A First Course in Mathematical Analysis

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A First Course in Mathematical Analysis

Mathematical Analysis (often called Advanced Calculus) is generally found by students to be one of their hardest courses in Mathematics. This text uses the so-called sequential approach to continuity, differentiability and integration to make it easier to understand the subject. Topics that are generally glossed over in the standard Calculus courses are given careful study here. For example, what exactly is a ‘continuous’ function? And how exactly can one give a careful definition of ‘integral’? This latter is often one of the mysterious points in a Calculus course – and it is quite tricky to give a rigorous treatment of integration! The text has a large number of diagrams and helpful margin notes, and uses many graded examples and exercises, often with complete solutions, to guide students through the tricky points. It is suitable for self study or use in parallel with a standard university course on the subject.

A First Course in Mathematical Analysis DAVID ALEXANDER BRANNAN Published in association with The Open University

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521864398 © The Open University 2006 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2006 eBook (EBL) ISBN-13 978-0-511-34857-0 ISBN-10 0-511-34857-6 eBook (EBL) ISBN-13 ISBN-10

hardback 978-0-521-86439-8 hardback 0-521-86439-9

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

To my wife Margaret and my sons David, Joseph and Michael

Contents

Preface

page ix

1

Numbers 1.1 Real numbers 1.2 Inequalities 1.3 Proving inequalities 1.4 Least upper bounds and greatest lower bounds 1.5 Manipulating real numbers 1.6 Exercises

1 2 9 14 22 30 35

2

Sequences 2.1 Introducing sequences 2.2 Null sequences 2.3 Convergent sequences 2.4 Divergent sequences 2.5 The Monotone Convergence Theorem 2.6 Exercises

37 38 43 52 61 68 79

3

Series 3.1 Introducing series 3.2 Series with non-negative terms 3.3 Series with positive and negative terms 3.4 The exponential function x 7! ex 3.5 Exercises

83 84 92 103 122 127

4

Continuity 4.1 Continuous functions 4.2 Properties of continuous functions 4.3 Inverse functions 4.4 Defining exponential functions 4.5 Exercises

130 131 143 151 161 164

5

Limits and continuity 5.1 Limits of functions 5.2 Asymptotic behaviour of functions 5.3 Limits of functions – using " and  5.4 Continuity – using " and  5.5 Uniform continuity 5.6 Exercises

167 168 176 181 193 200 203

vii

Contents

viii 6

Differentiation 6.1 Differentiable functions 6.2 Rules for differentiation 6.3 Rolle’s Theorem 6.4 The Mean Value Theorem 6.5 L’Hoˆpital’s Rule 6.6 The Blancmange function 6.7 Exercises

205 206 216 228 232 238 244 252

7

Integration 7.1 The Riemann integral 7.2 Properties of integrals 7.3 Fundamental Theorem of Calculus 7.4 Inequalities for integrals and their applications 7.5 Stirling’s Formula for n! 7.6 Exercises

255 256 272 282 288 303 309

8

Power series 8.1 Taylor polynomials 8.2 Taylor’s Theorem 8.3 Convergence of power series 8.4 Manipulating power series 8.5 Numerical estimates for p 8.6 Exercises

313 314 320 329 338 346 350

Appendix 1

Sets, functions and proofs

354

Appendix 2

Standard derivatives and primitives

359

Appendix 3

The first 1000 decimal places of

pffiffiffi 2, e and p

361

Appendix 4 Solutions to the problems Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8

363 363 371 382 393 402 413 426 443

Index

457

Preface

Analysis is a central topic in Mathematics, many of whose branches use key analytic tools. Analysis also has important applications in Applied Mathematics, Physics and Engineering, where a good appreciation of the underlying ideas of Analysis is necessary for a modern graduate. Changes in the school curriculum over the last few decades have resulted in many students finding Analysis very difficult. The author believes that Analysis nowadays has an unjustified reputation for being hard, caused by the traditional university approach of providing students with a highly polished exposition in lectures and associated textbooks that make it impossible for the average learner to grasp the core ideas. Many students end up agreeing with the German poet and philosopher Goethe who wrote that ‘Mathematicians are like Frenchmen: whatever you say to them, they translate into their own language, and forthwith it is something entirely different!’ Since 1971, the Open University in United Kingdom has taught Mathematics to students in their own homes via specially written correspondence texts, and has traditionally given Analysis a central position in its curriculum. Its philosophy is to provide clear and complete explanations of topics, and to teach these in a way that students can understand without much external help. As a result, students should be able to learn, and to enjoy learning, the key concepts of the subject in an uncluttered way. This book arises from correspondence texts for its course Introduction to Pure Mathematics, that has now been studied successfully by over ten thousand students. This book is therefore different from most Mathematics textbooks! It adopts a student-friendly approach, being designed for study by a student on their own OR in parallel with a course that uses as set text either this text or another text. But this is the text that the student is likely to use to learn the subject from. The author hopes that readers will gain enormous pleasure from the subject’s beauty and that this will encourage them to undertake further study of Mathematics! Once a student has grasped the principal notions of limit and continuous function in terms of inequalities involving the three symbols ", X and , they will quickly understand the unity of areas of Analysis such as limits, continuity, differentiability and integrability. Then they will thoroughly enjoy the beauty of some of the arguments used to prove key theorems – whether their proofs are short or long. Calculus is the initial study of limits, continuity, differentiation and integration, where functions are assumed to be well-behaved. Thus all functions continuous on an interval are assumed to be differentiable at most points in the interval, and so on. However, Mathematics is not that simple! For example, there exist functions that are continuous everywhere on R, but differentiable

Johann Wolfgang von Goethe (1749–1832) is said to have studied all areas of science of his day except mathematics – for which he had no aptitude.

ix

Preface

x nowhere on R; this discovery by Karl Weierstrass in 1872 caused a sensation in the mathematical community. In Analysis (sometimes called Advanced Calculus) we make no assumptions about the behaviour of functions – and the result is that we sometimes come across real surprises! The book has two principal features in its approach that make it stand out from among other Analysis texts. Firstly, this book uses the ‘sequential approach’ to Analysis. All too often students starting on the subject find that they cannot grasp the significance of both " and  simultaneously. This means that the whole underlying idea about what is happening is lost, and the student takes a very long time to master the topic – or, in many cases in fact, never masters the topic and acquires a strong dislike of it. In the sequential approach they proceed at a more leisurely pace to understand the notion of limit using " and X – to handle convergent sequences – before coming across the other symbol , used in conjunction with " to handle continuous functions. This approach avoids the conventional student horror at the perceived ‘difficulty of Analysis’. Also, it avoids the necessity to re-prove broadly similar results in a range of settings – for example, results on the sum of two sequences, of two series, of two continuous functions and of two differentiable functions. Secondly, this book makes great efforts to teach the "– approach too. After students have had a first pass at convergence of sequences and series and at continuity using ‘the sequential approach’, they then meet ‘the "– approach’, explained carefully and motivated by a clear ‘"– game’ discussion. This makes the new approach seem very natural, and this is motivated by using each approach in later work in the appropriate situation. By the end of the book, students should have a good facility at using both the sequential approach and the "– approach to proofs in Analysis, and should be better prepared for later study of Analysis than students who have acquired only a weak understanding of the conventional approach.

Outline of the content of the book In Chapter 1, we define real numbers to be decimals. Rather than give a heavy discussion of least upper bound and greatest lower bound, we give an introduction to these matters sufficient for our purposes, and the full discussion is postponed to Chapter 7, where it is more timely. We also study inequalities, and their properties and proofs. In Chapter 2, we define convergent sequences and examine their properties, basing the discussion on the notion of null sequence, which simplifies matters considerably. We also look at divergent sequences, sequences defined by recurrence formulas and particular sequences which converge to p and e. In Chapter 3, we define convergent infinite series, and establish a number of tests for determining whether a given series is convergent or divergent. We demonstrate the equivalence of the two definitions of the exponential function x 7! ex , and prove that the number e is irrational. In Chapter 4, we define carefully what we mean by a continuous function, in terms of sequences, and establish the key properties of continuous functions. We also give a rigorous definition of the exponential function x 7! ax .

 n We define ex as lim 1 þ nx n!1 1 P xn and as n! . n¼0

Preface In Chapter 5, we define the limit of a function as x tends to c or as x tends to 1 in terms of the convergence of sequences. Then we introduce the "– definitions of limit and continuity, and check that these are equivalent to the earlier definitions in terms of sequences. We also look briefly at uniform continuity. In Chapter 6, we define what we mean by a differentiable function, using difference quotients Q(h); this enables us to use our earlier results on limits to prove corresponding results for differentiable functions. We establish some interesting properties of differentiable functions. Finally, we construct the Blancmange function that is continuous everywhere on R, but differentiable nowhere on R. In Chapter 7, we give a careful definition of what we mean by an integrable function, and establish a number of related criteria for establishing whether a given function is integrable or not. Our integral is the so-called Riemann integral, defined in terms of upper and lower Riemann sums. We check the standard properties of integrals and verify a number of standard approaches for calculating definite integrals. Then we give a number of applications of integrals to limits of certain sequences and series and prove Stirling’s Formula. Finally, in Chapter 8, we study the convergence and properties of power series. The chapter ends with a marvellous proof of the irrationality of the number p that uses a whole range of the techniques that have been met in the previous chapters. For completeness and for students’ convenience, we give a brief guide to our notation for sets and functions, together with a brief indication of the logic involved in proofs in Mathematics (in particular, the Principle of Mathematical Induction) in Appendix 1. Appendix 2 contains a list of standard derivatives and primitives and Appendix 3 the first 1000 decimal places in the values of the pffiffiffi numbers 2, p and e. Appendix 4 contains full solutions to all the problems set during each chapter. Solutions are not given to the exercises at the end of each chapter, however. Lecturers/instructors may wish to use these exercises in homework assignments.

Study guide This book assumes that students have a fair understanding of Calculus. The assumptions on technical background are deliberately kept slight, however, so that students can concentrate on the newer aspects of the subject ‘Analysis’. Most students will have met some of the material in the early chapters previously. Although this means that they can therefore proceed fairly quickly through some sections, it does NOT mean that those sections can be ignored – each section contains important ideas that are used later on and most include something new or have a different emphasis. Most chapters are divided into five or six sections (each often further divided into sub-sections); sections are numbered using two digits (such as ‘Section 3.2’) and sub-sections using three digits (such as ‘Sub-section 3.2.4’). Generally a section is considered to be about one evening’s work for an average student. Chapter 7 on Integration is arguably the highlight of the book. However, it contains some rather complicated mathematical arguments and proofs.

xi

Stirling’s Formula says that, for large pffiffiffiffiffiffiffi ffi nn, n! is ‘roughly’ 2pn ne , in a sense that we explain.

xii Therefore, when reading Chapter 7, it is important not to get bogged down in details, but to keep progressing through the key ideas, and to return later on to reading the things that were left out at the earlier reading. Most students will require three or four passes at this chapter before having a good idea of most of it. We use wide pages with a large number of margin notes in which we place teaching comments and some diagrams to aid in the understanding of particular points in arguments. We also provide advice on which proofs to omit on a first study of the topic; it is important for the student NOT to get bogged down in a technical discussion or a proof until they have a good idea of the message contained in the result and the situations in which it can be used. Therefore clear encouragement is given on which portions of the text to leave till later, or to simply skim on a first reading. The end of the proof of a Theorem is indicated by a solid symbol ‘&’ and the end of the solution of a worked Example by a hollow symbol ‘&’. There are many worked examples within the text to explain the concepts being taught, together with a good stock of problems to reinforce the teaching. The solutions are a key part of the teaching, and tackling them on your own and then reading our version of the solution is a key part of the learning process. No one can learn Mathematics by simply reading – it is a ‘hands on’ activity. The reader should not be afraid to draw pictures to illustrate what seems to be happening to a sequence or a function, to get a feeling for their behaviour. A wise old man once said that ‘A picture is worth a thousand words!’. A good picture may even suggest a method of proof. However, at the same time it is important not to regard a picture on its own as a proof of anything; it may illustrate just one situation that can arise and miss many other possibilities! It is important NOT to become discouraged if a topic seems difficult. It took mathematicians hundreds of years to develop Analysis to its current polished state, so it may take the reader a few hours at several sittings to really grasp the more complex or subtle ideas.

Acknowledgements The material in the Open University course on which this book is based was contributed to in some way by many colleagues, including Phil Rippon, Robin Wilson, Andrew Brown, Hossein Zand, Joan Aldous, Ian Harrison, Alan Best, Alison Cadle and Roberta Cheriyan. Its eventual appearance in book form owes much to Lynne Barber. Without the forebearance of my family, the writing of the book would have been impossible.

Preface

It is important to read the margin notes! This signposting benefits students greatly in the author’s experience.

Tackling the problems is a good use of your time, not something to skimp.

1

Numbers

In this book we study the properties of real functions defined on intervals of the real line (possibly the whole real line) and whose image also lies on the real line. In other words, they map R into R. Our work will be from a very precise point of view in order to establish many of the properties of such functions which seem intuitively obvious; in the process we will discover that some apparently true properties are in fact not necessarily true! The types of functions that we shall examine include: exponential functions, such as x 7! ax ; where a; x 2 R; trigonometric functions, such as x 7! sin x; where x 2 R; pffiffiffi root functions, such as x 7! x; where x  0: The types of behaviour that we shall examine include continuity, differentiability and integrability – and we shall discover that functions with these properties can be used in a number of surprising applications. However, to put our study of such functions on a secure foundation, we need first to clarify our ideas of the real numbers themselves and their properties. In particular, we need to devote some time to the manipulation of inequalities, which play a key role throughout the book. In Section 1.1, we start by revising the properties of rational numbers and their decimal representation. Then we introduce the real numbers as infinite decimals, and describe the difficulties involved in doing arithmetic with such decimals. In Section 1.2, we revise the rules for manipulating inequalities and show how to find the solution set of an inequality involving a real number, x, by applying the rules. We also explain how to deal with inequalities which involve modulus signs. In Section 1.3, we describe various techniques for proving inequalities, including the very important technique of Mathematical Induction. The concept of a least upper bound, which is of great importance in Analysis, is introduced in Section 1.4, and we discuss the Least Upper Bound Property of R. Finally, in Section 1.5, we describe how least upper bounds can be used to define arithmetical operations in R. Even though you may be familiar with much of this material we recommend that you read through it, as we give the system of real numbers a more careful treatment than you may have met before. The material on inequalities and least upper bounds is particularly important for later on. In later chapters we shall define exactly what the numbers p and e are, and find various ways of calculating them. But, first, we examine numbers in general.

For example,pwhat exactly is ffiffiffi the number 2?

You may omit this section at a first reading.

1

1: Numbers

2

1.1

Real numbers

We start our study of the real numbers with the rational numbers, and investigate their decimal representations, then we proceed to the irrational numbers.

1.1.1

Rational numbers

We assume that you are familiar with the set of natural numbers N ¼ f1; 2; 3; . . .g;

Note that 0 is not a natural number.

and with the set of integers Z ¼ f. . .; 2; 1; 0; 1; 2; 3; . . .g: The set of rational numbers consists of all fractions (or ratios of integers)   p : p 2 Z; q 2 N : Q ¼ q Remember that each rational number has many different representations as a ratio of integers; for example 1 2 10 ¼ ¼ ¼ . . .: 3 6 30 We also assume that you are familiar with the usual arithmetical operations of addition, subtraction, multiplication and division of rational numbers. It is often convenient to represent rational numbers geometrically as points on a number line. We begin by drawing a line and marking on it points corresponding to the integers 0 and 1. If the distance between 0 and 1 is taken as a unit of length, then the rationals can be arranged on the line with positive rationals to the right of 0 and negative rationals to the left. negative –3

– 52

–2

positive –1

0

1 2

1

3 2

2

3

For example, the rational 32 is placed at the point which is one-half of the way from 0 to 3. This means that rationals have a natural order on the number-line. For 7 example, 19 22 lies to the left of 8 because 19 76 ¼ 22 88

and

7 77 ¼ : 8 88

If a lies to the left of b on the number-line, then a is less than b or b is greater than a; and we write a a:

For example 19 7 7 19 < or > : 22 8 8 22 We write a  b, or b  a, if either a < b or a ¼ b.

1.1

Real numbers Problem 1

3

Arrange the following rational numbers in order: 17 45 45 0; 1; 1; 17 20 ; 20 ; 53 ; 53:

Problem 2 Show that between any two distinct rational numbers there is another rational number.

1.1.2

Decimal representation of rational numbers

The decimal system enables us to represent all the natural numbers using only the ten integers 0; 1; 2; 3; 4; 5; 6; 7; 8 and 9; which are called digits. We now remind you of the basic facts about the representation of rational numbers by decimals. Definition

A decimal is an expression of the form a0  a1 a2 a3 . . .;

where a0 is a non-negative integer and a1, a2, a3, . . . are digits. If only a finite number of the digits a1, a2, a3, . . . are non-zero, then the decimal is called terminating or finite, and we usually omit the tail of 0s. Terminating decimals are used to represent rational numbers in the following way  a1 a2 an  a0  a1 a2 a3 . . . an ¼  a0 þ 1 þ 2 þ    þ n : 10 10 10 It can be shown that any fraction whose denominator contains only powers of 2 and/or 5 (such as 20 ¼ 22  5) can be represented by such a terminating decimal, which can be found by long division. However, if we apply long division to many other rationals, then the process of long division never terminates and we obtain a non-terminating or infinite decimal. For example, applying long division to 13 gives 0.333 . . ., and for 19 22 we obtain 0.86363 . . .. Problem 3 decimals.

Apply long division to 17 and

2 13

For example 0:8500 . . .; 13:1212 . . .; 1:111 . . .: For example, 0:8500 . . . ¼ 0:85: For example 8 5 0:85 ¼ 0 þ 1 þ 2 10 10 85 17 ¼ : ¼ 100 20

to find the corresponding

These non-terminating decimals, which are obtained by applying the long division process, have a certain common property. All of them are recurring; that is, they have a recurring block of digits, and so can be written in shorthand form, as follows: 0:333 . . .

¼ 0:3;

0:142857142857 . . . ¼ 0:142857 . . .; 0:86363 . . .

¼ 0:863:

It is not hard to show, whenever we apply the long division process to a fraction pq , that the resulting decimal is recurring. To see why we notice that there are only q possible remainders at each stage of the division, and so one of these remainders must eventually recur. If the remainder 0 occurs, then the resulting decimal is, of course, terminating; that is, it ends in recurring 0s.

Another commonly used notation is : : : 0:3 or 0:142857:

1: Numbers

4 Non-terminating recurring decimals which arise from the long division of fractions are used to represent the corresponding rational numbers. This representation is not quite so straight-forward as for terminating decimals, however. For example, the statement 1 3 3 3 ¼ 0:3 ¼ 1 þ 2 þ 3 þ    3 10 10 10 can be made precise only when we have introduced the idea of the sum of a convergent infinite series. For the moment, when we write the statement 13 ¼ 0:3 we mean simply that the decimal 0:3 arises from 13 by the long division process. The following example illustrates one way of finding the rational number with a given decimal representation.

We return to this topic in Chapter 3.

Example 1 Find the rational number (expressed as a fraction) whose decimal representation is 0:863: Solution First we find the fraction x such that x ¼ 0:63: If we multiply both sides of this equation by 102 (because the recurring block has length 2), then we obtain 100 x ¼ 63:63 ¼ 63 þ x: Hence 99x ¼ 63 ) x ¼

63 7 ¼ : 99 11

Thus 0:863 ¼

8 x 8 7 95 19 þ ¼ þ ¼ ¼ : 10 10 10 110 110 22

&

The key idea in the above solution is that multiplication of a decimal by 10k moves the decimal point k places to the right. Problem 4 Using the above method, find the fractions whose decimal representations are: (a) 0:231;

(b) 2:281.

The decimal representation of rational numbers has the advantage that it enables us to decide immediately which of two distinct positive rational numbers is the greater. We need only examine their decimal representations and notice the first place at which the digits differ. For example, to order 78 and 19 22 we write 7 ¼ 0:875 . . . 8

19 ¼ 0:86363 . . .; 22

and

and so #

#

0:86363 . . . < 0:875 )

19 7 < : 22 8

Problem 5 Find the first two digits after the decimal point in the deci45 mal representations of 17 20 and 53, and hence determine which of these two rational numbers is the greater. Warning Decimals which end in recurring 9s sometimes arise as alternative representations for terminating decimals. For example 1 ¼ 0:9 ¼ 0:999 . . .

and

1:35 ¼ 1:349 ¼ 1:34999 . . .:

Whenever possible, we avoid using the form of a decimal which ends in recurring 9s.

1.1

Real numbers

5

You may find this rather alarming, but it is important to realise that this is a matter of convention. We wish to allow the decimal 0.999 . . . to represent a number x, so x must be less than or equal to 1 and greater than each of the numbers 0:9; 0:99; 0:999; . . .: The only rational with these properties is 1.

1.1.3

Irrational numbers

One of the surprising mathematical discoveries made by the Ancient Greeks was that the system of rational numbers is not adequate to describe all the magnitudes that occur in geometry. For example, consider the diagonal of a square of side 1. What is its length? If the length is x, then, by Pythagoras’ Theorem, x must satisfy the equation x2 ¼ 2. However, there is no rational number which satisfies this equation. Theorem 1

x

1

x2 ¼ 12 þ 12 ¼ 2

There is no rational number x such that x2 ¼ 2.

Proof Suppose that such a rational number x exists. Then we can write x ¼ pq. By cancelling, if necessary, we may assume that p and q have no common factor. The equation x2 ¼ 2 now becomes p2 ¼ 2; q2

so

2

ð2rÞ ¼ 2q ;

This is a proof by contradiction.

p2 ¼ 2q2 :

Now, the square of an odd number is odd, and so p cannot be odd. Hence p is even, and so we can write p ¼ 2r, say. Our equation now becomes 2

1

so

2

2

q ¼ 2r :

For ð2k þ 1Þ2 ¼ 4k2 þ 4k þ 1 ¼ 4ðk2 þ kÞ þ 1:

Reasoning as before, we find that q is also even. Since p and q are both even, they have a common factor 2, which contradicts our earlier statement that p and q have no common factors. Arguing from our original assumption that x exists, we have obtained two contradictory statements. Thus, our original assumption must be false. In other & words, no such x exists. Problem 6 By imitating the above proof, show that there is no rational number x such that x3 ¼ 2. Since we want equations such as x2 ¼ 2 and x3 ¼ 2 to have solutions, we must introduce new We denote pffiffiffinumbers pffiffiffi which are not rational ffi2 ffiffiffi3 these pffiffinumbers. p new numbers by 2 and 3 2, respectively; thus 2 ¼ 2 and p3ffiffi2ffi p¼ ffiffiffiffiffi2. Of course, we must introduce many other new numbers, such as 3, 5 11, and m so on. Indeed, it can be shown ffiffiffi that, if m, n are natural numbers and x ¼ n has p m no integer solution, then n cannot be rational. A number which is not rational is called irrational. There are many other mathematical quantities which cannot be described exactly by rational numbers. For example, the number p which denotes the area of a disc of radius 1 (or half the length of the perimeter of such a disc) is irrational, as is the number e.

The case m ¼ 2 is treated in Exercise 5 for this section in Section 1.6.

Lambert proved that p is irrational in 1770.

1: Numbers

6 pffiffiffi It is natural to ask whether irrational numbers, such as 2 and p, can be represented as decimals. Using your calculator, you can check that (1.41421356)2 is p very approffiffiffi close to 2, and so 1.41421356 is a very pffiffigood ffi ximate value for 2. But is there a decimal which represents 2 exactly? If such a decimal exists, then it cannot be recurring, because all the recurring decimals correspond to rational numbers. In fact, it is possible to represent all the irrational numbers mentioned so far by non-recurring decimals. For example, there are non-recurring decimals such that pffiffiffi 2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .: It is also natural to ask whether non-recurring decimals, such as 0:101001000100001 . . .

and

In fact ð1:41421356Þ2 ¼1:9999999932878736:

pffiffiffi We prove that 2 has a decimal representation in Section 1.5.

0:123456789101112 . . .;

represent irrational numbers. In fact, a decimal corresponds to a rational number if and only if it is recurring; so a non-recurring decimal must correspond to an irrational number. We may summarise this as: recurring decimal , rational number non-recurring decimal , irrational number

1.1.4

The real number system

Taken together, the rational numbers (recurring decimals) and irrational numbers (non-recurring decimals) form the set of real numbers, denoted by R. As with rational numbers, we can determine which of two real numbers is greater by comparing their decimals and noticing the first pair of corresponding digits which differ. For example #

#

0:10100100010000 . . . < 0:123456789101112 . . .: We now associate with each irrational number a point on the number-line. For example, the irrational number x ¼ 0.123456789101112 . . . satisfies each of the inequalities 0:1 < x < 0:2 0:12 < x < 0:13 0:123 < x < 0:124 .. . We assume that there is a point on the number-line corresponding to x, which lies to the right of each of the (rational) numbers 0.1, 0.12, 0.123 . . ., and to the left of each of the (rational) numbers 0.2, 0.13, 0.124, . . ..

As usual, negative real numbers correspond to points lying to the left of 0; and the ‘number-line’, complete with both rational and irrational points, is called the real line.

When comparing decimals in this way, we do not allow either decimal to end in recurring 9s.

1.1

Real numbers

7

There is thus a one–one correspondence between the points on the real line and the set R of real numbers (or decimals). We now state several properties of R, with which you will be already familiar, although you may not have met their names before. These properties are used frequently in Analysis, and we do not always refer to them explicitly by name. Order Properties of R 1. Trichotomy Property If a, b 2 R, then exactly one of the following inequalities holds a < b or a ¼ b or a > b: 2. Transitive Property a a:

b < c ) a < c: If a 2 R, then there is a positive integer n

4. Density Property If a, b 2 R and a < b, then there is a rational number x and an irrational number y such that a < x < b and a < y < b:

Remark The Archimedean Property is sometimes expressed in the following equivalent way: for any positive real number a, there is a positive integer n such that 1n < a. The following example illustrates how we can prove the Density Property. Example 2

Find a rational number x and an irrational number y satisfying a < x < b and a < y < b;

where a ¼ 0:123 and b ¼ 0:12345 . . .. Solution

The two decimals #

a ¼ 0:1233 . . .

#

and

b ¼ 0:12345 . . .

differ first at the fourth digit. If we truncate b after this digit, we obtain the rational number x ¼ 0.1234, which satisfies the requirement that a < x < b. To find an irrational number y between a and b, we attach to x a (sufficiently small) non-recurring tail such as 010010001 . . . to give y ¼ 0.1234j010010001 . . .. It is then clear that y is irrational (because its decimal is non-recurring) and that & a < y < b. Problem 7 Find a rational number x and an irrational number y such that a < x < b and a < y < b, where a ¼ 0:3 and b ¼ 0:3401:

The first three of these properties are almost selfevident, but the Density Property is not so obvious.

1: Numbers

8 Theorem 2 Density Property of R If a, b 2 R and a < b, then there is a rational number x and an irrational number y such that a < x < b and

a < y < b:

You may omit the following proof on a first reading.

Proof For simplicity, we assume that a, b  0. So, let a and b have decimal representations a ¼ a0  a1 a 2 a3 . . .

and

b ¼ b0  b1 b2 b3 . . .;

where we arrange that a does not end in recurring 9s, whereas b does not terminate (this latter can be arranged by replacing a terminating representation by an equivalent representation that ends in recurring 9s). Since a < b, there must be some integer n such that

Here a0, b0 are non-negative integers, and a1, b1, a2, b2, . . . are digits.

a0 ¼ b0 ; a1 ¼ b1 ; . . .; an1 ¼ bn1 ; but an < bn : Then x ¼ a0  a1a2a3 . . . an 1bn is rational, and a < x < b as required. Finally, since x < b, it follows that we can attach a sufficiently small non& recurring tail to x to obtain an irrational number y for which a < y < b.

Remark One consequence of the Density Property is that between any two real numbers there are infinitely many rational numbers and infinitely many irrational numbers. Problem 8 Prove that between any two real numbers a and b there are at least two distinct rational numbers.

1.1.5

Arithmetic in R

We can do arithmetic with recurring decimals by first converting the decimals to fractions. However, it is not obvious how to perform arithmetical operations with non-recurring decimals. pffiffiffi Assuming that we can represent 2 and p by the non-recurring decimals pffiffiffi 2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .; pffiffiffi pffiffiffi can we also represent the sum 2 þ p and the product 2  p as decimals? Indeed, what do we mean by the operations of addition and multiplication when non-recurring decimals (irrationals) are involved, and do these operations satisfy the same properties as addition and multiplication of rationals? It would take many pages to answer these questions fully. Therefore, we shall assume that it is possible to define all the usual arithmetical operations with decimals, and that they do satisfy the usual properties. For definiteness, we now list these properties.

A proof of the previous remark would involve ideas similar to those involved in tackling Problem 8.

1.2

Inequalities

9

Arithmetic in R Addition A1 If a, b 2 R, then a þ b 2 R.

Multiplication M1 If a, b 2 R, then a  b 2 R.

A2

If a 2 R, then a þ 0 ¼ 0 þ a ¼ a.

M2

If a 2 R, then a  1 ¼ 1  a ¼ a.

A3

If a 2 R, then there is a number a 2 R such that a þ (a) ¼ ( a) þ a ¼ 0.

M3

If a 2 R  {0}, then there is a number a1 2 R such that a  a1 ¼ a1  a ¼ 1.

A4

If a, b, c 2 R, then (a þ b) þ c ¼ a þ (b þ c). If a, b 2 R, then a þ b ¼ b þ a.

M4

If a, b, c 2 R, then (a  b)  c ¼ a  (b  c). If a, b 2 R, then a  b ¼ b  a.

A5 D

M5

If a, b, c 2 R, then a  (b þ c) ¼ a  b þ a  c.

CLOSURE

IDENTITY

INVERSES

ASSOCIATIVITY

COMMUTATIVITY

DISTRIBUTATIVITY

To summarise the contents of this table: Properties A1–A5



R is an Abelian group under the operation of addition þ; R  {0} is an Abelian group under the operation of multiplication ;



These two group structures are linked by the Distributive Property.

Property D



It follows from the above properties that we can perform addition, subtraction (where a  b ¼ a þ (b)), multiplication and division (where ab ¼ a  b1) in R, and that these operations satisfy all the usual properties. Furthermore, we shall assume that the set R contains the nth roots and rational powers of positive real numbers, with their usual properties. In Section 1.5 we describe one way of justifying the existence of nth roots.

1.2

Properties M1–M5

Any system satisfying the properties listed in the table is called a field. Both Q and R are fields.

Inequalities

Much of Analysis is concerned with inequalities of various kinds; the aim of this section and the next section is to provide practice in the manipulation of inequalities.

1.2.1

Rearranging inequalities

The fundamental rule, upon which much manipulation of inequalities is based, is that the statement a < b means exactly the same as the statement b  a > 0. This fact can be stated concisely in the following way: Rule 1

For any a, b 2 R,

a < b , b  a > 0.

Put another way, the inequalities a < b and b  a > 0 are equivalent. There are several other standard rules for rearranging a given inequality into an equivalent form. Each of these can be deduced from our first rule above. For

Recall that the symbol ‘,’ means ‘if and only if’ or ‘implies and is implied by’.

1: Numbers

10 example, we obtain an equivalent inequality by adding the same expression to both sides. Rule 2

For any a, b, c 2 R,

a < b , a þ c < b þ c.

Another way to rearrange an inequality is to multiply both sides by a non-zero expression, making sure to reverse the inequality if the expression is negative. For example

Rule 3  

For any a, b 2 R and any c > 0, For any a, b 2 R and any c < 0,

a < b , ac < bc; a < b , ac > bc.

253 , 20530 ðc ¼ 10Þ; 253 , 20 > 30 ðc ¼ 10Þ:

Sometimes the most effective way to rearrange an inequality is to take reciprocals. However, in this case, both sides of the inequality should be positive, and the direction of the inequality has to be reversed. For example

Rule 4 (Reciprocal Rule) For any positive a, b 2 R, a5b , 1a > 1b : Some inequalities can be simplified only by taking powers. However, in order to do this, both sides must be non-negative and must be raised to a positive power. Rule 5 (Power Rule) For any non-negative a, b 2 R, and any p > 0,

1 2 < 4 , ð¼ 0:5Þ 2 1 > ð¼ 0:25Þ: 4 For example

a < b , a p < b p.

1

4 < 9 , 42 ð¼ 2Þ 1

For positive integers p, Rule 5 follows from the identity   bp  ap ¼ ðb  aÞ bp1 þ bp2 a þ    þ bap2 þ ap1 ; thus, since the right-hand bracket is positive, we have

< 92 ð¼ 3Þ: We shall discuss the meaning of non-integer powers in Section 1.5.

b  a > 0 , bp  ap > 0; which is equivalent to our desired result.

Remark There are corresponding versions of Rules 1–5 in which the strict inequality a < b is replaced by the weak inequality a  b.

For example, b3  a3 ¼ ðb  aÞ  2  b þ ba þ a2 :

Problem 1 State (without proof) the versions of Rules 1–5 for weak inequalities. We shall give one more rule for rearranging inequalities in Sub-section 1.2.3.

1.2.2

Solving inequalities

Solving an inequality involving an unknown real number x means determining those values of x for which the given inequality holds; that is, finding the solution set of the inequality. We can often do this by rewriting the inequality in an equivalent, but simpler form, using the rules given in the last sub-section.

The solution set is the set of those values of x for which the inequality holds.

1.2

Inequalities

11

In this activity we frequently use the usual rules for the sign of a product, and the fact that the square of any real number is non-negative. Also, we need to remember the difference between the logical statements: ‘implies’, ‘is implied by’ and ‘implies and is implied by’. Example 1



þ



þ

þ







þ

x3 Solve the inequality xþ2 xþ4 > 2x1 :

Solution We rearrange this inequality to give a somewhat simpler inequality, using Rule 1 xþ2 x3 xþ2 x3 > ,  >0 x þ 4 2x  1 x þ 4 2x  1 x2 þ 2x þ 10 >0 , ðx þ 4Þð2x  1Þ ,

It is a common strategy to bring all terms to one side. We bring everything to a common denominator.

ð x þ 1Þ 2 þ 9 > 0: ðx þ 4Þð2x  1Þ

Here we complete the square in the numerator, since we cannot factorise it.

Now, the numerator is always positive. The denominator vanishes when x ¼ 4 or x ¼ 12. By examining separately the sign of the denominator when x < 4, 4 < x < 12 and x > 12, we can deduce that the last fraction is positive precisely when x < 4 or x > 12. Hence the solution set of the original inequality is     xþ2 x3 > x: & ¼ ð1; 4Þ [ 12 ; 1 : x þ 4 2x  1

Example 2 Solution

Solve the inequality 2x21þ 2 < 14 : Since 2x2 þ 2 > 0, we have 1 1 < , 2x2 þ 2 > 4 þ2 4 , x2 þ 1 > 2

2x2

2

,x 1>0 , ðx  1Þðx þ 1Þ > 0:

ðby Rule 4Þ ðby Rule 3Þ ðby Rule 1Þ

This last inequality holds precisely when x < 1 or x > 1. It follows that the solution set of the original inequality is   1 1 x: 2 < ¼ ð1; 1Þ [ ð1; 1Þ: & 2x þ 2 4 Problem 2 Use each of the following expressions to write down an inequality with the given expression on its left-hand side which is equivalent to the inequality x > 2: (a) x þ 3;

Problem 3 (a)

This is because the final displayed inequality is equivalent to the inequality we are solving. The logical implication symbols between the displayed inequalities were all ‘implies and is implied by’.

4xx2 7 x2 1

(b) 2  x;

(c) 5x þ 2;

(d)

Solve the following inequalities:  3;

(b) 2x2  ðx þ 1Þ2 :

1 5xþ2.

Here we factorise the lefthand side of the inequality to examine the signs of its factors.

1: Numbers

12 Warning Great care is needed when solving inequalities which involve rational powers. In particular, when applying Rule 5 both sides of the inequality must be non-negative. pffiffiffiffiffiffiffiffiffiffiffiffiffi Example 3 Solve the inequality 2x þ 3 > x. pffiffiffiffiffiffiffiffiffiffiffiffiffi Solution The expression 2x þ 3 is defined only when 2x þ  3  0; that is, when x   32. Hence we need only consider those x in 3 , 1 . 2 We can obtain an equivalent inequality by squaring, provided that both pffiffiffiffiffiffiffiffiffiffiffiffiffi 2x þ 3 and x are non-negative. Thus, for x  0, we obtain pffiffiffiffiffiffiffiffiffiffiffiffiffi 2x þ 3 > x , 2x þ 3 > x2 ðby Rule 5; with p ¼ 2Þ

Note that, for the moment, we are examining only those x for which x  0.

, x2  2x  3 < 0 , ðx  3Þðx þ 1Þ < 0: So the part of the solution set in [0, 1) is [0, 3). pffiffiffiffiffiffiffiffiffiffiffiffiffi We now examine those x for which 32  x < 0. For such x, 2x þ 3  0 pffiffiffiffiffiffiffiffiffiffiffiffiffi and x < 0, so that 2x þ 3 ð 0Þ > x, for all such x. It follows that all these x, namely the set ½ 32 , 0Þ, belong to the solution set too. Combining these results, the solution set of the original inequality is

pffiffiffiffiffiffiffiffiffiffiffiffiffi  3 x : 2x þ 3 > x ¼  ; 0 [ ½0; 3Þ 2

3 & ¼  ;3 : 2 Problem 4

1.2.3

Solve the inequality

Notice the use of the Transitive Property here.

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x2  2 > x:

Inequalities involving modulus signs

We now turn our attention to inequalities involving the modulus, or absolute value, of a real number. Recall that, if a 2 R, then its modulus jaj is defined by  a; if a  0, j aj ¼ a; if a < 0.

y

y= x

It is often useful to think of jaj as the distance along the real line from 0 to a. x

For example j3j ¼ j3j ¼ 3:

In the same way, ja  bj is the distance along the real line from 0 to a  b, which is the same as the distance from a to b.

We sometimes write ja  bj ¼ dða; bÞ:

Notice also that ja þ bj ¼ ja  (b)j is the distance from a to b.

For example, the distance from 2 to 3 is jð2Þ  3j ¼ j5j ¼ 5:

1.2

Inequalities

13

We now list some basic properties of the modulus, which follow immediately from the definition: Properties of the modulus

For example, if a ¼ 2, b ¼ 1, then:

For any real numbers a and b:

1. jaj  0, with equality if and only if a ¼ 0; 2. –jaj  a  jaj;

1. j2j > 0; 2. j2j  2  j2j;

3. jaj2 ¼ a2;

3. j 2j2 ¼ ( 2)2; 4. j(2)  1j ¼ j1  (2)j;

4. ja  bj ¼ jb  aj; 5. jabj ¼ jaj  jbj.

5. j(2)  1j ¼ j 2j  j1j.

There is a basic rule for rearranging inequalities involving modulus signs: Rule 6

For any real numbers a and b, where b > 0: jaj < b , b < a < b.

Note that, in a similar way jaj  b , b  a  b:

Also, it is often possible, and sometimes easier, to use Rule 5 with p ¼ 2 than to use Rule 6. The following example illustrates the use of both rules. Example 4 Solution

Solve the inequality jx  2j < 1. Using Rule 6, we obtain

We take a ¼ x  2, b ¼ 1 in Rule 6.

jx  2j < 1 , 1 < x  2 < 1 , 1 < x < 3: So the solution set of the original inequality is fx: jx  2j < 1g ¼ ð1; 3Þ: Alternatively, using Rule 5 (with p ¼ 2), we obtain jx  2j < 1 , ðx  2Þ2 < 1

For jx  2j2 ¼ (x  2)2.

2

, x  4x þ 3 < 0 , ðx  1Þðx  3Þ < 0: Again, this shows that the required solution set is (1, 3). Example 5 Solution

&

Solve the inequality jx  2j  jx þ 1j. Using Rule 5 (with p ¼ 2), we obtain jx  2j  jx þ 1j , ðx  2Þ2  ðx þ 1Þ2 , x2  4x þ 4  x2 þ 2x þ 1

, 3  6x 1 ,  x: 2 So the solution set of the original inequality is

1 fx: jx  2j  jx þ 1jg ¼ ; 1 : 2

&

The inequalities in Examples 4 and 5 can easily be interpreted geometrically. In Example 4, the inequality jx 2j < 1 holds when the distance from x to 2 is strictly less than 1. So it holds for all points on either side of 2 at a distance less than 1 from 2  namely, in the open interval (1, 3).

1: Numbers

14 In Example 5, the inequality jx  2j  jx þ 1j holds when the distance from x to 2 is less than or equal to the distance from x to 1, since jx þ 1j ¼ jx  (1)j. The mid-point of 2 and 1 (that is, the point x where the distance from x to 2 equals the distance from x to 1) is 12. So the inequality holds when x lies in [12 , 1). Some good ideas when tackling problems involving inequalities of these types are:  

use your geometrical intuition, where possible, to give yourself an idea of the sets involved; test one or two values of x in your final solution set to see if they are valid – this often detects errors in manipulating inequality signs! Problem 5

Solve the following inequalities:

2

(a) j2x  13j < 5;

1.3

(b) jx  1j  2jx þ 1j.

Proving inequalities

In this section we show you how to prove inequalities of various types. We shall use the rules for rearranging inequalities given in Section 1.2, and also use other rules which enable us to deduce new inequalities from old. We have already met the first rule in Section 1.1, where it was called the Transitive Property of R. Transitive Rule

a 2. Solution (a) Suppose that jaj  1. The Triangle Inequality then gives     3 þ a3  j3j þ a3 ¼ 3 þ jaj3 3þ1

ðsince jaj  1Þ

Note the use of the Transitive Rule here.

¼ 4: (b) Suppose that jbj < 1. The ‘reverse form’ of the Triangle Inequality then gives   j3  bj  j3j  jbj   ¼ 3  jbj  3  jbj: Now jbj < 1, so that  jbj > 1. Thus 3  jbj > 3  1 ¼ 2;

Again, we use the Transitive Rule.

and we can then deduce from the previous chain of inequalities that & j3  bj > 2, as desired.

Remarks 1. The results of Example 1 can also be stated in the form: (a) j3 þ a3j  4, for jaj  1; (b) j3  bj > 2, for jbj < 1. 2. The reverse implications   3 þ a3  4 ) jaj  1

and

j 3  bj > 2 ) j bj < 1

are FALSE. For example, try putting a ¼ 32 and b ¼ 2! Problem 1

Use the Triangle Inequality to prove that:   (b) jbj < 1 ) b3  1 > 7.

(a) jaj  12 ) ja þ 1j  32;

2

8

1.3

Proving inequalities

1.3.2

17

Inequalities involving n

In Analysis we often need to prove inequalities involving an integer n. It is a common convention in mathematics that the symbol n is used to denote an integer (frequently a natural number). It is often possible to deal with inequalities involving n by using the rearrangement rules given in Section 1.2. Here is such an example. Example 2 Solution

Prove that 2n2  ðn þ 1Þ2 ;

for n  3:

Rearranging this inequality into an equivalent form, we obtain 2n2  ðn þ 1Þ2 , 2n2  ðn þ 1Þ2  0 , n2  2n  1  0

n

1

2

3

4

2n2 (n þ 1)2

2 4

8 9

18 16

32 25

, ðn  1Þ2  2  0 ðby ‘completing the square’Þ , ðn  1Þ2  2: This final inequality is clearly true for n  3, and so the original inequality & 2n2  (n þ 1)2 is true for n  3.

Remarks 1. In Problem 3 of Section 2x2  (x þ 1)2; pffiffiffiyou to solve pffiffiffi the inequality  1.2, we asked  its solution set was 1; 1  2 [ 1 þ 2; 1 . In Example 2, above, we found those natural numbers n lying in this solution set. 2. An alternative solution to Example 2 is as follows 

nþ1 2 2 2 ðby Rule 3Þ 2n  ðn þ 1Þ , 2  n pffiffiffi 1 1 ðby Rule 5, with p ¼ Þ; , 21þ n 2 and this final inequality certainly holds for n  3. Problem 2

1.3.3

Prove that n23nþ2 < 1; for n > 2.

More on inequalities

We now look at a number of inequalities and methods for proving inequalities that will be useful later on.  2 Example 3 Prove that ab  a þ2 b , for a, b 2 R. Solution We tackle this inequality using the various rearrangement rules and a chain of equivalent inequalities until we obtain an inequality that we know must be true 

aþb 2 a2 þ 2ab þ b2 ab  , ab  2 4 , 4ab  a2 þ 2ab þ b2 , 0  a2  2ab þ b2 , 0  ð a  bÞ 2 :

This has the following geometric interpretation: The area of a rectangle with sides of length a and b is less than or equal to the area of a square with sides of length aþb 2 . b a

1: Numbers

18 This final inequality is certainly true, since  2all squares are non-negative. It follows that the original inequality ab  aþb is also true, for a, b 2 R. & 2

Remark

pffiffiffiffiffi In the form ab  aþb 2 this inequality is sometimes called the Arithmetic–Geometric Mean Inequality for a, b.

A close examination of the above chain of equivalent statements shows that in  2 fact ab ¼ aþb if and only if a ¼ b. 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi  Problem 3 Prove that aþb a2 þ b2 ; for a, b 2 R. 2 pffiffiffi Problem 4 Suppose that a > 2. Prove the following inequalities:     2 (a) 12 a þ 2a < a; (b) 12 a þ 2a > 2: Hint: In part (b), use the result of Example 3 and the subsequent remark. Example 4

Prove that

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2  a þ b; for a; b  0:

Solution We tackle this inequality using the various rearrangement rules and a chain of equivalent inequalities until we obtain an inequality that we know must be true pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2  a þ b , a2 þ b 2  ð a þ bÞ 2

This has the following geometric interpretation: The length of the hypotenuse of a right-angled triangle whose other sides are of lengths a and b is less than or equal to the sum of the lengths of those two sides.

, a2 þ b2  a2 þ 2ab þ b2 a2 + b2

, 0  2ab: This final inequality is certainly true, since a, b  0. It follows that the pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi & original inequality a2 þ b2  a þ b is also true, for a, b  0. Problem p ffiffiffi pffiffiffi 5 Use the result of Example 4 to prove that c þ d; for c; d  0: pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Example 5 Prove that  a  b  ja  bj, for a, b  0.

a

pffiffiffiffiffiffiffiffiffiffiffi cþd 

Solution Notice first that interchanging the roles of a and b leaves the inequality unaltered. It follows that it is sufficient to prove the inequality under the assumption that a  b. pffiffiffi pffiffiffi So, assume that a  b. Then we know that a  b and ja  bj ¼ a  b: Hence pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi    a  b  ja  bj , a  b  a  b pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi , a  a  b þ b: This final inequality is certainly true, and is obtained from the result of Problem 5 by simply substituting a  and bffi in place of d. It pbffiffiffi in place pffiffiffi ofpcffiffiffiffiffiffiffiffiffiffiffiffiffi follows that the original inequality  a  b  ja  bj is also true, for & a, b  0. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi Problem 6 Prove that a þ b þ c  a þ b þ c; for a; b; c  0: We often use the Binomial Theorem and the Principle of Mathematical Induction (see Appendix 1) to prove inequalities. Example 6 Prove the following inequalities, for n  1: 1 (a) 2n  1 þ n; (b) 2n  1 þ 1n :

b

This will simplify the details of our chain of inequalities.

We avoid one modulus as a result of our simplifying assumption! Never be ashamed to utilise every tool at your disposal! (Why do the same work twice?)

n n

2 1þn

1

2

3

4

2 2

4 3

8 4

16 5

1.3

Proving inequalities

Solution

19 n

(a) By the Binomial Theorem for n  1 nð n  1Þ 2 x þ    þ xn ð1 þ xÞn ¼ 1 þ nx þ 2!  1 þ nx; for x  0: Then, if we substitute x ¼ 1 in this last inequality, we get

1 2

1

2n 1 þ 1n

3

4

2 1.41 1.26 1.19 2 1.5 1.33 1.25

We decrease the sum by omitting subsequent non-negative terms.

2n  1 þ n; for n  1: (b) We start by rewriting the required result in an equivalent form 

1 1 n 1 n 2 1þ ,2 1þ ðby the Power RuleÞ: n n Now, if we substitute x ¼ 1n in the Binomial Theorem for (1 þ x)n, we get 

   n 1 n 1 nð n  1Þ 1 2 1 1þ ¼1þn þ þ þ n n 2! n n  1 þ 1 ¼ 2:  n Since the inequality 2  1 þ 1n is true, it follows that the original 1 & inequality 2n  1 þ 1n, for n  1, is also true, as required. Problem 7

We decrease the sum by omitting all but the first two terms.

 n 1 Prove the inequality 1 þ 1n  52  2n ; for n  1.

Hint: consider the first three terms in the binomial expansion. Example 7 Solution

Prove that 2n  n2, for n  4. Let P(n) be the statement PðnÞ : 2n  n2 :

First we show that P(4) is true: 24  42. STEP 1 Since 24 ¼ 16 and 42 ¼ 16, P(4) is certainly true. STEP 2 We now assume that P(k) holds for some k  4, and deduce that P(k þ 1) is then true. So, we are assuming that 2k  k2. Multiplying this inequality by 2 we get 2kþ1  2k2 ; so it is therefore sufficient for our purposes to prove that 2k2  (k þ 1)2. Now 2k2  ðk þ 1Þ2 , 2k2  k2 þ 2k þ 1 , k2  2k  1  0 ðby ‘completing the square’Þ , ðk  1Þ2 2  0: This last inequality certainly holds for k  4, and so 2kþ1  (k þ 1)2 also holds for k  4. In other words: P(k) true for some k  4 ) P(k þ 1) true. It follows, by the Principle of Mathematical Induction, that 2n  n2 , for & n  4. Problem 8

Prove that 4n > n4, for n  5.

n n

2 n2

1

2

3

4

5

2 1

4 4

8 9

16 16

32 25

This assumption is just P(k).

Since P(k þ 1) is: 2kþ1  (k þ 1)2.

1: Numbers

20

Three important inequalities in Analysis Our first inequality, called Bernoulli’s Inequality, will be of regular use in later chapters. Theorem 1 Bernoulli’s Inequality For any real number x  1 and any natural number n, (1 þ x)n  1 þ nx.

Remark

The value of this result will come from making suitable choices of x and n for particular purposes.

In part (a) of Example 6, you saw that (1 þ x)n  1 þ nx, for x > 0 and n a natural number. Theorem 1 asserts that the same result holds under the weaker assumption that x  1. Proof

Let P(n) be the statement n

PðnÞ : ð1 þ xÞ  1 þ nx;

for x  1:

STEP 1 First we show that P(1) is true: (1 þ x)1  1 þ x. This is obviously true. STEP 2 We now assume that P(k) holds for some k  1, and prove that P(k þ 1) is then true. So, we are assuming that ð1 þ xÞk  1 þ kx; for x  1. Multiplying this inequality by (1 þ x), we get

We prove the result using Mathematical Induction.

This assumption is P(k). This multiplication is valid since ð1 þ xÞ  0.

ð1 þ xÞkþ1  ð1 þ xÞð1 þ kxÞ ¼ 1 þ ðk þ 1Þx þ kx2  1 þ ðk þ 1Þx: Thus, we have ð1 þ xÞkþ1  1 þ ðk þ 1Þx; in other words the statement P(k þ 1) holds. So, P(k) true for some k  1 ) P(k þ 1) true.

We decrease the expression if we omit the final nonnegative term.

It follows, by the Principle of Mathematical Induction, that ð1 þ xÞn  & 1 þ nx, for x  1, n  1: 1 Problem 9 By applying Bernoulli’s Inequality with x ¼ ð2nÞ , prove 1 1 that 2n  1 þ 2n1, for any natural number n.

You saw in part (b) of 1 Example 6 that 2n  1 þ 1n :

Our second inequality is of considerable use in various branches of pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi  Analysis. In Problem 3 you proved that aþb a2 þ b2 , for a, b 2 R. We 2 can rewrite this inequality in the equivalent form ða þ bÞ2  2ða2 þ b2 Þ or ða þ bÞ2  ða2 þ b2 Þð12 þ 12 Þ. The Cauchy–Schwarz Inequality is a generalisation of this result to 2n real numbers. Theorem 2 Cauchy–Schwarz Inequality For any real numbers a1 ; a2 ; . . .; an and b1 ; b2 ; . . .; bn ; we have ða1 b1 þ a2 b2 þ    þ an bn Þ2     a21 þ a22 þ    þ a2n b21 þ b22 þ    þ b2n :

We give the proof of Theorem 2 at the end of the sub-section.

1.3

Proving inequalities Problem 10

21

Use Theorem 2 to prove that  for any positive real numbers

a1 , a2 , . . ., an , then ða1 þ a2 þ    þ an Þ

1 a1

þ

1 a2

þ  þ

1 an

For example, with n ¼ 3

2

n :

ð1 þ 2 þ 3Þ



1 1 1 11 þ þ ¼6 1 2 3 6

Our final result also has many useful applications. In Example 3 you proved  2 that ab  a þ2 b , for a, b 2 R; it follows that, if a and b are positive, then

¼ 11  32 :

1

ðabÞ2  aþb 2 . The Arithmetic Mean–Geometric Mean Inequality is a generalisation of this result for two real numbers to n real numbers. Theorem 3 Arithmetic Mean–Geometric Mean Inequality For any positive real numbers a1, a2, . . ., an, we have 1 a1 þ a2 þ    þ a n : ð a1 a2 . . . an Þ n  n Problem 11 Use Theorem 3 with the n þ 1 positive numbers 1, 1 þ 1n, 1 þ 1n, . . ., 1 þ 1n to prove that, for any positive integer n 



nþ1 1 n 1 1þ  1þ : n nþ1

We give the proof of Theorem 3 at the end of the sub-section.

For example, with n ¼ 3 

1 3 64 ¼ 2:37 . . . 1þ ¼ 3 27 

5 1þ

1 4

4

¼

625 ¼ 2:44 . . . 256

Proofs of Theorems 2 and 3 You may omit these proofs at a first reading.

Theorem 2 Cauchy–Schwarz Inequality For any real numbers a1, a2, . . ., an and b1, b2, . . ., bn, we have    ða1 b1 þ a2 b2 þ    þan bn Þ2  a21 þ a22 þ    þ a2n b21 þ b22 þ    þ b2n : Proof If all the as are zero, the result is obvious; so we need only examine the case when not all the as are zero. It follows that, if we denote the Pn 2 2 2 2 sum k¼1 ak ¼ a1 þ a2 þ    þ an by A, then A > 0. Also, denote

Pn

2 2 2 2 k¼1 bk ¼ b1 þb2 þþbn

Pn

by B and k¼1 ak bk ¼ a1 b1 þ a2 b2 þþan bn by C. Now, for any real number l, we have that ðlak þ bk Þ2  0, so that n  n X  X l2 a2k þ 2lak bk þ b2k ¼ ðlak þ bk Þ2  0; k¼1

Note that we will use the  notation to keep the argument brief.

k¼1

which we may rewrite in the form l2 A þ 2lC þ B  0: But this inequality is equivalent to the inequality ðlA þ C Þ2 þ AB  C2 ; for any real number l: Since A is non-zero, we may now choose l ¼  CA. It follows from the last & inequality that AB  C2, which is exactly what we had to prove.

Remark If not all the as are zero, equality can only occur if

n P k¼1

ðlak þ bk Þ2 ¼ 0; that is,

if all the numbers ak are proportional to all the numbers bk, 1  k  n:

A is non-zero, by assumption, so that 1/A makes sense.

1: Numbers

22 Theorem 3 Arithmetic Mean–Geometric Mean Inequality For any positive real numbers a1, a2, . . ., an, we have 1 a1 þ a2 þ . . . þan ð a1 a2 . . . an Þ n  : n Proof

(2)

Since the ai are positive, we can rewrite (2) in the equivalent form 1

ð a1 a2 . . . an Þ n  1: ða1 þ a2 þ    þan Þ=n

(3)

Now, replacing each term ai by lai for any non-zero number l does not alter the left-hand side of the inequality (3). It follows that it is sufficient to prove the inequality (2) in the special case when the product of the terms ai is 1. Hence it is sufficient to prove the following statement P(n) for each natural number n: P(n): For any positive real numbers ai with a1a2 . . . an ¼ 1, then a1 þ a2 þ    þ an  n. First, the statement P(1) is obviously true. Next, we assume that P(k) holds for some k  1, and prove that P(k þ 1) is then true. Now, if all the terms a1, a2, . . ., akþ1 are equal to 1, the result P(k þ 1) certainly holds. Otherwise, at least two of the terms differ from 1, say a1 and a2, such that a1 > 1 and a2 < 1. Hence ða1  1Þ  ða2  1Þ  0; which after some manipulation we may rewrite as a1 þ a 2  1 þ a1 a2 :

(4)

We denote the typical term by ai rather than ak to avoid confusion with a different use of the letter k in the Mathematical Induction argument below.

We will prove this by Mathematical Induction.

The argument is exactly the same whichever two terms actually differ from 1. You should check this yourself.

We are now ready to tackle P(k þ 1). Then a1 þ a2 þ    þ akþ1  1 þ a1 a2 þ a3 þ a4 þ    þ akþ1

By (4).

 k þ 1; since we may apply the assumption that P(k) holds to the k quantities a1a2, a3, a4, . . ., akþ1. This last inequality is simply the statement that P(k þ 1) is indeed true. It follows by the Principle of Mathematical Induction that P(n) holds for all & natural numbers n, and so the inequality (2) must also hold.

Remark A careful examination of the proof of Theorem 3 shows that equality can only occur if all the terms ai are equal.

1.4 1.4.1

Least upper bounds and greatest lower bounds Upper and lower bounds

Any finite set {x1, x2, . . ., xn} of real numbers obviously has a greatest element and a least element, but this property does not necessarily hold for infinite sets.

That is ða1 a2 Þ a3 a4 . . . akþ1 ¼ 1 ) ða1 a2 Þ þ a3 þ a4 þ    þ akþ1  k:

1.4

Least upper bounds and greatest lower bounds

23

For example, the interval (0, 2] has greatest element 2, but neither of the sets N ¼ {1, 2, 3, . . .} nor [0, 2) has a greatest element. However the set [0, 2) is bounded above by 2, since all points of [0, 2) are less than or equal to 2. Definitions A set E R is bounded above if there is a real number, M say, called an upper bound of E, such that x  M; for all x 2 E: If the upper bound M belongs to E, then M is called the maximum element of E, denoted by max E. Geometrically, the set E is bounded above by M if no point of E lies to the right of M on the real line. For example, if E ¼ [0, 2), then the numbers 2, 3, 3.5 and 157.1 are all upper bounds of E, whereas the numbers 1.995, 1.5, 0 and 157.1 are not upper bounds of E. Although it seems obvious that [0, 2) has no maximum element, you may find it difficult to write down a formal proof. The following example shows you how to do this: Example 1 Determine which of the following sets are bounded above, and which have a maximum element: (a) E1 ¼ [0, 2);

(b) E2 ¼ f1n : n ¼ 1; 2; . . .g;

(c) E3 ¼ N.

Solution (a) The set E1 is bounded above. For example, M ¼ 2 is an upper bound of E1, since x  2;

for all x 2 E1 :

However, E1 has no maximum element. For each x in E1, we have x < 2, and so there is some real number y such that x < y < 2; by the Density Property of R. Hence y 2 E1, and so x cannot be a maximum element. (b) The set E2 is bounded above. For example, M ¼ 1 is an upper bound of E2, since 1  1; n

for all n ¼ 1; 2; . . .:

Also, since 1 2 E2 max E2 ¼ 1: (c) The set E3 is not bounded above. For each real number M, there is a positive integer n such that n > M, by the Archimedean Property of R. Hence M cannot be an upper bound of E3. & This also means that E3 cannot have a maximum element. Problem 1 Sketch the following sets, and determine which are bounded above, and which have a maximum element: (a) E1 ¼ (1, 1]; (b) E2 ¼ f1  1n : n ¼ 1; 2; . . .g; 2 (c) E3 ¼ {n : n ¼ 1, 2, . . .}.

For example, y can be of the form 1.99. . . 9 or y ¼ 12(x þ 2). 2 is not a maximum element, since 2 2 = E1.

1: Numbers

24 Similarly, we define lower bounds. For example, the interval (0, 2) is bounded below by 0, since 0  x;

for all x 2 ð0; 2Þ:

However, 0 does not belong to (0, 2), and so 0 is not a minimum element of (0, 2). In fact, (0, 2) has no minimum element. Definitions A set E R is bounded below if there is a real number, m say, called a lower bound of E, such that m  x;

for all x 2 E:

If the lower bound m belongs to E, then m is called the minimum element of E, denoted by min E. Geometrically, the set E is bounded below by m if no point of E lies to the left of m on the real line. Problem 2 Determine which of the following sets are bounded below, and which have a minimum element: (a) E1 ¼ (1, 1]; (b) E2 ¼ f1  1n : n ¼ 1; 2; . . .g; 2 (c) E3 ¼ {n : n ¼ 1, 2, . . .}. The following terminology is also useful: Definition below.

A set E R is bounded if it is bounded above and bounded

For example, the set E2 ¼ f1  1n : n ¼ 1; 2; . . .g is bounded, but the sets E1 ¼ (1, 1] and E3 ¼ {n2: n ¼ 1, 2, . . .} are not bounded. Similar terminology applies to functions. Definitions A function ƒ defined on an interval I R is said to:  be bounded above by M if f(x)  M, for all x 2 I; M is an upper bound of ƒ;  be bounded below by m if f(x)  m, for all x 2 I; m is a lower bound of ƒ;  have a maximum (or maximum value) M if M is an upper bound of ƒ and f(x) ¼ M, for at least one x 2 I;  have a minimum (or minimum value) m if m is a lower bound of ƒ and f(x) ¼ m, for at least one x 2 I.  Example 2 Let ƒ be the function defined by f (x) ¼ x2, x 2 12 , 3 . Determine whether ƒ is bounded above or below, and any maximum or minimum value of ƒ. Solution First, ƒ is increasing on the interval 12 , 3Þ, so that since 12  x < 3 it follows that 14 f ðxÞ < 9. Hence ƒ is bounded above and bounded below. Next, since f 12 ¼ 14 and 14 is a lower bound for ƒ on the interval 12 , 3Þ, it follows that ƒ has a minimum value of 14 on this interval. Finally, 9 is an upper bound for ƒ on the interval 12 , 3Þ but there is no point 1 x in 2 , 3Þ for which ƒ(x) ¼ 9. So 9 cannot be a maximum of ƒ on the interval. pffiffiffi pffiffiffi  However, if y is any number in (8, 9) there is a number x > y in y, 3

 pffiffiffi  1  2 2 2, 3 2 , 3 such that ƒ(x) ¼ x > y, so that no number in (8, 9) will serve

Strictly speaking, M and m are the upper bound and lower bound of the image set {ƒ(x): x 2 I}.

1 2

2√2

y

3

1.4

Least upper bounds and greatest lower bounds

25

as a maximum of ƒ on the interval. It follows that ƒ has no maximum value & on 12 , 3Þ. Problem 3 Let ƒ be the function defined by f ð xÞ ¼ x12 ; x 2 ½3; 2Þ. Determine whether ƒ is bounded above or below, and any maximum or minimum value of ƒ.

1.4.2

Least upper bounds and greatest lower bounds

We have seen that the interval [0, 2] has a maximum element 2, but [0, 2) has no maximum element. However, the number 2 is ‘rather like’ a maximum element of [0, 2), because 2 is an upper bound of [0, 2) and any number less than 2 is not an upper bound of [0, 2). In other words, 2 is the least upper bound of [0, 2). Definition A real number M is the least upper bound, or supremum, of a set E R if: 1. M is an upper bound of E; 2. if M0 < M, then M0 is not an upper bound of E. In this case, we write M ¼ sup E.

Part 1 says that M is an upper bound. Part 2 says that no smaller number can be an upper bound.

If E has a maximum element, max E, then sup E ¼ max E. For example, the closed interval [0, 2] has least upper bound 2. We can think of the least upper bound of a set, when it exists, as a kind of ‘generalised maximum element’. If a set does not have a maximum element, but is bounded above, then we may be able to guess the value of its least upper bound. As in the case E ¼ [0, 2), there may be an obvious ‘missing point’ at the upper end of the set. However it is important to prove that your guess is correct. We now show you how to do this. Example 3 Solution

Prove that the least upper bound of [0, 2) is 2. We know that M ¼ 2 is an upper bound of [0, 2), because x  2;

for all x 2 ½0; 2Þ:

To show that 2 is the least upper bound, we must prove that each number M0 < 2 is not an upper bound of [0, 2). To do this, we must find an element x in [0, 2) which is greater than M0 . But, if M0 < 2, then there is a real number x such that 0

M <x M .

GUESS CHECK

the value of M, then parts 1 and 2.

Notice that, if M is an upper bound of E and M 2 E, then part 2 is automatically satisfied, and so M ¼ sup E ¼ max E. Example 4

Determine the least upper bound of E ¼ f1  n12 : n ¼ 1; 2; . . .g.

Solution We guess that the least upper bound of E is M ¼ 1. Certainly, 1 is an upper bound of E, since 1 1  2  1; for n ¼ 1; 2; . . .: n To check part 2 of the strategy, we need to show that, if M0 < 1, then there is some natural number n such that 1 1  2 > M0: (1) n However 1 1 1  2 > M0 , 1  M0 > 2 n n 1 , < n2 ðsince 1  M 0 > 0Þ 1rffiffiffiffiffiffiffiffiffiffiffiffiffiffi M0 1 1 , >0 0Þ: We can certainly choose n so that this final inequality holds, by the Archimedean Property of R, and so we can choose n so that inequality (1) holds. & Hence 1 is the least upper bound of E.

That is, 1 ¼ sup E.

Remark Although we used double-headed arrows in this solution, the actual proof required only the implications going from right to left. In other words, the proof uses only the fact that rffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 0 1 2 >M ( < n: n 1  M0 Problem 4 Determine sup E, if it exists, for each of the following sets: (a) E1 ¼ ( 1, 1]; (b) E2 ¼ f1  1n : n ¼ 1, 2, . . .g; 2 (c) E3 ¼ {n : n ¼ 1, 2, . . .}. Similarly, we define the notion of a greatest lower bound. Definition A real number m is the greatest lower bound, or infimum, of a set E R if: 1. m is a lower bound of E; 2. if m0 > m, then m0 is not a lower bound of E. In this case, we write m ¼ inf E.

Part 1 says that m is a lower bound. Part 2 says that no larger number can be a lower bound.

1.4

Least upper bounds and greatest lower bounds

27

If E has a minimum element, min E, then inf E ¼ min E. For example, the closed interval [0, 2] has greatest lower bound 0. We can think of the greatest lower bound of a set, when it exists, as a kind of ‘generalised minimum element’. The strategy for establishing that a number is the greatest lower bound of a set is very similar to that for proving that a number is the least upper bound of a set. Strategy Given a subset E of R, to show that m is the greatest lower bound, or infimum, of E, check that: 1. x  m, for all x 2 E; 0

GUESS CHECK

0

2. if m > m, then there is some x 2 E such that x < m .

the value of m, then parts 1 and 2.

Notice that, if m is a lower bound of E and m 2 E, then part 2 is automatically satisfied, and so m ¼ inf E ¼ min E. Problem 5

Determine inf E, if it exists, for each of the following sets:

(a) E1 ¼ (1, 5];

(b) E2 ¼ fn12 : n ¼ 1; 2; . . .g.

Remarks 1. For any subset E of R, inf E  sup E. This follows from the fact that, for any x 2 E, we have inf E  x  sup E. 2. For any bounded interval I of R, let a be its left end-point and b its right end-point. Then inf I ¼ a and sup I ¼ b.

Least upper bounds and greatest lower bounds of functions Similar terminology applies to bounds for functions. Definitions Let ƒ be a function defined on an interval I R. Then: A real number M is the least upper bound, or supremum, of ƒ on I if:



1. M is an upper bound of ƒ(I); 2. if M0 < M, then M0 is not an upper bound of ƒ(I). In this case, we write M ¼ sup f or sup f or supf f ð xÞ : x 2 I g or sup f ð xÞ: I



x2I

A real number m is the greatest lower bound, or infimum, of ƒ on I if: 1. m is a lower bound of ƒ(I); 2. if m0 > m, then m0 is not a lower bound of ƒ(I). In this case, we write m ¼ inf f or inf f or inf f f ð xÞ : x 2 I g or inf f ð xÞ. I

There are similar definitions for the least upper bound and the greatest lower bound of f on a general set S in R. These are really the definitions for the least upper bound or the greatest lower bound of the set {ƒ(x): x 2 I}.

x2I

For sup f is the least upper bound,

Notice, for instance, that: 

if M is an upper bound for f on I, then sup f  M,



if m is a lower bound for f on I, then inf f  m.

I

I

I

and inf f is the greatest lower I

bound, for f on I.

1: Numbers

28 The strategies for proving that M is the least upper bound or m the greatest lower bound of ƒ on I are similar to the corresponding strategies for the least upper bound or the greatest lower bound of a set E. Strategies Let ƒ be a function defined on an interval I R. Then:  To show that m is the greatest lower bound, or infimum, of f on I, check that: 1. f(x)  m, for all x 2 I; 

2. if m0 > m, then there is some x 2 I such that f(x) < m0 . To show that M is the least upper bound, or supremum, of f on I, check that: 1. f(x)  M, for all x 2 I; 2. if M0 < M, then there is some x 2 I such that f(x) > M0 .

 Example 5 Let ƒ be the function defined by f ð xÞ ¼ x2 ; x 2 12 ; 3 . Determine 1  the least upper bound and the greatest lower bound of ƒ on 2 ; 3 .  Solution We have already seen that 9 is an upper bound for ƒ on 12 ; 3 , and that no smaller number will serve as  an upper bound. It follows that 9 must be the least upper bound of ƒ on 12 ; 3 .  Similarly, we have already seen that 14 is the minimum value of ƒ on 12 ; 3 ;  it follows that 14 is the greatest lower bound of ƒ on 12 ; 3 , and this is actually & attained ðat the point 12Þ.

You saw this in Example 2.

Example 2.

Problem 6 Let ƒ be the function defined by f ð xÞ ¼ x12 ; x 2 ½1; 4Þ: Determine the least upper bound and the greatest lower bound of ƒ on ½1, 4).

Remark For any interval I of R, inf f  sup f . This follows from the fact that, for I I any x 2 I, we have inf f  f ð xÞ  sup f . I I The least upper bound and the greatest lower bound of a function on an interval will be particularly significant in our later work on continuity and integrability of functions.

1.4.3

Chapters 4 and 7.

The Least Upper Bound Property

In the examples in the previous sub-section, it was easy to guess the values of sup E and inf E. At times, however, we shall meet sets for which these values are not so easy to determine. For example, if  

1 n E¼ 1þ : n ¼ 1; 2; . . . ; n then it can be shown that E is bounded above by 3, but it is not easy to guess the least upper bound of E. In such circumstances, it is reassuring to know that sup E does exist, even though it may be difficult to find. This existence is guaranteed by the following fundamental result.

We will study this set closely in Section 2.5.

1.4

Least upper bounds and greatest lower bounds

The Least Upper Bound Property of R Let E be a non-empty subset of R. If E is bounded above, then E has a least upper bound. We leave the proof of the Least Upper Bound Property of R to the next subsection. However, the Property itself is intuitively obvious. If the set E lies entirely to the left of some number M, then you can imagine moving M steadily to the left until you meet E. At this point, sup E has been reached. The Least Upper Bound Property of R can be used to show that pffiffiffi R does include decimals which represent irrational numbers such as 2, as we claimed in Section 1.1. In Sections 1.2 and 1.3 we have taken for granted the existence of rational powers and their properties, without giving formal definitions. pffiffiffi How can we supply these definitions? For example, how can we define 2 as a decimal? Consider the set

 E ¼ x 2 Q : x > 0; x2 < 2 :

29 The Least Upper Bound Property of R is an example of an existence theorem, one which asserts that a real number exists having a certain property. Analysis contains many such results which depend on the Least Upper Bound Property of R. While these results are often very general, and their proofs elegant, they do not always provide the most efficient methods of calculating good approximate values for the numbers in question.

This is the p setffiffiffi of positive rational numbers whose squares are less than 2. Intuitively, 2 lies on the number lineptoffiffiffi the right of the numbers in E, but ‘only just’! In fact, we should expect 2 to be the least upper bound of E. Certainly E has a least upper bound, by the Least Upper Bound Property, because E is bounded above, by 1.5 pffiffiffifor example. Thus it seems likely that sup E is the decimal representation of 2. But how can we prove that (sup E)2 ¼ 2? We shall prove this in Section 1.5, once we have described how to do arithmetic with real numbers (decimals). Finally, note that there is a corresponding result about lower bounds. The Greatest Lower Bound Property of R Let E be a non-empty subset of R. If E is bounded below, then E has a greatest lower bound.

1.4.4

Proof of the Least Upper Bound Property

You may omit this proof at a first reading.

We know that E is a non-empty set, and we shall assume for simplicity that E contains at least one positive number. We also know that E is bounded above. The following procedure gives us the successive digits in a particular decimal, which we then prove to be the least upper bound of E. Procedure to find a ¼ a0  a1a2 . . . ¼ sup E Choose in succession:  the greatest integer a0 such that a0 is not an upper bound of E; 

the greatest digit a1 such that a0  a1 is not an upper bound of E;



the greatest digit a2 such that a0  a1a2 is not an upper bound of E; .. . the greatest digit an such that a0  a1a2 . . . an is not an upper bound of E; .. .



For example, if E ¼ fx 2 Q : x > 0; x2 < 2g; then a0 ¼ 1; since 12 < 2 < 22 ; a0  a1 ¼ 1:4; since 1:42 < 2 < 1:52 ; a0  a1 a2 ¼ 1:41; since 1:412 < 2 < 1:422 ; .. .

1: Numbers

30 Thus, at the nth stage we choose the digit an so that:  

a0  a1a2 . . . an is not an upper bound of E; a0  a1 a2 . . . an þ 101n is an upper bound of E.

We now prove that the least upper bound of E is a ¼ a0  a1a2 . . .. First, we have to prove that a is an upper bound of E. To do this, we prove that, if x > a, then x 2 = E (this is equivalent to proving, that, if x 2 E, then x  a). We begin by representing x as a non-terminating decimal x ¼ x0  x1x2 . . .. Since x > a, there is an integer n such that a < x0  x1 x2 . . . xn : Hence x0  x1 x2 . . . xn  a0  a1 a2 . . . an þ

1 ; 10n

and so, by our choice of an, x ¼ x0  x1 x2 . . . xn is an upper bound of E. Since x > x0  x1x2 . . . xn, we have that x 2 = E, as required. Next, we have to show that, if x < a, then x is not an upper bound of E. Since x < a, there is an integer n such that x < a0  a1 a2 . . . an ; and so x is not an upper bound of E, by our choice of an. Thus we have proved that a is the least upper bound of E.

&

Remark Notice that this proof does not use any arithmetical properties of the real numbers but only their order properties, together with the arithmetical properties of rational numbers. In the next section, we use the Least Upper Bound Property to define some of the arithmetical operations on R.

1.5 1.5.1

Manipulating real numbers Arithmetic in R

At the end of Section 1.1 we discussed the decimals pffiffiffi 2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .; and asked whether it is possible to add and multiply these numbers to obtain another real number. We now explain how this can be done, using the Least Upper Bound Property of R. pffiffiffi A natural way to obtain a sequence of approximations to the sum 2 þ p is to truncate each of the above decimals, and form the sums of the truncations. If each of the decimals is truncated at the same decimal place, this gives a sequence of approximations which is increasing:

pffiffiffi Here we are assuming that 2 and p can be represented as decimals.

1.5

Manipulating real numbers

31

pffiffiffi 2

p

pffiffiffi 2þp

1 1.4

3 3.1

4 4.5

1.41

3.14

4.55

1.414 1.4142 .. .

3.141 3.1415 .. .

4.555 4.5557 .. .

pffiffiffi Intuitively, we should expect that the sum 2 þ p is greater than each of the numbers in the right-hand column, pffiffiffi but ‘only just’! To accord with our intuition, therefore, we define the sum 2 þ p to be the least upper bound of the set of numbers in the right-hand column; that is pffiffiffi 2 þ p ¼ supf4; 4:5; 4:55; 4:555; 4:5557; . . .g: To be sure that this definition makes sense, pffiffiwe ffi need to show that this set is bounded above. But all the truncations of 2 are less than 1.5, and all the truncations of p are less than, say, 4. Hence, all the sums in the right-hand column are less than 1.5 þ 4 ¼ 5.5. So, by the Least Upper Bound Property, the set of numbers pffiffiffi in the right-hand column does have a least upper bound, and we can define 2 þ p in this way. This method can be used to define the sum of any pair of positive real numbers. Let us check that this method of adding decimals gives the correct answer when we use it in a familiar case. Consider the simple calculation 1 2 þ ¼ 0:333 . . . þ 0:666 . . .: 3 3 Truncating each of these decimals and forming the sums, we obtain the set f0; 0:9; 0:99; 0:999; . . .g: The supremum of this set is, of course, the number 0.999. . . ¼ 1, which is the correct answer. Similarly, we canp define the product of any two positive real numbers. For ffiffiffi example, to define 2  p, we can form the sequence of products of their truncations: pffiffiffi 2

p

pffiffiffi 2p

1 1.4

3 3.1

3 4.34

1.41 1.414

3.14 3.141

4.4274 4.441374

1.4142 .. .

3.1415 .. .

4.4427093 .. .

We do not expect you to use this method to add decimals!

1: Numbers

32 pffiffiffi As before, we define 2  p to be the least upper bound of the set of numbers in the right-hand column. Similar ideas can be used to define the operations of subtraction and division. Thus we can define arithmetic with real numbers in terms of the familiar arithmetic with rationals, using the Least Upper Bound Property of R. Moreover, it can be proved that these operations in R satisfy all the usual properties of a field.

1.5.2

We omit the details.

These properties were listed in Sub-section 1.1.5.

The existence of roots

Just as we usually take for granted the basic arithmetical operations with real numbers, so we usually assume that,pgiven any positive real number a, there is ffiffiffi a unique positive real number b ¼ a such that b2 ¼ a. We now discuss the justification for this assumption. First, here is a geometrical justification. Given line segments of lengths 1 and a, we can construct a semi-circle with diameter a þ 1 as shown.

For each positive integer pffiffiffin, we can also construct n as follows: 1

1

1 1

b

5 6

a

1

4 3

1

2 7

Using similar triangles, we see that a b ¼ ; b 1 and so

1

1

b2 ¼ a: This shows that there should be a positive real number b such that b2 ¼ a, so that the length of the vertical line segment pffiffiffi pffiffiffi in the figure can be described exactly by the expression a. But does b ¼ a exist exactly as a real number? In fact it does, and a more general result is true. Theorem 1 For each positive real number a and each integer n > 1, there is a unique positive real number b such that bn ¼ a: ffiffiffi p n a. We also define We call this number b the nth root of a, and we write bp¼ffiffiffiffiffiffiffiffiffiffi p ffiffi ffi pffiffiffi n 0, since 0n ¼ 0, and if n is odd we define n ðaÞ ¼  n a, since 0p¼ ffiffi ffi n ð n aÞ ¼ a if n is odd. Let us illustrate Theorem 1 with the special case a ¼ 2 and n ¼ 2. In this case, Theorem 1 asserts the existence of a real number b such that b2 ¼ 2. In other pffiffiffi words, it asserts the existence of a decimal b which can be used to define 2 precisely. Here is a direct proof of Theorem 1 in this special case. We choose the numbers 1, 1.4, 1.41, 1.414, . . . to satisfy the inequalities

We shall prove Theorem 1 in Sub-section 4.3.3.

For example,

ffiffiffiffiffiffiffiffiffiffi p 3 ð8Þ ¼ 2:

1.5

Manipulating real numbers

33

12 < 2 < 22 ð1:4Þ2 < 2 < ð1:5Þ2 ð1:41Þ2 < 2 < ð1:42Þ2 ð1:414Þ2 < 2 < ð1:415Þ2 .. .

(1)

This process gives an infinite decimal b ¼ 1:414 . . .; and we claim that b2 ¼ ð1:414 . . .Þ2 ¼ 2: This can be proved using our method of multiplying decimals: b

b

b2

1

1

1

1.4

1.4

1.96

1.41 1.414 .. .

1.41 1.414 .. .

1.9881 1.999396 .. .

Notice that b ¼ 1:414 . . . is the decimal that we obtained as the least upper bound of the set fx 2 Q : x > 0; x2 < 2g in Sub-section 1.5.1.

We have to prove that the least upper bound of the set E of numbers in the righthand column is 2, in other words that sup E ¼ sup f1; ð1:4Þ2 ; ð1:41Þ2 ; ð1:414Þ2 ; . . .g ¼ 2: To do this, we employ the strategy given in Sub-section 1.4.2. First, we check that M ¼ 2 is an upper bound of E. This follows from the lefthand inequalities in (1). Next, we check that, if M0 < 2, then there is a number in E which is greater than M0 . To prove this, put x0 ¼ 1; x1 ¼ 1:4; x2 ¼ 1:41; x3 ¼ 1:414; . . .: Then, by the right-hand inequalities in (1), we have that 

1 2 xn þ n > 2: 10 Also 



1 2 2 1 1 xn þ n  xn ¼ n 2xn þ n 10 10 10 1 5 5 n ð2  2 þ 1Þ ¼ n ; 10 10 and so 

1 2 5 2 xn > xn þ n  10 10n 5 >2  n 10 ¼ 1:99 . . . 95: n digits

For example, if n ¼ 1, then 

1 2 ¼ 1:52 > 2: 1:4 þ 10

For example, if n ¼ 2, then ð1:41Þ2 > 1:95:

1: Numbers

34 So, if M0 < 2, then we can choose n so large that xn2 > M0 (while still having xn 2 E). This proves that the least upper bound of E is 2, and so (1.414. . .)2 ¼ 2. Thus we can define pffiffiffi 2 ¼ 1:414 . . .; pffiffiffi which justifies our earlier claim that 2 can be represented exactly by a decimal.

1.5.3

Rational powers

Having discussed nth roots, we are now in a position to define the expression ax, where a is positive and x is rational. Definition

If a > 0, m 2 Z and n 2 N, then  pffiffiffim m an ¼ n a :

pffiffiffi 1 For example, for a >p0,ffiffiffi with m ¼ 1 we have an ¼ n a, and with m ¼ 2 and 2 2 n ¼ 3 we have a3 ¼ ð 3 aÞ : This notation is particularly useful, because rational powers (or rational exponents) satisfy the following exponent laws (whose proofs depend on Theorem 1): Exponent Laws  If a, b > 0 and x 2 Q ,  

For example

then axbx ¼ (ab)x.

and x, y 2 Q , then ax ay ¼ axþ y. and x, y 2 Q , then ðax Þy ¼ axy .

If a > 0 If a > 0

If x and y are integers, these laws actually hold for all non-zero real numbers a and b. However, if x and y are not integers, then we must have a, b > 0. For 1 example, ð1Þ2 is not defined as a real number. m However, if a is a negative real number, then a n can be defined whenever m 2 Z, n 2 N and mn is reduced to its lowest terms with n odd, as follows m

an ¼

p ffiffiffim n a :

Finally, you may have wondered why we did not mention that each positive number has two nth roots when n is even. 22 ¼ (2)2 ¼ 4. We p ffiffiffiFor example, 1 n shall adopt the convention that, for a > 0, a and an always refer to the positive nth root of a. If we wish pffiffiffi to refer to both roots (for example, when solving equations), we write  n a.

1.5.4

Real powers

We conclude this section by briefly discussing the meaning of ax when a > 0 and x is an arbitrary real number. We have defined this expression when x is rational, but the same definition does not work if x is irrational. However, it pffiffiffipffiffi2 is common practice to write down expressions such as 2 , and even to apply the Exponent Laws to give equalities such as

1

1

1

22  32 ¼ 62 ; 1

1

5

22  23 ¼ 26 ;  1 13 1 22 ¼ 26 :

This extends our above m definition of a n ; for instance, 1 it defines an whenever n 2 N and n2 is odd. For example, ð8Þ3 ¼ 4.

1.6

Exercises  pffiffi p2ffiffi  pffiffi pffiffi   pffiffiffi 2 pffiffiffi 2 2 pffiffiffi 2 2 ¼ 2 ¼ 2 ¼ 2:

Can such manipulations be justified? In fact, it is possible to define ax, for a > 0 and x 2 R, using the Least Upper Bound Property of R, but it is then rather tricky to check that the Exponent Laws work. In Chapters 2 and 3 we shall explain how to define the expression ex, and in Chapter 4 we use ex to define the real powers in general and show that the Exponent Laws hold. For the time being, whenever the expression ax appears, you should assume that x is rational.

1.6

Exercises

Section 1.1 1. Arrange the following numbers in increasing order: (a)

7 3 1 7 11 36 ; 20 ; 6 ; 45 ; 60 ;

(b) 0:465, 0:465, 0:465, 0.4655, 0.4656. 2. Find the fractions whose decimal expansions are: (a) 0:481; (b) 0:481. 3. Let x ¼ 0:21 and y ¼ 0:2. Find x þ y and xy (in decimal form). 4. Find a rational number x and an irrational number y in the interval (0.119, 0.12). pffiffiffi 5. Prove that, if n is a positive integer which is not a perfect square, then n is irrational.  2 Hint: If p, q are positive integers such that pq ¼ n, and k is the positive integer such that k < pq < k þ 1 (why does such a positive integer k exist?), p show that 0 < p  kq < q and nqkp pkq ¼ q, and hence obtain a contradiction.

Section 1.2 1. Solve the following inequalities: pffiffiffiffiffiffiffiffiffiffiffiffiffi xþ1 (a) xx1 (b) 4x  3 > x; 2 þ4 < x2 4;   (c) 17  2x4   15; (d) jx þ 1j þ jx  1j < 4.

Section 1.3 1. Use the Triangle Inequality to prove that jaj  1 ) ja  3j  2: 2. Prove that (a2 þ b2)(c2 þ d2)  (ac þ ad)2, for any a, b, c, d 2 R. 3. Prove the inequality 3n  2n2 þ 1, for n ¼ 1, 2, . . .:

35

1: Numbers

36 (a) by using the Binomial Theorem, applied to (1 þ x)n with x ¼ 2; (b) by using the Principle of Mathematical Induction. 4. Use the Principle of Mathematical Induction to prove that, for n ¼ 1, 2, . . . : (a) 12 þ 22 þ 32 þ    þ n2 ¼ nðnþ1Þ6ð2nþ1Þ; qffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffi 5=4 135 ... ð2n1Þ 3=4 (b)   4nþ1 246 ... ð2nÞ 2nþ1. 2 5. Apply Bernoulli’s Inequality, first with x ¼ 2n and then with x ¼ ð3nÞ to prove that 2 2 1  3n  1 þ ; for n ¼ 1; 2; . . . : 1þ 3n  2 n 6. By applying the Arithmetic Mean–Geometric Mean Inequality to the n þ 1 positive numbers 1, 1  1n , 1  1n , 1  1n , . . . , 1  1n, prove that nþ1  n  1 1  1n  1  nþ1 ; for n ¼ 1; 2; . . . :

7. Use the Cauchy–Schwarz Inequality to prove that, if a1, a2, . . . , an are positive numbers with a1 þ a2 þ    þ an ¼ 1, then pffiffiffi pffiffiffi pffiffiffi pffiffiffi a1 þ a2 þ    þ an  n: 8. Use the Cauchy–Schwarz Inequality to prove that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi 3 4 cos x þ 4 1  cos x  5; for x 2 0; p2 :

Section 1.4 In Exercises 1–4, take E1 ¼ {x: x 2 Q , 0  x < 1} and E2 ¼ n ¼ 1; 2; . . .g:



2 1 þ n1 :

1. Prove that each of the sets E1 and E2 is bounded above. Which of them has a maximum element? 2. Prove that each of the sets E1 and E2 is bounded below. Which of them has a minimum element? 3. Determine the least upper bound of each of the sets E1 and E2. 4. Determine the greatest lower bound of each of the sets E1 and E2. 5. For each of the following functions, determine whether it has a maximum or a minimum, and determine its supremum and infimum: 1 (a) f ð xÞ ¼ 1þx (b) f ð xÞ ¼ 1  x þ x2 ; x 2 ½0; 2Þ. 2 ; x 2 ½0; 1Þ; 6. Prove that, for any two numbers a, b 2 R minfa; bg ¼ 12 ða þ b  ja  bjÞ and maxfa; bg ¼ 12 ða þ b þ ja þ bjÞ:

2

Sequences

This chapter deals with sequences of real numbers, such as 1 1 1 1 1 2 3 4 5 6

1; ; ; ; ; ; . . .;

Three dots are used to indicate that the sequence continues indefinitely.

0; 1; 0; 1; 0; 1; . . .; 1; 2; 4; 8; 16; 32; . . .: It describes in detail various properties that a sequence may possess, the most important of which is convergence. Roughly speaking, a sequence is convergent, or tends to a limit, if the numbers, or terms, in the sequence approach arbitrarily close to a unique real number, which is called the limit of the sequence. For example, we shall see that the sequence 1 1 1 1 1 2 3 4 5 6

1; ; ; ; ; ; . . .

is convergent with limit 0. On the other hand, the terms of the sequence 0; 1; 0; 1; 0; 1; . . . do not approach arbitrarily close to any unique real number, and so this sequence is not convergent. Likewise, the sequence 1; 2; 4; 8; 16; 32; . . . is not convergent. A sequence which is not convergent is called divergent. The sequence 0; 1; 0; 1; 0; 1; . . .; is a bounded divergent sequence. The sequence 1; 2; 4; 8; 16; 32; . . . is unbounded; its terms become arbitrarily large and positive, and we say that it tends to infinity. Intuitively, it seems plausible that some sequences are convergent, whereas others are not. However, the above description of convergence, involving the phrase ‘approach arbitrarily close to’, lacks the precision required in Pure Mathematics. If we wish to work in a serious way with convergent sequences, prove results about them and decide whether a given sequence is convergent, then we need a rigorous definition of the concept of convergence. Historically, such a definition emerged only in the late nineteenth century, when mathematicians such as Cantor, Cauchy, Dedekind and Weierstrass sought to place Analysis on a rigorous non-intuitive footing. It is not surprising, therefore, that the definition of convergence seems at first sight rather obscure, and it may take you a little time to master the logic that it involves. 37

2: Sequences

38 In Section 2.1 we show how to picture the behaviour of a sequence by drawing a sequence diagram. We also introduce monotonic sequences; that is, sequences which are either increasing or decreasing. In Section 2.2 we explain the definition of a null sequence; that is, a sequence which is convergent with limit 0. We then establish various properties of null sequences, and list some basic null sequences. In Section 2.3 we discuss general convergent sequences (that is, sequences which converge but whose limit is not necessarily 0), together with techniques for calculating their limits. In Section 2.4 we study divergent sequences, giving particular emphasis to sequences which tend to infinity or tend to minus infinity. We also show that convergent sequences are bounded; it follows that unbounded sequences are necessarily divergent. In Section 2.5 we prove the Monotone Convergence Theorem, which states that any increasing sequence which is bounded above must be convergent, and, similarly, that any decreasing sequence which is bounded below must be convergent. We use this theorem to study simple examples of sequences defined by recurrence formulas, and particular sequences which converge to p and e.

2.1 2.1.1

Introducing sequences What is a sequence?

Ever since learning to count you have been familiar with the sequence of natural numbers 1; 2; 3; 4; 5; 6; . . .: You have also encountered many other sequences of numbers, such as 2; 4; 6; 8; 10; 12; . . .; 1 1 1 1 1 1 ; ; ; ; ; ; . . .: 2 4 8 16 32 64

We begin our study of sequences with a definition and some notation. Definition

A sequence is an unending list of real numbers a1 ; a2 ; a3 ; . . .:

The real number an is called the nth term of the sequence, and the sequence is denoted by fan g: In each of the sequences above, we wrote down the first few terms and left you to assume that subsequent terms were obtained by continuing the pattern in an obvious way. It is sometimes better, however, to give a precise description of a typical term of a sequence, and we do this by stating an explicit formula for the nth term. Thus the expression {2n  1} denotes the sequence 1; 3; 5; 7; 9; 11; . . .;

Alternative notations are 1 fan g1 1 and fan gn¼1 :

2.1

Introducing sequences

39

and the sequence {an} defined by the statement an ¼ ð1Þn ; n ¼ 1; 2; . . .; has terms a1 ¼ 1; a2 ¼ 1; a3 ¼ 1; a4 ¼ 1; a5 ¼ 1; . . .: Problem 1 (a) Calculate the first five terms of each of the following sequences: (i) {3n þ 1};

(ii) {3n};

(iii) {( 1)nn}.

(b) Calculate the first five terms of each of the following sequences {an}: (i) an ¼ n!; n ¼ 1; 2; . . .;  n (ii) an ¼ 1 þ 1n ; n ¼ 1; 2; . . . (to 2 decimal places). Sequences often begin with a term corresponding to n ¼ 1. Sometimes, however, it is necessary to begin a sequence with some other value of n. We indicate this by writing, for example, fan g1 3 to represent the sequence

For example, the sequence   1 n!n cannot begin with n ¼ 1 or n ¼ 2.

a3 ; a4 ; a5 ; . . .:

Sequence diagrams It is often helpful to picture how a given sequence {an} behaves by drawing a sequence diagram; that is, a graph of the sequence in R 2. To do this, we mark the values n ¼ 1, 2, 3, . . . on the x-axis and, for each value of n, we plot the point example, the sequence diagrams for the sequences {2n  1},   (n,an). For n 1 n and {( 1 )} are as follows: In Figure (a), the points plotted all lie on the straight line y ¼ 2x  1. In Figure (b), they all lie on the hyperbola y ¼ 1x :

Problem 2 Draw a sequence diagram, showing the first five points, for each of the following sequences: n no  n (a) {n2}; (b) {3}; (c) 1 þ 1n g; (d) ð1Þ . n (In part (c), use the result of Problem 1, part (b).)

2.1.2

Monotonic sequences

Many of the sequences considered so far have the property that, as n increases, their terms are either increasing or decreasing. For example, the sequence {2n  1} has terms 1, 3, 5, 7, . . ., which are increasing, whereas the sequence

2: Sequences

40 1

has terms 1; 12 ; 13 ; 14 ; . . .; which are decreasing. The sequence {( 1)n} is neither increasing nor decreasing. All this can be seen clearly on the above sequence diagrams. We now give a precise meaning to these words increasing and decreasing, and introduce the term monotonic. n

Definition

A sequence {an} is:



constant, if

anþ1 ¼ an ;

for n ¼ 1; 2; . . .;

 

increasing, if decreasing, if

anþ1  an ; anþ1  an ;

for n ¼ 1; 2; . . .; for n ¼ 1; 2; . . .;



monotonic, if

{an} is either increasing or decreasing.

Remarks 1. Note that, for a sequence {an} to be increasing, it is essential that anþ1  an, for all n  1. However, we do not require strict inequalities, because we wish to describe sequences such as 1; 1; 2; 2; 3; 3; 4; 4; . . . and 1; 2; 2; 3; 4; 4; 5; 6; 6; . . . as increasing. One slightly bizarre consequence of the definition is that constant sequences are both increasing and decreasing! 2. A sequence {an} is said to be: strictly increasing, if anþ1 > an ;

for n ¼ 1; 2; . . .;

strictly decreasing, if anþ1 < an ; for n ¼ 1; 2; . . .; strictly monotonic, if {an} is either strictly increasing or strictly decreasing. 3. A diagram does NOT constitute a proof! In our first example we formally establish the monotonicity properties of our three sequences. Example 1 monotonic:

Determine which of the following sequences {an} are

(a) an ¼ 2n 1, n ¼ 1, 2, . . .; (b) an ¼ 1n , n ¼ 1, 2, . . .; (c) an ¼ ( 1)n, n ¼ 1, 2, . . . Solution (a) The sequence {2n  1} is monotonic because an ¼ 2n  1 and anþ1 ¼ 2ðn þ 1Þ  1 ¼ 2n þ 1; so that anþ1  an ¼ ð2n þ 1Þ  ð2n  1Þ ¼ 2 > 0; for n ¼ 1; 2; . . .: Thus {2n  1} is increasing.  (b) The sequence 1n is monotonic because 1 1 and anþ1 ¼ ; an ¼ n nþ1 so that

In fact, strictly increasing.

2.1

Introducing sequences anþ1  an ¼

41

1 1 n  ð n þ 1Þ 1  ¼ ¼ < 0; for n ¼ 1; 2; ...: nþ1 n ðn þ 1Þn ðn þ 1Þn

 Thus 1n is decreasing. Alternatively, since an > 0, for all n, and anþ1 n < 1; for n ¼ 1; 2; . . .; ¼ nþ1 an it follows that anþ1 < an ; for n ¼ 1; 2; . . .:

In fact, strictly decreasing.

Thus {an} is decreasing. (c) The sequence {(1)n} is not monotonic. In fact, a1 ¼ 1, a2 ¼ 1 and a3 ¼ 1. Hence a3 < a2, which means that {an} is not increasing. Also, a2 > a1, which means that {an} is not decreasing. Thus {(1)n} is neither increasing nor decreasing, and so is not mono& tonic.

A single counter-example is sufficient to show that {(1)n} is not increasing; similarly a (different!) single counter-example is sufficient to show that {(1)n} is not decreasing.

Example 1 illustrates the use of the following strategies: Strategy To show that a given sequence {an} is monotonic, consider the expression anþ1  an. 

If anþ1  an  0;

for n ¼ 1; 2; . . .;



If anþ1  an  0;

for n ¼ 1; 2; . . .;

then fan g is increasing. then fan g is decreasing.

If an > 0 for all n, it may be more convenient to use the following version of the strategy: Strategy To show that a given sequence of positive terms, {an}, is monotonic, consider the expression aanþ1 . n  

If aanþ1  1; n anþ1 If an  1;

for n ¼ 1; 2; . . .; for n ¼ 1; 2; . . .;

then fan g is increasing. then fan g is decreasing.

Problem 3 Show that the following sequences {an} are monotonic: (a) an ¼ n!; n ¼ 1; 2; . . .; (b) an ¼ 2n ; n ¼ 1; 2; . . .; (c) an ¼ n þ 1n ; n ¼ 1; 2; . . .: It is often possible to guess whether a sequence given by a specific formula is monotonic by calculating the first few terms. For example, consider the sequence {an} given by   1 n ; n ¼ 1; 2; . . .: an ¼ 1 þ n In Problem 1 you found that the first five terms of this sequence are approximately 2; 2:25; 2:37; 2:44 and 2:49:

We will study this important sequence in detail in Section 2.5.

2: Sequences

42 These terms suggest that the sequence {an} is increasing, and in fact it is. However, the first few terms of a sequence are not always a reliable guide to the sequence’s behaviour. Consider, for example, the sequence an ¼

10n ; n!

The first two terms certainly prove that the sequence is neither decreasing nor constant.

n ¼ 1; 2; . . .:

The first five terms of this sequence are approximately 10; 50; 167; 417 and 833; this suggests that the sequence {an} is increasing. However, calculation of more terms shows that this is not so, and the sequence diagram for {an} looks like this:

In fact a6 ’ 1389; a7 ’ 1984; a8 ’ 2480; a9 ’ 2756; a10 ’ 2756; a11 ’ 2505; a12 ’ 2088: In particular, the fact that a12 < a11 shows that the sequence {an} cannot be increasing.

Simplifying aanþ1 , we find that n  n anþ1 10nþ1 10 10 : ¼ ¼ nþ1 an ðn þ 1Þ! n! 10 But nþ1  1, for n ¼ 9, 10, . . ., so that aanþ1  1, for n ¼ 9, 10, . . .; it follows n that the sequence {an} is decreasing, if we ignore the first eight terms. In a situation like this, when a given sequence has a certain property provided that we ignore a finite number of terms, we say that the sequence  n eventually has the property. Thus we have just seen that the sequence 10n! is eventually decreasing. Another example of this usage is the following statement:

In fact, aa109 ¼ 1 and for n ¼ 10; 11; . . .:

anþ1 an

< 1;

the terms of the sequence {n2} are eventually greater than 100. This statement is true because n2 > 100, for all n > 10. Problem 4 Classify each of the following statements as TRUE or and justify your answers (that is, if a statement is TRUE, prove it; if a statement is FALSE, give a specific counter-example):

FALSE,

(a) The terms of the sequence {2n} are eventually greater than 1000. (b) The terms of the sequence {(1)n} are eventually positive. 1 (c) The terms of the n are eventually less than 0.025. n 4sequence o n (d) The sequence 4n is eventually decreasing.

Note this general approach to ‘TRUE’ and ‘FALSE’.

2.2

Null sequences

2.2

43

Null sequences

2.2.1

What is a null sequence?

In this sub-section we give a precise definition of a null sequence (that is, a sequence which converges to 0) and introduce some properties of null sequences. We shall frequently use the rules for rearranging inequalities which you met in Chapter 1, so you may find it helpful to reread that section quickly before starting here. We shall also use the following inequalities which were proved in Sub-section 1.3.3 2n  1 þ n;

Section 1.2

In fact, we use

for n ¼ 1; 2; . . .;

2n  n; for n ¼ 1; 2; . . .:

and n

2

2 n ;

for n  4:

Problem 1 For each of the following statements, find a number X such that the statement is true: 1 3 (a) 1n < 100 ; for all n > X; (b) 1n < 1000 ; for all n > X: Problem 2 For each of the following statements, find a number X such that the statement is true: n n Þ Þ 1 3 (a) ð1 (b) ð1 n2 < 100 ; for all n > X; n2 < 1000 ; for all n > X: The solutions of Problems 1 and 2 both suggest that the larger and larger n we o  Þn choose n, the closer and closer to 0 the terms of the sequences 1n and ð1 n2 become. We can express this in terms of the Greek letter " (pronounced ‘epsilon’), which we introduce to denote a positive number that may be as small as we please given o particular instance. In terms of the sequence diagrams for   innany ð1Þn 1 , this means that the terms of these sequences eventually lie n and n2 inside a horizontal strip in the sequence diagram from " up to ". However, the smaller we choose ", the further to the right we have to go before we can be sure that all the terms of the sequence from that point onwards lie inside the strip. That is, the smaller we choose " the larger we have to choose X if we wish to have ð1Þn 1 < "; for all n > X; or 2 < "; for all n > X: n n Definition

A sequence {an} is a null sequence if:

for each positive number ", there is a number X such that jan j < ";

for all n > X:

(1)

Using this terminology, n n o it follows from our previous discussion that the 1 Þ sequences n and ð1 are both null. n2 We can interpret our finding of a suitable number X for (1) to hold as an ‘"  X game’ in which player A chooses a positive number " and challenges player B to find some number X such that the property (1) holds.

Thus in Problems 1 and 2 we chose the particular examples 1 3 and 1000 in place of the of 100 ‘general’ positive number ".

ε X

n

–ε

Note that here X need not be an integer; any appropriate real number will serve. X does NOT depend on n, but does in general depend on ".

2: Sequences

44 n

Þ For example, consider the sequence {an} where an ¼ ð1 n2 ; n ¼ 1; 2; . . .. This sequence is null. For ð1Þn 1 jan j ¼ 2 ¼ 2 ; n n

so that, if we make the choice X ¼ p1ffiffi", it is certainly true that jan j < ";

for all n > X:

In terms of the "  X game, this simply means that whatever choice of " is made by player A, player B can always win by making the choice X ¼ p1ffiffin !

This is the case because, if n > p1ffiffi" ½¼ X, then we have n2 > 1" or " > n12 ; hence, for n > X, we have jan j ¼ n12 < X12 ¼ ":

Remarks

n no Þ 1. In the sequence ð1 , the signs of the terms made no difference to n2 whether the sequence was null, for they disappeared immediately we took the modulus of the terms in order to examine whether the definition (1) of a null sequence was satisfied. Indeed, in general, the signs of terms in a sequence make no difference to whether a sequence is null. We can express this formally as follows: A sequence {an} is null if and only if the corresponding sequence {janj} is null. 2. A null sequence {an} remains null if we add, delete or alter a finite number of terms in the sequence. Similarly, a non-null sequence remains non-null if we add, delete or alter a finite number of terms. 3. If one number serves as a suitable value of X for the inequality in (1) to hold, then any larger number will also serve as a suitable X. Hence, for simplicity in some proofs, we may assume if we wish that our initial choice of X in (1) is a positive integer.   Example 1 Prove that the sequence n13 is a null sequence. Solution

All that happens is that the definition (1) remains valid with a possibly different value of X having to be chosen. This is often expressed in the following memorable way: ‘a finite number of terms do not matter’. For example, if 7 þ p will serve as a suitable value for X, then so will 12 or 37; but 10 might not.

We have to prove that

for each positive number ", there is a number X such that 1 < "; for all n > X: n3

(2)

In1 order to find a suitable value of X for (2) to hold, we rewrite the inequality 3 < " in various equivalent ways until we spy a value for X that will suit n our purpose. Now 1 < " , 1 < " n3 n3 1 " 1 ffiffiffi : , n>p 3 "

, n3 >

So, let us choose X to be p13 ffiffi". With this choice of X, the above chain of equivalent inequalities shows us that, if n > X (the last line in the chain), 1  then  3 < " (in the first line). Thus, with this choice of X, (2) holds; so 13 is n n & indeed null.

The term inside the modulus is positive. Here we use the Reciprocal Rule for inequalities that you met in Sub-section 1.2.1. Here we use the Power Rule for inequalities that you met in Sub-section 1.2.1.

2.2

Null sequences

Example 2

45

Prove that the following sequence is not null  1; if n is odd; an ¼ 0; if n is even:

Solution To prove that the sequence is not null, we have to show that it does not satisfy the definition. In other words, we must show that the following statement is not true: for each positive number ", there is a number X such that jan j5 ", for all n 4X: So, what we have to show is that the following is true: for some positive number ", whatever X one chooses jan j5", 6 for all n4X: which we may rephrase as: there is some positive number ", such that whatever X one chooses jan j 5 6 ", for all n4X: 2 1 n –2

Sequence diagram 1 1/2 –1/2

n

Strip of half-width 12

So, we need to find some positive number " with such a property. The 1 sequence diagram provides  1 1the clue! If we choose " ¼ 2, then the point (n, an) lies outside the strip  2 ; 2 for every odd n. In other words, whatever X one then chooses, the statement jan j < "; for all n > X; & is false. It follows that the sequence is not a null sequence.

Note Notice that any positive value of " less than 1 will serve for our purpose here; there is nothing special about the number 12. These two examples illustrate the following strategy: Strategy for using the definition of null sequence 1. To show that {an} is null, solve the inequality janj < " to find a number X (generally depending on ") such that janj < ", for all n > X. 2. To show that {an} is not null, find ONE value of " for which there is NO number X such that janj < ", for all n > X.

However, any value of " greater than 1 provides no information!

2: Sequences

46 Problem 3 Use the above strategy to determine which of the following sequences are null: n o n no  1  Þn Þ (a) 2n1 ; (c) ðn1 ; (b) ð1 4 þ1 : 10 We now look at a number of Rules for ‘getting new null sequences from old’. Power Rule If  {an} is a null sequence, where an  0, for n ¼ 1, 2, . . ., and p > 0, then apn is a null sequence. 1 pffiffi Thus, for example, the 1sequence n is null – we simply apply the Power Rule to the sequence n that we saw earlier to be null, using the positive power p ¼ 12.

We shall prove these Rules later, in Sub-section 2.2.2. Recall that apn ¼ ðan Þp :

For p1ffiffin ¼

112 n

:

Remark Notice that the Power Rule also holds without the requirement that an  0, so long as apn is defined – for example, if p ¼ m1 where m is a positive integer. Combination Rules If {an} and {bn} are null sequences, then the following are also null sequences: Sum Rule {an þ bn}; Multiple Rule {lan}, for any real number l; Product Rule {anbn}.

Note that the number l has to be a fixed number that does not depend on n.

Thus, for example, we may use known examples of null sequences to verify that the following sequences are also null: 1 1     þ – by applying the Sum Rule to the null sequences 1n and n13 ; n47pn3 1 – by applying the Multiple Rule to the null sequence n3 with n3 l ¼ 47p;  1 fnð2n1 g – by applying the Product Rule to the null sequences 1n and  1Þ 2n1 : Problem 4 Use the above rules to show that the following sequences are null: n o n o 1 6ffiffi 5 1 p (a) þ ; (b) (c) 7 ; 3 1 : 5 ð2n1Þ

n

f

ð2n1Þ

3n4 ð2n1Þ3

g

Our next rule, the Squeeze Rule, also enables us to ‘get new null sequences from old’ – but in a slightly different way. To illustrate  1  this  rule, we look first at 1pffiffi p ffiffi the sequence diagrams of the two sequences and 1þ n . n

1

}√1n }

}1 +1√ n} ... ... n

In your solution, you may use any of the sequences that you have proved to be null so far in this sub-section.

2.2

Null sequences

47

  The points corresponding to the sequence 1þ1pffiffin are squeezed in between   the horizontal axis and the points corresponding to the null sequence p1ffiffin , since 1þ1pffiffin < p1ffiffin, for n ¼ 1, 2, . . .. Hence, if from some point onwards all the   points corresponding to p1ffiffin lie in a narrow strip in the sequence diagram of half-width " about the axis, then (from the same point onwards) all the points   corresponding to 1þ1pffiffin will also lie in the same strip. So, since " may be any positive number, it certainly looks from this sequence diagram argument that   the sequence 1þ1pffiffin must be a null sequence too. Squeeze Rule

If {bn} is a null sequence and

jan j  bn ; for n ¼ 1; 2; . . .; then {an} is a null sequence.

That is, {an} is dominated by {bn}.

The trick in using the Squeeze Rule to prove that a given sequence {an} is null is to think of a suitable sequence {bn} that  dominates {an}and is itself null. Thus, for example, since the sequence p1ffiffin is null and 1þ1pffiffin < p1ffiffin, it follows   from the Squeeze Rule that 1þ1pffiffin is also a null sequence. Proof of the Squeeze Rule We want to prove that {an} is null; that is: for each positive number ", there is a number X such that jan j < "; for all n > X: (3) We know that {bn} is null, so there is some number X such that jbn j < "; for all n > X: (4) We also know that janj  bn, for n ¼ 1, 2, . . ., and hence it follows from (4) that jan j ð< jbn jÞ < ";

And this is the X that we shall use to verify (3).

for all n > X:

Thus inequality (3) holds, as required.

&

Remark In applying the Squeeze Rule, it is often useful to remember the following fact: The behaviour of a finite number of terms does not matter: it is sufficient to check that janj  bn holds eventually.  n  is null. Example 3 Use the Squeeze Rule to prove that the sequence 12 1n  Proof We want to prove that 2 is null.   n What sequence can we find that dominates 12 ? Well, we saw at the start n of this sub-section that 2  n, for n ¼ 1, 2, . . ., and it follows from this, by the Reciprocal Rule for inequalities, that 21n  1n ; for n¼ 1 11;n 2; . . .: Thus the sequence n dominatesthe sequence , and is itself null. It 2  n  & follows from the Squeeze Rule that 12 is null. n 2 Problem 5 Use the inequality  12n  n , for n  4, and the Squeeze Rule is null. to prove that the sequence n 2  n Example 4 Use the Squeeze Rule to prove that the sequence 10n! is null.  n Solution We want to prove that 10n! is null. 10n  With a bit of linspiration, we guess that the sequence n! is eventually dominated by n , for some constant l.

That is, that {an} is eventually dominated by {bn}.

2: Sequences

48 n

Writing out the expression 10n! in full, we see that           10n 10 10 10 10 10 10 ...  ... ¼ 1 2 10 11 n1 n n! 10 < 3000  ; for all n > 10; n 30000 : ¼ n  n  In other words, 10n! is eventually dominated by ln , for l ¼ 30000.  This  latter 1 sequence is null, by the Multiple Rule applied to thenull sequence : n  It follows, fromthe  Squeeze Rule and the fact that 30000 is a null sequence, n n & that the sequence 10n! is also null. Problem 6 Prove that n the ofollowingnsequences o are null:  1  ð1Þn sin n2 (a) n2 þn ; ; (c) n2 þ2n : (b) n! We can now list a good number of generic types of null sequences, of which we have seen specific instances already in this sub-section. We call these basic null sequences, since we shall use them commonly together with the Combination Rules and other rules for null sequences in order to prove that particular sequences that we meet are themselves null. Basic null sequences   (a) n1p ; for p > 0; (b) fcn g; for jcj < 1; p n (c) fn c g; for p > 0 and jcj < 1;  n for any real c; (d) cn! ; np  for p > 0: (e) n! ;

2.2.2

Instances of these that we have seen are 

1 pffiffiffi ; n  n 1 ;  2 n 1 n ; 2  n 10 ; n!  10 n : n!

Proofs

We now give a number of proofs which were omitted from the previous subsection so as not to slow down your gaining an understanding of the key ideas there. First, recall the definition of a null sequence. Definition

A sequence {an} is null if

We suggest that you omit these proofs on a first reading, and return to them when you are confident that you understand the basic ideas.

for each positive number ", there is a number X such that jan j < ";

for all n > X:

Proofs of the Power Rule and the Combination Rules

Recall that X need not be an integer; any appropriate real number will serve.

In the previous sub-section we proved the Squeeze Rule, but did not prove the Power Rule or the Combination Rules. We now supply these proofs. Power Rule  If{an} is a null sequence, where an  0, for n ¼ 1, 2, . . ., and, if p>0, then apn is a null sequence.

Recall that anp ¼ (an)p.

2.2

Null sequences

Proof

49

  We want to prove that apn is null; that is:

for each positive number ", there is a number X such that apn < ";

for all n > X:

(5)

For janpj ¼ anp, since an  0.

(6)

Here we use the number "p in place of " in the definition of null sequence.

We know that {an} is null, so there is some number X such that 1

an < " p ;

1

for all n > X:

Taking the pth power of both sides of (6), we obtain the desired result (5), with & the same value of X.

Remark 1

Notice how we used the (positive) number "p in place of " in (6) in order to obtain " in the final result (5). In the proofs which follow, we again apply the definition of null sequence, using positive numbers which depend in some way on ", in order to obtain " in the inequality that we are aiming to prove. Sum Rule sequence. Proof

If {an} and {bn} are null sequences, then {an þ bn} is a null

We want to prove that the sum {an þ bn} is null; that is:

for each positive number ", there is a number X such that jan þ bn j < "; for all n > X:

(7)

We know that {an} and {bn} are null, so there are numbers X1 and X2 such that 1 jan j < "; for all n > X1 ; 2 and 1 jbn j < "; for all n > X2 : 2 Hence, if X ¼ max {X1, X2}, then both the two previous inequalities hold; so if we add them we obtain, by the Triangle Inequality, that jan þ bn j  jan j þ jbn j 1 1 < " þ " ¼ "; for all n > X: 2 2 Thus inequality (7) holds, with this value of X.

We use 12 " here rather than ", in order to end up with the symbol " on its own eventually in the desired inequality (7). You met the Triangle Inequality in Subsection 1.3.1.

&

Before going on, first a comment about the number 12 that appears several times in the above proof. It is used twice in expressions 12 " at the start of the proof in order to end up with a final inequality that says that some expression is ‘ X; where K is a positive real number that does not depend on " or X. Then {an} is a null sequence.

In this case, the result that a particular sequence is null. Loosely speaking, we may express this result as ‘K" is just as good as "’ in the definition of null sequence. For example, K might be 2 or p7 or 259, but it could not be 2n X or 259 .

2: Sequences

50 Proof

We want to prove that the sequence {an} is null; that is:

You may omit this proof on a first reading.

for each positive number ", there is a number X such that jan j < ";

for all n > X:

Now, whatever this positive number " may be, the number K" is also a positive number. It follows from the hypothesis stated in the Lemma that therefore there is some number X such that " jan j < K  K ¼ "; for all n > X: This is precisely the condition for {an} to be null.

&

From time to time we shall use this Lemma in order to avoid arithmetic complexity in our proofs. Multiple Rule If {an} is a null sequence, then {lan} is a null sequence for any real number l. Proof

We want to prove that the multiple {lan} is null; that is:

for each positive number ", there is a number X such that jlan j < ";

for all n > X:

(8)

If l ¼ 0, this is obvious, and so we may assume that l 6¼ 0. We know that {an} is null, so there is some number X such that j an j
X:

Thus the desired result (8) holds.

&

If {an} and {bn} are null sequences, then {anbn} is a null

Product Rule sequence. Proof

We use j1lj " here rather than " in order to end up eventually with the symbol " on its own in the desired inequality (8).

We want to prove that the product {anbn} is null; that is:

for each positive number ", there is a number X such that jan bn j < ";

for all n > X:

(9)

We know that {an} and {bn} are null, so there are numbers X1 and X2 such that pffiffiffi jan j < "; for all n > X1 ; and j bn j
X:

In terms of the "  X game, this simply means that whatever choice of " is made by player A, player B can always win by making the choice X ¼ 3"! I'll play X = 3ε

I'll play ANY positive

ε

This is the case because, if 3 ½¼ X; " then we have " > 3n or 3n < "; hence, for n > X, we have n>

jan  4j ¼

3 3 < ¼ ": n X

B wins A

B

3. If a sequence is convergent, then it has a unique limit. We can see this by using a sequence diagram. Suppose that the sequence {an} has two limits, ‘ and m, where ‘ 6¼ m. Then it is possible to choose a positive number " such that horizontal strips from ‘  " up to ‘ þ " and from m  " up to m þ " do not overlap. For example, we can take " ¼ 13 j‘  mj. m+ε m m–ε

The vertical distance between ‘ and m is j‘  mj, so that any choice of " that is less than half of this quantity will serve.

–m

+ε  –ε n

However, since ‘ and m are both limits of {an}, the terms an must eventually lie inside both strips, and this is impossible. A formal proof of this fact is given in the Corollary to Theorem 3, later in this section. 4. If a given sequence converges to ‘, then this remains true if we add, delete or alter a finite number of terms. This follows from the corresponding result for null sequences. 5. Not all sequences are convergent. For example, the sequence {(1)n} is not convergent. In this section we restrict our attention to sequences which do converge.

In other words ‘altering a finite number of terms does not matter’. See Subsection 2.2.1. We discuss non-convergent sequences in Section 2.4.

2.3

Convergent sequences

2.3.2

55

Combination Rules for convergent sequences

So far you have tested the convergence of a given sequence {an} by calculating an  ‘ and showing that {an  ‘} is null. This presupposes that you know in advance the value of ‘. Usually, however, you are not given the value of ‘. You are given only a sequence {an} and asked to decide whether or not it converges and, if it does, to find its limit. Fortunately many sequences can be dealt with by using the following Combination Rules, which extend the Combination Rules for null sequences:

Sub-section 2.2.1.

Theorem 1 Combination Rules If lim an ¼ ‘ and lim bn ¼ m, then: n!1

n!1

Sum Rule Multiple Rule Product Rule Quotient Rule

lim ðan þ bn Þ ¼ ‘ þ m;

n!1

lim ðlan Þ ¼ l‘;

n!1

for any real number l;

lim ðan bn Þ ¼ ‘m;  an ‘ lim ¼ ; provided that m 6¼ 0: n!1 bn m n!1

Remarks 1. In applications of the Quotient Rule, it may happen that some of the terms bn take the value 0, in which case abnn is not defined. However, we shall see (in Lemma 1) that this can happen only for finitely many bn (because m ¼ 6 0), and so {bn} is eventually non-zero. Thus the statement of the rule does make sense. 2. The following rule is a special case of the Quotient Rule: If lim an ¼ ‘ and ‘ 6¼ 0, then: n!1 1 1 Reciprocal Rule lim ¼ . n!1 an ‘

Corollary 1

We shall prove the Combination Rules at the end of this sub-section, but first we illustrate how to apply them.

Applying the Combination Rules Example 1 Show that each of the following sequences {an} is convergent, and find its limit: þ 1Þðn þ 2Þ (a) an ¼ ð2n3n ; 2 þ 3n

2

n

(b) an ¼ 2nn! þþ3n103 .

Solution Although the expressions for an are quotients, we cannot apply the Quotient Rule immediately, because the sequences defined by the numerators and denominators are not convergent. In each case, however, we can rearrange the expressions for an and then apply the Combination Rules. (a) In this case we divide both the numerator and denominator by n2 to give    2 þ 1n 1 þ 2n ð2n þ 1Þðn þ 2Þ ¼ : an ¼ 3n2 þ 3n 3 þ 3n

A finite number of terms do not matter.

2: Sequences

56 Since

1 n

is a basic null sequence, we find by the Combination Rules that ð2 þ 0Þð1 þ 0Þ 2 ¼ : lim an ¼ n!1 3þ0 3

(b) This time we divide both the numerator and denominator by n! to give 2

n

2n2 þ 10n 2nn! þ 10n! an ¼ ¼ 3 n! þ 3n3 1 þ 3nn!  3  2  n Since nn! , 10n! and nn! are all basic null sequences, we find by the Combination Rules that lim an ¼

n!1

0þ0 ¼ 0: 1þ0

&

See the list of basic null sequences (introduced in Sub-section 2.2.1) which is repeated in the margin below.

Remark We simplified each of the above sequences by dividing both numerator and denominator by the dominant term. In part (a), we divided by n2, which is the highest power of n in the expression. In part (b), the choice of dominant term was a little harder, but the choice of n! ensured that the resulting quotients in the numerator and denominator were all typical terms of convergent sequences. In choosing the dominant term the following ordering is often useful: Domination Hierarchy A factorial term n! dominates a power term cn for any c 2 R. A power term cn dominates a term np for p > 0, jcj < 1. The above examples illustrate the following general strategy: Strategy

Basic  null sequences: 1 ; for p > 0; np

To evaluate the limit of a complicated quotient:

1. Identify the dominant term, bearing in mind the basic null sequences. 2. Divide both the numerator and the denominator by the dominant term.

fcn g;

3. Apply the Combination Rules.

 n c ;  n!p n ; n!

Problem 3 Show that each of the following sequences {an} is convergent, and find its limit: 3

(a) an ¼ n

þ 2n2 þ 3 2n3 þ 1 ;

2

n

2 (b) an ¼ n3n þ þ n3 ;

n

ð1Þ (c) an ¼ n!þ 2n þ 3n! :

Proofs of the Combination Rules We prove the Sum Rule, the Multiple Rule and the Product Rule by using the corresponding Combination Rules for null sequences. Remember that lim an ¼ ‘

n!1

means that fan ! ‘g is a null sequence:

for jcj < 1;

fnp cn g; for p > 0; c > 1; for c 2 R; for p > 0:

Warning Notice that ‘3n!’ means ‘3  (n!)’ and not ‘(3n)!’.

You may omit these proofs at a first reading.

2.3

Convergent sequences

57

Sum Rule If lim an ¼ ‘ and lim bn ¼ m, then n!1 n!1 lim ðan þ bn Þ ¼ ‘ þ m: n!1

Proof

By assumption, {an  ‘} and {bn  m} are null sequences; since ðan þ bn Þ  ð‘ þ mÞ ¼ ðan  ‘Þ þ ðbn  mÞ

we deduce that {(an þ bn)  (‘ þ m)} is null, by the Sum Rule for null & sequences. Product Rule

If lim an ¼ ‘ and lim bn ¼ m, then n!1

n!1

lim ðan bn Þ ¼ ‘m:

n!1

Proof

The idea here is to express anbn  ‘m in terms of an  ‘ and bn  m an bn  ‘m ¼ ðan  ‘Þðbn  mÞ þ mðan  ‘Þ þ ‘ðbn  mÞ:

Since {an  ‘} and {bn  m} are null, we deduce that {anbn  ‘m} is null, by & the Combination Rules for null sequences.

Remark Note that the Multiple Rule is just a special case of the Product Rule in which the sequence {bn} is a constant sequence. To prove the Quotient Rule, we need to use the following lemma, which will also be needed in the next sub-section: Lemma 1

If lim an ¼ ‘ and ‘ > 0, then there is a number X such that n!1 1 an > ‘, for all n > X: 2

Proof Since 12 ‘ > 0, the terms an must eventually lie within a distance 12 ‘ of the limit ‘. {an} 3 2

|an – | < 12 

 1 2

1  < an < 32  2

X

n

In other words, there is a number X such that 1 jan  ‘j < ‘; for all n > X: 2 Hence 1 1  ‘ < an  ‘ < ‘; for all n > X; 2 2 and so the left-hand inequality gives 1 ‘ < an ; for all n > X; 2 as required.

&

Here we are taking " ¼ 12 ‘ in the definition of convergence.

2: Sequences

58

If lim an ¼ ‘ and lim bn ¼ m, then n!1  n!1  an ‘ lim ¼ , provided that m 6¼ 0: n!1 bn m

Quotient Rule

Proof We assume that m > 0; the proof for the case m < 0 is similar. Once again the idea is to write the required expression in terms of an  ‘ and bn  m an ‘ mðan  ‘Þ  ‘ðbn  mÞ :  ¼ bn m bn m Now, however, there is a slight problem: {m(an  ‘)  ‘(bn  m)} is certainly a null sequence, but the denominator is rather awkward. Some of the terms bn may take the value 0, in which case the expression is undefined. However, by Lemma 1, we know that that for some X we have 1 bn > m; for all n > X: 2 Thus, for all n > X an ‘ jmðan  ‘Þ  ‘ðbn  mÞj  ¼ b m bm n

n

jmðan  ‘Þ  ‘ðbn  mÞj  1 2 2m jmj  jan  ‘j þ j‘j  jbn  mj  : 1 2 2m

In particular, this implies that the terms of {bn} are eventually positive.

Here we use Lemma 1. Here we apply the Triangle Inequality to the numerator.

Since this last expression defines a null sequence, it follows by the Squeeze   & Rule that abnn  m‘ is null.

2.3.3

Further properties of convergent sequences

There are several other theorems about convergent sequences which will be needed in later chapters. The first is a general version of the Squeeze Rule. Theorem 2 If:

You met the Squeeze Rule for null sequences in Sub-section 2.2.1.

Squeeze Rule

1. bn  an  cn, for n ¼ 1, 2, . . ., 2. lim bn ¼ lim cn ¼ ‘, n!1

n!1

then lim an ¼ ‘: n!1

Proof

By the Combination Rules lim ðcn  bn Þ ¼ ‘  ‘ ¼ 0 ; n!1

so that fcn  bn g is a null sequence. Also, by condition 1 0  an  bn  cn  bn ; for n ¼ 1; 2; . . .; and so fan  bn g is null, by the Squeeze Rule for null sequences.

That is, fan  bn g is squeezed by fcn  bn g.

2.3

Convergent sequences

59

Now we write an in the form an ¼ ð an  bn Þ þ bn : Hence by the Combination Rules lim an ¼ lim ðan  bn Þ þ lim bn

n!1

n!1

n!1

&

¼ 0 þ ‘ ¼ ‘:

Remark In applications of the Squeeze Rule, it is sufficient to check that condition 1 applies eventually. This is because the values of a finite number of terms do not affect convergence.

A finite number of terms do not matter.

The following example and problem illustrate the use of the Squeeze Rule and the Binomial Theorem in the derivation of two important limits. Example 2 (a) Prove that, if c > 0, then 1 c ð1 þ c Þn  1 þ ; n (b) Deduce that

for n ¼ 1; 2; . . .:

We proved this inequality for the case c ¼ 1 in Example 6(b) of Sub-section 1.3.3.

1

lim an ¼ 1; for any positive number a:

n!1

Solution (a) Using the rules for inequalities, we obtain

1 c c n ð1 þ c Þn  1 þ , 1 þ c  1 þ ; since c > 0 : n n The right-hand inequality holds, because

c c n 1þ 1þn ¼ 1 þ c; n n by the Binomial Theorem. It follows that the required inequality also holds. (b) We consider the cases a > 1, a ¼ 1, a < 1 separately. If a > 1, then we can write a ¼ 1 þ c, where c > 0. Then, by part (a) 1 c 1 1  an ¼ ð1 þ cÞn  1 þ ; for n ¼ 1; 2; . . .: n   Since lim 1 þ nc ¼ 1, we deduce that

All the terms in the Binomial Theorem expansion of ð1 þ ncÞn are positive, so we get a smaller number if we ignore all the terms after the first two.

n!1

1

lim an ¼ 1;

n!1

by the Squeeze Rule. 1 If a ¼ 1, then an ¼ 1, for n ¼ 1, 2, . . ., so 1

lim an ¼ 1:

n!1

 1 If 0 < a < 1, then 1a > 1, so that lim 1a n ¼ 1 by the first case in part (b). n!1 Hence, by the Reciprocal Rule 1 1 1 & lim an ¼ 11n ¼ 1 ¼ 1: n!1 lim n!1 a

In this application of the Squeeze Rule, the ‘lower’ sequence is {1}.

2: Sequences

60 Problem 4 (a) Use the Binomial Theorem to prove that rffiffiffiffiffiffiffiffiffiffiffi 2 1 n n 1 þ ; for n ¼ 2; 3; . . .: n1 Þ 2 ð1 þ xÞn  nðn1 2! x ;

Hint:

for n  2; x  0:

(b) Use the Squeeze Rule to deduce that 1

lim nn ¼ 1:

n!1

Next we show that taking limits preserves weak inequalities. Theorem 3 Limit Inequality Rule If lim an ¼ ‘ and lim bn ¼ m, and also n!1

n!1

an  bn ;

for n ¼ 1; 2; . . .;

then ‘  m. Proof Suppose that an ! ‘ and bn ! m, where an  bn for n ¼ 1, 2, . . ., but that it is not true that ‘  m. Then ‘ > m and so, by the Combination Rules lim ðan  bn Þ ¼ ‘  m > 0:

n!1

Hence, by Lemma 1, there is an X such that 1 (2) an  bn > ð‘  mÞ; for all n > X: 2 However, we assumed that an  bn  0, for n ¼ 1, 2, . . ., so statement (2) is a contradiction. & Hence it is true that ‘  m.

Warning Taking limits does

NOT preserve strict inequalities. That is, if lim an ¼ ‘ and lim bn ¼ m, and also an < bn, for n ¼ 1, 2, . . ., then it follows

n!1

n!1

from Theorem 3 that ‘  m – but it does NOT necessarily follow that ‘ < m. 1 1 < 1n, for n ¼ 1, 2, . . ., but lim 2n ¼ lim 1n ¼ 0. For example, 2n n!1

n!1

We can now give the formal proof, promised earlier, that a convergent sequence has only one limit. Corollary 2

If lim an ¼ ‘ and lim an ¼ m, then ‘ ¼ m. n!1

n!1

Proof Applying Theorem 3 with bn ¼ an, we deduce that ‘  m and m  ‘. & Hence ‘ ¼ m. In Sub-section 2.2.1, we saw that a sequence {an} is null if and only if the sequence fj an jg is null. Our next result is a partial generalisation of this fact. Theorem 4

If lim an ¼ ‘, then lim jan j ¼ j‘j. n!1

n!1

This is a ‘proof by contradiction’.

2.4

Divergent sequences

Proof

Using the ‘reverse form’ of the Triangle Inequality, we obtain jan j  j‘j  jan  ‘j; for n ¼ 1; 2; . . .:

Since {an  ‘} is null, we deduce from the Squeeze Rule for null sequences that & fjan j  j‘jg is null, as required.

61 You met this in Sub-section 1.3.1, expressed in the form ja  bj  jaj  jbj . Here we are writing an in place of a and ‘ in place of b.

Remark Theorem 4 is only a partial generalisation of the earlier result about null sequences, because the converse statement does not hold. That is, if janj ! j‘j, it does NOT necessarily follow that an ! ‘. For example, consider the sequence an ¼ (1)n, for n ¼ 1, 2, . . .; in this case, janj ! 1, but {an} does not even converge.

2.4 2.4.1

Divergent sequences What is a divergent sequence?

We have commented several times that not all sequences are convergent. We now investigate the behaviour of sequences which do not converge. Definition

A sequence is divergent if it is not convergent.

Here are the sequence diagrams for fð1Þn g, {2n} and fð1Þn ng. Each of these sequences is divergent but, as you can see, they behave differently.

It is rather tricky to prove, directly from the definition, that these sequences are divergent. The aim of this section is to obtain criteria for divergence, which avoid having to argue directly from the definition. At the end of the section, we give a strategy involving two criteria which deal with all types of divergence. We obtain these criteria by establishing certain properties, which are necessarily possessed by a convergent sequence; if a sequence does not have these properties, then it must be divergent.

For example, to show that the sequence fð1Þn g is divergent, we have to show that fð1Þn g is not convergent; that is, for every real number ‘, the sequence fð1Þn  ‘g is not null.

2: Sequences

62

2.4.2

Bounded and unbounded sequences

One property possessed by a convergent sequence is that it must be bounded. Definition

A sequence {an} is bounded if there is a number K such that jan j  K;

for n ¼ 1; 2; . . .:

A sequence is unbounded if it is not bounded. Thus a sequence {an} is bounded if all the terms an lie on the sequence diagram in a horizontal strip from K up to K, for some positive number K. {an}

K

⏐an⏐ ≤ K

n

–K ≤ an ≤ K

–K

For example, the sequence fð1Þn g is bounded, because jð1Þn j  1;

for n ¼ 1; 2; . . .:

However the sequences {2n} and {n2} are unbounded, since, for each number K, we can find terms of these sequences whose absolute values are greater than K. Problem 1

Classify the following sequences as bounded or unbounded:    n  ; (d) 1  1n . (b) fð1Þn ng; (c) 2nþ1 n

(a) f1 þ ð1Þn g;

The sequence fð1Þn g shows that: a bounded sequence is not necessarily convergent. However we can prove that: a convergent sequence is necessarily bounded. Theorem 1 Boundedness Theorem If {an} is convergent, then {an} is bounded. Proof We know that an ! ‘, for some real number ‘. Thus {an  ‘} is a null sequence, and so there is a number X such that jan  ‘j < 1;

Take " ¼ 1 in the definition of a null sequence.

for all n > X:

For simplicity in the rest of the proof, we shall now assume that our initial choice of X is a positive integer. Now jan j ¼ jðan  ‘Þ þ ‘j  jan  ‘j þ j‘j; by the Triangle Inquality: It follows that jan j  1 þ j‘j; for all n > X: This is the type of inequality needed to prove that {an} is bounded, but it does not include the terms a1, a2, . . ., aX. To complete the proof, we let K be the maximum of the numbers ja1j, ja2j, . . ., jaXj, 1 þ j‘j.

K

+1  –1

X

n

–K

That is, K ¼ max{ja1j, ja2j, . . ., jaXj, 1 þ j‘j}.

2.4

Divergent sequences

63

It follows that jan j  K;

for n ¼ 1; 2; . . .;

&

as required.

From Theorem 1, we obtain the following test for the divergence of a sequence: If {an} is unbounded, then {an} is divergent.

Corollary 1

For example, the sequences {2n} and fð1Þn ng are both unbounded, so they are both divergent. Problem 2 Classify the following sequences as convergent or divergent, and as bounded unbounded: n or o    pffiffiffi n n2 þ n (a) f ng; (b) n2 þ 1 ; (c) ð1Þn n2 ; (d) nð1Þ :

2.4.3

Sequences which tend to infinity

Although the sequences {2n} and fð1Þn ng are both unbounded (and hence divergent), there is a marked difference in their behaviour. The terms of both sequences become arbitrarily large, but those of the sequence {2n} become arbitrarily large and positive. To make this precise, we must explain what we mean by ‘arbitrarily large and positive’. Definition

The sequence {an} tends to infinity if:

for each positive number K, there is a number X such that an > K;

for all n > X:

In this case, we write Often, we omit ‘as n ! 1’.

an ! 1 as n ! 1 :

Remarks 1. In terms of the sequence diagram for {an}, this definition states that, for each positive number K, the terms an eventually lie above the line at height K.

K {an}

X

n

2. If a sequence tends to infinity, then it is unbounded – and hence divergent, by Corollary 1.

2: Sequences

64 3. If a sequence tends to infinity, then this remains true if we add, delete or alter a finite number of terms. 4. In the definition we can replace the phrase ‘for each positive number K’ by ‘for each number K’; for, if the inequality ‘an > K, for all n > X’ holds for each number K, then in particular it certainly holds for each positive number K.

A finite number of terms do not matter.

There is a version of the Reciprocal Rule for sequences which tend to infinity. This enables us to use our knowledge of null sequences to identify sequences which tend to infinity. Theorem 2

Reciprocal Rule

(a) If the sequence {an} satisfies both of the following conditions: 1. {an} is eventually positive, 2. fa1n g is a null sequence, then an ! 1. (b) If an ! 1, then a1n ! 0. Proof of part (a)

To prove that an ! 1, we have to show that:

We prove only part (a): the proof of part (b) is similar.

for each positive number K, there is a number X such that an > K;

for all n > X:

(1)

Since {an} is eventually positive, we can choose a number X1 such that an > 0; Since

fa1n g

for all n > X1 :

is null, we can choose a number X2 such that 1 1 < ; for all n > X2 : a K n

Now, let X ¼ max {X1X2}; then 1 1 0 < < ; for all n > X: an K

Here we are taking " ¼ K1 in the definition of a null sequence. We make this choice of X so that BOTH of the preceding inequalities hold for all n > X.

This statement is equivalent to the statement (1), so an ! 1, as required. & Example 1 Use the Reciprocal Rule to prove that the following sequences tendnto o infinity: 3 (a) n2 ; (c) fn!  10n g (b) fn! þ 10n g;

Notice that, in parts (b) and (c), n! is the dominant term.

Solution

   3 (a) Each term of the sequence n2 is positive and n31=2 ¼ n23 . Now, n13 is a 2  3 basic null sequence and so n3 is null, by the Multiple Rule. Hence n2 tends to infinity, by the Reciprocal Rule. (b) Each term of the sequence fn! þ 10n g is positive and 1 1 0 ¼ 0; lim ¼ lim n! 10n ¼ n n!1 n! þ 10 n!1 1 þ 1 þ 0 n! by the Combination Rules. Hence fn! þ 10n g tends to infinity, by the Reciprocal Rule.

Alternatively, since 1  1 , the sequence n!  þ 101 n  n! n! þ 10n is null, by the Squeeze Rule for null sequences.

2.4

Divergent sequences

(c) First note that

65 

 10n n!  10 ¼ n! 1  ; for n ¼ 1; 2; . . .: n! 10n  n Since n! is a basic null sequence, we know that 10n! is eventually less n than 1, and so n!  10 is eventually positive. Also n

1 n!

lim

n!1 n!

1 0 ¼ 0; ¼ lim n ¼ 10 n n!1  10 10 1  n!

by the Combination Rules. Hence {n!  10n} tends to infinity, by the Reciprocal Rule.

This follows by taking " ¼ 1 in the definition of a null sequence.

&

There are also versions of the Combination Rules and Squeeze Rule for sequences which tend to infinity. We state these without proof. Theorem 3 Combination Rules If {an} tends to infinity and {bn} tends to infinity, then: Sum Rule Multiple Rule

{an þ bn} tends to infinity; {lan} tends to infinity, for l > 0;

Product Rule

{anbn} tends to infinity.

Theorem 4 Squeeze Rule If {bn} tends to infinity, and an  bn ;

for n ¼ 1; 2; . . .;

then {an} tends to infinity. Problem 3  n (a) 2n ;

For each of the following sequences {an}, prove thatnan ! 1: o   n  n 2 (b) 2n  n100 ; (c) 2n þ 5n100 ; (d) 2n10þþnn .

We can also define {an} tends to minus infinity. Definition The sequence {an} tends to minus infinity if an ! 1 as n ! 1: In this case, we write an ! 1

as n ! 1:

For example, the sequences {n2} and f10n  n!g both tend to minus infinity, because {n2} and fn!  10n g tend to infinity. Sequences which tend to minus infinity are unbounded, and hence divergent. However, the sequence fð1Þn ng shows that an unbounded sequence need not tend to infinity or to minus infinity.

2.4.4

Subsequences

We now give two useful criteria for establishing that a sequence diverges; both of them involve the idea of a subsequence. For example, consider the bounded divergent sequence fð1Þn g. This sequence splits naturally into two:

This follows from the fact that sequences which tend to infinity are unbounded.

2: Sequences

66 the even terms a2, a4, . . ., a2k, . . ., each of which equals 1;

{(–1)n}

the odd terms a1, a3, . . ., a2k 1, . . ., each of which equals 1.

{a2k}

1

Both of these are sequences in their own right, and we call them the even subsequence {a2k} and the odd subsequence {a2k  1}. In general, given a sequence {an} we may consider many different subsequences, such as:

1

2

3

4

5

7

n

{a2k – 1}

–1

{a3k}, comprising the terms a3, a6, a9, . . .; {a4kþ1}, comprising the terms a5, a9, a13, . . .;

6

{a2k!}, comprising the terms a2, a4, a12, . . .. Definition The sequence fank g is a subsequence of the sequence {an} if {nk} is a strictly increasing sequence of positive integers; that is, if n1 < n2 < n3 < . . .:

The sequence fank g is the sequence an1 ; an2 ; an3 ; . . .:

For example, the subsequence {a5kþ2} corresponds to the sequence of positive integers nk ¼ 5k þ 2; k ¼ 1; 2; . . .:

Note that the sequence {an} is a subsequence of itself.

The first term of {a5k þ 2} is a7, the second is a12, the third is a17, and so on. Notice that any strictly increasing sequence {nk} of positive integers must tend to infinity, since it can be proved by Mathematical Induction that nk  k;

for k ¼ 1; 2; . . .:

Problem 4 (a) Let an ¼ n2, for n ¼ 1, 2, . . .. Write down the first five terms of each of the subsequences fank g, where: (i) nk ¼ 2k; (ii) nk ¼ 4k  1; (iii) nk ¼ k2. (b) Write down the first three terms of the odd and even subsequences of n the following sequence: an ¼ nð1Þ ; n ¼ 1; 2; . . .: Our next theorem shows that certain properties of sequences are inherited by their subsequences. Theorem 5

Inheritance Property of Subsequences

For any subsequence fank g of {an}: (a) if an ! ‘ as n ! 1, then ank ! ‘

as k ! 1;

(b) if an ! 1

as k ! 1.

as n ! 1, then ank ! 1

A result similar to part (b) holds for sequences where an ! 1.

Proof of part (a) We want to show that for each positive number ", there is a number K such that jank  ‘j < ";

for all k > K:

(2)

However, since {an  ‘} is null, we know that there is a number X such that for all n > X: jan  ‘j < ";

You may omit this proof at a first reading.We prove only part (a): the proof of part (b) is similar.

2.4

Divergent sequences

For simplicity in the rest of the proof, we shall now assume that our initial choice of X is a positive integer. So, if we take K so large that nK  X ; then nk > nK  X; for all k > K; and so jank  ‘j < "; for all k > K; & which proves (2).

67 We will occasionally make this assumption if we want to refer to terms such as aX rather than mess about with nasty expressions such as a[X], where [X] is the integral part of X.

The following criteria for establishing that a sequence is divergent are immediate consequences of Theorem 5, part (a): Corollary 2 1. First Subsequence Rule The sequence {an} is divergent if it has two convergent subsequences with different limits. 2. Second Subsequence Rule The sequence {an} is divergent if it has a subsequence which tends to infinity or a subsequence which tends to minus infinity.

Though we do not prove the fact, ANY divergent sequence is of one (or both) of these two types.

We can now formulate the strategy promised at the beginning of this section. Strategy

To prove that a sequence {an} is divergent:

EITHER

1. show that {an} has two convergent subsequences with different limits OR

The fact that this strategy will always apply is simply a reformulation of the result mentioned in the above margin note.

2. show that {an} has a subsequence which tends to infinity or a subsequence which tends to minus infinity. For example, the sequence fð1Þn g has two convergent subsequences which have different limits: namely, the even subsequence with limit 1 and the odd subsequence with limit 1. So the sequence fð1Þn g is divergent, by the First Subsequence Rule.   Þn a subsequence (the even On the other hand, the sequence nð1  ð1Þhas n subsequence) which tends to infinity. So n is divergent by the Second Subsequence Rule.

Remark In order to apply the above strategy successfully to prove that a sequence is divergent, you need to be able to spot convergent subsequences with different limits or subsequences which tend to infinity or to minus infinity. It is not always easy to do this, and some experimentation may be required! If the formula for an involves the expression (1)n, it is a good idea to consider the odd and even subsequences, although this may not always work. It may be helpful to calculate the values of the first few terms in order to try to spot some subsequences whose behaviour can be identified. Problem 5 Use the above strategy to prove that each of the following sequences {an} is divergent:         (a) ð1Þn þ 1n ; (c) n sin 12 np . (b) 13 n  13 n ;

In part (b) the square brackets denote ‘the integer part’ function.

2: Sequences

68 We end this section by giving a result about subsequences which will be needed in later chapters. Theorem 6 If the odd and even subsequences of {an} both tend to the same limit ‘, then lim an ¼ ‘: n!1

We want to show that:

Proof

for each positive number ", there is a number X such that jan  ‘j < "; for all n > X: We know that there are integers K1 and K2 such that ja2k1  ‘j < "; for all k > K1 ; and ja2k  ‘j < "; for all k > K2 : So we now let

(3)

X ¼ maxf2K1  1; 2K2 g: With this choice of X, each n > X is either of the form 2k  1, with k > K1, or of the form 2k, with k > K2; it follows then that (3) holds with this value of X. &

2.5

The Monotone Convergence Theorem

2.5.1

Monotonic sequences

In Section 2.3, we gave various techniques for finding the limit of a convergent sequence. As a result, you may be under the impression that, if we know that a sequence converges, then there is some way of finding its limit explicitly. However, it is sometimes possible to prove that a sequence is convergent, even though we do not know its limit. For example, this situation occurs with a given sequence {an}, which has the following two properties: 1. {an} is an increasing sequence; 2. {an} is bounded above; that is, there is a real number M such that an  M; for n ¼ 1; 2; . . .: Likewise, if {an} is a sequence which is decreasing and bounded below, then {an} must be convergent. We combine these two results into one statement. Theorem 1

Monotone Convergence Theorem

If the sequence {an} is: EITHER

1. increasing and 2. bounded above OR

1. decreasing and 2. bounded below, then {an} is convergent.

{a2k1} and {a2k} are the odd and even subsequences of {an}.

2.5

The Monotone Convergence Theorem

69

Proof of the first case Let {an} be a sequence that is increasing and bounded above. Since {an} is bounded above, the set {an: n ¼ 1, 2, . . .} has a least upper bound, ‘ say; this is true by the Least Upper Bound Property of R. We now prove that lim an ¼ ‘. n!1 We want to show that:

We prove only the increasing version; the proof of the decreasing version is similar. You met this Property in Sub-section 1.4.3.

for each positive number ", there is a number X such that: (1)

+ε

We know that, if " > 0, then, since ‘ is the least upper bound of the set {an: n ¼ 1, 2, . . .}, there is an integer X such that aX > ‘  ":



jan  ‘j < ";

for all n > X:

X

Since {an} is increasing, an  aX, for n > X, and so an > ‘  ";

for all n > X: for all n > X;

which proves (1). Hence {an} converges to ‘.

n

This follows from the definition of least upper bound.

Thus jan  ‘j ¼ ‘  an < ";

–ε

jan  ‘j ¼ ‘  an, because an  ‘.

&

 The1Monotone Convergence Theorem tells us that a sequence such as must 1  n , which is increasing and bounded above (by 1, for example),   be convergent. In this case, of course, we know already that 1  1n is convergent (with limit 1) without using the Monotone Convergence Theorem. The Monotone Convergence Theorem is often used when we suspect that a sequence is convergent, but cannot find the actual limit. It can also be used to give precise definitions of numbers, such as p, about which we have only an intuitive idea, and we do this later in this section. For completeness, we point out that:

We will use the Monotone Convergence Theorem in precisely this way in Chapter 3. Sub-section 2.5.4.

If {an} is increasing but is not bounded above, then an ! 1 as n ! 1. Indeed, for any real number K, we can find an integer X such that aX > K; because {an} is not bounded above. Since {an} is increasing, an  aX for n > X, and so an > K; for all n > X: Hence an ! 1 as n ! 1. Similarly, we have the following result: If {an} is decreasing but is not bounded below, then an ! 1 as n ! 1. We summarise these results about monotonic sequences in the following useful theorem: Theorem 2

Monotonic Sequence Theorem

If the sequence {an} is monotonic, then: EITHER OR

{an} is convergent an ! 1.

The expression ‘an ! 1’ is shorthand for ‘either an ! 1 or an ! 1’.

2: Sequences

70 Next we note a consequence of the Monotone Convergence Theorem that is sometimes useful. Corollary 1 (a) If a sequence {an} is increasing and bounded above, then it tends to its least upper bound, sup {an: n ¼ 1, 2, . . .}. (b) If a sequence {an} is decreasing and bounded below, then it tends to its greatest lower bound, inf {an: n ¼ 1, 2, . . .}. Problem 1

Prove part (b) of the Corollary.

We meet applications of these results in the rest of this section, to functions defined recursively and to the study of the numbers e and p. But our first application is to a result of considerable value in Analysis, Topology and other parts of Mathematics.

The proof of part (a) is contained in the proof of Theorem 1 above.

This proof is similar to the earlier discussion.

We shall use it in Section 4.2.

The Bolzano–Weierstrass Theorem We have seen that if a sequence is convergent, then certainly it must be bounded. However if a sequence is bounded, it does not follow that it is necessarily convergent. For example, the sequence {an}, where an ¼ 1n, is bounded (by 1, since jan j ¼ j 1n j  1, 2, . . .) and is also convergent (to the n limit 0). Yet the sequence {an}, where an ¼ (1) , is also bounded (by 1, since n jan j ¼ ð1Þ ¼ 1, for n ¼ 1, 2, . . .) but is divergent (since its odd and even subsequences tend to different limits). Our next result shows that, if a sequence is bounded, then, even though it may diverge, it cannot behave ‘too badly’! Theorem 3

This was the Boundedness Theorem – Theorem 1, Sub-section 2.4.2.

Bolzano–Weierstrass Theorem

A bounded sequence must contain a convergent subsequence. That is, if a sequence {an} is such that janj  M, for all n, then there exist some number ‘ in [ M, M] and some subsequence fank g such that fank g converges to ‘ as k ! 1. For example, the sequence {sin n} is bounded. With the tools at our disposal, it is not at all clear what the behaviour of this sequence might be as n ! 1! However, the Bolzano–Weierstrass Theorem asserts that there is at least one number ‘ in [1, 1] such that some subsequence {sin nk} converges to ‘. This is itself a surprising result! In fact, however, using quite sophisticated mathematics we can prove that every number ‘ in [1, 1] has this property! The proof of the Bolzano–Weierstrass Theorem that we give is of interest in its own right; it depends on the notion of repeated bisection, which is a standard technique in many areas of Analysis. Proof We shall assume that the bounded sequence {an} has the property that janj  M, for all n. Then all the terms an lie in the closed interval [M, M], which we will denote as [A1, B1]. Next, denote by p the midpoint of the interval [A1, B1]. Then at least one of the two intervals [A1, p] and [p, B1] must contain infinitely many terms in the

For jsin nj  1.

Sadly, this is beyond the range of this book.

A1 ¼ M and B1 ¼ M. p ¼ 12 ðA1 þ B1 Þ.

2.5

The Monotone Convergence Theorem

sequence {an} – for otherwise the whole sequence would only contain finitely many terms. If only one of the two intervals has this property, denote that interval by the notation [A2, B2]; if both intervals have this property, choose the left one (that is, [A1, p]) to be [A2, B2]. In either case, we obtain: 1. ½A2 , B2  ½A1 , B1 ; 2. B2  A2 ¼ 12 ðB1  A1 Þ; 3. [A1, B1] and [A2, B2] both contain infinitely many terms in the sequence {an}.

71

If both intervals contain infinitely many terms in the sequence, it does not matter which choice we take; but we specify the left interval just for definiteness.

We now repeat this process indefinitely often, bisecting [A2, B2] to obtain [A3, B3], and so on. This gives a sequence of closed intervals {[An, Bn]} for which: 1. ½Anþ1 ; Bnþ1  ½An ; Bn , for each n 2 N;  n1 2. Bn  An ¼ 12 ðB1  A1 Þ, for each n 2 N; 3. each interval [An, Bn] contains infinitely many terms in the sequence {an}. Property 1 implies that the sequence {An} is increasing and bounded above by B1 ¼ M. Hence, by the Monotone Convergence Theorem, {An} is convergent; denote by A its limit. By the Limit Inequality Rule, we must have that A  M. Similarly, Property 1 implies that the sequence {Bn} is decreasing and bounded below by A1 ¼ M. Hence, by the Monotone Convergence Theorem, {Bn} is convergent; denote by B its limit. By the Limit Inequality Rule, we must have that B  M. We may then deduce, by letting n ! 1 in Property 2 and using the Combination Rules for sequences, that

Theorem 3, Sub-section 2.3.3.

B  A ¼ lim Bn  lim An ¼ lim ðBn  An Þ n!1 n!1 n!1  n1 1 ¼ lim ð B1  A 1 Þ n!1 2  n1 1 ¼ ðB1  A1 Þ lim ¼ 0: n!1 2 In other words, the sequences {An} and {Bn} both converge to a common limit. Denote this limit by ‘. Since ‘ ¼ A ¼ B, we must have M  ‘  M. We find a suitable subsequence of {an} as follows. Choose any an1 that lies in [A1, B1]; this is possible since [A1, B1] contains infinitely terms in {an}. Next, choose a term an2 in the sequence, with n2 > n1, such that an2 2 ½A2 ; B2 ; this is possible since [A2, B2] contains infinitely terms in {an}, by Property 3. And so on. In this way, we construct a subsequence fank g of {an}, with nk þ 1 > nk, for each k. Since Ak  ank  Bk , it follows from the Squeeze Rule for sequences & (that is, by letting k ! 1) that fank g converges to ‘, as required.

2.5.2

Sequences defined by recursion formulas

As we have seen, often sequences are defined by formulas; that is, for a given n, we substitute that value of n into a formula and obtain the term an in the sequence {an} Another way of specifying a sequence is to define its terms ‘inductively’, or ‘recursively’; here we specify the first term (or several terms) in the sequence, and then have a formula that enables us to calculate all successive terms.

That is, ‘ 2 [M, M].

Since nk þ 1 > nk, we must have nk  k, for each k. For Ak ! A ¼ ‘ and Bk ! B ¼ ‘.

2: Sequences

72 For example, the following sequences are defined recursively:   {an}, where a1 ¼ 2 and anþ1 ¼ 14 a2n þ 3 , for n  1; {an}, where a1 ¼ 2, a2 ¼ 4 and an ¼ 12 ðan1 þ an2 Þ, for n  3. Do these sequences converge? If they do converge, what are their limits? If we choose a different value for a1 (or for a1 and a2, in the second sequence), do their behaviours change? We can use the Monotone Convergence Theorem to answer many such questions. First, consider the sequence with recursion formula  1 anþ1 ¼ a2n þ 3 ; for n  1: (2) 4 If indeed {an} is convergent, what value could its limit take? If we let ‘ denote lim an and let n ! 1 in equation (2), we obtain ‘ ¼ 14 ð‘2 þ 3Þ. This can be n!1 rearranged as 4‘ ¼ ‘2 þ 3, and so as ‘2  4‘ þ 3 ¼ 0. Hence (‘  1)(‘  3) ¼ 0, so that ‘ = 1 or ‘ = 3. This suggests that we should look at the differences anþ1  1 and anþ1  3. Using equation (2), we deduce that  1 1 anþ1  1 ¼ a2n  1 ¼ ðan þ 1Þðan  1Þ; for n  1; (3) 4 4  1 1 anþ1  3 ¼ a2n  9 ¼ ðan þ 3Þðan  3Þ; for n  1: (4) 4 4

Of course we have not yet proved that {an} is convergent. This has been a ‘what if?’ discussion so far, but a useful one. Because 1 and 3 are the possible limits.

Next, it is useful to examine whether the sequence is monotonic by looking at the difference anþ1  an  1 1 anþ1  an ¼ a2n  4an þ 3 ¼ ðan  1Þðan  3Þ: (5) 4 4 So the sign of anþ1  an depends on where an lies on the number-line in relation to the numbers 1 and 3. For example, it follows from (5) that if 1 < an < 3, for some n  1, then anþ1  an < 0, so that anþ1 < an :

(6)

If an ¼ 1, for some number n  1, it follows from equation (3) that anþ1 ¼ 1. Consequently, if our initial term is a1 ¼ 1, then it follows that an ¼ 1, for all n  1; in other words, the sequence is simply the constant sequence {1}. Similarly, if an ¼ 3, for some number n  1, it follows from equation (3) that anþ1 ¼ 3. Consequently, if our initial term is a1 ¼ 3, then it follows that an ¼ 3, for all n  1; in other words, the sequence is simply the constant sequence {3}. Next, suppose that 1 < an < 3, for some number n  1. It follows from equation (3) that anþ1  1 is positive, so that 1 < anþ1. On the other hand, it follows from equation (4) that anþ1  3 is negative, so that anþ1 < 3. Thus, if 1 < an < 3, for some number n  1, then 1 < an þ 1 < 3. We can then prove by induction that, if 1 < a1 < 3, then 1 < an < 3, for all n > 1. (We omit the details.) Next, it follows, from this last fact and (6) above that, if 1 < a1 < 3, then the sequence {an} is strictly decreasing. Hence, if 1 < a1 < 3 the sequence {an} is strictly decreasing and is bounded below. Hence, by the Monotone Convergence Theorem, {an} is convergent.

We shall use this fact below.

In particular this discussion covers the case a1 ¼ 2 that we mentioned at the start of this sub-section. Now comes the coup de grace!

2.5

The Monotone Convergence Theorem

Since {an} is decreasing, whatever its limit might be (and we know that the only two possibilities for the limit are 1 and 3), its limit must be less than or equal to its first term a1. It follows that the limit of the sequence must be 1.

73 We saw that 1 and 3 were the only possible limits at the start of the discussion.

Problem2 Letthe sequence {an} be defined by the recursion formula anþ1 ¼ 14 a2n þ 3 , for n  1. (a) Prove that, if a1 > 3, then an ! 1 as n ! 1. (b) In the case that 0  a1 < 1, determine whether {an} is convergent and, if so, to what limit. (c) Describe the behaviour of the sequence {an} in the case that a1 < 0.

2.5.3

The number e

 n  We will define e to be the limit of the sequence 1 þ 1n . If we plot the first few terms of this sequence on a sequence diagram, then it seems that the sequence is increasing and converges to a limit, which is less than 3.

 n  is convergent, using the Monotone Convergence To show that 1 þ 1n Theorem, we prove that the sequence is increasing and is bounded above.  n  1. 1 þ 1n is increasing By the Binomial Theorem        n 1 n 1 nð n  1Þ 1 2 . . . 1 ¼1þn þ þ : þ 1þ n n 2! n n The general term in this expansion is   nð n  1Þ . . . ð n  k þ 1Þ 1 k k! n      1 1 2 k1 (7) 1 ¼ 1 ... 1  : k! n n n If k is fixed, then each of the factors      1 2 k1 1  ; 1  ;...; 1  n n n is increasing, and so the general term (7) increases  as n increases. Since this is true for each fixed k, the sequence ð1 þ 1nÞn is increasing.  n  2. 1 þ 1n is bounded above Here, note that the general term (7) satisfies the inequality      1 1 2 k1 1 1 1 ... 1   ; k! n n n k!

For example, to three decimal places the first two terms are 2 and 2.25, the sixth term is 2.521, and the hundredth term is 2.704.

2: Sequences

74 since each of the brackets is at most 1. Hence   1 n 1 1 1  1 þ 1 þ þ þ ... þ 1þ n 2! 3! n! 1 1 1  1 þ 1 þ 1 þ 2 þ . . . þ n1 ; 2 2 2 since k! ¼ k (k  1) . . . 2.1  2k1. Now  n  n   1  12 1 1 1 1 ¼2 1 1 þ 1 þ 2 þ þ n1 ¼ 1  1 2 2 2 12 2 1 ¼ 2  n1 ; 2 and so   1 n 1 1þ  3  n1 ; for n ¼ 1; 2; . . .: n 2  n  is bounded above, by 3. Thus the sequence 1 þ 1n

For the sum of the geometric progression a þn ar þ ar2 þ . . . þ arn1 is a 1r 1r , if r 6¼ 1; here we have a ¼ 1, r ¼ 12 :

 It follows, n  by the Monotone Convergence Theorem, that the sequence 1 þ 1n is convergent with limit at most 3. This allows us to make the following definition:  n Definition e ¼ lim 1 þ 1n : n!1

 n For larger and larger values of n, the terms 1 þ 1n give  better n  and better approximate values for e. Unfortunately, the sequence 1 þ 1n converges to e rather slowly, and we need to take very large integers n to get a reasonable approximation to e ¼ 2.71828. . .. Now we like in general to make the definition of ex as  would  x x n e ¼ lim 1 þ n , but first we need to know that this limit actually exists. n!1

For example   1 1000 ¼ 2:716 . . .: 1 þ 1000

Then e ¼ e1

Problem 3 Let x > 0.   (a) Prove that ð1 þ nxÞn is an increasing sequence, by adapting the method above where x ¼ 1. k 1 k (b) Verify  that x1 nþ  n  ð1 þ nÞ , for k ¼ 1, 2, . . .. Using this fact, prove that ð1 þ nÞ is bounded above.   (c) Deduce that ð1 þ nxÞn is convergent.

Problem 4 By considering the product   nof the first n terms of the , for n ¼ 1, 2, . . .. sequence ð1 þ 1nÞn , prove that n! > nþ1 e   We now examine the convergence of the sequence ð1 þ nxÞn , for x < 0. Recall that, when x < 0, then  x > 0; in particular, from the above discussion, the sequence ð1  nxÞn converges. We shall use Bernoulli’s inequality (1 þ c)n  1 þ nc, for c  1.   In investigating the convergence of the sequence ð1 þ nxÞn , we are only interested in what happens when n is large. So, we need consider 2 only the situation when n > x. Then n2 > x2, so that n12 < x12 ; thus nx2 < 1, x2 or  n2 > 1.

 n So lim 1 þ nx exists n!1 for x > 0. Note that the limit is at least 1, and so cannot be zero.

You met Bernoulli’s Inequality in Subsection 1.3.3.

Recall that  x > 0.

2.5

The Monotone Convergence Theorem

75

2

It follows that we may substitute  nx2 for c in Bernoulli’s Inequality. This gives that  n  2 x2 x 1 2 1þn  2 n n x2 ¼ 1  ; for n > x: n It follows that  n x2 x2 1  1  2  1  ; for n > x: (8) n n n o 2 Now the sequence 1  xn converges to the limit 1, by the Combination Rules. Hence, by applying the Squeeze Rule to inequality (8), the sequence n n o 2 1  nx2 converges to the limit 1 as n ! 1

Here we use the fact that f1ng is a basic null sequence.

But

 n  n 1 þ nx 1  nx x n  n ¼ 1þ n 1  nx

n 2 1  nx2 n : ¼  1  nx

(9)

We have just seen that the numerator is convergent, and we saw above that the denominator is convergent (to a non-zerolimit) forx > 0; it follows from (9), by the Quotient Rule, that the sequence ð1 þ nxÞn is convergent.   So whatever value we choose for the real number x, the sequence ð1 þ nxÞn is convergent. This means that we can now make the following legitimate definition: Definition

For any real value of x,

x n : ex ¼ lim 1 þ n!1 n

Recall that here we are considering the case x < 0. Note that when x ¼ 0, the sequence is simply the constant sequence {1}. The function x 7! ex , for x 2 R, is called the exponential function.

We can deduce more than this from equation (9), if we rewrite it in the convenient form  n

x n x n x2 1 ¼ 1 2 : 1þ n n n Letting n ! 1, we deduce from this last equation that ex ex ¼ 1: We have shown that this holds for x < 0; however, by simply interchanging x and x, it is clear that this holds for x > 0 also.

Here wealso usethe fact that n x2 lim 1  2 ¼ 1; n!1 n that we proved above.

Theorem 4 Inverse Property of ex For any real value of x, ex ex ¼ 1 The next property of the exponential function that we need to verify is that ex ey ¼ ex þ y for any real values of x and y. We shall prove this fact later.

In Sub-section 3.4.3.

2: Sequences

76

2.5.4

The number p

One of the oldest mathematical problems is to determine the area of a disc of radius r, and the length of its perimeter. It is well known that these magnitudes are pr2 and 2pr, respectively. But what exactly is p? Is there a real number p which makes these formulas correct? – and, if so, how is it formally defined? We now give a precise definition of p as the area of a disc of radius 1, using a method originally devised by Archimedes to calculate approximate values for the area of this disc. The idea is to calculate the areas of regular polygons inscribed in the disc. Archimedes found an easy way to calculate the areas of such regular polygons with 6 sides, 12 sides, 24 sides and so on. The results of parts (a) and (b) of the following problem help to give the first two of these areas. Problem 5

Here ‘inscribed’ means that all the vertices of the polygon lie on the circle, so that the inside of the polygon is contained in the inside of the circle.

Verify that the following triangles have the stated areas: Here p is used only as a symbol to represent angles; its value is not required.

2 tan 1#

Hint: In part (d), use the half-angle formula tan # ¼ 1tan22 1# : 2

Let sn denote the number of sides of the nth such inner polygon (so s1 ¼ 6, s2 ¼ 12, s3 ¼ 24, and, in general, sn ¼ 3  2n) and let an denote the area of the nth inner polygon. Then   1 2p an ¼ sn sin ; for n ¼ 1; 2; . . .: (10) 2 sn Geometrically, it is obvious that each time we double the number of sides of the inner polygon the area increases, and so a1 < a2 < a3 < . . . < an < anþ1 < . . .: Hence {an} is (strictly) increasing. But is {an} convergent?

For example, to three decimal places a1 ¼ 2:598, a2 ¼ 3, ... a6 ¼ 3:141:

2.5

The Monotone Convergence Theorem

77

Notice that each of the polygons lies inside a square of side 2, which has area 4. This means that an  4;

for n ¼ 1; 2; . . .;

and so {an} is increasing and bounded above (by 4). Hence, by the Monotone Convergence Theorem, {an} is convergent. Our intuitive idea of the area of the disc suggests that its area is greater than each of the areas an, but ‘only just’! Put another way, the area of the disc should be the limit of the increasing sequence {an}. This leads us to make the following definition: Definition

p ¼ lim an : n!1

We shall explain in a moment how to calculate the terms an without assuming a value for p. First, however, we describe how to estimate the area of the disc using outer polygons. Once again we start with a regular hexagon and repeatedly double the number of sides. The results of part (c) and (d) of the last problem help to give the first two such areas.

As before, let sn ¼ 3  2n, for n ¼ 1, 2, . . ., and let bn denote the area of the nth outer polygon. This nth outer polygon consists of sn isosceles triangles, each of height 1 and base 2 tanðspn Þ. Thus   p bn ¼ sn tan ; for n ¼ 1; 2; . . .: (11) sn

This will enable us to squeeze p between the area of the inner polygons and the area of the outer polygons.

For example, to three decimal places b1 ¼ 3:464;

Geometrically, it is obvious that each time we double the number of sides of the outer polygon the area decreases, and so b1 > b2 > b3 > > bn > bnþ1 > : So the sequence {bn} is (strictly) decreasing and bounded below (by 0, for example). Thus, by the Monotone Convergence Theorem, {bn} is also convergent. Intuitively, we expect that {bn} has the same limit as {an}, which we have defined to be p. But how can we prove this?

b2 ¼ 3:215; ... b6 ¼ 3:142:

2: Sequences

78 The terms an and bn can be calculated by using the following equations, which are known jointly as the Euclidean algorithm pffiffiffiffiffiffiffiffiffi anþ1 ¼ an bn ; for n ¼ 1; 2; . . .; (12) and 2anþ1 bn ; for n ¼ 1; 2; . . .: (13) anþ1 þ bn pffiffiffi pffiffiffi Starting with a1 ¼ 32 3 ¼ 2:598 . . . and b1 ¼p2ffiffiffiffiffiffiffiffiffi 3 ¼ 3:464 . . ., we use 2 b1 these equations iteratively to calculate first a2 ¼ a1 b1 , then b2 ¼ a2a2 þb , and 1 so on. Here are (approximations to) the first few values of each sequence obtained in this way: bnþ1 ¼

sn

6

12

an 2.598 3

24

48

96

Equations (12) and (13) can be verified from equations (10) and (11); we give the details in Subsection 2.5.5 below.

192

3.106 3.133 3.139 3.141

bn 3.464 3.215 3.160 3.146 3.143 3.142 It does appear that both sequences do converge to a common limit, namely p.

Remark

We prove this in Subsection 2.5.5 below.

We have defined p using the areas of approximating polygons. An alternative approach uses the perimeters of these polygons.

2.5.5

Proofs

You may omit this subsection at a first reading.

We now prove several of the results referred to in the discussion of p in the previous sub-section. First, we prove equations (12) and (13) pffiffiffiffiffiffiffiffiffi anþ1 ¼ an bn ; for n ¼ 1; 2; . . .; (12) and bnþ1

2anþ1 bn ¼ ; anþ1 þ bn

for n ¼ 1; 2; . . .:

(13)

Using the half-angle formulas for sin # and cos #, it is easy to check that rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 1 sin # tan # sin # ¼ 2 2 2 and 1 sin # tan # : tan # ¼ 2 sin # þ tan # Hence, since snþ1 ¼ 2sn   1 2p anþ1 ¼ snþ1 sin 2 snþ1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi     1 2p p sn sin ¼  sn tan 2 sn sn pffiffiffiffiffiffiffiffiffi ¼ an bn :

These comprise the Euclidean algorithm.

We omit the details in both cases.

This proves equation (12).

2.6

Exercises

79

Similarly

 bnþ1 ¼ snþ1 tan

p



snþ1



 2sn sin spn  sn tan spn



 ¼ sn sin spn þ sn tan spn ¼

2anþ1 bn : anþ1 þ bn

This proves equation (13).

Finally we prove that lim bn ¼ lim an . n!1 n!1 To do this, we rearrange equation (12) as follows bn ¼

a2nþ1 : an

Then, by the Combination Rules, we can deduce that

2 lim anþ1 p2 ¼ p: ¼ lim bn ¼ n!1 n!1 lim an p n!1

Then, since we defined p to be precisely lim an , it follows that lim bn ¼ lim an , n!1 n!1 n!1 as required. Definition Let an denote the area of the regular polygon with 3  2n sides inscribed in a disc of radius 1, and bn the area of the regular polygon with 3  2n sides circumscribing the disc. Then p is the common value of the limits lim an and lim bn . n!1

n!1

Remark It can be shown that every two applications of equations (12) and (13) give an extra decimal place in the decimal expansion for p. This decimal expansion is now known to many millions of decimal places, using sequences which converge to p much more rapidly than {an} or {bn}. Since p is irrational, there is no possibility of the decimal expansion of p recurring.

2.6

Exercises

Section 2.1 1. Calculate the first five terms of each of the following sequences, and draw a sequence diagram in each case: n nþ1 o    (b) ð1n!Þ (a) {n2  4n þ 4}; ; (c) sin 14 np .

We shall return in Section 8.5 to the question of approximating p, by the use of power series.

2: Sequences

80 2. Determine which of the following sequences are monotonic: n no  1  1 ð1Þ (a) nn þ ; (c) 2n . ; (b) þ2 n 3. Prove that the following sequences are eventually monotonic:  n   (a) 5n! ; (b) n þ 8n .

Section 2.2 1. For each of the following sequences {an} and numbers ", find a number X such that janj < ", for all n > X: n

Þ , " ¼ 0.001; (a) an ¼ ð1 n5

(b) an ¼ ð2n þ1 1Þ2 , " ¼ 0.002.

2. Use the definition of null sequence to prove that the two sequences in the previous exercise are null. 3. Prove that the following sequences are not null: n o n pffiffiffi (a) f ng; (b) 1 þ ð1n Þ .  4. Assuming that 1n is null, deduce that the following sequences are null. State which rules you use. n o n o n 1 o ð2nÞ n (b) nsin (c) ntan (a) p2ffiffin þ n37 ; 2 þ1 ; þ 1 þ sin n . pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 5. Prove that n þ 1  n is null. 2  b2 Hint: Use the fact that a  b ¼ aa þ b , for a þ b 6¼ 0. 6. Use the list of basic null sequences to prove that the following sequences are null. State which rules you use.    10   10 n  (b) 6nn! ; (c) n n!10 . (a) 43n þ 2n 3n ; 7. Prove that, if the sequence {an} of positive numbers is null, then null.

pffiffiffiffiffi an is

Section 2.3 1. Show that the following sequences converge to 1, by calculating an  1 in each case:  n2   1 (b) n2 þnþ1 . (a) nn  þ3 ; 2. Use the Combination Rules to find the limits of the following sequences: n o  þ 5n   10 2n  3 (a) 2n3 þn3n þ 4 ; ; (c) n5n! (b) 2nn þn100 100 þ n! . n

3. (a) Prove that, if the sequence{an}of positive numbers is convergent pffiffi with pffiffiffiffiffi ‘ . limit ‘, where ‘ > 0, then an is convergent with limit

You met this list in Subsection 2.2.1.

2.6

Exercises

81 2

2

b Hint: Use the fact that a  b ¼ aa þ b , for a þ b 6¼ 0. 1 (b) Prove that, if the sequence {ak} is convergent with limit ‘, ‘ 6¼ 0, and ank n 1o 1 is defined, where n 2 N, then ank is convergent with limit ‘n .

Hint: Use the identity aq  bq ¼ (a  b)(aq1 þ aq2b þ aq3b2 þ þ bq1) with suitable choices for a, b and q.

Section 2.4 1. Classify the following sequences as convergent or divergent, and as bounded or unbounded: n n no  1 (b) ð1Þn!100 . (a) n4 ; 2. Use the Reciprocal Rule to prove that the following sequences tend to infinity:         (a) nn!3 ; (b) n2 þ 2n ; (c) n2  2n ; (d) n!  n3  3n . 3. Use the Subsequence Rules to prove that the following sequences are divergent: n n 2o 1Þ n (a) fð1Þn 2n g; (b) ð2n . 2 þ1 4. Each of the following general statements concerning sequences {an} and {bn} is true or false. For each statement that is true, prove it. For each statement that is false, write down examples of sequences to illustrate your assertion: (a) If {an} and {bn} are divergent, then {an þ bn} is divergent. (b) If {an} and {bn} are divergent, then {anbn} is divergent. (c) If {an} is convergent and {bn} is divergent, then {anbn} is divergent.   (d) If a2n is convergent, then {an} is convergent.  2 (e) If an is convergent, and an > 0, then {an} is convergent. (f) If {bn} is convergent, where then {an} is convergent. n bn ¼ anþ1an, o anþ1 þanþ2 þ

þa2n ffiffi p (g) If an ! 0 as n ! 1, then is convergent. n 5. Prove that every sequence has a monotonic subsequence.

Section 2.5 1. Prove that, if {an} is increasing and has a subsequence fank g which is convergent, then {an} is convergent. 3an þ 1 2. Let fan g1 n¼1 be a sequence for which anþ1 ¼ an þ3 , for n  1. (a) Prove that, if a1 > 1, the sequence {an} converges. Hint: consider the cases 1 < a1 < 1, a1 ¼ 1 and a1 > 1 separately. (b) If a1  1, does the sequence converge or diverge?

The Power Rule in Subsection 2.2.1 and the subsequent remark there show that this result holds in the case that ‘ ¼ 0.

2: Sequences

82

 2 ¼ 12 an þ a‘ n , for n  1, where

3. Let fan g1 n¼1 be a sequence for which anþ1 a1 > 0 and ‘ > 0. (a) Prove that {an} is decreasing for n  2, and hence that an ! ‘ as n ! 1. (b) With ‘2 ¼ 2 and a1 ¼ 1.5, calculate the terms a2, a3, a4 and a5. 4. Obtain formulas for the perimeters of the inner and outer polygons introduced in Sub-section 2.5.4. Prove that these sequences tend to 2p as n ! 1. 5. Bernoulli’s Inequality states that (1 þ x)n  1 þ nx, for x  1, n ¼ 1, 2, . . .. 1 (a) Apply  Bernoulli’s   Inequality, with x ¼ n2 , to give an alternative proof 1 n is increasing. that 1 þ n n nþ1 o (b) Apply Bernoulli’s Inequality, with x ¼ n211, to prove that 1 þ 1n

is decreasing. (c) Deduce from parts (a) and (b) that     1 n 1 nþ1 1þ e 1þ ; n n

You met this in Subsection 1.3.3.

for n ¼ 1; 2; . . .:

6. Let f½an ; bn g1 n¼1 be a sequence of closed intervals with the property that ½anþ1 ; bnþ1   ½an ; bn , for n ¼ 1, 2, . . .. (a) Prove that there exists some point c that lies in all the intervals [an, bn], for n ¼ 1, 2, . . .. (b) Prove that, if bn  an ! 0 as n ! 1, then there exists only one such point c.

This result is sometimes called The Nested Intervals Theorem.

3

Series

The Ancient Greek philosopher Zeno of Elea proposed a number of paradoxes of the infinite, which are said to have had a profound influence on Greek mathematics. For example, Zeno claimed that it is impossible for an object to travel a given distance, since it must first travel half the distance, then half of the remaining distance, then half of what remains, and so on. There must always remain some distance left to travel, and so the journey cannot be completed. This paradox relies partly on the intuitive feeling that it is impossible to add up an infinite number of positive quantities and obtain a finite answer. However, the following illustration of the paradox suggests that such an infinite sum is plausible. 1 4

1 2

0

1 2

1 8 3 4

1 16 7 15 8 16

1

The distance from 0 to 1 can be split up into the infinite sequence of distances 12 , 14 , 18, . . ., and so it seems reasonable to write 1 1 1 1 þ þ þ þ    ¼ 1: 2 4 8 16 This chapter is devoted to the study of such expressions, which are called infinite series. The following example shows that infinite series need to be treated with care. Suppose that it is possible to add up 2, 4, 8, . . ., and that the answer is s 2 þ 4 þ 8 þ 16 þ    ¼ s: If we multiply through by 12, we obtain 1 1 þ 2 þ 4 þ 8 þ    ¼ s; 2 which we can rewrite in the form 1 1 þ s ¼ s: 2 It follows from this equation that s ¼ 2, which is obviously non-sensical. We can avoid reaching such absurd conclusions by performing arithmetical operations only with convergent infinite series. In Section 3.1 we define the concept of convergent infinite series in terms of convergent sequences, and give some examples. We also describe various properties which are common to all convergent series.

Because the left-hand side of the previous equation can be expressed as 1 þ ð2 þ 4 þ 8 þ   Þ ¼ 1 þ s:

83

3: Series

84 Section 3.2 is devoted to series with non-negative terms, and we give several tests for the convergence of such series. In Section 3.3 we deal with the much harder problem of determining whether a series is convergent when it contains both positive and negative terms. We also look at what can happen if we rearrange the terms of a convergent series – does the new series still converge? This section is quite long, so you might wish to tackle it in several sessions. In Section 3.4 we explain how ex can be represented as an infinite series of powers of x, and we use this representation to prove that the number e is irrational, and that ex ey ¼ exþy .

3.1 3.1.1

You first met ex in Section 2.5.

Introducing series What is a convergent series?

We begin by defining precisely what is meant by the statement that 1 1 1 1 þ þ þ þ    ¼ 1: 2 4 8 16 Let sn be the sum of the first n terms on the left-hand side. Then 1 s1 ¼ ; 2 1 1 3 s2 ¼ þ ¼ ; 2 4 4 1 1 1 7 s3 ¼ þ þ ¼ ; . . .; and so on: 2 4 8 8 In general, using the formula for the sum of a geometric progression, with a ¼ r ¼ 12, we obtain 1 1 1 1 1 1 1 1 sn ¼ þ þ þ    þ n ¼ þ 2 þ 3 þ    þ n 2 4 8 2 2 2 2 2  n  n 1 1  12 1 ¼  ¼1 : 1 2 2 12  n  is a null sequence, and so The sequence 12  n   1 lim sn ¼ lim 1  n!1 n!1 2  n 1 ¼ 1  lim ¼ 1: n!1 2 It is the precise mathematical statement that sn ! 1 as n ! 1 which justifies our earlier (intuitive) statement that 1 1 1 1 þ þ þ þ    ¼ 1: 2 4 8 16 We use this approach to define a convergent infinite series.

The sum of a geometric progression a þ ar þ n ar 2 þ    þ arn1 is a 1r 1r , if r 6¼ 1.

Here we use the Combination Rules for sequences.

3.1

Introducing series

85

Given a sequence {an} of real numbers, the expression

Definition

a1 þ a2 þ a3 þ    is called an infinite series, or simply a series. The nth partial sum of this series is sn ¼ a1 þ a2 þ a3 þ    þ an : The behaviour of the infinite series a1 þ a2 þ a3 þ    is determined by the behaviour of {sn}, its sequence of partial sums. Definition The series a1 þ a2 þ a3 þ    is convergent with sum s if the sequence {sn} of partial sums converges to s. In this case, we say that the series converges to s, and we write a1 þ a2 þ a3 þ    ¼ s: The series diverges if the sequence {sn} diverges. Thus we prove results about series by applying results for sequences proved in Chapter 2 to the sequence of partial sums {sn}. Example 1 For each of the following infinite series, calculate its nth partial sum, and determine whether the series is convergent or divergent: (a) 1 þ 1 þ 1 þ   ;

(b)

1 3

þ 312 þ 313 þ   ;

(c) 2 þ 4 þ 8 þ   .

Solution (a) In this case sn ¼ 1 þ 1 þ    þ 1 ¼ n: The sequence {sn} tends to infinity, and so this series is divergent. (b) Using the formula for summing a finite geometric progression, with a ¼ r ¼ 13, we obtain 1 1 1 1 þ þ þ  þ n 3 32 33 3  n  n   1 1  13 1 1 1  ¼  ¼ : 3 2 3 1  13

sn ¼

Since

1n  3

is a basic null sequence

1 lim sn ¼ ; 2 and so this series is convergent, with sum 12. (c) In this case n!1

s n ¼ 2 þ 4 þ 8 þ    þ 2n  n  2 1 ¼2 ¼ 2nþ1  2: 21 The sequence {2nþ1  2} tends to infinity, and so this series is divergent.&

3: Series

86

Sigma notation Next, we explain how to use the sigma notation to represent infinite series. Recall that a finite sum such as 1 1 1 1 1 1 1 1 þ þ þ  þ ¼ þ þ þ    þ 10 2 4 8 1024 21 22 23 2 can be represented using sigma notation as

 is the Greek capital letter ‘sigma’.

10 X 1 : 2n n¼1

This notation can be adapted to represent infinite series, as follows 1 X an ¼ a1 þ a2 þ a3 þ   :

For example

1 X 1 1 1 1 ¼ þ þ þ ; 2n 2 4 8 n¼1 1 X 1

n¼1

n n¼1

Convention When using the sigma notation to represent the nth partial sum sn of a series, we write n X ak : s n ¼ a1 þ a2 þ a3 þ    þ an ¼ k¼1

We have written k as the ‘dummy variable’ here, to avoid using n for two different purposes in the same expression. We could have used any letter (other n n P P than n) for the dummy variable; for example, the expressions ai and ar i¼1

r¼1

also stand for a1 þ a2 þ a3 þ    þ an. If we need to begin a series with a term other than a1, then we write, for example 1 1 X X an ¼ a0 þ a1 þ a2 þ    or an ¼ a3 þ a4 þ a5 þ   : n¼0

n¼3

For any series, the nth partial sum sn is obtained by adding the first n terms. For example, the nth partial sums of the above two series are sn ¼ a0 þ a1 þ a2 þ    þ an1 ¼

n1 X

ak

and

k¼0

sn ¼ a3 þ a4 þ a5 þ    þ anþ2 ¼

nþ2 X

1 1 1 ¼ 1þ þ þ þ : 2 3 4

For the above examples 1 1 1 1 sn ¼ þ þ þ   þ n 2 4 8 2 n X 1 ¼ ; k 2 k¼1 1 1 1 1 sn ¼ 1 þ þ þ þ   þ 2 3 4 n n X 1 ¼ : k k¼1 For example, we write 1 X 1 n! n n¼3 for the series 1 1 þ þ   : 3!  3 4!  4

ak :

k¼3

Problem 1 For each of the following infinite series, calculate the nth partial sum sn, and determine whether the series is convergent or divergent: 1  1 1   n P P P 1 n 13 ; (b) ð1Þn ; (c) (a) 2 : n¼1

n¼0

n¼1

Remark Notice that inserting into a series, or omitting or altering, a finite number of terms does not affect the convergence of the series, but may affect the sum. For example, since the series 12 þ 14 þ 18 þ    converges with limit 1, it follows that

Recall that the convergence or divergence of sequences was not affected by omitting or altering a finite number of terms in the sequence.

3.1

Introducing series

87

the series 8 þ 4 þ 2 þ 1 þ 12 þ 14 þ 18 þ    ¼ ð8 þ 4 þ 2 þ 1Þ þ 12 þ 14 þ 18 þ    converges with limit (8 þ 4 þ 2 þ 1) þ 1 ¼ 16.

3.1.2

For 12 þ 14 þ 18 þ    ¼ 1:

Geometric series

All the series considered so far are examples of geometric series. The standard geometric series with first term a and common ratio r is 1 X ar n ¼ a þ ar þ ar 2 þ   : n¼0

The following theorem enables us to decide whether any given geometric series is convergent or divergent. Theorem 1

Geometric Series 1 P a (a) If jrj < 1, then ar n is convergent, with sum 1r . n¼0 1 P n (b) If jrj  1 and a 6¼ 0, then ar is divergent. n¼0

Proof (a) If r 6¼ 1, then the nth partial sum sn is given by 1  rn : 1r n Now, if jr j < 1, then fr g is a basic null sequence, and so 1  rn a lim ð1  r n Þ ¼ lim sn ¼ lim a n!1 n!1 1  r 1  r n!1 a ; ¼ 1r 1 P by the Combination Rules for sequences. Thus, if jrj < 1, then ar n is a n¼0 convergent, with sum 1r. sn ¼ a þ ar þ ar 2 þ    þ ar n1 ¼ a

(b) Part (b) can be deduced immediately from the Non-null Test, which & appears later in this section. We defer the proof until then.

This value for sn is easily verified by Mathematical Induction.

Sub-section 3.1.5, Example 3.

Decimal representation of rational numbers Geometric series provide another interpretation of the decimal representation of rational numbers. Recall that rational numbers have terminating, or recurring, decimal representations. For example 3 1 ¼ 0:3 and ¼ 0:333 . . . ¼ 0:3: 10 3 Another way to interpret the symbol 0.333 . . . is as an infinite series 3 3 3 þ þ þ   : 101 102 103 3 1 and common ratio r ¼ 10 . This is a geometric series, with first term a ¼ 10 1 Since 0 < 10 < 1, this series is convergent, with sum 3 a 3 1 ¼ 10 1 ¼ ¼ : 1  r 1  10 9 3

Sub-section 1.1.3.

3: Series

88 Thus we have another method of finding the fraction which is equal to a given recurring decimal, by calculating the sum of the corresponding geometric series. Problem 2 Interpret the following decimals as infinite series, and hence represent them as fractions: (a) 0.111 . . .;

3.1.3

(b) 0.86363 . . .;

(c) 0.999 . . ..

You should at least read the solution to part (c), even if you do not tackle it first.

Telescoping series

Geometric series are easy to deal with because it is possible to find a formula for the nth partial sum sn. The next problem deals with another series for which we can calculate sn explicitly. Problem 3 Calculate the first four partial sums of the following series, giving your answers as fractions 1 X 1 1 1 1 ¼ þ þ þ   : n ð n þ 1 Þ 1  2 2  3 3  4 n¼1 The results obtained in Problem 3 suggest the general formula sn ¼

1 1 1 1 n þ þ þ  þ ¼ : 12 23 34 nð n þ 1Þ n þ 1

This formula can be proved by Mathematical Induction, or we can use the identity 1 1 1 ¼  ; for n ¼ 1; 2; . . .; nð n þ 1 Þ n n þ 1 which implies that 1 1 1 1 1 sn ¼ þ þ þ  þ þ 12 23 34 ðn  1Þn nðn þ 1Þ           1 1 1 1 1 1 1 1 1 1 ¼      þ þ þ  þ þ 1 2 2 3 3 4 n1 n n nþ1 1 ¼1 nþ1 n : ¼ nþ1 Since n 1 lim sn ¼ lim ¼ lim ¼ 1; n!1 n!1 n þ 1 n!1 1 þ 1 n we deduce that the given series is convergent, with sum 1 X 1 ¼ 1: n ð n þ 1Þ n¼1 Problem 4 fact that

Find the nth partial sum of the series

1 P n¼1

2 1 1 ¼  ; nðn þ 2Þ n n þ 2

for n ¼ 1; 2; . . .:

Deduce that this series is convergent, and find its sum.

1 nðnþ2Þ,

using the

This cancellation of adjacent terms explains why this series is said to be telescoping.

3.1

Introducing series

3.1.4

89

Combination Rules for convergent series

At the start of this chapter we saw that performing arithmetical operations on the divergent series 2 þ 4 þ 8 þ    can lead to absurd conclusions. However, the following result shows that there are Combination Rules for convergent series, which follow directly from the Combination Rules for convergent sequences: Theorem 2 Combination Rules 1 1 P P If an ¼ s and bn ¼ t, then: n¼1

Sum Rule

n¼1 1 P

Multiple Rule

n¼1 1 P

ðan þ bn Þ ¼ s þ t; lan ¼ ls, for l 2 R.

n¼1

Proof

Consider the sequences of partial sums {sn} and {tn}, where n n X X ak and tn ¼ bk : sn ¼ k¼1

k¼1

By assumption, sn ! s and tn ! t as n ! 1. Sum Rule 1 P The nth partial sum of the series ðan þ bn Þ is n¼1 n X ðak þ bk Þ ¼ ða1 þ b1 Þ þ ða2 þ b2 Þ þ  þ ðan þ bn Þ k¼1

¼ ða1 þ a2 þ  þ an Þ þ ðb1 þ b2 þ   þ bn Þ ¼ sn þ tn : By the Sum Rule for sequences lim ðsn þ tn Þ ¼ lim sn þ lim tn ¼ s þ t; n!1 n!1 1 P and so the sequence {sn þ tn} of partial sums of ðan þ bn Þ has limit s þ t. n¼1 Hence this series is convergent and 1 X ðan þ bn Þ ¼ s þ t: n!1

n¼1

Multiple Rule 1 P The nth partial sum of the series lan is n¼1 n X lak ¼ la1 þ la2 þ    þ lan k¼1

¼ lða1 þ a2 þ    þ an Þ ¼ lsn : By the Multiple Rule for sequences lim ðlsn Þ ¼ l lim sn ¼ ls;

n!1

n!1

We may rearrange the terms here, because this is the sum of a finite number of terms.

3: Series

90 1 P

and so the sequence {lsn} of partial sums of lan has limit ls. Hence this n¼1 series is convergent and 1 X lan ¼ ls: & n¼1

Example 2

Solution

Prove that the following series is convergent and calculate its sum  1 3 þ : 2n nð n þ 1Þ n¼1

1  X

1 P 1

We know that

n¼1

2n

is convergent, with sum 1, and that

1 P n¼1

1 nðnþ1Þ

is

convergent, with sum 1. Hence, by the Sum Rule and the Multiple Rule  1  X 1 3 þ is convergent; with sum 1 þ ð3  1Þ ¼ 4: & 2n nð n þ 1Þ n¼1 Problem 5 its sum

Prove that the following series is convergent and calculate 1  n X 3 n¼1

3.1.5

4



 2 : nð n þ 1Þ

The Non-null Test

For all the infinite series we have so far considered, it is possible to derive a simple formula for the nth partial sum. For most series, however, this is very difficult or even impossible. Nevertheless, it may still be possible to decide whether such series are convergent or divergent by applying various tests. The first test that we give arises from the following result: Theorem 3

If

1 P

an is a convergent series, then {an} is a null sequence.

n¼1

Proof

You met the second series in Sub-section 3.1.3.

Let sn ¼

n P

ak denote the nth partial sum of

1 P

an . Because

n¼1

k¼1

1 P n¼1

sn ¼ sn1 þ an ; and so an ¼ sn  sn1 : Thus, by the Combination Rules for sequences lim an ¼ lim ðsn  sn1 Þ ¼ lim sn  lim sn1 n!1

n!1

n!1

¼ s  s ¼ 0; and so {an} is a null sequence, as required.

1 X 1

n n¼1

&

1 1 ¼ 1 þ þ þ ; 2 3

and 1 X n¼1

an is

convergent, we know that {sn} is convergent. Suppose that lim sn ¼ s. n!1 We want to deduce that {an} is null. To do this, we write

n!1

For example, consider the series

n2 1 4 9 ¼ þ þ þ : þ 1 3 9 19

2n2

3.1

Introducing series

91

The following useful test for divergence is an immediate corollary of Theorem 3: Corollary divergent.

Non-null Test

If {an} is not a null sequence, then

1 P

an is

n¼1

The virtue of the Non-null Test is its ease of application. For example, it enables us to decide immediately that 1 1 1 X X X 1; ð1Þn and n are divergent; n¼1

n¼1

n¼1

because the corresponding sequences {1}, {(1n)} and {n} are not null. There are various ways to show that a given sequence {an} does not tend to zero. For example, we can show that an ! ‘ as n ! 1; where ‘ 6¼ 0, or that {an} tends to infinity or to minus infinity. More generally, we can use the result about subsequences which states that if an ! 0 as n ! 1; then ank ! 0 as k ! 1; for any subsequence fank g of {an}. This leads to the following strategy: Strategy EITHER

To show that

1 P

an is divergent, using the Non-null Test:

You met this Inheritance Property of subsequences in Sub-section 2.4.4, Theorem 5, for any limit ‘.

n¼1

1. show that {an} has a convergent subsequence with non-zero limit; OR

2. show that {an} has a subsequence which tends to infinity, or a subsequence which tends to minus infinity. This strategy can be used to prove part (b) of Theorem 1, as follows: 1 P Example 3 Prove that, if jrj  1 and a 6¼ 0, then ar n is divergent. n¼0

Solution We want to show that, if jrj  1 and a 6¼ 0, then {arn} is not a null sequence. Now, the even subsequence  2k    ¼ a; ar 2 ; ar 4 ; ar 6 ; . . . ar converges to a 6¼ 0 if jrj ¼ 1, and tends to infinity or to minus infinity if jrj  1. Thus, {ar2k} is not null, and so {arn} is not null. 1 P Hence, by the Non-null Test, ar n is divergent if jrj > 1 and a 6¼ 0. &

It tends to infinity if a > 0 and to minus infinity if a < 0.

n¼0

Warning

The converse of the Non-null Test is FALSE. In other words:

If {an} is a null sequence, it is not necessarily true that

1 P

an is convergent.

n¼1 1 P 1 1 1 For example, the sequence f1ng is a null sequence, but n ¼1 þ 2 þ 3 þ    n¼1 is divergent.

We shall prove this rather surprising fact in Subsection 3.2.1.

3: Series

92 The important thing to remember is that you can never use the Non-null Test to prove that a series is convergent, although you may be able to use it to prove that a series is divergent. 1 P n2 Problem 6 Prove that the series 2n2 þ 1 is divergent. n¼1

3.2

Series with non-negative terms

In this section we consider only series

1 P

an with non-negative terms. In other

n¼1

words we assume that an  0, for n ¼ 1, 2, . . .. 1 P It follows that the partial sums of an , which are given by n¼1

s 1 ¼ a1 ;

s 2 ¼ a1 þ a2 ;

s3 ¼ a1 þ a2 þ a3 ; . . .;

sn ¼ a1 þ a2 þ a3 þ    þ an ; . . .; form an increasing sequence {sn}. The fact that {sn} is increasing makes it easier to deal with series having non-negative terms. If we can prove that the sequence {sn} of partial sums is bounded above, then {sn} is convergent, by the Monotone Convergence 1 P Theorem, and so an is convergent. On the other hand, if we can prove

Sub-section 2.5.1, Theorem 1.

n¼1

that the sequence {sn} of partial sums is not bounded above, then {sn} cannot be convergent, and so we must have that sn ! 1 as n ! 1. We can rephrase these facts in the following convenient way: Boundedness Theorem for series

A series

1 P

an of non-negative terms is

n¼1

convergent if and only if its sequence {sn} of partial sums is bounded above. We shall use this, for example, to prove the Cauchy Condensation Test in Sub-section 3.2.1.

3.2.1

Tests for convergence

We now explore several tests for the convergence of series with non-negative terms. Problem 1 Use your calculator to find the first eight partial sums of each of the following series 1 1 X X 1 1 1 1 1 1 ¼ 1 þ þ þ   ; ¼ 1 þ þ þ    and 2 2 2 n 2 3 n 2 3 n¼1 n¼1 giving your answers to 2 decimal places. Plot your answers on a sequence diagram. Example 1

Prove that the series

1 P n¼1

1 n2

¼ 1 þ 212 þ 312 þ    is convergent.

This is essentially the same result as that of the Boundedness Theorem for sequences, Sub-section 2.4.2, Theorem 1.

3.2

Series with non-negative terms

93

Solution All the terms of the series are positive, so we shall use the Monotone Convergence Theorem. Let sn be the nth partial sum of the series. Then 1 1 1 1 þ þ þ  þ 2 22 32 42 n 1 1 1 1 þ þ þ  þ 1, k12
1þ þ ; 2 3 4 2 2     1 1 1 1 1 1 1 1 1 1 s8 ¼ 1 þ þ þ þ þ þ þ >1þ þ þ ; 2 3 4 5 6 7 8 2 2 2 .. . 1 1 1 1 s2k > 1 þ þ þ    þ ¼ 1 þ k: 2 2 2 2 |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} k terms

This series is often called the harmonic series, since its terms are proportional to the lengths of strings that produce harmonic tones in music. We think of n as being fairly large, in order to see the pattern. Here each bracket adds up to more than 12.

3: Series

94 Hence the sequence {sn} is increasing and not bounded above. Therefore it cannot be convergent, so that by the Monotone Convergence Theorem the & sequence {sn} is divergent. Hence the series itself is also divergent. We can extend this type of domination approach to test further series for convergence, in the following way: Theorem 1

Comparison Test 1 P

(a) If 0  an  bn , for n ¼ 1, 2, . . ., and

bn is convergent, then

n¼1

convergent.

(b) If 0  bn  an , for n ¼ 1, 2, . . ., and

1 P

an is

n¼1

bn is divergent, then

n¼1

divergent.

1 P 1 P

an is

n¼1

In the proof of part (a), in Subsection 3.2.2, we shall in fact see that 1 X

an 

n¼1

1 X

bn :

n¼1

Remark As with the Squeeze Rule for non-negative null sequences, it is sufficient to prove that the necessary inequalities in parts (a) and (b) hold eventually. In applications, we use this result in the following way: Strategy

To test a series

using the Comparison Test: EITHER

OR

1 P

an of non-negative terms for convergence

n¼1 1 P

find a convergent series dominate the an, find a divergent series

bn of non-negative terms where the bn

n¼1 1 P

That is, where an  bn.

bn of non-negative terms where the an

n¼1

dominate the bn. Example 3

That is, it is sufficient that 0  an  bn or 0  bn  an for all n > X, for some number X.

Prove that the series

1 P n¼1

That is, where bn  an. 1 n3

is convergent.

Solution We shall use the Comparison Test. Let bn ¼ n12 . Then 0  n13  bn , for n ¼ 1, 2, . . ., and we know that the series 1 P 1 n2 is convergent (this was Example 1 above). It follows from part (a) of the n¼1 1 P 1 & Comparison Test that the series n3 is convergent.

For n13  n12 :

n¼1

Example 4

Prove that the series

1 P n¼1

p1ffiffi n

is divergent.

Solution We shall use the Comparison Test. Let bn ¼ 1n. Then 0  bn  p1ffiffin, for n ¼ 1, 2, . . ., and we know that the series 1 P 1 2 above). It follows from part (b) of the n is divergent (this was Example 1 P n¼1 1 pffiffi is divergent. & Comparison Test that the series n

For 1n  p1ffiffin :

n¼1

1 P

ffiffi . Its terms are somewhat similar to pffiffi 1p Next, consider the series nþ3 4 nþ1 1 n¼1 P p1ffiffi, since we can express pffiffi 1p ffiffi in the form those of the series n nþ3 4 nþ1 n¼1

pffiffiffi In other words, the term ffiffiffi n p 4 dominates the term 3 n þ 1.

3.2 p1ffiffi n

Series with non-negative terms

1 1 1 , where the 1þ3n4 þn2 1 P 1 pffiffi is divergent, series n n¼1



95

second fraction tends to 1 as n ! 1. So, since the it seems likely that the series

1 P n¼1

pffiffi 1p4 ffiffi nþ3 nþ1

is also

By Example 4.

divergent. We can pin down the underlying idea here in the following useful result: Theorem 2

Limit Comparison Test 1 1 P P Suppose that an and bn have positive terms, and that n¼1

n¼1

an !L bn (a) If

1 P

(b) If

n¼1 1 P

as

n ! 1;

bn is convergent, then bn is divergent, then

n¼1

1 P

where L 6¼ 0: an is convergent.

In other words, bn behaves ‘rather like an’ for large n. Note that it is IMPORTANT that L is non-zero.

n¼1 1 P

an is divergent.

n¼1

So, take an ¼ pffiffinþ31p4 ffiffinþ1 and bn ¼ p1ffiffin, for n ¼ 1, 2, . . .. Then both an and bn are positive, and pffiffiffi pffiffiffi ð1=ð n þ 3 4 n þ 1ÞÞ an pffiffiffi ¼ bn ð1= nÞ pffiffiffi n ffiffiffi p ¼ pffiffiffi nþ34 nþ1 1 ¼ 1 1 ! 1 as n ! 1:  1 þ 3n 4 þ n2 It then follows from part (b) of the Limit Comparison Test that, since the 1 1 P P p1ffiffi is divergent, the series ffiffi is also divergent. pffiffi 1p series n nþ3 4 nþ1 n¼1

Since 1 6¼ 0. By Example 4.

n¼1

Example 5 Use the Limit Comparison Test to prove that the series 1 2 P n 3nþ4 5n4 n is convergent.

n¼1

2

1 Solution Set an ¼ n 5n3nþ4 4 n and bn ¼ n2 , for n ¼ 1, 2, . . .. Then both an and bn are positive, and

an n2  3n þ 4 n2  ¼ 5n4  n bn 1 4 3 2 n  3n þ 4n : ¼ 5n4  n 1  3n1 þ 4n2 1 as n ! 1: ¼ ! 3 5 5n 1 P 1 Since 15 6¼ 0 and the series n2 is known to be convergent, then the series 1 n¼1 P n2 3nþ4 & 5n4 n is also convergent.

n¼1

We make this choice for bn since, for large n, the 2 expression n 5n3nþ4 4 n is ‘more or n2 1 less the same as’ 5n 4 ¼ 5n2 – where the factor ‘5’ will not affect our argument. Dividing numerator and denominator by the dominant term n4. By Example 1.

3: Series

96 Problem 2 Use the Comparison Test or the Limit Comparison Test to determine the convergence of the following series: 1 1 1 1 P P P P cos2 ð2nÞ 1 1pffiffi nþ4 (a) (b) ; (c) (d) n3 : n3 þn; 2n3 nþ1; nþ n n¼1

n¼1

n¼1

n¼1

Our next test is motivated in part by the geometric series – recall that the 1 P ar n , where a 6¼ 0, is convergent if

geometric series a þ ar þ ar 2 þ    ¼

n¼0

jrj < 1 but divergent if jrj 1. The reason that this converges, if jrj < 1, is that the terms are then forced to tend to 0 so quickly that the partial sums sn increase so slowly as n increases that they remain bounded above. On the other hand, if jrj  1, the terms do not tend to 0, so that the series must diverge. Theorem 3

Ratio Test 1 P an has positive terms, and that abnþ1 ! ‘ as n ! 1. Suppose that n n¼1 1 P (a) If 0  ‘ < 1, then an is convergent. n¼1

(b) If ‘ > 1, then

1 P

By the Non-null Test. Note that with the Ratio Test, we concentrate on the series 1 P an itself and do not need to n¼1

‘think of’ some other series 1 P bn . n¼1

an is divergent.

Part (b) includes the case anþ1 an ! 1.

n¼1

Remark If ‘ ¼ 1, then the test gives us no information on whether the series converges. n For example, if an ¼ 1n, then aanþ1 ¼ nþ1 ¼ 1þ1 1 ! 1 as n ! 1; we have seen n n 1 1 P P 1 1 an ¼ that the series n diverges. On the other hand, if an ¼ n2 then 1 1 n¼1 n¼1 P P 1 anþ1 n2 1 an ¼ an ¼ ðnþ1Þ2 ¼ 1þ2þ 1 ! 1 as n ! 1; but the series n2 converges. n

Example 6 series: 1 P n (a) 2n ;

n2

n¼1

n¼1

Use the Ratio Test to determine the convergence of the following (b)

n¼1

1 P 10n n¼1

n!

.

Solution (a) Let an ¼ 2nn , n ¼ 1, 2, . . .. Then an is positive, and anþ1 n þ 1 2n ¼ nþ1  2 an n 1 1þn 1 as n ! 1: ! ¼ 2 2 Since 0 < 12 < 1, it follows from the Ratio Test that the series 1 1 P P n an ¼ 2n is convergent. n¼1

n¼1 n

(b) Let an ¼ 10n! , n ¼ 1, 2, . . .. Then an is positive, and anþ1 10nþ1 n!  ¼ an ðn þ 1Þ! 10n 10 ! 0 as n ! 1: ¼ nþ1

That is, the Ratio Test is inconclusive if ‘ ¼ 1. By Example 2. By Example 1.

3.2

Series with non-negative terms

97

It follows from the Ratio Test that the series convergent.

1 P

an ¼

n¼1

1 P n¼1

10n n!

is &

Problem 3 Use the Ratio Test to determine whether the following series are convergent 1 3 1 2 n 1 P P P ð2nÞ! n n 2 (a) (b) (c) n! ; n! ; nn . n¼1

n¼1

n¼1

We have now seen examples of many different convergent and divergent series, including examples of various generic types. We call these basic series, since we shall use them commonly together with various other tests in order to prove that particular series are themselves convergent or divergent. Basic series 1 P 1 (a) np ; (b)

n¼1 1 P

(c)

n¼1 1 P

(d)

n¼1 1 P n¼1

cn ; np c n ;

The following series are convergent: for p  2;

Instances of these that we have seen are 1 P 1 n2 ;

for 0  c < 1;

Geometric Series,

n¼1

1 P

for p > 0; 0  c < 1;

n¼1

cn n! ;

n 2n

¼

1   P n n 12 ;

1  P 1 n n¼1

2

;

n¼1

1 P 10n ;

for c  0.

n¼1

The following series is divergent: 1 P 1 (e) for p  1. np ;

1 P n¼1

n¼1

n!

p1ffiffi n

¼

1 P n¼1

1 . n1=2

Another test for convergence When we examined the convergence or divergence of the series

1 P 1 n, we found

n¼1

it convenient to group the terms into blocks that contained in turn 1, 2, 4, 8, . . . successive individual terms of the series. In fact, a similar technique applies to a wide range of series, and the following result is thus often invaluable in order to save repeating that same type of argument: Theorem 4

You will see this technique used in the proof of Theorem 4, in Subsection 3.2.2.

Cauchy Condensation Test

Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2n a2n , for n ¼ 1, 2, . . . 1 1 X X an is convergent if and only if bn is convergent: n¼1

n¼1

1 P 1 1 For example, consider the series n. Here, let an ¼ n, so that n¼1 1 bn ¼ 2n a2n ¼ 2n  n ¼ 1: 2 1 P 1 Then the Condensation Test tells us that the series n is convergent if and 1 1 n¼1 P P bn ¼ 1 is convergent. This is divergent, by the Nononly if the series 1 n¼1 n¼1 P 1 null Test, so that the original series n must itself have been divergent. n¼1

For, if an ¼ 1n, then a2n ¼ 21n .

3: Series

98 This use of the Condensation Test is much simpler than setting out to perform a grouping and estimation exercise on such occasions! The Condensation Test is also often useful when the individual terms of a series include expressions such as loge n. Example 7 Use the Condensation Test to determine the convergence of the 1 P 1 pffiffiffiffiffiffiffiffi series . n¼2 n

loge n

1 Solution Let an ¼ pffiffiffiffiffiffiffiffi , n ¼ 2, . . .. Then ann is positive; and, since o n loge n  pffiffiffiffiffiffiffiffiffiffiffiffi 1 n loge n is an increasing sequence, fan g ¼ pffiffiffiffiffiffiffiffi is a decreasing n

loge n

sequence. Next, let bn ¼ 2n a2n ; thus 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi loge ð2n Þ 1 1 1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffi  1 : n loge 2 loge 2 n2

bn ¼ 2n 

Since 1 P

1 P

1 1

n¼2 n2

2n

is a basic divergent series, it follows by the Multiple Rule that

bn must be divergent. Hence by the Condensation Test, the original series

n¼2 1 P n¼2 n

1 pffiffiffiffiffiffiffiffi must also be divergent.

&

loge n

Problem 4 Use the Condensation Test to determine the convergence of the following series: 1 1 P P 1 1 (a) ; (b) . n log n nðlog nÞ2 n¼2

3.2.2

e

n¼2

1

loge 1

loge 1 ¼ 0. Again we are using the fact that the convergence of a series is not affected if we add or alter a finite of terms.

e

Proofs

You may omit these proofs at a first reading.

In the previous sub-section we gave a number of tests for the convergence of series with non-negative terms. We now supply the proofs of these tests. Theorem 1

Note that here we only sum from n ¼ 2, since it makes no 1 sense to talk about pffiffiffiffiffiffiffiffi , as

Comparison Test

1 1 P P (a) If 0  an  bn, for n ¼ 1, 2, . . ., and bn is convergent, then an is n¼1 n¼1 convergent. 1 1 P P (b) If 0  bn  an, for n ¼ 1, 2, . . ., and bn is divergent, then an is n¼1 n¼1 divergent.

Proof (a) Consider the nth partial sums sn ¼ a1 þ a2 þ    þ an ;

for n ¼ 1; 2; . . .;

3.2

Series with non-negative terms

99

and t n ¼ b 1 þ b2 þ    þ bn ; We know that

for n ¼ 1; 2; . . .:

a1  b1 ; a2  b2 ; . . .; an  bn ; and so sn  tn ;

for n ¼ 1; 2; . . .: 1 P

We also know that

By adding up the previous inequalities.

bn is convergent, and so the increasing sequence

n ¼1

{tn} is convergent, with limit t, say. Hence sn  tn  t;

for n ¼ 1; 2; . . .;

and so the increasing sequence {sn} is bounded above by t. By the Monotone 1 P Convergence Theorem, {sn} is therefore also convergent, and so an is a n ¼1 convergent series. Note: Moreover, by the Limit Inequality Rule 1 1 X X an  bn : n ¼1

n¼1

1 P an is (b) We can deduce part (b) from part (a), as follows. Suppose that 1 P n ¼1 convergent. Then, by part (a), bn must also be convergent. However, 1 1 n ¼1 P P bn is assumed to be divergent, and so an must also be divergent.& n ¼1

You met this Rule in Sub-section 2.3.3, Theorem 3. This additional inequality will sometimes be useful. Such ‘proofs by contradiction’ can sometimes save a great deal of detailed arguments.

n¼1

Theorem 2

Limit Comparison Test 1 1 P P Suppose that an and bn have positive terms, and that n ¼1

n¼1

an ! L as n ! 1; where L 6¼ 0: bn 1 1 P P (a) If bn is convergent, then an is convergent. (b) If

n¼1 1 P

n ¼1

bn is divergent, then

n ¼1

1 P

an is divergent.

n¼1

Proof   (a) Because abnn is convergent, it must be bounded. Thus there is a constant K such that an  K; bn

for n ¼ 1; 2; . . .;

and so an  Kbn ;

for n ¼ 1; 2; . . .: 1 P By the Multiple Rule, Kbn is convergent. Hence, by the Comparison n¼1 1 P an is also convergent. Test, n¼1

See Sub-section 2.4.2, Theorem 1 (the Boundedness Theorem).

3: Series

100 (b) We can deduce part (b) from part (a), as follows. 1 1 P P Suppose that an is convergent. Then, by part (a), bn must also be n¼1

n¼1

convergent, because bn 1 as n ! 1; ! L an by the Quotient Rule for sequences. 1 1 P P However, bn is assumed to be divergent, and so an must also be n¼1 & divergent. n¼1

We are following a ‘proof by contradiction’ approach here. Remember that L 6¼ 0.

Theorem 3

Ratio Test 1 P Suppose that an has positive terms, and that aanþ1 ! ‘ as n ! 1. n n¼1 1 P (a) If 0  ‘ < 1, then an is convergent. n¼1 1 P

(b) If ‘ > 1, then

an is divergent.

Remember that part (b) includes the case aanþ1 ! 1. n

n¼1

Proof (a) Because ‘ < 1, we can choose " > 0 so that r ¼ ‘ þ " < 1: Hence there is a positive number X, which we may assume to be a positive integer, such that anþ1  r; for all n  X: an Thus, for n  X, we have      an an an1 aXþ1 ¼ ...  r nX ; aX an1 an2 aX since each of the n  X brackets is less than or equal to r. Hence an  aX r nX ;

for all n  X:

For example, take " ¼ 12 ð1  ‘Þ. Recall that we occasionally make the convenient assumption that X is an integer to avoid notational complications. 1 +ε  –ε

(1)

X

n

Now 1 X

aX r nX ¼ aX þ aX r þ aX r 2 þ   ;

n¼X

which is a geometric series with first term aX and common ratio r. Since 0 < r < 1, this series is convergent, and so, by inequality (1) and the 1 1 P P Comparison Test, an is also convergent. Hence an is convergent. n¼X

n¼1

(b) Since anþ1 ! ‘ as n ! 1; an and ‘ > 1, there is a positive number X, which we may assume to be a positive integer, such that anþ1  1; for all n  X: an

 1

X

n

3.2

Series with non-negative terms

101

(This also holds if aanþ1 ! 1 as n ! 1.) n Thus, for n  X, we have      an an an1 aXþ1 ¼ ...  1; aX an1 an2 aX since each of the brackets is greater than or equal to 1. Hence an  aX > 0;

for all n  X; 1 P

and so {an} cannot be a null sequence. Thus, by the Non-null Test, divergent. Hence

1 P

an is

n¼X

&

an is also divergent.

n¼1

Theorem 4 Cauchy Condensation Test Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2n a2n for n ¼ 1, 2, . . . 1 1 X X an is convergent if and only if bn is convergent: n¼1

n¼1

1 1 P P Proof Denote by sn and tn the nth partial sums of the series an and bn , n¼1 n¼1 respectively

s n ¼ a1 þ a2 þ    þ an

and tn ¼ 21 a2 þ 22 a4 þ    þ 2n a2n :

Since the terms an are decreasing and a1  0, it follows that The kth bracket contains 2k1 terms. The kth bracket contains 2k1 occurrences of a2k.

s2n ¼ a1 þ ða2 Þ þ ða3 þ a4 Þ þ    þ ða2n1 þ1 þ    þ a2n Þ  0 þ ða2 Þ þ ða4 þ a4 Þ þ    þ ða2n þ    þ a2n Þ   ¼ 0 þ ða2 Þ þ ð2a4 Þ þ    þ 2n1 a2n 1 ¼ tn : 2

ð2Þ

We may also write s2n in another way s 2 n ¼ a1 þ ð a2 þ a3 Þ þ ð a4 þ a5 þ a6 þ a7 Þ þ    þ ða2n1 þ    þ a2n 1 Þ þ a2n  a1 þ ð a2 þ a2 Þ þ ða4 þ a4 þ a4 þ a4 Þ þ    þ ða2n1 þ    þ a2n1 Þ þ a2n   ¼ a1 þ ð2a2 Þ þ ð4a4 Þ þ    þ 2n1 a2n1 þ a2n  a1 þ t n : ð3Þ Now suppose that

1 P

an is convergent. Then, by the Boundedness Theorem for

n¼1

series, its sequence of partial sums {sn} must be bounded above and so the sequence of {sn}. It therefore fs2ng must also be bounded above – since it is a subsequence  follows from inequality (2) that the sequence 12 tn must be bounded above, so the sequence {tn} must also be bounded above. Another application of the Boundedness 1 P Theorem shows that therefore the series bn must be convergent. n¼1

The kth bracket contains 2k terms.

The kth bracket contains 2k occurrences of a2k. In obtaining (3), we have replaced the term a2n in the previous line by a larger term 2n a2n. The Boundedness Theorem for series appeared in Section 3.2, above.

3: Series

102 Next, suppose that

1 P

an is divergent. Then, by the Boundedness Theorem

n¼1

for series, its sequence of partial sums {sn} must be unbounded above, and so the sequence fs2ng must also be unbounded above – since it is a subsequence of {sn}. It therefore follows from inequality (3) that the sequence {a1 þ tn} must be unbounded above, so the sequence {tn} must also be unbounded above. Another application of the Boundedness Theorem shows that therefore the 1 P & series bn must be divergent. n¼1

We end this section by proving the convergence/divergence of the basic series given earlier. The following series are convergent:

Basic series 1 P 1 (a) np ;

for p  2;

(b)

n¼1 1 P

c ;

for 0  c < 1;

(c)

n¼1 1 P

np c n ;

for p > 0; 0  c < 1;

(d)

n¼1 1 P

cn n! ;

for c  0.

n¼1

In Sub-section 3.2.1. In Chapter 7 we shall prove 1 P 1 that np is convergent, for n¼1

n

all p > 1.

The following series is divergent: 1 P 1 (e) for p  1. np ; n¼1

Proof (a) This series is convergent, by the Comparison Test, since, if p  2, then 1 1  2 ; for n ¼ 1; 2; . . .; p n n 1 P 1 and the series n2 is convergent.

Sub-section 3.2.1, Example 1.

n¼1

(b) The series

1 P

cn is a geometric series with common ratio c, and so it

n¼1

converges if 0  c < 1. p n

(c) Let an ¼ n c , for n ¼ 1, 2, . . .. Then, for c 6¼ 0   anþ1 ðn þ 1Þp cnþ1 1 p ¼ ¼ 1þ c: n an np c n Thus, if k is any integer greater than or equal to p, then   anþ1 1 k c  1þ c: n an Now, by the Product Rule for sequences   1 k 1þ ! 1 as n ! 1; n

Sub-section 3.1.2. If c ¼ 0, the series is clearly convergent.

We introduce the integer k here, so that we can use the Product Rule for sequences.

3.3

Series with positive and negative terms

103

and so, by the Squeeze Rule for sequences anþ1 ! c as n ! 1: an 1 P Since 0  c < 1, we deduce, from the Ratio Test, that np cn is n¼1 convergent. n

(d) Let an ¼ cn! , for n ¼ 1, 2, . . .. Then, for c 6¼ 0,

n anþ1 cnþ1 c c ¼ : ¼ an ðn þ 1Þ! n! n þ 1 1 n P c ! 0 as n ! 1, and we deduce from the Ratio Test that Thus aanþ1 n! is n n¼1 convergent. 1 P 1 (e) We saw earlier that the series n is divergent. It follows that the series 1 n¼1 P 1 1 1 np is also divergent, by the Comparison Test, since if p  1 then np  n,

If c ¼ 0, the series is clearly convergent.

In Sub-section 3.2.1, Example 2.

n¼1

&

for n ¼ 1, 2, . . ..

3.3

Series with positive and negative terms

The study of series

1 P

an , with an  0 for all values of n, is relatively straight-

n¼1

forward, because the sequence of partial sums {sn} is increasing. Similarly, if an  0 for all values of n, then {sn} is decreasing. However, it is harder to determine the behaviour of a series with both positive and negative terms, because {sn} is neither increasing nor decreasing. However, if the sequence {an} contains only finitely many negative terms, then the sequence {sn} is eventually increasing, and we can apply the methods of Section 3.2. Similarly, if {an} contains only finitely many positive terms, then the sequence {sn} is eventually decreasing, and we can again apply the methods of Section 3.2, after making a sign change. In this section we look at series such as 1 1 1 1 1 1 1 1 1 1 1  þ  þ  þ    and 1 þ 2  2 þ 2 þ 2  2 þ   ; 2 3 4 5 6 2 3 4 5 6 which contain infinitely many terms of either sign. We give several methods which can often be used to prove that such series are convergent.

3.3.1

For example, the convergence of 1þ2þ3

1 1 1     42 52 62

follows from that of

1 P

n¼1

Absolute convergence

Suppose that we want to determine the behaviour of the infinite series 1 X ð1Þnþ1 1 1 1 1 1 ¼ 1  2 þ 2  2 þ 2  2 þ   : 2 2 3 4 5 6 n n¼1 We know that the series 1 X 1 1 1 1 1 1 ¼ 1 þ 2 þ 2 þ 2 þ 2 þ 2 þ  2 n 2 3 4 5 6 n¼1

(1)

(2)

is convergent. Does this mean that the series (1) is also convergent? In fact it does, as we now prove.

The series (2) is a basic convergent series.

1 n2

:

3: Series

104 Consider the two related series 1 1 1 þ 0 þ 2 þ 0 þ 2 þ 0 þ  3 5 and 1 1 1 0 þ 2 þ 0 þ 2 þ 0 þ 2 þ : 2 4 6 Each of these series is dominated by the series (2), and so they are both convergent, by the Comparison Test. Applying the Multiple Rule with l ¼ 1, and the Sum Rule, we deduce that the series 1 1 1 1 1 1  2 þ 2  2 þ 2  2 þ    is convergent: 2 3 4 5 6 The argument just given is the basis for a concept called absolute convergence, which we now define. Definitions 1 1 P P A series an is absolutely convergent if jan j is convergent. A series

n¼1 1 P

n¼1

an that is convergent but not absolutely convergent is condi-

If the terms an are all nonnegative, then absolute convergence and convergence have the same meaning.

n¼1

tionally convergent. For example, the series (1) is absolutely convergent, because the series is convergent. However, the series 1 X ð1Þnþ1 n¼1

n

1 P n¼1

1 n2

We shall examine the behaviour of conditionally convergent series later, in Sub-sections 3.3.2 and 3.3.3.

1 1 1 1 1 ¼ 1  þ  þ  þ  2 3 4 5 6

1 P 1 is not absolutely convergent, because the series n is divergent; the series is, n¼1 in fact, conditionally convergent.

Theorem 1 Absolute Convergence Test 1 1 P P If an is absolutely convergent, then an is convergent. n¼1

You saw this in Example 2, Sub-section 3.2.1.

The proofs of the results in this sub-section appear in Sub-section 3.3.6.

n¼1

It follows that the series (1) is convergent (as we have already seen), and also that the series 1þ

1 1 1 1 1  þ þ  þ  22 32 42 52 62

is convergent, because the series

1 P

1 n2

is convergent. Indeed, no matter how   we distribute the plus and minus signs among the terms of n12 , the resulting series is convergent. However, the Absolute Convergence Test tells us nothing about the behaviour of the series n¼1

1 1 1 1 1 1  þ  þ  þ  2 3 4 5 6

(3)

1 1 1 1 1 1 1 1 1 þ  þ þ  þ þ  þ : 2 3 4 5 6 7 8 9

(4)

and

In fact, series (3) is convergent (by the Alternating Test, which we introduce in Subsection 3.3.2), and series (4) is divergent (see Exercise 2(a) for Section 3.3).

3.3

Series with positive and negative terms

The series

1 P n¼1

1 n

105

is divergent, and so the two series (3) and (4) are not

absolutely convergent. Example 1 Prove that the following series are convergent: 1 P ð1Þnþ1 1 1 (a) (b) 1 þ 12  14 þ 18 þ 16  32   . n3 ; n¼1

Solution (a) Let an ¼

1 P 1 for n ¼ 1, 2, . . .; then jan j ¼ n13 . We know that n3 is 1 n¼1 P ð1Þnþ1 convergent, so it follows that is absolutely convergent. Hence, n3 ð1Þnþ1 n3 ,

For

1 P n¼1

1 n3

is an example of a

basic convergent series.

n¼1

1 P ð1Þnþ1 is convergent. by the Absolute Convergence Test, n3 n¼1 1 P 1 (b) The series 2n is a convergent geometric series, so that the series n¼0

1 1 1 1 1  1þ  þ þ  2 4 8 16 32

is absolutely convergent. Then, by the Absolute Convergence Test, this & series is also convergent. Problem 1 Prove that the following series are convergent: 1 1 P P ð1Þnþ1 n cos n (a) ; (b) 3 2n . n þ1 n¼1

n¼1

P The Absolute Convergence Test states that, if jan j is convergent, then P an is also convergent, but it does not indicate P any explicitPconnection between the sums of these two series. Clearly, an is less than jan j if any of the terms an are negative. 1 1 P P ð1Þnþ1 1 1 1 1 For example, whereas ¼ 2n ¼ 2 þ 4 þ 8 þ    ¼ 1, 2n n¼1

1 2

n¼1

 14 þ 18     ¼ 13 : The following result relates the values of

P

an and

P

jan j:

Triangle Inequality (infinite form) 1  1 1 P  P P If an is absolutely convergent, then  an   jan j. n¼1

n¼1

n¼1

1 1 1 Problem 2 Show that the series 12 þ 14  18 þ 16 þ 32  64 þ    is convergent, and that its sum lies in [1,1]. (You do NOT need to find the sum of the series.)

3.3.2

The Alternating Test

Suppose that we want to determine the behaviour of the following infinite series, in which the terms have alternating signs 1 X ð1Þnþ1 n¼1

n

1 1 1 1 1 ¼ 1  þ  þ  þ   : 2 3 4 5 6

Recall that you met the Triangle Inequality for a1 ; a2 ; . . .; an   n n X  X   a ja j;  k   k¼1  k¼1 k in Sub-section 1.3.1.

3: Series

106 1 P

1 The Absolute Convergence Test does not help us with this series, because n n¼1 is divergent. 1 P ð1Þnþ1 In fact, the series is convergent. To see why, we first calculate n n¼1

some of its partial sums and plot them on a sequence diagram s1 ¼ 1; 1 s2 ¼ 1  ¼ 0:5; 2 1 1 s3 ¼ 1  þ ¼ 0:83; 2 3 1 1 1 s4 ¼ 1  þ  ¼ 0:583; 2 3 4 1 1 1 1 s5 ¼ 1  þ  þ ¼ 0:783; 2 3 4 5 1 1 1 1 1 s6 ¼ 1  þ  þ  ¼ 0:616: 2 3 4 5 6 The sequence diagram for {sn} suggests that s1  s3  s5  . . .  s2k1  . . . and s2  s4  s6  . . .  s2k  . . . ; for all k. In other words: the odd subsequence {s2k1} is decreasing and: the even subsequence {s2k} is increasing. Also, the terms of {s2k1} all exceed the terms of {s2k}, and both subsequences appear to converge to a common limit s, which lies between the odd and even partial sums. To prove this, we write the even partial sums {s2k} as follows s2k ¼

      1 1 1 1 1 þ  þ  þ  : 1 2 3 4 2k  1 2k

All the brackets are positive, and so the subsequence {s2k} is increasing. We can also write 

s2k

     1 1 1 1 1 1 1        : ¼1 2 3 4 5 2k  2 2k  1 2k

Again, all the brackets are positive, and so {s2k} is bounded above, by 1. Hence {s2k} is convergent, by the Monotone Convergence Theorem. Let lim s2k ¼ s: k!1

Since s2k ¼ s2k1  and

1 2k

is null, we have

1 2k

You met subsequences earlier, in Sub-section 2.4.4.

3.3

Series with positive and negative terms lim s2k1

k!1

107

    1 1 ¼ lim s2k þ ¼ lim s2k þ lim k!1 k!1 k!1 2k 2k ¼ s þ 0 ¼ s;

by the Combination Rules for sequences. Thus the odd and even subsequences of {sn} both tend to the same limit s, and so {sn} itself tends to s. Hence 1 X ð1Þnþ1

n

n¼1

1 1 1 1 1 ¼ 1  þ  þ  þ    is convergent, with sum s: 2 3 4 5 6

By Theorem 6 of Sub-section 2.4.4. In fact, s ¼ loge 2 ’ 0.69.

The same method can be used to prove the following general result: Theorem 2

This test is sometimes called the Leibniz Test.

Alternating Test

If an ¼ ð1Þnþ1 bn ;

n ¼ 1; 2; . . .;

where {bn} is a decreasing null sequence with positive terms, then 1 X an ¼ b1  b2 þ b3  b4 þ    is convergent: n¼1

When we apply the Alternating Test, there are a number of conditions which must be checked. We now describe these in the form of a strategy. Strategy To prove that check that

1 P

an is convergent, using the Alternating Test,

n¼1

an ¼ ð1Þnþ1 bn ;

n ¼ 1; 2; . . .;

where: 1. bn  0,

for n ¼ 1, 2, . . .;

To show that {bn} is null, use the techniques introduced in Sub-section 2.2.1.

2. {bn} is a null sequence; 3. {bn} is decreasing.

Remark It is often easiest to check that {bn} is decreasing by verifying that increasing.

1 bn

Here are some examples. Example 2 Determine which of the following series are convergent: 1 1 1 P P P ð1Þnþ1 ð1Þnþ1 pffiffi ; (a) (b) ; (c) ð1Þnþ1 . 4 n n n¼1

n¼1

n¼1

Solution n nþ1 o (a) The sequence ð1pÞffiffin has terms of the form an ¼ ð1Þnþ1 bn , where 1 bn ¼ pffiffiffi ; n

n ¼ 1; 2; . . .:

is

3: Series

108 Now: 1. bn ¼ p1ffiffin  0, for n ¼ 1, 2, . . .;   2. fbn g ¼ p1ffiffin is a basic null sequence;     pffiffiffi 3. fbn g ¼ p1ffiffin is decreasing, because b1n ¼ f ng is increasing. 1 P ð1Þnþ1 pffiffi is convergent. Hence, by the Alternating Test, n (b) The sequence

n

nþ1

ð1Þ n4

bn ¼

1 ; n4

o

n¼1

has terms of the form an ¼ ð1Þnþ1 bn , where

Alternatively, we can show that this series is convergent by the Absolute Convergence Test.

n ¼ 1; 2; . . .:

Now: 1. bn ¼ n14  0, for n ¼ 1, 2, . . .;   2. fbn g ¼ n14 is a basic null sequence;       3. fbn g ¼ n14 is decreasing, because b1n ¼ n4 is increasing. 1 P ð1Þnþ1 Hence, by the Alternating Test, is convergent. n4 n¼1

  (c) The sequence ð1Þnþ1 is not a null sequence. Hence, by the Non-null Test 1 X ð1Þnþ1 is divergent. & n¼1

Problem 3 (a)

1 P n¼1

3.3.3

ð1Þnþ1 n3

Notice that the fact that the nth term includes a factor (1)nþ1 does not imply at all that a series is convergent!

Determine which of the following series are convergent: ;

(b)

1 P n¼1

n ð1Þnþ1 nþ2 ;

(c)

1 P n¼1

ð1Þnþ1 . n1=3 þ n1=2

Rearrangement of a series

In the last sub-section we saw that the series 1 12 þ 13  14 þ 15  16 þ    is convergent. Let us denote by ‘ the sum of this series. If we temporarily ignore the need for careful mathematical proof and simply ‘move terms about’, look what we get 1 1 1 1 1 ‘ ¼ 1  þ  þ  þ  2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 ¼ 1   þ   þ   þ   þ  2 4 3 6 8 5 10 12 7 14 16       1 1 1 1 1 1 1 1   ¼ 1  þ  þ  2 4 3 6 8 5 10 12   1 1 1  þ  þ  7 14 16

In fact this series is conditionally convergent, since the harmonic series 1 þ 12 þ 13 þ 14 þ 15 þ 16 þ    is divergent.

(5)

This is the definition of ‘.

ð6Þ

The pattern in (6) is to follow the next available positive term by the two next available negative terms. Here we insert some brackets.

3.3

Series with positive and negative terms 1 1 1 1 1 1 1 1  þ  þ  þ  þ  2 4 6 8 10 12 14 16   1 1 1 1 1 1 1 ¼ 1  þ  þ  þ   2 2 3 4 5 6 8

¼

109 Here we insert the value of each bracket. Here we pull out a common factor of 12.

1 ¼ ‘: 2 It follows from the fact that ‘ ¼ 12 ‘ that ‘ ¼ 0. However this is impossible, since the partial sum s2 of the original series is 12, and the even partial sums s2k are increasing. What has gone wrong? We assumed that operations that are valid for sums of a finite number of terms also hold for sums of an infinite number of terms – we rearranged (infinitely many of) the terms in the series (5) to obtain the series (6) – without any justification. This rearrangement operation is not valid in general! We now give a precise definition of what we mean by a rearrangement. 1 P Loosely speaking, the series bn ¼ b1 þ b2 þ    is a rearrangement of the 1 n¼1 P an ¼ a1 þ a2 þ    if precisely the same terms appear in the series

Recall that, since the series converges to ‘, then s2k converges to ‘ also. In fact the series (6) does converge; we ask you to supply a careful proof of this in Exercise 2 on Section 3.3.

n¼1

sequences {bn} ¼ {b1, b2, . . .} and {an} ¼ {a1, a2, . . .}, though they may occur in a different order. 1 P Definition A series bn ¼ b1 þ b2 þ    is a rearrangement of the series 1 n¼1 P an ¼ a1 þ a2 þ    if there is a bijection ƒ such that n¼1

f: N !N bn 7! af ðnÞ:

Recall that a bijection is a one–one onto mapping.

1 P ð1Þnþ1 Assuming that the sum of the series ¼ 1  12 þ 13  14 þ n 1 n¼1 P 1 1 an ¼ 1 þ 13  12 þ 15 þ 17  14 þ 19 þ 5  6 þ    is loge 2, prove that the series

Example 2

1 11

 16 þ    converges to 32 loge 2.

n¼1

Solution Let sn and tn denote the nth partial sums of the series 1 1 þ 13  12 þ 15 þ 17  14 þ 19 þ 11  16 þ    and 1  12 þ 13  14 þ 15  16 þ   , respectively. We now introduce an extra piece of notation, Hn; we denote by Hn the nth n P 1 1 1 partial sum k ¼ 1þ 2 þ    þ n of the harmonic series. k¼1 1 P an come ‘in a natural way’ in threes, so it Now, the terms of the series

This will enable us to tackle things much more easily!

n¼1

seems sensible to look at the partial sums s3n 1 1 1 1 1 1 1 1 s3n ¼ 1 þ  þ þ  þ þ  þ    3 2 5 7 4 9 11 6 1 1 1 þ þ  4n  3 4n  1 2n

From the definition of s3n.

3: Series

110       1 1 1 1 1 1 1 1 ¼ 1þ  þ þ  þ þ  þ  3 2 5 7 4 9 11 6   1 1 1 þ  þ 4n  3 4n  1 2n     1 1 1 1 1 1 1  þ þ þ  þ ¼ 1þ þ þ    þ 3 5 4n  1 2 4 6 2n     1 1 1 1 1 1 1 1 ¼ 1þ þ þ    þ  þ þ þ  þ  Hn 2 3 4n 2 4 6 4n 2     1 1 1 1 1 1 1 1 ¼ 1þ þ þ    þ  1þ þ þ    þ  Hn 2 3 4n 2 2 3 2n 2 1 1 ¼ H4n  H2n  Hn : (7) 2 2

We insert some brackets in a finite sum, for convenience. We separate out the positive and negative terms. We add some terms to the contents of the first bracket, then subtract them again; the last bracket is simply 12Hn.

Furthermore 1 1 1 1 1 1 1  t2n ¼ 1 þ  þ  þ    þ  2 3 4 5 6  2n  1 2n  1 1 1 1 1 1 1 ¼ 1þ þ þ    þ  2 þ þ þ  þ 2 3 2n 2 4 6 2n   1 1 1 ¼ H2n  1þ þ þ    þ 2 3 n ¼ H2n  Hn :

By the definition of t2n. We insert some positive terms, then remove them again. By the definition of H2n.

(8)

We now eliminate the H’s from equations (7) and (8) as follows 1 1 s3n ¼ H4n  H2n  Hn 2 2 1 ¼ ðH4n  H2n Þ þ ðH2n  Hn Þ 2 1 ¼ t4n þ t2n : 2

(9)

But it follows from the hypotheses of the example that tn ! loge 2, and so t2n ! loge 2 and t4n ! loge 2, as n ! 1. Hence, letting n ! 1 in equation (9), we see that 1 s3n ! loge 2 þ loge 2 2 3 ¼ loge 2: 2 Next s3n1 ¼ s3n þ

1 2n

3 ! loge 2 2 and s3n2 ¼ s3n þ

1 1  2n 4n  1

3 ! loge 2 2

as n ! 1:

It follows from the above three results that sn ! 32 loge 2 as n ! 1:

&

By the definition of Hn.

3.3

Series with positive and negative terms Problem 4

Prove that the series

1 P n¼1

111 an ¼ 1 12  14 þ 13  16  18 þ 15 

1 10

1 1 1  12 þ 17  14  16 þ    converges to 12 loge 2. [You may assume that 1 P ð1Þnþ1 ¼1 12 þ 13  14 þ 15  16 þ    is loge 2.] the sum of the series n n¼1

In order to better understand the behaviour of series with positive and negative terms, we introduce some notation. For a conditionally convergent series 1 X an ¼ a1 þ a2 þ   ;

For example, for the series 1 X n¼1

n¼1  we define the quantities aþ n and an as follows   an ; if an  0, 0; if an  0,  aþ ¼ ¼ and a n n 0; if an < 0, an ; if an < 0.  þ In other words,    the sequence an picks out the non-negative terms and the sequence an picks out the non-positive terms (and discards their sign).  þ  In particular, an ¼ aþ n  an , and an  0 and an  0. 1 P Next, since the series an is convergent, an ! 0 as n ! 1 . It follows that n¼1

1 aþ n ¼ ðjan j þ an Þ ! 0 2

as n ! 1

1 1 1 an ¼ 1  þ  2 3 4 1 1 þ  þ   ; 5 6

we have 1 X n¼1

1 aþ n ¼10þ 0 3 1 þ  0þ 5

and 1 X

1 1 a n ¼0þ þ0þ 2 4 n¼1 1 þ 0 þ   : 6

and 1 a n ¼ ðjan j  an Þ ! 0 as n ! 1: 2 1 P Now, since an is conditionally convergent, it must contain infinitely 1 1 n¼1 P P many negative terms, since otherwise an and jan j would be the same n¼1

n¼1

apart from a finite number of terms. Since altering a finite number of terms does not affect the convergence of a series, it would then follow that the series 1 P jan j would be convergent – which we know is not the case.

For the series

1 P

an is

n¼1

assumed to be only conditionally convergent.

n¼1

1 P an . Similarly, there must be infinitely many positive terms in n¼1 1 P Finally, we show that, if the series an is conditionally convergent, then 1 1 1 n¼1 P P P aþ an ) and a the corresponding series n (the ‘positive part’ of n (the 1 n¼1 n¼1 n¼1 P an ) are both divergent. ‘negative part’ of n¼1 P 1 þ  For, suppose that aþ n were convergent. Then, since an ¼ an  an , we 1 n¼1 P þ have a a n ¼ an  an ; it follows that the series n must also be convergent. 1 n¼1 P  And then, since jan j ¼ aþ jan j would be n þ an , it would also follow that n¼1 convergent – which we know is not the case. 1 P A similar argument shows that a n cannot be convergent. 1 1 n¼1 P P aþ a In particular, it follows from the divergence of the series n and n, n¼1

For the series

1 P

an is

n¼1

assumed to be only conditionally convergent.

n¼1

whose terms are non-negative, that their partial sums must tend to 1 as n ! 1.

By the Boundedness Theorem for series.

3: Series

112 In general it is not true that any rearrangement of a conditionally convergent series converges. The situation is much more interesting!

Some rearrangements converge, some diverge.

Theorem 3

Riemann’s Rearrangement Theorem 1 P Let the series an ¼ a1 þ a2 þ    be conditionally convergent, with nth n¼1

partial sum sn. Then there are rearrangements of the series that have the following properties: 1. For any given number s, the rearranged series converges to s; 2. For any given numbers x and y, one subsequence of the sn converges to x and another subsequence of the sn converges to y;

This is not a complete list of all the possibilities that can occur!

3. The sequence sn converges to 1. 4. One subsequence of the sn converges to 1 and another subsequence of the sn converges to 1. We will not prove Theorem 3. Instead, we illustrate case (1) using the facts that we have already discovered about conditionally convergent series. Using similar ideas, we can prove the various parts of the theorem. 1 P ð1Þnþ1 Example 3 Find a rearrangement of the series ¼ 1 12 þ 13  14 þ n 1 1 n¼1 5  6 þ    that converges to the sum 3. 1 nþ1 P Solution Since the ‘positive’ part of the series ð1nÞ has partial sums that

Recall that the sum of this series is loge 2 or approximately 0.69.

n¼1

tend to 1, we start to construct the desired rearranged series as follows. Take enough of the ‘positive’ terms 1, 13 , 15 , . . ., 2N111 so that the sum 1 þ 13 þ 15 þ    þ 2N111 is greater than 3, choosing N1 so that it is the first integer such that this sum is greater than 3. Then these N1 terms will form the first N1 terms in our desired rearranged series. Next, take enough of the ‘negative’ terms 12 , 14 , 16 , . . ., 2N1 2 so that the expression 

   1 1 1 1 1 1 1 þ þ þ þ 1 þ þ þþ  3 5 2N1  1 2 4 6 2N2

is less than 3, choosing N2 so that it is the first integer such that this sum is less than 3. These N1 þ N2 terms form the first N1 þ N2 terms in our desired rearranged series. Now add in some more ‘positive’ terms 2N11þ1 , 2N11þ3 , . . ., 2N311 so that the sum     1 1 1 1 1 1 1 þ þ þ þ 1 þ þ þþ  3 5 2N1  1 2 4 6 2N2   1 1 1 þ þ þþ 2N1 þ 1 2N1 þ 3 2N3  1 is greater than 3, choosing N3 so that it is the first available integer such that this sum is greater than 3. Then these N2 þ N3 terms will form the first N2 þ N3 terms in our desired rearranged series.

This is possible since 1 1 1þ þ þ    ! 1; 3 5 as n ! 1.

3.3

Series with positive and negative terms

113

We then add in just enough ‘negative’ terms to make the next sum of N3 þ N4 terms less than 3, and so on indefinitely. In each set of two steps in this process we must use at least one of the ‘positive’ terms and one of the ‘negative’ terms, so that eventually all the ‘positive’ terms and all the ‘negative’ terms of the original series will be taken 1 P exactly once in the new series, which we denote by bn . So certainly the series

At each step there are always ‘positive’ terms or ‘negative’ terms left to choose, since there are infinitely many of each.

n¼1

1 P

bn is a rearrangement of the original series

n¼1

1 P ð1Þnþ1 n¼1

n

.

But how do we know that the sum of the rearranged series is 3? As we keep 1 P making the partial sums of bn swing back and forth from one side of 3 to the n¼1

other, we need the ‘radius’ of the swings to tend to 0. We achieve this by 1 P ð1Þnþ1 that we choose next, once always changing the sign of the terms from n n¼1 1 P the partial sum of bn has crossed beyond the value 3. For this ensures that n¼1

(from 2N1 2 onwards) the difference between a partial sum and 3 will always be less than the absolute value of the last term 2Nk11 or 2N1 k used, and we know that this last term tends to 0 as we go further out along the original series. 1 P It follows that the rearranged series bn converges to 3, as required. &

For clearly N1k  1k, since we use at least 1 term in the original series at each step of the rearrangement process.

n¼1

Problem 5 Find a rearrangement of the series 1 1 1 4 þ 5  6 þ    whose partial sums tend to 1.

1 P ð1Þnþ1 n

n¼1

¼1 12 þ 13 

The situation is very much simpler, though, in the case of absolutely convergent series. Let the series

Theorem 4

Then any rearrangement 1 P

bn ¼

n¼1

1 P

1 P

an ¼ a1 þ a2 þ    be absolutely convergent.

n¼1 1 P

bn of the series also converges absolutely, and

n¼1

an .

n¼1

For example, the series 1

1 1 1 1 1 1 1 1 þ  þ  þ  þ   22 32 42 52 62 72 82 9 2

For the series  1  1 P ð1Þnþ1  P 1  n2  ¼ n2 , which is n¼1

n¼1

a basic convergent series.

is absolutely convergent. It follows from Theorem 4 that the following series is also absolutely convergent – and to the same sum

3.3.4

1 |{z}

1 1  2 2 2 4 |fflfflfflfflfflffl{zfflfflfflfflfflffl}

1 1 1 þ 2þ 2þ 2 3 5 7 |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

1 term

2 terms

3 terms

1 1 1 1  2  2  2  2þ 6 8 10 12 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} : 4 terms

Multiplication of series

We now look at the question of how we might multiply together two infinite 1 1 1 P P P series an and bn to obtain a single series cn with the property that n¼0

n¼0

n¼0

In this sub-section we sum our series from n ¼ 0 rather than from n ¼ 1, simply because in many applications of Theorem 5 (below) this is the version that we shall require.

3: Series

114 1 X

cn ¼

n¼0

1 X

!

1 X



an

n¼0

! bn :

n¼0

1 P an bn ¼ Now, afirstthought might be that cn ¼ anbn, so that   n¼0 1 1 P P an  bn : However this is not the case! n¼0 n¼0  n  n For example, let an ¼ 12 and bn ¼ 13 . Then 1 1  n 1 1  n X X X X 1 1 3 an ¼ ¼ 2 and bn ¼ ¼ ; 2 3 2 n¼0 n¼0 n¼0 n¼0

but 1 X

an bn ¼

n¼0

1  n X 1 n¼0

6

These are both geometric series.

6 ¼ : 5

Note that 65 6¼ 2  32!

We can get a clue to what the correct formula for cn might be by, temporarily, abandoning our rigorous approach and simply ‘pushing symbols about’! If we multiply out both brackets and collect terms in a convenient way, we get 1 X n¼0

!

an 

1 X

!

bn

n¼0

¼ ða0 þ a1 þ a2 þ a3 þ  Þ  ðb0 þ b1 þ b2 þ b3 þ  Þ ¼ a0 ðb0 þ b1 þ b2 þ b3 þ   Þ þ a1 ðb0 þ b1 þ b2 þ b3 þ   Þ þ a2 ðb0 þ b1 þ b2 þ b3 þ   Þ þ a3 ðb0 þ b1 þ b2 þ b3 þ   Þ þ    ¼ a0 b0 þ a0 b1 þ a0 b2 þ a0 b3 þ    þ a1 b0 þ a1 b1 þ a1 b2 þ a1 b3 þ    þ a2 b0 þ a2 b1 þ a2 b2 þ a2 b3 þ    þ a3 b0 þ a3 b1 þ a3 b2 þ a3 b3 þ    ¼ a0 b0 þ ða0 b1 þ a1 b0 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ þ   :

Notice that every term that we would expect to be in such a product appears exactly once in this final expression. We have grouped the terms in the final expression in a particularly convenient way, so that we can see a pattern emerging from this non-rigorous argument. We can formalise this discussion as follows: Theorem 5

Product Rule 1 1 P P Let the series an and bn be absolutely convergent, and let n¼0

n¼0

cn ¼ a0 bn þ a1 bn1 þ a2 bn2 þ    þ an b0 ¼

n X

ak bnk :

k¼0

1 P

cn is absolutely convergent, and ! ! 1 1 1 X X X cn ¼ an  bn :

Then the series

n¼0

n¼0

n¼0

Let us now return to the two series

n¼0 1 P

an ¼

n¼0

we considered earlier, whose sums were 2 mula in Theorem 5 for cn, we have

1  P 1 n 2

n¼0 and 32,

and

1 P n¼0

bn ¼

1  P 1 n n¼0

3

that

respectively. With the for-

The pattern here is that in the kth bracket, the sum of the subscripts of each term is precisely k.

3.3

Series with positive and negative terms cn ¼

n X

ak bnk ¼

k¼0

115

n X 1 1  nk k 2 3 k¼0

n  k 1X 3 ¼ n 3 k¼0 2

 nþ1 1 1  32 ¼ n 3 1  32 1 ¼ n  ð2Þ  3

 nþ1 ! 3 3 2 ¼ n n: 1 2 2 3

It follows that 1 X

1 1 X X 1 1 2 n n 2 3 n¼0 n¼0   3 ¼322 ¼ 6  3 ¼ 3: 2

cn ¼ 3 

n¼0

Since 2  32 ¼ 3; we see that the sum of the series predicts it should be.

1 P

cn is what Theorem 5

n¼0

Remark Notice that the theorem requires that the series

1 P

an and

n¼0

1 P

bn are absolutely

n¼0

convergent. Without this assumption, the result may be false. 1 1 P P Þnþ1 pffiffiffiffiffiffi . Then both an and bn are converFor example, let an ¼ bn ¼ ð1 nþ1 n¼0

n¼0

gent, by the Alternating Test – but they are not absolutely convergent. Now, by the formula for cn, we have cn ¼

n X

n X ð1Þkþ1 ð1Þnkþ1 pffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kþ1 nkþ1 k¼0 n X 1 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi; ¼ ð1Þn k þ 1  nkþ1 k¼0

ak bnk ¼

k¼0

so that n X

1 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kþ1 nkþ1 k¼0 n X 1 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ 1:  n þ 1  nþ1 k¼0

jcn j ¼

1 P Since we therefore do not have cn ! 0 as n ! 1, it follows that the series cn n¼0 is divergent. We shall use Theorem 5 in Section 3.4 to give a definition of the exponential function x 7! ex in terms of series and to examine its properties, and in Section 8.4 on power series.

Yet again, this demonstrates that the hypotheses of our theorems really do matter!

3: Series

116

3.3.5

Overall strategy for testing for convergence

P We now give an overall strategy for testing a series an for convergence, as a flow chart. We suggest that, given a series which you wish to test for convergence or absolute convergence, you use the tests in the order indicated in the chart. Naturally, you may be able to short-circuit this strategy in certain cases. Is {an} null?

Σan is divergent (Non-null Test)

NO

For example, if you wish to examine a series whose terms alternate in sign, it is best to go straight to the Alternating Test.

YES Σ an is (absolutely) convergent

YES

Is Σ⏐an⏐ convergent? Try using: Basic series Ratio Test Limit Comparison Test Comparison Test Condensation Test

NO or CAN’T DECIDE Σ an is convergent (Alternating Test)

YES

Does Alternating Test apply? NO Try using First Principles

Remark We have labelled the final box ‘First Principles’, to indicate that, if the various tests do not produce a result, then it may be possible to work directly with the sequence of partial sums {sn}. Problem 6 Test the following series for convergence and absolute convergence: 1 1 1 P P P 1 5nþ2n 3 (a) n; (b) ; (c) n 2 2n3 1; 3 (d)

n¼1 1 P

(g)

n¼1 1 P n¼1

ð1Þnþ1 1

n3 2n ; n6

;

(e)

n¼1 1 P

(h)

n¼1 1 P n¼1

ð1Þnþ1 n2 n2 þ1 ;

(f)

n¼1 1 P

ð1Þnþ1 n n2 þ2 ;

(i)

n¼1 1 P

ð1Þnþ1 n n3 þ5 ; 1 3

n¼2 nðloge nÞ4

:

There is a variety of further tests for convergence or divergence which can be applied to series with non-negative terms. In this book we give only one further test, called the Integral Test. In particular it will enable us to prove that 1 X 1 is convergent; for all p > 1: np n¼1

You will meet the Integral Test in Section 7.4.

3.3

Series with positive and negative terms

3.3.6

117

Proofs

You may omit these proofs on a first reading.

We now supply the proofs omitted earlier in this section. Theorem 1 Absolute Convergence Test 1 1 P P If an is absolutely convergent, then an is convergent. n¼1

Proof

n¼1

We know that

1 P

jan j is convergent, and we want to prove that

n¼1

1 P

an

n¼1

is convergent. To do this, we define two new sequences   an ; if an  0, 0; if an  0, þ  and an ¼ an ¼ 0; if an < 0, an ; if an < 0.     Both the sequences aþ and a are non-negative, and n n  an ¼ aþ n  an ;

For example, if 1 X 1 1 an ¼ 1  2 þ 2 2 3 n¼1 1  2 þ   ; 4

for n ¼ 1; 2; . . .:

then 1 X

Also aþ n

 jan j;

for n ¼ 1; 2; . . .;

(10)

and a n  jan j;

for n ¼ 1; 2; . . .:

(11)

1 P

Since jan j is convergent, we deduce from (10) and (11) that n¼1 1 P a n are convergent, by the Comparison Test. Thus

1 P n¼1

aþ n ¼1þ0þ

n¼1

and 1 X

a n ¼ 0þ

n¼1

aþ n

and

þ

1 þ 0 þ  32

1 þ0 22

1 þ   : 42

n¼1

1 X

an ¼

n¼1

1  X

1 1 X  X  aþ aþ a n  an ¼ n  n

n¼1

n¼1

(12)

n¼1

is convergent, by the Combination Rules for series.

&

Sub-section 3.1.4.

Triangle Inequality (infinite form)   1 1 1 P  P P If an is absolutely convergent, then  an   jan j. n¼1

Proof

n¼1

n¼1

Let s n ¼ a1 þ a2 þ    þ an ;

n ¼ 1; 2; . . .;

and tn ¼ ja1 j þ ja2 j þ    þ jan j;

n ¼ 1; 2; . . .:

Then, by the Absolute Convergence Test 1 X an exists; lim sn ¼ n!1

n¼1

also lim tn ¼

n!1

1 X n¼1

jan j:

Alternatively, the infinite form of the Triangle Inequality can be deduced from statements (10), (11) and (12) in the proof of the Absolute Convergence Test.

3: Series

118 Now, by the Triangle Inequality

In Remark 3 following the proof of the Triangle Inequality in Sub-section 1.3.1, we commented that the Triangle Inequality for two numbers, a and b, can be extended in the obvious way to any finite sum of numbers.

jsn j ¼ ja1 þ a2 þ    þ an j  ja1 j þ ja2 j þ    þ jan j ¼ tn ; and so tn  sn  tn : Thus, by the Limit Inequality Rule for sequences

You met this in Subsection 2.3.3, Theorem 3.

 lim tn  lim sn  lim tn ; n!1

n!1

n!1

that is 

1 X

j an j 

n¼1

and so

1 X

an 

n¼1

1 X

jan j;

n¼1

   X X 1 1   an   ja j:   n¼1  n¼1 n

Theorem 2

&

Alternating Test

If an ¼ ð1Þnþ1 bn ;

n ¼ 1; 2; . . .;

where {bn} is a decreasing null sequence with positive terms, then 1 X an ¼ b1  b2 þ b3  b4 þ    is convergent: n¼1

Proof

We can write the even partial sums s2k of

1 P

an as follows

n¼1

s2k ¼ ðb1  b2 Þ þ ðb3  b4 Þ þ    þ ðb2k1  b2k Þ: Since {bn} is decreasing, all the brackets are non-negative, and so the even subsequence of partial sums, {s2k}, is increasing. We can also write the even partial sums s2k as s2k ¼ b1  ðb2  b3 Þ  ðb4  b5 Þ      ðb2k2  b2k1 Þ  b2k : Again, all the brackets are non-negative, and so s2k is bounded above, by b1. Hence {s2k} is convergent, by the Monotone Convergence Theorem. Now let lim s2k ¼ s:

k!1

Since s2k ¼ s2k1  b2k so that s2k1 ¼ s2k þ b2k , and {bn} is null, we have lim s2k1 ¼ lim ðs2k þ b2k Þ

k!1

k!1

¼ lim s2k þ lim b2k ¼ s; k!1

k!1

3.3

Series with positive and negative terms

119

by the Sum Rule for sequences. Thus the odd and even subsequences of {sn} both tend to the same limit s, and so {sn} tends to s. Hence 1 X

By Theorem 6, Subsection 2.4.4.

an ¼ b1  b2 þ b3  b4 þ    is convergent; with sum s: &

n¼1

Let the series

Theorem 4

1 P

an ¼ a1 þ a2 þ    be absolutely conver-

n¼1 1 P

gent. Then any rearrangement and

1 P

1 P

bn ¼

n¼1

bn of the series also converges absolutely,

n¼1

an :

n¼1

Proof First of all, we prove the result in the special case that ak  0 for all k. Choose any positive integer n. Then choose a positive integer N such that b1, b2, . . ., bn all occur among the terms a1, a2, . . ., aN of the original series. It follows that n X

bk 

k¼1



N X k¼1 1 X

ak ak :

(13)

k¼1 n P

Hence the partial sums

bk of the rearranged series

k¼1

1 P

bk are bounded

In particular, N  n. For the right-hand side is simply the left-hand side with possibly some more nonnegative terms inserted. For the partial sums of a series of non-negative terms are always less than or equal the sum of the whole series.

k¼1

above, so that the rearranged series must be convergent. It also follows from 1 1 P P bk  ak . (13) that k¼1

k¼1

By reversing the roles of the terms a1, a2, . . . and b1, b2, . . . in the above 1 1 P P argument, the same argument shows that ak  bk . It thus follows that k¼1 k¼1 1 1 P P ak and bk must in fact have the same sum. the two series k¼1

k¼1

To complete our proof, we now drop the condition that ak  0 for all k. þ   We then define the quantities aþ k , ak , bk and bk as follows aþ k bþ k

 ¼

0; 

¼

ak ; if ak  0, if ak < 0,

bk ; if bk  0, 0; if bk < 0,

a k b k

 ¼

if ak  0,

ak ; if ak < 0, 

¼

0;

0; if bk  0, ak ; if bk < 0.

1   1 P þ 1 a  and P a aþ k k k is convergent, since ak ¼ 2 ðjak j þ ak Þ and both 1 1 k¼1 P k¼1 k¼1 P þ þ converge. bk is a rearrangement of ak , and both series have non-

Now,

1 P

k¼1

k¼1

negative terms. It follows from the first part of the proof that both series converge, and have the same sum. 1 1 P P Similarly, a b k and k both converge, and have the same sum. k¼1

k¼1

By the Combination Rules for series.

3: Series

120    But bk  ¼ bþ k þ bk , and so the series

  b  converges, by the Sum Rule k 1 k¼1 P bk is absolutely convergent. for series. In other words, the rearranged series 1 P

k¼1

And, in fact 1   1 1 P b  ¼ P bþ þ P b . k k k k¼1

k¼1

k¼1

Finally 1 X

bk ¼

k¼1

¼ ¼

1 X k¼1 1 X k¼1 1 X

bþ k  aþ k 

1 X

b k

k¼1 1 X

a k

k¼1

ak :

&

k¼1

Theorem 5

Product Rule 1 1 P P Let the series an and bn be absolutely convergent, and let n¼0

n¼0

cn ¼ a0 bn þ a1 bn1 þ a2 bn2 þ    þ an b0 ¼

n X

ak bnk :

k¼0

1 P

cn is absolutely convergent, and ! ! 1 1 1 X X X cn ¼ an  bn :

Then the series

n¼0

n¼0

Proof

n¼0

n¼0

First, we introduce some notation, for n  0 n n n P P P sn ¼ ak ; tn ¼ bk ; un ¼ ck ; s0n

¼

s¼ s0 ¼

k¼0 n P

jak j;

k¼0 1 P

ak ;

k¼0 1 P k¼0

jak j;

tn0

k¼0 n P

¼

t¼ t0 ¼

jbk j;

k¼0 1 P

u0n

¼

k¼0 n P

jck j;

k¼0

bk ;

k¼0 1 P

jbk j:

k¼0

In particular, we know that sn ! s, tn ! t, sn0 ! s0 and tn0 ! t0 as n ! 1. Also, by the Product Rule for sequences, sn tn ! st and sn0 tn0 ! s0 t0 , as n ! 1. Hence, from the definition of limit, it follows that, for any positive number ", there is some integer N for which   1 1 jsn tn  stj < " and s0n tn0  s0 t0  < "; for all n  N: 3 3 It follows that, for n  N 0 0      s t  s0 t0  ¼  s0 t0  s0 t0  s0 t0  s0 t0  n n N N n n N N      s0 t 0  s0 t 0  þ s0 t 0  s0 t 0  n n

1 1 < "þ " 3 3 2 ¼ ": 3

N N

Here we use an integer N rather than a general number X and a weak inequality n  N rather than a strict inequality; this simplifies the notation in the rest of the argument a little.

3.3

Series with positive and negative terms

121

Now let n be such that n  2N, and consider the terms ai bj that occur in the expression un  sN tN. Every such term has i þ j  n, since un consists of all such terms; but none of these terms has both i  N and j  N. Hence, for every term ai bj in un  sN tN, a corresponding term jai bjj will occur in the expression sn0 tn0  sN0 tN0 – for this last expression consists of all terms jai bjj with i  n and j  n, but not with both i  N and j  N. (Note that we made the requirement that n  2N in order that every term in sN tN appears in un.) N

0 a0b0 a0b1 a0b2 a1b0 a1b1 a1b2

...

a2b0 ...

...

a2b1 a2b2

2N

For terms with both i  N and j  N are all in sN tN.

n

...

sNtN N

un – sNtN

2N

n

In the above diagram, we set out the terms ai bj in rows and columns, where the term ai bj occurs in the (i þ 1)th row and the ( j þ 1)th column. Then the small square contains all the terms ai bj with 0  i  N and 0  j  N; these add up to sN tN. Hence, for all n  2N, we have jun  stj ¼ jðun  sN tN Þ þ ðsN tN  stÞj  jun  sN tN j þ jsN tN  stj    s0 t0  s0 t0  þ jsN tN  stj n n

N N

2 1 < "þ " 3 3 ¼ "; it follows, from this inequality, that un ! st as n ! 1. Finally, we have u0n  s0n tn0  s0 t 0 :

By the Triangle Inequality.

By the discussions above concerning un  sN tN and sn0 tn0  sN0 tN0 . For the product sn0 tn0 contains more non-negative terms than does un0 ; and the sequences {sn0 } and {tn0 } are increasing.

3: Series

122 Hence, the increasing sequence {un0 } is bounded above, and so tends to a 1 P & limit as n ! 1. Thus the series cn is absolutely convergent. n¼0

3.4

The exponential function x j!ex

Earlier, we defined e ¼ 2.71828 . . . to be the limit   1 n e ¼ lim 1 þ ; n!1 n

In Sub-section 2.5.3.

and we defined ex to be the limit  x n ex ¼ lim 1 þ ; for any real x: n!1 n Here we show that the formula ex ¼

1 n X x n¼0

n!

;

for any real x;

is an equivalent definition of the quantity ex.

Remark The series

1 n P x n¼0

n!

converges for all values of x. For

This series is a basic series of type (d), in the case that x  0.

 nþ1   n   x  x  j xj     ðn þ 1Þ!  n!  ¼ n þ 1 !0

as n ! 1;

so that, by the Ratio Test, the series is absolutely convergent for all x, and so is convergent for all x.

3.4.1

The definition of ex as a power series, for x > 0

If we plot the partial sum functions of the infinite series of powers of x 1 n X x n¼0

n!

¼1þxþ

x2 x3 þ þ   ; 2! 3!

(1)

the resulting graph appears to be that of ex. (A series of multiples of increasing powers of x is called a power series.) In particular, when x ¼ 1, the sum of the series 1 X 1 1 1 ¼ 1 þ 1 þ þ þ  n! 2! 3! n¼0

is approximately 2.71828 . . ..

3.4

The exponential function x j! ex

123

y

y = ex

3

2

y = 1 + x + 2!x + 3!x 2

y = 1 + x + 2!x

y=1+x e y=1 –1

0

1

x

However the fact that the sequence of partial sums of the series (1) appears to converge to ex does not constitute a proof of this fact. Our first aim is to supply this proof in the particular case that x > 0. Theorem 1

If x > 0, then

1 n P x n¼0

n!

¼ ex .

We shall deal with the case x < 0 in Sub-section 3.4.3. Hence, for x > 0, the series 1 n P x n! gives an equivalent n¼0

definition of ex.

Proof

 n We defined ex to be ex ¼ lim 1 þ nx , and so we have to show that n!1

 x n ¼ lim 1 þ ; n n! n!1 n¼0

1 n X x

for x > 0:

The (n + 1)th partial sum of the above series is x2 x3 xn þ þ  þ : 2! 3! n! By the Binomial Theorem   x  nð n  1Þ  x  2  x n x n 1þ ¼1þn þ þ : þ n n 2! n n A typical term in this expansion is      nðn  1Þ . . . ðn  k þ 1Þ x k xk 1 2 k1 1 ¼ 1   1  k! n n n n k! xk  ; k! since each bracket is less than 1. Thus  x n x2 x3 xn 1þ  1 þ x þ þ þ  þ n 2! 3! n! ¼ snþ1 ; snþ1 ¼ 1 þ x þ

and so, by the Limit Inequality Rule for sequences  x n lim 1 þ  lim snþ1 ; n!1 n!1 n

You may omit this proof at a first reading.

3: Series

124 so that ex 

1 n X x n¼0

n!

:

(2)

On the other hand, for any integers m and n with m  n, we have   x  nð n  1Þ  x  2 x n nðn  1Þ . . . ðn  m þ 1Þ xm 1þn þ þ þ 1þ n n  2! n n    m!  x2 1 xm 1 2 m1 1 1 ¼1þxþ þ  þ 1 ... 1  : n n n n 2! m!

This follows since {sn} and {snþ1} both tend to the same limit, the sum of the infinite series.

Now keep m fixed and let n ! 1. By the Limit Inequality Rule for sequences, we obtain  x n x2 x3 xm lim 1 þ  1 þ x þ þ þ  þ ; n!1 n 2! 3! m! and so ex  smþ1 : Now let m ! 1; by the Limit Inequality Rule for sequences, we obtain ex  lim smþ1 ; m!1

so that ex 

1 n X x n¼0

(3)

n!

Combining inequalities (2) and (3), we obtain 1 n X x ; for x > 0: ex ¼ n! n¼0

&

Problem 1 Estimate e2 (to 3 decimal places) by calculating the seventh partial sum of the series (1) when x = 2.

3.4.2

Calculating e

The representation of e by the infinite series 1 X 1 1 1 ¼ 1 þ 1 þ þ þ  e¼ n! 2! 3! n¼0

(4)

provides a much more efficient way of calculating approximate values for e  n than the equation e ¼ lim 1 þ 1n . This is illustrated by the following table of n!1

approximate values: 

n

1 þ 1n n P 1

k¼0

k!

n

1

2

3

4

5

2

2.25

2.37

2.44

2.49

2

2.50

2.67

2.71

2.717

The calculation of e via the limit is a new calculation each time, whereas via the series involves only adding one extra term to the previous approximation.

3.4

The exponential function x j! ex

125

We can estimate how quickly the sequence of partial sums sn ¼

n1 X 1 1 1 1 ¼ 1 þ 1 þ þ þ  þ ; k! 2! 3! ð n  1Þ! k¼0

n ¼ 1; 2; . . .;

converges to e as follows. The difference between e and sn is given by 1 X 1 1 1 1 ¼ þ þ þ  e  sn ¼ k! n! ð n þ 1 Þ! ð n þ 2Þ! k¼n

1 1 1 1þ þ þ  ¼ n! ðn þ 1Þ ðn þ 1Þðn þ 2Þ

1 1 1 < 1 þ þ 2 þ  : n! n n The last expression in square brackets is a geometric series with first term n 1 and common ratio 1n, and so its sum is 11 1 ¼ n1 . Hence n 1 n 0 < e  sn <  n! n  1 1 1 (5)  ; for n ¼ 1; 2; . . .: ¼ ðn  1Þ! n  1

Here we replace various terms by larger terms (because their denominators are smaller); so the new sum is greater.

For example, this estimate shows that 0 < e  s6
0. Since, obviously, n! ¼ e when x ¼ 0, this completes the n¼0 1 P xn x proof that n! ¼ e , for all x 2 R.

Theorem 3

This was Theorem 1 in Sub-section 3.4.1.

¼ 1:

Since c0 ¼ 1  1 ¼ 1, the result then follows.

Proof

Here we have simply substituted mn for e in the previous expression.

n!

¼ ex .

For x < 0, the following chain of equalities holds , 1 n 1 X X x ðxÞn ¼1 ðby Lemma 1Þ n! n! n¼0 n¼0

Here we apply the Binomial Theorem to ðx þ ðxÞÞn .

In other words, the definition of ex as a power series is equivalent to its definition as a limit. Here the definition of ex on the right is that  x n . ex ¼ lim 1 þ n!1 n

3.5

Exercises

127

¼1

.

 xn lim 1 þ n!1 n

ðby Theorem 1; since x > 0Þ

¼ 1=ex ¼e

ðthis is the definition of ex Þ

x

ðby the Inverse Property of ex Þ:

&

This completes the proof.

In order to be crystal clear where we have reached, we now combine the results of Theorem 1 and Theorem 3 into the following: Theorem 4

For all x 2 R,

1 n P x n¼0

n!

 n ¼ lim 1 þ nx ¼ ex . n!1

Finally, we verify the Fundamental Property of the exponential function, which was left as unfinished business at the end of Sub-section 2.5.3. Theorem 5 Fundamental Property of the exponential function For all x, y 2 R, ex ey ¼ exþy .

Proof

By the Product Rule for series, ex  ey ¼

1 n P x n¼0

n!



1 n P y n¼0

n!

¼

1 P

cn ,

n¼0

where, for n  1, we have cn ¼ ¼

n X xk k¼0 n X

k!



ynk ðn  kÞ!

xk ynk k!ðn  kÞ! k¼0

¼

n 1X n! xk ynk n! k¼0 k!ðn  kÞ!

¼

ðx þ yÞn : n!

Since c0 ¼ 1  1 ¼ 1, the result then follows.

3.5

&

Exercises

Section 3.1 1. Prove that

1  P n¼1

34

n

is convergent, and find its sum.

2. Interpret 0:12 as an infinite series, and hence find the value of 0:12 as a fraction.

The Inverse Property of ex was Theorem 4 of Subsection 2.5.3.

3: Series

128 3. Prove that

  1 1 1 1  ¼ ; 4n2  1 2 2n  1 2n þ 1

for n ¼ 1; 2; . . .;

and deduce that 1 X

1 1 ¼ : 21 4n 2 n¼1

4. Determine whether the following series converge:   1   1  1   n  P P P 4 n 4 1 p1ffiffi  pffiffiffiffiffiffi (a) 1 þ 12 ; (b) . (c) 5 þ nðnþ2Þ ; n nþ1 n¼1

n¼1

n¼1

Section 3.2 1. Determine whether the following series converge: pffiffiffiffi 1 1 1 P P P cosð1=nÞ 2n n2 (b) (c) (a) 2n3 n; 2n2 þ3 ; 4n3 þnþ2; (d)

n¼1 1 P

n¼1

ðnþ1Þ5 2n ;

(e)

n¼1 1 P

n¼1

n2 3n n! ;

(f)

n¼1 1 P

n¼1

ðn!Þ2 ð2nÞ!.

1 n P 2 n!

1 n P 3 n! 2. (a) Use the Ratio Test to prove that nn converges, but that nn n¼1 n¼1 diverges. (b) For which positive values of c can you use the Ratio Test to prove 1 n P c n! that nn is convergent? n¼1

3. Prove that 1 1 1 pffiffiffi  pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffipffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ; for n ¼ 1; 2; . . .; n nþ1 n nþ1 nþ1þ n and use Exercise 4(c) on Section 3.1 to deduce that

1 P

1 3

n¼1 n2

is convergent.

4. Determine whether the following series converge: pffiffiffiffiffiffi 1 1 P P nþ1 1 (a) . ; (b) nðloge nÞðloge ðloge nÞÞ ð2n2 3nþ1Þðloge nþðloge nÞ2 Þ n¼3 n¼2

Section 3.3 1. Test the following series for convergence and absolute convergence: 1 1 1 1 P P P P ð1Þnþ1 ð1Þnþ1 n! sin n nþ2n pffiffi ; (a) (b) (c) (d) 3n þ5. n4 þ3 ; n2 ; 1þ n n¼1

n¼1

n¼1

n¼1

2. (a) Prove that 1 1 1 1 þ  > ; 3n  2 3n  1 3n 3n

for n ¼ 1; 2; . . .;

and deduce that 1 1 1 1 1 1 þ  þ þ  þ  2 3 4 5 6

is divergent:

3.5

Exercises

129

(b) Prove that   1 1     1 < 1; n 2n 2n þ 2 n2

for n ¼ 1; 2; . . .;

and deduce that 1 1 1 1 1 1 1 1 1 1 1 1  þ   þ   þ   þ    2 4 3 6 8 5 10 12 7 14 16 is convergent: (c) Prove that 1 1 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ; 2n  1 2n 2n

for n ¼ 1; 2; . . .;

and deduce that 1 1 1 1 1 1  þ pffiffiffi  þ pffiffiffi  þ    2 3 4 5 6 is divergent: 3. Prove that there exists a rearrangement of the series 1 4

1 P ð1Þnþ1 n¼1

n

¼ 1  12 þ 13 

þ 15  16 þ    for which one subsequence of its partial sums tends to 1 and another subsequence of its partial sums tends to 1. Alternatively, prove that no such rearrangement exists.

4

Continuity

The graphs of many functions that we come across look to be smooth curves, without any jumps.

y = ex

1

y = sin x

1 – 12 π

1 π 2

–1

On the other hand, the graphs of some functions that arise naturally, or that we construct for various purposes, do not appear to be smooth, but to contain jumps.

y = [x]

y = tan x

1

–2

–1

1

2

– 12 π

1π 2

–1

However there are some functions where the graphs appear to have an unreasonable number of jumps! How can we describe this complicated situation? The key underlying idea is that of continuity. Loosely speaking, a function is said to be continuous if we can draw its graph without taking our pencil off the page. This is the same as the graph ‘having no jumps’. The concept is an important one in Analysis since, in many situations, the crucial step in proving that a function has some property is to prove that the function is continuous. This is the first of two chapters that study continuous functions. In Section 4.1, we define the phrase: the function f is continuous at the point c, and we give a number of rules which state, for example, that various combinations and compositions of continuous functions are themselves continuous. 130

4.1

Continuous functions

131

Using these rules, together with a list of basic continuous functions, we can deduce that many functions are continuous at each point of their domains. For example, the functions x 7! x þ 1x and x 7! x sin 1x are continuous at each point of R  {0}, and the trigonometric and exponential functions are continuous at each point of their domains. Section 4.2 is devoted to the properties of continuous functions. The two fundamental properties of continuous functions are the Intermediate Value Theorem and the Extreme Value Theorem. We shall see a useful application of the Intermediate Value Theorem to finding the zeros of various functions. In Section 4.3, we discuss the Inverse Function Rule; this rule gives conditions under which a continuous function f has a continuous inverse function f 1. We then use the Inverse Function Rule to obtain the inverses of standard functions. Some of these inverse functions will be familiar to you already, but the Inverse Function Rule enables us to establish their properties by a rigorous argument. Finally, in Section 4.4, we shall provide a rigorous definition of the exponential function x 7! ax , where x 2 R and a > 0.

4.1

Continuous functions

4.1.1

What is continuity?

To accord with our intuitive idea of what we should mean by ‘the function f is continuous at the point c’, we wish to define this concept in such a way that the following two functions are continuous at the point c: 1.

y

y

2.

f (c) f (c)

c

x

c

x

On the other hand, we wish to formulate our definition so that the following two functions are not continuous at the point c: 3.

y

4.

f (c)

y

f (c)

c

x

c

x

In particular, to finding solutions of polynomial equations.

4: Continuity

132 So, our definition must say, in a precise way, that: if x tends to c, then f (x) tends to f (c). There are several ways of making this idea precise. In this chapter we adopt a definition which involves the convergence of sequences, as this will enable us to use the results about sequences that we met in Chapter 2.

In Chapter 5 we shall meet a different definition, and see that the two definitions are in fact equivalent.

Definitions A function f defined on an interval I that contains c as an interior point is continuous at c if: for each sequence {xn} in I such that xn ! c, then f (xn) ! f (c). If f is not continuous at the point c in I, then it is discontinuous at c. Thus, for example, in graph 3 above, if {xn} is a strictly decreasing sequence that tends to c as n ! 1, then {f (xn)} is a strictly decreasing sequence that tends to {f (c)}. On the other hand, if {xn} is a strictly increasing sequence that tends to c as n ! 1, then {f (xn)} is a strictly increasing sequence that tends to a limit as n ! 1 – but that limit is some number less than f (c). However, in graph 4, if {xn} is a strictly increasing sequence that tends to c as n ! 1, then {f (xn)} is a strictly increasing sequence that tends to f (c); but, if {xn} is a strictly decreasing sequence that tends to c as n ! 1, then the sequence {f (xn)} may not tend to any limit as n ! 1. On the other hand, in graphs 1 and 2, no matter how the sequence {xn} tends to c, then for sure we do have that {f (xn)} tends to f (c). For these examples, then, the definition agrees with what we believe the essence of continuity should be. Example 1

Prove that the function f (x) ¼ x3, x 2 R, is continuous at the point 12.

Solution Let {xn} be any sequence in R that converges to 12; that is, xn ! 12. Then, by the Combination Rules for sequences, it follows that f ðxn Þ ¼

x3n

 3 1 1 ! ¼ as n ! 1; 2 8

  while f 12 ¼ 18. In other words, {f(xn)} converges to f 12 as n ! 1. It follows that f is continuous at 12, as required.  Example 2

Prove that the function f ð xÞ ¼

y y = x3 1 8

x

1 2

y

&

2 1

1; x < 0; is discontinuous at 0. 2; x  0;

Solution Let {xn} be any sequence in R that converges to 0 from the right. By looking at the graph y ¼ f(x), it is clear that for such a sequence f(xn) ! 2 ¼ f(0). This will not help us to show that f is discontinuous at 0! However, if we let {xn} be any non-constant sequence in R that converges to 0 from the left, the situation will be very different. By looking at the graph y ¼ f(x), it is clear that for such a sequence f(xn) ! 1 6¼ f(0). We now make this precise.

1, x < 0, y = 2, x ≥ 0.

{

x

To prove that a function is discontinuous, all we need to do is to find just one sequence for which our definition of continuity at the relevant point does not hold.

4.1

Continuous functions

  Choose the sequence fxn g ¼  1n ; n ¼ 1; 2; . . .. For this sequence f ðxn Þ ¼ 1 ! 1 6¼ f ð0Þ as n ! 1: It follows that the function f cannot be continuous at 0.

133

&

Here we make a specific choice for the sequence {xn} to prove that f cannot be continuous at 0.

Problem 1 (a) Determine whether the function f ð xÞ ¼ x3  2x2 ; x 2 R, is continuous at 2. (b) Determine whether the function f ð xÞ ¼ ½ x; x 2 R, is continuous at 1. Problem 2 (a) Prove that the function f ð xÞ ¼ 1; x 2 R, is continuous on R. (b) Prove that the function f ð xÞ ¼ x; x 2 R, is continuous on R.

Here [] is the integer part function. By ‘continuous on R’, we mean ‘continuous at each point of R’.

However, as we know, not every function is defined on the whole of R – but we still want to be able to discuss the continuity of such functions. For example: pffiffiffi  f ð xÞ ¼ x, where the domain of f is the interval [0, 1);  

1

1

f ð xÞ ¼ x2 þ ð1  xÞ2 , where the domain of f is the half-closed interval (0, 1]; f ð xÞ ¼ 1x, where the domain of f is R  {0}.

How can we deal with these various different situations? The first two situations are dealt with by introducing the notion of one-sided continuity. Suppose that f is defined on some set S that contains an interval [c, c þ r) for some r > 0, but where f is not necessarily defined on any interval (c  r, c þ r) that contains c as an interior point. Then it seems reasonable to say that f is continuous on one side of c. In particular, that f is continuous on the right at the point c of S. Similarly, if f is defined on some set S that contains an interval (c  r, c] for some r > 0, but f is not necessarily defined on any interval (c  r, c þ r) that contains c as an interior point, then it seems reasonable to say that f is continuous on one side of c. In particular, that f is continuous on the left at the point c of S. The case of the function f ð xÞ ¼ 1x, where the domain of f is R – {0}, is different. The domain of f is the whole of R, with just one point omitted. That is, it is the union of two open intervals (1, 0) and (0, 1) – to both of which the definition of continuity applies in its originally stated form. So it makes sense to simply drop the original restriction that the domain of f must be an open interval in R. These considerations lead us to making the following less restrictive definition of continuity than our original definition:

y y=

1 x

x

Definitions A function f defined on a set S in R that contains a point c is continuous at c if: for each sequence {xn} in S such that xn ! c, then f(xn) ! f(c). If f is continuous on the whole of its domain, we often simply say that f is continuous (without explicit mention of the points of continuity). If f is not continuous at the point c in S, then it is discontinuous at c.

Here f is continuous on S.

4: Continuity

134 In addition, we have the following related definitions: Definitions A function f whose domain contains an interval [c, c þ r) for some r > 0 is continuous on the right at c if: for each sequence {xn} in [c, c þ r) such that xn ! c, then f(xn) ! f(c). A function f whose domain contains an interval (c  r, c] for some r > 0 is continuous on the left at c if: for each sequence {xn} in (c  r, c] such that xn ! c, then f(xn) ! f(c). A function f whose domain contains an interval I is continuous on I if it is continuous at each interior point of I, continuous on the right at the left endpoint of I (if this belongs to I), and continuous on the left at the right endpoint of I (if this belongs to I). The connection between the definitions of continuity and of one-sided continuity is rather obvious. Theorem 1 A function f whose domain contains an interval I that contains c as an interior point is continuous at c if and only if f is both continuous on the left at c and continuous on the right at c. Example 3 Determine whether the function f given by f ð xÞ ¼ continuous on fx : x 2 R; x  0g.

pffiffiffi x; x  0, is

Solution The domain of f is the interval I ¼ fx : x  0g. Thus, we have to show that for each c in I: pffiffiffi pffiffiffiffiffi for each sequence {xn} in I such that xn ! c, then xn ! c. if c ¼ 0, then we know already that, for any null sequence {xn} in I, pFirst, ffiffiffiffiffi xn is also a null sequence. let c > 0. We have to prove that if fxn  cg is a null sequence, then so Next, pffiffiffiffiffi pffiffiffi is xn  c . Now, from the identity

We omit a proof of this straight-forward result.

In fact, f is continuous on the right at 0, and continuous at all other points of its domain. y y = √x

x

0

pffiffiffiffiffi pffiffiffi xn  c xn  c ¼ pffiffiffiffiffi pffiffiffi ; xn þ c we see that, since c 6¼ 0 pffiffiffiffiffi pffiffiffi 0 xn  c ! pffiffiffi ¼ 0 2 c In other words, proof.

as n ! 1:

pffiffiffiffiffi pffiffiffi xn  c is indeed a null sequence. This completes the &

Problem 3

Prove that the nth root function  1 x  0; if n is even; n f ð xÞ ¼ x ; where n 2 N and x 2 R; if n is odd;

is continuous.

Recall that c 6¼ 0, so that p1ffiffic is indeed defined.

You may omit this Problem if you are short of time. 1

We defined an in Sub-section 1.5.3.

4.1

Continuous functions Hints: Use the result of Exercise 3(b) on Section 2.3, in Section 2.6, and the identity aq  bq ¼ ða  bÞðaq1 þ aq2 b þ aq3 b2 þ    þ bq1 Þ with suitable choices for a, b and q. In your solution, be careful not to use the letter n for two different purposes.

Example 4 continuous.

135

Example 3 above was the special case of Problem 3, where n ¼ 2.

Determine whether the function f ð xÞ ¼ 1x ; x 2 R  f0g, is

Solution The domain of f is the set R  {0}, the union of the two open intervals (1, 0) and (0, 1). Let c be any point of R  {0}, and let {xn} be any sequence in R  {0} that converges to c. Then, by the Quotient Rule for sequences, it follows that ff ðxn Þg ¼ x1n ! 1c as n ! 1; in other words, that f ðxn Þ ! f ðcÞ ¼ 1c as n ! 1. So f is continuous at c. Since c is an arbitrary point of R  {0}, it follows that f is continuous & on R  {0}.

Sub-section 2.3.2.

Our work so far on continuity illustrates the following general strategy: Strategy for continuity 

To prove that a function f : S ! R is continuous at a point c of S, prove that: for each sequence {xn} in S such that xn ! c, then f(xn) ! f(c).



To prove that a function f : S ! R is discontinuous at a point c of S: find one sequence {xn} in S such that xn ! c but f(xn) 6! f(c). Problem 4

Recall that just one such sequence suffices.

Prove that the following functions are continuous on R:

(a) f(x) ¼ c, x 2 R; (b) f(x) ¼ xn, x 2 R, n 2 N; (c) f(x) ¼ jxj, x 2 R. Problem 5 Determine the points of continuity and discontinuity of the signum function 8 < 1; x < 0, x ¼ 0, f ð xÞ ¼ 0; : 1; x > 0.

y 1

y = f(x) x –1

Remarks The definition of function that we are using involves a mapping that we call f (say) from a set in R, the domain A (say), to another set in R, the codomain B (say). 1. Let f be continuous at a point c in A, and assume that, for some set A0  A, c 2 A0 . Suppose that another function g has domain A0 on which g(x) ¼ f(x). Technically g is a different function from f, for sure. However if f is continuous at c, then it is a simple matter of some definition checking to

You can think of the mapping f as being a formula x 7! y that maps a point x of A onto to a point y of B, and we write y ¼ f (x) to indicate this.

4: Continuity

136 verify that g too is continuous at c. So, restriction of a function to a smaller domain does not affect its continuity at a point. 2. Let f and g be functions defined on sets in R that contain an open interval I, and c 2 I. Then, if f(x) ¼ g(x) for all x 2 I, f is continuous at c if g is continuous at c, and f is discontinuous at c if g is discontinuous at c. Again, we may simply ignore any difference between the domains of the functions when we are studying continuity at a point. 3. The underlying point here is that continuity at a point is a local property. It is only the behaviour of the function near that point that determines whether it is continuous at the point.

4.1.2

Rules for continuous functions

We have seen how to recognise whether a given function is continuous at a point. However, it would be tedious to have to go back to first principles to determine on each occasion whether a complicated function is continuous. As usual, we avoid such problems by having a set of rules that enable us to construct continuous functions. Combination Rules for continuous functions If f and g are functions that are continuous at a point c, then so are: Sum Rule f þ g; Multiple Rule

lf,

Product Rule Quotient Rule

f g,

for l 2 R;

fg; provided that f (c) 6¼ 0.

For example, any polynomial p(x) ¼ a0 þ a1x þ    þ anxn, x 2 R, is continuous at all points of R since we can build up the expression for p by successive applications of the Combination Rules for continuous functions. Similarly, any rational function rðxÞ ¼ pðxÞ qðxÞ, where p and q are polynomials and the domain of r is R minus the points where q vanishes, is continuous at all points of its domain, since we can build up the expression for r by successive applications of the Combination Rules for continuous functions. The Combination Rules above are all natural analogues of the corresponding results for sequences. However, we can combine functions in more ways than we can combine sequences – for example, we can compose functions f and g to obtain the function g  f . This approach too will often enable us to obtain new continuous functions. Composition Rule Let f be continuous on a set S1 that contains a point c, and g be continuous on a set S2 that contains the point f(c). Then g  f is continuous at c. For example, we know that thepfunction f(x) ¼ x2 þ 1, x 2 R, is continuous on R ffiffiffi and that the function gð xÞ ¼ x; x  0, is continuous on I ¼ {xp : xffiffiffiffiffiffiffiffiffiffiffiffi  0}.ffi It follows from the Composition Rule that the function g  f : x ! 7 x2 þ 1 is

For instance, the function pðxÞ ¼ x17 34x5 þ5x  8 is continuous on R. For instance, the function r ð xÞ ¼

1  2x ; x2  4

x 2 R  f 2g; is continuous on its domain.

Functions obtained in this way are called composite functions.

4.1

Continuous functions

137

continuous at all points c of R (the domain of f) whose image f(c) lies in I. But all such points f(c) lie in I; so it follows that in fact the composite g  f is continuous on R. 3

Problem 6 Prove that the function f given by f ð xÞ ¼ x2 ; x  0; is continuous. Finally, just as we had a Squeeze Rule for convergent sequences, we have a corresponding Squeeze Rule for continuous functions. Theorem 1

Squeeze Rule

Let the functions f, g and h be defined on an open interval I, and c 2 I. If: 1. g(x) f(x) h(x), for all x 2 I,

We state this rule only in the case of an interior point of an open interval. However more general versions also exist! y

y = h(x) y = f (x) y = g (x)

2. g(c) ¼ f(c) ¼ h(c), 3. g and h are continuous at c, I

then f is also continuous at c.



Example 5 Prove that the function f given by f ð xÞ ¼ is continuous at 0.

2

x sin 0;

1 x

c

;

x¼ 6 0, x ¼ 0,

Solution The diagram in the margin suggests that we should find functions g and h that squeeze f near 0. So, we define g(x) ¼ x2, x 2 R, and h(x) ¼ x2, x 2 R. With these two chosen, we check the conditions of the Squeeze Rule. The first condition is of course vital! Now we know that   1 1 sin

1; x It follows that x2 x2 sin

We shall study the trigonometric functions in detail in Sub-section 4.1.3. The only property that you need here is that, for any real number x, jsin xj 1. y y = x2 sin 1x

for any x 6¼ 0:

  1

x2 ; x

x

0

x

for any x 6¼ 0;

so that



 gð xÞ ¼ x2 f ð xÞ x2 ¼ hð xÞ;

for any x 2 R:

So condition 1 of the Squeeze Rule is satisfied. Next, the functions f, g and h all take the value 0 at the point 0. Thus condition 2 of the Squeeze Rule is satisfied. Finally, the functions g and h are polynomials, and so in particular they are continuous at 0. So condition 3 of the Squeeze Rule is satisfied. It follows then from the Squeeze Rule that f is continuous at 0, as required. & To test your understanding of these techniques, try the following problems. Problem 7 Prove that the following function is continuous on R, stating each rule or fact about continuity that you are using f ð xÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 

5x ; x 2 R: 1 þ x2

These inequalities are obviously true for x ¼ 0, as well as for x 6¼ 0.

Recall that all polynomials are continuous on R.

4: Continuity

138 Problem 8 Using the elementary properties of the trigonometric functions, determine whether the following functions are continuous at 0:   x sinð1xÞ; x 6¼ 0, sinð1xÞ; x 6¼ 0, (a) f ðxÞ ¼ (b) f ðxÞ ¼ 0; x = 0; 0; x = 0. Proofs We now give the proofs of the Combination, Composition and Squeeze Rules. First, recall the Combination Rules. Combination Rules for continuous functions If f and g are functions that are continuous at a point c, then so are: Sum Rule

f þ g;

Multiple Rule Product Rule

lf, for l 2 R; fg;

Quotient Rule

f g,

provided that f(c) 6¼ 0.

The proofs of all these are very similar, and depend on the corresponding results for sequences. We prove only the Sum Rule. Proof of the Sum Rule We want to prove that f þ g is continuous at c. Suppose that f and g have domains S1 and S2, respectively. Then the domain of f þ g is S1 \ S2, and this set contains the point c. Thus, we have to show that: for each sequence {xn} in S1 \ S2 such that xn ! c, then f ðxn Þ þ gðxn Þ ! f ðcÞ þ gðcÞ: We know that the sequence {xn} lies in S1 and in S2, and that both functions f and g are continuous at c. Hence f ðxn Þ ! f ðc Þ

and gðxn Þ ! gðcÞ;

and it follows from the Sum Rule for Sequences that f(xn) þ g(xn) ! & f(c) þ g(c), as required. Next, recall the Composition Rule. Composition Rule Let f be continuous on a set S1 that contains a point c, and g be continuous on a set S2 that contains the point f(c). Then g  f is continuous at c. Proof We want to prove that g  f is continuous at c. Now, we know that g  f is certainly defined on the set S ¼ {x : x 2 S1 and f(x) 2 S2}, and this set contains the point c. Thus, we have to show that: for each sequence {xn} in S such that xn ! c, then gðf ðxn ÞÞ ! gðf ðcÞÞ. We know that the sequence {xn} lies in S1, and that f is continuous at c. Hence, we have that f(xn) ! f(c).

You may omit the rest of this Sub-section at a first reading.

4.1

Continuous functions

139

We also know that {f(xn)} lies in S2, and that g is continuous at f(c). It & follows that gðf ðxn ÞÞ ! gðf ðcÞÞ, as required. Finally, recall the Squeeze Rule. Theorem 1

Squeeze Rule

Let the functions f, g and h be defined on an open interval I, and c 2 I. If: 1. g(x) f(x) h(x), for all x 2 I, 2. g(c) ¼ f(c) ¼ h(c), 3. g and h are continuous at c, then f is also continuous at c. Proof We have to show that f is continuous at c. Thus, we have to prove that: for each sequence {xn} in the domain of f such that xn ! c, then f (xn) ! f (c). Now, since xn ! c there is some number X such that xn 2 I, for all n > X. Hence, by condition 1 we have that gðxn Þ f ðxn Þ hðxn Þ;

for all n > X:

(1)

This follows from the definition of a sequence converging to c.

So, if we now let n ! 1 and use conditions 2 and 3, we get that lim gðxn Þ ¼ gðcÞ ¼ f ðcÞ and

n!1

lim hðxn Þ ¼ hðcÞ ¼ f ðcÞ:

n!1

(2)

It follows, from (1), (2) and the Squeeze Rule for sequences, that

Sub-section 2.3.3.

lim f ðxn Þ ¼ f ðcÞ;

n!1

&

as required.

4.1.3

Trigonometric functions and the exponential function

Trigonometric functions For the moment we are assuming that you have a knowledge of the trigonometric functions arising from your study of trigonometry. We will prove that the trigonometric functions are continuous on the whole of their domains. But, first, we need a basic inequality for the sine function. Lemma 1

0 sin x x,

for 0 x p2 :

Proof If x ¼ 0, then sin 0 ¼ 0; so there is an equality. Suppose next that 0 < x p2, and consider the following diagram, which represents a quarter circle, centred at the origin, with radius 1.

4: Continuity

140

Since the circle has radius 1, the arc AB has length x and the perpendicular AC has length sin x. Hence 0 < sin x x;

p for 0 < x : 2

&

For the shortest distance from the point A to the line BC is the perpendicular from A to BC.

We can now extend the inequality in Lemma 1 to obtain a more general result. Theorem 2

y

The Sine Inequality jsin xj j xj; for x 2 R:

y = ⎜x ⎜ 1

Proof We saw in Lemma 1 that the inequality holds for 0 x p2 : For x > p2, we have

–π ⎜sin

y = ⎜sin x ⎜

0 x ⎜≤ ⎜x ⎜, for x ∈

π x

1 jsin xj 1 < p < x ¼ j xj; 2 and so the desired inequality is also true in this case. Finally, the inequality also holds for x < 0, since jsinðxÞj ¼ jsin xj

and jxj ¼ j xj:

&

This is the key tool that we need to prove the continuity of the trigonometric functions. Theorem 3 continuous.

The trigonometric functions (sine, cosine and tangent) are

Recall that this means they are continuous at each point of their domains.

Proof To prove that the sine function is continuous at each point c 2 R, we need to show that: for each sequence fxn g in R such that xn ! c; then sin xn ! sin c:

(3)

We use the formula sin xn  sin c ¼ 2 cos it follows that

    1 1 ðxn þ cÞ sin ðxn  cÞ ; 2 2

This is a standard trigonometric formula.

4.1

Continuous functions

141

    1 1 jsin xn  sin cj ¼ 2 cos ðxn þ cÞ  sin ðxn  cÞ 2 2  1

2 sin ðxn  cÞ 2 1

2 ðxn  cÞ ¼ jxn  cj: 2

By taking moduli. For cos 12 ðxn þ cÞ 1: Here we use the Sine Inequality, Theorem 2.

Hence, if {xn  c} is null, then {sin xn  sin c} is also null, by the Squeeze Rule for null sequences. In other words, the result (3) holds. The continuity of the cosine and tangent functions now follows from the formulas   1 sin x and tan x ¼ ; cos x ¼ sin x þ p 2 cos x

&

using the Combination Rules and the Composition Rule.

Problem 9 Prove that the following function is continuous on R, stating each rule or fact about continuity that you are using   f ð xÞ ¼ x2 þ 1 þ 3 sin x2 þ 1 ; for x 2 R: Problem 10 on R.

Prove that the function f ð xÞ ¼ sin

p

2 cos x



is continuous

The exponential function x 7! ex We now prove that the exponential function is continuous on R. But first we start with some inequalities that we will need to do this. Theorem 4

The Exponential Inequalities

x

for x > 0;

(a) e > 1 þ x, x

1 1x ;

(b) e

1 (c) 1 þ x ex 1x ; Proof

and

ex  1 þ x,

for x  0;

for 0 x < 1; for j xj < 1:

We prove inequalities (a) and (b) using the exponential series

x2 x3 ex ¼ 1 þ x þ þ þ   ; for x  0: 2! 3! 2 3 (a) For x > 0, we have x2! > 0, x3! > 0, and so on. Hence ex > 1 þ x; for x > 0: It follows from this that ex  1 þ x, for x  0, since e0 ¼ 1 2 3 (b) For x  0, we have x2! x2 , x3! x3 , and so on. Hence ex 1 þ x þ x2 þ x3 þ   : 1 The series on the right is a geometric series; it converges to the sum 1x , for 0 x 1. Hence 1 ; for 0 x < 1: ex

1x

You met this series in Sub-section 3.4.1.

4: Continuity

142 (c) We have just seen that these inequalities hold for 0 x < 1. For 1 < x < 0, we have that 0 < x < 1, so that, by parts (a) and (b) 1 ; for 1 < x < 0: 1 þ ðxÞ ex

1  ðxÞ By taking reciprocals and reversing the inequalities, we may reformulate this in the form 1 1 þ x ex

; for 1 < x < 0: 1x & This completes the proof of the desired result.

Remark Notice that, when x 6¼ 0, the results of Theorem 4 hold with strict inequalities. We are now able to prove the continuity of the exponential function. Theorem 5

The exponential function x 7! ex, x 2 R, is continuous.

Proof To prove that the exponential function is continuous at each point c 2 R, we need to show that: for each sequence fxn g in R such that xn ! c; then exn ! ec :

(4)

We use the formula exn  ec ¼ ec ðexn c  1Þ. If we apply Theorem 4, part (c), with xn  c in place of x, we obtain 1 1 þ ðxn  cÞ exn c

; for jxn  cj < 1: 1  ðx n  c Þ Thus, if {xn  c} is null, then jxn  cj < 1 eventually, and so exn c ! 1 as n ! 1, by the Squeeze Rule for sequences. Hence exn ! ec , so that the desired & result (4) holds.

Remark Later we shall give a rigorous definition of the general exponential function x 7! ax, for x 2 R and a > 0, and we shall show that all exponential functions are continuous on R.

Section 4.4.

Problem 11 Prove that the following function is continuous on R, stating each rule or fact about continuity that you are using   2 f ð xÞ ¼ cos x5  5x2 þ 7ex : We end this section by listing the various types of functions that we have found to be continuous on their domains. For example

Basic continuous functions

The following functions are continuous:



polynomials and rational functions;



modulus function; nth root function;

  

trigonometric functions (sine, cosine and tangent); the exponential function.

f ð xÞ ¼ 3x7  4x2 þ 5; 3x f ð xÞ ¼ 4 ; x 6 f ð xÞ ¼ jxj; 1

f ð xÞ ¼ x2 ; x  0; 1

f ð xÞ ¼ x3 ; x 2 R; f ð xÞ ¼ sin x; cos x; tan x; f ð xÞ ¼ ex :

4.2

Properties of continuous functions

4.2 4.2.1

143

Properties of continuous functions The Intermediate Value Theorem

In Section 4.3 we shall prove that the function f(x) ¼ x5 þ x  1, x 2 R, is a one-function on R. From the graph of f it certainly looks as though f maps R onto R; how can we prove this? For example, is there a value of x for which f(x) ¼ 0? In other words, is there a root of the equation x5 þ x  1 ¼ 0? The shape of the graph y ¼ x5 þ x  1 certainly suggests that such a number x exists; since f(0) ¼ 1 and f(1) ¼ 1, we would expect there to be some number x in the interval (0, 1) such that f(x) ¼ 0. However we do not have a formula for solving the equation to find x. Now, we introduced the concept of continuity on the grounds that it would enable us to pin down precisely the idea that the graph of a ‘well-behaved’ function does not have ‘gaps’ or ‘jumps’, and this is the key to proving that such a number x exists. Theorem 1 Intermediate Value Theorem Let f be a continuous function on [a, b], and let k be any number lying (strictly) between f(a) and f(b). Then there exists a number c in (a, b) such that f(c) ¼ k. This result is illustrated below in the two possible cases:

As the graph above on the right shows, there may be more than one possible value of c such that f(c) ¼ k. All that we claim is that there is at least one such point c. The requirement in Theorem 1 that f be continuous at each point of [a, b] is essential. For example, the function 81  1 x < 0, <x; x ¼ 0, f ðxÞ ¼ 0; :1 ; 0 < x 1, x is continuous on [1, 1] except at 0 – where it is discontinuous. For this function, f(1) ¼ 1 and f(1) ¼ 1, but there is no number c in (1, 1) such that f ðcÞ ¼ 12 (for example). The following example shows a typical application of the Intermediate Value Theorem.

This is one of the main existence theorems in Analysis. Note that f(a) 6¼ f(b) and a < c < b.

4: Continuity

144 Example 1

Prove that there is a number c in (0, 1) such that c5 þ c  1 ¼ 0.

Solution Consider the function f(x) ¼ x5 þ x  1 on the interval [0, 1]. Then f is continuous on [0, 1] since it is a basic continuous function, and f(0) ¼ 1 and f(1) ¼ 1. Since f(0) < 0 < f(1), it then follows from the Intermediate Value Theorem that there is a number c in (0, 1) such that f(c) ¼ 0; that is, such that & c5 þ c  1 ¼ 0. If f is a function and c is a real number such that f(c) ¼ 0, then c is called a zero of the function f. We often show that an equation has a solution by proving that a related continuous function has a zero (by using the Intermediate Value Theorem with k ¼ 0). Problem 1 cos c ¼ c.

For f is a polynomial. The function f is strictly increasing on [0, 1], and so the number c must be unique in this case.

We do not consider the possibility of complex zeros in this book.

Prove that there is a real number c in (0, 1) such that

Problem 2 Let the function f: [0, 1] ! [0, 1] be continuous. Prove that there is a real number c in [0, 1] such that f(c) ¼ c. Proof of Theorem 1 We use the method of repeated bisection. We shall assume that f(a) < f(b). If, in fact, f(a) > f(b), the proof is very similar. First, denote the closed interval [a, b] as [A1, B1]. Then, denote by p the midpoint of the interval [A1, B1]. Notice that, if f(p) ¼ k, then the proof is complete, since we can take c ¼ p. Otherwise, we define one of the two intervals [A1, p] and [p, B1] to be [A2, B2] in the following way  ½A1 ; p; if f ( p) > 0, ½ A2 ; B2  ¼ ½p; B1 ; if f ( p) < 0. In either case, we obtain:

For example   f ð xÞ ¼ 13 þ 12 x sin p2 x : You should at least skim this proof on a first reading, as the method is an important one. p ¼ 12 ðA1 þ B1 Þ:

1. [A2, B2] [A1, B1]; 2. B2  A2 ¼ 12 ðB1  A1 Þ; 3. f(A2) < k < f(B2). We now repeat this process indefinitely often, bisecting [A2, B2] to obtain [A3, B3], and so on. If, at any stage, we encounter a bisection point p such that f(p) ¼ k, then the proof is complete. Otherwise, we obtain a sequence of closed intervals {[An, Bn]} with the following properties: 1. [Anþ1, Bnþ1] [An, Bn], for each n 2 N; 1n1 2. Bn  An ¼ 2 ðB1  A1 Þ, for each n 2 N; 3. f(An) < k < f(Bn),

for each n 2 N.

Property 1 implies that the sequence {An} is increasing and bounded above by B1 ¼ b. Hence by the Monotone Convergence Theorem, {An} is convergent; denote by A its limit. By the Limit Inequality Rule for sequences, we must have that A b. Similarly, Property 1 implies that the sequence {Bn} is decreasing and bounded below by A1 ¼ a. Hence by the Monotone Convergence Theorem,

Theorem 3, Sub-section 2.3.3.

4.2

Properties of continuous functions

145

{Bn} is convergent; denote by B its limit. By the Limit Inequality Rule for sequences, we must have that B  a. We may then deduce, by letting n ! 1 in Property 2 and using the Combination Rules for sequences, that B  A ¼ lim Bn  lim An ¼ lim ðBn  An Þ n!1 n!1 n!1  n1 ¼ lim 12 ðB1  A1 Þ n!1  n1 ¼ ðB1  A1 Þ lim 12 ¼ 0: n!1

In other words, the sequences {An} and {Bn} both converge to a common limit. Denote this limit by c; then we must have a c b. Now we use the fact that f is continuous at c. It follows that lim f ðAn Þ ¼ f ðcÞ

n!1

and

That is, c 2 [a, b].

lim f ðBn Þ ¼ f ðcÞ:

n!1

Then, by Property 3, f(An) < k, for n ¼ 1, 2, . . .; so that, by letting n ! 1, we obtain f(c) k by the Limit Inequality Rule for sequences. Similarly, we can deduce from the fact that k < f(Bn) that f(c)  k. It follows that f(c) ¼ k. Finally, notice that, since f(a) < k < f(b), we cannot have either c ¼ a or & c ¼ b; so that, in fact, a < c < b as required.

Recall that taking limits ‘flattens inequalities’. That is, c 2 (a, b).

The method of repeated bisection is very powerful, and of wide application in Mathematics. Now, we saw above that the continuous function f(x) ¼ x5 þ x  1, x 2 [0, 1], 1 1 has a zero in (0, 1). Now, since f 2 ¼ 32 þ 12  1 ¼  15 ¼ 1 > 0, 32 < 0 andf(1)  1 we can make the stronger statement that f must have a zero in 2 ; 1 , by the Intermediate Value Theorem. Problem 3 Use the method of repeated bisection to find an interval of length 18 that contains a zero of the function f(x) ¼ x5 þ x  1, x 2 [0, 1].

Antipodal points Two points on the surface of the Earth are called antipodal points if the line between them passes through the centre of the Earth. (In this sense, antipodal points are ‘opposite each other’.) The following result is a rather interesting application of the Intermediate Value Theorem! It makes the (physically reasonable) assumption that temperature is a continuous function of position on the Earth’s surface. Theorem 2

Antipodal Points Theorem

There is always a pair of antipodal points on the Equator of the Earth at which the temperature is the same. To prove this, we must set up the situation as a mathematical problem. Let f() denote the temperature at a point on the Equator at an angle  radians East of Greenwich, for 0  < 2p, and extend f to be defined on [0, 2p] by

In fact the same result holds for any ‘Great Circle’ on the Earth’s surface – that is, the intersection of a plane through the Earth’s centre with the Earth’s surface. This process is often called Mathematical modelling.

4: Continuity

146 requiring that f(2p) ¼ f(0). Then Theorem 2 can be rephrased in the following equivalent way: Theorem 20 Antipodal Points Theorem Let f : [0, 2p] ! R be continuous, with f(0) ¼ f(2p). Then there exists a number c in [0, p] such that f(c) ¼ f(c þ p).

Proof of Theorem 20 Notice, first, that if f(0) ¼ f(p) then we can take c ¼ 0. We now prove the result under the assumption that f(0) < f(p). (The proof in the case that f(0) > f(p) is very similar.) Next, define the function g as follows gðÞ ¼ f ðÞ  f ð þ pÞ;

for  2 ½0; p:

Then, since f is continuous on [0, 2p], it follows that g is continuous on [0, p], by the Combination Rules. But gð0Þ ¼ f ð0Þ  f ðpÞ < 0

For, if f(c) ¼ f(c þ p), then c and c þ p are antipodal points with the same temperature.

Quite often in mathematics we obtain our results by applying a standard result to some cunningly chosen auxiliary function!

and gðpÞ ¼ f ðpÞ  f ð2pÞ ¼ f ðpÞ  f ð0Þ > 0: It then follows from the Intermediate Value Theorem, with k ¼ 0, that there exists some number c in (0, p) such that g(c) ¼ 0; in other words, such that & f(c)  f(c þ p) ¼ 0. This completes the proof.

4.2.2

Zeros of polynomials

You will have already met a standard method for solving a polynomial equation of degree 2, and possibly equations of degrees 3 and 4 too. However there exists no method of solving a general polynomial equation of degree 5 or higher by means of formulas. How many zeros can a polynomial equation have? In fact, a polynomial equation of degree n can have at most n roots in R. Theorem 3 Fundamental Theorem of Algebra Let p(x) ¼ anxn þ an1xn1 þ    þ a1x þ a0, x 2 R, where an 6¼ 0. Then the equation p(x) ¼ 0 has at most n roots in R.

Remark In fact, in its most general form the Fundamental Theorem of Algebra states that, if p(x) ¼ anxn þ an1xn1 þ    þ a1x þ a0, where the coefficients ak may

These are called quadratic, cubic and quartic equations, respectively.

This is a quite difficult result to prove! We do not prove this result, which requires methods beyond the scope of this book.

4.2

Properties of continuous functions

147

be real or complex numbers and an 6¼ 0, then the equation p(x) ¼ 0 has exactly n roots in C, the set of all complex numbers. Now, it is sometimes straight-forward to locate zeros of a polynomial if we have some idea of where to look for them in the first place. Problem 4 Let pð xÞ ¼ x6  4x4 þ x þ 1, x 2 R. Prove that p has a zero in each of the intervals (1, 0), (0, 1) and (1, 2). However to follow this approach for finding zeros (if any) of a given polynomial, we need first to have an inkling where to look for them. We usually start by applying the following result, which gives an interval in which the zeros must lie: Theorem 4

Zeros Localisation Theorem

Let pð xÞ ¼ x þ an1 x þ    þ a1 x þ a0 ; x 2 R, be a polynomial. Then all the zeros of p (if there are any) lie in the open interval ( M, M), where n

n1

M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg:

Notice that the coefficient of xn, the leading coefficient of p, has been set as 1. The proof appears at the end of the sub-section.

Example 2 Prove that the polynomial pð xÞ ¼ x4  2x2  x þ 1, x 2 R, has at least two zeros in R. Solution We will apply the Zeros Localisation Theorem to p, since its leading coefficient is 1. Since M ¼ 1 þ maxfj2j; j1j; j1jg ¼ 3; it follows that all the zeros of p lie in (3, 3). We now compile a table of values of p(x), for x ¼ 3, 2, 1, 0, 1, 2, 3: x p(x)

3 67

2

1

0

1

2

3

11

1

1

1

7

61

We find that p(0) and p(1) have opposite signs, as do p(1) and p(2), so p must have a zero in each of the intervals (0, 1) and (1, 2), by the Intermediate Value Theorem. & Thus we have proved that p has at least two zeros in R.

Often, it is not necessary to compute the values of p(x) for all integers x in [M, M].

In fact this polynomial has exactly two zeros in R.

Problem 5 Prove that the polynomial pð xÞ ¼ x5 þ 3x4  x  1, x 2 R, has at least three zeros in R. Although we cannot at this stage prove the Fundamental Theorem of Algebra, nevertheless we can use the Zeros Localisation Theorem and the Intermediate Value Theorem to prove the following: Theorem 5

Every real polynomial of odd degree has at least one zero in R.

Thus, for example, for sufficiently large x the polynomial pð xÞ ¼ xn þ an1 xn1 þ    þ a1 x þ a0 , x 2 R is essentially dominated by its leading term xn, and so is positive for large positive values of x and is negative for large negative values of x.

For example, pð xÞ ¼ x5 þ 3x4  x  1: Since n is odd.

4: Continuity

148 No corresponding result holds for polynomials of even degree. Thus, for instance, the polynomial pð xÞ ¼ x2 þ 1 has no zeros in R. Proofs of Theorems 4 and 5 We supply these proofs which were omitted earlier, so as not to disturb the flow of the text. Theorem 4 Zeros Localisation Theorem Let pð xÞ ¼ xn þ an1 xn1 þ    þ a1 x þ a0 ; x 2 R, be a polynomial. Then all the zeros of p (if there are any) lie in the open interval (M, M), where M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg: Proof In order to concentrate on the dominating term xn, we define the function r as follows pð x Þ an1 a1 a0 r ð xÞ ¼ n  1 ¼ þ    þ n1 þ n ; x 2 R  f0g: x x x x Then, by using the Triangle Inequality, we obtain that, for jxj > 1 a a1 a0 n1 þ    þ n1 þ n jr ð xÞj ¼ x x a x a1 a0 n1

þ    þ n1 þ n x x x ! 1 1 1 þ    þ n1 þ n

maxfjan1 j; . . .; ja1 j; ja0 jg  j xj j xj j xj ! 1 1 1 <M þ    þ n1 þ n þ    j xj j xj j xj ¼M

1 j xj

1  j1xj

¼

M : j xj  1

You may omit these at a first reading.

Recall that the coefficient of xn, the leading coefficient of p, is 1.

Here the modulus of each coefficient is at most the maximum value over all coefficients. jxj > 1, so that j1xj < 1:

By summing the geometric series.

It follows that, if j xj  M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg, then jr ð xÞj < 1: Now, from the definition of r(x) we see that pð xÞ ¼ xn ð1 þ r ð xÞÞ;

for j xj  M:

But since jr ð xÞj < 1, we certainly have that 1 þ r ð xÞ > 0. It follows from the above expression for p(x) in terms of r(x) that p(x) must have the same sign as xn, for j xj  M: & It follows that any zero of p must lie in (M, M). Theorem 5

jr ð xÞj51 , 15r ð xÞ51:

Every polynomial of odd degree has at least one zero in R.

Proof By dividing the polynomial by its leading coefficient, we may assume that the polynomial is of the form pð xÞ ¼ xn þ an1 xn1 þ    þ a1 x þ a0 ; x 2 R, where n is odd. We can then define M and r(x) as in Theorem 4, and all the statements in the proof of Theorem 4 then hold here too. Since n is odd, xn is positive for x  M and negative for x M. It follows from the arguments in the proof of Theorem 4 that p(x) must be positive for x  M and negative for x M. In particular, p(M) > 0 and p(M) < 0. Then, by the Intermediate Value Theorem, p must have a zero in (M, M). &

So we have certainly got good value from the arguments in the previous proof!

4.2

Properties of continuous functions

4.2.3

149

The Extreme Values Theorem

We now look at whether continuous functions on closed intervals are bounded or unbounded. It turns out that they are bounded, and we use this fact a great deal throughout Analysis! Theorem 6

The Extreme Values Theorem

A continuous function on a closed interval possesses a maximum value and a minimum value on the interval. In other words, if f is a function that is continuous on a closed interval [a, b], then there exist points c and d in [a, b] such that f ðcÞ f ð xÞ f ðdÞ;

We discussed bounds, suprema, infima, maxima and minima of functions earlier, in Sub-section 1.4.2. We prove this at the end of the sub-section.

for x 2 ½a; b:

If f is continuous on an interval that is not closed, then it may not be bounded. For example, the function f ð xÞ ¼ 1x, x 2 ð0; 1, is continuous on (0,1] but is not bounded above (and so it is not bounded), but it is bounded below (by 1). Similarly, if a function is not continuous on a closed interval, then it may not be bounded. For example, the function  1; x ¼ 0; f ðxÞ ¼ 1 0 < x 1; x; takes the values 1 when x ¼ 0 and 1x when 0 < x 1. f is bounded below, by 0, but is not bounded above (and so it is not bounded). Problem 6 (a) Determine the maximum and the minimum of the function f ð xÞ ¼ x2 , x 2 ½1; 2, on [1, 2]. Specify all points in [1, 2] where these are attained. (b) Determine the maximum and the minimum of the function gð xÞ ¼ sin x, x 2 ½0, 2p, on [0, 2p]. Specify all points in [0, 2p] where these are attained. Since a function is bounded if and only it is both bounded above and bounded below, we sometimes use the following version of Theorem 6 in applications: Corollary 1

The Boundedness Theorem

A continuous function on a closed interval is bounded. In other words, if f is a function that is continuous on the closed interval [a, b], then there exist a number M such that jf ð xÞj M; for x 2 ½a; b: On other occasions, we shall find the following consequence of Theorem 6 and the Intermediate Value Theorem useful: Corollary 2

The Interval Image Theorem

The image of a closed interval under a continuous function is a closed interval.

For it will often be sufficient for our purposes.

4: Continuity

150 Proof of Theorem 6 Without some systematic approach to studying continuous functions, we would never be able to prove this theorem, and would be reduced to much hand-waving arguments and loose assertions. In this case we shall use our earlier work on sequences, and the following related result: Lemma 1

Let f be a function defined on an interval I.

(a) If f is bounded above on I and supff ð xÞ: x 2 I g ¼ M, then there exists some sequence {xn} in I such that f ðxn Þ ! M as n ! 1. (b) If f is not bounded above on I, then there exists some sequence {xn} in I such that f ðxn Þ ! 1 as n ! 1. Proof (a) If there is some point, c say, in I for which f(c) ¼ M, then the constant sequence {c} has the desired property. Now suppose that no such point c exists. Then, since supff ð xÞ: x 2 Ig ¼ M, it follows from the definition of supremum that there is some point, x1 say, in I for which

Notice that this is a result about sup for any function f on an interval; no assumption of continuity is involved. A similar result holds for functions that are bounded below on I or are unbounded below on I.

This is a proof by contradiction.

f ðx1 Þ > M  1: Next, since f(x1) < M (from our assumption that there is no point where f takes the value M), choose a point, x2 say, in I for which 

1 f ðx2 Þ > max f ðx1 Þ; M  ; 2 notice, in particular, that this last inequality ensures that x2 6¼ x1. Continuing this process indefinitely, we obtain a sequence {xn} of distinct points in I for which 

1 f ðxn Þ > max f ðx1 Þ; f ðx2 Þ; . . .; f ðxn1 Þ; M  : n   1 Since the sequence M  n converges to M, it follows, by the Squeeze Rule for sequences, that f ðxn Þ ! M as n ! 1. (b) It follows from the definition of ‘unbounded above’ that there is some point, x1 say, in I for which f(x1) > 1. Then, for each n > 1, we can construct a sequence {xn} of distinct points in I for which f ðxn Þ > maxff ðx1 Þ; f ðx2 Þ; . . .; f ðxn1 Þ; ng: Since the sequence {n} tends to 1, it follows, by the Squeeze Rule for & sequences, that f(xn) ! 1 as n ! 1. We are now in a position to prove Theorem 6. Theorem 6 The Extreme Values Theorem A continuous function on a closed interval possesses a maximum value and a minimum value on the interval. In other words, if f is a function that is continuous on the closed interval [a, b], then there exist points c and d in [a, b] such that f ðcÞ f ð xÞ f ðd Þ; for x 2 ½a; b:

So f ðx2 Þ > f ðx1 Þ and f ðx2 Þ > M  12. Also, f ðx2 Þ < M:

For M  1n < f ðxn Þ < M:

4.3

Inverse functions

151

Proof First, we shall assume that f is not bounded above, and verify that this assumption leads to a contradiction with known facts about f. It follows from part (b) of Lemma 1 that there exists some sequence {xn} in [a, b] such that f(xn) ! 1 as n ! 1. Then, by the Bolzano–Weierstrass Theorem, {xn} must contain a convergent subsequence fxnk g; denote by d the limit of fxnk g. Since all the xnk lie in [a, b], it follows, by the Limit Inequality Rule for sequences, that d 2 [a, b]. Since f is continuous at d and the sequence fxnk g converges to d, it follows that f ðxnk Þ ! f ðdÞ. However, ff ðxnk Þg is a subsequence of the sequence {f(xn)} which tends to 1, so that f ðxnk Þ ! 1: This is a contradiction. So f must be bounded above on [a, b] after all. Next, denote by M the number supff ð xÞ: x 2 ½a; bg. It follows, from part (a) of Lemma 1, that there exists some sequence {xn} in [a, b] such that f(xn) ! M as n ! 1. Then, by the Bolzano–Weierstrass Theorem, {xn} must contain a convergent subsequence fxnk g; denote by d the limit of fxnk g. Since all the xnk lie in [a, b], it follows, by the Limit Inequality Rule for sequences, that d 2 [a, b]. Since f is continuous at d and the sequence fxnk g converges to d, it follows that f ðxnk Þ ! f ðdÞ. Therefore, since ff ðxnk Þg is a subsequence of the sequence {f(xn)} which tends to M, we have M ¼ f(d). This complete the proof that there exists a point d in [a, b] such that f ð xÞ f ðd Þ; for all x 2 ½a; b: The proof of the existence of a point c in [a, b] such that f(c) f(x), for all & x 2 [a, b], is similar; we omit it.

4.3 4.3.1

Inverse functions Existence of an inverse function

Let f be the function f(x) ¼ 2x, x 2 R. Then, given any number y in R, we can find a unique number x ¼ 12 y in the domain of f such that y ¼ f(x) ¼ 2x. The inverse function f 1, defined by f 1 ð yÞ ¼ 12 y; y 2 R, undoes the ‘effect’ of f; that is, f 1( f(x)) ¼ x, x 2 R. Also, f undoes the effect of f 1; that is, f ( f 1(y)) ¼ y, y 2 R. Not every function has an inverse function! For example, consider the function gð xÞ ¼ x2 ;

x 2 R:

Since g(2) ¼ 4 ¼ g(2), we cannot define g1(4) uniquely. Thus g fails to have an inverse function, because it is not one–one. However the function hð xÞ ¼ x2 ; x 2 ½0; 1Þ; is one–one and has an inverse function pffiffiffi h1 ð yÞ ¼ y; y 2 ½0; 1Þ: In general, if f : A ! R is one–one, then, for each point y in the image f(A), there is a unique point x in A such that f(x) ¼ y. Thus f is a one–one correspondence between A and f(A); and so we can define the inverse function f 1 by f 1( y) ¼ x, where y ¼ f(x).

It is also possible to prove Theorem 6 by the method of repeated bisection. Theorem 3, Sub-section 2.5.1.

xnk 2 ½a; b , a xnk b:

Inheritance Property of Subsequences, Theorem 5, Sub-section 2.4.4. M must exist, since we have shown that the set ff ð xÞ: x 2 ½a; bg is bounded.

4: Continuity

152 Definition Let f : A ! R be a one–one function. Then the inverse function f 1 has domain f(A) and is specified by f 1 ð yÞ ¼ x;

where y ¼ f ð xÞ; x 2 A:

For some functions f, we can find the inverse function f 1 directly, by solving the equation y ¼ f(x) algebraically to obtain x in terms of y. 1 Example 1 Prove that the function f defined by f ð xÞ ¼ 1x ; x 2 ð1; 1Þ, has an inverse function defined on (0, 1). 1 Solution First, we solve the equation y ¼ 1x to give x in terms of y. By simple manipulation, we obtain 1 1 ,x¼1 : y¼ 1x y 1 Now, for each x 2 (1, 1), we have x < 1, and so f ð xÞ ¼ 1x > 0; thus f ðð1,1ÞÞ  ð0,1Þ. Also, for each y 2 (0, 1), we have 1 x ¼ 1  2 ð1; 1Þ; y

and so f is a one–one correspondence between (1, 1) and (0, 1). Hence 1 f 1 ð yÞ ¼ 1  ; y 2 ð0; 1Þ: & y

Remark Usually, when defining a function we write x for the domain variable. To conform with this practice, we may rewrite the inverse function f 1 in Example 1 as follows 1 f 1 ð xÞ ¼ 1  ; x 2 ð0; 1Þ: x The graph y ¼ f 1 (x) is obtained by reflecting the graph y ¼ f(x) in the line y ¼ x. This reflection interchanges the x- and y-axes.

Proving that a function f is one–one We have seen that, if f: A ! R is one–one, then f has an inverse function f 1 with domain f(A). For the function f considered in Example 1, it is possible to determine f 1 explicitly by solving the equation y ¼ f (x) to obtain x in terms of y. Unfortunately, it is generally not possible to solve the equation y ¼ f (x) in this way. Nevertheless, it may still be possible to prove that f has an inverse function f 1 by showing that f is one–one in some other way. For example, f is one–one if it is either strictly increasing or strictly decreasing; that is, if f is strictly monotonic. A function f defined on an interval I is: increasing on I if x1 5x2 ) f ðx1 Þ f ðx2 Þ; for x1 ; x2 2 I; strictly increasing on I if x1 5x2 ) f ðx1 Þ5 f ðx2 Þ; for x1 ; x2 2 I; decreasing on I if x1 5x2 ) f ðx1 Þ  f ðx2 Þ; for x1 ; x2 2 I;

Definitions   

y

1 1–x

1

(– ∞, 1)

0

y=

1

x

4.3

Inverse functions

153 x1 5x2 ) f ðx1 Þ > f ðx2 Þ; for x1 ; x2 2 I;



strictly decreasing on I if



monotonic on I if

f is either increasing on I or decreasing on I;



strictly monotonic on I if

f is either strictly increasing on I or strictly decreasing on I.

The most powerful technique for proving that a function f is strictly monotonic is to compute the derivative f 0 of f and examine the sign of f 0 (x). We shall not study differentiation in detail until Chapter 6, so here we consider only functions which can be proved to be strictly monotonic by manipulating inequalities rather than by Calculus. For example, if n 2 N, then the function f(x) ¼ xn, x 2 [0, 1), is strictly increasing; and, if n is odd, the function f(x) ¼ xn, x 2 R, is strictly increasing. Similarly, if n 2 N, then the function f(x) ¼ xn, x 2 (0, 1), is strictly decreasing. Prove that the function f ð xÞ ¼ x5 þ x  1; x 2 R; is one–one.

Example 2 Solution

If x1 < x2, then x51 < x52 . Hence x51 þ x1  1 < x52 þ x2  1;

so f is strictly increasing, and thus one–one. Problem 1

Prove that the following functions are one–one:

(a) f ð xÞ ¼ x4 þ 2x þ 3; 2

(b) f ð xÞ ¼ x 

4.3.2

&

1 x;

x 2 ½0; 1Þ;

x 2 ð0; 1Þ:

The Inverse Function Rule

If the function f: A ! R is strictly monotonic, then f is one–one, and so f has an inverse function f 1 with domain f(A). However, it is not always easy to determine f(A). However, if f is known to be continuous on A, then the following result simplifies the problem immediately. Theorem 1

Inverse Function Rule

Let f : I ! J, where I is an interval and J is the image f(I), be a function such that: 1. f is strictly increasing on I; 2. f is continuous on I. Then J is an interval, and f has an inverse function f 1: J ! I such that: 10 . f 1 is strictly increasing on J; 20 . f 1 is continuous on J.

Remarks 1. The interval I may be any type of interval: open or closed, half-open, bounded or unbounded. 2. There is a similar version of the Inverse Function Rule with ‘strictly increasing’ replaced by ‘strictly decreasing’.

We prove the Inverse Function Rule in Sub-section 4.3.4.

4: Continuity

154 Example 3 Prove that the function f ð xÞ ¼ x5 þ x  1; x 2 R; has a continuous inverse function, with domain R. Solution The domain of f is R, which is an interval. We have already seen, in Example 2, that f is strictly increasing on R and so has an inverse function f1. Since f is a polynomial, it is a basic continuous function on R. Thus f satisfies the hypotheses of the Inverse Function Rule. Then it follows from the Rule that the image J ¼ f(R) is an interval, and that the inverse function f1: J ! R is strictly increasing and continuous on J. It remains to check that the image J is the whole of R. Notice that J ¼ f(R) contains each of the numbers f ðnÞ ¼ n5 þ n  1;

n 2 Z:

Now, J contains each of the intervals [f (n), f (n)]; and f ðnÞ ¼ n5  n 1 ! 1 as n ! 1, while f ðnÞ ¼ n5 þ n  1 ! 1 as n ! 1. It follows that, in fact, J ¼ f (R) must be (1, 1) ¼ R. & Thus f has a continuous inverse function f 1: R ! R. As the above example shows, when we apply the Inverse Function Rule the hardest step is to determine the image J ¼ f (I). Since J is an interval, it is sufficient to determine the end-points of J, which may be real numbers or one of the symbols 1 and 1. We must also determine whether or not these endpoints belong to J. The following diagrams illustrate two examples: y

y

d

d = f (b) y = f (x)

J = [c, d )

J = (c, d ]

ð0; 1 has end-points 0 and 1; ½1; 1Þ has end-points 1 and 1: Do not let this use of the symbol 1 tempt you to think that 1 is a real number. y

y = f (x)

c

c = f (a)

For example:

d y = f (x)

J = [c, d )

a

I = [a, ∞)

x

a

b

x

I = (a, b]

Notice that, if a is an end-point of I and a 2 I, then c ¼ f(a) is the corresponding end-point of J and c 2 J. On the other hand, if a is an end-point of I and a 2 = I (this includes the possibility that a may be 1 or 1), then it is a little harder to find the corresponding end-point of J. However, it can be shown that, if {xn} is a monotonic sequence in I and xn ! a, then f(xn) ! c, and c 2 = J: (We prove this in Sub-section 4.3.4.) Example 4 Prove that the function f ð xÞ ¼ x4 þ 2x þ 3; x 2 ½0; 1Þ, has a continuous inverse function with domain [3, 1). Solution The domain of f is [0, 1), which is an interval. Also, we know that f is strictly increasing and continuous on [0, 1), and so conditions 1 and 2 of the Inverse Function Rule hold. It follows that the image J ¼ f([0, 1)) is an interval, and f has a continuous inverse function f 1: J ! [0, 1) which is strictly increasing on J. It remains to check that the image J is [3, 1). For the end-point 0 of I, we have 0 2 I, so the corresponding end-point of J is f(0) ¼ 3, and 3 2 J.

c = f (a) c∈J

a a∈I

x

I = [a, ∞)

y d = f (b) J = (c, d ]

{ f (an)}

y = f (x)

c c∉J

{an} a a∉I

I = (a, b]

b

x

You saw this in Problem 1(a), above.

4.3

Inverse functions

155

The other end-point of I is 1, so to find the corresponding end-point of J we choose the monotonic sequence {n}, which lies in I and tends to infinity. Then f ðnÞ ¼ n4 þ 2n þ 3 ! 1 as n ! 1. It follows that the corresponding end& point of J is 1. Thus, J ¼ [3, 1), as required. We now summarise the strategy for establishing that a given continuous function f has a continuous inverse function. Strategy To prove that f : I ! J, where I is an interval with end-points a and b, has a continuous inverse f 1: J ! I: 1. show that f is strictly increasing on I; 2. show that f is continuous on I; 3. determine the end-point c of J corresponding to the end-point a of I as follows: 

if a 2 I, then f(a) ¼ c (and c 2 J);



if a 2 = I, then f(xn) ! c (and c 2 = J),

where {xn} is a monotonic sequence in I such that xn ! a; 4. determine the end-point d of J corresponding to the end-point b of I, similarly. Problem 2 Use the above strategy to prove that the function f ð xÞ ¼ x2  1x ; x 2 ð0; 1Þ, has a continuous inverse function with domain R. Hint: Use the result of Problem 1(b) in Sub-section 4.3.1.

4.3.3

Inverses of standard functions

We now use the Inverse Function Rule and the above strategy to define continuous inverse functions for certain standard functions. Although you will be familiar with these inverse functions already, we can now prove that they exist and are continuous. We also remind you of some properties of these inverse functions. For each function, we give brief remarks on the four steps of the strategy. In each case, the continuity of the function f follows directly from the results of Section 4.1.

The nth root function We asserted the existence of the nth root function in Sub-section 1.5.2, Theorem 1. We can at last provide the proof of that assertion! The nth root function For any positive integer n  2, the function f ð xÞ ¼ xn ; x 2 ½0; 1Þ; pffiffiffi has a strictly increasing continuous inverse function f 1 ð xÞ ¼ n x with domain ½0; 1Þ, called the nth root function.

There is a corresponding version of this strategy for the case that f is decreasing on I.

4: Continuity

156 In this case, the strategy is easy to apply: 1. f is strictly increasing on [0, 1); 2. f is continuous on [0, 1); 3. f(0) ¼ 0; 4. f(k) ¼ kn ! 1 as k ! 1. It follows, from the Inverse Function Rule, that f has a strictly increasing continuous inverse function f 1: [0, 1) ! [0, 1).

Remark If n is odd, then the nth root function can be extended to a continuous function whose domain is the whole of R.

Inverse trigonometric functions The function sin1

The function   1 1 f ð xÞ ¼ sin x; x 2  p; p ; 2 2

has a strictly increasing continuous inverse function with domain [1, 1], called sin1. In this case: 1. the geometric  definition of f(x) ¼ sin x shows that f is strictly increasing on 12 p; 12 p ;

 2. f is continuous on 12 p; 12 p ;   3. sin 12 p ¼ 1;   4. sin 12 p ¼ 1. 

 It follows that f 12 p; 12 p ¼ ½1; 1. Hence, by the Inverse Function Rule, f has increasing continuous inverse function f 1:

1 a 1strictly  ½1; 1 !  2 p; 2 p : The decreasing version of the strategy can be applied similarly to prove that the cosine function has an inverse, if we restrict its domain suitably. The function cos1 The function f ð xÞ ¼ cos x; x 2 ½0; p; has a strictly decreasing continuous inverse function with domain [1, 1], called cos1. The domain [0, p] of f is chosen here by convention, so that f is a strictly monotonic restriction of the cosine function. Similarly, to form an  inverse of the tangent function we must restrict its domain to  12 p; 12 p , since the tangent function is strictly increasing and continuous on this interval.

f is a basic continuous function. We use {k} rather than {n} here, to avoid using n for two different purposes in the same expression.

4.3

Inverse functions

157

The function tan1

The function   1 1 f ð xÞ ¼ tan x; x 2  p; p ; 2 2

has a strictly increasing continuous inverse function with domain R, called tan1.   In this case, the image set f  12 p; 12p is R, because (for example) if {xn} is a monotonic sequence in  12 p; 12 p and xn ! 12 p as n ! 1, then sin xn f ðxn Þ ¼ tan xn ¼ ! 1 as n ! 1: cos xn

Remark Some texts use arc sin, arc cos and arc tan instead of sin1, cos1 and tan1, respectively. Problem 3   pffiffiffi (a) Determine the values of sin1 ðp1ffiffi2Þ, cos1 12 and tan1 3 .   (b) Prove that cos 2 sin1 x ¼ 1  2x2 , for x 2 ½1; 1. Hint:

Let y ¼ sin1 x:

The function loge We now discuss one of the most important inverse functions. The function loge The function f ð xÞ ¼ ex ; x 2 R; has a strictly increasing continuous inverse function f 1 with domain (0, 1), called loge. In this case: 1. f is strictly increasing on R, since x1 < x2 ) x2  x1 > 0 ) ex2 x1 > 1 ) e x2 > e x1 ;

ðsince ex  1 þ x > 1; for x > 0Þ

2. f is continuous on R; 3. f ðnÞ ¼ en ! 1 as n ! 1; 4. f ðnÞ ¼ en ! 0 as n ! 1: It follows that the image of R under f is f ðR Þ ¼ ð0; 1Þ. Hence, by the Inverse Function Rule, f has a strictly increasing continuous inverse function f 1: ð0; 1Þ ! R. Problem 4 Prove that loge x þ loge y ¼ loge ðxyÞ, for x; y 2 ð0; 1Þ. Hint: Let a ¼ loge x and b ¼ loge y.

4: Continuity

158

Inverse hyperbolic functions The function sinh1

The function 1 f ð xÞ ¼ sinh x ¼ ðex  ex Þ; x 2 R; 2 has a strictly increasing continuous inverse function f 1 with domain R, called sinh1.

In this case: 1. f is strictly increasing on R, since x1 < x2 ) ex1 <ex2 ) ex1 < ex2 ) ex1  ex1 < ex2  ex2 ) sinh x1 < sinh x2 ; 2. f is continuous on R, by the Combination Rules; 3. f ðnÞ ¼ 12 ðen  en Þ ! 1 as n ! 1; 4. f ðnÞ ¼ 12 ðen  en Þ ! 1 as n ! 1. It follows that the image of R under f is f (R) ¼ R. Hence, by the Inverse Function Rule, f has a strictly increasing continuous inverse function f 1: R ! R. The function cosh1

The function 1 f ð xÞ ¼ cosh x ¼ ðex þ ex Þ; x 2 ½0; 1Þ; 2 has a strictly increasing continuous inverse function f 1 with domain [1, 1), called cosh1.

In this case: 1. f is strictly increasing on [0, 1), since x1 < x2 ) sinh x1 < sinh x2  1  1 ) 1 þ sinh2 x1 2 < 1 þ sinh2 x2 2 ) cosh x1 < cosh x2 ; since cosh2 x ¼ 1 þ sinh2 x; 2. f is continuous on [0, 1), by the Combination Rules; 3. f ð0Þ ¼ 1; 4. f ðnÞ ¼ 12 ðen þ en Þ ! 1 as n ! 1. It follows that the image of [0, 1) under f is f ([0, 1)) ¼ [1, 1). Hence, by the Inverse Function Rule, f has a strictly increasing continuous inverse function f 1: [1, 1) ! [0, 1). The strategy can be applied in a similar way to show that f(x) ¼ tanh x is strictly increasing and continuous on R, with f (R) ¼ (1, 1). We omit the details.

4.3

Inverse functions

159

The function tanh1

The function sinh x ; x 2 R; f ð xÞ ¼ tanh x ¼ cosh x has a strictly increasing continuous inverse function f1 with domain (1, 1), called tanh1.

The inverse hyperbolic functions can be expressed in terms of loge, as the following example shows. pffiffiffiffiffiffiffiffiffiffiffiffiffi  Example 5 Prove that sinh1 x ¼ loge x þ x2 þ 1 , for x 2 R. Solution

Let y ¼ sinh1 x, for x 2 R. Then 1 x ¼ sinh y ¼ ðey  ey Þ; 2

and so e2y  2xey  1 ¼ 0: This is a quadratic equation in ey, so that pffiffiffiffiffiffiffiffiffiffiffiffiffi e y ¼ x x 2 þ 1: Since ey > 0, we must choose the þ sign here, so that  pffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ loge x þ x2 þ 1 :

Problem 5

4.3.4

&

 pffiffiffiffiffiffiffiffiffiffiffiffiffi Prove that cosh1 x ¼ loge x þ x2  1 , for x 2 ½1; 1Þ.

Proof of the Inverse Function Rule

We now prove the Inverse Function Rule, and justify our strategy for determining the domains of inverse functions. Theorem 1 Inverse Function Rule Let f: I ! J, where I is an interval and J is the image f (I), be a function such that: 1. f is strictly increasing on I; 2. f is continuous on I. Then J is an interval, and f has an inverse function f 1: J ! I such that: 10 . f 1 is strictly increasing on J; 20 . f 1 is continuous on J. Proof The proof is in four parts: J ¼ f(I) is an interval. Let y1, y2 2 f(I), with y1 < y2, and let y 2 (y1, y2). Now y1 ¼ f (x1) and y2 ¼ f(x2), for some x1, x2 2 I, with x1 < x2 since f is strictly increasing on I. It follows, by the Intermediate Value Theorem, that there is some number x 2 (x1, x2) for which f(x) ¼ y. Hence y 2 f(I). It follows that f(I) is an interval.

You may omit the proofs in this sub-section at a first reading, but you should at least read the statement of Theorem 2 below.

4: Continuity

160 The inverse function f 1: J ! I exists. The function f is strictly increasing and is therefore one–one; also, f maps I onto J, so the function f 1: J ! I exists, by definition. f 1 is strictly increasing on J. We have to show that y1 < y2 ) f 1 ðy1 Þ < f 1 ðy2 Þ;

for y1 ; y2 2 J:

Notice that     f 1 ðy1 Þ  f 1 ðy2 Þ ) f f 1 ðy1 Þ  f f 1 ðy2 Þ ) y1  y2 : Hence y1 < y2 ) f 1 ðy1 Þ < f 1 ðy2 Þ, as required. f 1 is continuous on J. Let y 2 J, and (for simplicity) assume that y is not an end-point of J. Then y ¼ f(x), for some x 2 I, and we want to prove that yn ! y ) f 1 ðyn Þ ! f 1 ð yÞ ¼ x: Thus, we want to deduce that: for each " > 0, there is some number X such that x  " < f 1 ðyn Þ < x þ "; for all n > X: Since f is strictly increasing, we know that f ðx  "Þ < f ð xÞ < f ðx þ "Þ;

(1)

also, since yn ! y ¼ f(x), there is some number X such that f ðx  "Þ < yn < f ðx þ "Þ;

for all n > X:

If we then apply the strictly increasing function f 1 to these inequalities, we obtain (1), as required. This completes the proof of the Inverse Function Rule.

&

In fact, a careful check of the first part of the above proof shows that a slight change enables us to prove the following result that is of interest in its own right. Theorem 2 an interval.

The image of an interval under a continuous function is also

Proof Let f be a continuous function on an interval I. We shall assume that f is non-constant on I, since otherwise the result is trivial. Let y1, y2 2 f(I), with y1 < y2, and let y 2 (y1, y2). Now y1 ¼ f(x1) and y2 ¼ f(x2), for some x1, x2 2 I, and x1 6¼ x2. Let I0 denote the interval with endpoints x1 and x2. It follows, by applying the Intermediate Value Theorem to the function f on I0 , that there is some number x 2 I0 for which f(x) ¼ y. Hence y 2 f(I). & It follows that f(I) is an interval. Finally, we justify our strategy for determining the end-points of J ¼ f(I).

In Corollary 2, Sub-section 4.2.3, we showed that the image of a closed interval under a continuous function is a closed interval. We must introduce the symbol I0 since we do not know in general whether x1 < x2 or x2 < x1.

4.4

Defining exponential functions

161

Strategy for finding the end-points of J If a is an end-point of the interval I, then we can find the corresponding end-point c of J as follows: 

if a 2 I, then f (a) ¼ c (and c 2 J);



if a 2 = I, then f(xn) ! c (and c 2 = J), where {xn} is a monotonic sequence in I such that xn ! a.

Proof

Here we consider only the end-point a of I; a similar result holds for the other end-point b of I.

For simplicity, we shall suppose that a is the left end-point of I. y

y

d

d = f (b)

J = [c, d )

J = (c, d ]

y = f (x)

{ f(xn)} y = f (x)

c = f (a) c∉J

c∈J a

I = [a, ∞)

{xn}

x

a a∉I

a∈I

I = (a, b]

b

x

If a 2 I, then c ¼ f (a) 2 J and f ð xÞ  f ðaÞ ¼ c;

for x 2 I;

and so c is the corresponding left end-point of J. On the other hand, if a 2 = I, then we select any decreasing sequence {xn} in I such that xn ! a as n ! 1. Then {f(xn)} is also decreasing; it therefore follows, by the Monotonic Sequence Theorem for sequences, that f ðxn Þ ! c as n ! 1;

(2)

where c is a real number or 1. Now, f(xn) 2 J for n ¼ 1, 2, . . .; therefore, since J is an interval, it follows that ½f ðxn Þ; f ðx1 Þ  J;

This was Theorem 2 in Subsection 2.5.1: if the sequence {an} is monotonic, then either {an} is convergent or an ! 1.

for n ¼ 1; 2; . . .:

Hence, by (2), we have that 1 [ ðc; f ðx1 Þ ¼ ½f ðxn Þ; f ðx1 Þ  J: n¼1

To deduce, finally, that c is the left end-point of J, we need to show that c 2 = J. Suppose that in fact c 2 J. Then c ¼ f(x), for some x 2 I; it follows that f ð xÞ ¼ c < f ðxn Þ; ¼) x < xn ; ¼) x a:

for n ¼ 1; 2; . . . for n ¼ 1; 2; . . .

Thus x 2 = I. This contradiction completes the proof.

4.4 4.4.1

&

Taking limits ‘flattens’ inequalities.

Defining exponential functions The definition of ax

pffiffiffi Earlier, we looked at the definition of the irrational number 2 ¼ 1:4142 . . .. We have also defined ax for a > 0 when x is rational, but we have not yet defined ax when x is irrational.

Sub-section 1.1.1. Sub-section 1.5.3.

4: Continuity

162 pffiffi 2

One possible method for pffiffiffidefining the irrational powerp2ffiffiffi involves the decimal representation of 2. Each of the truncations of 2 ¼ 1:4142 . . . is a rational number, and the corresponding rational numbers 21 ; 21:4 ; 21:41 ; 21:414 ; 21:4142 ; . . .

(1)

are defined, and form an increasing sequence which is bounded above by 22 ¼ 4 (for example). Hence, by the Monotone Convergence Theorem for sequences, the sequence (1) ispconvergent, and the limit of this sequence can ffiffi be taken as the definition of 2 2 : We can define ax for a > 0 and x 2 R similarly. However, with this definition it is difficult to establish the properties of ax, such as the Exponential Laws. It is more convenient to define ax by using the exponential function x 7! ex, whose properties we have already discussed. Recall that 1 n  x n X x ; for x 2 R; ex ¼ lim 1 þ ¼ n!1 n n! n¼0 ex ¼ ðex Þ1 ;

for x 2 R;

and exþy ¼ ex  ey ;

for x; y 2 R:

(2)

x

Recall too that the function x 7! e is strictly increasing and continuous; and hence, by the Inverse Function Rule, it has a strictly increasing continuous inverse function x 7! loge x, x 2 (0, 1). Thus we have (3) loge ðex Þ ¼ x; for x 2 R; and eloge x ¼ x; x

for x 2 ð0; 1Þ: y

Now let a ¼ e and b ¼ e . Then, by equation (2), we have ab ¼ ex ey ¼ exþy ; so that, by equation (3), we obtain loge(ab) ¼ logea þ logeb, for a, b 2 (0, 1). We deduce that, if a > 0 and n 2 N, then loge ðan Þ ¼ n loge a; and so an ¼ en loge a : With a little more manipulation, we can show that the equation ax ¼ ex loge a is true for each rational number x. This suggests that we define ax, for a > 0 and x irrational, by means of this equation. Definition

If a > 0, then ax ¼ ex loge a ;

for x 2 R:

For example, 2x ¼ ex loge 2 for x 2 R, so that the graph y ¼ 2x is obtained from the graph y ¼ ex by a scaling in the x-direction with scale factor log1 2. e

Sub-section 2.5.1.

Section 3.4.

Sub-section 3.4.3.

4.4

Defining exponential functions

163

This relationship between the graphs y ¼ ex and y ¼ 2x suggests that the function x 7! 2x must also be continuous. In fact, we can deduce the continuity of the function x 7! 2x ¼ ex loge 2 ; x 2 R, from the continuity of the function x 7! ex, by using the Multiple Rule and the Composition Rule for continuity.

Remark Since x 7! 2x is continuous, we deduce that the sequence 21 ; 21:4 ; 21:41 ; 21:414 ; 21:4142 ; . . .; where 1, 1.4, 1.41,pffiffi1.414, 1.4142, . . . are truncations of pffiffi does converge to 2 2 , and so both definitions of 2 2 agree.

pffiffiffi 2 ¼ 1:4142 . . .,

In general, we have the following result. Theorem 1 continuous.

If a > 0, then the function x 7! ax ¼ ex loge a ; x 2 R, is

Prove that the following functions are continuous:

Problem 1

(a) f(x) ¼ x , where x 2 (0, 1) and  2 R; (b) f(x) ¼ xx, where x 2 (0, 1). 

4.4.2

Further properties of exponentials

Our definition of ax enables us to give straight-forward proofs of the following Exponent Laws. Exponent Laws x x x  If a, b > 0 and x 2 R, then a b ¼ (ab) .  

If a > 0 and x, y 2 R, then a a ¼ a . If a > 0 and x, y 2 R, then ðax Þy ¼ axy . x y

xþy

You first met these laws in Sub-section 1.5.3, but there x, y 2 Q .

4: Continuity

164 For example, to prove the final Exponent Law, notice that, from the definition of ax, we have loge(ax) ¼ x loge a; so that x ðax Þy ¼ ey loge ða Þ ¼ exy loge a ¼ axy : Thus manipulations such as  pffiffi p2ffiffi pffiffiffi 2 pffiffiffiðp2ffiffipffiffi2Þ pffiffiffi2 2 ¼ 2 ¼ 2 ¼2 are indeed justified. Problem 2

Prove that, if a > 0 and x, y 2 R , then axay ¼ ax þ y.

Finally, our definition of ax enables us to prove the rule for rearranging inequalities by taking powers. Rule 5

For any non-negative a, b 2 R , and any p > 0, a < b , ap < bp.

You met this Rule in Subsection 1.2.1.

If a ¼ 0, the result is obvious. In general, since the functions x 7! loge x and x 7! ex are strictly increasing, we have a < b , loge a < loge b , p loge a < p loge b ,e

p loge a

<e

ðsince p > 0Þ

p loge b

, ap < bp : The following example illustrates the use of Rule 5. Example 1 Solution

Determine which of the numbers ep and pe is greater. We use the inequality x

e > 1 þ x;

for x > 0:

 Applying this inequality with x ¼ pe  1, we obtain     p p 1 e ð e Þ1 > 1 þ e p ¼ ¼ pe1 ; e p then, by multiplying through by the positive factor e, we obtain that e e > p. & Finally, by applying Rule 5 with p ¼ e, we obtain ep > pe. Problem 3

4.5

Prove that ex > xe, for x > e.

Exercises

Section 4.1 1. Use the appropriate rules, together with the list of basic continuous functions, to prove that the following functions are continuous:

This was one of the Exponential Inequalities: part (a) of Theorem 4 in Sub-section 4.1.3.

4.5

Exercises

165

(a) f ð xÞ ¼ expðsinðx2 þ 1ÞÞ; x 2 R; pffiffi x 2 ½0; 1Þ: (b) f ð xÞ ¼ e x þ x5 ; 2. Determine  whether the following functions are continuous at 0: sin x sinð1xÞ; x 6¼ 0, (a) f ð xÞ ¼ 0; x ¼ 0;  1 1 sin x ; x 6¼ 0, (b) f ð xÞ ¼ x 0; x ¼ 0: 3. Prove that each of the following sequences is convergent, and determine its limit: n  1 o n  1 o (b) cos 21n 2 : (a) sin en  1 ; 4. Let f be defined on an open interval I, and c 2 I. Prove that, if f is continuous at c and f(c) 6¼ 0, then there is an open interval J  I such that c 2 J and f(x) 6¼ 0, for any x 2 J.  1; x ¼ 0; 1; 5. Determine the points where the function f ð xÞ ¼ is x þ ½2x; 0 < x < 1; (a) continuous on the left; (b) continuous on the right; (c) continuous. 6. Prove that the following function is continuous on R 8 x  12 p; < 1; f ð xÞ ¼ sin x;  12 p < x < 12 p; : 1; x  12 p: 7. Determine at which points the following function is continuous  x; 1 x < 0; f ð xÞ ¼ ex ; 0 x 1: 8. Write down examples of functions with the following properties: (a) f and g are discontinuous on R, but g  f is continuous on R; (b) f and g are discontinuous on R, but f þ g and fg are continuous on R;   (c)  f is continuous on R  1; 12 ; 13 ; 14 ; . . . but discontinuous at  1; 12 ; 13 ; 14 ; . . . :

Section 4.2 1. The function f is continuous on (0, 1), and takes every real value at most once. Use the Intermediate Value Theorem to prove that f is strictly monotonic on (0, 1). 2. Give examples of functions f continuous on the half-open interval [0, 1) in R, to show that f ([0,1)) can be open, closed or half-open. 3. Prove that each of the following polynomials has the stated number of (real) zeros: (a) pð xÞ ¼ x4  4x3 þ 3x2 þ 2x  1; 4 zeros; (b) pð xÞ ¼ 3x3  8x2 þ x þ 3;

3 zeros:

2 4. Prove that  the function f ð xÞ ¼ x  sin x  3 p; x 2 R, has a zero in 2 5 3 p; 6 p .

We shall use this result in Chapter 6.

4: Continuity

166 5. Using the Zeros Localisation Theorem and the Extreme Values Theorem, prove that every polynomial of even degree n pð xÞ ¼ an xn þ an1 xn1 þ    þ a1 x þ a0 ;

x 2 R;

where an 6¼ 0, has a minimum value on R. 6. Write down an example of a function f(x), x 2 (0, 1), that is continuous on (0, 1) and for which f ((0, 1)) ¼ (0, 1), but such that there is no point c in (0, 1) for which f (c) ¼ c.

This result is closely related to Problem 2 in Subsection 4.2.1, so you might like to look back at that problem and compare the two.

Section 4.3 1. For each of the following functions, prove that it has a continuous inverse function and determine the domain of that inverse function: (a) f ð xÞ ¼ x3 þ 1  x12 ;

x 2 ð0; 1Þ;

1 ð1þx3 Þ2

(b) f ð xÞ ¼ ; x 2 ð1; 1Þ: 2. Determine whether each of the following statements is true:   (a) sin sin1 x ¼ x, for x 2 ½1; 1; (b) sin1 ðsin xÞ ¼ x, for x 2 R.  

3. (a) Prove that tan1 x þ tan1 y ¼ tan1   tan1 x þ tan1 y lies in  12 p; 12 p . (b) Use the result in part (a) to evaluate

x þy 1xy

, provided that



tan1 12

þ



tan1 13

.

This is known as the Addition Formula for tan1.

5

Limits and continuity

In Chapter 4 we made some progress in pinning down the idea of ‘a wellbehaved function’ in precise terms. Using our earlier work on sequences we defined what was meant by a function being continuous: roughly speaking, its graph has no jumps or gaps. However, a number of functions arise quite naturally in mathematics where we need to handle functions that are already defined near some particular point, but that are not defined at the point itself. For example, the function sin x f ð xÞ ¼ ; x 6¼ 0; x which arises when we examine the question of whether the sine function is differentiable. Clearly from their graphs, the behaviour of f differs significantly from the behaviour of the function   1 gð xÞ ¼ sin ; x 6¼ 0; x that you have already met. It is possible to assign a value (namely, 1) to f at 0 so that this extension of the domain of f makes f continuous on a domain that is an interval. On the other hand, it is not possible to assign a value to g at 0 so that this extension of the domain of g makes g continuous on a domain that is an interval. In Section 5.1, we discuss limits of functions, and show that the existence of this helpful value 1 for f can be stated as lim sinx x ¼ 1. The concept of a limit of a x!0 function is closely related to that of a continuous function, and many of the rules for calculating limits are similar to those for continuity. We also discuss one-sided limits. In Section 5.2, we discuss the behaviour of functions near asymptotes of their graphs. In particular, we prove that, if n 2 Z, then lim xn ex ¼ 0:

1 y = sin 1x

–1

x!1

Next, in Section 5.3, we introduce a slightly different definition of the limit lim f ð xÞ of a function f at a point c. Instead of using sequences tending to c to x!c define the limit of f at c, we define lim f ð xÞ directly in terms of inequalities x!c involving x and f (x), and we verify that this new definition is completely equivalent to the earlier definition. We also illustrate the changes to proofs of results about limits that the new definition involves. In Section 5.4, we introduce a definition for the continuity of a function f at a point c in terms of inequalities involving x and f (x), rather than in terms of the behaviour of f on sequences that tend to c. We verify that this new definition is completely equivalent to the earlier definition of continuity, and illustrate the changes to proofs of results about continuity that the new definition involves. Finally, in Section 5.5, we introduce a concept that will be extremely powerful in your further study of Analysis; namely, that of uniform continuity.

At first sight this new definition will seem more complicated. However throughout the rest of the book we shall see just how powerful this new definition turns out to be!

167

5: Limits and continuity

168 Before starting on Sections 5.3 and 5.4, you may find it useful to quickly revise the material in Sections 1.2 and 1.3 on inequalities.

5.1 5.1.1

‘may’ means ‘will’!

Limits of functions What is a limit of a function?

The graph of the function sin x ; x 6¼ 0; f ð xÞ ¼ x shows that, if x takes values which are ‘close to’ but distinct from 0, then f(x) takes values which are ‘close to’ 1. The closer that x gets to 0, the closer f(x) gets to 1. We now pin down this idea precisely. First, we need the following fact. Theorem 1

Proof

If {xn} is a null sequence whose terms are non-zero, then sin xn ! 1 as n ! 1: xn

First we deduce from the inequality p sin x  x; for 0 < x < ; 2

sin xn n!1 xn

That is, lim

¼ 1.

Lemma 1, Sub-section 4.1.3.

that sin x p  1; for 0 < x < : x 2 Next, we need to use the formula for the area of a sector of a disc of radius 1 in Figure (a), below. Compare the area of this sector with the area of the triangle in Figure (b), below.

We find that p for 0 < x < : 2 sin x Since tan x ¼ cos x and cos x > 0 for 0 < x < p2, we obtain sin x p ; for 0 < x < : cos x  x 2 Combining this inequality with our earlier upper estimate for sinx x, we find that sin x p  1; for 0 < x < : cos x  x 2 x  tan x;

We discussed the area and perimeter of a disc of radius 1 in Sub-section 2.5.4 and in Exercise 4 on Section 2.5 in Section 2.6.

5.1

Limits of functions

169

In fact, this inequality holds for 0 < jxj < p2, since sinðxÞ sin x ¼ : ðxÞ x Now, if {xn} is any null sequence with non-zero terms, then the terms xn must eventually satisfy the inequality jxn j < p2, and so there is some number X such that sin xn cos xn   1; for n > X: xn But cos xn ! 1 as n ! 1, since the cosine function is continuous at 0 and cos 0 ¼ 1. Hence, by the Squeeze Rule for sequences sin xn ! 1: & xn cosðxÞ ¼ cos x and

Take " ¼ p2 in the definition of null sequence.

sin x This behaviour of cos x near 0 is an example of a function f tending to a limit as x tends to a point c. To define this concept, we need to ensure that the function f is defined near the point c, but not necessarily at the point c itself. We first introduce the idea of a punctured neighbourhood of c.

Definitions A neighbourhood of a point c of R is an open interval that contains the point c, and a punctured neighbourhood of a point c of R is a neighbourhood of c from which the point c itself has been deleted. For example, the sets (0, 9), (1, 1) and R are neighbourhoods of the point 2, and (1, 2) [ (2, 5) and (1, 2) [ (2, 4) are punctured neighbourhoods of the point 2. In general, a neighbourhood of c is an interval of the form (c  r, c þ s), for some r, s > 0, and a punctured neighbourhood of c is the union of a pair of intervals (c  r, c) [ (c, c þ s), for some r, s > 0. In practice, we often choose as a neighbourhood of c an open interval (c  r, c þ r), r > 0, with centre at c; and as a punctured neighbourhood of c the union (c  r, c) [ (c, c þ r) of two open intervals of equal length. We now define the limit of a function in terms of limits of sequences. Definition Let the function f be defined on a punctured neighbourhood N of a point c. Then f (x) tends to the limit ł as x tends to c if: for each sequence fxn g in N such that xn ! c; then f ðxn Þ ! ‘: (1)

So the ‘puncture’ is at c.

These choices are simply matters of convenience!

Note that xn 6¼ c, for any n.

We write this as either ‘lim f ð xÞ ¼ ‘’ or ‘f (x) ! ‘ as x ! c’. x!c

Let us check that this definition holds for f ðxÞ ¼ sinx x at 0. This function f is defined on the domain R  {0}, and so in particular on every punctured neighbourhood of 0. We have just seen that the statement (1) holds; it follows then that sin x ¼ 1: lim x!0 x Since the definition of the limit of a function involves the limit of sequences, we can use our various Combination Rules for sequences to determine the limits of many functions.

For example, f is defined on (1, 0) [ (0, 1).

5: Limits and continuity

170 Example 1 Prove that each of the following functions tends to a limit as x tends to 2, and determine these limits: 2 3 3x  2 (a) f ð xÞ ¼ xx 24; (b) f ð xÞ ¼ xx2   3x þ 2 : Solution (a) First, notice that f is defined on every punctured neighbourhood of 2. Next, notice that x2  4 ¼ x þ 2; for x 6¼ 2: x2 Thus, if {xn} is any sequence that lies in some punctured neighbourhood N of 2 and xn ! 2, then f ð xÞ ¼

f ðxn Þ ¼ xn þ 2 ! 4

as n ! 1;

by the Sum Rule for sequences. It follows that x2  4 ¼ 4: x!2 x  2 (b) The function

We can cancel x  2, since x 6¼ 2. For example, N ¼ (1, 2) [ (2, 3); any other punctured neighbourhood of 2 would serve our purpose equally well.

lim

f ð xÞ ¼

x3  3x  2 ðx  2Þðx2 þ 2x þ 1Þ ¼ x2  3x þ 2 ð x  2Þ ð x  1Þ

has domain R  {1, 2}, and so f is defined on the punctured neighbourhood N ¼ (1, 2) [ (2, 3) of 2. Next, notice that f ð xÞ ¼

x3  3x  2 x2 þ 2x þ 1 ; ¼ x2  3x þ 2 x1

Of course any smaller punctured neighbourhood of 2 would serve our purpose equally well.

for x 2 N ¼ ð1; 2Þ [ ð2; 3Þ: Thus, if {xn} is any sequence that lies in N ¼ (1, 2) [ (2, 3) and xn ! 2, then x2n þ 2xn þ 1 4þ4þ1 ¼ 9 as n ! 1; ! xn  1 21 by the Combination Rules for sequences. It follows that f ðxn Þ ¼

lim

x!2

x3  3x  2 ¼ 9: x2  3x þ 2

&

Later in this section we give several further techniques for calculating limits. Our next example illustrates how to prove that a limit does not exist.

Sub-sections 5.1.2 and 5.1.3.

Example 2 Prove that each of the following functions does not tend to a limit as x tends to 0:   pffiffiffi (a) f ðxÞ ¼ 1x ; x 6¼ 0; (c) f ð xÞ ¼ x; x  0: (b) f ðxÞ ¼ sin 1x ; x 6¼ 0; Solution (a) The function f is defined on  the  punctured neighbourhood N ¼ (2,  0) [ (0, 2) of 0. The null sequence 1n lies in N and tends to 0, but f 1n ¼ n ! 1. Hence f does not tend to a limit as x tends to 0. (b) The function f is defined on the punctured neighbourhood N ¼ (2, 0) [ (0, 2) of 0.

Any other punctured neighbourhood of 0 would serve equally well here, of course.

5.1

Limits of functions

171

To prove that f(x) does not tend to a limit as x tends to 0, we choose two null sequences {xn} and {x0n} that lie in N, such that   f ðxn Þ ! 1 and f x0n ! 1:     Since sin 2np þ 12 p ¼ 1 and sin 2np þ 32 p ¼ 1 for n 2 Z, we can choose 1 1 xn ¼ and x0n ¼ ; for n ¼ 0; 1; 2; . . .: 2np þ 12 p 2np þ 32 p Hence f does not tend to a limit as x tends to 0. pffiffiffi (c) The function f ð xÞ ¼ x has domain [0, 1), and so f is not defined on any punctured neighbourhood of zero. Hence f does not tend to a limit as x & tends to zero.

In particular, the terms xn and x n0 will be non-zero.

We collect these techniques together in the form of a strategy. Strategy To show that lim f ð xÞ does not exist: x!c 1. Show that there is no punctured neighbourhood N of c on which f is defined; OR

2. Find two sequences {xn} and {xn0 } (in some punctured neighbourhood N of c) which tend to c, such that {f (xn)} and {f (xn0 )} have different limits; OR

3. Find a sequence {xn} (in some punctured neighbourhood N of c) which tends to c, such that f (xn) ! 1 or f (xn) ! 1. Problem 1 Determine whether the following limits exist: 2 (a) lim x xþx ; (b) lim jxxj : x!0

5.1.2

x!0

Limits and continuity

Consider the function 1; f ð xÞ ¼ 0;

x 6¼ 0; x ¼ 0:

Does this function tend to a limit as x tends to zero; and, if it does, what is the limit? Certainly, f is defined on any punctured neighbourhood of zero, since the domain of f is R. Also, if {xn} is any null sequence with non-zero terms, then f (xn) ¼ 1, for n ¼ 1, 2, . . ., so that f (xn) ! 1 as n ! 1. It follows that lim f ð xÞ ¼ 1. x!0

This example serves to emphasise that the value of a limit lim f ð xÞ has nothing x!c to do with the value of f (c) – even if f happens to be defined at the point c. However, if f is defined at c, and f is also continuous at c, then the only possible value for lim f ð xÞ is f (c). x!c

We put these observations together in the following result. Theorem 2 Then

Let the function f be defined on an open interval I, with c 2 I. f is continuous at c , lim f ð xÞ ¼ f ðcÞ: x!c

Notice that I is a neighbourhood of c.

5: Limits and continuity

172 Using Theorem 2 and our knowledge of continuous functions, we can evaluate many limits rather easily.   For example, to determine lim 3x5  5x2 þ 1 notice that the function x!2

f (x) ¼ 3x  5x þ 1 is defined on R and is continuous at 2, since f is a polynomial. Hence, by Theorem 2   lim 3x5  5x2 þ 1 ¼ f ð2Þ ¼ 77: 5

2

Polynomials are basic continuous functions: see Sub-section 4.1.3.

x!2

We saw earlier that the following functions are continuous at 0 2  1  x sin x ; x 6¼ 0; x sin 1x ; x 6¼ 0; and f ð xÞ ¼ f ð xÞ ¼ 0; x ¼ 0; 0; x ¼ 0:

Problem 8 and Example 5, Sub-section 4.1.2, respectively.

It therefore follows from Theorem 2 that     1 1 2 lim x sin ¼ 0 and lim x sin ¼ 0: x!0 x!0 x x On the other hand, we saw in part (b) of Example 2 that   1 lim sin ¼ 0 does not exist: x!0 x

Sub-section 5.1.1.

It follows from   Theorem 2 that, no matter how we try to extend the domain of f ðxÞ ¼ sin 1x to include x ¼ 0, we can never obtain a continuous function.

Remark If f is defined on an open interval I, with c 2 I, and lim f ð xÞ exists but x!c

lim f ð xÞ 6¼ f ðcÞ, then f is said to have   a removable discontinuity at c. For x!c x sin 1x ; x 6¼ 0; has a removable discontinuity example, the function f ð xÞ ¼ 3; x ¼ 0; at 0.

This means that, if we redefine the value of f just at c itself, the resulting function is then continuous at c.

Problem 2 Use Theorem 1 to determine the following limits: pffiffiffiffiffiffiffiffiffi pffiffiffi ex (a) lim x; (c) lim 1þx (b) limp sin x; x!2

x!1

x!2

In the remainder of this chapter we shall frequently use Theorem 2. When the function f is one of our basic continuous functions, however, we shall not always refer to the theorem explicitly. For example, f (x) ¼ x2 þ 1 and g(x) ¼ sin x are basic continuous functions, and so we can write limðx2 þ 1Þ ¼ 5 and x!2 lim sin x ¼ 0 without further explanation. x!0

Basic continuous functions:  polynomials and

rational functions;  modulus function;  nth root function;  trigonometric functions

(sine, cosine and tangent);

5.1.3

 the exponential function.

Rules for limits

As you might expect from your experience with sequences, series and continuous functions, we often find limits by using various rules. First, we state the Combination Rules. Theorem 3

Combination Rules

If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ m, then: x!c

x!c

Sum Rule

limð f ð xÞ þ gð xÞÞ ¼ ‘ þ m;

Multiple Rule

lim lf ð xÞ ¼ l‘;

x!c x!c

for l 2 R;

5.1

Limits of functions

173

Product Rule

lim f ð xÞgð xÞ ¼ ‘m;

Quotient Rule

lim gf ððxxÞÞ ¼ m‘ ; provided that m 6¼ 0.

x!c x!c

For example, since lim sinx x ¼ 1 and limðx2 þ 1Þ ¼ 1; we have x!0 x!0    2  sin x þ 2 x þ 1 ¼ 1 þ 2  1 ¼ 3: lim x!0 x The proofs of these rules are all simple consequences of the corresponding results for sequences. We illustrate this by proving just one rule. Proof of the Sum Rule Let the functions f and g be defined on a punctured neighbourhood N of a point c. We want to show that: for each sequence fxn g in N such that xn ! c, then f ðxn Þ þ gðxn Þ ! ‘ þ m:

There are no new ideas in the proof. All that we have to do is to set up things so that we can use the Sum Rule for sequences.

Now, since lim f ð xÞ ¼ ‘, we know that, for any sequence {xn} in N for which x!c

xn ! c, then f (xn) ! ‘. Also, since lim gð xÞ ¼ m, we know that, as xn ! c, then x!c

g(xn) ! m. It follows by the Sum Rule for sequences that, as xn ! c, then & f (xn) þ g(xn) ! ‘ þ m. This completes the proof. We can also use the following Composition Rule. Theorem 4 Composition Rule If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ L, then lim gð f ð xÞÞ ¼ L, provided that: x!c

x!c

x!‘

f (x) 6¼ ‘ in some punctured neighbourhood of c; g is defined at ‘ and is continuous at ‘.

EITHER OR

Note that the second limit is a limit as x ! ‘, not as x ! c.

Remarks on the Composition Rule (a) In any particular case, the Composition Rule may be FALSE if we do not ensure that one or other of the two provisos holds! For example, if f ð xÞ ¼ 0; x 2 R;

sin x and gð xÞ ¼

x

0;

;

x 6¼ 0; x ¼ 0;

Here we take c ¼ 0, ‘ ¼ 0 and L ¼ 1.

then lim f ð xÞ ¼ 0 and

x!0

lim gð xÞ ¼ 1; but

x!0

lim gð f ð xÞÞ ¼ limð0Þ ¼ 0:

x!0

x!0

So, in this case, lim gð f ð xÞÞ 6¼ L:

x!0

(b) Suppose the first proviso ‘f (x) 6¼ ‘ in some punctured neighbourhood of c’ in Theorem 4 holds. Then we know that, for each sequence {xn} in a punctured neighbourhood N of c for which xn ! c, f (xn) does not take the value ‘ but must lie in a punctured neighbourhood of ‘. In this case, the desired result will follow from the facts that (i) the sequence {f (xn)} converges to ‘, (ii) f (xn) lies in a punctured neighbourhood of ‘, and (iii) lim gð xÞ ¼ L: x!‘

(c) Suppose the second proviso ‘g is defined at ‘ and is continuous at ‘’ in Theorem 4 holds. In this case, the desired result will follow from

We omit the details of the proof in this case, which are now straight-forward to write down.

5: Limits and continuity

174 the facts that (i) the sequence {f (xn)} converges to ‘, and (ii) g is continuous at ‘. The following example illustrates the use of the Composition Rule. Example 3 Determine the following limits:  2 ðsin xÞ (a) lim sinsin (b) lim 1þ sinx x : x ; x!0

x!0

Solution (a) Let f (x) ¼ sin x, x 2 R, and gðxÞ ¼ sinx x , x 6¼ 0: Then sin x ¼ 1: x!0 x!0 x!0 x!0 x Also, f (x) ¼ sin x ¼ 6 0 in the punctured neighbourhood ðp; 0Þ [ ð0; pÞ of 0 (for example). It follows, by the Composition Rule, that sinðsin xÞ lim gð f ð xÞÞ ¼ lim ¼ 1: x!0 x!0 sin x lim f ð xÞ ¼ lim sin x ¼ 0

(b) Let f ðxÞ ¼

and

lim gð xÞ ¼ lim

x 6¼ 0, and gð xÞ ¼ 1 þ x2 , x 2 R: Then   sin x ¼ 1 and lim 1 þ x2 ¼ 2: lim x!0 x x!1

Here we have c ¼ 0, ‘ ¼ 0 and L ¼ 1.

sin x x ,

Here we have c ¼ 0, ‘ ¼ 1 and L ¼ 2.

Also, g is defined and continuous at 1. It follows, by the Composition Rule, that   ! sin x 2 lim gð f ð xÞÞ ¼ lim 1 þ ¼ 2: & x!0 x!0 x Problem 3 Use the Combination Rules and the Composition Rule to determine the following limits:  1 sinðx2 Þ sin x (b) lim x2 ; (c) lim sinx x 2 : (a) lim 2xþx 2 ; x!0

x!0

x!0

For the functions 8 x ¼ 0; 0. Then f(x) tends to the limit ł as x tends to c from the right if:

Note that f need not be defined at the point c.

for each sequence {xn} in (c, c þ r) such that xn ! c, then f (xn) ! ‘. We write this either as ‘ limþ f ð xÞ ¼ ‘’ or as ‘f (x) ! ‘ as x ! c þ ’.

For example, limþ x!0

x!c

pffiffiffi x ¼ 0:

There is a corresponding definition for limits as x tends to c from the left. Definition Let f be defined on (c  r, c), for some r > 0. Then f(x) tends to the limit ł as x tends to c from the left if: for each sequence {xn} in (c  r, c) such that xn ! c, then f(xn) ! ‘. We write this either as ‘ lim f ð xÞ ¼ ‘’ or as ‘f ð xÞ ! ‘ as x ! c ’. x!c

Sometimes both left and right limits exist. Even when this happens, the two values need not be equal, as the following example shows. Example 4 Prove that the function f ð xÞ ¼ jxxj , x 6¼ 0, tends to different limits as x tends to 0 from the right and from the left. Solution The function f is defined on (0, 1), and f (x) ¼ 1 on this interval. Thus, if {xn} is a null sequence in (0, 1), then lim f ðxn Þ ¼ lim ð1Þ ¼ 1: n!1

n!1

So limþ f ð xÞ ¼ 1: x!0

Similarly, f is defined on (1, 0), and f(x) ¼ 1 on this interval. Thus, if {xn} is a null sequence in (1, 0), then lim f ðxn Þ ¼ lim ð1Þ ¼ 1:

n!1

So lim f ð xÞ ¼ 1: x!0

n!1

&

Again, f need not be defined at the point c.

5: Limits and continuity

176 Problem 6 Write down a function f defined on the interval [1, 3] for which lim f ð xÞ does not exist but limþ f ð xÞ ¼ 1. Verify that f has the x!0

x!0

specified properties. The relationship between one-sided limits and ‘ordinary’ limits is given by the following result. Theorem 7 Then

Let f be defined on a punctured neighbourhood of the point c.

We omit a proof of this result.

lim f ð xÞ ¼ ‘ x!c

, limþ f ð xÞ x!c

and

lim f ð xÞ both exist and equal ‘:

x!c

Remark If f is defined on an open interval I that contains the point c, and lim f ð xÞ and lim f ð xÞ both exist; but limþ f ð xÞ 6¼ lim f ð xÞ;

x!cþ

x!c

x!c

x!c

then f is said to have a jump discontinuity at c. Analogues of the Combination Rules, Composition Rule and Squeeze Rule can also be used to determine one-sided limits. In the statements of all these rules, we simply replace lim by limþ or lim , and replace the open interval I containx!c

x!c

x!c

ing c by (c, c þ r) or (c  r, c), as appropriate. Problem 7 Prove pffiffiffithat: (a) limþ sinx x þ x ¼ 1; x!0

Problem 8 prove that

pffiffi (b) limþ sinpð ffiffix xÞ ¼ 1: x!0

1 ; for jxj < 1, to Use the inequalities 1þ x  ex  1x

limþ

x!0

ex 1 ex 1 ¼ lim ¼ 1: x!0 x x

Deduce that ex 1 ¼ 1: x!0 x lim

5.2

Asymptotic behaviour of functions

In this section we define formally a number of statements that you will have already met in your study of Calculus, such as 1 ! 1 as x ! 0þ and ex ! 1 as x ! 1; x and describe the relationship between them.

5.2.1

Functions which tend to infinity

Just as we defined the statement lim f ð xÞ ¼ ‘ in terms of the convergence of x!c sequences, so we can define the statement f (x) ! 1 as x ! c.

We verified these inequalities as part (c) of Theorem 4 (‘The Exponential Inequalities’) in Sub-section 4.1.3.

5.2

Asymptotic behaviour of functions

177

Definition Let f be defined on a punctured neighbourhood N of the point c. Then f (x) ! 1 as x ! c if: for each sequence {xn} in N such that xn ! c, then f (xn) ! 1. The statement ‘f (x) ! 1 as x ! c’ is defined similarly, with 1 replaced by 1. As with sequences, there is a version of the Reciprocal Rule which relates the behaviour of functions which tend to infinity and functions which tend to 0.

Theorem 2, Sub-section 2.4.3.

Theorem 1 Reciprocal Rule (a) If the function f satisfies the following two conditions: 1. f (x) > 0 for all x in some punctured neighbourhood of c, and 2. f (x) ! 0 as x ! c, then f ð1xÞ ! 1 as x ! c. (b) If f (x) ! 1 as x ! c, then f ð1xÞ ! 0 as x ! c. For example, lim x2 ¼ 0.

1 x2

! 1 as x ! 0, since f (x) ¼ x2 > 0, for x 2 R  {0}, and

x!0

The statements ‘f ð xÞ ! 1 or 1 as x ! cþ ’ and

‘f ð xÞ ! 1 or 1 as x ! c ’

are defined similarly, with the punctured neighbourhood of c being replaced by open intervals (c, c þ r) or (c  r, c), as appropriate, for some r > 0. The Reciprocal Rule can also be applied with ‘x ! c’ replaced by ‘x ! cþ’ or ‘x ! c’, and with the punctured neighbourhood of c being replaced by open intervals (c, c þ r) or (c  r, c), as appropriate, for some r > 0. For example, 1x ! 1 as x ! 0þ, since f(x) ¼ x > 0, for x 2 ð0; 1Þ, and limþ x ¼ 0:

x!0

Problem 1 Prove the following: 1 (a) jxj ! 1 as x ! 0; (b) 1 1 x3 ! 1 as x ! 1þ ; (c)

sin x x3

! 1 as x ! 0:

Remark There are also versions of the Combination Rules and Squeeze Rule for functions which tend to 1 or 1 as x tends to c, cþ or c; here we state only the Combination Rules for functions which tend to 1 as x tends to c. Theorem 2 Combination Rules If f(x) ! 1 and g(x) ! 1 as x ! c, then: Sum Rule f(x) þ g(x) ! 1; Multiple Rule l f(x) ! 1, for l > 0; Product Rule f(x) g(x) ! 1.

These are similar to results stated for sequences in Theorems 3 and 4, Sub-section 2.4.3.

5: Limits and continuity

178 Problem 2 Prove that the following statements are false: (a) If f(x) ! 1 and g(x) ! 1 as x ! 0, then f(x)  g(x) ! 1; (b) If f(x) ! 1 and g(x) ! 0 as x ! 0, then f(x) g(x) ! 0 or f(x) g(x) ! 1.

5.2.2

Behaviour of f(x) as x tends to 1 or 1

Finally, we define various types of behaviour of real functions f (x) as x tends to 1 or 1. To avoid repetition, here we allow ‘ to denote either a real number or one of the symbols 1 and 1.

We adopt this convention for simplicity ONLY in this sub-section!

Definition Let f be defined on an interval (R, 1), for some real number R. Then f (x) ! ‘ as x ! 1 if: for each sequence {xn} in (R, 1) such that xn ! 1, then f (xn) ! ‘. The statement ‘f (x) ! ‘ as x ! 1’ is defined similarly, with 1 replaced by 1, and (R, 1) by (1, R). In practice, we usually prove that f (x) ! ‘ as x ! 1 by showing that f (x) ! ‘ as x ! 1. When ‘ is a real number, we also use the notation lim f ð xÞ ¼ ‘ and

lim f ð xÞ ¼ ‘:

x!1

x!1

Once again, we can use versions of the Combination Rules and Reciprocal Rule to obtain statements about the behaviour of given functions as x tends to 1 or 1. In the statements of these rules, we need only replace c by 1 or 1, and the punctured neighbourhood of c by (R,1) or (1, R), as appropriate. For example, for any positive integer n xn ! 1 as x ! 1 and lim xn ¼ 0: x!1

More generally, we have the following result for the behaviour of polynomials as x ! 1. If a0, a1, . . ., an1 are real numbers and

Theorem 3

n

pð xÞ ¼ x þ an1 x

n1

þ    þ a1 x þ a0 ;

x 2 R;

then pð xÞ ! 1 as x ! 1

and

We ask you to prove the first part of this result in Section 5.6. The second part then follows at once.

1 ! 0 as x ! 1: pð x Þ

There are also versions of the Squeeze Rule for functions as x tends to 1, which have some important applications. Theorem 4 Squeeze Rule Let the functions f, g and h be defined on some interval (R, 1). (a) If f, g and h satisfy the following two conditions: 1. g(x)  f(x)  h(x), for all x in (R, 1), and 2. lim gð xÞ ¼ lim hð xÞ ¼ ‘; x!1

x!1

then lim f ð xÞ ¼ ‘: x!1

We omit the proof of this theorem, which is straightforward.

5.2

Asymptotic behaviour of functions

179 y

(b) If f and g satisfy the following two conditions: 1. f(x)  g(x), for all x in (R, 1), and

l y = f (x)

2. g(x) ! 1 as x ! 1, then f (x) ! 1 as x ! 1.

y = g(x)

x2 xn xnþ1 þ  þ þ þ  2! n! ðn þ 1Þ! for x  0. Since x  0, all the terms on the right here are non-negative, and so ex ¼ 1 þ x þ

ex xn

xnþ1 ; ðn þ 1Þ!

(R, ∞)

R

An important application of this Squeeze Rule is to show that ex tends to 1 faster than any power of x, as x tends to 1. We prove this in the next example. ex Example 1 Prove that for each n ¼ 0, 1, 2, . . ., xn ! 1 as x ! 1. Solution We use the power series expansion

ex 

y = h(x)

y

x

y = f (x) y = g(x) R

(R, ∞)

x

See Section 3.4.

for x  0:

It follows that ex x ; for x > 0:  xn ðn þ 1Þ! x Since ðnþ1Þ! ! 1 as x ! 1, it follows from part (b) of the Squeeze Rule that & ! 1 as x ! 1.

Remark We deduce from Example 1 and the Product Rule that, for any integer n, xnex ! 1 as x ! 1. Thus, by the Reciprocal Rule, for any integer n, xnex ! 0 as x ! 1. Problem 3 Determine the behaviour of the following functions as x ! 1: 3 (a) f ð xÞ ¼ 2xx3þx ; (b) f ð xÞ ¼ sinx x. In our next example, we describe the behaviour of loge x as x ! 1. Example 2

Prove that loge x ! 1 as x ! 1.

Solution We prove this from first principles. First, note that the function f (x) ¼ loge x is defined on (0, 1). Next, we have to prove that: for each sequence {xn} in (0, 1) such that xn ! 1, then loge xn ! 1. To prove that loge xn ! 1, we need to show that: for each positive number K, there is a number X such that loge xn > K; for all n > X:

(1)

However, since we know that xn ! 1, we can choose X such that xn > eK ;

for all n > X:

Since the function loge is strictly increasing, the statement (1) is therefore & true. It follows that loge x ! 1 as x ! 1, as required.

There is no easy way to prove this result by using the Squeeze Rule, since loge x tends to 1 ‘rather reluctantly’; that is, more slowly than any function that we have considered so far.

5: Limits and continuity

180

5.2.3

Composing asymptotic behaviour

Earlier we gave a Composition Rule for limits. We now describe a more general Composition Rule which permits the composition of the different types of asymptotic behaviour that we have now met. For example, since 1 f ðxÞ ¼ ! 1 as x ! 0þ and gðxÞ ¼ ex ! 1 as x ! 1; x we should expect that 1

Sub-section 5.1.3, Theorem 4.

as x ! 0þ :

gð f ð x Þ Þ ¼ e x ! 1

This result is true, as the following Composition Rule shows. To avoid repetition, we again allow ‘ and L to denote either a real number or one of the symbols 1 and 1.

We again adopt this convention for simplicity ONLY in this sub-section!

Theorem 5 Composition Rule If: 1. f (x) ! ‘ as x ! c (or cþ, c, 1 or 1), and

We omit the proof of Theorem 5.

2. g (x) ! L as x ! ‘, then

If ‘ denotes 1 or 1, then the first proviso is automatically satisfied. Note that we must have conditions 1 and 2 and the first proviso satisfied or conditions 1 and 2 and the second proviso satisfied if we are to make any application of this result.

as x ! c ðor cþ ; c ; 1 or 1; respectivelyÞ;

gð f ð xÞÞ ! L provided that: EITHER

OR

f (x) ¼ 6 ‘ in some punctured neighbourhood of c (or in (c, c þ r), (c  r, c), (R, 1) or (1, R), respectively, for some r > 0) ‘ is finite, and g is defined at ‘ and is continuous at ‘.

Example 3 Prove that x 1 (b) xex ! 1 as x ! 0þ ; (a) ex2 ! 1 as x ! 1; 1 (c) x sin x ! 1 as x ! 1: Solution x

(a) Let f ð xÞ ¼ 2x , x 2 R, and gð xÞ ¼ ex , x 2 R  f0g, so that x

g ð f ð x ÞÞ ¼

e2 x 2

x

¼

2e2 , x

for x 2 R  f0g, and so in

particular for x 2 ð0; 1Þ: Now, by the Multiple Rule, f(x) ! 1 as x ! 1; and, by Example 1, g(x) ! 1 as x ! 1. It follows, by the Composition Rule, that x

2e2 gð f ð x Þ Þ ¼ ! 1 as x ! 1, x x so that; by the Multiple Rule; we have ex2 ! 1 as x ! 1: x

(b) Let f ð xÞ ¼ 1x , x 2 R  f0g, and gð xÞ ¼ ex , x 2 R  f0g, so that 1

g ð f ð x ÞÞ ¼

ex 1 x

1

¼ xex ,

for x 2 R  f0g, and so in

particular for x 2 ð0; 1Þ:

5.3

Limits of functions – using " and 

181

Now, f (x) ! 1, as x ! 0þ; and, by Example 1, g(x) ! 1 as x ! 1. It follows, by the Composition Rule, that 1

gð f ð xÞÞ ¼ xex ! 1

as x ! 0þ :

(c) Let f ð xÞ ¼ 1x , x 2 R  f0g; and gð xÞ ¼ sinx x ; x 2 R  f0g, so that    sin 1x 1 gð f ð xÞÞ ¼ 1 ¼ x sin ; for x 2 R  f0g, and so x x

In Theorem 5, we have ‘ ¼ 0, L ¼ 1.

for x 2 ð0; 1Þ: Now f ð xÞ ! 0 as x ! 1; gð xÞ ! 1 as x ! 0; and

This is condition 1.

f ð xÞ 6¼ 0; for x 2 ð0; 1Þ:

So the first proviso is satisfied.

It follows, by the Composition Rule, that   1 gð f ð xÞÞ ¼ x sin ! 1 as x ! 1: x

This is condition 2.

&

Problem 4 Prove that: 1 (b) xex ! 0 as x ! 0 . (a) loge (loge x) ! 1 as x ! 1; Hint for part (b): Use the fact that 1x ! 1 as x ! 0 . Problem 5 Give examples of functions f and g (and a specific value for each of ‘ and m) for which 1. f(x) ! ‘ as x ! 1 and 2. g(x) ! m as x ! ‘, but for which gð f ð xÞÞ6! m as x ! 1.

5.3

Limits of functions – using e and d

Earlier we gave definitions of the limit of a sequence, continuity of a function and limit of a function, and strategies for using these definitions. In each case, the strategy was in two parts: GUESS GUESS

that the definition HOLDS: prove that it holds for all cases. that the definition FAILS: find ONE counter-example.

Each definition is particularly convenient if we wish to prove that the definition FAILS, but it is not always easy to work with it when we wish to prove that the definition HOLDS. For example, when we wish to prove that a given function f is discontinuous at a point c, then we have to find ONE sequence in a punctured neighbourhood of c such that xn ! c

BUT

f ðxn Þ 6! f ðcÞ:

5: Limits and continuity

182 On the other hand, if we wish to use the definition to prove that f is continuous at c, then we have to show that: for each sequence {xn} in a punctured neighbourhood of c such that xn ! c; then f ðxn Þ ! f ðcÞ : As it is sometimes tricky to determine the behaviour of every such sequence {xn}, this definition can be very inconvenient to use. In this section we introduce a definition of the limit of a function that is equivalent to our earlier definition, but which does not use sequences.

5.3.1

In Section 5.4 we shall introduce a definition of continuity which does not use sequences.

The e – d definition of limit of a sequence

We motivate our discussion by returning to the definition of a convergent sequence. Definition The sequence {an} is convergent with limit ł, or converges to the limit ł, if {an  ‘} is a null sequence. In other words, if: for each positive number ", there is a number X such that jan  ‘j < "; for all n > X:

Sub-section 2.3.1.

(1)

{an} l+ε

|an – l| < ε ⇔ l – ε < an < l + ε

l l−ε

n

X

The condition (1) means that, from some point on, the terms of the sequence all lie in the shaded strip between ‘  " and ‘ þ ", and thus lie ‘close to’ ‘. If we choose a smaller number ", then the shaded strip in the diagram becomes narrower, and we may need to choose a larger number X in order to ensure that the inequality (1) still holds. But, whatever positive number " we choose, we can always find a number X for which (1) holds. n

Þ Problem 1 Let an ¼ ð1 n2 , n ¼ 1, 2, . . .. How large must we take X in order that: (a) |an  0| < 0.1, for all n > X?

(b) |an  0| < 0.01, for all n > X? (c) |an  0| < ", for all n > X, where " is a given positive number? Earlier we described a ‘game’, based on the above definition, in which player A chooses a positive number " and challenges player B to find a number X such that (1) holds. If the sequence in question converges, then such a number X always exists, and player B can always win. If the sequence does not converge, then, for SOME choices of ", NO number X exists such that (1) holds, and so player A can always win.

Sub-section 2.2.1.

5.3

Limits of functions – using " and 

183

For example, consider the null sequence an ¼ p1ffiffin , n ¼ 1, 2, . . .. In this case, the game might proceed as follows:

I'll play X = 100

I'll play ε = 0.1

B wins A

B

Here player B wins, since, for n > 100, we have 1 1 pffiffiffi  0 ¼ p1ffiffiffi < pffiffiffiffiffiffiffi ffi ¼ 0:1: n n 100

Here X ¼ 100.

Player A tries again:

I'll play ε = 0.01

I'll play X = 10 000

B wins A

B

Player B again wins, since, for n > 10 000, we have 1 1 pffiffiffi  0 ¼ p1ffiffiffi < pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:01: n n 10 000

Here X ¼ 10 000.

But now player B has figured out a winning strategy, and challenges player A to do his worst!

I'll play X = 1/ε 2

I'll play ANY positive

ε

B wins A

B

This is indeed a winning strategy, since 1 pffiffiffi 1 1 1 for n > 2 , we have n > , and so pffiffiffi  0 ¼ pffiffiffi < " : " " n n The reason for introducing " in the definition of convergent sequence is to formalise the idea of ‘closeness’. The statement (1) means that we can make the terms an of the sequence as close as we please to ‘ by choosing X large enough.

Here X ¼ "12 .

5: Limits and continuity

184 Put another way, we can think of a sequence {an} as a function with domain R n 7! ðn; an Þ: The number " formalises the idea of closeness of an to ‘ (that is, in the codomain). The condition ‘for all n > X’ restricts the values n in the domain to values for which condition (1) holds.

In general, the smaller the value of ", the larger the X that we have to choose.

{an} l+ε l

{

an close to l

|an – l| < ε for all n > X

l–ε

n

X for all n > X

5.3.2

The e – d definition of limit of a function

The concept of the limit of a function as x ! c also involves the idea of closeness. To define f ð xÞ ! ‘

as x ! c;

(2)

we require the statement j f ð xÞ  ‘j < "; this means that the values of f (x) and ‘ are within a distance " of each other. Earlier, we formalised the statement (2) by defining the limit of a function in terms of limits of sequences.

Sub-section 5.1.1.

Definition Let the function f be defined on a punctured neighbourhood N of a point c. Then f (x) tends to the limit ł as x tends to c if: for each sequence fxn g in N such that xn ! c, then f ðxn Þ ! ‘: (3)

Note that xn 6¼ c, for any n.

Intuitively, this means that, as xn gets close to c, so f (xn) gets close to ‘. More precisely, we can make f (xn) as close as we please to ‘ by choosing xn sufficiently close to c; we can ensure this by considering only xns for sufficiently large n. But the sequence {xn} can be any sequence of points in N that converges to c, so what we are really saying is that we can make f (x) as close as we please to ‘ by choosing x sufficiently close to c (but not equal to c). Thus we now need to specify not only closeness in the codomain j f ð xÞ  ‘j < "; but also closeness in the domain. To do this, we introduce another small positive number  (which depends on ") and specify closeness in the domain by a statement of the form 0 < jx  cj < :

Actually it is the requirement j x  c j <  that specifies the closeness in the domain. The requirement 0 < j x  c j is made simply to ensure that x¼ 6 c.

5.3

Limits of functions – using " and  y

185 y

y = f (x)

y = f (x)

{ f (xn)}

{

{

f (xn) close l to l

f (x) close l to l

{xn} c

c

x

x

x close to c

xn close to c

Problem 2 Let f(x) ¼ 2x þ 3, x 2 R. How small must we choose  in order that: (a) j f (x)  5j < 0.1, for 0 < jx  1j < ? (b) j f (x)  5j < 0.01, for 0 < jx  1j < ? (c) j f (x)  5j < ", number?

for 0 < jx  1j < , where " is a given positive

In general, the statement ‘f (x) ! ‘ as x ! c’ means the following: for EACH positive number ", there is a  such that j f ð xÞ  ‘j < ";

CORRESPONDING

positive number

for all x satisfying 0 < jx  cj < :

(4)

The following diagram illustrates how the numbers " and  are used to formalise the idea of closeness. y

y y = f (x)

y = f (x) l+ε

{

f (x) close l to l

l l–ε

y

c

The same number  may serve for various different values of "; but, in general, the value of  depends on the value of ".

Condition (4) means that, for all x with 0 < jx  cj < , the points (x, f (x)) of the graph lie in the heavily shaded rectangle, and thus lie ‘close to’ (c, ‘).

c–δ c c+δ x

x close to c

This leads us to the following definition of the limit of a function. Definition Let the function f be defined on a punctured neighbourhood N of a point c. Then f (x) tends to the limit ł as x tends to c if: for each positive number ", there is a positive number  such that j f ð xÞ  ‘j < ";

for all x satisfying 0 < jx  cj <  :

(5)

Thus f ðxÞ is ‘close to’ ‘ whenever x is ‘close to’ c:

5: Limits and continuity

186

Remarks 1. We assume that  is chosen sufficiently small in (5), so that the interval (c  , c þ ) lies within N. We also require that 0 < jx  cj in (5), since we are not concerned with the value of f at c, or even whether f is defined at c. 2. There are similar definitions of limits from the left and limits from the right at c, where we simply replace ‘0 < jx  cj < ’ by ‘c   < x < c’ and ‘c < x < c þ ’, respectively; and similar definitions of lim f ð xÞ and x!1

lim f ð xÞ.

For otherwise the values f (x) may be undefined. The existence and value of a limit is a local property, but the behaviour of the function AT the point in question is irrelevant.

x!1

We now formally state the equivalence of the two definitions of ‘limit of a function’. Theorem 1 The ‘"   definition’ and the ‘sequential definition’ of the statement lim f ð xÞ ¼ ‘ are equivalent. x!c

Proof Let the function f be defined on a punctured neighbourhood N of a point c. We have to show that: for each sequence {xn} in N such that xn ! c, then f ðxn Þ ! ‘

(6)

, for each positive number ", there is a positive number  such that (7) j f ð xÞ  ‘j < "; for all x satisfying 0 < jx  cj <  :

You may omit this proof at a first reading.

We shall assume, for convenience, that N ¼ ðc  r; cÞ [ ðc; c þ r Þ ¼ fx : 0 < jx  cj < r g:

First, let us assume that (6) holds. Then, for each sequence {xn} in N and each positive number ", there is a number X such that ( ) j f ðxn Þ  ‘j < "; for all n > X:

Here we simply restate (6).

We will now use a ‘proof by contradiction’. So, suppose that (7) does not hold. Then there must exist some positive number " for which there is no positive number  such that

Here we write down what is meant by ‘(7) does not hold’.

j f ð xÞ  ‘j < ";

for all x satisfying 0 < jx  cj < :

(8)

In particular, we can always assume that  < r, so that {x : 0 < jx  cj < } N. It follows that, for each positive integer n, there is some number xn (say) in N such that 1 (9) j f ðxn Þ  ‘j  "; where 0 < jxn  cj < : n But, from our assumption (6), the sequence {xn} must satisfy (*). Hence, in addition, in particular, for the positive number " that we have chosen for (*), there is a number X such that jf (xn)  ‘j < ", for all n > X. But this is inconsistent with (9), so that we have a contradiction. It follows, therefore, that (7) must hold after all! Next, let us assume that (7) holds. That is, for each positive number ", there is a positive number  such that j f ð xÞ  ‘j < ";

for all x satisfying 0 < jx  cj < :

(10)

Recall that N ¼ fx : 0 < jx  cj < r g: This is possible since (8) does not hold for any , and since we may take as successive choices of  the values 1; 12 ; 13 ; . . ..

For all n > X, f (xn) now has to satisfy the two inconsistent inequalities j f ðxn Þ  ‘j  " and j f ðxn Þ  ‘j < ": Here we simply restate (7).

5.3

Limits of functions – using " and 

187

Now let {xn} be any sequence in N such that xn ! c. In particular, for the positive number  that we have chosen for (10), there is a number X such that (0 < )jxn  cj <  for all n > X. It follows from (10) that jf(xn)  ‘j < " for all n > X. Since this holds for each positive number ", we deduce that f (xn) ! ‘, so that & (6) holds, as required.

Remark Depending on the particular circumstances, we can therefore use whichever definition of limit is the more convenient for our purposes. We can think of the choice of  in terms of " in the definition of limit of a function as a game with players A and B, just as we did with the choice of X in the definition of convergent sequence. Thus, player A chooses some value for ", and then player B has to try to find a value of  such that j f ð xÞ  ‘j < "; for all x satisfying 0 < jx  cj < : For example, consider how we might show that f (x) ¼ x2 ! 0 as x ! 0. Here, for the function f (x) ¼ x2, x 2 R, and c ¼ 0, player B has to find a number  such that j f ð xÞ  0j ¼ 4x2 < "; for all x satisfying 0 < j xj < : The game might proceed as follows: I'll play δ = 0.1

I'll play ε = 0.2

B wins A

B

Here player B wins, since

for 0 < j xj < 0:1; we have that 4x2  0:04 < 0:2:

So player A tries again: I'll play δ = 0.1

I'll play ε = 0.02

A wins A

B

Here player A wins, since x ¼ 0:09 satisfies 0 < j xj < 0:1

BUT

4  0:092 ¼ 0:0324 6< 0:02:

Player B lost because he did not choose  sufficiently small; so he tries again.

We have 0 < jxn  cj since xn lies in a punctured neighbourhood of c.

5: Limits and continuity

188

I'll play δ = 0.01

I'll play ε = 0.02

B wins A

B

This time player B wins, since

for 0 < j xj < 0:01; we have that 4x2  0:0004 < 0:02:

But now player B has figured out a winning strategy! He challenges player A to do his worst!

I'll play ANY positive

I'll play δ = 12√ε

ε

B wins A

B

This is indeed a winning strategy, since

  2 1 pffiffiffi 1 pffiffiffi 2 " ; we have that 4x < 4  " ¼ ": for 0 < j xj < 2 2

In general, if f (x) ! ‘ as x ! c, then such a number  always exists, and player B can always win. On the other hand, if f ðxÞ 6! ‘ as x ! c, then for SOME choices of " no corresponding positive number  exists, and so player A can always win. Our strategy for using the ‘"   definition’ of lim f ð xÞ in particular cases is x!c as follows. Strategy for using the ‘e  d definition’ of limit of a function 1. To show that f (x) ! ‘ as x ! c, find an expression for  in terms of " such that: for each positive number " j f ð xÞ  ‘j < ";

for all x satisfying 05jx  cj5: (11)

2. To show that f ðxÞ 6! ‘ as x ! c, find: positive number " for which there is NO positive number  such that

ONE

j f ð xÞ  ‘j < ";

for all x satisfying 0 < jx  cj < :

Whatever value of " is chosen by player A, playerpBffiffiffineed only specify  ¼ 12 ".

Of course the choice of  may depend on c too.

5.3

Limits of functions – using " and 

Example 1

189

Using the strategy, prove that

(a) limð2x þ 3Þ ¼ 5; x!1   (c) lim x sin 1x ¼ 0; x!0

(b) limð2x3 Þ ¼ 0; x!0   (d) lim sin 1x does not exist. x!0

Solution (a) The function f (x) ¼ 2x þ 3 is defined on every punctured neighbourhood of 1. We must prove that: for each positive number ", there is a positive number  such that jð2x þ 3Þ  5j < "; for all x satisfying 0 < jx  1j <  ; that is j2ðx  1Þj < ";

for all x satisfying 0 < jx  1j < :

(12)

We choose  ¼ 12 "; then the statement (12) holds, since

We could equally well choose  to be any positive number less than 12 ", since the statement (12) would still hold.

for 0 < jx  1j < 12 ", we have j2ðx  1Þj < ": Hence, limð2x þ 3Þ ¼ 5. x!1

(b) The function f (x) ¼ 2x3 is defined on every punctured neighbourhood of 0. We must prove that: for each positive number ", there is a positive number  such that 3 2x  0 < "; for all x satisfying 0 < jx  0j <  ; that is

3 2x < ";

for all x satisfying 0 < j xj <  : p ffiffi ffi We choose  ¼ 12 3 "; then the statement (13) holds, since: pffiffiffi for 0 < j xj < 12 3 ", we have that   3 pffiffiffi 3 2x ¼ 2j xj3 < 2  1 3 " ¼ 1 " < ": 2 4

(13) Notice that many other choices of  would also have served our purpose.

Hence, limð2x3 Þ ¼ 0. x!0

(c) The function f ð xÞ ¼ x sin 1x is not defined at 0, but it is defined on every punctured neighbourhood of 0. We must prove that: for each positive number ", there is a positive number  such that x sin 1  0 < "; for all x satisfying 0 < jx  0j <  ; x that is

x sin 1 < "; x

for all x satisfying 0 < j xj < :

Now, sin 1x  1, for all non-zero x, so that

(14)

5: Limits and continuity

190 x sin 1  j xj; x

for all non-zero x:

We therefore  choose   ¼ "; then the statement (14) holds. Hence, lim x sin 1x ¼ 0. x!0

(d) Suppose that the limit exists; call it ‘. Take " ¼ 12, say. Then there is some positive number , and points xn and yn of the form 1 1   ; where n 2 N; and yn ¼  xn ¼  1 2n þ 2 p 2n þ 32 p such that 0 < jxn j < 

and

0 < jyn j < 

and

1 j f ðxn Þ  ‘j < ; 2 1 j f ðyn Þ  ‘j < : 2

and :

Hence, for sufficiently large n j f ðxn Þ  f ðyn Þj ¼ jð f ðxn Þ  ‘Þ  ð f ðyn Þ  ‘Þj 1 1 þ ¼ 1: 2 2 But f (xn)  f (yn) ¼ 1  (1) ¼ 2, which contradicts the previous inequal  ity. It follows that no such numbers ‘ and  exist, so that in fact lim sin 1x x!0 & does not exist.  j f ðxn Þ  ‘j þ j f ðyn Þ  ‘j


X; where X ¼ ; n  but

y

1 0

For example, the points of the sequence {xn}, where xn ¼  1n, for all n  1, since then f (xn) ¼ 1. We make a suitable choice of " such that no corresponding  exists that satisfies the requirement (3).

  1  f ð0Þ j f ðxn Þ  f ð0Þj ¼ f  n

¼10 1 ¼ 1 6< ¼ ": 2 Thus, with this choice of ", no value of  exists such that j f ð xÞ  f ð0Þj < ";

This was requirement (3).

for all x satisfying jx  0j < :

This proves that f is discontinuous at 0.

&

Problem 3 Use the Strategy to prove that the following function is discontinuous at 2 8 < x  12 ; if x > 2; f ð xÞ ¼ 1; if x ¼ 2; : x  1; if x < 2: Finally, we demonstrate a nice application of the "   approach to continuity in order to obtain a result that will be useful later on.

x

5.4

Continuity – using " and 

Theorem 2 Let f be continuous at an interior point c of an interval I, and f ðcÞ 6¼ 0. Then there exists a neighbourhood N ¼ (c  r, c þ r) of c on which: (a) f(x) has the same sign as f(c), and

197

Here r > 0.

(b) j f ð xÞj > 12 j f ðcÞj. Proof Since f(c) 6¼ 0, we may take 12 j f ðcÞj as the positive number " in the definition of continuity at c. It follows that there exists some positive number r such that 1 (4) j f ð xÞ  f ðcÞj < j f ðcÞj; for all x satisfying jx  cj < r: 2 We now rewrite (4) in the following equivalent form 1 1  j f ðcÞj < f ð xÞ  f ðcÞ < j f ðcÞj; for all x 2 ðc  r; c þ r Þ: (5) 2 2

For convenience, here we use the symbol r rather than  in the definition of continuity – this makes no real difference.

Suppose, first, that f (c) > 0. It follows from the left inequality in (5) that, for x 2 (c  r, c þ r), we have 12 f ðcÞ < f ð xÞ  f ðcÞ – in other words, that 1 2 f ðcÞ < f ð xÞ. Thus, in this case, both (a) and (b) hold for x 2 (c  r, c þ r). Suppose, next, that f (c) < 0. It follows from the right inequality in (5) that, for x 2 (c  r, c þ r), we have f ð xÞ  f ðcÞ < 12 j f ðcÞj ¼ 12 f ðcÞ – in other words, that f ð xÞ < 12 f ðcÞ. Thus, in this case too, both (a) and (b) hold for x 2 (c  r, c þ r). & This completes the proof of the theorem.

5.4.2

The Dirichlet and Riemann functions

We can use our new "   approach to continuity to tackle two very strange functions that were devised in the second half of the nineteenth century. We start with the Dirichlet function. Definition

Dirichlet’s function is defined on R by the formula 1; if x is rational; f ð xÞ ¼ 0; if x is irrational:

y f (xn)

1

The graph of f looks rather like two parallel lines, but each line has ‘infinitely many gaps in it’.

f (yn) c

Theorem 3

The Dirichlet function is discontinuous at each point of R.

Proof Let c be any point of R.Then, for each  n = 1, 2, . . ., by the Density Property of R, the open interval c  1n , c þ 1n contains a rational number, xn say, and an irrational number, yn say. Then xn ! c and yn ! c as n ! 1, but f ðxn Þ ¼ 1 and f ðyn Þ ¼ 0: Thus f (x) does not tend to a limit as x tends to c, whether c is rational or irrational. & It follows that f is discontinuous at each point c of R.

xn

yn

c + 1n

x

Recall that the Density Property of R asserts that, between any two unequal real numbers, there exists at least one rational and one irrational number – Sub-section 1.1.4.

5: Limits and continuity

198 Our next function possesses even stranger behaviour. Definition

Riemann’s function is defined on R by the formula 1

f ð xÞ ¼

q; 0;

if x is a rational number if x is irrational:

p

q ; in

lowest terms; with q > 0;

The expression ‘in lowest terms’ means that p and q have no common factor.

It is not clear from the graph of f whether f is continuous at any point of R. Theorem 4 Riemann’s function is discontinuous at each rational point of R, and continuous at each irrational point of R. Proof First, let c ¼ pq be any rational point of R, where q > 0 and pq is in its lowest terms. Then, Property of R, for each n = 1, 2, . . ., the open inter by1 the Density  val c  n , c þ 1n contains an irrational number xn, for which f ðxn Þ ¼ 0 6¼ 1q ¼ f ðcÞ. Hence as n ! 1

In this part of the proof, we use a sequential approach. Note that xn 6¼ c.

xn ! c but f ðxn Þ 6! f ðcÞ: Thus f is discontinuous at any rational point c. Next, suppose that c is any irrational point of R. We must prove that: for each positive number ", there is a positive number  such that j f ð xÞ  f ðcÞj < "; for all x satisfying jx  cj < :

In this part of the proof, we use an "   approach.

(6)

Since c is irrational, f (c) ¼ 0; also f (x)  0 for all x. It follows that condition (6) becomes f ð xÞ < ";

for all x satisfying jx  cj < :

(7)

Our task is therefore to find a value of  such that (7) holds. Now, let N be any positive integer such that N > 1", and let SN be the set of rational numbers pq (in lowest terms) in the interval (c  1, c þ 1) for which 0 < q < N – that is

p p SN ¼ : c   1; 0 < q < N : q q Since there are only a finite number of points in SN and c 2 = SN, we can define a positive number  as follows

Thus, in particular, N1 < ". We choose (c  1, c þ 1) simply so that we are only looking at points ‘near to c’.

c2 = SN since c is irrational.

5.4

Continuity – using " and 

199

 ¼ minfjx  cj : x 2 SN g: Thus there are NO rational numbers (c  , c þ  ). It follows that, if jx  cj < , then: EITHER OR

p q,

with 0 < q < N, that lie in the interval

x is irrational, so that f(x) ( ¼ 0) < "; x is rational, so that x ¼ pq, with q  N, and f ð xÞð¼ 1qÞ  N1 < ".

In either case, f(x) < ". Hence we have proved that (7) holds, and so that f is continuous at c.

5.4.3

 is the distance from c to the nearest point of SN.

&

Proofs

Since the definition of continuity introduced in this section is equivalent to the earlier sequential definition, the rules for continuity can also be proved using the "   approach. To give you a flavour of what this involves, we first prove one of the Combination Rules. Example 3 (Sum Rule) Let f and g be defined on an open interval I containing the point c. Then, if f and g are continuous at c, so is the sum function f þ g. Solution The functions f, g and f þ g are certainly defined on some neighbourhood (c  r, c þ r) I of c, where r > 0. We want to prove that: for each positive number ", there is a positive number  such that jð f ð xÞ þ gð xÞÞ  ð f ðcÞ þ gðcÞÞj < "; for all x satisfying jx  cj < 

(8)

We know that, since f is continuous at c, there is a positive number 1 (which we may assume is  r) such that 1 (9) j f ð xÞ  f ðcÞj < "; for all x satisfying jx  cj < 1 ; 2 similarly, since g is continuous at c, there is a positive number 2 (which we may assume is also  r) such that 1 (10) jgð xÞ  gðcÞj < "; for all x satisfying jx  cj < 2 : 2 We now choose  ¼ min{1, 2}. Then both statements (9) and (10) hold for all x satisfying jx  cj < , so that jð f ð xÞ þ gð xÞÞ  ð f ðcÞ þ gðcÞÞj ¼ jð f ð xÞ  f ðcÞÞ þ ðgð xÞ  gðcÞÞj  j f ð xÞ  f ðcÞj þ jgð xÞ  gðcÞj 1 1 < " þ " ¼ "; 2 2 so that the statement (8) holds. Hence f þ g is continuous at c.

&

Problem 4 Prove that if f is defined on an open interval I containing the point c at which it is continuous, and f (c) 6¼ 0, then 1f is defined on some neighbourhood of c and is also continuous at c. Hint: Use the result of Theorem 2 in Sub-section 5.4.1.

You might like to compare this solution with that of Example 3 in Subsection 5.3.2.

5: Limits and continuity

200 Finally, we give a proof of the Composition Rule for continuity; this is a particularly pleasing example of the "   approach. Example 4 (Composition Rule) Prove that if f is continuous at c and g is continuous at f (c), then g  f is continuous at c. Solution

We must prove that:

for each positive number ", there is a positive number  such that jgð f ð xÞÞ  gð f ðcÞÞj < ";

for all x satisfying jx  cj < :

We do not bother to mention appropriate neighbourhoods of c and f (c) on which f and g are defined, respectively, just to simplify the statement of the Rule.

(11)

We know that, since g is continuous at f(c), there is a positive number 1 such that: jgð f ð xÞÞ  gð f ðcÞÞj < ";

for all f ðxÞ satisfying j f ð xÞ  f ðcÞj < 1 :

(12)

g

f (c) – δ1

f (c) f (x)

f (c) + δ1

g(f (c)) – ε

g( f (c)) g( f (x)) g( f (c)) + ε

Also, since f is continuous at c, there is a positive number  such that j f ð x Þ  f ð c Þ j < 1 ;

for all x satisfying jx  cj < :

(13)

f

c–δ

c

x c+δ

f (c) – δ1

f (c) f (x)

f (c) + δ1

Combining (12) and (13), we deduce that: for all x satisfying jx  cj < , then j f ð xÞ  f ðcÞj < 1 ,

By (13).

so that jgð f ð xÞÞ  gð f ðcÞÞj < ": Hence g  f is continuous at c.

5.5

By (12).

&

Uniform continuity

We start by reminding you of the "   definition of continuity at points of an interval I. Definition A function f defined on an interval I that contains a point c is continuous at c if: for each positive number ", there is a positive number  such that j f ð xÞ  f ðcÞj < ";

for all x in I satisfying jx  cj < :

f is said to be continuous on I if it is continuous at each point c of I.

Here the choice of  depends, in general, on both " and c.

5.5

Uniform continuity

201

Earlier we pointed out that, for a given value of the positive number ", we may need to choose different values of  for different points c. This suggests the following related definition. Definition on I if:

Sub-section 5.4.1, Remark 6.

A function f defined on an interval I is uniformly continuous

for each positive number ", there is a positive number  such that (1) j f ð xÞ  f ðcÞj < "; for all x and c in I satisfying jx  cj < :

Here the choice of  depends on ", the same choice whatever c may be.

ONLY

Since the definition of uniform continuity is more restrictive than that of continuity, it follows that uniform continuity implies continuity. Theorem 1 If a function f is uniformly continuous on an interval I, then it is continuous on I. However the converse result is not true in general. Example 1 Prove that the function f ð xÞ ¼ 1x , 0 < x  1; is not uniformly continuous on (0, 1]. Solution

Since f is a rational function, it is a basic continuous function, and so is continuous on (0, 1].

To prove that f is NOT uniformly continuous on (0,1], we need to

find ONE positive number " for which there is NO positive number  such that j f ð xÞ  f ðcÞj < "; for all x and c in ð0; 1 satisfyingjx  cj5:

(2)

We will take as our choice of " the number 1, and prove that there is corresponding positive number  such that j f ð xÞ  f ðcÞj < 1; for all x and c in ð0;1 satisfying jx  cj < :

For the statement (2) asserts that statement (1) is not valid.

NO

(3)

Suppose on the contrary that some number  did exist such that (3) was valid. We will show that this assumption leads us to a contradiction. Choose a positive integer N with 1 < : N Now each point of (0,1] belongs to at least one interval of the form           1 3 k kþ2 2N  2 2N 2N  1 0 2 ; ; ; ;1 ; ; ; . . .; ; . . .; ; ; 2N 2N 2N 2N 2N 2N 2N 2N 2N (4) and, in addition, all points x and c in any given such interval satisfy the inequality   2 1 ¼ < ; jx  cj < 2N N Hence, by (3) all points x and c in ð0; 1 in any interval in ð4Þ satisfy j f ð xÞ  f ðcÞj < 1: (5) Now, from the definition of f, we have   2N 2N  1 f ð xÞ < ¼ K; say; for all x in ;1 : 2N  1 2N

This is what we now set out to do.

For, successive intervals in (4) overlap.

In particular, f (x) < f (c) þ 1.

For, f is strictly decreasing  2N and f 2N1 . ¼ 2N1 2N

5: Limits and continuity

202     Then, since the intervals 2N2N1 ; 1 and 2N2N2 ; 22NN overlap, there is a point c in 2N 2 2N  f (c) < K. It follows from (5) that in the second last 2N ; 2N for which  2N 2 2N interval 2N ; 2N in (4) we have   2N  2 2N ; : f ð xÞ < K þ 1; for all x in 2N 2N Working backwards through the 2N intervals, we can check that   0 2 ; f ð xÞ < K þ ð2N  1Þ; for all x in : 2N 2N

For f is strictly decreasing.

It follows then, that in fact f(x) < K þ (2N  1), for all x in (0,1]. In other words, f is bounded in (0,1]. But this is not the case, since f (x) ! 1 as x ! 0þ. & We have thus reached the desired contradiction. Problem 1 Let f be the function f(x) ¼ x2, x 2 [2, 3]. For any given positive number ", find a formula for  in terms of " such that (1) holds. This proves directly from the definition that f is uniformly continuous on [2, 3]. Hint:

Use the fact that x2  c2 ¼ (x þ c)(x  c).

Notice, though, that the function in Problem 1 is a continuous function on a closed interval, and recall that we saw earlier that continuous functions on closed intervals have ‘nice properties’. It turns out that closed intervals are important too for uniform continuity. Theorem 2 If a function f is continuous on a closed interval [a, b], then it is uniformly continuous on [a , b].

Section 4.2. Example 1 shows that, in general, f may NOT be uniformly continuous if its interval of continuity is a halfclosed interval.

Proof of Theorem 2 We use the Bolzano–Weierstrass Theorem, and a proof by contradiction. So, suppose that, in fact, f is not uniformly continuous on [a, b]. Then for some choice of ", there is NO corresponding positive number  such that

You may omit this proof on a first reading.

j f ð xÞ  f ð yÞj < ";

(6)

We use y rather than c in (6), since it fits in better with the notation that we will use later in the proof.

(7)

The choice of x and y will depend, in general, on the choice of .

for all x and y in ½a, b satisfying jx  yj < :

We can reformulate this statement (6) in the following convenient way: for any positive number , there are two points x, y in [a, b] such that j f ð xÞ  f ð yÞj  "

and jx  yj < : 1 n, n

¼ 1, 2, . . ., in turn, we can then define By applying (7) for the choices n ¼ two sequences {xn} and {yn} in [a,b] such that 1 (8) j f ðxn Þ  f ðyn Þj  " and jxn  yn j < ; for n 2 N: n Now, the sequence {xn} is bounded, so that by the Bolzano–Weierstrass Theorem it contains a subsequence fxnk g that converges to some point c of [a, b] as k ! 1. Then, since jxnk  ynk j < n1k , it follows that the corresponding subsequence fynk g must also converge to c as k ! 1. Since f is continuous at c, we must have f ðxnk Þ ! f ðcÞ and f ðynk Þ ! f ðcÞ; as k ! 1: (9)

" is now fixed from here on in this proof.

Sub-section 2.5.1, Theorem 3. From (8).

5.6

Exercises

203

Now, it follows from (8) that j f ðxnk Þ  f ðynk Þj  ";

for each k 2 N:

(10)

So, if we now let k ! 1 in (10) and use the limits in (9), we deduce that 0  ";

Here we use the Limit Inequality Rule, Theorem 6 of Sub-section 5.1.3.

which is absurd. This contradiction shows that (7) and so (6) cannot be valid. It follows & that f must be uniformly continuous on [a, b] after all. We shall use Theorem 2 later, to prove that a function continuous on a closed interval [a, b] is integrable on [a, b]. Theorem 2 is one of the important theorems in Analysis beyond the scope of this book.

5.6

Exercises

Section 5.1 1. Determine whether the following limits exist: 3 3 1 1 (a) lim xx1 ; (b) lim jxx1 (c) lim esin x . j; x!1

x!1

x!0

2. Determine limits:  the following  x (a) lim sin x þ e x1 ; (b) lim

ex 1 ; x!0 sin x

x!0

(c) lim

x!0

ejxj 1 j xj ;

3

1 (d) limþ jxx1 j. x!1

3. Write out the proof of section 5.1.3. 4. For the functions 8 < 0; f ð xÞ ¼ 1; :

2;

Theorem 4 (the Composition Rule) in Sub-

x ¼ 0; x ¼ 1; and x 6¼0; 1;

gð xÞ ¼

0; 1 þ j xj;

x ¼ 0; x 6¼ 0;

determine f ðgð0ÞÞ; f lim gð xÞ and lim f ðgðxÞÞ. x!0

x!0

Section 5.2 1. Prove that (a) x14 ! 1 as x ! 0; (c) expðex  xÞ ! 1 as x ! 1; (e) x þ sin x ! 1 as x ! 1;

(b) cot x ! 1 as x ! 0þ ; (d) loge x ! 1 as x ! 0þ ; (f) xx ! 1 as x ! 1.

2. Let pð xÞ ¼ xn þ an1 xn1 þ    þ a1 x þ a0 , where a0 ; a1 ; . . .; an1 2 R. Prove that pð xÞ ! 1 as x ! 1.    Hint: Write pð xÞ ¼ xn 1 þ r 1x , for a suitable polynomial r. 3. For a > 0 and n 2 Z, prove that: (b) eax xn ! 0 as x ! 1. (a) eax xn ! 1 as x ! 1;

In Theorem 3, Subsection 7.2.2.

5: Limits and continuity

204 4. For a < 0 and n 2 Z, prove that: (a) eax xn ! 0 as x ! 1; 1; as x ! 1; (b) eax xn ! 1; as x ! 1;

if n is odd; if n is even:

Section 5.3 1. Use the strategy based on the "   definition of limit to prove the following statements: pffiffiffiffiffiffiffiffiffiffiffiffiffi (a) limð3x  4Þ ¼ 5; (b) limþ x2  4 ¼ 0. x!3

x!2

2. Use the strategy based on the "   definition of limit to prove the following statements:  x (a) lim sin (b) lim cos 1x does not exist. j xj does not exist; x!0

x!0

3. Use the strategy based on the "   definition of limit to prove that, if lim f ð xÞ ¼ ‘, where ‘ 6¼ 0, then lim f ð1xÞ exists and equals 1‘ . x!c

x!c

Section 5.4 1. Use the strategy based on the "   definition of continuity to prove that: (a) f ð xÞ ¼ x5 is continuous at 0; (b) f ð xÞ ¼ 1x is continuous at 2. 2. Use the strategy based on the "   definition of continuity to prove that the function sin 1x ; if x 6¼ 0; f ð xÞ ¼ 0; if x ¼ 0; is discontinuous at 0. 3. Use the "   definition of continuity to prove that, if f and g are continuous at c, then so is the product fg.

Section 5.5 1. A function f is defined on a bounded interval I, on which it satisfies the following inequality for some given number K 2 R: j f ð xÞ  f ð yÞj  K jx  yj; for all x; y in I: Prove that f is (a) bounded on I, and (b) uniformly continuous on I. 2. Give an example of a function f that is continuous and bounded on R but is not uniformly continuous on R, and verify your assertions. 3. Prove that, if a function f is continuous on a half-closed interval (a, b], then it is uniformly continuous on (a, b] if and only if limþ f ð xÞ exists (and is x!a finite).

6

Differentiation

The family of all functions is so large that there is really no possibility of finding many interesting properties that they all possess. In the last two chapters we concentrated our attention on the class of all continuous functions, and we found that continuous functions share some important properties – for example, they satisfy the Intermediate Value Theorem, the Extreme Values Theorem and the Boundedness Theorem. However, many of the most interesting and powerful properties of functions are obtained only when we further restrict our attention to the class of all differentiable functions. You will have already met the idea of differentiating a given function f; that is, finding the slope of the tangent to the graph y ¼ f(x) at those points of the graph where a tangent exists. The slope of the tangent at the point (c, f(c)) is called the derivative of f at c, and is written as f 0 (c). But when does a function have a derivative? Geometrically, the answer is: whenever the slope of the chord through the point (c, f(c)) and an arbitrary point (x, f(x)) of the graph approaches a limit as x ! c. In this chapter we investigate which functions are differentiable, and we discuss some of the important properties that all differentiable functions possess. In Section 6.1 we give a strategy for determining whether a given function f is differentiable at a given point c. In particular, we prove that, if f is differentiable at c, then it is continuous at c; whereas a function f may be continuous at c, but not differentiable at c. We also consider functions that possess higher derivatives; that is, functions which can be differentiated more than once. In Section 6.2 we obtain various standard derivatives and rules for differentiation, including the Inverse Function Rule. In Sections 6.3 and 6.4 we study the properties of functions that are differentiable on an interval, and establish some powerful results about derivatives which are easy to describe geometrically. In Section 6.5 we meet an important and useful result, called l’Hoˆpital’s Rule, which enables us to find limits of the form

In fact there exist functions that are continuous everywhere on R, but differentiable nowhere on R. This discovery by Karl Weierstrass in 1872 caused a sensation in Analysis.

f ð xÞ ; x!c gð xÞ

lim

in some of the awkward cases when f (c) ¼ g(c) ¼ 0. In Section 6.6 we construct the Blancmange function, a function that is continuous everywhere on R, but differentiable nowhere on R. It is related to certain types of fractals.

In this case, the Quotient Rule for limits of functions fails.

205

6: Differentiation

206

6.1 6.1.1

Differentiable functions What is differentiability?

Differentiability arises from the geometric concept of the tangent to a graph. The tangent to the graph y ¼ f (x) at the point (c, f (c)) is the straight line through the point (c, f (c)) whose direction is the limiting direction of chords joining the points (c, f (c)) and (x, f (x)) on the graph, as x ! c. The following three examples illustrate some of the possibilities that can occur when we try to find tangents in particular instances.

The function f (x) ¼ x2, x 2 R, is continuous on R, and its graph has a tangent at each point; for example, the line y ¼ 2x  1 is the tangent to the graph at the point (1, 1). On the other hand, although the function g(x) ¼ jx  1j, x 2 R, is continuous on R, its graph does not have a tangent at the point (1, 0); no line through the point (1, 0) is a tangent to the graph. Finally, the signum function ( 1; x < 0, x ¼ 0, kð xÞ ¼ 0; 1; x > 0, is discontinuous at 0, and no line through the point (0, 0) is a tangent to the graph. We now make these ideas more precise, using the concept of limit to pin down what we meant above by ‘limiting direction’. We define the slope of the graph at (c, f (c)) to be the limit, as x tends to c, of the slope of the chord through the points (c, f (c)) and (x, f (x)). The slope of this chord is f ð xÞ  f ðcÞ ; xc and this expression is called the difference quotient for f at c. Thus the slope of the graph of f at (c, f (c)) is f ð xÞ  f ðcÞ ; (1) x!c xc provided that this limit exists. Sometimes it is more convenient to use an equivalent form of the difference quotient, particularly when we are examining a specific function for lim

6.1

Differentiable functions

207

differentiability at a specific point. If we replace x by c þ h, then ‘x ! c’ in (1) is equivalent to ‘h ! 0’. Thus we can rewrite the difference quotient for f at c as f ð c þ hÞ  f ð c Þ ; QðhÞ ¼ h and the slope of the graph of f at (c, f (c)) is then lim QðhÞ; (2) h!0

provided that this limit exists. We say that f is differentiable at c if the graph y ¼ f (x) has a tangent at the point (c, f (c)), and that the derivative of f at c is the limit of the difference quotient given by expression (1) or expression (2). To formalise this concept, we need to ensure that f is defined near the point c, and so we assume that c belongs to some open interval I in the domain of f. Definitions Let f be defined on an open interval I, and c 2 I. Then the derivative of f at c is f ð xÞ  f ðcÞ lim or lim QðhÞ; x!c h!0 xc provided that this limit exists. In this case, we say that f is differentiable at c. The derivative of f at c is denoted by f 0 (c), and the function f 0: x 7! f 0 ð xÞ is called the derived function. The operation of obtaining f 0 (x) from f (x) is called differentiation.

Note that the difference quotient depends on the choice of c; for difference choices of c, the difference quotient is generally different. Sometimes f 0 is denoted by Df and f 0 (x) by Df (x).

For future reference, we reformulate the definition in terms of " and  as follows. Definition Let f be defined on an open interval I, and c 2 I. Then f is differentiable at c with derivative f 0 (c) if: for each positive number ", there is a positive number  such that   f ðxÞf ðcÞ   xc  f 0 ðcÞ < ", for all x satisfying 0 < jx  cj < .

Remarks 1. In ‘Leibniz notation’, f 0 (x) is written as dy dx, where y ¼ f (x). As we shall see later, this notation is sometimes useful; however, it is important to recall that the symbol dy dx is purely notation and does NOT mean some quantity dy ‘divided by’ another quantity dx. 2. The existence of the derivative f 0 (c) is not quite equivalent to the existence Þf ðcÞ of a tangent to the graph y ¼ f (x) at the point (c, f (c)). For, if lim f ðxxc x!c

exists (and so is some real number), then the graph has a tangent at the point (c, f (c)), and the slope of the tangent is the value of this limit. On the other hand, the graph may have a vertical tangent at the point (c, f (c)). In this Þf ðcÞ Þf ðcÞ ! 1 or 1 as x ! c, so that lim f ðxxc does not exist; thus case, f ðxxc x!c f is not differentiable at c. 3. Since the concept of a derivative is defined in terms of a limit, we shall use many results obtained for limits of functions to prove analogous results for derivatives.

This is a simple reformulation, with nothing else taking place. We shall use the definition occasionally in this form, especially when proving general results as distinct from looking at specific functions and specific points.

For instance, in the statement of the Composition Rule in Sub-section 6.2.2.

6: Differentiation

208 Example 1 Prove that the function f (x) ¼ x3, x 2 R, is differentiable at any point c, and determine f 0 (c). Solution

At the point c QðhÞ ¼

f ðc þ hÞ  f ðcÞ ðc þ hÞ3  c3 ¼ h h 3c2 h þ 3ch2 þ h3 ¼ h ¼ 3c2 þ 3ch þ h2 :

Recall that h 6¼ 0.

Thus, Q(h) ! 3c2 as h ! 0. It follows that f is differentiable at c, and that & f (c) ¼ 3c2. 0

Thus, the derived function of f is f 0 (x) ¼ 3x2, x 2 R.

For comparison, we now prove the same result using the "   definition of differentiability; for simplicity, though, we shall assume that c ¼ 2 and we shall simply check that f 0 (2) ¼ 12. We think that you will see why the Q(h) method is often preferred! Solution We have to show that for each positive number ", there is a positive number  such that  3  x  8    < "; for all x satisfying 0 < jx  2j < ;  12 x2  that is

 2  x þ 2x  8 < ";

for all x satisfying 0 < jx  2j < :

Now, x2 þ 2x  8 ¼ (x þ 4)(x  2); so, for 0 < jx  2j < 1, we have x 2 (1, 3) and jx þ 4j < 7.   Next, choose  ¼ min 1; 17 " . With this choice of , it follows that, for all x satisfying 0 < jx  2j < , we have  2  x þ 2x  8 ¼ jx þ 4j  jx  2j 1 < 7  " ¼ ": 7 It follows that f is differentiable at 2, with derivative f 0 (2) ¼ 12.

&

To prove that a function is not differentiable at a point, we use the strategy for limits that applies in these situations. Strategy

Here we use the fact that x3  8 ¼ ðx  2Þðx2 þ 2x þ 4Þ:

To show that lim QðhÞ does not exist:

We arrange for x 2 (1, 3), in order to concentrate simply on values of x near to 2. The bound 7 for jx þ 4j is simply used in order to keep some given bound (not necessarily a small bound) for this term. It is now clear why we made that particular choice for ; it arose from trying various values and then adjusting the choice appropriately in order to end up with a neat ‘"’. Alternatively, we could have used a different value for  in terms of ", and the K " Lemma.

h!0

1. Show that there is no punctured neighbourhood of c on which Q is defined; OR

  2.  Find non-zero null sequences {hn} and h0n such that {Q(hn)} and  two   Q h0n have different limits; OR

3. Find a null sequence {hn} such that Q(hn) ! 1 or Q(hn) ! 1. Example 2 Prove that the modulus function f(x) ¼ jxj, x 2 R, is not differentiable at 0.

For convenience, we rephrase this strategy in terms of lim QðhÞ rather than lim f ð xÞ.

h!0

x!c

6.1

Differentiable functions

209

Solution The graph suggests that the slopes of chords joining the origin to points on the graph of f to the left and to the right do not tend to the same limit as these points approach the origin. This suggests that we investigate whether case 2 of the above strategy may be useful. At the point 0 f ð0 þ hÞ  f ð0Þ jhj  j0j ¼ h h jhj ¼ : h  0 Now let {hn} and hn be two null sequences, where hn ¼ 1n and h0n ¼  1n. Then   1 1 Q ð hn Þ ¼ Q ¼ n1 n n

This is motivation for the approach to the solution rather than part of the solution itself. This is the start of the solution.

Q ð hÞ ¼

y y = ⏐x⏐ –1 n

0

1 n

x

1

¼ n1 ! 1

as n ! 1;

n

and    1  0  1 Q hn ¼ Q  ¼ n1 n n ¼

1 n

1n

! 1

as n ! 1:

It follows that f is not differentiable at 0.

&

Problem 1 (a) Prove that the function f (x) ¼ xn, x 2 R, n 2 N, is differentiable at any point c, and determine f 0 (c). (b) Prove that the constant function on R is differentiable at any point c, with derivative zero.

In particular, the derivative of the identity function f (x) ¼ x, x 2 R, is the constant function f 0 (x) ¼ 1.

Problem 2 Prove that the function f ð xÞ ¼ 1x ; x 2 R  f0g, is differentiable at any point c 6¼ 0, and determine f 0 (c). Problem 3 Use the "   definition of differentiability to prove that the function f (x) ¼ x4 is differentiable at 1, with derivative f 0 (1) ¼ 4. Problem 4 Prove that the following functions f are not differentiable at the given point c: 1

(a) f ð xÞ ¼ j xj2 ; x 2 R, c ¼ 0;

(b) f(x) ¼ [x], x 2 R, c ¼ 1.

Looking back at Example 2, it looks as though chords joining the origin to points (h, f (h)) have slopes that tend to a limit 1 if h ! 0þ, whereas they have slopes that tend to a limit 1 if h ! 0. This suggests the concept of one-sided derivatives that will be useful in our work later on.

Here [x] denotes the integer part of x.

6: Differentiation

210

y

slope fL′(c)

y

y = f (x)

y = f (x)

c

x

slope fR′(c) c

x

Definitions Let f be defined on an interval I, and c 2 I. Then the left derivative of f at c is f ð xÞ  f ðcÞ or fL0 ðcÞ ¼ lim QðhÞ; x!c h!0 xc provided that this limit exists. In this case, we say that f is differentiable on the left at c. Similarly, the right derivative of f at c is f ð x Þ  f ðc Þ or fR0 ðcÞ ¼ limþ QðhÞ; fR0 ðcÞ ¼ limþ x!c h!0 xc provided that this limit exists. In this case, we say that f is differentiable on the right at c. A function f whose domain contains an interval I is differentiable on I if it is differentiable at each interior point of I, differentiable on the right at the left end-point of I (if this belongs to I ) and differentiable on the left at the right end-point of I (if this belongs to I ). fL0 ðcÞ ¼ lim

All these definitions are analogous to definitions for continuity that you met in Sub-section 4.1.1.

With this notation, the function f in Example 2 is differentiable on the left at 0 and fL0 ð0Þ ¼ 1; it is also differentiable on the right at 0, and fR0 ð0Þ ¼ 1. The connection between the definitions of differentiability and of one-sided differentiability is rather obvious. Theorem 1 A function f whose domain contains an interval I that contains c as an interior point is differentiable at c if and only if f is both differentiable on the left at c and differentiable on the right at c AND fL0 ðcÞ ¼ fR0 ðcÞ: Example 3

We omit a proof of this straight-forward result. The common value is simply f 0 (c).

Determine whether the function x þ x2 ; 1  x < 0, f ð xÞ ¼ sin x; 0  x  2p, y

is differentiable at the points c ¼ 1, 0 and 2p, and determine the corresponding derivatives when they exist. Solution We investigate the behaviour of the difference quotient Q(h) at each point in turn. At 1, the function is not defined to the left of 1; and, for 0 < h < 1, we have QðhÞ ¼

f ð1 þ hÞ  f ð1Þ h

y = f (x) –1

π

2π x

6.1

Differentiable functions

211

n o n o ð1 þ hÞ þ ð1 þ hÞ2  ð1Þ þ ð1Þ2 ¼

h h þ h2 ¼ h ¼ 1 þ h ! 1 as h ! 0þ : It follows that f is differentiable on the right at 1, and fR0 ð1Þ ¼ 1. At 0, the function is defined on either side of 0, but by a different formula; we therefore examine each side separately. At 0, for 0 < h < 2p, we have f ðhÞ  f ð0Þ sin h  sin 0 ¼ Q ð hÞ ¼ h h sin h ! 1 as h ! 0þ : ¼ h Also at 0, for 1 < h < 0, we have   h þ h2  fsin 0g f ð hÞ  f ð 0Þ Q ð hÞ ¼ ¼ h h h þ h2 ¼ h ¼ 1 þ h ! 1 as h ! 0 : It follows that f is differentiable on the right at 0 and fR0 ð0Þ ¼ 1, and that f is differentiable on the left at 0 and fL0 ð0Þ ¼ 1. Since the left and right derivatives at 0 are equal, it follows that f is differentiable at 0 and f 0 ð0Þ ¼ 1. Finally, at 2p, the function is not defined to the right of 2p; and, for 2p < h < 0, we have Q ð hÞ ¼

f ð2p þ hÞ  f ð2pÞ sinð2p þ hÞ  sin 2p ¼ h h sin h ! 1 as h ! 0 : ¼ h

It follows that f is differentiable on the left at 2p, and fL0 ð2pÞ ¼ 1. Problem 5

&

Determine whether the function 8 x2 ; 2  x > < 4 x ; 0  x < 1, f ð xÞ ¼ 3 > x ; 1  x  2, > : 0; x > 2,

is differentiable at the points c ¼ 2, 0, 1 and 2, and determine the corresponding derivatives when they exist. Hint:

Sketch the graph y ¼ f(x) first.

Remarks Just as with continuity, the definition of differentiability involves a function f defined on a set in R, the domain A (say), that maps A to another set in R, the codomain B (say).

These remarks are analogous to similar remarks for continuity in Sub-section 4.1.1.

6: Differentiation

212 1. Let f and g be functions defined on open intervals I and J, respectively, where I  J; and let f (x) ¼ g(x) on J. Technically g is a different function from f. However, if f is differentiable at an interior point c of J, it is a simple matter of some definition checking to verify that g too is differentiable at c. Similarly, if f is non-differentiable at c, g too is nondifferentiable at c. 2. The underlying point here is that differentiability at a point is a local property. It is only the behaviour of the function near that point that determines whether it is differentiable at the point.

6.1.2

Differentiability and continuity

All the functions that we have met so far that are differentiable at a particular point c are also continuous at that point. In fact, this property holds in general.

differentiable ) continuous

Theorem 2 Let f be defined on an open interval I, and c 2 I. If f is differentiable at c, then f is also continuous at c.

There is a similar result for one-sided derivatives.

Proof

If f is differentiable at c, then there is some number f 0 (c) such that

fð xÞ  fðcÞ ¼ f 0 ðcÞ: x!c xc It follows that lim



f ð xÞ  f ðcÞ  ðx  cÞ limf f ð xÞ  f ðcÞg ¼ lim x!c x!c xc

n o f ð xÞ  f ðcÞ ¼ lim  limðx  cÞ x!c x!c xc

By the Product Rule for Limits, Sub-section 5.1.3.

¼ f 0 ðcÞ  0 ¼ 0: Hence, by the Sum and Multiple Rules for limits lim f ð xÞ ¼ f ðcÞ:

Sub-section 5.1.3

x!c

Thus f is continuous at c.

&

In fact this gives us a useful test for non-differentiability. Corollary 1

If f is discontinuous at c, then f is not differentiable at c.

For example, the signum function ( 1; x < 0, x ¼ 0, f ð xÞ ¼ 0; 1; x > 0, is discontinuous at c, and so cannot be differentiable at c. It is often worth checking whether a function is even continuous at a point before setting out on a complicated investigation of difference quotients to determine whether it is differentiable at that point.

Problem 5, Sub-section 4.1.1.

6.1

Differentiable functions Problem 6

213

Prove that the function sin 1x ; x 6¼ 0, f ð xÞ ¼ 0; x ¼ 0,

is not differentiable at 0. Example 4

Show that the function x sin 1x ; x 6¼ 0, f ð xÞ ¼ 0; x ¼ 0,

is continuous at 0, but not differentiable at 0. Solution At 0

We proved earlier that f is continuous at 0.

Sub-section 4.1.2, Problem 8 (a).

 f ð0 þ hÞ  f ð0Þ h sin 1h  0 ¼ Q ð hÞ ¼ h  h 1 ¼ sin : h o 1   n and x0n ¼ 2nþ1 1 p ; Thus, if we define two null sequences fxn g ¼ np ð 2Þ n 2 N, we have       0 1 1 Qðxn Þ ¼ sinðnpÞ ¼ 0 and Q xn ¼ sin 2n þ p ¼ sin p ¼ 1; 2 2   0  so that {Q (xn)} and Q xn tend to different limits as n ! 1. It follows that & f cannot be differentiable at 0. Problem 7

y y = x sin x

y

Prove that the function x2 sin 1x ; x 6¼ 0, f ð xÞ ¼ 0; x ¼ 0,

y = x2 sin

1 x

x

is differentiable at 0. Given that, for x 6¼ 0; f 0 ð xÞ ¼ 2x sin 1x  cos 1x, is f 0 continuous at 0?

6.1.3

1 x

The sine, cosine and exponential functions

In order to study the differentiability of each of the sine, cosine and exponential functions, we need to use the following three standard limits. Theorem 3

Three standard limits

(a) lim sinx x ¼ 1; x!0

x (b) lim 1cos ¼ 0; x x!0

x

(c) lim e x1 ¼ 1: x!0

Proof The limits (a) and (c) have been verified earlier, so we have only to verify (b) now. Using the half-angle formula for cosine, we have   2 sin2 12 x 1  cos x ¼ lim lim x!0 x!0 x x

(a) was proved in Subsection 5.1.1; (c) in Sub-section 5.1.4, Problem 8. For cos x ¼ 1  2 sin2

1  2x .

6: Differentiation

214     ! sin 12 x 1 ¼ lim  sin x 1 x!0 2 x 2 1 !    sin 2 x 1 ¼ lim  lim sin x 1 x!0 x!0 2 2x ¼ 1  0 ¼ 0:

By the Product Rule for limits.

&

Since sine is continuous at 0.

With this tool, we can now tackle the various functions. Theorem 4

The function f (x) ¼ sin x is differentiable on R, and f 0 (x) ¼ cos x.

Let c be any point in R. At c, the difference quotient for f is

Proof

sinðc þ hÞ  sin c h sin c cos h þ cos c sin h  sin c ¼ h sin h 1  cos h  sin c  ; ¼ cos c  h h

QðhÞ ¼

so that sin h 1  cos h  sin c  lim h!0 h h ¼ cos c  1  sin c  0

lim QðhÞ ¼ cos c  lim

h!0

h!0

¼ cos c: It follows that f is differentiable at c, and f 0 (c) ¼ cos c.

&

Problem 8 Prove that the function f (x) ¼ cos x is differentiable on R, and f 0 (x) ¼ sin x. Finally we find the derivative of the exponential function. This is an extremely important result, as the function f (x) ¼ lex, l an arbitrary constant, is the only function f that satisfies the differential equation f 0 (x) ¼ f (x) on R. Theorem 5 Proof

The function f (x) ¼ ex is differentiable on R, and f 0 (x) ¼ ex.

Let c be any point in R. At c, the difference quotient for f is ecþh  ec QðhÞ ¼ h eh  1 c ; ¼e  h

so that eh  1 h!0 h c ¼e 1 ¼ ec :

lim QðhÞ ¼ ec  lim

h!0

It follows that f is differentiable at c, and f 0 (c) ¼ ec.

&

That is, f is its own derived function.

6.1

Differentiable functions

6.1.4

215

Higher-order derivatives

In the previous sub-section, we saw that sin0 ¼ cos and cos0 ¼ sin. An exactly similar argument shows that (sin)0 ¼ cos and (cos)0 ¼ sin. Thus sin and cos are all differentiable on R, and so clearly we can differentiate the functions sin and cos as many times as we please on R. In general, when we differentiate any given differentiable function f, we obtain a new function f 0 (whose domain may be smaller than that of f ). The notion of differentiability can then be applied to the function f 0 , just as before, yielding another function f 00 ¼ ( f 0 )0 , whose domain consists of those points where f 0 is differentiable. Definitions Let f be differentiable on an open interval I, and c 2 I. If f 0 is differentiable at c, then f is called twice differentiable at c, and the number f 00 (c) is called the second derivative of f at c. The function f 00 is called the second derived function of f. Provided that the derivatives exist, we can define f 000 or f (3), f (4), . . ., (n) f , . . .. The functions f 00 , f (3), f (4), . . ., f (n), . . . are called the higher-order derived functions of f, whose values f 00 (x), f (3)(x), f (4)(x), . . ., f (n)(x), . . . are called the higher-order derivatives of f.

f 00 ¼ (f 0 )0 is sometimes written as f (2).

However, not all derived functions are themselves differentiable. You have already seen, for example, that the function x2 sin 1x ; x 6¼ 0, f ð xÞ ¼ 0; x ¼ 0, is differentiable at 0. In fact, it has as its derived function 2x sin 1x  cos 1x ; x 6¼ 0, 0 f ð xÞ ¼ 0; x ¼ 0, which is not even continuous at 0. Example 5

Prove that the function 1 2  x ; x < 0, f ð xÞ ¼ 1 22 x  0, 2x ;

is differentiable on R, but that its derived function is not differentiable at 0. For c > 0, the difference quotient for f at c is 1 ðc þ hÞ2  12 c2 Q ð hÞ ¼ 2 h 1 ¼ ð2c þ hÞ ! c as h ! 0; 2 so that f is differentiable at c and f 0 (c) ¼ c. When c ¼ 0 and h > 0, a similar argument shows that f is differentiable on the right at 0 and fR0 ð0Þ ¼ 0. For c < 0, the difference quotient for f at c is

Solution

1 ðc þ hÞ2 þ 12 c2 Q ð hÞ ¼ 2 h 1 ¼  ð2c þ hÞ ! c 2

as h ! 0;

We assume that h is sufficiently small that jhj < c, so that c þ h > 0; and hence that we are using the correct value for f at c þ h. We assume that h is sufficiently small that jhj < c, so that c þ h < 0; and hence that we are using the correct value for f at c þ h.

6: Differentiation

216 so that f is differentiable at c and f 0 (c) ¼ c. When c ¼ 0 and h < 0, a similar argument shows that f is differentiable on the left at 0 and fL0 ð0Þ ¼ 0. Since f has fL0 ð0Þ ¼ fR0 ð0Þ ¼ 0, it follows that f is differentiable at 0 and f (0) ¼ 0. Thus the derived function for f is given by f 0 ð xÞ ¼ j xj;

x 2 R:

This is the modulus function, which as we showed earlier is not differentiable & at 0. It follows that f 0 is not differentiable at 0. Problem 9

Example 2, Sub-section 6.1.1.

Prove that the function 2 x ; x < 0; f ð xÞ ¼ x3 ; x  0;

is differentiable on R. Is f 0 differentiable at 0?

6.2

Rules for differentiation

In the last section we showed that the functions sin, cos and exp are differentiable on R, by appealing directly to the definition of differentiability. However, it would be very tedious if, every time that we wished to prove that a given function is differentiable and to determine its derived function, we had to use the difference quotient method described in Section 6.1. Sometimes we do need to use that method, but usually we can avoid the algebra involved in the difference quotient method by using various rules for differentiation. In this section, we introduce the Combination Rules, the Composition Rule and the Inverse Function Rule for differentiable functions. These are similar to the rules for continuous functions that you met earlier. Note that each of the rules for differentiation supplies two pieces of information:

In Sections 4.1 and 4.3.

1. a function of a certain type is differentiable; 2. an expression for the derivative.

6.2.1

The Combination Rules

The Combination Rules for differentiable functions are a consequence of the Combination Rules for limits. Theorem 1 Combination Rules Let f and g be defined on an open interval I, and c 2 I. Then, if f and g are differentiable at c, so are: ( f þ g)0 (c) ¼ f 0 (c) þ g0 (c);

Sum Rule

f þ g,

Multiple Rule

l f, for l 2 R,

and

(l f ) (c) ¼ l f (c);

Product Rule

fg,

and

( fg)0 (c) ¼ f 0 (c)g (c) þ f (c)g0 (c);

and

0

0

We differentiate each term in turn. We differentiate one term at a time.

6.2

Rules for differentiation f g,

Quotient Rule

217

provided that g(c) 6¼ 0; and  0 gðcÞf 0 ðcÞf ðcÞg0 ðcÞ f : g ðcÞ ¼ ðgðcÞÞ2

Now, we saw in Section 6.1 that, for any n 2 N, the function x 7! xn ; x 2 R; is differentiable on R, and that its derived function is x 2 R:

x 7! nxn1 ;

We can use this fact, together with the Combination Rules, to prove that any polynomial function is differentiable on R, and that its derivative can be obtained by differentiating the polynomial term-by-term. Corollary 1 Let p(x) ¼ a0 þ a1x þ a2x2 þ    þ anxn, x 2 R, where a0, a1, a2, . . ., an 2 R. Then p is differentiable on R, and its derivative is p0 ð xÞ ¼ a1 þ 2a2 x þ    þ nan xn1 ;

x 2 R:

Since a rational function is a quotient of two polynomials, it follows from Corollary 1 and the Quotient Rule that a rational function is differentiable at all points where the denominator does not vanish (that is, the denominator does not take the value 0). 3

Example 1 Prove that the function f ð xÞ ¼ x2x1 ; x 2 R  f1g, is differentiable on its domain, and find its derivative. Solution The function f is a rational function whose denominator does not vanish on R  {1}; hence f is differentiable on this set (the whole of its domain). By Corollary 1, the derived function of x 7! x3 is x 7! 3x2 , and the derived function of x 7! x2  1 is x 7! 2x. It follows from the Quotient Rule that the derivative of f is ðx2  1Þ  3x2  x3  ð2xÞ f 0 ð xÞ ¼ ð x 2  1Þ 2 x4  3x2 & ¼ : ð x 2  1Þ 2 Problem 1 Find the derivative of each of the following functions: (a) f (x) ¼ x7  2x4 þ 3x3  5x þ 1, x 2 R; 2

x 2 R  f1g; (b) f ð xÞ ¼ xx3 þ1 1 ; (c) f (x) ¼ 2 sin x cos x, x 2 R; x

e (d) f ð xÞ ¼ 3þsin x2 cos x ;

Problem 2 x 2 R.

x 2 R:

Find the third order derivative of the function f (x) ¼ xe2x,

In the last section we found the derived functions for sin, cos and exp. We now ask you to find the derived functions for the remaining trigonometric functions and the three most common hyperbolic functions.

In particular,  0 g0 ðcÞ 1 g ðcÞ ¼  ðgðcÞÞ2 :

6: Differentiation

218 Find the derivative of each of the following functions:   (a) f ð xÞ ¼ tan x; x 2 R   12 p;  32 p;  52 p; . . . ; (b) f ð xÞ ¼ cosec x; x 2 R  f0; p; 2p; . . .g;   (c) f ð xÞ ¼ sec x; x 2 R   12 p;  32 p;  52 p; . . . ; (d) f ð xÞ ¼ cot x; x 2 R  f0; p; 2p; . . .g: Problem 3

Problem 4

Find the derivative of each of the following functions

(a) f (x) ¼ sinh x, x 2 R; (c) f (x) ¼ tanh x, x 2 R.

6.2.2

(b) f(x) ¼ cosh x, x 2 R;

The Composition Rule

In the last sub-section we extended our stock of differentiable functions to include all rational, trigonometric and hyperbolic functions. However, to differentiate many other functions we need to differentiate composite functions, such as the function f (x) ¼ sin(cos x), x 2 R, which is the composite of two differentiable functions – namely, f ¼ sin cos. The Composition Rule tells us that the composite of two differentiable functions is itself differentiable. Theorem 2 Composition Rule Let g and f be defined on open intervals I and J, respectively, and let c 2 I and gðI Þ J. If g is differentiable at c and f is differentiable at g(c), then f g is differentiable at c and ð f gÞ0 ðcÞ ¼ f 0 ðgðcÞÞg0 ðcÞ:

This rule is sometimes known as the Chain Rule.

We give the proof of the Rule in Sub-section 6.2.4.

Remarks 1. When written in Leibniz notation, the Composition Rule has a form that is easy to remember: if we put u ¼ gð xÞ and y ¼ f ðuÞ ¼ f ðgð xÞÞ then dy dy du ¼  : dx du dx 2. We can extend the Composition Rule to a composite of three (or more) functions; for example ð f g hÞ0 ð xÞ ¼ f 0 ½gðhðxÞÞg0 ðhð xÞÞh0 ð xÞ: In Leibniz notation, if we put v ¼ hð xÞ; u ¼ gðvÞ and

y ¼ f ðuÞ ¼ f ½gðhð xÞÞ;

We frequently use this extension of the Composition Rule without mentioning it explicitly.

6.2

Rules for differentiation

219

then we obtain the chain dy dy du dv ¼   : dx du dv dx Example 2 Prove that each of the following composite functions is differentiable on its domain, and find its derivative: (a) k(x) ¼ sin (cos x 2 R; (b) k(x) ¼ cosh (e2x), x 2 R;  x), 1 x2 (c) kð xÞ ¼ tan 4 e þ sin px ; x 2 ð1; 1Þ: Solution (a) Here k(x) ¼ sin (cos x), so let gð xÞ ¼ cos x and

f ð xÞ ¼ sin x;

for x 2 R;

then g0 ð xÞ ¼ sin x

and f 0 ð xÞ ¼ cos x;

for x 2 R:

By the Composition Rule, k ¼ f g is differentiable on R, and k0 ð xÞ ¼ f 0 ðgð xÞÞg0 ð xÞ ¼ cosðcos xÞ  ð sin xÞ ¼ cosðcos xÞ  sin x: (b) Here k(x) ¼ cosh (e2x), so let hð xÞ ¼ 2x; gð xÞ ¼ ex and f ð xÞ ¼ cosh x; then h, g and f are differentiable on R, and

for x 2 R;

h0 ð xÞ ¼ 2; g0 ð xÞ ¼ ex and f 0ð xÞ ¼ sinh x; for x 2 R: By the (extended form of the) Composition Rule, k ¼ f g h is differentiable on R, and k0 ð xÞ ¼ f 0 ½gðhð xÞÞg0 ðhð xÞÞh0 ð xÞ   ¼ sinh e2x  e2x  2 2x 2x ¼  2e2 sinh e : (c) Here kð xÞ ¼ tan 14 ex þ sin px ; x 2 ð1; 1Þ, so let 1  x2 e þ sin px ; x 2 ð1; 1Þ; and gð x Þ ¼ 4   1 1 f ð xÞ ¼ tan x; x 2  p; p : 2 2

Now, when x 2 (1, 1), we have 1  x2 e þ jsin pxj j gð x Þ j  4 1  ðe þ 1Þ 4 0, n 2 N; (b) f ð xÞ ¼ tan x; x 2  12 p; 12 p ; (c) f (x) ¼ ex, x 2 R.

I

x

6: Differentiation

222 Solution (a) The function f ð xÞ ¼ xn ; x > 0; is continuous and strictly increasing on (0, 1), and f ((0, 1)) ¼ (0, 1). Also, f is differentiable on (0, 1), and its derivative f 0 (x) ¼ nxn1 is nonzero there. So f satisfies the conditions of the Inverse Function Rule. Hence f 1 is differentiable on (0, 1); and, if y ¼ f (x), then  1 0 1 1 ¼ ð yÞ ¼ 0 f f ð xÞ nxn1 1 1 1n ¼ x1n ¼ y n : n n If we now replace the domain variable y by x, we obtain  1 0 1 1 ð xÞ ¼ xn1 ; x 2 ð0; 1Þ: f n (b) The function   1 1 f ð xÞ ¼ tan x; x 2  p; p ; 2 2  1 1   1 1  is continuous and strictly increasing R.  1 1on 2 p; 2 p , and f 0 2 p; 2 p ¼ Also, f is differentiable on 2 p; 2 p , and its derivative f ð xÞ ¼ sec2 x is non-zero there. So f satisfies the conditions of the Inverse Function Rule. Hence f 1 ¼ tan1 is differentiable on R; and, if y ¼ f (x), then  1 0 1 1 f ¼ ð yÞ ¼ 0 f ð xÞ sec2 x 1 1 ¼ : ¼ 1 þ tan2 x 1 þ y2 If we now replace the domain variable y by x, we obtain  1 0 1 ð xÞ ¼ ; x 2 R: tan 1 þ x2 (c) The function f ð xÞ ¼ ex ; x 2 R; is continuous and strictly increasing on R, and f (R) ¼ (0, 1). Also, f is differentiable on R, and its derivative f 0 (x) ¼ ex is non-zero there. So f satisfies the conditions of the Inverse Function Rule. Hence f  1 ¼ loge is differentiable on (0, 1); and, if y ¼ f (x), then  1 0 1 1 ¼ f ð yÞ ¼ 0 f ð xÞ ex 1 ¼ : y If we now replace the domain variable y by x, we obtain 1 & ðloge Þ0 ð xÞ ¼ ; x 2 ð0; 1Þ: x Problem 6 For each of the following functions f, show that f1 is differentiable and determine its derivative: (a) f(x) ¼ cos x, x 2 (0, p);

(b) f(x) ¼ sinh x, x 2 R.

1

y ¼ xn, so that x ¼ yn :

6.2

Rules for differentiation

223

Most equations of the form y ¼ f (x) cannot be solved explicitly to give x as some formula involving y alone. The Inverse Function Rule is often used in such situations to solve problems that would otherwise be intractable. The following problem illustrates this type of application. Problem 7 Prove that the function f (x) ¼ x5 þ x  1, x 2 R, has an inverse function f1 which is differentiable on R. Find the values of ( f1)0 (d ) at those points d corresponding to the points c ¼ 0, 1 and 1, where d ¼ f (c).

Exponential functions Earlier we defined the number ax, for a > 0, by the formula ax ¼ expðx loge aÞ: Since the functions exp and log are differentiable on R and (0, 1), respectively, it follows that we can use this formula to determine the derivatives of functions such as x 7! x ; x 7! ax and x 7! xx . Example 4 Prove that, for  2 R, the power function f (x) ¼ x, x 2 (0, 1), is differentiable on its domain, and that f 0 (x) ¼ x1. Solution

By definition f ð xÞ ¼ expð loge xÞ:

The function x 7!  loge x is differentiable on (0, 1), with derivative x . By the Composition Rule, f is differentiable on (0, 1) with derivative   f 0 ð xÞ ¼ expð loge xÞ  x   1 ¼x  ¼ x ; x 2 ð0; 1Þ: & x

Remark In the case when  is a rational number,  ¼ mn say, the result of Example 4 can also be proved by applying the Composition Rule to the functions x 7! xm and 1 1 x 7! xn . The function x 7! xn is differentiable on (0, 1), by the Inverse Function Rule, as it is the inverse of the function x 7! xn . Example 5 Prove that, for a > 0, the function f (x) ¼ ax, x 2 R, is differentiable on its domain, and that f 0 (x) ¼ ax loge a. Solution

By definition f ð xÞ ¼ expðx loge aÞ:

The function x 7! x loge a is differentiable on R, with derivative loge a. By the Composition Rule, f is differentiable on R, with derivative f 0 ð xÞ ¼ expðx loge aÞ  loge a ¼ ax loge a;

x 2 R:

&

Problem 8 Prove that the function f (x) ¼ xx, x 2 (0, 1), is differentiable on its domain, and find its derivative.

Section 4.4. Section 6.1.

6: Differentiation

224 We end this sub-section with a list of basic differentiable functions. Basic differentiable functions The following functions are differentiable on their domains: polynomials and rational functions

nth root function trigonometric functions (sine, cosine and tangent)



exponential function



hyperbolic functions (sinh, cosh and tanh).



6.2.4

Proofs

You may omit these proofs at a first reading.

We now supply the proofs of the Combination Rules and the Composition Rule, which we omitted earlier. We also illustrate how to prove such results using the "   method. Theorem 1

Combination Rules

Let f and g be defined on an open interval I, and c 2 I. Then, if f and g are differentiable at c, so are: Sum Rule Multiple Rule

f þ g, and ( f þ g)0 (c) ¼ f 0 (c) þ g0 (c); l f, for l 2 R, and (l f )0 (c) ¼ l f 0 (c);

Product Rule

fg,

Quotient Rule

f g, provided that g (c) 6¼ 0;

and

0 ( fg)0 (c) ¼ f0 (c) g(c) þ f (c) g (c);

and

f g

0

0

0

ðcÞg ðcÞ ðcÞ ¼ gðcÞf ððcgÞf : ðcÞÞ2

Proof

In this proof we use the Combination Rules for limits given in Sub-section 5.1.

Sum Rule Let F ¼ f þ g. Then F ð xÞ  F ðcÞ f f ð xÞ þ gð xÞg  f f ðcÞ þ gðcÞg ¼ lim lim x!c x!c xc xc f ð x Þ  f ðc Þ gð xÞ  gðcÞ ¼ lim þ lim x!c x!c xc xc ¼ f 0 ðcÞ þ g0 ðcÞ: Thus F is differentiable at c, with derivative f 0 (c) þ g0 (c). Multiple Rule This is just a special case of the Product Rule, with g(x) ¼ l. Product Rule Let F ¼ fg. Then F ð xÞ  F ðcÞ f ð xÞgð xÞ  f ðcÞgðcÞ ¼ lim x!c x!c xc xc

lim

6.2

Rules for differentiation

225

f f ð xÞ  f ðcÞggð xÞ þ f ðcÞfgð xÞ  gðcÞg xc f ð xÞ  f ðcÞ gð xÞ  gðcÞ ¼ lim gð xÞ þ lim f ðcÞ x!c x!c xc xc f ð xÞ  f ðcÞ gð xÞ  gðcÞ ¼ lim  lim gð xÞ þ f ðcÞ  lim x!c x!c x!c xc xc ¼ f 0 ðcÞgðcÞ þ f ðcÞg0 ðcÞ;

¼ lim x!c

since f and g are differentiable at c, and since g is continuous at c, so that g(x) ! g(c) as x ! c. Thus F is differentiable at c, with derivative f 0 (c)g(c) þ f (c)g0 (c).

For, differentiable ) continuous.

Quotient Rule Let F ¼ gf . Recall, first, that, since g is continuous at c (as it is differentiable there) and g (c) 6¼ 0, there exists an open interval J containing c on which g(x) 6¼ 0. Thus the domain of F contains the open interval J, and c 2 J. Then f ð xÞ f ðcÞ F ð xÞ  F ðcÞ gð xÞ  gðcÞ ¼ lim x!c x!c x  c xc f ð xÞgðcÞ  f ðcÞgð xÞ ¼ lim x!c ðx  cÞ  gð xÞgðcÞ ff ð xÞ  f ðcÞggðcÞ  f ðcÞfgð xÞ  gðcÞg ¼ lim x!c ðx  cÞ  gð xÞgðcÞ f ð x Þ  f ðc Þ f ðcÞ gð xÞ  gðcÞ  lim  ¼ lim x!c ðx  cÞ  gð xÞ x!c gðcÞ ðx  cÞ  gð xÞ f ð x Þ  f ðc Þ 1 f ðc Þ gð x Þ  gð c Þ 1 ¼ lim  lim   lim  lim x!c x!c gð xÞ gðcÞ x!c ðx  cÞ x!c gð xÞ xc f 0 ðcÞ f ðcÞg0 ðcÞ f 0 ðcÞgðcÞ  f ðcÞg0 ðcÞ  2 ¼ ; ¼ gð c Þ g ðcÞ g2 ð c Þ

lim

since f and g are differentiable at c, and g is continuous at c. 0 ðcÞg0 ðcÞ Thus F is differentiable at c, with derivative f ðcÞgðcgÞf : 2 ð cÞ

&

As an illustration of how such results may be proved using the "   method that we introduced in Section 5.4, we prove just the Sum Rule. Notice that in our proofs we avoid many complications by judicious use of the ‘K" Lemma’.

The ‘K" Lemma’ first appeared in Sub-section 2.2.2.

Proof of the Sum Rule Let F ¼ f þ g. In view of the K" Lemma, we must prove that:

As you work through this proof, compare it with the earlier proof using limits.

for each positive number ", there is a positive number  such that   F ð x Þ  F ðcÞ  0 0   < K" for all x satisfying  f ð c Þ þ g ð c Þ f g  xc  0 < jx  cj < :

(1)

First, we write the expression on the left-hand side of (1) in a convenient form as

Recall that K must NOT depend on x or ".

6: Differentiation

226 ff ð x Þ þ gð x Þ g  ff ð c Þ þ gð c Þ g  ff 0 ðcÞ þ g0 ðcÞg xc ff ð xÞ  f ðcÞg þ fgð xÞ  gðcÞg  ff 0 ðcÞ þ g0 ðcÞg ¼ x  c



f ð xÞ  f ðcÞ gð xÞ  gðcÞ  f 0 ðcÞ þ  g0 ð c Þ : ¼ xc xc

(2)

We now use the information that we already have to tackle each of the terms in (2) in turn. Since f is differentiable at c, there exists some positive number 1 such that   f ð x Þ  f ðc Þ 0     x  c  f ðcÞ < "; for all x satisfying 0 < jx  cj < 1 : (3) Also, since g is differentiable at c, there exists some positive number 2 such that    gð x Þ  gð c Þ  0    x  c  g ðcÞ < "; for all x satisfying 0 < jx  cj < 2 : (4) Now let  ¼ min{1, 2}. It follows that, for all x satisfying 0 < jx  cj < , both (3) and (4) hold. Hence, if we apply the Triangle Inequality to (2) and then use both (3) and (4), we find that, for all x satisfying 0 < jx  cj <    f f ð xÞ þ gð xÞg  f f ðcÞ þ gðcÞg  0 0    f ð c Þ þ g ð c Þ f g   xc      f ð xÞ  f ðcÞ   gð x Þ  gð c Þ   f 0 ðcÞ þ   g0 ðcÞ   xc xc

By the Triangle Inequality. By (3) and (4).

< " þ " ¼ 2": But this is just the statement (1) that we set out to prove, with K ¼ 2.

This expression appears in (2).

&

Remark Note that our use of the ‘K" Lemma’ meant that we did not need to know at the steps (3) and (4) to use expressions like 12 " rather than " in order to end up with a final conclusion that ‘some expression is 0.

6.3

Rolle’s Theorem

229

When we wish to locate local extrema of a differentiable function ƒ, instead of using the above definition we usually use the following result, which gives a connection between local extrema of a function ƒ and the points where f 0 vanishes. Theorem 1

Local Extremum Theorem

Let ƒ be defined on an interval [a, b]. If ƒ has a local extremum at c, where a < c < b, and if ƒ is differentiable at c, then f 0 ðcÞ ¼ 0: Proof Suppose that ƒ has a local maximum at c. Since a < c < b, it follows from the definition of local maximum that there exists a neighbourhood N of c with N ½a, b such that f ðxÞ  f ðcÞ;

for x 2 N:

We now choose numbers r, s > 0 such that N ¼ (c  r, c þ s). First, looking to the left of c, we have f ðxÞ  f ðcÞ  0

and

x  c < 0;

for c  r < x < c;

so that f ðxÞ  f ðcÞ  0; xc

for c  r < x < c:

Thus f ðxÞ  f ðcÞ  0: xc Next, looking to the right of c, we have fL0 ðcÞ ¼ lim

(1)

x!c

f ðxÞ  f ðcÞ  0

and

x  c > 0;

for c < x < c þ s;

so that

Thus

f ðxÞ  f ðcÞ  0; xc

for c < x < c þ s:

f ðxÞ  f ðcÞ  0: (2) xc Since ƒ is differentiable at c, the left and right derivatives at c exist and are equal. Hence, in view of inequalities (1) and (2), their common value, f 0ðcÞ, & must be 0. fR0 ðcÞ ¼ limþ x!c

We prove only the local maximum version: the proof of the local minimum version is similar.

6: Differentiation

230

Remarks 1. The Local Extremum Theorem applies only if the function is differentiable at a local extremum. For example, the function f ðxÞ ¼ jxj; x 2 ½1; 1; has a local minimum 0 at 0, but ƒ is not differentiable at 0. 2. The Local Extremum Theorem does not assert that a point where the derivative vanishes is necessarily a local extremum. For example, the function f ðxÞ ¼ x3 ; x 2 ½1; 1; does not have a local extremum at 0, although f 0 (0) ¼ 0. 3. The Local Extremum Theorem does not make any assertion about a local extremum that occurs at a point c that is one of the end-points of the interval [a, b]. Find the local extrema of the function 1 1 f ð xÞ ¼ x4  x3 ; x 2 ½1; 2: 4 3 Clearly any extremum of ƒ on [a, b] that occurs at a point other than a or b must be a local extremum. It follows from Theorem 1 that such a point c must be a point where f 0 (c) ¼ 0. Thus an immediate consequence of the Local Extremum Theorem is the following criterion for finding all the extrema of ‘well-behaved functions’ on closed intervals. Problem 1

Corollary 1 Let ƒ be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then the extrema of ƒ on [a, b] can occur only at a, at b, or at points c in (a, b) where f 0 (c) ¼ 0. We now reformulate Corollary 1 as a strategy for locating minima and maxima. Strategy To determine the maximum and the minimum of a function ƒ which is continuous on [a, b] and differentiable on (a, b): 1. Determine the points c1, c2, . . . in (a, b) where f 0 vanishes; 2. Compare the values of f (a), f (b), f (c1), f (c2), . . .; the least is the minimum, and the greatest is the maximum.

Problem 2 Use the above Strategy to determine the minimum

 and the maximum of the function f ð xÞ ¼ sin2 x þ cos x, for x 2 0; 12 p :

6.3.2

Rolle’s Theorem

In the previous sub-section we saw that, if a function ƒ is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then the extrema of ƒ can occur only at a, at b, or at points c in (a, b) where f 0 (c) ¼ 0.

6.3

Rolle’s Theorem

231

Now the function 

 1 f ðxÞ ¼ sin px ; 2

  2 2 x2  ; ; 3 3

shows that it can happen that no interior point of (a, b) corresponds to a maximum or a minimum of ƒ, and that f 0 need not vanish at any interior point at all. However, the situation is quite different for the function   1 f ðxÞ ¼ sin px ; x 2½2; 2: 2

That is, the extrema may not occur at interior points of (a, b).

Here f(2) ¼ f(2) ¼ 0; on [2, 2] the function f has a maximum at 1 and a minimum at 1, and at both these interior points f 0 vanishes. This is a special case of Rolle’s Theorem which asserts that, if f(a) ¼ f(b), then there is at least one point c strictly between a and b at which f 0 vanishes. Theorem 2 Rolle’s Theorem Let ƒ be defined on the closed interval [a, b] and differentiable on the open interval (a, b). If f(a) ¼ f(b), then there exists some point c, with a < c < b, for which f 0 (c) ¼ 0.

This is an existence theorem. Often it is difficult to evaluate c explicitly.

Remarks 1. This apparently simple result is one of the most important results in Analysis. 2. In geometric terms, Rolle’s Theorem means that, if the line joining the points (a, f(a)) and (b, f(b)) on the graph of ƒ is horizontal, then so is the tangent to the graph at some point c in (a, b). 3. There may be more than one point c in (a, b) at which f 0 vanishes (as in the diagram in the margin). Rolle’s Theorem simply asserts that at least one such point c exists. Proof If ƒ is constant on [a, b], then f 0 (x) ¼ 0 everywhere in (a, b); in this case, we may take c to be any point of (a, b). If ƒ is non-constant on [a, b], then either the maximum or the minimum (or both) of ƒ on [a, b] is different from the common value f (a) ¼ f (b). Since one of the extrema occurs at some point c with a < c < b, the Local Extremum & Theorem applied to the point c shows that f 0 (c) must be zero. Example 1 function

Verify that the conditions of Rolle’s Theorem are satisfied by the f ð xÞ ¼ 3x4  2x3  2x2 þ 2x;

x 2 ½1; 1;

and determine a value of c in (1, 1) for which f 0 (c) ¼ 0. Solution Since f is a polynomial function, f is continuous on [1, 1] and differentiable on (1, 1). Also, f (1) ¼ f (1) ¼ 1. Thus f satisfies the conditions of Rolle’s Theorem on [1, 1]. It follows that there exists a number c 2 (1, 1) for which f 0 (c) ¼ 0. Now

Since ƒ is continuous on [a, b], ƒ must have both a maximum and a minimum on [a, b], by the Extreme Values Theorem.

6: Differentiation

232 f 0 ð xÞ ¼ 12x3  6x2  4x þ 2    1 1 2 x ; ¼ 12 x  3 2 so that f 0 vanishes at the points p1ffiffi3,  p1ffiffi3 and 12 in (1, 1). Any of these three & numbers will serve for c. Problem 3 Verify that the conditions of Rolle’s Theorem are satisfied by the function f ð xÞ ¼ x4  4x3 þ 3x2 þ 2; x 2 ½1; 3; and determine a value of c in (1,3) for which f 0 (c) ¼ 0. Problem 4 For each of the following functions, state whether Rolle’s Theorem applies for the given interval: (a) f ð xÞ ¼ tan x; x 2 ½0; p; (b) f ð xÞ ¼ x þ 3jx  1j; x 2 ½0; 2; (c) f ð xÞ ¼ x  9x17 þ 8x18 ; x 2 ½0; 1;

 (d) f ð xÞ ¼ sin x þ tan1 x; x 2 0; 12 p :

6.4

The Mean Value Theorem

Here we continue to study the geometric properties of functions that are differentiable on intervals, and describe some of their applications.

6.4.1

The Mean Value Theorem

First, recall the geometric interpretation of Rolle’s Theorem: Under suitable conditions, if the chord joining the points (a, f (a)) and (b, f (b)) of the graph of f is horizontal, then so is the tangent at some point c of (a, b). If you imagine pushing the chord (as shown in the margin), always parallel to itself, until it is just about to lose contact with the graph of f, it looks as though at this point the chord becomes a tangent to the graph. Similarly, the ‘chord-pushing’ approach suggests that, even if the original chord is not horizontal (that is, if f(a) 6¼ f(b)), there must still be some point c of (a, b) at which the tangent is parallel to the chord. Example 1

Consider the function f ð xÞ ¼ 3  3x þ x3 ; x 2 ½1; 2:

Find a point c of (1, 2) such that the tangent to the graph of f is parallel to the chord joining (1, f (1)) to (2, f (2)). Solution Since f(1) ¼ 3  3 þ 1 ¼ 1 and f(2) ¼ 3  6 þ 8 ¼ 5, the slope of the chord joining the endpoints of the graph is f ð 2Þ  f ð 1Þ 5  1 ¼ ¼ 4: 21 21

6.4

The Mean Value Theorem

233

Now, since f is a polynomial, it is differentiable on (1, 2) and its derivative is qffiffi f 0 (x) ¼ 3 þ 3x2; hence f 0 (c) ¼ 4 when 3c2 ¼ 7, or c ¼ 73 ’ 1:53. Thus, at the point (c, f(c)) the tangent to the curve is parallel to the chord joining the & end-points. We now generalise Rolle’s Theorem and assert that there is always a point where the tangent to the graph is parallel to the chord joining the end-points. This result is known as the Mean Value Theorem, so-called since f ð bÞ  f ð aÞ ba can be thought of as the mean value of the derivative between a and b. Theorem 1

Mean Value Theorem

Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists a point c in (a, b) such that f ð bÞ  f ð aÞ f 0 ðcÞ ¼ : ba

Again, this is an existence theorem. Note that when f(a) ¼ f(b), the Mean Value Theorem simply reduces to Rolle’s Theorem.

The idea of the proof is as follows. We define h(x) to be the vertical distance from the chord to the curve; then h(a) and h(b) are both 0; in fact, h satisfies all the conditions of Rolle’s Theorem. Applying Rolle’s Theorem to h, we obtain the desired result. The slope of the chord joining the points (a, f (a)) and (b, f (b)) is f ðbÞ  f ðaÞ m¼ ; ba and so the equation of the chord is y ¼ mðx  aÞ þ f ðaÞ: It follows that the vertical height, h(x), between points with ordinate x on the graph and those on the chord is given by hð xÞ ¼ f ð xÞ  ½mðx  aÞ þ f ðaÞ: Now h(a) ¼ h(b) ¼ 0, and h is continuous on [a, b] and differentiable on (a, b). Thus h satisfies all the conditions of Rolle’s Theorem. It follows from Rolle’s Theorem that there exists some point c in (a, b) for which h0 (c) ¼ 0. But, since h0 (c) ¼ f 0 (c)  m, it follows that f ð bÞ  f ð aÞ : & f 0 ðcÞ ¼ m ¼ ba Proof

Example 2 Verify that the conditions of the  Mean Value Theorem are 7 satisfied by the function f ð xÞ ¼ x1 ; x 2 1; xþ1 2 ; and find a value for c that satisfies the conclusion of the theorem. Solution whose denominator

 The function f is a rational

function   is non-zero  on 1; 72 ; so f is continuous on 1; 72 and differentiable on 1; 72 : Thus f satisfies the conditions of the Mean Value Theorem. Now  5 f 72  f ð1Þ 2 90 ¼ ; ¼ 7 5 9 21 2

y y = f (x)

5 9

1

7 2

x

6: Differentiation

234 and f 0 ð xÞ ¼

2 ðx þ 1Þ2

;

  so the Mean Value Theorem asserts that there exists some point c in 1; 72 for which f 0 (c) ¼ 29. Thus 2 2 ¼ ; 2 9 ð c þ 1Þ and so ðc þ 1Þ2 ¼ 9. It follows that c ¼ 2.

&

This is a quadratic equation with roots c ¼ 2 and 4; we ignore the solution c¼ 4,  since it lies outside 1; 72 .

Problem 1 For each of the following functions, verify that the conditions of the Mean Value Theorem are satisfied, and find a value for c that satisfies the conclusion of the theorem: (a) f ð xÞ ¼ x3 þ 2x; x 2 ½2; 2; (b) f ð xÞ ¼ ex ; x 2 ½0; 3:

6.4.2

Positive, negative and zero derivatives

We now study some consequences of the Mean Value Theorem for functions whose derivatives are always positive, always negative, or always zero. First, we prove a crucial result about monotonic functions which you use regularly to sketch the graph of a function f. It concerns the behaviour of f on a general interval I, so here we denote the interior of I (the set of all interior points of I ) by Int I. Theorem 2

For example, if I ¼ [0, 1), then Int I ¼ (0, 1).

Increasing-Decreasing Theorem

Let f be continuous on an interval I and differentiable on Int I. (a) If f 0 (x)  0 on Int I, then f is increasing on I. (b) If f 0 (x)  0 on Int I, then f is decreasing on I. Proof Choose any two points x1 and x2 in I, with x1 < x2. The function f satisfies the conditions of the Mean Value Theorem on the interval [x1, x2], so there exists a point c in (x1, x2) such that f ðx2 Þ  f ðx1 Þ ¼ f 0 ðcÞ: x2  x1 It follows that f (x2)  f(x1) must have the same sign as f 0 (c). (a) If f 0 (x)  0 on Int I, then f(x2)  f(x1)  0, so that f(x2)  f(x1). Thus f is increasing on I. (b) If f 0 (x)  0 on Int I, then f(x2)  f(x1)  0, so that f(x2)  f (x1). Thus f is & decreasing on I.

Remark If the inequalities in the statement of Theorem 2 are replaced by strict inequalities, the conclusions of the Theorem become the following: (a) If f 0 (x) > 0 on Int I, then f is strictly increasing on I; (b) If f 0 (x) < 0 on Int I, then f is strictly decreasing on I.

The proofs of these assertions are similar to the proofs in Theorem 2.

6.4

The Mean Value Theorem

235

Problem 2 For each of the following functions f, determine whether f is increasing, strictly increasing, decreasing or strictly decreasing: 4 (a) f ð xÞ ¼ 3x3  4x; x 2 ½1; 1Þ; (b) f ð xÞ ¼ x  loge x; x 2 ð0; 1: Two useful consequences of Theorem 2 are the following corollaries. Corollary 1

Zero Derivative Theorem

Let f be continuous on an interval I and differentiable on Int I. If f 0 (x) ¼ 0 for all x in Int I, then f is constant on I. Proof Cases (a) and (b) of Theorem 2 both apply, so that f is both increasing and decreasing on I. & Hence f is constant on I. As an illustration of the use of this important result, we can now prove the claim made earlier that the function f (x) ¼ lex, l an arbitrary constant, is the only function f that satisfies the differential equation f 0 (x) ¼ f (x) on R. For, if f 0 (x) ¼ f (x), then d x ðe f ð xÞÞ ¼ ex f 0 ð xÞ  ex f ð xÞ dx ¼ ex ð f 0 ð xÞ  f ð xÞÞ ¼ 0; it then follows from Corollary 1 that ex f(x) is just some constant l, say, so that f (x) ¼ lex.

Sub-section 6.1.3.

Corollary 2 Let f and g be continuous on an interval I and differentiable on Int I. If f 0 (x) ¼ g0 (x) for all x in Int I, then f(x) ¼ g(x) þ c for all x in I, for some constant c. Proof This follows immediately by applying Corollary 1 to the function & h ¼ f  g, since h0 (x) ¼ 0 for all x in the interior of I. Example 3 Solution

pffiffiffiffiffiffiffiffiffiffiffiffiffi  Prove that sinh1 x ¼ loge x þ x2 þ 1 , for all x 2 R. Let f ð xÞ ¼ sinh1 x; x 2 R;  pffiffiffiffiffiffiffiffiffiffiffiffiffi gð xÞ ¼ loge x þ x2 þ 1 ;

x 2 R:

Then f and g are continuous on R and differentiable on R. Now 1 f 0 ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R; 2 x þ1 also, by the Composition Rule for Derivatives     1 1 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi  1 þ  2x x2 þ 1 2 g0 ð x Þ ¼ 2 x þ x2 þ 1 x ffi 1 þ pffiffiffiffiffiffiffiffi x2 þ 1 pffiffiffiffiffiffiffiffiffiffiffiffi ffi ¼ x þ x2 þ 1 1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R: x2 þ 1

For h is continuous on I and differentiable on Int I.

6: Differentiation

236 Hence f 0 (x) ¼ g0 (x) for all x in R. It follows from Corollary 2 that  pffiffiffiffiffiffiffiffiffiffiffiffiffi sinh1 x ¼ loge x þ x2 þ 1 þ c; for some constant c. Putting x ¼ 0 in the above identity, we obtain 0 ¼ loge(1) þ c, & so that c ¼ 0. Problem 3 Use Corollary 1 to prove the following identities: (a) sin1 x þ cos1 x ¼ 12 p, for x 2 ½1; 1;  (b) tan1 x þ tan1 1x ¼ 12 p, for x > 0. Next, note that second derivatives can often be used to identify whether a point c where f 0 (c) ¼ 0 is a local maximum or a local minimum of a function f. Suppose that f is defined on a neighbourhood of c, and that f 0 ðcÞ ¼ 0 and

f 00 ðcÞ > 0:

It can be shown that there is some punctured neighbourhood N ¼ ðc  r; cÞ [ ðc; c þ sÞ of c such that f 0 ð xÞ  f 0 ðcÞ > 0; xc

for x 2 N;

In Exercise 6 on this subsection, in Section 6.7, we ask you to verify a result that implies this assertion.

that is f 0 ð xÞ > 0; for x 2 N: xc We can rewrite this inequality in the form f 0 ð xÞ < 0; f 0 ð xÞ > 0;

for x 2 ðc  r; cÞ; for x 2 ðc; c þ sÞ:

It follows from the Increasing–Decreasing Theorem that f has a local minimum at c. A similar argument shows that, if f 0 ðcÞ ¼ 0

and

f 00 ðcÞ < 0;

then f has a local maximum at c. Thus we have proved the following result. Theorem 3 Second Derivative Test Let f be defined on a neighbourhood of c, and f 0 (c) ¼ 0. (a) If f 00 (c) > 0, then f (c) is a local minimum of f. (b) If f 00 (c) < 0, then f (c) is a local maximum of f.

Remark The theorem gives us no information in the case that f 00 (c) ¼ 0. Problem 4 For the function f(x) ¼ x3  3x2 þ 1, x 2 R, determine those points c where f 0 (c) ¼ 0. Using the Second Derivative Test, determine whether these correspond to local maxima, local minima or neither.

The following diagrams are helpful in remembering this result.

6.4

The Mean Value Theorem

237

Inequalities We now demonstrate how the Increasing–Decreasing Theorem can be used to prove certain inequalities involving differentiable functions. Prove that, for  > 1,

Example 4



ð1 þ xÞ  1 þ x;

This is a generalisation of Bernoulli’s Inequality

for x  1:

ð1 þ xÞn  1 þ nx;

Solution

Let f ð xÞ ¼ ð1 þ xÞ ð1 þ xÞ;

for x   1, n 2 N, that you met in Sub-section 1.3.3.

for x 2 ½1; 1Þ:

The function f is continuous on [1, 1) and differentiable on (1, 1), and f 0 ð xÞ ¼ ð1 þ xÞ1   h i ¼  ð1 þ xÞ1 1 : Firstly, if 1 < x < 0, then 0 < 1 þ x < 1: Since  > 1, we can then take the (  1)th power of each side of the inequality 1 þ x < 1 to obtain ð1 þ xÞ1 < 1;

for 1 < x < 0;

Since   1 > 0, we can use Rule 5 for inequalities, which you met in Sub-section 1.2.1.

so that f 0 ð xÞ < 0;

for 1 < x < 0:

Next, if x > 0, then 1 þ x > 1; so that ð1 þ xÞ1 > 1;

Since   1 > 0, we can again use Rule 5 for inequalities.

for x > 0:

Hence f 0 ð xÞ > 0;

for x > 0:

Bringing together these two arguments, we have f 0 ð xÞ < 0; f 0 ð xÞ > 0;

for 1 < x < 0; for x > 0:

Also, f (0) ¼ 0. Hence, by the Increasing–Decreasing Theorem f is decreasing on ½1; 0; f is increasing on ½0; 1Þ; so we have f ð xÞ  f ð0Þ ¼ 0;

for x 2 ½1; 0;

f ð xÞ  f ð0Þ ¼ 0;

for x 2 ½0; 1Þ:

It follows that ð1 þ xÞ  ð1 þ xÞ  0; as required.

for x 2 ½1; 1Þ;

&

6: Differentiation

238 Example 4 illustrates the following general strategy. To prove that g(x)  h(x) on [a, b]:

Strategy 1. Let

f ð xÞ ¼ gð xÞ  hð xÞ; and show that f is continuous on [a, b] and differentiable on (a, b); 2. Prove that: and f 0 ð xÞ  0 on ða; bÞ; and f 0 ð xÞ  0 on ða; bÞ;

f ðaÞ  0; f ðbÞ  0;

EITHER OR

There is a corresponding version of the Strategy in which the weak inequalities are replaced by strict inequalities.

This also works if b is 1. This also works if a is 1.

Prove the following inequalities:

 (a) x  sin x, for x 2 0; 12 p ; Problem 5 (b)

6.5

2 3x

2

þ 13  x3 ,

for x 2 [0; 1].

L’Hoˆpital’s Rule

In order to differentiate the functions sin and exp, we needed to use the following results sin x ¼1 x!0 x lim

and

ex  1 ¼ 1: x!0 x lim

Each of these limits is the limit as x ! 0 of a quotient in which the numerator and denominator take the value 0 when x is 0. The evaluation of these limits was not trivial and required considerable care. In Analysis and in Mathematical Physics we often need to evaluate limits of the form lim

f ð xÞ

x!c gð xÞ

;

x!2

cos 3x sin x  ecos x

and

lim

x

x!0 cosh x

The Quotient Rule for limits of functions states that f ð xÞ f ð xÞ lim ; ¼ x!c x!c gð xÞ lim gð xÞ

lim

2

1

exist? If they do, what are their values? Do we have to evaluate all such limits by a direct argument, or is there a handy rule which we can apply? We shall find that there is, called l’Hoˆpital’s Rule.

6.5.1

See Sub-section 5.1.1 and Problem 8 in Sub-section 5.1.4, respectively.

where f ðcÞ ¼ gðcÞ ¼ 0:

Such limits cannot be evaluated by the Quotient Rule for limits of functions, because it does not apply in this situation. For example, do the limits limp

Sub-section 6.1.3.

x!c

provided that these last two limits exist.

Cauchy’s Mean Value Theorem

Recall that the Mean Value Theorem asserts that, under certain conditions, the graph of a function f(x), for x 2 [a, b], has the property that at some intermediate point c the tangent to the graph is parallel to the chord joining the endpoints of the graph. In other words, there exists a point c in (a, b) such that

Theorem 1, Sub-section 6.4.1.

6.5

L’Hoˆpital’s Rule f 0 ðcÞ ¼

239

f ð bÞ  f ð aÞ : ba

(1)

The key tool that we shall need in Sub-section 6.5.2 in the proof of l’Hoˆpital’s Rule is the following result. Theorem 1

Cauchy’s Mean Value Theorem

Let f and g be continuous on [a, b] and differentiable on (a, b). Then there is some point c in (a, b) for which f 0 ðcÞ

 fgðbÞ  gðaÞg ¼ g0 ðcÞ  ff ðbÞ  f ðaÞg:

(2)

0

In particular, if g(b) 6¼ g(a) and g (c) 6¼ 0, this equation can be written in the form f 0 ðcÞ f ðbÞ  f ðaÞ : (3) ¼ g0 ð c Þ gð bÞ  gð aÞ

This form is easier to remember.

Notice that the Mean Value Theorem is simply the special case when g(x) ¼ x. For then g0 (c) ¼ 1 and g(b)  g(a) ¼ b  a, and equation (3) reduces to equation (1). However, Theorem 1 is NOT a simple consequence of the Mean Value Theorem. For, if we apply the Mean Value Theorem separately to the functions f and g, we establish the existence of two points c1 and c2 in (a, b) for which f 0 ðc1 Þ ¼

f ðbÞ  f ðaÞ ba

and

g0 ð c 2 Þ ¼

gð bÞ  gð aÞ : ba

However, since c1 and c2 are usually unequal, we cannot deduce the existence of a single point c satisfying the statement (2). Our proof of Theorem 1 is similar to that of the Mean Value Theorem: we choose a suitable ‘auxiliary’ function h on [a, b] for which the conditions of Rolle’s Theorem are satisfied. Proof

Consider the function hð xÞ ¼ f ð xÞ  fgðbÞ  gðaÞg  gð xÞ  ff ðbÞ  f ðaÞg; for x 2 ½a; b:

By the Combination Rules for continuity and differentiability, h is continuous on [a, b] and differentiable on (a, b). Also hðaÞ ¼ f ðaÞ  fgðbÞ  gðaÞg  gðaÞ ¼ f ðaÞgðbÞ  gðaÞ f ðbÞ

 f f ðbÞ  f ðaÞg Here two terms cancel.

6: Differentiation

240 and

 fgð bÞ  gð aÞ g  gð bÞ ¼ f ðaÞgðbÞ  gðaÞf ðbÞ;

hð bÞ ¼ f ð bÞ

 f f ð bÞ  f ð aÞ g Again two terms cancel.

so that h(a) ¼ h(b). Thus h satisfies the conditions of Rolle’s Theorem on [a, b]. Therefore there exists a point c in (a, b) for which h0 ðcÞ ¼ 0 ;

The common value of h(a) and h(b) does not matter; the important thing is that the two are equal.

that is, these exists a point c in (a, b) for which f 0 ðcÞ

 fgðbÞ  gðaÞg ¼ g0 ðcÞ  f f ðbÞ  f ðaÞg:

This is precisely equation (2). Equation (3) follows immediately from equation (2).

&

By applying Cauchy’s Mean Value Theorem to the functions 1 f ð xÞ ¼ x6  1 and gð xÞ ¼ x4 þ 3x3 þ 3x  3 on ½1; 2; 2 5 3 2 prove that the equation 3x  2x  9x ¼ 3 has at least one root in (1, 2).

Example 1

Solution Since f and g are both polynomials, they are continuous on [1, 2] and differentiable on (1, 2). It follows that Cauchy’s Mean Value Theorem applies to the functions f and g on [1, 2]. Now, gð2Þ ¼ 12  16 þ 3  8 þ 3  2  3 ¼ 8 þ 24 þ 6  3 ¼ 35 and gð1Þ ¼ 12 þ 3 þ 3  3 ¼ 3 12, so that g(2) 6¼ g(1). Also, f 0 (x) ¼ 6x5 and g0 (x) ¼ 2x3 þ 9x2 þ 3 6¼ 0 on (1, 2). It therefore follows from Cauchy’s Mean Value Theorem that there exists at least one point c in (1, 2) for which 5

6c 63  0 ¼ 2c3 þ 9c2 þ 3 35  3 12 63 ¼ 1 ¼ 2: 31 2 By cross-multiplying, we see that c satisfies the equation 6c5 ¼ 4c3 þ 18c2 þ 6, or 3c5  2c3  9c2 ¼ 3: In other words, the equation 3x5  2x3  9x2 ¼ 3 has at least one root in & (1, 2). Problem 1

By applying Cauchy’s Mean Value Theorem to the functions f ð xÞ ¼ x3 þ x2 sin x and gð xÞ ¼ xcos x  sin x on ½0;p;

prove that the equation 3x ¼ (p2  2) sin x  xcos x has at least one root in (0, p).

6.5.2

l’Hoˆpital’s Rule

We are now in a position to state the key result that we need to evaluate certain types of limits in a fairly routine way.

Here we use equation (3).

6.5

L’Hoˆpital’s Rule

241

Theorem 2 l’Hoˆpital’s Rule Let f and g be differentiable on a neighbourhood of the point c, at which f(c) ¼ g(c) ¼ 0. Then f ð xÞ f 0 ð xÞ exists and equals lim 0 ; x!c gð xÞ x!c g ð xÞ

lim

provided that this last limit exists. We assume that f 0 ð xÞ lim 0 exists and equals ‘: (4) x!c g ð xÞ Hence there is some punctured neighbourhood N ¼ (c  r, c) [ (c, c þ s) of c 0 on which gf 0 is defined and g0 (x) 6¼ 0. We now prove that our assumption (4) implies that f ð xÞ exists and equals ‘: (5) lim x!c gðxÞ Let y be any specific point in N for which y > c. The functions f and g are continuous on [c, y] and differentiable on (c, y), so they satisfy the conditions of Cauchy’s Mean Value Theorem on [c, y]. Now g(y)  g(c) 6¼ 0; for otherwise we would have g(c) ¼ 0 (from our assumptions) and so g(y) ¼ 0; this would imply, by Rolle’s Theorem, that g0 would vanish somewhere in (c, y), which it does not. It follows from the conclusion (3) of Cauchy’s Mean Value Theorem that there exists some point z in (c, y) for which f 0 ðzÞ f ð y Þ  f ðc Þ ¼ g0 ðzÞ gð yÞ  gðcÞ f ð yÞ ¼ ; since f ðcÞ ¼ gðcÞ ¼ 0: g ð yÞ Now let y ! cþ. Since c < z < y, it follows that z ! cþ too. But we know that f 0 ðzÞ limþ 0 exists and has the value ‘: z!c g ðzÞ It follows that f ð yÞ exists and has the value ‘: lim y!cþ gð yÞ This is exactly the same as the statement that f ð xÞ lim exists and has the value ‘: x!cþ gð xÞ A similar argument shows that f ð xÞ lim exists and has the value ‘: x!c gð xÞ & Combining the last two statements, we obtain the desired result (5). Proof

We now show how we can use l’Hoˆpital’s Rule to evaluate the two limits that we mentioned at the start of this section.

You may omit this proof at a first reading.

Theorem 1, Sub-section 6.5.1.

This is just a special case of (4), with z in place of x and z ! cþ in place of x ! c.

Here x is simply a ‘dummy variable’: it does not matter what letter we assign to the variable in the limit. We simply repeat the whole argument, starting with a specific point y in N for which y < c.

6: Differentiation

242 cos 3x sin x  ecos x exists and determine its value.

Example 2

Solution

Prove that limp

(6)

x!2

Let f ð xÞ ¼ cos 3x

and gð xÞ ¼ sin x  ecos x ;

for x 2 R:

Then f and g are differentiable on R, and  p p f ¼g ¼ 0; 2 2 so that f and g satisfy the conditions of l’Hoˆpital’s Rule at p2. Now f 0 ð xÞ 3 sin 3x ¼ : g0 ð xÞ cos x þ sin x  ecos x Then, by l’Hoˆpital’s Rule, the limit (6) exists and equals f 0 ð xÞ 3 sin 3x ¼ limp ; x!2 g0 ð xÞ x!2 cos x þ sin x  ecos x

limp

provided that this last limit exists. 0 By the Quotient Rule for continuous functions, we know that gf 0 is continuous at p2, so that  f 0 ð xÞ f 0 p2 ¼ 0  p limp 0 x!2 g ð xÞ g 2 3 ¼ ¼ 3: 1 ˆ It follows, from l’Hopital’s Rule, that the original limit (6) must also exist, & and that its value is 3. x2 x!0 cosh x  1 exists and determine its value.

Example 3

Solution

Here we are using the fact (see Remark 7, Sub-section 5.4.1) that, if f is continuous at c, then lim f ð xÞ ¼ f ðcÞ: x!c

(In future, we shall not mention this fact explicitly in this particular connection.)

(7)

Prove that lim

Let

f ð xÞ ¼ x2 and gð xÞ ¼ cosh x  1; Then f and g are differentiable on R, and

for x 2 R:

f ð0Þ ¼ gð0Þ ¼ 0; so that f and g satisfy the conditions of l’Hoˆpital’s Rule at 0. Now, the derivatives of f and g are f 0 ð xÞ ¼ 2x

and g0 ð xÞ ¼ sinh x;

for x 2 R:

It follows from l’Hoˆpital’s Rule that the limit (7) exists and equals 2x (8) lim x!0 sinh x provided that this last limit exists. Now, both f 0 and g0 are differentiable on R, and f 0 (0) ¼ g0 (0) ¼ 0; thus f 0 and 0 g satisfy the conditions of l’Hoˆpital’s Rule at 0. Since f 00 ð xÞ ¼ 2 and g00 ð xÞ ¼ cosh x; for x 2 R;

We cannot assert that f 0 ð xÞ f 0 ð0Þ ¼ 0 ; x!0 g0 ð xÞ g ð0Þ lim

since f 0 (0) ¼ g0 (0) ¼ 0.

6.5

L’Hoˆpital’s Rule

243

it follows from l’Hoˆpital’s Rule that the limit (8) exists and equals 2 ; (9) lim x!0 cosh x provided that this last limit exists. The function cosh is continuous on R, and cosh 0 ¼ 1, so that lim cosh x ¼ 1. x!0

It follows, from the Quotient Rule for limits, that the limit (9) does exist and that its value is 2 2 ¼ ¼ 2: cosh 0 1 Working backwards, we conclude that the limit (8) exists, and equals 2. Working further backwards, we conclude that the limit (7) also exists and & equals 2. Before applying a theorem, it is important to check that its conditions are satisfied; for if they are not satisfied you cannot use the theorem! For instance, a thoughtless application of l’Hoˆpital’s Rule can give an incorrect answer! For example, consider the problem of evaluating 2x2  x  1 : (10) x!1 x2  x If we put f(x) ¼ 2x2  x  1 and g(x) ¼ x2  x, we might be tempted to evaluate (10) as follows f ð xÞ f 0 ð xÞ 4x  1 ¼ lim 0 ¼ lim lim (11) x!1 gð xÞ x!1 g ð xÞ x!1 2x  1 f 00 ð xÞ 4 ¼ lim 00 ¼ lim x!1 g ð xÞ x!1 2

Note that the carefully set out logic of the argument here is essential, since it is a consequence of what we know from Theorem 2.

lim

¼ 2: In fact, the value of the limit (10) is 3; so let us review the above argument carefully! The line (11) in the calculation is valid, since f and g are differentiable on R and f(1) ¼ g(1) ¼ 0; so the conditions of l’Hoˆpital’s Rule are satisfied for the first application of the Rule. However, f 0 (1) ¼ 3 and g0 (1) ¼ 1, so the conditions are not satisfied for the second application of l’Hoˆpital’s Rule! In fact, we should have concluded directly from line (11) that f 0 ð xÞ 4x  1 ¼ 3: lim 0 ¼ lim x!1 g ð xÞ x!1 2x  1 The moral is that you must apply l’Hoˆpital’s Rule with care, particularly to make the proviso ‘provided that this last limit exists’ at the appropriate points. At the end, you then work backwards through the chain of applications of the Rule to reach your conclusion about the limit that you set out originally to examine.

Remark Notice that we cannot apply l’Hoˆpital’s Rule to evaluate the limits sin x ex  1 ¼ 1 and lim ¼ 1; lim x!0 x x!0 x because we used these limits to find the derivatives of sin x and ex, respectively!

Note that f(1) ¼ g(1) ¼ 0.

Here the careful arguments that we gave in Examples 2 and 3 have been abandoned in favour of thoughtless ‘formula-pushing’!

6: Differentiation

244 Problem 2

Prove that the following limits exist, and evaluate them. 1

2x (a) lim sinh sin 3x ; x!0

(c) lim

x!0

6.6

sinðx2 þsin x2 Þ 1cos 4x ;

(b) lim

x!0

1

ð1þxÞ5 ð1xÞ5 2 2 ð1þ2xÞ5 ð12xÞ5

;

cos x (d) lim sin xx . x3 x!0

The Blancmange function

We now meet a function, called the Blancmange function, which is continuous at each point of R but is differentiable nowhere on R. The construction of the first function with these properties by Karl Weierstrass caused a huge excitement among mathematicians! You saw earlier that, if a function is differentiable at a point c, then it is also continuous at c. The Blancmange function shows, in a very striking way, that the converse result is false!

6.6.1

You may omit this section, apart from the next two paragraphs, at a first reading.

Sub-section 6.1.2.

What is the Blancmange function?

We give the formal definition first, and then look at the underlying geometry of the graph of the function. Definition

Let f be the function x  ½x; f ðxÞ ¼ 1  ðx  ½xÞ;

0  x  [x]  12, 1 2 < x  [x] < 1,

where [x] denotes the integer part of x. Then the Blancmange function is defined on R by the formula 1 1 1 BðxÞ ¼ f ðxÞ þ f ð2xÞ þ f ð4xÞ þ f ð8xÞ þ    2 4 8 1 X 1 ¼ f ð2n xÞ: 2n n¼0 So, what does the graph of B look like? And can we obtain any idea why the function has its strange properties? The function f is continuous on R, but is not differentiable at the points 12 n, for any integer n.

This series converges by the Comparison Test, since jf ð2n xÞj  12 ; for all x 2 R:

6.6

The Blancmange function

245

Next, we construct the graphs y ¼ 12 f ð2xÞ; y ¼ 14 f ð4xÞ; y ¼ 18 f ð8xÞ; . . . ; we obtain each graph by scaling the previous graph by a factor of 12 in both the x-direction and the y-direction. Each graph has twice as many peaks in any interval as its predecessor, and each of these peaks is half the height of the peaks in its predecessor. Thus the number of points in the interval where the function is not differentiable doubles (approximately) at each stage.

We obtain the Blancmange function B by adding together all these functions to form an infinite series Bð xÞ ¼ f ð xÞ þ 12 f ð2xÞ þ 14 f ð4xÞ þ 18 f ð8xÞ þ   . Thus, for example   B 12 ¼ f 12 þ 12 f ð1Þ þ 14 f ð2Þ þ 18 f ð4Þ þ          ¼ 12 þ 12  0 þ 14  0 þ 18  0 þ    ¼ 12 ; and B

1 4

1 1 1 1 Þ þ 18 f ð2Þ þ    4 þ 2 f 2 þ 4 f ð1    ¼ 14 þ 12  12 þ 14  0 þ 18  0 þ    ¼f

¼ 14 þ 14 ¼ 12 : To get an idea of the shape of the graph of B, we look at the graphs of successive partial sum functions of B. To help you follow the construction, we also draw in the graph at the previous stage (in light dashes) and the function being added to it (in heavy dashes).

The first few of these graphs are included in the diagrams below.

246

6: Differentiation

Eventually, we obtain the following graph of B:

Here the dashes indicate the graphs at the early stages and also the graphs of functions occurring in the summation.

It looks as though B is continuous. However, it is NOT true in general that the sum of infinitely many continuous function is itself continuous, so a proof of the continuity of B is necessary. Similarly, it looks as though B might not be differentiable. For example, we have marked the points on the above graphs corresponding    to x ¼ 13. At the first stage, 13 lies in the interval 0; 12 , and so the point 13 ; f 13

 lies on a line segment of slope 1. At the next stage, 13 lies in the interval 14 ; 12 ,     and so the point 13 ; f 13 þ 12 f 23 lies on a line segment of slope 0. And so on. At successive stages in the construction of the graph of B, the point corresponding to 13 lies alternately on line segments of slope 0 and 1. Thus it seems    plausible that the slopes of chords joining the points 13 ; B 13 and (x, B(x)) do not tend to any fixed value as x tends to 13, so that B would not be differentiable at 13. Notice in the last diagram above that however closely we look at the graph of the Blancmange function B, it seems to have small blancmanges growing on it everywhere. At the nth stage, each horizontal line segment has one or two mini-blancmanges growing on it, and each sloping line segment has one or two ‘sheared’ mini-blancmanges growing on it. Finally, notice that the Blancmange function B is periodic, with period 1. This occurs because f is periodic, with period 1.

We give this in Sub-section 6.6.2. We prove this in Sub-section 6.6.3.

6.6

The Blancmange function

6.6.2

247

Continuity of the Blancmange function

To verify that B is continuous on R, we must use the power of the "   definition of continuity. Theorem 1 Proof

The Blancmange function B is continuous at each point c 2 R.

We must show that

for each positive number ", there is a positive number  such that jBðxÞ  BðcÞj < ";

for all x satisfying jx  cj < :

(1)

Now, it follows, from the definition of B, that Bð xÞ  BðcÞ ¼ 1 P 1 n n 2n ð f ð2 xÞ  f ð2 cÞÞ; hence, by the infinite form of the Triangle Inequality

Sub-section 3.3.1.

n¼0

j Bð x Þ  Bð c Þ j 

1 X 1 j f ð2n xÞ  f ð2n cÞj: n 2 n¼0

(2)

 For all x and c, both of the numbers f(2nx) and f(2nc) lie in 0; 12 , so that the modulus of their difference is at most 12; that is 1 j f ð 2n x Þ  f ð 2n c Þ j  : 2 Now we choose an integer N such that 21N < 12 ". (This choice is possible, since   the sequence 21n is null.) It follows that 1 1 X X 1 1 1 n n f ð 2 x Þ  f ð 2 c Þ   j j n 2 2n 2 n¼N n¼N 1 2N 1 (3) < ": 2 Next, since each of the functions x 7! f(2nx), n ¼ 0, 1, . . ., N  1, is continuous, it follows that ¼

for each n ¼ 0, 1, . . ., N  1, there is a positive number n such that n

n

j f ð2 xÞ  f ð2 cÞj
X:  y x n

c

yn

x

(5)

n

Since f is differentiable at c, we know that, for each positive number ", there exists some positive number  such that    f ð x Þ  f ðc Þ  1 0   < "; for all x satisfying 0 < jx  cj < ;  f ð c Þ  xc  4 so that 1 (6) j f ð xÞ  f ðcÞ  f 0 ðcÞðx  cÞj  "jx  cj; for all x satisfying jx  cj < : 4 Now, since xn ! c and yn ! c as n ! 1, it follows that there are numbers X1 and X2 such that jxn  cj <  for all n > X1 ; and jyn  cj <  for all n > X2 ;

This is the definition of differentiability at c, but with 14 " in place of ", in order to obtain an " in our final result (5). In fact, this is a strict inequality if x 6¼ c. However later in the argument we will need to allow the possibility that x ¼ c, so we must write (6) with a weak inequality sign.

so, if we set X ¼ max{X1, X2}, we certainly have jxn  cj < 

and jyn  cj < 

for all n > X:

It now follows from (6) that, for all n > X, we have 1 j f ðxn Þ  f ðcÞ  f 0 ðcÞðxn  cÞj  "jxn  cj and 4 1 0 j f ðyn Þ  f ðcÞ  f ðcÞðyn  cÞj  "jyn  cj: 4

We put x ¼ xn and then x ¼ yn into (6).

6.6

The Blancmange function

249

Hence, we can apply the Triangle Inequality to obtain j f ðyn Þ  f ðxn Þ  f 0 ðcÞðyn  xn Þj ¼ jf f ðyn Þ  f ðcÞ  f 0 ðcÞðyn  cÞg  ff ðxn Þ  f ðcÞ  f 0 ðcÞðxn  cÞgj  j f ðyn Þ  f ðcÞ  f 0 ðcÞðyn  cÞj þ j f ðxn Þ  f ðcÞ  f 0 ðcÞðxn  cÞj 1 1  "jyn  cj þ "jxn  cj 4 4 1 1  "jyn  xn j þ "jyn  xn j 4 4 1 ¼ "jyn  xn j; for all n > X: 2

We first insert some terms and take them away again. We use the two inequalities that we have just verified for n > X. Since xn  c  yn, both jyn  cj and jxn  cj are  |yn  xn|.

Since we know that xn 6¼ yn, we can divide this inequality by the non-zero term yn  xn to obtain, for n > X   f ðy n Þ  f ðx n Þ  1 0   f ðcÞ  " < ":  y x 2 n n

&

This is the result (5) that we set out to prove.

Remark The hypothesis that xn and yn must lie on opposite sides of c cannot generally be omitted. For example, consider the function  x2 sin 1x ; x 6¼ 0, f ðxÞ ¼ 0; x ¼ 0, that you saw earlier is differentiable at 0, with f 0 (0) ¼ 0. If we set 1 1  ; for n ¼ 0; 1; 2; . . .; and yn ¼  xn ¼ ð2n þ 1Þp 2n þ 12 p then

 f ðy n Þ  f ðx n Þ ¼ yn  xn

1 2 ð2nþ12Þ p2

1 ð2nþ12Þp







 ð 0Þ 

1 ð2nþ1Þp

y = x2

y



2ð2n þ 1Þ  ¼ 2n þ 12 p 2 ! as n ! 1: p However, the value of this limit is not 0, the value of f 0 (0). We are now ready to prove the principal result in this sub-section. Theorem 2 c 2 R.

Problem 7, Sub-section 6.1.2.

The Blancmange function B is not differentiable at any point

Proof In order to apply the lemma, we construct two sequences {xn} and {yn} converging to c such that the corresponding sequence of difference quotients is not convergent. We use the method of repeated bisection to construct {xn} and {yn}.

y =f (x) xn yn x

y = – x2

6: Differentiation

250 Since B is periodic with period 1, we assume for simplicity that c 2 [0, 1]. [x0, y0] ¼ [0, 1]. Then, since c lies in one of the intervals 

We  start by defining 0; 12 and 12 ; 1 , we can define

 1 0; ; if c 2 0; 12 ; 1  ½x1 ; y1  ¼ 1 2  if c 2 2 ; 1 : 2; 1 ; With this definition of [x1, y1], we have that: 1. ½x1 ; y1  ½x0 ; y0 ; 2. y1  x1 ¼ 12 ; 3. x1  c  y1; 4. x1 ¼ 12 p1 and y1 ¼ 12 ð p1 þ 1Þ, for some integer p1. We can then repeat this process, bisecting the interval [x1, y1] to obtain [x2, y2], and so on. In this way, we obtain a sequence of closed intervals [xn, yn], for n ¼ 1, 2, . . ., such that: 1. ½xnþ1 ; ynþ1  ½xn ; yn ;  n 2. yn  xn ¼ 12 ; 3. xn  c  yn; 4. xn ¼ 21n pn and yn ¼ 21n ð pn þ 1Þ, for some integer pn. Properties 2 and 3 imply that both the sequences {xn} and {yn} converge to c, but we shall show that the sequence of difference quotients {Qn}, where Bð y n Þ  Bð x n Þ yn  xn n ¼ 2 ðBðyn Þ  Bðxn ÞÞ;

Qn ¼

We use Property 2 for the value of yn  xn.

is not convergent. It will then follow, from the lemma, that B is not differentiable at c. To prove that the sequence {Qn} is divergent, it is sufficient to prove that Qnþ1 ¼ Qn  1;

for n ¼ 0; 1; 2; . . .;

(7)

since it then follows that {Qn} cannot converge. To prove (7), we put 1 1 zn ¼ ðxn þ yn Þ ¼ nþ1 ð2pn þ 1Þ: 2 2 Note that ½xnþ1 ; ynþ1  ¼

½xn ; zn ; ½zn ; yn ;

if c 2 ½xn ; zn ; if c 2 ðzn ; yn :

We now claim that, for n ¼ 0, 1, 2, . . . 1 1 Bðzn Þ ¼ ðBðxn Þ þ Bðyn ÞÞ þ nþ1 : 2 2 It would then follow from (8) that, if [xnþ1, ynþ1] ¼ [xn, zn], then Qnþ1 ¼ 2

nþ1

n

ðBðzn Þ  Bðxn ÞÞ

¼ 2 ðBðyn Þ  Bðxn ÞÞ þ 1 ¼ Qn þ 1;

(8)

The structure of the proof is as follows

(8) ) (7) ) Theorem 2.

6.6

The Blancmange function

251

whereas, if [xnþ1, ynþ1] ¼ [zn, yn], then Qnþ1 ¼ 2nþ1 ðBðyn Þ  Bðzn ÞÞ ¼ 2n ðBðyn Þ  Bðxn ÞÞ  1 ¼ Qn  1: In either case, the required result (7) would then hold. Thus, in order to complete the proof that (7) holds, it is sufficient to prove that (8) holds. To verify (8), note that, for k ¼ 0, 1, 2, . . ., we have 9 1 > 2k xn ¼ nk pn ; > > > 2 > = 1 k (9) 2 yn ¼ nk ðpn þ 1Þ; > 2 > > > > 1 2k zn ¼ nkþ1 ð2pn þ 1Þ; ; 2

We shall shortly use the 1  k  P 1 expression 2k f 2 x for k¼0

B(x), hence some ks now appear.

and f ( p) ¼ 0 for any integer p. It then follows from the definition of B as an infinite series that Bðxn Þ þ Bðyn Þ ¼ ¼

Bðzn Þ ¼ ¼

1 X   1  k  f 2 x n þ f 2k y n k 2 k¼0 n1 X   1  k  f 2 x n þ f 2k y n ; k 2 k¼0

For, f (2kxn) ¼ f (2kyn) ¼ 0 if k > n  1.

and

1 X 1  k  f 2 zn 2k k¼0 n X 1  k  f 2 zn : 2k k¼0

For, f (2kzn) ¼ 0 if k > n.

We deduce from (9) that, for k ¼ 0, 1, 2, . . ., n  1, the terms 2kxn, 2kyn and 2 zn all lie in the interval  

k  1 1 k 2 xn ; 2 yn ¼ nk pn ; nk ðpn þ 1Þ ; 2 2

1  and so in some interval of the form 2 qk ; 12 ðqk þ 1Þ , where qk is an integer. Now, the restriction of f to such an interval is linear, so that we have        f 2k zn ¼ 12 f 2k xn þ f 2k yn ; k ¼ 0; 1; 2; . . .; n  1; k

hence n1 n1 X   1  k  1X 1  k  f 2 z f 2 x n þ f 2k y n : ¼ n k k 2 2 2 k¼0 k¼0

(10)

Finally

  1 1 1 n f ð 2 z Þ ¼ f p þ n n 2n 2n 2 1 ¼ nþ1 : 2 If we then add (10) and (11), we obtain (8), as required. This completes the proof of Theorem 2.

(11)

&

6: Differentiation

252

6.7

Exercises

Section 6.1 1. Determine whether each of the following functions f is differentiable at the specified point c; if it is, evaluate the derivative f 0 (c). x2 ; x  0, (a) f ð xÞ ¼ c ¼ 0; x2 ; x > 0, 1; x < 1, c ¼ 1; (b) f ð xÞ ¼ x2 ; x  1, tan x;  12 p < x < 13 p; c ¼ 13 p; (c) f ð xÞ ¼ x2 ; x  13 p; 1  x; x < 1, (d) f ðxÞ ¼ c ¼ 1; x  x2 ; x  1, 1 x sin x2 ; x 6¼ 0; c ¼ 0; (e) f ð xÞ ¼ 0; x ¼ 0; 1 sin x sin x ; x 6¼ 0; (f) f ð xÞ ¼ c ¼ 0: 0; x ¼ 0; 2. Write down an expression for a function f with domain (1, 2] and the following properties: (a) fL0 exists at 1, and fL0 (1) ¼ 1; (b) fR0 exists at 1, and fR0 (1) ¼ 2. Verify that f has the properties (a) and (b).

Section 6.2 1. Use the rules for differentiation to verify that the function 2 f ð xÞ ¼ loge ð1 þ xÞ þ ex ; x 2 ð1; 1Þ, is differentiable on its domain, and determine its derivative. 2. Write down (without justification) the derivatives of the following functions: 2 þ1 (a) f ð xÞ ¼ xx1 ; x 2 ð1; 1Þ; (b) f ð xÞ ¼ loge ðsin xÞ;

x 2 ð0; pÞ;

(c) f ð xÞ ¼ loge ðsec x þ tan xÞ;  cos xþsin x (d) f ð xÞ ¼ cos x 2 0; xsin x ;

 x 2 12 p;  1 4p :



1 2p

;

3. Prove that the function f(x) ¼ tanh x, x 2 R, has an inverse function f 1 that is differentiable on (1, 1), and find an expression for (f 1)0 (x).   4. Prove that the function f(x) ¼ tan x þ 3x, x 2 12 p; 12 p , has an inverse function f 1 that is differentiable on R, and find the value of ( f 1)0 (0). 5. Determine the derivatives of the following functions, assuming that they are differentiable on their domains: (a) f ð xÞ ¼ coth x; x 2 R  f0g; (b) f ð xÞ ¼ ðloge xÞloge x ; x 2 ð1; 1Þ:

6.7

Exercises

253

Section 6.3 1. The function f has domain [2, 2] and its graph consists of four line segments, as shown.

Identify (a) the local minima of f, (b) the minima of f, (c) the local maxima of f, and (d) the maxima of f, and state where these occur, 2. Let f be the function f ð xÞ ¼ 2x3  3x2 ; x 2 ½0; 2: Find (a) the local minima and minima of f, (b) the local maxima and maxima of f and state where these occur. 3. Prove that, of all rectangles with given perimeter, the square has the greatest area. 4. Verify that the conditions of Rolle’s Theorem are satisfied by the function f ð xÞ ¼ 1 þ 2x  x2 ;

x 2 ½0; 2;

and determine a value of c in (0,2) for which f 0 (c) ¼ 0. 5. Use Rolle’s Theorem to prove that, if p is a polynomial and l1, l2, . . ., ln are distinct zeros of p, then p0 has at least (n 1) zeros. 6. Use Rolle’s Theorem to prove that, for any real number l, the function 3 f ð xÞ ¼ x3  x2 þ l; x 2 R; 2 never has two distinct zeros in [0, 1]. Hint: Assume that f has two distinct zeros in [0, 1], and obtain a contradiction. 7. The function f is twice differentiable on an interval [a, b]; and, for some point c in (a, b), f(a) ¼ f(b) ¼ f(c). Prove that there exists some point d in (a, b) with f 00 (d) ¼ 0. 8. The function f is differentiable on a neighbourhood N of a point c; f 0 is continuous at c and f 0 (c) > 0. Prove that f is increasing on some neighbourhood of c. Harder: Give an example of a function f that is differentiable on R with the properties that (a) f 0 (0) > 0 and (b) there is NO open neighbourhood of 0 on which f is increasing. 9. Let the function f : ½0; 1 7! ½0; 1 be continuous on [0, 1] and differentiable on (0, 1). We know that there is at least one point c in [0, 1] for which f (c) ¼ c. Use Rolle’s Theorem to prove that, if f 0 ð xÞ 6¼ 1 in (0, 1), then there is exactly one such point c. Hint: Consider the function hð xÞ ¼ f ð xÞ  x; x 2 ½0; 1; assume that two such points c exist, and obtain a contradiction.

You saw this in Problem 2 of Sub-section 4.2.1.

6: Differentiation

254

Section 6.4 1. For each of the following functions, verify that the conditions of the Mean Value Theorem are satisfied, and determine a value of c that satisfies the conclusion of the theorem: (a) f ð xÞ ¼ 2x1 (b) f ð xÞ ¼ x3 þ 2x2 þ x; x 2 ½0; 1: x2 ; x 2 ½1; 1; 2. Use the Mean Value Theorem to prove that jsin b  sin aj  jb  aj; for a; b 2 R: 3. Use the Zero Derivative Theorem to prove that  pffiffiffiffiffiffiffiffiffiffiffiffiffi cosh1 x ¼ loge x þ x2  1 ; for x  1: 4. For the function f(x) ¼ x3  2x2 þ x, x 2 R, determine those points c where f 0 (c) ¼ 0. Using the Second Derivative Test, determine whether these correspond to local maxima, local minima, or neither. 5. Prove the following inequalities: (a) loge x  1  1x ; for x 2 ½1; 1Þ; 1

(b) 4x4  x þ 3; for x 2 ½0; 1; (c) loge ð1 þ xÞ > x  12 x2 ; for x 2 ð0; 1Þ: 6. Let f be defined on a neighbourhood of c, with f 0 (c) > 0. Prove that there is some punctured neighbourhood N of c such that f ð xÞ  f ðcÞ > 0; for x 2 N: xc pffiffiffi 7. By applying the Mean Value Theorempto the function f ð xÞ ¼ x on the ffiffiffiffiffiffiffi ffi 1 1 interval [100, 102], prove that 10 11 < 102 < 10 10 : 8. Let f be differentiable on a closed interval [a, b]. (a) Prove that, if the minimum of f on [a, b] occurs at a, then fR0 (a)  0; and that, if the minimum of f on [a, b] occurs at b, then fL0 (b)  0. (b) Prove that, if fR0 (a) < 0 and fL0 (b) > 0, then f 0 (x) ¼ 0 for some x 2 (a, b). Hint: Check that the minimum of f on [a, b] occurs at some point in (a, b). (c) By considering the function g (x) ¼ f(x)  kx, x 2 [a, b], prove that, if fR0 (a) < k < fL0 (b), then f 0 (x) ¼ k for some x 2 (a, b).

Section 6.5 1. Verify that the following functions satisfy the conditions of Cauchy’s Mean Value Theorem on [0, 2], and determine a value of c that satisfies the conclusion of the theorem f ð xÞ ¼ x4 þ 2x2

and

gð xÞ ¼ x3 þ 3x;

for x 2 ½0; 2:

2. Use l’Hoˆpital’s Rule to prove that the following limits exist, and evaluate the limits: 1 1 x þ sin xÞ (a) lim sinhðsin ; x

ðxþ3Þ (b) lim ð5xþ3Þx1 ;

x ; (c) lim 1xcos 2

xx (d) lim sinh sinð3x3 Þ :

x!0 x!0

3

x!1

x!0

2

This result is known as Darboux’s Theorem, and is an intermediate value theorem for f 0 .

7

Integration

We have already used the idea of the area of a set in the plane; for example, to define the number p and to prove that sinx x ! 1 as x ! 0. However, our earlier discussions begged the question of what exactly we mean by ‘area’. For simplicity, we shall restrict our attention in this book to defining only the area between a graph and the x-axis. In Section 7.1 we use the idea of lower and upper estimates to give a rigorous definition of such an area. This is done by trapping the desired area between underestimates and overestimates, each of which is the sum of the areas of suitably chosen rectangles.

Then the area between the graph y ¼ f (x), x 2 [a, b], and the segment [a, b] of the x-axis is defined to be A if: the least upper bound of the underestimates ¼ A, and: the greatest lower bound of the overestimates ¼ A. We call A the integral of f over [a, b], and denote it by Z b Z b f or f ð xÞdx: a

a

In practice, it would be inconvenient if we had to revert to the definition in order to tell whether a given function was integrable or not, and we shall verify several criteria for integrability. In Section 7.2, we identify some large classes of integrable functions, and verify the standard rules for integrable functions. In Section 7.3 we meet the Fundamental Theorem of Calculus which enables us to avoid use of the definition of integrability in order to evaluate many integrals. For instance, we verify the usual methods for integration by parts and integration by substitution.

It follows from the Fundamental Theorem of Calculus that we can think of integration as the operation inverse to differentiation.

255

7: Integration

256 Often it is not possible to evaluate an integral explicitly, and the best that we can do is to obtain upper and lower estimates for its value. In Section 7.4, we meet a range of inequalities for integrals, including the Triangle Inequality for integrals Z b  Z b    f   j f j; where a < b:  a

a

We use these inequalities to prove Wallis’s Formula for p, namely that   2 2 4 4 6 6 2n 2n p : : : : : :  : : ¼ ; lim n!1 1 3 3 5 5 7 2n  1 2n þ 1 2 and to establish the Maclaurin Integral Test, which enables us to determine the 1 1 P P 1 1 convergence or divergence of series such as np , for p > 0, and n log n. n¼1

n¼2

e

Finally, in Section 7.5, we discuss the evaluation of n!. It is easy to evaluate n! for values of n up to 10, say, by direct multiplication; but for n ¼ 100 or n ¼ 200, the number n! cannot be evaluated by a standard scientific calculator. Surprisingly, we can use integration techniques to obtain an excellent estimate for n!, called Stirling’s Formula pffiffiffiffiffiffiffiffinn n!  2pn as n ! 1: e Before starting to read the chapter, we recommend that you refresh your memory of greatest lower bounds and least upper bounds of functions. Recall that, if ƒ is a function defined on an interval I  R, then: 

A real number m is the greatest lower bound, or infimum, of ƒ on I if: 1. m is a lower bound of ƒ(I); 2. if m0 > m, then m0 is not a lower bound of ƒ(I).



At least not in 2005, when these words are being written.

We shall also explain the precise meaning of the symbol tilda: ‘’. You met sup and inf in Subsection 1.4.2. We often denote m by inf { f(x) : x 2 I}, inf f ð xÞ, x2I

inf f or simply by inf f. I

M is the least upper bound, or supremum, of ƒ on I if: 1. M is an upper bound of ƒ(I); 2. if M 0 < M, then M 0 is not an upper bound of ƒ(I).

In Sections 7.1 and 7.2 we shall discuss some new properties of infimum and supremum that are needed for our work on integrability. Also, we shall recommend that you omit working your way through quite a number of detailed proofs during your first reading of this chapter. This is not necessarily because they are particularly difficult, but simply in order to guide you through the key ideas first before you return to study some of the ‘gory details’ later on once you have grasped the overall picture.

7.1

We often denote M by sup { f(x) : x 2 I}, sup f ð xÞ, x2I

sup f or simply by sup f. I

The Riemann integral

Earlier we defined the number p to be the area of the unit disc. We obtained lower and upper estimates for p by trapping the area of the disc between the area of a 3  2n-sided inner polygon and the area of a 3  2n-sided outer polygon, and letting n ! 1.

Sub-section 2.5.4.

7.1

The Riemann integral

257

For simplicity, in this book we do not consider general areas in the plane, but restrict our attention to the area between a graph y ¼ f(x), x 2 [a, b], and the interval [a, b] of the x-axis. For a continuous function f, we certainly require such a definition to agree with our intuitive notion of area! However, it is not obvious that we can always say that the region between a graph and the x-axis has an area. For example, how can we define such an area for the following functions?

(a) f(x) =

x2, 0 ≤ x ≤ 1, 2, 1 < x ≤ 2.

{

(b) f(x) =

–2, 0 ≤ x < 1, 3, x = 1.

{

(c) f(x) =

1, 0 ≤ x ≤ 1, x rational, 0, 0 ≤ x ≤ 1. x irrational.

{

Our approach is to find lower and upper estimates for the area, if such a thing exists, that we can assign to each region by splitting it up into smaller regions with areas which we can approximate by rectangles, and use the fact that Area of a rectangle ¼ base  height. We now illustrate the underlying idea of these estimates by considering the situation when the function f is positive on [a, b] (see the diagrams below). First, we divide up the interval [a, b] into a family of smaller intervals, called a partition of [a, b]. Then we approximate the area by finding two sequences of rectangles each with one of the subintervals as base. In one sequence, we choose rectangles as large as possible so that the sum of their individual areas forms an underestimate for the ‘area’ of the region; in the other, we choose rectangles as small as possible so that the sum of their individual areas forms an overestimate for the ‘area’. Then, if there is a real number A with the properties: the least upper bound of the underestimates ¼ A,

When defining p, we used triangles rather than rectangles.

7: Integration

258 and: the greatest lower bound of the overestimates ¼ A, we define A to be the area between the graph and the x-axis.

We find that we can define such an area for the graphs (a) and (b) above, but not for the graph (c).

7.1.1

The Riemann integral and integrability

We now start on a process leading to the formal definition of the integral. Definitions A partition P of an interval [a, b] is a family of a finite number of subintervals of [a, b] P ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . . ; ½xi1 ; xi ; . . . ; ½xn1 ; xn g;

(1)

where

For brevity, we sometimes shorten this expression for P simply to ½xi1 ; xi gni¼1 or f½xi1 ; xi : 1  i  ng:

a ¼ x0 < x1 < x2 <    < xi1 < xi <    < xn1 < xn ¼ b: The points xi, 0  i  n, are called the partition points in P. The length of the ith subinterval is denoted by xi ¼ xi  xi1, and the mesh of P is the quantity kPk ¼ max fxi g.

a = x0 x1 .... xi–1 xi .... xn = b

1in

A standard partition is a partition with equal subintervals. For example, consider the partition P of [0, 1], where

     1 1 3 3 3 3 P¼ 0; ; ; ; ; ; ; 1 : 2 2 5 5 4 4

Equivalently P ¼ f½0; 0:5 ; ½0:5; 0:6 ; ½0:6; 0:75 ; ½0:75; 1 g:

Here 1 1 3 1 1 3 3 3  0 ¼ ; x2 ¼  ¼ ; x3 ¼  ¼ 2 2 5 2 10 4 5 20 3 1 x4 ¼ 1  ¼ ; 4 4 and the mesh of P is

 1 1 3 1 1 ¼ : kPk ¼ max ; ; ; 2 10 20 4 2 x1 ¼

and

P is not a standard partition of [0, 1], since not all its subintervals are of equal length.

7.1

The Riemann integral

259

Next, we introduce some notation, mi and Mi, associated with the height of the graph of a bounded function f on its subintervals. Since we are intending to develop a definition of integral that will apply to discontinuous functions as well as to continuous functions, we use the concepts of greatest lower bound and least upper bound of f on subintervals rather than minimum and maximum of f on the subintervals. We also introduce some quantities, L( f, P) and U( f, P), that correspond to underestimates and overestimates of the possible area.

We need to do this, since, if f is not continuous on a typical subinterval [xi1, xi], it may not possess a minimum or a maximum on the subinterval.

Definitions Let f be a bounded function on [a, b], and P a partition of [a, b] given by P ¼ {[xi1, xi] : 1  i  n}. We denote by mi and Mi the quantities mi ¼ inf f f ð xÞ : x 2 ½xi1 ; xi g and Mi ¼ supf f ð xÞ : x 2 ½xi1 ; xi g: Then the corresponding lower and upper Riemann sums for f on [a, b] are n n X X mi xi and U ð f , PÞ ¼ Mi xi : Lð f , PÞ ¼ i¼1

You met infimum and supremum earlier, in Subsection 1.4.2. The quantities mi and Mi exist, since f is bounded.

i¼1

y

y

y = f (x) m2

m1 x0

x1

m3 x2

x3

...

M2

mn

xn–1 xn x

M3

x0

Example 1 Let f(x) ¼ x, x 2 [0, 1], and let P ¼ partition of [0, 1]. Evaluate L( f, P) and U( f, P).

y

y = f (x)

M1

Mn x1

x2

x3

...

1 y=x

xn–1 xn x

1 1 1 1  0; 5 ; 5 ; 2 ; 2 ; 1 be a

Solution In this case, the function f is increasing and continuous. Thus, on each subinterval in [0, 1], the infimum of f is the value of f at the left end-point of the subinterval and the supremum of f is the value of f at the right end-point of the subinterval. Hence, on the three subintervals in P, we have   1 1 1 1 m1 ¼ f ð0Þ ¼ 0; x1 ¼  0 ¼ ; M1 ¼ f ¼ ; 5 5 5 5     1 1 1 1 1 1 3 M2 ¼ f x2 ¼  ¼ ; m2 ¼ f ¼ ; ¼ ; 5 5 2 2 2 5 10   1 1 1 1 M3 ¼ f ð1Þ ¼ 1; m3 ¼ f x3 ¼ 1  ¼ : ¼ ; 2 2 2 2 It then follows, from the definitions of L( f, P) and U( f, P), that 3 X L ð f , PÞ ¼ mi xi ¼ m1 x1 þ m2 x2 þ m3 x3 i¼1

1 1 3 1 1 ¼0 þ  þ  5 5 10 2 2 3 1 31 ; ¼0þ þ ¼ 50 4 100

0

1 5

1 2

1

x

We use the fact that, if a function is continuous on a closed interval, then it attains its infimum and supremum there, by the Extreme Values Theorem in Sub-section 4.2.3.

7: Integration

260

U ð f , PÞ ¼

3 X

Mi xi ¼ M1 x1 þ M2 x2 þ M3 x3

i¼1

1 1 1 3 1 ¼  þ  þ1 5 5 2 10 2 1 3 1 69 þ þ ¼ : ¼ 25 20 2 100

&

Problem 1 Evaluate L ( f, P) and U ( f, P) for the following function and partition of [0, 1]: 8 1 1 > > < 2x; 0  x < 2 ; 2 < x < 1; f ð xÞ ¼ 0; x ¼ 1 ; > > 2 : 1; x ¼ 1;    x 2 ½0; 1 ; and P ¼ 0; 13 ; 13 ; 34 ; 34 ; 1 : Hint: Be careful over the values of m2 and M3; you may find it helpful to sketch the graph of f. Example 2 Evaluate L( f, Pn) and U( f, Pn) for the following function and standard partition of [0, 1]

Notice how a careful laying out of the calculation for the various terms makes it straight-forward to calculate these two sums.

It is quite important that you tackle this problem, to make sure that you understand the notation being used. You should also read its solution carefully.

y 1

f ð xÞ ¼ x; x 2 ½0; 1 ; and

     1 1 2 i1 i 1 ; ;...; 1  ;1 ; Pn ¼ 0; ; ; ; . . . ; n n n n n n

y=x

and determine lim Lð f ; Pn Þ and lim U ð f ; Pn Þ, if these exist. n!1

n!1

Solution In this case, the function f is increasing and continuous. Thus, on each subinterval in [0, 1], the infimum of f is the value of f at the left end-point of the subinterval and the supremum of f is the value of f at the right end-point of the subinterval.

 i Hence, on the ith subinterval i1 n ; n in Pn, for 1  i  n, we have     i1 i1 i i ; Mi ¼ f mi ¼ f ¼ ¼ ; and n n n n xi ¼

i i1 1  ¼ : n n n

It then follows, from the definitions of L( f, Pn) and U( f, Pn), that n n X X i1 1  Lð f ; Pn Þ ¼ mi xi ¼ n n i¼1 i¼1 1 ¼ 2 n ¼

(

n X i¼1

i

n X

) 1

i¼1

 1 nð n þ 1Þ n1  n ¼ ; 2 n 2 2n

0

1 n

2 n

...

n –1 n

1

We follow the general structure of the solution to Example 1.

x

7.1

The Riemann integral U ð f ; Pn Þ ¼

n X

261 Mi xi ¼

i¼1

n X i i¼1

n



1 n

¼

n 1X i n2 i¼1

¼

1 nð n þ 1Þ n þ 1 ¼ :  n2 2 2n

:

It follows that n1 1 ¼ and 2n 2 nþ1 1 lim U ð f ; Pn Þ ¼ lim ¼ : n!1 n!1 2n 2 lim Lð f ; Pn Þ ¼ lim

n!1

n!1

&

Problem 2 Evaluate L( f, Pn) and U( f, Pn) for the following function and standard partition of [0, 1] f ð xÞ ¼ x2 ; x 2 ½0; 1 ; and

     1 1 2 i1 i 1 ; ; . . .; 1  ; 1 ; Pn ¼ 0; ; ; ; . . .; n n n n n n and determine lim Lð f , Pn Þ and lim U ð f , Pn Þ, if these exist. n!1

n!1

Properties of Riemann sums In all the examples that we have seen so far, the lower Riemann sum for f over an interval has been less than or equal to the corresponding upper Riemann R1 sum. Also, in Example 2 we found that the value that ‘we hope for’ for 0 xdx, namely 12, is greater than all the lower Riemann sums and less than all the upper Riemann sums; it seems reasonable to ask whether such a property holds in general. So we now examine some of the key properties of Riemann sums. We shall need to use some properties of least upper bounds and greatest lower bounds.

Here we are assuming that our rigorous treatment of integration will give the same answers as those that you obtained in your initial Calculus course!

We will set these out carefully as we need them.

Theorem 1 For any function f bounded on an interval [a, b] and any partition P of [a, b], L( f, P)  U( f, P). Proof Let P ¼ {[xi1, xi] : 1  i  n}. Then, on each subinterval [xi1, xi], 1  i  n, we have inf{ f(x) : x 2 [xi1, xi]}  sup { f(x):x 2 [xi1, xi]} – in other words, mi  Mi. It follows that n n X X mi xi  Mi xi ; i¼1

i¼1

in other words, L( f, P)  U( f, P), as required.

&

Next, we need some techniques for comparing the Riemann sums for different partitions on the interval [a, b]; this will enable us shortly to verify that, for two partitions P and P0 of [a, b], we must have L( f, P)  U( f, P0 ).

For each term in the left-hand sum is less than or equal to the corresponding term in the right-hand sum.

Theorem 3, below.

7: Integration

262 Notice that this fact does NOT follow from Theorem 1, which deals only with one partition at a time. Definition Let P ¼ {[x0, x1], [x1, x2], . . ., [xi1, xi], . . ., [xn1, xn]} be a partition of an interval [a, b]. Then a partition P0 of [a, b] is a refinement of P if the partition points of P0 include the partition points of P. A partition Q of [a, b] is the common refinement of two partitions P and P0 of [a, b] if the partition points of Q comprise the partition points of P together with the partition points of P0 .      For example, the partition P0 ¼ 0, 12 , 12 , 35 , 35 , 34 , 34 , 1 of [0, 1] is    a refinement of the partition P ¼ 0, 12 , 12 , 34 , 34 , 1 , since it simply has one additional partition point 35 as compared with P. Similarly,      P00 ¼ 0, 13 , 13 , 12 , 12 , 35 , 35 , 34 , 34 , 1 is also a refinement of P. However the     partition Q ¼ 0, 15 , 15 , 12 , 12 , 1 of [0, 1] is not a refinement of P, since its partition points do not include all the partition points of P.     Also, the common refinement of the partitions P ¼ 0, 12 , 12 , 34 , 34 , 1





  and P0 ¼ 0, 13 , 13 , 23 , 23 , 1 is

      1 1 1 1 2 2 3 3 0; ; ; ; ; ; ; ; ; 1 : 3 3 2 2 3 3 4 4

The point 34 is missing from Q.

We shall need the following crucial result in our work on refinements. I

Lemma 1 For any bounded function f defined on intervals I and J, with I  J, we have inf f  inf f x2J

Proof

x2I

and

sup f  sup f : x2I

x2J

By definition of greatest lower bound, we know that inf f  f ð xÞ, for x2J

J

Loosely speaking, the larger interval gives the function more space to get smaller and more space to get larger.

x 2 J. It follows, from the fact that I  J, that inf f  f ð xÞ; for x 2 I: x2J

For inf f is the greatest lower

Hence inf f is a lower bound for f on I, so that inf f  inf f . x2J

x2J

The proof that sup f  sup f is similar, so we omit it. x2I

x2I

x2I

&

bound of f on I.

x2J

Problem 3 Prove that, for any bounded function f defined on intervals I and J where I  J; sup f  sup f . x2I

x2J

Lemma 2 Let f be a bounded function on an interval [a, b]. Let P and P0 be partitions of [a, b], where P0 is a refinement of P that contains just one additional partition point. Then Lð f , PÞ  Lð f , P0 Þ Proof

and

U ð f , P0 Þ  U ð f , PÞ:

Let P be the partition P ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . . ; ½xi1 ; xi ; . . . ; ½xn1 ; xn g

of [a, b], and suppose that P0 contains an additional partition point c in the particular subinterval [, ] of P.

Loosely speaking, the addition of one partition point increases L and decreases U.

a = x0 x1 ... xi–1 xi ... xn = b

7.1

The Riemann integral

263

Note that c must be an interior point of [, ]. For we are assuming that all the partition points of partitions are distinct; and, if c were an end-point  or , then P0 would contain that point twice in its set of partition points. Since [, c] [, ], it follows from Lemma 1 that inf f  inf f

x2½;

and

x2½;c

inf f  inf f :

x2½;

(2)

x2½c;

Now, the terms in the lower sums L ( f, P) and L ( f, P0 ) are the same, except that the contribution to L ( f, P) associated with the interval [, ] is

We use  and  here rather than xi1 and xi simply in order to avoid sub-subscripts.

α

β c

inf f  ð  Þ;

x2½;

whereas the contribution to L( f, P0 ) associated with the intervals [, c] and [c, ] is inf f  ðc  Þ þ inf f  ð  cÞ:

x2½; c

x2½c; 

It then follows, from the inequalities (2), that inf f  ðc  Þ þ inf f  ð  cÞ  inf f  ðc  Þ þ inf f  ð  cÞ

x2½; c

x2½c; 

x2½; 

x2½; 

¼ inf f  fðc  Þ þ ð  cÞg x2½; 

¼ inf f  ð  Þ: x2½; 

Since the lower sums L ( f, P) and L ( f, P0 ) are the same, apart from these contributions to each, it follows that L ( f, P0 )  L ( f, P), as required. & The proof that U ( f, P0 )  U ( f, P) is similar, so we omit it. Lemma 2 shows that the addition of just one point to a partition increases the lower Riemann sum and decreases the upper Riemann sum. By applying this fact a finite number of times, we deduce the following general result. Theorem 2 Let f be a bounded function on an interval [a,b], and let P and P0 be partitions of [a,b], where P0 is a refinement of P. then 0

Lð f ; P Þ  L ð f ; P Þ

0

and U ð f ; P Þ  U ð f ; PÞ:

We now compare the Riemann sums for two different partitions, and discover that all the lower Riemann sums of a bounded function on an interval are less than or equal to all the upper Riemann sums. This is a significant improvement on the result of Theorem 1 that, for a given partition P of [a, b], L( f, P)  U( f, P). Theorem 3 Let f be a bounded function on an interval [a, b], and let P and P0 be partitions of [a, b]. Then Lð f , PÞ  U ð f , P0 Þ: Proof Let Q be the common partition of P and P0 . Then, since Q is a refinement of both P and P0 , we have: L( f, P)  L( f, Q), L( f, Q)  U( f, Q),

by Theorem 2, by Theorem 1,

U( f, Q)  U( f, P0 ),

by Theorem 2.

Loosely speaking, refining a partition increases L and decreases U.

7: Integration

264 It follows from this chain of inequalities that L( f, P)  U( f, P0 ), as & required. This result is exactly what makes our definitions of lower and upper Riemann sums useful – whatever the integral of f over an interval [a, b] might be, if it even exists, we are certainly using lower Riemann sums, all of which provide underestimates, and upper Riemann sums, all of which provide overestimates. Definitions define:  

Let f be a bounded function on an interval [a, b]. Then we

the lower integral of f on [a, b] to be

Sometimes written as Rb a f ð xÞdx.

Rb

a f ¼ sup Lð f , PÞ,



P R the upper integral of f on [a, b] to be ab f ¼ inf U ð f , PÞ, P

where P denotes partitions of [a, b]. Further, if the R b lower and upper integrals are equal, we define the integral of f on [a, b], a f , to be their common value; that is Z b Z b Zb f ¼ f ¼ f: a



a

Sometimes written as Rb a f ð xÞdx. Often written as

Rb a

f ðxÞdx.

a

It is all too easy to be lulled into a sense of false security by a firmly stated definition! So we now prove the following result that assures us that the above definitions make sense. Let f be a bounded function on an interval [a, b]. Then: Rb R (a) The lower integral a f and the upper integral ab f both exist; Rb R  (b) a f  ab f .

Theorem 4



Proof (a) Since f is bounded on [a, b], there is some number M such that j f (x)j  M on [a, b]; in particular, f ð xÞ  M;

for x 2 ½a; b :

Thus for any partition P ¼ {[xi 1, xi]: 1  i  n} of [a, b], we have f ð xÞ  M;

for x 2 ½xi1 ; xi and 1  i  n;

in particular, we have For, inf f ðxÞ  f ðxi Þ  M:

mi ¼ inf f ð xÞ  M:

½xi1 ; xi

½xi1 ; xi

It follows that Lð f ; P Þ ¼

n X i¼1

mi xi 

n X

Mxi

i¼1

¼M

n X

xi ¼ M ðb  aÞ:

i¼1

Since all the lower sums L ( f, P) are bounded above by M (b  a), it follows that the greatest lower bound of the lower sums, sup Lðf ; PÞ, must exist. Rb P This greatest lower bound is precisely the lower integral a f . 

Also,

Rb 

a

f  M ðb  aÞ.

7.1

The Riemann integral

265

The proof of the existence of the upper integral we omit it here.

Rb a

f is very similar, so

(b) Let P and P0 be any two partitions of [a, b]. We know, from Theorem 3, that Lð f , PÞ  U ð f , P0 Þ: If we fix P0 for the moment, then we know that U( f, P0 ) serves as an upper bound for all the lower Riemann sums L( f, P), whichever partition P may be. It follows, then, that the least upper bound of the L( f, P), the lower Rb integral a f , must satisfy the inequality  Z b f  U ð f , P0 Þ: (3) 

a



a

Rb It follows from the inequality (3) that a f serves as a lower bound for all  the upper Riemann sums U( f, P0 ), whichever partition P0 may be. It follows,  then, that the greatest lower bound of the U( f, P0 ), the upper Rb integral a f , must satisfy the inequality Z b Z b f  f: & a

Remark It follows, from the definition of the integral (when it exists) as Z b f ¼ sup Lð f , PÞ ¼ inf U ð f , PÞ; P

P

a

that, for any partition P of [a, b] Z b Lð f ; P Þ  f  U ð f ; PÞ: a

Now, already seen that, if f(x) ¼ x, x 2 [0, 1], and  have  we

 Pn ¼ 0; 1n ; 1n ; 2n ; . . . ; 1  1n ; 1 is a standard partition of [0, 1], then lim Lð f , Pn Þ ¼

n!1

1 2

and

1 lim U ð f , Pn Þ ¼ : 2

n!1

(4)

R1 Since f is bounded on [0, 1], we know that the lower integral 0 f exists. But  R1 Lð f , Pn Þ  0 f , so that, by the Limit Inequality Rule for sequences, it follows 

from (4) that 1  2

Z

1

f: 

0

(5) R1 0

Similarly, since f is bounded on [0, 1], we know that the upper integral f R1 exists. But 0 f  U ð f , Pn Þ, so that, by the Limit Inequality Rule for sequences, it follows from (4) that Z 1 1 (6) f  : 2 0

Example 2, above.

7: Integration

266 We then observe that Z 1 Z 1 1 f   f , in light of (5) and (6), and 2 Z 01 Z 1  0 f  f , by part (b) of Theorem 4. 0

0

R1 R1 It follows that we must have 0 f ¼ 0 f ¼ 12, so that f is integrable on [0, 1]  R1 and 0 f ¼ 12. 

In other words,

R1 0

xdx ¼ 12.

Problem 4 Use the result of Problem 2 to R 1prove that the function f(x) ¼ x2 is integrable on [0, 1], and evaluate 0 f . Problem 5 Let f be the constant  function x 7! k on [0, 1], and Pn ¼ 0; 1n ; 1n ; 2n ; . . . ; 1  1n ; 1 a standard partition of [0, 1]. Calculate the Riemann sums L( f, Pn)Rand U( f, Pn). Hence prove that f is 1 integrable on [0, 1], and evaluate 0 f . However, not all bounded functions defined on closed intervals are integrable! Example 3

Prove that the function

1; 0  x  1; f ð xÞ ¼ 0; 0  x  1;

y

x rational, x irrational,

y = 1, x ∈ 1

is not integrable on [0, 1]. Solution Let P ¼ {[x0, x1], [x1, x2], . . . , [xi1, xi], . . . , [xn1, xn]} be any partition of [0, 1]. Then n n X X Lð f , PÞ ¼ mi xi ¼ 0  xi i¼1

i¼1

¼ 0; and U ð f , PÞ ¼

n X

Mi xi ¼

i¼1

¼

n X i¼1 n X

y = 0, x ∉ 0

xi–1 xi

1

x

For, each subinterval [xi1, xi] contains both rational and irrational points.

1  xi xi ¼ 1:

i¼1

Since all the lower Riemann sums are 0, their least upper bound

R1 

0

f is also

zero. Similarly, since all the upper Riemann sums are 1, their greatest lower R1 bound 0 f is also 1. R1 R1 & Since 0 f 6¼ 0 f , it follows that f is not integrable on [0, 1]. 

7.1.2

Criteria for integrability

It would be tedious to have to go through the ab initio discussion of integrability on each occasion that we wished to determine whether a given function was integrable on a given closed interval. Therefore we now meet three criteria that we often use to avoid that process.

We suggest that at a first reading you omit ALL the proofs in this sub-section but read the rest of the text.

7.1

The Riemann integral

267

The first criterion is of particular interest in that its statement says nothing about the value of the integral itself: it mentions only the difference between the upper and the lower Riemann sums. Theorem 5 Riemann’s Criterion for integrability Let f be a bounded function on an interval [a, b]. Then

y area = U( f, P) – L( f, P)

f is integrable on ½a; b if and only if:

y = f (x)

for each positive number ", there is a partition P of [a, b] for which U ð f ; PÞ  Lð f ; PÞ < ":

(7) a

Proof

b

Recall that L( f, P)  U( f, P), for any partition P.

sup Lð f ; PÞ ¼ inf U ð f ; PÞ; over all partitions P of ½a; b ; P

P

denote by I the common value of these two quantities. It follows that, for any given positive number ", there are partitions Q and Q0 of [a, b] for which 1 1 (8) Lð f ; QÞ > I  " and U ð f ; Q0 Þ < I þ ": 2 2 Now let P be the common refinement of Q and Q0 . It follows from (8) and Theorem 2 that 1 Lð f ; PÞ  Lð f ; QÞ > I  "; 2 1 0 U ð f ; PÞ  U ð f ; Q Þ < I þ ": 2 Hence, we may deduce from subtracting these inequalities that



 1 1 U ð f ; P Þ  Lð f ; P Þ < I þ "  I  " 2 2

That is, that refining a partition increases L and decreases U.

¼ ": This is precisely the assertion (7). Suppose next that the statement (7) holds; that is, that for each positive number ", there is some partition P of [a, b] for which U ð f ; PÞ  Lð f ; PÞ < ": Since

x

Suppose, first, that f is integrable on [a, b]. Then

Rb a

f  U ðf ; PÞ and

from (9) that Z b a

f

Z

Rb 

a

(9)

f  Lðf ; PÞ, whatever partition P is, it follows

b

f  U ð f ; P Þ  Lð f ; P Þ a

< ":

(10)

Since the left-hand side of (10) is some non-negative number independent of ", Rb Rb it follows that a f  a f ¼ 0: In other words, f is integrable on [a, b], as 

required.

&

By (9).

7: Integration

268 Problem 6 Use Riemann’s Criterion to determine the integrability of the following functions on [0, 1]:

2; 0  x < 1, (a) f ð xÞ ¼ 3; x ¼ 1;

1; 0  x  1, x rational, (b) f ð xÞ ¼ 0; 0  x  1, x irrational. Our next criterion for integrability is phrased in terms of a sequence of partitions and its statement involves a value for the integral. Theorem 6

Common Limit Criterion

Let f be a bounded function on an interval [a, b]. (a) If f is integrable on [a, b], then there is a sequence {Pn} of partitions of [a, b] such that Z b Z b f and lim U ðf ; Pn Þ ¼ f: (11) lim Lð f ; Pn Þ ¼ n!1

a

n!1

a

(b) If there is a sequence {Pn} of partitions of [a, b] such that lim Lð f ; Pn Þ

n!1

and

lim U ð f ; Pn Þ both exist and are equal;

n!1

(12)

then R b f is integrable on [a, b] and the common value of these two limits is a f . Proof

Rb (a) Suppose first that f is integrable on [a, b], and let I denote a f . Then, corresponding to each integer n  1, there exist some partitions of [a, b], Qn and Qn0 say, for which   1 1 and U f ; Q0n < I þ : (13) Lð f ; Qn Þ > I  n n Now let Pn be the common refinement of Qn and Qn0 . It follows that 1 I  < Lð f ; Qn Þ ðby ð13ÞÞ n  Lð f ; Pn Þ ðsince Pn is a refinement of Qn Þ  U ð f ; Pn Þ    U f ; Q0n ðsince Pn is a refinement of Q0n Þ 1 ðby ð13ÞÞ m, then there is some x 2 I such that f (x) < m0 . 

To show that M is the least upper bound, or supremum, of f on I, check that: 1. f (x)  M, for all x 2 I; 2. if M0 < M, then there is some x 2 I such that f (x) > M0 . Problem 1 Let f be a bounded function on an interval I. Prove that, for any constant k inf fk þ f ð xÞg ¼ k þ inf f f ð xÞg and x2I

When using these strategies, we shall often use the numbers m þ " and M  " in place of m0 and M0 , to fit in with our increased use of " as a positive number that can be as small as we please.

x2I

supfk þ f ð xÞg ¼ k þ supf f ð xÞg: x2I

x2I

In fact you have already met one property of inf f and sup f in your study of integration.

Lemma 1, Sub-section 7.1.1.

Lemma 1 For any bounded function f defined on intervals I and J where I  J, we have inf f  inf f and sup f  sup f :

Loosely speaking, the larger interval gives the function more space to get smaller and more space to get larger.

x2J

x2I

x2I

x2J

A key tool in our work will be the following innocuous-looking result. Lemma 2 Let f be a bounded function on an interval I. If, for some number K, f (x)  f (y)  K for all x and y in I, then sup f  inf f  K. I

Proof

I

Note that K must be independent of the choice of x and y in I.

Since f (x)  f (y)  K for all x and y in I, we have

f ð xÞ  K þ f ð yÞ; for all x and y in I: (1) It follows from (1) that, for any choice whatsoever of y in I, then K þ f (y) serves as an upper bound for the set of all possible values of f (x) for x in I. It follows that For supff ð xÞ : x 2 I g is simply the least upper bound. (See also the Remark above.)

supf f ð xÞ : x 2 I g  K þ f ð yÞ; which we may rewrite in the form supf f ð xÞ : x 2 I g  K  f ð yÞ (2) In a similar way, it follows from (2) that supf f ð xÞ : x 2 I g  K serves as a lower bound for the set of all possible values of f(y) for y in I. It follows that supf f ð xÞ : x 2 I g  K  inf f f ð yÞ : y 2 I g:

(3)

We may rearrange the inequality (3) in the form supff ð xÞ : x 2 I g inf f f ð yÞ : y 2 I g  K. This is precisely the required result, since the letters & x and y in this last inequality are simply ‘dummy variables’.

Remarks 1. If f is bounded on I and f (x)  f (y) < K for all x and y in I, then it may not be true that sup f  inf f < K. For example, if f is the identity function on I

I ¼ (0, 1), then

I

For inf ff ð yÞ: y 2 I g is simply the greatest lower bound. (See also the Remark above.)

7: Integration

274 f ð xÞ  f ð yÞ < 1, for all x and y in ð0,1Þ, but sup f  inf f ¼ 1  0 ¼ 1: ½0,1

½0,1

2. The conclusion of Lemma 2 also holds if we know that, for some number K, j f (x)  f (y)j  K, for all x and y in I, since f ð xÞ  f ð yÞ  j f ð xÞ  f ð yÞj: In some situations, the following result related to Lemma 2 is also useful. Let f be a bounded function on an interval I. Then sup ff ð xÞ  f ð yÞg ¼ sup f  inf f :

Lemma 3

x; y2I

I

I

Proof We use the strategy for supremum mentioned earlier. Let x and y be any numbers in I. Then, in particular f ð xÞ  sup f ;

for all x 2 I ;

I

and, since f ð yÞ  inf f , for all y 2 I I

 f ð yÞ  inf f ;

for all y 2 I:

I

If we add these inequalities, we obtain f ð xÞ  f ð yÞ  sup f  inf f ; I

I

for all x; y 2 I;

so that sup f f ð xÞ  f ð yÞg  sup f  inf f : x;y2I

By the earlier Remark.

I

I

To prove the desired result, we now need to prove that for each positive number ", there are X and Y in I for which   f ð X Þ  f ðY Þ > sup f  inf f  ": I

I

By the strategy for supremum.

(4)

Now, by the definition of infimum and supremum on I, we know that, since " > 0, there exist X and Y in I such that

1 2

1 f ð X Þ > sup f  " 2 I

1 and f ðY Þ < inf f þ ": I 2

It follows from these two inequalities that  1   1  f ð X Þ  f ðY Þ > sup f  "  inf f þ " I 2 2 I ¼ sup f  inf f  ": I

I

&

This completes the proof.

Problem 2 Let f (x) ¼ x2 on the interval I ¼ (2, 3]. Determine inf f; sup f; inf f f ð xÞ  f ð yÞg and sup f f ð xÞ  f ð yÞg: I

I

x;y2I

x; y2I

7.2

Properties of integrals

275

Theorem 1 Combination Rules Let f and g be bounded functions on an interval I. Then: inf ð f þ gÞ  inf f þ inf g and supð f þ gÞ  sup f þ sup g;

Sum Rule

 I I I I  I Multiple Rule inf ðlf Þ ¼ l inf f and supðlf Þ ¼ l sup f ; for l > 0; I I I I   Negative Rule inf ðf Þ ¼  sup f and supðf Þ ¼  inf f . I

I

I

I

I

Proof We will prove only the first part of each Rule, for the proofs of the second parts are similar. Sum Rule For any x and y in I f ð xÞ  inf f

gð xÞ  inf g;

and

I

I

so that, by adding these two inequalities, we obtain f ð xÞ þ gð xÞ  inf f þ inf g: I

I

Since inf f þ inf g is a lower bound for f(x) þ g(x) on I, it follows that I

I

inf ðf þ gÞ  inf f þ inf g: I

I

I

Multiple Rule For any x in I, f ð xÞ  inf f so that I

lf ð xÞ  l inf f ;

for all x 2 I:

I

Since l > 0.

Thus inf ðlf ð xÞÞ  l inf f : x2I I  To prove that inf ðlf ð xÞÞ ¼ l inf f , we now need to prove that: x2I

I

for each positive number ", there is some X in I for which   lf ð X Þ < l inf f þ ": I

(5)

Now, since " > 0 and l > 0, we have l" > 0. It follows, from the definition of infimum of f on I, that there exists an X in I for which " f ð X Þ < inf f þ : I l Multiplying both sides by the positive number l, we obtain the desired result (5). Negative Rule For any x in I, f ð xÞ  sup f so that I

 f ð xÞ   sup f;

for all x 2 I :

I

Thus

inf ðf ð xÞÞ  sup f : x2I

I

7: Integration

276 To prove that inf ðf ð xÞÞ ¼ sup f , we now need to prove that: x2I

I

for each positive number ", there is some X in I for which  f ð X Þ <  sup f þ ":

(6)

I

Now, since " > 0 it follows, from the definition of supremum of f on I, that there exists an X in I for which f ð X Þ > sup f  " I

so that

 f ð X Þ < sup f þ ": I

&

This is precisely the inequality (6).

Problem 3 Prove that if f is a bounded function on an interval I, then: (a) supðlf Þ ¼ l sup f ; for l > 0; (b) supðf Þ ¼ inf f . I

I

I

I

We will now use these results on greatest lower bounds and least upper bounds to address the main topic of this section: integration!

7.2.2

Monotonic and continuous functions

We now prove that bounded functions on an interval [a, b] are integrable if they are either monotonic on [a, b] or continuous on [a, b]. This provides us with very many integrable functions! Neither of these two classes of functions includes the other. For example, the function f ð xÞ ¼ x  x2 ; x 2 ½0; 1 ; is continuous on [0,1] but is not monotonic on [0, 1]; while the function

x; 0  x < 1; f ð xÞ ¼ ; x 2 ½0; 1 ; 2; x ¼ 1;

You should sketch the graphs of these two functions to appreciate these statements.

is monotonic on [0,1] but is not continuous on [0, 1]. Theorem 2 Let f be a bounded function on an interval [a, b]. If f is monotonic on [a, b], then it is integrable on [a, b]. Proof We shall assume that f is increasing on [a, b]; if it is decreasing, the proof is similar. If f is constant on [a, b], it is certainly integrable on [a, b]. So, we shall assume, in addition, that f is also non-constant on [a, b]; it follows, in particular, that f (a) 6¼ f (b). We will prove that: for each positive number ", there is a partition P of [a, b] for which U ð f ; PÞ  Lð f ; PÞ < ": It then follows from the Riemann Criterion for integrability that f is integrable on [a, b]. For any given " > 0, let P be any partition of [a, b] with mesh jjPjj such that " kPk < f ðbÞf ðaÞ. Then, with the usual notation for Riemann sums

Theorem 5, Sub-section 7.1.2.

7.2

Properties of integrals

277

U ðf ; PÞ  Lðf ; PÞ ¼

n X

ðMi  mi Þxi ¼

i¼1

n X

ðf ðxi Þ  f ðxi1 ÞÞxi

Since f is increasing on [a, b].

i¼1

 kPk 

n X

ð f ðxi Þ  f ðxi1 ÞÞ

i¼1

¼ kPk  ð f ðbÞ  f ðaÞÞ < ":

&

This completes the proof.

Notice that the brevity of the above proof hides the fact that within the proof we are using quite a lot of work needed to prove the Riemann Criterion! Theorem 3 Let f be a bounded function on an interval [a, b]. If f is continuous on [a, b], then it is integrable on [a, b]. Proof

We will prove that:

for each positive number ", there is a partition P of [a, b] for which U ð f ; PÞ  Lð f ; PÞ < ": It then follows from the Riemann Criterion for integrability that f is integrable on [a, b]. Our key new tool here is the fact that, since f is continuous on an interval [a, b], which is a closed interval, it is therefore uniformly continuous on [a, b]. " It follows that, for any given " > 0; 2ðba Þ > 0, so that there is a positive number  for which " ; for all x and y in ½a, b satisfying jx  yj <  : (7) j f ðxÞ  f ð yÞj < 2ðb  aÞ Now, let P ¼ f½xi1 , xi gni¼1 be any partition of [a, b] with mesh jjPjj such that jjPjj < . Then, for each i we have, for all x and y in [xi1, xi]

You met uniform continuity in Section 5.5; this assertion is Theorem 2 there. Here the choice of  depends ONLY on ", the same choice whatever x and y may be. We " choose 2ðba Þ rather than ", for convenience later on.

jx  yj  xi  xi1  kPk < : It follows, from (7), that " ; for all x and y in ½xi1 ; xi ; 2ð b  aÞ so that, by Remark 2 following Lemma 2 " sup f ð xÞ  inf f ð xÞ  : 2 ð b  aÞ x2 ½ x ;x i1 i x2½xi1 ;xi j f ð x Þ  f ð y Þj
0 Lðl f ; Pn Þ ¼

n X

inf ðlf Þ  xi

½x ; x i¼1 i1 i n X

¼l

By the Combination Rules for inf and sup, Theorem 1, Sub-section 7.2.1.

inf f  xi

i¼1

½xi1 ; xi

¼ lLðf ; Pn Þ ! l

Z

b

f

as n ! 1;

a

while, if l < 0 Lðlf ; Pn Þ ¼

n X

inf ðlf Þ  xi

½x ; x i¼1 i1 i n X

¼l

By the Combination Rules for inf and sup.

sup f  xi

i¼1 ½xi1 ; xi

¼ lU ðf ; Pn Þ ! l

Z

b

f a

It follows that, for all real l, Lðlf ; Pn Þ ! l

as n ! 1: Rb a

f as n ! 1.

Rb

A similar argument shows that, for all real l, U ðlf ; Pn Þ ! l a f as n ! 1. Since the limits of the two sequences of Riemann sums are equal, it follows from part (b) of the Null Partitions Criterion for integrability, that lf is integrable on [a, b] and Z b Z b ðlf Þ ¼ l f : a

For, if l ¼ 0, the result is trivial.

Theorem 7, Sub-section 7.1.2.

a

Product Rule Since f and g are bounded on [a, b], there is some number M, say, such that j f ð x Þj  M

and

jgð xÞj  M;

for all x 2 ½a; b :

Now let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as n ! 1. It follows that, if x, y 2 [xi1, xi] for some i, then j f ð xÞgðxÞ  f ð yÞgðyÞj ¼ jff ð xÞgðxÞ  f ð xÞgð yÞg þ ff ðxÞgð yÞ  f ð yÞgðyÞgj ¼ j f ð xÞfgðxÞ  gð yÞg þ gð yÞff ð xÞ  f ð yÞgj  M jgðxÞ  gðyÞj þ M jf ðxÞ  f ð yÞj ( ) ( ) M

sup g  inf g ½xi1 ;xi

½xi1 ;xi

þM

sup f  inf f ; ½xi1 ;xi

½xi1 ;xi

Pn ¼ f½xi1 ; xi gni¼1 :

7: Integration

280 so that, by Lemma 2 in Sub-section 7.2.1 and Remark 2 following that Lemma (

sup ð fgÞ  inf ð fgÞ  M ½xi1 ;xi

)

(

sup g  inf g

½xi1 ;xi

½xi1 ;xi

)

þM

½xi1 ;xi

(8)

sup f  inf f : ½xi1 ;xi

½xi1 ;xi

In terms of the Riemann sums for fg, it follows!that U ð fg; Pn Þ  Lð fg; Pn Þ ¼

n X i¼1

M

n X i¼1

By definition of U and L.

sup ð fgÞ  inf ð fgÞ xi ½xi1 ;xi

½xi1 ;xi

(

)

sup g  inf g xi þ M

n X

½xi1 ;xi

½xi1 ;xi

i¼1

(

) sup f  inf f xi ½xi1 ;xi

½xi1 ;xi

By (8).

¼ M fU ðg; Pn Þ  Lðg; Pn Þg þ M fU ðf ; Pn Þ  Lðf ; Pn Þg ! 0 as n ! 1:

It then follows, from part (b) of the Null Partitions Criterion for integrability, that fg is integrable on [a, b].

Modulus Rule Let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as n ! 1. Then, if x, y 2 [xi  1, xi] for some i, we have   j f ð xÞj  j f ð yÞj  j f ð xÞ  f ð yÞj; so that

  j f ð xÞj  j f ð yÞj  sup f  inf f

By part (a) of the Null Partitions Criterion for integrability (Theorem 7, Sub-section 7.1.2), since f and g are integrable on [a, b]. Pn ¼ f½xi1 ; xi gni¼1 : This follows from the ‘reverse form’ of the Triangle Inequality, that you met in Sub-section 1.3.1.

½xi1 ;xi

½xi1 ;xi

and therefore sup j f j  inf j f j  sup f  inf f : ½xi1 ;xi

½xi1 ;xi

(9)

½xi1 ;xi

½xi1 ;xi

In terms of the Riemann sums for j f j, it follows that U ðj f j; Pn Þ  Lðj f j; Pn Þ ¼

n X



!

sup j f j  inf j f j xi ½xi1 ;xi

i¼1

½xi1 ;xi

By definition of U and L.

!

n X

sup f  inf f xi ½xi1 ;xi

i¼1

½xi1 ;xi

¼ U ð f ; Pn Þ  Lð f ; Pn Þ !0

By Remark 2 following Lemma 2, Sub-section 7.2.1.

as n ! 1:

It then follows, from part (b) of the Null Partitions Criterion for integrability, & that jfj is integrable on [a, b].

By (9). By part (a) of the Null Partitions Criterion for integrability (Theorem 7, Sub-section 7.1.2), since f is integrable on [a, b].

Our next result is often overlooked as obvious – which it is not! a

Theorem 5

The Sub-interval Theorem

(a) Let f be integrable on [a, Rb], and Rlet a c; it then follows from the above equation, using the Triangle Inequality for integrals, that    Z   F ð xÞ  F ðcÞ 1  x    f ðcÞ ¼ ff ðtÞ  f ðcÞgdt  xc  xc c Z x 1  ðsince x > cÞ: j f ðtÞ  f ðcÞjdt xc c But, since f is continuous at c, we know that there must exist some positive number  such that 1 j f ðtÞ  f ðcÞj < "; for all x satisfying jx  cj < : 2 It follows that, for x satisfying c < x < c þ , we have   Z x F ð xÞ  F ðcÞ  1   1  f ð c Þ "dt  xc  xc 2 c

1 ¼ " 2 < ": A similar argument applies in the case that x < c; so that, for all x satisfying 0 < jx  cj < , we have   F ð xÞ  F ðcÞ    < ":  f ð c Þ  xc  This completes the proof.

7.4.2

We omit the details.

&

Wallis’s Formula

In this sub-section we use the method of Reduction of Order together with various inequalities between integrals to establish a remarkable formula for the number p.

Reduction of Order method Quite often we need to evaluate an integral In that involves a non-negative integer n. A common approach to such integrals is to relate the value of In to the value of In1 or In2 by a reduction formula, using integration by parts. Rp Example 4 Let In ¼ 02 sinn xdx, n ¼ 0, 1, 2, . . .. (a) Evaluate I0 and I1. (b) Prove that In ¼ n1 n In2 , for n  2.

This is just another name for a recurrence formula.

7: Integration

294 (c) Deduce from part (b) that, for n  1 I2n ¼

1:3: . . . :ð2n  3Þð2n  1Þ p  2:4: . . . :ð2n  2Þð2nÞ 2

I2nþ1 ¼

and

2:4: . . . :ð2n  2Þð2nÞ : 3:5: . . . :ð2n  1Þð2n þ 1Þ

Solution (a) We can evaluate the first two integrals easily Z p 2 p I0 ¼ 1dx ¼ ; and 2 0 Z p 2 p I1 ¼ sin xdx ¼ ½ cos x 02 ¼ 1: 0

(b) We first write In in the form In ¼ parts, we find that, for n  2

Rp 2

0

sin x sinn1 xdx. Using integration by

Z p 2

p In ¼ ðcos xÞ sinn1 x 02  ðcos xÞðn  1Þ sinn2 x cos xdx ¼ 0 þ ð n  1Þ ¼ ð n  1Þ

Z

Z

0

p 2

2

cos x sinn2 xdx

0 p 2



 1  sin2 x sinn2 xdx

0

¼ ðn  1ÞfIn2  In g: We can then rewrite this result in the form nIn ¼ (n  1)In2, so that In ¼

n1 In2 : n

(c) If we replace n in the formula for In in part (b) by 2n, 2n  2, 2n  4, . . ., in turn, we obtain I2n ¼

2n  1 I2n2 ; 2n

I2n2 ¼

2n  3 I2n4 ; 2n  2

I2n4 ¼

Hence I2n ¼

2n  1 I2n2 2n

¼

ð2n  3Þð2n  1Þ I2n4 ð2n  2Þð2nÞ

¼

ð2n  5Þð2n  3Þð2n  1Þ I2n6 ; ð2n  4Þð2n  2Þð2nÞ

2n  5 I2n6 ; . . .: 2n  4

We will integrate sin x and differentiate sinn1 x.

7.4

Inequalities for integrals

295

and so on. Continuing this process, we obtain I2n ¼ ¼

1:3: . . . :ð2n  3Þð2n  1Þ I0 2:4: . . . :ð2n  2Þð2nÞ 1:3: . . . :ð2n  3Þð2n  1Þ p  : 2:4: . . . :ð2n  2Þð2nÞ 2

For I0 ¼ p2 :

Similarly 2n I2n1 2n þ 1 ð2n  2Þð2nÞ I2n3 ¼ ð2n  1Þð2n þ 1Þ

I2nþ1 ¼

.. .

Problem 4

¼

2:4: . . . :ð2n  2Þð2nÞ I1 3:5: . . . :ð2n  1Þð2n þ 1Þ

¼

2:4: . . . :ð2n  2Þð2nÞ : 3:5: . . . :ð2n  1Þð2n þ 1Þ

Let In ¼

R1 0

&

For I1 ¼ 1.

ex xn dx, n ¼ 0, 1, 2, . . ..

(a) Evaluate I0. (b) Prove that In ¼ e  nIn1, for n  1. (c) Deduce the values of I1, I2, I3 and I4.

Wallis’s Formula We now use the various results that we have just proved for the integral Rp In ¼ 02 sinn xdx to verify some surprising results. Theorem  5 Wallis’s Formula  2n 2n (a) lim 21 : 23 : 43 : 45 : 65 : 67 : . . . : 2n1 : 2nþ1 ¼ p2 : n!1 pffiffiffi n!Þ2 22n pffiffi ¼ p: (b) lim ðð2n Þ! n

Each of the two limits is called Wallis’s Formula.

n!1

In the next problem, we ask you to establish a number of relationships between the terms of the two sequences in the statement of Theorem 5. Problem 5

2 2n

n!Þ 2 pffiffi, Let an ¼ 21 : 23 : 43 : 45 : 65 : 67 : . . . : 2n2n 1 : 2n2nþ 1 and bn ¼ ðð2n Þ! n

n  1. (a) Evaluate an and bn, for n ¼ 1, 2 and 3. (b) Verify that b2n ¼ 2nþ1 n an , for n ¼ 1, 2 and 3. 2 2nþ1 (c) Prove that bn ¼ n an , for all n  1.

You may omit tackling this problem and the following proof at a first reading.

7: Integration

296 Rp Proof of Theorem 5 Let In ¼ 02 sinn xdx, for all n  0, and let the sequences {an} and {bn} be as given in Problem 5. (a) Using the formulas for I2n and I2nþ1 in part (c) of Example 4 above, we obtain I2n 1:3:3:5:5: . . .:ð2n  1Þð2n  1Þð2n þ 1Þ p  ; ¼ 2:2:4:4:6: . . .:ð2n  2Þð2nÞð2nÞ 2 I2nþ1

The numerator of I2nþ1 is the same as the denominator of I2n.

which we arrange in the form an ¼

I2nþ1 p  : 2 I2n

Notice that, in order to complete the proof of part (a), it is sufficient to show that I2nþ1 ! 1 as n ! 1: (2) I2n We do this as follows.

 Since 0  sin x  1 for x 2 0; p2 , we have h pi sin2n x  sin2nþ1 x  sin2nþ2 x; for x 2 0; : 2 It follows, from the Inequality Rule for integrals, that I2n  I2nþ1  I2nþ2 : Thus

Note that

I2nþ1 I2nþ2  1 I2n I2n 2n þ 1 : ¼ 2n þ 2

I2nþ2 ¼

(3)

2n þ 1 I2n ; 2n þ 2

by the reduction formula for In.

By letting n ! 1 in (3) and using the Squeeze Rule for sequences, we deduce that the limit (2) holds, as desired. (b) We know, from part (c) of Problem 5, that 2n þ 1 an ; for all n  1: n We also know, from the proof of part (a) of this theorem, that p as n ! 1: an ! 2 b2n ¼

It follows, by applying the Product Rule for sequences to the above formula for bn2, that p b2n ! 2  ¼ p as n ! 1: 2 Hence,pby the continuity of the square root function, we conclude that ffiffiffi & bn ! p as n ! 1. Unfortunately the sequences in Theorem 5 converge rather slowly as n pffiffiffi increases, so they are of little value in determining p or p to any reasonable degree of approximation. In the next chapter, we shall meet practical ways of estimating p.

Section 8.5.

7.4

Inequalities for integrals

7.4.3

297

Maclaurin Integral Test

We now introduce a method for determining the convergence or divergence of 1 P f ðnÞ, where the terms are positive and decrease to certain series of the form 1 n¼1 P 1 zero. In particular, we show that the series np converges if p > 1 and n¼1 diverges if 0 < p  1. We use the fact that each term in such a series can be regarded as a contribution to a lower Riemann sum or upper Riemann sum for a suitable integral. Theorem 6 Maclaurin Integral Test Let f be positive and decreasing on [1, 1), and let f(x) ! 0 as x ! 1. Then: 1 R n  P (a) f ðnÞ converges if the sequence 1 f: n 2 N is bounded above; (b)

n¼1 1 P

f ðnÞ diverges if the sequence

R n

n¼1

1



f: n 2 N tends to 1 as n ! 1.

You saw in Sub-section 3.2.1 that this series converges if p  2.

In these results, the number 1 may be replaced by any convenient positive integer.

Before proving the theorem, we illustrate the underlying ideas. Example 5

Let Pn1 be the standard partition of [1, n] with n  1 subintervals f½1; 2 ; ½2; 3 ; . . .; ½i; i þ 1 ; . . .; ½n  1; n g:

(a) Determine the lower and upper Riemann sums, L( f,Pn1) and U( f,Pn1), for the function f ð xÞ ¼ x12 , x 2 ½1; 1Þ. 1 1 P P 1 1 (b) Deduce that the series n2 converges, and that 1  n2  2. n¼1

n¼1

Solution (a) Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n  1, m i ¼ f ð i þ 1Þ ¼ Mi ¼ f ðiÞ ¼

1 ð i þ 1Þ 2

;

1 : i2

Also, each subinterval in the partition has length 1.

In2 fact, the sum of this series is p 6 , but to prove this goes outside the scope of this book.

7: Integration

298 Hence the lower and upper Riemann sums for f are n1 X

Lð f ; Pn1 Þ ¼

mi  1 ¼

i¼1 n1 X

1 1 1 þ 2 þ  þ 2 ; 2 2 3 n

1 1 1 þ 2 þ  þ : 2 1 2 ðn  1Þ2 i¼1 1 P 1 (b) Let sn denote the nth partial sum of the series n2 ; thus U ð f ; Pn1 Þ ¼

Mi  1 ¼

n¼1

1 1 1 1 sn ¼ 2 þ 2 þ    þ þ 2: 2 1 2 n ð n  1Þ It follows that Lð f ; Pn1 Þ ¼ sn  1

and

U ð f ; Pn1 Þ ¼ sn 

1 : n2

Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have Z n Lð f ; Pn1 Þ  f  U ðf ; Pn1 Þ; 1

so that sn  1 

Z 1

n

dx 1  sn  2 : 2 x n

(4)

Now, the sequence {sn} is increasing, since the series has positive terms. Also, from (4), we have Z n dx sn  þ1 2 1 x  n 1 ¼  þ1 x 1   1 1 (5) þ 1 ¼ 2   2: ¼ 1 n n Thus {sn} is bounded above. Hence, by the Monotone Convergence Theorem for sequences, {sn} is convergent, so that 1 X 1 is convergent: 2 n n¼1 Since s1 ¼ 1 and {sn} is increasing, the sum of the series is at least 1. Finally, we deduce from (5), using the Limit Inequality Rule for sequences, that lim sn  2; so the sum of the series is at most 2. Hence n!1

1

Example 6

1 X 1  2: n2 n¼1

&

Let Pn1 be the standard partition of [1, n] with n  1 subintervals f½1; 2 ; ½2; 3 ; . . .; ½i; i þ 1 ; . . .; ½n  1; n g:

7.4

Inequalities for integrals

299

(a) Determine the lower and upper Riemann sums, L( f, Pn1) and U( f, Pn1), for the function f ð xÞ ¼ 1x ; x 2 ½1; 1Þ. 1 P 1 (b) Deduce that the series n diverges. n¼1

Solution (a) Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n  1 1 ; m i ¼ f ð i þ 1Þ ¼ iþ1 1 Mi ¼ f ðiÞ ¼ : i Also, each subinterval in the partition has length 1. Hence the lower and upper Riemann sums for f are n1 X

Lð f ; Pn1 Þ ¼

mi  1 ¼

i¼1

1 1 1 þ þ  þ ; 2 3 n

n1 X

1 1 1 : Mi  1 ¼ þ þ    þ 1 2 n1 i¼1 1 P 1 (b) Let sn denote the nth partial sum of the series n; thus U ð f ; Pn1 Þ ¼

n¼1

1 1 1 1 þ : sn ¼ þ þ    þ 1 2 ð n  1Þ n It follows that 1 and U ð f ; Pn1 Þ ¼ sn  : n Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have Z n Lð f ; Pn1 Þ  f  U ð f ; Pn1 Þ; Lð f ; Pn1 Þ ¼ sn  1

1

so that sn  1 

Z 1

n

dx 1  sn  ; x n

or 1 sn  1  loge n  sn  : (6) n In particular, sn  loge n. Then, since loge n ! 1 as n ! 1, it follows, from the Squeeze Rule for sequences that tend to infinity, that sn ! 1 as n ! 1. 1 P 1 & Hence, the series n diverges. n¼1

Remarks We can in fact get more information from the above argument than just the result of the example. 1. If we define the sequence { n} by the expression 1 1 1 n ¼ 1 þ þ þ    þ  loge n 2 3 n

7: Integration

300 then 1 nþ1  loge nþ1 n Z nþ1 1 dx ¼  nþ1 x n  Z nþ1  1 1  dx  0; ¼ nþ1 x n

nþ1  n ¼

so that { n} is decreasing. Also, it follows from (6), using the fact that  n ¼ sn  loge n, that 1 n  ; n thus the sequence { n} is bounded below by 0. Hence the sequence { n} tends to a limit,  say, as n ! 1. The value of  clearly lies between 1 (for  1 ¼ 1) and 0. This number is called Euler’s constant, and occurs often in Analysis. In fact,  ¼ 0:57721 56649 01532 86060 65120 90082 40243 . . .: 1 2. Although the sequence sn ¼ 1 þ 12 þ    þ n1 þ 1n tends to infinity, we can make more precise statements than that, arising from (6). We can rewrite the inequalities in (6) in the following form 1 þ loge n  sn  1 þ loge n; n if we then divide throughout by logen and let n ! 1, we obtain, by the Limit Inequality Rule for sequences, that sn ! 1 as n ! 1: loge n We can write this last limit in an equivalent form as ‘sn  logen as n ! 1’ – in other words, ‘the behaviour of sn and logen are essentially the same for large n’.

Problem 6 Use the Maclaurin Integral Test to determine the behaviour 1 P 1 of the series np , for p > 0, p 6¼ 1. n¼1

R Show that xðlogdx xÞ2 ¼  log1 x, for x > 1, and hence prove e 1 e P 1 that the series 2 converges. nðlog nÞ

Problem 7

n¼2

e

R dx ¼ loge ðloge xÞ, for x > 1, and hence Problem 8 Show that x log ex 1 P 1 prove that the series n log n diverges. n¼2

e

Sequences revisited We can apply many ideas similar to those in Example 5 to obtain useful information about certain types of convergent sequences.

By definition of the sequence { n}. For loge n  sn  1n :

We give here just the first 35 decimal places.

We shall address this  notation more carefully in the next section.

7.4

Inequalities for integrals

Example 7

301

1 Prove that n þ1 1 þ n þ1 2 þ    þ 2n ! loge 2 as n ! 1.

 

 1 Solution Let f ð xÞ ¼ 1þx , x 2 [0, 1]; and let Pn ¼ 0; 1n ; 1n ; 2n ; . . .; n1 n ;1 be the standard partition of [0, 1] into n sub-intervals of equal length 1n.

 i Since f is decreasing on [0, 1], it follows that on the ith sub-interval i1 n ; n we have   i 1 : mi ¼ f ¼ n 1 þ ni

Recall that

 i mi ¼ inf f ð xÞ : x 2 i1 n ;n :

Thus the lower Riemann sum for f on Pn is n X 1 1 i n 1 þ n i¼1 n X 1 1 1 1 ¼ ¼ þ þ  þ : nþi nþ1 nþ2 2n i¼1

L ð f ; Pn Þ ¼

Since f is decreasing on [0, 1], it is integrable on [0, 1]. Hence, as n ! 1, we have Z 1 Z 1 1 1 1 dx þ þ    þ ¼ Lð f ; P n Þ ! f¼ nþ1 nþ2 2n 1 þx 0 0 ¼ ½loge ð1 þ xÞ 10 ¼ loge 2: & Example 7 illustrates a general technique that is often useful. Strategy n ! 1.

If f is positive and decreasing on [0, 1], then: 1n

Problem 9

Prove that n



1 n2 þ12

n   R1 P f ni ! 0 f as

i¼1

The trick in applying this strategy is to make a good choice of f.

 1 1 þ n2 þ2 ! p4 as n ! 1. 2 þ    þ n2 þn2

You have already met the corresponding ideas for divergent sequences in Example 6 and Remark 2 after that example, so we state without comment the general strategy illustrated there. Strategy If f is positive and decreasing on [1, 1), f(x) ! 0 as x ! 1, and n Rn Rn P f ðiÞ  1 f as n ! 1. 1 f ! 1 as n ! 1, then i¼1

Recall that this means that n P

f ðiÞ R n ! 1 as n ! 1: 1 f

i¼1

Remark In fact under the hypotheses of this strategy

n P i¼1

Rn f ðiÞ  ð 1 f Þ þ c, for any fixed

The reason for this will be explained in Section 7.5.1.

number c; in particular situations we may choose c in any convenient way. Problem 10

pffiffiffi Prove that 1 þ p1ffiffi2 þ p1ffiffi3 þ    þ p1ffiffin  2 n as n ! 1.

Here you will use both the strategy and the subsequent remark.

7: Integration

302 Proof of the Maclaurin Integral Test Theorem 6 Maclaurin Integral Test Let f be positive and decreasing on [1, 1), and let f(x) ! 0 as x ! 1. Then: 1 R n  P (a) f ðnÞ converges if the sequence 1 f : n 2 N is bounded above; (b)

n¼1 1 P

f ðnÞ diverges if the sequence

R n

n¼1

1

 f : n 2 N tends to 1 as n ! 1.

Rn Proof Let In denote the integral 1 f , let sn ¼ f(1) þ f(2) þ    þ f(n) denote the nth partial sum of the series, and let Pn1 be the standard partition of [1, n] with n  1 subintervals f½1; 2 ; ½2; 3 ; . . .; ½i; i þ 1 ; . . .; ½n  1; n g:

Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n  1 m i ¼ f ð i þ 1Þ

and Mi ¼ f ðiÞ:

Also, each subinterval in the partition has length 1. Hence the lower and upper Riemann sums for f are Lð f ; Pn1 Þ ¼

n1 X

m i  1 ¼ f ð 2 Þ þ f ð 3Þ þ    þ f ð n Þ

i¼1

¼ sn  f ð1Þ; U ð f ; Pn1 Þ ¼

n1 X

M i  1 ¼ f ð 1 Þ þ f ð 2Þ þ    þ f ð n  1 Þ

i¼1

¼ sn  f ðnÞ: Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have Lð f ; Pn1 Þ  In  U ð f ; Pn1 Þ; so that sn  f ð1Þ  In  sn  f ðnÞ:

(7)

Case 1: {In} is bounded We are now assuming that, for some M, In  M, for all n. It follows from (7) that sn  f ð1Þ þ M; for all n:

7.5

Stirling’s Formula for n!

303

Thus the increasing sequence {sn} is bounded above, and so, by the Monotone Convergence Theorem, it is convergent. 1 P f ðnÞ is convergent. Hence the series n¼1

Case 2: {In} is not bounded

R nþ1 The sequence {In} is increasing, since Inþ1  In ¼ n f  0. Since we are now assuming that {In} is not bounded, it follows that In ! 1 as n ! 1. Now, from (7), sn  In; so, by the Squeeze Rule for sequences which tend to infinity, sn ! 1 as n ! 1: Hence the series

1 P

&

f ðnÞ is divergent.

n¼1

7.5

Stirling’s Formula for n!

For small values of n, we can evaluate n! directly by multiplication or by using a scientific calculator. Problem 1 n

Complete the following table of values of n! n!

n

n!

n

n!

1

1

6

720

20

2 3

2 6

7 8

5040 40 320

30 40



4

24

9

362 880

50

5

120

10

3 628 800

60

As n increases, n! grows very quickly; for instance: around 14! seconds have elapsed since the birth of Christ; around 18! seconds have elapsed since the formation of the Earth. A calculator soon becomes useless for evaluating n!; thus the author’s current scientific calculator (2005) gives that 69! ’ 1.7  1098, but an error message when asked the value of 70!. Stirling’s Formula gives us a way of estimating n! for large values of n. In order to state the formula, though, we must first introduce some notation that enables us to compare the behaviour of positive functions of n for large values of n.



14! ’ 8:7  1010



18! ’ 6:4  1015

7.5.1

The tilda notation

For many numbers that arise in the normal way, we often use the symbol ‘’’ to pffiffiffi denote ‘is approximately equal to’. For example, 2 ¼ 1:4142135623730950 4880 . . . (where we have given only the first 20 decimal places); for many pffiffiffi pffiffiffi purposes it is sufficient to use estimates such as 2 ’ 1:414 or even 2 ’ 1:4,

Often an estimate is sufficient for our purposes.

7: Integration

304 since the error involved is small. But what do we mean by ‘the error involved is small’? Sometimes we mean that the error, the difference between the exact value and the approximation, is a small number; sometimes we mean that the percentage error, namely actual error  100; percentage error ¼ exact value is a small number. To handle the behaviour of positive functions of n for large values of n, we introduce a very precise mathematical notation. Notation

For positive functions f and g with domain N, we write f ðnÞ  gðnÞ as n ! 1;

to mean that gf ððnnÞÞ ! 1

as n ! 1. 2

þ 10 ! 1 as For example, n2 þ 1000n þ 10  n2 as n ! 1, since n þ1000n n2 1 1 sinð1nÞ n ! 1; and sin n  n as n ! 1, since 1 ! 1 as n ! 1. n

Notice that, if f(n)  g(n) as n ! 1, then g(n)  f(n) as n ! 1. Also, if f(n)  g(n) as n ! 1, g(n) ! 1 and c is a given number, then f(n) þ c  g(n) as n ! 1, since, as n ! 1

‘For example, the error is less than 1’. For example, ‘the percentage error is less than 1%’.

The notation is ONLY used for positive functions. We say ‘f tilda g’ or ‘f twiddles g’. You can think of this as saying that the percentage error involved by replacing f(n) by g(n) is small for large n. For, lim sinx x ¼ 1: x!0

For, gf ððnnÞÞ ! 1 , gf ððnnÞÞ ! 1.

f ð nÞ þ c f ð nÞ c ¼ þ ! 1 þ 0 ¼ 1: gðnÞ gðnÞ gðnÞ Notice, however, that the statement f(n)  g(n) as n ! 1 does NOT mean that f(n)  g(n) ! 0 or even that f(n)  g(n) is bounded. For example, n2 þ 1000n þ 10  n2 as n ! 1, yet  2    n þ 1000n þ 10  n2 ¼ 1000n þ 10 ! 1 as n ! 1: In situations like this example, ‘’ compares the sizes of the dominant terms in f and g. Problem 2 Find two pairs from the following functions such that fi(n)  fj(n) as n ! 1    f1 ðnÞ ¼ sinðn2 Þ; f2 ðnÞ ¼ sin n12 ; f3 ðnÞ ¼ 1  cos 1n ; f4 ðnÞ ¼ n22 ; f5 ðnÞ ¼ n12 ; f6 ðnÞ ¼ 1  1n; f7 ðnÞ ¼ 2n12 :

Remark If ‘ is finite and positive, then the statements ‘f ðnÞ ! ‘ as n ! 1’ and

‘f ðnÞ  ‘ as n ! 1’

are equivalent, since each is equivalent to the statement ‘f ðnÞ ! 1 as n ! 1’: ‘ We can also, in this situation, legitimately say that f(n)  ‘ for large n. The Combination Rules for handling tilda are direct consequences of the corresponding Combination Rules for sequences.

For example, we may write   1 n  e as n ! 1; 1þ n since  1þ

 1 n ! e as n ! 1: n

Note that we only use  for positive funtions.

7.5

Stirling’s Formula for n!

305 For example, if

Combination Rules

f1 ðnÞ ¼ n2 þ n; g1 ðnÞ ¼ n2 ; f2 ðnÞ ¼ n3 þ n; g2 ðnÞ ¼ n3 ;

Let f1(n)  g1(n) and f2(n)  g2(n) as n ! 1. Then: Sum Rule

f1(n) þ f2(n)  g1(n) þ g2(n);

Multiple Rule Product Rule

lf1(n)  lg1(n), for any number l > 0; f1(n)  f2(n)  g1(n)  g2(n);

Reciprocal Rule

1 f1 ðnÞ

Example 1 as n ! 1.

then  2    n þ n þ n3 þ n  n2 þ n3 ;  2  2 5 n þ n  5n ðl ¼ 5Þ;  2    n þ n  n3 þ n  n2  n3 ¼ n5 ;

 g11ðnÞ.

1 1  : n2 þ n n2 1

1

Prove that, if f(n)  g(n) as n ! 1, then ð f ðnÞÞn  ðgðnÞÞn

Since f(n)  g(n) as n ! 1, we have

Solution

f ð nÞ !1 gð nÞ

as n ! 1;

so that loge

f ð nÞ ! 0 as n ! 1; gð nÞ

and therefore 1 f ð nÞ log !0 n e gðnÞ But

Here we use the fact that the function loge is continuous at 1.

as n ! 1:

 1 1 f ð nÞ f ð nÞ n loge ¼ loge n gðnÞ gð nÞ 1

¼ loge

ð f ðnÞÞn 1

;

ðgðnÞÞn

so that 1

loge

ð f ð nÞ Þ n 1

ð gð nÞ Þ n

! 0 as n ! 1:

We now use the fact that the exponential function is continuous at 0; it follows from the previous limit that 1

ð f ðnÞÞn 1

ðgðnÞÞn

! 1 as n ! 1; 1

1

which is precisely the statement that ð f ðnÞÞn  ðgðnÞÞn as n ! 1.

&

Problem 3 Give specific examples of functions f and g to show that 1 1 ðf ðnÞÞn  ðgðnÞÞn as n ! 1 does not imply that f(n)  g(n) as n ! 1. Hint:

7.5.2

Try f(n) ¼ n2 and g(n) ¼ n.

Stirling’s Formula

Stirling’s Formula was discovered in the eighteenth century, in an analysis of a gambling problem!

7: Integration

306 Theorem 1

Stirling’s Formula pffiffiffiffiffiffiffiffinn n!  2pn as n ! 1: e

pffiffiffiffiffiffiffiffi n Problem 4 Use your calculator to evaluate 2pn ne for n ¼ 5. For this value of n does the expression approximate n! to within 1%? For small values of n, Stirling’s Formula gives reasonable approximations to n!, and the percentage error quickly decreases as n increases: n 10

n!

Stirling’s approximation

3 628 800

3 598 696 18

Error 0.83% (1 in 120)

18

20 52

2.433  10 8.066  1067

2.423  10 8.053  1067

0.42% (1 in 240) 0.16% (1 in 620)

100

9.333  10157

9.325  10157

0.09% (1 in 1170)

We illustrate the use of Stirling’s Formula as follows. If 200 coins are tossed, then the probability of there being exactly 100 heads and 100 tails is    200 1 200! 200  ¼ : 100 2 ð100!Þ2 2200 It follows from Stirling’s Formula that this probability is pffiffiffiffiffiffiffiffiffiffi 200200 400p  200! ’ pffiffiffiffiffiffiffiffiffiffi   e 2 2 200 100 ð100!Þ 2 200p  100  2200 e 1 pffiffiffi 10 p 1 : ¼ 17:724 . . . In other words, the probability of there being exactly 100 heads and 100 tails is about 1 in 18 – rather higher than you might expect. ¼

Problem 5 Hint:

 1n

nn n!1 n!

Prove that lim

¼ e.

Use the result of Example 1.

Problem 6 Use Stirling’s Formula to estimate the following numbers to two significant figures:    300  300 1 300! 1 300 (a)  ; (b)  : 3 150 2 3 ð100!Þ Problem 7

Use Stirling’s Formula to determine a number l such that    2n 4n  l22n as n ! 1: n 2n

This fact is easily proved using Probability Theory.

Here we substitute n ¼ 200 and n ¼ 100 into Stirling’s Formula.

7.5

Stirling’s Formula for n!

7.5.3

307

Proof of Stirling’s Formula

Theorem 1

Stirling’s Formula pffiffiffiffiffiffiffiffinn n!  2pn as n ! 1: e

We divide up the proof into a number of steps, for clarity.

Proof

Step 1: Setting things up We consider the function f ðxÞ ¼ loge x;

x 2 ½1; n ;

and the standard partition Pn1 of the interval [1, n] with n  1 subintervals f½1; 2 ; ½2; 3 ; . . .; ½i; i þ 1 ; . . .; ½n  1; n g: We consider also the sequence of numbers fcn g1 n¼2 , where cn is the total area between the concave curve y ¼ loge x, for x 2 [1, n], and the polygonal line with vertices (1, 0), (2, loge 2), (3, loge 3), . . ., (n, loge n), as illustrated below. This consists of n  1 small slivers. y y = loge x

total area of silvers = cn

y = loge x

contribution to cn

(i + 1,loge(i + 1)) (i, logei) 1

2

3 ... i i + 1

...

n

x

Step 2: Calculating areas The area between the curve y ¼ loge x and the x-axis, for x 2 [1, n], is Z n loge xdx ¼ ½x loge x  x n1 1

¼ n loge n  ðn  1Þ:

(1)

Next, the area between the polygonal line and the x-axis is 1 2fLð f , Pn1 Þ

þ U ð f ; Pn1 Þg;

(2)

and, since f is increasing, we have Lð f , Pn1 Þ ¼ loge 1 þ loge 2 þ    þ loge ðn  1Þ ¼ loge ðn  1Þ! ¼ loge n!  loge n

(3)

and U ð f , Pn1 Þ ¼ loge 2 þ    þ loge n ¼ loge n!:

(4)

Substituting from (3) and (4) into (2), we find that the area between the polygonal line and the x-axis is

7: Integration

308 loge n!  12 loge n: It follows from (1) and (5) that cn ¼ n loge n  ðn  1Þ  loge n! þ 12 loge n

(5)

1

nnþ2 ¼ loge n1 : e n!

(6)

Step 3: Behaviour of the sequence {cn} It is obvious from the definition of cn that the sequence {cn} is positive and increasing. In order to apply the Monotone Convergence Theorem to the sequence, we must prove next that {cn} is bounded above. So, let i be an integer with 1  i  n  1, and let A ¼ ði; loge iÞ; B ¼ ði þ 1; loge ði þ 1ÞÞ; and C ¼ ði þ 1; loge iÞ: Then, as illustrated in the diagram in the margin BC ¼ loge ði þ 1Þ  loge i   1 ¼ loge 1 þ : i Next, let the tangent at A to the curve meet the line BC at D. Since AC ¼ 1 and the slope of the line AD is 1i , it follows that CD ¼ 1i . Hence BD ¼ CD  BC   1 1 1 ¼  loge 1 þ  2: i i 2i Hence the contribution to the area cn of this sliver between the lines x ¼ i and x ¼ i þ 1 is at most 1 area of  ABD ¼ AC  BD 2 1 1 1  1 2 ¼ 2: 2 2i 4i If we now sum such areas over i ¼ 1, 2, . . ., n  1, we obtain n1 1X 1 cn  4 i¼1 i2 

Here we use the inequality loge ð1 þ xÞ > x  12x2 ; for x > 0; that you met in Section 6.7, Exercise 5(c) on Section 6.4.

1 1X 1 : 4 i¼1 i2

Since this infinite series converges, it follows that the sequence {cn} is bounded above. It follows from the Monotone Convergence Theorem that the sequence {cn} is convergent.

Step 4: Properties of the sequence {an}, where an ¼ ecn Since {cn} is convergent and the exponential function is continuous, it follows that, if we set an ¼ ecn , for n ¼ 2, 3, . . ., then the sequence {an} is also convergent. Thus, using the expression (6) for cn, we have 1 nnþ2 (7) an ¼ n1 ! L; as n ! 1; e n! for some non-zero number L.

The actual value of the bound does not matter here.

We now introduce {an} so that we can use Wallis’s Formula in the next Step.

L 6¼ 0 since the exponential does not take the value 0.

7.6

Exercises

309

It follows from the formula for an in (7) that a2n n2nþ1 e2n1  ð2nÞ! ¼  1 2 a2n e2n2 ðn!Þ ð2nÞ2nþ2 1

¼

ð2nÞ!n2 e ðn!Þ2 22nþ2 1

;

(8)

after some cancellation.

Step 5: Proving Stirling’s Formula We can rewrite (8) in the form 1

ð2nÞ!n2 a2n e ¼  pffiffiffi : a2n ðn!Þ2 22n 2

(9)

But, by Wallis’s Formula, the first quotient on the right-hand side of (9) tends to p1ffiffi as n ! 1. It follows that, if we let n ! 1 on both sides of (9), we obtain p

L2 1 e ¼ pffiffiffi  pffiffiffi ; L p 2 so that

e L ¼ pffiffiffiffiffiffi : 2p

Knowing the value of L, we can then rewrite (7) in the form 1

nnþ2 e ! pffiffiffiffiffiffi ; n1 e n! 2p by the Reciprocal Rule for sequences, it follows that pffiffiffiffiffiffi 2p en1 n! pffiffiffi ! : n e n n . By the definition of tilda, this limit is exactly equivalent to the relation pffiffiffiffiffiffiffiffinn n!  2pn as n ! 1; e & that we set out to prove.

Remark It is quite remarkable how many of the techniques that we have met so far in the book are needed in order to prove this apparently straight-forward result.

7.6

Exercises

Section 7.1 1. Sketch the graph of the function

1  j xj; 1 < x < 1; f ð xÞ ¼ 1; x ¼ 1: Determine the minimum, maximum, infimum and supremum of f on [1, 1].

You met Wallis’s Formula in Theorem 5 of Subsection 7.4.2. ðn!Þ2 22n pffiffiffi pffiffiffi ¼ p: n!1 ð2nÞ! n lim

7: Integration

310 2. Let f be the function

j xj; f ð xÞ ¼ 1 2;

 1 < x < 1; x ¼ 1:

Evaluate L( f, P) and U( f, P) for each of the following partitions P of [1, 1]:     (a) P ¼ 1;  12 ;  12 ; 0 ; 0; 12 ; 12 ; 1 ;    (b) P ¼ 1;  14 ;  14 ; 13 ; 13 ; 1 : 3. Let f be the function

1  x; 0  x < 1; f ð xÞ ¼ 2; x ¼ 1: (a) Using the standard partition Pn of [0, 1] with n equal subintervals, evaluate L( f, Pn) and U( f, Pn). R1 (b) Deduce that f is integrable on [0, 1], and evaluate 0 f : 4. Let f be the function f(x) ¼ x3, x 2 [0, 1]. (a) Using the standard partition Pn of [0, 1] with n equal subintervals, evaluate L( f, Pn) and U( f, Pn). R1 (b) Deduce that f is integrable on [0, 1], and evaluate 0 f :

 5. Let f be the function f (x) ¼ sin x, x 2 0; p2 :

 (a) Using the standard partition Pn of 0; p2 with n equal subintervals, evaluate L ( f, Pn) and U ( f, Pn).

 Rp (b) Deduce that f is integrable on 0; p2 , and evaluate 02 f : Hint:

Use the formula sin A þ sin þ ðn  1ÞBÞ 1 ðA þ BÞþ    þ sinðA  sin 2 nB n1   sin A þ ¼ B ; B 6¼ 0: 2 sin 12 B

6. Prove that the function

1 þ x; f ð xÞ ¼ 1  x;

0  x  1; 0  x  1;

This formula can be proved by Mathematical Induction.

x rational; x irrational;

is not integrable on [0, 1]. 7. Prove that Dirichlet’s function

1 ; if x is a rational number pq, in lowest terms, with q > 0, f ð xÞ ¼ q 0; if x is irrational. is integrable on [0, 1]. Hint: Verify that there are at most 12 nðn þ 1Þ points x in [0, 1] for which f ð xÞ > 1n :

Section 7.2 1. For bounded functions f and g on an interval I, prove that supð f þ gÞ  sup f þ sup g: I

I

I

For the interval I ¼ [1, 2], write down functions f and g for which strict inequality holds in this result.

You met this function in Subsection 5.4.2.

7.6

Exercises

311

2. Write down functions f and g and an interval I for which it is not true that supð fgÞ ¼ sup f  sup g: I

I

I

3. Write down functions f and g on [0, 1] that are not integrable on [0, 1] but such that f þ g is integrable on [0, 1]. 4. (a) Write down functions f and g on [0, 1] such that f is integrable, g is not integrable, and fg is integrable. (b) Write down a function f on [0, 1] such that jfj is integrable but f is not integrable on [0, 1]. 5. Prove that, if the functions f and g are integrable on [a, b], then so is the function max{f, g}. Hint:

Use the formula maxfa; bg ¼ 12 fa þ b þ ja  bjg, for a, b 2 R.

6. Prove that, if f and g are integrable on [a, b], then Z b 2 Z b Z b  fg  f2 g2 : a

a

a



Hint:

This is known as the Cauchy–Schwarz Inequality for integrals.

Use the Cauchy-Schwarz Inequality

n P

2 ai bi

i¼1

 

n P i¼1

a2i



n P

i¼1

 b2i :

Section 7.3 1. Write down a primitive of each of the following functions: pffiffiffiffiffiffiffiffiffiffiffiffiffi (a) f ð xÞ ¼ x2  9; x 2 ð3; 1Þ; (b) f ð xÞ ¼ sinð2x þ 3Þ  4 cosð3x  2Þ;

x 2 R;

(c) f ð xÞ ¼ e sinð3xÞ; x 2 R: 2. Using the result of part (c) of Exercise 1, write down a primitive F of the function f ð xÞ ¼ e2x sinð3xÞ, x 2 R, for which F(p) ¼ 0. 2x

3. Show that the following   functions are all primitives of the function f(x) ¼ sech x, x 2  p2 ; p2 : (b) F2 ð xÞ ¼ 2 tan1 ðex Þ;   (a) F1 ð xÞ ¼ tan1 ðsinh xÞ; 1 (c) F3 ð xÞ ¼ sin ðtanh xÞ; (d) F4 ð xÞ ¼ 2 tan1 tanh 12 x : Re 4. Let In ¼ 1 xðloge xÞn dx, for n  0: (a) Prove that In ¼ 12 e2  12 nIn1 , for n  1. (b) Evaluate I0, I1, I2 and I3. 5. Evaluate each of the following integrals, using the suggested substitution where given: Rp R 1 ðtan1 xÞ2 (a) 02 tanðsin xÞ cos xdx; (b) 0 1þx2 dx ðtry u ¼ tan1 xÞ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R1  (c) 0 x5 þ 2x2 x6 þ 4x3 þ 4dx;  1  R p dx 1  tan2 ð12xÞ 2 (d) 0 2þcos x dx try u ¼ tan 2 x and use the identity cos x ¼ 1 þ tan2 1x ; ð2 Þ R e2 loge ðloge xÞ Re 7 (f) e dx; (e) 1 8x loge xdx; x R p sinð2xÞ R 4 dx pffiffiffi 2 (g) 0 1þ3 cos2 x dx; (h) 1 ð1þxÞpffiffix ðtry u ¼ xÞ:

You met this in Subsection 1.3.3.

7: Integration

312

Section 7.4 1. Prove the following inequalities: R 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 1 x4 1 ffi dx  10 (a) 0 x3 2ð1 þ x99 Þdx  12 ; (b) 0 pffiffiffiffiffiffiffiffiffiffi : 1þ3x97 R 30 1 2. Prove that 12  0 1þx 2x99 dx  2: R 2   2  Þ sinð99xÞ 3. Prove that  0 x ðx3 dx   4: 20 1þx 1 R dx P 12 1 4. Show that , and deduce that is 3 ¼ 2ðloge xÞ 3 xðloge xÞ2 nðloge nÞ2 n¼2 convergent. 1 R dx P 1 1 2 5. Show that 1 ¼ 2ðloge xÞ , and deduce that 1 is divergent. xðloge xÞ2 n¼2 nðloge nÞ2   1 1 6. Prove that lim n ðnþ1 þ ðnþ2 þ    þ ð2n1Þ2 ¼ 12 : Þ2 Þ2 n!1

7. Determine the convergence or divergence of the following series: 1 1 P P 1 1 (a) (b) loge n ; loge ðloge nÞ : ðlog nÞ n¼2

8. Prove that

e

n¼2

n P loge k k

k¼2

ðloge nÞ

 12 ðloge nÞ2 , as n ! 1.

9. Prove that, if f is integrable on [a, b], where a < b, and f(x) > 0, for all x 2 [a, b], Rb then a f > 0: Hint: Use the following steps: Rb (a) Assume that the result is false, so that in fact a f ¼ 0; Rd (b) Verify that, if ½c; d  ½a; b , then c f ¼ 0; (c) Prove that there is a partition P of [a, b] with U( f, P)  b  a, and deduce that there is some subinterval [a1, b1] of [a, b] with a1 < b1 and sup f  1; ½a1 ,b1

Rb (d) Using the fact that a11 f ¼ 0, prove that there is a sequence 1 f½an , bn g1 n ¼ 1 of intervals with ½anþ1 , bnþ1 ½an , bn and sup f  n ; ½an ,bn

(e) Prove that there is some point c in [a, b] such that an ! c as n ! 1, and that c 2 [an, bn], for all n  1; (f) Verify that f(c) ¼ 0.

Section 7.5 

ð3nÞ! 3n n!1 n

1. Prove that lim

1n

¼ 27 e3 :

2. Use Stirling’s Formula to estimate each of the following numbers to two significant figures: pffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1400  400! ð800pÞ 400  ; (b) : (a) 200 2 ð100!Þ4 4400 3. Use Stirling’s Formula to determine the number l such that ð8nÞ! ðð2nÞ!Þ

4

l

216n 3

n2

as n ! 1:

8

Power series

The evaluation of given functions at given points in their domains is of great importance. If we are dealing with a polynomial function, the calculation of the function’s values presents no problem: it is simply a matter of arithmetic. For example, if 1 1 1 1 f ð xÞ ¼ 1þ x  x2  x3 þ x4 ; 2 2 6 4 then 1 1 1 1 13 f ð1Þ ¼ 1þ   þ ¼ : 2 2 6 4 12 On the other hand, the sine function is rather different: there is no way of calculating its values precisely merely by the use of arithmetic. It is important to be able to estimate values of functions that cannot be evaluated exactly, and to know how close the estimates are to the actual values of the function. In this chapter we are primarily concerned with a procedure for calculating approximate values of functions, like the sine function, whose values cannot be calculated easily at all points in their domains. We see how, in principle, we can use a certain sequence of polynomials to calculate the values of the sine function, for example, to any desired degree of accuracy; and how we can represent sin x, for any x, as the sum of a series. We shall see, for instance, that the polynomial pð xÞ ¼ x  16 x3 approximates f (x) ¼ sin x to within 105 for all x in the interval [0, 0.1], and that, in general

Section 8.2, Example 1, and Theorem 3.

x3 x5 x7 þ  þ   ; for x 2 R: 3! 5! 7! In Section 8.1 we define the Taylor polynomial Tn(x) and discuss some particular examples of functions for which the Taylor polynomials appear to provide useful approximations. In Section 8.2 we investigate how closely Taylor polynomials approximate a given function in the neighbourhood of a point a in its domain, and we establish a criterion for when we can say that sin x ¼ x 

f ð xÞ ¼ lim Tn ð xÞ ¼ n!1

1 X

an ð x  aÞ n :

n¼0

In this case, we say that f is the sum function of the power series

1 P

an ðx  aÞn .

n¼0

In Sections 8.3 and 8.4 we look at the behaviour of power series in their own right; that is, we consider functions which are defined by power series. In 1 P particular, in Section 8.3 we see that a power series an ðx  aÞn behaves in n¼0

one of three ways: it converges only for x ¼ a, or it converges for all x, or it

313

8: Power series

314 converges in an interval (a  R, a þ R), where R is a positive number called the radius of convergence. In Section 8.4 we discuss various rules for power series, including the Sum, Multiple and Product Rules. We also find that it is valid to differentiate or integrate a given power series term-by-term, and that this does not affect the radius of convergence. Finally, in Section 8.5 we look briefly at various methods for estimating the number p, and prove that p is irrational. Some of the proofs in this chapter are not particularly illuminating, and you may wish to leave them till a second reading of the chapter. You will need a calculator handy while you work through Sections 8.1 and 8.2.

8.1

Taylor polynomials

8.1.1

What are Taylor polynomials?

Let f be a continuous function defined on an open interval I containing the point a. It follows from the definition of continuity that

That is, I is a neighbourhood of a.

lim f ð xÞ ¼ f ðaÞ:

x!a

Thus we can write f ð xÞ ’ f ðaÞ;

for x near a:

In geometrical terms, this means that we can approximate the graph y ¼ f(x) near a by the horizontal line y ¼ f(a) through the point (a, f(a)). For most continuous functions, this does not give a very good approximation. However, if the function is differentiable on I, then we can obtain a better approximation by using the tangent line through (a, f(a)) instead of the horizontal line. The tangent to the graph at (a, f(a)) has equation y  f ð aÞ ¼ f 0 ð aÞ xa or y ¼ f ðaÞ þ f 0 ðaÞðx  aÞ; so, for x near a, we can write f ð xÞ ’ f ðaÞ þ f 0 ðaÞðx  aÞ: This approximation is called the tangent approximation to f at a. Notice that the function f and the approximating polynomial f ð aÞ þ f 0 ð aÞ ð x  aÞ have the same value at a and the same first derivative at a. Example 1 Solution

Determine the tangent approximation to the function f(x) ¼ ex at 0. Here f ð xÞ ¼ ex ;

f ð0Þ ¼ 1;

f 0 ð xÞ ¼ ex ;

f 0 ð0Þ ¼ 1:

We can think of the tangent at (a, f(a)) as the line of best approximation to the graph near a.

8.1

Taylor polynomials

315

Hence the tangent approximation to f at 0 is ex ’ f ð0Þ þ f 0 ð0Þðx  0Þ ¼ 1 þ x:

&

Problem 1 Determine the tangent approximation to each of the following functions f at the given point a: (a) f ð xÞ ¼ ex ;

a ¼ 2;

(b) f ð xÞ ¼ cos x;

a ¼ 0:

So far we have seen two approximations to f(x) for x near a: f ð xÞ ’ f ðaÞ

ða constant functionÞ; 0

f ð xÞ ’ f ðaÞ þ f ðaÞðx  aÞ ða linear functionÞ: If the function f is twice differentiable on a neighbourhood I of a, then we can find an even better approximation to f(x) by considering the expression f ð x Þ  f f ð aÞ þ f 0 ð aÞ ð x  aÞ g ð x  aÞ 2

:

Let F ð xÞ ¼ f ð xÞ  ff ðaÞ þ f 0 ðaÞðx  aÞg; 2

for x 2 I; for x 2 I:

G ð x Þ ¼ ð x  aÞ ; Then F and G are differentiable on I, and F 0 ð xÞ ¼ f 0 ð xÞ  f 0 ðaÞ; G0 ð xÞ ¼ 2ðx  aÞ: Also, F ðaÞ ¼ GðaÞ ¼ 0: Hence, by l’Hoˆpital’s Rule, it follows that F ð xÞ x!a Gð xÞ lim

exists and equals

Theorem 2, Sub-section 6.5.2.

F 0 ð xÞ ; x!a G0 ð xÞ lim

provided that this latter limit exists. Now F 0 ð xÞ f 0 ð x Þ  f 0 ð aÞ ; ¼ lim x!a G0 ð xÞ x!a 2ðx  aÞ lim

and, since the function f is twice differentiable at a, this limit exists and equals 1 00 2 f ðaÞ. Hence lim

x!a

f ð xÞ  f f ðaÞ þ f 0 ðaÞðx  aÞg ð x  aÞ 2

exists and equals

1 00 f ðaÞ: 2

We can reformulate this result as follows: for x near a 1 f ð xÞ ’ f ðaÞ þ f 0 ðaÞðx  aÞ þ f 00 ðaÞðx  aÞ2 ða quadratic functionÞ: 2 Notice that the function f and the approximating polynomial f ðaÞ þ f 0 ðaÞ ðx  aÞ þ 12 f 00 ðaÞðx  aÞ2 have the same value at a and the same first and second derivatives at a.

8: Power series

316 This suggests that, if the function is n-times differentiable on I, then we may be able to find a better approximating polynomial of degree n whose value at a and whose first n derivatives at a are equal to those of f. This leads to the following definition. Definition Let f be n-times differentiable on an open interval containing the point a. Then the Taylor polynomial of degree n for f at a is the polynomial f 0 ð aÞ ð x  aÞ 1! 00 f ð aÞ f ðnÞ ðaÞ þ ð x  aÞ 2 þ    þ ðx  aÞn : 2! n!

Tn ð x Þ ¼ f ð a Þ þ

Strictly speaking, we should use more complicated notation to indicate that the Taylor polynomial Tn(x) depends on n, a and f.

Notice that Tn ðaÞ ¼ f ðaÞ; Tn0 ðaÞ ¼ f 0 ðaÞ; . . .; TnðnÞ ðaÞ ¼ f ðnÞ ðaÞ; that is, f and Tn have the same value at a and have equal derivatives at a of all orders up to and including n. Example 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x) for the function f(x) ¼ sin x at each of the following points: (a) a ¼ 0; Solution

(b) a ¼ p2 Here

p ¼ 1; 2 p f 0 ð0Þ ¼ 1; f0 ¼ 0; f 0 ð xÞ ¼ cos x; 2 p f 00 f 00 ð xÞ ¼ sin x; f 00 ð0Þ ¼ 0; ¼ 1; 2 p f 000 ð xÞ ¼ cos x; f 000 ð0Þ ¼ 1; f 000 ¼ 0: 2

f ð xÞ ¼ sin x;

f ð0Þ ¼ 0;

f

Hence: 3

(a) T1 ð xÞ ¼ x; T2 ð xÞ ¼ x and T3 ð xÞ ¼ x  x3! ;  2  2 (b) T1 ð xÞ ¼ 1; T2 ð xÞ ¼ 1  12 x  p2 and T3 ð xÞ ¼ 1  12 x  p2 : & Problem 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x) for each of the following functions f at the given point a: (a) f ð xÞ ¼ ex ; a ¼ 2;

(b) f ð xÞ ¼ cos x; a ¼ 0:

Problem 3 Determine the Taylor polynomial of degree 4 for each of the following functions f at the given point a: (a) f ð xÞ ¼ 7  6x þ 5x2 þ x3 ; a ¼ 1; (c) f ð xÞ ¼ loge ð1 þ xÞ; a ¼ 0;

1 (b) f ð xÞ ¼ 1x ; a ¼ 0; (d) f ð xÞ ¼ sin x; a ¼ p4 ;

(e) f ð xÞ ¼ 1 þ 12 x  12 x2  16 x3 þ 14 x4 ; a ¼ 0: Problem 4 Determine the percentage error involved in using the Taylor polynomial of degree 3 for the function f (x) ¼ tan x at 0 to evaluate tan 0.1. (Use your calculator to evaluate tan 0.1.)

We do not usually multiply out such brackets, since that would make the results less clear.

8.1

Taylor polynomials

8.1.2

317

Approximation by Taylor polynomials

We now look at some specific examples of functions to investigate the assertion that Taylor polynomials provide good approximations for a large class of functions.

The function f ðx Þ ¼ 1 þ 21 x  21 x 2  61 x 3 þ 41 x 4 It follows from the result of Problem 3(e) above that the Taylor polynomials of degrees 1, 2, 3 and 4 for f at 0 are: 1 1 1 T2 ð xÞ ¼ 1þ x  x2 ; T1 ð xÞ ¼ 1þ x; 2 2 2 1 1 1 T3 ð xÞ ¼ 1þ x  x2  x3 ; 2 2 6

1 1 1 1 and T4 ð xÞ ¼ 1þ x  x2  x3 þ x4 : 2 2 6 4

Since f ðnÞ ð0Þ ¼ 0 for n  5, it follows that for n  5 the Taylor polynomial of degree n for f at 0 is just the same as the Taylor polynomial of degree 4 for f at 0. The graphs of these Taylor polynomials are as follows:

In this case, the polynomials T2(x) and T3(x) provide good approximations to f(x) near 0, and Tn(x) ¼ f(x) for all n  4.

The function f(x) ¼ sin x

Tn ð xÞ ¼ f ð xÞ

By calculating higher derivatives of the function f(x) ¼ sin x at 0, we can show that the following are Taylor polynomials for f at 0 T1 ð xÞ ¼ T2 ð xÞ ¼ x; T5 ð xÞ ¼ T6 ð xÞ ¼ x 

x3 x5 þ ; 3! 5!

In general, if f is a polynomial of degree N, then

T 3 ð xÞ ¼ T 4 ð xÞ ¼ x 

x3 ; 3!

T 7 ð xÞ ¼ T 8 ð xÞ ¼ x 

x3 x5 x7 þ  : 3! 5! 7!

for all n  N:

8: Power series

318 The following graphs illustrate how the approximation to f(x) given by Tn(x) gets better as n increases.

For example, the  pgraph  of T5 appears to be very close to the graph of sine p over the interval 2 , 2 , so that sin x and T5(x) do not differ by very much for values of x in this interval. Thus T5(x) seems to be a good approximation to sin x in this interval. It appears that, as the degree of the Taylor polynomial increases, so its graph becomes a good approximation to that of sine over more and more of R. For instance, in the above diagrams the shaded area covers the interval of the x-axis on which the Taylor polynomial Tn(x) agrees with sin x to three decimal places. 1 The function f ðx Þ ¼ 1x

By repeated differentiation it is easy to verify that, for k ¼ 1, 2, . . . k! ; f ð k Þ ð xÞ ¼ ð1  xÞkþ1 thus, in particular, f ðkÞ ð0Þ ¼ k!. Hence the Taylor polynomial of degree n for f at 0 is Tn ð x Þ ¼ ¼

n X f ðkÞ ð0Þ k¼0 n X

k!

xk

xk ¼ 1 þ x þ x2 þ    þ xn :

k¼0

The following diagram shows the graphs of the Taylor polynomials for f at 0 of degrees 1, 2, 4 and 7.

8.1

Taylor polynomials

319

The graphs show that the nature of the approximation is very different from the previous examples. For sine, the interval over which the approximation is good seems to expand indefinitely as the degree of the polynomials increases; 1 the interval of good approximation always seems to be but for f ðxÞ ¼ 1x contained in the interval ( 1, 1). Notice, however, that for this function f, the Taylor polynomials Tn(x) are 1 P xk ; this series converges just the nth partial sums of the geometric series k¼0

1 for jxj < 1, and diverges for jxj  1. It follows that, if jxj < 1, then with sum 1x

Tn ð xÞ ! f ð xÞ as n ! 1; so that, if jxj < 1, the polynomials Tn(x) do provide better and better approximations to f(x) as n increases.

Which functions can be approximated by Taylor polynomials? In the next section we obtain a criterion for determining those functions f for which the Taylor polynomials provide useful approximating polynomials, and

For jxj  1, the sequence {Tn(x)} does not converge, and so cannot provide an approximation to f(x).

8: Power series

320 the intervals on which the approximation occurs. We also introduce certain basic power series which correspond to the functions in the following problem. Problem 5 Determine the Taylor polynomial of degree n for each of the following functions at 0: 1 (a) f ð xÞ ¼ 1x ; (d) f ð xÞ ¼ sin x;

8.2

(b) f ð xÞ ¼ loge ð1 þ xÞ; (e) f ð xÞ ¼ cos x:

(c) f ð xÞ ¼ ex ;

Taylor’s Theorem

8.2.1

Taylor’s Theorem and approximation

In Section 8.1 we demonstrated how to find the Taylor polynomial Tn(x) of degree n for a function f at a point a. This polynomial and its first n derivatives agree with f and its first n derivatives at a, and the polynomial appears to approximate f at points near a. But how good an approximation is it? What error is involved if we replace f by a Taylor polynomial at a? The answer to these questions is given by the following important result. Theorem 1 Taylor’s Theorem Let f be (n þ 1)-times differentiable on an open interval containing the points a and x. Then f ð xÞ ¼ f ðaÞ þ f 0 ðaÞðx  aÞ f 00 ðaÞ f ðnÞ ðaÞ ð x  aÞ 2 þ    þ ðx  aÞn þ Rn ð xÞ; þ 2! n! where Rn ð x Þ ¼

f ðnþ1Þ ðcÞ ðx  aÞnþ1 ; ðn þ 1Þ!

and c is some point between a and x.

Remarks 1. When n ¼ 0, Taylor’s Theorem reduces to the assertion f ð xÞ ¼ f ðaÞ þ f 0 ðcÞðx  aÞ; which we can rewrite (for x 6¼ a) in the form f ð x Þ  f ð aÞ ¼ f 0 ðcÞ; for some c between a and x: xa But this is just the Mean Value Theorem! It follows that Taylor’s Theorem can be considered as a generalisation of the Mean Value Theorem. 2. The result of Theorem 1 can be expressed in the form f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ; where Rn(x) is thought of as a ‘remainder term’ or ‘error term’ involved in approximating f(x) by the estimate Tn(x).

Strictly speaking, we should use more complicated notation to indicate that the remainder term Rn(x) depends on n, a and f.

8.2

Taylor’s Theorem

321

Proof For simplicity, we assume that x > a; the proof is similar if x < a. We use the auxiliary function hðtÞ ¼ f ðtÞ  Tn ðtÞ  Aðt  aÞnþ1 ;

t 2 ½a; x;

You may omit this proof at a first reading.

(1)

where Tn is the Taylor polynomial of degree n for f at a, and A is a constant chosen so that hðaÞ ¼ hð xÞ:

(2)

Now f ðaÞ ¼ Tn ðaÞ; f 0 ðaÞ ¼ Tn0 ðaÞ; . . .;

and f ðnÞ ðaÞ ¼ TnðnÞ ðaÞ;

so that hðaÞ ¼ 0; h0 ðaÞ ¼ 0; . . .;

and hðnÞ ðaÞ ¼ 0:

The function h is continuous on the closed interval [a, x] and differentiable on the open interval (a, x); also, h(a) ¼ h(x). Hence, by Rolle’s Theorem, there exists some number c1 between a and x for which h0 ðc1 Þ ¼ 0: Next, we apply Rolle’s Theorem to the function h0 on the interval [a, c1]. The function h0 is continuous on [a, c1] and differentiable on (a, c1); also, h0 (a) ¼ h0 (c1) ¼ 0. Hence, by Rolle’s Theorem, there exists some number c2 between a and c1 for which h00 ðc2 Þ ¼ 0: In turn, we apply Rolle’s Theorem to the functions h00 ; h000 ; . . .; hðnÞ on the intervals ½a; c2 ; ½a; c3 ; . . . ; ½a; cn ; where c2 > c3 > c4 >    > cn > a: At the last stage, we find that there exists some point c between a and cn for which hðnþ1Þ ðcÞ ¼ 0:

(3)

By differentiating equation (1) (n þ 1) times, we obtain hðnþ1Þ ðtÞ ¼ f ðnþ1Þ ðtÞ  Aðn þ 1Þ!:

(4)

From (3) and (4), we deduce that 0 ¼ f ðnþ1Þ ðcÞ  Aðn þ 1Þ!; so that A¼

f ðnþ1Þ ðcÞ : ðn þ 1Þ!

(5)

Finally, it follows from equations (1) and (2), with h(a) ¼ 0, that f ð xÞ ¼ Tn ð xÞ þ Aðx  aÞnþ1 : Hence, by equation (5) f ð xÞ ¼ Tn ð xÞ þ as required.

f ðnþ1Þ ðcÞ ðx  aÞnþ1 ; ðn þ 1Þ!

&

This choice of A is made so that we can apply Rolle’s Theorem to h on [a, x].

8: Power series

322 Problem 1 Obtain an expression for R1(x) when Taylor’s Theorem is 1 applied to the function f ð xÞ ¼ 1x at a ¼ 0. Calculate the value of c 3 when x ¼ 4. Problem 2 What can you say about Rn(x) when f is a polynomial of degree at most n? Problem 3 By applying Taylor’s Theorem to the function f (x) ¼ cos x at a ¼ 0, prove that cos x ¼ 1  12 x2 þ R3 ð xÞ, x 2 R, where 1 4 x : jR3 ð xÞj  24 In most applications of Taylor’s Theorem, we do not know the value of c explicitly. However, for many purposes this does not  matter, since we can show that jRn(x)j is small by finding an estimate for  f ðnþ1Þ ðcÞ which is valid for all c between a and x, and then applying the following result. Corollary 1

Remainder Estimate

Let f be (n+1)-times differentiable on an open interval containing the points a and x. If  ðnþ1Þ  f ðcÞ  M; for all c between a and x, then f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ; where jRn ð xÞj 

Proof

M jx  ajnþ1 : ðn þ 1Þ!

From Taylor’s Theorem, we have R n ð xÞ ¼

f ðnþ1Þ ðcÞ ðx  aÞnþ1 ; ðn þ 1Þ!

where c is some point between a and x. It follows that  ðnþ1Þ  f ðcÞ jx  ajnþ1 jRn ð xÞj ¼ ðn þ 1Þ! M  jx  ajnþ1 : ðn þ 1Þ!

&

Example 1 By applying the Remainder Estimate to the function f (x) ¼ sin x, with a ¼ 0 and n ¼ 3, calculate sin 0.1 to four decimal places. Solution

Here f ð xÞ ¼ sin x;

f ð0Þ ¼ 0;

f 0 ð xÞ ¼ cos x;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ sin x;

f 00 ð0Þ ¼ 0;

f 000 ð xÞ ¼ cos x;

f 000 ð0Þ ¼ 1:

Strictly speaking, we should use more complicated notation to indicate that the upper bound M depends on n, a and f.

8.2

Taylor’s Theorem

323

Hence the Taylor polynomial of degree 3 for f at 0 is T3 ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ

f 00 ð0Þ 2 f 000 ð0Þ 3 x þ x 2! 3!

1 ¼ x  x3 : 6 Also, f ð4Þ ð xÞ ¼ sin x, so that  ð4Þ   f ðcÞ ¼ jsin cj  1;

for c 2 R:

Taking M ¼ 1 in the Remainder Estimate, we deduce that 1 jR3 ð0:1Þj   ð0:1Þ4 4! 1  104 ¼ 24 1 <  105 : 2 Now sin 0:1 ’ T3 ð0:1Þ 1 ¼ 0:1   103 6 ¼ 0:1  0:0001666 . . . ¼ 0:0998333 . . .: It follows from the Remainder Estimate that T3(0.1) gives an estimate for sin 0.1 to four decimal places. Hence sin 0.1 ¼ 0.0998 (to four decimal places). & Problem 4 By applying the Remainder Estimate to the function f ð xÞ ¼ loge ð1 þ xÞ, with a ¼ 0 and n ¼ 2, calculate loge 1:02 to four decimal places. In many practical situations we do not know how many terms of the power series are needed in order to calculate the value of a given function to a prescribed number of decimal places. In such cases, we can use the Remainder Estimate to determine how many terms are needed. Example 2 By applying the Remainder Estimate to the function f (x) ¼ ex, with a ¼ 0, calculate the value of e to three decimal places. Solution

Since f ðkÞ ð xÞ ¼ ex ; f

ðk Þ

ð0Þ ¼ 1;

for k ¼ 0; 1; . . .; we have

for k ¼ 0; 1; . . .:

It follows that, for each n Tn ð xÞ ¼ 1 þ x þ

x2 xn þ  þ : 2! n!

Also, f ðnþ1Þ ð xÞ ¼ ex , for all x, so that, for all c 2 (0,1), we have  ðnþ1Þ  f ðcÞ  e < 3: It follows from the Remainder Estimate, with x ¼ 1 and M ¼ 3, that j Rn ð 1 Þ j 

3  1nþ1 : ðn þ 1Þ!

We proved that e < 3 in Subsection 2.5.3.

8: Power series

324 To calculate e to three decimal places, we must choose n so that 3 < 2  104 ; or 15;000 < ðn þ 1Þ!: ðn þ 1Þ! Since 7! ¼ 5,040 and 8! ¼ 40,320, we may safely choose n ¼ 7. It follows that e ’ T7 ð1Þ 1 1 1 1 1 1 1 ¼1þ þ þ þ þ þ þ 1! 2! 3! 4! 5! 6! 7! ¼ 1 þ 1 þ 0:5 þ 0:16 þ 0:1416 þ 0:0083 þ 0:00138 þ 0:000198412 . . . ¼ 2:7182 . . .:

&

Hence, e ¼ 2.718 (to three decimal places).

Problem 5 By applying the Remainder Estimate to the function f(x) ¼ cos x, with a ¼ 0, calculate cos 0.2 rounded to four decimal places. Our next example illustrates how to obtain an approximation valid over an interval. Example 3

1 Calculate the Taylor polynomial T3(x) for f ð xÞ ¼ xþ2 at 1.

Show that T3(x) approximates f (x) with an error less than 5  103 on the interval [1, 2]. Solution

Here f ð xÞ ¼

1 ; xþ2

f 0 ð xÞ ¼

1 f ð 1Þ ¼ ; 3

1

;

1 f 0 ð 1Þ ¼  ; 9

3

;

f 00 ð1Þ ¼

4

;

f 000 ð1Þ ¼ 

ðx þ 2Þ2

f 00 ð xÞ ¼

2 ð x þ 2Þ

f 000 ð xÞ ¼

6

2 ; 27 2 : 27

ð x þ 2Þ Hence the Taylor polynomial of degree 3 for f at 1 is 1 1 1 1 T 3 ð x Þ ¼  ð x  1Þ þ ð x  1Þ 2  ð x  1Þ 3 : 3 9 27 81 24 Also, f ð4Þ ð xÞ ¼ ðxþ2 , so that Þ5  ð4Þ  24  f ðcÞ  ; for c 2 ð1; 2Þ: 35 in the Remainder Estimate, we deduce that Taking M ¼ 24 35 24 ð2  1Þ4  35 4! 1 ¼ 5 ¼ 0:0041 . . .; for x 2 ½1; 2: 3 Since the remainder term is less than 0.005, it follows that T3(x) approximates & f (x) with an error less than 5  103 on [1, 2]. jR3 ð xÞj 

8.2

Taylor’s Theorem

325

Problem 6 Calculate the Taylor polynomial T4(x) for f (x) ¼ cos x at p. Show that approximates f (x) with an error less than 3  103 on the 3 T4(x)  5 interval 4 p; 4 p .

8.2.2

Taylor’s Theorem and power series

From Taylor’s Theorem, we know that, if a function f can be differentiated as often as we please on an open interval containing the points a and x, then, for any n f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ n X f ðkÞ ðaÞ ðx  aÞk þ Rn ð xÞ; ¼ k! k¼0 where Rn ð x Þ ¼

f ðnþ1Þ ðcÞ ðx  aÞnþ1 ; ðn þ 1Þ!

for some c between a and x. It follows that, if Rn(x) ! 0 as n ! 1, then we can express f (x) as a power series in (x  a). Theorem 2 Let f have derivatives of all orders on an open interval containing the points a and x. If Rn(x) ! 0 as n ! 1, then 1 ðnÞ X f ð aÞ ð x  aÞ n : f ð xÞ ¼ n! n¼0 We call f the sum function of the power series

1 ð nÞ P f ðaÞ n¼0

n!

ðx  aÞn , and we call

this power series the Taylor series for f at a.

Warning A dramatic example of what happens when the remainder term does not tend to zero is given by the function 1 2 x f ð xÞ ¼ e ; x 6¼ 0; 0; x ¼ 0: For this function, f (0) ¼ 0, f 0 (0) ¼ 0, f 00 (0) ¼ 0, . . .. Thus the Taylor polynomial Tn(x) is identically zero for each n, although the function f is not identically zero. In this case, Rn(x) ¼ f (x) for all n, and so the Taylor series for f at 0 1 ðnÞ X f ð0Þ n x ¼ 0 þ 0x þ 0x2 þ   ; n! n¼0 converges to f (x) only at 0! We can use Theorem 2 to obtain the following basic power series. Theorem 3 (a)

1 1x

Basic power series

¼1 þ x þ x2 þ x3 þ    2

3

¼

(b) loge ð1 þ xÞ ¼ x  x2 þ x3     ¼

1 P n¼0 1 P n¼1

xn ;

for j xj < 1; n

ð1Þnþ1 xn ;

forj xj < 1;

y 1 2

y = e –1/x –3

–2

–1 0

1

2

3 x

We indicate a proof of this assertion in Section 8.6, Exercise 6 on Section 8.2.

8: Power series

326 2

3

(c) ex ¼ 1 þ x þ x2! þ x3! þ    ¼ 3

5

(d) sin x ¼ x  x3! þ x5!     ¼ 2

1 n P x n¼0 1 P n¼0 1 P

4

(e) cos x ¼ 1  x2! þ x4!     ¼

n¼0

for x 2 R;

n! ; 2nþ1

for x 2 R;

2n

for x 2 R:

x ð1Þn ð2nþ1 Þ!; x ð1Þn ð2n Þ!;

Proof (a) Here we use the fact that, for x 6¼ 1 1 xnþ1 ¼ 1 þ x þ x2 þ x3 þ    þ xn þ : 1x 1x n nþ1 o But the sequence x1x is null, if jxj < 1. Thus, by letting n ! 1, we

This identity is easily proved by multiplying both sides by 1  x:

obtain 1 X 1 ¼ xn ; 1  x n¼0

for j xj < 1:

(b) Similarly 1 ð1Þnþ1 tnþ1 ¼ 1 t þ t2     þ ð1Þn tn þ : 1þt 1þ t Integrating both sides from 0 to x, where jxj < 1, we obtain ! Z x Z x nþ1 nþ1 dt ð 1 Þ t ¼ 1 t þ t2     þ ð1Þn tn þ dt; 1þt 0 1þt 0 so that

 nþ1 x t2 t3 n t ½loge ð1 þ tÞ0 ¼ t  þ     þ ð1Þ 2 3 nþ1 0 Z x nþ1 t dt: þ ð1Þnþ1 0 1þt x

Hence x2 x3 xnþ1 þ     þ ð1Þn 2 3 nþ1 Z x nþ1 t nþ1 dt: þ ð1Þ 0 1þt

loge ð1 þ xÞ ¼ x 

When 0  x < 1, we have, by the Inequality Rule for integrals  Z x nþ1  Z x nþ1 Z x   t t ð1Þnþ1 ¼ dt dt  tnþ1 dt   0 1þt 0 1þt 0

nþ2 x t ¼ nþ2 0 ¼

xnþ2 1 ! 0 as n ! 1:  nþ2 nþ2

When 1 < x < 0, we put T ¼ t and X ¼ x, so that 0 < T < 1. Then

Theorem 2, Sub-section 7.4.1.

8.2

Taylor’s Theorem  Z  ð1Þnþ1 

0

x

327

 Z X nþ1 Z X tnþ1  T 1 dt ¼ dT  T nþ1 dT 1X 0 1þt 0 1T ¼

X nþ2 1 !0  ð1  X Þðn þ 2Þ ð1  X Þðn þ 2Þ

For, 0 < X < 1 and 1 1T

1  1X

for T 2 [0, X].

as n ! 1:

Combining these two results, we deduce that 1 X xn ð1Þnþ1 ; for 1 < x < 1: loge ð1 þ xÞ ¼ n n¼1 (c) This was one of our definitions of the exponential function.

See Sub-section 3.4.3.

(d) Let f (x) ¼ sin x. Then, by Taylor’s Theorem, we have sin x ¼ Tn ð xÞþ Rn ð xÞ, where Rn ð x Þ ¼

f ðnþ1Þ ðcÞ nþ1 x ; ðn þ 1Þ!

for some number c between 0 and x. We saw earlier that ðnþ1Þ

 cos c;   so that, in particular, we can be sure that  f ðnþ1Þ ðcÞ  1. It follows from the Remainder Estimate, with M ¼ 1, that f

ðcÞ ¼  sin c

jRn ð xÞj 

Problem 5(d), Sub-section 8.1.2.

or

1 j xjnþ1 ! 0 as n ! 1; ðn þ 1Þ!

so that, in particular, Rn(x) ! 0 as n ! 1. Hence, by letting n ! 1 in the equation sin x ¼ Tn ð xÞ þ Rn ð xÞ, we obtain sin x ¼ x 

1 X x3 x5 x7 x2nþ1 þ  þ  ¼ ; ð1Þn 3! 5! 7! ð2n þ 1Þ! n¼0

(e) The proof is similar to that of part (d), so we omit it.

for x 2 R:

&

Remark Probably the first definitions of sin x and cos x that you met were expressed in terms  ofpa right-angled triangle, but those definitions only make sense for x 2 0; 2 . We have now shown that sin x and cos x can be represented by the power series in parts (d) and (e) of Theorem 3, and we can use these power series to define sin x and cos x for all x 2 R. Finally, we use Taylor’s Theorem to prove an interesting limit which is a generalisation of two limits that you met earlier

 1 n x n lim 1 þ ¼ e and lim 1 þ ¼ ex ; for any x 2 R: n!1 n!1 n n Example 4 (a) Calculate T1(x) and R1(x) for the function f ð xÞ ¼ loge ð1 þ xÞ; x 2 ð1; 1Þ, at 0. (b) By replacing x in part (a) by x , where jxj > jj, prove that, for any real numbers  and 

Sub-section 2.5.3.

8: Power series

328  x lim 1 þ ¼ e : x!1 x (c) Deduce from part (a) that

1 n loge 1 þ ! 1 as n ! 1: n Solution (a) For the function f ð xÞ ¼ loge ð1 þ xÞ, x 2 ð1; 1Þ, we have f ð xÞ ¼ loge ð1 þ xÞ; f ð0Þ ¼ 0; f 0 ð xÞ ¼

1 ; 1þx

f 00 ð xÞ ¼

1 ð1 þ x Þ2

f 0 ð0Þ ¼ 1; :

Hence T1 ð xÞ ¼ f ð0Þ þ f 0 ð0Þx ¼ x

and R1 ð xÞ ¼

f 00 ðcÞ 2 x2 x ¼ ; 2! 2ð1 þ cÞ2

for some number c between 0 and x. (b) Replacing x by x in the equation x2 ; for j xj < 1; loge ð1 þ xÞ ¼ x  2ð1 þ cÞ2 we obtain    2 loge 1 þ ; for j xj > jj: ¼  x x 2ð 1 þ c Þ 2 x 2 Hence   2  ; for j xj > jj: ¼   x loge 1 þ x 2ð 1 þ c Þ 2 x Since 1x ! 0 as x ! 1, it follows that   ¼ ; lim x loge 1 þ x!1 x so that  x ¼ : lim loge 1 þ x!1 x Since the exponential function is continuous on R, and

  x x ; ¼ 1þ exp loge 1 þ x x it follows, from the Composition Rule for limits, that  x ¼ e : lim 1 þ x!1 x x2 , (c) For x 2 (1, 1), we have, from part (a), that loge ð1 þ xÞ ¼ x  2ð1þc Þ2 1 for some number c between 0 and x. Putting x ¼ n, we obtain

1 1 1 loge 1 þ ; ¼  n n 2n2 ð1 þ cn Þ2 for some number cn between 0 and 1n.

8.3

Convergence of power series

329

Hence

1 1 n loge 1 þ ¼1 n 2nð1 þ cn Þ2 ! 1 as n ! 1; since 0 and

8.3

1 2n

1 2nð1 þ cn Þ

2



1 2n

&

is a null sequence.

Convergence of power series

8.3.1

The radius of convergence

In the previous section you saw that certain standard functions can be expressed as the sum functions of power series; for example 1 X 1 ¼ xn 1  x n¼0

and

loge ð1 þ xÞ ¼

1 X

xn ð1Þnþ1 ; n n¼1

for j xj < 1:

These power series are the Taylor series for the given functions at 0. Conversely, power series can be used to define functions. Thus, for instance, we defined the exponential function x 7! ex, x 2 R, to be the sum function for 1 n P x the power series n! .

Section 3.4.

n¼0

But there are many other functions which are defined as the sum functions of power series (as distinct from a power series obtained as the Taylor series for a given function). For example, the Bessel function  2n 1 X ð1Þn 2x J0 ð xÞ ¼ ; for x 2 R; ðn!Þ2 n¼0

See Exercise 6 on Section 8.4, in Section 8.6.

arises in analyses of the vibrations of a circular drum and of the radiation from certain types of radio antenna. However, all such uses of power series depend on knowledge of those 1 P numbers x for which a power series an ðx  aÞn converges. In each of the n¼0

above examples, the series converges on an interval; indeed, all power series have this property. Theorem 1

Radius of Convergence Theorem 1 P an ðx  aÞn , precisely one of the following

For a given power series possibilities occurs:

n¼0

(a) The series converges only when x ¼ a; (b) The series converges for all x;

We give the proof of Theorem 1 in Subsection 8.3.3.

8: Power series

330 (c) There is a number R > 0, called the radius of convergence of the power series, such that the series converges if jx  aj < R and diverges if jx  aj > R. For example: 1 P (a) n!xn converges only when x ¼ 0; (b)

n¼0 1 P

(c)

n¼0 1 P

xn n!

For all non-zero x, the series 1 P n!xn diverges, by the Nonn¼0   nð1xÞn o null Test, as n!x1 n ¼ n! is

converges for all x;

xn converges if jxj < 1 and diverges if jxj > 1 – its radius of conver-

a basic null sequence.

n¼0

gence is 1.

Remarks 1. Sometimes we abuse our notation by writing R ¼ 0, if a series converges only for x ¼ a, or R ¼ 1, if a series converges for all x. 2. It is important to remember that Theorem 1, part (c), makes no assertion about the convergence or divergence of the power series at the end-points a  R, a þ R of the interval (a  R, a þ R). For example, we shall see shortly that the three power series 1 X n¼1

xn ;

1 n X x n¼1

n

and

Example 2.

1 n X x n¼1

n2

all have radius of convergence 1. However, the intervals on which these series converge are, respectively, ( 1, 1), [ 1, 1) and [ 1, 1]. The interval of convergence of the power series is the interval (a  R, a þ R), together with any end-points of the interval at which the series converges. The following diagram illustrates the various types of interval of conver1 P gence of an ðx  aÞn : n¼0

Theorem 1 is not concerned with the problem of evaluating the radius of convergence of a power series. However, a power series is simply a particular type of series, and so all the convergence tests for series can be applied to power series. In practice, we can tackle most commonly arising power series by using the following version of the Ratio Test. Ratio Test for Radius of Convergence 1 P Suppose that an ðx  aÞn is a given power series, and that

Sections 3.1–3.3.

Theorem 2

n¼0

a   nþ1    ! L; an

We give the proof of Theorem 2 in Subsection 8.3.3.

as n ! 1:

(a) If L is 1, the series converges only for x ¼ a.

R ¼ 0.

8.3

Convergence of power series

331

(b) If L ¼ 0, the series converges for all x.

R ¼ 1.

(c) If L > 0, the series has radius of convergence

1 L:

R ¼ L1 

Example 1 Determine the radius of convergence of each of the following power series: 1 n 1 P P n ðxþ1Þn ðx2Þn ; (b) (a) n! n! : n¼1

n¼1

Solution n (a) Here an ¼ nn! for all n, so

a  ðn þ 1Þnþ1 n! 1 n  nþ1  ¼ 1þ  ;  ¼ ðn þ 1Þ! n an nn Thus

for all n:

a   nþ1    ! e as n ! 1: an

Hence, by the Ratio Test, the radius of convergence is 1e  (b) Here an ¼ n!1 , for all n, so a  n! 1  nþ1  ¼ ; ¼  ðn þ 1Þ! n þ 1 an Thus

for all n:

a   nþ1    ! 0 as n ! 1: an

Hence, by the Ratio Test, the power series

1 P ðx2Þn n¼1

n!

converges for all x. &

Problem 1 Determine the radius of convergence of each of the following power series: 1 1 P P ðn!Þ2 n (a) ð2n þ 4n Þxn ; (b) ð2nÞ! x ; (c)

n¼0 1 P

n

ðn þ 2n Þðx  1Þ ;

(d)

n¼1

n¼1 1 P

xn 1: n ð n! n¼1 Þ

Hint for part (d): Use Stirling’s Formula and the result of Example 1, Sub-section 7.5.1. If we wish to find the interval of convergence (as opposed to the radius of convergence), then we may need to use some of the other tests for series in order to determine the behaviour at the end-points of the interval. Example 2 Determine the interval of convergence of each of the following power series: 1 1 n 1 n P P P x x (a) xn ; (b) (c) n; n2 . n¼1

n¼1

n¼1

Solution (a) Here an ¼ 1, for all n, so a   nþ1    ¼ 1; for all n: an

8: Power series

332 Hence, by the Ratio Test, the radius of convergence is 1; in other words 1 X xn converges for 1 < x < 1: n¼1

Next, we consider the behaviour of the power series at the end-points of this interval, namely 1 and 1.1Since the sequences f1n gand fð1Þn g are P n x diverges when x ¼ 1, by the Nonboth non-null, it follows that n¼1 null Test. 1 P Hence the interval of convergence of xn is ( 1, 1). n¼1 1 (b) Here an ¼ n, for all n, so a  n 1  nþ1  ¼ ; for all n:  ¼ n þ 1 1 þ 1n an Thus

a   nþ1   ! 1 as n ! 1:  an Hence, by the Ratio Test, the radius of convergence is 1; in other words 1 n X x converges for 1 < x < 1: n n¼1 1 1 P P ð1Þn 1 But we know that n converges. It follows that the n diverges and n¼1 n¼1 1 n P x interval of convergence of the power series n is ½1; 1Þ. (c) Here an ¼ n12 , for all n, so a  n2 1  nþ1  ¼ ; ¼  2 an ðn þ 1Þ ð1 þ 1nÞ2

n¼1

By Example 2 of Sub-section 3.2.1, and at the start of Subsection 3.3.2, respectively.

for all n:

Thus

a   nþ1   ! 1 as n ! 1:  an Hence, by the Ratio Test, the radius of convergence is 1; in other words 1 n X x converges for 1 < x < 1: 2 n n¼1 1 P 1 But we know that n2 converges; so, by the Absolute Convergence Test, n¼1 1 n P ð1Þ the series n2 also converges. It follows that the interval of convern¼1 1 n P x & gence of the power series n2 is ½1; 1. n¼1

Now, the Radius of Convergence Theorem tells us that, if a power series 1 P an ðx  aÞn has radius of convergence R, then it converges for jx  aj < R. In n¼0

fact, we can say more than this – namely, that, for all points x with jx  aj < R, the power series is actually absolutely convergent! Theorem 3

Absolute Convergence Theorem 1 P Let the power series an ðx  aÞn have radius of convergence R. Then it is n¼0

absolutely convergent for all x with jx  aj < R.

Theorem 1, Sub-section 3.3.1.

8.3

Convergence of power series

333

Proof Let x be a number such that jx  aj < R, and chose a number X such that jx  aj < jX  aj < R. 1 P Now, the series an ðX  aÞn is convergent, so that an ðX  aÞn ! 0 as n¼0

n ! 1. In particular, there exists some number N such that jan ðX  aÞn j < 1, for all n > N. It follows that  x  a n   n jan ðx  aÞn j ¼   jan ðX  aÞ j X  a  x  a n     ; for n > N: Xa 1  1  P  x  a n is convergent, so that the series P jan ðx  aÞn j is also But the series Xa n¼0

n¼0

convergent, by the Comparison Test for series. 1 P an ðx  aÞn is absolutely convergent, as It follows that the power series n¼0 & required.

Such a choice is always possible – for example, choose X to be the midpoint of x and the nearest end-point of the interval (a  R, a þ R).

For

1   P  x  a n n¼0

Xa

is a geometric

series.

Problem 2 Determine the interval of convergence of each of the following power series: 1 1 P P 1 n (a) nxn ; (b) n3n x : n¼0

n¼1

Determine the radius of convergence of the power series 1 X  ð   1Þ 2  ð   1Þ . . . ð   n þ 1Þ n 1 þ x þ x þ  ¼ x ; 2! n! n¼0

Problem 3

where  6¼ 0; 1; 2; . . .:

8.3.2

Abel’s Limit Theorem

We can sometimes use the ideas of power series to sum interesting and commonly arising series such as 1 1 X X ð1Þnþ1 1 1 1 ð1Þn 1 1 1 ¼1  þ  þ    and ¼1  þ  þ   : 2 3 4 3 5 7 n 2n þ 1 n¼1 n¼0 A major tool in this connection in the following result. Theorem 4

Abel’s Limit Theorem

1 P

Let f (x) be the sum function of the power series an xn , which has radius of 1 n¼0 P convergence 1; and let an be convergent. Then n¼0 1 X

lim f ð xÞ ¼

x!1

an :

n¼0

Proof This proof is a wonderful illustration of the sheer power and magic of the " !  method for proving results in Analysis! We use the standard approach for proving that a limit exists as x ! 1 . 1 P Let sn ¼ a0 þ a1 þ    þ an1 be the nth partial sum of the series an ; and n¼0 1 P let s denote the sum an . n¼0

A similar result holds for power series of the form 1 P an ðx  aÞn ; a 6¼ 0. n¼0

Equivalently 1

1 P P n an x ¼ an : lim x!1

n¼0

n¼0

You may omit this proof at a first reading.

8: Power series

334 Now, a0 ¼ s1 and an ¼ snþ1  sn; for all n  1:

(1)

It follows that, for j xj51 ð1  x Þ

1 X

  snþ1 xn ¼ ð1  xÞ s1 þ s2 x þ s3 x2 þ    þ snþ1 xn þ   

n¼0

¼ s1 þ s2 x þ s3 x2 þ s4 x3    þ snþ1 xn þ  s1 x  s2 x2  s3 x3      sn xn  snþ1 xnþ1     ¼ s1 þ ðs2  s1 Þx þ ðs3  s2 Þx2 þ    þ ðsnþ1  sn Þxn þ    1 X an xn ¼ f ðxÞ: ¼

Using (1).

n¼0

We now study closely the identity ð1  xÞ

1 X

snþ1 xn ¼ f ðxÞ:

(2)

n¼0

Next, we want to prove that: for each positive number ", there is a positive number  such that j f ð xÞ  sj < ",

This is simply the definition of lim f ð xÞ ¼ s:

for all x satisfying 1   < x < 1.

x!1

Now, we use equation (2) in the following way to get a convenient expression for f (x)  s f ð xÞ  s ¼

1 X

an x n 

n¼0

1 X

Here we use the definition of s, equation (2), and the fact that the sum of the series 1 P 1 x n is 1x :

an

n¼0

¼ ð1  xÞ ¼ ð1  xÞ

1 X n¼0 1 X

snþ1 x n  s

n¼0

snþ1 x n  ð1  xÞ

n¼0

¼ ð1  xÞ

1 X

1 X

sx n

n¼0

ðsnþ1  sÞx n :

(3)

n¼0

Next, choose a number N such that jsnþ1  sj < 12 " for all n > N. We can then apply the Triangle Inequality to equation (3), for x 2 ð0; 1Þ, to see that   N 1 X  X  n n ðsnþ1  sÞx þ ðsnþ1  sÞx  j f ð x Þ  sj ¼ ð 1  x Þ  n¼0  n¼Nþ1     X  X   N 1   n n ðsnþ1  sÞx  þ ð1  xÞ ðsnþ1  sÞx   ð 1  x Þ  n¼0 n¼Nþ1   1 X 1 xn  ð1  x Þ jsnþ1  sjx þ ð1  xÞ  "  2 n¼0 n¼Nþ1 N X

n

Such a choice of N is possible, since snþ1 ! s as n ! 1.

We split the infinite sum into two parts. Using the Triangle Inequality. For jsnþ1  sj < 12 ", for all n < N; also, we use that 1 P 1 0 < x < 1 so that xn ¼ 1x . n¼0

8.3

Convergence of power series

 ð1  xÞ

335

N X

1 jsnþ1  sjxn þ " 2 n¼0

N X

1 (4) jsnþ1  sj þ " 2 n¼0 N P Finally, since the linear function x 7!ð1  xÞ jsnþ1  sj is continuous on  ð1  xÞ

Again we use that 0 < x < 1.

n¼0

R, and takes the value 0 at 1, it follows that we can choose a positive number  (with  2 (0, 1)) such that N X 1 ð1  x Þ jsnþ1  sj < "; for all x satisfying 1   < x < 1: 2 n¼0 N P If we substitute this upper bound 12 " for ð1  xÞ jsnþ1  sj into the

We add in the extra requirement that  < 1 in order to ensure that we are only considering values of x in (0, 1).

n¼0

inequality (4), we find that we have proved that

for each positive number ", there is a positive number  such that j f ð xÞ  sj < "; for all x satisfying 1   < x < 1:

&

This is what we set out to prove. As an application of Abel’s Limit Theorem, we use it to evaluate 1 X ð1Þnþ1 1 1 1 ¼ 1  þ  þ   : 2 3 4 n n¼1 Let f ð xÞ ¼

1 P ð1Þnþ1 n¼1

n

xn . We have seen already that this power series converges

in (1, 1) to the sum loge ð1 þ xÞ. Further, we saw earlier that the series 1 P ð1Þnþ1 is convergent, by the Alternating Test. It follows, from Abel’s n

Part (b) of Theorem 3, Subsection 8.2.2.

n¼1

Limit Theorem, that 1 X ð1Þnþ1 n¼1

n

¼ lim x!1

1 X ð1Þnþ1 n¼1

n

xn ¼ lim loge ð1 þ xÞ ¼ loge 2: x!1

Remark It is always necessary to check that the conditions of Abel’s Theorem apply before using it. For example, a thoughtless application to the identity 1 X 1 ; wherej xj < 1; 1  x þ x2  x3 þ    ¼ ð1Þn xn ¼ 1 þ x n¼0 of taking the limit as x ! 1, would give the following absurd conclusion 1 X 1 1  1 þ 1  1 þ  ¼ ð1Þn ¼ ! 2 n¼0

We did not check that the 1 P ð1Þn xn was series n¼0

convergent at 1!

Problem 4 Use Abel’s Limit Theorem to evaluate 1 P ð1Þn 1 1 1 2nþ1 ¼ 1  3 þ 5  7 þ   .

n¼0

3

5

7

Hint: Use the facts that tan1 x ¼ x  x3 þ x5  x7 þ    for jxj < 1, and that the latter series has radius of convergence 1.

We prove these facts in Subsection 8.4.1.

8: Power series

336

8.3.3

Proofs

You may omit this subsection at a first reading.

We now supply the proofs that we omitted from Sub-section 8.3.1. Theorem 1

Radius of Convergence Theorem 1 P For a given power series an ðx  aÞn , precisely one of the following n¼0 possibilities occurs: (a) The series converges only when x ¼ a; (b) The series converges for all x; (c) There is a number R > 0 such that the series converges if jx  aj < R and diverges if jx  aj > R. Proof For simplicity, we shall assume that a ¼ 0. Clearly the possibilities (a) and (b) are mutually exclusive; we shall therefore assume that for a given power series neither possibility occurs, and then prove that possibility (c) must occur. First, let ( ) 1 X n S ¼ x: an x converges :

For the general proof, replace x throughout by (x  a).

n¼0

Since the possibility (b) has been excluded, we deduce that S must be bounded. 1 P For, if an X n diverges, then, in view of the Absolute Convergence Theorem n¼0 1 P for power series, an xn cannot converge for any x with jxj > jXj, since n¼0 1 P otherwise an X n would then have to be convergent – which is not the case.

Theorem 3, Sub-section 8.3.1.

n¼0

Also, S is non-empty, since the power series

1 P

an xn converges at 0.

n¼0

It then follows, from the Least Upper Bound Property of R, that the set S 1 P an xn is must have a least upper bound, R say. We now prove that the series n¼0

convergent if jxj < R and divergent if jxj > R. First, notice that R > 0. For, since possibility (a) has been excluded, there must 1 P an xn1 is convergent; hence

be at least one non-zero value of x, x1 say, such that

n¼0

we have x1 2 S. It follows, from the Absolute Convergence Theorem, that 1 P an xn must converge for those x with jxj < jx1j, so that ðjx1 j; jx1 jÞ lies in S. n¼0

Now, choose any x for which jxj < R. Then by the definition of R as sup S, 1 P there exists some number x2 with jxj < x2 < R such that an xn2 converges. It n¼0 1 P an xn converges. follows, from the Absolute Convergence Theorem, that n¼0

Finally, choose any x for which j xj > R ¼ sup S. It follows that x 2 = S, so that 1 P n & an x must be divergent. n¼0

Here we identify a number R that will be the desired radius of convergence.

8.3

Convergence of power series

337

In view of the Absolute Convergence Theorem, we can slightly strengthen the conclusion of the Radius of Convergence Theorem, as follows. 1 P

If the power series

Corollary

an ðx  aÞn has radius of convergence

n¼0

R > 0, then it is absolutely convergent if jx  aj < R. If the series converges for all x, then it is absolutely convergent for all x.

Theorem 3, Sub-section 8.3.1.

This result is sometimes also called the Absolute Convergence Theorem.

Finally we prove the Ratio Test for Radius of Convergence of power series. Theorem 2

Ratio Test for Radius of Convergence 1 P Suppose that an ðx  aÞn is a given power series, and that n¼0 a   nþ1   ! L as n ! 1:  an (a) If L is 1, the series converges only for x = a. (b) If L ¼ 0, the series converges for all x. (c) If L > 0, the series has radius of convergence L1 : Proof

For simplicity, we shall assume that a ¼ 0.

(a) Suppose that jaanþ1 j ! 1 as n ! 1. Then, for any non-zero value of x, the n sequence

ja a xx j ! 1, so that the sequence {anxn} is unbounded. It nþ1

n

nþ1 n

follows, from the Non-null Test, that the series

1 P

an xn must be divergent.

n¼0

anþ1 an

(b) Suppose that j j ! 0 as n ! 1. Then, for any non-zero value of x   anþ1 xnþ1  anþ1     a xn  ¼  a   j xj n

n

! 0  j xj ¼ 0;

1 P

an xn is absolutely convergent, and so is convergent.     (c) Suppose that  aanþ1  ! L as n ! 1, where L > 0. n so that

n¼0

First, suppose that j xj > L1. Then   anþ1 xnþ1  anþ1     a xn  ¼  a   j xj n

n

! L  j xj > 1; so that

1 P

an xn is not absolutely convergent. It follows, from the above

n¼0

Corollary to the Radius of Convergence Theorem, that the radius of 1 P an xn must be less than or equal to L1. convergence of n¼0

Next, suppose that j xj < L1. Then   anþ1 xnþ1  anþ1     a xn  ¼  a   j xj n

n

! L  j xj < 1;

For the general proof, replace x throughout by (xa).

8: Power series

338 1 P

an xn is absolutely convergent, and so is convergent. 1 P an xn must be at least equal to L1. Hence the radius of convergence of so that

n¼0

n¼0

Combining these two facts, it follows that the desired radius of conver& gence is exactly L1.

8.4

Manipulating power series

It would be tedious to have to apply Taylor’s Theorem every time that we wished to determine the Taylor series of a given function. While sometimes this really has to be done, in many commonly arising situations we can use standard rules for power series and the list of basic power series to avoid most of the effort. We now set out to establish the rules for manipulating power series.

8.4.1

Theorem 3, Sub-section 8.2.2.

Rules for power series

Many of the rules for manipulating power series are similar to the corresponding rules for manipulating ‘ordinary’ series. Theorem 1

Combination Rules

Let f ð xÞ ¼ gð xÞ ¼

1 X n¼0 1 X

an ð x  aÞ n ;

for jx  aj < R;

bn ð x  aÞ n ;

for jx  aj < R0 :

and

n¼0

Then: Sum Rule

ðf þ gÞð xÞ ¼

1 P

ðan þ bn Þðx  aÞn ;

for jx  aj < r;

n¼0

where r ¼ minfR; R0 g; 1 P lan ðx  aÞn ; Multiple Rule lf ð xÞ ¼

for jx  aj < R; where l 2 R:

Thus, if the power series for both f and g converge at a particular point x, then so does the series for f þ g.

n¼0

We do not supply a proof, as Theorem 1 is a simple restatement of the Sum and Multiple Rules for ‘ordinary’ series. We can use the Combination Rules to find the Taylor series at 0 for the function cosh x. We start with the power series for the exponential function 1 n X x2 x3 x ; for x 2 R: ex ¼ 1 þ x þ þ þ    ¼ 2! 3! n! n¼0 2

3

It follows that ex ¼ 1  x þ x2!  x3! þ    ¼

1 P n¼0

ð1Þn xn! ; n

Theorem 2, Sub-section 3.1.4.

Theorem 3, Sub-section 8.2.2.

for x 2 R. We

may then use the Sum Rule to obtain

1 X x2 x4 x2n x x ; e þ e ¼ 2  1 þ þ þ  ¼ 2  2! 4! ð2nÞ! n¼0

for x 2 R;

For the odd-powered terms cancel.

8.4

Manipulating power series

339

so that, by using the Multiple Rule with l ¼ 12, we obtain 1 X 1 x2 x4 x2n ; for x 2 R: cosh x ¼ ðex þ ex Þ ¼ 1 þ þ þ    ¼ 2 2! 4! ð2nÞ! n¼0 Problem 1 Find the Taylor series at 0 for each of the following functions, and state its radius of convergence: (a) f ð xÞ ¼ sinh x; x 2 R; (b) f ð xÞ ¼ loge ð1  xÞ þ 2ð1  xÞ1 ;

j xj < 1:

Problem 2 Find the Taylor series at 0 for each of the following functions, and state its radius of convergence:   (a) f ð xÞ ¼ sinh x þ sin x; x 2 R; (b) f ð xÞ ¼ loge 1þx j xj < 1; 1x ; 1 x 2 R: (c) f ð xÞ ¼ 1þ2x 2;

Remark In Theorem 1, the radius of convergence of the power series for f þ g may be larger than r ¼ minfR; R0 g. For example, we can use the basic power series and the Combination Rules to verify that the Taylor series at 0 for the functions 1 1 and gðxÞ ¼ 1x þ 11 x are f ð xÞ ¼ 1x 2 1 X xn and f ð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 þ    ¼ n¼0

1 X 1 3 2 7 3 15 4 1 n gð x Þ ¼  x  x  x  x     ¼ 1 þ n x ; 2 4 8 16 2 n¼0 each with radius of convergence 1. It follows, from Theorem 1, that the power series for the function ð f þ gÞð xÞ ¼ 11 x is 2

1 X 1 1 1 1 1 n ð f þ gÞð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 þ    ¼ x ; n 2 4 8 16 2 n¼0

We simply add the power series term-by-term.

and that this power series has radius of convergence at least 1. In fact, it has radius of convergence 2. x We can find the Taylor series at 0 for the function f ð xÞ ¼ 11 þ  x, for jxj < 1, by

writing

1þx 1x

function

2 1x

2ð1xÞ 1x

2 ¼ 1x  1. Since the Taylor series at 0 for the 1 P 2 3 is 2 þ 2x þ 2x þ 2x þ    ¼ 2xn , with radius of conver-

in the form

n¼0

gence 1, it follows that the Taylor series for f at 0 is 1 þ 2x þ 2x2 þ 1 P 2x3 þ    ¼ 1þ 2xn , with radius of convergence 1. n¼1

However we could obtain the same result by multiplying together the Taylor 1 , since series for the functions x 7! 1 þ x and x 7! 1x     ð1 þ xÞ 1 þ x þ x2 þ x3 þ    ¼ 1 þ x þ x2 þ x3 þ    þ x 1 þ x þ x2 þ x3 þ    ¼ 1 þ 2x þ 2x2 þ 2x3 þ   : In fact, we can justify such multiplication together of power series to obtain further power series.

We now ‘collect together’ the multiples of successive various powers of x.

8: Power series

340

Theorem 2 Let

Product Rule f ð xÞ ¼ gð x Þ ¼

1 X n¼0 1 X

an ð x  aÞ n ;

for jx  aj < R;

bn ð x  aÞ n ;

for jx  aj < R0 :

and

n¼0

Then ð fgÞð xÞ ¼

1 X

cn ðx  aÞn ; for jx  aj < r; where r ¼ minfR; R0 g;

n¼0

and c 0 ¼ a0 b0 ; c 1 ¼ a0 b1 þ a1 b0

As in Theorem 1, the radius of convergence of the product 1 P series cn ðx  aÞn may be n¼0

and

greater than r.

cn ¼ a0 bn þ a1 bn1 þ    þ an1 b1 þ an b0 : Theorem 2 is an immediate consequence of the Product Rule for series, applied 1 1 P P to the two series an ðx  aÞn and bn ðx  aÞn , both of which are absolutely n¼0

n¼0

convergent for jx  aj < r – by the Absolute Convergence Theorem. Example 1

Theorem 5, Sub-section 3.3.4.

Theorem 3, Sub-section 8.3.1.

1þx Determine the Taylor series at 0 for the function f ð xÞ ¼ ð1x . Þ2

Solution We use our knowledge of the power series at 0 for the functions x x 7! 1 1 x and x 7! 11 þ  x, both of which have radius of convergence 1. Thus, by the Product Rule, we have 1þx 1 1þx  ¼ 2 1x 1x ð1  xÞ     ¼ 1 þ x þ x2 þ x3 þ x4 þ     1 þ 2x þ 2x2 þ 2x3 þ 2x4 þ    1 X ¼ cn xn ; n¼0

where c0 ¼ 1  1 ¼ 1, c1 ¼ 1  2 þ 1  1 ¼ 3 and cn ¼ 1  2 þ 1  2 þ . . . þ 1  2 þ 1  1 ¼ 2n þ 1. 1 P & Thus the required power series for f at 0 is 1 þ x 2 ¼ ð2n þ 1Þxn : ð1  x Þ

Problem 3

Determine the Taylor series at 0 for:

(a) f ð xÞ ¼ ð1 þ xÞ loge ð1 þ xÞ; (b) f ð xÞ ¼

n¼0

1þx ð1xÞ3

;

for j xj < 1;

Also, its radius of convergence is at least 1. (In fact it is easy to check that its radius of convergence is exactly 1.)

for j xj < 1:

Next, we have seen already that the hyperbolic functions have the following Taylor series at 0 x3 x5 x2 x4 sinh x ¼ x þ þ þ    and cosh x ¼ 1 þ þ þ    ; 3! 5! 2! 4! each with radius of convergence 1. Notice that the derivative of the function sinh x is cosh x, and that term-by-term differentiation of the series 3 5 2 4 x þ x3! þ x5! þ    gives the series 1 þ x2! þ x4! þ   . It looks as though we can

At the start of this sub-section.

8.4

Manipulating power series

341

obtain the Taylor series for the derivative of a function f simply by differentiating the Taylor series for f itself. Our next result states that we can differentiate or integrate the Taylor series of a function fRterm-by-term to obtain the Taylor series of the corresponding function f 0 or f , respectively. Theorem 3

Differentiation and Integration Rules 1 P Let f ð xÞ ¼ an ðx  aÞn ; for jx  aj < R. Then: n¼0

Differentiation Rule f 0 ð xÞ ¼ R

Integration Rule

1 P n¼1

f ð xÞdx ¼

nan ðx  aÞn1 ; 1 P n¼0

for jx  aj < R;

nþ1

aÞ an ðx  n þ 1 þ constant;

To find the constant, put x ¼ a.

for jx  aj < R:

They may, however, behave differently at the end-points of their intervals of convergence.

All three series have the same radius of convergence. For example, consider the Taylor series at 0 for the function tan1 . We know that 1 ¼ 1  x2 þ x4  x6 þ    1 þ x2 1 X ¼ ð1Þn x2n ; with radius of convergence 1;

This is a geometric series.

n¼0

and that d 1 tan1 x ¼ ; for x 2 R: dx 1 þ x2 It follows, from the Integration Rule, that 3

5

7

x x x þ  þ  þc; for j xj < 1;and some constant c: 3 5 7 Substituting x ¼ 0 into this equation, we find that c ¼ tan1 0 ¼ 0. It follows that tan1 x ¼ x 

tan1 x ¼ x  Problem 4

x3 x5 x7 þ  þ   ; 3 5 7

for j xj < 1:

Find the Taylor series at 0 for the following functions:

(a) f ð xÞ ¼ ð1  xÞ2 ; for j xj < 1; (c) f ð xÞ ¼ tanh1 x; for j xj < 1:

(b) f ð xÞ ¼ ð1  xÞ3 ; for j xj < 1;

Problem 5 Find the first three non-zero terms in the Taylor series at 0 for the function f (x) ¼ ex(1  x)2, jxj < 1. State its radius of convergence. 3

5

x x Problem 6 Let f be the function f ð xÞ ¼ x þ 1:3 þ 1:3:5 þ þ x2nþ1 þ   ; x 2 R. 1:3::ð2nþ1Þ Determine the Taylor series at 0 for each of the following functions: (b) f 0 ð xÞ  xf ð xÞ: (a) f 0 ð xÞ;

Problem 7 Determine the Taylor series at 0 for the function 2 f ð xÞ ¼ ex , x 2 R. R1 2 Þn 1 1 Deduce that 0 ex dx ¼ 1  13 þ 10  42 þ    þ ð2nð1 þ 1Þn! þ   :

By the Integration Rule, the final Taylor series must have the same radius of convergence as the original Taylor series, namely 1.

8: Power series

342 We have now found a whole variety of techniques for identifying Taylor series: Taylor’s Theorem;

Combination, Product, Differentiation and Integration Rules.

But how do we know that different techniques will always give us the same expression as the Taylor series? The following result states that there is only one Taylor series for a function f at a given point a – any valid method will give the same coefficients. Theorem 4 Uniqueness Theorem 1 1 X X If an ðx  aÞn ¼ bn ðx  aÞn ; for jx  aj < R; then an ¼ bn : n¼0

Proof

n¼0

Let f ð xÞ ¼

1 P

an ðx  aÞn and gð xÞ ¼

n¼0

1 P

bn ðx  aÞn , for jx  aj R. Let c be any number such that R < c < R0 . Then 1 P nan cn1 is absolutely convergent, by the Absolute Convergence Theorem. n¼1

For example, c ¼ 12 ðR þ R0 Þ.

But   c jan cn j ¼ nan cn1    n  c  nan cn1 : Hence, by the Comparison Test for series, the series

1 P

an cn is absolutely

n¼1

convergent, and so is convergent. This contradicts the definition of R, and thus shows that R0 6> R. It follows that R ¼ R0 . Next, we show that f is differentiable on (R, R), and that f 0 is of the stated form. So, choose any point c 2 (R, R). Then choose a positive number r < R such that c 2 (r, r). Then, for all non-zero h such that jhj < r  jcj 1 1   f ð c þ hÞ  f ð c Þ X 1X  nan cn1 ¼ an ðc þ hÞn  cn hncn1 : h h n¼1 n¼1

(3)

Now, we apply Taylor’s Theorem to the function x 7! xn on the interval with end-points c and c þ h. We obtain 1 ðc þ hÞn ¼ cn þ nhcn1 þ nðn  1Þh2 cn2 n ; 2

For example, r ¼ 12 ðjcj þ RÞ.

8: Power series

346 where cn is some number between c and c þ h (and, in particular, with jcn j  r). Now   ðc þ hÞn cn  nhcn1   1 nðn  1Þh2 r n2 : (4) 2 It follows from (3) and (4) and the Triangle Inequality that    f ð c þ hÞ  f ð c Þ X  1 X 1 1  n1   nan c   jhj nðn  1Þjan jr n2 : (5)    h 2 n¼1 n¼2 1 P Since the series nan xn1 has radius of convergence R, it follows that so n¼1 1 1 P P does the series nðn  1Þan xn2 ; we conclude that the series nð n  1Þ n¼2

an r n2 is (absolutely) convergent. Hence, it follows from (5) and the Limit Inequality Rule that 1 f ðc þ hÞ  f ðcÞ X ¼ lim nan cn1 : h!0 h n¼1 The series 1 X an ð x  aÞ n f ð xÞ ¼

n¼2

&

Integration Rule

n¼0

and

1 X an ðx  aÞnþ1 F ð xÞ ¼ n þ 1 n¼0

have the same radius of convergence, R say. Also, f is integrable on (a  R, a þ R), and Z f ð xÞdx ¼ F ð xÞ: Proof The two power series have the same radius of convergence, by the Differentiation Rule for power series, applied to F. It also follows, from the Differentiation Rule for power series, that F is & differentiable and F0 ¼ f, so that F is a primitive of f.

8.5

Numerical estimates for p

One of the most interesting problems in the history of mathematics throughout the last two thousand years has been the determination to any pre-assigned pffiffiffi degree of accuracy of naturally arising irrational numbers such as 2, e and p. Here we look at various way of estimating p, and prove that p is irrational.

8.5.1

For we have shown that termby-term differentiation does not alter the radius of convergence of a power series.

Calculating p

It is easy to check that p is slightly larger than 3 by wrapping a string round a circle of radius r and measuring the length of the string corresponding to one circumference; for this length 2pr is just slightly larger than 6r.

This is the other part of Theorem 3, Sub-section 8.4.1.

That is, F is a primitive of f on (a  R, a þ R)

8.5

Numerical estimates for p

347

Over hundreds of years mathematicians devised more sophisticated methods for estimating the value of p. Some of their results are as follows: the Babylonians

c. 2000 BC

p ¼ 3 18 ¼ 3:125

the Egyptians the Old Testament

c. 2000 BC c. 550 BC

p ¼ 3 13 81 ’ 3:160 p¼3

Archimedes

c. 300–200 BC

between 3 17 and 3 10 71, p ’ 3:141

the Chinese the Hindus

c. 400–500 AD c. 500–600 AD

p ’ 3:1415926 pffiffiffiffiffi p ’ 10 ’ 3:16

1 Kings vii, 23; 2 Chronicles iv, 2.

With the development of the Calculus in the seventeenth century, new formulas for estimating p were discovered, including: Wallis’s Formula

p 2

2n 2n ¼ 21  23  43  45  65  67      2n1  2nþ1  

Leibniz’s Series

tan

Gregory’s Series

p 4

1

x¼x

x3 3

þ

x5 5



x7 7

þ 

¼ 1  13 þ 15  17 þ   

However, neither Wallis’s Formula nor Gregory’s Series is very useful for calculating p beyond a few decimal places, as they converge far too slowly. However, we can use the Taylor series for tan1 effectively for calculating p by choosing values of x closer to 0; in fact, the smaller the value of x, the faster the series converges, and so the fewer the number of terms we need to calculate p to a given degree of accuracy. For example, if we choose x ¼ p1ffiffi3 in Leibniz’s Series, we obtain p 1 1 1 1 3 1 1 5 1 1 7 pffiffiffi þ pffiffiffi  pffiffiffi þ    ¼ tan1 pffiffiffi ¼ pffiffiffi  6 5 7 3 3 3 3 3 3

1 1 1 1 þ  : ¼ pffiffiffi 1  þ  9 45 189 3 This formula can be used to calculate p to several decimal places without much effort; we gain about one extra place for every two extra terms. To obtain series that are more effective for calculating p, we can use the Addition Formula for tan1

1 1 1 x þ y tan x þ tan y ¼ tan ; 1  xy   provided that tan1 x þ tan1 y lies in the interval  p2 ; p2 . Problem 1

Theorem 5, Sub-section 7.4.2. Just before Problem 4, Subsection 8.4.1. Problem 4, Sub-section 8.3.2.

This is known as Sharp’s Formula.

Problem 3(a) on Section 4.3, in Section 4.5.

Use the Addition Formula for tan1 to prove that p 1 1 1 1 1 2 tan þ tan þ tan ¼ : 3 4 9 4

Similar applications of the Addition Formula give further expressions for p4, such as

1 p 1 1 1 1 1 6 tan þ 2 tan þ tan ¼ ; 8 57 239 4

1 1 p 4 tan1  tan1 ¼ : 5 239 4

This is known as Machin’s Formula.

8: Power series

348 Such formulas have been used to calculate p to great accuracy: 1 million decimal places were achieved in 1974. Recently, improved methods of numerical analysis have been used to calculate p correct to several million decimal places. In 1770, Lambert showed that p is irrational; that is, p is not the solution of a linear equation a0 þ a1x ¼ 0, where the coefficients a0 and a1 are integers. Then, in 1882, Lindemann showed that p is transcendental; that is, p is not a solution of any polynomial equation a0 þ a1x þ a2x2 þ    þ anxn ¼ 0, where all the coefficients are integers. We end with a couple of mnemonics that can be used to recall the first few digits for p; the word lengths give the successive digits: May I have a large container of coffee? 3: 1 4 1 5 9 2 6

For interest only, we list the first 1000 decimal places of p in Appendix 3.

This solved the ancient Greek problem of finding a ruler and compass method to construct a square with an area equal to that of a given circle.

and

8.5.2

How 3:

I 1

need 4

a 1

drink; alcoholic of 5 9 2

after 5

all those formulas 3 5 8

involving 9

course; 6

tangent 7

functions! 9

Proof that p is irrational

You may omit this at a first reading.

We now prove that p is irrational. Our proof is rather intricate; and, surprisingly, it uses many of the properties of integrals that you met in Chapter 7! It depends on the properties of a suitably chosen integral that we examine in Lemmas 1–3.   R1 n Lemma 1 Let In ¼ 1 ð1  x2 Þ cos 12 px dx, n ¼ 0, 1, 2, . . .. Then: (a) In > 0;

(b) In  2.

Proof   n (a) The function f ð xÞ ¼ ð1  x2 Þ cos 12 px , for x 2 [1, 1], is non-negative, for n ¼ 0, 1, 2, . . .. It follows, from the Inequality Rule for integrals, that In  0, for each n. To prove that In > 0 we need to examine the integral in more detail. By the Inequality Rule for integrals, we have Z 1 Z 1 Z 1 Z 1 ! 2 2 f ð xÞdx ¼ þ þ f ð xÞdx 1

1



Z

1 2

12

12

1 2

f ð xÞdx:

     n   n Now, if x 2 12 ; 12 , then ð1  x2 Þ cos 12 px  34 cos 14 p , so that Z 1 Z 1 n 2 3 1 f ð xÞdx  cos p dx 4 1 1 4 2 n 3 1 ¼ cos p > 0; 4 4 as required.

R 1 For 12 f ð xÞdx  0 and R1 1 f ð xÞdx  0: 2

By the Inequality Rule for integrals. For the length of the interval of integration is 1.

8.5

Numerical estimates for p

349

  n (b) For x 2 ½1; 1, ð1  x2 Þ cos 12 px  1  1 ¼ 1, so that Z 1 Z 1 f ð xÞdx  1dx ¼ 2: 1

Problem 2

&

1

By the Inequality Rule for integrals.

Prove that pI0 ¼ 4 and p3 I1 ¼ 32.

Next, we obtain a reduction formula for In, using integration by parts. Lemma 2

The integral In satisfies the following reduction formula p2 In ¼ 8nð2n  1ÞIn1  16nðn  1ÞIn2 :

Proof Using integration by parts twice, we obtain



1

Z     1 2 1 1 2 n2 2 n1 sin px  nð2xÞ 1  x sin px dx In ¼ 1  x p 2 2 1 p 1

Z 1  n1 4n 1 ¼ x 1  x2 sin px dx p 1 2



1  4n  2 1 2 n1 ¼ x 1x  cos px p p 2 1

Z 1n  n1  n2 o 8n 1 þ 2 1  x2  2x2 ðn  1Þ 1  x2 cos px dx p 1 2

Z 1  n2 8n 16 1 ¼ 2 In1  2 nðn  1Þ x2 1  x2 cos px dx p p 2 1

Z 1n     o 8n 16 1 2 n2 2 n1 ¼ 2 In1  2 nðn  1Þ 1x  1x cos px dx p p 2 1 ¼

The value of the first term is zero.

The value of the first term is zero.

For  n2 x2 1  x2    n2 1  x2 ¼ 1  1  x2  n2  n1 ¼ 1  x2  1  x2 :

8n 16 In1  2 nðn  1ÞfIn2  In1 g: p2 p

Multiplying both sides by p2, we obtain p2 In ¼ 8nIn1  16nðn  1ÞIn2 þ 16nðn  1ÞIn1 ¼ 8nð2n  1ÞIn1  16nðn  1ÞIn2 :

&

We now use the result of Lemma 2 to prove the crucial tool in our proof that p is irrational. Lemma 3 that

Proof

For n ¼ 0, 1, 2, . . ., there exist integers a0, a1, a2, . . ., an such ! n X p2nþ1 In ¼ a0 þ a1 p þ a2 p2 þ    þan pn ¼ ak p k : n! k¼0 2nþ1

For simplicity, let Jn ¼ p n! In . It follows from Problem 2 that J0 ¼ pI0 ¼ 4

and

J1 ¼ p3 I1 ¼ 32:

(1)

8: Power series

350 Next, we rewrite the reduction formula in Lemma 2 in terms of Jn as p2nþ1 In Jn ¼ n! 8nð2n  1Þ 2n1 16nðn  1Þ 2n1 p p ¼ In1  In2 n! n! ¼ 8ð2n  1ÞJn1  16p2 Jn2 : (2) The desired result now follows by Mathematical Induction. We know that it holds for n ¼ 0 and for n ¼ 1, by the statements (1) above. Using the reduction formula (2) we can prove that the statement of the Lemma holds & for all n  2. Finally, we can use the fact that, for any integer p, the sequence null, to prove that p is irrational. Theorem 1

n

p2nþ1 n!

o

is

The sequence

n ðp 2 Þ

p is irrational.

where the coefficients ak are integers. If we now multiply both sides by the expression q2nþ1 , we find that n X p2nþ1 In ¼ ak pk q2nþ1k ; n! k¼0 so that p2nþ1 In is an integer: n!

(3)

However, we know that the sequence

n

p

o 2nþ1 n!

is a null sequence; and we know,

from Lemma 1, that jInj 2. It follows, from the Squeeze Rule for sequences, that p2nþ1 In ! 0 n!

as n ! 1:

n!

basic null sequence.

Proof Suppose, on the contrary, that p is rational; so that p ¼ pq, for p, q 2 N. Then, the conclusion of Lemma 3 can be written in the form n X p2nþ1 pk I ¼ a ; n k q2nþ1 n! qk k¼0

(4)

2nþ1

It follows from (4) that eventually p n! In < 1; and so, from (3), that in fact In must equal 0. This contradicts the result of Lemma 1, part (a). This contra& diction proves that, in fact, p must be rational.

8.6

We omit ‘the gory details’!

Exercises

Section 8.1 1. Determine the tangent approximation to the function f(x) ¼ 2  3x þ x2 þ ex at the given point a: (a) a ¼ 0; (b) a ¼ 1.

This is a proof by contradiction.

is a

8.6

Exercises

351

2. Determine the Taylor polynomial of degree 3 for each of the following functions f at the given point a: (a) f(x) ¼ loge(1 þ x), a ¼ 2; (b) f ð xÞ ¼ sin x, a ¼ p6 ; (c) f ð xÞ ¼ ð1 þ xÞ2 , a ¼ 12 ; (d) f ð xÞ ¼ tan x, a ¼ p4 : 3. Determine the Taylor polynomial of degree 4 for each of the following functions f at the given point a: (a) f(x) ¼ cosh x, a ¼ 0; (b) f(x) ¼ x5, a ¼ 1. 4. Determine the percentage error involved in using the Taylor polynomial of degree 3 for the function f(x) ¼ ex at 0 to evaluate e0.1.

Section 8.2 1. Obtain an expression for the remainder term R1(x) when Taylor’s Theorem is applied to the function f(x) ¼ ex at 0. Show that, when x ¼ 1 and n ¼ 1, then the value of c in the statement of Taylor’s Theorem is approximately 0.36. 2. By applying Taylor’s Theorem to the function f(x) ¼ sin x with a ¼ p4, prove that

 1 p 1  p2 sin x ¼ pffiffiffi 1 þ x   x þ R2 ð xÞ; x 2 R; 4 2 4 2   3 where jR2 ð xÞj  16 x  p4 : 3. By applying the Remainder Estimate to the function f(x) ¼ sinh x with a ¼ 0, calculate sinh 0.2 to four decimal places. x 4. Calculate the Taylor polynomial T3(x) for the function f ð xÞ ¼ xþ3 at 2. 4 Show that T3(x) approximates f(x) to within 10 on the interval 2; 52 :

5. (a) Determine the Taylor polynomial of degree n for the function f(x) ¼ loge x at 1. (b) Write down the remainder term Rn(x) in Taylor’s Theorem for this function, and show that Rn(x) ! 0 as n ! 1 if x 2 (1, 2). (c) By using Theorem 2 in Section 8.2, determine a Taylor series for f at 1 which is valid when x 2 (1, 2).  12 x 6. Let f ð xÞ ¼ e ; x 6¼ 0; 0; x ¼ 0: (a) Prove that f 0 ð0Þ ¼ 0: (b) Prove that, for x 6¼ 0, f 0 (x) is of the form 1 (a polynomial of degree at most 3 in 1x)e x2 . (n) (c) Prove that, for x 6¼ 0, f (x) is of the form 1

(a polynomial of degree at most 3n in 1x)e x2 . (d) Prove that, for any positive integer n, f (n)(0) ¼ 0.

Section 8.3 1. Determine the radius of convergence of each of the following power series: 1 1 n  P P ð3nÞ! n 1 n (a) ð x þ 1Þ n ; (b) n! x  e ; ðn!Þ2 n¼1

n¼1

For this function, the Taylor polynomial Tn(x) is identically zero for each n, although the function f is not identically zero. In this case, Rn(x) ¼ f(x), for all n; and so the Taylor series for f 1 ðnÞ P at 0, f n!ð0Þ xn , converges to n¼0

f(x) only at 0.

8: Power series

352 (c)

1 P

ðn þ 1Þn xn ;

(d)

n¼1

1 2x

2 1:3:5 3 þ 1:3 2:5 x þ 2:5:8 x þ   ;

aðaþ1Þ:bðbþ1Þ 2 (e) 1 þ a:b 1:c x þ 1:2:c ðcþ1Þ x þ   ,

where a, b, c > 0:

2. Determine the Taylor series for the function f(x) ¼ loge(2 þ x) at the given point a; in each case, indicate the general term and state the radius of convergence of the series: (a) a ¼ 1;

This series is often called the hypergeometric series.

(b) a ¼ 1.

Hint: Use the Taylor series at 0 for the function x 7! loge ð1 þ xÞ, with t ¼ x  1 and t ¼ x þ 1, respectively. 3. Determine the interval of convergence of each of the following power series: 1 n 1 P P n 2 n (a) (b) ð1Þn 2n ðx  1Þn : n! x ; n¼1

n¼1

4. Give an example (if one exists) of each of the following: 1 P an xn which diverges at both x ¼ 1 and x ¼ 2; (a) a power series n¼0

(b) a power series

1 P

(c) a power series

n¼0 1 P

an xn which diverges at x ¼ 1 but converges at x ¼ 2; an xn which converges at x ¼ 1 but diverges at x ¼ 2.

n¼0

5. Give an example (if one exists) of each of the following: 1 P (a) a power series an xn which converges only if 3 < x < 3; (b) a power series

n¼0 1 P

(c) a power series

n¼0 1 P

(d) a power series

n¼0 1 P

an xn which converges only if 3  x < 3; an xn which converges only if 3 < x  3; an xn which converges only if 3  x  3.

n¼0

Section 8.4 1. Determine the Taylor series for each of the following functions at 0; in each case, indicate the general term and state the radius of convergence of the series:   3

x (a) f ð xÞ ¼ loge ð1 þ x þ x2 Þ ¼ loge 1x (b) f ð xÞ ¼ cosh 1x ; 1x : 2. Determine the first three non-zero terms in the Taylor series for each of the following functions at 0; and state the radius of convergence:

(a) f(x) ¼ cos(ex  1); (b) f(x) ¼ loge(1 þ sin x); (c) f(x) ¼ ex sin x. 2:4:6 7 5 3. For the function f ð xÞ ¼ x þ 23 x3 þ 2:4 3:5 x þ 3:5:7 x þ   , for jxj < 1, prove that   1  x2 f 0 ð xÞ  xf ð xÞ ¼ 1: 4. By using the Integration Rule for power series, prove that sinh1 x ¼ x 

1 x3 1:3 x5 1:3:5 x7 þ  þ   ; 2 3 2:4 5 2:4:6 7

for j xj < 1:

1

xffi . In fact, f ð xÞ ¼ psinffiffiffiffiffiffiffi 1x2

8.6

Exercises

353

5. By using R 1 the Integration Rule for power series, find an infinite series with sum 0 sinðt2 Þdt. 6. The Bessel function J0 is defined by the power series J0 ð xÞ ¼ 1 

x2 22 :ð1!Þ

þ 2

x4 24 :ð2!Þ

 2

x6 26 :ð3!Þ

þ  þ 2

ð1Þn x2n 22n :ðn!Þ2

þ  ; for x 2 R:

(a) Determine power series for J00 ð xÞ and J000 ð xÞ, indicating the general term in each. (b) Prove that xJ000 ð xÞ þ J00 ð xÞ þ xJ0 ð xÞ ¼ 0.

J0 is not the Taylor series at 0 of any familiar function.

Appendix 1: Sets, functions and proofs

Sets A set is a collection of objects, called elements. We use the symbol 2 to mean ‘is a member of’ or ‘belongs to’; thus ‘x 2 A’ means ‘the element x is a member of the set A’. Similarly, we use the symbol 62 to mean ‘is not a member of’ or ‘does not belong to’; thus ‘x 62 A’ means ‘the element x is not a member of the set A’. We often use curly brackets (or braces) to list the elements of a set in some way. Thus {a, b, c} denotes the set whose three elements are a, b and c. Similarly, {x : x2 ¼ 2} denotes the set whose elements x are such that x2 ¼ 2; we would read this symbol in words as ‘the set of x such that x2 ¼ 2’. When we define a set in this latter way, the symbol x is a dummy variable; that is, if we replace that symbol x by any other symbol, such as y, the set is exactly the same set; thus, for instance, the sets {x : x2 ¼ 2} and {y : y2 ¼ 2} are identical. Two sets are equal if they contain the same elements. We say that a set A is a subset of a set B, if all the elements of A are elements of B, and we denote this by writing ‘A  B’. We may wish to indicate specifically the possibility that the subset A is equal to B by writing ‘A  B’, or that A is a proper subset of B (that is, A is a subset of B but A 6¼ B), by writing ‘A  B’.

Thus the colon ‘:’ means ‘such that’.

6¼

Notice that, to show that two sets A and B are equal, it is necessary to prove that A is a subset of B and that B is a subset of A. In symbols A ¼ B is equivalent to the two properties both holding: A  B and B  A: The empty set is the set that contains no elements; it is denoted by the symbol ‘Ø’. It may seem strange to define such a thing; but, in practice, it is often a convenient set to use. The union of two sets A and B consists of the set of elements that belong to at least one of A and B; in symbols A [ B ¼ fx : x 2 A or x 2 Bg: The intersection of two sets A and B consists of the set of elements that belong to both A and B; in symbols A \ B ¼ fx : x 2 A and x 2 Bg: We denote the set of elements that belong to A but not to B by A  B; that is A  B ¼ fx : x 2 A; x 2 = Bg: There are some special symbols used that are used to denote commonly arising sets of real numbers:

N, Z, Q,

the set of all natural numbers; thus N ¼ {1, 2, 3, . . .}; the set of all integers; thus Z ¼ {0, 1, 2, 3, . . .}; the set of all rational numbers; that is, numbers of the form pq, where p, q 2 Z but q 6¼ 0; R, the set of all real numbers; R þ, the set of all positive real numbers. On the real line R, an interval I is a set of real numbers such that, if a and b both lie in I, then all numbers between a and b also lie in I.

354

Here we also allow x to belong to both A and B.

Sets, functions and proofs

355

There are nine types of intervals in R: ½a; b ¼ fx : x 2 R; a  x  bg; ða; bÞ ¼ fx : x 2 R; a < x < bg; ½a; bÞ ¼ fx : x 2 R; a  x < bg;

The first four types of intervals are bounded intervals.

ða; b ¼ fx : x 2 R; a < x  bg; ½a; 1Þ ¼ fx : x 2 R; x  ag; ða; 1Þ ¼ fx : x 2 R; x > ag; ð1; b ¼ fx : x 2 R; x  bg; ð1; bÞ ¼ fx : x 2 R; x < bg; R. An interval is said to be closed if it contains both of its end-points, open if it contains neither of its end-points, and half-closed or half-open if it is neither closed nor open.

Functions A function is a mapping of some element of R onto another element of R. Thus a function f is defined by specifying: a set A, called the domain of f; a set B, called the codomain of f; a rule x 7! f ð xÞ that associates with each element x of A a unique element f(x) of B.

Symbolically, we write this in the following way: f: A!B x 7! f ð xÞ: The element f(x) is called the image of x under f, and the set f (A) ¼ {y : y ¼ f (x) for some x 2 A} the image of A under f. A particularly simple function is the identity function that maps each element of a set to itself. Thus, the identity function on a set A, iA, is the function iA : A ! B x 7! x: Sometimes every point of the codomain is in the image. We say that a function f : A ! B is an onto function if f (A) ¼ B. Sometimes every point of the image is the image of precisely one point of the domain. We say that a function f : A ! B is a one–one function if: whenever f(x) ¼ f(y) for elements x, y of A, then necessarily x ¼ y. Notice that this does NOT mean that f is a one–one mapping of A onto B. A given function f : A ! B may be onto, one–one, both onto and one–one, or neither onto nor one–one. A function f : A ! B is called a bijection if it is both onto and one–one. If a function f : A ! B is one–one, then it has an inverse function f1 : f (B) ! A with the defining rule f 1 ð yÞ ¼ x;

where y ¼ f ðxÞ:

Notice that a function f only has an inverse function f 1 if it is bijective. Sometimes we want to ‘change the domain’ of a function to a larger set or a smaller set.

The final five types of intervals are unbounded intervals.

Appendix 1

356 If f : A ! B, and C is a subset of A, we say that the function g : C ! B is the restriction of f to C if gðxÞ ¼ f ð xÞ; for all x 2 C: Similarly, if f : A ! B, and A is a subset of C, we say that the function g : C ! B is the extension of f to C if gðxÞ ¼ f ð xÞ;

for all x 2 A:

Finally, the composite function or composite g f of two functions f :A!B

and

Sometimes we use the term extension to denote a function g : C ! D where A  C, B  D, and g(x) ¼ f(x), for all x 2 A.

g : C ! D;

where f (A)  C, is the function g f : A!D x 7! gð f ð xÞÞ: Proofs Logical implications are so important in Mathematics that we use some special symbols, namely ), ( and ,, to denote implications. Thus, if we are discussing two statements P and Q, we write P ) Q to mean: ‘if P is true, then Q is true’ or ‘P implies Q’; P ( Q to mean: ‘P is implied by Q’, or ‘if Q is true, then P is true’ or ‘Q implies P’; P , Q to mean: ‘P is true if and only if Q is true’; this is equivalent to the two separate statements P ) Q and Q ) P: In this case, P and Q are equivalent statement. The implications P ) Q and Q ) P are called converse implications. In order to prove that an assertion ‘P ) Q’ is true, we need to verify that, in all situations where the statement P holds, then the statement Q also holds. To prove that an assertion ‘P ) Q’ is not true, all that we need do is to find one example of a situation where the statement P holds but the statement Q does not. Such a situation is called a counter-example to the assertion. Sometimes we prove assertions P by simply checking all possible cases; generally, though, we devise logical arguments that deal with all cases at the same time. Sometimes our logical arguments appear slightly convoluted to non-mathematicians; two examples of these are:

Thus, to prove that P , Q, we need to prove that P ) Q and Q ) P.

This is sometimes called disproof by counter-example. This approach is sometimes called proof by exhaustion.

Proof by contradiction: In order to show that a statement P holds, we start by

assuming that P is false; we then look at the consequences of that assumption, and identify some specific consequence that is untrue or contradictory. It then follows that P must hold, after all. Proof by contraposition: In order to show that an implication P ) Q is true, it is

sufficient to prove that if Q does not hold, then P does not hold. Sometimes mathematical proofs are long and complicated, and involve the use of a variety of approaches to prove the desired result. Notice that not all approaches to proving a result are necessarily valid! Some examples of such approaches are: Proof by picture: Here you make a false claim in your argument because it appears

to be true in the particular diagram that you have drawn. Proof by example: For instance, you prove that some property holds for n ¼ 1 and

n ¼ 2, and then assert that it therefore holds for all positive integers.

This fact is sometimes (humorously) called proof by perspiration.

Sets, functions and proofs

357

Proof by omission: For instance, you prove one or two special cases of a result, and

claim that the general proof is ‘similar’. Proof by superiority: Here you claim that the result is ‘obvious’, rather than sit down

to construct a logical proof of it. Proof by mumbo-jumbo: Here you write down a jumble of formulas and relevant

words, ending up with the claim that you have proved the result.

Principle of Mathematical Induction This is a standard method of proving statements involving an integer n, generally a positive integer. Principle of Mathematical Induction Let P(n) denote a statement involving a positive integer n. If the following two conditions are satisfied: 1. the statement P(1) is true, 2. whenever the statement P(k) is true for a positive integer k, then the statement P(k þ 1) is also true, then the statement P(n) is true for all positive integers n. Sometimes we need to use an equivalent version of the Principle. (Second) Principle of Mathematical Induction Let P(n) denote a statement involving a positive integer n. If the following two conditions are satisfied: 1. the statement P(1) is true, 2. whenever the statements P(1), P(2), . . ., P(k) are true for a positive integer k, then the statement P(k þ 1) is also true, then the statement P(n) is true for all positive integers n. We have stated the Principle when P(n) applies to all integers n  1; analogous results hold when P(n) applies to all integers n  N, whatever integer N may be. We end with some useful results that can be proved using the Principle of Mathematical Induction. Sums n X

1 ¼ n;

k¼1 n X k¼1 n X

k2 ¼

nðn þ 1Þð2n þ 1Þ ; 6

n X

nðn þ 1Þ ; 2 k¼1   n X nðn þ 1Þ 2 k3 ¼ ; 2 k¼1 k¼

ð2k þ 1Þ ¼ ðn þ 1Þ2 ;

k¼0

sin A þ sinðA þ BÞ þ    þ sinðA þ ðn  1ÞBÞ     sin 12 nB n1 1  sin A þ ¼ B ; B 6¼ 0: 2 sin 2 B Arithmetic progression   a þ ða þ d Þ þ ða þ 2d Þ þ    þ ða þ ðn  1Þd Þ ¼ n a þ n1 2 d :

Thus, in order to use the Principle, we have a two stage strategy:

1. check P(1), 2. check that P(k) ) P(k þ 1).

Appendix 1

358 Geometric progression n a þ ar þ ar2 þ    þ ar n1 ¼ a 1r r 6¼ 1: 1r ; Binomial Theorem n   X nðn  1Þ 2 n k x þ    þ xn ; ð1 þ xÞn ¼ x ¼ 1 þ nx þ k 2! k¼0 ða þ bÞn ¼

n   X n nk k a b k k¼0

¼ an þ nan1 b þ

nðn  1Þ n2 2 a b þ    þ bn : 2!

It follows that the sum of the 1 P geometric series ar k , for k¼0

a jrj < 1,is 1r  . n n! Here, ¼ k!ðnk Þ! and k 0! ¼ 1.

Appendix 2: Standard derivatives and primitives

f(x)

f 0 (x)

Domain

k

0

R

x xn, n 2 Z  {0}

1 nxn1

R R

x,  2 R a x, a > 0

x1 ax loge a

(0, 1) R

xx

xx (1 þ loge x)

(0, 1)

sin x

cos x

R

cos x tan x

sin x sec2 x

R    R  n þ 12 p : n 2 Z

cosec x

cosec x cot x

sec x cot x

sec x tan x cosec2 x 1 ffi pffiffiffiffiffiffiffi 1x2 1 ffi pffiffiffiffiffiffiffi 2

R  {np : n 2 Z}    R  n þ 12 p : n 2 Z R  {np : n 2 Z}

sin1 x cos1 x

(1, 1) (1, 1)

tan1 x

1x 1 1þx2

ex

ex

R

loge x

1 x

(0, 1)

sinh x cosh x

cosh x sinh x

R R

tanh x sinh1 x

sech2 x 1 ffi pffiffiffiffiffiffiffi 2

R R

cosh1 x tanh

1

x

1þx 1 ffi pffiffiffiffiffiffiffi x2 1 1 1x2

R

(1, 1) (1, 1)

359

Appendix 2

360

f (x)

Primitive F(x)

xn, n 2 Z  {1}

nþ1

Domain

x nþ1 xþ1 þ1 ax loge a

R

sin x

cos x

R

cos x tan x

sin x loge (sec x)

R  1 1   2 p; 2 p

ex

ex loge x

R (0, 1)

loge x

loge jxj x loge x  x

(1, 0) (0, 1)

sinh x

cosh x

R

cosh x tanh x

R R

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2  x 2 , a > 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2  a2 , a > 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ x 2 , a > 0 ax e sin bx, a,b 6¼ 0

sinh x loge (cosh x) aþx 1 2a loge ax  1 1 x a tan a (  sin1 ax  cos1 ax  ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi loge x þ x2  a2  1 x cosh a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi loge x þ a2 þ x2  sinh1 ax pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 2  x2 þ 1 a2 sin1 x 2 xpa  a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 2  a2  1 a2 log x þ x 2  a2 2 xpx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 21 2 e  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 2 2 2 2 x a þ x þ 2 a loge x þ a þ x eax a2 þb2 ða sin bx  b cos bxÞ

eax cos bx, a,b 6¼ 0

eax a2 þb2

x ,  2 R  {1} 

a x, a > 0

1 x 1 x

1 a2 x2 , 1 a2 þx2 ,

a>0 a>0

1 pffiffiffiffiffiffiffiffiffi , a2 x2

a>0

1 pffiffiffiffiffiffiffiffiffi , x2 a2

a>0

1 pffiffiffiffiffiffiffiffiffi , a2 þx2

a>0

(0, 1) R

ða cos bx þ b sin bxÞ

(a, a) R (a, a) (a, a) (a, 1) (a, 1) R R (a, a) (a, 1) R R R

Appendix 3: The first 1000 decimal places of pffiffiffi 2, e and p More than a million decimal places of these numbers, and various similar commonly arising numbers, are easily available on the internet. pffiffiffi 2 ¼ 1. 4142135623 7379907324 9248360558 9505011527 6408891986 6872533965 8355752203 0305440277 0130156185 0146824720 0758474922 8246468157 2636881313 5738108967 4823728997 0249441341 6581726889

7309504880 7846210703 5073721264 8206057147 0955232923 4633180882 3753185701 9031645424 6898723723 7714358548 6572260020 0826301005 7398552561 5040183698 1803268024 7285314781 4197587165

1688724209 8850387534 4121497099 0109559971 0484308714 9640620615 1354374603 7823068492 5288509264 7415565706 8558446652 9485870400 1732204024 6836845072 7442062926 0580360337 8215212822

6980785696 3276415727 9358314132 6059702745 3214508397 2583523950 4084988471 9369186215 8612494977 9677653720 1458398893 3186480342 5091227700 5799364729 9124859052 1077309182 9518488472

7187537694 3501384623 2266592750 3459686201 6260362799 5474575028 6038689997 8057846311 1542183342 2264854470 9443709265 1948972782 2269411275 0607629969 1810044598 8693147101 ...

8073176679 0912297024 5592755799 4728517418 5251407989 7759961729 0699004815 1596668713 0428568606 1585880162 9180031138 9064104507 7362728049 4138047565 4215059112 7111168391

e ¼ 2. 7182818284 6277240766 8174135966 8298807531 7614606680 5517027618 7093287091 4637721112 9316368892 6680331825 3012381970 7825098194 5988885193 4841984443 3043699418 7683964243 1718986106

5904523536 3035354759 2904357290 9525101901 8226480016 3860626133 2744374704 5238978442 3009879312 2886939849 6841614039 5581530175 4580727386 6346324496 4914631409 7814059271 8739696552

0287471352 4571382178 0334295260 1573834187 8477411853 1384583000 7230696977 5056953696 7736178215 6465105820 7019837679 6717361332 6738589422 8487560233 3431738143 4563549061 1267154688

6624977572 5251664274 5956307381 9307021540 7423454424 7520449338 2093101416 7707854499 4249992295 9392398294 3206832823 0698112509 8792284998 6248270419 6405462531 3031072085 9570350354

4709369995 2746639193 3232862794 8914993488 3710753907 2656029760 9283681902 6996794686 7635148220 8879332036 7646480429 9618188159 9208680582 7862320900 5209618369 1038375051 ...

9574966967 2003059921 3490763233 4167509244 7744992069 6737113200 5515108657 4454905987 8269895193 2509443117 5311802328 3041690351 5749279610 2160990235 0888707016 0115747704

Numerical Analysts enjoy adding more digits to such decimals, using ever-more sophisticated computing techniques! For most practical purposes pffiffiffi 2 ’ 1:414:

For most practical purposes e ’ 2:718: A useful fact sometimes is that e3 ’ 20.

361

Appendix 3

362 p ¼ 3. 1415926535 5923078164 0938446095 6446229489 2712019091 7245870066 0113305305 0921861173 8912279381 1907021798 0005681271 2249534301 8640344181 0597317328 2619311881 5982534904 1712268066

8979323846 0628620899 5058223172 5493038196 4564856692 0631558817 4882046652 8193261179 8301194912 6094370277 4526356082 4654958537 5981362977 1609631859 7101000313 2875546873 1300192787

2643383279 8628034825 5359408128 4428810975 3460348610 4881520920 1384146951 3105118548 9833673362 0539217176 7785771342 1050792279 4771309960 5024459455 7838752886 1159562863 6611195909

5028841971 3421170679 4811174502 6659334461 4543266482 9628292540 9415116094 0744623799 4406566430 2931767523 7577896091 6892589235 5187072113 3469083026 5875332083 8823537875 2164201989

6939937510 8214808651 8410270193 2847564823 1339360726 9171536436 3305727036 6274956735 8602139494 8467481846 7363717872 4201995611 4999999837 4252230825 8142061717 9375195778 ...

5820974944 3282306647 8521105559 3786783165 0249141273 7892590360 5759591953 1885752724 6395224737 7669405132 1468440901 2129021960 2978049951 3344685035 7669147303 1857780532

For most practical purposes p ’ 3:1416:

A useful fact sometimes is that p2 ’ 10.

Appendix 4: Solutions to the problems

Chapter 1 Section 1.1 45 45 17 1. 1 <  17 20 <  53 < 0 < 53 < 20 < 1:

2. Let a and b be two distinct rational numbers, where a < b. Let c ¼ 12ða þ bÞ; then c is rational, and 1 c  a ¼ ðb  aÞ > 0; and 2 1 b  c ¼ ðb  aÞ > 0; 2 so that a < c < b. 3.

1 7

2 ¼ 0:142857142857 . . .; 13 ¼ 0:153846153846 . . ..

4. (a) Let x ¼ 0:231. Multiplying both sides by 103, we obtain 1000x ¼ 231:231 ¼ 231 þ x: Hence 999x ¼ 231 ) x ¼

231 77 ¼ : 999 333

(b) Let x ¼ 0:81. Multiplying both sides by 102, we obtain 100x ¼ 81:81 ¼ 81 þ x: Hence 99x ¼ 81 ) x ¼

81 9 ¼ : 99 11

Thus 2:281 ¼ 2 þ 5.

17 20

2 9 251 þ ¼ : 10 110 110

45 17 ¼ 0:85 and 45 53 ¼ 0:84 . . ., so that 53 < 20.

6. Suppose that there exists a rational number x such that x3 ¼ 2. Then we can write x ¼ pq. By cancelling, if necessary, we may assume that p and q have no common factor. The equation x3 ¼ 2 now becomes p3 ¼ 2q3 : Now, the cube of an odd number is odd, because ð2k þ 1Þ3 ¼ 8k3 þ 12k2 þ 6k þ 1   ¼ 2 4k3 þ 6k2 þ 3k þ 1; and so p must be even. Hence we can write p ¼ 2r, say. Our equation now becomes

363

Appendix 4

364 ð2r Þ3 ¼ 2q3 ; so that q3 ¼ 4r 3 : Hence q is also even, so that p and q do have a common factor 2, which contradicts our earlier statement that p and q have no common factors. Thus, no such number x exists. 7. We may take, for example, x ¼ 0.34 and y ¼ 0.34001000100001. . .. 8. Let a and b have decimal representations a ¼ a0  a1 a2 a3 . . .

and

Here a0, b0 are non-negative integers, and a1, b1, a2, b2, . . . are digits.

b ¼ b0  b1 b2 b3 . . .;

where we arrange that a does not end in recurring 9s, whereas b does not terminate (this latter can be arranged by replacing a terminating representation by an equivalent representation that ends in recurring 9s). Since a < b, there must be some integer n such that a0 ¼ b0 ; a1 ¼ b1 ; . . . ; an1 ¼ bn1 ; but an < bn : Then x ¼ a0  a1a2a3 . . . an  1bn is rational, and a < x < b as required. Next, let c ¼ 12 ðx þ bÞ, so that x < c < b. By repeating the same procedure, this time to the interval between the numbers c and b, we can find a rational number y with c < y < b. We have thus constructed two distinct rational numbers between a and b, as required.

Section 1.2 1. Rule 1 Rule 2 Rule 3

For any a, b 2 R, For any a, b, c 2 R,

a  b , b  a  0. a  b , a þ c  b þ c.

 For any a, b 2 R and any c > 0,  For any a, b 2 R and any c < 0,

a  b , ac  bc; a  b , ac  bc.

Remark Note that the following results are NOT true in general:  For any a, b 2 R and any c  0,  For any a, b 2 R and any c  0,

a  b , ac  bc; a  b , ac  bc.

For, if c ¼ 0, then we can make no assertion as to whether a  b or a  b from the information that ac  bc or ac  bc. To see this, take in turn a ¼ 2, b ¼ 3 and a ¼ 2, b ¼ 1. Rule 4 (Reciprocal Rule) For any positive a, b 2 R, a  b , 1a  1b. Rule 5 (Power Rule) For any non-negative a, b 2 R, and any p > 0, a  b , a p  b p. 2. (a) x þ 3 > 5. (b) 2  x < 0. (c) 5x þ 2 > 12. (d)

1 ð5x þ 2Þ

> 1 12 .

3. (a) Rearranging the inequality, we obtain 4x  x2  7 4x  x2  7  3 , 30 x2  1 x2  1 x2  x þ 1 0 , x2  1

4x  4x2  4 0 x2  1   2 x  12 þ 34 ,  0: x2  1 ,

Note that x does not end in recurring 9s, from the way in which it was constructed.

Solutions to the problems

365

 2 Since x  12 þ 34 > 0, for all x, the inequality holds if and only if x2  1 < 0. Hence the solution set is  x:

 4x  x2  7  3 ¼ ð1; 1Þ: x2  1

(b) Rearranging the inequality, we obtain 2x2  ðx þ 1Þ2 , 2x2  x2 þ 2x þ 1 , x2  2x  1  0 , ðx  1Þ2  2: Hence the solution set is n o n pffiffiffio n pffiffiffio x : 2x2  ðx þ 1Þ2 ¼ x : x  1   2 [ x : x  1  2  pffiffiffii h pffiffiffi  ¼ 1; 1  2 [ 1 þ 2; 1 : 4. We can obtain an equivalent inequality by squaring, provided that both sides are nonnegative. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Now, 2x2  2 is defined when 2x2  2  0; that is, for x2  1. So, 2x2  2 is defined and non-negative if x lies in (1, 1] [ [1, 1). Hence, for x  1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x2  2 > x , 2x2  2 > x2 , x2 > 2: So the part of the solution set in [1, 1) is Suppose, next, that x  1. Then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x2  2  0 > x;

pffiffiffi  2; 1 .

so that the whole of (1, 1] lies in the solution set. Hence the complete solution set is n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi o pffiffiffi  x : 2x2  2 > x ¼ ð1; 1 [ 2; 1 : 5. (a) 2x2  13 < 5 , 5 < 2x2  13 < 5 ðby Rule 6Þ , 8 < 2x2 < 18 , 4 < x2 < 9 , 2 < j xj < 3: Hence the solution set is

2  x : 2x  13 < 5 ¼ ð3; 2Þ [ ð2; 3Þ: (b) jx  1j  2jx þ 1j , ðx  1Þ2  4ðx þ 1Þ2 , x2  2x þ 1  4x2 þ 8x þ 4 , 0  3x2 þ 10x þ 3 , 0  ð3x þ 1Þðx þ 3Þ: Hence the solution set is

1 fx : jx  1j  2jx þ 1jg ¼ ð1; 3 [  ; 1 : 3

Appendix 4

366

Section 1.3 1. (a) Suppose that jaj  12. The Triangle Inequality gives

Hence

ja þ 1j  jaj þ 1 1  þ 1 ðsince jaj  12Þ 2 3 ¼ : 2 3 ja þ 1j  : 2

(b) Suppose that jbj < 12. The ‘reverse form’ of the Triangle Inequality gives 3 b  1  jjb3 j  1j ¼ jbj3 1  1  jbj3 : Now jbj
; 2 8 8

so, from the previous chain of inequalities 3 b  1 > 7: 8 2. Rearranging the inequality, we obtain 3n < 1 , 3n < n2 þ 2 n2 þ 2 , 0 < n2  3n þ 2 , 0 < ðn  1Þðn  2Þ; and this final inequality certainly holds for n > 2. 3. The result is true for all those a and b for which a þ b  0. We now consider those a and b for which a þ b > 0. Rearranging the given inequality, we obtain a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ bÞ2 pffiffiffi  a2 þ b2 ,  a2 þ b2 2 2 , a2 þ 2ab þ b2  2a2 þ 2b2 , 0  a2  2ab þ b2 , 0  ða  bÞ2 ; and this final inequality certainly holds. This completes the proof. 4. (a) Using the rules for rearranging inequalities, we obtain 

1 2 1 2 1 aþ 2a (as a2 > 2); so, by ( ), it follows that  

2 1 2 2 aþ a < ; a 2 a which gives the required inequality.   Alternatively, use a direct argument after squaring the expression 12 a þ 2a : 5. Since c, d  0, we can choose non-negative numbers a and b with c ¼ a2 and d ¼ b2. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Substituting into the result a2 þ b2  a þ b, for a, b  0, of Example 4, we obtain pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi c þ d  c þ d: 6. In Problem 5 we saw that pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi cþd  cþ d;

for numbers c; d  0:

( )

Following the good advice in the margin note before the Problem, we use this! For a, b, c  0, we deduce from ( ) that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ b þ c ¼ a þ ðb þ c Þ pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi  aþ bþc ðby an application of ð ÞÞ pffiffiffi pffiffiffi pffiffiffi  aþ bþ c ðby a further application of ð ÞÞ: 7. If we substitute x ¼ 1n in the Binomial Theorem for (1 þ x)n, when n  2 we get 

   n 1 n 1 nðn  1Þ 1 2 1 ¼1þn þ þ 1þ þ n n 2! n n n1 1þ1þ 2n 1 1 5 1 ¼2þ  ¼  : 2 2n 2 2n When n ¼ 1, the inequality also holds. This completes the proof. 8. Let P(n) be the statement PðnÞ : 4n > n4 : STEP 1 First we show that P(5) is true: 45  54. Since 45 ¼ 1024 and 54 ¼ 625, P(5) is certainly true. STEP 2 We now assume that P(k) holds for some k  5, and deduce that P(k þ 1) is then true. So, we are assuming that 4k > k4. Multiplying this inequality by 4 we get

This assumption is just P(k).

4kþ1 > 4k4 ; so it is therefore sufficient for our purposes to prove that 4k4  (k þ 1)4. Now 

1 4 4k4  ðk þ 1Þ4 , 4  1 þ : ( ) k As k increases, the expression 1 þ 1k decreases. Since k  5, 1 þ 1k  1 þ 15 ¼ 65 ; it follows that

Since P(k þ 1) is: 4kþ 1 > (k þ 1)4.

Appendix 4

368 

 4 1 4 6 1þ  k 5 ¼

1296 ¼ 2:0736 < 4: 625

Thus the inequality ( ) certainly holds for k  5, and so it follows that 4kþ 1  (k þ 1)4 also holds for k  5. In other words: P(k) true for some k  5 ) P(k þ 1) true. It follows, by the Principle of Mathematical Induction, that 4n > n4, for n  5. 1 9. Substituting x ¼ ð2nÞ into Bernouilli’s Inequality (1 þ x)n  1 þ nx (which we may 1 do since ð2nÞ  1 for all natural numbers n), we obtain 

1 n 1 1n 1 2n 2n 1 ¼ : 2

If we take the nth root of both sides of this final inequality (which is permissible, by the Power Rule), we find that 1

1 1  1; 2n 2n

so that 1

2n 

1 2n ¼ 1 2n  1 1  2n 1 : ¼ 1þ 2n  1

pffiffiffiffiffi ffi 10. If we apply the Cauchy–Schwarz Inequality, with ak in place of ak and p1ffiffiffi ak in place of bk, we have 

1 1 1 ða1 þ a2 þ    þ an Þ þ þ  þ a1 a2 an 

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 1 1 1 2  a1  pffiffiffiffiffi þ a2  pffiffiffiffiffi þ    þ an  pffiffiffiffiffi a1 a2 an ¼ ð1 þ 1 þ    þ 1Þ2

ðwith n terms in the bracketÞ

2

¼n : 11. Applying Theorem 3 to the n þ 1 positive numbers 1; 1 þ 1n ; 1 þ 1n ; . . . ; 1 þ 1n, we obtain 

1 

1 n nþ1 1 1 1þ   nþ1þn n nþ1 n ¼

nþ2 nþ1

¼ 1þ

1 ; nþ1

taking the (n þ 1)th power of this last inequality, by the Power Rule, we deduce that 



nþ1 1 n 1  1þ : 1þ n nþ1

Solutions to the problems

369

Section 1.4 1. (a) E1 is bounded above. For example, M ¼ 1 is an upper bound of E1, since x  1; for all x 2 E1 : Also, max E1 ¼ 1, since 1 2 E1. (b)

E2 is bounded above. For example, M ¼ 1 is an upper bound of E2, since 1 1   1; for n ¼ 1; 2; . . .: n However, E2 has no maximum element. If x 2 E2, then x ¼ 1  1n for some 1 positive integer n; and so there is another element of E2, such as 1  nþ1 , with 

1 1 1 1 since > : 1 M (for instance, take n > M, which implies that n2 > n > M). Hence M cannot be an upper bound of E3. 2. (a) The set E1 ¼ (1, 1] is not bounded below, and so it cannot have a minimum element. For each number m, there is a (negative) number x such that x < m. Since x 2 E1, m cannot be a lower bound of E1. (b) E2 is bounded below by 0, since 1

1  0; n

for n ¼ 1; 2 . . .:

Also, 0 2 E2, and so min E2 ¼ 0. (c) E3 is bounded below by 1, since n2  1;

for n ¼ 1; 2 . . .:

Also, 1 2 E3, and so min E3 ¼ 1. 3. ƒ is increasing on theinterval [3, 2). Since 3  x < 2 so that x2 2 (4, 9], we have f ð xÞ ¼ x12 2 19 , 14 . Hence ƒ is bounded above and bounded below. Next, since f ð3Þ ¼ 19 and 19 is a lower bound for ƒ on the interval [3, 2), it follows that ƒ has a minimum value of 19 on this interval. Finally, 14 is an upper bound for ƒ on the interval [3, 2) but there is no point x in [3, 2) for which f ðxÞ ¼ 14. So 14 cannot be a maximum of f on the interval.   However, if y is any number in 19 ; 14 , there is a number x >  p1ffiffiy in      p1ffiffiy ; 2 ½3; 2Þ such that f ð xÞ ¼ x12 > y, so that no number in 19 , 14 will serve as a maximum of f on the interval. It follows that f has no maximum value on [3, 2).

Appendix 4

370 4. (a) The set E1 ¼ (1, 1] has a maximum element 1, and so sup E1 ¼ max E1 ¼ 1: (b) We know that E2 ¼ f1  1n : n ¼ 1, 2, . . .g is bounded above by 1, since 1

1  1; n

for n ¼ 1; 2 . . .:

To show that M ¼ 1 is the least upper bound of E2, we prove that, if M0 < 1, then there is an element 1  1n of E2 such that 1 1  > M0: n However, since M0 < 1, we have 1

1 1 > M0 , 1  M0 > n n 1 ,n> 1  M0

ðsince 1  M 0 > 0Þ:

1 1 0 Choosing a positive integer n so large that n > 1M 0 , we obtain 1  n > M , as required. Hence 1 is the least upper bound of E2.

(c) The set E3 ¼ {n2 : n ¼ 1, 2, . . .} is not bounded above, and so it cannot have a least upper bound. 5. (a) The set E1 ¼ (1, 5] is bounded below by 1, since 1  x;

for all x 2 E1 :

To show that m ¼ 1 is the greatest lower bound of E1, we prove that, if m0 > 1, then there is an element x in E1 which is less than m0 . Since m0 > 1, there is a number x of the form x ¼ 1.00 . . . 01 such that 1 < x < m0 ; and clearly x 2 E1. Hence 1 is the greatest lower bound of E1. (b) The set E2 ¼ fn12 : n ¼ 1, 2, . . .g is bounded below by 0, since 0
0, then there is an element n12 in E2 such that n12 < m0 . Since m0 > 0, we have 1 1 < m0 , n2 > 0 n2 m 1 , n > pffiffiffiffiffi : m0 ffi, we obtain Choosing a positive integer n so large that n > p1ffiffiffi m0

1 n2

< m0 , as

required. Hence 0 is the greatest lower bound of E2. 6. ƒ is decreasing on the interval [1, 4). Since 1  x < 4 so that x2 2 [1, 16), we have 1  1 , 1 . Hence ƒ is bounded above by 1 and bounded below by 16 . f ð xÞ ¼ x12 2 16 Next, since f(1) ¼ 1 and 1 is an upper bound for ƒ on the interval [1, 4), it follows that ƒ has least upper bound 1 on [1, 4). 1 Finally, 16 is a lower bound for ƒ on the interval [1, 4) but there is no point x in 1 . [1, 4) for which f ðxÞ ¼ 16 1  However, if y is any number in 16 , 1 , there is a number x < p1ffiffiy in [1, 4) such that   1 , 1 will serve as a lower bound of ƒ on [1, 4). f ð xÞ ¼ x12 < y, so that no number in 16 1 as its greatest lower bound on [1, 4). It follows that ƒ has 16

Solutions to the problems

371

Chapter 2 Section 2.1 1. (a)

(i) 4, 7, 10, 13, 16; (ii)

1 1 1 1 1 3 , 9 , 27 , 81 , 243 ;

(iii) 1, 2, 3, 4, 5. (b) (i) a1 ¼ 1, a2 ¼ 2, a3 ¼ 6, a4 ¼ 24, a5 ¼ 120; (ii) a1 ¼ 2, a2 ¼ 2.25, a3 ¼ 2.37, a4 ¼ 2.44, a5 ¼ 2.49. 2. (a)

(b)

(c)

(d)

3. (a) n! is monotonic, because an ¼ n!

and

anþ1 ¼ ðn þ 1Þ!;

Appendix 4

372 so that anþ1  an ¼ ðn þ 1Þ!  n! ¼ n  n! > 0; for n ¼ 1; 2; . . .: Thus {n!} is increasing. Alternatively, an > 0, for all n, and anþ1 ðn þ 1Þ! ¼ n þ 1  1; ¼ n! an

for n ¼ 1; 2; . . .;

so that {n!} is increasing. (b) {2n} is monotonic, because an ¼ 2n and anþ1 ¼ 2ðnþ1Þ ; so that 1 1  2nþ1 2n 

1 1 ¼ n  1 < 0; 2 2

anþ1  an ¼

for n ¼ 1; 2; . . .:

Thus {2n} is decreasing. Alternatively, an > 0, for all n, and anþ1 2n 1 ¼ nþ1 ¼ < 1; 2 an 2

for n ¼ 1; 2; . . .;

so that {2n} is decreasing.

 (c) n þ 1n is monotonic, because an ¼ n þ

1 1 and anþ1 ¼ n þ 1 þ ; n nþ1

so that



1 1  nþ nþ1 n nðn þ 1Þ  1 ¼ > 0; for n ¼ 1; 2; . . .: nðn þ 1Þ

anþ1  an ¼

 nþ1þ

 Thus n þ 1n is increasing. 4. (a) (b)

TRUE:

2n > 1000, for n > 9, since {2n} is increasing and 210 ¼ 1024.

FALSE:

All the terms a1, a3, a5, . . . are negative.

1 < 0:025, for all n > 0:025 ¼ 40:   anþ1 1 nþ1 4 (d) TRUE: an > 0 for all n, and an ¼ 4 n :

(c)

1 TRUE: n

Now





1 nþ1 4 nþ1 4 1, 4 4 n n 1 pffiffiffi ,  21 n

So anþ1  1; an

for n > 2:

anþ1  an ;

for n > 2;

Hence

so that

n4  4n

is eventually decreasing.

1 1  44 n 1 , n  pffiffiffi ’ 2:414: 21

,1þ

Solutions to the problems

373

Section 2.2 1. (a)

1 1 n < 100 , n > 100, by the Reciprocal Rule (for positive numbers). Hence we may take X ¼ 100.

Any value for X greater than 100 will also be valid.

1 n

Any value for X greater than 333 will also be valid.

3 < 1000 , n > 1000 3 ¼ 333:3333 . . ., by the Reciprocal Rule (for positive numbers). Hence we may take X ¼ 333. 2. (a) ð1Þn 1 1 1 n2 < 100 , n2 < 100

(b)

, n2 > 100 , n > 10: Hence we may take X ¼ 10. ð1Þn 3 1 3 n2 < 1000 , n2 < 1000 1000 , n2 > 3 rffiffiffiffiffiffiffiffiffiffi 1000 ’ 18:25: ,n> 3 Hence we may take X ¼ 18. n o 3. (a) The sequence 2n 1 1 is a null sequence.

Any value for X greater than 10 will also be valid.

(b)

Any value for X greater than 18 will also be valid.

To prove this, we want to show that: for each positive number ", there is a number X such that 1 2n  1 < "; for all n > X:

( )

We know that 1 1 2n  1 < " , 2n  1 < " ðsince 2n  1 > 0Þ 1 , 2n  1 >  " 1 , n > 12 1 þ ; "   n o 1 1 it follows that ( ) holds if we take X ¼ 2 1 þ 1" . Hence 2n1 is null. n no Þ is not a null sequence. (b) The sequence ð1 10 To prove this, we must find a positive value of " such that the sequence does not eventually lie in the horizontal strip on the sequence diagram from " up 1 to ". We can take " ¼ 20 , as the following sequence diagram shows:

Appendix 4

374 There is NO value of X such that the following statement holds ð1Þn 1 10 < 20 ; for all n > X: n no Þ is a null sequence. (c) The sequence ðn1 4 þ1 To prove this, we want to show that: for each positive number ", there is a number X such that ð1Þn n4 þ 1 < "; for all n > X:

( )

We know that ð1Þn 1 n4 þ 1 < " , n4 þ 1 < " 1 , n4 þ 1 > " 1 4 , n >  1: " 1 Now, if "  1, then "  1  0, so that 1 n4 >  1; for n ¼ 1; 2; . . .; " hence ( ) holds with X ¼ 1. On the other hand, if 0 < " < 1, then 1"  1 > 0, so that 

14 1 1 4 n > 1,n> 1 ; " " 1 14 hence ( ) holds with X ¼ "  1 . n no Þ Thus ( ) holds in either case. Hence nð1 is null. 4 þ1 n o n o 1 1 4. (a) We know that 2n1 is null, and so ð2n1 is also null, by the Power Rule. 3 n o nÞ o n o

1 1 5 (b) The sequences n and 2n1 are null, so p56ffiffin and ð2n1 are also null, by Þ7 the Power Rule and Multiple Rule. o n 5 Hence the sequence p56ffiffin þ ð2n1 is null, by the Sum Rule. Þ7   o n

1

 1 1 are null, so n14 and are also null, by (c) The sequences n and 2n1 1 ð2n1Þ3 the PowerRule.  1 Hence is also null, by the Product Rule and Multiple Rule. 1 4 3n ð2n1Þ3

5. Here, using the Hint, we have  n 1 1 ¼n n n 2 2 1 1 n 2 ¼ : n n

1

 n  Thus, the sequence n dominates the sequence n 12 , and is itself null. It

 n  is null. follows, from the Squeeze Rule, that the sequence n 12 n o

1 1 6. (a) We guess that n2 þn is dominated by n . To check this, we have to show that 1 1  ; for n ¼ 1; 2; . . .; n2 þ n n this certainly holds, because n2 þ n  n; for n ¼ 1; 2; . . .: n o

 Since 1n is null, we deduce that n21þn is null, by the Squeeze Rule.

Solutions to the problems (b) We guess that

n

ð1Þ n!

o n

375 is dominated by

1 n .

To check this, we have to show that ð1Þn 1 n!  n ; for n ¼ 1; 2; . . . ; this certainly holds, because ð1Þn 1 n! ¼ n! ; for n ¼ 1; 2; . . .; and n!  n; for n ¼ 1; 2; . . .:

Þn  is null, by the Squeeze Rule. Since n is null, we deduce that ð1 n! n o

 2 n is dominated by n12 . (c) We guess that nsin 2 þ 2n

1

To check this, we have to show that sin n2 1 n2 þ 2n  n2 ; for n ¼ 1; 2; . . .; this certainly holds, because sin n2  1; for n ¼ 1; 2; . . .; and

Since

1 n2

n2 þ 2n  n2 ; for n ¼ 1; 2; . . .: n o n2 is null, we deduce that nsin is null, by the Squeeze Rule. 2 þ 2n

Section 2.3 1. (a)

The sequence appears to converge to 1. (b) Since bn ¼ an  1 ¼ n þn 1  1 ¼ 1n, it follows that fbn g ¼ 2. (a)

1 n

is a null sequence.

Appendix 4

376 2

1 2 Here an  1 ¼ nn2  þ 1  1 ¼ n2 þ 1. We know that

n

o

2 n2 þ 1

is a null sequence, so it

follows that {an  1} is also a null sequence. Hence {an} converges to 1. (b)

n

3

n

Þ Þ Here an  12 ¼ n þ2nð1  12 ¼ ð1 3 2n3 . We know that

 follows that an  12 is also a null sequence. Hence {an} converges to 12.

n

ð1Þn 2n3

3. (a) The dominant term is n3, so we write an as n3 þ 2n2 þ 3 2n3 þ 1 1 þ 2n þ n33 : ¼ 2 þ n13



 Since 1n and n13 are basic null sequences an ¼

lim an ¼

n!1

1þ0þ0 1 ¼ ; 2þ0 2

by the Combination Rules. (b) The dominant term is 3n, so we write an as an ¼ ¼

n2 þ 2n 3n þ n3 2n n2 3n þ 3

: 3 1 þ n3n n 2 o    n 3o n and n3n are basic null sequences Since n3n , 23 lim an ¼

n!1

0þ0 ¼ 0; 1þ0

by the Combination Rules. (c) The dominant term is n!, so we write an as an ¼

n! þ ð1Þn 2n þ 3n! n

1 þ ð1Þ ¼ 2n n! : n! þ 3 n no

n Þ and 2n! are basic null sequences Since ð1 n! lim an ¼

n!1

1þ0 1 ¼ ; 0þ3 3

by the Combination Rules.

o

is a null sequence, so it

Solutions to the problems

377

4. (a) We know that

rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi!n 2 2 ,n 1þ n 1þ : n1 n1 qffiffiffiffiffiffi 2 , we obtain Using the hint, with x ¼ n1 rffiffiffiffiffiffiffiffiffiffiffi!n rffiffiffiffiffiffiffiffiffiffiffi!2 2 nðn  1Þ 2 1þ  n1 2! n1 nðn  1Þ 2  ¼ n; ¼ 2 n1 as required. 1 n

1

(b) Since n  1, we have nn  1. Combining this inequality with that in part (a), we obtain rffiffiffiffiffiffiffiffiffiffiffi 2 1 n ; for n  2: 1n 1þ n  1 nqffiffiffiffiffiffiffiffio1 2 Now, n  1 n¼2 is a null sequence, by the Power Rule, so that rffiffiffiffiffiffiffiffiffiffiffi! 2 ¼ 1: lim 1 þ n!1 n1  1 Hence, by the Squeeze Rule, lim nn ¼ 1: n!1

Section 2.4 1. (a) {1 þ ( 1)n} is bounded, since the terms 1 þ (1)n take only the values 0 and 2. Hence j1 þ ð1Þn j  2; for n ¼ 1; 2; . . .: n

(b) {( 1) n} is unbounded. Given any number K, there is a positive integer n such that jð1Þn nj > K; for instance, n ¼ jK j þ 1.

2n þ 1 is bounded, since 2n nþ 1 ¼ 2 þ 1n so that n 2n þ 1 1 n ¼ 2 þ n  3; for n ¼ 1; 2; . . .:

 n  is bounded, since (d) 1  1n 1 0  1   1; for n ¼ 1; 2; . . .; n so that 

n 

n 1  1 ¼ 1  1  1; for n ¼ 1; 2; . . .: n n pffiffiffi 2. (a) fn ng o is unbounded, and hence divergent, by Corollary 1. 2 n (b) nn2 þ þ 1 is convergent (with limit 1), and hence bounded, by Theorem 1. In fact n2 þ n 1 þ 1n 1 ¼  1 þ  2; for n ¼ 1; 2; . . .: 1 2 n þ 1 1 þ n2 n (c)

(c) {( 1)nn2} is unbounded, and hence divergent, by Corollary 1.

n (d) The terms of the sequence nð1Þ are 1 1 1; 2; ; 4; ; 6; . . .; 3 5 so the sequence is unbounded: given any number K, there is an even positive integer 2n such that 2n > K. Hence the sequence is divergent, by Corollary 1.

Appendix 4

378

n

 n 3. (a) Each term of 2n is positive, and 2nn is a basic null sequence. Hence 2n ! 1, by the Reciprocal Rule. (b) First, note that



n100 2n  n100 ¼ 2n 1  n ; for n ¼ 1; 2; . . .; 2 n 100 o 100 and that n2n is a basic null sequence. It follows that n2n is eventually less

than 1, so that 2n  n100 is eventually positive. Also 1 1 0 n ¼ 0; ¼ lim 2 n100 ¼ lim n!1 2n  n100 n!1 1  n 1  0 2 by the Combination Rules. Hence 2n  n100 ! 1, by the Reciprocal Rule. n

(c) We know that 2n ! 1, by part (a), and that 2n 2n þ 5n100  ; for n ¼ 1; 2; . . .: n n n Hence 2n þ 5n100 ! 1, by the Squeeze Rule.

Remark You could have used the Reciprocal Rule or the Sum and Multiple Rules. n n 2o (d) Each term of the sequence 2n10þþnn is positive, and  n

1 2 þ n2 n10 þ n ¼ lim n lim 10 n!1 n þ n n!1 2 þ n2 ¼ lim

n10 2n

þ 2nn 2

1 þ n2n 0þ0 ¼ 0; ¼ 1þ0 n!1

by the Combination Rules. n 2 Hence 2n10þþnn ! 1, by the Reciprocal Rule. 4. (a)

(i) a2 ¼ 4, a4 ¼ 16, a6 ¼ 36, a8 ¼ 64, a10 ¼ 100; (ii) a3 ¼ 9, a7 ¼ 49, a11 ¼ 121, a15 ¼ 225, a19 ¼ 361;

(iii) a1 ¼ 1, a4 ¼ 16, a9 ¼ 81, a16 ¼ 256, a25 ¼ 625. (b) a1 ¼ 1; a3 ¼ 13; a5 ¼ 15; a2 ¼ 2, a4 ¼ 4, a6 ¼ 6. 5. (a) If an ¼ ð1Þn þ 1n, for n ¼ 1, 2, . . ., then 1 1 a2k ¼ 1 þ and a2k1 ¼ 1 þ ; 2k 2k  1 so that lim a2k ¼ 1; whereas lim a2k1 ¼ 1: k!1

k!1

Hence {an} is divergent, by the First Subsequence Rule.   (b) If an ¼ 13n  13n , for n ¼ 1, 2, . . ., then 1 a3k ¼ 0 and a3kþ1 ¼ ; for k ¼ 1; 2; . . .; 3 so that lim a3k ¼ 0; whereas lim a3kþ1 ¼ 13: k!1

k!1

Hence {an} is divergent, by the First Subsequence Rule.

Solutions to the problems

379

  (c) If an ¼ n sin 12np , for n ¼ 1, 2, . . ., then Now

a1 ¼ 1; a2 ¼ 0; a3 ¼ 3; a4 ¼ 0; a5 ¼ 5; a6 ¼ 0; . . .: 

1 a4kþ1 ¼ ð4k þ 1Þ sin 2kp þ p 2 ¼ 4k þ 1;

for k ¼ 1; 2; . . .;

so that a4kþ1 ! 1. Hence {an} is divergent, by the Second Subsequence Rule.

Section 2.5 1. Suppose that the sequence {an} is decreasing and bounded below, so that it is necessarily convergent. We show that the limit of {an} is the number m ¼ inf {an : n ¼ 1, 2, . . .}. To see this, let ‘ denote lim an . By the Limit Inequality Rule, since an  m we know n!1 that ‘  m. Now assume that in fact ‘ > m. Since ‘ > m, it follows that ‘ > 12 ð‘ þ mÞ > m. It follows, from the definition of greatest lower bound, that there then exists some integer X such that aX < 12 ð‘ þ mÞ; and so, since {an} is decreasing, that 1 an < ð‘ þ mÞ; for all n > X: 2 We deduce from the Limit Inequality Rule that ‘  12 ð‘ þ mÞ. We may rearrange this inequality as 2‘  ‘ þ m, or ‘  m. This contradicts our assumption that ‘ > m. It follows that, in fact, ‘ ¼ m. 2. We use the ideas in the discussion prior to the problem.

We shall show that this assumption leads to a contradiction.

By letting n ! 1. For we cannot have ‘ < m and ‘  m!

(a) Let a1 > 3. Then, from equation (5) in the discussion before the statement of the problem 1 a2  a1 ¼ ða1  1Þða1  3Þ 4 > 0; so that a2 > a1. This suggests that, in general, it might be that anþ1 > an in the case that a1 > 3; we check this, using Mathematical Induction. Let P(n) be the statement: If a1 > 3, then an þ 1 > an. We have just seen that the statement P(1) is true. Now assume that the statement P(k) is true; that is, that akþ1 > ak for some k  1. It follows, from equation (5) in the discussion before the statement of the problem, that 1 akþ1  ak ¼ ðak  1Þðak  3Þ 4 > 0; so that ak þ 1 > ak; in other words, the statement P(k þ 1) is true. This proves that P(n) holds for all n  1. Thus, if a1 > 3, the sequence {an} is increasing. It follows, from the Monotone Convergence Theorem, that either {an} converges to some (finite) limit ‘ or tends to infinity as n ! 1. But the only possible limits of the sequence are 1 and 3, so that since a1 > 3 and the sequence is increasing, clearly {an} cannot tend to a limit. It follows that an ! 1 as n ! 1. (b) Let 0  a1 < 1. Then, from equation (5) in the discussion before the statement of the problem,

We need a discussion of this type in order to identify the result that we wish to verify using Mathematical Induction.

Appendix 4

380 1 a2  a1 ¼ ða1  1Þða1  3Þ 4 > 0; so that a2 > a1. This suggests that, in general, it might be that anþ1 > an in the case that 0  a1 < 1; we check this, using Mathematical Induction. Let P(n) be the statement: If 0  a1 < 1, then anþ1 > an. We have just seen that the statement P(1) is true. Now assume that the statement P(k) is true; that is, that akþ1 > ak for some k  1. It follows, from equation (5) in the discussion before the statement of the problem, that 1 akþ1  ak ¼ ðak  3Þðak  1Þ 4 > 0; so that ak þ 1 > ak; in other words, the statement P(k þ 1) is true. This proves that P(n) holds for all n  1. Thus, if 0  a1 < 1, the sequence {an} is increasing. It follows, from the Monotone Convergence Theorem, that either {an} converges to some (finite) limit ‘ or tends to infinity as n ! 1. Now, we can use equation (3) before the problem, with n ¼ 1, to see that the assumption a1 < 1 implies that a2  1 ¼

 1 2 a  1 < 0; 4 1

so that a2 < 1. We can then prove, by Mathematical Induction, using an argument similar to that in part (a), that, if a1 < 1, then an < 1 for all n  1. But the only possible limits of the sequence are 1 and 3, so that since an < 1 for all n  1, clearly {an} cannot tend to 3 or to infinity. It follows that an ! 1 as n ! 1.   (c) Let a1 < 0. Since anþ1 ¼ 14 a2n þ 3 , it is clear that the behaviour of the sequence as n ! 1 depends only on the magnitude of a1 and not on its sign. It follows from this observation that: if 1 < a1 < 0, if a1 ¼ 1, if 3 < a1 < 1, if a1 ¼ 3, if a1 < 3,

then an ! 1 then an ! 1 then an ! 1 then an ! 3 then an ! 1

as n ! 1; as n ! 1; as n ! 1; as n ! 1; as n ! 1.

3. (a) The Binomial Theorem gives  x nðn  1Þ x2  x n x n þ 1þ ¼1þn þ þ n n  2! n n 1 1 2 xn 1  x þ  þ n : ¼1þxþ 2! n n The general term in this expansion is 





nðn  1Þ . . . ðn  k þ 1Þ xk 1 1 2 k1 k 1 1 ... 1  x: ¼ k! n k! n n n If k and x are fixed, n  this general term increases as n increases. Hence the sequence 1 þ nx is increasing if x > 0. (b) By the Binomial Theorem 

 1 k 1 k ¼1þ ; 1þk 1þ n n n

for k ¼ 1; 2; . . .:

( )

Solutions to the problems

381

 n  is bounded above, we choose an integer k  x and use To prove that 1 þ nx the inequality ( ), as follows 

 x n k n 1þ  1þ n n 

!n 

k 1 k 1 n  1þ ¼ 1þ n n



  1 n

 ek ;

 n  is increasing with limit e. Hence 1 þ nx is since the sequence 1 þ n k bounded above by e .

 n  (c) Since 1 þ nx is increasing and bounded above, it must be convergent, by the Monotone Convergence Theorem.

 n  4. The first n terms of the sequence 1 þ 1n are  1  2  3 

2 3 4 nþ1 n ; ; ;...; : 1 2 3 n Taking the product of these terms, each of which is less than e, we obtain 21 32 43 . . . ðn þ 1Þn < en ; 11 22 33 . . . nn it follows, by cancellation, that ðn þ 1Þn < en : n! Hence n! >

ðn þ 1Þn ¼ en



nþ1 n ; e

for n ¼ 1; 2; . . .:

5. We use the formulas:

pffiffi pffiffi   (a) Area ¼ 12  1  1  sin 13p ¼ 12  23 ¼ 43.   (b) Area ¼ 12  1  1  sin 16p ¼ 12  12 ¼ 14.   (c) Area ¼ 12  2 tan 16p  1 ¼ 12  2  p1ffiffi3 ¼ p1ffiffi3. 1  1  (d) Area ¼ 12  2 tan 12 p  1 ¼ tan 12 p . 1  To determine tan 12p , we use the formula 1   2 tan 12 p 1 1 : tan p ¼ 6 1  tan2 12 p 1  1ffiffi p Since tan 6p ¼ 3, we obtain 

pffiffiffi   1 1 tan2 p  1 ¼ 0; p þ 2 3 tan 12 12 so that pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffi2 ffi 

2 3  2 3 þ4 1 p ¼ tan 12 2 pffiffiffi ¼  3  2:

Appendix 4

382 pffiffiffi 1  1  Since tan 12 p > 0, we must have tan 12 p ¼ 2  3; it follows that pffiffiffi Area ¼ 2  3:

Chapter 3 Section 3.1 1. (a) Using the formula for summing a geometric series, with a ¼ r ¼  13, we obtain   2  3  n 1 1 1 1 þ  sn ¼  þ  þ þ  3 3 3 3  1n 1   1  3 ¼  3 1   13  n  1 1 1  : ¼ 4 3

 n  Since  13 is a basic null sequence 1 lim sn ¼  ; n!1 4 and so

1  X 1 n 1  is convergent; with sum  : 3 4 n¼1 (b) Here sn ¼ 1 þ ð1Þ1 þð1Þ2 þ    þ ð1Þn1 ¼ 1  1 þ 1  1 þ    þ ð1Þn1 ; so that

 sn ¼

1; n odd; 0; n even:

Hence s2kþ1 ! 1 and s2k ! 0 as k ! 1, so that {sn} is divergent, by the First 1 P Subsequence Rule. Thus the series ð1Þn is divergent. n¼0

(c) Using the formula for summing a geometric series with a ¼ 2 and r ¼ 12, we obtain   2  n3 1 1 1 sn ¼ 2 þ 1 þ þ þ þ 2 2 2  n 1  12  ¼2 1  12  n  1 : ¼4 1 2

 n  Since 12 is a basic null sequence, lim sn ¼ 4, and so n!1 1  n X 1 is convergent; with sum 4: 2 n¼1 2. (a) We interpret 0.111. . . as 1 1 1 þ þ þ   : 101 102 103

Solutions to the problems

383

1 1 This is a geometric series with a ¼ 10 and r ¼ 10 . Since convergent with sum

1 10

< 1, the series is

1 a 1 ¼ 10 1 ¼ ; 1  r 1  10 9

hence 1 0:111 . . . ¼ : 9 (b) We interpret 0.86363. . . as 

8 1 63 63 þ þ    : þ 10 10 1001 1002 63 1 The series in the bracket is a geometric series with a ¼ 100 and r ¼ 100 . Since 1 < 1, this series is convergent with sum 100 63 a 63 7 ¼ 100 1 ¼ ¼ ; 1  r 1  100 99 11

hence 0:86363 . . . ¼

8 7 19 þ ¼ : 10 110 22

(c) We interpret 0.999. . . as 9 9 9 þ þ þ   : 101 102 103 9 1 and r ¼ 10 . Since This is a geometric series with a ¼ 10 convergent with sum

1 10

< 1, this series is

9 a ¼ 10 1 ¼ 1; 1  r 1  10

hence 0:999 . . . ¼ 1: 1 12 1 s2 ¼ 12 1 s3 ¼ 12 1 s4 ¼ 12 4. Since 3. s1 ¼

1 ; 2 1 1 1 2 þ ¼ þ ¼ ; 23 2 6 3 1 1 2 1 3 þ þ ¼ þ ¼ ; 23 34 3 12 4 1 1 1 3 1 4 þ þ þ ¼ þ ¼ : 23 34 45 4 20 5 ¼



1 1 1 1 ¼  ; nðn þ 2Þ 2 n n þ 2 we have

for n ¼ 1; 2; . . .;



1 X 1 1 1 1 ¼  ; nðn þ 2Þ n¼1 2 n n þ 2 n¼1

1 X

and so 1 1 1 1 1 þ þ þ þ   þ 13 24 35 46 nðn þ 2Þ 









 1 1 1 1 1 1 1 1 1 1 ¼ 1 þ  þ  þ  þ þ  : 2 3 2 4 3 5 4 6 n nþ2

sn ¼

Appendix 4

384 Most of the terms in alternate brackets cancel out, leaving 

1 1 1 1 sn ¼ 1þ   2 2 nþ1 nþ2 3 1 1  : ¼  4 2ðn þ 1Þ 2ðn þ 2Þ n o n o Since n þ1 1 and n þ1 2 are null sequences, lim sn ¼ 34, and so n!1

1 X

1 3 is convergent, with sum : n ð n þ 2 Þ 4 n¼1 1   P n 3 3 5. The series 4 is a geometric series, with a ¼ r ¼ 4. Hence, it is convergent, with n¼1

3

sum 14 3 ¼ 3. 4

The series

1 P n¼1

1 nðn þ 1Þ

is convergent, with sum 1 (cf. Sub-section 3.1.3). Hence, by

the Combination Rules

1  n X 3 2 is convergent, with sum 3  ð2  1Þ ¼ 1:  4 nðn þ 1Þ n¼1 6. By the Combination Rules for sequences n2 1 1 ¼ lim ¼ ; þ 1 n!1 2 þ n12 2 n 2 o so that the sequence 2n2n þ 1 is not null. 1 P n2 Hence, by the Non-null Test, 2n2 þ 1 is divergent. lim

n!1 2n2

n¼1

Section 3.2 1. Let sn ¼ 1 þ 212 þ 312 þ    þ n12 and tn ¼ 1 þ 12 þ 13 þ    þ 1n. The values of these partial sums are as listed below:

n

1

2

3

4

5

6

7

8

sn 1

1.25

1.36

1.42

1.46

1.49

1.51

1.53

tn

1.50

1.83

2.08

2.28

2.45

2.59

2.72

1

2. (a) We use the Comparison Test. Since n3 þ n  n3 ;

for n ¼ 1; 2; . . .;

Solutions to the problems

385

we have 1 1  ; n3 þ n n3 Since

1 P n¼1

1 n3

for n ¼ 1; 2; . . .:

is convergent, we deduce, from the Comparison Test, that 1 X

1 is convergent: 3þn n n¼1

(b) Let an ¼

1 pffiffiffi nþ n

1 bn ¼ ; n

and

for n ¼ 1; 2; . . .;

so that lim

an

n!1 bn

n pffiffiffi þ n 1 ¼ lim n!1 1 þ p1ffiffi ¼ lim

n!1 n

n

¼ 1 6¼ 0: 1 P 1 Since n is divergent, we deduce, from the Limit Comparison Test, that n¼1 1 X 1 pffiffiffi is divergent: n þ n n¼1 (c) Let an ¼

nþ4 2n3  n þ 1

and

bn ¼

1 ; n2

for n ¼ 1; 2; . . .;

so that an n3 þ 4n2 ¼ lim 3 n!1 bn n!1 2n  n þ 1 1 þ 4n ¼ lim n!1 2  12 þ 13 n n 1 ¼ 6¼ 0: 2 is convergent, we deduce, from the Limit Comparison Test, that 1 X nþ4 is convergent: 3nþ1 2n n¼1 lim

Since

1 P n¼1

1 n2

(d) We use the Comparison Test. Since 0  cos2 ð2nÞ  1; for n ¼ 1; 2; . . .; we have cos2 ð2nÞ 1  3 ; for n ¼ 1; 2; . . .: n3 n is convergent, we deduce, from the Comparison Test, that 0

Since

1 P n¼1

1 n3

1 X cos2 ð2nÞ

n3

n¼1

is convergent:

3

3. (a) Let an ¼ nn! , for n ¼ 1, 2, . . ., so that !  anþ1 ðn þ 1Þ3 n!  3 ¼ n an ðn þ 1Þ! ðn þ 1Þ2 n3 2 n þ 2n þ 1 1 2 1 ¼ ¼ þ 2þ 3: n3 n n n ¼

Appendix 4

386 Hence, by the Combination Rules for sequences anþ1 lim ¼ 0; n!1 an 1 3 P n it follows, from the Ratio Test, that n! is convergent. n¼1

2 n

(b) Let an ¼ n n!2 , for n ¼ 1, 2, . . ., so that ! 

anþ1 ðn þ 1Þ2 2nþ1 n! ¼  2 n n 2 an ðn þ 1Þ! 2ðn þ 1Þ 2 n

1 1 ¼2 þ 2 : n n

¼

Hence, by the Combination Rules for sequences anþ1 ¼ 0; lim n!1 an 1 2 n P n 2 it follows, from the Ratio Test, that n! is convergent. n¼1

Þ! (c) Let an ¼ ð2n nn , for n ¼ 1, 2, . . ., so that ! 

anþ1 ð2n þ 2Þ! nn ¼  an ð2nÞ! ðn þ 1Þnþ1

¼

ð2n þ 2Þð2n þ 1Þnn

ðn þ 1Þnþ1 2ð2n þ 1Þnn ¼ ðn þ 1Þn 4n þ 2 n : ¼ 1 þ 1n   n n o   ð1þ1nÞ 1 n 1 We know that lim 1 þ n ¼ e and that 4n þ 2 is null, so that 4n þ 2 is null, n!1

by the Product Rule for sequences. It follows, from the Reciprocal Rule for sequences, that anþ1 ! 1 as n ! 1: an 1 P ð2nÞ! It follows, from the Ratio Test, that nn is divergent. n¼1

Remark Notice that ð2nÞ!  nn

 



2n 2n  1 nþ1     n n n

 1; it follows, from the Non-null Test, that

1 P ð2nÞ! n¼1

nn

is divergent.

4. (a) Let an ¼ n log1 n, n ¼ 2, .n. .. Then o an is positive; and, since {nlogen} is an increase

ing sequence, fan g ¼

1 n loge n

is a decreasing sequence.

Next, let bn ¼ 2n a2n ; thus 1 2n loge ð2n Þ 1 1 1 ¼ ¼  : n loge 2 loge 2 n

bn ¼ 2n 

Solutions to the problems

387

1 P

1 Since n is a basic divergent series, it follows by the Multiple Rule that 1 n¼2 P bn must be divergent. Hence, by the Condensation Test, the original series n¼2 1 P n¼2

1 n loge n

must also be divergent.

, n ¼ 2, . . .. Then an is positive; and, since {n(loge n)2} is an n o increasing sequence, fan g ¼ nðlog1 nÞ2 is a decreasing sequence.

(b) Let an ¼ nðlog1

e

nÞ2

e

Next, let bn ¼ 2n a2n ; thus bn ¼ 2n  ¼

1 2n ðloge ð2n ÞÞ2

1 ðn loge 2Þ2

¼

1 ðloge 2Þ2



1 : n2

1 P 1 Since n2 is a basic convergent series, it follows by the Multiple Rule that 1 n¼2 P bn must be convergent. Hence by the Condensation Test, the original series n¼2 1 P n¼2

1 nðloge nÞ2

must also be convergent.

Section 3.3 nþ1

Þ n 1. (a) Let an ¼ ð1 n3 þ 1 , for n ¼ 1, 2, . . . , so that

jan j ¼

n ; n3 þ 1

for n ¼ 1; 2; . . .:

Now n n 1  ¼ ; n3 þ 1 n3 n2 and

1 P n¼1

1 n2

for n ¼ 1; 2; . . .;

is a basic convergent series. Hence, by the Comparison Test 1 X n is convergent: 3þ1 n n¼1

It follows, from the Absolute Convergence Test, that 1 X ð1Þnþ1 n is convergent: n3 þ 1 n¼1 n (b) Let an ¼ cos 2n , for n ¼ 1, 2, . . .; then 1 jan j  n ; for n ¼ 1; 2; . . .; 2

since jcos nj  1. 1 P 1 Now 2n is a basic convergent series (in fact, a convergent geometric n¼1

series). Hence, by the Comparison Test 1 X cos n n¼1

2n

is absolutely convergent:

It follows, from the Absolute Convergence Test, that 1 X cos n n¼1

2n

is convergent:

Appendix 4

388 2. The series 1 1 1 1 1 1 þ  þ þ  þ  2 4 8 16 32 64 is absolutely convergent, and hence convergent, by the Comparison Test, since the 1 P 1 series 2n is a convergent geometric series. n¼1

By the infinite form of the Triangle Inequality, we have 1 1 1 1 þ  þ þ 1  1 þ     1 þ 1 þ 1 þ 1 þ 1 þ 1 þ    2 4 8 16 32 64 2 4 8 16 32 64 ¼ 1: It follows that the sum of the original series lies in the interval [1, 1]. n nþ1 o 

is of the form ð1Þnþ1 bn , where 3. (a) The sequence ð1nÞ3 bn ¼

1 ; n3

for n ¼ 1; 2; . . .:

Now: 1 n3  0, for n ¼ 1, 2, . . .;

 2. n13 is a basic null sequence;

 3. n13 is decreasing, because {n3} is increasing. 1 P ð1Þnþ1 Hence, by the Alternating Test, is convergent. n3 n o n¼1 n is is not a null sequence, since (b) The sequence ð1Þnþ1 nþ2 n ¼ 1; lim n!1 n þ 2 and so the odd subsequence tends to 1. Hence, by the Non-null Test 1 X n ð1Þnþ1 is divergent: n þ 2 n¼1   n o Þnþ1 is of the form ð1Þnþ1 bn , where (c) The sequence ð1 1 1

1.

n3 þn2

bn ¼ Now: 1. 2.

1 1 3

n þn 

1 2

1 1 3

1

n þ n2

for n ¼ 1; 2; . . .:

;

 0, for n ¼ 1, 2, . . .;  is a null sequence, by the Squeeze Rule, since 1

1 1 n3 þ n2

0

1 1 3

1 2



1 1 ; n2

n þn n o where 11 is a basic null sequence;  n2 n 1 o 1 1 is decreasing, because n3 þ n2 is increasing. 3. 1 1 n3 þ n2

Hence, by the Alternating Test,

1 P ð1Þnþ1 1

1

n¼1 n3 þ n2

is convergent.

4. Let sn and tn denote the nth partial sums of the series 1 1 1 1 1 1 1 1 1 1 1 1 1 1   þ   þ   þ    and 1  þ  þ  þ    ; 2 4 3 6 8 5 10 12 2 3 4 5 6 n P 1 1 1 respectively. Denote by Hn the nth partial sum k ¼ 1þ 2 þ    þ n of the harmonic k¼1 series.

Solutions to the problems The terms of the series

389 1 P

an come ‘in a natural way’ in threes, so it seems

n¼1

sensible to look at the partial sums s3n 1 1 1 1 1 1 1 1 1 1 1 s3 n ¼ 1   þ   þ   þ    þ   2 4 3 6 8 5 10 12 2n  1 4n  2 4n 







1 1 1 1 1 1 1 1 1 1 1 þ þ þ  þ ¼ 1        2 4 3 6 8 5 10 12 2n  1 4n  2 4n 



1 1 1 1 1 1 1 1 1  þ þ þ þ  þ þ ¼ 1 þ þ þ  þ 3 5 2n  1 2 4 6 8 4n  2 4n 



1 1 1 1 1 1 1 1 þ  þ þ  þ ¼ 1 þ þ þ þ  þ 2 3 4 2n  1 2n 2 4 2n 

1 1 1 1 1 þ þ þ  þ  2 2 3 2n 1 1 ¼ H2n  Hn  H2n 2 2 1 ¼ ðH2n  Hn Þ: 2 But as we saw in Example 2 t2n ¼ H2n  Hn ; so that 1 s3n ¼ t2n : 2 By assumption, tn ! loge 2 and so t2n ! loge 2 as n ! 1. It follows, by the Multiple Rule for sequences, that 1 s3n ! loge 2 as n ! 1: 2 Finally 1 1 s3n1 ¼ s3n þ ! loge 2 as n ! 1; 4n 2 and 1 1 1 s3n2 ¼ s3n þ þ ! loge 2 as n ! 1: 4n  2 4n 2 It follows from these three results that sn ! 12 loge 2 as n ! 1. 5. We follow the pattern of the solution to Example 3, by finding a subsequence of partial 1 P ð1Þnþ1 sums of the series ¼1  12 þ 13  14 þ 15  16 þ    that are greater than 2, 3, . . . n n¼1

in turn and ensuring that the other partial sums are not much less than these values. 1 P ð1Þnþ1 First, note that all terms of the series are at most 1 in modulus. n 1 n¼1 P ð1Þnþ1 Next, since the ‘positive’ part of the series has partial sums that tend to n n¼1

1, we start to construct the desired rearranged series as follows. Take enough of the ‘positive’ terms 1; 13 ; 15 ; . . .; 2N11 1 so that the sum 1 þ 13 þ 15 þ    þ 2N11 1 is greater than 2, choosing N1 so that it is the first integer such that this sum is greater than 2. Then these N1 terms will form the first N1 terms in our desired rearranged series. Next, take one ‘negative’ term 12 and consider the expression 

 1 1 1 1 :  1 þ þ þ þ 3 5 2N1  1 2 It is certainly less than the partial sum tN1 ¼ 1 þ 13 þ 15 þ    þ 2N11 1 of the rearranged series, for which tN1 > 2. But, since all terms are at most 1 in magnitude, it follows that tN1 þ1 > 1.

Appendix 4

390 Now add in some more ‘positive’ terms 2N11þ1 ; 2N11þ3 ; . . .; 2N211 so that the sum 

 

1 1 1 1 1 1 1 þ 1 þ þ þ  þ  þ þ  þ 3 5 2N1  1 2 2N1 þ 1 2N1 þ 3 2N2  1 is greater than 3, choosing N2 so that it is the first available integer such that this sum is greater than 3. Then these N2 þ 1 terms will form the first N2 þ 1 terms in our desired rearranged series. We then add in just one ‘negative’ term; this makes the next partial sum of the rearranged series satisfy tN2 þ2 > 2. Then we add in sufficient positive terms to make the partial sum tN3 þ2 > 4. And so on indefinitely. In each set of two steps in this process we must use at least one of the ‘positive’ terms and one of the ‘negative’ terms, so that eventually all the ‘positive’ terms and all the ‘negative’ terms of the original series will be taken exactly once in the new 1 1 P P series, which we denote by bn . So certainly the series bn is a rearrangement of n¼1 n¼1 1 P ð1Þnþ1 the original series n . n¼1

From our construction, we have ensured that: 1 P for all n > Nk, for any k, the nth partial sums of bn are >k. n¼1 1 P bn tend to infinity as n ! 1. It follows, then, that the nth partial sums of n¼1 1

1  P 1 6. (a) The sequence 2 n is not null; hence, by the Non-null Test, the series 2 n is n¼1 divergent. 1 P 1 Thus 2 n is neither convergent nor absolutely convergent. n¼1

(b) We have

 n  n 5n þ 2n 1 2 ¼ 5n þ ; for n ¼ 1; 2; . . .: 3n 3 3 1   1   P P n 2 n Now, n 13 and are both basic convergent series; hence, by the 3 n¼1

n¼1

Combination Rules for series 1 X 5n þ 2n n¼1

3n

(c) The sequence fan g ¼

n

is convergent ðand absolutely convergentÞ: o

3 2n3  1

is null (so the Non-null Test is inappropriate),

and contains only positive terms. The Ratio Test fails, since anþ1 2n3  1 ¼ lim ¼ 1: n!1 an n!1 2ðn þ 1Þ3 1 lim

Instead, we use the Limit Comparison Test, with bn ¼

1 ; n3

for n ¼ 1; 2; . . .:

Then 3 an 3 ¼ lim 2n 11 n!1 bn n!1 n3

lim

3n3 n!1 2n3  1 3 ¼ 6¼ 0: 2

¼ lim

Since all terms are non-negative.

Solutions to the problems 1 P

Since

n¼1

1 n3

391

is a basic convergent series, we deduce, from the Limit Comparison

Test, that 1 X

3 is convergent: 31 2n n¼1

Since the terms in the series are non-negative, it follows that it is also absolutely convergent. n nþ1 o 1 P 1 (d) The sequence ð1Þ1 is null, but 1 is a basic divergent series. Hence 3 n n¼1 n3 1 nþ1 P ð1Þ is not absolutely convergent. 1 n¼1 n3 n nþ1 o n o is of the form ð1Þnþ1 bn , where However, ð1Þ1 n3

bn ¼

1 1

n3

;

for n ¼ 1; 2; . . . :

Now: 1 for n ¼ 1, 2, . . .; 1  0, n3 n o 2. 11 is a null sequence; 3 nn o

1 3. 11 is decreasing, because n3 is increasing.

1.

n3

Hence, by the Alternating Test, (e) The sequence an ¼ lim

ð1Þnþ1 n2 n2 þ 1 2

n!1 n2

1 P ð1Þnþ1 1

n¼1

n3

is convergent.

; n ¼ 1; 2; . . ., is not null, since

n 1 ¼ lim ¼ 1; þ 1 n!1 1 þ n12

so that the odd subsequence tends to 1. 1 1 P P an are divergent, by the Non-null Test. It follows that Hence, jan j and n¼1

n¼1

1 X ð1Þnþ1 n2 n¼1

n2 þ 1

is neither convergent nor absolutely convergent:

nþ1

Þ n (f) Let an ¼ ð1 n3 þ5 ; n ¼ 1; 2; . . .; then n ; n ¼ 1; 2; . . .; jan j ¼ 3 n þ5 so that n 1 jan j  3 ¼ 2 ; for n ¼ 1; 2; . . .: n n 1 1 P P an It follows, by the Comparison Test, that jan j is convergent; that is, that n¼1 n¼1 is absolutely convergent. 1 P Hence, by the Absolute Convergence Test, an is convergent.

n6 

n¼1

(g) Since 2n is a basic null sequence of positive terms, it follows, from the Reciprocal Rule for sequences, that 2n ! 1 as n ! 1: 6

2n  n Hence n6 is not a null sequence; it follows, from the Non-null Test, that 1 n X 2 is divergent: n6 n¼1 1 n P 2 Thus, n6 is neither convergent nor absolutely convergent. n¼1

Since

1 P n¼1

1 n2

is a basic

convergent series.

Appendix 4

392 nþ1

Þ n (h) Let an ¼ ð1 n2 þ2 , n ¼ 1, 2, . . ., so that n ; for n ¼ 1; 2; . . .: jan j ¼ 2 n þ2

Now we try the Limit Comparison Test, with 1 bn ¼ ; n

for n ¼ 1; 2; . . .:

We obtain n jan j 2 ¼ lim n 1þ2 n!1 bn n!1 n

lim

n2 ¼ 1 6¼ 0: þ2 1 1 1 P P P 1 Thus, an is not jan j is divergent, since n is divergent; it follows that ¼ lim

n!1 n2

n¼1

n¼1

absolutely convergent. n o However,

ð1Þ n n2 þ 2

bn ¼

n¼1

n

nþ1

o

is of the form ð1Þnþ1 bn , where

n ; n2 þ 2

for n ¼ 1; 2; . . .:

Now: n n2 þ 2

 0, for n ¼ 1, 2, . . .; o 2. n2 nþ 2 is a null sequence, since 1  n ¼  n 2  ! 0 as n ! 1; n2 þ 2 1 þ n2 n o n2 o 3. n2 nþ 2 is decreasing, since n nþ2 is increasing, because 



ðn þ 1Þ2 þ2 n2 þ 2 2 2  ¼ nþ1þ  nþ nþ1 n nþ1 n 2 ¼1  0: nðn þ 1Þ 1 P ð1Þnþ1 n Hence, by the Alternating Test, n2 þ 2 is convergent. n¼1 n o 3 1 (i) Let an ¼ nðloge nÞ4 is an 3 , n ¼ 2, . . .. Then an is positive; and, since o n nðloge nÞ4 1 is a decreasing sequence. increasing sequence, fan g ¼ 3 1.

n

nðloge nÞ4

Next, let bn ¼ 2n a2n ; thus

1

bn ¼ 2n  ¼

3

2n ðloge 2n Þ4 1 3

ðn loge 2Þ4 1 1 ¼ 3  3 : ðloge 2Þ4 n4 1 P 1 Since 3 is a basic divergent series, it follows (for example, by the Ratio Test) n¼2 n4 1 P that bn must be divergent. Hence, by the Condensation Test, the original n¼2 1 P 1 series 3 must also be divergent. Hence the series is not convergent, and, n¼2 nðloge nÞ4

since its terms are non-negative, is not absolutely convergent.

Solutions to the problems

393

Section 3.4 1. Substituting x ¼ 2 into the power series for ex, we find that the seventh partial sum of the power series for e2 gives 2 22 23 24 25 26 e2 ’ 1 þ þ þ þ þ þ 1! 2! 3! 4! 5! 6! 4 8 16 32 64 þ ¼1þ2þ þ þ þ 2 6 24 120 720 4 2 4 4 ¼1þ2þ2þ þ þ þ 3 3 15 45 16 ¼ 7 þ ¼ 7:355: 45 Hence this method gives as an estimate for e2 to 3 decimal places the number 7.355.

Remark In fact, e2 ¼ 7.389, to 3 decimal places. 2. From equation (5) we know that 1 1  ; 0 < e  sn < ðn  1Þ! n  1 so that we want n to satisfy the requirement that 1 1 1  < 5  1011 ; or ðn  1Þ!  ðn  1Þ >  1011 ¼ 2  1010 : ðn  1Þ! n  1 5 But 13!  13 ’ 8.095  1010, so n ¼ 14 is sufficient. In other words, 14 terms will suffice.

Chapter 4 Section 4.1 1. (a) Let {xn} be any sequence in R that converges to 2. Then, by the Combination Rules for sequences, we have f ðxn Þ ¼ x3n  2x2n ! 23  2  22 ¼ 8  8 ¼ 0

as n ! 1:

But f(0) ¼ 0, so that f(xn) ! f(0) as n ! 1. Thus f is continuous at 2. (b) Consideration of the graph of f near 1 suggests that f is not continuous at 1. So, let {xn} be any sequence in (0, 1) that tends to 1. Then f(xn) ¼ 0, for all n, so that f ðxn Þ ! 0 as n ! 1: But f(1) ¼ 1, so that f(xn) 6! f(1) as n ! 1. Thus f is not continuous at 1.

Remark If we had chosen {xn} to be any sequence in (1, 2) that tends to 1, we would have found that f(xn) ! f(1) as n ! 1. However this does NOT prove that f is continuous at 1, since the definition of continuity requires that f(xn) ! f(1) for ALL sequences that tend to 1. 2. (a) Let c be any point in R. Let {xn} be any sequence in R that converges to c. Then f(xn) ¼ 1, for all n, so that f ðxn Þ ! 1 as n ! 1:

1 For example, xn ¼ 1  2n .

Appendix 4

394 But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c. Since c is an arbitrary point of R, it follows that f is continuous on R. (b) Let c be any point in R. Let {xn} be any sequence in R that converges to c. Then f(xn) ¼ xn, for all n, so that f ðxn Þ ! c as n ! 1: But f(c) ¼ c, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c. Since c is an arbitrary point of R, it follows that f is continuous on R. 3. The domain of f is the interval I ¼ {x : x  0} if n is even and R if n is odd. Thus, we have to show that for each c in I: 1

1

for each sequence {xk} in I such that xk ! c, then xnk ! cn . First, let c ¼ 0. We have already seen (in the Power Rule for null and the n sequences o 1 subsequent Remark) that, for any null sequence {xk} in I, xnk is also a null sequence. In other words, f(xk) ! f(0) as k ! 1. Hence f is continuous at 0. n Next, olet c 6¼ 0. We have to prove that if {xk  c} is a null sequence, then so is 1

1

xnk  cn .

Now, using the hint, we have 1

xk  c

1

xnk  cn ¼

n1 n

n2 n

n3

1 n

2

xk þ xk c þ xk n cn þ    þ c

n1 n

:

By the result of part (b) of Exercise 3 on Section 2.3, we obtain n1

n2

n3

1

2

xkn þ xkn cn þ xkn cn þ    þ c !c

n1 n

n2

1

n3

2

n1 n

þ c n cn þ c n cn þ    þ c

n1 n

as k ! 1

n1 n

¼ nc ; since c 6¼ 0, we deduce that 1 0 1 xnk  cn ! n1 nc n

as k ! 1

¼ 0: o 1 In other words, xk  cn is null, as required. It follows that f is continuous at c. n

1 n

4. (a) Let d be any point in R. Let {xn} be any sequence in R that converges to d. Then f(xn) ¼ c, for all n, so that f ðxn Þ ! c

as n ! 1:

But f(d) ¼ c, so that f(xn) ! f(d) as n ! 1. Thus f is continuous at d. Since d is an arbitrary point of R, it follows that f is continuous on R. (b) Let c be any point in R. Let {xk} be any sequence in R that converges to c. Then f(xk) ¼ xkn, for all k, so that, by the Product Rule for sequences f ðxk Þ ! cn

as k ! 1:

n

But f(c) ¼ c , so that f (xk) ! f (c) as k ! 1. Thus f is continuous at c. Since c is an arbitrary point of R, it follows that f is continuous on R. (c) Let c be any point in R. Let {xn} be any sequence in R that converges to c. Then, by Theorem 4 of Sub-section 2.3.3, we have f ðxn Þ ¼ jxn j ! jcj

as n ! 1:

But f(c) ¼ jcj, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c. Since c is an arbitrary point of R, it follows that f is continuous on R.

Solutions to the problems

395

5. First, let c ¼ 0. Consideration of the graph of f near 0 suggests that f is not continuous at 0. So, let {xn} be any sequence in (0, 1) that tends to 0. Then f(xn) ¼ 1, for all n, so that f ðxn Þ ! 1

as n ! 1:

But f(0) ¼ 0, so that f(xn) 6! f(0) as n ! 1. Thus f is not continuous at 0. Next, let c > 0. We will show that f is continuous at c. So, let {xn} be any sequence that tends to c. Since c > 0, it follows that eventually xn > 0; thus xn > 0 for all n > N, for a suitable choice of N. Then f(xn) ¼ 1, for all n > N, so that f ðxn Þ ! 1

as n ! 1:

But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c. Finally, let c < 0. We will show that f is continuous at c. So, let {xn} be any sequence that tends to c. Since c < 0, it follows that eventually xn < 0; thus xn < 0 for all n > N, for a suitable choice of N. Then f(xn) ¼ 1, for all n > N, so that f ðxn Þ ! 1 as n ! 1: But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c. It follows that f is continuous on R  {0}, and discontinuous at 0. 6. We have already seen, in Example 3 of Sub-section 4.1.1, that the function 1

g : x 7! x2 ;

for x  0;

is continuous on [0, 1). Also, it follows from Problem 4 that the function h : x 7! x3 ;

for x 2 R;

is continuous on R. We may apply the Composition Rule to the functions g and h, since g([0, 1)) R; it follows that  1 3 3 h g : x 7! x2 ¼ x2 ; for x  0; is continuous on [0, 1). 7. By the Sum and Product Rules, and by using the fact that the constant function and the identity function are continuous on R, we see that g : x 7! x2 þ x þ 1;

Equivalently, all polynomials are continuous on R.

x 2 R;

is continuous on R. Since g(R) (0, 1) and the square root function h is continuous on R, it follows, by the Composition Rule, that the function pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h g : x 7! x2 þ x þ 1; x 2 R; is continuous on R. Next, by the Sum, Multiple and Product Rules, and by using the fact that the constant function and the identity function are continuous on R, we see that the functions k : x 7! 5x

and

l : x 7!1 þ x2 ;

Equivalently, all polynomials are continuous on R.

x 2 R;

are continuous on R. Since l(x) 6¼ 0, it follows from the Product and Quotient Rules, that 5x m : x 7! ; x 2 R; 1 þ x2 is continuous on R. It then follows, by applying the Sum Rule to h g and m, that the function pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5x ; x 2 R; f ð xÞ ¼ x2 þ x þ 1  1 þ x2 is continuous on R.

Equivalently, rational functions are continuous on their domains.

Appendix 4

396 8. (a) The graph of f suggests that we should find functions g and h that squeeze f near 0. y y = x sin

1 x

x

So, we define g(x) ¼  jxj, x 2 R, and h(x) ¼ jxj, x 2 R. With these two chosen, we check the conditions of the Squeeze Rule. Now we know that  1  1  sin  1; for any x 6¼ 0: x It follows that

 1  j xj  x sin  jxj; x

for any x 6¼ 0;

so that gð xÞ½¼ jxj  f ð xÞ  ½j xj ¼ hð xÞ;

for any x 2 R:

So condition 1 of the Squeeze Rule is satisfied. Next, the functions f, g and h all take the value 0 at the point 0. Thus condition 2 of the Squeeze Rule is satisfied. Finally, the functions g and h are both continuous at 0. It then follows, from the Squeeze Rule, that f is continuous at 0. (b) Consideration of the graph of f near 0 suggests that f is not continuous at 0. 1 y = sin 1x

–1

So, let {xn} be the sequence ( ) 1   : fxn g ¼ 2n þ 12 p Then xn ! 0 as n ! 1, and 

1 f ðxn Þ ¼ sin 2n þ p 2  1 ¼ sin p ¼ 1 2 for all n, so that f ðxn Þ ! 1 as n ! 1: But f(0) ¼ 0, so that f(xn) 6! f(0) as n ! 1. Thus f is not continuous at 0.

Solutions to the problems

397

9. Since the constant function and the identity function are continuous on R, it follows from the Combination Rules that the function 2

gðxÞ ¼ x þ 1;

x 2 R;

Equivalently, all polynomials are continuous on R.

is continuous on R. Next, since the sine function is continuous on R, it follows by the Composition Rule, that   sin gð xÞ ¼ sin x2 þ 1 ; x 2 R; is continuous on R. It then follows from the Multiple and Sum Rules, that the function   f ð xÞ ¼ x2 þ 1 þ 3 sin x2 þ 1 ; x 2 R; is continuous on R. 10. The cosine function is continuous on R, so that, by the Multiple Rule, the function p x 7! cos x; x 2 R; 2 is continuous on R. Then, since the sine function, also, is continuous on R, we deduce, from the Composition Rule, that p  f ð xÞ ¼ sin cos x ; x 2 R; 2 is continuous on R. 11. Since the identity function is continuous on R, it follows, from the Combination Rules, that the function gð xÞ ¼ x5  5x2 ;

Equivalently, all polynomials are continuous on R.

x 2 R;

is continuous on R. Then, since the cosine function is continuous on R, we deduce, from the Composition Rule, that   hð xÞ ¼ cos x5  5x2 ; x 2 R; is continuous on R. Next, since the identity function is continuous on R, it follows, from the Combination Rules, that the function 2

kð xÞ ¼ x ;

x 2 R;

is continuous on R. Then, since the exponential function is continuous on R, we deduce, from the Composition Rule, that 2

lðxÞ ¼ ex ;

x 2 R;

is continuous on R. Finally, it follows, from the Combination Rules, that   2 f ðxÞ ¼ cos x5  5x2 þ 7ex ; x 2 R; is continuous on R.

Section 4.2 1. Consider the function f ðxÞ ¼ cos x  x;

x 2 ½0; 1:

We shall show that f has a zero c in (0, 1).

Equivalently, all polynomials are continuous on R.

Appendix 4

398

Certainly, f is continuous on [0, 1], and f ð0Þ ¼ cos 0  0 ¼ 1 > 0; f ð1Þ ¼ cos 1  1< 0: Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that f (c) ¼ 0, and so such that cos c ¼ c: 2. If f (0) ¼ 0 or f (1) ¼ 1, we can take c ¼ 0 or c ¼ 1, respectively. Otherwise, we have f (0) > 0 and f (1) < 1, since the image of [0, 1] under f lies in [0, 1]. We now define the auxiliary function gðxÞ ¼ f ð xÞ  x;

x 2 ½0; 1;

and show that g has a zero c in (0, 1). Certainly, g is continuous on [0, 1], and gð0Þ ¼ f ð0Þ  0 > 0; gð1Þ ¼ f ð1Þ  1< 0: Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that g(c) ¼ 0, and so such that f ðcÞ ¼ c: 3. For the function f (x) ¼ x5 þ x  1, x 2 [0, 1], we have f ð0Þ ¼ 1 < 0; f ð1Þ ¼ 1 þ 1  1 ¼ 1 > 0;  1 1 1 15 f ¼ þ 1¼ : 2 32 2 32

and

1  It follows, 1  by the Intermediate Value Theorem applied to 2 ; 1 , that f has a zero in 2 ; 1 .   3  243 13 Next, f 34 ¼ 1024 þ 34  1 ¼  1024 < 0. Thus, since 3 f 4 < 0 and f(1) > 0it3 fol lows, by the Intermediate Value Theorem applied to 4 ; 1 , that f has a zero in 4 ; 1 . 7 16;807 1 12;711 3 7  Finally, f 8 ¼ 32;768  8 ¼ 32;768 > 0. Thus, since f 4 < 0 and f 8 > 0 it fol    lows, by the Intermediate Value Theorem applied to 34 ; 78 , that f has a zero in 34 ; 78 . 3 7 This interval 4 ; 8 is of length 18, as required. 4. We compile the following table of values for the continuous function p:

1

0

1

2

p(x) 3

1

1

3

x

Since p(1) < 0 and p(0) > 0, it follows, from the Intermediate Value Theorem, that p has a zero in (1, 0).

Solutions to the problems

399

Since p(0) > 0 and p(1) < 0, it follows, from the Intermediate Value Theorem, that p has a zero in (0, 1). Since p(1) < 0 and p(2) > 0, it follows, from the Intermediate Value Theorem, that p has a zero in (1, 2). 5. For the polynomial p(x) ¼ x5 þ 3x4  x  1, x 2 R, that is continuous on R, we have M ¼ 1 þ maxfj3j; j1j; j1jg ¼ 4; so that, by the Zeros Localisation Theorem, all the zeros of p lie in (4, 4). We now compile a table of values of p(x), for x ¼ 4, 3, . . ., 4:

x

4

p(x)

253

3 2 1 2

17

0 1

2

3

4

2 1 2 77 482 1,787

By applying the Intermediate Value Theorem to p on each of the three intervals [4, 3], [1, 0] and [0, 1], we deduce that p has a zero in each of the intervals ð4; 3Þ; ð1; 0Þ

and

ð0; 1Þ:

6. (a) First, we look separately at the two closed subintervals [1, 0] and [0, 2], on each of which f is strictly monotonic. The function f is continuous on each of these subintervals, and so has a maximum and a minimum on each. Since f is strictly decreasing on [1, 0], it follows that maxf f ð xÞ : x 2 ½1; 0g ¼ f ð1Þ ¼ 1; and this is attained in [1, 0] only at  1; also minf f ð xÞ : x 2 ½1; 0g ¼ f ð0Þ ¼ 0; and this is attained in [1, 0] only at 0. Since f is strictly increasing on [0, 2], it follows that maxf f ð xÞ : x 2 ½0; 2g ¼ f ð2Þ ¼ 4; and this is attained in [0, 2] only at 2; also minf f ð xÞ : x 2 ½0; 2g ¼ f ð0Þ ¼ 0; and this is attained in [0, 2] only at 0. Combining these results, we see that maxf f ð xÞ : x 2 ½1; 2g ¼ 4; and this is attained in [ 1, 2] only at 2; also minf f ð xÞ : x 2 ½1; 2g ¼ 0; and this is attained in [1, 2] only at 0.

    (b) First, we look separately at the three closed subintervals 0; 12p , 12p; 32p and 3  2p; 2p on each of which f is strictly monotonic. The function f is continuous on each of these subintervals, and so has a maximum and a minimum on each.   Since f is strictly increasing on 0; 12p , it follows that

    max f ð xÞ : x 2 0; 12p ¼ f 12p ¼ 1;   and this is attained in 0; 12p only at 12 p; also

  min f ðxÞ : x 2 0; 12p ¼ f ð0Þ ¼ 0;   and this is attained in 0; 12p only at  0.  Since f is strictly decreasing on 12p; 32p , it follows that

    max f ð xÞ : x 2 12p; 32p ¼ f 12p ¼ 1;   and this is attained in 12p; 32p only at 12p; also

    min f ðxÞ : x 2 12p; 32p ¼ f 32p ¼ 1;

Appendix 4

400 1



only at 32p.   Next, since f is strictly increasing on 32p; 2p , it follows that

  max f ð xÞ : x 2 32p; 2p ¼ f ð2pÞ ¼ 0;   and this is attained in 32p; 2p only at 0; also

    min f ð xÞ : x 2 32p; 2p ¼ f 32p ¼ 1;   and this is attained in 32p; 2p only at 32p. Combining these results, we see that and this is attained in

3 2p; 2p

maxf f ð xÞ : x 2 ½0; 2pg ¼ 1; and this is attained in [0, 2p] only at 12p; also minf f ð xÞ : x 2 ½0; 2pg ¼ 1; and this is attained in [0, 2p] only at 32p.

Section 4.3 1. (a) If 0  x1 < x2, then 2x1 < 2x2 and x14 < x24. Hence x41 þ 2x1 þ 3 < x42 þ 2x2 þ 3; so that f is strictly increasing on R , and is thus one–one on R . (b) If 0 < x1 < x2, then x12 < x22 and x11 > x12 , so that  x11 <  x12 : Hence 1 1 x21  < x22  ; x1 x2 it follows that f is strictly increasing on (0, 1), and is thus one–one on (0, 1). 2. We perform the three steps of the strategy. 1. We showed, in Problem 1(b) above, that f is strictly increasing on (0, 1). 2. The function 1 x3  1 ¼ ; x 2 R  f0g; x x is a rational function, and therefore is continuous on its domain R  {0}. In particular, f is continuous on (0, 1). x 7! x2 

3. Choose the increasing sequence {n}, which tends to 1, the right-hand end-point of (0, 1). Then 1 f ðnÞ ¼ n2  ! 1 as n ! 1; n by the Reciprocal Rule for sequences. Thus the right-hand end-point of J ¼ f ðð0; 1ÞÞ is 1.

 4. Choose the decreasing sequence 1n , which tends to 0, the left-hand end-point of (0, 1). Then  1 1 f ¼ 2  n ! 1 as n ! 1; n n by the Reciprocal Rule for sequences. Thus the left-hand end-point of J ¼ f ðð0; 1ÞÞ is 1. Hence f has a continuous inverse f 1: R ! (0, 1), by the Inverse Function Rule.       3. (a) Since sin 14p ¼ p1ffiffi2 and 14p lies in 12p; 12p , we have sin1 p1ffiffi2 ¼ 14p. 1  1 2  1 2 Since cos 3p ¼ 2, we have cos 3p ¼ 2, and 3p lies in [0, p], so that   1 2 1 cos 2 ¼ 3p.   pffiffiffi   pffiffiffi Since tan 13p ¼ 3 and 13p lies in 12p; 12p , we have tan1 3 ¼ 13p.

Solutions to the problems

401

(b) Following the Hint, we put y ¼ sin1x. Then   cos 2 sin1 x ¼ cosð2yÞ ¼ 1  2 sin2 y ¼ 1  2x2 ; since x ¼ sin y. 4. Following the Hint, we put a ¼ logex and b ¼ logey. Then x ¼ ea and y ¼ eb, so that   loge ðxyÞ ¼ loge ea eb   ¼ loge eaþb ¼aþb ¼ loge x þ loge y: 5. Let y ¼ cosh1 x, where x  1. Then x ¼ cosh y ¼ 12ðey þ ey Þ; so that e2y  2xey þ 1 ¼ 0: This is a quadratic equation in ey, and so pffiffiffiffiffiffiffiffiffiffiffiffiffi ey ¼ x  x2  1: Both choices of þ or  give a positive expression on the right, but recall that y  0 and so ey  1. Since  pffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffi x þ x2  1  x  x2  1 ¼ 1; pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi we choose the þ sign, because x þ x2  1  1, whereas x  x2  1  1. Hence  pffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ cosh1 x ¼ loge x þ x2  1 :  pffiffiffiffiffiffiffiffiffiffiffiffiffi (The value y ¼ loge x  x2  1 gives the negative solution of the equation cosh y ¼ x.)

Section 4.4 1. (a) For x > 0, f ð xÞ ¼ x ¼ e loge x . Now, the functions x 7! loge x; x 2 ð0; 1Þ;

and

x 7!ex ; x 2 R;

are continuous; hence, by the Multiple Rule and the Composition Rule, f is continuous. (b) For x > 0, f ð xÞ ¼ xx ¼ ex loge x . Now, the functions x 7! loge x;

x 2 ð0; 1Þ;

and x 7! ex ;

x 7! x;

x 2 R;

x 2 R;

are continuous; hence, by the Product Rule and the Composition Rule, f is continuous. 2. Since ax ¼ ex loge a and ay ¼ ey loge a , we have ax ay ¼ ex loge a ey loge a ¼ ex loge aþy loge a ¼ eðxþyÞ loge a ¼ axþy :

Appendix 4

402 x 3. Weuse  the inequality e > 1 þ x, for x > 0. Applying this inequality with x replaced by xe  1, we obtain x  x eðeÞ1 > 1 þ  1 ; for x > e; e x ¼ ¼ xe1 : e It follows that x

ee > x; so that, by Rule 5 with p ¼ e ex > xe :

Chapter 5 Section 5.1 1. (a) The function f ð xÞ ¼

x2 þ x x

has domain R  {0}, so that f is defined on any punctured neighbourhood of 0. Next, notice that f (x) ¼ x þ 1, for x 6¼ 0. It follows that, if {xn} lies in R  {0} and xn ! 0 as n !1, then f(xn) ! 1 as n ! 1. Hence lim f ð xÞ ¼ 1: x!0

(b) First, notice that f ð xÞ ¼

j xj ¼ x



1; 1;

x > 0, x < 0.

The domain of f is R  {0}, so that f is defined on any punctured neighbourhood of 0.



 However, the two null sequences 1n and  1n both have non-zero terms, but   1 1 ¼ 1; whereas lim f  ¼ 1: lim f n!1 n!1 n n Hence, lim f ð xÞ does not exist. x!0 pffiffiffi 2. (a) The function f ð xÞ ¼ x is defined on [0, 1), which contains the point 2; it is also continuous at 2. Hence, by Theorem 2 pffiffiffi pffiffiffi lim x ¼ 2: x!2 pffiffiffiffiffiffiffiffiffi (b) The function f ð xÞ ¼ sin x is defined on [0, p], which contains the point p2; it is also continuous at p2, by the Composition Rule for continuous functions. Hence, by Theorem 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  p ffi pffiffiffiffiffiffiffiffiffi limp sin x ¼ sin ¼ 1: x!2 2 x

(c) The function f ð xÞ ¼ 1 eþ x is defined on ( 1, 1), which contains the point 1; it is also continuous at 1, by the Composition Rule for continuous functions. Hence, by Theorem 2 ex 1 lim ¼ e: x!1 1 þ x 2

Solutions to the problems

403

3. (a) Since lim

x!0

sin x ¼ 1 and x

1 1 ¼ ; þx 2

lim

x!0 2

we deduce from the Product Rule for limits that sin x 1 ¼ : þ x2 2

lim

x!0 2x

(b) Let f(x) ¼ x2 and gð xÞ ¼ sinx x; then lim x2 ¼ 0 and

x!0

lim

x!0

sin x ¼ 1: x

Also, f(x) ¼ x2 6¼ 0 in R  {0}; hence, by the Composition Rule sinðx2 Þ ¼ 1: x!0 x!0 x2 p ffiffi ffi (c) Let f ð xÞ ¼ sinx x and gð xÞ ¼ x; then f is defined on the punctured neighbourhood ( p, 0) [ (0, p) of 0, and, by the Quotient Rule for limits 

sin x 1 lim f ð xÞ ¼ lim ¼ 1: x!0 x!0 x lim gð f ð xÞÞ ¼ lim

Also, g is defined on (0, 1), and is continuous at 1; hence, by Theorem 2 and the Composition Rule  x 12 ¼ gð1Þ ¼ 1: lim gð f ð xÞÞ ¼ lim x!0 x!0 sin x 4. Here



f ð0Þ; f ð1 þ xÞ; 8 > < f ð0Þ; ¼ f ð0Þ; > : f ð1 þ xÞ; 8 0;
0 for x 6¼ 0, and, since f is continuous at 0 limj xj ¼ 0:

x!0

Hence, by the Reciprocal Rule 1 1 ¼ ! 1 as x ! 0: f ð xÞ j xj (b) Let f(x) ¼ x3  1. Then f(x) > 0 for x 2 (1, 1), and, since f is continuous at 1   limþ x3  1 ¼ 0: x!1

Hence, by the Reciprocal Rule 1 1 ¼ !1 f ð xÞ x3  1

as x ! 1þ ;

so that 1 1 ¼ ! 1 as x ! 1þ : f ð xÞ 1  x3  1 1  x3 (c) Let f ð xÞ ¼ sin x. Then f(x) > 0 for x 2  2 p; 2 p , and  2 x3 x ¼ lim sin x lim x!0 sin x x!0 x   lim x2 x!0  ¼ lim sinx x 

x!0

¼

0 ¼ 0; 1

by the Quotient Rule for limits.

Appendix 4

406 Hence, by the Reciprocal Rule 1 sin x ¼ 3 !1 f ð xÞ x

as x ! 0:

2. (a) Let f ð xÞ ¼

1 ; x 2 R  f0g; x2

and

gð xÞ ¼

1 ; x 2 R  f0g: x2

Then f(x) ! 1 as x ! 0, and g(x) ! 1 as x ! 0; but f ð xÞ  gðxÞ ¼ 0 ! 0 as x ! 0: (b) Let f ð xÞ ¼

1 ; x 2 R  f0g; x2

and

gð xÞ ¼ x2 ; x 2 R  f0g:

Then f(x) ! 1 as x ! 1, and g(x) ! 0 as x ! 1; but f ð xÞgð xÞ ¼ 1 ! 1 as x ! 0: 3. (a) By the Sum Rule for limits as x ! 1, we have 

2x3 þ x 1 ¼ lim 2 þ lim x!1 x!1 x3 x2  1 ¼ lim ð2Þ þ lim 2 x!1 x!1 x ¼ 2 þ 0 ¼ 2: (b) Since  1  sin x  1 for x 2 R, we have 

1 sin x 1   ; x x x

for x 2 ð0; 1Þ:

Also 1 gð xÞ ¼  ! 0 x

as x ! 1

and 1 ! 0 as x ! 1: x Hence, by the Squeeze Rule for limits as x ! 1 , hð xÞ ¼

f ð xÞ ¼

sin x ! 0 as x ! 1: x

4. (a) Let f(x) ¼ log ex and g(x) ¼ log ex, for x 2 (0, 1). Then f ð xÞ ! 1

as x ! 1 and

gð xÞ ! 1 as x ! 1:

Hence, by the Composition Rule for limits gð f ð xÞÞ ¼ log e ðlog e xÞ ! 1 as x ! 1: x

(b) Let f ð xÞ ¼ 1x and gð xÞ ¼ ex , for x 2 (0, 1). Then 1

gð f ð xÞÞ ¼ xex ;

for x 2 ð0; 1Þ:

Now f ð xÞ ! 1

as x ! 0 :

In order to use the Composition rule for limits, we need to determine the behaviour of g(x) as x ! 1; that is, the behaviour of g(x) as x ! 1.

Solutions to the problems

407

Now gðxÞ ¼ 

ex 1 ¼ x xe x

and xex ! 1 as x ! 1: It follows, by the Reciprocal Rule, that gðxÞ ! 0 as x ! 1: Hence, by the Composition Rule for limits, 1

gðf ð xÞÞ ¼ xex ! 0 as x ! 0 : 5. Let for all x 2 R;

f ðxÞ ¼ 1; and

 gðxÞ ¼

2; 3;

x ¼ 1, x 6¼ 1.

Then f ðxÞ ! 1 as x ! 1 ðso that ‘ ¼ 1Þ; and gð xÞ ! 3 as x ! 1

ðso that m ¼ 3Þ;

however as x ! 1, we have gðf ð xÞÞ ¼ gð1Þ ¼ 2 ! 2 6¼ 3ð¼ mÞ:

Section 5.3 1. (a)

jan  0j < 0:1 n Þ , ð1 < 0:1 n2 ,

1 n2 2

, n

< 0:1 > 10:

It follows that j an  0j < 0.1 for all n > X, so long as X  (b)

pffiffiffiffiffi 10 ’ 3:16:

ja  0j < 0:01 n n ð1Þ , n2 < 0:01 , n12 , n2 , n

< 0:01 > 100 > 10:

It follows that jan  0j < 0.01 for all n > X, so long as X  10. (c)

ja  0j < " n n ð1Þ , n2 < " , n12 , n2 , n

p1ffiffi" :

It follows that jan  0 j < " for all n > X, so long as X  p1ffiffi".

Appendix 4

408 2. (a)

j f ð xÞ  5j < 0:1 , j2x  2j < 0:1 , jx  1j < 0:05: It follows that j f (x)  5j < 0.1 whenever 0 < jx  1j < , so long as   0.05.

(b)

j f ð xÞ  5j < 0:01 , j2x  2j < 0:01 , jx  1j < 0:005: It follows that j f (x)  5j < 0.01 whenever 0 < jx  1j < , so long as   0.005.

(c)

j f ð xÞ  5j < " , j2x  2j < " 1 , jx  1j < ": 2 It follows that j f (x)  5j < " whenever 0 < jx  1j < , so long as   12".

3. (a) The function f (x) ¼ 5x  2 is defined on every punctured neighbourhood of 3. We must prove that: for each positive number ", there is a positive number  such that jð5x  2Þ  13j < ";

for all x satisfying 0 < jx  3j < ;

that is 1 jx  3j < "; for all x satisfying 0 < jx  3j < : 5 Choose  ¼ 15 " ; then the statement ( ) holds. Hence

( )

limð5x  2Þ ¼ 13:

x!3

(b) The function f (x ) ¼ 1  7x3 is defined on every punctured neighbourhood of 0. We must prove that: for each positive number ", there is a positive number  such that   1  7x3  1 < "; for all x satisfying 0 < jx  0j < ; that is

3 7x < "; for all x satisfying 0 < j xj < : qffiffiffiffiffi Choose  ¼ 3 17 " (so that 73 ¼ "); then the statement ( ) holds. Hence   lim 1  7x3 ¼ 1:

( )

x!0

(c) The function f ð xÞ ¼ x2 cos We must prove that:

1 x3

is defined on every punctured neighbourhood of 0.

for each positive number ", there is a positive number  such that  2 x cos 1  0 < "; for all x satisfying 0 < jx  0j < ; x3 that is

 2 x cos 1 < "; 3 x

for all x satisfying 0 < jxj < :

( )

Solutions to the problems

409

  pffiffiffi Choose  ¼ ". Then, since cos x13  1 for all non-zero x, it follows that, for 0 < jxj <   2 x cos 1  x2 x3 < 2 ¼ "; so that the statement ( ) holds. Hence 

1 lim x2 cos 3 ¼ 0: x!0 x (d) We consider the cases that x < 0 and x > 0 separately; that is, the two one-sided limits. Since jxj ¼  x when x < 0, we have lim

x!0

x þ j xj 0 ¼ lim ¼ 0; x!0 x x

also, since jxj ¼ x when x > 0, we have lim

x!0þ

x þ j xj 2x ¼ limþ ¼ limþ ð2Þ ¼ 2: x!0 x!0 x x

Since the two one-sided limits are not equal, it follows that lim xþxjxj does not x!0 exist. 3 4. The function f(x) ¼ x , x 2 R, is defined on every punctured neighbourhood of 1. We must prove that: for each positive number ", there is a positive number  such that jx3  1j < " for all x satisfying 0 < jx  1j < ; that is 2 x þ x þ 1  jx  1j < "; for all x satisfying 0 < jx  1j < :

( )

Choose  ¼ minf1; 17 "g, so that: (i) for 0 < jx  1j < , we have x 2 (0, 2), and so 2 x þ x þ 1 ¼ x2 þ x þ 1 < 22 þ 2 þ 1 ¼ 7; (ii) jx  1j < 17 ": It follows that 2 x þ x þ 1  jx  1j < 7  1" ¼ "; for all x satisfying 0 < jx  1j < ; 7 that is, the statement ( ) holds. Hence   lim x3 ¼ 1: x!1

5. Since the limits for f and g exist, the functions f and g must be defined on some punctured neighbourhoods (c  r1, c) [ ( c, c þ r1) and (c  r2, c) [ (c, c þ r2) of c, it follows that the function fg is certainly defined on the punctured neighbourhood (c  r, c) [ ( c, c þ r), where r ¼ min{r1, r2}. We want to prove that: for each positive number ", there is a positive number  such that jðf ð xÞgð xÞÞ  ð‘mÞj < "; for all x satisfying 0 < jx  cj < :

( )

Appendix 4

410 In order to examine the first inequality in ( ), we write f ð xÞgð xÞ  ‘m

as ð f ðxÞ  ‘ÞðgðxÞ  mÞ þ mð f ðxÞ  ‘Þ þ ‘ðgð xÞ  mÞ;

so that j f ð xÞgðxÞ  ‘mj ¼ jð f ð xÞ  ‘Þðgð xÞ  mÞ þ mð f ð xÞ  ‘Þ þ ‘ðgð xÞ  mÞj  j f ð xÞ  ‘j  jgð xÞ  mj þ jmjj f ð xÞ  ‘j þ j‘jjgð xÞ  mj:

(1)

We know that, since lim f ð xÞ ¼ ‘, there is a positive number 1 such that x!c

j f ð xÞ  ‘j < ";

for all x satisfying 0 < jx  cj < 1 ;

(2)

and a positive number 2 such that j f ð xÞ  ‘j < 1;

for all x satisfying 0 < jx  cj < 2 ;

(3)

similarly, since lim gð xÞ ¼ m, there is a positive number 3 such that x!c

jgð xÞ  mj < ";

for all x satisfying 0 < jx  cj < 3 :

(4)

We now choose  ¼ min {1, 2, 3}. Then the statements (2), (3) and (4) hold for all x satisfying 0 < jx  cj < , so that, from (1), we deduce j f ð xÞgðxÞ  ‘mj  j f ðxÞ  ‘j  jgð xÞ  mj þ jmjj f ð xÞ  ‘j þ j‘jjgð xÞ  mj  1  " þ jmj" þ j‘j" ¼ ð1 þ jmj þ j‘jÞ"; for all x satisfying 0 < jx  cj < . This is equivalent to the statement (*), in which the number " has been replaced by K", where K ¼ 1 þ jmj þ j‘j. Since K is a constant, this is sufficient, by the K" Lemma. It follows that lim f ð xÞgð xÞ ¼ ‘m, as required. x!c

6. Since lim f ð xÞ ¼ ‘ and ‘ 6¼ 0, we may take " ¼ 12 j‘j in the definition of limit; it x!c

follows that there is a positive number  such that 1 j f ð xÞ  ‘j < j‘j for all x satisfying 0 < jx  cj < ; 2 in other words, for all x satisfying 0 < jx  cj < , we have 1 1  j‘j < f ð xÞ  ‘ < j‘j; 2 2 which we may rewrite in the form 1 1 ‘  j‘j < f ð xÞ < ‘ þ j‘j: 2 2

( )

It follows, from (*), that, if ‘ > 0, then 1 f ð xÞ > ‘  j‘j 2 1 1 ¼ ‘  ‘ ¼ ‘ > 0; 2 2 and, if ‘ < 0, then 1 f ð xÞ < ‘ þ j‘j 2 1 1 ¼ ‘  ‘ ¼ ‘ < 0: 2 2 In either case, we obtain that f(x) has the same sign as ‘ on the punctured neighbourhood {x: 0 < jx  cj < } of c.

Solutions to the problems

411

Section 5.4 1. The function f (x) ¼ x3, x 2 R, is defined on R. We must prove that: for each positive number ", there is a positive number  such that 3 x  1 < "; for all x satisfying jx  1j < ; that is

2 x þ x þ 1  jx  1j < ";

 Choose  ¼ min 1; 17 " , so that

for all x satisfying jx  1j < :

( )

(i) for jx  1j < , we have x 2 (0, 2), and so 2 x þ x þ 1 ¼ x2 þ x þ 1 < 22 þ 2 þ 1 ¼ 7; (ii) jx  1j < 17 ": It follows that 2 x þ x þ 1  jx  1j < 7  1 " ¼ "; 7

for all x satisfying jx  1j < ;

that is, the statement (*) holds. Hence f is continuous at 1. pffiffiffi pffiffiffi 2. The function f ð xÞ ¼ x is defined on [0, 1), and f ð4Þ ¼ 4 ¼ 2. We must prove that: for each positive number ", there is a positive number  such that pffiffiffi x  2 < "; for all x satisfying jx  4j < : Now

pffiffiffi x  2 ¼ pxffiffiffi 4 x þ 2 jx  4j ¼ pffiffiffi xþ2 1  jx  4j; 2

since

ð Þ

pffiffiffi x  0:

So, choose  ¼ ". It follows that, for all x satisfying jx  4j < , we have pffiffiffi x  2  1 jx  4j 2 1 <  2 1 ¼ " 2 < ": That is, the statement ( ) holds. Hence f is continuous at 4. 3. The graph of f suggests that, while f is ‘well behaved’ to the left of 2, we should examine the behaviour of f to the right of 2. If we choose " to be any positive number less than 14, say, then there will always be points x (with x > 2) as close as we please to 2 where f (x) is not within a distance " of f (2) ¼ 1. So, take " ¼ 14. Then, if f were continuous at 2, there would be some positive number  such that

Appendix 4

412 1 j f ð xÞ  1j < ; for all x satisfying jx  2j < : 4 Now let xn ¼ 2 þ 1n, for n ¼ 1, 2, . . .. Then 

1 1 xn ¼ 2 þ 2 ð2; 2 þ Þ; for all n > X; where X ¼ ; n  but

( )



1 1  1 j f ðxn Þ  1j ¼ 2 þ  n 2

1 1 þ 2 n 1 5 ¼ ": 6 4 Thus, with this choice of ", no value of  exists such that requirement (*) holds. This proves that f is discontinuous at 2. ¼

4. Since f is continuous at an interior point c of I and f (c) 6¼ 0, it follows, by Theorem 2 of Sub-section 5.4.1, that there is a neighbourhood N ¼ ( c  r, c þ r) of c on which 1 j f ð xÞj > j f ðcÞj: 2 Since f (c) 6¼ 0, it follows from the inequality (1) that

(1)

1 j f ð x Þj > j f ðc Þj 2 > 0; for all x with jx  cj < r: In particular, f (x) 6¼ 0 on N, so that 1f is defined on N. To prove that 1f is continuous at c, we must prove that, for each positive number ", there is a positive number  such that 1 1  f ðxÞ f ðcÞ < "; for all x satisfying jx  cj < : Now

ð Þ

1 1 f ðcÞ  f ð xÞ  ¼ f ð xÞ f ðcÞ f ð xÞf ðcÞ ¼

j f ð x Þ  f ðc Þj : j f ð x Þj  j f ðc Þj

(2)

Since f is continuous at c, we know that, for the given value of " in (*), there is a positive number 1 such that j f ð xÞ  f ðcÞj < ";

for all x satisfying jx  cj < 1 :

(3)

Now choose  ¼ min{r, 1}, so that both (1) and (3) hold, for all x satisfying jx  cj < . It then follows, from (1), (2) and (3), that 1 1 jf ð xÞ  f ðcÞj  f ð xÞ f ðcÞ ¼ j f ð xÞj  j f ðcÞj " h > 0, the difference quotient for f at 2 is f ð2 þ hÞ  f ð2Þ h ð2 þ hÞ2 þ 4 ¼ h ¼4h ! 4 as h ! 0þ :

QðhÞ ¼

Hence f is differentiable on the right at 2, and fR0 (2) ¼ 4. f is not differentiable at 2, since it is not defined to the left of 2. (b) For 1 < h < 1, the difference quotient for f at 0 is QðhÞ ¼ ¼

f ðhÞ  f ð0Þ ( 2h h 0 h ; h < 0; 

¼

h4 0 h ;

h; h3 ;

h > 0; h < 0; h > 0:

Thus lim QðhÞ ¼ 0 and

h!0

lim QðhÞ ¼ 0;

h!0þ

so that fL0 (0) and fR0 (0) both exist and equal 0. It follows that f is differentiable at 0, and f 0 (0) ¼ 0.

Appendix 4

416 (c) The difference quotient for f at 1 is f ð1 þ hÞ  f ð1Þ QðhÞ ¼ h 8 < ð1þhÞ4 1 ; 1 < h < 0, h ¼ 3 ð 1þh Þ 1 : ; 0 < h < 1, h  4 þ 6h þ 4h2 þ h3 ; 1 < h < 0, ¼ 3 þ 3h þ h2 ; 0 < h < 1. Thus lim QðhÞ ¼ 4 and limþ QðhÞ ¼ 3; h!0

h!0

so that fL0 (1) ¼ 4 and fR0 (1) ¼ 3. It follows that f is not differentiable at 1. (d) The difference quotient for f at 2 is f ð2 þ hÞ  f ð2Þ QðhÞ ¼ h ( ¼ ( ¼

ð2þhÞ3 8 ; h 08 h ;

1 < h < 0, 0 < h < 1,

12 þ 6h þ h2 ;

1 < h < 0,

 8h ;

0 < h < 1.

Since limþ QðhÞ does not exist, it follows that f is not differentiable at 2. h!0 However f is differentiable on the left at 2, and fL0 (2) ¼ 12. 6. For n 2 N

! 

1 1  p ¼ sin 2n þ f  2 2n þ 12 p ¼ 1 ! 1 as n ! 1:

Hence, although

1 ð2nþ12Þp

f 

! 0 as n ! 1 !

1  2n þ 12 p

6! 0 ¼ f ð0Þ:

It follows that f is not continuous at 0; and so, by Corollary 1, f is not differentiable at 0. 7. For any non-zero h, the difference quotient Q(h) at 0 is f ðhÞ  f ð0Þ QðhÞ ¼ h  h2 sin 1h ¼ h 1 ¼ h sin h !0

as h ! 0:

It follows that f is differentiable at 0, and that f 0 (0) ¼ 0. Now, for x 6¼ 0, f 0 ð xÞ ¼ 2x sin 1x  cos 1x. Hence, if n 2 N 

1 1 f0 ¼ sinð2npÞ  cosð2npÞ 2np np ¼01 ¼ 1; as n ! 1: Since f 0 (0) 6¼ 1, it follows that f 0 is not continuous at 0.

Solutions to the problems

417

8. Let c be any point in R. At c, the difference quotient for f is cosðc þ hÞ  cos c h cos c cos h  sin c sin h  cos c ¼ h cos h  1 sin h  sin c  ¼ cos c  h h ! cos c  0  sin c  1; as h ! 0;

QðhÞ ¼

¼  sin c: It follows that f is differentiable at c, and f 0 (c) ¼  sin c. Since c is an arbitrary point of R, it follows that f is differentiable on R. 9. For c > 0, the different quotient for f at c is 3

We assume that h is sufficiently small that jhj < c, so that c þ h > 0; and hence that we are using the correct value for f at c þ h.

3

ðc þ hÞ  c h  1 2 ¼ 3c h þ 3ch2 þ h3 h ¼ 3c2 þ 3ch þ h2 ! 3c2

QðhÞ ¼

as h ! 0;

2

0

so that f is differentiable at c and f (c) ¼ 3c . Next, for c < 0, the different quotient for f at c is 2

We assume that h is sufficiently small that jhj <  c, so that c þ h < 0; and hence that we are using the correct value for f at c þ h.

2

ðc þ hÞ  c h  1 ¼ 2ch þ h2 h ¼ 2c þ h ! 2c

QðhÞ ¼

as h ! 0;

0

so that f is differentiable at c and f (c) ¼ 2c. Finally, the difference quotient for f at 0 is QðhÞ ¼ ¼

f ðhÞ  f ð0Þ ( 2 h h 0 h < 0, h ; 

h3 0 h ;

h; ¼ h2 ; ! 0; as

h > 0, h < 0, h > 0, h ! 0:

It follows that f is differentiable at 0, and that f 0 (0) ¼ 0. Therefore, the function f 0 is given by the following formula ( 2x; x < 0, f 0 ð xÞ ¼

0; 3x2 ;

x ¼ 0, x > 0.

It follows that f 0 is continuous at 0, since lim f 0 ð xÞ ¼ limþ f 0 ð xÞ ¼ f 0 ð0Þ ¼ 0: x!0

Section 6.2 1. (a) f 0 ð xÞ ¼ 7x6  8x3 þ 9x2  5; (b) f 0 ð xÞ ¼

3

x 2 R:

2

ðx  1Þ2x  ðx þ 1Þ3x2

ðx3  1Þ2 x  3x2  2x ¼ ; ðx3  1Þ2 4

x 2 R  f1g:

x!0

Appendix 4

418 (c) f 0 ð xÞ ¼ 2 cos2 x  2 sin2 x ¼ 2 cos 2x; x 2 R: ð3 þ sin x  2 cos xÞex  ex ðcos x þ 2 sin xÞ (d) f 0 ð xÞ ¼ ð3 þ sin x  2 cos xÞ2 ex ð3  sin x  3 cos xÞ ¼ ; x 2 R: ð3 þ sin x  2 cos xÞ2 2.

f 0 ð xÞ ¼ e2x þ 2xe2x ¼ e2x ð1 þ 2xÞ; f 00 ð xÞ ¼ 2e2x ð1 þ 2xÞ þ 2e2x ¼ e2x ð4 þ 4xÞ; f 000 ð xÞ ¼ 2e2x ð4 þ 4xÞ þ 4e2x ¼ e2x ð12 þ 8xÞ:

3. In each case, we use the Quotient Rule and the derivatives of sine and cosine. sin x (a) Here f ð xÞ ¼ cos x, so that

cos x cos x  sin xðsin xÞ cos2 x 1 ¼ cos2 x ¼ sec2 x;

f 0 ð xÞ ¼

on the domain of f. (b) Here f ð xÞ ¼ sin1 x, so that cos x sin2 x ¼ cosec x cot x;

f 0 ð xÞ ¼ 

on the domain of f. (c) Here f ð xÞ ¼ cos1 x, so that sin x cos2 x ¼ sec x tan x;

f 0 ð xÞ ¼

on the domain of f. x (d) Here f ð xÞ ¼ cos sin x , so that

sin xð sin xÞ  cos x cos x sin2 x 1 ¼ 2 sin x ¼ cosec2 x;

f 0 ð xÞ ¼

on the domain of f. 4. (a) Here f ð xÞ ¼ 12 ðex  ex Þ; x 2 R; so that 1 f 0 ð xÞ ¼ ðex þ ex Þ 2 ¼ cosh x; x 2 R: (b) Here f ð xÞ ¼ 12 ðex þ ex Þ; x 2 R; so that 1 f 0 ð xÞ ¼ ðex  ex Þ 2 ¼ sinh x; x 2 R: (c) Here f ð xÞ ¼

sinh x ; cosh x

x 2 R;

Solutions to the problems

419

so that cosh x cosh x  sinh x sinh x cosh2 x 1 ¼ cosh2 x ¼ sech2 x; x 2 R:

f 0 ð xÞ ¼

5. (a) Here   f ð xÞ ¼ sinh x2 ; so that

x 2 R;

  f 0 ð xÞ ¼ 2x cosh x2 ;

x 2 R:

(b) Here f ð xÞ ¼ sinðsinhð2xÞÞ;

x 2 R;

so that f 0 ð xÞ ¼ cosðsinhð2xÞÞ2 coshð2xÞ;

x 2 R:

(c) Here 

cos 2x f ð xÞ ¼ sin ; x2

x 2 ð0; 1Þ;

so that



 2

cos 2x x 2 sin 2x  2x cos 2x f ðxÞ ¼ cos  ; x2 x4 0

x 2 ð0; 1Þ:

6. (a) The function f ð xÞ ¼ cos x;

x 2 ð0; pÞ;

is continuous and strictly decreasing on (0, p), and f ðð0; pÞÞ ¼ ð1; 1Þ: Also, f is differentiable on (0, p), and its derivative f 0 (x) ¼ sin x is non-zero there. So f satisfies the conditions of the Inverse Function Rule. Hence f 1 ¼ cos1 is differentiable on (1, 1); and, if y ¼ f (x), then  1 0 1 1 : ¼ ð yÞ ¼ 0 f f ð xÞ sin x Since sin x > 0 on (0, p), and sin2 x þ cos2x ¼ 1, it follows that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi sin x ¼ 1  cos2 x ¼ 1  y2 : Hence



0 1 f 1 ðyÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi : 1  y2

If we now replace the domain variable y by x, we obtain  1 0 1 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 ð1; 1Þ: f 1  x2 (b) The function f ð xÞ ¼ sinh x;

x 2 R;

is continuous and strictly increasing on R, and f ðR Þ ¼ R:

Appendix 4

420 Also, f is differentiable on R, and its derivative f 0 (x) ¼ cosh x is non-zero there. So f satisfies the conditions of the Inverse Function Rule. Hence f 1 ¼ sinh1 is differentiable on R; and, if y ¼ f (x), then  1 0 1 1 : ¼ ð yÞ ¼ 0 f f ðxÞ cosh x Since cosh x > 0 on R, and cosh2 x ¼ 1 þ sinh2 x, it follows that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi cosh x ¼ 1 þ sinh2 x ¼ 1 þ y2 : Hence 

0 1 f 1 ð yÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi : 1 þ y2

If we now replace the domain variable y by x, we obtain  1 0 1 ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R: f 1 þ x2 7. If x1 < x2, then x15 < x25; it follows that f (x1) < f (x2), so that f is strictly increasing on R. Since f is a polynomial function, it is continuous and differentiable on R. Also f 0 ð xÞ ¼ 5x4 þ 1 6¼ 0 on R: Thus, f satisfies the conditions of the Inverse Function Theorem. Now, f (0) ¼ 1, f (1) ¼ 1 and f (1) ¼ 3. Hence, by the Inverse Function Rule  1 0 1 f ¼ 1; ð1Þ ¼ 0 f ð0Þ  1 0 1 1 f ¼ ; ð1Þ ¼ 0 f ð1Þ 6 and 

0 f 1 ð3Þ ¼

1 1 ¼ : f 0 ð1Þ 6

8. By definition, f ð xÞ ¼ xx ¼ expðx loge xÞ: The functions x 7! x and x 7! loge x are differentiable on (0, 1), and exp is differentiable on R. It follows, by the Product Rule and the Composition Rule, that f is differentiable on (0, 1), and that 

1 0 f ð xÞ ¼ expðx loge xÞ  loge x þ x  x ¼ xx ðloge x þ 1Þ:

Section 6.3 1. Since f is a polynomial function, f is continuous on [1, 2] and differentiable on (1, 2); also f 0 ð xÞ ¼ x3  x2 ¼ x2 ðx  1Þ: Thus f 0 vanishes at 0 and 1. First, we consider the behaviour of f near 0. For x 2 (1, 1), for example 

1 4 f ð xÞ ¼ x3 x  4 3 has the opposite sign to that of x, since x  43 < 0; thus

Solutions to the problems f ðxÞ > 0 for x 2 ð1; 0Þ and f ðxÞ < 0 for x 2 ð0; 1Þ: Since f (0) ¼ 0, it follows that 0 is not a local extremum of f. Next, we consider the behaviour of f near 1. Now



 1 4 1 3 1 1 x  x   f ð xÞ  f ð1Þ ¼ 4 3 4 3  1 3  1 4 ¼ x 1  x 1 4 3   1   1 ¼ ðx  1Þ x3 þ x2 þ x þ 1  ðx  1Þ x2 þ x þ 1 4 3   1 ¼ ðx  1Þ 3x3  x2  x  1 12   1 ¼ ðx  1Þ2 3x2 þ 2x þ 1 : 12 It follows that, for x 2 (0, 2), for example f ðxÞ  f ð1Þ  0; 1 so that f has a local minimum at 1, with value f ð1Þ ¼  12 .

2. Since the sine and cosine functions are continuous and differentiable on R, so also is f. Now f 0 ð xÞ ¼ 2 sin x cos x  sin x ¼ sin xð2 cos x  1Þ;   thus f 0 vanishes in 0; 12p only when cos x ¼ 12; that is, when x ¼ 13 p. Since f (0) ¼ 1, f 12 p ¼ 1 and  

1 1 pffiffiffi 2 1 3 1 5 p ¼ 3 þ ¼ þ ¼ ; f 3 2 2 4 2 4  1  it follows that, on 0; 2 p : the minimum of f is 1, and occurs when x ¼ 0 and x ¼ 12 p; the maximum of f is 54, and occurs when x ¼ 13 p. 3. Since f is a polynomial function, f is continuous on [1, 3] and differentiable on (1, 3). Also, f (1) ¼ 2 and f (3) ¼ 2, so that f (1) ¼ f (3). Thus f satisfies the conditions of Rolle’s Theorem on [1, 3]. Now   f 0 ð xÞ ¼ 4x3  12x2 þ 6x ¼ 2x 2x2  6x þ 3 ; pffiffiffi  so that f 0 (x) ¼ 0 when x ¼ 0 and x ¼ 12 3  3 (the roots of the quadratic equation 0 2x2  6x þ p 3¼ ffiffiffi0). Thus the only value  ofpffiffixffi in (1, 3) such that f (x) ¼ 0 is 1 1 x ¼ 2 3 þ 3 , so we must take c ¼ 2 3 þ 3 ’ 2:37. NO:

f is not defined at 12 p.

(b)

NO:

f is not differentiable at 1, and f (0) 6¼ f (2).

(c)

YES:

4. (a)

All the conditions are satisfied.   (d) NO: f ð0Þ 6¼ f 12 p .

Section 6.4 1. (a) Since f is a polynomial function, f is continuous on [2, 2] and differentiable on (2, 2). Thus f satisfies the conditions of the Mean Value Theorem on [2, 2].

421

Appendix 4

422 Now f 0 ð xÞ ¼ 3x2 þ 2; so that c satisfies the conclusion of the theorem when f ð2Þ  f ð2Þ 12  ð12Þ ¼ 2  ð2Þ 2  ð2Þ ¼ 6; qffiffi that is, when 3c2 ¼ 4 so that c ¼  43 ’ 1:15. Thus there are two possible values of c. 3c2 þ 2 ¼

(b) The function exp is continuous on [0, 3] and differentiable on (0, 3). Thus f satisfies the conditions of the Mean Value Theorem on [0, 3]. Now, f 0 (x) ¼ ex, so that c satisfies the conclusion of the theorem when  f ð3Þ  f ð0Þ 1  3 ¼ e 1 ; 30 3   that is, when c ¼ loge 13 ðe3  1Þ ’ 1:85.   1 1 2. (a) Here f 0 ð xÞ ¼ 4x3  4 ¼ 4 x3  1 , so that f 0 (x) > 0 on (1, 1). Hence, by the Increasing-Decreasing Theorem, f is strictly increasing on [1, 1). ec ¼

(b) Here f 0 ðxÞ ¼ 1  1x, so that f 0 (x) < 0 on (0, 1). Hence, by the IncreasingDecreasing Theorem, f is strictly decreasing on (0, 1]. 3. (a) Let f ð xÞ ¼ sin1 x þ cos1 x;

x 2 ½1; 1:

Then f is continuous on [1, 1] and differentiable on (1, 1). Now 1 1 f 0 ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0; for x 2 ð1; 1Þ: 2 1x 1  x2 It follows, from Corollary 1, that sin1 x þ cos1 x ¼ c;

for x 2 ½1; 1;

for some constant c. Putting x ¼ 0, we obtain 0 þ 12 p ¼ c, so that c ¼ 12 p. Hence 1 sin1 x þ cos1 x ¼ p; x 2 ½1; 1: 2 (b) Let 1 f ð xÞ ¼ tan1 x þ tan1 ; x

x 2 ð0; 1Þ:

Then f is continuous and differentiable on (0, 1). Now 1 1 x2 f 0 ð xÞ ¼   2 ¼ 0; for x 2 ð0; 1Þ: 2 1þx 1þ 1 x

It follows, from Corollary 1, that 1 tan1 x þ tan1 ¼ c; for x 2 ð0; 1Þ; x for some constant c. Putting x ¼ 1, we obtain 14 p þ 14 p ¼ c, so that c ¼ 12 p. Hence 1 1 tan1 x þ tan1 ¼ p; x 2 ð0; 1Þ: x 2

Solutions to the problems

423

4. We have f 0 ð xÞ ¼ 3x2  6x ¼ 3xðx  2Þ;

for x 2 R:

0

Thus f (x) ¼ 0 when x ¼ 0 and x ¼ 2, so that c ¼ 0, 2. Now f 00 ð xÞ ¼ 6x  6;

for x 2 R;

so that f 00 (0) ¼ 6 < 0 and f 00 (2) ¼ 6 > 0. It follows, from the Second Derivative Test, that f has a local maximum that occurs at 0 and a local minimum that occurs at 2. The value of the local maximum is f (0) ¼ 1, and the value of the local minimum is f (2) ¼ 3. 5. (a) Let  1 x 2 0; p : 2

f ð xÞ ¼ x  sin x;  Then f is continuous on 0; Now



1 2p

 and differentiable on 0;

f 0 ð xÞ ¼ 1  cos x > 0;

for x 2



1 2p

.



1 0; p ; 2

and f (0) ¼ 0.   It follows that f (x)  0 for x 2 0; 12 p , so that  1 x  sin x; for x 2 0; p : 2 (b) Let 2 1 2 f ð xÞ ¼ x þ  x3 ; 3 3

x 2 ½0; 1:

Then f is continuous on [0, 1] and differentiable on (0, 1). Now 2 2 1 f 0 ð xÞ ¼  x3 3 3  2 1 1  x3 < 0; for x 2 ð0; 1Þ; ¼ 3 and f ð1Þ ¼ 23 þ 13  1 ¼ 0. It follows that f (x)  0 for x 2 [0, 1], so that 2 1 2 x þ  x3 ; for x 2 ½0; 1: 3 3

Section 6.5 1. Here f ðxÞ ¼ x3 þ x2 sin x

and

gð xÞ ¼ x cos x  sin x on ½0; p:

Since polynomial functions and the sine and cosine functions are continuous and differentiable on R, it follows, by the Combination Rules, that f and g are continuous on [0, p] and differentiable on (0, p). It follows that Cauchy’s Mean Value Theorem applies to the functions f and g on [0, p]. Now, g(0) ¼ 0  1  0 ¼ 0 and g(p) ¼ pcosp  sinp ¼  p, so that g(p) 6¼ g(0). Also f 0 ð xÞ ¼ 3x2 þ 2x sin x þ x2 cos x ¼ x2 ð3 þ cos xÞ þ 2x sin x

Appendix 4

424 and g0 ð xÞ ¼ cos x  x sin x  cos x ¼ x sin x on (0, p). In particular, g0 (x) 6¼ 0 on (0, p). It therefore follows from Cauchy’s Mean Value Theorem that there exists at least one point c in (0, p) for which c2 ð3 þ cos cÞ þ 2c sin c ðp3 Þ  ð0Þ ¼ ; c sin c ðpÞ  ð0Þ so that cð3 þ cos cÞ þ 2 sin c ¼ p2 :  sin c By cross-multiplying, we obtain cð3 þ cos cÞ þ 2 sin c ¼ p2 sin c; so that

  3c ¼ p2  2 sin c  c cos c:

Hence the equation 3x ¼ (p2  2)sinx  xcosx has at least one root in (0, p). 2. (a) Let f ð xÞ ¼ sinh 2x

and

gð xÞ ¼ sin 3x;

for x 2 R:

Then f and g are differentiable on R, and f ð0Þ ¼ gð0Þ ¼ 0; so that f and g satisfy the conditions of l’Hoˆpital’s Rule at 0. Now f 0 ð xÞ ¼ 2 cosh 2x and g0 ðxÞ ¼ 3 cos 3x; so that f 0 ð xÞ 2 cosh 2x : ¼ g0 ð xÞ 3 cos 3x It follows, from l’Hoˆpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals x!0 f 0 ð xÞ 2 cosh 2x ; ¼ lim lim x!0 g0 ð xÞ x!0 3 cos 3x provided that this last limit exists. But, by the Combination Rules for continuous functions, the function x 7!

2 cosh 2x 3 cos 3x

is continuous at 0, so that f 0 ðxÞ f 0 ð0Þ ¼ 0 x!0 g0 ð xÞ g ð0Þ 2 ¼ : 3 lim

It follows, from l’Hoˆpital’s Rule, that the original limit exists, and that its value is 23. (b) Let 1

1

f ð xÞ ¼ ð1 þ xÞ5  ð1  xÞ5 and 2

2

gð xÞ ¼ ð1 þ 2xÞ5  ð1  2xÞ5 ;  1 1 for x 2  2 ; 2 : 

Solutions to the problems

425

  Then f and g are differentiable on  12 ; 12 , and f ð0Þ ¼ gð0Þ ¼ 0; so that f and g satisfy the conditions of l’Hoˆpital’s Rule at 0. Now 4 4 1 1 f 0 ð xÞ ¼ ð1 þ xÞ5 þ ð1  xÞ5 5 5 and 3 3 4 4 g0 ð xÞ ¼ ð1 þ 2xÞ5 þ ð1  2xÞ5 ; 5 5 so that 4

4

1 ð1 þ xÞ5 þ 15 ð1  xÞ5 f 0 ð xÞ ¼ 5 : 0 g ð xÞ 4 ð1 þ 2xÞ35 þ 4 ð1  2xÞ35 5 5

It follows, from l’Hoˆpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals x!0

0

lim

f ð xÞ

x!0 g0 ð xÞ

¼ lim

þ xÞ

45

þ

þ 2xÞ

35

þ

1 5 ð1

x!0 4 ð1 5

1 5 ð1 4 5 ð1

 xÞ

45 3

 2xÞ5

;

provided that this last limit exists. But, by the Combination Rules and the Power Rule for continuous functions, the functions f 0 and g0 are continuous at 0; hence, by the Quotient Rule f 0 ð0Þ g0 ð0Þ 1 þ1 ¼ 54 54 5þ5 1 ¼ : 4 It follows, from l’Hoˆpital’s Rule, that the original limit exists, and equals 14. (c) Let    f ð xÞ ¼ sin x2 þ sin x2 and gð xÞ ¼ 1  cos 4x; for x 2 R: lim

f 0 ð xÞ

x!0 g0 ð xÞ

¼

Then f and g are differentiable on R, and f ð0Þ ¼ gð0Þ ¼ 0; so that f and g satisfy the conditions of l’Hoˆpital’s Rule at 0. Now       f 0 ð xÞ ¼ cos x2 þ sin x2  2x þ 2x cos x2 and g0 ð xÞ ¼ 4 sin 4x; so that f 0 ð xÞ cosðx2 þ sinðx2 ÞÞ  ð2x þ 2x cosðx2 ÞÞ ¼ : 4 sin 4x g0 ð xÞ It follows, from l’Hoˆpital’s Rule, that the limit lim gf ððxxÞÞ exists and equals f 0 ð xÞ lim 0 ; x!0 g ð xÞ

x!0

( )

provided that this last limit ( ) exists. Next, f 0 (0) ¼ 0 and g0 (0) ¼ 0, and f 0 and g0 are differentiable on R. Thus f 0 and 0 g satisfy the conditions of l’Hoˆpital’s Rule at 0. 0 It follows, from l’Hoˆpital’s Rule, that the limit lim gf 0ððxxÞÞ exists and equals x!0

Appendix 4

426 f 00 ðxÞ ; x!0 g00 ð xÞ

( )

lim

provided that this last limit ( ) exists. Now      2 f 00 ð xÞ ¼ sin x2 þ sin x2  2x þ 2x cos x2         þ cos x2 þ sin x2  2 þ 2 cos x2  4x2 sin x2 and g00 ð xÞ ¼ 16 cos 4x: But, by the Combination Rules for continuous functions, the functions f 00 and g00 are continuous at 0, so that f 00 ðxÞ f 00 ð0Þ ¼ 00 lim 00 x!0 g ð xÞ g ð0Þ 4 1 ¼ ¼ : 16 4 It follows, from l’Hoˆpital’s Rule, that the limit ( ) exists, and that its value is 14. It then follows, from a second application of l’Hoˆpital’s Rule, that the original limit exists, and that its value is 14. (d) Let f ð xÞ ¼ sin x  x cos x

and

gð xÞ ¼ x3 ;

for

x 2 R:

Then f and g are differentiable on R, and f ð0Þ ¼ gð0Þ ¼ 0; so that f and g satisfy the conditions of l’Hoˆpital’s Rule at 0. Now f 0 ð xÞ ¼ x sin x

and

g0 ð xÞ ¼ 3x2 ;

so that f 0 ð xÞ x sin x ¼ g0 ð xÞ 3x2 1 sin x : ¼  3 x It follows, from l’Hoˆpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals x!0 

f 0 ð xÞ 1 sin x lim ¼ lim  x!0 g0 ð xÞ x!0 3 x 

1 sin x ; ¼  lim 3 x!0 x provided that  this  last limit exists. 0 But lim sinx x ¼ 1, so that the limit lim gf 0ððxxÞÞ does exist and equals 13. It follows, x!0

x!0

from l’Hoˆpital’s Rule, that the original limit also exists, and that its value is 13.

Chapter 7 Section 7.1 1. In this case, the function f is continuous, except at 12 and 1.       On the three subintervals 0; 13 , 13 ; 34 and 34 ; 1 in P, we have

Solutions to the problems

427  1 2 ¼ ; 3 3  3 3 M2 ¼ f ¼ ; 4 2   M3 ¼ 2 ¼ lim f ð xÞ ;

1 1 0¼ ; 3 3 3 1 5 x2 ¼  ¼ ; 4 3 12 3 1 m3 ¼ f ð1Þ ¼ 1; x3 ¼ 1  ¼ : x!1 4 4 It then follows, from the definitions of L( f, P) and U( f, P), that 3 X mi xi ¼ m1 x1 þ m2 x2 þ m3 x3 Lð f ; PÞ ¼ i¼1 1 5 1 ¼0 þ0 þ1 3 12 4 1 1 ¼0þ0þ ¼ ; 4 4 3 X U ðf ; PÞ ¼ Mi xi ¼ M1 x1 þ M2 x2 þ M3 x3 i¼1 2 1 3 5 1 ¼  þ  þ2 3 3 2 12 4 2 5 1 97 ¼ þ þ ¼ : 9 8 2 72 m1 ¼ f ð0Þ ¼ 0;  1 ¼ 0; m2 ¼ f 2

M1 ¼ f

x1 ¼

2. In this case, the function f is increasing and continuous on [0, 1]. Thus, on each subinterval in [0, 1], the infimum of f is the value of f at the left end-point of the subinterval and the supremum of f is the value of f at the right end-point of the subinterval.   i Hence, on the ith subinterval i1 n ; n in Pn, for 1  i  n, we have 



  2 i1 i1 2 i i ¼ ¼ ; Mi ¼ f ; and mi ¼ f n n n n i i1 1 ¼ : xi ¼  n n n It then follows, from the definitions of L( f, Pn) and U( f, Pn), that

n n  X X i1 2 1 Lð f ; Pn Þ ¼ mi xi ¼  n n i¼1 i¼1 n  1 X 2 i  2i þ 1 ¼ 3 n i¼1   1 nðn þ 1Þð2n þ 1Þ nðn þ 1Þ 2 þn ¼ 3 n 6 2   1 ðn þ 1Þð2n þ 1Þ ¼ 2  ðn þ 1 Þ þ 1 n 6   1  ¼ 2 2n2 þ 3n þ 1  6ðn þ 1Þ þ 6 6n  1 1 1

1 ¼ 2 2n2  3n þ 1 ¼  þ 2 ; 6n 3 2n 6n n n  2 X X i 1 Mi xi ¼  U ð f ; Pn Þ ¼ n n i¼1 i¼1 n X 1 ¼ 3 i2 n i¼1 1 nðn þ 1Þð2n þ 1Þ  n3 6 2n2 þ 3n þ 1 1 1 1 ¼ ¼ þ þ 2: 2 6n 3 2n 6n ¼

y 2

y = f (x) 1

0

1 1 3 2

3 4

x

1

y 1 y = f (x)

1

x

Appendix 4

428 It follows that



1 1 1 1  þ 2 ¼ n!1 3 2n 6n 3

lim Lð f ; Pn Þ ¼ lim

n!1

and

 lim U ð f ; Pn Þ ¼ lim

n!1

n!1

1 1 1 þ þ 3 2n 6n2

1 ¼ : 3

3. By definition of least upper bound, we know that f ð xÞ  sup f , for x 2 J. It follows, x2J from the fact that I J, that f ð xÞ  sup f ;

for x 2 I :

x2J

Hence sup f is an upper bound for f on I, so that sup f  sup f . x2J

x2I

x2J

4. In Problem 2, we showed that for the function f (x) ¼ x2, x 2 [0, 1], and the partition

        i 1 Pn ¼ 0; 1n ; 1n ; 2n ; . . .; i1 of [0, 1] n ; n ; . . .; 1  n ; 1 1 1 and lim U ð f ; Pn Þ ¼ : lim Lð f ; Pn Þ ¼ n!1 n!1 3 3 It follows that

Z 

1

0

Z 1

1 f  3

and

f 

0

1 : 3

But Z 

1

f 

Z 1 f ;

by part ðbÞ of Theorem 4:

0

0

R1 R1 that we must have 0 f ¼ 0 f ¼ 13, so that f is integrable on [0, 1] and RIt1follows  1 0 f ¼ 3. 5. Here the function f (x) = k, x 2 [0, 1], isthe constant function on [0, 1]. i Hence, on the ith subinterval i1 n ; n in Pn, for 1  i  n, we have i i1 1 ¼ : mi ¼ k; Mi ¼ k; and xi ¼  n n n It then follows, from the definitions of L( f, Pn) and U( f, Pn), that n n X X 1 Lð f ; Pn Þ ¼ m i  xi ¼ k n i¼1 i¼1 n X1 ¼ k; ¼k n i¼1 U ð f ; Pn Þ ¼

n X

Mi  xi ¼

i¼1

n X

k

i¼1 n X

¼k

i¼1

1 n

1 ¼ k: n

It follows that Z But



f k

and

0

Z 

1

Z 1 f  k: 0

1

f  0

Z 1 f;

by part ðbÞ of Theorem 4:

0

R1 R1 RIt1follows that we must have 0 f ¼ 0 f ¼ k, so that f is integrable on [0, 1] and 0 f ¼ k:

Solutions to the problems

429

6. (a) Let Pn ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . .; ½xi1 ; xi ; . . .; ½xn1 ; xn g be a standard partition of [0, 1]. The function f is constant on the interval [0, 1), so that, for this partition Pn, we have

2

mi ¼ 2 for all i;  2; 1  i  n  1; Mi ¼ 3; i ¼ n; 1 xi ¼ : n

1 1 –1

It follows that Lð f ; Pn Þ ¼

n X

mi xi ¼

i¼1

–2

n X

1 ð2Þ  n

i¼1

¼ 2

n X 1 i¼1

n

¼ 2;

and U ðf ; Pn Þ ¼

n X

Mi xi ¼

i¼1

n1 X

Mi xi þ Mn xn

i¼1

¼

n1 X i¼1

¼ 2

1 1 ð2Þ þ 3 n n n1 X 1 i¼1

n

þ

3 n

n1 3 5 þ ¼ 2 þ : ¼ 2 n n n Thus, in particular 5 U ð f ; Pn Þ  Lð f ; Pn Þ ¼ : n Now let " be any given positive number. It follows that, if we choose n such that 5 5 n < " (that is, if we choose n > " ), then 

5 < ": U ð f ; Pn Þ  Lð f ; Pn Þ ¼ n It follows, from Riemann’s Criterion for integrability, that f is integrable on [0, 1]. Alternatively, let P ¼ {[x0, x1], [x1, x2], . . ., [xi 1, xi], . . ., [xn 1, xn]} be any partition of [0, 1] . Then, since Mi ¼ mi for 1  i  n 1, we have U ð f ; PÞ  Lð f ; PÞ ¼

n X i¼1

y 3

ðMi  mi Þxi ¼ ðMn  mn Þxn ¼ ð3  ð2ÞÞxn ¼ 5xn : ( )

Now let " be any given positive number, and choose P to be any partition of [0, 1] such its mesh, jjPjj, is less than 5"; then, in particular, we have xn < 5". It then follows, from the inequality ( ), that U ð f ; PÞ  Lð f ; PÞ ¼ 5xn " < 5  ¼ ": 5 It follows, from Riemann’s Criterion for integrability, that f is integrable on [0, 1].

x

Appendix 4

430 (b) Let P ¼ {[x0, x1], [x1, x2], . . ., [xi 1, xi], . . ., [xn 1, xn]} be any partition of [0, 1]. y y = 1, x ∈ 1 y = 0, x ∉ xi–1 xi

0

1

x

Then, for the partition P, we have mi ¼ 0 and

Mi ¼ 1;

for all i. It follows that Lð f ; PÞ ¼

n X

mi xi ¼

i¼1

n X

0  xi ¼ 0;

i¼1

and U ð f ; PÞ ¼

n X

Mi xi ¼

i¼1

¼

n X i¼1 n X

1  xi xi ¼ 1:

i¼1

Thus, for ALL partitions P of [0, 1], we have U ð f ; PÞ  Lð f ; PÞ ¼ 1: In particular, it follows that, for any positive " for which 0 < " < 1, there is NO partition P of [0, 1] for which U ð f ; PÞ  Lð f ; PÞ < ": It follows, from Riemann’s Criterion for integrability, that f is not integrable on [0, 1].

Section 7.2 1. First, we want to use the Strategy to prove that inf fk þ f ð xÞg ¼ k þ inf f f ð xÞg: x2I

x2I

Since f is bounded on I, inf ff ðxÞg exists. In particular x2I

f ð xÞ  inf ff ð xÞg; x2I

for all x 2 I:

It follows that k þ f ð xÞ  k þ inf f f ð xÞg; x2I

for all x 2 I;

so that k þ inf ff ð xÞg is a lower bound for k þ f (x) on I. x2I

Next, let m0 be any number greater than k þ inf f f ð xÞg. It follows that x2I

m0  k > inf f f ð xÞg; x2I

Solutions to the problems

431

so that, in particular, there is some number x in I such that m0  k > f ð xÞ: We may reformulate this fact as: there is some number x in I such that m0 > k þ f ð xÞ; so that m0 is not a lower bound for k þ f (x) on I. It follows that the greatest lower bound for k þ f (x) on I is k þ inf ff ð xÞg; in other x2I words inf fk þ f ðxÞg ¼ k þ inf f f ðxÞg: x2I

x2I

Next, we want to use the Strategy to prove that supfk þ f ð xÞg ¼ k þ supf f ð xÞg: x2I

x2I

Since f is bounded on I, supf f ð xÞg exists. In particular x2I

f ðxÞ  supf f ð xÞg;

for all x 2 I:

x2I

It follows that k þ f ð xÞ  k þ supf f ð xÞg;

for all x 2 I;

x2I

so that k þ supff ð xÞg is an upper bound for k þ f (x) on I. x2I

Next, let M0 be any number less than k þ supf f ðxÞg. It follows that x2I

M 0  k < supf f ð xÞg; x2I

so that, in particular, there is some number x in I such that M 0  k < f ð xÞ: We may reformulate this fact as: there is some number x in I such that M 0 < k þ f ð xÞ; so that M0 is not an upper bound for k þ f (x) on I. It follows that the least upper bound for k þ f (x) on I is k þ supf f ðxÞg; in other x2I words supfk þ f ð xÞg ¼ k þ supf f ð xÞg: x2I

x2I

2. The function f is decreasing on the interval (2, 0], so that inf f ¼ f ð0Þ ¼ 0 and

x2ð2;0

sup f ¼ lim þ f ð xÞ ¼ 4;

x2ð2;0

x!2

f is increasing on the interval [0, 3], so that inf f ¼ f ð0Þ ¼ 0 and

sup f ¼ f ð3Þ ¼ 9:

x2½0;3

x2½0;3

Since I ¼ (2, 0] [ [0, 3], it follows that inf f ¼ f ð0Þ ¼ 0 and I

n o sup f ¼ max f ð3Þ; lim þ f ð xÞ ¼ 9: x!2

I

Since 0  f (x)  9 and 0  f (y)  9, so that 9   f (y)  0, we have  9  f ðxÞ  f ð yÞ  9: We will prove that inf f f ð xÞ  f ð yÞg ¼ 9 and sup f f ð xÞ  f ð yÞg ¼ 9. x;y2I

x;y2I

First, since f (x)  f (y)  9, 9 is a lower bound for f (x)  f (y), for x, y 2 I. To prove that 9 is the greatest lower bound, we now need to prove that: for each positive number ", there are X and Y in I = (2, 3] for which f ðX Þ  f ðY Þ < 9 þ ":

Appendix 4

432 Now, by the definition of infimum and supremum on I, we know that, since 12 " > 0, there exist X and Y in (2, 3] such that f ð X Þ < 12 "

For example, for any number pffiffiffi X in (2, 3] with j X j  12 ", we have

f ðY Þ > 9  12 ":

and

It follows from these two inequalities that     f ð X Þ  f ðY Þ < 12 " þ 9 þ 12 " ¼ 9 þ ":

f ðXÞ ¼ X 2 

we could, pffiffiffi for instance, choose X ¼ 12 ".

It follows that inf f f ð xÞ  f ð yÞg ¼ 9. x;y2I

Finally, since f (x)  f (y)  9, 9 is an upper bound for f (x)  f (y), for x, y 2 I. To prove that 9 is the least upper bound, we now need to prove that: for each positive number ", there are X and Y in I ¼ (2, 3] for which f ð X Þ  f ðY Þ > 9  ": Now, by the definition of infimum and supremum on I, we know that, since 12 " > 0, there exist X and Y in (2, 3] such that f ð X Þ > 9  12 "

f ðY Þ < 12 ":

and

It follows from these two inequalities that     f ð X Þ  f ðY Þ > 9  12 " þ 12 " ¼ 9  ": It follows that sup ff ð xÞ  f ð yÞg ¼ 9. x;y2I

3. (a) For any x in I, f ð xÞ  sup f , so that, since l > 0 I

lf ð xÞ  l sup f ;

for all x 2 I:

I

Thus supðlf ð xÞÞ  l sup f : x2I

I

  To prove that supðlf ð xÞÞ ¼ l sup f , we now need to prove that: x2I

I

for each positive number ", there is some X in I for which   lf ð X Þ > l sup f  ":

( )

I " l

Now, since " > 0 and l > 0, we have > 0. It follows, from the definition of supremum of f on I, that there exists an X in I for which " f ð X Þ > sup f  : l I Multiplying both sides by the positive number l, we obtain the desired result ( ). (b) For any x in I, f ðxÞ  inf f so that I

 f ð xÞ   inf f ;

for all x 2 I:

I

Thus supðf ð xÞÞ   inf f : I

x2I

To prove that supðf ð xÞÞ ¼  inf f , we now need to prove that: I

x2I

for each positive number ", there is some X in I for which  f ð X Þ >  inf f  ": I

1 "; 4

( )

Solutions to the problems

433

Now, since " > 0 it follows, from the definition of infimum of f on I, that there exists an X in I for which f ðX Þ < inf f þ "; I

so that  f ð X Þ >  inf f  ": I

This is precisely the desired result ( ).

Section 7.3

 pffiffiffiffiffiffiffiffiffiffiffiffiffi1  1 1 1. (a) F 0 ð xÞ ¼ x þ x2  4  1 þ x2  4 2 ð2xÞ 2  pffiffiffiffiffiffiffiffiffiffiffiffiffi1  1  1   x2  4 2  x2  4 2 þx ¼ x þ x2  4  1 ¼ x2  4 2 : 

12 1 1 pffiffiffiffiffiffiffiffiffiffiffiffi2ffi 1 1 1 1 4  x þ x  4  x2 2 ð2xÞ þ 2 1  x2 (b) F 0 ð xÞ ¼ 2 2 2 4 2 

1     1 1  4  x2  x2 þ 2 ¼ 4  x2 2 2 2   1  2 2 2 ¼ 4x 4x 1   ¼ 4  x2 2 : 2. (a) From the Fundamental Theorem of Calculus and the Table of Standard Primitives, we obtain  Z 4   2 1 1 9  1  4 1  x þ 9 2 dx ¼ x x2 þ 9 2 þ loge x þ x2 þ 9 2 2 2 0 0 9 9 ¼ 10 þ loge 9  loge 3 2 2 9 ¼ 10 þ loge 3: 2 (b) From the Fundamental Theorem of Calculus and the Table of Standard Primitives, we obtain Z e loge xdx ¼ ½x loge x  xe1 1

¼ ðe  eÞ  ð0  1Þ ¼ 1: 3. Using the Table of Standard Primitives and the Combination Rules, we obtain the following primitives: x ; (a) F ð xÞ ¼ 4ðx loge x  xÞ  tan1 2 2 1 2x (b) F ð xÞ ¼ 3 loge ðsec 3xÞ þ 5e ð2 sin x  cos xÞ: 4. (a) Here we use integration by parts. Let f ð xÞ ¼ loge x

and

1

g0 ð xÞ ¼ x3 ;

so that f 0 ð xÞ ¼

1 x

and

4

gð xÞ ¼ 34x3 :

Appendix 4

434 It follows that Z

Z 3 4 3 4 1 x3 loge x ¼ x3 loge x  x1 x3 dx 4 Z 4 3 4 3 1 3 x3 dx ¼ x loge x  4 4 3 4 9 4 ¼ x3 loge x  x3 : 4 16 (b) Here we use integration by parts, twice. On each occasion we differentiate the power function and integrate the trigonometric function. Hence Z p Z p 2 2  p x2 cos xdx ¼ x2 sin x 02  2x sin xdx 0 Z p 0 2 1 2 ¼ p 2 x sin xdx; 4 0 and Z p Z p 2

0

p

x sin xdx ¼ ½x  ðcos xÞ02  p 2

2

ðcos xÞdx

0

¼ 0 þ ½sin x0 ¼ 1: It follows that Z

p 2

0

1 x2 cos xdx ¼ p2  2: 4

5. (a) We follow the strategy just before Example 3, with x in place of t and u in place of x. Let u ¼ g(x) ¼ 2sin 3x, x 2 R . Then du ¼ 6 cos 3x; dx Hence

Z

so that du ¼ 6 cos 3xdx: Z

1 sin udu 6 1 ¼  cos u 6 1 ¼  cosð2 sin 3xÞ: 6

sinð2 sin 3xÞ cos 3xdx ¼

(b) We follow the strategy just before Example 3, with x in place of t and u in place of x. Let u ¼ gðxÞ ¼ ex , x 2 R. The function g is one–one on R. Then du ¼ ex ; so that du ¼ ex dx; dx also when x ¼ 0; then u ¼ 1; when x ¼ 1; Hence

Z 0

1

ex ð1 þ

ex Þ

then u ¼ e:

dx ¼ 2

Z

e

du

ð1 þ uÞ2  1 e ¼ 1þu 1 1 1 þ ¼ 1þe 2 e1 : ¼ 2ð1 þ eÞ 1

Solutions to the problems

435

6. (a) We follow the strategy just before Example 4. 1 Let t ¼ gðxÞ ¼ ðx  1Þ2 ; x 2 ð1; 1Þ. The function g is one–one on (1, 1). Then t2 ¼ x 1, so that x ¼ t2 þ 1. It follows that dx ¼ 2t; so that dx ¼ 2tdt: dt Hence Z Z dx 2tdt ¼ 3 1 3 3t þ ðt2 þ 1Þt 3ðx  1Þ2 þxðx  1Þ2 Z 2dt ¼ 4t2 þ 1 ¼ tan1 ð2tÞ   1 ¼ tan1 2ðx  1Þ2 : (b) We follow the strategy justffi before Example 4. pffiffiffiffiffiffiffiffiffiffiffiffi Let t ¼ gðxÞ ¼ 1 þ ex , x 2 [0, 1). The function g is one–one on [0, 1). Then t2 ¼ 1 þ ex, so that ex ¼ t2 1 and x ¼ loge (t2 1). It follows that: dx 2t 2t ¼ ; so that dx ¼ 2 dt; dt t2  1 t 1 also: pffiffiffi when x ¼ 0, then t ¼ 2, when x ¼ loge 3, then t ¼ 2. Hence Z loge 3 Z 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi   ex 1 þ ex dx ¼ pffiffi t2  1  t  0

2

¼

Z

¼

2 pffiffi 2

23 t 3

2t2 dt 2 pffiffi 2

pffiffiffi 16  4 2 ¼ : 3

Section 7.4 1. Since

 x sin

1 x10

 x;

for x 2 ½1; 3;

it follows, from part (a) of Theorem 2, that  Z 3 Z 3 1 x sin 10 dx  xdx x 1 1 3 1 ¼ x2 2 1 1 ¼ ð9  1Þ ¼ 4: 2 2. Since 1 2 ex  1  x2  1 4 1  ¼ ; for x 2 0; ; 1 3 2 1 4

2t dt t2  1

Appendix 4

436 it follows, from part (b) of Theorem 2, that 

Z 1 2 4 1 2 0 ex dx  3 2 0 2 ¼ : 3 (Remark We could obtain a smaller upper estimate for the integral by applying 2 1 part (a) of Theorem 2 to the inequality ex  1x 2 .) Next, since

 1 e  1 þ x  1; for x 2 0; ; 2 it follows, from part (b) of Theorem 2, that 

Z 1 2 1 2 0 ex dx  1  2 0 1 ¼ : 2 3. (a) Since  sin 1  1; for x 2 ½1; 4; x x2

and 2 þ cos

2

 1  1; x

it follows that sin1 x    1; 2 þ cos 1x

for x 2 ½1; 4;

for x 2 ½1; 4:

It follows, from Theorem 3, that Z 4 sin1 x    1  ð4  1Þ 1 2 þ cos 1x ¼ 3: (b) Since

and

h pi tan x  0 for x 2 0; ; 4   3  sin x2  2;

h pi for x 2 0; ; 4

it follows that tan x 1  tan x; 3  sinðx2 Þ 2

h pi for x 2 0; : 4

It follows, from Theorems 2 and 3, that Z p Z p 4 4 tan x tan x dx  dx 2 2 0 3  sinðx Þ 0 3  sinðx Þ Z p 4 1 tan xdx  0 2 p4 1 ¼ loge ðsec xÞ 2 0   p 1 ¼ loge sec  loge 1 2 4 pffiffiffi 1 1 ¼ loge 2 ¼ loge 2: 2 4

Solutions to the problems 4. (a)

I0 ¼ ¼

Z

437

1

ex dx

0 ½ex 10

¼ e  1:

(b) Using integration by parts, we obtain Z 1 In ¼ ex xn dx 0 Z 1 ¼ ½ex xn 10  ex nxn1 dx 0

¼ e  nIn1 : (c) Using the result of part (b), with n ¼ 1, 2, 3 and 4 in turn, we obtain I1 ¼ e  I0 ¼ e  ðe  1Þ ¼ 1; I2 ¼ e  2I1 ¼ e  2; I3 ¼ e  3I2 ¼ e  3ðe  2Þ ¼ 6  2e; I4 ¼ e  4I3 ¼ e  4ð6  2eÞ ¼ 9e  24: 5. (a)

2 2 4  ¼ ; 1 3 3 2 2 4 4 64 a2 ¼    ¼ ; 1 3 3 5 45 2 2 4 4 6 6 256 a3 ¼      ¼ ; 1 3 3 5 5 7 175

a1 ¼

ð1!Þ2 22 pffiffiffi ¼ 2; 2! 1 ð2!Þ2 24 4 pffiffiffi 2; b2 ¼ pffiffiffi ¼ 3 4! 2

b1 ¼

b3 ¼ (b)

ð3!Þ2 26 16 pffiffiffi pffiffiffi ¼ 3: 15 6! 3

b21 ¼ 4

32 9 256 2 b3 ¼ 75 (c) We know that b22 ¼

b2n ¼

and and and

ðn!Þ4 24n ðð2nÞ!Þ2 n

3a1 ¼ 4; 5 32 a2 ¼ ; 2 9 7 256 a3 ¼ : 3 75

:

We now express an in a similar way, tackling the numerator and denominator separately. The numerator is a product of 2n even numbers; so, taking a factor 2 from each term, we obtain 2:2:4:4: . . . :ð2nÞ:ð2nÞ ¼ 22n ð1:1:2:2: . . . :n:nÞ ¼ 22n ðn!Þ2 : The denominator of an cannot be tackled in quite the same way, as all its factors are odd. However, we can relate it to factorials by introducing the missing even factors:

Appendix 4

438 1:3:3:5:5: . . . :ð2n  1Þ:ð2n þ 1Þ ¼ ¼

1:2:2:3:3:4:4: . . . :ð2n  1Þ:ð2nÞ:ð2nÞ:ð2n þ 1Þ 2:2: 4:4: . . . :ð2nÞ:ð2nÞ ðð2nÞ!Þ2 ð2n þ 1Þ 22n ðn!Þ2

:

It follows that an ¼

22n ðn!Þ2 ðð2nÞ!Þ2 ð2nþ1Þ 22n ðn!Þ2

24n ðn!Þ4

¼

ðð2nÞ!Þ2 ð2n þ 1Þ n ¼ b2n ; 2n þ 1 so that b2n ¼

2n þ 1 an : n

6. Let 1 ; xp

f ð xÞ ¼

for x 2 ½1; 1Þ;

where p > 0 and p 6¼ 1. Then f is positive and decreasing, and f ð xÞ ! 0 as x ! 1: Also,

Z

n

f ¼

1

Z

n 1

dx xp

¼

n x1p 1p 1

¼

n1p  1 : 1p

( )

Now, if p > 1, then 1  p < 0, so that Z n 1 n1p 1 < f ¼  : p  1 p  1 p  1 1

R n  Since the set 1 f : n 2 N is bounded above, it follows, from the Maclaurin Integral Test, that the series converges. Finally, if 0 < p < 1, then 1  p > 0, so that n1p ! 1

as n ! 1:

It follows, from the Maclaurin Integral Test, that the series diverges. 7. Let t ¼ g(x) ¼ loge x, x 2 (1, 1). The function g is one–one on (1, 1). Then x ¼ et; also dt 1 ¼ dx x Hence

Z

and so dt ¼

dx : x

Z dt ¼ 2 t xðloge xÞ 1 1 ¼ ¼ : t loge x dx

2

Solutions to the problems

439

Now let f ð xÞ ¼

1 xðloge xÞ2

x 2 ½2; 1Þ:

;

Then f is positive and decreasing on [2, 1), and f ðxÞ ! 0 as x ! 1: Also

Z

n

f ¼

Z

n

dx

xðloge xÞ2  1 n ¼  loge x 2 1 1  ¼ loge 2 loge n 1 :  loge 2

R n  Since the set 2 f : n 2 N is bounded above, it follows, from the Maclaurin Integral Test, that the series converges. 2

2

8. Let t ¼ g(x) ¼ loge x, x 2 (1, 1). The function g is one–one on (1, 1). Then x ¼ et; also dt 1 ¼ dx x Hence

Z

and so dt ¼

dx : x

Z dx dt ¼ x loge x t ¼ loge t ¼ loge ðloge xÞ:

Now let f ð xÞ ¼

1 ; x loge x

x 2 ½2; 1Þ:

Then f is positive and decreasing on [2, 1), and f ðxÞ ! 0 as x ! 1: Also

Z

n 2

Z

n

dx x loge x ¼ ½loge ðloge xÞn2 ¼ loge ðloge nÞ  loge ðloge 2Þ ! 1 as n ! 1:

f ¼

2

Hence, by the Maclaurin Integral Test, the series diverges. 1 9. Let f ð xÞ ¼ 1þx 2 , x 2 [0, 1]. Then f is positive and decreasing on [0, 1]; it follows, from the strategy, that  Z 1 n 1X i f f: ! n i¼1 n 0

Now

 n n n X 1X i 1X 1 1 ¼ f  i 2 ¼ n 2 þ i2 n i¼1 n n i¼1 1 þ n i¼1 n 

1 1 1 ; ¼n 2 þ þ    þ n þ 12 n2 þ 22 n2 þ n2

Appendix 4

440 and Z

1

0

Z

1

1 dx 1 þ x2 0  1 1 ¼ tan x 0

f ¼

1 ¼ tan1 1  tan1 0 ¼ p: 4 It follows that



1 1 1 þ þ  þ 2 n2 þ 12 n2 þ 22 n þ n2

n

1 ! p as n ! 1: 4

10. Let f ð xÞ ¼ p1ffiffix, x 2 [1, 1). Then f is positive and decreasing on [1, 1), and f (x) ! 0 as x ! 1. Now n n X X 1 pffi f ðiÞ ¼ i i¼1 i¼1 1 1 1 ¼ 1 þ pffiffiffi þ pffiffiffi þ    þ pffiffiffi ; n 3 2 and

Rn 1

f

R n 1 x 2 dx h1 1 in ¼ 2x2 ¼

1

1

¼ 2n2  2 ! 1

as n ! 1:

It follows, from the strategy, that 1 1 1 1 1 þ pffiffiffi þ pffiffiffi þ    þ pffiffiffi  2n2  2 as n ! 1; n 3 2 and so, using the remark before the problem, that 1 1 1 1 1 þ pffiffiffi þ pffiffiffi þ    þ pffiffiffi  2n2 as n ! 1: n 3 2

Section 7.5 1. 20! ’ 2:4329  1018 ; 30! ’ 2:6525  1032 ; 40! ’ 8:1592  1047 ; 50! ’ 3:0414  1064 ; 60! ’ 8:3210  1081 :

  2. First, f2(n)  f5(n); that is, sin n12  n12 as n ! 1. We know that sin x ! 1 as x ! 0; x

1 since the sequence n2 tends to 0 as n ! 1, it follows that   sin n12 ! 1 as n ! 1: 1 2

n   In other words, sin n12  n12 as n ! 1.   Second, f3(n)  f7(n); that is, 1  cos 1n  2n12 as n ! 1.   We can use the formula cos x ¼ 1  2 sin2 12 x to obtain

Solutions to the problems

441

  1  cos x 2 sin2 12x ¼ x2   x2   sin 12 x 1 sin 12 x ¼  1  1 : 2 2x 2x Then, since 12 x ! 0 as x ! 0, we deduce, from the fact that sinx x ! 1 as x ! 0, that 1  cos x 1 as x ! 0: ! x2 2

1 Then, since the sequence n tends to 0 as n ! 1, it follows that  1  cos 1n 1 as n ! 1; ! 1 2 2 n so that 1  cos

1 n

1 2n2

in other words, 1  cos

! 1 as n ! 1;

1 n

 2n12 as n ! 1. 2

3. Using the Hint, let f (n) ¼ n and g(n) ¼ n. Then  1 1 ð f ðnÞÞn ¼ n2 n  1 2 ¼ nn ! ð1Þ2 ¼ 1

as n ! 1;

and 1

1

ðgðnÞÞn ¼ ðnÞn !1

as n ! 1;

1 n

1

1

so that ðf ðnÞÞ1n ! 1 as n ! 1. In other words, ð f ðnÞÞn  ðgðnÞÞn as n ! 1. ðgðnÞÞ

However f ðnÞ n2 ¼ gðnÞ n ¼ n 6! 1

as n ! 1;

so that f (n) 6 g(n) as n ! 1.

pffiffiffiffiffiffiffiffi n 4. First, using a calculator, we find that the value of the expression 2pn ne when n ¼ 5 is approximately 118.019. Now 5! ¼ 120, so that the error in the Stirling’s Formula approximation is about 120  118.019 ¼ 1.981. The percentage error in this approximation is thus about 1:981  100 ’ 1:651%: 120 This approximation is therefore not within 1% of the exact value. 5. Using Stirling’s Formula, we obtain nn nn  pffiffiffiffiffiffiffiffinn n! 2pn e en ¼ pffiffiffiffiffiffi ; 2pn it follows from Example 1 that  n 1n n e  1 1 n! p2n ð2nÞ2n

as n ! 1:

Appendix 4

442 We have seen previously that, for any positive number a 1

1

an ! 1 and

nn ! 1;

as n ! 1:

It follows that 1

p2n ! 1 and It follows that  n 1n n e n!

1

ð2nÞ2n ! 1;

as n ! 1:

as n ! 1;

that is

6. (a)

(b)

7. Now

so that

 n 1n n ! e as n ! 1: n! ! 300 1 300! ¼ 300 150!  150!  2300 150 2 pffiffiffiffiffiffiffiffiffiffi300300 600p e ’  300 300p 150 2300 e rffiffiffi 1 6 ¼ 30 p ¼ 0:046 ðto two significant figuresÞ: pffiffiffiffiffiffiffiffiffiffi300300 600p e 300! 1  ’ 300 3 ð100!Þ3 3300 ð200pÞ2 100 3300 e pffiffiffi 3 ¼ 200p ¼ 0:0028 ðto two significant figuresÞ:



4n 2n

¼



ð4nÞ! ðð2nÞ!Þ

2

and

2n n



4n 2n ð4nÞ!ðn!Þ2  ¼ : 2n ðð2nÞ!Þ3 n

It follows, from Stirling’s Formula, that  4n pffiffiffiffiffiffiffiffi4n4n  2n 2n 8pn e 2pn ne   3  6n 2n ð4pnÞ2 2n e n 44n ¼ pffiffiffi 2  26n 22n ¼ pffiffiffi : 2 pffiffiffi Hence l ¼ 1 2.

¼

ð2nÞ! ðn!Þ2

;

Solutions to the problems

443

Chapter 8 Section 8.1 1. The tangent approximation to f at a is f ðxÞ ’ f ðaÞ þ f 0 ðaÞðx  aÞ: f ð xÞ ¼ ex ; f 0 ð xÞ ¼ ex ;

(a)

f ð2Þ ¼ e2 ; f 0 ð2Þ ¼ e2 :

Hence the tangent approximation to f at 2 is ex ’ e2 þ e2 ðx  2Þ: f ð xÞ ¼ cos x;

(b)

f ð0Þ ¼ 1;

0

f ð xÞ ¼ sin x;

f 0 ð0Þ ¼ 0:

Hence the tangent approximation to f at 0 is cos x ’ 1 þ 0ðx  0Þ ¼ 1: 2. (a)

f ð xÞ ¼ ex ;

f ð2Þ ¼ e2 ;

f 0 ð xÞ ¼ ex ;

f 0 ð2Þ ¼ e2 ;

f 00 ð xÞ ¼ ex ; f 000 ðxÞ ¼ ex ;

f 00 ð2Þ ¼ e2 ; f 000 ð2Þ ¼ e2 :

Hence f 0 ð2Þ ðx  2Þ 1! ¼ e2 þ e2 ðx  2Þ;

T1 ð xÞ ¼ f ð2Þ þ

f 0 ð2Þ f 00 ð2Þ ðx  2Þ þ ðx  2Þ2 1! 2! 1 ¼ e2 þ e2 ðx  2Þ þ e2 ðx  2Þ2 ; 2 f 0 ð2Þ f 00 ð2Þ f 000 ð2Þ ðx  2Þ þ ðx  2Þ2 þ ðx  2Þ3 T3 ð xÞ ¼ f ð2Þ þ 1! 2! 3! 1 1 ¼ e2 þ e2 ðx  2Þ þ e2 ðx  2Þ2 þ e2 ðx  2Þ3 : 2 6 T2 ð xÞ ¼ f ð2Þ þ

f ð xÞ ¼ cos x;

(b)

0

f ð0Þ ¼ 1;

f ð xÞ ¼ sin x;

f 0 ð0Þ ¼ 0;

f 00 ð xÞ ¼ cos x;

f 00 ð0Þ ¼ 1;

f 000 ðxÞ ¼ sin x;

f 000 ð0Þ ¼ 0:

Hence f 0 ð0Þ x ¼ 1; 1! 0 f ð0Þ f 00 ð0Þ 2 1 xþ x ¼ 1  x2 ; T2 ð xÞ ¼ f ð0Þ þ 1! 2! 2 f 0 ð0Þ f 00 ð0Þ 2 f 000 ð0Þ 3 1 T3 ð xÞ ¼ f ð0Þ þ xþ x þ x ¼ 1  x2 : 1! 2! 3! 2 T1 ð xÞ ¼ f ð0Þ þ

Appendix 4

444 3. The Taylor polynomial of degree 4 for f at a is f 0 ða Þ f 00 ðaÞ f 000 ðaÞ f ð4Þ ðaÞ ðx  aÞ þ ðx  aÞ2 þ ðx  aÞ3 þ ðx  aÞ4 : T4 ð xÞ ¼ f ðaÞ þ 1! 2! 3! 4! (a)

f ð xÞ ¼ 7  6x þ 5x2 þ x3 ;

f ð1Þ ¼ 7;

f 0 ð xÞ ¼ 6 þ 10x þ 3x2 ;

f 0 ð1Þ ¼ 7;

f 00 ð xÞ ¼ 10 þ 6x;

f 00 ð1Þ ¼ 16;

f 000 ð xÞ ¼ 6;

f 000 ð1Þ ¼ 6;

f ð4Þ ð xÞ ¼ 0;

f ð4Þ ð1Þ ¼ 0:

Hence T4 ð xÞ ¼ 7 þ 7ðx  1Þ þ 8ðx  1Þ2 þðx  1Þ3 : (b)

f ð xÞ ¼ ð1  xÞ1 ;

f ð0Þ ¼ 1;

2

f ð xÞ ¼ ð1  xÞ ;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ 2ð1  xÞ3 ;

f 00 ð0Þ ¼ 2;

f 000 ð xÞ ¼ 3!ð1  xÞ4 ;

f 000 ð0Þ ¼ 3!;

f ð4Þ ð xÞ ¼ 4!ð1  xÞ5 ;

f ð4Þ ð0Þ ¼ 4!:

0

Hence T4 ð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 : (c)

f ð xÞ ¼ loge ð1 þ xÞ; 1

f ð0Þ ¼ 0;

f 0 ð xÞ ¼ ð1 þ xÞ ;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ ð1 þ xÞ2 ;

f 00 ð0Þ ¼ 1;

f 000 ð xÞ ¼ 2ð1 þ xÞ3 ;

f 000 ð0Þ ¼ 2;

f ð4Þ ð xÞ ¼ 3!ð1 þ xÞ4 ;

f ð4Þ ð0Þ ¼ 3!:

Hence 1 1 1 T 4 ð xÞ ¼ x  x2 þ x3  x4 : 2 3 4 p 1 ¼ pffiffiffi ; f ð xÞ ¼ sin x; f 4 2 p 1 f0 ¼ pffiffiffi ; f 0 ð xÞ ¼ cos x; 4 2   1 00 00 p ¼  pffiffiffi ; f ð xÞ ¼  sin x; f 4 2 p 1 ¼  pffiffiffi ; f 000 f 000 ð xÞ ¼  cos x; 4 2   1 ð4Þ ð4Þ p f ¼ pffiffiffi : f ð xÞ ¼ sin x; 4 2

(d)

Hence 

 1 p 1  p2 1  p3 1  p 4  x x :  x þ T4 ð xÞ ¼ pffiffiffi 1 þ x  4 2 4 6 4 24 4 2

Solutions to the problems (e)

445

1 1 1 1 f ð xÞ ¼ 1 þ x  x2  x3 þ x4 ; 2 2 6 4 1 1 f 0 ð xÞ ¼  x  x2 þ x3 ; 2 2 00 f ð xÞ ¼ 1  x þ 3x2 ;

1 f 0 ð0Þ ¼ ; 2 00 f ð0Þ ¼ 1;

f 000 ðxÞ ¼ 1 þ 3!x;

f 000 ð0Þ ¼ 1;

f

ð4Þ

f ð0Þ ¼ 1;

f ð4Þ ð0Þ ¼ 3!:

ð xÞ ¼ 3!;

Hence 1 1 1 1 T4 ðxÞ ¼ 1 þ x  x2  x3 þ x4 : 2 2 6 4 f ð xÞ ¼ tan x;

4.

0

f ð0Þ ¼ 0;

f ð xÞ ¼ sec x;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ 2 sec2 x tan x;

f 00 ð0Þ ¼ 0;

f 000 ðxÞ ¼ 4 sec2 x tan2 x þ 2 sec4 x;

f 000 ð0Þ ¼ 2:

2

Hence 1 T3 ðxÞ ¼ x þ x3 ; 3 so that 1 T3 ð0:1Þ ¼ 0:1 þ  0:001 ¼ 0:1003: 3 Since (by calculator) tan 0.1 ¼ 0.10033467. . ., the percentage error involved is about 0:10033467  0:10033333  100 ’ 0:001%: 0:10033467 f ð xÞ ¼ ð1  xÞ1 ;

5. (a)

f ð0Þ ¼ 1;

f 0 ð xÞ ¼ ð1  xÞ2 ;

f 0 ð0Þ ¼ 1; 3

00

f ð xÞ ¼ 2  ð1  xÞ ;

f 00 ð0Þ ¼ 2;

.. . f ðnÞ ð xÞ ¼ ðn!Þ  ð1  xÞn1 ;

f ðnÞ ð0Þ ¼ n!:

Hence 1 00 1 f ð0Þx2 þ    þ f ðnÞ ð0Þxn 2! n! ¼ 1 þ x þ x2 þ    þ xn :

Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ

f ð xÞ ¼ loge ð1 þ xÞ;

(b)

0

1

f ð0Þ ¼ 0;

f ð xÞ ¼ ð1 þ xÞ ;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ ð1Þ  ð1 þ xÞ2 ;

f 00 ð0Þ ¼ 1;

f 000 ðxÞ ¼ ð1Þð2Þ  ð1 þ xÞ3 ;

f 000 ð0Þ ¼ 2;

.. . f ðnÞ ð xÞ ¼ ð1Þnþ1 ðn  1Þ!  ð1 þ xÞn ;

f ðnÞ ð0Þ ¼ ð1Þnþ1 ðn  1Þ!:

Hence 1 00 1 f ð0Þx2 þ    þ f ðnÞ ð0Þxn 2! n! 1 2 1 3 ð1Þnþ1 n x : ¼ x  x þ x   þ n 2 3

Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ

Appendix 4

446 f ð xÞ ¼ ex ;

(c)

f ð0Þ ¼ 1;

and, for each positive integer k f ðk Þ ð xÞ ¼ ex ;

f ðkÞ ð0Þ ¼ 1:

Hence 1 00 1 f ð0Þx2 þ    þ f ðnÞ ð0Þxn 2! n! x2 x3 xn ¼ 1 þ x þ þ þ  þ : 2! 3! n!

Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ

f ð xÞ ¼ sin x;

(d)

f ð0Þ ¼ 0;

0

f ð xÞ ¼ cos x;

f 0 ð0Þ ¼ 1;

f 00 ð xÞ ¼ sin x;

f 00 ð0Þ ¼ 0;

f 000 ð xÞ ¼ cos x;

f 000 ð0Þ ¼ 1;

f ð4Þ ð xÞ ¼ sin x;

f ð4Þ ð0Þ ¼ 0;

and, in general, for each positive integer k 8 < 0; if k is even, ðkÞ 1; if k  1 (mod 4), f ð0Þ ¼ : 1; if k  3 (mod 4). Hence T n ð xÞ ¼ x 

x3 x5 xn þ þ    þ ð0 or 1 or 1Þ ; 3! 5! n!

where the coefficient of xk is f f ð xÞ ¼ cos x;

(e)

0

ðkÞ

ð0Þ k! ,

and the value of f (k)(0) is as specified above.

f ð0Þ ¼ 1;

f ð xÞ ¼ sin x;

f 0 ð0Þ ¼ 0;

f 00 ð xÞ ¼ cos x;

f 00 ð0Þ ¼ 1;

f 000 ð xÞ ¼ sin x;

f 000 ð0Þ ¼ 0;

f

ð4Þ

ð xÞ ¼ cos x;

f ð4Þ ð0Þ ¼ 1;

and, in general, for each positive integer k 8 < 0; if k is odd; f ðkÞ ð0Þ ¼ 1; if k  0 ðmod 4Þ; : 1; if k  2 ðmod 4Þ: Hence T n ð xÞ ¼ 1 

x2 x4 xn þ     þ ð0 or 1 or  1Þ ; 2! 4! n!

where the coefficient of xk is above.

f ðkÞ ð0Þ k! ,

and the value of f (k)(0) is as specified

Section 8.2 1. Here f ð xÞ ¼

1 1 ; f 0 ð xÞ ¼ 1x ð1  x Þ2

Hence R1 ð xÞ ¼

f 00 ðcÞ 2 x2 : x ¼ 2! ð1  cÞ3

and

f 00 ð xÞ ¼

2 ð1  xÞ3

:

Solutions to the problems

447

To calculate the value of c, we use Taylor’s Theorem, with n ¼ 1 f ðxÞ ¼ f ðaÞ þ f 0 ðaÞðx  aÞ þ R1 ð xÞ: This gives f that is

and so

  3 3 3 ¼ f ð0Þ þ f 0 ð0Þ þ R1 ; 4 4 4

32 3 4 4¼1þ þ ; 4 ð1  cÞ3 32 9 4 ¼ : 4 ð1  c Þ3

It follows that ð1  cÞ3 ¼ 14 ; so that 1  c ¼

113 4

’ 0:630. Thus c ’ 0:370:

2. When f is a polynomial of degree n or less f ðnþ1Þ ðcÞ ¼ 0; 3.

f ð xÞ ¼ cos x; 0

so that Rn ð xÞ ¼ 0: f ð0Þ ¼ 1;

f ð xÞ ¼ sin x;

f 0 ð0Þ ¼ 0;

f 00 ð xÞ ¼ cos x;

f 00 ð0Þ ¼ 1;

f 000 ðxÞ ¼ sin x;

f 000 ð0Þ ¼ 0;

f ð4Þ ð xÞ ¼ cos x: Hence, by Taylor’s Theorem with a ¼ 0, f (x) ¼ cos x and n ¼ 3, we have 1 cos x ¼ 1  x2 þ R3 ð xÞ; 2 where f ð4Þ ðcÞ 4 x : R3 ð xÞ ¼ 4! Thus jcos cj 4 x jR3 ðxÞj  24 1  x4 : 24 4. f ð xÞ ¼ loge ð1 þ xÞ; f ð0Þ ¼ 0; 1 ; f 0 ð0Þ ¼ 1; f 0 ð xÞ ¼ 1þx 1 ; f 00 ð0Þ ¼ 1; f 00 ð xÞ ¼ ð1 þ x Þ2 2 : f 000 ðxÞ ¼ ð1 þ x Þ3 Hence, by Taylor’s Theorem with x ¼ 0.02 loge 1:02 ¼ 0 þ ð1  0:02Þ 



1  0:022 þ R2 ð0:02Þ 2

¼ 0:0198 þ R2 ð0:02Þ: Now, for c 2 (0, 0.02)

2 j f ðc Þj ¼  2: ð1 þ cÞ3 000

Appendix 4

448 Hence, by the Remainder Estimate with M ¼ 2, we obtain 2 jR2 ð0:02Þj   0:023 ¼ 0:0000026: 3! It follows that loge 1:02 ¼ 0:0198 ðto four decimal placesÞ: 5. For any positive integer n f ðnÞ ð xÞ ¼  cos x hence

ðnÞ f ðcÞ  1;

or

 sin x;

for all c 2 R;

so that M ¼ 1. It follows that the remainder term in the Remainder Estimate satisfies the inequality 1 ð0:2Þnþ1 : jRn ð0:2Þj  ðn þ 1Þ! Now ð0:2Þ4 ð0:2Þ5 ¼ 0:00006 and ¼ 0:0000026 . . .; 4! 5! so we should try n ¼ 4 in the Remainder Estimate, to be safe. Here f ð xÞ ¼ cos x;

f ð0Þ ¼ 1;

0

f ð xÞ ¼ sin x;

f 0 ð0Þ ¼ 0;

f 00 ð xÞ ¼ cos x;

f 00 ð0Þ ¼ 1;

f 000 ð xÞ ¼ sin x;

f 000 ð0Þ ¼ 0;

f ð4Þ ð xÞ ¼ cos x;

f ð4Þ ð0Þ ¼ 1

f ð5Þ ð xÞ ¼ sin x: Hence, by Taylor’s Theorem with f (x) ¼ cos x, a ¼ 0, x ¼ 0.2 and n ¼ 4, we have 1 1 f ð xÞ ¼ 1  x2 þ x4 þ R4 ð xÞ; 2 24 in other words 1 1 cos 0:2 ¼ 1  ð0:2Þ2 þ ð0:2Þ4 þR4 ð0:2Þ 2 24 ¼ 1  0:02 þ 0:00006 þ R4 ð0:2Þ ¼ 0:9801 ðrounded to four decimal placesÞ: 6. Using the same function f as in Problem 5, we find that f ðpÞ ¼ 1;

f 0 ðpÞ ¼ 0;

f 00 ðpÞ ¼ 1;

f 000 ðpÞ ¼ 0;

f ð4Þ ðpÞ ¼ 1:

It follows that 1 1 T4 ð xÞ ¼ 1 þ ðx  pÞ2  ðx  pÞ4 : 2 24 The difference between T4(x) and f (x) is f ð5Þ ðcÞ ðx  p Þ5 5!  sin c ¼ ðx  pÞ5 ; 120

R4 ð xÞ ¼

Solutions to the problems

449

for some number c between p and x. Hence, for x 2 1 p5 jR4 ðxÞj  120 4 ¼ 0:00249 . . . < 3  103 :

3



5 4 p; 4 p

, we have

Section 8.3 1. (a) Applying the Ratio Test with an ¼ 2n þ 4n, we obtain anþ1 2nþ1 þ 4nþ1 2 þ 2nþ2 a ¼ 2n þ 4n ¼ 1 þ 2n n 1n þ2 ¼ 2 21n 2 þ1 ! 2  2 ¼ 4 as n ! 1: Hence the series has radius of convergence 14 : 2

ðn!Þ (b) Applying the Ratio Test with an ¼ ð2nÞ! , we obtain    1 1 anþ1 ¼ ðn þ 1Þðn þ 1Þ ¼ 1 þ n1 þ n an ð2n þ 1Þð2n þ 2Þ 2 þ 1n 2 þ 2n 1 as n ! 1: ! 4 Hence the series has radius of convergence 4.

(c) Applying the Ratio Test with an ¼ n þ 2n, we obtain anþ1 n þ 1 þ 2n1 1 þ 1n þ n21nþ1 ¼ a ¼ n þ 2n 1 þ n21n n ! 1 as n ! 1: Hence the series has radius of convergence 1. (d) Applying the Ratio Test with an ¼ 1 1n ; we obtain ðn!Þ pffiffiffiffiffiffiffiffi1n n 1 n anþ1 2pn e ð n! Þ ¼ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1   a nþ1 n ððn þ 1Þ!Þ 2pðn þ 1Þ nþ1 nþ1 e  1 12 pffiffiffiffiffiffi1n nn 2p n ¼ pffiffiffiffiffiffi 1  1  1 n þ 1 nþ1 2 2p ððn þ 1Þnþ1 Þ ! 1 as n ! 1: (Here we have used Stirling’s Formula to estimate n! and (n þ 1)!.) Hence the series has radius of convergence 1. 2. (a) Applying the Ratio Test with an ¼ n, we obtain anþ1 n þ 1 a ¼ n ! 1 as n ! 1: n Hence the series has radius of convergence 1. Thus the series converges if x 2 (1, 1), and diverges if jxj > 1. When x ¼ 1, the sequence {nxn} is non-null; hence, by the Non-null Test, the series diverges. Hence the interval of convergence of the series is (1, 1). (b) Applying the Ratio Test with an ¼ n31n , we obtain

Appendix 4

450 anþ1 n3n 1 an ¼ ðn þ 1Þ3nþ1 ¼ 1 þ 13 n 1 ! as n ! 1: 3 Thus the series converges if x 2 (3, 3), and diverges if jxj > 3. 1 P 1 When x ¼ 3, the series is n, which is divergent. n¼1 1 P

When x ¼ 3, the series is

n¼1

ð1Þn n ,

which is convergent.

Hence the interval of convergence of the series is [3, 3). 3. Applying the Ratio Test with an ¼ ð1Þ: ...n!:ðnþ1Þ, we obtain anþ1   n n  1 ¼ ¼ a n þ 1 1 þ 1 n n ! 1 as n ! 1: Hence the series has radius of convergence 1. 4. From the Hint, we know that tan1 x is the sum function of the series 1 X ð1Þn n¼0

2n þ 1

x2nþ1 ¼ x 

x3 x5 x7 þ  þ   ; 3 5 7

for j xj < 1:

( )

From the Hint, we also know that the radius of convergence of this power series is 1. It follows that, by applying Abel’s Limit Theorem to the power series ( ), we may obtain the following 1   X ð1Þn lim tan1 x ¼ ; x!1 2n þ 1 n¼0 provided that this last series is convergent; it is indeed convergent, by the Alternating Test for series. Since the function x 7! tan1 x is continuous at 1, it follows that 1 X   ð1Þn ¼ lim tan1 x x!1 2n þ 1 n¼0 p ¼ tan1 1 ¼ : 4

Section 8.4 1. (a) We know that ex ¼ 1 þ x þ

x2 x3 xn þ þ    þ þ   ; 2! 3! n!

for x 2 R;

and so ex ¼ 1  x þ

x2 x3 xn  þ    þ ð1Þn þ   ; 2! 3! n!

for x 2 R:

Hence, by the Sum and Multiple Rules for power series 1 sinh x ¼ ðex  ex Þ 2 x3 x2nþ1 þ   ; ¼ x þ þ  þ 3! ð2n þ 1Þ! This series converges for all x.

for x 2 R:

For x 7! tan1 x is continuous on its domain R.

Solutions to the problems

451

(b) We know that loge ð1  xÞ ¼ x 

x2 x3 xn          ; 2 3 n

for jxj < 1;

and 1 ¼ 1 þ x þ x2 þ x3 þ    þ xn þ   ; 1x

for j xj < 1:

Hence, by the Sum and Multiple Rules for power series 2 loge ð1  xÞ þ 1  x 

  x2 x3 xn ¼ x            þ 2 þ 2x þ 2x2 þ 2x3 þ    þ 2xn þ    2 3 n 3 2 5 3 2n  1 n x þ   ; for jxj < 1: ¼ 2 þ x þ x þ x þ  þ 2 3 n The radius of convergence of this series is 1. 2. (a) We know that x3 x2nþ1 þ  þ þ   ; 3! ð2n þ 1Þ!

for x 2 R;

x3 ð1Þn x2nþ1 þ  þ þ   ; 3! ð2n þ 1Þ!

for x 2 R:

sinh x ¼ x þ and that sin x ¼ x 

Hence, by the Sum and Multiple Rules for power series sinh x þ sin x 



x3 x2nþ1 x3 ð1Þn x2nþ1 þ  þ x  þ  þ þ  ¼ x þ þ  þ ð2n þ 1Þ! ð2n þ 1Þ! 3! 3! 

x5 x4nþ1 þ    ; for x 2 R: ¼ 2 x þ þ  þ ð4n þ 1Þ! 5! This series converges for all x. (b) We know that loge ð1 þ xÞ ¼ x 

x2 x3 xn þ     þ ð1Þnþ1 þ   ; 2 3 n

for j xj < 1;

and loge ð1  xÞ ¼ x 

x2 x3 xn          ; 2 3 n

for jxj < 1:

Hence, by the Sum and Multiple Rules for power series 

1þx loge 1x ¼ loge ð1 þ xÞ  loge ð1  xÞ 



x2 x3 xn x2 x3 xn ¼ x  þ     þð1Þnþ1 þ     x            2 3 n 2 3 n 

x3 x2nþ1 ¼ 2 x þ þ  þ þ    ; for jxj < 1: 3 2n þ 1 The radius of convergence of this series is 1. (c) We know that 1 ¼ 1 þ x þ x2 þ x3 þ    þ xn þ    ; 1x

for jxj < 1:

Appendix 4

452 It follows, by replacing x by 2x2, that    2  3  n 1 ¼ 1 þ 2x2 þ 2x2 þ 2x2 þ    þ 2x2 þ    1 þ 2x2 ¼ 1  2x2 þ 22 x4  23 x6 þ    þ ð1Þn 2n x2n þ   ; for j2x2j < 1; that is, for j xj < p1ffiffi2. The radius of convergence of this series is p1ffiffi2. 3. (a) We know that loge ð1 þ xÞ ¼ x 

x2 x3 xn þ     þ ð1Þnþ1 þ    ; 2 3 n

for jxj < 1:

Hence, by the Product Rule ð1 þ xÞ loge ð1 þ xÞ 

x2 x3 xn ¼ x  þ     þð1Þnþ1 þ    2 3 n 

3 4 x x xn þ x2  þ     þ ð1Þn þ  2 3 n1 n x2 x3 x þ    ; for jxj < 1: ¼ x þ      þ ð1Þn 2 6 nðn  1Þ The radius of convergence of this series is 1. (b) We know that 1 ¼ 1 þ x þ x2 þ x3 þ    þ xn þ    ; 1x

for j xj < 1;

and (from Example 1) that 1þx ¼ 1 þ 3x þ 5x2 þ    þ ð2n þ 1Þxn þ    ; ð1  x Þ2

for j xj < 1:

Hence, by the Product Rule, we obtain 1þx ð1  x Þ3 1 1þx  ¼ 1  x ð1  x Þ2   ¼ 1 þ x þ x2 þ x3 þ    þ xn þ       1 þ 3x þ 5x2 þ    þ ð2n þ 1Þxn þ    ¼ 1 þ ð3 þ 1Þx þ ð5 þ 3 þ 1Þx2 þ ð7 þ 5 þ 3 þ 1Þx3 þ    þ ðð2n þ 1Þ þ ð2n  1Þ þ    þ 1Þxn þ    ¼ 1 þ 4x þ 9x2 þ 16x3 þ    þ ðn þ 1Þ2 xn þ    ;

for j xj < 1:

(We can prove, by Mathematical Induction, that n X 1 þ 3 þ 5 þ    þ ð2n þ 1Þ ¼ ð2k þ 1Þ ¼ ðn þ 1Þ2 ;

for n 2 N:Þ

k¼0

The radius of convergence of this series is 1. 4. (a) We know that ð1  xÞ1 ¼ 1 þ x þ x2 þ x3 þ    þ xn þ    ;

for j xj < 1:

Hence, by the Differentiation Rule, we obtain ð1  xÞ2 ¼ 1 þ 2x þ 3x2 þ 4x3 þ    þ nxn1 þ    ; The radius of convergence of this series is 1.

for j xj < 1:

Solutions to the problems

453

(b) If we differentiate the series in part (a), by the Differentiation Rule for power series, we obtain 2ð1  xÞ3 ¼ 2 þ 6x þ 12x2 þ    þ nðn  1Þxn2 þ   ;

for jxj < 1:

Hence, by the Multiple Rule, we have nðn  1Þ n2 x þ   ; 2 The radius of convergence of this series is 1. (c) Notice that, for the function f (x) ¼ tanh1 x, j x j < 1 , we have ð1  xÞ3 ¼ 1 þ 3x þ 6x2 þ    þ

for j xj < 1:

1 ¼ 1 þ x2 þ x4 þ    þ x2n þ   : 1  x2 Hence, by the Integration Rule for power series, the Taylor series for f at 0 is f 0 ð xÞ ¼

x3 x5 x2nþ1 þ þ  þ þ   ; 3 5 2n þ 1 since f (0) ¼ 0, it follows that c ¼ 0. Hence tanh1 x ¼ c þ x þ

tanh1 x ¼ x þ

x3 x5 x2nþ1 þ þ  þ þ   ; 3 5 2n þ 1

for j xj < 1;

for j xj < 1:

The radius of convergence of this series is 1. 5. We know that ex ¼ 1 þ x þ

x2 x3 xn þ þ    þ þ   ; 2! 3! n!

for x 2 R;

and we know, from part (a) of Problem 4, that ð1  xÞ2 ¼ 1 þ 2x þ 3x2 þ 4x3 þ    þ ðn þ 1Þxn þ   ;

for j xj < 1:

Hence, by the Product Rule for power series, we have 

  x2 x3 2 x e ð1  xÞ ¼ 1 þ x þ þ þ     1 þ 2x þ 3x2 þ 4x3 þ    2 6 

1 ¼ 1 þ ð2 þ 1Þx þ 3 þ 2 þ x2 þ    2 11 2 ¼ 1 þ 3x þ x þ    ; for jxj < 1: 2 The radius of convergence of this series is 1. 6. We are given that f ð xÞ ¼ x þ

x3 x5 x2nþ1 þ þ  þ þ   ; 1:3 1:3:5 1:3: ::: :ð2n þ 1Þ

for x 2 R:

(a) By the Differentiation Rule, we can differentiate the power series term-by-term; this gives f 0 ð xÞ ¼ 1 þ

x2 x4 x2n þ þ  þ þ   ; 1 1:3 1:3: ::: :ð2n  1Þ

for x 2 R:

(b) It follows, from the definition of f, that xf ð xÞ ¼ x2 þ

x4 x6 x2nþ2 þ þ  þ þ   ; 1:3 1:3:5 1:3: ::: :ð2n þ 1Þ

Hence, by the Combination Rules for power series, f 0 ð xÞ  xf ð xÞ ¼ 1; since all the other terms cancel out.

for x 2 R:

Appendix 4

454 7. Since ex ¼ 1 þ x þ

x2 x3 xn þ þ    þ þ   ; 2! 3! n!

for x 2 R;

we may deduce that 2

ex ¼ 1  x2 þ

x4 x6 x2n  þ    þ ð1Þn þ   ; 2! 3! n!

for x 2 R:

It follows, from the Integration Rule for power series, that 1 Z 1 x3 x5 x7 x2nþ1 2  þ    þ ð1Þn þ  ex dx ¼ x  þ 3 5  2! 7  3! ð2n þ 1Þ  n! 0 0 1 1 1 1 n þ   : ¼ 1  þ  þ    þ ð1Þ 3 10 42 ð2n þ 1Þ  n! 8. By the General Binomial Theorem 1 1 X 1 4 xn ; ð1 þ 6xÞ4 ¼ n n¼0 where 1 4

n

¼

for j6xj < 1;

1 3 7 1  4 4 4 ... 4  n þ 1 : n!

It follows that

 1 3 7  34 4 4 ð6xÞ2 þ 4 ð6xÞ3 þ    1 2 6 3 27 189 3 1 x    ; for jxj < : ¼ 1 þ x  x2 þ 2 8 16 6 The radius of convergence of this series is 16. 1

ð1 þ 6xÞ4 ¼ 1 þ

1 4

1

ð6xÞ þ

4

9. (a) By the General Binomial Theorem

1  X 1  12 ðxÞn ; ð1  xÞ2 ¼ n n¼0 where 

for j xj < 1;



 1 3 5  1 2 2 2 ... 2  n þ 1  12 : ¼ n n!

It follows that 1 1 3 ð1  xÞ2 ¼ 1 þ x þ x2 þ    þ ð1Þn 2 8



 12 n x þ   ; for j xj < 1: n

The radius of convergence of this series is 1. (b) We know that d 1 sin1 x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; dx 1  x2

for x 2 ð1; 1Þ:

Then, by the result of part (a), with x replaced by x2, we obtain  1 1 1 3  2 2n pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1 þ x2 þ x4 þ    þ ð1Þn x þ   ; for j xj < 1: 2 n 2 8 1x Hence, by the Integration Rule for power series 1 3 sin1 x ¼ c þ x þ x3 þ x5 þ    6 40  1 2nþ1 2 x þ   ; for j xj < 1: þ ð1Þn n 2n þ 1

Solutions to the problems

455

Putting x ¼ 0 into this equation, we see that c ¼ 0. It follows that 1 3 sin1 x ¼ x þ x3 þ x5 þ    6 40  1 2nþ1 2 x þ   ; þ ð1Þn n 2n þ 1

for j xj < 1:

The radius of convergence of this series is 1.

Section 8.5 1. We use the Addition Formula tan1 x þ tan1 y ¼ tan1



xþy ; 1  xy

( )

  which holds provided that tan1 x þ tan1 y lies in the interval  p2 ; p2 . First, we deduce, from the Addition Formula ( ), that !   1 1 1 1 1 1 1 3þ4 þ tan ¼ tan tan 3 4 1  13  14 

4þ3 ¼ tan1 12  1  7 : ¼ tan1 11 This equation holds, since   1 1 þ tan1 ’ 0:3218 þ 0:2450 ¼ 0:5668; tan1 3 4   and 0:5668 2  p2 ; p2 ’ ð1:5708; 1:5708Þ: Next, we deduce, from the Addition Formula ( ), that      1 1 2 7 2 þ tan1 þ tan1 ¼ tan1 þ tan1 tan1 3 4 9 11 9 ! 7 2 11 þ 9 ¼ tan1 7 1  11  29 

63 þ 22 ¼ tan1 99  14 p 1 ¼ tan ð1Þ ¼ : 4 This equation holds, since   7 2 þ tan1 ’ 0:5667 þ 0:2187 ¼ 0:7854; tan1 11 9   and 0:7854 2  p2 ; p2 ’ ð1:5708; 1:5708Þ: 2. First



1 px dx 2 1 

1 2 1 4 ¼ : sin px ¼ p 2 p 1

I0 ¼

Z

1

cos

456 It follows that p I0 ¼ 4. Next, using integration by parts twice, we obtain 

Z 1   1 2 1  x cos px dx I1 ¼ 2 1 

1 Z 1 

  2 1 2 1 ¼ 1  x2 sin px  ð2xÞ sin px dx p 2 p 2 1 1 

Z 4 1 1 ¼ x sin px dx p 1 2  

1 

Z  4 2 1 4 1 2 1 x  cos px cos px dx   ¼ p p 2 p 2 1 p 1 

1 8 2 1 ¼ 2 sin px p p 2 1 32 ¼ 3: p It follows that p 3I1 ¼ 32.

Appendix 4

Index

ax, definition and continuity, 161 Abel’s limit theorem, 333 absolute convergence test, 104 absolute convergence theorem, 332, 337 absolute value, 12 absolutely convergent series, 104 addition formula for tan1, 166 alternating test, 107 antipodal points theorem, 145 approximation by Taylor polynomials, 317 Archimedean property of R, 7 arithmetic in R, 8, 30 arithmetic mean – geometric mean inequality, 18, 21 asymptotic behaviour of functions, 176 basic continuous functions, 142 differentiable functions, 224 null sequences, 48 power series, 325 series, 97 Bernoulli’s inequality, 20, 237 Bessel function, 329 Bijection, 355 binomial theorem, 358, 342 blancmange function, 244 Bolzano–Weierstrass theorem, 70 bound, greatest lower, 26 bound, least upper, 25 boundedness theorem, 62, 92, 149 Cauchy condensation test, 97 Cauchy’s mean value theorem, 239 Cauchy–Schwarz inequality, 20, 311 Chain Rule, 218 combination rules continuous functions, 136 convergent sequences, 55 convergent series, 89 differentiable functions, 216 functions which tend to 1, 177 inequalities, 14 infimum, 275 integrable functions, 278 limits of functions, 172 null sequences, 46 power series, 338

primitives, 284 sequences which tend to 1, 65 series, 89 supremum, 275 tilda, 305 common limit criterion, 268 common refinement, 262 comparison test, 94 composition rule asymptotic behaviour of functions, 180 continuous functions, 136 differentiable functions, 218 limits of functions, 173 conditionally convergent series, 104 continuity continuity, 132, 194 limits of functions, 171 one-sided, 134, 194 uniform, 201 continuous functions basic, 142 boundedness theorem, 149 combination rules, 136 composition rule, 136 extreme values theorem, 169 integrability, 277 intermediate value theorem, 143 inverse function rule, 153 squeeze rule, 137 convergence, interval of, 330 radius of, 330 convergent sequence, 53 combination rules, 55 quotient rule, 55 squeeze rule, 58 convergent series, 85 combination rules, 89 cos1, 156 cosh, power series, 339 cosh1, 158 cosine function derived function, 214 power series, 326 Darboux’s theorem, 254 decimal representation of numbers, 3, 87 density property of R, 7 derivatives, 207 higher order, 215

one-sided, 210 standard, 359 difference quotient, 206 differentiability of f(x) ¼ e x, 214 f(x) ¼ a x, 223 f(x) ¼ x, 223 f(x) ¼ x x, 223 trigonometrical functions, 214 differentiable functions combination rules, 216 composition rules, 218 inequalities involving, 237 inverse function rule, 221 differentiation, 210 rule for power series, 341 Dirichlet’s function, 197, 310 discontinuity, removable, 172 divergent sequence, 61 series, 85 domination hierarchy, 56 e, definition, 74 irrationality, 175 "   game, 43, 187 Euclidean algorithm, 78 even subsequence, 66 ex, definition, 75, 122 fundamental property, 127 inverse property, 75 exponent laws, 163 exponential function continuity, 141 derived function, 214 differentiability, 223 inequalities, 141 power series, 326 extreme values theorem, 149 extremum, 228 field, 9 first subsequence rule, 67 function, nowhere differentiable, 244 fundamental inequality for integrals, 288 fundamental theorem of Algebra, 146 of Calculus, 283, 292 general binomial theorem, 342 geometric series, 87

457

Index

458 Goethe, ix greatest lower bound, 26, 272 property of R, 29 Gregory’s series for estimating p, 346 harmonic series, 93, 109 higher-order derivatives, 215 hyperbolic functions, differentiability, 224 hypergeometric series, 352 increasing–decreasing theorem, 234 inequalities, power rule, 10 rules for integrals, 289 infimum, 26, 272 infinite series, 85 inheritance property of subsequences, 66 Int I, 234 Integrability, 264 combination rules, 278 continuous functions, 277 integral test, 297 modulus rule, 278 monotonic functions, 276 Riemann’s criterion, 267 integral, 264 additivity, 280 inequality rules, 289 lower, 264 upper, 264 integration by parts, 285 by substitution, 286 integration rule for power series, 341 integration, reduction of order method, 293 intermediate value theorem, 143 interval image theorem, 149 interval of convergence, 230 inverse function rule for continuous functions, 153 differentiable functions, 221 inverse hyperbolic functions, 158 inverse trigonometric functions, 156 irrational numbers, 5 K" lemma, 49, 193 least upper bound, 25, 262 property of R, 29 Leibniz notation, 207 Leibniz test, 107 Leibniz’s series for estimating p, 346 l‘Hoˆpital’s rule, 241 limit comparison test, 95 limit inequality rule for limits, 175 sequences, 60 limit of a function, 169, 177, 185 as x ! 1, 177 one-sided, 175 limit of a sequence, 53, 182 limits of functions and continuity, 171

limits of functions, combination rules, 172 composition rule, 173 one-sided, 175 squeeze rule, 174 local extremum theorem, 229 loge ð1 þ xÞ, power series for, 325 lower integral, 264 lower Riemann sum, 259 Maclaurin integral test, 297 mathematical induction, 357 maximum, 23 mean value theorem, 233 minimum, 24 modulus function, continuity, 135 modulus rule for integrable functions, 278 modulus, 12 monotone convergence theorem, 68 monotonic function, 153 integrability, 276 monotonic sequence theorem, 69 multiplication of series, 114 n!, Stirling’s formula for, 306 neighbourhood, 169 non-null test, 91 nth partial sum of series, 85 nth root function, continuity, 155 nth root of a positive real number, 32, 155 null partition criterion, 269 null sequence, 43 odd subsequence, 66 one-one function, 152, 355 one-sided derivative, 210 one-sided limit, 175 onto function, 355 order properties of R, 7 p, 76 irrationality of, 348 numerical estimates, 346 partial sum of series, 85 partition of [a, b], 258 standard, 258 power series absolute convergence, 332 basic, 325 combination rules, 338 differentiation rule, 341 integration rule, 341 uniqueness theorem, 342 primitives, 282 combination rules, 278 scaling rule, 278 standard, 360 uniqueness theorem, 284

radius of convergence theorem, 329 ratio test, 96 for radius of convergence, 330 rational function, continuity, 136 differentiability, 217 rational power, 34 rearrangement of series, 109 reciprocal rule for functions which tend to 1, 177 sequences which tend to 1, 64 tilda, 305 recursion formula, 71 reduction of order method, 293 refinement, 262 remainder estimate, 322 removable discontinuity, 172 Riemann’s rearrangement theorem, 112 Riemann’s function, 198 Riemann’s-criterion for integrability, 267 Rolle’s theorem, 231 scaling rule for primitives, 284 second derivative test, 236 second subsequence rule, 67 sequences, 38 basic null, 48 combination rules, 55 convergent, 53 divergent, 61 limit inequality rule, 60 monotonic, 40 null, 43 unbounded, 62 which tend to 1, 63 which tend to 1, 65 squeeze rule, 58 series, 85 absolutely convergent, 104 basic, 97 convergent, 85 divergent, 85 geometric, 87 harmonic, 93, 109 hypergeometric, 352 integral test, 297 multiplication, 114 product rule, 114 Taylor, 325 telescoping, 88 sine function derived function, 214 power series, 326 sine inequality, 140 sin1, 156 sinh, 158 power series, 339 square root function, continuity, 134 squeeze rule as x ! 1, 178 continuous functions, 137

Index convergent sequences, 58 limits of functions, 174 null sequences, 47 sequences which tend to 1, 179 standard derivatives, 359 standard partition, 258 standard primitives, 360 Stirling’s formula, 306 strategy for testing for convergence, 116 sub-interval theorem, 280 subsequence, 66 sum function, 325 supremum, 35 tan1, 157 power series, 341

459 tanh1, 159 power series, 344 tangent approximation, 314 Taylor polynomial, 316 Taylor series, 325 Taylor’s theorem, 320 tilda notation, 300, 304 combination rules, 305 transitive property of R, 7 transitive rule for inequalities, 291 triangle inequality, 15, 16 backwards form, 15 infinite form, 105 integrals, 291 trichotomy property of R, 7 trigonometric functions

continuity, 139 differentiability, 213 unbounded sequence, 62 uniform continuity, 201 uniqueness theorem, power series, 342 primitives, 284 upper integral, 264 upper Riemann sum, 259 Wallis’s formula, 293 Weierstrass, K., x zero derivative theorem, 235 zero of polynomial, 295 zeros localisation theorem, 147