of Gravity In this subsection, we define the centre of gravity of a material line and a material planar region enclosed by certain curves. First let us consider material particles on the plane at points (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ) with masses m 1 , m 2 , . . . m n , respectively. Then the centre of gravity of the system of these particles is at the point (x0 , y0 ), where n xi m i , x0 := i=1 n i=1 m i

n yi m i y0 := i=1 . n i=1 m i

Centre of gravity of a material line in the plane We would like to find a formula for the centre of gravity of a material line L in the plane. If M(X, r ) is the mass of an arc of the line containing the point X with length r , then the point density of this line at the point X is defined by γ (X ) := lim

r →0

M(X, r ) . r

Now, suppose L is given by the equation y = f (x), x ∈ [a, b].

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3 Definite Integral

In order to find the centre of gravity of L, we first consider a partition P : a = x0 < x1 < . . . < xk of [a, b], and points X i := (ξi , f (ξi )) for i = 1, . . . , n, on L, where ξi ∈ [xi−1 , xi ], i = 1, . . . , n. Then the centre of gravity of the system of material points X 1 , . . . , X n is the point (x0 (P), y0 (P)), where n ξi m i , x0 (P) = i=1 n i=1 m i

n f (ξi )m i n y0 (P) := i=1 , i=1 m i

(∗)

where, m i is the mass of the material point X i , that is, m i = γi si with γi as the point density at the point X i and si is the length of the arcs joining (xi−1 , yi−1 ) to (xi , yi ). Here yi = f (xi ) for i = 1, . . . , n. Note that m i = γi si is the approximate mass of the arc joining (xi−1 , yi−1 ) to (xi , yi ). We may consider the centre of gravity of L is at (x0 , y0 ), where x0 := lim x0 (P), μ(P)→0

y0 := lim y0 (P). μ(P)→0

Now, suppose that the function x → γ (X ) := γ (x, f (x)) is continuous on [a, b]. Then, in view of the formula for x0 (P) and y0 (P) in (∗), we have   2 b dy n xγ (x, y) 1 + dx a dx i=1 ξi m i = lim x0 (P) = lim n ,   2 μ(P)→0 μ(P)→0 b i=1 m i dy dx a γ (x, y) 1 + dx 

 2 dy n yγ (x, y) 1 + dx a dx i=1 f (ξi )m i n lim y0 (P) = lim = ,   2 μ(P)→0 μ(P)→0 b i=1 m i dy dx a γ (x, y) 1 + dx b

Definition 3.5.12 Suppose that the material line L is given by the equation y = f (x), x ∈ [a, b], and for each point X := (x, f (x)) on L, γ (x, f (x)) is the point density such that the function x → γ (X ) := γ (x, f (x)) is continuous on [a, b]. Then the mass of L is defined by b

 γ (x, y) 1 +

M := a

 dy 2 dx

dx

3.5 Some Applications

235

and the centre of gravity of the line L is defined to be at the point (x0 , y0 ), where 1 x0 = M



b

xγ (x, y) 1 +

 dy 2 dx

dx,

a

1 x0 = M



b

yγ (x, y) 1 +

 dy 2 dx

dx.

♦

a

Example 3.5.17 We find the centre of gravity of the semi-circular arc x 2 + y 2 = a 2 , y ≥ 0, assuming that the density of the material is constant. In this case, y = f (x) := so that it follows that

#

 1+

dy dx

2

 a2 − x 2,

a =√ . 2 a − x2

Since γ (x, y) is constant,

xC = 0



 2 dy y 1 + dx −a dx 2a yC = = .    2 π a dy 1 + dx dx −a a

♦

Centre of gravity of a material planar region Next we consider the centre of gravity of a material planar region  bounded by two curves y = f (x),

y = g(x) with f (x) ≤ g(x) a ≤ x ≤ b.

Suppose that the density of the material at the point X is γ (X ). This density is defined as follows: Suppose M(X, r ) is the mass of the circular region in  with centre at X and radius r > 0, and α(X, r ) is the area of the same circular region. Then the density of the material at the point x is defined by γ (X ) := lim

r →0

M(X, r ) . α(X, r )

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3 Definite Integral

Now, in order to find the centre of gravity of , first consider a partition P : a = x0 < x1 < · · · < xn = b of of the interval [a, b]. Let ξi be the mid-point of the interval [xi−1 , xi ], that is, ξi = xi−12+xi for i = 1, . . . , n. Consider the rectangular boxes Ri : xi−1 ≤ x ≤ xi , f (ξi ) ≤ y ≤ g(ξi ) for i = 1, . . . , n. Note that, for each i ∈ {1, . . . , n},  f (ξi ) + g(ξi )  , X i = ξi , 2 is the mid-point of the rectangular box Ri . Of course, if f (ξi ) = g(ξi ) for some i ∈ {1, . . . , n}, then the corresponding Ri is a line segment. Assuming that the mass of the rectangular box Ri is concentrated at its mid-point, we can assume that this mass is approximately equal to m i = γ (X i )[g(ξi ) − f (ξi )]xi , where γ (X i ) is the density of the material at X i and xi = xi − xi−1 for i = 1, . . . , n. Now, the centre of gravity of the system of material points at X 1 , X 2 , . . . , X n is x0,P

n ξi m i := i=1 , n i=1 m i

n y0,P :=

f (ξi )+g(ξi ) mi 2

n

i=1

i=1

mi

.

Then, the centre of gravity of  may be defined as x0 = lim x0,P , μ(P)→0

Denoting

y0 = lim y0,P . μ(P)→0

 f (ξi ) + g(ξi )  as γ˜ (ξi ), γ (X i ) = γ ξi , 2

we have n ξi γ˜ (ξi )[g(ξi ) − f (ξi )]xi , x0,P = i=1 n i=1 γ˜ (ξi )[g(ξi ) − f (ξi )]x i n γ˜ (ξi )[ f (ξi ) + g(ξi )][g(ξi ) − f (ξi )]xi 1 i=1 n y0,P = 2 i=1 γ˜ (ξi )[g(ξi ) − f (ξi )]x i n 1 i=1 γ˜ (ξi )[(g(ξi ))2 − ( f (ξi ))2 ]xi n = . 2 i=1 γ˜ (ξi )[g(ξi ) − f (ξi )]x i

3.5 Some Applications

237

Thus, b

x γ˜ (x)[g(x) − f (x)] dx x0 = lim x0,P = a b , μ(P)→0 a γ˜ (x)[g(x) − f (x)] dx b 1 a γ˜ (x)[(g(x))2 − ( f (x)2 ] dx y0 = lim y0,P = . b μ(P)→0 2 γ˜ (x)[g(x) − f (x)] dx a

b

x γ˜ (x)[g(x) − f (x)] dx , x0 = a b a γ˜ (x)[g(x) − f (x)] dx b 1 a γ˜ (x)[(g(x))2 − ( f (x))2 ] dx y0 = . b 2 γ˜ (x)[g(x) − f (x)] dx a

Example 3.5.18 We find the coordinates of the centre of gravity of the planar region bounded by the parabola y 2 = a x cut off by the straight line x = a, assuming that the density to constant. In this case √ √ f (x) = − a x, g(x) = a x, 0 ≤ x ≤ a. Hence the coordinates of the centre of gravity are b b √ 2 0 x a x dx 3 0 x[g(x) − f (x)] dx x0 =  b = b √ = a. 5 0 [g(x) − f (x)] dx 0 2 a x dx 1 y0 = 2

b 0

[ f (x) + g(x)][g(x) − f (x)] dx = 0. b 0 [g(x) − f (x)] dx

♦

Moment of Inertia Suppose there are n material points in the plane. Let their masses be m 1 , m 2 , . . . m n respectively. Suppose that these points are at distances d1 , . . . , dn from a fixed point O. Then the moment of inertia (M.I) of the system of these points with respect to the point O is defined by the quantity: I O :=

n  i=1

di2 m i .

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3 Definite Integral

If O is the origin, and (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ) are the points, then n  (xi2 + yi2 )m i .

I O :=

i=1

M.I. of a material line in the plane Suppose a curve L is given by the equation y = f (x), a ≤ x ≤ b. We assume that this curve is a material line. Suppose the density of the material at the point X = (x, y) is γ (X ). Now, in order to find the moment of inertia of L, we first consider a partition P : a = x0 < x1 < . . . < xk = b, and take points ξi = [xi−1 , xi ], i = 1, . . . , n. Then we consider the moment of inertia of the system of material points at (ξ1 , ηi ), i = 1, . . . , n. Here, ηi = f (ξi ), i = 1, . . . , n. I O,P :=

n  (ξi2 + ηi2 )m i . i=1

Thus, I O,P :=

n  (ξi2 + ηi2 )γi si . i=1

Here, si is the length of the arcs joining (xi−1 , yi−1 ) to (xi , yi ), and γi is the density at the point (ξi , ηi ). Note that γi si is the approximate mass of the arc joining (xi−1 , f (xi−1 )) to (xi , f (xi )). Now, assuming that the functions f (x) and γ˜ (x) := γ (x, f (x)) are continuous on [a, b], the moment of inertia of L with respect to O is n  I O = lim I O,P = lim (ξi2 + ηi2 )γi si . μ(P)→0

μ(P)→0

i=1

Thus,  IO =

b

 (x + y )γ (x, y) 1 + 2

a

2

 dy 2 dx

dx.

M.I. of a circular arc with respect to the centre Suppose the given curve is a circular arc: ρ = a, α ≤ θ ≤ β. Following the arguments in the above paragraph, we compute the moment of inertia using polar coordinates:

3.5 Some Applications

239

The moment of inertia, in this, case is given by I O := lim

μ(P)→0

n 

di2 m i ,

i=1

where di = a, m i = γi aθi , for i = 1, . . . , n, with γ (θ ) being the point density. Hence, n  I O = lim a 2 γi [aθi ]. μ(P)→0

i=1

Thus,  IO = a

β

3 α

γ (θ )dθ.

If γ (θ ) = γ , a constant, then β IO = a

γ (θ )dθ = (β − α)γ a 3 .

3 α

In particular, M.I of the circle ρ = a, 0 ≤ θ ≤ 2π , is I O = 2π γ a 3 . M.I. of inertia of a material sector in the plane The region is R : 0 ≤ ρ ≤ a, α ≤ θ ≤ β with constant density γ . To find the M.I. of R, we partition it by rays and circular arcs: P : α = θ0 < θ1 < θ2 < . . . < θn = β, Q : 0 = ρ0 < ρ1 < ρ2 < . . . < ρm = a. Consider the elementary region obtained by the above partition: Ri j : ρ j−1 ≤ ρ ≤ ρ j , θi−1 ≤ θ ≤ θi .

240

3 Definite Integral

Assume that the mass of this region Ri j is concentrated at the point (ρˆ j , θˆi ), where ρˆ j ∈ [ρ j−1 , ρ j ], θˆi ∈ [θi−1 , θi ]. Then the M.I. of the material point at (ρˆ j , θˆi ) is m i j di2j where m i j is the mass of the region Ri j which is approximately equal to [ρˆ j θi ρ j ]γ , and di j = ρˆ j . Thus the M.I. of the subsector θi−1 ≤ θ ≤ θi is defined by lim

μ(Q)→0

n  j=1

m i j di2j = lim

μ(Q)→0

= lim

μ(Q)→0

n  [ρˆ j θi ρ j ]γ ρˆ 2j j=1 n   3  γ ρˆ j ρ j θi j=1

⎛ a ⎞  = γ ⎝ ρ 3 dρ ⎠ θi 0

γ a4 θi . = 4 From this, it follows that, the moment of inertia of the sector α ≤ θ ≤ β is given by M.I =

(β − α)γ a 4 . 4

In particular, moment of inertia of a circular disc is Ma 2 π γ a4 = , 2 2 where M = πa 2 γ is the mass of the disc. Exercise 3.5.1 If M is the mass of a right circular homogeneous cylinder with base 2 radius a, then show that its moment of inertia is Ma . 2

3.6 Appendix Theorem 3.6.1 A bounded function f : [a, b] → R is Riemann integrable if and only if there exists γ such that for every ε > 0, there exists a partition P of [a, b] satisfying |S(P, f, T ) − γ | < ε for every tag T on P, and in that case γ =

b a

f (x)dx.

3.6 Appendix

241

Proof Let ε > 0 be given. Suppose f is Riemann integrable. Then, by Theorem 3.1.3, there exists a partition P of [a, b] such that U (P, f ) − L(P, f ) < ε. Since L(P, f ) ≤ S(P, f, T ) ≤ U (P, f ) for every tag T on P, we obtain |S(P, f, T ) − γ | < ε for every tag T on P. Conversely, suppose there exists γ such that for every ε > 0, there exists a partition P of [a, b] satisfying |S(P, f, T ) − γ | < ε for every tag T on P. So, assume that such a γ ∈ R exists. For a given ε > 0, let P be a partition of [a, b] such that γ − ε < S(P, f, T ) < γ + ε

(1)

holds. Let  : i = 1, . . . , k}, Tn = {ti,n

 Tn = {ti,n : i = 1, . . . , k}

be tags on P : a = x0 < x1 < . . . < xk = b such that  ) → mi , f (ti,n

 f (ti,n ) → Mi as n → ∞.

Then S(P, f, Tn ) → L(P, f ),

S(P, f, Tn ) → U (P, f ) as n → ∞.

(2)

Also, from (1), we have γ − ε < S(P, f, Tn ) < γ + ε,

γ − ε < S(P, f, Tn ) < γ + ε

∀ n ∈ N.

Now, taking limit and using (2), we have γ − ε < L(P, f ) < γ + ε,

γ − ε < U (P, f ) < γ + ε.

Therefore, U (P, f ) − L(P, f ) < ε. By Theorem 3.1.3, f is integrable.



As a consequence of the above theorem we have the following. Corollary 3.6.2 Suppose f : [a, b] → R is a bounded function. If (Pn ) is a sequence of partitions on [a, b] such that {S(Pn , f, Tn )} converges for every tag Tn on Pn for each n ∈ N, then f is Riemann integrable, and b f (x)dx = lim S(Pn , f, Tn ). n→∞

a

Knowing a sequence (Pn ) of partitions, how to assert the convergence of {S(Pn , f, Tn )} for every tag Tn of Pn for each n ∈ N?

242

3 Definite Integral

In this regard, analogous to Theorem 3.1.16, we have the following result (See [6]) for its proof). Theorem 3.6.3 Suppose f : [a, b] → R is a Riemann integrable (bounded) function. If (Pn ) is a sequence of partitions on [a, b] such that μ(Pn ) → 0 as n → ∞, then b S(Pn , f, Tn ) → f (x)dx as n → ∞ a

for every tag Tn on Pn for each n ∈ N. Now, we specify a large class of functions of practical importance which are Riemann integrable. Theorem 3.6.4 Suppose f : [a, b] → R is bounded and piecewise continuous, i.e., there are at most a finite number of points in [a, b] at which f is discontinuous. Then f is integrable. Proof Let ε > 0 be given. We have to show that there exists a partition P such that U (P, f ) − L(P, f ) < ε. Suppose that c ∈ (a, b) such that f is continuous on [a, c) and (c, b]. Let δ > 0 be such that c + δ < b and c − δ > a. Let f 1 , f 2 , f 3 be restrictions of f to the intervals [a, c − δ], [c + δ, b] and [c − δ, c + δ], respectively. Since f is continuous on [a, c − δ] and [c + δ, b], f is integrable on these intervals, so that there exist partitions P1 on [a, c − δ] and P2 on [c + δ, b] such that U (P1 , f 1 ) − L(P1 , f 1 )
0 is so small that 2(M − m)δ < ε/3, we have U (P3 , f 3 ) − L(P3 , f 3 )
0, there exists a partition P such that U (P, f ) − L(P, f ) < ε, so that f is integrable. The case of more than one (but finite number of) points of discontinuity, including discontinuity at the endpoints, can be handled analogously. 

3.7 Additional Exercises 1. Let f be a bounded function on [a, b]. Prove that, if there is a partition P of [a, b] such that L(P, f ) = U (P, f ), then f is a constant function. 2. If f and g are bounded functions on [a, b] such that f ≤ g, then prove that b b a f (x)dx ≤ a g(x)dx. 3. Let f : [a, b] → R be a bounded function. If P and Q are partitions of [a, b] such that Q is a refinement of P, then show that L(P, f ) ≤ L(Q, f ) ≤ U (Q, f ) ≤ U (P, f ) for all n ∈ N. 4. For a bounded function f : I → R, let m f := inf x∈I f (x) and M f := supx∈I f (x). Prove that, if f and g are bounded functions on I , then m f + m g ≤ m f +g ≤ M f +g ≤ M f + Mg . 5. Let f : [a, b] → R be a bounded function. Show that L(P, f ) + L(P, g) ≤ L(P, f + g) ≤ U (P, f + g) ≤ U (P, f ) + U (P, g) for any partition P of [a, b]. 6. Suppose f : [a, b] → R is Riemann integrable. Justify the following statements: (a) For every ε > 0, there exists a partition P of [a, b] such that b f (x)dx − L(P, f ) < ε.

0≤ a

(b) For every ε > 0, there exists a partition P of [a, b] such that b f (x)dx < ε.

0 ≤ U (P, f ) − a

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3 Definite Integral

(c) For every ε > 0, there exists a partition P of [a, b] such that





b



0 ≤

S(P, f, T ) − f (x)dx

< ε



a

for every tag T on P. 7. If f is either monotonically increasing or monotonically decreasing function on [a, b], then prove that f is integrable. [Hint: Use Corollary 3.1.4.] 8. Prove that every piecewise constant function on [a, b] is integrable. 9. Given a bounded function f and a partition P on [a, b], find piecewise constant functions g and h such that b L(P, f ) =

b g(x)dx, U (P, f ) =

a

h(x)dx. a

10. Prof Theorem 3.1.7 and Theorem 3.1.8 using Corollary 3.1.4. 11. Prove that every bounded piecewise continuous function on [a, b] is integrable. 12. Let f be a bounded function on [a, b] such that | f | is integrable. Is it necessary that f is integrable? 13. Let f be a bounded function on [a, b]. Prove the following: (a) If f is continuous on (a, b), then f is integrable on [a, b]. (b) If g is a bounded function on (a, b) such that g(x) = f (x) for all x ∈ [a, b], b b then g is integrable on [a, b] and a g(x)dx = a f (x)dx. (c) If h is a bounded function on [a, b] which is continuous  b (c, b] for  b on [a, c) and some c ∈ (a, b), then h is integrable on [a, b] and a h(x)dx = a f (x)dx. x 14. Let f be integrable on [a, b] and g(x) = a f (t)dt. Show the following: (a) If f (x) ≥ 0 for all x ∈ [a, b], then g is monotonically increasing. (b) If f (x) ≤ 0 for all x ∈ [a, b], then g is monotonically decreasing. 15. Suppose f is integrable on [a, b]. Prove the following.  b  t (a) lim f (x)dx = f (x)dx. t→b

a

a

(b) If (an ) and (bn ) are in [a, b] which converge to a and b, respectively, then b b limn→∞ ann f (x)dx = a f (x)dx. 16. Using Newton-Leibnitz formula, evaluate the following integrals.

3.7 Additional Exercises

245

√ 1/  2

(a)

√ 0 2

(c)

1

dx 1−

x2

,

(b)

√ 3x + 1dx,

0 2π/ω 

(x 3 + x + 1)dx, (d) 1

cos2 (ωt)dt, ω > 0. 0

17. Verify the conclusions in Theorem 3.3.2 for the following functions.  −1, −1 ≤ x ≤ 0, (a) f (x) = |x|, x ∈ [−1, 1], (b) f (x) =   1, 0 < x ≤ 1 x, 0 ≤ x ≤ 1, x, 0 ≤ x ≤ 1, (c) f (x) = , (d) f (x) = 1, 1 < x ≤ 2 x − 1, 1 < x ≤ 2 18. Antiderivative of a Riemann integrable function is Riemann integrable, and difference of any two antiderivatives of a given function is a constant -- Why? 19. Find the following limits by interpreting the sequence as a Riemann sum and then applying Newton-Leibnitz formula. 1

[sin(π/n) + sin(2π/n) + · · · + sin(nπ/n)] .  1 1 1 + + ··· + . (b) lim n→∞ n 1 3n  2 n+  1 22 (n − 1)2 (c) lim . + + · · · + n→∞ n 3 n3 n3 ) ( 1 1 1 + ··· +  (d) lim √ + √ . n→∞ n2 + n n2 n 2 + (n − 1)n [Hint: Theorem 3.1.16 together with Theorem 3.3.1] (a) lim

n→∞  n

20. Let g : [0, 1] → R be defined by  g(x) =

sin(1/x), 0 < x ≤ 1, 0, x = 0.

Show that g is differentiable at every point in (0, 1), but g  does not have a Riemann integrable extension to [0, 1].  x 2 sin(1/x), x = 0, 21. Let f : [0, 1] → R be defined by f (x) = Show that f 0, x = 0. is differentiable in (0, 1) and f  has Riemann integrable extension to [0, 1]. Find b  a f (x)dx. 22. If f is continuous and bounded on an open interval (a, b), then it has Riemann integrable extension to [a, b]. Why? 23. Suppose f is defined on (a, b). If f 1 and f 2 are Riemann integrable extensions b b of a function f to [a, b], then a f 1 (x)dx = a f 2 (x)dx. Why?

246

3 Definite Integral

24. Justify the statement: Given a continuous function f on [a, b] and c ∈ R, there exists a function g which is continuous on [a, b] and differentiable on (a, b) such that g  = f on (a, b) and g(a) = c. x d 25. State the required properties of the function f : [a, b] → R such that dx ( a f (t) x d dt) = a dt f (t)dt. 26. Find explicit expressions for the following. d (a) dx (a)

d dx

ex

d (b) dx

ln t dt 1 x 2

1

1 d dt (b) 1 + t3 dx

sin x √ 0 x

dt 1 − t2

" 1 + t2

x2

27. Evaluate the following integral by using appropriate change of variable, and properties of integrals. 1  (a) 1 − x 2 dx

1 (b)

0

(d) 1

xe dx 0

4

4

x2

dx √ 1 + x3

x2

(e)

√ sin x √ dx x

1 (c)

2

x 3 e x dx 0

(f)

1

1 √ x 1 + x 2 dx 0

28. Prove the following: 1

1 x (1 − x) dx = m

(a) 0

b

b f (x)dx =

(b) a

x n (1 − x)m dx for any m, n ∈ N.

n

0

f (a + b − x)dx. a

29. Suppose f is continuous on [a, b] and ϕ and ψ are differentiable on an open interval J such that ϕ(x), ψ(x) ∈ (a, b) for all x ∈ J . Prove that the func ϕ(x) tion g, defined by g(x) = ψ(x) f (t)dt, x ∈ J , is differentiable and g  (x) =  f (ϕ(x))ϕ  (x)  x− f (ψ(x))ψ (x) for all x ∈ J . 30. Let f (x) = 1 (1/t)dt for x > 0. Justify the following statements: (a) f (x) < 0 for 0 < x < 1 and f (x) > 0 for x > 1, (b) f is continuous and strictly increasing on (0, ∞), (c) f is differentiable and f  (x) = 1/x for x > 0.  x dt 31. Let f (x) = 0 √1−t for −1 < x < 1. Show that 2

3.7 Additional Exercises

247

(a) f is continuous and strictly increasing √ on (−1, 1), (b) f is differentiable and f  (x) = 1/ 1 − x 2 on (−1, 1), (c) f (sin x) = x for all x ∈ (−1, 1). 32. Show that, if f is twice differentiable and f  and f  are continuous in an open interval containing [a, ], then b

b



f (x)dx + a

x f  (x)dx = [b f  (b) − a f  (a)].

a

 ϕ(x) f (t)dt = 33. Drive Theorem 3.4.2 using Theorem 3.4.5. [Hint: Observe that a  ϕ(x)  ϕ(α) f (t)dt + ϕ(α) f (t)dt.] a 34. Suppose f and ψ are as in Theorem 3.4.5. Assume that ψ  (x) = 0 for all x ∈ (α, β). If ϕ is the inverse of ψ defined on the range of ψ and if ϕ  exists, then show that ψ(β) β  f (ψ(t))dt = f (x)ϕ  (x)dx. α

ψ(α)

[Hint: Write f (x) = ( f ◦ ψ)(ϕ(x)) and use Theorem 3.4.5.] 35. Use Problem 34 to evaluate the following integrals. 4 (a) 1

√

t √ dt (b) 1+ t

3 1

dt √ t 1+t

Geometric and mechanical applications 36. Find the area of the portion of the circle x 2 + y 2 = 1 which lies inside the parabola y 2 = 1 − x. [Hint: Area enclosed by the circle in the second and third quadrant and the area enclosed by the parabola in the first and fourth quadrant. The required area is 1√ π + 2 0 1 − x dx. Ans: π2 + 43 . ] 2 37. Find the area common to the cardioid ρ = a(1 + cos θ ) and the circle ρ = 3a . 2 [Hint: The points of intersections of the given curves are given by 1 + cos θ = 23 ,   π/3 π i.e., for θ = ± π3 . Hence the required area is 2 21 0 ( 3a )2 dθ + 21 π/3 a 2 (1 + 2  √ cos θ )2 dθ . Ans: 47 π − 9 8 3 . ] 38. For a, b > 0, find the area included between the parabolas given by y 2 = 4a(x + a) and y 2 = 4b(b − x).

248

3 Definite Integral

[Hint: Points of intersection√of the curves is given by a(x + a) = b(b − x), i.e.,   b−a √ 2 −a 2 x = ba+b = b − a; y = 2 ab. The required area is 2 × −a 4a(x + a) +  √ b √ 4b(b − x) dx . Ans: 83 ab(a + b).] b−a 39. Find the area of the loop of the curve r 2 cos θ = a 2 sin 3θ [Hint:r 0 for θ = 0 and θ = π/3, and r is maximum for θ = π/6. The area  π/3 = 2 is 0 r2 dθ . ] 40. Find the area of the region bounded by the curves x − y 3 = 0 and x − y = 0. [Hint:  1 Points of intersections of the curves are at x = 0, 1, −1. The area is 2 0 (x 1/3 − x)dx. Ans: 1/2 ] 41. Find the area of the region that lies inside the circle r = a cos θ and outside the cardioid r = a(1 − cos θ ). [Hint: Note that the circle is the one with centre at (0, a/2)  π/3 and radius a/2. The curves intersect at θ = ±π/3. The required area is −π/3 (r12 − r22 )dθ , where √ 3 r1 = a cos θ , r2 = a(1 − cos θ ). Ans: a3 (3 3 − π ) ] 42. Find the area of the loop of the curve x = a(1 − t 2 ), y = at (1 − t 2 ) for −1 ≤ t ≤ 1. [Hint: y = 0 for t ∈ {−1, 0, 1}, and y negative for −1 ≤ t ≤ 0 and positive for 0 ≤ t ≤ 1. Also, y 2 = x 2 (a − x)/a so that the curve is symmetric w.r.t. the a 0 x-axis. Area is 2 0 ydx = 2 1 y(t)x  (t)dt. Ans: 8a 2 /15 ] 43. Find the length of an arch of the cycloid x = a(t − sin t), y = a(1 − cos t). [Hint: The curve cutsthe  x-axis at x = a and x = 2πa for t = 0 and t = 2π , 2π [x  (t)]2 + [y  (t)]2 dt. Ans: 8a. ] respectively. (C) = 0 44. For a > 0, find the length of the loop of the curve 3a y 2 = x(x − a)2 . [Hint: The curve cuts the x-axis at x = a, and the curve is symmetric w.r.t. a " the x-axis. Thus the required area is 2 0 1 + ( dy )2 dx. Since 6ayy  = (x − dx a)(3x − a), 1 + y 2 =

(3x+a)2 . 12ax

4a √ . 3 2 , 1+cos √θ

Ans:

]

0 ≤ θ ≤ π/2. 45. Find the length of the curve r = √  π/4 3 [Hint: (C) = 2 0 sec θ dθ . Ans: 2 + ln( 2 + 1). ] 46. Find the volume of the solid obtained by revolving the curve y = 4 sin 2x, 0 ≤ x ≤ π/2, about y-axis. [Hint: Writing y = 4 sin 2x for 0 ≤ x ≤ π/4 and y = 4 sin 2u for π/4 ≤ u ≤  π/2  π/4 4 π/2, V = 0 (u 2 − x 2 )dy = π π/4 u 2 (8 cos 2u)du − π 0 x 2 (8 cos 2x)dx. Also, note that the curve is symmetric w.r.t. the line x = π/4. Hence, the volume  π/4 is π 0 [( π4 − x)2 − x 2 ]dy. Ans: 2π 2 .] 47. Find the area of the surface obtained by revolving a loop of the curve 9ax 2 = y(3a − y)2 about y-axis.  3a " [Hint: x = 0 iff y = 0 or y = 3a. A = 2π 0 x 1 + ( dx )2 dx. Ans: 3πa 2 . ] dy 48. Find the area of the surface obtained by revolving about x-axis, an arc of the catenary y = c cosh(x/c) x = −a and x = a for a > 0.   a between a [Hint: The area is 2π −a y 1 + y 2 dx = 2π c −a cosh2 xc dx. Ans: π c [2a + c sinh 2ac ]. ]

7 Additional Exercises

249

49. The lemniscate ρ 2 = a 2 cos 2θ revolves about the line θ = π4 . Find the area of the surface of the solid generated.  π/4  [Hint: The required surface area is 4π −π/4 h ρ 2 + ρ 2 dθ , where h := ρ sin √  π 2 − θ , ρ = a cos 3θ so that ρ 2 + ρ 2 = cosa 2θ . Ans: 4πa 2 . ] 4 50. Find the volume of revolution of the cardioid ρ = a(1 + cos θ ) about the initial line. [Ans: 83 . ]

Chapter 4

Improper Integrals

In Chap. 3, we defined the concept of definite integral, more precisely, the Riemann integral, only for those functions which are bounded and defined on closed and bounded intervals. In this chapter, we extend the notion of the integral in a natural way when some of the above requirements are not satisfied. The resulting integrals are known as improper integrals.

4.1 Definitions 4.1.1 Integrals over Unbounded Intervals First we consider integrals of functions defined over intervals of the forms [a, ∞), (−∞, b] and (−∞, ∞). Recall that the Riemann integral was defined for bounded functions defined on intervals of the form [a, b] where a, b ∈ R with a < b. Definition 4.1.1 (Integral over [a, ∞) ) Suppose f is a real valued function defined on [a, ∞) for some a ∈ R and f is integrable on [a, t] for all t > a. If t f (x) dx

lim

t→∞ a

exists as a real  ∞ number, then we say that the improper integral of f over [a, ∞), denoted by a f (x)dx, exists and it is defined by

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_4

251

252

4 Improper Integrals

∞

t f (x)dx = lim

f (x) dx.

t→∞

a

♦

a

Definition 4.1.2 (Integral over (−∞, b] ) Suppose f is a real valued function defined on (−∞, b] for some b ∈ R and f is integrable on [t, b] for all t < b, . If b f (x) dx

lim

t→−∞ t

exists as a real number, then we say that the improper integral of f over (−∞, b], b denoted by −∞ f (x) dx, exists and it is defined by b

b f (x) dx = lim

f (x) dx.

t→−∞

−∞

♦

t

∞ c Suppose f : R → R is such that the improper integrals −∞ f (x) dx and c f (x) dx in the sense of Definitions 4.1.1 and 4.1.2, respectively, exist for some c ∈ R. Then, d it can be shown that (Exercise), for any d ∈ R, the improper integrals −∞ f (x) dx ∞ and d f (x) dx exist in the sense of Definitions 4.1.1 and 4.1.2, respectively, and c

∞ f (x) dx +

−∞

d f (x) dx =

∞ f (x) dx +

−∞

c

f (x) dx. d

c ∞ Thus, the sum −∞ f (x) dx + c f (x) dx is independent of the choice of c ∈ R. In view of this, we have the following definition (Figs. 4.1 and 4.2). c Definition 4.1.3 Suppose f : R → R is such that the improper integrals −∞ f (x) dx ∞ and c f (x) dx exist in the sense of Definitions 4.1.1 and 4.1.2, respectively, for some  ∞c ∈ R. Then we say that the improper integral of f over (−∞, ∞), denoted by −∞ f (x) dx, exists and it is defined by ∞

c f (x) dx =

−∞

∞ f (x) dx +

−∞

f (x) dx. c

♦

4.1 Definitions

253

Fig. 4.1 Improper integral over [a, ∞)

Fig. 4.2 Improper integral over (−∞, a]

Remark 4.1.1 We may observe that the existence of t f (x) dx

lim

t→∞ −t

does not, in general, imply the existence of function f : R → R defined by

∞

−∞

f (x) dx. To see this, consider the

f (x) = x, x ∈ R. c t Then we have −t f (x) dx = 0 for every t ∈ R, but the integrals −∞ f (x) dx and ∞ ♦ c f (x) dx do not exist for any c ∈ R.  lim

t

t→∞ −t

 f (x) dx exists 

Remark 4.1.2 If lim

t

t→∞ −t

⇒

∞

−∞

f (x)dx

f (x) dx exists, then its value is called the Cauchy principal

value of the integral of f . This particular integral will be useful while studying integrals of complex valued functions of a complex variable. ♦ Next we consider integrals of functions defined over infinite integrals of the form (a, ∞) and (−∞, b).

254

4 Improper Integrals

Definition 4.1.4 Suppose f is a real valued function defined on (a, ∞) and the ∞ improper integral t f (x)dx exists in the sense of Definition 4.1.1 for all t > a. If ∞ f (x) dx

lim

t→a+ t

exists, then we say that the improper integral of f over (a, ∞), denoted by ∞ f (x) dx, exists, and it is defined by a ∞

∞ f (x) dx = lim+

f (x) dx.

t→a

a

♦

t

Definition 4.1.5 Suppose f is a real valued function defined on (−∞, b) and the t improper integral −∞ f (x)dx exists in the sense of Definition 4.1.2 for all t < b. If t lim

t→b− −∞

f (x) dx

exists, then we say that the improper integral of f over (−∞, b), denoted by b −∞ f (x) dx, exists, and it is defined by b

t f (x) dx = lim−

−∞

t→b

f (x) dx.

♦

−∞

Remark 4.1.3 In the case of (a, ∞), the function may not be defined at the point a or may be unbounded on (a, a + δ) for some δ > 0 so that we cannot talk about the Riemann integral over [a, a + δ] for δ > 0. Analogous remark holds for functions defined on (−∞, b). ♦

4.1.2 Improper Integrals over Bounded Intervals Now, we consider improper integral of a function f which is defined on a bounded interval of the form either (a, b] or [a, b) and use them to define for the case of [a, b] \ J0 , where J0 is a finite subset of [a, b] (Figs. 4.3 and 4.4).

4.1 Definitions

255

Fig. 4.3 Improper integral over (a, b]

Fig. 4.4 Improper integral over [a, b)

Definition 4.1.6 Suppose f : (a, b] → R is such that it is integrable on [t, b] for every t ∈ (a, b). If b lim+ f (x) dx t→a

t

exists, then we say that the improper integral of f over (a, b], denoted by exists, and it is defined by b

a

f (x) dx,

b f (x) dx = lim+

f (x) dx.

t→a

a

b

♦

t

Definition 4.1.7 Suppose f : [a, b) → R is such that it is integrable on [a, t] for every t ∈ (a, b). If

256

4 Improper Integrals

t f (x) dx

lim

t→b− a

exists, then we say that the improper integral of f over [a, b), denoted by exists and it is defined by b

b a

f (x) dx,

t f (x) dx = lim−

f (x) dx.

t→b

a

♦

a

Definition 4.1.8 Suppose f is a real valued function defined on the set [a, c) ∪ (c, b] c b for some c with a < c < b. If the improper integrals a f (x) dx and c f (x) dx exist in the sense of Definitions 4.1.7 and 4.1.6, respectively, then we say that the improper b integral of f over [a, b], denoted by a f (x) dx, exists and it is defined by b

c f (x) dx =

a

b f (x) dx +

a

f (x) dx.

♦

c

Definition 4.1.9 Suppose f is a real valued function defined on (a, b). If for some c b c ∈ (a, b), the improper integrals a f (x) dx and c f (x) dx exist in the sense of Definitions 4.1.6 and 4.1.7, respectively, then we say that the improper integral of b f over (a, b), denoted by a f (x) dx, exists and it is defined by b

c f (x) dx =

a

b f (x) dx +

a

f (x) dx.

♦

c

More generally we have the following definition. Definition 4.1.10 Suppose f is defined on [a, b] \ J0 , where J0 is a finite subset of [a, b], say J0 = {x1 , . . . xk }. Suppose f is integrable over every closed and bounded interval contained in [a, b] \ J0 . Assume further that the improper integrals xi f (x)dx, i = 1, . . . , k + 1, xi−1

4.1 Definitions

257

exist, where x0 = a and xk+1 = b. Then we say that the improper integral of f over b [a, b], denoted by a f (x) dx, exists and it is defined by b

k+1  

xi

f (x) dx =

f (x) dx.

♦

i=1 x

a

i−1

Terminology: If an improper integral exists, then we also say that the improper integral converges; otherwise we say that the improper integral diverges. Remark 4.1.4 In the case of improper integrals over a non-closed bounded interval of the form (a, b], the function may not be defined at the point a or may be unbounded on (a, a + δ) for some δ > 0. If the function is not defined at a, then we can assign some value to it at a, say f (a) = α for some α ∈ R. Now, if the redefined function is not bounded in [a, b], then we cannot talk about the Riemann integral over [a, b]. Analogous statement holds for the case of improper integrals over [a, b) or (a, b) or ♦ [a, b] \ J0 , where J0 is a finite set. Remark 4.1.5 Suppose f is defined on a set of the form [a, b] \ J0 , where J0 is a finite set. In case f is bounded on [a, b] \ J0 and can be extended to [a, b] such that the extended function f˜ is integrable over [a, b], then it can be shown that (Exercise) b the improper integral f (x) dx exists and a

b

b f (x) dx =

a

f˜ (x) dx,

a

where the integral on the right-hand side is the Riemann integral of f˜ .

♦

Let us illustrate the above remark by an example. Example 4.1.1 Let f (x) =

sin x , x

0 < x ≤ 1. We know that f˜ defined by

f˜ (x) =



f (x), x = 0, . 1, x = 0

is a continuous extension of f to [0, 1], and hence, is whether, 1 lim

t→0+ t

sin x dx = x

1 0

1 0

f˜ (x)dx exists. Now, the question

f˜ (x)dx.

258

4 Improper Integrals

Since f˜ (x) =

sin x x

for t ≤ x ≤ 1 for any t ∈ (0, 1), we have 1

f˜ (x)dx −

1

sin x dx = x

t

0

t

f˜ (x)dx.

0

But, since |f˜ (t)| ≤ 1 for all t ∈ [0, 1],   t      f˜ (x)dx ≤ t → 0 as t → 0.     0

Thus, we have shown that 1 lim

t→0+

sin x dx = x

t

Thus, the improper integral

1 0

sin x dx x

1 0

1

f˜ (x)dx.

0

exists and

sin x dx = x

1

f˜ (x)dx.

♦

t

4.1.3 Typical Examples Example 4.1.2 Consider the improper integral t

∞ 1

1 x

dx. Note that

1 dx = [ln x]t1 = ln t → ∞ as t → ∞. x

1

Hence,

∞ 1

1 x

♦

dx diverges.

Example 4.1.3 Consider the improper integral t 1

∞ 1

1 x2

dx. Note that

  1 1 t 1 dx = − = 1 − → 1 as t → ∞. x2 x 1 t

4.1 Definitions

Hence,

∞ 1

1 x2

259

dx converges and ∞

1 dx = 1. x2

♦

1

∞ Example 4.1.4 For p = 1, consider the improper integral 1 we have  −p+1 t t 1 x t −p+1 − 1 dx = = . xp −p + 1 1 −p + 1

1 xp

dx. In this case,

1

Note that, p>1

1 t −p+1 − 1 → as t → ∞, −p + 1 p−1

⇒

and p 1, diverges for p ≤ 1,

1

and ∞ 1

 1

∞

1 1 for p > 1. dx = p x p−1

1 dx exsits iff p > 1 and in that case xp

 1

∞

1 1 dx = p x p−1

Example 4.1.5 Note that t 0



t e−x dx = −e−x 0 = 1 − e−t → 1 as t → ∞.

♦

260

Hence,

4 Improper Integrals

∞ 0

e−x dx converges and ∞

e−x dx = 1.

♦

0



∞

Example 4.1.6 The integral 0

t 0

1 dx converges: Note that for t > 0, 1 + x2

 t 1 −1 dx = tan (x) = tan−1 (t) − tan−1 (0) = tan−1 (t). 0 1 + x2

Since tan−1 (t) → π/2 as t → ∞ we obtain ∞ 1

1 dx = lim t→∞ 1 + x2

Example 4.1.7 The integral

∞

0 −t

1 −∞ 1+x2 dx

t

1 π dx = . 2 1+x 2

1

♦

converges: Note that for t > 0.

1 dx = 1 + x2

t

1 dx. 1 + x2

0

Hence, as in last example, 0 lim

t→∞ −t

Thus, both

1 dx = lim t→∞ 1 + x2

t 0

1 π dx = . 2 1+x 2

0

∞ 1 1 −∞ 1+x2 dx and 0 1+x2 dx exist, and each of them is equal to π/2. Hence, ∞ −∞

1 dx = 1 + x2

0 −∞

1 dx + 1 + x2

∞ 0

1 dx = π. 1 + x2

♦

4.1 Definitions

261

1 Example 4.1.8 (i) The integral 0 1x dx is improper as the function f (x) = 1/x, 0 < x ≤ 1 is unbounded. We note, for 0 < t < 1, 1

1 dx = [log x]1t = log 1 − log t x

t

= − log t = log

Thus lim+ t→0

1 t

1 dx x

= ∞, so that

1

1 0 x

1 . t

dx diverges.

b Since 1x dx exists as a Riemann integral for any a > 0, b > a, we can conclude b 1 a that 0 x dx diverges for every b > 0. 1 (ii) Clearly, for p ≤ 0, the integral 0 x1p dx exists as a Riemann integral. Next, 1 consider the case p > 0. In this case, 0 x1p dx is an improper integral, as the function f (x) = 1/xp , 0 < x ≤ 1 is unbounded. We observe that, for 0 < t < 1, 1 t

 −p+1 1 x 1 1 − t −p+1 . dx = = xp −p + 1 t −p + 1

Note that, p>1

1 − t −p+1 t −p+1 − 1 = → ∞ as t → 0, −p + 1 p−1

⇒

and p 0, b > a and p = 0, we b can conclude that, for every b > 0, 0 x1p dx converges for p < 1 and diverges for p > 1. ♦  0

1

1 dx exists iff p < 1 and in that case xp



1

1 1 dx = xp 1−p

0

Example 4.1.9 Let a < b and p > 0. We shall discuss the convergence of the improper integrals b b dx dx and . (b − x)p (x − a)p a

a

(i) We observe that t lim

t→b− a

dx exists ⇐⇒ lim t→b− (b − x)p

b−a b−t

b−a ⇐⇒ lim

ε→0

 b−a ds sp ε→0 ε

We know that lim

ds exists sp

ε

does not exist for p = 1. Thus,

ds exists sp b

dx a (b−x)p

diverges if p = 1.

Next, let p = 1 and 0 < ε < b − a. Then, we have b−a ε

 1−p b−a ε1−p ds s (b − a)1−p − . = = sp 1−p ε 1−p 1−p

Since lim ε → 0ε1−p /(1 − p) exists as a real number if and only if p < 1, t lim

t→b− a

and in that case

b a

dx exists ⇐⇒ p < 1, (b − x)p

dx (b − a)1−p . = (b − x)p 1−p

(ii) Modifying the arguments above, it can also be seen that, for p > 0, the  b dx improper integral a (x−a) ♦ p converges if and only if p < 1.

4.1 Definitions

263

Remark 4.1.6 For the computation of improper integrals, whenever they exist, we essentially used the fundamental theorem of integration over intervals over which the integral exist as Riemann integrals and then took limits. This motivates us in considering the following procedure: Suppose f is defined on an open interval (a, b), where a can be −∞ and b can be ∞. Suppose f is integrable on every closed and bounded interval contained in (a, b) and there exists a function g which is differentiable on (a, b) and g  (x) = f (x) for all x ∈ (a, b). Assume further that β := lim g(x) and α := lim g(x) exist. x→a+

x→b−

Then, it can be shown that the improper integral

b a

f (x)dx is given by

b f (x)dx = β − α. a

For example, to compute

b a

first note that



dx for p < 1, (b − x)p

(b − x)−p+1 dx . = g(x) := − p (b − x) −p + 1

Since g(x) → β := 0 as x → b−1 and g(x) → α := g(a) as x → a+ , we obtain b a

dx (b − a)1−p for p < 1. = β − α = (b − x)p 1−p

♦

Exercise 4.1.1 Suppose f : [a, ∞) → R be such that it is integrable on [a, b] for ∞ every b > a. Show that a f 9x)dx converges if and only if for ε > 0, there exists β β0 > a such that | α f (x) dx| < ε for all α, β ≥ β0 .

264

4 Improper Integrals

4.2 Tests for Integrability 4.2.1 Integrability by Comparison We state a result which will be useful in asserting the existence of certain improper integrals by comparing it with certain other improper integrals. Notation 4.2.1 Suppose J is an interval of either finite or infinite length. Suppose f is defined on J , except possibly at a finite number of points in J . We denote the improper integral of f over J by  f (x) dx.

♦

J

For example, if J = [a, b], then f may not be defined at a or at b or at some point c ∈ (a, b), and then the corresponding improper integrals, by definition, are b

t f (x) dx, lim

lim

t→a

f (x) dx,

t→b

t

a

t

b f (x) dx + lim+

lim

t→c−

f (x) dx,

t→c

a

t

respectively. We omit the proof of the following two theorems as they can be proved using the corresponding results for definite integrals and then taking limits. Theorem 4.2.1 (Comparison test) Let J be an interval, f and g be defined on J \ J0 ,  where J0 is a finite set. If 0 ≤ f (x) ≤ g(x) for all x ∈ J \ J0 and if J g(x) dx exists, then J f (x) dx exists and   f (x)dx ≤ J

g(x)dx. J

Theorem and f be defined on J \ J0 , where J0 is a finite   4.2.2 Let J be an interval set. If J |f (x)| dx exists, then J f (x) dx exists. and       f (x) dx ≤ |f (x)| dx. J

J

Definition 4.2.1 Let J be an interval and f be definedon J \ J0 , where J0 is a finite set. Then f is said to be absolutely integrable over J , if J |f (x)| dx is convergent. ♦

4.2 Tests for Integrability

265

∞ Example 4.2.1 We have already seen that the improper integral 0 verges. This can also be seen as follows: 1 1 We know that the Riemann integral 0 1+x 2 dx exists. Also, since

1 dx 1+x2

con-

1 1 ≤ 2 for all x ≥ 1, 2 1+x x ∞ 1 ∞ 1 and since 1 x2 dx converges, by Theorem 4.2.1, 1 1+x 2 dx also converges. Sim0 1 ilarly, we can assert the convergence of the improper integrals −∞ 1+x 2 dx and ∞ 1 ♦ −∞ 1+x2 dx. Example 4.2.2 Let us check the convergence of ∞ 1

We have already seen that

∞ dx 1

x2

dx . 1 + ex

converges and its value is 1. Now, for x ≥ 1 we have 1 2 ≤ 2 x 1+e x

so that by Theorem 4.2.1

∞ 1

dx 1+ex

also converges.

♦

Example 4.2.3 Let p > 1. Since      sin x    ≤ 1 ,  cos x  ≤ 1  xp  xp xp xp for x > 0, it follows from Example 4.1.4 and Theorem 4.2.2 that the improper integrals ∞ ∞ sin x cos x dx and dx p x xp 1

converge for all p > 1. ∞ ∞ In fact 1 sinxp x dx and 1 example.

1

♦ cos x xp

dx converge for all p > 0 as we see in the next

266

4 Improper Integrals

Example 4.2.4 Let p > 0. Then for t > 0, t 1

 t t sin x 1 1 dx = p (− cos x) − p cos x dx p p+1 x x x 1 1

  t cos x cos t = cos 1 − p − p dx. t xp+1 1

By the result in Example 4.2.3,

∞ cos x 1

as t → ∞. Hence, ∞

xp+1

dx converges for all p > 0. Also,

sin x dx xp

converges for all

cos x dx xp

converges for all p > 0.

cos t tp

→0

p > 0.

1

Similarly, we see that ∞

♦

1

1 1 Example 4.2.5 Clearly, for p ≤ 0, 0 sinxp x dx and 0 grals. So, let p > 0. Then the relations

cos x xp

dx exist as Riemann inte-

     sin x   sin x  1 1 1  cos x   =  ≤ , p  ≤ p  xp   x  xp−1 p−1 x x x and the results in Example 4.1.8 and Theorem 4.2.1 imply that 1

sin x dx converges for all p < 2 xp

0

and

1 0

cos x dx converges for all p < 1. xp

♦

4.2 Tests for Integrability

267

∞

sin x dx converges for all p > 0 xp

1

∞

cos x dx converges for all p > 0 xp

1

∞

sin x dx converges for 0 < p < 2 xp

0

∞

cos x dx converges for 0 < p < 1 xp

0

Example 4.2.6 Since

sin x x

is decreasing on (0, 1],

sin x 1 sin 1 sin x = ≥ p−1 ∀x ∈ (0, 1]. p p−1 x x x x 1 1 Also, since 0 xp−1 dx diverges for p − 1 ≥ 1, i.e., for p ≥ 2, by comparison test (Theorem 4.2.1), we have 1

sin x dx diverges for all p ≥ 2. xp

♦

0

1

sin x dx converges if and only if p < 2 xp

0

4.2.2 Integral Test for Series of Numbers Recall that, in Example 1.2.9, we have proved the convergence of the series for p > 1 by using the fact that the sequence (an ) with n an = 1

dx , n ∈ N, xp

∞

1 n=1 np

268

4 Improper Integrals

converges. Now, we obtain a general procedure for showing the convergence or divergence of a series of positive terms by using the convergence or divergence of an appropriate improper integral. First we observe the following result (Exercise): Suppose f is defined on [a, ∞) with values in [0, ∞). Then n only if limn→∞ a f (x)dx exists.

∞ a

f (x)dx converges if and

Theorem 4.2.3 (Integral test) Let f be a continuous positive and decreasing function defined on [1, ∞). Then ∞ 

∞ f (n) converges ⇐⇒

n=1

f (x) dx converges. 1

Proof We observe that, for each k ∈ N, k ≤x ≤k +1 ⇒ ⇒

f (k + 1) ≤ f (x) ≤ f (k) k+1 f (x)dx ≤ f (k). f (k + 1) ≤ k

Hence, n 

k+1 n  n   f (k + 1) ≤ f (x)dx ≤ f (k),

k=1

i.e., n  k=1

k=1 k

k=1

n+1 n  f (k + 1) ≤ f (x)dx ≤ f (k), 1

k=1

Hence, by comparison we ∞  test for sequences, and the statement preceding the theorem f (n) converges if and only if the improper integral f (x) dx obtain, the series ∞ n=1 1 converges.  Remark 4.2.1 From the above theorem it follows that if f is a continuous positive ∞ and decreasing function defined on [1, ∞) such that 1 f (x) dx converges, then f (n) → 0 as n → ∞. In fact, under the above assumption, it can also shown that f (x) → ∞ as x → ∞. Can we conclude the same if we drop the assumption that f is a decreasing? The answer is in the negative. To see this consider the continuous function f : [1, ∞) → R defined by  1 − n2 |x − 2n| if |x − 2n| ≤ 1/n2 , n ∈ N , f (x) = 0 otherwise.

4.2 Tests for Integrability

269

We observe that |x − 2n| ≤ 1/n2 if and only if 2n − 1/n2 ≤ x ≤ 2n + 1/n2 so that ∞

2

f (x) dx =

n=1

1

Thus, n ∈ N.

∞ 1

2n+1/n  ∞ 

f (x) dx =

2n−1/n2

∞  1 . n2 n=1

f (x) dx converges, but f (x) → 0 as x → ∞, since f (2n) = 1 for all ♦

Exercise 4.2.1 Let f : [1, ∞) → R be defined by  f (x) =

n − n4 |x − 2n| if |x − 2n| ≤ 1/n3 , n ∈ N , 0 otherwise.

Show that f is a continuous function such that as x → ∞.

∞ 1

f (x) dx converges and f (x) → ∞

4.2.3 Integrability Using Limits Now we prove a result which facilitates the assertion of convergence and divergence of improper integrals. Theorem 4.2.4 Suppose 1. f (x) ≥ 0, g(x) > 0 for all x ∈ [a, ∞), b b 2. a f (x)dx and a g(x)dx exist for every b > a, f (x) 3. g(x) →  as x → ∞ for some  ≥ 0. Then the following hold. ∞ ∞ (i) If  > 0, then a f (x)dx converges ⇐⇒ a g(x)dx converges. ∞ ∞ (ii) If  = 0 and a g(x)dx converges, then a f (x)dx converges. ∞ ∞ f (x) → ∞ as x → ∞ and a f (x)dx converges, then a g(x)dx conFurther, if g(x) verges. f (x) →  as x → ∞ for some  ≥ 0. g(x) (i) Suppose  = 0. Then  > 0, and for ε > 0 with  − ε > 0, there exists x0 ≥ a such that f (x) <  + ε ∀ x ≥ x0 . −ε < g(x)

Proof Suppose

Hence ( − ε)g(x) < f (x) < ( + ε)g(x) ∀ x ≥ x0 .

270

4 Improper Integrals

∞ ∞ x Consequently, x0 f (x)dx converges iff x0 g(x)dx converges. Also, a 0 f (x)dx and  x0 a g(x)dx exist, by assumption. (ii) Suppose  = 0. Then for ε > 0, there exists x0 ≥ a such that f (x) < ε ∀ x ≥ x0 . g(x) ∞ Thus, f (x) < εg(x) for all x ≥ x0 . Hence, convergence of x0 g(x)dx implies the ∞ convergence of x0 f (x)dx. f (x) Next, suppose that → ∞ as x → ∞. Then for any M > 0, there exists g(x) x0 ≥ a such that f (x) ≥ M ∀ x ≥ x0 . g(x) Hence

1 f (x) ∀ x ≥ x0 . M ∞ ∞ x Consequently, if x0 f (x)dx converges, then x0 g(x)dx converges. As a 0 f (x)dx x and a 0 g(x)dx exist, the proof is over.  0 ≤ g(x) ≤

Example 4.2.7 The integral ∞ 0

1+x dx converges: 1 + x3

To see this, first note that 1 + x + x2 + x3 (1 + x)(1 + x2 ) = → 1 as x → ∞. 1 + x3 1 + x3 Hence, by Theorem 4.2.4, ∞ 0

1+x dx converges ⇐⇒ 1 + x3

We have already seen that

∞ 0

1 dx 1+x2

∞ 0

1 dx converges. 1 + x2

converges. Hence,

∞ 0

1+x dx 1+x3

converges.

♦

Exercise 4.2.2 Suppose f and g are non-negative continuous functions on J . Then, prove that   f (x)dx exists ⇐⇒ g(x)dx exists J

J

4.2 Tests for Integrability

271

in the following cases: f (x) 1. J = (a, b] and lim g(x) =  and  > 0. x→a

f (x) 2. J = [a, b) and lim g(x) =  and  > 0. x→b

3. J = [a, ∞) and lim

f (x)

x→∞ g(x)

4. J = (−∞, b] and

=  and  > 0.

lim f (x) x→−∞ g(x)

=  and  > 0.   In 1–4 above, prove that, if  = 0 and J g(x)dx exists, then J f (x)dx exists.



4.3 Gamma and Beta Functions Gamma and beta functions are certain improper integrals which appear in many applications.

4.3.1 Gamma Function We show that for x > 0, the improper integral ∞ (x) :=

t x−1 e−t dt

0

converges. The function (x), x > 0, is called the gamma function. Let x > 0. Note that t x−1 e−t ≤ t x−1 ∀ t > 0 and by the result in Example 4.1.8(ii), the improper integral Hence, by comparison test (Theorem 4.2.1), 1

1 0

t x−1 dt converges.

t x−1 e−t dt converges.

0

Also, we observe that

t x−1 e−t → 0 as t → ∞, t −2 ∞ and by the result Example 4.1.3, the integral 1 t −2 dt converges. Hence, by  ∞inx−1 Theorem 4.2.4, 1 t e−t dt converges. Thus,

272

4 Improper Integrals

∞ (x) :=

t

x−1 −t

e

1 dt =

0

t

x−1 −t

e

∞ dt +

0

t x−1 e−t dt

1

converges for every x > 0. Recall from Example 4.1.5 that we have (1) = 1.(1)

∞ 0

e−t dt = 1. Hence,

Also, using integration by parts, we obtain (x + 1) = x(x) ∀ x > 1.(2) Indeed, for x > 1, we have   t x e−t dt = −t x e−t + xt x−1 e−t dt. Since t x e−t → 0 as t → 0 and also as t → ∞, we obtain ∞ (x + 1) =

x −t

t e

∞ dt = x

0

t x−1 e−t dt = x(x).

0

Combining (1) and (2), we obtain (n + 1) = n! ∀ n ∈ N. We may also observe that ∞ (1/2) =

t −1/2 e−t dt.

0

Using the change of variable, y = t 1/2 , we have dy = ∞

t −1/2 e−t dt = 2

0

∞

1 dt, 2t 1/2

it can be seen that

e−y dy. 2

0

It is known (can be proved using the method of calculus of two variables) that ∞ 0

e−x dx = 2

√

π . 2

4.3 Gamma and Beta Functions

273

Hence, we have (1/2) =

√

π.

This, together with (2) gives 

2n + 1  2

=

1 × 3 × · · · × (2n − 1) √ π. 2n

4.3.2 Beta Function We show that for x > 0, y > 0, the improper integral 1 t x−1 (1 − t)y−1 dt

β(x, y) := 0

converges. The function β(x, y) for x > 0, y > 0 is called the beta function. Clearly, the above integral is proper for x ≥ 1 and y ≥ 1. In order to treat the remaining cases, we consider the integrals 1/2 t x−1 (1 − t)y−1 dt, 0

1 t x−1 (1 − t)y−1 dt. 1/2

We note that if 0 < t ≤ 1/2, then (1 − t)y−1 ≤ 21−y so that t x−1 (1 − t)y−1 ≤ t x−1 21−y ≤ 2t x−1 for 0 < t
0, y > 0 0

converges for every x > 0, y > 0. Beta function and gamma function are related by the relation, β(x, y) =

(x)(y) (x + y)

the proof of which is too much involved (see, e.g., [6]).

4.4 Additional Exercises 1. Justify the following statements: ∞ ∞ (a) If a f (x) dx exists, then c f (x) dx exists for any c > a. c b (b) If −∞ f (x) dx exists, then −∞ f (x) dx exists for any c < b. (c) If f is defined on on every closed and bounded ∞  c(−∞, ∞) and integrable interval, and if −∞ f (x) dx and c f (x) dx exist for some c ∈ R, then ∞ d −∞ f (x) dx and d f (x) dx exist for any d ∈ R. 2. Suppose f : R → R is an even function, that  ∞is, f (−x) = f (x) for every x ∈ R. ∞ Prove that −∞ f (x)dx exists if and only if 0 f (x)dx exists, and in that case ∞

∞ f (x)dx = 2

−∞

f (x)dx. 0

3. Suppose f : R → R is an odd function, that is, f (−x) = −f (x) for every x ∈ R. 0 ∞ Prove that −∞ f (x)dx exists if and only if 0 f (x)dx exists, and in that case ∞ f (x)dx = 0. −∞

4. Justify the following statements:

∞ (a) Suppose f is defined on [a, ∞)with values in [0, ∞). Then a f (x)dx n converges if and only if limn→∞ a f (x)dx exists. b (b) Suppose f is defined on (−∞, b] with values in [0, ∞). Then −∞ f (x)dx b converges if and only if limn→∞ −n f (x)dx exists.

4.4 Additional Exercises

275

5. Justify the following statements:

∞ (a) Suppose f is defined on [a, ∞). Then a f (x)dx converges if and only if b for any sequence (bn ) in R, bn → ∞ implies limn→∞ a n f (x)dx exists. b (b) Suppose f is defined in (−∞, b]. Then −∞ f (x)dx converges if and only if b for any sequence (an ) in R, an → −∞ implies limn→∞ an f (x)dx exists.

6. Suppose f is absolutely integrable over (−∞, ∞). Show that ∞

t f (x)dx = lim

f (x)dx.

t→∞

−∞

−t

7. Suppose f is defined on (a, b] or [a, b) and it has an integrable extension f˜ b b to [a, b]. Prove that the improper integral a f (x) dx exists and a f (x) dx = b ˜ a f (x) dx. 8. For p < 1, let fp (x) := sinxp x , 0 < x ≤ 1. Show (without using the previous problem) that fp has a continuous extension f˜p to [0, 1] and the improper integral 1 1 is equal to 0 f˜p (x)dx. 0 fp (x)dx 2  2 2 −1 dx = 1 (x + 1)dx. 9. Justify: 1 xx−1 t 10. Suppose f ≥ 0 on [a, b) and the integral a f (x)dx exists for every t ∈ [a, b). b If lim (b − x)α f (x) converges for some α < 1, then show that a f (x)dx also x→b converges.

11. 12. 13. 14. 15. 16. 17. 18.

19.

[Hint: Observe that for any ε > 0, there exists x0 ∈ [a, b) such that the number β+ε β := limx→b (b − x)α f (x) satisfies 0 ≤ f (x) ≤ (b−x) α for all x ∈ [x0 , b).]  b dx For a < b and p < 1, show that the improper integral a (x−a) p converges.  ∞ cos x Show that the improper integral 1 xp dx converges for all p > 0. 1 p Show that, for p > 1, 0 xdx 1/p = p−1 .  ∞ dx  1 dx Show that, for p > 1, 0 x1/p − 1 x1/p = 1. ∞ 1 Show that 1 dx → 0 as p → ∞ and 0 dx → 1 as p → 0. xp xp  1 dx Show that limt→1− 0 xp = ∞.    ∞   Does 1 sin x12 dx converge? [ Hint: Note that sin x12  ≤ x12 .]  ∞ cos x Does 2 x(log x)2 dx converge?    cos x  ≤ x(log1 x)2 and use the change of variable t = log x.] [Hint: Observe  x(log x)2  ∞ 2 Does 0 sinx2 x dx converge? [Hint: Observe sion on [0, 1].]

sin2 x x2

≤

1 x2

for x ≥ 1 and

sin2 x ,0 x2

< x ≤ 1 has a continuous exten-

276

4 Improper Integrals

20. Does

1 0

sin x dx x2

converge?

  [Hint: Observe sinx2 x = sinx x 1x ≥ sin1 1 .] t 21. Suppose f ≥ 0 on [a, b) and the integral a f (x)dx exists for every t ∈ [a, b). b If lim (b − x)α f (x) converges for some α < 1, then show that a f (x)dx also x→b converges. [Hint: Observe that for any ε > 0, there exists x0 ∈ [a, b) such that the number β+ε β := limx→b (b − x)α f (x) satisfies 0 ≤ f (x) ≤ (b−x) α for all x ∈ [x0 , b).]  ∞ b f (x)dx exists implies a f (x)dx → 0 as a, b → ∞. 22. Does a0

b a b [Hint: Observe: a f (x)dx = a0 f (x)dx − a0 f (x)dx for a0 > 0 and take limits.] ∞ 2 23. Does 0 e−x dx converge? [Hint: e−x ≤ x12 for 1 ≤ x ≤ ∞.] ∞ x) dx converge? 24. Does 2 sin(log x 2

[Hint: Change of variable t = log x, and the fact that 1 25. Does 0 ln xdx converge? [Hint: Change of variable t = log x.]

∞ log 2

sin t dt diverges.]

Chapter 5

Sequence and Series of Functions

Suppose we have real valued functions f1 , f2 , . . . defined on an interval I . Then, for each x ∈ I , we have a sequence (fn (x)) of real numbers. Suppose (fn (x)) converges at each x ∈ I to say, f (x). Some of the natural questions that one would like to ask are the following: (i) If each fn is continuous, then is f continuous? (ii) If each fn is differentiable on (a, b) ⊆ I , then is f differentiable on (a, b), and if so, is it true that f  (x) = limn→∞ fn (x) for each x ∈ (a, b)? (iii) If each fn is Riemann integrable on [a, b] ⊆ I , then is f Riemann integrable on [a, b], b b and if so, is it true that a f (x) dx = limn→∞ a fn (x) dx? ∞ n  If ∞ n=1 fn (x) converges for each x ∈ I to g(x) = n=1 fj (x) and if gn (x) = j=1 fj (x) for n ∈ N and for each x ∈ I , then questions (i), (ii), (iii) above can be asked about gn and g in place of fn and f , respectively. The purpose of this chapter is to discuss the above questions and many other related issues. We shall also deal with the convergence of a special case of the series of functions, namely the power series.

5.1 Sequence of Functions 5.1.1 Pointwise Convergence and Uniform Convergence Analogous to the concepts of sequence and series of numbers, we have the concepts of sequence and series of functions. Definition 5.1.1 For each n ∈ N, let fn be a (real valued) function defined on an ♦ interval I . Then, we say that (fn ) is a sequence of functions defined on I .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_5

277

278

5 Sequence and Series of Functions

Definition 5.1.2 Let (fn ) be a sequence of functions defined on an interval I . We say that (1) (fn ) converges at a point x0 ∈ I , if the sequence (fn (x0 )) of real numbers converges; ♦ (2) (fn ) converges pointwise on I , if it converges at every x ∈ I . Suppose (fn ) converges pointwise on I . Then, we can define a function f : I → R by f (x) = lim fn (x), x ∈ I . n→∞

Definition 5.1.3 Suppose (fn ) converges pointwise on I . Then, the function f defined by f (x) = lim fn (x), x ∈ I , n→∞

is called the pointwise limit of (fn ), and we write this fact as fn → f

pointwise on I .

♦

Exercise 5.1.1 Let (fn ) be a sequence of functions defined on an interval I such that  fn → f and fn → g pointwise for some functions f and g on I . Then f = g. We may observe that, for a given sequence (fn ) of functions defined on an interval I, fn → f pointwise on I if and only if for every ε > 0, and for each x ∈ I , there exists N ∈ N (in general, depends on both ε and x) such that |fn (x) − f (x)| < ε for all n ≥ N .

For any given ε > 0, if we are able to find an N ∈ N which does not vary as x varies over I such that |fn (x) − f (x)| < ε for all n ≥ N and for all x ∈ I , then we say that (fn ) converges uniformly to f on I (Figs. 5.1 and 5.2). More precisely, we have the following definition. Definition 5.1.4 Let (fn ) be a sequence of functions defined on an interval I . We say that (fn ) converges to f uniformly on I if for every ε > 0 there exists N ∈ N such that

Fig. 5.1 Graph of fn lies in a strip containing the graph of f

5.1 Sequence of Functions

279

Fig. 5.2 |fn (x) − f (x)| < ε for all x ∈ I

|fn (x) − f (x)| < ε

∀ n ≥ N and ∀ x ∈ I ,

and in that case we write fn → f

uniformly on

I.

♦

Remark 5.1.1 It is to be remarked that the Definitions 5.1.1–5.1.4 will also make sense if we take the domains of definition of the functions as any subset of R instead of an interval I . However, for the sake of simplicity of presentation and also for illustrating with examples, we have taken the domains of definition of the functions as intervals. ♦ Remark 5.1.2 We observe from the definition that if (fn ) converges uniformly to f ♦ on I , then (fn ) converges uniformly to f on every subinterval I0 ⊆ I . Exercise 5.1.2 Verify the statement in the above remark.



It is obvious that uniform convergence implies pointwise convergence. However, the converse is not true. To see this, look at the following example. Example 5.1.1 For each n ∈ N, let fn (x) = xn , x ∈ (0, 1). We have fn (x) → 0 for every x ∈ (0, 1). Thus, (fn ) converges pointwise to the zero function on (0, 1). However, this convergence is not uniform. To see this, let ε > 0 be given. If the convergence is uniform, then there exists N ∈ N such that |xn | < ε for all x ∈ (0, 1) and for all n ≥ N ; in particular, |xN | < ε for all x ∈ (0, 1), which is not possible if we had chosen ε < 1. This is illustrated in Fig. 1.1 below. ♦ In the next example we give two uniformly convergent sequence of functions (Fig. 5.3).

280

5 Sequence and Series of Functions

Fig. 5.3 fn (x) = xn , 0 0, there exists N ∈ N such that αn < ε for all n ≥ N . Hence, |fn (x) − f (x)| < ε ∀ n ≥ N , ∀ x ∈ I . (ii) Suppose (fn ) converges uniformly to f on I . Then for every ε > 0, there exists N ∈ N such that |fn (x) − f (x)| < ε ∀ n ≥ N , ∀ x ∈ I . In particular, for any sequence (xn ) in I , we have |fn (xn ) − f (xn )| < ε ∀ n ≥ N . Thus, |fn (xn ) − f (xn )| → 0.



In order to prove (fn ) does not converge uniformly to f on I , it is enough to find a sequence (xn ) in I such that |fn (xn ) − f (xn )| → 0

Let (fn ) be a sequence of functions defined on I , and let f : I → R. By Theorem 5.1.1 (i), sup |fn (x) − f (x)| → 0 ⇒ fn → f uniformly . x∈I

In fact, the converse is also true.

282

5 Sequence and Series of Functions

Theorem 5.1.2 Let (fn ) be a sequence of functions defined on I , and let f : I → R. Then sup |fn (x) − f (x)| → 0 ⇐⇒ fn → f uniformly . x∈I

Proof We have already observed that if sup |fn (x) − f (x)| → 0, then fn → f unix∈I

formly. Conversely, suppose fn → f uniformly. Let ε > 0 be given. Then we know that there exists N ∈ N such that |fn (x) − f (x)| < ε ∀ n ≥ N, ∀ x ∈ I . In particular, sup |fn (x) − f (x)| < ε ∀ n ≥ N. x∈I

Hence, sup |fn (x) − f (x)| → 0.



x∈I

(fn ) converges uniformly to f ⇐⇒ supx∈I |fn (x) − f (x)| → 0. Let us use Theorem 5.1.1 to verify uniform convergence and non-uniform convergence of some sequences. Example 5.1.3 For each n ∈ N, let fn (x) =

2nx , 1 + n4 x2

x ∈ [0, 1].

Using the relation a2 + b2 ≥ 2ab for any real numbers a and b, we have 0 ≤ fn (x) =

2nx 1 ≤ . 4 2 1+n x n

Thus, by Theorem 5.1.1 (i), fn → 0 uniformly.

♦

Example 5.1.4 For each n ∈ N, let fn (x) =

1 log(1 + n4 x2 ), n3

Then we have 0 ≤ fn (x) ≤ where g(t) :=

1 t3

x ∈ [0, 1].

1 log(1 + n4 ) = g(n) ∀ n ∈ N, n3

log(1 + t 4 ) for t > 0. Note that, by L’Hospital’s rule, we have lim g(t) = limt→∞

t→∞

4t 3 = 0. 3t 2 (1 + t 4 )

5.1 Sequence of Functions

283

In particular, limn→∞ g(n) = 0. Hence, by Theorem 5.1.1 (i), (fn ) converges uniformly to the zero function. ♦ Example 5.1.5 For each n ∈ N, let fn (x) = xn , x ∈ (0, 1). Taking f as the zero function on (0, 1), we have seen in Example 5.1.1 that fn → f pointwise, but not uniformly. Now, the non-uniform convergence of (fn ) also follows by Theorem 5.1.1 (ii), by taking xn = n/(n + 1) for n ∈ N, since |fn (xn ) − f (xn )| = fn (xn ) =

 n n 1 → . n+1 e

♦

Example 5.1.6 For each n ∈ N, let fn (x) =

nx , 1 + n2 x2

x ∈ [0, 1].

Note that fn (0) = 0, and for x = 0, fn (x) → 0 as n → ∞. Hence, (fn ) converges pointwise to the zero function. Since fn (1/n) = 1/2 for all n ∈ N, it follows from ♦ Theorem 5.1.1 (ii) that (fn ) does not converge uniformly on [0, 1]. Example 5.1.7 For each n ∈ N, let fn (x) = tan−1 (nx),

x ∈ R.

Note that fn (0) = 0, and for x = 0, fn (x) → f (x) as n → ∞, where ⎧ ⎨ π/2, x > 0, 0, x = 0 f (x) = ⎩ −π/2, x < 0. Hence, (fn ) converges pointwise to f . However, it does not converge uniformly to f on any interval containing 0. To see this, let I be an interval containing 0 and let xn = 1/n for n ∈ N. Then we have fn (xn ) = π/4 and f (xn ) = π/2 ∀ n ∈ N. Hence, by Theorem 5.1.1 (ii), (fn ) does not converge uniformly.

♦

Exercise 5.1.3 Show the non-uniform convergence of the sequence (fn ) in Example 5.1.7 by using ε − N arguments. 

284

5 Sequence and Series of Functions

Example 5.1.8 Let {r1 , r2 , . . .} be an enumeration of rational numbers in [0, 1]. For each n ∈ N, let fn : [0, 1] → R be defined by fn (x) =

0, x ∈ {r1 , . . . , rn }, 1, x ∈ / {r1 , . . . , rn }.

We observe that for each x ∈ [0, 1], fn (x) → f (x) :=

0, x rational, 1, x irrational

Thus, fn → f pointwise. Now, note that fn (rn+1 ) = 1 and f (rn+1 ) = 0 for all n ∈ N so that |fn (rn+1 ) − f (rn+1 )| → 0. Hence, by Theorem 5.1.1 (ii), (fn ) does not converge to f uniformly. ♦ Example 5.1.9 For each n ∈ N, let fn (x) =

1 , x ∈ (0, 1). nx

Clearly, fn (x) → 0 for every x ∈ (0, 1]. Thus, taking f = 0, the zero function, fn → f pointwise. But, (fN ) does not converge uniformly. This follows from Theorem ♦ 5.1.1 (ii), since fn (1/n) = 1 for all n ∈ N. We observe that the sequence (fn ) of functions in Example 5.1.9 is such that f is bounded, whereas every fn is unbounded. Such situation never arise for a uniformly convergent sequence (fn ) of functions, as Theorem 5.1.3 below shows. Before stating the theorem we introduce the following definition. Definition 5.1.5 A sequence (fn ) of functions is said to be eventually uniformly bounded on I if there exists M > 0 and N ∈ N such that |fn (x)| ≤ M for all x ∈ I and for all n ≥ N . ♦ Theorem 5.1.3 Suppose fn → f uniformly on I . Then, f is bounded if and only if (fn ) is eventually uniformly bounded. Proof Let ε > 0 be given. Since fn → f uniformly on I , there exists N ∈ N such that |fn (x) − f (x)| < ε for all x ∈ I and n ≥ N . Now, suppose f is bounded on I . Then there exists M > 0 such that |f (x)| ≤ M for all x ∈ I . Hence, |fn (x)| ≤ |fn (x) − f (x)| + |f (x)| < ε + M ∀ x ∈ I , ∀n ≥ N , and thus, (fn ) is eventually uniformly bounded. Conversely, suppose (fn ) is eventually uniformly bounded on I . Then there exists M > 0 and N ∈ N such that |fn (x)| ≤ M for all x ∈ I and for all n ≥ N . Hence,

5.1 Sequence of Functions

285

|f (x)| ≤ |f (x) − fn (x)| + |fn (x)| ≤ ε + M ∀x ∈ I , ∀n ≥ N , 

and thus, f is bounded on I .

Before closing this subsection, let us observe the following simple properties of the uniform convergence. Theorem 5.1.4 Let (fn ) and (gn ) be sequences of functions defined on I . If fn → f and gn → g uniformly for some functions f and g defined on I , then fn + gn → f + g

uniformly.

Proof Note that for every n ∈ N and for all x ∈ I , |(fn (x) + gn (x)) − (f (x) + g(x))| ≤ |fn (x) − f (x)| + |gn (x) − g(x)|.

(∗)

Now, let ε > 0 be given. Since fn → f and gn → g uniformly, there exist N1 , N2 ∈ N such that |fn (x) − f (x)| < ε ∀ x ∈ I , ∀ n ≥ N1 , |gn (x) − g(x)| < ε ∀ x ∈ I , ∀ ≥ N2 . Hence, from (∗), we obtain |(fn (x) + gn (x)) − (f (x) + g(x))| ≤ |fn (x) − f (x)| + |gn (x) − g(x)| < 2ε for all n ≥ N := max{N1 , N2 } and for all x ∈ I . Thus, fn + gn → f + g uniformly.

5.1.2 Uniform Convergence and Continuity We have seen in Examples 5.1.5 and 5.1.7 that the limit function f is not continuous, although every fn is continuous. The following theorem shows that such a situation will not arise if the convergence is uniform. Theorem 5.1.5 Suppose (fn ) is a sequence of continuous functions defined on an interval I such that it converges uniformly to a function f . Then f is continuous on I . Proof Suppose x0 ∈ I . Then for any x ∈ I and for any n ∈ N, |f (x) − f (x0 )| ≤ |f (x) − fn (x)| + |fn (x) − fn (x0 )| + |fn (x0 ) − f (x0 )|.

(∗)

Let ε > 0 be given. Since (fn ) converges to f uniformly, there exists N ∈ N such that

286

5 Sequence and Series of Functions

|fn (x) − f (x)| < ε ∀ n ≥ N , ∀ x ∈ I . Since fN is continuous, there exists δ > 0 such that |fN (x) − fN (x0 )| < ε whenever |x − x0 | < δ. Hence from (∗), we have |f (x) − f (x0 )| ≤ |f (x) − fN (x)| + |fN (x) − fN (x0 )| + |fN (x0 ) − f (x0 )| < 3ε whenever |x − x0 | < δ. Thus, f is continuous at x0 . This is true for all x0 ∈ I . Hence, f is a continuous function on I .  If (fn ) is a sequence of continuous functions defined on an interval I which converges pointwise to a discontinuous function f on I , then (fn ) does not converge uniformly to f

5.1.3 Uniform Convergence and Integration Suppose fn → f pointwise on an interval I . If each fn is Riemann integrable on [a, b] ⊆ I , then can we say that f is integrable? If f is integrable, do we have the convergence

b

b fn (x)dx → f (x)dx? a

a

The answer to the above question need not be in the affirmative as the following examples show. Example 5.1.10 Let (fn ) and f be as in Example 5.1.8. That is, if {r1 , r2 , . . .} is an enumeration of rational numbers in [0, 1], then for each n ∈ N, fn : [0, 1] → R is defined by 0, x ∈ {r1 , . . . , rn }, fn (x) = 1, x ∈ / {r1 , . . . , rn }. Then, for each n ∈ N, fn is continuous except at the points r1 , . . . , rn so that fn is integrable. Note that (fn ) converges pointwise to f : [0, 1] → R defined by f (x) :=

0, x rational, 1, x irrational,

5.1 Sequence of Functions

1

0 fn (x)dx

287

= 1 for all n ∈ N, but f is not integrable.

♦

Example 5.1.11 For each n ∈ N, let fn (x) = nx(1 − x2 )n ,

0 ≤ x ≤ 1.

Then we see that lim fn (x) = 0 ∀ x ∈ [0, 1].

n→∞

Indeed, fn (0) = 0 = fn (1) and for each x ∈ (0, 1),  n+1 fn+1 (x) = (1 − x2 ) → (1 − x2 ) as n → ∞. fn (x) n Since (1 − x2 ) < 1 for x ∈ (0, 1), we obtain limn→∞ fn (x) = 0 for every x ∈ [0, 1]. We note that,

1 1 n → as n → ∞. fn (x)dx = 2n + 2 2 0

Thus, the limit of the integrals is not the integral of the limit.

♦

We observe that in the above two examples, the convergence of (fn ) is not uniform: (a) The non-uniform convergence of the sequence (fn ) in Example 5.1.10 has been shown in Example 5.1.8 (b) Let (fn ) be as in Example 5.1.11. Note that,  1 n 1 n  1 n  1+ . fn (1/n) = 1 − 2 = 1 − n n n We know that (see Example 2.3.23)   1 n 1 n = exp(−1), limn→∞ 1 + = exp(1). lim 1 − n→∞ n n Hence, we have limn→∞ fn (1/n) = 1. In particular, with f as the zero function, |fn (1/n) − f (1/n)| → 1 as n → ∞. Hence, by Theorem 5.1.1 (ii), (fn ) does not converge to f uniformly. Now, we show that the answer to the question raised in the beginning of this subsection is in the affirmative if the convergence of (fn ) to f is uniform. Theorem 5.1.6 Let (fn ) be a sequence of integrable functions defined on an interval [a, b]. Suppose (fn ) converges uniformly on [a, b] and f is its limit. Then f is integrable, and if

288

5 Sequence and Series of Functions

x gn (x) =

x fn (t)dt, g(x) =

a

f (t)dt, n ∈ N a

for x ∈ [a, b] and n ∈ N, then gn → g uniformly on [a, b]. In particular,

b

b fn (x)dx →

a

f (x)dx. a

Proof We know that αn := supx∈[a,b] |fn (x) − f (x)| → 0 (Theorem 5.1.2). Since fn (x) − αn ≤ f (x) ≤ fn (x) + αn for all x ∈ [a, b] and for all n ∈ N, by Theorem 3.1.14, f is integrable. Next, let ε > 0 be given. Let N ∈ N be such that |fn (t) − f (t)| < ε ∀ n ≥ N ∀ t ∈ [a, b]. Then we have

x |gn (x) − g(x)| ≤

|fn (t) − f (t)|dt ≤ ε(b − a) a

for all n ≥ N and for all x ∈ [a, b]. Hence, gn → g uniformly. In particular, gn (b) → g(b), that is,

b

b fn (x)dx → f (x)dx. a

a



Thus, the proof is complete.

The following example shows that the uniform convergence of (fn ) to f in the above theorem is not necessary for its conclusion. Example 5.1.12 For n ∈ N, let fn (x) =

nx , x ∈ [0, 1]. 1 + nx

Note that fn (0) = 0 for every n ∈ N, and fn (x) → 1 for every x ∈ (0, 1]. Thus, (fn ) converges pointwise to the function f defined by f (x) =

0, x = 0, 1, 0 < x ≤ 1.

Since each fn is continuous and f is not continuous on [0, 1], the convergence of (fn ) to f is not uniform. This can also be seen by noting that

5.1 Sequence of Functions

289

f (1/n) − fn (1/n) =

1 ∀ n ∈ N. 2

However (cf. Example 2.2.17 and Example 2.3.28),

1

1 (f (x) − fn (x))dx =

0

0

1 dx = log(1 + n) → 0. 1 + nx n

Thus, the conclusion in Theorem 5.1.6 holds, though the sequence (fn ) does not converge uniformly to f . ♦

5.1.4 Uniform Convergence and Differentiation Suppose fn → f pointwise on an open interval I and each fn is differentiable on I . Then the following questions are natural: 1. Is f differentiable on I ? 2. If f is differentiable on I , then do we have d d f (x) = lim fn (x) n→∞ dx dx for every x ∈ I ? The following examples show that the answers need not be in the affirmative even if the convergence of (fn ) to f is uniform. Example 5.1.13 For each n ∈ N, let fn (x) = Note that

sin(nx) √ , n

x ∈ R.

1 |fn (x)| ≤ √ , ∀ n ∈ N, n

and hence (fn ) converges to the zero function uniformly. Since fn (x) = we have

√

fn (0) =

n cos(nx), x ∈ R, n ∈ N, √

n → ∞ as n → ∞.

Therefore, (fn ) does not converge pointwise.

♦

290

5 Sequence and Series of Functions

Example 5.1.14 For each n ∈ N, let

fn (x) =

1 + x2 for x ∈ R. n2

Then we see that, limn→∞ fn (x) = |x| for all x ∈ R. Note that the function f (x) = |x|, x ∈ R, is not differentiable at 0. Thus, pointwise limit of a sequence of differentiable functions need not be even differentiable. In this case, with f (x) = |x| for x ∈ R, the convergence (fn ) to f is uniform. Indeed, for every x ∈ R,

( 1 + x2 ) − x2 1 1/n2 2 − |x| = n2  + x = n2 1 1 + x2 + |x| + x2 + |x| n2 n2

so that |fn (x) − f (x)| ≤ 1/n for all x ∈ R and for all n ∈ N. Hence, (fn ) converges to f uniformly. ♦ Example 5.1.15 For n ∈ N, let fn (x) = Since |fn (x)| =

x , 1 + n2 x2

x ∈ (−1, 1).

|x| 1 ≤ ∀ x ∈ (−1, 1), ∀ n ∈ N, 1 + n2 x2 2n

(fn ) converges uniformly to the zero function f ≡ 0. Note that fn (x) =

(1 + n2 x2 ) − x(2n2 x) 1 − n2 x2 = . (1 + n2 x2 )2 (1 + n2 x2 )2

Note that fn (0) = 1, and for x = 0, fn (x) → 0 as n → ∞. Hence, the sequence (fn ) converges pointwise to the function g defined by g(x) =

1 if x = 0, 0 if 0 < |x| < 1,

Thus, we have shown that fn → f ≡ 0 uniformly and fn → g pointwise, but, ♦ f  = g. In Examples 5.1.13, 5.1.13 and 5.1.15 we have seen that (fn ) is a sequence of differentiable functions converging to a function f uniformly, but either (a) f is not differentiable or (b) f may be differentiable, but (fn ) does not converge pointwise, or

5.1 Sequence of Functions

291

(c) f is differentiable and (fn ) converges pointwise to a function g, but f  is not equal to g. In view of the above discussion, the following theorem is important. Theorem 5.1.7 Suppose (fn ) is a sequence of differentiable functions defined on an open interval I such that each fn is continuous, (fn (x0 )) converges for some x0 ∈ I , and (fn ) converges uniformly on I . Then (fn ) converges uniformly to a differentiable function f on I and lim fn (x) = f  (x) ∀ x ∈ I . n→∞

Proof Let g(x) := limn→∞ fn (x) for x ∈ I , and α := limn→∞ fn (x0 ). Since the convergence of (fn ) to g is uniform, by Theorem 5.1.5, the function g is continuous. Now, let

x

x  ϕn (x) = fn (t)dt, ϕ(x) = g(t)dt for x ∈ I . x0

x0

Then, using the arguments as in Theorem 5.1.6, (ϕn ) converges to ϕ uniformly. Since g is continuous, by Theorem 3.3.2, ϕ is differentiable and ϕ  (x) = g(x) for x ∈ I , and by Theorem 3.3.1,

x ϕn (x) =

fn (t)dt = fn (x) − fn (x0 ).

x0

Hence, we have lim fn (x) = lim [ϕn (x) + fn (x0 )] = ϕ(x) + α ∀ x ∈ I .

n→∞

n→∞

Let f (x) = ϕ(x) + α, x ∈ I . Since ϕn → ϕ uniformly, it follows from the above that fn → f uniformly, and f is differentiable. Note that g(x) = ϕ  (x) = f  (x) ∀ x ∈ I . Thus, (fn ) converges to f  .



Let us illustrate the above theorem using the sequence of functions in Example 5.1.15, but defined on a different interval. Example 5.1.16 For n ∈ N, let fn (x) =

x , 1 + n2 x2

As in Example 5.1.15, we see that

x ∈ (1, 2).

292

5 Sequence and Series of Functions

(i) |fn (x)| ≤

1 1 − n2 x2 and (ii) fn (x) = 2n (1 + n2 x2 )2

for all x ∈ (1, 2) and for all n ∈ N. The inequality (i) implies that fn → 0 uniformly on (1, 2), and by (ii), we have |fn (x)| ≤

1 + n2 x2 1 1 ≤ ≤ 2 2 2 2 2 (1 + n x ) 1+n x 1 + n2

for all x ∈ (1, 2) and for all n ∈ N. Hence, fn → 0 uniformly on (1, 2). Thus, in this case, we have we have limn→∞ fn (x) = f  (x) for all x ∈ (1, 2), as in the conclusion of Theorem 5.1.7. We may observe that, in the above, in place of the interval (1, 2), we could have taken any open interval (a, b) with 0 ∈ / [a, b]. ♦ Remark 5.1.3 In Theorem 5.1.7, the requirement of uniform convergence of (fn ) is not a necessary condition. To see this, for each n ∈ N, consider fn (x) =

xn , x ∈ (0, 1). n

Note that (fn ) converges uniformly to the zero function f = 0 and fn (x) = xn−1 ∀ x ∈ (0, 1), ∀ n ∈ N. In this case we have limn→∞ fn (x) = f  (x) for every x ∈ (0, 1). Note that the conver♦ gence of (fn ) is not uniform.

5.2 Series of Functions Series of functions is defined as in the case of series of numbers. Definition 5.2.1 By a series of functions on an interval I , we mean an expression of the form ∞ ∞   fn or fn (x), n=1

n=1

where (fn ) is a sequence of functions defined on I .  Definition 5.2.2 Given a series ∞ n=1 fn of functions on an interval I , let sn (x) :=

n  i=1

fi (x), x ∈ I .

♦

5.2 Series of Functions

293

Then sn is called the nth partial sum of the series

∞

n=1 fn .

♦

Definition 5.2.3 Let (fn ) be a sequence of functions defined on an interval I , and let sn be the nth partial sum of the series ∞ n=1 fn . Then we say that the series ∞ (1) n=1 fn converges at a point x0 ∈ I if (sn ) converges at x0 , (2) ∞ n=1 fn converges pointwise on I if (sn ) converges pointwise on I , (sn ) converges uniformly on I , (3) ∞ n=1 fn converges uniformly on I if  ∞ ∞ f converges absolutely on I if (4) n=1 n n=1 |fn (x)| converges for every x ∈ I . ♦ Recall from Chap. 1 that if (an ) is a sequence of real numbers, then convergence ∞  |a | implies the convergence of a . of ∞ n=1 n n=1 n From this, it follows that, absolute convergence of a series of functions implies its pointwise convergence. The proof of the following theorem is easy and left as an exercise. Theorem 5.2.1 (Comparison test) Suppose (fn ) and (gn ) are sequences of functions defined on an interval I such that |fn (x)| ≤ |gn (x)| ∀ x ∈ I , ∀ n ∈ N. If the series lutely on I .

∞ n=1

gn (x) converges absolutely on I , then

∞

n=1 fn (x)

converges abso-

The first theorem below is a consequence of Theorems 5.1.5 and 5.1.6, and the second theorem follows from Theorem 5.1.7.  Theorem 5.2.2 Suppose (fn ) is a sequence of continuous functions on I . If ∞ n=1 fn (x) converges uniformly on I , say to f (x), then f is continuous on I , and for [a, b] ⊆ I ,

b f (x)dx = a

∞



b

fn (x)dx.

n=1 a

Theorem 5.2.3 Suppose ∞ functions (fn ) is a sequence of continuously differentiable f (x) converges uniformly on I , and if on an open interval I . If ∞ n=1 n  n=1 fn (x) conf (x) converges uniformly to a differentiable verges at some point x0 ∈ I , then ∞ n n=1 function on I , and ∞ ∞   d  fn (x) = fn (x). dx n=1 n=1 Exercise 5.2.1 Show that, in Theorem 5.2.2, the continuity of the functions fn can  be replaced by Riemann integrability of fn and f .  Definition 5.2.4 Given a series ∞ n=1 fn of functions ∞  defined on an open interval I , are differentiable on I , then the series if f n n=1 fn is called the derived series of ∞ f . ♦ n=1 n

294

5 Sequence and Series of Functions

5.2.1 Dominated Convergence Now, we consider a useful sufficient condition to check uniform convergence. First a definition.  Definition 5.2.5 We say that ∞ n=1 fn is a dominated series if there exists a sequence (αn ) of positive real numbers such that |fn (x)| ≤ αn ∀ x ∈ I , ∀ n ∈ N, and the series

∞ n=1

αn converges.

♦

By Theorem 5.2.1, every dominated series converges absolutely. Theorem 5.2.4 (Weierstrass test) Every dominated series converges uniformly and absolutely.  Proof Let ∞ n=1 fn be a dominated series defined on an interval I , and let (αn ) be asequence of positive reals such that |fn (x)| ≤ αn for all n ∈ N and for all x ∈ I , ∞ n=1 αn converges.   Let sn (x) = ni=1 fi (x) and σn = nk=1 αk for n ∈ N. Then for n > m,  n  n n       fi (x) ≤ |fi (x)| ≤ αi = σn − σm . |sn (x) − sm (x)| =    i=m+1

i=m+1

i=m+1

 Since ∞ n=1 αn converges, the sequence (σn ) is a Cauchy sequence. Now, let ε > 0 be given, and let N ∈ N be such that |σn − σm | < ε ∀n, m ≥ N . Hence, from the relation |sn (x) − sm (x)| ≤ σn − σm , we have |sn (x) − sm (x)| < ε ∀n, m ≥ N , ∀x ∈ I . This, in particular, implies that (sn (x)) is a Cauchy sequence at each x ∈ I . Hence, (sn (x)) converges for each x ∈ I . Let f (x) = limn→∞ sn (x), x ∈ I . Then, we have |f (x) − sm (x)| = lim |sn (x) − sm (x)| ≤ ε ∀m ≥ N , ∀x ∈ I . n→∞

∞

Thus, the series n=1 fn converges uniformly to f on I . By Theorem 5.2.1, it converges absolutely as well.  Remark 5.2.1 Theorem 5.2.4 is usually known as Weierstrass M-test, as in standard books on Calculus, the numbers αn in Theorem 5.2.4 are usually denoted by ♦ Mn .

5.2 Series of Functions

295

Example 5.2.1 The series

∞ n=1

cos nx n2

and

∞ n=1

sin nx n2

are dominated series, since

   cos nx  1 1  sin nx     2  ≤ 2 ,  2  ≤ 2 ∀n ∈ N n n n n ∞

is convergent. ♦ ∞ n Example 5.2.2 The series n=0 xis a dominated series on [−ρ, ρ] for 0 < ρ < 1, n ♦ since |xn | ≤ ρ n for all n ∈ N, and ∞ n=0 ρ is convergent. ∞ x Example 5.2.3 Consider the series n=1 n(1+nx2 ) on R. Note that and

1 n=1 n2

|x| 1 1  ≤ √ , 2 n(1 + nx ) n 2 n ∞

converges. Thus, the given series is dominated series. ♦ ∞ Example 5.2.4 Consider the series n=1 1+nx2 x2 for x ∈ [c, ∞), c > 0. Note that

and

1 n=1 n3/2

1 x 1 x ≤ 2 2 ≤ 2 ≤ 2 2 2 1+n x n x n x n c  1 and ∞ n=1 n2 converges. Thus, the given series is a dominated series on [c, ∞) for any c > 0. ♦ ∞ Example 5.2.5 The series n=1 (xe−x )n is dominated on [0, ∞): To see this, note that n  n! xn xn = n xe−x = nx ≤ n e (nx) /n! n and the series

∞

n! n=1 nn

♦

converges (Fig. 5.4).

 n−1 is not uniformly convergent on (0, 1), in parExample 5.2.6 The series ∞ n=1 x ticular, not dominated on (0, 1). This is seen as follows: Note that sn (x) :=

n  k=1

xk−1 =

1 1 − xn → f (x) := as n → ∞. 1−x 1−x

Also, we observe that f (x) − sn (x) =

xn . 1−x

Taking xn := n/(n + 1), we have xnn → 1/e and 1 − xn → 0 so that f (xn ) − sn (xn ) =

xnn →→ ∞. 1 − xn

296

Fig. 5.4 sn (x) =

5 Sequence and Series of Functions

n

k=1 x

k−1 ,

0 1. Also, the series diverges at x = 1 and at x = −1. Hence, its radius of convergence is 1, and its domain of convergence is the open interval (−1, 1). Note that in this case the function 1 , |x| < 1, f (x) = 1−x represents the given power series.

♦

5.3 Power Series

301

 xn Example 5.3.2 Consider the power series ∞ n=0 n . In this case, we know that the series converges at x = −1 and diverges at x = 1. Hence, its radius of convergence is 1, and its domain of convergence is [−1, 1). ♦  xn Example 5.3.3 Consider the power series ∞ n=0 n2 . We know that this series converges at x = 1 and x = −1. Since n2 |xn+1 /(n + 1)2 | = |x| → |x| as n → ∞, |xn /n2 | (n + 1)2   xn    by ratio test the series ∞ n=0 n2 converges for x with |x| < 1 and diverges for x with |x| > 1. Therefore, the radius of convergence is 1, and the domain of convergence is [−1, 1]. ♦ ∞ xn Example 5.3.4 Consider the power series n=0 n! . Since |x| |xn+1 /(n + 1)!| = → 0 as n → ∞, n |x /n!| n+1 by ratio test the series converges at every x ∈ R. Hence, the radius of convergence is ∞, and the domain of convergence is R. Note that in this case the function f (x) = ex , x ∈ R, ♦

represents the given power series. Example 5.3.5 Consider the power series

∞ n=0

n!xn . Since

|(n + 1)!xn+1 | = (n + 1)|x| → ∞ as n → ∞, |n!xn | by ratio test, the series diverges at every nonzero x ∈ R. Hence, the radius of convergence is 0, and the domain of convergence is the singleton set {0}. ♦ For finding the radius of convergence and the domain of convergence, Theorem 5.3.2 below will be useful. CONVENTION: We shall use the convention, that if a ∈ {0, ∞}, then a1 := ∞ if a = 0, 0 if a = ∞.  n Theorem 5.3.2 Consider the power series ∞ n=0 an (x − c) , and let R be its radius of convergence.     (i) If limn→∞  aan+1  = L ∈ [0, ∞], then R = 1/L. n (ii) If limn→∞ |an |1/n = ∈ [0, ∞], then R = 1/ .

302

5 Sequence and Series of Functions

Proof Taking un (x) = an (x − c)n , n ∈ N0 , the proofs follow from Abel’s theorem (Theorem 5.3.1) and applying ratio test and root test.  Example 5.3.6 Let us find out the radius of convergence and domain of convergence of some power series ∞ ∞   xn xn . and 2n − 1 n4n n=1 n=1  xn (i) ∞ n=1 2n−1 : In this case an = 1/(2n − 1) and |an+1 /an | → L = 1. Hence R = 1/L = 1. Note that the series diverges at x = 1 and, by using Leibnitz’s theorem, the series converges at x = −1. Hence, the domain of convergence is [−1, 1).  xn n : (ii) ∞ n n=1 n4 In this case an = 1/(n4 ) and |an+1 /an | → L = 1/4. Hence, R = 1/L = 4. Note that the series diverges at x = 4 and, by using Leibnitz’s theorem, the series converges at x = −4. Hence, the domain of convergence is [−4, 4). ♦ Suppose (an )n∈N0 and (bn )n∈N0 are sequences of numbers such that |an | ≤ |bn | for all n ∈ N0 . Then, using the comparison test for the convergence of series of functions, and by Abel’s theorem, we can assert the following:  ∞ n 1. If ∞ n=0 bn (x − c) converges for |x − c| < r for some r > 0, then n=0 an (x − c)n converges for |x − c| < r. ∞ n n 2. If the radii of convergence of ∞ n=0 an (x − c) and n=0 bn (x − c) are R1 and R2 , respectively, then R1 ≥ R2 . Exercise 5.3.2 Let (akn )n∈N0  be a subsequence of a sequence (an ) of numbers. ∞ n (i) If the power series n=0 an (x − c) has radius of convergence R, then ∞ kn n=0 akn (x − c) is a power series and its radius of convergence is at least R— Why? ∞ x n  n (ii) Using (i), and knowing the radii of convergence of ∞ n=1 x and n=1 n , find ∞ xn! ∞ n2  the radii of convergence of n=1 x and n=1 n! . Remark 5.3.2 Although the domain of convergence of a power series is an interval, the domain of the function to which the power series converges can be larger than the domain of convergence of the series. For instance, the domain of convergence  n of the power series ∞ n=0 x is (−1, 1), whereas the function f (x) := 1/(1 − x) to which the series converges has domain of definition R \ {0}. Moreover, the function has a power series representation around any point c = 1. Indeed, if c = 1, 1 1 = 1−x (1 − c) − (x − c) 1 1   = (1 − c) 1 − x−c 1−c

=

1 (1 − c)

∞   n=0

x − c n 1−c

5.3 Power Series

303

whenever |x − c| < |1 − c|. Thus, ∞

 1 = an (x − c)n for 1−x n=0

|x − c| < |1 − c|,

where an := 1/(1 − c)n+1 for n ∈ N0 .

♦

At this point one may ask whether a function can have different power series expansion around the same point or not. We shall see in the next subsection that, that is not possible.

5.3.2 Term by Term Differentiation and Integration In this subsection, consider, without loss of generality, that a power series  we shall n a x with domain of convergence and radius of convergence are is of the form ∞ n n=0 D and R, respectively. Let f (x) =

∞ 

an xn , x ∈ D.

n=0

Now, we address the following questions on the power series

∞ n=0

an xn .

1. Is f continuous on D? 2. Is f integrable on every interval [a, b] ⊆ D? 3. If f is integrable on [a, b] ⊆ D, then do we have the equality

b f (x)dx =

∞  n=0

a

b xn dx ?

an a

4. Is f differentiable on (−R, R)? 5. If f is differentiable on (−R, R), and if g = f  , then do we have the equality g(x) =

∞ 

nan xn−1 on (−R, R)?

n=1

We show that the answers to all the above questions are in the affirmative. In this regard the following theorem is very crucial.  n Theorem 5.3.3 Let R > 0 be the radius of convergence of a power series ∞ n=0 an x . Then the series converges uniformly on [−ρ, ρ] for any ρ with 0 < ρ < R, and the function f defined by

304

5 Sequence and Series of Functions

f (x) :=

∞ 

an xn , x ∈ (−R, R)

n=0

is continuous on the interval (−R, R). Proof Let 0 < ρ < R, and r be such that ρ < r < R. Then for every x with |x| ≤ ρ, we have  x n  ρ n | ≤ |an r n | . |an xn | ≤ |an r n | r r  n n Since the series ∞ n=0 an r is convergent, an r → 0 as n → ∞, and hence the n n N, for some  M > 0. Thus, sequence (an r ) is bounded,  say |annr | ≤ M for all n ∈ ∞ ρ n (ρ/r) converges, since < 1, so that |an xn | ≤ M (ρ/r)n with ∞ n=0 n=0 an x is a ∞ r n dominated series on [−ρ, ρ]. Hence the series n=0 an x is uniformly convergent on [−ρ, ρ], and by Theorem 5.2.4, the function f defined by f (x) :=

∞ 

an xn

n=0

is continuous on [−ρ, ρ]. Since, this is true for any ρ < R, the function f is continuous at every x ∈ (−R, R). Indeed, for any x0 ∈ (−R, R), we may take ρ such that |x0 | < ρ < R.  ∞ Definition 5.3.4 Given a power series n=0 an xn , the power series ∞ 

nan xn−1

n=1

is called the derived series of

∞ n=0

♦

an xn .

Next theorem is about the derived series

∞ n=1

nan xn−1 .

Theorem 5.3.4 The power series and its derived series have the same radius of convergence.  n Proof Let R > 0 be the radius of convergence of a power series ∞ n=0 an x . Let x0 ∈ (−R, R), and let ρ be such that |x0 | < ρ < R. Then, we have |nan x0n−1 | = n|an ρ n−1 |

 |x | n−1 0 ρ

∀n ∈ N.

 n n−1 Since ∞ | → 0 so that it is bounded, say |an ρ n−1 | ≤ M n=0 an ρ converges, |an ρ for all n ∈ N. Thus, 

|nan x0n−1 |

|x0 | ≤ Mn ρ

n−1 ∀n ∈ N.

5.3 Power Series

305

∞

 n−1 converges, the series ∞ n=1 |nan x0 | also converges. Thus, for every x ∈ (−R, R).  n−1 . Suppose It remains to show that R is the radius of convergence of ∞ n=1 nan x ∞ n−1 na x converges at some point y with |y | > R. Then taking r with |y0 | > n 0 0 n=1  n−1 na r converges (by Abel’s theorem applied to this series). r > R, we see that ∞ n n=1  n But, |nan r n−1 | ≥ |an r n |/r so that by comparison test ∞ n r converges. This is n=1 a n−1 is R.  not possible since r > R. Thus, the radius of convergence of ∞ n=1 nan x ∞ Corollary 5.3.1 Let R > 0 be the radius of n=0 an xn . Then for any ∞of convergence n−1 converges uniformly on [−ρ, ρ] ρ with 0 < ρ < R, the derived series n=1 n an x and it represents a continuous function on the interval (−R, R).  Proof By Theorem 5.3.4, the radius of convergence of the derived series ∞ n=1 n an  ∞ xn−1 is R. Hence, Theorem 5.3.3 applied to the series n=1 n an xn−1 gives the result. n−1 n=1 n(|x0 |/ρ) n−1 converges n=1 nan x

Since ∞

 Theorem 5.3.5 Suppose R > 0 is the radius of convergence of a power series ∞ n a x n=0 n , and let ∞  an xn for |x| < R. f (x) := n=0

Then the following hold. (i) f is a continuous function on (−R, R) and for [a, b] ⊆ (−R, R),

b f (x)dx = a

∞  an [bn+1 − an+1 ]. n + 1 n=0

(ii) f is differentiable on (−R, R) and ∞

 d f (x) = nan xn−1 ∀ x ∈ (−R, R). dx n=1  n Proof (i) Let [a, b] ⊆ (−R, R). By Theorem 5.3.3, the series ∞ n=0 an x converges uniformly on [a, b]. Hence, by Theorem 5.2.2, the series can be integrated term by term over [a, b], i.e.,

b f (x)dx = a

∞  n=0

b

∞  an [bn+1 − an+1 ]. x dx = n + 1 n=0 n

an a

 n−1 (ii) By Corollary 5.3.1, the derived series ∞ converges uniformly on n=1 nan x any interval of the form [−ρ, ρ], where 0 < ρ < R. Hence, by Theorem 5.2.3, we have

306

5 Sequence and Series of Functions

f  (x) =

∞ 

nan xn−1

(∗)

n=1

for every x ∈ (−ρ, ρ), where 0 < ρ < R. Now, if x0 ∈ (−R, R), we may take ρ such that x0 ∈ (−ρ, ρ), so that the equation (∗) holds for x = x0 . This is true for any  x0 ∈ (−R, R). Hence, (∗) holds for all x ∈ (−R, R). Every power series can be integrated and differentiated term by term in its interval of convergence

In view of Theorems 5.3.5 and 5.3.4, we have the following. Theorem 5.3.6 Let R > 0 be the radius of convergence of the power series (x − c)n and let ∞  an (x − c)n , |x − c| < R. f (x) =

∞ n=0

an

n=0

Then f is infinitely differentiable on (c − R, c + R) and an =

f (n) (c) ∀ n ∈ N. n!

Proof By Theorems 5.3.5 and 5.3.4, the function f is infinitely differentiable on (−R, R) and for any k ∈ N, f (k) (x) =

∞ 

n(n − 1) · · · (n − k + 1)an (x − c)n−k ∀ x ∈ (−R, R).

n=k

Hence, f (k) (c) = k!ak so that ak = f (k) (c)/k! for all k ∈ N.



In view of the above theorem, we can assert the following:

If

∞  n=0

an (x − c) = n

∞ 

bn (x − c)n for x in a neighbourhood of the point c,then

n=0

an = bn for all n ∈ N ∪ {0}

Definition 5.3.5 If f is infinitely differentiable in a neighbourhood of a point c, and if the series ∞  f (n) (c) (x − c)n n! n=0

5.3 Power Series

307

has positive radius of convergence, then this series is called the Taylor series of f . If c = 0, then the Taylor series of f is called the Maclaurin series of f . ♦ Example 5.3.7 It can be easily seen that the Maclaurin series of sin x and cos x are given by ∞ ∞   (−1)n x2n+1 (−1)n x2n sin x = , cos x = , (2n + 1)! (2n)! n=0 n=0 respectively, for all x ∈ R.

♦

Remark 5.3.3 It is to be remarked that any infinitely differentiable function in a neighbourhood of a point need not have the Taylor series expansion around that point. To see this consider the function f (x) =

e−x , x = 0, 0, x = 0. 2

It can be easily seen that f is infinitely differentiable on R and f (n) (0) = 0 for all n ∈ N. Hence, ∞  f (n) (0) n f (x) = ♦ x for x = 0. n! n=0 Example 5.3.8 Recall that for |x| < 1, ∞

 1 = (−1)n xn . 1+x n=0

(1)

Integrating term by term, we get log(1 + x) =

∞  xn+1 . (−1)n n+1 n=0

(2)

Since the above series converges at x = 1, we obtain log 2 =

∞  (−1)n n=0

n+1

=1−

1 1 1 (−1)n + − + ··· + + ··· , 2 3 4 n+1

which we have observed in Sect. 1.2.4. Observe that the series in (1) and (2) have the same radii of convergence 1, but different domains of convergence, namely (−1, 1) and (−1, 1], respectively. ♦ Example 5.3.9 Recall that for |x| < 1,

308

5 Sequence and Series of Functions ∞

 1 = (−1)n x2n . 2 1+x n=0

(1)

Integrating term by term, we get

tan

−1

x x= 0

∞

2n+1  dt n x = (−1) , |x| < 1. 1 + t2 2n + 1 n=0

(2)

Note that the above series converges at x = 1 and also at x = −1. Hence, taking x = 1, we obtain ∞  (−1)n π = . 4 2n + 1 n=0 which is the Madhava–Nilakantha series for π/4 (cf. Sect. 1.2.4). Observe that the series in (1) and (2) have the same radii of convergence 1, but different domains of convergence, namely (−1, 1) and [−1, 1], respectively. ♦ Example 5.3.10 Let |x| < 1. Then we have ∞

 1 (−1)n xn . = 1+x n=0 Observe that

1 1 d . =− (1 + x)2 dx (1 + x)

Hence, term by term differentiation of the series for 1/(1 + x) gives ∞ ∞   1 n n−1 = − (−1) nx = (−1)n+1 nxn−1 . (1 + x)2 n=1 n=1

♦

Remark  5.3.4 nWe know that a series of functions need not be a power series of the form ∞ n=0 an x . But, in certain cases, it may be possible to convert them into this form after some change of variable, and obtain their domain of convergence. For example, consider the series ∞ ∞   (i) an (x − x0 )n (ii) an x3n n=0 ∞ 

n=0

∞  an (iii) (iv) an sinn x n x n=0 n=0 In each of these cases, we may take a new variable as follows: In (i) y = x − x0 , in (ii) y = x3 , in (iii) y = 1x , and in (iv) y = sin x. Suppose the radius of convergence n of ∞ n=0 an y is R. Then

5.3 Power Series

(i) (ii) (iii) (iv)

|y| < R |y| < R |y| < R |y| < R

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

309

|x − x0 | < R, |x|3 < R ⇐⇒ |x| < R1/3 1/|x| < R ⇐⇒ |x| > 1/R. | sin x| < R.

Of course one can apply some of the tests for convergence for the series of numbers and obtain the values of x for which the series in (1)–(iv) converge or diverge. ♦

5.4 Additional Exercises In the following I denotes an interval. 1. Prove that, if (fn ) converges pointwise on I , then there exists a unique function f : I → R such that fn → f pointwise. 2. Prove that, for every x ≥ 0, lim xe−nx = 0,

n→∞

lim n2 x2 e−nx = 0.

n→∞

3. Let fn (x) = (1 − x)xn for x ∈ [0, 1]. Show that (fn ) converges uniformly. [Hint: Check sup{|fn (x) − f (x)| : x ∈ [0, 1]} → 0.] 4. Suppose fn → f uniformly on [a, b]. Show that, if f is continuous on [a, b], then for every (xn ) in [a, b], xn → x implies fn (xn ) → f (x). 5. Let fn for every n ∈ N and f be continuous functions on [a, b]. Justify the statement: fn → f uniformly if and only if for every sequence (xn ) in [a, b], |fn (xn ) − f (xn )| → 0. 6. Justify the statement: If (fn ) converges uniformly to f on I , then (fn2 ) need not converge uniformly to f 2 . [Hint: Consider fn (x) = x + 1/n on [0, ∞).] 7. Justify the statement: If (fn ) is a sequence of continuous functions on an interval I which converges pointwise to a continuous function f on I , then the convergence need not be uniform. [Hint: Consider fn (x) = x/n on [0, ∞.] 8. For n ∈ N, let fn : [−1, 1] → R be defined by fn (x) =

n(1 − n|x|), 0 < |x| < 1/n, 0, x = 0 & |x| ≥ 1/n

and (fn ) converges pointwise to for x ∈ [−1, 1]. Show that each    fn is integrable 1 1 an integrable function f , but −1 fn (x)dx does not converge to −1 fn (x)dx. 9. For n ∈ N, let fn (x) = n2 x(1 − x2 )n , x ∈ [0, 1]. Show that fn → 0 pointwise , but



1 −1 fn (x)dx

 does not converge to 0.

310

5 Sequence and Series of Functions

10. For n ∈ N, let fn (x) =

log(1 + n3 x2 ) 2nx , gn (x) = , x ∈ [0, 1]. n2 1 + n3 x2

Show that gn → 0 uniformly on [0, 1]. Using this fact, show also that fn → 0 uniformly on [0, 1]. 11. For n ∈ N, let ⎧ 2 0 ≤ x ≤ 1/n, ⎨ n x, fn (x) = 2n − n2 x, 1/n ≤ x ≤ 2/n, ⎩ 0, 2/n ≤ x ≤ 1. Show that (fn ) does not converge uniformly of [0, 1]. [Hint: Use the relation between uniform convergence and integrals.] 

12. For n ∈ N, let fn (x) =

x2 + 1n , n ∈ N. Show the following.

(a) (fn ) converges uniformly on (−1, 1). (b) Each fn is continuously differentiable on (−1, 1). (c) (fn ) converge point wise, but not uniformly. 13. Let I be an open interval and for n ∈ N, let fn (x) =

14. 15. 16. 17.

18. 19.

2x , x ∈ I. 1 + n3 x2

Find all possible intervals I such that fn → f and fn → f  uniformly for some differentiable function f on I . [Hint: Refer example 5.1.16.]  ∞ an x2n Suppose (an ) is such that ∞ n=1 an is absolutely convergent. Show that n=1 1+x2n is a dominated series on R.  xn Show that for each p > 1, the series ∞ n=1 np is convergent on [−1, 1] and the limit function is continuous.  2 n+1 − n2 xn }(1 − x) converges to a continShow that the series ∞ n=1 {(n + 1) x uous function on [0, 1], but it is not dominated.  1 1 Show that the series ∞ n=1 1+(k+1)x − 1+kx is convergent on [0, 1], but it is not a dominated series. Show also that the series can be integrated term by term over the interval [0, 1].  n−1 converges pointwise, but not uniformly, on Show that the series ∞ n=1 nx (−1, 1). Find the interval of convergence of the following power series. ∞ ∞   (−1)n (−1)n 3n (i) (ii) nxn (4n − 1)xn n=1 n=1 ∞ ∞  n(x + 5)n  2n sinn x (iii) (iv) 3 (2n + 1) n2 n=1 n=1

5.4 Additional Exercises

311

[Answers: (i): (−∞, −1) ∪ [1, ∞); (ii): (−∞, −3) ∪ [3, ∞); (iii): [−6, −4]; (iv): [− π6 + kπ, π6 + kπ ], k ∈ Z.]  (n−1)! n 20. Find the radius of convergence of ∞ n=0 nn x . 21. Find the radius of convergence and interval of convergence of the following series: ∞ ∞   2 (i) α n xn for α > 0 (ii) α n xn for α > 0 n=0 ∞ 

(iii)

n=0

xn!

(iv)

∞  sin(nπ/6)

∞  (−1)n

n

n=0

n=0

xn(n+1) ∞  (−i)n

(x − 1) (vi) x2n n n nα 2 4 n=0 n=0 22. Using the limit of known series find the Maclaurin series for the functions (v)

n

(i)

x2 1 + x3

(ii)

x2 . (1 + x3 )2

23. The functions |x|xk has no power series expansion around 0 for any value k ∈ N. Why?

Chapter 6

Fourier Series

In the final section of the last chapter we have considered a special case of series of functions, namely, the power series. In this chapter we consider another special case of series of functions, namely, the Fourier series which is in many respects similar to power series. In fact, while studying the heat conduction problem in the year 1804, Fourier found it necessary to use a special type of function series associated with certain functions f , later known as Fourier series of f . As it is a special case, all the results known for general series of functions are true for these series as well. In addition, we obtain many more interesting and important results.

6.1 Fourier Series of 2π-Periodic Functions Definition 6.1.1 Let (an ) and (bn ) be sequences of real numbers. Then a series of functions of the form c0 +

∞ 

(an cos nx + bn sin nx), x ∈ R,

♦

n=1

is called a trigonometric series. The trigonometric series in Definition 6.1.1 is also written as c0 + a1 cos x + b1 sin x + a2 cos 2x + b2 sin 2x + · · · . If an = 0 and bn = 0 for all n > k for some k ∈ N ∪ {0}, then the resulting trigonometric series takes the form

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0_6

313

314

6 Fourier Series

c0 +

k  (an cos nx + bn sin nx). n=1

Definition 6.1.2 Given c0 , an , bn ∈ R for n = 1, . . . , k, a function of the form c0 +

k 

(an cos nx + bn sin nx), x ∈ R,

♦

n=1

is called a trigonometric polynomial of degree at most k. We observe that trigonometric polynomials are 2π -periodic on R, i.e., if f (x) is a trigonometric polynomial, then f (x + 2π ) = f (x) ∀ x ∈ R. This, in particular, implies that for any x ∈ R, f (x + 2nπ ) = f (x) ∀ n ∈ Z. From this, we can infer that, if the trigonometric series ∞  (an cos nx + bn sin nx) c0 + n=1

converges at a point x ∈ R, then it has to converge at x + 2π as well; and hence at x + 2nπ for all integers n. Indeed, if we write f k (x) = c0 +

k  (an cos nx + bn sin nx) n=1

for k ∈ N, then each f k (x) is a trigonometric polynomial, and hence lim f k (x) = lim f k (x + 2π ).

k→∞

k→∞

This shows that we can restrict the discussion of convergence of a trigonometric series to an interval of length 2π . Definition 6.1.3 A function f : R → R is said to be periodic with period T > 0 if f (x + T ) = f (x) for all x ∈ R, and in that case we also say that f is a T -periodic function. ♦ Remark 6.1.1 Suppose f is a function defined on an interval [a, b] with f (a) = f (b). Then f can be extended to all of R as a T -periodic function f˜ with T = b − a, by defining

6.1 Fourier Series of 2π -Periodic Functions

315

f˜(x + nT ) = f (x), x ∈ I, n ∈ Z. Indeed, corresponding to each y ∈ R, there exists a unique x ∈ I and n ∈ Z such that y = x + nT , so that f˜ is defined at all points in R and its restriction to [a, b] is f. In case, the value of f is not specified at one or both of the end-points, then we may assign the value(s) such that f (a) = f (b). In the due course, we shall denote the extended function f˜ by the same notation f . Thus, if we say that f is a T -periodic function with its values specified on an interval I , then we mean the T -periodic extension of f : I → R. ♦ We know that, if a trigonometric series converges, then the limit function is 2π periodic. What about the converse? Does every 2π -periodic function have a trigonometric series representation?

Suppose, for a moment, that f is a 2π -periodic function and that we can write f (x) = c0 +

∞  (an cos nx + bn sin nx) n=1

for all x ∈ R. Then what should be the coefficients c0 , an , bn ? Before, discussing this question, let us consider a simpler problem: Suppose f is a trigonometric polynomial, say f (x) = c0 +

k  (an cos nx + bn sin nx) n=1

for some k ∈ N. Then, we have π

  k    an cos nx dx + bn sin nx dx f (x) dx = 2c0 π + n=1

−π

π

π

−π

−π

= 2c0 π, since

π

−π

cos nx dx = 0 =

π

−π

sin nx dx. Thus,

1 c0 = 2π

π f (x) dx. −π

(∗)

316

6 Fourier Series

Also, for n, m ∈ N, we have

π

−π

π

cos nx sin mxdx = 0 and 

π cos nx cos mxdx =

−π

sin nx sin mxdx = −π

0, if n = m π, if n = m.

Therefore, from (∗) we obtain π

π f (x) cos mx dx = am π,

−π

f (x) sin mx dx = bm π −π

for m = 1, 2 . . . , k. Thus, for n = 1, . . . , k, we have 1 an = π

π −π

1 f (x) cos nx dx, bn = π

π f (x) cos nx dx. −π

 If a trigonometric series a20 + ∞ n=1 (an cos nx + bn sin nx) converges pointwise to a function f , which is integrable over [−π, π ], and if the series can be integrated term by term over [−π, π ], then the above procedure can be adopted to obtain the coefficients c0 , a1 , b1 , a2 , b2 , . . .. Motivated by the above consideration, we have the following definition. Definition 6.1.4 If f is a 2π -periodic function integrable over [−π, π ], then the Fourier series of f is the trigonometric series ∞

a0  + (an cos nx + bn sin nx), 2 n=1 where 1 an = π

π −π

1 f (x) cos nxdx, bn = π

π f (x) sin nxdx, −π

and this fact is written as ∞

f (x) ∼

a0  + (an cos nx + bn sin nx). 2 n=1

The numbers an and bn are called the Fourier coefficients of f .

♦

Remark 6.1.2 In the above, and in what follows, by integrability of a function f over an interval [a, b] we mean that it is Riemann integrable over [a, b]. The notion of Fourier series also can be considered in a more general context of Lebesgue

6.1 Fourier Series of 2π -Periodic Functions

317

integration (see, e.g., Nair [9]), so as to be applicable to functions that are not necessarily Riemann integrable. In this book, we shall deal only Riemann integrable functions. ♦ We have already observed that if a trigonometric series converges pointwise to a function f , which is integrable over [−π, π ], and if the series can be integrated term by term over [−π, π ], then it is the Fourier series of f . For instance, if a trigonometric series is uniformly convergent to a function f in [−π, π ], then f is continuous on [−π, π ] and the series can be integrated term by term. Here is a sufficient condition for the uniform convergence of the trigonometric series, which follows from Theorem 5.2.4 (Weierstrass test):  ∞ If the series ∞ n=0 an and n=0 bn are absolutely convergent, then the trigonometric series ∞ c0 + n=1 (an cos nx + bn sin nx) is uniformly convergent.

If f is a trigonometric polynomial, then its Fourier series is itself The following two theorems show that there is a large class of functions which can be represented by their Fourier series. Interested readers may look for their proofs in books on Fourier series; for example Bhatia [2]. Theorem 6.1.1 Suppose f is a 2π -periodic function which is bounded and monotonic on [−π, π ]. Then the Fourier series of f converges, and the limit function f˜(x) is given by f˜(x) =



f (x) if f is continuous at x, f (x−) + f (x+)] if f is not continuous at x.

1 [ 2

Theorem 6.1.2 (Dirichlet’s theorem) Let f : R → R be a 2π -periodic function which is piecewise differentiable on (−π, π ). Then the Fourier series of f converges, and the limit function f˜(x) is given by f˜(x) =



f (x) if f is continuous at x, f (x−) + f (x+)] if f is not continuous at x.

1 [ 2

In Theorem 6.1.2 we used the terminology piecewise differentiable as per the following definition. Definition 6.1.5 A function f : [a, b] → R is said to be piecewise differentiable if f exists and is piecewise continuous on [a, b] except possibly at a finite number of points. ♦ Remark 6.1.3 It is known that there are continuous functions f defined on [−π, π ] whose Fourier series does not converge pointwise to f . Its proof relies on concepts from advanced mathematics (cf. [2, 8]) ♦

318

6 Fourier Series

Recall that, in the case of a power series, the partial sums are polynomials and the limit function is infinitely differentiable. In the case of a Fourier series, the partial sums are infinitely differentiable, but, the limit function, if exists, need not be even continuous at certain points. This fact is best illustrated by the following example. Example 6.1.1 Let f be a 2π -periodic function defined on (−π, π ] by  f (x) =

0, −π < x ≤ 0, 1, 0 < x ≤ π.

Note that this function satisfies the conditions in Dirichlet’s theorem (Theorem 6.1.2). Hence, its Fourier series converges to f (x) for every x = 0, and at the point 0, the series converges to 1/2. Note that, for n ∈ N ∪ {0}, 1 an = π



π cos nxdx =

1, n = 0, 0, n = 0,

0

and for n ∈ N, 1 bn = π

π sin nxdx = 0

Thus,

1 π



1 − cos nπ n

=

1 π



1 − (−1)n . n

⎧ 2 ⎪ ⎨ , n odd, bn = π n ⎪ ⎩ 0, n even.

Thus, the Fourier series of f is ∞ 1 2  sin(2n + 1)x + . 2 π n=0 (2n + 1)

In particular, for x = π/2, 1=

∞ ∞ 1 2  sin[(2n + 1)π/2] 1 2  (−1)n + = + 2 π n=0 (2n + 1) 2 π n=0 (2n + 1)

which leads to the Madhava–Nilakantha series ∞

 (−1)n π = . 4 (2n + 1) n=0

♦

6.2 Best Approximation Property

319

6.2 Best Approximation Property Let f be a 2π -periodic function which is integrable on [−π, π ]. We know that the Fourier series of f need not converge to f pointwise. However, we have the following theorem on the best approximation property for the truncations of the Fourier series. Theorem 6.2.1 Let f be a 2π -periodic function which is integrable on [−π, π ], and for k ∈ N, let a0  + (an cos nx + bn sin nx), 2 n=1 k

f k (x) =

where a0 , a1 , a2 , . . . and b1 , b2 , . . . are the Fourier coefficients of f . Then π

π | f (x) − f k (x)| dx ≤

| f (x) − g(x)|2 dx

2

−π

−π

for any trigonometric polynomial g of the form g(x) = c0 +

k  (cn cos nx + dn sin nx) n=1

with c0 , c1 , . . . , ck , d1 , . . . , dk are in R.  Proof Let g(x) = c0 + kn=1 (cn cos nx + dn sin nx) for some numbers c0 , c1 , . . . , ck , d1 , . . . , dk in R. Then π

π | f (x) − g(x)| dx =

|( f (x) − f k (x)) + ( f k (x) − g(x))|2 dx

2

−π

−π π

=

π | f (x) − f k (x)| + 2

−π

| f k (x) − g(x)|2 dx

−π

π +2

( f (x) − f k (x))( f k (x) − g(x))dx.

−π

Note that f k (x) − g(x) =

k  a0 − c0 + [(an − cn ) cos nx + (bn − dn ) sin nx)]. 2 n=1

320

6 Fourier Series

Multiplying by f (x) − f k (x) and integrating, and observing the facts that f k (x))d = 0 and π

π −π

( f (x) −

π ( f (x) − f k (x)) cos nx dx = 0,

−π

( f (x) − f k (x)) sin nx dx = 0 −π

for n = 1, . . . , k, we obtain π ( f (x) − f k (x))( f k (x) − g(x))dx = 0. −π

Hence, we have π

π | f (x) − g(x)| dx =

−π

π | f (x) − f k (x)| +

2

2

−π π

≥

| f k (x) − g(x)|2 dx

−π

| f (x) − f k (x)|2 . −π



Thus, the proof is complete. The conclusion in the above theorem can also be written as π

π | f (x) − f k (x)| dx = inf

| f (x) − g(x)|2 dx,

2

g∈Tk

−π

−π

where Tk is the set of all trigonometric polynomials of degree at most k. In fact, in advanced courses (see, e.g., Rudin [13] or Nair [8], one learns that π | f (x) − f k (x)|2 dx → 0 as k → ∞. −π

6.3 Fourier Series for Even and Odd Functions Definition 6.3.1 Suppose f is a function defined on an interval containing (−, ) for some  > 0. Then on the interval (−, ), f is said to be an (i) even function, if f (−x) = f (x) for all x ∈ (−, ),

6.3 Fourier Series for Even and Odd Functions

321

(ii) odd function, if f (−x) = − f (x) for all x ∈ (−, ) .

♦

An even or odd function f defined on (−, ) can be extended to all of R as a 2-periodic function by assigning the same value for f at − and , and defining f˜(x + 2n) = f (x), − ≤ x ≤ . Note that if f is an odd function on (−, ), then for the extended 2-periodic function f˜ to be odd on R, we require f˜() = 0 = f˜(−), since f˜() = f˜(−) = − f˜(). This shows that, for an odd function on (−, ) to have an odd 2-periodic extension, the values at − and  cannot be assigned arbitrarily. However, if f is an even function on (−, ), then we can assign any value at  and then assign the same value at − so that the 2-periodic extension remains even. Now, let us consider the cases of even and odd on (−π, π ) separately: Case(i): f is even on (−π, π ) and integrable function on [−π, π ]. In this case, f (x) cos nx is an even function and f (x) sin nx is an odd function. Hence bn = 0 for all n ∈ N, so that in this case ∞

a0  2 + f (x) ∼ an cos nx with an := 2 π n=1

π f (x) cos nxdx. 0

Note that at the points x = 0 and x = π , the above series takes the forms ∞

∞

a0  a0  + + an and (−1)n an , 2 2 n=0 n=1 respectively. Case(ii): f is odd on (−π, π ) and integrable on [−π, π ]. In this case, f (x) sin nx is an even function and f (x) cos nx is an odd function. Hence an = 0 for all n ∈ N ∪ {0}, so that ∞ 

2 f (x) ∼ bn sin nx with bn := π n=1 At the point x = π/2, the series takes the form

π f (x) sin nxdx, 0

322

6 Fourier Series ∞ 

(−1)n b2n+1 .

n=0

Let us illustrate the above observations by some examples. Example 6.3.1 Consider the 2π -periodic function f defined on [−π, π ] by f (x) = |x|, x ∈ [−π, π ]. Note that f is an even function. Hence, bn = 0 for n = 1, 2, . . ., and ∞

f (x) ∼

a0  + an cos nx, x ∈ [−π, π ] 2 n=1

with 2 a0 = π

π x dx = π 0

and for n = 1, 2, . . ., ⎫ ⎧

π sin nx ⎬ 2 ⎨ sin nx π x dx x cos nxdx = − ⎭ π⎩ n n 0 0 0

2  cos nx π 2 (−1)n − 1 = = π n2 0 π n2

2 an = π

π

Thus, a2n = 0, a2n+1 = so that |x| ∼

−4 , n = 0, 1, 2, . . . π(2n + 1)2

∞ π 4  cos(2n + 1)x − , x ∈ [−π, π ]. 2 π n=0 (2n + 1)2

Taking x = 0 (using Dirichlet’s theorem), we obtain ∞

 π2 1 = . 8 (2n + 1)2 n=0 Example 6.3.2 Consider the 2π -periodic function f defined on (−π, π ] by f (x) = x, x ∈ (−π, π ].

♦

6.3 Fourier Series for Even and Odd Functions

323

Note that f is an odd function on (−π, π ). Hence, an = 0 for n ∈ N ∪ {0} and the Fourier series is ∞  bn sin nx, x ∈ (−π, π ] n=1

with 2 bn = π

π 0

⎫ ⎧ π cos nx ⎬ 2 ⎨ cos nx π −x dx x sin nx dx = + ⎭ π⎩ n n 0 0

2 cos nπ  (−1)n+1 2 = −π = . π n n Thus the Fourier series is 2

∞  (−1)n+1

n

n=1

sin nx.

In particular (using Dirichlet’s theorem), with x = π/2 we have ∞

∞

 (−1)n+1  (−1)n nπ π = sin = , 4 n 2 2n + 1 n=1 n=0 ♦

the Madhava–Nilakantha series. Example 6.3.3 Consider the 2π -periodic function f defined on (−π, π ] by  f (x) =

−1, −π < x < 0, 1, 0 ≤ x ≤ π.

Note that f is an odd function on (−π, π ). Hence, an = 0 for n ∈ N ∪ {0} and the Fourier series is ∞  bn sin nx, n=1

with 2 bn = π

π sin nx dx =

2 2 (1 − cos nπ ) = [1 − (−1)n ]. nπ nπ

0

Thus f (x) ∼

∞ 4  sin(2n + 1)x . π n=0 2n + 1

324

6 Fourier Series

Taking x = π/2, we have again the Madhava–Nilakantha series ∞

 (−1)n π = . 4 2n + 1 n=0

♦

Example 6.3.4 Consider the 2π -periodic function f defined on [−π, π ] by f (x) = x 2 , x ∈ [−π, π ]. Note that f is an even function. Hence, bn = 0 for n = 1, 2, . . ., and the Fourier series is ∞

a0  2 + an cos nx, x ∈ [−π, π ), an = 2 π n=1

π x 2 cos nx dx. 0

It can be see that a0 = 2π 2 /3, and an = (−1)n 4/n 2 . Thus x2 ∼

∞  (−1)n cos nx π2 +4 , x ∈ [−π, π ]. 3 n2 n=1

Taking x = 0 and x = π (using Dirichlet’s theorem), we have ∞

 (−1)n+1 π2 = , 12 n2 n=1

∞

 1 π2 = 6 n2 n=1

respectively (Fig. 6.1).

♦

6.4 Sine and Cosine Series Expansions We have seen that the Fourier series of an odd function involves only sine functions, whereas Fourier series of an even function involves only cosine functions. The above observation points to the following: Suppose a function f is defined on [0, π ]. In order to get sine series expansion for f , we may extend it to [−π, π ] so that the extended function is an odd 2π -periodic function. Similarly, to get cosine series expansion for f , we may extend it to [−π, π ] so that the extended function is an even 2π -periodic function.

6.4 Sine and Cosine Series Expansions

325

Fig. 6.1 Fourier approximation of f (x) = x/|x|, 0 < |x| ≤ 3

For obtaining such 2π -periodic functions based on a given integrable function f : [0, π ] → R, we first define an odd function f od and even function f ev on (−π, π ) as follows:  f (x) if 0 ≤ x < π, f od (x) = , − f (−x) if − π < x < 0,  f (x) if 0 ≤ x < π, f ev (x) = f (−x) if − π < x < 0. Clearly, f od (−x) = − f od (x),

f ev (−x) = f ev (x)

for all x ∈ (−π, π ), so that f od is an odd function and f ev is an even function on (−π, π ). These functions f od and f ev can be extended to 2π -periodic functions with their values at −π and π coincide with f (π ), and thus we obtain the corresponding get corresponding Fourier series on [0, π ] as f (x) ∼

∞ 

∞

bn sin nx and f (x) ∼

n=1

with 2 an = π

π 0

2 f (x) cos nx dx, bn = π

a0  + an cos nx 2 n=1 π f (x) sin nx dx. 0

326

6 Fourier Series

Definition 6.4.1 Let f : [0, π ] → R be an integrable function. Then the series f (x) ∼

∞ 

∞

bn sin nx and f (x) ∼

n=1

with 2 an = π

π 0

2 f (x) cos nx dx, bn = π

a0  + an cos nx 2 n=1 π f (x) sin nx dx 0

are called, respectively, the sine series expansion and the cosine series expansion of f . ♦ Example 6.4.1 Consider the function f (x) = x 2 , x ∈ [0, π ]. Then  f od (x) =

x 2 , if 0 ≤ x < π, −x 2 , if − π < x < 0,

f ev (x) = x 2 , x ∈ (−π, π ). Hence, the cosine series expansion of f on [0, π ] is (cf. Example 6.3.4) x2 ∼

∞  (−1)n cos nx π2 +4 , x ∈ [0, π ]. 3 n2 n=1

Its sine series expansion is f (x) ∼

∞ 

bn sin nx, x ∈ [0, π ],

n=1

where 2 bn = π

π 0

⎫ ⎧ ⎨ π π cos nx ⎬ cos nx 2 −x 2 dx . x 2 sin nx dx = + 2x ⎭ π⎩ n n 0 0

Note that  cos nx π cos nπ (−1)n+1 −x 2 = π2 , = −π 2 n n n 0

π π 2 sin nx π 2 sin nx cos nx d= x dx 2x − n n n n n 0 0 0

2  cos nx π 2 (−1)n − 1 = 2 . = 2 n n n n 0

6.4 Sine and Cosine Series Expansions

Thus,

327

4 (−1)n+1 + bn = 2π n π



(−1)n − 1 . n3

♦

Example 6.4.2 Consider the function f (x) = x, x ∈ [0, π ]. Note that f od (x) = x, x ∈ (−π, π ) 

and f ev (x) =

x, if 0 ≤ x < π, −x, if − π < x < 0.

Thus, f ev (x) = |x|, x ∈ (−π, π ). From Examples 6.3.2 and 6.3.1, we obtain x ∼2

∞  (−1)n+1

n

n=1

and |x| ∼

sin nx, x ∈ [0, π ]

∞ π 4  cos(2n + 1)x − , x ∈ [0, π ]. 2 π n=0 (2n + 1)2

Example 6.4.3 Let us consider sine and cosine expansions of the function  f (x) =

0, if 0 ≤ x < π/2, 1, if π/2 ≤ x < π.

Then the sine series of f is given by f (x) ∼

∞ 

bn sin nx, x ∈ [0, π ],

n=1

where 2 bn = π

π sin nx dx = − π/2



2  cos nx π 2 cos nπ/2 − cos nπ . = π n π n π/2

♦

328

6 Fourier Series

Note that b2n−1 =

2 and (2n − 1)π b2n =

2 [(−1)n − 1] = 2nπ



2 if n odd, − nπ 0 if n even.

Thus, for x ∈ [0, π ], we have π sin x sin 2x sin 3x sin 5x sin(4n − 3)x f (x) ∼ − + + + ··· + 2 1 1 3 5 4n − 3 sin(4n − 1)x sin(4n + 1)x sin(4n − 2)x + + + ··· . − 4n − 2 4n − 1 4n + 1

♦

6.5 Fourier Series of 2-Periodic Functions Suppose f is a T -periodic function. We may write T = 2. Then we may consider the change of variable t = π x/ so that the function f (x) = f (t/π ), as a function of t, it is 2π -periodic. Hence, its Fourier series is ∞

a0  + (an cos nt + bn sin nt) 2 n=1 where 1 an = π

π −π

1 bn = π

π −π

1 f (t/π) cos ntdt = 

1 f (t/π) sin ntdt = 

In particular, on the interval (−, ), • f even implies bn = 0 and an = • f odd implies an = 0 and bn =

2 

2 





 f (x) cos

nπ x dx, 

f (x) sin

nπ x dx. 

−

 −

nπ x dx,  0  nπ x 0 f (x) sin  dx. f (x) cos

6.5 Fourier Series of 2-Periodic Functions

329

Example 6.5.1 Consider the function f (x) = 1 − |x|, −1 ≤ x ≤ 1. Here,  = 1, so that 1 an =

1 (1 − |x|) cos nπ x dx = 2

−1

(1 − x) cos nπ x dx, 0

1 bn =

(1 − |x|) sin nπ x dx = 0. −1

Note that

1 cos nπ x dx =

sin nπ x nπ

1

0

1



sin nπ x x cos nπ x dx = x nπ

= 0, 0

1

0

1 −

0

0

 cos nπ x 1 sin nπ x dx = nπ n2π 2 0

(−1) − 1 = . n2π 2 n

Hence, 1 an = 2 0

2 (1 − x) cos nπ x dx = 2 2 [1 − (−1)n ] = n π

Thus, f (x) ∼

∞ 4  cos(2n + 1)π x . π 2 n=0 (2n + 1)2



0, neven, 4/n 2 π 2 , n odd.

♦

6.6 Fourier Series on Arbitrary Intervals Suppose a function f is defined in an interval [a, b] such that f (a) = f (b). We can obtain Fourier expansion of it as follows: . Let Method 1: Let us consider a change of variable as y = x − a+b 2

330

6 Fourier Series

 a + b , ϕ(y) := f (x) = f y + 2

where

−≤ y ≤

with  = (b − a)/2. We can extend ϕ as a 2-periodic function and obtain its Fourier series as ∞ a0  nπ nπ ϕ(y) ∼ + y + bn sin y) (an cos 2   n=1 where 1 an = 

 −

nπ y 1 dy, bn = ϕ(y) cos  

 ϕ(y) sin −

nπ y dy. 

Method 2: Considering the change of variable as y = x − a and  := b − a, we define ϕ(y) := f (x) = f (y + a) where 0 ≤ y < . We can extend ϕ as a 2-periodic function in any manner and obtain its Fourier series. Here are two specific cases: (a) Extending ϕ to (−, ) as an even function, we obtain ∞

ϕ(y) ∼

a0  nπ an cos + y 2  n=1

where  = (b − a)/2 and 1 an = 

 ϕ(y) cos −

nπ y dy. 

(b) Extending ϕ to (−, ) as an odd function, we obtain ϕ(y) ∼

∞ 

bn sin

n=1

where 1 bn = 

 ϕ(y) sin −

nπ y 

nπ y dy. 

From the series of ϕ we can recover the corresponding series of f on [a, b] by writing y = x − a.

6.7 Additional Exercises

331

6.7 Additional Exercises 1. Find the Fourier of the 2π - period function f such that:  series 1, − π2 ≤ x < π2 (a) f (x) = 0, π2 < x < 3π . 2  x, − π2 ≤ x < π2 (b) f (x) = π − x, π2 < x < 3π . 2  2x 1 + π , −π ≤ x ≤ 0 (c) f (x) = 1 − 2x , 0 ≤ x ≤ π. π x2 (d) f (x) = 4 , −π ≤ x ≤ π. 2. Using the Fourier series in Exercise 1, find the sum of the following series: 1 1 1 1 1 1 (a) 1 − + − + . . ., (b) 1 + + + + . . .. 3 5 7 4 9 16 1 1 1 1 1 1 + . . ., (d) 1 + 2 + 2 + 2 + . . .. (c) 1 − + − 4  9 16 3 5 7 π sin x, 0 ≤ x ≤ 4 3. If f (x) = , then show that cos x, π4 ≤ x < π2 π 8 f (x) ∼ cos π 4



sin x sin 3x sin 10x + + + ... . 1.3 5.7 9.11

4. Show that for 0 < x < 1, x − x2 =

8 π2



sin xπ sin 3π x sin 5π x + + + . . . . 13 33 53

5. Show that for 0 < x < π , sin x +

sin 5x π sin 3x + + ... = . 3 5 4

6. Show that for −π < x < π , x sin x = 1 −

2 2 2 1 cos x − cos 2x + cos 3x − cos 4x + . . . , 2 1.3 2.4 3.5

and find the sum of the series 1 1 1 1 − + − + .... 1.3 3.5 5.7 7.9

332

6 Fourier Series

7. Show that for 0 ≤ x ≤ π ,

cos 2x cos 4x cos 6x π2 − + + + ... , x(π − x) = 6 12 22 32

sin 3x sin 5x 8 sin x + + + ... . x(π − x) = π 13 33 53 8. Find the Fourier series of the 2-periodic functions f given below defined on [−1, 1) by  −1, −1 ≤ x < 0, (a) f (x) = 1, 0 ≤ x < 1,  0, x 2 ≤ x < 0, (b) f (x) = 1, 0 ≤ x < 1, 9. Find the cosine series and sine series for the functions f given below:  1, 0 ≤ x < 1, (a) f (x) = 0, 1 ≤ x < 2. (b) f (x) = x, 0 ≤ x < 1. 10. Justify the statement: If f is an odd 2-periodic function, then f (−) = f (0) = f () = 0. 11. If f : [0, ] → R, then show that  f (x), 0 ≤ x ≤ , is an even func(a) g : [−, ] → R defined by g(x) = f (−x), − ≤ x < 0, tion on [−, ], and ⎧ f (x), 0 < x < , ⎨ (b) h : [−, ] → R defined by g(x) = − f (−x), − < x < 0, is an odd ⎩ 0, x ∈ {0, , −} function on [−, ]. 12. If f is (Riemann) integrable on [−π, π ], then show that ∞

a02  2 1 + (an + bn2 ) ≤ 2 π n=1

π [ f (x)]2 dx. −π

13. If f is (Riemann) integrable on [−π, π ], then justify the following: π

π f (x) cos nxdx → 0 and

−π

as n → ∞.

f (x) sin nxdx → 0 −π

6.7 Additional Exercises

333

14. Assuming that the Fourier series of f converges uniformly on [−π, π ], show that π ∞ 1 a2  2 [ f (x)]2 dx = 0 + (an + bn2 ). π 2 n=1 −π

15. Using Exercises 7 and 14 show that ∞ (−1)n−1 ∞ 1 π4 π2 , (b) (a) = = n=1 n 4 n=1 90 n2 12 ∞ (−1)n−1 ∞ 1 π6 π3 (d) (c) = = n=1 n 6 n=1 (2n − 1)3 945 32

References

1. R.G. Bartle, D.R. Sherbert, Introduction to Real Analysis, 3rd edn (Wiley & Sons, Inc., 2000) 2. R. Bhatia, Fourier Series, TRIM, 2nd edn, (Hindustan Book Agnency, New Delhi 1993) (2003) 3. K.G. Binmore, Mathematical Analysis: A Straight Forward Approach (Cambridge University Press, 1991) 4. C.G. Delninger, Elements of Real Analysis (Jones & Martlett Learning, 2011) 5. P.M. Fitzpatric, Advanced Calculus (American Mathematical Society, 2006) 6. S. Ghorpade, B.V. Limaye, A Course in Calculus and Analysis, 2nd edn (Springer, Berlin, 2018) 7. G.G. Joseph, Indian Mathematics: Engaging with the World from Ancient to Modern Times (World Scientific, 2016) 8. M.T. Nair, Functional Analysis: A First Course. Prentice-Hall of India, New Delhi, 2002 (Fourth Print, 2014) 9. M.T. Nair, Measure and Integration: A First Course (Taylor & Francis, CRC Press, 2019) 10. D. Perkins, ϕ, π, e, I (MAAPRESS, The Mathematical Association of America, 2017) 11. N. Piskunov, Differential Integral Calculus, vol. I & vol. II (Mir Publishers, Moscow, 1974) 12. C.H. Pugh, Real Mathematical Analysis (Springer, Berlin, 2004) 13. W. Rudin, Principles of Mathematical Analysis, 3rd edn (McGraw-Hill, International Student Edition, 1964)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0

335

Index

Symbols 2π -periodic, 314 δ-neighbourhood, 72 k th -derivative, 154 k th -root, 102, 112 n th term, 2

A Abel’s theorem, 298 Absolute convergence, 62, 298 Absolutely convergent, 62 Absolutely integrable, 264 Alternating series, 57, 58 Antiderivative, 200 Archimdean property, 7 Arc length, 220 Area, 215 Area of the region, 215, 216, 219

B Beta function, 273 Binary expansion, 50 Bolzano-Weierstrass theorem, 35 Bounded, 21, 26 above, 21, 25 below, 21, 26 function, 78 on a subset, 78 Boundedness test, 20, 78 Bounded set, 104

C Caratheodory, 124 Cartesian coordinates, 215

Cauchy criterion, 40, 41 Cauchy integral reminder formula, 213 Cauchy principal value, 253 Cauchy product, 114 Cauchy sequence, 40 Cauchy’s generalized mean value theorem, 142 Cauchy’s root test, 54 Centre of gravity, 233, 236 Change of variable, 211 Closed subset, 42 Comparison test, 48, 49, 264 Composition, 81 Composition rule, 130 Concave function , 164 Conditionally convergent, 62 Constant sequence, 8 Continuous, 95 Continuous at a point, 95 Convergent sequence, 3 Convergent series , 44 Converges pointwise, 278, 293 uniformly, 278, 293 Convex function , 164 Cosine series expansion, 324, 326 Critical point, 136 Curve determined by a function, 163

D D’Alembert’s test, 54 Decimal expansion, 50 Decreasing, 24, 110 strictly, 110 Definite integral, 173 Deleted δ-neighbourhood, 72

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 M. T. Nair, Calculus of One Variable, https://doi.org/10.1007/978-3-030-88637-0

337

338 Deleted neighbourhood, 72 Dense, 67 Density, 235 Derivative, 120 Derived series, 293, 304 Differentiable at x0 , 120 Differential quotient, 121 Dirichlet’s theorem, 317 Divergent sequence, 3 Divergent series, 45 Diverges to −∞, 11 Diverges to ∞, 11 Domain of convergence, 299 Dominated series, 294

E Equidistant points, 183 Euler, 32 Even function, 320 Eventually uniformly bounded, 284 Eventually constant sequence, 8 Exponential, 112 function, 113 Extension, 201 Extension of a function, 201 Extremum, 105

F Fibonacci sequence, 19, 20 Fourier coefficients, 316 Fourier series, 316 Fourier series of 2-periodic functions, 328 2π -periodic functions, 313 functions on arbitrary intervals, 329 Fundamental theorem, 199, 205

G Gamma function, 271 Generalized Mean-Value Theorem (GMVT), 142, 198 Geometric and mechanical applications, 247 Geometric series, 45 Global maximum, 136 Global minimum, 136 Golden ratio, 19 Golden rectangle, 20 Greatest lower bound property, 27

Index H Hemachandra-Fibonacci sequence, 20 Higher derivatives, 153

I Improper integral, 254 converges, 257 diverges, 257 Improper integral of f over (−∞, ∞), 252 over (−∞, b), 254 over (−∞, b], 252 over (a, ∞), 254 over (a, b), 256 over [a, ∞), 251 over [a, b], 256 over (a, b], 255 over [a, b), 256 Increasing, 24, 110 strictly, 110 Indefinite integral, 200 Infimum, 27 Infinitely differentiable, 154 Initial point, 215 Integrable, 177 Integral, 177 Integral test, 197, 268 Interior point, 125 Intermediate Value Theorem (IVT), 106 Interval of convergence, 300 Inverse, 109 Inverse Function Theorem (IFT), 109, 111

J Jyeshthadeva, 203

L Lagrange’s mean value theorem, 140 Least upper bound, 26, 27 Least upper bound property, 25, 27 Left derivative, 126 differentiability, 125 limit, 88 Leibnitz-Gregory series, 60 Leibnitz’s theorem, 58 Length of the curve, 220, 224, 225 L’Hospital’s rule, 144, 147, 149–151, 155 Limit of a sequence, 4 Limit of f , 73 Limit as x → −∞, 90

Subject Index

339

Limit as x → ∞, 90 Limit point, 71 Local extremum, 136, 143 strict, 136 Local maxima, 135 Local minima, 135 Local minimum, 135 strict, 135 Logarithm, 52, 112, 118 function, 118 Lower bound, 26 Lower integral, 179 Lower sum, 176

Pingala, 20 Pointwise limit, 278 Polar coordinates, 218, 224 Power series, 298 Primitive, 200 Product, 81 Product formula, 208, 209 Product rule, 129

M Maclaurin series, 159, 307 Madhava, 60 Madhava–Nilakantha, 323, 324 Madhava–Nilakantha series, 60, 308, 318 Maximum, 105 Mean-Value Theorem (MVT), 140, 198 generalized, 198 Mesh, 192 Minimum, 105 Moment of inertia, 237 Monotonic, 110 Monotonically decreasing, 24, 110 increasing, 24, 110 Monotonic convergence theorem, 25 Monotonic sequences, 24

R Radius of convergence, 300 Ratio test, 15, 54 Rearrangement, 64 Refinement, 178 Riemann sum, 181 Right differentiability, 125 differentiable, 126 limit, 88 Rolle’s theorem, 138 Root function, 111 Root test, 54

N Natural logarithm, 117 Natural logarithm function, 118 Neighbourhood, 72 Newton-Leibnitz formula, 200 Nilakantha, 60, 61 Norm, 192

O Odd function, 321 Open set, 43

P Partial sum, 44, 293 Partition, 175 Periodic, 314 Piecewise continuous, 195 Piecewise differentiable, 317

Q Quotient rules, 129

S Sandwich theorem, 12, 84 Second derivative, 153 Sequence, 1 alternating, 11 strictly decreasing, 24 strictly increasing, 24 unbounded, 21 Series, 44 of functions, 292 Series of functions, 292 Signum function, 127 Sine series expansion, 324, 326 Slope of the tangent, 120 Stationary points, 136 Strictly decreasing, 110 Strictly increasing, 110 Strictly monotonic, 110 Subsequence, 32 Sum, 81 Supremum, 26 Surface of revolution, 230, 231

340 T Tags, 181 Taylor series, 159, 307 Taylor’s formula, 156, 157 Taylor’s polynomial, 158 Terminal point, 215 Ternary expansion, 50 Trigonometric polynomial, 314 Trigonometric series, 313 Twice differentiable, 153 U Unbounded, 26 function, 78 on a subset, 78 Unbounded set, 104 Uniform continuity, 193 Uniform continuous, 193 Uniform convergence, 278

Index Uniformly bounded eventually , 284 Upper bound, 25 Upper integral, 179 Upper sum, 176

V Volume of a solid, 226 of revolution, 227 Volume of the solid of revolution, 227, 228

W Weierstrass M-test, 294

Y Yuktibhasha, 203