TEXTBOOKS VOL A Radical Approach to Real Analysis

A Radical Approach to Real AnalysisSecond Edition

AMS / MAA TEXTBOOKS VOL 10

David Bressoud

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book3 MAAB001/Bressoud October 20, 2006 4:18

A Radical Approach toReal Analysis

10.1090/text/010

© 2007 byThe Mathematical Association of America (Incorporated)

Library of Congress Catalog Card Number 2006933946

Hardcover ISBN: 978-0-88385-747-2 (out of print)

Paperback ISBN: 978-1-93951-217-8

Electronic ISBN: 978-1-61444-623-1

Printed in the United States of AmericaCurrent Printing (last digit):

10 9 8 7 6 5 4 3

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A Radical Approach toReal Analysis

Second Edition

David M. BressoudMacalester College

®

Published and Distributed byThe Mathematical Association of America

Committee on BooksFrank Farris, Chair

MAA Textbooks Editorial BoardZaven A. Karian, Editor

George ExnerThomas Garrity

Charles R. HadlockWilliam Higgins

Douglas B. MeadeStanley E. SeltzerShahriar Shahriari

Kay B. Somers

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to the memory of my mother

Harriet Carnrite Bressoud

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Preface

The task of the educator is to make the child’s spiritpass again where its forefathers have gone, mov-ing rapidly through certain stages but suppressingnone of them. In this regard, the history of sciencemust be our guide.

—Henri Poincare

This course of analysis is radical; it returns to the roots of the subject. It is not a historyof analysis. It is rather an attempt to follow the injunction of Henri Poincare to let historyinform pedagogy. It is designed to be a first encounter with real analysis, laying out itscontext and motivation in terms of the transition from power series to those that are lesspredictable, especially Fourier series, and marking some of the traps into which even greatmathematicians have fallen.

This is also an abrupt departure from the standard format and syllabus of analysis.The traditional course begins with a discussion of properties of the real numbers, moveson to continuity, then differentiability, integrability, sequences, and finally infinite series,culminating in a rigorous proof of the properties of Taylor series and perhaps even Fourierseries. This is the right way to view analysis, but it is not the right way to teach it. Itsupplies little motivation for the early definitions and theorems. Careful definitions meannothing until the drawbacks of the geometric and intuitive understandings of continuity,limits, and series are fully exposed. For this reason, the first part of this book follows thehistorical progression and moves backwards. It starts with infinite series, illustrating thegreat successes that led the early pioneers onward, as well as the obstacles that stymiedeven such luminaries as Euler and Lagrange.

There is an intentional emphasis on the mistakes that have been made. These highlightdifficult conceptual points. That Cauchy had so much trouble proving the mean valuetheorem or coming to terms with the notion of uniform convergence should alert us to thefact that these ideas are not easily assimilated. The student needs time with them. The highlyrefined proofs that we know today leave the mistaken impression that the road of discoveryin mathematics is straight and sure. It is not. Experimentation and misunderstanding havebeen essential components in the growth of mathematics.

ix

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x Preface

Exploration is an essential component of this course. To facilitate graphical and numericalinvestigations, Mathematica and Maple commands and programs as well as investigativeprojects are available on a dedicated website at www.macalester.edu/aratra.

The topics considered in this book revolve around the questions raised by Fourier’strigonometric series and the restructuring of calculus that occurred in the process of an-swering them. Chapter 1 is an introduction to Fourier series: why they are important andwhy they met with so much resistance. This chapter presupposes familiarity with partialdifferential equations, but it is purely motivational and can be given as much or as littleemphasis as one wishes. Chapter 2 looks at the background to the crisis of 1807. Weinvestigate the difficulties and dangers of working with infinite summations, but also theinsights and advances that they make possible. More of these insights and advances aregiven in Appendix A. Calculus would not have revolutionized mathematics as it did if ithad not been coupled with infinite series. Beginning with Newton’s Principia, the physicalapplications of calculus rely heavily on infinite sums. The chapter concludes with a closerlook at the understandings of late eighteenth century mathematicians: how they saw whatthey were doing and how they justified it. Many of these understandings stood directly inthe way of the acceptance of trigonometric series.

In Chapter 3, we begin to find answers to the questions raised by Fourier’s series. Wefollow the efforts of Augustin Louis Cauchy in the 1820s to create a new foundation to thecalculus. A careful definition of differentiability comes first, but its application to many ofthe important questions of the time requires the mean value theorem. Cauchy struggled—unsuccessfully—to prove this theorem. Out of his struggle, an appreciation for the natureof continuity emerges.

We return in Chapter 4 to infinite series and investigate the question of convergence.Carl Friedrich Gauss plays an important role through his complete characterization ofconvergence for the most important class of power series: the hypergeometric series. Thischapter concludes with a verification that the Fourier cosine series studied in the first chapterdoes, in fact, converge at every value of x.

The strange behavior of infinite sums of functions is finally tackled in Chapter 5. Welook at Dirichlet’s insights into the problems associated with grouping and rearranginginfinite series. We watch Cauchy as he wrestles with the problem of the discontinuity ofan infinite sum of continuous functions, and we discover the key that he was missing. Webegin to answer the question of when it is legitimate to differentiate or integrate an infiniteseries by differentiating or integrating each summand.

Our story culminates in Chapter 6 where we present Dirichlet’s proof of the validity ofFourier series representations for all “well behaved” functions. Here for the first time weencounter serious questions about the nature and meaning of the integral. A gap remains inDirichlet’s proof which can only be bridged after we have taken a closer look at integration,first using Cauchy’s definition, and then arriving at Riemann’s definition. We conclude withWeierstrass’s observation that Fourier series are indeed strange creatures. The functionrepresented by the series

cos(πx) + 1

2cos(13πx) + 1

4cos(169πx) + 1

8cos(2197πx) + · · ·

converges and is continuous at every value of x, but it is never differentiable.

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Preface xi

The material presented within this book is not of uniform difficulty. There are computa-tional inquiries that should engage all students and refined arguments that will challenge thebest. My intention is that every student in the classroom and each individual reader strikingout alone should be able to read through this book and come away with an understanding ofanalysis. At the same time, they should be able to return to explore certain topics in greaterdepth.

Historical Observations

In the course of writing this book, unexpected images have emerged. I was surprisedto see Peter Gustav Lejeune Dirichlet and Niels Henrik Abel reveal themselves as thecentral figures of the transformation of analysis that fits into the years from 1807 through1872. While Cauchy is associated with the great theorems and ideas that launched thistransformation, one cannot read his work without agreeing with Abel’s judgement that“what he is doing is excellent, but very confusing.” Cauchy’s seminal ideas required twoand a half decades of gestation before anyone could begin to see what was truly importantand why it was important, where Cauchy was right, and where he had fallen short ofachieving his goals.

That gestation began in the fall of 1826 when two young men in their early 20s, GustavDirichlet and Niels Henrik Abel, met to discuss and work out the implications of whatthey had heard and read from Cauchy himself. Dirichlet and Abel were not alone in thisundertaking, but they were of the right age to latch onto it. It would become a recurringtheme throughout their careers. By the 1850s, the stage was set for a new generationof bright young mathematicians to sort out the confusion and solidify this new vision formathematics. Riemann and Weierstrass were to lead this generation. Dirichlet joined Gaussas teacher and mentor to Riemann. Abel died young, but his writings became Weierstrass’sinspiration.

It was another twenty years before the vision that Riemann and Weierstrass had graspedbecame the currency of mathematics. In the early 1870s, the general mathematical com-munity finally understood and accepted this new analysis. A revolution had taken place.It was not an overthrow of the old mathematics. No mathematical truths were discred-ited. But the questions that mathematicians would ask and the answers they would accepthad changed in a fundamental way. An era of unprecedented power and possibility hadopened.

Changes to the Second Edition

This second edition incorporates many changes, all with the aim of aiding students whoare learning real analysis. The greatest conceptual change is in Chapter 2 where I clarifythat the Archimedean understanding of infinite series is the approach that Cauchy and themathematical community has adopted. While this chapter still has a free-wheeling style inits use of infinite series—the intent being to convey the power and importance of infiniteseries—it also begins to introduce rigorous justification of convergence. A new sectiondevoted entirely to geometric series has been added. Chapter 4, which introduces tests ofconvergence, has been reorganized.

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xii Preface

I have also trimmed some of the digressions that I found led students to lose sight of myintent. In particular, the section on the Newton–Raphson method and the proof of Gauss’stest for convergence of hypergeometric series have been taken out of the text. Because Ifeel that this material is still important, though not central, these sections and much moreare available on the web site dedicated to this book.

Web Resource: When you see this box with the designation “Web Resource”,more information is available in a pdf file, Mathematica notebook, or Mapleworksheet that can be downloaded at www.macalester.edu/aratra. The box isalso used to point to additional information available in Appendix A.

I have added many new exercises, including many taken from Problems in MathematicalAnalysis by Kaczor and Nowak. Problems taken from this book are identified in Appendix C.I wish to acknowledge my debt to Kaczor and Nowak for pulling together a beautifulcollection of challenging problems in analysis. Neither they nor I claim that they are theoriginal source for all of these problems.

All code for Mathematica and Maple has been removed from the text to the website.

Exercises for which these codes are available are marked with the symbol��

��M&M . The

appendix with selected solutions has been replaced by a more extensive appendix of hints.I considered adding a new chapter on the structure of the real numbers. Ultimately,

I decided against it. That part of the story properly belongs to the second half of thenineteenth century when the progress described in this book led to a thorough reappraisalof integration. To everyone’s surprise this was not possible without a full understandingof the real numbers which were to reveal themselves as far more complex than had beenthought. That is an entirely other story that will be told in another book, A Radical Approachto Lebesgue’s Theory of Integration.

Acknowledgements

Many people have helped with this book. I especially want to thank the NSA and theMAA for financial support; Don Albers, Henry Edwards, and Walter Rudin for their earlyand enthusiastic encouragement; Ray Ayoub, Allan Krall, and Mark Sheingorn for helpfulsuggestions; and Ivor Grattan-Guinness who was extremely generous with his time andeffort, suggesting historical additions, corrections, and references. The epilogue is amongthe additions that were made in response to his comments. I am particularly indebtedto Meyer Jerison who went through the manuscript of the first edition very carefully andpointed out many of the mathematical errors, omissions, and questionable approaches in theearly versions. Some was taken away and much was added as a result of his suggestions. Itake full responsibility for any errors or omissions that remain. Susan Dziadosz assisted withthe exercises. Her efforts helped weed out those that were impossible or incorrectly stated.Beverly Ruedi helped me through many aspects of production and has shepherded this booktoward a speedy publication. Most especially, I want to thank the students who took thiscourse at Penn State in the spring of 1993, putting up with a very preliminary edition andhelping to identify its weaknesses. Among those who suggested improvements were RyanAnthony, Joe Buck, Robert Burns, Stephanie Deom, Lisa Dugent, David Dunson, Susan

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Preface xiii

Dziadosz, Susan Feeley, Rocco Foderaro, Chris Franz, Karen Lomicky, Becky Long, EdMazich, Jon Pritchard, Mike Quarry, Curt Reese, Brad Rothenberger, Chris Solo, RandyStanley, Eric Steel, Fadi Tahan, Brian Ward, Roger Wherley, and Jennifer White.

Since publication of the first edition, suggestions and corrections have come from manypeople including Dan Alexander, Bill Avant, Robert Burn, Dennis Caro, Colin Denis,Paul Farnham II, Julian Fleron, Kristine Fowler, Øistein Gjøvik, Steve Greenfield, MichaelKinyon, Mary Marion, Betty Mayfield, Mi-Kyong, Helen Moore, Nick O’Neill, DavidPengelley, Mac Priestley, Tommy Ratliff, James Reber, Fred Rickey, Wayne Roberts,Cory Sand, Karen Saxe, Sarah Spence, Volker Strehl, Simon Terrington, and Stan Wagon.I apologize to anyone whose name I may have forgotten.

David M. [email protected] 20, 2006

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Contents

Preface ix

1 Crisis in Mathematics: Fourier’s Series 1

1.1 Background to the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Difficulties with the Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Infinite Summations 9

2.1 The Archimedean Understanding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Calculating π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Logarithms and the Harmonic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.5 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.6 Emerging Doubts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3 Differentiability and Continuity 57

3.1 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.2 Cauchy and the Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.3 Continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.4 Consequences of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3.5 Consequences of the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4 The Convergence of Infinite Series 117

4.1 The Basic Tests of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

4.2 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

xv

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16 Contents

4.3 The Convergence of Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

4.4 The Convergence of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

5 Understanding Infinite Series 171

5.1 Groupings and Rearrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

5.2 Cauchy and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

5.3 Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

5.4 Verifying Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

6 Return to Fourier Series 217

6.1 Dirichlet’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

6.2 The Cauchy Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

6.3 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

6.4 Continuity without Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

7 Epilogue 267

A Explorations of the Infinite 271

A.1 Wallis on π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

A.2 Bernoulli’s Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

A.3 Sums of Negative Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

A.4 The Size of n! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

B Bibliography 303

C Hints to Selected Exercises 305

Index 317

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Appendix A

Explorations of the Infinite

The four sections of Appendix A lead to a proof of Stirling’s remarkable formula for thevalue of n!:

n! = nne−n√

2πn eE(n), (A.1)

where limn→∞ E(n) = 0, and this error term can be represented by the asymptotic series

E(n) ∼ B2

1 · 2 · n+ B4

3 · 4 · n3+ B6

5 · 6 · n5+ · · · , (A.2)

where B1, B2, B3, . . . are rational numbers known as the Bernoulli numbers. Note that wedo not write equation (A.2) as an equality since this series does not converge to E(n) (seesection A.4).

In this first section, we follow John Wallis as he discovers an infinite product that is equalto π . While his formula is a terrible way to approximate π , it establishes the connectionbetween n! and π , explaining that curious appearance of π in equation (A.1). In section 2,we show how Jacob Bernoulli was led to discover his mysterious and pervasive sequenceof numbers by his search for a simple formula for the sum of kth powers. We continuethe applications of the Bernoulli numbers in section 3 where we follow Leonhard Euler’sdevelopment of formulæ for the sums of reciprocals of even powers of the positive integers.It all comes together in section 4 when we shall prove this identity discovered by AbrahamdeMoivre and James Stirling.

A.1 Wallis on π

When Newton said, “If I have seen a little farther than others it is because I have stood onthe shoulders of giants,” one of those giants was John Wallis (1616–1703). Wallis taught

271

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272 Appendix A. Explorations of the Infinite

at Cambridge before becoming Savilian Professor of Geometry at Oxford. His ArithmeticaInfinitorum, published in 1655, derives the rule (found also by Fermat) for the integral of afractional power of x:

∫ 1

0xm/n dx = 1

1 + m/n= n

m + n. (A.3)

We begin our development of Wallis’s formula for π with the observation that π is thearea of any circle with radius 1. If we locate the center of our circle at the origin, then thequarter circle in the first quadrant has area

π

4=

∫ 1

0(1 − x2)1/2 dx. (A.4)

This looks very much like the kind of integral Wallis had been studying. Any means ofcalculating this integral will yield a means of calculating π . Realizing that he could notattack it head on, Wallis looked for similar integrals that he could handle. His genius isrevealed in his decision to look at ∫ 1

0(1 − x1/p)q dx.

When q is a small positive integer, we can expand the integrand:∫ 1

0(1 − x1/p)0 dx =

∫ 1

0dx

= 1,∫ 1

0(1 − x1/p)1 dx =

∫ 1

0(1 − x1/p) dx

= 1 − p

p + 1

= 1

p + 1,

∫ 1

0(1 − x1/p)2 dx =

∫ 1

0(1 − 2x1/p + x2/p) dx

= 1 − 2p

p + 1+ p

p + 2

= 2

(p + 1)(p + 2),

∫ 1

0(1 − x1/p)3 dx =

∫ 1

0(1 − 3x1/p + 3x2/p − x3/p) dx

= 1 − 3p

p + 1+ 3p

p + 2− p

p + 3

= 6

(p + 1)(p + 2)(p + 3).

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A.1 Wallis on π 273

A pattern is emerging, and it requires little insight to guess that∫ 1

0(1 − x1/p)4 dx = 4!

(p + 1)(p + 2)(p + 3)(p + 4),

∫ 1

0(1 − x1/p)5 dx = 5!

(p + 1)(p + 2)(p + 3)(p + 4)(p + 5),

...

where 4! = 4 · 3 · 2 · 1 = 24, 5! = 5 · 4 · 3 · 2 · 1 = 120, and so on.These numbers should look familiar. When p and q are both integers, we get reciprocals

of binomial coefficients,

q!

(p + 1)(p + 2) · · · (p + q)= 1(

p+q

q

) ,where (

p + q

q

)=

(p + q

p

)= (p + q)!

p! q!.

This suggested to Wallis that he wanted to work with the reciprocals of his integrals:

f (p, q) = 1

/∫ 1

0(1 − x1/p)q dx = (p + 1)(p + 2) · · · (p + q)

q!. (A.5)

We want to find the value of f (1/2, 1/2) = 4/π . Our first observation is that we canevaluate f (p, q) for any p so long as q is a nonnegative integer. We use induction on q toprove equation (A.5). As shown above, when q = 0 we have f (p, q) = 1/

∫ 10 dx = 1. In

exercise A.1.2, you are asked to prove that

∫ 1

0

(1 − x1/p

)qdx = q

p + q

∫ 1

0

(1 − x1/p

)q−1dx. (A.6)

With this recursion and the induction hypothesis that

f (p, q − 1) = (p + 1)(p + 2) · · · (p + q − 1)

(q − 1)!,

it follows that

f (p, q) =(

p + q

q

)(p + 1)(p + 2) · · · (p + q − 1)

(q − 1)!= (p + 1)(p + 2) · · · (p + q)

q!.

We also can use this recursion to find f (1/2, 3/2) in terms of f (1/2, 1/2):

f (1/2, 3/2) = 1/2 + 3/2

3/2f (1/2, 1/2) = 4

3f (1/2, 1/2).

We can now prove by induction (see exercise A.1.3) that

f

(1

2, q − 1

2

)= (2q)(2q − 2) · · · 4

(2q − 1)(2q − 3) · · · 3f

(1

2,

1

2

). (A.7)

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Table A.1. Values of f (p, q) in terms of � = 4/π .

f (p, q)

p↓ q → −1/2 0 1/2 1 3/2 2 5/2 3 7/2 4

−1/2 ∞ 1 12� 1

213� 3

84

15� 516

835� 35

128

0 1 1 1 1 1 1 1 1 1 1

1/212� 1 � 3

243� 15

885� 35

166435� 315

128

1 12 1 3

2 2 52 3 7

2 4 92 5

3/213� 1 4

3� 52

83� 35

86415� 105

1612821 � 1155

128

2 38 1 15

8 3 358 6 63

8 10 998 15

5/24

15� 1 85� 7

26415� 63

812815 � 231

1651235 � 3003

128

3 516 1 35

16 4 10516 10 231

16 20 42916 35

7/28

35� 1 6435� 9

212821 � 99

851235 � 429

161024

35 � 6435128

4 35128 1 315

128 5 1155128 15 3003

128 35 6435128 70

What if p is a nonnegative integer or half-integer? The binomial coefficient is symmetricin p and q, (

p + q

q

)=

(p + q

p

)= (p + q)!

p! q!.

It is only a little tricky to verify that for any values of p and q, we also have that f (p, q) =f (q, p) (see exercise A.1.4). This can be used to prove that when p and q are positiveintegers (see exercise A.1.5),

f

(p − 1

2, q − 1

2

)= 2 · 4 · 6 · · · (2p + 2q − 2)

3 · 5 · · · (2p − 1) · 3 · 5 · · · (2q − 1)f

(1

2,

1

2

). (A.8)

Wallis could now construct a table of values for f (p, q), allowing � to stand forf (1/2, 1/2) = 4/π (see Table A.1.).

We see that any row in which p is a positive integer is increasing from left to right, andit is reasonable to expect that the row p = 1/2 is also increasing from left to right (seeexericse A.1.6 for the proof). Recalling that � = 4/π , this implies a string of inequalities:

1 <4

π<

3

2<

16

3π<

15

8<

32

5π<

35

16<

256

35π<

315

128.

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A.1 Wallis on π 275

These, in turn, yield inequalities for π/2:

4

3<

π

2< 2,

64

45<

π

2<

16

9,

256

175<

π

2<

128

75,

16384

11025<

π

2<

2048

1225.

It is easier to see what is happening with these inequalities if we look at our string ofinequalities in terms of the ratios that led us to find them in the first place:

1 <4

π<

3

2<

4

3· 4

π<

3

2· 5

4<

4

3· 6

5· 4

π<

3

2· 5

4· 7

6< · · · .

In general, we have that

3 · 5 · 7 · · · (2n − 1)

2 · 4 · 6 · · · (2n − 2)<

4 · 6 · 8 · · · (2n)

3 · 5 · 7 · · · (2n − 1)

4

π<

3 · 5 · 7 · · · (2n + 1)

2 · 4 · 6 · · · (2n). (A.9)

This yields a general inequality for π/2 that we can make as precise as we want by takingn sufficiently large:

22 · 42 · 62 · · · (2n)2

1 · 32 · 52 · · · (2n − 1)2 · (2n + 1)<

π

2<

22 · 42 · 62 · · · (2n − 2)2 · (2n)

1 · 32 · 52 · · · (2n − 1)2. (A.10)

As n gets larger, these bounds on π/2 approach each other. Their ratio is 2n/(2n + 1)which approaches 1. Wallis therefore concluded that

π

2= 2

1· 2

3· 4

3· 4

5· 6

5· 6

7· · · . (A.11)

Note that this product alternately grows and shrinks as we take more terms.

Exercises

The symbol��

��M&M indicates that Maple and Mathematica codes for this problem are

available in the Web Resources at www.macalester.edu/aratra.

A.1.1.��

��M&M Consider Wallis’s infinite product for π/2 given in equation (A.11).

a. Show that the product of the kth pair of fractions is 4k2/(4k2 − 1) and therefore

π = 2∞∏

k=1

4k2

4k2 − 1.

b. How many terms of this product are needed in order to approximate π to 3-digitaccuracy?

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c. We can improve our accuracy if we average the upper and lower bounds. Prove that thisaverage is

22 · 42 · · · (2n) · (2n + 1/2)

1 · 32 · · · (2n − 1)2 · (2n + 1).

How large a value of n do we need in order to approximate π to 3-digit accuracy?

A.1.2. Prove equation (A.6).

A.1.3. Use equation (A.6) to prove equation (A.7) by induction on q.

A.1.4. Prove that

∫ 1

0

(1 − x1/p

)qdx =

∫ 1

0

(1 − x1/q

)pdx, (A.12)

and therefore f (p, q) = f (q, p).

A.1.5. Using the results from exercises A.1.3 and A.1.4, prove equation (A.8) when p andq are positive integers.

A.1.6. Prove that if p is positive and q1 > q2, then(1 − x1/p

)q1<

(1 − x1/p

)q2,

for all x between 0 and 1, and therefore f (p, q1) > f (p, q2). Prove that if p is negativeand q1 > q2, then f (p, q1) < f (p, q2).

A.1.7.��

��M&M We have seen that as long as p or q is an integer, we have that f (p, q) =(

p+q

q

). This suggests a way of defining binomial coefficients when neither p nor q are

integers. We would expect the value of( 1

1/2

)to be f (1/2, 1/2) = 4/π . Using your favorite

computer algebra system, see what value it assigns to( 1

1/2

). How should we define

(a

b

)for

arbitrary real numbers a and b?

A.1.8. When p and q are integers, we have the relationship of Pascal’s triangle,

f (p, q) = f (p, q − 1) + f (p − 1, q). (A.13)

Does this continue to hold true when p and q are not both integers? Either give an examplewhere it does not work or prove that it is always true.

A.1.9. Show that if we use the row p = −1/2 to approximate π/2:

1 >4

2π>

1

2>

4

3π>

3

8>

16

15π>

5

16>

32

35π>

35

128> · · · ,

then we get the same bounds for π/2.

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A.2 Bernoulli’s Numbers 277

A.1.10. What bounds do we get for π/2 if we use the row p = 3/2?

A.1.11. What bounds do we get for π/2 if we use the diagonal:

1 <4

π< 2 <

32

3π< 6 <

512

15π< 20 <

4094

35π< 70 < · · ·?

A.1.12. As far as you can, extend the table of values for f (p, q) into negative values of p

and q.

A.1.13.��

��M&M Use the method of this section to find an infinite product that approaches

the value of

f (2/3, 1/3) = 1

/∫ 1

0

(1 − x3/2

)1/3dx .

Compare this to the value of( 1

1/3

)given by your favorite computer algebra system.

A.2 Bernoulli’s Numbers

Johann and Jacob Bernoulli were Swiss mathematicians from Basel, two brothers whoplayed a critical role in the early development of calculus. The elder, Jacob Bernoulli, diedin 1705. Eight years later, his final masterpiece was published, Ars Conjectandi. It laid thefoundations for the study of probability and included an elegant solution to an old problem:to find formulas for the sums of powers of consecutive integers. He bragged that with it hehad calculated the sum of the tenth powers of the integers up to one thousand,

110 + 210 + 310 + · · · + 100010,

in “half a quarter of an hour.”The formula for the sum of the first k − 1 integers is

1 + 2 + 3 + 4 + · · · + (k − 1) = k2

2− k

2. (A.14)

The proof is simple:

1 + 2 + · · · + (k − 2) + (k − 1)

+ (k − 1) + (k − 2) + · · · + 2 + 1= k + k + · · · + k + k = (k − 1)k.

No one knows who first discovered this formula. Its origins are lost in the mists of time.Even the formula for the sum of squares is ancient,

12 + 22 + 32 + 42 + · · · + (k − 1)2 = k3

3− k2

2+ k

6. (A.15)

It was known to and used by Archimedes, but was probably even older. It took quite a bitlonger to find the formula for the sum of cubes. The earliest reference that we have to thisformula connects it to the work of Aryabhata of Patna (India) around the year 500 .. The

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formula for sums of fourth powers was discovered by ibn Al-Haytham of Baghdad about1000 .. In the early 14th century, Levi ben Gerson of France found a general formulafor arbitary kth powers, though the central idea which draws on patterns in the binomialcoefficients can be found in other contemporary and earlier sources: Al-Bahir fi’l Hisab(Shining Treatise on Calculation) written by al-Samaw’al in 1144 in what is now Iraq,Siyuan Yujian (Jade Mirror of the Four Unknowns) written by Zhu Shijie in 1303 in China,and Ganita Kaumudi written by Narayana Pandita in 1356 in India.

Web Resource: To see a derivation and proof of this historic formula for thesum of kth powers based on properties of binomial coefficients, go to Binomialcoefficients and sums of kth powers.

Jacob Bernoulli may have been aware of the binomial coefficient formula, but that didnot stop him from finding his own. He had a brilliant insight. The new integral calculusgave efficient tools for calculating limits of summations that today we call Riemann sums.Perhaps it could be turned to the task of finding formulas for other types of sums.

The Bernoulli Polynomials

Jacob Bernoulli looked for polynomials, B1(x), B2(x), . . . , for which

1 + 2 + · · · + (k − 1) =∫ k

0B1(x) dx,

12 + 22 + · · · + (k − 1)2 =∫ k

0B2(x) dx,

13 + 23 + · · · + (k − 1)3 =∫ k

0B3(x) dx,

...

1n + 2n + · · · + (k − 1)n =∫ k

0Bn(x) dx.

Such a polynomial must satisfy the equation

∫ k+1

k

Bn(x) dx = kn. (A.16)

In fact, for each positive integer n, there is a unique monic1 polynomial of degree n thatsatisfies this equation for all values of k, not only when k is a positive integer. It is easiestto see why this is so by means of an example. We shall set

B3(x) = x3 + a2x2 + a1x + a0

and show that there exist unique values for a2, a1, and a0.

1 The coefficient of xn is 1.

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A.2. Bernoulli’s Numbers 279

Substituting this polynomial for B3(x) in equation (A.16), we see that

k3 =∫ k+1

k

(x3 + a2x

2 + a1x + a0)dx

= 1

4(k + 1)4 + a2

3(k + 1)3 + a1

2(k + 1)2 + a0(k + 1) − 1

4k4 − a2

3k3 − a1

2k2 − a0k

= k3 +(

3

2+ a2

)k2 + (1 + a2 + a1)k +

(1

4+ a2

3+ a1

2+ a0

). (A.17)

The coefficients of the different powers of k must be the same on both sides:

0 = 3

2+ a2, (A.18)

0 = 1 + a2 + a1, (A.19)

0 = 1

4+ a2

3+ a1

2+ a0. (A.20)

These three equations have a unique solution: a2 = −3/2, a1 = 1/2, a0 = 0, and so

B3(x) = x3 − 3x2

2+ x

2. (A.21)

Integrating this polynomial from 0 to k, we obtain the formula for the sum of cubes:

13 + 23 + · · · + (k − 1)3 =∫ 1

0

(x3 − 3x2

2+ x

2

)dx

= k4

4− k3

2+ k2

4.

The first two Bernoulli polynomials are

B1(x) = x − 1

2, (A.22)

B2(x) = x2 − x + 1

6. (A.23)

We now make an observation that will enable us to construct Bn+1(x) from Bn(x). If wedifferentiate both sides of equation (A.16) with respect to k, we get:

Bn(k + 1) − Bn(k) = nkn−1. (A.24)

This implies that

n[0n−1 + 1n−1 + 2n−1 + · · · + (k − 1)n−1

]= [Bn(1) − Bn(0)] + [Bn(2) − Bn(1)] + [Bn(3) − Bn(2)]

+ · · · + [Bn(k) − Bn(k − 1)]

= Bn(k) − Bn(0), (A.25)

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and therefore

Bn(k) − Bn(0)

n= 1n−1 + 2n−1 + · · · + (k − 1)n−1 =

∫ k

0Bn−1(x) dx. (A.26)

Our recursive formula is

Bn(x) = n

∫ x

0Bn−1(t) dt + Bn(0). (A.27)

Given that B4(0) = −1/30, we can find B4(x) by integrating B3(x), multiplying by 4, andthen adding −1/30:

B4(x) = 4

(x4

4− x3

2+ x2

4

)− 1

30= x4 − 2x3 + x2 − 1

30.

If we know the constant term in each polynomial: B1(0) = −1/2, B2(0) = 1/6, B3(0) =0, . . . , then we can successively construct as many Bernoulli polynomials as we wish.These constants are called the Bernoulli numbers:

B1 = −1

2, B2 = 1

6, B3 = 0, B4 = −1

30, . . .

A Formula for Bn(x )

We can do even better. Recalling that B1(x) = x + B1 and repeatedly using equation (A.27),we see that

B2(x) = 2∫ x

0(t + B1) dt + B2

= x2 + 2B1x + B2,

B3(x) = 3∫ x

0(t2 + 2B1t + B2) dt + B3

= x3 + 3B1x2 + 3B2x + B3,

B4(x) = 4∫ x

0(t3 + 3B1t

2 + 3B2t + B3) dt + B4

= x4 + 4B1x3 + 6B2x

2 + 4B3x + B4,...

A pattern is developing. Our coefficients are precisely the coefficients of the binomialexpansion. Pascal’s triangle has struck again. Once we see it, it is easy to verify byinduction (see exercise A.2.3) that

Bn(x) = xn + nB1xn−1 + n(n − 1)

2!B2x

n−2 + n(n − 1)(n − 2)

3!B3x

n−3

+ · · · + nBn−1x + Bn. (A.28)

The only problem left is to find an efficient means of determining the Bernoulli numbers.

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A Recursive Formula for Bn

We turn to equation (A.24), set k = 0, and assume that n is larger than one:

Bn(1) − Bn(0) = n · 0n−1 = 0. (A.29)

We use equation (A.28) to evaluate Bn(1):

0 = Bn(1) − Bn

=[

1 + nB1 + n(n − 1)

2!B2 + · · · + n(n − 1)

2!Bn−2 + nBn−1 + Bn

]− Bn, (A.30)

Bn−1 = −1

n

(1 + nB1 + n(n − 1)

2!B2 + · · · + n(n − 1)

2!Bn−2

). (A.31)

It follows that

B5 = −1

6

(1 + 6 · −1

2+ 15 · 1

6+ 20 · 0 + 15 · −1

30

)= 0,

B6 = −1

7

(1 + 7 · −1

2+ 21 · 1

6+ 35 · 0 + 35 · −1

30+ 21 · 0

)

= 1

42.

Continuing, we obtain

B7 = 0, B8 = −1

30, B9 = 0,

B10 = 5

66, B11 = 0, B12 = −691

2730.

Bernoulli’s Calculation

Equipped with equation (A.26) and the knowledge of B1, B2, . . . , B10, we can find theformula for the sum of the first k − 1 tenth powers:

k−1∑i=1

i10 = 1

11[B11(k) − B11]

= 1

11(k11 + 11B1k

10 + 55B2k9 + 165B3k

8 + 330B4k7

+ 462B5k6 + 462B6k

5 + 330B7k4 + 165B8k

3

+ 55B9k2 + 11B10k)

= 1

11k11 − 1

2k10 + 5

6k9 − k7 + k5 − 1

2k3 + 5

66k. (A.32)

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Since it is much easier to take powers of 1000 = 103 than of 1001, let us add 100010 =1030 to the sum of the tenth powers of the integers up to 999:

110 + 210 + 310 + · · · + 100010

= 1030 + 1

111033 − 1

21030 + 5

61027 − 1021 + 1015 − 1

2109 + 5

66103.

This is a simple problem in arithmetic:

1 00000 00000 00000 00000 00000 00000 .00+ 90 90909 09090 90909 09090 90909 09090 .90− 50000 00000 00000 00000 00000 00000 .00+ 83 33333 33333 33333 33333 33333 .33− 10 00000 00000 00000 00000 .00+ 1 00000 00000 00000 .00− 5000 00000 .00+ 75 .75

91 40992 42414 24243 42424 19242 42500

Seven and a half minutes is plenty of time.

Fermat’s Last Theorem

The Bernoulli numbers will make appearances in each of the next two sections. Once theywere discovered, mathematicians kept finding them again, and again, and again. One of themore surprising places that they turn up is in connection with Fermat’s last theorem.

After studying Pythagorean triples, triples of positive integers satisfying

x2 + y2 = z2,

Pierre de Fermat pondered the question of whether such triples could exist when theexponent was larger than 2. He came to the conclusion that no such triples exist, but nevergave a proof. It should be noted that if there is no solution to

xn + yn = zn,

then there can be no solution to

xmn + ymn = zmn,

because if x = a, y = b, z = c were a solution to the second equation, then x = am,y = bm, z = cm would be a solution to the first. If we want to prove Fermat’s statement,then it is enough to prove that there are no solutions when n = 4 and no solutions when n

is an odd prime.The case n = 4 can be handled by methods described by Fermat. Euler essentially proved

the case n = 3 in 1753. His proof was flawed, but his approach was correct. Fermat’s“theorem” for n = 5 came in pieces. Sophie Germain (1776–1831), one of the first womento publish mathematics, showed that if a solution exists, then either x, y, or z must bedivisible by 5. Gustav Lejeune Dirichlet made his mark on the mathematical scene when,in 1825 at the age of 20, he proved that the variable divisible by 5 cannot be even. In thesame year, Adrien Marie Legendre, then in his 70’s, picked up Dirichlet’s analysis and

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carried it forward to show the general impossibility of the case n = 5. Gabriel Lame settledn = 7 in 1839.

In 1847, Ernst Kummer proved that there is no solution in positive integers to

xp + yp = zp

whenever p is a regular prime. The original definition of a regular prime is well outsidethe domain of this book, but Kummer found a simple and equivalent definition:

An odd prime p is regular if and only if it does not divide the numerator of anyof the Bernoulli numbers: B2, B4, B6, . . . , Bp−3.

The prime 11 is regular. Up to p − 3 = 8, the numerators are all 1. The prime 13 is regular.So is 17. And 19. And 23. Unfortunately, not all primes are regular. The prime 37 is not,nor is 59 or 67. Methods using Bernoulli numbers have succeeded in proving Fermat’s lasttheorem for all primes below 4,000,000. The proof by Andrew Wiles and Richard Tayloruses a very different approach.

Exercises

The symbol��


available in the Web Resources at www.macalester.edu/aratra.

A.2.1.��

��M&M Find the polynomials B5(x), B6(x), B7(x), and B8(x).

A.2.2.��

��M&M Graph the eight polynomials B1(x) through B8(x). Describe the symmetries

that you see. Prove your guess about the symmetry of Bn(x) for arbitrary n.

A.2.3. Prove equation (A.28) by induction on n.

A.2.4. Prove that

Bn(1 − x) = (−1)nBn(x) (A.33)

provided n ≥ 1.

A.2.5. Prove that

B2n+1 = 0 (A.34)

provided that n ≥ 1.

A.2.6. Show how to use Bernoulli polynomials and hand calculations to find the sum

18 + 28 + 38 + · · · + 10008.

A.2.7. Show how to use Bernoulli polynomials and hand calculations to find the sum

110 + 210 + 310 + · · · + 1,000,00010.

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A.2.8.��

��M&M Explore the factorizations of the numerators and of the denominators of

the Bernoulli numbers. What conjectures can you make?

A.2.9.��

��M&M Find all primes less than 100 that are not regular. Show that 691 is not a

regular prime.

A.3 Sums of Negative Powers

Jacob Bernoulli and his brother Johann were also interested in the problem of summingnegative powers of the integers. The first such case is the harmonic series,

1 + 1

2+ 1

3+ · · · + 1

k,

which by then was well understood. The next case involves the sums of the reciprocals ofthe squares:

1 + 1

22+ 1

32+ 1

42+ · · · .

We observe that this seems to approach a finite limit. The sum up to 1/1002 is 1.63498. Upto 1/10002 it is 1.64393. In fact, the Bernoullis knew that it must converge because

1

n2<

1

n(n − 1)= 1

n − 1− 1

n,

and so

N∑n=1

1

n2< 1 +

N∑n=2

(1

n − 1− 1

n

)

= 1 +(

1 − 1

2

)+

(1

2− 1

3

)

+ · · · +(

1

N − 2− 1

N − 1

)+

(1

N − 1− 1

N

)

= 2 − 1

N.

The sum of the reciprocals of the squares must converge and it must converge to somethingless than 2. What is the actual value of its limit?

It was around 1734 that Euler discovered that the value of this infinite sum is, in fact,π2/6. His proof stretched even the credulity of his contemporaries, but it is worth giving toshow the spirit of mathematical discovery. The fact that π2/6 = 1.64493 . . . is very closeto the expected value was convincing evidence that it must be correct.

Consider the power series expansion of sin(x)/x:

sin x

x= 1 − x2

3!+ x4

5!+ · · · . (A.35)

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A.3 Sums of Negative Powers 285

We know that this function has roots at ±π , ±2π , ±3π , . . . , and so we should be able tofactor it:

sin x

x=

(1 − x

π

) (1 + x

π

) (1 − x

2π

) (1 + x

2π

). . .

=(

1 − x2

π2

)(1 − x2

22π2

)(1 − x2

32π2

). . . . (A.36)

We compare the coefficient of x2 in equations (A.35) and (A.36) and see that

−1

6= −

(1

π2+ 1

22π2+ 1

32π2+ · · ·

),

or equivalently,

π2

6= 1 + 1

22+ 1

32+ 1

42+ · · · . (A.37)

Comparing the coefficients of x4 and doing a little bit of work, we can also find theformula for the sum of the reciprocals of the fourth powers:

1

120= 1

12 · 22π4+ 1

12 · 32π4+ 1

12 · 42π4+ · · · + 1

22 · 32π4+ 1

22 · 42π4+ · · ·

=∑

1≤j<k<∞

1

j 2k2π4,

π4

120=

∑1≤j<k<∞

1

j 2k2. (A.38)

If we square both sides of equation (A.37) and separate the pieces of the resulting product,we see that

π4

36=

(1 + 1

22+ 1

32+ · · ·

)(1 + 1

22+ 1

32+ · · ·

)

=∑

1≤j<k<∞

1

j 2k2+

∑1≤j=k<∞

1

j 2k2+

∑∞>j>k≥1

1

j 2k2

= 2∑

1≤j<k<∞

1

j 2k2+

∞∑k=1

1

k4.

Using the result from equation (A.38), we obtain our formula:

π4

36= 2

π4

120+

∞∑k=1

1

k4,

π4

90= 1 + 1

24+ 1

34+ 1

44+ · · · . (A.39)

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One can—and Euler did—continue this to find formulas for the sums of the reciprocalsof the other even powers,

∞∑k=1

1

k6,

∞∑k=1

1

k8,

∞∑k=1

1

k10, . . . .

In 1740, Euler discovered a formula that covered all of these cases.

A Generating Function

A problem that is not unreasonable at first glance is that of finding a power series expansionfor 1/(1 − ex). It looks as if it should be quite straightforward. Expand as a geometricseries, use the power series for the exponential function, then rearrange (note that we needto define 0! = 1):

1

1 − ex= 1 + ex + e2x + e3x + · · ·

= 1 +∞∑

k=1

ekx

= 1 +∞∑

k=1

(1 + kx + (kx)2

2!+ (kx)3

3!+ · · ·

)

= 1 +∞∑

k=1

∞∑n=0

(kx)n

n!

= 1 +∞∑

n=0

xn

n!

∞∑k=1

kn . . . whoa!

Something is wrong. We are getting infinite coefficients. We need to back up.The constant term in the power series expansion should be the value of the function at

x = 0. There is our problem. If we set x = 0 in our original function, we get a zero in thedenominator. We can get rid of the zero in the denominator if we multiply the numeratorby x. The function we should try to expand is

x

1 − ex.

We check what happens as x approaches 0 and see that we get −1. So far so good. It wouldbe nice if the constant were +1 instead, so we change the sign of the denominator. We arelooking for the coefficients in the power series expansion:

x

ex − 1= 1 + a1x + a2x

2 + a3x3 + · · · . (A.40)

The fact that we have multiplied by −x is not going to make our original argument work,but this power series should exist. There is little choice but to compute the coefficients, thean, by brute force. We could do it by using Taylor’s formula, but those derivatives quicklybecome very messy. Instead, we shall multiply both sides of equation (A.40) by ex − 1,expanded as a power series, and then equate the coefficients of comparable powers of x to

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solve for the an:

x = (ex − 1)(1 + a1x + a2x2 + a3x

3 + · · · )

=(

x + x2

2!+ x3

3!+ x4

4!+ · · ·

)(1 + a1x + a2x

2 + a3x3 + · · · )

= x +(

1

2!+ a1

)x2 +

(1

3!+ a1

2!+ a2

)x3 +

(1

4!+ a1

3!+ a2

2!+ a3

)x4 + · · · .

(A.41)

We obtain an infinite sequence of equations which we can solve for an:

0 = 1

2+ a1 =⇒ a1 = −1

2,

0 = 1

6+ a1

2+ a2 =⇒ a2 = 1

12,

0 = 1

24+ a1

6+ a2

2+ a3 =⇒ a3 = 0.

We continue in this manner:

a4 = −1

720, a5 = 0, a6 = 1

30240, a7 = 0, a8 = −1

1209600, . . . .

If you do not yet see what is happening, try multiplying each an by n!:

1 · a1 = −1

2, 2! · a2 = 1

6, 3! · a3 = 0, 4! · a4 = −1

30,

5! · a5 = 0, 6! · a6 = 1

42, 7! · a7 = 0, 8! · a8 = −1

30.

The Bernoulli numbers!Once we see them, it is not hard to prove that they are really there. Equation (A.41)

implies that

0 = 1

n!+ a1

(n − 1)!+ a2

(n − 2)!+ a3

(n − 3)!+ · · · + an−2

2!+ an−1

1!. (A.42)

We multiply both sides by n!:

0 = 1 + n a1 + n(n − 1)

2!(2! · a2) + n(n − 1)(n − 2)

3!(3! · a3)

+ · · · + n(n − 1)

2![(n − 2)! · an−2] + n [(n − 1)! · an−1]. (A.43)

This is precisely the recursion that we saw in equation (A.30) for the Bernoulli numbers.We have proven that

x

ex − 1= 1 +

∞∑n=1

Bn

xn

n!. (A.44)

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Euler’s Analysis

Once he had realized equation (A.44), Euler was off and running. One of the things thatit shows is that x/(ex − 1) is almost an even function: Bn is zero whenever n is odd andlarger than 1. If we add x/2 to both sides, we knock out the single odd power of x andobtain an even function:

1 +∞∑

m=1

B2m

x2m

(2m)!= x

ex − 1+ x

2= 2x + xex − x

2(ex − 1)= x(ex + 1)

2(ex − 1). (A.45)

We replace x by 2t :

1 +∞∑

m=1

B2m

(2t)2m

(2m)!= t(e2t + 1)

e2t − 1

= tet + e−t

et − e−t

= t coth t, (A.46)

where coth t is the hyperbolic cotangent of t . Euler knew that

sin z = eiz − e−iz

2i= −i sinh iz, (A.47)

cos z = eiz + e−iz

2= cosh iz, (A.48)

and so he saw that

z cot z = zcosh iz

−i sinh iz

= iz coth iz

= 1 +∞∑

m=1

B2m

(2iz)2m

(2m)!

= 1 +∞∑

m=1

(−1)mB2m

(2z)2m

(2m)!. (A.49)

Euler knew of another expansion for z cot z. Recognizing that the denominator of cot z =cos z/ sin z is zero whenever z = kπ , k any integer, he had found an infinite partial fractiondecomposition of the cotangent (see exercise A.3.16):

cot z = · · · + 1

z + 2π+ 1

z + π+ 1

z+ 1

z − π+ 1

z − 2π+ 1

z − 3π+ · · ·

= 1

z+

∞∑k=1

(1

z + kπ+ 1

z − kπ

)

= 1

z+ 2

∞∑k=1

z

z2 − k2π2

= 1

z− 2

∞∑k=1

z

k2π2 − z2. (A.50)

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If we multiply by z, we get an alternate expression for z cot z:

z cot z = 1 − 2∞∑

k=1

z2

k2π2 − z2

= 1 − 2∞∑

k=1

z2/k2π2

1 − z2/k2π2

= 1 − 2∞∑

k=1

(z2

k2π2+ z4

k4π4+ z6

k6π6+ · · ·

)

= 1 − 2∞∑

k=1

∞∑m=1

z2m

k2mπ2m

= 1 − 2∞∑

m=1

z2m

π2m

(1 + 1

22m+ 1

32m+ 1

42m+ · · ·

). (A.51)

Comparing the coefficients of z2m in equations (A.49) and (A.51), we see that

(−1)mB2m22m

(2m)!= −2

π2m

(1 + 1

22m+ 1

32m+ 1

42m+ · · ·

), (A.52)

or equivalently

1 + 1

22m+ 1

32m+ 1

42m+ · · · = (−1)m+1 (2π )2m

2 · (2m)!B2m. (A.53)

Euler had them all, provided the exponent was even:

∞∑n=1

1

n2= (2π )2

4· 1

6= π2

6, (A.54)

∞∑n=1

1

n4= (2π )4

2 · 24· 1

30= π4

90, (A.55)

∞∑n=1

1

n6= (2π )6

2 · 720· 1

42= π6

945, (A.56)

...

The function∑

n−s which Euler had shown how to evaluate when s is a positive eveninteger would come to play a very important role in number theory. Today it is called thezeta function:

ζ (s) =∞∑

n=1

1

ns, s > 1.

It can be defined for all complex values of s except s = 1. When Riemann laid out hisprescription for a proof that the number of primes less than or equal to x is asymptoticallyx/ ln x, he conjectured that all of the nonreal roots of ζ (s) lie on the line of s’s with real

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part 1/2. This is known as the Riemann hypothesis. It says a great deal about the error thatis introduced when x/ ln x is used to approximate the prime counting function. It is stillunproven.

If the Exponent is Odd?

If the exponent is odd, it appears that there is no simple formula. The most that can be said,and this was only proved in 1978 by Roger Apery, is that

∞∑n=1

1

n3

is definitely not a rational number.

Exercises

The symbol��


available in theWeb Resources at www.macalester.edu/aratra.

A.3.1.��

��M&M Calculate

100∑n=1

1

n2and

1000∑n=1

1

n2.

The first differs from π2/6 by about 1/100, the second from π2/6 by about 1/1000. Thissuggests that

1

N+

N∑n=1

1

n2

should be a pretty good approximation to π2/6. Test this hypothesis for different values ofN , including at least N = 5000 and N = 10000.

A.3.2.��

��M&M Calculate

N∑n=1

1

n4and

N∑n=1

1

n6

for N = 100, 500, and 1000. Compare your results with the predicted values of π4/90 andπ6/945, respectively. Are you willing to believe Euler’s result? In each case, what is theapproximate size of the error?

A.3.3. Prove that if k is larger than 1, then

∫ ∞

N+1

dx

xk<

∞∑n=N+1

1

nk<

∫ ∞

N

dx

xk. (A.57)

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Use these bounds to prove that∑N

n=1 1/nk differs from∑∞

n=1 1/nk by an amount that liesbetween

1

(k − 1)(N + 1)k−1and

1

(k − 1)Nk−1.

A.3.4. Set x = π/2 in equation (A.36) and see what identity you get. Does it look familiar?It should.

A.3.5. Set x = π/3 in equation (A.36) and see what identity you get. What happens if youset x = π/4?

A.3.6. Comparing the coefficients of x6 in equations (A.35) and (A.36) tells us that

π6

7!=

∑1≤j<k<l<∞

1

j 2k2l2. (A.58)

Use this fact together with equations (A.37) and (A.38) to prove that

∞∑k=1

1

k6= π6

945.

A.3.7. Consider the aborted derivation on page 286. Remember that any equality involvinginfinite series must, in general, carry a restriction on those x’s for which it is valid. Whatare the restrictions that need to go with each equality? Where precisely does the argumentgo wrong?

A.3.8.��

��M&M Graph the polynomials

y = 1 +N∑

n=1

Bn

xn

n!

for N = 4, 6, 8, 10, and 12. Compare these to the graph of x/(ex − 1). Describe what yousee. Where does it appear that this series converges?

A.3.9. We observe that∑∞

n=1 n−2m = 1 + 2−2m + · · · is always larger than 1. Use this factand equation (A.53) to prove that

|B2m| >2 · (2m)!

(2π )2m. (A.59)

Evaluate this lower bound for B20, B40, and B100. Do these numbers stay small or do theyget large? Express the lower bound in scientific notation with six digits of accuracy.

A.3.10. Show that limn→∞ ζ (n) = 1. Use this fact and the formula for Bn implied byequation (A.53) to find the interval of convergence of the series 1 + ∑∞

n=1 Bnxn/n!. Explain

your analysis of convergence at the endpoints.

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A.3.11.��

��M&M Taylor’s theorem tells us that Bn must be the nth derivative of x/(ex − 1)

evaluated at x = 0. Verify that this is correct when n = 1, 2, 3, and 4 by finding thederivatives.

A.3.12. Use the power series expansions of ex , cos x, and sin x to prove that

eix = cos x + i sin x. (A.60)

A.3.13. Use equation (A.60) to prove equations (A.47) and (A.48).

A.3.14.��

��M&M Graph the polynomials

y = 1 +N∑

m=1

(−1)mB2m

(2z)2m

(2m)!

for N = 2, 4, and 6. Compare these to the graph of z cot z. Describe what you see. Estimatethe radius of convergence for this series.

A.3.15. Determine the interval of convergence for the series in exercise A.3.14. Show thework that supports your answer.

A.3.16. We assume that cot z has a partial fraction decomposition. This means that thereare constants, ak , such that

cot z =∑

−∞<k<∞

ak

z − kπ.

To find the values of the ak , we multiply both sides by sin z,

cos z =∑

−∞<k<∞ak

sin z

z − kπ,

and then take the limit as z approaches mπ . Show that

sin z

z − kπ

approaches 0 if m = k and that it approaches cos mπ = (−1)m if m = k. Finish the proofthat am = 1.

A.3.17.��

��M&M Graph the functions

RN (z) = 1

z+ 2

N∑k=1

z

z2 − k2π2

for N = 3, 6, 9, and 12. Compare these to the graph of cot z. Describe what you see. Wheredoes it appear that this series converges? Plot the differences cot(z) − RN (z) for variousvalues of N and find a reasonable approximation, in terms of N and z, to this error function.Test the validity of your approximation for N = 1000.

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A.4 The Size of n! 293

A.3.18. What are the exact values of

∞∑n=1

1

n8and

∞∑n=1

1

n10,

expressed as a power of π times a rational number?

A.3.19. Prove that B2m and B2m+2 always have opposite sign.

A.3.20.��

��M&M Apery proved that

∑n−3 is not a rational number. We still do not know

if it can be written as π3 times a rational number. Calculate

N∑n=1

1

n3

for large values of N (at least 1000) and estimate the size of your error (see exer-cise A.3.3).

A.4 The Size of n!

An accurate approximation to n! was discovered in 1730 in a collaboration between Abra-ham de Moivre (1667–1754) and James Stirling (1692–1770). de Moivre was a FrenchProtestant. He and his parents had fled to London after the revocation of the Edict of Nantesin 1685. Despite his brilliance, he was always a foreigner and never obtained an academicappointment. He struggled throughout his life to support himself on the meager incomeearned as a tutor. Stirling was a Jacobite and in 1716, a year after the Jacobite rebellion,was expelled from Oxford for refusing to swear an oath of allegiance to the king. Becauseof his politics, he too was denied an academic position.

Even though it was a joint effort, the formula that we will find is called Stirling’s formula.This is primarily de Moivre’s own fault. When he published his result he gave Stirling creditfor finding the constant, but his language was sufficiently imprecise that the attribution ofthe constant to Stirling could easily be misread as crediting him with the entire identity. Inany event, Stirling’s name does deserve to be attached to this identity because it was thefruit of both their efforts.

Our first task is to turn n! into a summation so we can use an integral approximation.This is easily accomplished by taking the natural logarithm:

ln(n!) =n∑

k=1

ln k.

We can bound this above and below by integrals:

∫ n

1ln x dx <

n∑k=1

ln k <

∫ n−1

0ln(x + 1) dx + ln n,

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1.5

1

0.5

00 1 2 3 4 5 6 7

x

FIGURE A.1. Graphs of ln(x + 1) and ln x bounding the step function lnx.

where the value of∑n

k=1 ln k is represented by the area under the staircase in Figure A.1.Evaluating these integrals, we see that

n ln n − n + 1 < ln(n!) < n ln n − n + 1 + ln n,

e(n

e

)n

< n! < ne(n

e

)n

.

This gives us a pretty good idea of how fast n! grows, but because the summands areincreasing, our upper and lower bounds get further apart as n increases, unlike the situationwhen we estimated the rate of the growth of the harmonic series.

A Trick for Approximating Summations

To get a better approximation, we use a trick that is part of the repertoire of number theorywhere there are many summations that need to be approximated. We rewrite ln k as theintegral of 1/x from x = 1 to x = k and then interchange the integral and the summation:

n∑k=1

ln k =n∑

k=1

∫ k

1

dx

x

=∫ n

1

∑nk=x+1 1

xdx

=∫ n

1

n − xx

dx. (A.61)

We now split this integral into two pieces:∫ n

1

n − xx

dx =∫ n

1

n − x + 1/2

xdx +

∫ n

1

x − x − 1/2

xdx

= n ln n − n + 1 + 1

2ln n +

∫ n

1

x − x − 1/2

xdx. (A.62)

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x

108642

0.2

0.1

0

−0.1

−0.2

−0.3

−0.4

−0.5

FIGURE A.2. Graph of (x − x − 1/2)/x.

The integrand in the last line has a graph that oscillates about and approaches the x-axis(see Figure A.2). The limit of this integral as n approaches infinity exists because∣∣∣∣

∫ ∞

n

x − x − 1/2

xdx

∣∣∣∣ <

∣∣∣∣∫ n+1/2

n

x − x − 1/2

xdx

∣∣∣∣=

∣∣∣∣∫ 1/2

0

x − 1/2

n + xdx

∣∣∣∣= −1

2+

(n + 1

2

)(ln

(n + 1

2

)− ln n

),

(A.63)

which approaches 0 as n goes to infinity (see exercise A.4.1).We have proven that

ln(n!) = n ln n − n + 1

2ln n + 1 +

∫ ∞

1

x − x − 1/2

xdx + E(n), (A.64)

where

E(n) = −∫ ∞

n

x − x − 1/2

xdx

approaches 0 as n approaches infinity. Equivalently,

n! = C(n

e

)n √n eE(n) (A.65)

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where

ln C = 1 +∫ ∞

1

x − x − 1/2

xdx.

What is the value of C?

Evaluating C

Wallis’s formula comes to our aid in a very slick evaluation. Stirling’s formula implies that

(2n)!

n! · n!= C(2n/e)2n

√2n eE(2n)

C2(n/e)2nn e2E(n)

= 22n

C

√2

neE(2n)−2E(n). (A.66)

We solve for C and then do a little rearranging:

C = 22n(n!)2

(2n)!

√2√n

eE(2n)−2E(n)

= (2 · 4 · 6 · · · 2n)2

1 · 2 · 3 · · · 2n

2√2n

eE(2n)−2E(n)

= 2 · 4 · 6 · · · 2n

1 · 3 · 5 · · · (2n − 1)

2√2n

eE(2n)−2E(n)

= 2 · 4 · 6 · · · (2n − 2) · √2n

1 · 3 · 5 · · · (2n − 1)2 eE(2n)−2E(n). (A.67)

Looking back at Wallis’s work, we see from equation (A.10) that

2 · 4 · 6 · · · (2n − 2) · √2n

1 · 3 · 5 · · · (2n − 1)

approaches√

π/2 as n gets large. That means that the right side of equation (A.67)approaches 2

√π/2 = √

2π as n approaches infinity. Since the left side is independent ofn, the constant C must actually equal

√2π . We have proven Stirling’s formula:

n! = nne−n√

2πn eE(n), (A.68)

where E(n) is an error that approaches 0 as n gets large.

The Asymptotic Series for E (n)

Not long after deMoivre and Stirling published their formula for n! in 1730, both LeonardEuler and Colin Maclaurin realized that something far more general was going on, a formulafor approximating arbitrary series that today is called the Euler–Maclaurin formula. Eulerwrote to Stirling in 1736 describing this general formula. Stirling wrote back in 1738saying that Colin Maclaurin had also discovered this result. Euler’s proof was publishedin 1738, Maclaurin’s in 1742. Because it takes very little extra work, we shall develop theasymptotic series for E(n) in the more general context of the Euler-Maclaurin formula.

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We want to find a formula for∑n

k=1 f (k) where f is an analytic function for x > 0. Weset

S(n) =n∑

k=1

f (k)

and assume that S also can be defined for all x > 0 so that it is analytic. By Taylor’sformula, we have that

S(n + x) = S(n) + S ′(n)x + S ′′(n)

2!x2 + S ′′′(n)

3!x3 + · · · .

We set x = −1 and observe that

f (n) = S(n) − S(n − 1) = S ′(n) − S ′′(n)

2!+ S ′′′(n)

3!− S(4)(n)

4!+ · · · .

We want to invert this and write S ′(n) in terms of f and its derivatives at n. In principle,this is doable because

f ′(n) = S ′′(n) − S ′′′(n)

2!+ S(4)(n)

3!− S(5)(n)

4!+ · · ·

f ′′(n) = S ′′′(n) − S(4)(n)

2!+ S(5)(n)

3!− S(6)(n)

4!+ · · ·

f ′′′(n) = S(4)(n) − S(5)(n)

2!+ S(6)(n)

3!− S(7)(n)

4!+ · · ·

...

In other words, we want to find the constants a1, a2, a3, . . . such that

S ′(n) = f (n) + a1f′(n) + a2f

′′(n) + a3f′′′(n) + · · · . (A.69)

We substitute the expansions of the derivatives of f in terms of the derivatives of S intoequation (A.69). This tells us that

S ′(n) =∞∑

j=1

(−1)j−1 S(j )(n)

j !+

∞∑k=1

ak

∞∑j=k+1

(−1)j−k−1 S(j )(n)

(j − k)!

= S ′(n) +∞∑

j=2

(−1)j−1S(j )(n)

(1 +

j−1∑k=1

(−1)kak

(j − k)!

). (A.70)

This will be true if and only if

1 − a1

(j − 1)!+ a2

(j − 2)!− a3

(j − 3)!+ · · · + (−1)j−1 aj−1

1!= 0, j ≥ 2.

This equation should look familiar. Except for the sign changes, it is exactly the equalitythat we saw in equation (A.42) on page 287, an equality uniquely satisfied by the Bernoullinumbers divided by the factorials. In our case,

ak = (−1)kBk

k!.

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Since we know that B2m+1 = 0 for m ≥ 1, we have shown that

S ′(x) = f (x) +∞∑

k=1

(−1)kBk

k!f (k)(x)

= f (x) + 1

2f ′(x) +

∞∑m=1

B2m

(2m)!f (2m)(x). (A.71)

We do need to keep in mind that these have all been formal manipulations, what Cauchyreferred to as “explanations drawn from algebraic technique.” This derivation should beviewed as suggestive. In no sense is it a proof. In particular, there is no guarantee that thisseries converges.

Nevertheless, even when the series does not converge, it does provide useful approxi-mations. If we now integrate each side of equation (A.71) from x = 1 to x = n and thenadd S(1) = f (1), we get the Euler–Maclaurin formula.

Theorem A.1 (Euler–Maclaurin Formula). Let f be an analytic function for x > 0,then, provided the series converge, we have that

n∑k=1

f (k) =∫ n

1f (x) dx + 1

2f (n) +

∞∑m=1

B2m

(2m)!f (2m−1)(n)

+ 1

2f (1) −

∞∑m=1

B2m

(2m)!f (2m−1)(1). (A.72)

When we set f (x) = ln x and use the fact that we know that the constant term is ln(2π )/2,this becomes Stirling’s Formula:

ln(n!) =n∑

k=1

ln k = n ln n − n + 1

2ln n + 1

2ln(2π ) + E(n), (A.73)

where E(n) can be approximated by the asymptotic series,

E(n) ∼∞∑

m=1

B2m

(2m)(2m − 1)n2m−1. (A.74)

Difficulties

Does the fact that the constant term is ln(√

2π ) mean that

1 −(

B2

1 · 2+ B4

3 · 4+ B6

5 · 6+ · · ·

)= ln

(√2π

)= .9189385. . . ?

Hardly. If we try summing this series, we find that it does not approach anything. The firstfew Bernoulli numbers are small, but as we saw in the last section, they start to grow. They

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Table A.2. Partial sums of de Moivre’s series.

N 1 − ∑Nm=1 B2m/2m(2m − 1)

1 0 .9166667

2 0 .9194444

3 0 .9186508

4 0 .9192460

5 0 .9184043

6 0 .9203218

7 0 .9139116

8 0 .9434622

9 0 .763818

10 2 .15625

11 −11 .2466

12 145 .602

13 −2047 .5

14 34061 .3

15 −657411 .0

grow faster than 2(2m)!/(2π )2m. Table A.2. lists the partial sums of

1 −N∑

m=1

B2m

2m(2m − 1).

The first few values look good—up to N = 4 they seem to be approaching ln√

2π—butthen they begin to move away and very quickly the series is lurching out of control.

What about the error function:

E(n) ∼ B2

1 · 2n+ B4

3 · 4n3+ B6

5 · 6n5+ · · · ;

does it converge? In exercises A.4.2 and A.4.3, the reader is urged to experiment with thisseries. What you should see is that no matter how large n is, eventually this series will startto oscillate with increasing swings. But that does not mean that it is useless. If you take thefirst few terms, say the first two, then

nne−n√

2πn e1/(12n)−1/(360n3)

is a better approximation to n! than just

nne−n√

2πn.

Something very curious is happening. As we take more terms, the approximation keepsgetting better up to some point, and then it starts to get worse as the series moves into itsuncontrolled swings. This is what we mean by an asymptotic series. Even though it does

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not converge, it does give an approximation to the quantity in question. How many termsof the asymptotic series should you take? That depends on n. As n gets larger, you can gofarther. Infinite series do strange things.

Exercises

The symbol��


available in theWeb Resources at www.macalester.edu/aratra.

A.4.1. Show that

−1

2+

(n + 1

2

)(ln

(n + 1

2

)− ln n

)

= −1

2+ 1

2ln

(1 + 1

2n

)+ 1

2ln

([1 + 1

2n

]2n)

.

Use this identity to prove (see equation (A.63)) that

limn→∞

∫ ∞

n

x − x − 1/2

xdx = 0.

A.4.2.��

��M&M Evaluate

nne−n√

2πn e1/(12n)−1/(360n3)

for n = 5, 10, 20, 50, and 100 and compare it to n!.

A.4.3.��

��M&M To see how many terms of the asymptotic series we should take, find the

summand in the asymptotic series that is closest to zero and stop at that term. For each ofthe values n = 5, 10, 20, 50, and 100, find which summand is the smallest in absolute value.Estimate the function of n that describes how many terms of the asymptotic series shouldbe taken for any given n. How accurately does this approximate n! when the number ofterms is chosen optimally?

A.4.4. Use the approximation

|B2m| ≈ 2(2m)!

(2π )2m

to check your estimate from exercise A.4.3.

A.4.5.��

��M&M Using the Euler–Maclaurin formula with f (x) = 1/x gives us an ap-

proximation for the harmonic series. Show that the constant term of the Euler–Maclaurinformula is

1

2+

∞∑m=1

B2m

2m.

Determine how useful this is in approximating the value of Euler’s γ .

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A.4.6.��

��M&M Use the Euler–Maclaurin formula to show that

n∑k=1

1

k= ln n + 1

2n+ γ − H (n)

where H (n) can be approximated by the asymptotic series

H (n) ∼∞∑

m=1

B2m

2m n−2m.

For each of the values n = 5, 10, 20, 50, and 100, find which summand is the smallest inabsolute value. Estimate the function of n that describes how many terms of the asymptoticseries should be taken for any given n. How accurately does this approximate the harmonicseries when the number of terms is chosen optimally?

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Appendix B

Bibliography

Birkhoff, Garrett, A Source Book in Classical Analysis, Harvard University Press, Cambridge, MA,1973.

Bonnet, Ossian, “Remarques sur quelques integrales definies,” Journal de Mathematiques Pures etAppliquees, vol. 14, August 1849, pages 249–256.

Borwein, J. M., P. B. Borwein, and D. H. Bailey, “Ramanujan, Modular Equations, and Approxima-tions to Pi or How to Compute One Billion Digits of Pi,” The American Mathematical Monthly,vol. 96, no. 3, March 1989, pages 201–219.

Cauchy, Augustin-Louis, Cours d’Analyse de l’Ecole Royale Polytechnique, series 2, vol. 3 in Œuvrescompletes d’Augustin Cauchy, Gauthier-Villars, Paris, 1897.

Cauchy, Augustin-Louis, Lecons sur le calcul differentiel, series 2, vol. 4 in Œuvres completesd’Augustin Cauchy, Gauthier-Villars, Paris, 1899.

Cauchy, Augustin-Louis, Resume des Lecons donnees a l’Ecole Royale Polytechnique sur le calculinfinitesimal, series 2, vol. 4 in Œuvres completes d’Augustin Cauchy, Gauthier-Villars, Paris,1899.

Dijksterhuis, E. J., Archimedes, translated by C. Dikshoorn, Princeton University Press, Princeton,1987.

Dirichlet, G. Lejeune, Werke, reprinted by Chelsea, New York, 1969.Dunham, William, Journey through Genius: the great theorems of mathematics, John Wiley & Sons,

New York, 1990.Edwards, C. H., Jr., The Historical Development of the Calculus, Springer–Verlag, New York, 1979.Euler, Leonhard, Introduction to Analysis of the Infinite, books I & II, translated by John D. Blanton,

Springer–Verlag, New York, 1988.Gauss, Carl Friedrich, Werke, vol. 3, Koniglichen Gesellschaft der Wissenschaften, 1876.Grabiner, Judith V., The Origins of Cauchy’s Rigorous Calculus, MIT Press, Cambridge, MA, 1981.Grattan-Guinness, Ivor, Convolutions in French Mathematics, 1800–1840, vols. I, II, III, Birkhauser

Verlag, Basel, 1990.Grattan-Guinness, Ivor, The Development of the Foundations of Mathematical Analysis from Euler

to Riemann, MIT Press, Cambridge, MA, 1970.

303

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304 Appendix B. Bibliography

Grattan-Guinness, Ivor, Joseph Fourier, 1768–1830, MIT Press, Cambridge, MA, 1972.Hawkins, Thomas, Lebesgue’s theory of integration: its origins and development, 2nd edition,

Chelsea, New York, 1975.Hermite, Charles and Thomas Jan Stieltjes, Correspondance d’Hermite et de Stieltjes, B. Baillaud

and H. Bourget, eds., Gauthier-Villars, Paris, 1903–1905.Kaczor, W. J., and M. T. Nowak, Problems in Mathematical Analysis, vols. I, II, III, Student

Mathematical Library vols. 4, 12, 21, American Mathematical Society, Providence, RI, 2000–2003.

Kline, Morris, Mathematical Thought from Ancient to Modern Times, Oxford, 1972.Lacroix, S. F., An Elementary Treatise on the Differential and Integral Calculus, translated by

Babbage, Peacock, and Herschel with appendix and notes, J. Deighton and Sons, Cambridge,1816.

Lacroix, S. F., Traite Elementaire de Calcul Differentiel et de Calcul Integral, 4th edition, Bachelier,Paris, 1828.

Medvedev, Fyodor A., Scenes from the History of Real Functions, translated by Roger Cooke,Birkhauser Verlag, Basel, 1991.

Olsen, L., A new proof of Darboux’s theorem, American Mathematical Monthly, vol. 111 (2004),pp. 713–715.

Poincare, Henri, “La Logique et l’Intuition dans la Science Mathematique et dans l’Enseignement,”L’Ensiegnement mathematique, vol. 1 (1889), pages 157–162.

Preston, Richard, “The Mountains of Pi,” The New Yorker, March 2, 1992, pages 36–67.Riemann, Bernhard, Gesammelte Mathematische Werke, reprinted with comments by Raghavan

Narasimhan, Springer–Verlag, New York, 1990.Rudin, Walter, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, New York, 1976.Serret, J.-A., Calcul Differentiel et Integral, 4th edition, Gauthier-Villars, Paris, 1894.Struik, D. J., A Source Book in Mathematics 1200–1800, Princeton University Press, Princeton, 1986.Truesdell, C., “The Rational Mechanics of flexible or elastic bodies 1638–1788,” Leonardi Euleri

Opera Omnia, series 2, volume 11, section 2, Orell Fussli Turici, Switzerland, 1960.Van Vleck, Edward B., “The influence of Fourier’s series upon the development of mathematics,

Science, N.S. vol. 39, 1914, pages 113–124.Weierstrass, Karl Theodor Wilhelm, Mathematische werke von Karl Weierstrass, 7 volumes, Mayer

& Muller, Berlin, 1894–1927.Whittaker, E. T., and G. N. Watson, A Course of Modern Analysis, 4th ed., Cambridge University

Press, Cambridge, 1978.

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Appendix C

Hints to Selected Exercises

Exercises which can also be found in Kaczor and Nowak are listed at the start of eachsection following the symbol

��

��KN . The significance of 3.1.2 = II:2.1.1 is that exercise

3.1.2 in this book can be found in Kaczor and Nowak, volume II, problem 2.1.1.

2.1.6 Use the fact that 1 + x + x2 + · · · + xk−1 = (1 − xk)/(1 − x).

2.1.8 If you stop at the kth term, how far away are the partial sums that have moreterms?

2.2.1 Use the fact that 1 + x + x2 + · · · + xk−1 = (1 − xk)/(1 − x).

2.2.4 Take the first 3k + 3 terms and rewrite this finite summation as (1 + 2−3 + 2−6 +· · · + 2−3k) + (2−1 + 2−4 + 2−7 + · · · + 2−(3k+1)) − (2−2 + 2−5 + 2−8 + · · · +2−(3k+2)).

2.2.6 Use the work from exercise 2.2.5.

2.2.8 Find an expression in terms of r and s for a partial sum of a rearranged series thatuses the first r positive summands and the first s negative summands. Show thatyou can get as close as desired to the target value provided only that r and s aresufficiently close, regardless of their respective sizes.

2.3.4 Take pairs of terms and assume that regrouping of the summands is allowed.

2.3.5 Take the tangent of each side and use the formula

tan(x + y) = tan x + tan y

1 − tan x tan y.

2.3.8 Explain what happens when you take a = −1 in equation (2.20).

305

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306 Appendix C. Hints to Selected Exercises

2.4.10 Begin by separating the summands according to the total number of digits in thedenominator: (

1

1+ 1

2+ · · · + 1

8

)

+(

1

10+ 1

11+ · · · + 1

18+ 1

20+ · · · + 1

88

)

+(

1

100+ 1

101+ · · · + 1

888

)

+ · · · +(

1

10k+ 1

10k + 1+ · · · + 1

8(10k+1 − 1)/9

)+ · · · .

a. There are 8 summands in the first pair of parentheses. Show that there are72 = 8 · 9 in the second, 648 = 8 · 92 in the third, 5832 = 8 · 93 in the fourth,and that in general there are 8 · 9k in the k + 1st. Hint: what digits are youallowed to place in the first position? in the second? in the third?

b. Each summand in a given pair of parentheses is less than or equal to the firstterm. Show that the sum of the terms in the k + 1st parentheses is strictly lessthan 8 · 9k/10k , and thus our series is bounded by

8

1+ 8 · 9

10+ 8 · 92

102+ 8 · 93

103+ · · · .

c. Evaluate the geometric series given above.

2.4.14 Show that

1 + 1

3+ · · · + 1

2n − 1=

(1 + 1

2+ 1

3+ · · · + 1

2n

)− 1

2

(1 + 1

2+ 1

3+ · · · + 1

n

).

2.4.16 Show that∞∑

m=n

1

m2<

1

n2+

∫ ∞

n

dx

x2.

2.4.18 Work with the fraction of the road that you have covered. The first step takes you1/2000th of the way, the next step 1/4000th, the third 1/6000th.

2.5.3 Integration by parts.

2.5.15 Is c the same for all values of n?

2.6.5 How can you use the fact that e−1/x2has all of its derivatives equal to 0 at x = 0?

��

��KN 3.1.2 = II:2.1.1, 3.1.3 = II:2.1.2, 3.1.4 = II.2.1.3, 3.1.5 = II.2.1.4, 3.1.6 = II:2.1.5,

3.1.15 = II:2.1.8, 3.1.16 = II:2.1.10b, 3.1.17 = II:2.1.9b, 3.1.18 = II:2.1.12,3.1.19 = II:2.1.13, 3.1.20 = II:2.1.13.

3.1.2 For those functions with |x|, consider x > 0 and x < 0 separately. Use the definitionof the derivative at x = 0. For functions with x, consider x ∈ Z separately. Usethe definition of the derivative at x ∈ Z.

3.1.3 logx a = (ln a)/(ln x).

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Appendix C. Hints to Selected Exercises 307

3.1.4 Consider the transition points: Is the function continuous there? If it is, rely on thedefinition of the derivative.

3.1.15 xf (a) − af (x) = (x − a)f (a) − a(f (x) − f (a)). The same trick will work inpart (b).

3.1.16 Rewrite the fraction asf (x)ex − f (0)e0

x − 0÷ f (x) cos x − f (0) cos 0

x − 0.

3.1.19 (b) Consider f (x) = x2 sin(1/x), x = 0.

3.1.20 Rewrite

f (xn) − f (zn)

xn − zn

= f (xn) − f (a)

xn − a· xn − a

xn − zn

+ f (zn) − f (a)

zn − a· a − zn

xn − zn

.

Show that this must lie betweenf (xn) − f (a)

xn − aand

f (zn) − f (a)

zn − a. Why doesn’t

this approach work when xn and zn lie on the same side of a?

3.2.9 Show that there is a k between 0 and �x for which

f (x0 + 2�x) − 2f (x0 + �x) + f (x0)

�x2= f ′(x0 + 2k) − f ′(x0 + k)

k.

Define g(h) = f ′(x0 + k + h), so that

f (x0 + 2�x) − 2f (x0 + �x) + f (x0)

�x2= g(k) − g(0)

k.

Use the generalized mean value theorem a second time.��

��KN 3.3.4 = II:1.2.1, 3.3.5 = II:1.2.2, 3.3.6 = II:1.2.3, 3.3.7 = II:1.2.4, 3.3.8 = II:1.3.3,

3.3.9 = II:1.3.4, 3.3.10 = II:1.3.7, 3.3.11 = II:1.3.10, 3.3.12 = II:1.3.11, 3.3.13 =II:1.3.12, 3.3.14 = II:1.2.6, 3.3.15 = II:1.2.7, 3.3.34 = II:2.1.23.

3.3.3 What fractions in (√

2 − 1,√

2 + 1) have denominators ≤ 5?

3.3.4 Where is sin x = 0?

3.3.6 For rational numbers, f (p/q) = p/(q + 1). What is the difference between p/q

and f (p/q)?

3.3.8 Apply the intermediate value theorem to the function g defined by g(x) = f (x) − x.

3.3.11 Consider g(x) = f (x + 1) − f (x), 0 ≤ x ≤ 1.

3.3.12 f (2) − f (0) = (f (2) − f (1)) + (f (1) − f (0)).

3.3.13 Start by explaining why f (i + 1) − f (i) cannot be strictly positive for all integervalues of i ∈ [0, n − 1].

3.3.14 Consider separately the cases x2 ∈ N, x2 ∈ N.

3.3.15 Consider separately the cases x ∈ N, x ∈ N.

3.3.17

| sin(x + h) − sin x| = |(sin x)(cos h − 1) + (cos x)(sin h)|≤ | sin x| · | cos h − 1| + | cos x| · | sin h|≤ | cos h − 1| + | sin h|.

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Graph | cos h − 1| + | sin h| and find an interval containing h = 0 where this func-tion is less than 0.1.

3.3.18

|(x + h)2 − x2| = |2xh + h2|= |h| · |2x + h|≤ |h| · |2 + h|.

3.3.22 Use the power series for ln(1 + x) to show that ln(1 + x) < x for x > 0, andtherefore if a > b > 0 then

ln a − ln b = ln

(1 + a − b

b

)<

a − b

b.

3.3.27 Since f is continuous on any interval that does not contain 0, you only need toprove that if c1 ≤ 0 ≤ c2 and if A is between f (c1) and f (c2), then there is some c,c1 < c < c2, for which f (c) = A.

3.3.28 When does a small change in x result in a change in f (x) that cannot be madearbitrarily small?

3.3.33f (x)

g(x)− f (c)

g(c)= f (x)

g(x)· g(c) − g(x)

g(c)+ f (x) − f (c)

g(c).

��

��KN 3.4.6 = I:1.1.7–12, 3.4.23 = II:2.1.24, 3.4.24 = II:2.1.25, 3.4.25 = II:2.1.26,

3.4.26 = II:2.1.27, 3.4.27 = II:2.1.28, 3.4.28 = II:2.1.29, 3.4.29 = II:1.2.17, 3.4.30= II:2.2.1.

3.4.1 The function cannot be continuous.

3.4.2 The domain cannot be a closed, bounded interval.

3.4.4 Prove the contrapositive. Explain why if A and B have opposite signs, then|A − B| ≥ |A|.

3.4.5 What exactly is the technical statement that corresponds to this condition? Forevery pair (ε, δ), what must exist? What happens for very large values of ε? Doesthis technical statement of existence make sense as the definition of a verticalasymptote?

3.4.6 (h) If you hold n constant, what value of m, 1 ≤ m ≤ 2n − 1 maximizes thisexpression? (j) How close can this expression get to 1? (n) Find the minimumvalue of x/y + 4y/x in the first quadrant. (o) Set m = kn and find the values of k

that maximize, minimize the resulting expression. (r) Find the maximum value ofxy/(1 + x + y) in the first quadrant.

3.4.11 Given the sequences x1 ≤ x2 ≤ · · · ≤ xk ≤ · · · < · · · ≤ yk ≤ · · · ≤ y2 ≤ y1, let c

be the least upper bound of {x1, x2, x3, . . .}. Prove that c ∈ [xk, yk] for every k.

3.4.12 Let S be the set of all x for which a ≤ x < x2 and g(x) ≥ g(x2). If S is not empty,then it is bounded and so has a least upper bound, call it B ≤ x2. Note that B mayor may not be in S.

a. Use the continuity of g to prove that g(B) ≥ g(x2).b. Use the fact that we can make |g′(x2) − (g(x2) − g(x))/(x2 − x)| as small as we

wish by taking x sufficiently close to x2 to prove that B < x2.

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c. Use the fact that B < x2, g(B) ≥ g(x2), and g′(x) ≥ 0 to prove that there areelements of S that are strictly larger than B. This implies that B is not an upperbound and so S must be the empty set.

3.4.13 Assume that we can find a pair (x1, x2), a ≤ x1 < x2 ≤ b, for which f (x1) > f (x2).It follows that there is a positive number α such that

f (x2) − f (x1)

x2 − x1< −α < 0.

3.4.16 If | sin(1/c) − c−1 cos(1/c)| > 1, then c cannot be in the range of g.

3.4.19 Let c = e(−8n+1)π/4, n ∈ N, and try to find an x for which sin(ln c) + cos(ln c) =sin(ln x). What are other values of c ∈ (0, 1) that do not correspond to any value ofx?

3.4.20 Recall Theorem 3.4.

3.4.22 Start by proving that between any two real roots of P there must be at leastone real root of P ′. If a polynomial P has a root of order n > 1 at x = a, thenP (x) = (x − a)nQ(x) where Q is a polynomial, Q(a) = 0. The derivative P ′(x) =n(x − a)n−1Q(x) + (x − a)nQ′(x) has a root of order n − 1 at x = a.

3.4.24 If f has a local maximum at x = c, then f ′−(c) ≥ 0 (why?). Let d = sup{x | f (x) >

f (c)/2}. Show that f ′−(d) ≤ 0. Complete the proof.

3.4.25 Use the idea that helped us prove the mean value theorem.

3.4.26 Use the result of exercise 3.4.25.

3.4.27 Use the result of exercise 3.4.26.

3.4.28 Let c = inf{x ∈ (a, b) | f (x) = 0}. Why is this set non-empty?

3.4.30 Consider f (x)eαx .��

��KN 3.5.2 = II2.3.6, 3.5.3 = II:2.3.7, 3.5.4 = II:2.2.11, 3.5.17 = II:2.3.8, 3.5.18 =

II:2.3.34.

3.5.1 Prove the contrapositive.

3.5.2 Consider negative as well as positive values of x.

3.5.4 Consider derivatives.

3.5.11 Rewrite the limit as

limx→0

e−1/x2

x= lim

x→0

x−1

e1/x2 .

3.5.17 Differentiate each side of

[f (x) − f (0)] g′(θ (x)) = [g(x) − g(0)] f ′(θ (x))

with respect to x, collect the terms that involve θ ′(x) on one side, divide both sidesby x, and then take the limit of each side as x → 0+.

3.5.18 Rewrite f (x)g(x) as eg(x) ln(f (x)).��

��KN 4.1.16 = I:3.4.10a.

4.1.1 In this case we know that the partial sum to n terms differs from the value of theseries by exactly (1/2)n/(1 − 1/2) = 1/2n−1.

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4.1.3 For an alternating series with summands whose absolute values are decreasingtoward zero, the partial sum approximation differs from the target value by at mostthe absolute value of the next term.

4.1.7 A function f is even if and only if f (−x) = f (x).

4.1.14 Are the hypotheses of the Alternating Series Test satisfied?

4.1.15 Combine consecutive summands with the same sign.

4.1.16 Write out enough terms that you get a feel for εn. Combine consecutive summandswith the same sign.

��

��KN 4.2.4 = I:2.2.50, 4.2.5 = I:3.2.1, 4.2.6 = I:3.4.1, 4.2.7 = I:3.4.13, 4.2.29 = I:3.2.17.

4.2.1 How do you know that for all n sufficiently large, |an| < 1?

4.2.2 Use the definition of convergence. To what value does this series converge? Showthat given any ε > 0, there is some N so that all of the partial sums past the N thdiffer from this value by less than ε.

4.2.4 (a) The arctangent function is bounded. (d) To test for absolute convergence, com-bine consecutive pairs of terms. (f) Show that n/(n + 1)2 > 1/(n + 3).

4.2.5 (a)√

n2 + 1 − 3√

n3 + 1 = n(1 + 1/n2)1/2 − n(1 + 1/n3)1/3. Use a Taylor poly-nomial approximation. (b) Show that limn→∞(n/(n + 1))n+1 = limn→∞(1 +1/n)−n−1 = 1/e. (c) Use a Taylor polynomial approximation. (f) Use the roottest.

4.2.6 (b) When does the rational function of a have absolute value less than 1? (c) Usethe root test.

4.2.8 Show that if n ≥ N , then |an| ≤ |aN | αn−N and so

n√

|an| ≤ αn√

|aN |/αN .

4.2.22 Prove and then use the fact that for k ≥ 2:

1

k ln(k ln 2)>

1

2k ln k.

4.2.24 Show that n1+(ln ln n+ln ln ln n)/ ln n = n(ln n)(ln ln n).

4.2.28 Use Stirling’s formula in place of the factorials.

4.2.29 (a) 2n/2 > 2n for n > 8.��

��KN 4.3.1 = I:3.3.2, 4.3.2 = I:3.3.3, 4.3.3 = I:3.3.6, 4.3.4 = I:3.3.7, 4.3.18 = I:2.4.11,

4.3.20 = I:2.4.15, 4.3.21 = I:2.4.19, 4.3.22 = I:2.4.20, 4.3.23 = I:2.4.26, 4.3.24 =II:1.2.18, 4.3.25 = II:1.2.19, 4.3.26 = II:1.2.20.

4.3.1 (a) Using the limit ratio test, we have absolute convergence if

1 > limn→∞

(n + 1)3|x|n+1

n3|x|n = |x|.

Check for convergence at x = +1 and at x = −1. (d) Using the lim sup root test,we have absolute convergence when

1 > limn→∞

∣∣(2 + (−1)n)x∣∣ = 3 |x|.

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Check for convergence at x = 1/3 and at x = −1/3. (f) Rewrite this summationso that the power of x is the index of summation:

∑∞n=1 2nxn2 = ∑∞

m=1 amxm

where am = 2√

m if m is a perfect square, am = 0 if m is not a perfect square.Now use the lim sup root test. (h) Use the lim sup root test and remember thatlimn→∞(1 + 1/n)n = e.

4.3.2 (b) Use the lim sup root test. This converges absolutely if

1 > limn→∞

(n

n + 1

)1/n ∣∣∣∣2x + 1

x

∣∣∣∣ =∣∣∣∣2x + 1

x

∣∣∣∣ .Check what happens when |(2x + 1)/x| = 1, i.e. when 2x + 1 = x and when 2x +1 = −x.

4.3.3 (a) This implies that limn→∞ |anxn|1/n = |x| limn→∞ L1/nn−α/n.

4.3.4 (a) Since the radius of convergence is R, we know that limn→∞ n√|anxn| = 1/R. It

follows that limn→∞ n√|2nanxn| = 2/R. (c) Use Stirling’s formula.

4.3.6 Do the summands approach 0 when |x| equals the radius of convergence?

4.3.7 Use Stirling’s formula.

4.3.8 Use Stirling’s formula.

4.3.11 Use the ratio test.

4.3.12 This is a hypergeometric series.

4.3.13 Show that

1 · 3 · 5 · · · (2n − 1) = (2n)!

2 · 4 · 6 · · · 2n= (2n)!

2n · n!.

If we ignore F (n) in equation (4.15), how close is this approximation whenn = 10? = 20? = 100?

4.3.14 Either use the result of exercise 4.3.13 together with Stirling’s formula, or use thefact that limk→∞(1 + 1/k)k = e.

4.3.18 (a) Let α = p/q where gcd(p, q) = 1. The answer is in terms of q.

4.3.20 First show that it is enough to prove the last two inequalities. Use the equivalentdefinition on the lim sup found in exercise 4.3.19.

4.3.21 Use the result from exercise 4.3.20.

4.3.23 Show that it is enough to prove the last inequality. Let

A = limn→∞

an+1

an

.

choose an ε > 0 and a response N such that for all n ≥ N , an+1/an < A + ε. Showthat for n ≥ N , an < AN (A + ε)n−N . Take the limit as n approaches infinity of thenth root of this upper bound.

4.4.2 Use equation (4.27).

4.4.4 At x = 1/2,∣∣∣∣∣n∑

k=1

(−1)k−1 cos[(2k − 1)π/4]

∣∣∣∣∣ =∣∣∣∣1 − (−1)n cos(πn/2)

2 cos(π/4)

∣∣∣∣ ≤√

2.

By equation (4.23), |Tn − Tm| ≤ 2√

2/(2m + 1).

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4.4.12 (c) Use Dirichlet’s test with bk = ckRk and ak = (x/R)k = eikθ .

5.1.1 The regrouped series has initial term 2/3 and ratio 1/9.

5.1.10 Choose any ten terms from the original series to be the first ten terms of therearranged series. Can the remaining terms be arranged so that the resulting seriesconverges to the target value? Does it matter in what order we put the first tenterms?

5.2.1 Show that the function represented by this series is not continuous.

5.2.4 For all x ∈ [−π, π ], this is an alternating series and therefore the sum of the firstN terms differs from the value of the series by an amount whose absolute value isless than |x|2N+1/(2N + 1)!.

5.2.7

−1 + 1

4− 1

9+ 1

16− 1

25+ 1

36− · · ·

= −(

1 + 1

4+ 1

9+ 1

16+ · · ·

)+ 2

(1

4+ 1

16+ 1

36+ · · ·

).

5.2.8 What is the power series expansion of ln(1 − x)?

5.2.9 Use the partial sums

Sn(x) =n∑

k=1

xk

k2

and the fact that

|Li2(1) − Li2(x)| ≤ |Li2(1) − Sn(1)| + |Sn(1) − Sn(x)| + |Sn(x) − Li2(x)|.��

��KN 5.3.2 = II:3.2.29.

5.3.1 Consider functions for which f ′k(x) = 0 for all k and all x.

5.3.2 Show that it converges at x = 0. Explain why it is enough to show that for any N

and any x,

∞∑n=N+1

1

n2 + x2≤

∞∑n=N+1

1

n2,

and then explain why this is true.

5.3.4 Use equation (4.28) from page 166.

5.3.6 Show that

x2

(1 + kx2)(1 + (k − 1)x2)= 1

1 + (k − 1)x2− 1

1 + kx2.

5.3.7 Show that |G(x) − Gn(x)| = | sin x|/(1 + nx2). Given ε > 0, find x0 > 0 so that|x| ≤ x0 implies that | sin x|/(1 + nx2) ≤ | sin x| ≤ | sin x0| < ε. For this valueof x0, find an N so that n ≥ N and |x| ≥ x0 implies that | sin x|/(1 + nx2) ≤1/(1 + nx2) ≤ 1/(nx2

0 ) < ε. Explain why this proves that the convergence isuniform.

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��

��KN 5.4.1 = II:3.2.2, 5.4.12 = II:3.2.14.

5.4.1 (a) For each n, find the supremum of {n2x2e−n2|x| | x ∈ R}. (c) Show that forany n, there is an x > 0 for which the nth summand is equal to 2n and all ofthe summands beyond the nth are ≥ 0. Explain why you cannot have uniformconvergence if this is true. (f ) Show that arctan x + arctan x−1 = π/2, and thereforethe nth summand is equal to arctan(1/(n2(1 + x2))). Explain why this is less thanor equal to 1/(n2(1 + x2)).

5.4.2 Use the fact that a1 + 2a2x + 3a3x2 + · · · converges uniformly and absolutely on

(0, R).

5.4.5 Find the values of N that are responses to ε at x = a, over the open interval (a, b),and at x = b.

5.4.6 Consider summands that are not continuous at x = a or x = b.

5.4.10 Show that for any n ≥ 2:

∞∑k=2

sin kπ/n

ln k≥

2n∑k=2

sin kπ/n

ln k

= − sin π/n

ln(n + 1)+

n∑k=2

sin(kπ/n)

(1

ln k− 1

ln(k + n)

)

≥ − sin π/n

ln(n + 1)+

n∑k=2

sin(kπ/n)ln 2

(ln n)(ln 2n)

= f (n) ln 2

(ln n)(ln 2n)− sin(π/n)

(1

ln(n + 1)+ ln 2

(ln n)(ln 2n)

),

where

f (n) = sin(π/n) + sin(2π/n) + sin(3π/n) + · · · + sin(nπ/n).

Use equation (5.64) to show that

f (n) = sin(π/n)

1 − cos(π/n).

5.4.12 (c) Show that 2 sin(n2x) sin(nx) = cos(n(n − 1)x) − cos(n(n + 1)x).(d) Rewrite the summation as

∞∑n=1

sin(nx)

n

(arctan(nx) − π

2

)+ π

2

∞∑n=1

sin(nx)

n.

In the first sum, let bn(x) = π/2 − arctan(nx). (e) Rewrite the summation as∞∑

n=1

(−1)n+1

nx−a/2n−a/2.

5.4.13 Consider∑∞

k=1 xk − xk−1 on [0, 1].

6.1.4 Show that f (x) = f (−x), g(x) = −g(−x).

6.1.5 If f is even and g is odd, then F (−x) = f (x) − g(x).

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6.1.6 If the Fourier series converges at x = 0, then∑∞

k=1 ak converges, and therefore thepartial sums of

∑∞k=1 ak are bounded.

6.1.9 Find an algebraic expression for this function on (−1, 1).

6.1.10 Uniform convergence means that you are allowed to interchange integration andinfinite summation.

6.1.18 Change variables using t = α + β − u and let h(u) = g(α + β − u). Show that h

is nonnegative and increasing on [α, β].

6.2.2 Where is the graph of y = x3 − 2x2 + x increasing? Where is it decreasing? Whereis the slope steepest?

6.2.4 Fix ε > 0. Put a bound on the error contributed by using an approximating sum overthe interval [0, ε]. Use the fact that sin(1/x) is continuous on the interval [ε, 1].

6.2.6 Use the mean value theorem.

6.2.7 We need a bounded, differentiable function whose derivative is not bounded.

6.2.11 Use the definition of differentiability. You must show that

limx→x0

∣∣∣∣∣∫ x

x0f (t) dt

x − x0− f (x0)

∣∣∣∣∣ = 0.

6.2.12 Consider Theorem 3.14.��

��KN 6.3.8 = III:1.1.7, 6.3.9 = III:1.1.6, 6.3.10 = III:1.1.14.

6.3.2 Show that given any σ > 0, there is a response δ so that for any partition withsubintervals of length < δ, the variation is less than σ .

6.3.7 Fix a variation σ . Can we limit the sum of the lengths of the intervals on which thevariation exceeds σ?

6.3.8 Where is this function discontinuous? How large is the variation at the points ofdiscontinuity?

6.3.10 (d) 1/(n + k) = (1/2n)(2/(1 + k/n). (f ) First show that the function of n is equalto e

∑nk=1(1/n) ln(1+k/n).

6.3.11 The summation is an approximation using a partition with infinitely many intervalsof the form [qn+1, qn]. Show that for any ε > 0, we can find a Riemann sum withintervals of length less than 1 − q that differs from our infinite summation by lessthan ε.

6.3.13∞∑

n=1

1

(2n − 1)2=

∞∑n=1

1

n2−

∞∑n=1

1

(2n)2.

6.3.20 Note that at points of discontinuity, the function decreases. Otherwise, it is anincreasing function. Show that if we approximate f (x) with

∑100n=1((nx))/n2, then

we are within 1/200 of the correct value. Now explain why it follows that if0 ≤ x < y ≤ 1, then

f (y) − f (x) <

100∑n=1

ny − nx

n2+ 1

100< (y − x)( ln(101) + γ ) + 1

100.

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6.3.16 Write f (x) = fN (x) + RN (x) where

RN (x) =∞∑

n=N+1

((nx))

n2.

Given an ε > 0, the task is to show how to find a response δ such that for 0 < ν < δ,

|fN (x + ν) + RN (x + ν) − fN (x + 0) − RN (x + 0)| < ε.

6.3.19 Let σ (k) be the sum of the divisors of k, and set k = 2ak1 where k1 is odd. Showthat ψ(k) = (2a+1 − 3)σ (k1). It is known that

limσ (k)

k ln ln k= eγ .

This is Gronwall’s Theorem, published in 1913.

6.3.20 Show that

g(1/5) = 1

5

∞∑n=0

1

5n + 1+ 2

5

∞∑n=0

1

5n + 2− 2

5

∞∑n=0

1

5n + 3− 1

5

∞∑n=0

1

5n + 4

=∞∑

n=0

125n2 + 125n + 26

5(5n + 1)(5n + 2)(5n + 3)(5n + 4).

6.3.21 Let x = p/q, gcd(p, q) = 1. Let m = (q − 1)/2. Show that

g(x) =m∑

k=1

((kx))

k+

m∑k=−m

((kx))∞∑

n=1

1

qn + k

=m∑

k=1

((kx))

(1

k− 2k

∞∑n=1

1

q2n2 − k2

).

6.4.1 Consider Theorem 3.14.

6.4.10 Use the fact that αm is an integer that is odd when m is even and even when m isodd.

A.1.2 Start with∫ 1

0

(1 − x1/p

)q =∫ 1

0

(1 − x1/p

)q−1dx −

∫ 1

0

(1 − x1/p

)q−1x1/p dx

=∫ 1

0

(1 − x1/p

)q−1dx

+ p

∫ 1

0

(1 − x1/p

)q−1(−x(1−p)/p

p

)x dx,

and then use integration by parts on the second integral.

A.1.4 Use the substitution u = (1 − x1/p)q .

A.1.8 Using equations (A.6) and (A.12), we see that

f (p, q) + p + q

qf (p, q − 1) = p + q

pf (p − 1, q).

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A.1.12 The values are undefined when p or q is a negative integer, but it is defined forother negative values of p and q.

A.1.13 Show that f (2/3, k) = (5 · 8 · 11 · · · (3k + 2))/(3 · 6 · 9 · · · (3k)). Show thatf (2/3, 1/3 + k) = f (2/3, 1/3) (6 · 9 · 12 · · · (3k + 3))/(4 · 7 · 10 · · · (3k + 1)).

A.2.4 Use the fact that equation (A.16) defines Bn(x). Show that∫ k+1k

Bn(1 − x) dx = (−1)n∫ k+1k

Bn(x) dx.

A.2.5 Use equation (A.33) from the previous exercise and equation (A.29).

A.3.6 Use the fact that∞∑

j=1

1

j 2

∞∑k=1

1

k4=

∑j =k

1

j 2 k4+

∞∑k=1

1

k6.

Find the coefficients of the summmations on the right side: ∞∑

j=1

1

j 2

3

= ? ×∑

1≤j<k<l<∞

1

j 2k2l2+ ? ×

∑j =k

1

j 2 k4+ ? ×

∞∑k=1

1

k6.

A.3.9 Rather than trying to evaluate 100!, find

A = ln

(2 · (2m)!

(2π )2m

)=

(2m∑n=1

ln n

)− (2m − 1) ln 2 − (2m) ln π.

Use the observation that

2 · (2m)!

(2π )2m= eA = 10A/ ln 10.

A.3.10 Show that ζ (n) < 1 + ∫ ∞1 x−ndx = 1 + 1/(n − 1).

A.3.19 Use equation (A.53).

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Index

Abel, Niels Henrik, xi, 160–161, 163, 169, 174,182, 203, 209, 244

Abel’s lemma (theorem 4.16), 161, 209, 244absolute convergence, 125, 175, 220, 248absolute convergence theorem (corollary 4.4),

126absolute uniform convergence (corollary 5.10),

204addition of series (theorem 5.4), 178al-Samaw’al, 278algebraic numbers, 263alternating harmonic series, 13, 16alternating series, 126alternating series test (corollary 4.5), 126Ampere, Andre Marie, 161, 258analytic function, 54Apery, Roger, 290Archimedean principle, 12Archimedean understanding, 12, 18Archimedes of Syracuse, 9–11, 19, 22, 237, 277arctangent

series expansion, 23Aryabhata, 277associative, 12asymptotic series, 299

Babbage, Charles, 54Baire, Rene Louis, 268

Berkeley, George, 50–52Bernoulli numbers, 118, 128, 271, 280–283,

287, 297Bernoulli polynomials, 269, 278–280Bernoulli’s identity, 40Bernoulli, Daniel, 4, 52–54, 248Bernoulli, Jacob, 118, 271, 277–282,

284Bernoulli, Johann, 40, 52, 109, 277, 284binomial series, 25, 129

convergence, 122, 146, 153, 206d’Alembert’s investigation of convergence,

41–43Bolzano, Bernhard, 57, 78, 81, 84, 90, 258Bolzano–Weierstrass theorem, 84Bonnet, Ossian, 72, 102, 220, 244Bonnet’s lemma (lemma 6.9), 244Bonnet’s mean value theorem (lemma 6.5), 231,

233, 244Borel, Emile, 268

Cantor, Georg, 84, 268Cauchy, Augustin Louis, ix–xi, 11, 12, 19–20,

55, 57, 71–76, 81, 84, 96, 102, 105, 123,135, 137, 152, 160, 181–185, 218, 220,237–242, 248–249, 252, 298

Cauchy criterionfor integrability, 240, 249

317

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318 Index

Cauchy criterion (theorem 4.2), 123, 129, 208,210

Cauchy integral, 238Cauchy sequence, 123Cauchy series, 123Cauchy’s condensation test (theorem 4.11), 135Cauchy’s remainder theorem (theorem 3.11),

107characteristic function, 268Charles X, 57Collins, John, 39commutative, 12comparison test (theorem 4.6), 130completeness, 125completeness (theorem 4.3), 125condensation test, 135conditional convergence, 126, 177conditions for Riemann integrability

(theorem 6.10), 251continued fractions, 83continuity, 81

and boundedness, 95and differentiability, 91, 258and integrability, 241, 268Lacroix’s definition, 78of composition, 90of power series, 205of product, 89of reciprocal, 89of sum, 88on an interval, 81piecewise, 227uniform, 228

continuity and uniform convergence impliesconvergence at endpoints (theorem 5.14),208

continuity of infinite series (theorem 5.6), 187continuity of integral (corollary 6.8), 243continuous implies bounded (theorem 3.6), 95continuous implies bounds achieved

(theorem 3.8), 98continuous implies integrable (theorem 6.6), 241continuous on [a, b] implies uniform continuity

(lemma 6.3), 229contrapositive, 122convergence

absolute, 125, 149, 175, 220, 248alternating series, 126

binomial series, 146Cauchy criterion, 123, 129, 208comparison test, 130condensation test, 135conditional, 126d’Alembert’s definition, 128Dirichlet’s test, 164Gauss’s test, 152improper integral, 138in norm, 145infinite series, 117integral test, 137limit ratio test, 131limit root test, 132of binomial series, 153of exponential series, 146of Fourier series, 158p-test, 137pointwise, 145radius of, 147ratio test, 130root test, 132uniform, 185, 188, 197, 203, 208, 254, 259

converse, 122cosine

series expansion, 44covering, 268Cp function, 54

d’Alembert, Jean Le Rond, 41–43, 52, 53, 122,128, 129, 248

Darboux’s theorem (theorem 3.14), 112Darboux, Jean Gaston, 111decreasing function, 87Dedekind, Julius Wilhelm Richard, 84, 175definite integral, 219de Moivre, Abraham, 40, 271, 293–296derivative

and continuity, 91, 258Lagrange’s definition, 54, 55of infinite series, 63–65, 195of power series, 205one-sided, 92

Diderot, Denis, 41, 52differentiable implies continuous (theorem 3.5),

91differentiation

of series, 6

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Index 319

Dijksterhuis, E. J., 11dilogarithm, 187Dirichlet, Peter Gustav Lejeune, x–xi, 7, 82, 88,

160–161, 163, 174, 175, 182, 203,217–227, 248, 267, 269, 270, 282

Dirichlet kernel, 223Dirichlet’s test (corollary 4.17), 164, 166Dirichlet’s test for uniform convergence

(theorem 5.16), 211Dirichlet’s theorem (theorem 6.1), 227discontinuity, 84distributive law for series (theorem 5.5), 179divergence, 18, 52–53, 149

comparison test, 130condensation test, 135d’Alembert’s definition, 128Gauss’s test, 152integral test, 137limit ratio test, 131limit root test, 132of binomial series, 153p-test, 137ratio test, 130root test, 132

divergence theorem (theorem 4.1), 122divergence to infinity, 29dominated uniform convergence (theorem 5.9),

203

Eisenstein, Ferdinand Gotthold Max, 174envelope, 185ε–δ, 61Euclid, 73Eudoxus of Cnidus, 11Euler, Leonhard, ix, 4, 17–18, 31, 38, 39, 43,

52–54, 138, 150, 151, 172, 248, 271, 282,284–289, 296

Euler’s constant, 31Euler–Maclaurin formula (theorem A.1), 139,

296, 298existence of radius of convergence

(theorem 4.14), 149exponential function

series expansion, 44

Fermat, Pierre de, 99, 272, 282Fermat’s last theorem, 160, 270, 282–283Fermat’s theorem on extrema (theorem 3.9), 100

floor, 30Fourier, Jean Baptiste Joseph, 1–7, 22, 43, 53,

54, 160, 161, 171, 197, 199, 217–220, 222,248, 267

Fourier series, 5–7, 63, 145, 166, 171, 182, 191,197, 199, 218, 248, 269

convergence, 158, 217Dirichlet’s test, 164

Dirichlet’s theorem, 227uniform convergence, 210uniqueness, 267

frequency, 53

γ , see Euler’s constantGauss, Carl Friedrich, x, 57, 149, 151, 152, 174Gauss’s test (theorem 4.15), 152, 187generalized mean value theorem (theorem 3.2),

75geometric series, 17Germain, Sophie, 217, 282Gilbert, Phillipe, 259Grattan-Guinness, Ivor, 18greatest lower bound, 97Gregory, James, 23, 35, 39, 40Gudermann, Christof, 203

Hachette, Jean Nicholas Pierre, 160Hadamard, Jacques, 270Hankel, Hermann, 259, 267Hardy, Godfrey Harold, 259harmonic series, 121, 284

partial sums, 33harmonics, 53Hawkins, Thomas, 259nHeine, Heinrich Eduard, 84, 267Hermite, Charles, 269Herschel, John, 54Houel, Guillaume Jules, 259Holmboe, Bernt Michael, 160hypergeometric series, 150

Gauss’s test for convergence, 152

ibn Al-Haytham, 278improper integral, 252

unbounded domain, 138value, 138

increasing function, 87infimum, 97

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320 Index

infinite limit, 109infinite series, 9, 171

addition of two series, 178alternating, 126differentiation, 63–65, 195divergent, 18, 52–53integration, 197multiplication by constant, 179of continuous functions, 182, 187rearranging, 13, 175, 177, 220regrouping, 13, 173

infinite summation, see infinite seriesintegral

as an area, 219as inverse of derivative, 219Cauchy, 238

with Cauchy criterion, 240definite, 219improper, 138Lebesgue, 268, 269of infinite series, 197of power series, 205Riemann, 249, 267

necessary and sufficient conditions forexistence, 251with Cauchy criterion, 249

integral form of the mean value theorem(theorem 6.7), 243

integral of Dirichlet kernel (lemma 6.4), 231integral test (theorem 4.13), 137integration

of series, 5intermediate value property, 73, 78, 79, 85intermediate value theorem (theorem 3.3), 85inverse, 122

Jacobi, Carl Gustav Jacob, 174

Kepler, Johann, 99Kummer, Ernst Eduard, 283

Lacroix, Sylvestre Francois, 4, 78, 161Lagrange, Joseph Louis, ix, 4, 6, 43, 53, 54, 57,

166, 217, 248Lagrange remainder, 43–47Lagrange’s remainder theorem (theorem 2.1), 44Lame, Gabriel, 283Laplace, Pierre Simon, 4, 57, 161

least upper bound, 97, 210Lebesgue, Henri Leon, 1, 220, 263, 268Lebesgue integral, 268–269Legendre, Adrien Marie, 160, 270, 282Leibniz, Gottfried, 23, 24, 40, 50, 99, 172Levi ben Gerson, 278L’Hospital, Guillaume Francois Antoine de, 109L’Hospital’s rule

0/0 (theorem 3.12), 109∞/∞ (theorem 3.13), 110

L’Huillier, Simon Antoine Jean, 52lim inf, 148lim sup, 148limit

at infinity, 109d’Alembert’s definition, 52from the left, 92from the right, 92infinite, 109interchanging, 171lower, 148one-sided, 92upper, 148

limit ratio test (corollary 4.8), 131limit root test (corollary 4.10), 132Liouville, Joseph, 217logarithm

series expansion, 28lower limit, 148

Machin, John, 23Maclaurin, Colin, 138, 296Madhava, 23mean value theorem

Bonnet’s, 231, 233, 244Bonnet’s proof, 72, 74, 102Cauchy’s first proof, 72–73Cauchy’s second proof, 75, 77generalized, 75integral form, 243

mean value theorem (theorem 3.1), 58, 72–77measure, 263, 268Medvedev, Fyodor A., 259nMeray, Charles, 84Mercator, Nicolaus, 28method of exhaustion, 11modified converse to intermediate value theorem

(theorem 3.4), 87

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Index 321

Monge, Gaspard, 4monotonic function, 87

Napoleon I, 217Narayana Pandita, 278Navier, Claude Louis Marie Henry, 217nested interval principle, 32Newton, Isaac, 23–26, 28, 40, 50, 99, 271Nilakantha, 23

Olsen, Lars, 111one-sided derivative, 92one-sided limit, 92Oresme, Nicole, 17

p-test (corollary 4.12), 137Peacock, George, 54π

calculations of, 22–26Wallis’s formula for, 24

piecewise continuous, 227piecewise monotonic, 87, 227, 262Poincare, Henri, ix, 269pointwise convergence, 145Poisson, Simeon Denis, 4, 161, 217,

220power series, 39, 145, 204–205

binomial, see binomial seriescontinuity, 205differentiation, 205expansion of e−1/x2

, 68hypergeometric, see hypergeometric seriesintegration, 205uniform convergence, 204, 209

primescounting function, 270, 289in arithmetic progression, 270

Pythagorean triples, 282

Q.E.D., 73quadrature of the parabola, 9

Raabe, Joseph Ludwig, 152, 258radius of convergence, 147

existence of, 149for complex-valued power series, 169

Ramanujan, S., 26ratio test (theorem 4.7), 130, 151

rational function, 38rearranging convergent series (theorem 5.2),

175rearranging infinite series, 13, 175, 177, 220refinement, 239regrouping infinite series (theorem 5.1), 13,

173regular prime, 283Riemann, Georg Friedrich Bernhard, xi, 4, 58,

174–175, 220, 228, 248–255, 258, 267,269, 270, 289

Riemann hypothesis, 290Riemann integral, 249, 267

necessary and sufficient conditions forexistence, 251

Riemann rearrangement theorem (theorem 5.3),177

Riemann’s lemma (lemma 6.2), 228, 242Rolle, Michel, 100Rolle’s theorem (theorem 3.10), 100root test (theorem 4.9), 132Russell, Bertrand Arthur William, 1

Saigey, Jacques Frederic, 160Seidel, Phillip, 182series, see infinite seriesSerret, Joseph Alfred, 72, 102sine

series expansion, 44Steiner, Jakob, 174Stieltjes, Thomas Jan, 269Stirling, James, 271, 293–296Stirling’s formula, 45, 118, 133, 146, 271,

293–298Stirling’s series, 118Stokes, George, 182Sturm, Charles Francois, 217supremum, 97Swineshead, Richard, 17Sylvester, James Joseph, 41n

Taylor series, 40Cauchy remainder, 107–109Lagrange remainder, 43–47, 71, 105–109

Taylor, Brook, 40Taylor, Richard, 270, 283term-by-term differentiation (theorem 5.7), 6,

195

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322 Index

term-by-term integration (theorem 5.8), 5, 197trigonometric series, 145, 218, 254

uniform continuity, 228uniform convergence, 185, 188, 197, 203, 208,

254, 259, 267Cauchy criterion, 202Dirichlet’s test, 211in general, 267of power series, 209Weierstrass M-test, 203, 204

uniform convergence of power series, I(corollary 5.13), 204

uniform convergence of power series, II(theorem 5.15), 209

uniformly bounded, 210upper bound implies least upper bound

(theorem 3.7), 98upper limit, 148

Vallee Poussin, Charles de la,270

variation, 250variation on dominated uniform convergence

(corollary 5.11), 204vibrating string problem, 4, 53–54

Wallis, John, 23, 271–275Wallis’s formula, 24, 275, 296Weierstrass, Karl Theodor Wilhelm, xi, 4, 58,

91, 182, 203, 258–263, 267Weierstrass M-test (corollary 5.12), 203, 204,

259Whitehead, Alfred North, 1Wiles, Andrew, 270, 283

ζ , see zeta functionzeta function, 36, 289Zhu Shijie, 278

In this second edition of the MAA classic, exploration continues to be an

essential component. More than 60 new exercises have been added, and

the chapters on Infinite Summations, Differentiability and Continuity, and

Convergence of Infinite Series have been reorganized to make it easier to

identify the key ideas.

A Radical Approach to Real Analysis is an introduction to real analysis,

rooted in and informed by the historical issues that shaped its develop-

ment. It can be used as a textbook, as a resource for the instructor who

prefers to teach a traditional course, or as a resource for the student who

has been through a traditional course yet still does not understand what

real analysis is about and why it was created.

The book begins with Fourier’s introduction of trigonometric series

and the problems they created for the mathematicians of the early 19th

century. It follows Cauchy’s attempts to establish a firm foundation for

calculus and considers his failures as well as his successes. It culminates

with Dirichlet’s proof of the validity of the Fourier series expansion and

explores some of the counterintuitive results Riemann and Weierstrass

were led to as a result of Dirichlet’s proof.

AMS / MAA TEXTBOOKS

TEXTBOOKS VOL A Radical Approach to Real Analysis

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