Excursions in classical analysis: pathways to advanced problem solving and undergraduate research 9780883857687, 9780883859353, 0883857685, 0883859351

Citation preview

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Excursions in Classical Analysis Pathways to Advanced Problem Solving and Undergraduate Research

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Chapter 11 (Generating Functions for Powers of Fibonacci Numbers) is a reworked version of material from my same name paper in the International Journal of Mathematical Education in Science and Technology, Vol. 38:4 (2007) pp. 531–537. www.informaworld.com Chapter 12 (Identities for the Fibonacci Powers) is a reworked version of material from my same name paper in the International Journal of Mathematical Education in Science and Technology, Vol. 39:4 (2008) pp. 534–541. www.informaworld.com Chapter 13 (Bernoulli Numbers via Determinants)is a reworked version of material from my same name paper in the International Journal of Mathematical Education in Science and Technology, Vol. 34:2 (2003) pp. 291–297. www.informaworld.com Chapter 19 (Parametric Differentiation and Integration) is a reworked version of material from my same name paper in the International Journal of Mathematical Education in Science and Technology, Vol. 40:4 (2009) pp. 559–570. www.informaworld.com

c 2010 by the Mathematical Association of America, Inc.

Library of Congress Catalog Card Number 2010924991 Print ISBN 978-0-88385-768-7 Electronic ISBN 978-0-88385-935-3 Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1

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Excursions in Classical Analysis Pathways to Advanced Problem Solving and Undergraduate Research

Hongwei Chen Christopher Newport University

Published and Distributed by The Mathematical Association of America

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Committee on Books Frank Farris, Chair Classroom Resource Materials Editorial Board Gerald M. Bryce, Editor Michael Bardzell William C. Bauldry Diane L. Herrmann Wayne Roberts Susan G. Staples Philip D. Straffin Holly S. Zullo

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CLASSROOM RESOURCE MATERIALS Classroom Resource Materials is intended to provide supplementary classroom material for students—laboratory exercises, projects, historical information, textbooks with unusual approaches for presenting mathematical ideas, career information, etc.

101 Careers in Mathematics, 2nd edition edited by Andrew Sterrett Archimedes: What Did He Do Besides Cry Eureka?, Sherman Stein The Calculus Collection: A Resource for AP and Beyond, edited by Caren L. Diefenderfer and Roger B. Nelsen Calculus Mysteries and Thrillers, R. Grant Woods Conjecture and Proof, Mikl´os Laczkovich Counterexamples in Calculus, Sergiy Klymchuk Creative Mathematics, H. S. Wall Environmental Mathematics in the Classroom, edited by B. A. Fusaro and P. C. Kenschaft Excursions in Classical Analysis: Pathways to Advanced Problem Solving and Undergraduate Research, by Hongwei Chen Exploratory Examples for Real Analysis, Joanne E. Snow and Kirk E. Weller Geometry From Africa: Mathematical and Educational Explorations, Paulus Gerdes Historical Modules for the Teaching and Learning of Mathematics (CD), edited by Victor Katz and Karen Dee Michalowicz Identification Numbers and Check Digit Schemes, Joseph Kirtland Interdisciplinary Lively Application Projects, edited by Chris Arney Inverse Problems: Activities for Undergraduates, Charles W. Groetsch Laboratory Experiences in Group Theory, Ellen Maycock Parker Learn from the Masters, Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson, and Victor Katz Math Made Visual: Creating Images for Understanding Mathematics, Claudi Alsina and Roger B. Nelsen Ordinary Differential Equations: A Brief Eclectic Tour, David A. S´anchez Oval Track and Other Permutation Puzzles, John O. Kiltinen A Primer of Abstract Mathematics, Robert B. Ash Proofs Without Words, Roger B. Nelsen Proofs Without Words II, Roger B. Nelsen She Does Math!, edited by Marla Parker Solve This: Math Activities for Students and Clubs, James S. Tanton Student Manual for Mathematics for Business Decisions Part 1: Probability and Simulation, David Williamson, Marilou Mendel, Julie Tarr, and Deborah Yoklic

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Student Manual for Mathematics for Business Decisions Part 2: Calculus and Optimization, David Williamson, Marilou Mendel, Julie Tarr, and Deborah Yoklic Teaching Statistics Using Baseball, Jim Albert Visual Group Theory, Nathan C. Carter Writing Projects for Mathematics Courses: Crushed Clowns, Cars, and Coffee to Go, Annalisa Crannell, Gavin LaRose, Thomas Ratliff, Elyn Rykken

MAA Service Center P.O. Box 91112 Washington, DC 20090-1112 1-800-331-1MAA FAX: 1-301-206-9789

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Contents Preface 1

xi . . . .

1 1 8 12 15

2

A New Approach for Proving Inequalities Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 22 23

3

Means Generated by an Integral Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 30 32

4

The L’Hˆopital Monotone Rule Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 37 38

5

Trigonometric Identities via Complex Numbers 5.1 A Primer of complex numbers . . . . . . . . 5.2 Finite Product Identities . . . . . . . . . . . . 5.3 Finite Summation Identities . . . . . . . . . . 5.4 Euler’s Infinite Product . . . . . . . . . . . . 5.5 Sums of inverse tangents . . . . . . . . . . . 5.6 Two Applications . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . .

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39 39 41 43 45 48 49 51 53

Special Numbers 6.1 Generating Functions 6.2 Fibonacci Numbers . 6.3 Harmonic numbers . 6.4 Bernoulli Numbers .

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55 55 56 58 61

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Two Classical Inequalities 1.1 AM-GM Inequality . . . . 1.2 Cauchy-Schwarz Inequality Exercises . . . . . . . . . . . . References . . . . . . . . . . . .

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Contents

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69 72

On a Sum of Cosecants 7.1 A well-known sum and its generalization 7.2 Rough estimates . . . . . . . . . . . . . 7.3 Tying up the loose bounds . . . . . . . . 7.4 Final Remarks . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . .

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73 73 74 76 79 79 81

8

The Gamma Products in Simple Closed Forms Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 89 91

9

On the Telescoping Sums 9.1 The sum of products of arithmetic sequences . . . . . . . . 9.2 The sum of products of reciprocals of arithmetic sequences 9.3 Trigonometric sums . . . . . . . . . . . . . . . . . . . . . 9.4 Some more telescoping sums . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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93 94 95 97 101 106 108

10 Summation of Subseries in Closed Form Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

109 117 119

11 Generating Functions for Powers of Fibonacci Numbers Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121 128 130

12 Identities for the Fibonacci Powers Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

131 140 142

13 Bernoulli Numbers via Determinants Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

143 149 152

14 On Some Finite Trigonometric Power Sums 14.1 Sums involving sec2p .k=n/ . . . . . . 14.2 Sums involving csc2p .k=n/ . . . . . . 14.3 Sums involving tan2p .k=n/ . . . . . . 14.4 Sums involving cot2p .k=n/ . . . . . . Exercises . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . .

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153 154 156 157 159 162 164

15 Power Series of .arcsin x/2 15.1 First Proof of the Series (15.1) . . . . . . . . . . . . . . . . . . . . . . . 15.2 Second Proof of the Series (15.1) . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

165 165 167 171

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Contents

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Six Ways to Sum .2/ 16.1 Euler’s Proof . . . . . . . . . . . 16.2 Proof by Double Integrals . . . . 16.3 Proof by Trigonometric Identities 16.4 Proof by Power Series . . . . . . 16.5 Proof by Fourier Series . . . . . 16.6 Proof by Complex Variables . . . Exercises . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . .

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175 176 177 181 182 183 183 184 187

17 Evaluations of Some Variant Euler Sums Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

189 197 199

18 Interesting Series Involving Binomial Coefficients 18.1 An integral representation and its applications . 18.2 Some Extensions . . . . . . . . . . . . . . . . 18.3 Searching for new formulas for  . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .

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201 202 208 210 212 215

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19 Parametric Differentiation and Integration Example 1 . . . . . . . . . . . . . . . . . . . Example 2 . . . . . . . . . . . . . . . . . . . Example 3 . . . . . . . . . . . . . . . . . . . Example 4 . . . . . . . . . . . . . . . . . . . Example 5 . . . . . . . . . . . . . . . . . . . Example 6 . . . . . . . . . . . . . . . . . . . Example 7 . . . . . . . . . . . . . . . . . . . Example 8 . . . . . . . . . . . . . . . . . . . Example 9 . . . . . . . . . . . . . . . . . . . Example 10 . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . .

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217 217 218 219 220 220 222 223 224 225 226 228 230

20 Four Ways to Evaluate the Poisson Integral 20.1 Using Riemann Sums . . . . . . . . . . 20.2 Using A Functional Equation . . . . . . 20.3 Using Parametric Differentiation . . . . 20.4 Using Infinite Series . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . .

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231 232 233 234 235 235 239

21 Some Irresistible Integrals 21.1 Monthly Problem 10611 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Monthly Problem 11206 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Monthly Problem 11275 . . . . . . . . . . . . . . . . . . . . . . . . . . .

241 241 243 244

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21.4 Monthly Problem 11277 . 21.5 Monthly Problem 11322 . 21.6 Monthly Problem 11329 . 21.7 Monthly Problem 11331 . 21.8 Monthly Problem 11418 . Exercises . . . . . . . . . . . References . . . . . . . . . . .

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245 246 247 249 250 252 254

Solutions to Selected Problems

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Index

297

About the Author

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Preface This book grew out of my work in the last decade teaching, researching, solving problems, and guiding undergraduate research. I hope it will benefit readers who are interested in advanced problem solving and undergraduate research. Math students often have trouble the first time they confront advanced problems like those that appear in Putnam competitions and in the journals of the Mathematical Association of America. And while many math students would like to do research in mathematics, they don’t know how to get started. They aren’t aware of what problems are open, and when presented with a problem outside of the realm of their courses, they find it difficult to apply what they know to this new situation. Historically, deep and beautiful mathematical ideas have often emerged from simple cases of challenging problems. Examining simple cases allows students to experience working with abstract ideas at a nontrivial level — something they do little of in standard mathematics courses. Keeping this in mind, the book illustrates creative problem solving techniques via case studies. Each case in the book has a central theme and contains kernels of sophisticated ideas connected to important current research. The book seeks to spell out the principles underlying these ideas and to immerse readers in the processes of problem solving and research. The book aims to introduce students to advanced problem solving and undergraduate research in two ways. The first is to provide a colorful tour of classical analysis, showcasing a wide variety of problems and placing them in historical contexts. The second is to help students gain mastery in mathematical discovery and proof. Although one proof is enough to establish a proposition, students should be aware that there are (possibly widely) various ways to approach a problem. Accordingly, this book often presents a variety of solutions for a particular problem. Some reach back to the work of outstanding mathematicians like Euler; some connect to other beautiful parts of mathematics. Readers will frequently see problems solved by using an idea that might at first have seemed inapplicable, or by employing a specific technique that is used to solve many different kinds of problems. The book emphasizes the rich and elegant interplay between continuous and discrete mathematics by applying induction, recursion, and combinatorics to traditional problems in classical analysis. Advanced problem solving and research involve not only deduction but also experimentation, guessing, and arguments from analogy. In the last two decades, computer alxi

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Preface

gebra systems have become a useful tool in mathematical research. They are often used to experiment with various examples, to decide if a potential result leans in the desired direction, or to formulate credible conjectures. With the continuing increases of computing power and accessibility, experimental mathematics has not only come of age but is quickly maturing. Appealing to this trend, I try to provide in the book a variety of accessible problems in which computing plays a significant role. These experimentally discovered results are indeed based on rigorous mathematics. Two interesting examples are the WZ-method for telescoping in Chapter 9 and the Hermite-Ostrogradski Formula for searching for a new formula for  in Chapter 18. In his classic “How to Solve It”, George P´olya provides a list of guidelines for solving mathematical problems: learn to understand the problem; devise a plan to solve the problem; carry out that plan; and look back and check what the results indicate. This list has served as a standard rubric for several generations of mathematicians, and has guided this book as well. Accordingly, I have begun each topic by categorizing and identifying the problem at hand, then indicated which technique I will use and why, and ended by making a worthwhile discovery or proving a memorable result. I often take the reader through a method which presents rough estimates before I derive finer ones, and I demonstrate how the more easily solved special cases often lead to insights that drive improvements of existing results. Readers will clearly see how mathematical proofs evolve — from the specific to the general and from the simplified scenario to the theoretical framework. A carefully selected assortment of problems presented at the end of the chapters includes 22 Putnam problems, 50 MAA Monthly problems, and 14 open problems. These problems are related not solely to the chapter topics, but they also connect naturally to other problems and even serve as introductions to other areas of mathematics. At the end of the book appear approximately 80 selected problem solutions. To help readers assimilate the results, most of the problem solutions contain directions to additional problems solvable by similar methods, references for further reading, and to alternate solutions that possibly involve more advanced concepts. Readers are invited to consider open problems and to consult publications in their quest to prove their own beautiful results. This book serves as a rich resource for advanced problem solving and undergraduate mathematics research. I hope it will prove useful in students’ preparations for mathematics competitions, in undergraduate reading courses and seminars, and as a supplement in analysis courses. Mathematicians and students interested in problem solving will find the collection of topics appealing. This book is also ideal for self study. Since the chapters are independent of one another, they may be read in any order. It is my earnest hope that some readers, including working mathematicians, will come under the spell of these interesting topics. This book is accessible to anyone who knows calculus well and who cares about problem solving. However, it is not expected that the book will be easy reading for many math students straight out of first-year calculus. In order to proceed comfortably, readers will need to have some results of classical analysis at their fingertips, and to have had exposure to special functions and the rudiments of complex analysis. Some degree of mathematical maturity is presumed, and upon occasion one is required to do some careful thinking.

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Preface

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Acknowledgments I wish to thank Gerald Bryce and Christopher Kennedy. Both read through the entire manuscript with meticulous and skeptical eyes, caught many of my errors, and offered much valuable advice for improvement. This book is much improved because of their dedication and I am deeply grateful to them. Naturally all possible errors are my own responsibility. I also wish to thank Don Albers at the Mathematical Association of America and the members of the Classroom Resource Materials Editorial Board for their kindness, thoughtful and constructive comments and suggestions. I especially wish to thank Susan Staples for carefully reviewing the book and suggesting its current extended title. Elaine Pedreira and Beverly Ruedi have helped me through many aspects of production and have shepherded this book toward a speedy publication. It is an honor and a privilege for me to learn from all of them. My colleagues Martin Bartelt, Brian Bradie and Ron Persky have read the first draft of several chapters and lent further moral support. Much of the work was accomplished during a sabbatical leave in the fall of 2007. My thanks go to Christopher Newport University for providing this sabbatical opportunity. Finally, I wish to thank my wife Ying and children Alex and Abigail for their patience, support and encouragement. I especially wish to thank Alex, who drew all of the figures in the book and commented on most of chapters from the perspective of an undergraduate. Comments, corrections, and suggestions from readers are always welcome. I would be glad to receive these at [email protected]. Thank you in advance.

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1 Two Classical Inequalities The heart of mathematics is its problems.

— P. R. Halmos

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. — N. H. Abel

Mathematical progress depends on a stream of new problems, but most new problems emerge simply from well-established principles. For students who want to master existing principles and build their own knowledge of mathematics, the moral is simple: “Learn from the masters!” Often, historical developments of a particular topic provide the best way to learn that topic. In this chapter, by examining various proofs of the arithmetic mean and geometric mean (AM-GM) inequality and the Cauchy-Schwarz inequality, we see the motivation necessary to work through all the steps of a rigorous proof. Some of these proofs contain brilliant ideas and clever insights from famous mathematicians. Personally, I still remember how exciting and stirring these proofs were when I first encountered them. We begin with the AM-GM inequality.

1.1 AM-GM Inequality Let a1 ; a2 ; : : : ; an be positive real numbers. Then p a1 C a2 C


C an n a1 a2


an  ; n with equality if and only if all ak .1  k  n/ are equal.

(1.1)

First Proof. The following proof is based on induction. Let H.n/ stand for the hypothesis that inequality (1.1) is valid for n. For n D 2, we have p a1 C a2 a1 a2  : (1.2) 2 p p p This is equivalent to a1 C a2 2 a1 a2  0 or . a1 a2 /2  0, which is always true. Moreover, equality in (1.2) holds if and only if a1 D a2 . Next, we show that H.2/ and H.n/ together imply H.n C 1/. Let a1 C a2 C


C an C anC1 AnC1 D : nC1 1

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Excursions in Classical Analysis

Then 1 Œ.n C 1/AnC1 C .n 1/AnC1  2n 1 D Œ.a1 C


C an / C .anC1 C AnC1 C


C AnC1 / „ ƒ‚ … 2n n 1 terms   q p 1 n n 1 n n a1 a2


an C n anC1 AnC1  2n q  2n a1 a2


an anC1 AnnC11 ;

AnC1 D

where the first inequality follows from H.n/, while the second inequality follows from H.2/ in the form p aCb ab  2 with q p a D n a1 a2


an ; b D n anC1 AnnC11 : Now, H.n C 1/ follows from

n 1 A2n nC1  a1 a2


anC1 AnC1 :

As before, equality holds if and only if a1 D a2 D


D an and anC1 D AnC1 , or equivalently, if and only if a1 D a2 D


D anC1 . Following the above induction proof is not difficult. The challenges lie in how to regroup the terms appropriately and then where to apply the induction hypothesis. It is natural to ask whether we can find methods that will help us meet these challenges and help us in investigating more complicated questions later in the book and in the future. To this end, we present five more proofs of the AM-GM inequality, chosen specifically to emphasize different and creative proof techniques. By relating the proofs to simple mathematical facts, we will see that the basic ingredients of each proof are few and simple. We start with a non-standard induction proof that is attributed to Cauchy. The proof consists of two applications of induction; one, a “leap-forward” induction, leads to the desired result for all powers of 2 while the other, a “fall-back” induction, from a positive integer to the preceding one, together with the leap-forward induction enables us to establish the result for all positive integers. Second Proof. Again let H.n/ stand for the hypothesis that inequality (1.1) is valid p p a2 /2  0 as noted in the first proof. Moreover, for n. H.2/ follows directly from . a1 H.2/ is self-generalizing: for example, H.2/ can be applied twice to get .a1 a2 a3 a4 /1=4 

.a1 a2 /1=2 C .a3 a4 /1=2 a1 C a2 C a3 C a4  : 2 4

This confirms H.4/, and then H.4/ can be used again with H.2/ to establish .a1 a2


a8 /1=8 

.a1


a4 /1=4 C .a5


a8 /1=4 a1 C a2 C


C a8  : 2 8

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This is precisely H.8/. Continuing in this way k times yields H.2k / for all k  1. Now, to prove H.n/ for all n between the powers of 2, let n < 2k . The plan is to extend .a1 ; a2 ; : : : ; an / into a longer sequence .b1 ; b2; : : : ; b2k / to which we can apply H.2k /. To see this, take the case n D 3 first. Set bi D ai .i D 1; 2; 3/ and let b4 be the value that yields the equality b1 C b2 C b3 C b4 a1 C a2 C a3 D : 4 3 This leads to a1 C a2 C a3 b4 D : 3 Applying H.4/ gives s   a1 C a2 C a3 a1 C a2 C a3 C .a1 C a2 C a3 /=3 a1 C a2 C a3 4  D a1 a2 a3 3 4 3 which then simplifies to the desired result H.3/: p a1 C a2 C a3 3 a1 a2 a3  : 3 In general, for n < 2k , similar to the case n D 3, set ( ai ; i D 1; 2; : : : ; n; bi D a1 C


Can D A; i D n C 1; : : : ; 2k : n Applying H.2k / to .b1 ; b2; : : : ; b2k /, we obtain 

k

a1


an A2

n

1=2k



a1 C


C an C .2k 2k

n/A

D

2k A D A; 2k

which simplifies to, after eliminating A on the left-hand side and then raising both sides to the power of 2k =n, a1 C a2 C


C an .a1 a2


an /1=n  : n This proves H.n/ as desired. The condition for equality is derived just as easily. Cauchy’s proof is longer than the first proof, but because of the self-generalizing nature of H.2/, almost without help, we see how to regroup the terms and where to apply the induction hypothesis. Moreover, we find that the above leap-forward fall-back induction approach is actually equivalent to verifying the following two steps: (I) H.n/ H) H.n

1/,

(II) H.n/ and H.2/ H) H.2n/. This leads to a short instructive proof of (1.1) as follows: To prove (I), set a1 C a2 C


C an 1 : AD n 1

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Excursions in Classical Analysis

Then H.n/ yields n Y1

kD1

and hence

!

ak A  n Y1

kD1

For (II), we have

Pn

1 kD1

ak  An

2n Y

kD1

1

ak D  

ak C A n

D



!n

D



.n

1/A C A n

a1 C a2 C


C Can n 1

n Y

kD1

1

n

n

D An

1

:

1 ! 0 2n Y ak @ ak A kDnC1

!n 0 2n n X X ak @ n

kD1

kDnC1

2n X ak 2n

kD1

!2n

1n ak A n

;

where in the first inequality we use the induction hypothesis H.n/, and in the second inequality we use H.2/. Let us now turn to an alternate approach to prove (1.1). This new approach easily avoids the hurdle of how to regroup the terms and where to apply the induction hypothesis. Introduce new variables ak Ak D p ; n a a


a 1 2 n

.1  k  n/

which are normalized in the sense that n Y

kD1

Ak D 1:

Now, the AM-GM inequality says that if A1 A2


An D 1 then A1 C A2 C


C An  n;

(1.3)

with equality if and only if A1 D A2 D


D An D 1. This observation yields the third induction proof. Third Proof. Let H.n/ stand for the hypothesis that, subject to A1 A2


An D 1, inequality (1.3) is valid for n. For n D 2, if A1 A2 D 1, we have p p 2 A1 C A2  2 () A1 A2  0;

which is always true. Assume that A1 A2


AnC1 D 1 and set Bk D Ak .1  k  n

1/; Bn D An AnC1 :

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Since B1B2


Bn D 1, H.n/ implies that B1 C B2 C


C Bn D A1 C


C An

1

C An AnC1  n:

We have to show that A1 C


C An C AnC1  n C 1. To see this, it suffices to prove that An C AnC1  An AnC1 C 1; which is equivalent to .An 1/.AnC1 1/  0. While this does not generally solely follow from the assumption that A1 A2


AnC1 D 1, due to the symmetry of the Ak ’s, we may order the numbers A1 ; A2 ; : : : ; AnC1 in advance such that An  Ak  AnC1 for all 1  k  n 1. Thus, we have nC1 AnnC1  A1 A2


AnC1  AnC1

and so An  1  AnC1 , which implies that .An 1/.AnC1 desired result H.n C 1/. The equality case means that A1 D A2 D


D An AnC1 D 1

and

.An

1/  0. This establishes the 1/.AnC1

1/ D 0;

which implies A1 D


D An D AnC1 D 1. The additive inequality (1.3) yields a quite different proof of (1.1). Indeed, it also connects to a larger theme, where normalization and orderings give us a systematic way to pass from an additive inequality to a multiplicative inequality. As another example of this process, we present Hardy’s elegant proof of (1.1), which is based on a special ordering designed by a smoothing transformation. Let f .a1 ; a2 ; : : : ; an / be a symmetric function, i.e., for all .a1 ; a2 ; : : : ; an /, f .a1 ; a2 ; : : : ; an / D f ..a1 ; a2 ; : : : ; an //; where .a1 ; a2 ; : : : ; an / is a permutation of .a1 ; a2 ; : : : ; an /. The smoothing transformation states that if the symmetric function f .a1 ; a2 ; : : : ; an / becomes larger as two of the variables are made closer in value while the sum is fixed, then f .a1 ; a2 ; : : : ; an / is maximized when all variables are equal. Hardy’s following proof provides one of the best applications of this principle. Fourth Proof.

Let GD

p a1 C a2 C


C an n a1 a2


an ; A D : n

(1.4)

If a1 D a2 D


D an , then A D G D a1 . Hence (1.1) holds with equality. Suppose that a1 ; a2 ; : : : ; an are not all equal. Without loss of generality, we may assume that a1 D minfa1 ; a2 ; : : : ; an g;

a2 D maxfa1 ; a2 ; : : : ; an g:

Then it is clear that a1 < G < a2 ; a1 < A < a2

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and a1 C a2

A > a1 > 0. Now, consider a new set of n positive numbers fA; a1 C a2

(1.5)

A; a3; : : : ; an g

and let G1 and A1 be the corresponding geometric and arithmetic means. Since A.a1 C a2

A/

a1 a2 D .A

a1 /.a2

A/ > 0;

we have A.a1 C a2

A/a3


an > a1 a2


an ;

from which it follows that G1 > G. On the other hand, nA1 D A C .a1 C a2

A/ C a3 C


C an D nA;

i.e., A1 D A. Thus, we have shown that the transformation defined by (1.5) does not change the arithmetic mean while the geometric mean is increased. In particular, if all numbers in (1.5) are the same then we have A D A1 D G1 > G; which proves (1.1). Otherwise we continue to apply the transformation to the latest numbers. Notice that under the transformation, at least one more element in the underlying set of n positive numbers is replaced with the arithmetic mean. Thus, after applying the transformation up to k times, where k  n 1, we arrive at a vector with n equal coordinates. Now, we have Gk D Ak and A D A1 D


D Ak D Gk > Gk

1

>


> G1 > G:

This finally proves (1.1) as desired. As an example of applying the smoothing transformation above, consider .a1 ; a2; a3 ; a4 / D .1; 7; 5; 3/: The transformation successively yields the sets of positive numbers .1; 7; 5; 3/ H) .4; 4; 5; 3/ H) .4; 4; 4; 4/: In Hardy’s proof, observe that the key ingredient is that the double inequalities a1 < A < a2 are put together to build the simple quadratic inequality .A a1 /.a2 A/ > 0. In fact, to prove symmetric inequalities, without loss of generality, we can always assume that a1  a2 


 an . Taking this monotonicity as an additional bonus, Alzer obtained the following stunning proof of (1.1). Fifth Proof. Let a1  a2 


 an , G and A be given by (1.4). Then a1  G  an and so there exists one k with ak  G  akC1 . It follows that k Z X i D1

G ai



1 t

1 G



dt C

Z n X

i DkC1

ai G



1 G

 1 dt  0 t

(1.6)

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or equivalently, n Z X i D1

ai G

n

Z

X 1 dt  G i D1

Here the left-hand side gives

n X ai i D1

G G

Dn



ai

1 dt: t

G

 1 ;

A G

while the right-hand side becomes n X

n Y

ln G/ D ln

.ln ai

i D1

ai

i D1

!

n ln G D 0:

Thus, A=G 1  0, hence A  G, which gives our required inequality. In the case of equality, all integrals in (1.6) must be zero, which implies a1 D a2 D


D an D G. The final proof is due to George P´olya, who reported that the proof came to him in a dream. As a matter of fact, his proof yields the following stronger inequality: Weighted Arithmetic Mean and Geometric Mean Inequality Let a1 ; a2 ; : : : ; an and p1 ; p2 ; : : : ; pn be positive real numbers such that p1 C p2 C


C pn D 1. Then a1p1 a2p2


anpn  p1 a1 C p2 a2 C


C pn an ;

(1.7)

with equality if and only if all ak .1  k  n/ are equal. It is well known that

P´olya’s Proof.

x  exp.x

1/ for all x 2 R:

For 1  k  n, using this inequality repeatedly yields ak  exp.ak

1/

p

and

ak k  exp.pk ak

 exp

n X

and so p p a1 1 a2 2




anpn

pk ak

kD1

pk /

!

(1.8)

1 :

To obtain the desired arithmetic mean bound, the normalization process is used again. Let bk D

ak A

with

A D p1 a1 C p2 a2 C


C pn an :

Replacing ak in (1.8) by bk yields  a p1  a p2 1

1

A

A






 a pn n

A

 exp

n X

kD1

ak pk A

!

1 D 1:

This is equivalent to (1.7). In the case of equality, noticing that ak ak < exp. A A

1/

unless

ak D 1; A

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we see that (1.8) is strict unless ak D A for all k D 1; 2; : : : ; n. Thus, one has equality in (1.7) if and only if all ak ’s are equal. Similar to the AM-GM inequality, the Cauchy-Schwarz inequality is another one of the most fundamental mathematical inequalities. It also turns out to have a remarkable number of proofs, some just as charming and informative as the proofs of the AM-GM inequality. To highlight this, we present four proofs of the Cauchy-Schwarz inequality.

1.2 Cauchy-Schwarz Inequality Let ak ; bk .1  k  n/ be real numbers. Then n X

ak bk

kD1

!2



n X

ak2

kD1

!

n X

bk2

kD1

!

(1.9)

;

with equality if and only if there is a t 2 R such that ak D tbk .1  k  n/. First, we demonstrate that the Cauchy-Schwarz inequality is a consequence of the AMGM inequality. First Proof. Let H.n/ represent the hypothesis that inequality (1.9) is valid for n. For n D 2, by the AM-GM inequality, we have .a1 b1 C a2 b2 /2 D a12 b12 C 2.a1 b2 /.a2 b1 / C a22 b22  a12 b12 C a12 b22 C a22 b12 C a22 b22

D .a12 C a22 /.b12 C b22 /;

which confirms H.2/. For the case of equality, appealing to the equality condition in the AM-GM inequality, we have a1 b2 D a2 b1 , or equivalently, ai D tbi .i D 1; 2/ with t D a1 =a2 D b1 =b2. Now, we prove that H.2/ and H.n/ together imply H.n C 1/. Indeed, using the AM-GM inequality and H.n/ yields

2

n X

!

ak bk anC1 bnC1

kD1

v u u 2  2 tanC1 

2 anC1

n X

n X

kD1

bk2

kD1

!

!v u u 2 t 2 b b

nC1

k

C

n X

2 bnC1

ak2

kD1 n X

ak2

kD1

!

!

and hence nC1 X

kD1

ak bk

!2

D 

n X

ak bk

kD1 n X

kD1

ak bk

!2

!2

C2 C

n X

kD1

2 anC1

!

2 2 ak bk anC1 bnC1 C anC1 bnC1 n X

kD1

bk2

!

C

2 bnC1

n X

kD1

ak2

!

2 2 C anC1 bnC1

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1. Two Classical Inequalities

 D

n X

kD1 nC1 X

kD1

ak2

!

ak2

n X

bk2

kD1

!

nC1 X

kD1

!

C

2 anC1

n X

bk2

kD1

!

!

n X

2 C bnC1

ak2

kD1

!

2 2 C anC1 bnC1

bk2 :

For the case of equality, again, the equality condition in the AM-GM inequality implies and

ak bnC1 D bk anC1

for all k D 1; 2; : : : ; n:

ak D tbk ;

Therefore, ak D tbk for all k D 1; 2; : : : ; n; n C 1. The next proof is based on normalization. Second Proof. If all ak or bk are zero, we have equality in (1.9). Suppose that ai ¤ 0 and bj ¤ 0 for some 1  i; j  n. Set Ak D P n

ak


2 1=2 kD1 ak

and

kD1

A2k

D1

bk

kD1

which are normalized in the sense that n X

Bk D P n

and

n X

bk2


1=2 ;

Bk2 D 1:

kD1

(1.10)

With these assignments, the inequality (1.9) is equivalent to (1.11)

A1 B1 C A2 B2 C


C An Bn  1: To see this, adding the well-known elementary inequalities Ak Bk 

1 2 .A C Bk2 / 2 k

.() .Ak

Bk /2  0/

for k from 1 to n, in view of (1.10), we have 1 A1 B1 C A2 B2 C


C An Bn  2

n X

A2k

kD1

C

n X

Bk2

kD1

!

D 1:

Moreover, subject to (1.10), it is easy to see that (1.11) is equivalent to .A1

B1 /2 C .A2

B2 /2 C


C .An

Bn /2  0:

Thus, the equality in (1.9) occurs if and only if Ak D Bk for all 1  k  n; that is, if and only if !1=2 n n X X 2 2 ak D tbk for all 1  k  n with t D ak = bk : kD1

kD1

Once again, in terms of normalization, we see how to pass from a simple additive inequality to a multiplicative inequality. In the following, we show how to pass from a simple identity to the famous Lagrange’s identity, which provides a “one-line proof” of the Cauchy-Schwarz inequality.

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Third Proof.

For n D 2, the Cauchy-Schwarz inequality just says .a1 b1 C a2 b2 /2  .a12 C a22 /.b12 C b22/:

This is a simple assertion of the elementary identity .a12 C a22 /.b12 C b22 /

.a1 b1 C a2 b2 /2 D .a1 b2

a2 b1 /2 :

In general, let Dn D

n X

ak2

kD1

!

n X

kD1

bk2

!

n X

ak bk

kD1

!2

:

We have already seen that D2 is a perfect square. In view of the symmetry, we can rewrite Dn as n n n X n X 1 XX 2 2 Dn D .ai bj C aj2 bi2 / ai bi aj bj : 2 i D1 j D1

i D1 j D1

This quickly yields the well-known Lagrange’s identity n n 1 XX Dn D .ai bj 2

aj bi /2 :

i D1 j D1

Now the Cauchy-Schwarz inequality follows from Lagrange’s identity immediately. Finally, we present Schwarz’s proof,which is based on one simple fact about quadratic functions. Let x D .a1 ; a2 ; : : : ; an /; y D .b1 ; b2 ; : : : ; bn /:

Recall the inner product on Rn is defined by

.x; y/ D a1 b1 C a2 b2 C


C an bn p and the norm is given by kxk D .x; x/. In vector form, the Cauchy-Schwarz inequality becomes .x; y/2  kxk2kyk2: Schwarz observed that the quadratic function f .t/ D kxt C yk2 D kxk2t 2 C 2.x; y/t C kyk2 is nonnegative for all t 2 R and hence the discriminant .x; y/2

kxk2kyk2  0:

This provides another short proof of the Cauchy-Schwarz inequality. For readers interested in learning more about the Cauchy-Schwarz inequality, please refer to Michael Steele’s fascinating book The Cauchy-Schwarz Master Class [5]. With the Cauchy-Schwarz inequality as the initial guide, this lively, problem-oriented book will coach one towards mastery of many more classical inequalities, including those of H¨older, Hilbert, and Hardy. Analysis abounds with applications of these two fundamental inequalities. Let us single out two applications.

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1. Two Classical Inequalities

Prove that sequence

Application 1. Proof.

n

1C


1 n n

o

is monotonic increasing and bounded.

By the AM-GM inequality, we have s      1 n 1 1 1 nC1 1C
1 < n 1C C1 D 1C : n nC1 n nC1

This is equivalent to    nC1 1 n 1 < 1C ; 1C n nC1


n which shows that 1 C 1n is monotonic increasing as required. To find an upper bound of the sequence, for n > 5, using the AM-GM inequality once more, we have s    6 1 5 n 5 nC1 n 5
1 < 6  C .n 5/
1 D ; 6 nC1 6 nC1 and so .5=6/6 < .n=.n C 1//nC1 , or equivalently, .1 C 1=n/nC1 < .6=5/6 . Therefore,      6 1 n 1 nC1 6 1C < 1C < < 3; n n 5

.for n > 5/:

In view of the monotonicity of the sequence, for all n  1, we obtain 1 C well known that the sequence converges to the natural base e.


1 n n

< 3. It is

(Monthly Problem 11430, 2009). For real x1 ; : : : ; xn , show that

Application 2.

p x1 x2 xn C C


C < n: 2 2 2 2 2 1 C x1 1 C x1 C x2 1 C x1 C


C xn Solution Notice that x1 xn jx1 j jxn j C


C  C


C : 1 C x12 1 C x12 C


C xn2 1 C x12 1 C x12 C


C xn2 Without lose of generality, we only consider the case when x1 ; : : : ; xn are all nonnegative real numbers. Let S denote the left-hand side of the required inequality. Applying the Cauchy-Schwarz inequality yields 2 x1 x2 xn C1
C


C1
1 C x12 1 C x12 C x22 1 C x12 C


C xn2 " 2  2  2 # x1 x2 xn C C


C :  n
1 C x12 1 C x12 C x22 1 C x12 C


C xn2

 S D 1
2

For k D 1, we have



x1 1 C x12

2



x12 D1 1 C x12

1 ; 1 C x12

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and xk 2 1 C x1 C


C xk2

!2

 D

xk2 .1 C x12 C


C xk2 1C

x12

1 C


C xk2

1

/.1 C x12 C


C xk2/ 1C

1

for 2  k  n. Combining these inequalities results in 2  2  x1 xn C


C 1 1 C x12 1 C x12 C


C xn2

1

x12

C x12

1 ; C


C xk2

1 < 1: C


C xn2

Thus, we have S 2 < n, which implies the desired inequality. We now conclude this chapter by introducing the Gauss arithmetic-geometric mean based on the AM-GM inequality. Let a and b be two distinct positive numbers. Define the successive arithmetic and geometric means by a0 D a; anC1 D

p an C bn I b0 D b; bnC1 D an bn ; 2

.n D 1; 2; : : :/:

These sequences were mentioned by Lagrange in 1785 and six years later came to the attention of the 14-year old Gauss. By inequality (1.1) we see that an  bn for n D 1; 2; : : : : It follows that p aCb ab 


 bn  bnC1  anC1  an 


 2 and

1 .an bn /: 2 Hence the sequences fan g and fbn g both converge to the same limit, which is called the Gauss arithmetic-geometric mean and denoted by agm.a; b/. In 1799 Gauss successfully p found agm.1; 2/ to great precision, but it wasn’t until 1818 that he finally established ! 1 Z dx 2 =2 ; agm.a; b/ D p  0 a2 sin2 x C b 2 cos2 x anC1

bnC1 

which later was used to found the branch of mathematics called the “theory of elliptic functions.”

Exercises 1. Prove the AM-GM inequality by constructing a smoothing transformation in which the geometric mean is invariant while the arithmetic mean is decreased. 2. For positive real numbers a; b; x and y, show that .a C b/2 a2 b2  C : xCy x y Then use this self-generalizing inequality to prove the Cauchy-Schwarz inequality.

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3. Let ak > 0 for all 1  k  n and a1 a2


an D 1. Prove the following inequalities. (a) (1) .1 C a1 /.1 C a2 /


.1 C an /  2n . 1 a1

C a12 C


C a1n for any positive integer m  n 1. P 4. Let ak > 0 for all 1  k  n and nkD1 ak D 1. Prove the following inequalities. (b) (2) a1m C a2m C


C anm 

(a)

(b)

a12 a22 an2 1 C C


C  . a1 C a2 a2 C a3 an C a1 2

n X

kD1

(c)

n X

kD1

(d)

n X

kD1

(e) 1 

2

ak n  . ak 2n 1

akm  1 ak .n

1 1/nm

2

, where m > 1.

n X ak p 1 p ak . p 1 ak n 1 kD1

n X

kD1

ak p 1 C a0 C


C ak

  , where a0 D 0. p 2 ak C


C an 1

5. For all nonnegative sequences a1 ; a2 ; : : : ; and positive integers n and k, use leapforward fall-back induction to prove   a1 C a2 C


C ak n a1n C a2n C


C akn  : k k This is a special case of the power mean inequality, which we will explore more in Chapter 3. 6. Let ak .k D 1; 2; : : : ; n/ be positive real numbers. Let A and G be the arithmetic and geometric means of ak ’s respectively, and H be the harmonic mean of ak ’s defined by H D

n : 1=a1 C 1=a2 C


C 1=an

Prove that nA C H  .n C 1/G: In addition, if AH  G n ; n  3, prove that n X

kD1

7. Let a1 < a2 <


< an and n  2, n X kD1

Pn

kD1

1 2 ak C x 2

akn  n G n.n

1/

:

1=ak  1: Prove that for any real number x and !2



1 2.a1 .a1

1/ C x 2 /

:

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Excursions in Classical Analysis

8. Let fan g be a positive real sequence with ai Cj  ai C aj for all i; j D 1; 2; : : :. For each n 2 N, show that a2 an a1 C C


C  an : 2 n P P 9. Let a1  a2 


 an > 0. Let kiD1 ai  kiD1 bi for all 1  k  n. Show that Pn Pn 2 2 (a) i D1 ai  i D1 bi , Pn P n 3 2 (b) i D1 ai  i D1 ai bi .

10. The Schwab-Schoenberg Mean. Let a and b be positive numbers with a < b. Define a0 D a; b0 D b, and for n  0, anC1 D

p an C bn ; bnC1 D anC1 bn : 2

Show that fan g and fbn g approach to a common limit and determine the value of the limit. 11. Putnam Problem 1964-A5. Prove that there is a positive constant K such that the following inequality holds for any sequence of positive numbers a1 ; a2 ; a3 ; : : :: 1 X

nD1

1 X n 1 K : a1 C a2 C


C an a nD1 n

12. Monthly Problem 11145 [2005, 366; 2006, 766]. Find the least c such that if n  1 and a1 ; : : : ; an > 0 then n X

kD1

k Pk

j D1

1=aj

c

n X

ak :

kD1

Remark. The inequality above can be generalized as: for any positive p, !p   1 1 X k pC1 p X p  ak : Pk p 1=aj kD1

j D1

kD1

13. Putnam Problem 1978-A5. Let 0 < xk <  for k D 1; 2; : : : ; n and set xD Prove that

x1 C x2 C


C xn : n

  n Y sin xk sin x n  : xk x

kD1

14. Putnam Problem 2003-A2. Let a1 ; a2 ; : : : ; an and b1 ; b2 ; : : : ; bn be nonnegative real numbers. Show that .a1 a2


an /1=n C .b1 b2


bn /1=n  Œ.a1 C b1 /.a2 C b2 /


.an C bn /1=n :

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1. Two Classical Inequalities

15. Carleman’s Inequality. Let a1 ; a2 ; : : : ; be a positive sequence with 0 < 1. Prove that 1 1 X X .a1 a2


an /1=n  e an : nD1

P1

nD1

an
0. In view of the algebraic identity ak2 x C

 bk2 p D ak x x

we define fk .x/ D ak2 x C

bk p x

2

C 2ak bk ;

bk2 ; E D .0; C1/: x

Now, fk .x/ has minimum value of 2ak bk , which occurs at xkc D bk =ak . Furthermore, we find that ! Pn n n X X b2 2 f .x/ D fk .x/ D ak x C kD1 k x kD1

kD1

P P has a unique critical number at xc D . nkD1 bk2 = nkD1 ak2 /1=2 , and f .x/ has minimum value !1=2 n n X X 2 2 f .xc / D 2 ak bk : kD1

kD1

Applying (2.1) yields n X

kD1

n X

ak bk 

ak2

kD1

n X

bk2

kD1

!1=2

;

which is equivalent to (2.4). By Lemma 1, we have that equality holds in (2.4) if and only if all xkc D xc . Setting t D xc yields ak D tbk for all 1  k  n. We further show how to use (2.1) to prove the Triangle inequality, which is usually deduced via the Cauchy-Schwarz inequality. Theorem 2.3. Let ak ; bk .1  k  n/ be real numbers. Then n X

kD1

.ak C bk /2

!1=2



n X

kD1

ak2

!1=2

C

n X

bk2

kD1

!1=2

;

(2.5)

with equality if and only if there is a t 2 R such that ak D tbk .1  k  n/. Proof. Since ak C bk  jak j C jbk j, without loss of generality, we may assume that ak ; bk > 0. Modifying the functions used in the proof of (2.4), we define fk .x/ D .ak C bk /2 x C

ak2 ; E D .0; 1/: x

By completing the square, we find that minE ffk .x/g D f .xkc / D 2ak .ak C bk /, where xkc D ak =.ak C bk /. Similarly, we obtain that ! Pn n n X X a2 2 f .x/ D fk .x/ D .ak C bk / x C kD1 k x kD1

kD1

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Excursions in Classical Analysis

P P has a minimum value at xc D . nkD1 ak2 /1=2 =. nkD1 .ak C bk /2 /1=2 . In particular, f .xc / D 2

n X

kD1

ak2

!1=2

n X

kD1

2

.ak C bk /

!1=2

:

Inequality (2.1) thus ensures n X

2

kD1

Since 2

ak .ak C bk /  2

n X

kD1

(2.6) becomes n X

kD1

ak2

C

n X

kD1

ak .ak C bk / D

2

.ak C bk /

2

n X

ak2

kD1

!1=2

n X

ak2 C

n X

ak2

kD1

kD1

n X

kD1

!1=2

n X

2

kD1

.ak C bk /

.ak C bk /2

n X

kD1

!1=2

n X

(2.6)

:

bk2 ;

kD1

2

.ak C bk /

!1=2



n X

bk2 :

kD1

Completing the square on the left-hand side, we have 8 !1=2 92 !1=2 n n n = < X X X bk2 :  .ak C bk /2 ak2 ; : kD1

kD1

kD1

Taking the square root on both sides, we finally obtain !1=2 !1=2 n n X X 2 2 .ak C bk / ak  kD1

kD1

n X

kD1

bk2

!1=2

;

which is equivalent to (2.5). By using Lemma 1 one more time, we see the equality holds in (2.5) if and only if all xkc D xc . Set t D xc . Then all ak D tbk . Further exploration along these lines suggests that (2.1) can be adapted to establish some more general classical inequalities. For example, replacing fk .x/ by pk .ak ln x/ in the proof of Theorem 2.1 gives the following weighted arithmetic-geometric mean inequality Pn n P Y pk = n kD1 pk ak iD1 pi ak  P : n kD1 pk kD1

Moreover, for p; q > 1; 1=p C 1=q D 1, replacing fk .x/ by ak x p =p C bk x proof of Theorem 2.2 leads to H¨older’s inequality !1=p n !1=q n n X X X q p ak bk  ak bk : kD1

kD1

q

=q in the

kD1

Observe that inequality (2.1) can be extended to n X

kD1

inf fk .x/  inf

x2E

x2E

n X

kD1

fk .x/ 

n X

fk .y/

(2.7)

kD1

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2. A New Approach for Proving Inequalities

for all y 2 E. The relaxed right-hand side bound becomes more convenient for both proving and discovering inequalities. To demonstrate this, we present a new elementary proof of the Ky Fan inequality. P Theorem 2.4. Let ak 2 .0; 1=2; pk > 0 .1  k  n/ with nkD1 pk D 1. Define ! 1 n n n X Y X pk AD pk ak ; G D ak ; H D pk =ak kD1

kD1

kD1

and 0

A D

n X

0

pk .1

ak /; G D

kD1

n Y

pk

.1

ak /

kD1

n X

0

; H D

pk =.1

!

1

ak /

kD1

:

Then

G A H  0  0: 0 H G A with equality if and only if all ak ’s are equal.

(2.8)

Proof. We prove the first inequality and leave the proof of the second as an exercise. Motivated by the proof of Theorem 2.3 in [5], take   1 x 1 x x fk .x/ D pk C ln ; E D .0; 1=2: ak 1 ak x Since fk0 .x/ D pk



1 ak .1

1 ak /

x.1

x/



; fk00 .x/ D

pk .1 x 2 .1

2x/ > 0; x/2

fk .x/ is strictly convex and has a unique critical point at x D ak 2 E. In light of (2.7), we have     y 1 y 1 y 1 ak pk fk .ak / D ln  fk .y/ D pk C ln ak ak 1 ak y and ln

n  Y 1

kD1

ak ak

Thus, ln

pk



n  Y 1

kD1

n X pk ak

kD1

ak ak

!

pk

n X

y



kD1

y H

1

pk 1 ak y

H0

!

C ln

.1

1

y/ C ln

y y

1

y y

:

:

Taking y D H=.H C H 0 / 2 E leads to ln

G0 H0  ln ; G H

which is equivalent to the proposed first inequality. Since all fk are strictly convex on E, by Lemma 1, we conclude that equality in (2.8) holds if and only if all ak ’s are equal. This completes the proof.

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Excursions in Classical Analysis

Bellman observed that the fundamental results of mathematics are often inequalities rather than equalities. If he is right, then the ability to solve an inequality, especially those that differ from any previously encountered as exercises, is a valuable skill for anyone who does mathematics. Those whose appetites have been whetted by the approach presented here may want to examine other inequalities. To check recent advances in inequalities, the free online journal Journal of Inequalities in Pure and Applied Mathematics (jipam.vu.edu.au) is always a good resource.

Exercises 1. Let E  R. Prove that inf f .x/ C inf g.x/  inf .f .x/ C g.x//:

x2E

x2E

x2E

Moreover, if f and g are convex in E, show that equality holds if and only if f; g and f C g achieve the infimum at same point. 2. Minkowski’s Inequality. Let ak ; bk  0 for all 1  k  n and p > 1. Use (2.1) to prove that n X

kD1

p

.ak C bk /

!1=p



n X

p ak

kD1

!1=p

n X

C

p bk

kD1

!1=p

:

3. Let ak > 0 for all 1  k  n. Use (2.1) to prove that  Pn

kD1

ak

n

PnkD1 ak



n Y

a

ak k :

kD1

4. Let ak > 0 and bk  0 for all 1  k  n. Use (2.1) to prove that   Pn PnkD1 bk n  Y bk b k kD1 bk  Pn : ak kD1 ak

kD1

5. Monthly Problem 11189 [2005, 929; 2007, 645]. Let a1 ; a2 ; : : : ; an be positive. Let anC1 D a1 and p > 1. Prove that n X

kD1

n akp p X  ak ak C akC1 2 kD1

p

1

2p=.p

1/

n X

kD1

.ak C akC1 /1=.p

1/

:

6. Under the conditions stated in Theorem 2.4, prove that G A  0: 0 G A 7. An Interpolation Inequality. Under the conditions stated in Theorem 2.4, prove that A0 1 A   : G0 G C G0 G

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2. A New Approach for Proving Inequalities

8. Let ak 2 .0; 1=2 .1  k  n/, which do not all coincide. Define f .x/ D

n Y

kD1

n  X 1 1 Cx ak ai i D1

1 ak



1

!

1=n

; x 2 .0; 1=n/:

Show that f .x/ is continuous, strictly decreasing, and therefore 1

H G D f .1=n/  f .x/  f .0/ D 0 ; x 2 .0; 1=n/: H G

References [1] E. Beckenbach and R. Bellman, An Introduction to Inequalities, The Mathematical Association of America, 1961. [2] P. S. Bullen, D. S. Mitrinovics and P. M. Vasic, Means and their Inequalities, Reidel, Dordrecht, 1988. [3] H. Chen, A unified elementary approach to some classical inequalities, Internat. J. Math. Ed. Sci. Tech., 31 (2000) 289–292. [4] J. Sandor and V. Szabo, On an inequality for the sum infimums of functions, J. Math. Anal. Appl., 204 (1996) 646–654. [5] W. L. Wang, Some inequality involving means and their converses, J. Math. Anal. Appl., 238 (1999) 567–579.

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3 Means Generated by an Integral For a pair of distinct positive numbers, a and b, a number of different quantities M.a; b/ are known as means: 1. the arithmetic mean: A.a; b/ D .a C b/=2 p 2. the geometric mean: G.a; b/ D ab 3. the harmonic mean: H.a; b/ D 2ab=.a C b/ 4. the logarithmic mean: L.a; b/ D .b

a/=.ln b ln a/ p 5. the Heronian mean: N.a; b/ D .a C ab C b/=3 6. the centroidal mean: T .a; b/ D 2.a2 C ab C b 2 /=3.a C b/ These are all positively homogeneous, in the sense that M.a; b/ D M.a; b/

for all  > 0;

and symmetric, in the sense that M.a; b/ D M.b; a/. Moreover, all of the named means satisfy a third property, called intermediacy, as well: min.a; b/  M.a; b/  max.a; b/: This inequality, which ensures that M.a; b/ lies between a and b, is the essential feature of means. In the following discussion we assume that means satisfy these three properties. The arithmetic, geometric and harmonic means were studied by the early Greeks largely within the broader context of the theory of ratio and proportion, which in turn was motivated in part by its application to the theory of music. Pythagoras is generally regarded as the first Greek mathematician to investigate how these three means relate to musical intervals. Pappus is credited as the first to associate these three means with Figure 3.1. By manipulating the picture, one can conclude that H.a; b/  G.a; b/  A.a; b/. Since then, both for their beauty and importance, the study of means has become a cornerstone of the theory of inequalities. There are numerous new proofs, refinements and 25

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Excursions in Classical Analysis

A

B¢ O

B

b

a

Figure 3.1. Pappus’s construction: OA D A.a; b/; AB D G.a; b/; BB 0 D H.a; b/ variants on these means [4, pp. 21–48]. Howard Eves in [3] showed how some of these means occur in certain geometrical figures; for example, besides demonstrating three appearances of the harmonic mean, he also ranked most of the foregoing means in terms of the lengths of vertical segments of a trapezoid in Figure 3.2. Shannon Patton created an animated version of this figure using Geometer’s Sketchpad. The detail is available at Mathematical Magazine’s website: www.maa.org/pubs/mm supplements/index.html.

a

b

H

G

N A

T

Figure 3.2. The means in a trapezoid In this chapter, we show how to find functions that are associated to these means. To see this, we introduce [2] R b t C1 x dx f .t/ D aR b : (3.1) t dx x a

This encompasses all the means at the beginning of the chapter: particular values of t in (3.1) give each of the means on our list. Indeed, it is easy to verify that f . 3/ D H.a; b/; f.

1 / D N.a; b/; 2

f.

3 / D G.a; b/; 2

f .0/ D A.a; b/;

f . 1/ D L.a; b/; f .1/ D T .a; b/:

Moreover, upon showing that f .t/ is strictly increasing, as an immediate consequence we deduce that H.a; b/  G.a; b/  L.a; b/  N.a; b/  A.a; b/  T .a; b/

(3.2)

with equality if and only if a D b.

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3. Means Generated by an Integral

To prove that f .t/ is strictly increasing for 0 < a < b, we show that f 0 .t/ > 0. By the quotient rule, R b t C1 Rb R b t C1 Rb ln x dx a x t dx dx a x t ln x dx 0 a x a x f .t/ D : (3.3) Rb . a x t dx/2 Thus, it suffices to prove that the numerator for this quotient is positive. Since the bounds of the definite integrals are constants, the numerator for this quotient can be written as Z b Z b Z b Z b x t C1 ln x dx x t dx x t C1 dx x t ln x dx a

a

D D

Z

a

b

x t C1 ln x dx a

Z bZ a

b

x t y t .x

a

Z

Z

b

y t dy a

b

y t C1 dx a

Z

b

x t ln x dx a

y/ ln x dxdy:

a

Substituting in a different manner, we can write the same numerator as Z b Z b Z b Z b t C1 t t C1 x ln x dx x dx x dx x t ln x dx a

a

D D

Z

a

b

y

t C1

ln y dy

a

b

t

x dx

a

a

Z bZ

a

Z

b

x t y t .y

Z

b

x

t C1

dx

a

Z

b

y t ln y dy a

x/ ln y dxdy:

a

Averaging the two equivalent expressions shows that the numerator equals Z Z 1 b b t t x y .x y/.ln x ln y/ dxdy > 0: 2 a a With expression (3.3), this implies that f 0 .t/ > 0. Therefore, f .t/ is strictly increasing as desired. Note that the above proof is based on double integrals. A related single variable proof for the positivity of the numerator starts with Z u Z u Z u Z u F .u/ D x t C1 ln x dx x t dx x t C1 dx x t ln x dx: a

a

Clearly F .a/ D 0 and for u  a, 0

t

F .u/ D u

Z

a

a

u

x t .u a

x/.ln u

ln x/ dx  0;

so F .b/  0 as long as b > a. Next, notice that f .t/ is a continuous monotonic increasing mean. By comparing with the classic means given in (3.2), we can sharpen some inequalities. For example, since f . 2/ D

ab.ln b ln a/ G 2 .a; b/ D ; b a L.a; b/

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Excursions in Classical Analysis

the monotonicity of f .t/ implies that f . 2/ separates H and G which results in the following well-known interpolation inequality: H.a; b/ 

G 2 .a; b/ < G.a; b/: L.a; b/

Furthermore, defining the power mean by Mp .a; b/ D



ap C b p 2

1=p

;

we get the equalities M1 .a; b/ D A.a; b/; M0 .a; b/ D lim Mp .a; b/ D G.a; b/; M 1.a; b/ D H.a; b/: p!0

Observing that
1 G.a; b/ C A.a; b/ ; 2
1 N.a; b/ D G.a; b/ C 2A.a; b/ ; 3

M1=2 .a; b/ D

we challenge the reader to choose values of t to show that L.a; b/ < M1=3 .a; b/
ˇ. Then c.a; bI ˛; ˇ/ D



ˇ.b ˛ ˛.b ˇ

a˛ / aˇ /

1=.˛

ˇ/

D

Rb

a Rb a

x˛

1

dx

xˇ

1

dx

!1=.˛

ˇ/

:

(3.4)

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3. Means Generated by an Integral

Since a < c.a; bI ˛; ˇ/ < b, and positively homogeneous and symmetry are also obvious, it follows that c constitutes a continuous mean. Thus, f .t/ D c.a; bI t C 2; t C 1/; is just one member of the general mean family of (3.4). In particular, setting ˇ D 1 in (3.4), we obtain the Stolarsky mean S˛ .a; b/ D



b ˛ a˛ ˛.b a/

1=.˛

1/

;

.˛ ¤ 0; 1/

which is monotonic increasing in ˛. Note that S

1 .a; b/

D G.a; b/; S2.a; b/ D A.a; b/:

The limiting cases give S0 .a; b/ D lim S˛ .a; b/ D L.a; b/ ˛!0

and S1.a; b/ D lim S˛ .a; b/ D e ˛!1

1

.b b =aa /1=.b

a/

D I.a; b/:

Here I.a; b/ is called the identric mean and also arises from the Putnam Problem 1979-B2. Like the Heronian mean, the identric mean is trapped between L and A as well. Thus L.a; b/  I.a; b/  A.a; b/: It is also interesting to see that   A.a; b/ L.a; b/ I.a; b/ ln D : G.a; b/ L.a; b/ Finally, to gain a more general perspective on the topic of means, we state a set of axioms, which, if satisfied by a class of functions, entitle those functions to be called “means”. The axioms will be chosen by abstracting the most important properties of f .t/ in (3.1). We say function F .a; b/ defines a mean for a; b > 0 when 1. F .a; b/ is continuous in each variable, 2. F .a; b/ is strictly increasing in each variable, 3. F .a; b/ D F .b; a/, 4. F .ta; tb/ D tF .a; b/ for all t > 0, 5. a < F .a; b/ < b for 0 < a < b. We leave to the reader to show that a necessary and sufficient condition for a mean is that for 0 < a  b, F .a; b/ D b f .a=b/

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Excursions in Classical Analysis

where f .s/ is positive, continuous and strictly increasing for 0 < s  1, and satisfies s < f .s/  1;

for 0 < s < 1:

In particular, if  is a positive continuous function on .0; 1 and we let R1

f .s/ D f .s/ D Rs 1 s

x.x/ dx .x/ dx

;

then f satisfies these conditions and

Rb

F .a; b/ D bf .a=b/ D Ra b

x.x=b/ dx

a

defines a mean. Moreover, if ing, then f < f on .0; 1/.

.x=b/ dx

is positive continuous on .0; 1 and = is strictly increas-

The means generated by (3.1) craft continuous analogs to some discrete means, thereby providing a powerful method to study the means systematically. As the chapters progress, we will often translate between discrete and continuous mathematics in search of the effective approach to solve problems.

Exercises 1. Let f .t/ be defined by (3.1). Find another proof of its monotonicity. 2. Putnam Problem 1957-B3. For f .x/ a positive, monotonic decreasing function defined in 0  x  1 prove that R1

xf 2 .x/ dx

0

R1 0

xf .x/ dx

R1

 R0 1 0

f 2 .x/ dx f .x/ dx

:

3. Putnam Problem 1979-B2. Let 0 < a < b. Evaluate lim

t !0

Z

1 0

Œbx C a.1

t

x/ dx

 1=t

:

4. Slope mean: Define S.a; b/ as the slope of the line which bisects the angle formed by the lines y D ax and y D bx. Find an explicit form for S.a; b/ and then prove that S.a; b/ is a mean and satisfies the inequality H.a; b/  S.a; b/  A.a; b/: 5. We see that M2=3 .a; b/  I.a; b/. Is it possible to find 2=3 < t < 1 such that I.a; b/ < M t .a; b/? 6. For 1 < ˛ < 1=2 or 2 < ˛, show that S˛ .a; b/  M.˛C1/=3 .a; b/:

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3. Means Generated by an Integral

7. Show that  1 2 .2A.a; b/ C G.a; b//  I.a; b/  A.a; b/ C 1 3 e

2 e



G.a; b/:

8. Monthly Problem 11347 [2008, 167]. Determine all ordered 4-tuples .˛; ˇ; ; ı/ of positive numbers with ˛ > ˇ and > ı such that for all distinct positive a and b, I >

p ˛A C ˇG > .A G ı /1=. Cı/ > AG: ˛Cˇ

9. Let agm.a; b/ be the Gauss arithmetic-geometric mean. Prove that L.a; b/  agm.a; b/ 

A.a; b/ C G.a; b/  N.a; b/: 2

10. Let 0 < a < b and x; y  1. Prove that b xCy bx

axCy xCy  A.a; b/y x a x

where equality holds if and only if x D y D 1.

R ln b 11. Let 0 < a < b. Applying the midpoint and trapezoidal approximations to ln a e x dx (see Figure 3.3), prove that G.a; b/ < L.a; b/ < A.a; b/. Furthermore, prove that L.a; b/ < M1=3 .a; b/ via Simpson’s Three-Eighths Rule [1, p. 195].

Figure 3.3. Areas in the figure 12. (Some series involving means).Let x D .b P 1 2k (a) ln.A=G/ D 1 kD1 2k x , P1 1 (b) ln.I=G/ D kD1 2kC1 x 2k , P1 1 (c) ln.A=I / D kD1 2k.2kC1/ x 2k .

a/=.b C a/. Prove that

13. Open Problem. Notice that

ln I.a; b/ D

Z

1 0

lnŒta C .1

t/b dt:

Using various numerical methods to approximate the integral enables us to obtain some new inequalities. For example,

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(a) For a ¤ b, show that

(b) For a ¤ b, show that


1=3 A2 .a; b/G.a; b/ < I.a; b/; I 2 .a; b/
q > 0; x  1/ (4.1) x 1 in this way, we see the numerator of the derivative is .p

q/x pCq

1

px p

1

C qx q

1

;

where the positivity is not immediately evident. In general, the derivative of a quotient is quite messy and the process to show monotonicity can be tedious. In this chapter, we introduce a general method for proving the monotonicity of a large class of quotients. Because of the similarity to the hypotheses to those of l’Hˆopital’s rule, we refer to the following theorem as the “L’Hˆopital Monotone Rule.” L’Hˆopital’s Monotone Rule (LMR) Let f; g W Œa; b ! R be continuous functions that are differentiable on .a; b/ with g0 ¤ 0 on .a; b/. If f 0 =g0 is increasing (decreasing) on .a; b/, then the functions f .x/ g.x/

f .b/ g.b/

and

f .x/ g.x/

f .a/ g.a/

are likewise increasing (decreasing) on .a; b/. Proof. We may assume that g0 .x/ > 0 and f 0 .x/=g0 .x/ is increasing for all x 2 .a; b/. By the mean value theorem, for a given x 2 .a; b/ there exists c 2 .a; x/ such that f .x/ g.x/

f .a/ f 0 .c/ f 0 .x/ D 0  0 ; g.a/ g .c/ g .x/

and so f 0 .x/ g.x/


g.a/

g0 .x/ f .x/


f .a/  0:

33

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Therefore, 

f .x/ g.x/

f .a/ g.a/

0

D

f 0 .x/.g.x/

g.a// .g.x/

g0 .x/.f .x/ g.a//2

f .a//

 0:

This shows that .f .x/ f .a//=.g.x/ g.a// is increasing on .a; b/ as desired. Along the same lines, we can show that .f .x/ f .b//=.g.x/ g.b// is increasing. Here we assume that a and b are finite. Along the same path, the rule can be extended easily to the case where a or b is infinity. LMR first appeared in Gromov’s work [1] for volume estimation in differential geometry. Since then, LMR and its variants have been used in approximation theory, quasi-conformal theory and probability (see [3] and the references cited therein). But, for most analysis students, LMR is not as well known as it should be. To popularize and promote this monotonicity rule, we present four examples within the realm of elementary analysis. The reader can see the wide applicability of LMR and find more applications in the exercises. Our first example is to confirm the monotonicity of (4.1) by using LMR. To see this, let f .x/ D x p 1 and g.x/ D x q 1. Consider G.x/ D



f .x/=g.x/; p=q;

if x ¤ 1; if x D 1:

We have f 0 .x/=g0 .x/ D .p=q/x p q , which is increasing on .1; 1/ as long as p > q > 0. Hence, by LMR, G.x/ is increasing on .1; 1/. In particular, G.x/ > G.1/ D p=q. This yields the following bonus inequality: xq 1 xp 1 > ; for x > 1; p > q > 0: p q Our second example comes from Wilker’s inequality, which appears in [5]. Wilker asked for a proof that 

sin x x

2

C

tan x > 2; x

.0 < x
0;

cos4 x/ C sin x



 cos x C 2x tan2 x sec2 x

x 1 sin x cos4 x

f 00 .x/=g00 .x/ is increasing on .0; =2/. Thus, using LMR twice and noticing that f .0/ D f 0 .0/ D g.0/ D g0 .0/ D 0, we deduce that f 0 .x/=g0 .x/ and so F .x/ D f .x/=g.x/ is increasing on .0; =2/. Our third example is related to the well-known Jordan’s inequality: 2 x  sin x  x; 

.0  x 

 /: 2

Now, by using LMR, we improve Jordan’s inequality as follows. If x 2 Œ0; =2, then 2 x

C

 2 x. 2

2x/

 sin x 

2 x

C

2 x. 2

2 x 

C

1 x. 2 3

4x 2 /

 sin x 

2 x 

C

 2 x. 2 2

4x 2/;

2 x 

C

1 x. 4 2 5

 sin x 

2 x 

C

 2 x. 4 5

16x 4/;

16x 4/

2x/;

where the coefficients are all the best possible. We show the third inequality in the list. The other two can be proven similarly. It is easy to see that the third inequality is equivalent to sin x 2 1  2 x    : 2 5  4 16x 4 5

Let f1 .x/ D sin x=x; g1 .x/ D 16x 4 ; f2 .x/ D sin x

(4.3)

x cos x; g2.x/ D x 5 . Then

1 sin x x cos x 1 f2 .x/ f10 .x/ D D ; 0 5 g1 .x/ 64 x 64 g2 .x/ f20 .x/ 1 sin x D : 0 g2 .x/ 5 x3

Since sin x=x 3 is decreasing on .0; =2/, we have f20 .x/=g20 .x/ is decreasing on .0; =2/. By LMR, f2 .x/=g2 .x/ is decreasing on .0; =2/ and so is f10 .x/=g10 .x/. Applying LMR once more, we have sin x  f1 .x/ f1 .=2/ 2 D 4x g1 .x/ g1 .=2/  16x 4 decreasing on .0; =2/. Thus, the desired inequality (4.3) follows from sin x x x!0  4

lim

 2

16x 4

D



2 5

;

sin x x x!=2  4

lim

 2

16x 4

D

1 : 2 5

Our final example is Monthly Problem 11369 [2], which was recently posed by Donald Knuth: Prove that for all real t, and all ˛  2, e ˛t C e

˛t

2  .e t C e t /˛

2˛ :

(4.4)

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First, observe that the inequality is invariant when t is replaced by sufficient to show the inequality holds for t  0. Next, set x D e t . Hence (4.4) is equivalent to .x 2 C 1/˛ x 2˛ x˛

To apply LMR, let f .x/ D .x 2 C 1/˛

x 2˛

1

 2˛

t. Hence, it is

(4.5)

2:

1 and g.x/ D x ˛ . Then

f 0 .x/ 2x.x 2 C 1/˛ D g0 .x/ x˛

1

2x 2˛

1

1

:

Define

  f 0 .x/ 1 ˛ 1 F .x/ D 0 D 2x x C 2x ˛ : g .x/ x To show that f 0 .x/=g0 .x/ is increasing for x > 1, it suffices to prove that F 0 .x/  0 for x > 1. Indeed,     2.˛ 1/ 1 ˛ 1 F 0 .x/ D 2 x C ˛ 2˛x ˛ 1 : x x2 C 1

Using Bernoulli’s inequality

.1 C t/p  1 C pt

for t  0; p  1;

which can be proved via LMR by considering .1 C t/p =.1 C pt/ as well, we have     1 ˛ 1 1 ˛ 1 xC D x˛ 1 1 C 2  x ˛ 1 C .˛ 1/x ˛ 3 : x x Therefore, for x > 1 and ˛  2, F 0 .x/ 

2.˛

Now, LMR deduces that f .x/ g.x/

1/.˛

2/x ˛ x2 C 1

3

.x 2

f .1/ .x 2 C 1/˛ x 2˛ D g.1/ x˛ 1

1/

 0:

2˛ C 1

is increasing for x > 1. In particular, applying l’Hˆopital’s rule yields lim

x!1

f .x/ g.x/

f .1/ .x 2 C 1/˛ x 2˛ D lim x!1 g.1/ x˛ 1

2˛ C 1

D 2˛

2:

Thus,

f .x/ f .1/ .x 2 C 1/˛ x 2˛ 2˛ C 1 D  2˛ g.x/ g.1/ x˛ 1 This implies the desired (4.5) and therefore proves (4.4).

2:

To end this chapter, we give an example to show that f 0 .x/=g0 .x/ is not monotonic but f .x/=g.x/ is still monotonic. Consider Z x f .x/ D x .1 C cos.1=t// dt; g.x/ D x: 0

Then f .x/=g.x/ is increasing on .0; 1/, whereas f 0 .x/=g0 .x/ is not monotonic on .0; 1/. This also shows that the converse of LMR is false.

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4. The L’H^ opital Monotone Rule

Exercises 1. For x >

1, prove that F .x/ D

x ln.1 C x/ x ln.1 C x/

is increasing. 2. Prove that f .x/ D .1 C x/1=x is decreasing but g.x/ D .1 C x/1C1=x is increasing on .0; 1/. 3. For 0 < x < =2, prove that
3 (a) sinxx > cos x. (b)

2x sin x

C

x tan x

> 3.

Remark. Using t 2  2t

1 with t D x= sin x in the second inequality yields  x 2 x C > 2: sin x tan x

4. Monthly Problem 11009 [2003, 341; 2005, 92–93]. Let a > b  c > d > 0. Prove that ax b x f .x/ D x c dx is increasing and convex for all x. Remark. One may show that f .x/ is increasing based on LMR. Without loss of generality, assume that d D 1, otherwise divide the ratio by d x . Now, let y D c x ; ˛ D ln b= ln c; ˇ D ln a= ln c. Then ˇ > ˛  1. One can rewrite f .x/ as F .y/ WD

yˇ y

y˛ 1

y 2 .0; 1/ [ .1; 1/:

5. In Wilker’s inequality, prove that
sin x 2 C tanx x x x 3 tan x

2



6. Kober’s inequality claims that cos x  1 following refinement: 1

2  2 xC x.  2

16 ; .0 < x < =2/: 4 2x= for 0 < x < =2. Prove the

2x/  cos x  1

2 2 x C 2 x.  

2x/:

7. For 0 < x < 1, prove that x   4 x x < tan < : 2  1 x 2 2 1 x2

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8. Monthly Problem 11308 [2007, 640; 2009, 183–184]. Let n be a positive integer. For 1  i  n, let xi be a real number in .0; =2/ and ai be a real number in Œ1; 1/. Prove that 2ai Y ai n  n  Y xi xi C > 2: sin xi tan xi i D1

i D1

9. For =n  x  

=n, prove that sin nx  n sin x

10. Prove that Pn (a) PkD1 n (b) kD1

sin kx k sin kx k

> 0; 

1 6

1 : 3

for all 0 < x < ; n D 1; 2; : : : .

sin x;

for all =n < x < 

=n; n D 1; 2; : : :.

References [1] M. Gromov and M. Taylor, Finite propagation speed, kernel estimates for functions of the Laplace operator, and the geometry of complete Riemannian manifolds, J. Diff. Geom., 17 (1982) 15–53. [2] D. Knuth, Problem 11369, Amer. Math. Monthly, 115 (2008) 567. [3] I. Pinelis, On L’Hˆopital-type rules for monotonicity, J. Inequal. Pure Appl. Math. 7 (2006), Article 40. Available at jipam.vu.edu.au/article.php?sid=657 [4] J. S. Sumner, A. A. Jagers and J. Anglesio, Inequalities involving trigonometric functions, Amer. Math. Monthly, 98 (1991) 264–267. [5] J. B. Wilker, Problem E3306, Amer. Math. Monthly, 96 (1989) 55.

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5 Trigonometric Identities via Complex Numbers The shortest path between two truths in the real domain passes through the complex domain. — J. Hadamard

Trigonometric identities often arise in a variety of branches of mathematics, including geometry, analysis, number theory and applied mathematics. The classical standard tables [4] and [5] have collected many of these identities, such as   n Y1 k n sin D n 1: (5.1) n 2 kD1

However, the proofs for these identities are widely scattered throughout books and journals. In this chapter, we present a comprehensive study on proving trigonometric identities via complex numbers. To keep the chapter to a reasonable page limit, we only examine several families of identities in [4, 5]. But, we emphasize that the methods used here can be applied to prove many identities. As applications of these identities, we explicitly evaluate the Poisson integral as well as establish some more sophisticated finite trigonometric identities of the sort   n X1 .n 1/.n C 1/ k csc2 D : (5.2) n 3 kD1

5.1 A Primer of complex numbers Historically, complex numbers were first introduced by Cardano in the coursep of solving quadratic and cubic equations. He noticed that if the 1 were p “new pnumbers” 1 ˙ treated as ordinary numbers with the added rule that 1
1 D 1, they did solve the equation x 2 C 2x C 2 D 0. However, p for a long period of time, it was felt that no meaning could actually be assigned to i D 1 and it was therefore termed “imaginary”. In 1748, Euler discovered the amazing identity e i D cos  C i sin :

(5.3)

Such a close connection between i and trigonometric functions was quite startling and perhaps indicated that some meaning should be attached to these “imaginary” numbers. 39

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Finally, the work of Gauss and others clarified the meaning of complex numbers. In the complex domain, Gauss eventually established the fundamental theorem of algebra, which claims that any polynomial equation has at least one solution. In the middle of the nineteenth century, starting to extend the methods of the differential and integral calculus by including complex variables, Cauchy and others founded the branch of mathematics called the Theory of Complex Analysis. It has been acclaimed as one of the most harmonious theories in the abstract sciences. In the following, we review some selected basic facts on complex numbers.

5.1.1 Multiplication and De Moivre’s Formula Recall the geometric representation of a complex number. Let  denote the angle between the vector z D x C i y and the positive real axis, where 0   < 2 . The angle  is called the argument and is denoted  D arg z: Thus, we have z in a polar coordinate representation: z D r .cos  C i sin /; (5.4) where r D jzj D

p

x 2 C y 2 ; tan  D y=x:

(5.4) gives us a geometric method for performing complex multiplication: for any complex numbers z1 and z2 , jz1 z2 j D jz1 jjz2 j and arg.z1 z2 / D arg z1 C arg z2

.mod 2/:

(5.5)

In general, this leads to the well-known De Moivre’s formula: for a positive integer n, .cos  C i sin /n D cos n C i sin n:

(5.6)

Now, appealing to Euler’s formula (5.3), we get the following two popular identities: 1 C e ix D 1 C .e ix=2/2 D 2 cos2 .x=2/ C 2i sin.x=2/ cos.x=2/ D 2 cos.x=2/e ix=2 (5.7) and 1 e ix D 1 .e ix=2/2 D 2 sin2 .x=2/ 2i sin.x=2/ cos.x=2/ D 2i sin.x=2/e ix=2: (5.8)

5.1.2 Roots of Unity The solutions of the equation z n 1 D 0 are called the nth roots of unity. De Moivre’s formula can be reversed to find all roots of unity, namely f1; !; ! 2 ; : : : ; ! n

1

g;

where ! D e i.2=n/ is called the nth primitive root of unity. Geometrically, these numbers are the vertices of a regular n-gon inscribed in the unit circle jzj D 1, and have many nice properties that interconnect algebra, geometry, combinatorics and number theory. Since zn

1 D .z

1/.z n

1

C zn

2

C


C 1/;

(5.9)

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5. Trigonometric Identities via Complex Numbers

clearly, the factor z

1 accounts for the root 1, and so the other roots satisfy zn

1

C zn

2

(5.10)

C


C 1 D 0;

which yields zn

1

C zn

2

C


C 1 D .z

!n

!/


.z

1

(5.11)

/:

Now with the stage appropriately set, we show how several families of trigonometric identities that appeared in [4, 5] fall into place.

5.2 Finite Product Identities We begin with identity (5.11). Setting z D 1 and then taking modulus gives nD

n Y1

kD1

j1

! k j:

(5.12)

In the complex plane, this shows that the product of the distance from one vertex on the unit circle to other vertices is the constant n. In view of (5.8), 1

k

! D1

e

i.2k=n/



k D 2i sin n



e i k=n;

we see that (5.1) follows from (5.12) immediately. Next, for z D 1 Œ1 2

. 1/n  D . 1/n

n Y1

1

kD1

1, (5.11) gives

.1 C ! k /:

By (5.7), we have k

1C! D1Ce and so

n Y1

kD1

Since



k D 2 cos n

ˇ  ˇ ˇ ˇ ˇcos k ˇ D 1 ˇ n ˇ

n Y1

 k sin D 2n kD1   n Y k sin D 2n C 1 kD1   n Y k cos D 2n C 1

kD1





e i k=n;

. 1/n : 2n

     k .n k/ k D sin ; cos D n n n appealing to (5.12) and (5.13), we find that sin



i.2k=n/

cos

(5.13) 

.n

k/ n



;

p n ; n 2 1 p 2n C 1 ; 2n 1 : 2n

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Excursions in Classical Analysis

Next, notice that n Y1

kD0

  nY1 k 1  i.xCk=n/ sin x C D e n 2i kD0

D

n 1 1 Y i k=n e .2i /n

k

D !n

k

ix

kD0

1 D e i.n .2i /n Since !

i.xCk=n/

e

1/=2 i nx

 e 2xi n Y1

kD0



e i.



2k=n/

e 2xi

e i.



2k=n/



:

, setting z D e 2xi in (5.9) we have n Y1

kD0

 e 2xi

e i.

2k=n/



D e 2nxi

1;

so that n Y1

kD0

  1 k D .e  i=2 /n sin x C n .2i /n

1

.e nxi

e

nxi

/D

1 sin nx 2n 1

or equivalently sin nx D 2n

1

kD0

Replacing x by x C =2 in (5.14) yields sin nx D . 1/n=2 2n

1

n Y1

cos nx D . 1/

2

  k sin x C : n

  k cos x C ; n

kD0

.n 1/=2 n 1

n Y1

n Y1

kD0

(5.14)

when n is even;

  k cos x C ; n

when n is odd:

Finally, notice that the equation z 2n 1 D 0 has 2n roots evenly spaced in angle around the unit circle, with a separation of =n radians. Setting  D exp.i =n/, the factorization of z 2n 1 yields n Y1 z 2n 1 D .z  k /: kD n

Combining the conjugate factors gives z

2n

1 D .z

2

1/

n Y1

kD1

 z2



k 2z cos n





C1 :

Dividing by z n on both sides yields zn

z

n

D .z

z

1

/

n Y1

kD1

 z

2 cos



k n



Cz

1



:

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5. Trigonometric Identities via Complex Numbers

Setting z D e ix and appealing to De Moivre’s formula, we obtain sin nx D 2n sin x

1

n Y1

 cos x

kD1

Similarly, using the factorizations of z 2nC1 other product identities: .z

n  Y z2 1/



k n



(5.15)

:

1 and z 2n C 1 respectively, we can establish



2k 2z cos 2n C 1





C 1 D z 2nC1

1;

   n  Y 2k C 1 D z 2nC1 z 2 C 2z cos 2n C 1

1;

kD1

.z C 1/

cos

kD1

n Y1

kD0

 z2



.2k C 1/ 2z cos 2n





C 1 D z 2n C 1:

5.3 Finite Summation Identities We have illustrated how to obtain the finite product identities from the factorizations. In the following, we show how to establish the finite summation identities based on partial fractions. Since ! k D ! n k , we see that 1

f1; ! are also roots of z n

2

;!

; :::; !

.n 1/

g

1. Thus, zn D

1

n Y1

.1

zn

1

D

n X1

kD0

1

z n has the form of Ak ; z! k

where Ak are constants to be determined. Multiplying by 1 n X1

Ak

kD0

Since lim z! !

k

(5.16)

kD0

from which the partial fraction decomposition of 1 1

z! k /;

z n on both sides yields

1 zn D 1: 1 z! k

 1 zn 0; if i ¤ k; D i n; if i D k; 1 z!

we obtain Ak D 1=n and so 1 1

zn

D

n 1 1 X 1 n 1 z! k kD0

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or equivalently n 1

zn

D

In view of the identity

n 1 X

kD0

1 : 1 z! k

(5.17)

1 a ib a ib D D 2 ; aCib .a C i b/.a i b/ a C b2 setting z D xe i ˛ in (5.17) and then extracting the real part gives n X1 n.1 x n cos n˛/ 1 x cos.˛ C 2k=n/ D : n 2n 1 2x cos n˛ C x 1 2x cos.˛ C 2k=n/ C x 2

(5.18)

kD0

Replacing x by 1=x in (5.18) yields n X1 nx n .x n cos n˛/ x 2 x cos.˛ C 2k=n/ D : 1 2x n cos n˛ C x 2n 1 2x cos.˛ C 2k=n/ C x 2

(5.19)

kD0

Now, subtracting (5.19) from (5.18) we get n X1 n.1 x 2n / D 2x n cos n˛ C x 2n 1

1 x2 : 2x cos.˛ C 2k=n/ C x 2

(5.20)

n X1 n.1 x n cos n˛/ 1 x cos.˛ C 2k=n/ D : 1 2x n cos n˛ C x 2n 1 2x cos.˛ C 2k=n/ C x 2

(5.21)

1

kD0

Alternately, adding (5.19) and (5.20) we have

kD0

For ˛ D 0, we may specify (5.20) and (5.21), respectively, as n X1

kD0

and

1 n X1

kD0

n.1 C x n / 1 D ; 2x cos.2k=n/ C x 2 .1 x n /.1 x 2 /

(5.22)

1 x cos.2k=n/ n D : 2x cos.2k=n/ C x 2 1 xn

(5.23)

1

Another way of deriving a finite summation identity is to manipulate an existing finite product identity. For example, if we take the logarithm of both sides of (5.14), we obtain ln sin nx D .n

1/ ln 2 C

n X1

kD0

  k ln sin x C ; n

and then differentiate, the result is seen to be n 1 X

kD0

  k cot x C D n cot nx: n

(5.24)

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5. Trigonometric Identities via Complex Numbers

Replacing x by x C =2 in (5.24) gives n 1 X

kD0

    k n 1 tan x C D n tan nx C  : n 2

(5.25)

Differentiating (5.24) yields n X1

kD0

  k csc2 x C D n2 csc2 nx: n

This establishes (5.2) as follows:   n 1 X 2 k D lim .n2 csc2 nx csc x!0 n kD1

csc2 x/ D

.n

1/.n C 1/ : 3

Similarly, differentiating (5.25) gives n X1

kD0

    k n 1 sec2 x C D n2 sec2 nx C  : n 2

This identity will allow us to derive some more sophisticated trigonometric identities (see Exercise 14).

5.4 Euler’s Infinite Product One of the highlights in Euler’s 1748 work, Introduction to Analysis of the Infinite, is the representation of the sine as an infinite product  1  Y x2 sin x D x 1 : (5.26) k2  2 kD1

This result is the key to many other expansions and identities. It opens up the theory of infinite products and partial fraction decompositions of transcendental functions. We will see some of its remarkable corollaries in the next chapter, which includes the partial fractions of trigonometric functions and the summation of the Riemann zeta function for all even positive integers. We present here an elementary derivation of (5.26). The first step is to establish the following fact: for an odd natural number n, sin nx D Pn .sin x/; where Pn .t/ is an nth degree polynomial. For n D 3, we see the required polynomial is just P3 .t/ D 3t 4t 3 since sin 3x D 3 sin x 4 sin3 x. In general, for any positive integer m, we start with De Moivre’s formula (5.6) cos mx C i sin mx D .cos x C i sin x/m and take its imaginary part, which is     m m sin mx D sin x cosm 1 x sin2 x cosm 1 3

3

xC

  m sin4 x cosm 5

5

 x C


:

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In particular, if m D 2n C 1, appealing to cos2 x D 1 sin2 x, we arrive at  .2n C 1/2 12 2 sin.2n C 1/x D .2n C 1/ sin x 1 sin x 3Š C

Œ.2n C 1/2

12 Œ.2n C 1/2 5Š

32 



4

sin x C


:

That is, sin.2n C 1/x D .2n C 1/ sin x
P .sin2 x/;

(5.27)

where P .t/ is an nth degree polynomial with P .0/ D 1. Note that if r1 ; r2; : : : ; rn are roots of P .t/, then in factored form  n  Y t P .t/ D 1 : rk kD1

This is self evident, since substituting t D 0 gives P .0/ D 1, just as substituting t D rk yields P .rk / D 0 for k D 1; 2; : : : ; n. Next, due to (5.27), if x is a solution of sin.2n C 1/x D 0, but sin x ¤ 0, then sin2 x is a solution of P .t/ D 0. Consider xD

k ; 2n C 1

for k D 1; 2; : : : ; n:

For each of these values we have .2n C 1/x D k , and thus sin.2n C 1/x D 0, while 0 < x < =2 implies that we get n distinct positive roots for P .t/, namely r1 D sin2 .=.2n C 1//; r2 D sin2 .2=.2n C 1//; : : : ; rn D sin2 .n=.2n C 1//: Thus, in factored form, sin.2n C 1/x D .2n C 1/ sin x

n Y

kD1

Replacing x by x=.2n C 1/ yields  Y  n x sin x D .2n C 1/ sin 1 2n C 1 kD1

1

sin2 x sin2 .k=.2n C 1//

!

! sin2 .x=.2n C 1// ; sin2 .k=.2n C 1//

:

(5.28)

which is true for all natural numbers n. For a reader who is familiar with the uniformly convergence theorem for infinite products, which justifies that interchanging the product and the limit is valid, it is easy to see that (5.26) follows from (5.28) by letting n ! 1. To avoid using such an advanced result, n n we rewrite sin x as AK BK , where .K C 1/ > jxj; n > K and !   Y K 2 x sin .x=.2n C 1// n AK D .2n C 1/ sin 1 ; 2n C 1 sin2 .k=.2n C 1// kD1 ! n Y sin2 .x=.2n C 1// n BK D 1 : sin2 .k=.2n C 1// kDKC1

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For fixed K, 1  k  K, we have

and



x lim .2n C 1/ sin n!1 2n C 1 lim

n!1

Thus,

Dx

sin2 .x=.2n C 1// x2 : D k2  2 sin2 .k=.2n C 1//

n lim AK Dx

n!1



K  Y 1

kD1

x2 k2  2



def

D AK :

n n n Since BK D sin x=AK , we see that limn!1 BK D BK exists. To estimate BK , using the well-known inequality 2  < sin  < ; for 0 <  < =2;  we have x x2 sin2 < 2n C 1 .2n C 1/2 and 4 k2  2 k > 2
.k D K C 1; : : : ; n/: sin2 2n C 1  .2n C 1/2 Thus, sin2 .x=.2n C 1// x2 1 > 1 4k 2 sin2 .k=.2n C 1// and   n Y x2 n 1 > BK > 1 : 4k 2 kDKC1 P 2 Since x =.4k 2 / converges, we see that the infinite product   1 Y x2 1 4k 2 kDKC1

converges to 1 as K goes to infinity and so

lim BK D 1:

K!1

Thus, we finally establish Euler’s infinite product (5.26) by  1  Y x2 sin x D lim AK BK D x 1 : K!1 k2  2 kD1

As a corollary, from cos x D sin 2x=.2 sin x/, we find that  1  Y 4x 2 cos x D 1 : .2k 1/2  2

(5.29)

kD1

Moreover, for x D =2, (5.26) leads to the famous Wallis’s formula, namely  1  Y 2 1 1
3
3
5
5
7
7


D 1 D : 2  4k 2
2
4
4
6
6
8




(5.30)

kD1

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5.5 Sums of inverse tangents The evaluation of inverse tangent sums of the form 1 X

arctan f .k/

kD1

for a real function f has appeared in the Mathematical Association of America journal problem sections from time to time. For example, in a brief search in the Monthly one will find Raisbeck’s E930 back in 1950 and Borwein and Baily’s 11438 in 2009. There are very good chances that such problems can be solved by using the addition and difference formulas for inverse tangents. But the principle of the argument for complex multiplication provides an equally simple method. Q Let zk D ak Ci bk .1  k  n/ with ak ; bk 2 R. Consider Pn D nkD1 zk . Appealing to (5.5), we have   n n X X bk arg.Pn / D arg.zk / D arctan .mod 2 /: (5.31) ak kD1

kD1

Q Similarly, for a convergent infinite product P D 1 kD1 zk ;   1 X bk arg.P / D arctan .mod 2 /: ak

(5.32)

kD1

Of course, throughout arctan x will always denote the principal value. It seems that (5.31) and (5.32) are not as widely known as they deserve to be. To help spread the word, we demonstrate two examples. Example 1 (Monthly Problem E930 [1950,483; 1951, 262–263]). If n Y

kD1

show that

n X

kD1

.x C rk / D

arctan rk D arctan

n X

aj x n

j

(5.33)

;

j D0

a1 a0

a3 C a5 a2 C a4




:




The published solution by Apostol [8] involves the addition formula for arctan x, the elementary symmetric function of the n variables and mathematical induction. Here we present a short proof based on (5.31). For x D 1= i , (5.33) becomes n Y

kD1

.1 C i rk /= i D

n X

aj i j

n

;

j D0

which is equivalent to n Y

kD1

.1 C i rk / D

n X

j D0

aj i j D .a0

a2 C a4




/ C i .a1

a3 C a5




/:

Now, the required identity follows from applying (5.31) to this equation.

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Example 2.

Replacing x by  z in (5.26) yields 1  Y sin  z D  z 1 kD1

z2 k2



(5.34)

:

For z D x C i y, sin  z D sin  x cosh y C i cos  x sinh y: Appealing to (5.32), we recover an elegant formula in [3] by Glasser and Klamkin, namely     1 y X 2xy 1 tanh y D arctan tan .mod 2 /; arctan k2 x2 C y2 x tan  x kD1

p which, for x D y D 1= 2, becomes   1 X 1  arctan D k2 4 kD1

p ! tanh = 2 arctan p tan = 2

.mod 2 /:

(5.35)

This solves the Monthly Problem 3379, 1990 posed by Chapman.

5.6 Two Applications In contrast to Fourier series, identities such as (5.20) and (5.22) are finite sums over the angles equally dividing the plane. They often crop up in discrete Fourier series and number theory. We conclude this chapter with two applications of (5.20) and (5.22). Our first application shows how to use (5.22) to evaluate the following Poisson integral Z 2 d ID ; .jxj < 1/ 1 2x cos  C x2 0 via the limit of a Riemann sum. This integral is usually evaluated by using either the residue theorem or Fourier series. Since 1 2x cos  C x 2  .1 jxj/2 > 0 for jxj < 1; the integrand is continuous and integrable. Partition the interval Œ0; 2  into n equal subintervals by the partition points   2k k D W 1kn : n By (5.22), we get 0 n 1 2 X @ I D lim n!1 n kD0 1

2x cos

2 n.1 C x n / n!1 n .1 x n /.1 x 2 / 2 D : 1 x2 D lim

1 

2k n



C x2

1 A

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The second application uses (5.20) and (5.22) to deduce some more sophisticated finite trigonometric identities like (5.2). Regrouping (5.22) as n 1 X

kD1

2x cos

1

1 

2k n



C x2

D

n.1 C x n / .1 x n /.1 x 2 /

1 x/2

.1

;

and letting x ! 1, we re-prove (5.2) as n X1

kD1

csc2



k n





D 4 lim

x!1

n.1 C x n / .1 x n /.1 x 2 /

1 .1

x/2



D

.n

1/.n C 1/ ; 3

where l’H¨opital’s rule has been used to calculate the limit. Similarly, we assume that n is odd first. Setting ˛ D  in (5.20), we have n X1

kD0

1 C 2x cos

Letting x ! 1, we obtain n X1

sec2

kD0



1 



k n

2k n



C x2

D 4 lim

x!1

D

n.1 x 2n / : .1 C x n /2 .1 x 2 /

n.1 x 2n / D n2 ; .1 C x n /2 .1 x 2 /

(5.36)

where l’H¨opital’s rule has been used again. If n is even, setting ˛ D  in (5.20) and then regrouping yields n X1

kD0;k¤n=2

1 C 2x cos

Letting x ! 1, we have

1 

n 1 X

2k n



sec

kD0;k¤n=2

2

C

x2



k n

D



n.1 x 2n / .1 C x n /2 .1 x 2 /

D

.n

1 .1

1/.n C 1/ : 3

x/2

:

(5.37)

Appealing to the trigonometric identities 1 C tan2 x D sec2 x and 1 C cot2 x D csc2 x, by (5.35)–(5.37), we find that   n 1 X k .n 1/.n 2/ cot2 D n 3 kD1

and

n X1

kD0

2

tan



k n



D

(

n.n

1/;

.n 1/.n 2/ 3

if n is odd; if n is even, k ¤ n=2.

(5.38)

If we replace the power 2 in these identities by an arbitrary positive even power, finding an explicit closed form becomes more difficult. We will investigate these problems systematically by using the generating functions in Chapter 14. Remarkably, the identity (5.22) also provides us an experimental method to search for more trigonometric identities (see Exercises 15 and 16).

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Exercises 1. Let ! D e i.2=n/ and n be a positive integer. Prove that Q 1 (a) nkD1 j1 ! k j2k D nn .
n.nC1/=2 n=2 Q 1   k k (b) nkD1 sin n D 12 n . ˇ  ˇ
n.nC1/=2 Q 1 ˇ ˇk (c) nkD1 D 12 Œ1 . 1/n n=2. ˇcos k n ˇ

2. Prove that the sum of the squared distance from any point P on the unit circumcircle of a regular n-gon to its vertices is 2n. 3. Determine the closed forms of n X

r k sin.kx/

n X

and

kD1

r k cos.kx/:

kD0

Find their limits as n ! 1 if jr j < 1. 4. Let n  3. Find

n Y1

kD1

.1 C 2 cos.k=n// :

5. Define Pn .x/ by 2 n  X n x 2k .1 Pn .x/ D k

x/2.n

k/

:

kD0

Show that for 0  x  1,



2n n

sin2 .x=2kC1 /

Qk

Pn .x/  6. Prove that   Q 1 (a) nkD1 sin .2kC1/ D 2n   Q 1 (b) nkD1 sin .3kC1/ D 3n

1 22n



:


1 n 1 . 2 p 3 . 2n

7. Prove that

Q1



(a)

sin x x

D

(b)

sin x x

D cos2 .x=2/ C

kD1

cos

x 2k



: P1

kD1

mD1

cos.x=2m /.

8. Putnam Problem 1949-A6. For every x, prove that 1 Y sin x 1 C 2 cos.2x=3n / D : x 3 nD1

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Excursions in Classical Analysis

Comment. A new approach to this problem is to apply the Fourier transform to the product expansion Exercise 7(a), then use the convolution and delta distributions. Based on this method, one can generate a family of similar identities. For example,   1 Y sin x 1 x 3x 5x D 2 cos k C 2 cos k C 2 cos k : x 6 6 6 6 kD1

The details and further examples can be found in [7]. 9. Monthly Problem 11383 [2008, 757]. Show that ! p p 1 X 1 C n2 C 2n n2 C 4n C 3  D : arccos .n C 1/.n C 2/ 3 nD1

Remark. The answer seems to be =6. 10. Monthly Problem 5486 [1967, 447; 1968, 421–422]. Prove that n X

csc

kD0

2



.2k C 1/ 2n



2

2n X

and

Dn

csc

2

kD1



k 2n C 1



D

4 n.n C 1/: 3

11. Prove that sinh x D x

 1  Y x2 1C 2 2 ; n  nD1

cosh x D

1  Y 1C

nD1

 4x 2 : .2n 1/2  2

Use the latter result to prove that lim n

n!1

n  Y 1

1 5 C 2 k 4k

kD1



D

cosh. / : 

12. Prove that 1 X

arctan

kD1

13. Prove that



.2k

1 X

kD1

2xy 1/2 x 2 C y 2





 y  x D arctan tan tanh 2 2

x . 1/ arctan 2n C 1 k



.mod 2 /:

 x  D arctan tanh ; 4

where it is assumed that =2  arctan t  =2. 14. Putnam Problem 1986-A3. Evaluate 1 X

kD1

arccot.k 2 C k C 1/;

where arccotx for x  0 denotes the number  in the interval 0 <   =2 with cot  D x.

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15. Prove that n X1

kD0

    k n 1 2 2 tan x C D n sec nx C  n 2 2

n:

In general, let Sm .x/ D Prove that SmC1 .x/ D

n X1

kD0

1 0 S .x/ m m

  k tan xC : n m

Sm

for m D 2; 3; : : : :

1;

16. In view of (5.22), define Fp .x/ D

n X1

kD0

.1

1 x2 : 2x cos.2k=n/ C x 2 /p

If n ¤ 0 .mod 4/, show that n X1

kD0

secp .2k=n/ D 2p

1

. i /p .Fp .i //:

(5.39)

Moreover, prove the following recursive relation Fp .x/ x d F1Cp .x/ D C 1 x2 p dx



Fp .x/ 1 x2



:

This enables us to compute Fp .x/ in succession. Comment. Using the sampling theorems with second-order discrete eigenvalue problems, Hassan in [6] derives some trigonometric identities such as (5.39). With the help of Mathematica or Maple, Exercises 15 and 16 provide us an elementary experimental method to search for this kind of trigonometric identities. In the course of doing this, one should not miss     n 1 n 1 X X 2n.n C 1/ 2 .2k C 1/ 2 2 .2k C 1/ sec D 2n ; sec D : 4n 4n C 2 3 kD0

kD0

References [1] H. Chen, On a new trigonometric identity, Internat. J. Math. Ed. Sci. Tech., 37 (2006) 215–223. [2] W. Dunham, Euler: The Master of Us All, The Mathematical Association of America, 1990. [3] M. Glasser and M. Klamkin, On some inverse tangent summations, Fibonacci Quarterly, 14 (1976) 385–388. [4] I. Gradshteyn and I. Ryzhik, Table of Integrals, Series, and Product, 6th edition, edited by A. Jeffrey and D. Zwillinger, Academic Press, New York, 2000. [5] E. Hansen, A Table of Series and Products, Prentice Hall, New Jersey, 1975.

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[6] H. Hassan, New trigonometric sums by sampling theorem, J. Math. Anal. Appl., 339 (2008) 811–827. [7] K. Morrison, Cosine products, Fourier transforms, and random sums, Amer. Math. Monthly, 102 (1995) 716–724. [8] G. Raisbeck, Problem E930, Amer. Math. Monthly, 58 (1951) 262–263.

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6 Special Numbers Numbers constitute the only universal language.

— N. West

All numbers are not created equal. Some sequences of numbers in analysis seem quite complicated and mysterious, yet they have played an important role and have made many unexpected appearances in analysis and number theory. A large body of literature exists on these special numbers but much of it in widely scattered books and journals. In this chapter we investigate three of these special sets of numbers, the Fibonacci, harmonic and Bernoulli numbers. We derive some basic results that are readily accessible to those with a knowledge of elementary analysis. These materials will serve as a brief primer on these special numbers and lay the foundation for many of the latter chapters. We start with the Fibonacci numbers, which are defined by a recursive relation. Then we examine the harmonic numbers and some of their important features. Finally, we investigate the Bernoulli numbers, which are attributed to Jacob Bernoulli in the discussion of closed forms for the sums of kth powers of consecutive integers.

6.1 Generating Functions Special numbers are often defined by recurrences. But applications may well require explicit closed forms of the general terms. Generating functions provide one of the best routes to finding these formulas. In the preface of his fascinating book [10], H. Wilf writes “The full beauty of the subject of generating functions emerges from tuning in on both channels: the discrete and the continuous.” He further observes that “A generating function is a clothesline on which we hang up a sequence of numbers for display.” Thus, roughly speaking, generating functions will transform problems about sequences into problems about functions. This is useful because we have piles of mathematical machinery for manipulating functions. Definition 1 The ordinary generating function for the infinite sequence fa0 ; a1 ; a2 ; : : :g is the power series G.x/ D a0 C a1 x C a2 x 2 C


: Now it is clear that the unknown numbers an are arranged neatly on the clothesline: indexing from 0, an is the coefficient of x n in the generating function. Therefore, once 55

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we identify the function G.x/, we will be able to find the explicit formula for the an ’s by expanding G.x/ as a power series, and vice versa. Recall that the sum of an infinite geometric series, for jxj < 1, 1 C x C x2 C x3 C


D

1 1

x

:

This indicates that the generating function of the sequence f1; 1; 1; : : :g is 1=.1 the other hand, if G.x/ D ln.1 x/, we have

x/. On

G .n/ .0/ 1 D : nŠ n Therefore

ln.1

x/ generates the sequence f1; 1=2; 1=3; : : :g.

Sometimes a sequence fan g has a generating function whose properties are quite complicated, while the related sequence fan =nŠg has a generating function that is quite simple. This leads to beneath definition: Definition 2 The exponential generating function for the infinite sequence fa0 ; a1 ; a2 ; : : :g is the power series a2 a3 a1 G.x/ D a0 C x C x 2 C x 3 C


: 1Š 2Š 3Š Thus the series for e x is the exponential generating function for f1; 1; 1; : : :g, and the series for cos x and sin x, for the periodic sequences f1; 0; 1; 0 : : :g and f0; 1; 0; 1; : : :g, respectively. The magic of generating functions is that we can carry out all sorts of manipulations on sequences by performing mathematical operations on their associated generating functions. For more detailed theories and applications of generating functions, please refer to [10]. In the following, using Fibonacci, harmonic and Bernoulli numbers as case studies, we explore how generating functions lead to simple, direct proofs of basic properties of these sets of numbers.

6.2 Fibonacci Numbers Fibonacci numbers have a long and glorious mathematical history. They are named after the Italian mathematician Leonardo of Pisa, who was known as Fibonacci (a contraction of filius Bonaccio, “son of Bonaccio.”). In 1202, Fibonacci published Liber Abbaci, or “Book of Calculation,” a highly influential book of practical mathematics. In this book, he introduced and promoted the mighty Hindu-Arabic numerals to Europeans, who were still doing calculations with Roman numbers. His book also contained the following curious problem: “How many pairs of rabbits can be produced from a single pair in a year if every month each begets a new pair which from the second month on becomes productive?” In the course of answering this rabbit problem, the Fibonacci numbers 1; 1; 2; 3; 5; 8; 13; 21; : : : and the formula FnC2 D Fn C FnC1 for n  1; with F1 D F2 D 1 are revealed, where Fn denotes the nth Fibonacci number.

(6.1)

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The formula (6.1), which is often called a recursive formula, does not directly give the value of Fn . Instead it gives a rule telling us how to compute Fn from the previous numbers. To find the exact value of Fn , we now turn to the method of generating functions. Let the generating function be G.x/ D

1 X

Fn x n :

nD1

In view of (5.1), then G.x/ D x C x 2 C F3 x 3 C F4 x 4 C F5 x 5 C




D x C x 2 C .F1 C F2 /x 3 C .F2 C F3 /x 4 C .F3 C F4 /x 5 C




D x C x.F1 x C F2 x 2 C F3 x 3 C


/ C x 2.F1 x C F2 x 2 C F3 x 3 C


/ D x C xG.x/ C x 2 G.x/

and so

x : 1 x x2 To determine the coefficient of x n in G.x/, we factor G.x/ D

1

x 2 D .1 ˛x/.1 ˇx/; p 5/=2. Using the method of partial fractions and the

x

p where ˛ D .1 C 5/=2; ˇ D .1 geometric series expansion yields

x ˛x/.1 ˇx/   1 1 1 Dp 5 1 ˛x 1 ˇx 1 1 X n .˛ ˇ n /x n : Dp 5 kD1

G.x/ D

.1

Thus, we obtain Binet’s formula 1 Fn D p .˛ n 5

ˇn/

(6.2)

as an explicit expression for the Fibonacci number Fn . The formula (6.2) also can be obtained from the exponential generating function. To see this, let 1 X Fn n Ge .x/ D x : nŠ kD1

Differentiating Ge .x/ successively gives

F3 2 x C 2Š F4 2 Ge 00 .x/ D F2 C F3 x C x C 2Š Ge 0 .x/ D F1 C F2 x C

F4 3 x C 3Š F5 3 x C 3Š

F5 4 x C 4Š F6 4 x C 4Š

F6 5 x C


; 5Š F7 5 x C


: 7Š

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It follows that, from (6.1) and F1 D F2 D 1, Ge .x/ satisfies the differential equation Ge 00 .x/

Ge 0 .x/

Ge .x/ D 0:

This, together with Ge .0/ D 0; Ge 0 .0/ D F1 D 1, yields 1 Ge .x/ D p .e ˛x 5

e ˇx /:

Finally, the formula (6.2) is an immediate consequence of the power series expansion of e ax . In light of (6.2), Fibonacci numbers grow very fast. In fact, F45 D 1;134;903;170. That is, in less than four years, one will have more than one billion pairs of rabbits! Moreover, one can determine exactly how fast Fibonacci numbers grow. In particular, considering the ratios of successive Fibonacci numbers yields a particularly striking result: lim

n!1

FnC1 D lim .Fn /1=n D ; n!1 Fn

(6.3)

p where  D .1 C 5/=2 D 1:618034 : : : was called the golden ratio by the ancient Greeks. Recently, Viswanath in [8] took a fresh look at Fibonacci numbers by studying the Fibonacci-like recurrence: F nC2 D ˙FnC1 ˙ Fn ;

F0 D 0; F1 D 1;

where each ˙ sign is chosen randomly with probability 1=2. He unexpectedly found that lim .jFn j/1=n D 1:13198824 : : : :

n!1

This constant may explain the case of rabbits randomly allowed to prey on each other, and enables us closer to simulate the real world. Since the thirteenth century, Fibonacci numbers have intrigued us to the present day. There is a specialized journal entitled Fibonacci Quarterly that was founded in 1962 and is devoted to Fibonacci numbers and their generalizations. Recently, Fibonacci numbers even appeared in a popular novel, Dan Brown’s bestseller The Da Vinci Code. Although Fibonacci numbers were originally represented as an unrealistic model of a population of rabbits, scientists did find them turning up in many models of the natural world. For example, the flowers of surprisingly many plants have a Fibonacci number of petals. Lilies have 3 petals, buttercups have 5, sunflowers often have 55, 89 or 144. Similar numerical patterns involving successive Fibonacci numbers also occur in pine cones and pineapples. However, the precise reasons for Fibonacci numerology in plant life are still unknown. There’s still plenty of room for mathematical exploration and experimentation in a problem that began centuries ago.

6.3 Harmonic numbers The harmonic numbers fHn g are defined by Hn D 1 C

1 1 1 C C


C : 2 3 n

(6.4)

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These numbers satisfy the recurrence relation Hn D Hn

1

1 ; n

C

.for n > 1/

and H1 D 1. Hence 1 X

nD1

1 X

Hn x n D x C

nD2 1 X

DxC

nD2 1 X

Dx

Hn x n  Hn

Hn x

n

nD1

1 C

!

C

 1 xn n

1 X xn : n nD1

Since it is valid to interchange the order of summation and integration for a convergent power series, for jxj < 1, 1 Z 1 X 1 n X x n x D t n 0

nD1

nD1

1

dt D

Z

x

. 0

1 X

t

n 1

nD1

/dt D

Z

x 0

dt 1

t

D

ln.1

x/:

Thus, we find that the generating function of fHn g is 1 X

nD1

ln.1 x/ : 1 x

Hn x n D

(6.5)

Unlike the Fibonacci numbers, a closed form of Hn for general n does not exist. Moreover, even as limn!1 1=n D 0, Hn itself is divergent! Modern calculus textbooks prove this fact either by establishing the estimates H2n  1 C 21 n or by comparing Hn with R1 1 dx=x. But the following argument is equally simple and insightful. Suppose that limn!1 Hn D S . Then   1 1 1 1 1 1 1 C C


C C


D 1C C


C C


D S 2 4 2n 2 2 n 2 and so the sum of the odd numbered terms, 1C

1 1 C


C C


3 2n C 1

must be the other half of S . But this is impossible since 1 2n

1

>

1 2n

for every positive integer n. On the other hand, define the alternate signed harmonic numbers by Hn0 D 1

1 1 . 1/nC1 C C


C : 2 3 n

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0 H2n exhibits the following particular beautiful form — Catalan’s identity:

1 1 1 1 C


C 2 3 2n 1 2n 1 1 1 D H2n 2 C C


C 2 4 2n 1 1 1 D H2n Hn D C C


C ; nC1 nC2 2n

0 H2n D1

and

0 lim H2n D lim

n!1

n!1

Z 1 n 1 dx 1 X D D ln 2: n 1 C k=n 0 1Cx kD1

Now we study the asymptotic behavior of Hn as n ! 1. Define   1 nC1 an D ln : n n Since

Z

1



1 1 an D n n C x 0 a simple estimation on the integral yields Z 0 < an
1, Z

N

f .x/ dx D

0

N X1 j D0

N X1

N X1 Z j C1 j D0

j

f .x/ dx D

N X1 Z 1

N

j D0

j D0

1/

.j C 1/

f .2i

1/

f .x C j / dx;

1 .f .0/ C f .N //; 2

X 1 .f .j / C f .j C 1// D f .j / 2

.f .2i

0

j D0

.j // D f .2i

1/

f .2i

.N /

1/

.0/:

Thus we establish the Euler-Maclaurin summation formula: Z

N 0

f .x/ dx D

N X

1 .f .0/ C f .N // 2

f .j /

j D0

f

.2i 1/

1 .0// C .2k/Š

Z

0

k X B2i .f .2i .2i /Š

1/

.N /

i D1

1 N X1 j D0

B2k .x/f .2k/ .x C j /dx:

(6.19)

Equivalently, we can write this as a relation between a finite sum and an integral N X

j D0

f .j / D

Z

N 0

f .x/ dx C

1 .f .0/ C f .N // 2

k X B2i C .f .2i .2i /Š i D1

1/

.N /

f .2i

1/

.0// C RN ;

(6.20)

where RN D D

1 .2k/Š

Z

1 .2k/Š

Z

0

1 N X1 j D0

B2k .x/f .2k/ .x C j /dx

N

B2k .x 0

bxc/f .2k/ .x/dx:

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Here bxc denotes the largest integer  x. The formulas (6.19) and (6.20) enable us to approximate integrals by sums or, conversely, to estimate the values of certain sums by means of integrals. For example, let f .x/ D ln x. Since f .n/ .x/ D . 1/n 1 .n 1/Šx n , by (6.20), we get n X i D1

ln k D

Z

n

1 ln x dx C .ln 1 C ln n/ 2 0    1 1 1 1 2 C B2 C B4 2Š n 1 4Š n3

2 13



C

1 B6 6Š



24 n5

24 15



C


:

By using the known Bernoulli numbers,we have ln nŠ D n ln n C

1 1 ln n C 2 12n

1 1 C C Rn : 360n3 1250n5

Now exponentiating both sides yields nŠ D nn e

n

p Rn ne exp



1 12n

 1 1 C C


: 360n3 1250n5

Appealing to the power series of e x , we obtain   p 1 1 nŠ D nn e n ne Rn 1 C C C


: 12n 288n2

(6.21)

A careful analysis shows that (see Exercise 13) p as n ! 1: e Rn ! 2 This, together with (6.21), yields the famous Stirling’s formula   p 1 1 nŠ D 2 n nn e n 1 C C C


: 12n 288n2

(6.22)

6.4.4 Power series in terms of Bk Appealing to (6.10), we rewrite (6.9) as 1 X x x B2n 2n coth D x : 2 2 .2n/Š nD0

Substituting x by 2xi and noticing that coth.ix/ D x cot x D

1 X

nD0

. 1/n

22n B2n 2n x D1 .2n/Š

1 2 x 3

i cot x, we find that 1 4 x 45

2 6 x 945




:

(6.23)

Furthermore, by the trigonometric identities tan x D cot x

2 cot.2x/; csc x D cot x C tan.x=2/;

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we obtain tan x D

1 X

. 1/nC1

nD1

x csc x D 1 C

1 X

22n .22n 1/B2n 2n x .2n/Š

. 1/nC1

2.22n

nD1

1

(6.24)

;

1

1/B2n 2n x : .2n/Š

(6.25)

There are many applications for these power series expansions, the most impressive of which links to the values of the Riemann zeta function at even positive integers. It is well known that .2/ D  2=6. Now, we conclude this chapter by deriving Euler’s amazing formula:for any positive integer k, .2k/ D

1 X 1 . 1/k 1 22k D .2k/Š n2k nD1

Recall Euler’s infinite product (5.26), namely 1  Y sin x D 1 x nD1

1

B2k

 2k :

(6.26)

 x2 : n2  2

Taking the logarithm of both sides and then differentiating yields cot x

1

X 2x 1 D ; 2 2 x  n x2 nD1

which results in Euler’s elegant partial fraction expansion: 1 1  X X 1 2x 1 1 cot x D D 2 n2 2 x  x x n x nD1 nD1

1 n C x



:

(6.27)

With the previous established power series expansion (6.23), we now have proven that 1

2

1 X

nD1

1

X . 1/k 22k B2k x2 D 1 C x 2k :  2 n2 x 2 .2k/Š

(6.28)

kD1

In terms of a geometric series expansion, 1 X x2 x 2 =. 2 n2 / x 2k D D : 2 2 2 2 2 2 2k  n x 1 x =. n /  n2k kD1

Thus (6.28) becomes 1

2

1 X

nD1

1 X

kD1

x 2k  2k n2k

!

D1C

1 X . 1/k 22k B2k 2k x : .2k/Š

kD1

Interchanging the order of summation on the left yields ! 1 1 1 X X X 1 . 1/k 22k B2k 2k 2k 1C 2 x D 1 C x : .2k/Š  2k n2k nD1 kD1

kD1

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Now (6.26) follows from equating the coefficients of x 2k on both sides. In particular, for k D 2; 3; 4, we get 1 X 1 4 D ; n4 90 nD1

1 X 1 6 D ; n6 945 nD1

1 X 1 8 D : n8 9450 nD1

In light of (6.26), we see that .2k/ is a rational multiple of  2k , and hence irrational. In spite of such success, Euler’s original proof, as it emerged from (6.28), was geared toward even powers of x only, and thus unable to attain a formula for .2k C 1/ analogous to (6.26). The question whether .2k C 1/ is irrational has been asked since the time of Euler. For almost 200 years there had been no progress until 1978 when R. Apery proved the irrationality of .3/ by using only results known at the time of Euler. Despite considerable effort we still know very little about the irrationality of .2k C 1/ for k  2.

Exercises 1. Let Fn be the Fibonacci numbers and an D F0 C F1 C F2 C


C Fn : Find a generating function for an and then prove that an D FnC2

1:

2. Let a0 D 1. For n  1, define an by an D Show that an D F2n

n X1 i D1

.n

i /ai C 1:

1.

3. Monthly Problem 11258p[2006, 939; 2008, 949]. Let Fn denote the nth Fibonacci number, and let i denote 1. Prove that 1 X F3k 2F1C3k 1 D i C .1 F3k C i F2
3k 2

p 5/:

kD0

4. Let Hn be the nth harmonic number. For all n  2, prove that n 1 X

kD1

and Hn2

Hk D nHn

n



H3 Hn H2 2 C C


C 2 3 n

5. Prove that 2

ln .1



:

1 X Hn n x/ D 2 x n C1 nD1

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and 3

ln .1

x/ D 6

6. Let

n X

an D

1 X

n 1 X Hk nC2 kC1

nD1

kD1

. 1/k


C k

1 1 C 2 3

1

kD1

!

x nC2 :

!

1

ln 2 :

Does the sequence fan g converge? If so, to what value? 7. Monthly Problem 11382 [2008, 665]. Show that if p is prime and p > 5, then p 1

p 1 X H2 X Hk k  k k2

kD1

.mod p 2 /:

kD1

Remark. Two rational numbers are congruent modulo d if their difference can be expressed as a reduced fraction of the form da=b with b relatively prime to a and d . A good entrance to this problem is Terence Tao’s “Solving Mathematical Problems,” Section 2.3. P n 8. Prove that if f .x/ D 1 nD1 an x converges, then 1 X

nD1

an Hn x n D

Z

1

0

f .x/ f .tx/ dt: 1 t

In particular, show that 1 X

nD1

9. Let Rn D Hn

1 n2n

H D .2/; 1 n

1 X 1 Hn n2 nD1

1

D .3/:

ln.n C 1/ and be Euler’s constant. Prove that

Rn D

where 1 1 a1 D ; ak D 2 k

1 X

nD1

Z

ak ; .n C 1/.n C 2/


.n C k/

1

x.1 0

x/


.k

1

x/dx .k > 1/:

For the reader familiar with Stirling numbers of the first kind s.k; i /, explicitly, ak D

k . 1/kC1 X s.k; i / : k i C1 i D1

10. Monthly Problem 10949 [2002, 569; 2004, 264–265]. Show that, for each positive integer n, .n/  Hn C 2 ln.Hn /e Hn ; P with equality only for n D 1, where .n/ D djn d denotes the sum of the divisors of n.

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6. Special Numbers

11. Prove that

(b) (c)

P1

1 nD 1 nC1=2 x . P . 1/n  sec  x D 1 nD 1 nC1=2Cx . P 2 1 D 1 nD 1 .xCn/2 . sin2 x

(a)  tan  x D

12. Define b0 D 1; b0.x/ D 1 and

Z

bn .x/ D n

x

bn

where the constant bn is chosen so that and bn D Bn . 13. Show that lim

n!1 nn

14. Show that

D

n X 1 k

1 .t/ dt

C bn ;

0

R1

bn .t/ dt D 0. Show that bn .x/ D Bn .x/

0

nŠ p ne

D

n

p 2:

m

X B2k 1 1 C C


: 2n 2k n2k

ln n

kD1

kD1

15. Show that, for n  1,  n 1  X 1 B2kC2 2n D : 2k .2k C 2/ .2n C 1/.2n C 2/

kD0

16. Monthly Problem E 3237 [1987, 995; 1989, 364–365]. Prove that Bm D

1 n.1

nm /

m X1 kD0

k

n



m k



Bk

n X1

jm

j

;

j D1

for any positive integer m; n > 1. 17. Prove that 1 X

nD0

2kC1 . 1/n kC1 .2 / D . 1/ B2kC1 .1=4/: 2.2k C 1/Š .2n C 1/2kC1

18. Prove Raabe’s multiplication identity   m 1 1 X k Bn x C Dm m m

n

Bn .mx/:

kD0

Comment. Lehmer in [4] proves that the nth Bernoulli polynomial Bn .x/ is the unique monic polynomial of degree n which satisfies this identity. Therefore this identity provides another definition of the Bernoulli polynomials.

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References [1] P. Chandra and E. Weisstein, “Fibonacci Number.” From MathWorld—A Wolfram Web Resource. mathworld.wolfram.com/FibonacciNumber.html [2] T. Koshy, Fibonacci and Lucas Numbers with Applications, New York, John Wiley, 2001 [3] J. Lagarias, An elementary problem Amer. Math. Monthly, 109 (2002) 534–543.

equivalent

to

the

Riemann

Hypothesis,

[4] D. H. Lehmer, A new approach to Bernoulli polynomials, Amer. Math. Monthly, 95 (1988) 905–911. [5] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, research.att.com/ njas/sequences/ [6] Sondow, Jonathan and Weisstein, Eric W. “Harmonic Number.” From MathWorld—A Wolfram Web Resource. mathworld.wolfram.com/HarmonicNumber.html [7] K. R. Stromberg, Introduction to classical real analysis, Wadsworth, Belmont, California, 1981. [8] D. Viswanath, Random Fibonacci sequences and the number 1:13198824: : :, Math. Comp., 69 (2000) 1131–1155. [9] Wikipedia, “Bernoulli Numbers,” available at en.wikipedia.org/wiki/Bernoulli numbers [10] H. Wilf, Generatingfunctionology, A. K. Peters, Ltd., 1994. It can be downloaded from math.upenn.edu/ wilf/DownldGF.html

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7 On a Sum of Cosecants Mathematics is the study of analogies between analogies. —G. C. Rota

Undergraduate research has been a longstanding practice in the experimental sciences. Only recently, however, have significant numbers of undergraduates begun to participate in mathematical research. While many math students are interested in research, they often don’t know what problems are open or how to get started. Finding appropriate problems becomes of fundamental importance. Joseph Gallian has run an excellent undergraduate mathematics research program [4] at the University of Minnesota, Duluth since 1977. He offers the following criteria for successful undergraduate research problems:  not much background reading is required;  partial results are probable;  the problem has been posed recently;  new results will likely be publishable. How does one actually find an appropriate problem? One popular way is to generalize or specialize an existing result and see what develops. As Zeilberger in [6] suggested: at first, leave the exact form of the generalization or specialization blank, and as you move forward, see what kind of generalization or specialization would be required to make the proof work. Keep ‘guessing and erasing’ until you get it done, just like doing a crossword puzzle. This chapter is intended   to illustrate this strategy by studying a trigonometric sum in P 1 the form of nkD1 csc k . n

7.1 A well-known sum and its generalization The well-known sum we have in mind is n X1

kD1

sin



k n



D cot

  : 2n

(7.1) 73

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There are many ways to derive (7.1). Here we present a simple proof using complex numbers. Indeed, we derive a more general formula: n X1

kD1

sin.kx/ D

sinŒ.n

1/x=2 sin.nx=2/ : sin.x=2/

(7.2)

To see this, appealing to 1 C r C r2 C


C rn

1

D

rn 1 r 1

(7.3)

and De Moivre’s formula, we get n X1

kD1

sin.kx/ D Im

n 1 X

e

i kx

kD1

D Im e i nx=2

!

e i.n

D Im e 1/x=2

e ix=2

ix e

e e

i.n 1/x

e ix i.n 1/x=2 ix=2

1 1 !

!

sinŒ.n 1/x=2 Im.e i nx=2 / sin.x=2/ sinŒ.n 1/x=2 sin.nx=2/ ; D sin.x=2/ D

as proposed. Now (7.1) follows from (7.2) by setting x D =n. In analogy to (7.1) we seek to compute in closed form In D

n 1 X

kD1

  n X1 1 k D csc : sin.k=n/ n

(7.4)

kD1

To begin with, based on the identity (5.2) and other favorable results [2] on the power sums Pn 1 2p kD1 csc .k=n/, it seemed to me that there ought to be a closed form for In . I spent many hours reading related journal papers, using Mathematica to explore and test potential results, and even consulting experts in the field via email. As a matter of fact, no closed form of In exists [1]. In lieu of this, could we have a ‘nice’ estimate for In and eventually establish an asymptotic formula for (7.4)? Well, searching for these answers offers a ‘nice’ problem. As we will see, this problem is hard enough that the results will be publishable yet accessible enough to be understood by analysis students. The investigation also indicates that a certain generalization of an existing result can produce quite a different problem, one of interest on its own. To show how to proceed in such an investigation, we take the reader through some rough estimates before deriving the best possible one. This demonstrates how special cases, that often are easier to solve, lead to insights and improvements of the existing results.

7.2 Rough estimates Some rough estimates come from playing with the expression of (7.4) directly. We start with collecting some appropriate tools. The selection of the following results is pretty straightforward.

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Jordan’s inequality: For 0 < x  =2, 2x  sin x < x:  Clearly, this is equivalent to 1  < csc x  : x 2x

(7.5)

Young’s inequality: Let Hn be the nth harmonic number defined by (6.4) and let be Euler’s constant defined by (6.7). Then 1 < Hn 2.n C 1/

ln n


0 such that jan j  M bn for all n. Now we proceed to the estimates for In . Since     k .n k/ csc D csc ; n n the second inequality of (7.5) implies In  2

bn=2c X

csc

kD1

Notice that 1 n D 2bn=2c





1; 1C

k n



n

bn=2c X kD1

1 : k

if n D 2m; if n D 2m C 1:

1 ; 2m

This, together with the second inequality of (7.6), yields that, for n  3,   1 In  n.ln n C ln 2/ C 1 C O : n Next, to estimate In from below, for n  3, we rewrite 8   P 1/=2c ˆ 0. Since .ln.tan.x=2///0 D csc x;

(7.12)

from (7.11), integrating (7.12) provides ln.tan.x=2//

ln.x=2/ D

1 X ai 2i x : 2i

(7.13)

i D1

On the other hand, by (7.11) we have Œn=2 1 X 2n X 1 In  C2 ai .=n/2i  k i D1

kD1

Œn=2 1

X

k 2i

1

:

(7.14)

kD1

Since n=Œn=2  3, along the same lines we used in the rough estimate, we get Œn=2 2n X 1 2n 3 < .ln n C C ln 2/ C :  k   kD1

To estimate the double sum in (7.14), noticing Œn=2 X

k 2i

1

kD1


l /;

1 1 D : .a C kd /


Œa C .k C m/d  md.a C d /.a C 2d /


.a C md /

(9.15)

9.3 Trigonometric sums In this section, we show how to telescope various sums involving the trigonometric functions.

9.3.1 Angles in arithmetic sequences First, similar to Gauss’s pairing trick, we derive the formula
n X sin x C n2  sin   S WD sin.x C k/ D sin 2 kD0

nC1  2




:

(9.16)

Write S twice in the form of

S D sin x C sin.x C / C


C sin.x C n/;

S D sin.x C n/ C sinŒx C .n

1/ C


C sin x:

Summing the terms vertically in pairs and using the sum to product formula sin ˛ C sin ˇ D 2 sin

˛Cˇ ˛ ˇ cos ; 2 2

we get n   n X 2S D 2 sin x C  cos k 2 kD0

n   : 2

To eliminate the cosine sum, we multiply both sides by sin =2 and then apply the product to sum formula 1 sin ˛ cos ˇ D Œsin.˛ C ˇ/ sin.ˇ ˛/ (9.17) 2 to obtain     n   n X nC1 nC1 2S sin =2 D sin x C  sin .k C 1/  sin k  : 2 2 2 kD0

Telescoping the sum yields     nC1 nC1 2S sin =2 D 2 sin x C  sin  ; 2 2 thereby implying (9.16) as required.

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Next, differentiating (9.16) with respect to x gives an analogous formula for the summation of cosines, namely

n X cos x C n2  sin nC1  2   cos.x C k/ D : (9.18) sin 2 kD0

Furthermore, combining (9.16) and (9.18), we get another interesting formula,  sin x C sin.x C / C


C sin.x C n/ n  D tan x C  : cos x C cos.x C / C


C cos.x C n/ 2

Along the same lines, one can establish the following alternating sums which are perhaps less well known:

n X . C  / sin x C n2 . C  / sin nC1 2 k   . 1/ sin.x C k/ D ; cos 2 kD0

n n X sin nC1 2 . C  / cos x C 2 . C  / k   . 1/ cos.x C k/ D :  cos kD0 2 Finally, we derive a closed form for

Ts WD

n X1

r k sin k

kD1

in the same manner as we handled the sum in (9.1). Notice that .1

2r cos  C r 2 /Ts D r sin  C r 2 sin 2 C


C r n 2

2r sin  cos  n

2r sin.n n

C r sin.n






2r

2

sin.n

n 1

sin.n

2/ C r n

1/ cos  r sin  C


C r n

2/ C r

sin.n

1

sin.n

Appealing to (9.17), we see that all Œr k  vanish for 2  k  n simplified to sin n. Therefore r sin 

1/

3/

1/:

Let Œr k  denote the coefficient of r k . Combining the like terms yields 8 ˆ k D 2; 0.

kD1

where the value is independent of a. Differentiating this with respect to a yields 1 X

kD1

1 4.2k C a/ D 2 : 4 4 C .2k C a/ a C 2a C 2

In particular, for a D 0 and a D 1, unexpectedly, we get 1 X

kD1

2k 1 D 1 C 4k 4 2

and

1 X

kD1

4.2k C 1/ 1 D : 4 C .2k C 1/4 5

9.4 Some more telescoping sums In this section, we explore some other methods to form telescoping sums.

9.4.1 Sums of some famous families of numbers We first examine the sums involving the Fibonacci numbers in which the telescoping technique can be utilized. Since Fk D FkC2 FkC1 ; the series fFk g is telescoping and n X

kD1

Fk D FnC2

1:

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Moreover, since Fk

1 1 D F F 1 kC1 k 1 Fk

1 ; Fk FkC1

we find n X

kD2 n X

kD2

Fk

1 D1 1 FkC1

1 ; Fn FnC1

Fk

Fk D2 1 FkC1

1 Fn

1 : FnC1

In general, for any sequence fak g with akC1 D ˛ak C ak

1;

.k  3/

where ˛ is a nonzero constant, we have n X

kD2

ak D

1 .anC1 C an ˛

a1

a2 /:

Based on this fact, a large class of series related to special numbers can be readily computed.

9.4.2 Apéry-like formulas For x ¤ 0 and any sequence fak g, let b1 D

1 a1 a2


ak 1 ; bk D x x.x C a1 /


.x C ak

Then bk

bkC1 D

The telescoping sum results in n X

kD1

.3/ D

1 1 X 1 5 X D k3 2

kD1

.k  2/:

a1 a2


ak 1 : .x C a1 /


.x C ak /

a1 a2


ak 1 1 D .x C a1 /


.x C ak / x

This, along with

1/

kD1

a1 a2


an : x.x C a1 /


.x C an /

(9.23)

. 1/kC1  ; 2k 3 k k

has been made famous by Ap´ery’s proof [1] of the irrationality of .3/. Appealing to the identities (9.23) and (9.8), we establish an analogous formula for .2/: 1 1 X X 1 .2/ D D3 k2 kD1

kD1

1 k2



2k k

:

(9.24)

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To prove (9.24), we set x D n and ak D k in (9.23) thereby giving n X

kD1

.k 1/Š 1 D .n C 1/.n C 2/


.n C k/ n

nŠ : n.n C 1/


.2n/

This leads to n X 1 .k 1/Š .n 1/Š D C n2 n.n C 1/


.n C k/ n.n C 1/


.2n/ kD1

D

n X1

kD1

.k 1/Š .n 1/Š C 2 n.n C 1/


.n C k/ n .n C 1/


.2n

1/

:

Therefore, 1 1 X n 1 X X X 1 .k 1/Š .n 1/Š D C 2 2 n n.n C 1/


.n C k/ n .n C 1/


.2n

nD1

nD1 kD1

nD1

1/

(9.25)

:

For the double series in (9.20), we rearrange the series and sum in columns obtaining "1 # 1 X n 1 X X X 1 .k 1/Š D .n 1/Š n.n C 1/


.n C k/ .k C 1/.k C 2/


.n C k C 1/ nD1 kD1

nD1

D

1 X

nD1

kDn

.n 1/Š ; n.n C 1/


.2n/

where we have used the following telescoping sum from the identity (9.8): 1 X

kDn

1 .k C 1/.k C 2/


.n C k C 1/ 1  1 X 1 D n .k C 1/.k C 2/


.n C k/ kDn

D

1 .k C 2/.k C 3/


.n C k C 1/



1 : n.n C 1/


.2n/

Hence the right-hand side of (9.25) becomes 1 X

nD1

.n

1/Š



1 1 C 2 n.n C 1/


.2n/ n .n C 1/


.2n

1/



D3 D3

1 X

nD1 1 X

nD1

.n 1/Š n.n C 1/


.2n/ 1 n2



2n n



and (9.24) follows as desired.

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9.4.3 A variant of telescoping sums Sometimes a series itself is not telescoping, but it can be decomposed into the sum of a telescoping series and a well-known series. Let us illustrate this approach by two examples. Example 1. Monthly Problem E 3352 (1989). Show that 1 X

kD0

kŠ.k 4

1 e D : 2 C k C 1/ 2

(9.26)

Since .k 2

k C 1/.k 2 C k C 1/ D k 4 C k 2 C 1;

partial fractions gives

k4

2 kC1 D 2 2 Ck C1 k CkC1

k k2

1 : kC1

Thus, 1 1 D kŠ.k 4 C k 2 C 1/ 2

1 D 2

Setting bk D

k 1 2ŒkŠ.k 2 kC1/

kŠ.k 4

 

kC1 kŠ.k 2 C k C 1/

k

1 k C 1/

kŠ.k 2

k .k C 1/Š.k 2 C k C 1/

k

kŠ.k 2



 1 1 C : k C 1/ .k C 1/Š

yields 1 D bkC1 C k 2 C 1/

bk C

1 1 : 2 .k C 1/Š

This transforms the left-hand side of (9.26) into 1 1 1 1 X 1 1 X 1 e C D D : 2 2 .k C 1/Š 2 kŠ 2 kD0

kD0

Example 2. Transform .3/ into a faster converging series. We start with SD

1 X

kD2

.k

1 ; 1/k.k C 1/

which telescopes and, by (9.12) sums to 1=4 after shifting the index of summation. Therefore,  1  1 5 X 1 1 5 X 1 .3/ D D : 4 k3 k k3 4 k 3 .k 2 1/ kD2

Here the nth term is approximately n

kD2

5

whereas before it was n

3

.

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9.4.4 Summation by computer — Gosper’s algorithm In 1977, in conjunction with his work on the development of one of the first symbolic algebra programs, Gosper in [2] discovered a beautiful way to sum fak g whenever akC1 =ak is a rational function of k. As one of the landmarks in the history of computerization of the problem of closed form summation, Gosper’s algorithm answers the following question affirmatively: P If akC1 =ak is rational in k, is the sum nkD1 ak telescoping? Gosper’s algorithm proceeds in two steps, each of which is fairly straightforward. Step 1 is to express the ratio in the special form akC1 p.k C 1/ q.k/ D ; ak p.k/ r .k/ where p; q; and r are polynomials satisfying gcd.q.k/; r .k C h// D 1; for all integers h  0: Step 2 is to find a nonzero polynomial solution c.k/ of q.k/c.k C 1/

r .k

1/c.k/ D p.k/;

(9.27)

whenever possible. If c.k/ exists, we find bk D

r .k

1/c.k/ ak ; ak D bkC1 p.k/

bk :

Let us use Gosper’s algorithm to show that the sum Sn D

n X .4k C 1/
kŠ .2k C 1/Š

kD0

is telescoping. The term ratio akC1 4k C 5 D ak 2.4k C 1/.2k C 3/ is rational in k as expected. The choices p.k/ D 4k C 1; q.k/ D 1; r .k/ D 2.2k C 3/ clearly satisfy the conditions in Step 1. Thus, equation (9.27) becomes c.k C 1/

2.2k C 1/c.k/ D 4k C 1;

and the constant polynomial c.k/ D 1 is a solution. Hence bk D

2.2k C 1/ .4k C 1/
kŠ kŠ D 2 4k C 1 .2k C 1/Š .2k/Š

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satisfies ak D bkC1

bk and Sn D bnC1

b0 D 2

nŠ : .2n C 1/Š

(9.26) is also a good application of Gosper’s algorithm. Rewriting (9.26) in the equivalent form  1  X 1 1 D 0; kŠ.k 4 C k 2 C 1/ 2 kŠ kD0

Gosper’s algorithm finds that bk D expected.

k2 . 2ŒkŠ.k 2 kC1/

This yields the sum as b1

b0 D 0 as

Here we need not rely so much on cleverness and luck, as calculations can be done on the computer. Quite often, when the summand involves binomial coefficients, factorials, products of rational functions, there are very good chances that such a problem can be solved by Gosper’s algorithm or by its generalization — the WZ method. The interested reader can find more details in Petkovsek-Wilf-Zeilberger’s elegant book [3] and additional examples in their excellent article [4]. The latter contains 27 Monthly problems over the period of 1978–1997 solved by a computer. The related programs are available at www.math.rutgers.edu/zeilberg/programs.html In summary, we have given a partial account of everything you always wanted to know about the telescoping sums! It is interesting to notice that there are problems we may not be able to solve by this method, even though they may appear to be candidates. A good P 2 example related to our discussion in 9.3.3 is to find 1 nD1 arctan.1=n / in closed form. The answer turns out to be a curious identity (5.34), which cannot be derived by telescoping methods.

Exercises 1. Find the following sums in closed form: (a) (b)

12 1
3

C

Pn

kD1

22 3
5

C


C   n 2 k : k

n2 .2n 1/.2nC1/ .

2. Evaluate

n   X n cos kx sin.n k

k/x:

kD0

3. Prove that

n X

sec kx sec.k

1 X

1 nC1

kD1

4. Prove that 1C

nD1



1/x D

tan nx : sin x

1
3


.2n 1/ 2
4


2n

2

D

4 : 

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9. On the Telescoping Sums

5. Evaluate

2n X

. 1/k

kD0

    4n 2n
2k k

1

:

6. Let n be a positive integer. Prove that  1 1 1 p C p C


C p 0, prove that  n  X 2n . a/j D .1 C a/n cos.2n tan 2j

1

p a/:

j D0

5. Let the generating function of Fibonacci numbers be given by 1 X

nD1

Fn x n D

x : x x2

1

Find the generating functions for F2n and F2nC1 respectively via (10.2). 6. Determine the real function of x whose power series is 1 X

kD0

x 4kC3 x3 x7 x 11 D C C C


.4k C 3/Š 6 5040 39916800

7. Find

1 X

nD0

in closed form.

1 x 4nC1 .4n C 1/.4n C 3/

8. Monthly Problem 2991 [1922, 356; 1924, 51–52]. Sum the infinite series S2 .x/ D 1 C

3x 2 4z 4 6z 6 C C C


; 2Š 4Š 6Š

where the numerators of the coefficients form a series of numbers whose third differences are all equal to 2. 9. It is well-known that e x is bounded for x < 0. Determine all possible proper subseries of e x which remain bounded for x < 0. 10. Open Problem on Ramanujan q-series. Define A.q/ D

1 Y

.1

nD1

qn/ D 1

q

q2 C q5 C q7

q 12

q 15 C q 22 C


:

Now, we trisect the series into A0 D 1

A1 D

A2 D

q 12

q 15 C q 51 C q 57

q C q 7 C q 22

q 2 C q 5 C q 26

q 40

q 35

q 117

q 126 C


;

q 70 C q 100 C q 145 C


;

q 77 C q 92 C q 155 C


:

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10. Summation of Subseries in Closed Form

Look for algebraic relations among A0 ; A1; A2 by using Mathematica or Maple. For example, Berndt and Hart recently found that A2 A20 C A0 A21 C A1 A22 D 0. For an introduction on this topic, see Michael Simos’s “A multisection of q-series”, which is available at cis.csuohio.edu/ somos/multiq.html

References [1] J. Borwein, D. Bailey and R. Girgensohn, Experimentation in Mathematics, A. K. Peters, Massachusetts, 2004. [2] A. Chen, Accelerated series for  and ln 2, Pi Mu Epsilon J., 13 (2008) 529–534. [3] H. Chen, Solution to Problem 1147, Pi Mu Epsilon J., 12 (2007) 433-434. [4] A. DeMorgan, The differential and integral calculus, Robert Baldwin, London, 1942. [5] H. W. Gould, Combinatorial identities, West Virginia University, Revised edition, 1972. [6] E. R. Hansen, A Tables of Series and Products, Englewood Cliffs, Prentice-Hall, 1976. [7] K. Hardy and K. Williams, The Green Book — 100 practice problems for undergraduate mathematics competitions, Integer Press, Ottawa, 1985. [8] R. Honsberger, Mathematical Gems III, Washington, DC, MAA, 210–214, 1985. [9] G. R. Ierley and O. G. Ruehr, Analytic and numerical solutions of a nonlinear boundary layer problem, Stud. Appl. Math., 75 (1986) 1–36. [10] D. H. Lehmer, Interesting series involving Amer. Math. Monthly, 92 (1985) 449-457.

the

central

binomial

coefficients,

[11] C. Roussean, Problem 1147, Pi Mu Epsilon J., 12 (2007) 363. [12] L. Rubel and K. Stolarsky, Subseries of the Power Series for ex , Amer. Math. Monthly, 87 (1980) 371–376. [13] T. Trif, Combinatorial sums and series involving inverses of binomial coefficients, Fibonacci Quarterly, 38 (2000) 79–84.

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11 Generating Functions for Powers of Fibonacci Numbers In this chapter, using Binet’s formula and the power-reduction formulas, we derive explicit formulas of generating functions for powers of Fibonacci numbers. The corresponding results are extended to the famous Lucas and Pell numbers. As we did in Chapter 6, by FnC2 D FnC1 C Fn , we found that the generating function for Fibonacci numbers is 1 X x Fn x n D : (11.1) 1 x x2 nD1

In general, for any positive integer k, let the generating function of fFnk g be Gk .x/ D

1 X

Fnk x n :

nD1

It is natural to ask whether there is an explicit formula for Gk .x/. When k D 2, there is a recursive relation 2 2 2 FnC3 D 2FnC2 C 2FnC1 Fn2 : (11.2) To prove this, using Binet’s formula

where ˛ D .1 C (11.3) to get

1 Fn D p .˛ n ˇ n /; 5 p 1=˛ D .1 5/=2; appealing to ˛ˇ D

p 5/=2; ˇ D Fn2 D

1 ..˛ 2 /n 5

(11.3) 1, we square

2. 1/n C .ˇ 2 /n /:

This shows that Fn2 is a linear combination of the nth power of 1; ˛ 2; and ˇ 2 . Thus, the corresponding characteristic equation becomes .t C 1/.t

˛ 2/.t

ˇ2/ D t 3

2t 2

2t C 1 D 0

from which (11.2) follows. 121

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Another more accessible way to prove (11.2) relies on knowing in advance that such a formula as (11.2) exists. We search for it by setting 2 2 2 FnC3 D aFnC2 C bFnC1 C cFn2 ;

where a; b and c are constants to be determined. Putting n D 1; 2; 3, respectively, yields the system of equations 32 D 22 a C 12b C 12 c;

52 D 32 a C 32b C 12 c;

82 D 52 a C 32b C 22 c:

Solving for a; b; c leads to (11.2). Once (11.2) has been proved, we have G2 .x/ 2xG2.x/ 2x 2G2 .x/ x 3 G2 .x/

D F12 x CF22 x 2 CF32 x 3 CF42 x 4 D 2F12 x 2 C2F22 x 3 C2F32 x 4 D 2F12 x 3 C2F22 x 4 D F12 x 4

C


; C2F42 x 5 C


; C2F32 x 5 C2F42 x 6 C


; CF22 x 5 CF32 x 6 CF42 x 7 C


:

Summing the terms vertically and then appealing to (11.2), we obtain .1

2x

and so G2 .x/ D Similarly, we have

1 X

nD1

2x 2 C x 3 /G2 .x/ D x Fn2 x n D

x.1 .1 C x/.1

3 3 3 FnC4 D 3FnC3 C 6FnC2

x2; x/ : 3x C x 2/

3 3FnC1

(11.4)

Fn3

(11.5)

which results in G3 .x/ D

1 X

nD1

Fn3 x n D

x.1 2x .1 C x x 2 /.1

x2/ : 4x x 2 /

(11.6)

Theoretically, repeating this process, we can find recurrence relations for Fn4 ; Fn5 ; : : :, and thereby determine the corresponding generating functions. However, this process becomes tedious and not feasible for larger k. As a matter of fact a proof of a general explicit formula in print has not been found. In 1962, Riordan [2], instead of giving an explicit form, found that Gk .x/ satisfies the recurrence relation .1

ak x C . 1/k x 2 /Gk .x/ D 1 C kx

bk=2c

X

. 1/i aki Gk

2i ..

1/i x/= i;

i D1

where the aki have a rather complicated structure. Recently, Mansour [1] gave an explicit formula in terms of determinants of two k  k matrices, which is difficult to calculate even for k D 3; 4. Naturally, we ask whether there exists an elementary method for finding explicit formulas Gk .x/? In this chapter, we give an affirmative answer. First, we derive generating

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functions for fFk n g and the Lucas numbers fLk n g based on Binet’s formulas. Then, we introduce the power-reduction formula, which enables us to show that Fnk is a linear combination of either fLk n g or fFk n g, from which we establish an explicit formula for Gk .x/. Finally, we extend our results to more general sequences including Lucas and Pell numbers. We start with establishing the generating functions for fFk n g and fLk n g, where the Lucas numbers fLn g are defined by L1 D 1; Ln D Ln

1

C Ln

L2 D 3; (for n  3).

2

Similar to (11.3), we have Binet’s formula for Ln , namely Ln D ˛ n C ˇ n :

(11.7)

We now prove the generating functions for fFk n g and fLk n g are 1 X

Fk n x n D

nD0 1 X

Lk n x n D

nD0

To see this, appealing to ˛ˇ D 1 X

nD0

1

Fk x ; Lk x C . 1/k x 2

(11.8)

2

1

Lk x : Lk x C . 1/k x 2

(11.9)

1, by (11.3) and (11.7), we have

1 X ˛k n ˇk n n Fk n x D p x 5 nD0 n

1 D p 5

1 X

˛

kn

x

n

nD0

1 X

ˇ

kn

nD0

x

n

!

  1 1 1 D p 5 1 ˛k x 1 ˇk x 1 .˛ k ˇ k /x D p 5 1 Lk x C . 1/k x 2 Fk x ; D 1 Lk x C . 1/k x 2 as desired. (11.9) can be proved along the same lines. Next, we establish an explicit formula of Gk .x/ for any positive integer k. In order to motivate the idea, we study two special cases, namely k D 2 and k D 3. This process not only reveals how to approach the general case but also leads to new proofs for G2.x/ and G3 .x/. For k D 2, by (11.9), 1 X 2 3x L2n x n D : 1 3x C x 2 nD0

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In terms of L2n , we have Fn2 D

1 ..˛ 2 /n 5

1 .L2n 5

2. 1/n C .ˇ 2 /n / D

2. 1/n / ;

(11.10)

and so 1 X

Fn2 x n

nD1

1 D 5 

1 X

L2n x

n

2

1 X

n n

. 1/ x

nD0

nD0

2 3x 1 3x C x 2

2 1Cx

D

1 5

D

x.1 .1 C x/.1

x/ : 3x C x 2/

!



For k D 3, by (11.8) and (11.1), we have 1 X

nD0

F3n x n D

2x ; 4x x 2

1

1 X

nD0

Fn . x/n D

3x : 1 C x x2

Again, appealing to Fn3 D

˛ 3n

3˛ 2n ˇ n C 3˛ n ˇ 2n p 5 5

ˇ 3n

D

1 .F3n 5

3. 1/n Fn / ;

(11.11)

we find that 1 X

nD1

1 Fn3 x n D 5 1 D 5 D

1 X

F3n x n

3

nD0



1

1 X

Fn . x/n

nD0

2x 3x C 2 4x x 1 C x x2

x.1 2x .1 C x x 2 /.1

x2/ : 4x x 2 /

!



These cases are not isolated examples. In fact, they essentially reframe the problem as follows: once expressing Fnk as a linear combination of Fnk and Lnk , we can find an explicit formula of the underlying Gk .x/ by using the generating functions of Fnk and Lnk . To this end, we show that (11.10) and (11.11) are the base case of the following generalizations. Power-Reduction Formula. For each positive integer k, 5

k

Fn2k

 k 1  X 2k D . 1/i.nC1/ L2.k i

i /n

i D0

5

k

Fn2kC1

 k  X 2k C 1 D . 1/i.nC1/ F.2.k i

k.nC1/

C . 1/

i /C1/n :

  2k ; k

(11.12)

(11.13)

i D0

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To prove (11.12), starting with the binomial theorem, we have 5k Fn2k D .˛ n D

k 1 X i D0

C Replacing i by 2k

2k X

ˇ n /2k D . 1/i 2k X



i D0



2k i

. 1/i

i DkC1

. 1/i





˛ i n ˇ .2k

2k i





2k i i /n

˛ i n ˇ .2k

i /n



2k k

C . 1/k

˛ i n ˇ .2k

i /n



˛ k nˇk n

:

i in the last sum and appealing to     2k 2k D ; i 2k i

we obtain 2k X

i

. 1/

i DkC1



2k i



i n .2k i /n

˛ ˇ

D

k 1 X

i

. 1/

i D0



2k i



˛ .2k

i /n i n

ˇ :

Hence 5

k

Fn2k

D

k 1 X

i

. 1/

i D0



2k i



in

in

.˛/ .ˇ/ .˛

2.k i /n

Cˇ

2.k i /n

k

/ C . 1/



2k k



.˛ˇ/k n :

By ˛ˇ D 1 and (11.7), this proves (11.12). Since L1 D 1, we may rewrite (11.12) as !    k 1  1 X 2k 2k i.nC1/ k.nC1/ 2k . 1/ L2.k i /n C . 1/ L1 ; Fn D k i k 5 i D0

which indicates that Fn2k is a linear combination of Lucas numbers. Similarly, we can prove the formula (11.13). Finally, we are ready to derive an explicit formula for Gk .x/. Using the power-reduction formulas (11.12) and (11.13) respectively, we obtain 5k G2k .x/ D

k 1 X

. 1/i

 X 1 2k L2.k i nD0

i D0

i n k i /n .. 1/ x/ C . 1/

 X 1 2k .. 1/k x/n k nD0

    k 1 X 2 L2.k i /. 1/i x 2k 2k i k D . 1/ C . 1/ i 2 i k 1 1 L2.k i / . 1/ x C x i D0 5 G2kC1 .x/ D

k X

  1 2k C 1 X . 1/ F.2.k i

D

k X

  2k C 1 i 1

k

i D0

i D0

i

i /C1/n ..

1 . 1/k x

;

1/i x/n

nD0

F2.k i /C1 x L2.k i /C1 . 1/i x

x2

:

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In particular, we find that the generating functions for G4 ; G5; G6 are as follows: G4.x/ D

1 25



2 1

7x 7x C x 2

4.2 C 3x/ 6 C 2 1 C 3x C x 1 x



x.1 4x 4x 2 C x 3 / ; .1 x/.1 7x C x 2 /.1 C 3x C x 2 /   1 5x 10x 10x C C G5.x/ D 25 1 11x x 2 1 x x2 1 C 4x x 2 D

x.1 7x 16x 2 C 7x 3 C x 4 / ; .1 C 4x x 2 /.1 x x 2 /.1 11x x 2/  1 2 18x 15.2 3x/ 6.2 C 7x/ C G6.x/ D 125 1 18x C x 2 1 3x C x 2 1 C 7x C x 2 D

D

20 1Cx

x.1 12x 53x 2 C 53x 3 C 12x 4 x 5 / : x/.1 18x C x 2 /.1 3x C x 2 /.1 C 7x C x 2 /

.1



When k D 2; 3, it is interesting to see that the homogeneous recursive relations (11.2) and (11.5) are used to derive the generating functions G2 and G3. Conversely, Gk .x/ can be employed to find new homogeneous recursive relations. For example, G4 can be used to prove that 4 4 4 FnC5 D 5FnC4 C 15FnC3

4 15FnC2

4 5FnC1 C Fn4 :

The details will be studied in the next chapter. Generally, the ideas used to extract an explicit formula for Gk .x/ actually work for any power of numbers that satisfy a second-order recursive relation. In the following, we sketch the derivations of closed forms for the generating functions for powers of Lucas and Pell numbers.We leave the details and the general case for the reader. Similar to (11.12) and (11.13), we have the following power-reduction formulas: L2k n

 k 1  X 2k D . 1/i n L2.k i

i /n

i D0

L2kC1 n

kn

C . 1/

 k  X 2k C 1 D . 1/i n L.2.k i



2k k



;

i /C1/n :

i D0

Let the generating function of fLkn g be Hk .x/. Then, H2k .x/ D

   k 1  X 2 L2.k i / . 1/i x 2k 2k C i k 1 1 L2.k i /. 1/i x C x 2 i D0

1 . 1/k x

;

 k  X 2 L2.k i /C1 . 1/i x 2k C 1 H2kC1 .x/ D : i 1 L2.k i /C1 . 1/i x x 2 i D0

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Explicitly, H1 .x/ D H2 .x/ D H3 .x/ D H4 .x/ D

1 X

nD1 1 X

nD1 1 X

nD1 1 X

nD1

2

x

Ln x n D

1

L2n x n D

4 7x x 2 ; .1 C x/.1 3x C x 2 /

L3n x n D

8 13x 24x 2 C x 3 ; .1 C x x 2 /.1 4x x 2 /

L4n x n D

16 79x 164x 2 C 76x 3 C x 4 : .1 x/.1 C 3x C x 2 /.1 7x C x 2 /

x2

x

;

Next, recall that the Pell numbers are defined by Pn D 2Pn 1 CPn 2 with P1 D 1; P2 D 2 and the Pell-Lucas numbers satisfy the same recursive relation but with Q1 D 2; Q2 D 6. Setting p p s D 1 C 2; t D 1 2; we are able to find:

Binet-type Formulas 1 Pn D p .s n 2 2

t n /; Qn D s n C t n I

the generating functions of fPk n g and fQk n g: 1 X

nD0 1 X

nD0

Pk n x n D

1

Pk x ; Qk x C . 1/k x 2

Qk n x n D

1

2 Qk x I Qk x C . 1/k x 2

and the power-reduction formulas:  k 1  X 2k 23k Pn2k D . 1/i.nC1/ Q2.k i i D0

23k Pn2kC1 D

k.nC1/ i /n C . 1/

 k  X 2k C 1 . 1/i.nC1/ P.2.k i



2k k



;

i /C1/n :

i D0

P1

Pnk x n ; we have   k 1 X 2 Q2.k i / . 1/i x 2k 3k i 2 G2k .x/ D . 1/ i 1 Q2.k i / . 1/i x C x 2 i D0   1 2k C. 1/k ; k 1 . 1/k x

Finally, setting Gk .x/ D

nD0

 k  X 2k C 1 2 G2kC1 .x/ D i 1 3k

i D0

P2.k i /C1 x Q2.k i /C1 . 1/i x

x2

:

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In particular, G1.x/ D G2.x/ D G3.x/ D G4.x/ D

1 X

nD1 1 X

nD1 1 X

nD1 1 X

nD1

x 2x

Pn x n D

1

Pn2 x n D

x.1 .1 C x/.1

Pn3 x n D

x.1 4x .1 C 2x x 2 /.1

Pn4 x n D

x2

; x/ ; 6x C x 2/ x2/ 14x

x2/

;

x.1 C x/.1 14x C x 2 / x/.1 C 6x C x 2/.1 34x

.1

x2/

:

Exercises 1. Let  be the golden ratio. For positive integer n  2, prove that  n D Fn  C Fn

1:

2. Define u0 D a; u1 D b and un D pun

qun

1

2

with p 2 ¤ 4q. Find the generating function for fun g. 3. Let Gk .x/ D

1 X

ukn x n :

nD0

Prove that .Gm

qGm

1 /G1 .x/

D Œa C .b

pa/x

1 X

GmCn x n :

nD0

4. Monthly Problem 3801 [1936, 580; 1938, 636–637]. Show that cot

1

1 D cot

1

2 C cot

1

5 C cot

1

13 C cot

1

34 C


;

where these integers constitute every other term of the Fibonacci series and satisfy the recurrence unC1 D 3un un 1 . 5. A Problem of Euler/Monthly Problem 3674 [1934, 269; 1935, 518–521]. In how many different ways can a convex n-gon be divided into triangles by nonintersecting diagonals? Comment. Drawing two nonintersecting diagonals in a convex pentagon divides the interior into three triangles. This can be done in five distinct ways. Let an be the

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11. Generating Functions for Powers of Fibonacci Numbers

required number of ways for an .n C 2/-gon. Thus a3 D 5. In general, show that an satisfies a recurrence relation anC1 D

n X

ak an

k:

kD0

Use the generating function to show that an D .2n/Š=.nŠ.n C 1/Š/. 6. Find an explicit formula for an if an D nan

1

C .n C 1/Š for n  1, and a0 D 0.

7. Let a1 D a2 D 1, and let an D 1

n 2 1X .n n

k

1/kak ;

if n > 2:

kD1

Find an explicit formula for an . 8. For n  2, show that n X Fn FnC1 Fk2  FnC2 FnC2 Fk 1

1:

kD1

9. Prove that Fk n D Lk n D

F.k 5F.k

1/n Ln

C L.k 1/n Fn ; 2 1/n Ln C L.k 1/n Fn : 2

10. Prove that Fk n

 k  X k D Fni Fnk 1i Fi ; i i D0

Lk n D

 k  X k Fni Fnk 1i Li : i i D0

11. Monthly Problem 10765 [1999, 864; 2001, 978–979]. Fix positive integers k and n with n > 2k C 1. Prove that p p p b k Fn c b k Fn k C k fn 2k c is 0 unless Fn is a kth power, and then it is 1.

12. Monthly Problem 10825 [2000, 752; 2002, 762–763]. Given real numbers x and y, define Sk .x; y/ for k 2 N by S0 .x; y/ D x; S1.x; y/ D y, and the recurrence Sn .x; y/ D Sn 1 .x; y/ C Sn 2 .x; y/ for all n 2 N. Show that r jx 2 C xy y 2 j inf jSn .x; y/j  ; n2N 5 and determine when equality holds.

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13. Putnam Problem 1999-A3. Consider the power series expansion

1

1 2x

x3

D

1 X

an x n :

nD0

Prove that, for each integer n  0, there is an integer m such that 2 an2 C anC1 D am :

14. Open Problem. Let fn satisfy that fn D afn

1

C bfn

2

C cfn

3

with arbitrary initial values f0 ; f1 and f2 . Use Mathematica or Maple to find an explicit formula for second powers. Is there a similar formula for third powers? Also, what about formulas for fn D afn

1

C bfn

2

C cfn

3

C dfn

4‹

References [1] T. Mansour, A formula for the generating functions of powers of Horsdam’s sequence, Australasian J. Combinatorics, 30 (2004) 207–212. [2] J. Riordan, Generating functions for powers of Fibonacci numbers, Duke Math. J., 29 (1962) 5–12. [3] E. Weisstein, Fibonacci Number, Mathworld — A Wolfram Web Resource, available at mathworld.wolfram.com/FibonacciNumber.html [4] E. Weisstein, Pell Numbers, Mathworld — A Wolfram Web Resource, available at mathworld.wolfram.com/PellNumber.html

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12 Identities for the Fibonacci Powers In this chapter, using generating functions, we establish the following identities for any positive integers n and k, k FnCkC1 D

k X

k ai .k/ FnCk i;

i D0

where the ai .k/ are given explicitly in terms of Fibonomial coefficients (see (12.7) below). Along the way, we will focus on how to derive identities, instead of merely focusing on verification. Recall that Fibonacci numbers fFn g are defined by F1 D F2 D 1; FnC2 D FnC1 C Fn

(for n  1).

(12.1)

In the previous chapter, we proved that for any n  1, 2 2 2 FnC3 D 2FnC2 C 2FnC1

3 FnC4

D

3 3FnC3

C

3 6FnC2

Fn2 ; 3 3FnC1

(12.2) Fn3 :

(12.3)

In these identities, a power of a Fibonacci number is expressed as a linear combination of the same power of successive Fibonacci numbers. Naturally, we ask whether there are any more such curious and pretty identities as (12.1)–(12.3). Specifically, for any positive integers k and n, we look for a class of identities in the form k FnCkC1 D

k X

k ai .k/ FnCk i;

(12.4)

i D0

where ai .k/ are integers independent of n. For the fourth degree we make the reasonable guess that the formula has the form 4 4 4 4 4 FnC5 D aFnC4 C bFnC3 C cFnC2 C dFnC1 C eFn4 :

(12.5) 131

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Start with 0 B B B B B @

F54 F64 F74 F84 F94

F44 F54 F64 F74 F84

F34 F44 F54 F64 F74

F24 F34 F44 F54 F64

F14 F24 F34 F44 F54

1 0 1 0 41 a F6 C Bb C B F 4 C CB C B 7C C B C B 4C C B c C D B F8 C C B C B 4C A @d A @ F9 A 4 e F10

which results from setting n D 1; 2; 3; 4; 5 in (12.5) respectively. Solving this system, for example, by using Mathematica, a[n_] = Fibonacci[n]^4; b[n_] = a[n + 5] - a*a[n + 4] - b*a[n + 3] - c*a[n + 2] d*a[n + 1] - e*a[n]; Solve[{b[1] == 0, b[2] == 0, b[3] == 0, b[4] == 0, b[5] == 0}, {a, b, c, d, e}] {{a -> 5, b -> 15, c -> -15, d -> -5, e -> 1}} we obtain 4 4 4 FnC5 D 5FnC4 C 15FnC3

4 15FnC2

4 5FnC1 C Fn4 :

Along the same lines, we get 5 5 5 FnC6 D 8FnC5 C 40FnC4

5 60FnC3

5 5 40FnC2 C 8FnC1 C Fn5 I

6 6 6 FnC7 D 13FnC6 C 104FnC5

6 260FnC4

7 7 7 FnC8 D 21FnC7 C 273FnC6

7 1092FnC5

7 7 7 1820FnC4 C 1092FnC3 C 273FnC2

8 8 8 FnC9 D 34FnC8 C 714FnC7

8 4641FnC6

8 8 8 12376FnC5 C 12376FnC4 C 4641FnC3

6 21FnC1

6 6 6 260FnC3 C 104FnC2 C 13FnC1

Fn6 I

8 714FnC2

8 34FnC1 C Fn8 I

9 9 9 FnC10 D 55FnC9 C 1870FnC8 9 C 85085FnC4

9 19635FnC3

10 10 10 FnC11 D 89FnC10 C 4895FnC9 10 C 1514513FnC5

9 19635FnC7

Fn6 I

9 9 85085FnC6 C 136136FnC5

9 9 1870FnC2 C 55FnC1 C Fn9 I

10 83215FnC8

10 582505FnC4

10 10 582505FnC7 C 1514513FnC6

10 10 10 83215FnC3 C 4895FnC2 C 89FnC1

Fn10 :

The coefficients of the above identities are listed below: 0 B B B B B B B B B B B B B B B @

1 1 2 2 3 6 5 15 8 40 13 104 21 273 34 714 55 1870 89 4895

1 3 15 60 260 1092 4641 19635 83125

1 5 1 40 8 1 260 104 13 1820 1092 273 12376 12376 4641 85085 136136 85085 582505 1514513 1514513

1

1 21 714 19635 582505

1 34 1 1870 55 1 83215 4895 89

1

C C C C C C C C C: C C C C C C A

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This matrix reveals a wealth of patterns:  The first column is the Fibonacci numbers.  There is an alternating pattern of signs by double columns.  There is a symmetry similar to Pascal’s triangle up to sign. A similar table was obtained by Brousseau [1] via a different approach. However, it seems that there are no analytical formulas for determining the general ai .k/ in (12.4). To recognize the underlying pattern of the ai .k/, we apply Sloane’s On-Line Encyclopedia of Integer Sequences [5] to these results. This enables us to conjecture that the coefficients ai .k/ in (12.4) are: a0 .k/ D FkC1 D Fibonomial.k C 1; 1/;

a1 .k/ D FkC1 Fk D Fibonomial.k C 1; 2/; :: : ai .k/ D . 1/i Ci.i C1/=2 Fibonomial.k C 1; i C 1/I :: : ak .k/ D . 1/kCk.kC1/=2 ;

(12.6)

where a Fibonomial coefficient is defined by Fibonomial .n; k/ D

Fn Fn 1


Fn kC1 : Fk Fk 1


F1

(12.7)

When we look back over the discussion above, we find that the whole story is built on a hunch that a formula in the form of (12.4) exists. In the previous chapter, we saw how to use the identity (12.2) to derive the generating functions for Fibonacci squares, namely G2 .x/ D

1 X

nD0

Fn2 x n D

x.1 .1 C x/.1

x/ : 3x C x 2/

(12.8)

It is natural to ask about the converse of this derivation: Can we establish (12.2) based on the generating function (12.8)? Indeed, we can. To see this, rewrite (12.8) as .1

2x

2x 2 C x 3 /G2 .x/ D x

x2:

Now if G2 .x/ is the generating function of fFn2 g then xG2 .x/ is the generating function for the shifted sequence fFn2 1 g, and consequently we have .1

2x

2x 2 C x 3 /G2 .x/ D x

x2 C

and so Fn2

2Fn2

1

2Fn2

2

1 X

.Fn2

2Fn2

0

for all n  3;

1

2Fn2

nD3

C Fn2

3

2

C Fn2 3 /x n ;

which is equivalent to (12.2).

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It is no coincidence that the G2.x/ is a rational function if and only if (12.2) holds. In fact, the relationship between G2 .x/ and (12.2) is the base case of the following generalization. Theorem 12.1. A sequence fun g satisfies the recursive formula un D a1 un

1

C a2 un

2

C


C ak un

(12.9)

k

if and only if its generating function, G.x/, has the form G.x/ D

1

a1 x

g.x/ a2 x 2




ak x k

for some polynomial g.x/. Proof.

By the definition of a generating function, G.x/ D aj x j G.x/ D aj

1 X

un

jx

n

P1

nD0

un x n , we have

nDj

for each j D 1; 2; : : : ; k. Hence G.x/Œ1

a1 x

a2 x 2






ak x k 

a1 xG.x/ a2 x 2 G.x/


ak x k G.x/ 1 X D g.x/ C .un a1 un 1 a2 un 2


ak un k /x n ; D G.x/

nDk

where g.x/ is a polynomial. Thus, fun g satisfies (12.9) if and only if G.x/ has the asserted rational form. Therefore a sequence defined by a homogeneous linear recursive formula always has a rational generating function and vice versa. Moreover, the numerator contains information about the first few terms of the sequence, while the denominator describes the linear recursive relation. Based on Theorem 12.1, we can show the existence of (12.4) and also confirm the formulas stated in (12.6). Theorem 12.2. For any positive integers k and n, there exists an identity in the form of (12.4) with the coefficients ai .k/ determined by (12.6). To show the existence of (12.4), by Theorem 12.1, we need p p Lemma 12.1. Let ˛ D .1 C 5/=2; ˇ D .1 5/=2: If the generating function of fFnk g is in the form of 1 X Pk .x/ Gk .x/ D Fnk x n D ; (12.10) Qk .x/ nD0

then

Qk .x/ D

k Y

.1

˛ i ˇ k i x/

(12.11)

i D0

and the degree of Pk .x/ is less than the degree of Qk .x/.

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Proof.

By Binet’s formula and the binomial theorem, we have 1 1 k   X X X 1 k Fnk x n D . 1/k i ˛ ni ˇ n.k i 5k=2 nD0

nD0

D D

1 5k=2 1 5k=2

k X i D0

k X i D0

i D0

  k . 1/k i

i

1 X

i/

˛ ni ˇ n.k

i/

!

xn

xn

nD0

  . 1/k i k : i 1 ˛i ˇk i x

If k is odd, grouping the i -th term and the .k i /-th term together, we get ! .k 1/=2   1 X X 1 . 1/k i . 1/i k k n C Fn x D k=2 i 5 1 ˛i ˇk i x 1 ˛k i ˇi x nD0 i D0 D D

1 5k=2 1 5k=2

.kX 1/=2 i D0

.kX 1/=2 i D0

   k . 1/i i 1

1 ˛i ˇk i x

1 ˛k i ˇi x

1



  .˛ k i ˇ i ˛ i ˇ k i /x k . 1/i : i .1 ˛ k i ˇ i x/.1 ˛ i ˇ k i x/

If k is even, in view of the fact that ˛ˇ D 1, grouping the i -th term and the .k term together, with the exception of the middle term, we get 1 X

i /-th

Fnk x n

nD0

D D D

1 5k=2 1 5k=2 1 5k=2

k=2 1

X k  i i D0

. 1/k i . 1/i C 1 ˛i ˇk i x 1 ˛k i ˇi x

k=2 1

 X k  . 1/i i 1 i D0

1 C i ˛ ˇk i x 1

!



 . 1/k=2 k C k=2 1 . 1/k=2 x

1 ˛k i ˇi x



C



 . 1/k=2 k k=2 1 . 1/k=2 x

k=2 1

  X k  2 .˛ k i ˇ i C ˛ i ˇ k i /x . 1/k=2 k i . 1/ C : i k=2 1 . 1/k=2 x .1 ˛ k i ˇ i x/.1 ˛ i ˇ k i x/ i D0

This proves (12.11) as desired and confirms that the degree of Pk .x/ is less than the degree of Qk .x/. Appealing to Theorem 12.1 again, we see the coefficients in (12.4) are determined by the expansion of Qk .x/. To perform the expansion comfortably, we show Lemma 12.2. Let k Y

.1

i D0

Then, for i D 1; 2; : : : ; k; k C 1, bi D . 1/i r i.i

1/=2

r i x/ D 1 C

kC1 X

bi x i :

(12.12)

i D1

.r kC1 1/.r k .r 1/.r 2

1/


.r k 1/


.r i

i C2

1/

1/

:

(12.13)

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Proof.

Set f .x/ D

k Y

r i x/:

.1

i D0

Then f satisfies the functional equation .1

r kC1 x/;

x/f .r x/ D f .x/.1

which is equivalent to .1

x/ 1 C

kC1 X

i

bi r x

i

i D1

!

D 1C

kC1 X

bi x

i

i D1

!

.1

r kC1 x/:

Equating the coefficients of x i .i D 1; 2; : : : ; k; k C 1/ on both sides, we have r i bi

ri

1

bi

1

D bi

r kC1 bi

1;

and so

r kC1 r i 1 r i 1 .r k i C2 1/ b D bi 1 : i 1 1 ri ri 1 Iterating this relationship to write bi 1 in terms of bi 2 and continuing until bi is expressed as a multiple of b1 , we have bi D

bi D . 1/2

ri

1 i 2

.r k i C2 1/.r k i C3 .r i 1/.r i 1 1/

r

1/

bi

2

:: : D . 1/i

1

ri

1 i 2

r




1.r k i C2 .r i 1/.r i

1/.r k i C3 1/


.r k 1 1/


.r 2 1/

Since b1 D .1 C r C r 2 C


C r k / D

1/

b1 :

r kC1 1 ; r 1

we obtain bi D . 1/i

ri

1 i 2

r




1.r kC1 1/.r k 1/


.r k .r 1/.r 2 1/


.r i 1/

i C2

1/

;

which is equivalent to (12.13) and this ends the proof of Lemma 12.2. Now, we are ready to prove Theorem 12.2. Set Qk .x/ D 1

k X

ai .k/x i C1 :

(12.14)

i D0

Multiplying (12.10) by (12.14), we have Pk .x/ D

1 X

nD0

Fnk x n

1 X k X

ai .k/Fnk x nCi C1 :

nD0 i D0

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Equating the coefficients of x nCkC1 on both sides gives k X

k 0 D FnCkC1

k ai .k/FnCk i:

i D0

This proves (12.4) as desired. Thus, it remains to verify the formulas stated in (12.6). Replacing r by ˛=ˇ and x by ˇ k x in (12.12), in view of the exponents of ˇ in (12.13), namely k.i C 1/

1 i.i C 1/ 2

1 .i C 1/.i C 2/ 2

1 .i C 1/.2k 2

i C 2/ D

1 i.i C 1/ 2

and ˛ˇ D 1, we have ai .k/ D

bi C1 ˇ k.i C1/

D . 1/i . 1/i.i C1/=2

.˛ kC1

ˇ kC1 /.˛ k ˇ k /


.˛ k i C1 ˇ k .˛ ˇ/.˛ 2 ˇ 2 /


.˛ i C1 ˇ i C1 /

i C1

/

:

By the definition of Fibonomial coefficients (12.7), we get ai .k/ D . 1/i . 1/i.i C1/=2 Fibonomial.k C 1; i C 1/: This proves (12.6) as required. To demonstrate the power of the both theorems, we highlight two applications. 1. Let the sequence fUn g satisfy UnC2 D aUnC1 C bUn with a2 C 4b > 0 and the generating function for the kth power be given by 1 X

nD0

Unk x n D

pk .x/ : qk .x/

Based on the proof of Lemma 12.1, we find that qk .x/ D

k Y

.1

s i t k i x/;

i D0

where s and t are roots of x 2 ax b D 0. Furthermore, the degree of pk .x/ is less than the degree of qk .x/. Thus, similar to Theorem 12.2, we have k UnCkC1 D

k X

k ci .k/ UnCk i;

i D0

where the coefficients ci ; i D 0; 1; 2; : : : ; k; are given by ci D . 1/i . b/i.i C1/=2

.s kC1

t kC1 /.s k t k /


.s k i C1 t k .s t/.s 2 t 2 /


.s i C1 t i C1 /

i C1

/

:

In particular, for Lucas numbers fLn g, since LnC2 D LnC1 C Ln and Ln D ˛ n C ˇ n ;

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we observe that the denominator of the generating function of fLkn g is the same as that of fFnk g. Thus, Theorem 1 implies that identity (12.4) holds for Lucas numbers: LknCkC1 D

k X

ai .k/ LknCk i ;

i D0

where ai .k/ are given by (12.6). Explicitly, for any positive integer n, L2nC3 D 2L2nC2 C 2L2nC1

L2n ;

L3nC4 D 3L3nC3 C 6L3nC2

3L3nC1

L4nC5 D 5L4nC4 C 15L4nC3

L3n ;

15L4nC2

5L4nC1 C L4n :

For the Pell numbers fPn g, which are defined by PnC2 D 2PnC1 C Pn ;

we have a D 2; b D 1; s D 1 C k PnCkC1

D

k X

. 1/i Ci.i C1/=2

p 2; t D 1 .s kC1

p 2. Thus, Theorem 1 implies that

t kC1 /.s k t k /


.s k i C1 t k .s t/.s 2 t 2 /


.s i C1 t i C1 /

i D0

i C1

/

k PnCk i:

Explicitly, for any positive integer n, 2 2 2 PnC3 D 5PnC2 C 5PnC1

Pn2 ;

3 3 3 PnC4 D 12PnC3 C 30PnC2

3 12PnC1

4 4 4 PnC5 D 29PnC4 C 174PnC3

Pn3;

4 174PnC2

4 29PnC1 C Pn4:

2. Define ˇ ˇ Fnk ˇ ˇ k ˇ FnC1 Dk .Fn / D ˇˇ : ˇ :: ˇ ˇ Fk nCk

k FnC1 k FnC2 :: : k FnCkC1









:: :




k FnCk k FnCkC1 :: : k FnC2k

ˇ ˇ ˇ ˇ ˇ ˇ: ˇ ˇ ˇ ˇ

We show that Theorem 12.2 can be used to simplify the computation of Dk .Fn / as Dk .FnC1 / D . 1/nk.kC1/=2 Dk .F1 /: Indeed, since ˇ k ˇ FnC1 ˇ ˇ k ˇ FnC2 Dk .FnC1 / D ˇˇ :: ˇ : ˇ ˇ Fk nCkC1

k FnC2 k FnC3 :: : k FnCkC2









:: :




k FnCk k FnCkC1 :: : k FnC2k

ˇ k ˇ FnCkC1 ˇ ˇ k FnCkC2 ˇ ˇ; :: ˇ ˇ : ˇ k ˇ F nC2kC1

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applying (12.4) on the last column and then repeatedly using the column operations, we have ˇ ˇ k k k ˇ FnC1 FnC2


FnCk ak .k/Fnk ˇˇ ˇ ˇ ˇ k k k FnC3


FnCkC1 ak .k/Fnk ˇ ˇ FnC2 ˇ Dk .FnC1 / D ˇˇ :: :: :: :: ˇ :: ˇ ˇ : : : : : ˇ ˇ k k k ˇ ˇ Fk F


F a .k/F k n nCkC1 nCkC2 nC2k ˇ ˇ k k ˇ Fnk ˇ FnC1


FnCk ˇ ˇ ˇ k ˇ k k F F


F ˇ nC1 nC2 nCkC1 ˇˇ k ˇ D . 1/ ak .k/ ˇ : :: :: ˇ :: ˇ :: ˇ : : : ˇ ˇ k k ˇ Fk FnCkC1


FnC2k ˇ nCk D . 1/k ak .k/ Dk .Fn /:

Iterating this relationship, we get Dk .FnC1 / D . 1/2k ak2 .k/ Dk .Fn

1/

D


D . 1/nk aknk .k/ Dk .F1 / D . 1/nk.kC1/=2 Dk .F1 /:

In particular, for k D 1, we have ˇ ˇ D1 .Fn / D ˇˇ

ˇ FnC1 ˇˇ FnC2 ˇ

Fn FnC1

D . 1/n

1

D . 1/n

1

D1 .F1 / ˇ ˇ F2 n 1 ˇ F1 D . 1/ ˇ F2 F3 ˇ ˇ ˇ ˇ n 1 ˇ 1 1 ˇ D . 1/ ˇ 1 2 ˇ

That is, Fn FnC2

2 FnC1 D . 1/n

1

ˇ ˇ ˇ ˇ

:

, which is the well-known Cassini’s identity.

It is interesting to see that symbolic software provides a powerful tool for verifying identities. At the beginning of the chapter, we exhibited how to use Mathematica to verify the identity of Fibonacci fourth powers, but such a proof is not very attractive since it does not reveal why there is such an identity. A sophisticated application of a symbolic software should not only yield a deeper understanding but also offer new formulas. In [2], Dobbs presented an algorithmic method to manipulate trigonometric identities. This method was later simplified by Klamkin in [4]. Roughly speaking, they proved that all trigonometric identities can be derived from the basic identity sin2  C cos2  D 1. Thus, it is well worth knowing if the process of manipulating Fibonacci identities is also algorithmic. This should be a nice project for undergraduate research.

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Exercises 1. For n  2, prove that 5.Fn2

1

2 C FnC1 /  2. 1/nC1 .mod .Fn

1

C FnC1 //:

2. Show that Fn4

Fn

2 Fn 1 FnC1 FnC2

D 1:

3. Use a computer to show that 256F9n D 625Fn9 C 4500Fn7 L2n C 3150Fn5 L4n C 420Fn3 L6n C 6Fn L8n : Can you find a similar formula for L9n ? p 1 D i . Prove that 4. Let Fn D

n Y1

kD1

 1

k 2i cos n

n Y1

kD1

 1

D

b.n 1/=2c

Y

kD1

  2k 3 C 2 cos n

p ! p ! ni 1 C 5 2 n n 1C 5 ln p D p i sinh ln p ; 2 2 1 1 5 5 5

2 D p i nC1 sin 5 Ln D



2i cos

D 2i n i nC1 cos

 b.nY  2/=2c  .2k C 1/ .2k C 1/ D 3 C 2 cos 2n n kD1 ! p p ! ni 1 C 5 n 1C 5 n nC1 ln D 2i i cosh ln : p p 2 2 1 1 5 5

Comment: This should strengthen our conviction that manipulating Fibonacci identities is algorithmic. 5. Pascal’s triangle exhibits some interesting patterns of divisibility by various numbers. Rewrite (12.4) as kC1 X i D0

k . 1/i.i C1/=2 Fibonomial .k C 1; i C 1/ FnCk

i

D 0:

The triangle of Fibonomial coefficients begins with (see Sloane A010048 [5]) 1 1 1 1 1 1

2 3

5

1 1

1 2

6 15

1 3

15

1 5

1

Show that this triangle also has some nearly identical patterns of divisibility by the same numbers.

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12. Identities for the Fibonacci Powers

6. Putnam Problem 1986-A6. Let a1 ; a2 ; : : : ; an be real numbers, and let b1 ; b2; : : : ; bn be distinct positive integers. Suppose there is a polynomial f .x/ satisfying the identity 1 X .1 x/n f .x/ D 1 C ai x bi : i D1

Find a simple expression (not involving any sums) for f .1/ in terms of b1 ; b2; : : : ; bn and n ( but independent of a1 ; a2 ; : : : ; an .)

7. Putnam Problem 1993-A2. Let fxn g be a sequence of nonzero real numbers such that xn2 xn 1 xnC1 D 1 for n D 1; 2; 3 : : : : Prove there exists a real number a such that xnC1 D axn

xn

1

for all n  1.

8. Define u0 D 0; u1 D 1

and

un D aun

1

C bun

2:

Also, for any nonnegative integer m, define (   1; if j D 0; m D um


um j C1 : ; if j D 1; : : : m j u uj


u1 Prove that kC1 X

i.i C1/=2 i.i 1/=2

. 1/

b

i D0

  kC1 ukn i u

i

In particular, for k D 2,

u2nC3 D .a2 C b/u2nC2 C b.a2 C b/u2nC1

9. Let aij D



n

i j



ai Cj

n n j

b

D 0: b 3 u2n :

; 0  i; j  n

and AnC1 D .aij / be a matrix of order n C 1. Find all the eigenvalues of AnC1 and prove that uk.nC1/ tr.AknC1 / D : uk 10. Define the q–Fibonacci polynomials as f .n; a; b/ D af .n

1; a; b/ C q n

2

bf .n

2; a; b/

with initial values f .0; a; b/ D 0; f .1; a; b/ D 1: Prove that n X1 n 1 j  2 f .n; a; b/ D q j an 1 2j b j ; j j D0

where

  .1 q n /.1 q n 1 /


.1 q n k n D k .1 q/.1 q 2 /


.1 q k / denotes a q–binomial coefficient.

1

/

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References [1] A. Brousseau, A sequence of power formulas, Fibonacci Quarterly, 6 (1968) 81–83. [2] D. E. Dobbs, Proving trigonometric identities to fresh persons, MATYC J., 14 (1980) 39–42. [3] A. F. Horadam, Generating functions for powers of a certain generalized sequence of numbers, Duke Math. J., 32 (1965) 437–446. [4] M. S. Klamkin, On proving trigonometric identities, Math. Mag., 56 (1983) 215–220. [5] N. J. A. Sloane, The On-line Encyclopedia of Integer Sequences, available at www.research.att.com/ njas/sequences/

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13 Bernoulli Numbers via Determinants In this chapter, we represent Bernoulli numbers via determinants based on their recursive relations and Cramer’s rule. These enable us to evaluate a class of determinants involving factorials where the evaluation of these determinants by row and column manipulation is either quite challenging or almost impossible [3]. Recall the recursive relation for the Bernoulli numbers Bk (see (6.12)): B0 D 1;

n X1

kD0



n k



(13.1)

Bk D 0:

Let bk D Bk = kŠ. Rewriting (13.1) explicitly yields the following system of linear equations: 8 1 b1 C 2Š D 0; ˆ ˆ ˆ ˆ b ˆ 1 1 ˆ C b2 C 3Š D 0; ˆ 2Š ˆ ˆ ˆ ˆ b b 1 1 2 < C 2Š C b3 C 4Š D 0; 3Š

or equivalently

ˆ





ˆ ˆ ˆ ˆ b1 ˆ 1 ˆ C .nb22/Š C


C bn 1 C nŠ D 0; ˆ .n 1/Š ˆ ˆ ˆ : b1 1 C .nb21/Š C


C bn C .nC1/Š D 0; nŠ 8 1 b1 D 2Š ; ˆ ˆ ˆ ˆ ˆ b 1 1 ˆ ˆ 2Š C b2 D 3Š ; ˆ ˆ ˆ ˆ < b 1 C b 2 C b3 D 3Š 2Š

1 ; 4Š

ˆ





ˆ ˆ ˆ ˆ b1 ˆ ˆ C .nb22/Š C


C bn ˆ .n 1/Š ˆ ˆ ˆ : b1 C .nb21/Š C


C bn D nŠ

1

D

1 ; nŠ

1 : .nC1/Š

143

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Solving for bn by means of Cramer’s rule, we find ˇ ˇ 1 0 0 ˇ ˇ 1 ˇ 1 0 ˇ 2Š ˇ 1 1 ˇ 1 2Š ˇ 3Š ˇ bn D ˇ :: :: :: ˇ : : : ˇ ˇ 1 1 1 ˇ .n 1/Š .n 2/Š .n 3/Š ˇ ˇ 1 1 1 ˇ nŠ .n 1/Š .n 2/Š

1 2Š 1 3Š 1 4Š










::

:: :

:

1 nŠ






1 .nC1/Š






ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ: ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ

Using the basic properties of the determinants, factoring out the coefficient 1 and then permutating the columns, we obtain ˇ ˇ 1 ˇ 1 0


0 ˇˇ ˇ 2Š ˇ ˇ 1 1 ˇ 1


0 ˇˇ 2Š ˇ 3Š ˇ ˇ 1 1 1 ˇ


0 ˇˇ 4Š 3Š 2Š ˇ ˇ Bn D nŠbn D . 1/n nŠ ˇˇ :: :: :: : ˇ ; for all n 2 N: (13.2) :: ˇ : :: ˇ : : : ˇ ˇ ˇ ˇ 1 1 1 ˇ ˇ nŠ


1 .n 1/Š .n 2/Š ˇ ˇ ˇ 1 1 1 ˇˇ 1 ˇ .nC1/Š


nŠ .n 1/Š 2Š

In particular, appealing to the well-known fact (see (6.10))

B2nC1 D 0; n D 1; 2; 3; : : : : we get an added bonus from (13.2), namely ˇ 1 ˇ 1 0


ˇ 2Š ˇ 1 1 ˇ 1


2Š ˇ 3Š ˇ 1 1 1 ˇ


3Š 2Š ˇ 4Š ˇ :: :: :: ˇ :: ˇ : : : : ˇ ˇ 1 1 1 ˇ nŠ


.n 1/Š .n 2/Š ˇ ˇ 1 1 1 ˇ .nC1/Š


nŠ .n 1/Š

ˇ 0 ˇˇ ˇ 0 ˇˇ ˇ 0 ˇˇ ˇ :: ˇ D 0; : ˇˇ ˇ 1 ˇ ˇ 1 ˇˇ 2Š

for n D 3; 5; 7; : : : :

This was the proposed problem 3784 which appeared in the American Mathematical Monthly, 1936. Furthermore, using B0 D 1; B1 D 1=2 and B2nC1 D 0 for n  1, we may rewrite x ex as

1

D1

  x2 x3 xD xC C C


1 2Š 3Š

1 X 1 B2n 2 xC x n 2 .2n/Š nD1

 x C b2 x 2 C b4 x 4 C b6 x 6


: 2

(13.3)

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13. Bernoulli Numbers via Determinants

Applying the Cauchy product of power series on the right hand, and then equating the coefficients of x 2kC1 for k D 1; 2; : : : n in (13.3), we have 8 1 1 b2 C 3Š D 0; ˆ 2.2Š/ ˆ ˆ ˆ ˆ b2 1 1 ˆ ˆ C b4 C 5Š D 0; ˆ 3Š 2.4Š/ ˆ ˆ ˆ ˆ 1 < b 2 C b 4 C b6 C 1 D 0; 5Š 3Š 7Š 2.6Š/ ˆ ˆ





ˆ ˆ ˆ ˆ b2 ˆ ˆ ˆ .2n 3/Š C ˆ ˆ ˆ : b2 C .2n 1/Š

b4 .2n 5/Š

C


C b2.n

b4 .2n 3/Š

C


C b2n C

1/

C

1 .2n 1/Š

1 .2nC1/Š

1 2Œ2.n 1/Š

1 2Œ.2n/Š

D 0;

D 0;

or equivalently 8 1 ; b2 D 2 .3Š/ ˆ ˆ ˆ ˆ ˆ b2 3 ˆ ˆ ˆ 3Š C b4 D 2 .5Š/ ; ˆ ˆ ˆ ˆ < b 2 C b 4 C b6 D 5 ; 5Š 3Š 2 .7Š/ ˆ





ˆ ˆ ˆ ˆ ˆ b2 ˆ ˆ C ˆ .2n 3/Š ˆ ˆ ˆ : b2 .2n 1/Š C

(13.4)

b4 .2n 5/Š

C


C b2.n

b4 .2n 3/Š

C


C b2n D

1/

D

2n 3 ; 2Œ.2n 1/Š

2n 1 2 Œ.2nC1/Š ;

where we have used the identity 1 2 Œ.2k/Š

1 2k 1 D : .2k C 1/Š 2 Œ.2k C 1/Š

Therefore, using Cramer’s rule for the system of linear equations (13.4) and recalling B2n D .2n/Š b2n , we find

n 1

B2n D . 1/

ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ .2n/Š ˇˇ 2 ˇˇ ˇ ˇ ˇ ˇ ˇ ˇ

1 3Š 3 5Š 5 7Š

1

0

1 3Š 1 5Š






1 1 3Š






:: :

:: :

:: :

2n 3 .2n 1/Š

1 .2n 3/Š

1 .2n 5/Š

2n 1 .2nC1/Š

1 .2n 1/Š

1 .2n 3/Š




::

:









ˇ 0 ˇˇ ˇ 0 ˇˇ ˇ 0 ˇˇ ˇ :: ˇ : : ˇˇ ˇ 1 ˇ ˇ 1 ˇˇ 3Š

On the other hand, equating the coefficients of x 2k for k D 2; 3; : : : ; n C 1 in (13.3), we

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have 8 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i C1

Prove that

. 1/nC1 .22n 2/B2n : .2n/Š

det.Aij / D

7. Prove the Vandermonde determinant evaluation Y det .xij 1 / D 1i;j n

.xj

xi /:

1i 0; Factor[%] (-1+n) n n 1 X

2

tan

kD1



k n



D n.n

(14.18)

1/I

D[T[t],{t,4}]/4!/.t->0; Factor[%] 1/3 (-1 +n)n(-3+ n+n^2) Thus n X1

tan4

kD1



k n



D

n.n

1/ 3

.n2 C n

(14.19)

3/I

Tc[t_] :=n*(1 - t*Cot[n*ArcTan[t]])/(1 + t^2); D[Tc[t],{t,2}]/2!; Limit[%, t->0] 1/3 (-2 + n)(-1 + n) Thus n X1

tan2

kD1;k¤n=2



k n



D

1 .n 3

1/.n

(14.20)

2/I

D[Tc[t],{t,4}]/4!; Factor[ Limit[%, t->0]] 1/45 (-2 + n)(-1 + n)(-13 + 3n + n^2) Thus n X1

tan4

kD1;k¤n=2

14.4



k n



D

.n

1/.n 45

2/

.n2 C 3n

13/:

(14.21)

Sums involving cot2p .k=n/

Let Cp .n/ D

n 1 X

2p

cot

kD1



k n



and its generating function be A.n; t/ D

1 X

Cp .n/t 2p :

pD0

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Then 1 X

A.n; t/ D

n 1 X

pD0 n X1

D

kD1 n X1

D

kD1

cot

kD1



k n

!

t 2p

1 cot2 .k=n/

1

t2

1

1 cos2 .k=n/ .1 C t 2 / cos2 .k=n/

n 1 C t2

D

2p

n 1 t2 X 1=.1 C t 2 / : 1 C t2 cos2 .k=n/ 1=.1 C t 2 / kD0

Substituting t D tan  yields n X1

kD0

n 1 X 1=.1 C t 2 / cos2  D : cos2 .k=n/ 1=.1 C t 2 / cos2 .k=n/ cos2  kD0

Therefore,

n .1 t cot.n arctan t// D T  .n; t/: 1 C t2 As an immediate consequence, (14.20) and (14.21) deduce that   n X1 .n 1/.n 2/ 2 k cot D ; n 3 A.n; t/ D

(14.22)

kD1 n 1 X

4

cot

kD1



k n



D

.n

1/.n 45

2/

.n2 C 3n

13/:

(14.23)

We now have established a variety of finite trigonometric power sum identities. The reader should be aware that different approaches to the explicit evaluations of finite trigonometric sums often yield distinctly different types of closed form evaluations. Notably, our preceding method yields evaluation in terms of binomial coefficients (see Exercise 9), while the comprehensive paper [1] offers a different approach and yields sums in terms of Bernoulli numbers. Clearly, as the powers of the trigonometric functions increase, explicit computations become increasingly tedious. Consequently, it is reasonable to use Mathematica to perform such calculations. For completeness, we provide additional power sum identities below.

A table of more power sums 1. Power sums of secant When n is odd: n X1

sec

6

kD0 n X1

kD0

sec8



k n



D

n2 .2n4 C 5n2 C 8/; 15



k n



D

n2 .17n6 C 56n4 C 98n2 C 144/; 315

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14. On Some Finite Trigonometric Power Sums n X1

sec

10

kD0



k n



D

n2 .62n8 C 255n6 C 546n4 C 820n2 C 1152/: 2835

When n is even: n 1 X

sec6



k n



D

.n2 1/ .2n4 C 23n2 C 191/; 945

sec8



k n



D

.n2 1/ .3n6 C 43n4 C 337n2 C 2497/; 14175

sec10



k n



D

.n2 1/ .2n8 C 35n6 C 321n4 C 2125n2 C 14797/: 93555

kD0;k¤n=2 n 1 X

kD0;k¤n=2 n X1

kD0;k¤n=2

2. Power sums of cosecant n 1 X

csc

6

csc

8

kD1 n 1 X

kD1 n X1

csc10

kD1



k n



D

.n2 1/ .2n4 C 23n2 C 191/; 945



k n



D

.n2 1/ .3n6 C 43n4 C 337n2 C 2497/; 14175



k n



D

.n2 1/ .2n8 C 35n6 C 321n4 C 2125n2 C 14797/: 93555

3. Power sums of tangent When n is odd: n 1 X

tan

6

tan

8

kD1 n X1

kD1 n X1

kD1

tan10



k n



D

n.n 1/ .2n4 C 2n3 15



k n



D

n.n 1/ .17n6 C 17n5 315

95n4



k n



D

n.n 1/ .62n8 C 62n7 2385

448n6

2232n2

8n2

8n C 15/; 95n3 C 213n2 C 213n

315/:

48n5 C 1358n4 C 1358n3

2232n C 2835/:

When n is even: n 1 X

kD1;k¤n=2

6

tan



k n



D

.n

1/.n 945

2/

.2n4 C 6n3

28n2

96n C 251/;

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n X1

8

tan

kD1;k¤n=2

n X1



k n



D

.n

1/.n 2/ .3n6 C 9n5 14175 C1761n

10

tan

kD1;k¤n=2



k n



D

.n

59n4

195n3 C 457n2

52n6

168n5 C 546n4

3551/;

1/.n 2/ .2n8 C 6n7 93555 C 1974n3

3068n2

13152n C 22417/:

4. Power sums of cotangent n X1

cot6



k n



D

cot8



k n



D

kD1 n X1

kD1 n X1

kD1

.n

.n

1/.n 945

2/

.2n4 C 6n3

1/.n 2/ .3n6 C 9n5 14175

59n4

  k .n 1/.n 2/ cot D .2n8 C 6n7 n 93555 10

3068n2

28n2

96n C 251/; 195n3 C 475n2 C 1761n

52n6

3551/;

168n5 C 546n4 C 1974n3

13152n C 22417/:

Exercises 1. Prove that n 1 X

kD0

sin x n sin nx D : cos. C 2k=n/ cos nx cos n

cos x

2. Let ! be the nth primitive root of unity and 0  j  n X i ¤j

j! i

!j j

2

D

1. Prove that

1 n.n2 12

1/:

3. Let ! be the nth primitive root of unity and let gN .n/ D

n X1

.! k

1/

N

:

kD1

Show that gN .n/ is a polynomial with rational coefficients and degree at most N, and determine gN .n/ for N D 1; 2; 3: 4. Evaluate

n X cos N.2k 1/=n ; 2 sin .2k 1/=.2n/ kD1

in terms of the integers N and n, 0  N  n.

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5. Let a and b be positive integers such that 0 < ab < n. Then for any x; y 2 R, n X1

kD0

sin2a



 nX1     bk bk n 2a Cx D cos2a C y D 2a : a n n 2 kD0

6. Monthly Problem 4152 [1945, 163; 1946, 344–346]. For N < x < N , show the following trigonometric expansion for the binomial coefficient N 2N X  m N m x cos cos D N mD1 N N

7. Prove that

N Cx 2

NŠ
N x
: Š 2 Š

(a) If n is even, then n X1

kD0

. 1/k cos  cos2 

cos2 . k / n

D n csc n csc :

(b) If n is odd, then n X1

kD0

. 1/k cos.k=n/ cos2 

/ cos2 . k n

D n csc n csc :

8. For odd integer n, find the generating function for the alternating sums of sec2pC1 .k=n/. 9. If n is odd, prove that 2p 1

Sp .n/ D n

X

. 1/pCk

kD1



p

1 C kn 2p 1

 2p X1  j Dk

2p j C1



:

10. Let n; p be positive integers and let Sn;p D

    2p nX1 k csc2p : 2n 2n kD1

Find a closed form expression for Sn;p . 11. Evaluate SD

N X1 NX1 NX1 i D1 j D1 kD1

sin.kj=N / sin.ij=N / : 1 cos.j=N /

12. Let

i j .i j / sin cos ; i; j D 1; 2; : : : ; n n n n Find the eigenvalues of the matrix A D .aij /. aij D sin

1:

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References [1] B. Berndt and B. Yeap, Explicit evaluations and reciprocity theorems for finite trigonometric sums, Advances in Appl. Math., 29 (2002) 358–385. [2] H. Chen, On some trigonometric power sums, Int. J. Math. Math. Anal., 30 (2002) 185–191. [3] W. Chu, Summations on trigonometric functions, Appl. Math. Comp., 141 (2003) 161–176. [4] K. R. Stromberg, Introduction to Classical Real Analysis, Wadsworth, Belmont, 1981.

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15 Power Series of .arcsin x/2 The power series expansion .arcsin x/2 D

1 1 X 2 nD1

.2x/2n  ; 2n 2 n n

jxj  1

(15.1)

plays an important role in the evaluation of series involving the central binomial coefficient. Some classical examples [1] are 1 X

nD1

1 n2



1 X

nD1

1 n4



2 1 D .2/; 18 3

. 1/nC1 2   D .3/; 5 2n n3 n

nD1 1 X

2n n

 D

2n n

 D

17 4 17  D .4/; 3456 36

where .x/ is the Riemann zeta function. Since Euler, the series (15.1) has been established in a variety of ways, for example, by using Euler’s series transformation [2] or by using the Gregory series for arctan x [1]. There are also a number of elegant and clever elementary proofs. In this chapter, we present two such proofs and display some applications. The first proof is based on the power series solution to a differential equation while the second is based on the evaluation of a parametric integral.

15.1

First Proof of the Series (15.1)

Let y.x/ D .arcsin x/2 . We begin with deriving a differential equation for y.x/. Differentiating y.x/ gives p 1 x 2 y 0 .x/ D 2 arcsin x: (15.2) 165

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Differentiating (15.2) and then multiplying by nary differential equation

p 1

x 2 /y 00 .x/

.1

x 2 shows that y.x/ satisfies the ordi-

xy 0 .x/ D 2:

(15.3)

Since this is a differential equation with variable coefficients, it is natural to seek a solution via the power series representation y.x/ D

1 X

an x n :

(15.4)

nD0

Termwise differentiation of (15.4) yields y 0 .x/ D a1 C 2a2 x C 3a3 x 2 C


D

1 X

nan x n

1

nD1

D

1 X

nD0

.n C 1/anC1 x n

and y 00 .x/ D 2a2 C 6a3 x C


D

1 X

nD1

n.n C 1/anC1 x n

1

D

1 X

nD0

.n C 2/.n C 1/anC2 x n :

Since a0 D y.0/ D 0; a1 D y 0 .0/ D 0, substituting the series of y 0 .x/ and y 00 .x/ into (15.3) yields 2a2 C 6a3 x C

1 X 

nD2

.n C 2/.n C 1/anC2

 n2 an x n D 2:

Equating coefficients of x n successively, we obtain a2 D 1; a3 D 0 and, in general, anC2 D

n2 an ; .n C 2/.n C 1/

for n  2:

For n  2, this determines a2nC1 D 0; a2n D

Œ2n 1 .n 1/Š2 1 D .2n/.2n 1/


4
3 2

22n  ; 2n 2 n n

where we have used the fact that   .2n/Š 2n : D n .nŠ/2 Thus, we find that y.x/ D

1 1 X .2x/2n  ; 2 nD1 2 2n n n

which converges for jxj  1 by the Ratio Test. This proves (15.1) as desired.

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15.2

167

Second Proof of the Series (15.1)

The second proof uses two different evaluations of the parametric integral f .x/ D

Z

=2

d ; x sin 

1

0

for jxj < 1:

(15.5)

For the first evaluation, setting t D tan.=2/, we have Z

=2

d D2 x sin 

1

dt 2 / C .t x x/2 0  ˇ 2 t x ˇ1 Dp arctan p ˇ 1 x2 1 x2 0 ! r 2 1Cx arctan Dp ; 1 x 1 x2

1

0

Z

.1

where we have used the identity arctan ˛ C arctan ˇ D arctan



 ˛Cˇ : 1 ˛ˇ

Next, appealing to the identity p 2 arctan s D arcsin



s 1 s C1



C

 ; 2

which can be verified by showing that both sides have the same derivatives, and setting s D .1 C x/=.1 x/, our original function becomes f .x/ D

 arcsin x p Cp : 2 2 1 x 1 x2

(15.6)

The second evaluation of (15.5) comes from expanding the integrand of (15.5) into a geometric series and then integrating term by term. This gives ! Z =2 1 X n f .x/ D sin  d x n : 0

nD0

To proceed, for n  2, we integrate by parts to get Z

1 n sin n

n

sin  d D

1

 cos  C

n

1 n

Z

=2

sinn

2

 d:

0

Evaluating at 0 and =2 yields Z Since

=2 n

sin  d D

0

Z

=2

1 n

 sin  d D ; 2 0

0

n

Z

Z

=2

sinn

2

 d:

0

=2 0

sin  d D 1;

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by induction, we obtain the following famous Wallis’s formulas: Z and

Z

=2

sin 0

2k

1
3


.2k 1/    d D D 2kC1 2
4


.2k/ 2 2

=2 0

sin2kC1  d D

  2k k

(15.7)

22k

2
4


.2k/ D 1
3


.2k C 1/

 : 2k .2k C 1/ k

Therefore,   1 1 X  X 1 2n 2n x C f .x/ D 2 nD0 22n n nD0

22n   x 2nC1 : 2n .2n C 1/ n

(15.8)

Recalling the binomial series expansion p 1

1 x2

D

  1 X 1 2n x 2n ; 22n n

(15.9)

nD0

and comparing (15.6) and (15.8), we obtain 1 X arcsin x p D 1 x2 nD0

22n   x 2nC1 : 2n .2n C 1/ n

(15.10)

Finally, integrating (15.10) yields 1 X 1 .arcsin x/2 D 2 nD0

22n

 x 2nC2 ; 2n .2n C 1/.2n C 2/ n 

that is, after replacing n C 1 by n, 2

.arcsin x/ D

1 X

nD1

22n .2n

1



2.n 1/ 1/.2n/ n 1

 x 2n ;

which is equivalent to (15.1). Here, in addition to the required series, we get (15.10) as a bonus. To demonstrate the power of (15.1), let us give some applications within the realm of elementary analysis. 1. The series expansion (15.1) yields the .2n/th derivative .arcsin2 .0//.2n/ D 22n

1

Œ.n

1/Š2; for all n  1:

The direct calculation is quite tedious.

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169

2. Integrating (15.9) gives arcsin x D

Z

x 0

dt

p 1

t2

D

1 X

nD0

1 2n 2 .2n C 1/

  2n 2nC1 x : n

In view of the Cauchy product of arcsin x and itself, equating the coefficients of x 2n , we obtain the combinatorial identity n X

kD0

1 .2k C 1/.2n

2k

   2k 2.n k 1/ D n k 1 1/ k

24n 3  : 2n n2 n

Furthermore, the Cauchy product of arcsin x and .arcsin x/2 leads to !   nX1 1 X 3Š 1 2n .arcsin x/3 D x 2nC1 : .2n C 1/22n n .2k C 1/2 nD1

kD0

In particular, for x D 1, we get 1 X

nD1

3Š .2n C 1/22n

!   nX1 3 1 2n D : 2 n .2k C 1/ 8 kD0

3. Letting x D sin t in (15.1), we have t2 D

1 X

nD1

22n 1   sin2n t; 2n n2 n

for jtj  =2:

Integrating with respect to t from 0 to =2 gives 1 X 3 D 24 nD1

22n 1   2n n2 n

Z

=2

sin2n t dt: 0

Applying Wallis’s formula (15.7), we get the well-known Euler formula 1

X 1 2 D D .2/: 6 n2 nD1 4. Dividing (15.1) by x and integrating from 0 to x, we have Z

x 0

1 .arcsin t/2 1 X .2x/2n dt D  : t 4 nD1 3 2n n n

(15.11)

This function is viewed as a “higher transcendent” by Lehmer [1]. A symbolic evaluation by Mathematicagives

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Sum[(2*x)^(2*n)/(n^3*Binomial[2 n, n]), {n, 1, Infinity}] 2 x^2 HypergeometricPFQ[{1, 1, 1, 1}, {3/2, 2, 2}, x^2] That is, 1 X .2x/2n 3   D 2x 2 Hypergeometric4 F3 Œf1; 1; 1; 1g; f ; 2; 2g; x 2: 2 2n nD1 n3 n

Mathematica also deduces 1 X

nD1

22n   D  2 ln 2 2n n3 n

7 .3/: 2

(15.12)

Moreover, replacing x by sin x and then setting t D sin  in (15.11), we have Z x 1 X 22n 2 (15.13)  2 cot  d D   sin2n x: 2n 0 nD1 n3 n Setting x D =2 and manipulating (15.12) and (15.13) yields an integral representation for .3/: Z 2 2 8 =2 2 .3/ D ln 2  cot  d: 7 7 0 Integrating by parts converts this into Z 2 2 16 =2 .3/ D ln 2 C  ln sin  d: 7 7 0 5. Using the power series expansion of arcsin x, Ewell in [3] rediscovered the Euler formula ( ) 1 X .2n/ 2 .3/ D 1 4 : (15.14) 7 .2n C 1/.2n C 2/22n nD1

In the following, we give a different proof of (15.14). Indeed, recall (see (6.23) and (6.26)) 1 X .2n/ 2n I (15.15)  cot  D 1 2  2n nD1 by (15.13), we get 1 2 x 2

1 X

nD0

1 X .2n/ 2.nC1/ x D .n C 1/ 2n nD1

22n 2   sin2n x: 2n n3 n

(15.16)

For x D =2, appealing to (15.12), we obtain 1 2  8

1 2 X .2n/ 1 D . 2 ln 2 4 nD1 .n C 1/22n 4

7 .3//: 2

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171

Thus, 2 2 .3/ D 7

ln 2

By the identity

1

1 X .2n/ C 2 nD1 .n C 1/22n

 1 1 D2 nC1 2n C 1

!

(15.17)

:

 1 ; .2n C 1/.2n C 2/

and appealing to the existing sum (see Exercise 10) 1 X

nD1

.2n/ 1 D .1 .2n C 1/22n 2

ln 2/

(15.18)

we see that Euler’s formula (15.14) follows from (15.17) immediately. Furthermore, integrating (15.16) we obtain 1 3 x 6

1 X

nD0

1 X .2n/ 2nC3 x D .n C 1/.2n C 3/ 2n

nD1

22n 2   2n 3 n n

Z

x

sin2n t dt: 0

For x D =2, this gives another series representation for .3/: ( ) 1 X .2n/ 2 .3/ D  1 2 : .2n C 2/.2n C 3/22n nD1

(15.19)

We have seen a variety of consequences of (15.1). With the aid of these expressions (15.1) and (15.9)–(15.19), the interested reader is encouraged to find additional applications.

Exercises 1. Show that

Z

b a

f .x/ cot x dx D 2

Using this result verifies (5.12) directly. p 2. Setting x D 2=2 in (15.10) yields D

1 Z X

nD1

b

f .x/ sin 2nx dx: a

1 X .nŠ/2 2nC1 : .2n C 1/Š

nD0

Can you give another proof? 3. Using the power series expansion of (15.9) and Z x 1 arcsin t 2 .arcsin x/ D p dt; 2 0 1 t2 prove that 2 1 1 1 D 1C 2 C 2 C


C C


: 8 3 5 .2k C 1/2

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4. Prove that (n 1 ) X 1 i2

1 3 X .arcsin x/ D 2 nD1

.2x/2n  ; 2n 2 i D1 n n 8 9 1 0; u C v < =2g whenever .x; y/ 2 D. See Figure 16.2.

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y

v p/2

1

D

S 1

x

p/2

u

Figure 16.2. Transformed region of integration The Jacobian determinant is 0 sin u sin v cos u B cos v cos2 v det B @ sin u sin v cos v 2 cos u cos u

1

sin2 u sin2 v D1 cos2 u cos2 v

C CD1 A

x2y2 :

Thus (16.4) is a bijective transformation between D and S . Hence J D

ZZ

S

1 dudv D Area of S D

2 1   2 D 2 2 8

as required. Recently, Noam Elkies in [5] generalized (16.4) to the n-variable transformation x1 D

sin u1 sin u2 ; x2 D ; : : : ; xn cos u2 cos u3

1

D

sin un 1 sin un ; xn D cos un cos u1

(16.5)

and proved the following two properties: 1. The Jacobian determinant of the transformation (16.5) is @.x1 ; x2; : : : ; xn / D 1 ˙ .x1 x2


xn /2 ; @.u1 ; u2 ; : : : ; un / the sign

and C chosen according to whether n is even or odd.

2. The transformation (16.5) maps the polytope …n WD f.u1 ; u2 ; : : : ; un / W ui > 0; ui C ui C1 < =2 .1  i  n/g one-to-one into the open unit cube Sn WD f.x1 ; x2 ; : : : ; xn / W 0 < xi < 1 .1  i  n/g: In particular, he found that the volume of the polytope …n is S.n/ D

1 X

kD0

1 X . 1/nk 1 D : n .2k C 1/ .4k C 1/n kD 1

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16.3

Proof by Trigonometric Identities

After these two proofs via surprising and unexpected transformations, we can not resist the temptation to present two completely elementary proofs. The first one begins with the identity 1 1 D 2 2 sin x 4 sin .x=2/ cos2 .x=2/   1 1 1 D C 4 sin2 .x=2/ cos2 .x=2/   1 1 1 D C : 4 sin2 .x=2/ sin2 .. C x/=2/ Letting x D =2 and repeatedly applying this identity yields   1 1 1 1 1D D C 4 sin2 .=4/ sin2 .=2/ sin2 .3 /=4   1 1 1 1 1 D C C C 16 sin2 .=8/ sin2 .3 /=8 sin2 .5=8/ sin2 .7 /=8 n

2 1 1 1 X D n 2 4 sin ..2k C 1/=2nC1 / kD0

D

2 4n

1 1 2nX

kD0

1 : sin2 ..2k C 1/=2nC1 /

In the range of 0 < t < =2, we have sin t < t < tan t and so 1 1 1 > 2 > cot2 t D t sin2 t sin2 t

1:

Thus, for N D 2n and t D .2k C 1/=.2N /, we have 1>

N=2 1 8 X 1 >1 2 .2k C 1/2 kD0

1 : N

Taking the limit N ! 1 yields 1D

1 8 X 1 ; 2  .2k C 1/2 kD0

which is equivalent to (16.3). The second one is based on the trigonometric identity N X

kD1

cot2



k 2N C 1



D

N.2N 3

1/

:

(16.6)

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This identity initially appeared in a sequence of exercises in the problem book by the twin brothers Akiva and Isaak Yaglom, whose Russian original edition was published in 1954. Versions of closely related proofs were rediscovered and presented by F. Holme (1970), T. Apostol (1973) and by Ransford (1982). Indeed, Identity (16.6) is also given by (14.22). In view of 1 cot2 t < 2 < csc2 t; t using (16.6) yields N.2N 3 that is

1/

 N  X 2N C 1 2 2N.N C 1/ < < ; k 3 kD1

N X  2 2N 2N 1 1  2 2N 2N C 2 < < : 2 6 2N C 1 2N C 1 k 6 2N C 1 2N C 1 kD1

Both the left-hand and the right-hand side converge to  2=6 for N ! 1, which ends the proof of (16.1).

16.4

Proof by Power Series

As we have seen in Chapter 15, the power series .arcsin x/2 together with Wallis’s formula is used to derive (16.1). In the following, we present Boo Rim Choe’s proof (Monthly, 1987), which uses only the power series of arcsin x. Note that arcsin x D

Z

x 0

dt

p 1

t2

D

1 X

nD0

1 22n .2n C 1/

  2n 2nC1 x ; n

(16.7)

which is valid for jxj  1. Putting x D sin t we get   1 X 1 2n tD sin2nC1 t 22n .2n C 1/ n nD0

for jtj  =2. Integrating from 0 to =2 and using Wallis’s formula Z =2 2
4


.2n/ sin2nC1 t dt D 3
5


.2n C 1/ 0 yields (16.3) again. Indeed, Choe’s proof can be traced back to the following original proof of Euler. Let Z 1 tk Ik WD p dt: 0 1 t2 An integration by parts shows that IkC2 D

kC1 Ik kC2

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from which Ik can be calculated for all k. In particular, we have I2nC1 D

2
4


.2n/ : 3
5


.2n C 1/

In view of 1 .arcsin x/2 D 2

Z

x 0

(16.8)

arcsin t p dt; 1 t2

setting x D 1, then replacing arcsin t in the integrand by its power series expansion (16.7) and making use of (16.8), yields 2 1 1 1 D 1C 2 C 2 C


C C


8 3 5 .2n C 1/2 which is again equivalent to (16.1). Euler tried to push this method for .2k/ for k  2 but did not succeed. However, in this attempt, he found the following rapidly convergent formula .2/ D 3

16.5

1 1 X X Œ.n 1/Š2 D3 .2n/Š nD1 nD1

1  : 2n n2 n

(16.9)

Proof by Fourier Series

Let f .x/ D x.1

x/. Consider the Fourier cosine series of f .x/ on Œ0; 1. Notice that a0 D 2 an D 2

Z

1

f .x/ dx D

0

Z

0

1 ; 3

1

f .x/ cos n x dx D 2

1 C . 1/n : n2  2

Therefore f .x/ D D

1

a0 X C an cos n x 2 nD1 1 6

1 X cos 2 nx :  2n2

nD1

Since f is continuous, of bounded variation on Œ0; 1 and f .0/ D f .1/, the Fourier series of f converges to f pointwise. Setting x D 0 we again get .2/ D  2=6.

16.6

Proof by Complex Variables

The final proof by Russell in [7] uses only Euler’s formula e ix D cos x C i sin x

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and the well-known logarithmic series. It begins with the definite integral I WD Now 2 cos x D e ix C e

ix

Z

=2

ln.2 cos x/ dx: 0

D e ix .1 C e

2ix

/ implies

ln.2 cos x/ D ix C ln.1 C e

2ix

/:

In terms of the logarithmic series expansion, we have I D

Z

! 1 X . 1/n 2nix ix e dx n nD1 ! 1 2 X 1 : 8 .2k C 1/2

=2 0

Di

kD0

But I is real, forcing 1 X 2 1 D 8 .2k C 1/2 kD0

and thereby yielding an added bonus Z

=2

ln.cos x/ dx D

0

 ln 2: 2

In general, the calculus of residues from complex analysis provides another tool to evaluate sums. A full account of this technique can be found in any introductory text on complex analysis. For example, see Marsden and Hoffman’s Basic Complex Analysis. We have seen a variety of proofs of Euler’s famous formula for .2/. Some proofs appear in historical contexts. Some proofs have provided interesting connections to other topics. It is interesting to see how wide a range of mathematical subjects appeared in these proofs. Although one proof is enough to establish the truth of the formula, many generations of mathematicians have amused themselves by coming up with alternative proofs. It should come as no surprise if there are still more new proofs to come. The interested reader is encouraged to consult Robin Chapman’s survey article, which compiles 14 different proofs and is available at www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf.

Exercises 1. Prove that

.3/ D

8 7

1 X

1 ; 2
2n n nD1  ! 1 .ln 2/3 X 1 ln 2 1 C C 3 : 3 2n n2 n

.2/ D .ln 2/2 C 2

nD1

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2. Another proof of Euler’s formula by double integral. Prove that Z 1 Z 1 2 tanh2k x D t 2k dt D (a) dx, 2 2k C 1 1 1 cosh x (b)

(c)

1 X tanh2k x tanh2k y 1 D 2 2 cosh.x C y/ cosh.x cosh x cosh y kD0 1 X

kD0

1 1 D 2 .2k C 1/ 4

Z

1Z 1 1

y/

1 cosh.x C y/ cosh.x

1

;

y/

dxdy D

2 : 8

3. Recall that (for example, see (5.28)) 

x sin x D .2n C 1/ sin 2n C 1 Let uk D

 Y n

1

kD1

! sin2 .x=.2n C 1// : sin2 .k=.2n C 1//

sin2 .x=.2n C 1// : sin2 .k=.2n C 1//

Show that Pn kD1 uk P 1 1 C nkD1 uk

n X sin x uk  : .2n C 1/ sin.x=.2n C 1// 1 uk kD1

Based on this inequality, give a proof of Euler’s formula. 4. Monthly Problem 3996 [1941, 341; 1942, 483–485]. Sum the series 1 X Œ.n 1/kŠ ; .nk/Š nD1

where k is any integer greater than unity. 5. Monthly Problem 4318 [1948, 586; 1950, 345–346]. Show that 1 X

nD1

1 2 D G .n sinh n /2 3

11 .2/; 30

where G is Catalan’s constant, which is defined by GD

1 X

nD0

. 1/n : .2n C 1/2

6. Show that 1 n   X 1 X n . 1/kC1 D 2 ln 2 k 2n k nD1 kD1

and

1 n   X 1 X n . 1/kC1 D .2/: k 2n k2 nD1 kD1

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Comment. Recall Euler’s Series Transformation, which states that if f .t/ D P1 P1 n nD0 an t and nD0 an converges, then   X  ! 1 n  X 1 t n n f D t ak : k 1 t 1 t nD0

kD0

One may establish that  1 n  1 X X 1 X xn n xk D 2 : k 2n kp np nD1 nD1 kD1

7. (H. Wilf) For every integer n  0, show that 3

n X i D1

1 

i2

C

2i i

1 nŠ4 X k2 2 D : .2n/Š Œ.k C n C 1/Š2 6 kD0

Remark. This gives us a family of series which all converge to  2=6. When n D 0 we have the usual series for .2/, and when n ! 1 we have the faster convergent series (16.9). 8. Show that .3/ D

Z

D

Z

1 0

Z 1Z 1 ln.xy/ 1 dxdydz D dxdy 1 xyz 1 xy 0 0 0 0 Z =2 ln sin x ln cos x ln.1 x/ ln x dx D 4 dx: x sin x cos x 0

Z

1 0

1

Z

1

9. Show that .3/ D

Z

1 2

Use this identity to show that

An D

1 X

i1 D1






1 X

in D1

An D . 1/n

Bn D

1 X

i1 D1






ln.1 x/ ln x dx: x.1 x/

i D1 j D1

Show that

11. Define

0

1 1 1 XX 2

.3/ D

10. Define

1

1 X

Z

in D1

ij 2

i Cj j

1

:

1 : i1 i2


in .i1 C i2 C


C in / 1

0

1 

lnn x dx D nŠ.n C 1/: 1 x

. 1/i1 Ci2 C


Cin : i1 i2


in .i1 C i2 C


C in /

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Show that Bn D . 1/n n

D . 1/

Z

1 0

lnn .1 C x/ dx x

nŠ.n C 1/

1 X n X n.n

n lnnC1 2 nC1

kD1 i D1

1/


.n i C 1/ n ln 2k k i C1

i

!

2 :

12. Open Problem. Motivated by 1 X 1 X

i D1j D1

Z

. 1/i D ij.i C j /

define Cn D

1 X

i1 D1






1 X

in D1

1

ln.1

0

x/ ln.1 C x/ 5 dx D .3/; x 8

. 1/i1 Ci2 C


Cim : i1 i2


in .i1 C i2 C


C in /

For 0 < m < n, determine Cn in the perspective of Exercises 10 and 11 above. This is equivalent to evaluating the integrals I.n; m/ D

Z

1

lnm .1

0

x/ lnn x

m

.1 C x/

dx:

13. Let 0 < ˛ < 1=2. Prove that .p C 1/ D

1 X

nD0

.1

 n  X 1 . ˛/n k n : nC1 k ˛/ .k C 1/pC1 kD0

In particular, .p C 1/ D 2

 1 X n  X . 1/n k 2k n : k .k C 1/pC1

nD0 kD0

References [1] T. M. Apostol, A proof that Euler missed: Evaluating .2/ the easy way, Math. Intelligence, 5 (1983) 59–60. [2] A. Ayoub, Euler and the Zeta Function, Amer. Math. Monthly, 81 (1974) 1067–1086. [3] J. Bernoulli, Tractatus de seriebus infinitis, 1689. Available at www.kubkou.se/pdf/mh/jacobB.pdf P 2 2 [4] B. R. Choe, An elementary proof of 1 kD1 1=n D  =6, Amer. Math. Monthly, 94 (1987) 662–663. P n [5] N. D. Elkies, On the sums 1 kD 1 .4k C 1/ , Amer. Math. Monthly, 110 (2003) 561–573. [6] D. Kalman, Six ways to sum a series, College Math. J., 24 (1993) 402–421.

[7] D. C. Russell, Another Eulerian-type proof, Math. Magazine, 64 (1991) 349. [8] E. Weisstein, Riemann Zeta Function zeta(2), From MathWorld—A Wolfram Web Resource. Available at mathworld.wolfram.com/RiemannZetaFunctionZeta2.html

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17 Evaluations of Some Variant Euler Sums In the mathematical world Euler is akin to the likes of Shakespeare and Mozart — universal, richly detailed, and inexhaustible. His work, dating back to the early eighteenth century, is still very much alive and generating intense interest. In response to a letter from Goldbach, for integers m  1 and n  2, Euler considered sums of the form   1 1 X 1 1 m X 1 1 1 C C


C D H m; (17.1) S.m; n/ WD .k C 1/n 2 k .k C 1/n k kD1

kD1

and was successful in obtaining several explicit values of these sums in terms of the Riemann zeta function .k/. Indeed, he established the beautiful formula S.1; 2/ D

1 X

kD1

1 Hk D .3/ .1 C k/2

as well as the more general relations: for all n > 2, S.1; n/ D

n .n C 1/ 2

n 2 1X .k C 1/.n 2

k/:

(17.2)

kD1

Two slight variant sums of Euler’s definition, which are known nowadays as Euler sums, are defined as 1 X 1 .m/ .m/ S.m; n/ D H ; with Hk D 1 C 1=2m C


C 1= k m ; kn k

(17.3)

kD1

and

  1 X 1 1 1 m T .m; n/ D 1C C


C : kn 2 k

(17.4)

kD1

It is easy to see that

T .2; n/ D S.2; n/ C 2S.1; n C 1/ C .n C 2/: 189

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Over the centuries progress on evaluating S.m; n/ and T .m; n/ for m  2 has been minimal. In 1948, as Monthly Problem 4305, Sandham proposed to prove that 1 X 1 17 T .2; 2/ D H2 D .4/; k2 k 4

(17.5)

kD1

which apparently remained unnoticed until 1993 when Enrico Au-Yeung, an undergraduate at the University of Waterloo, numerically rediscovered (17.5). Shortly thereafter it was rigorously proven true by Borwein and Borwein in [2]. This empirical result launched a fruitful search for Euler sums through a profusion of methods: combinatorial, analytic, and algebraic. Based on extensive experimentation with computer algebra systems, a large class of Euler and related sums have been explicitly evaluated in terms of the Riemann zeta values. Some typical evaluations include 1 X

kD1 1 X

1 Hk D .3/ 2k k 2

2 ln 2; 12

1 7 .2/ Hk D .4/; 2 k 4

kD1 1 X

kD1

7 1 Hk2 D .5/ 3 k 2

.2/.3/;

and 1 X . 1/k k2

kD1

1

Hk D

5 .3/: 8

In particular, Borwein and Bradley in [3] recently collected thirty-two beautiful proofs of S.1; 2/ D .3/. Motivated by these results, replacing Hk by hk D 1 C

1 1 C


C ; 3 2k 1

(17.6)

we study the following new variant Euler sums 1 X

ak hk

kD1

where the ak are relatively simple functions of k. We begin by deriving some series involving hk . Since Z x 1 X dt xk ln.1 x/ D D ; t k 0 1 kD1

replacing x by x gives ln.1 C x/ D

1 X . 1/k 1 x k : k

kD1

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Averaging these two series yields   X 1 1 1Cx 1 ln D x 2k 2 1 x 2k 1

1

(17.7)

:

kD1

In terms of the Cauchy product and partial fractions, we have 1 2 ln 4



1Cx 1 x



D

1  X

kD1

1 .2k

1/
1

C

1 .2k

3/
3

C


C

1 1
.2k

1/



x 2k

      1 X 1 1 1 1 1 1 1 D C C C C


C C x 2k 2k 2k 1 1 2k 3 3 1 2k 1 kD1

D

 2k 1  X 1 x 1 : 1C C


C 3 2k 1 k

kD1

Appealing to (17.6), we have   1 X hk 2k 1 2 1Cx x D ln : k 4 1 x

(17.8)

kD1

This enables us to evaluate a wide variety of interesting series via specialization, differentiation and integration. First, setting x D 1=2, we find 1 X

kD1

p For x D 2=2,

1 hk D ln2 3: 4 22k k

(17.9)

1 X p hk 1 D ln2 .3 C 2 2/: k 4 2 k

(17.10)

kD1

p Putting x D . 5

N the reciprocal of the golden ratio, we get 1/=2 D , 1 X p hk 2k 1 N D ln2 .2 C 5/: k 4

(17.11)

kD1

p p Furthermore, for any ˛  2, putting x D . 5 C 1/=2˛ and x D . 5 respectively, we get 1 X

kD1

hk ˛ 2k k

and 1 X

kD1

hk ˛ 2k k

!2k p 5C1 1 D ln2 2 4

.2˛ C 1/ C

!2k p 5 1 1 D ln2 2 4

.2˛

.2˛

1/

1/ C

.2˛ C 1/

1/=2˛ in (17.8)

p ! 5 p 5

(17.12)

p ! 5 p : 5

(17.13)

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Recalling the Fibonacci numbers, which are defined by F1 D 1; F2 D 1; Fk D Fk

1

C Fk

2

for k  2

and Binet’s formula 0 p !k 5C1 1 Fk D p @ 2 5

and combining (17.12) and (17.13), we find p  2 1 X hk 5 ˛ C˛ F D ln 2k 2k 20 ˛2 ˛ ˛ k

1 1

kD1

In particular, for ˛ D 2, 1 X

kD1

hk 22k k

F2k D

p !k 1 5 1 A; 2 

! p ˛2 C ˛ 5 C 1 ln p : ˛2 ˛ 5 C 1

p p 5 ln 5 ln.9 C 4 5/: 4

(17.14)

(17.15)

Another step along this path is changing variables. Setting x D cos  in (17.8) leads to 1 X hk cos2k  D ln2 .cot.x=2// : k

(17.16)

kD1

Integrating both sides from 0 to  , and using   Z   2k 2k cos  d D 2k k 2 0 and

Z

we find



ln2 .cot.x=2// d D

0

1 X

kD1

3 ; 4

  2 2k D : 4 22k k k hk

(17.17)

Next, for 0 < x < 1, differentiating (17.8), then multiplying both sides by x, we obtain 1 X

kD1

hk x 2k D

Setting x D 1=2, we get

x 2.1

x2/

ln



1Cx 1 x

1 X hk 1 D ln 3: 2k 3 2



:

(17.18)

(17.19)

kD1

For x D

p 2=2,

p 1 X p hk 2 D ln.3 C 2 2/: 2 2k

(17.20)

kD1

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We can now deduce a result similar to (17.15): p 1 X p p hk 5 F D .10 ln.5 C 2 5/ C 3 5 ln 5 2k 50 22k

5 ln 5/:

(17.21)

kD1

Finally, for 0 < x  1, dividing both sides of (17.9) by x and integrating from 0 to x, we obtain   Z 1 X 1 x 1 2 1Ct hk 2k x D ln dt: (17.22) k2 2 0 t 1 t kD1

Using the substitution u D .1 Z 1 1 X hk 2k x D k2 .1

kD1

D

x/=.1 C x/ and integration by parts, we get ln2 u du 1 u2

x/=.1Cx/

1 ln x ln2 2





1 x 1Cx

In view of (17.7), we have   Z 1 ln u 1 u ln du D 1Cu .1 x/=.1Cx/ u Since

we find

Z

u2k ln u du D

C

2

Z

1 .1 x/=.1Cx/

1 X

kD0

1 2k C 1

1 u2kC1 ln u 2k C 1

  ln u 1 u ln du: u 1Cu Z

1

u2k ln u du: .1 x/=.1Cx/

1 u2kC1 C C; .2k C 1/2

    1   1 X hk 2k 1 1 x 1 x X 1 1 x 2kC1 2 x D ln x ln C 2 ln k2 2 1Cx 1Cx .2k C 1/2 1 C x

kD1

kD0

C2

1 X

kD0

1 .2k C 1/3

2

1 X

kD0

1 .2k C 1/3



1 x 1Cx

2kC1

:

(17.23)

In order to proceed comfortably, we employ the classical polylogarithm function Lin .x/ D

1 X xn ; kn

kD1

which is the generating function of the sequence f1= k ng. The special case n D 2 is the dilogarithm introduced by Euler. Clearly, .2/ D Li2 .1/ and 1 X

kD0

xn 1 D .Lin .x/ .2k C 1/n 2

Lin . x//:

Appealing to (17.23) and 1 X

kD0

1 X 1 1 D .2k C 1/3 k3 kD0

1 X

kD0

1 7 D .3/; .2k/3 8

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we finally obtain   1 X hk 2k 7 1 1 x 2 x D .3/ C ln x ln k2 4 2 1Cx kD1       1 x 1 x x 1 Li2 Li2 C ln 1Cx 1Cx 1Cx      1 x x 1 Li3 Li3 : 1Cx 1Cx

Taking the limit as x approaches 1, we get

1 X hk 7 D .3/: k2 4

(17.24)

kD1

For x D 1=3,

1 X

kD1

7 hk D .3/ 8 32k k 2 C

1 ln 3 ln3 2 2

1 3 ln 2 C ln 2 Li2 . 1=2/ C Li3 . 1=2/; 3

where we have used Li2 .1=2/ D

2 12

1 2 ln 2I 2

1 2 7 .3/ C ln3 2 ln 2: 8 6 12 Moreover, manipulating (17.22) yields   Z 1 X hk 1 1 1 2 1Ct 2k .1 x / D ln dt: k2 2 x t 1 t Li3 .1=2/ D

kD1

Appealing to

hk D we have

k Z X i D1

1

x

2.i 1/

0

dt D

Z

1 0

k X

x

2.i 1/

i D1

Z 1 1 1 X X h2k hk 1 x 2k D dx k2 k2 0 1 x2

kD1

!

dt D

Z

1 0

1 x 2k dx 1 x2

kD1

D

1 2

Z

1 0



1 1

x2

Z

1 x

1 2 ln t



1Ct 1 t



dt



dx:

Interchanging the order of the integration, we get  Z t  Z  1 X h2k 1 1 1 2 1Ct 1 D ln dx dt: k2 2 0 t 1 t x2 0 1 kD1   Z 1 1 1 3 1Ct D ln dt: 4 0 t 1 t

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17. Evaluations of Some Variant Euler Sums

Using the substitution x D .1 Z

1 0

t/=.1 C t/ and the well-known fact that x k ln3 x dx D

we find

1 X h2k D k2

1 2

kD1

Z

1 0

6 ; .k C 1/3

ln3 x dx 1 x2

1 Z 1 X 1 2k 3 x ln x dx 2 0

D

kD0

D3

1 X

kD0

1 45 D .4/: 4 .2k C 1/ 16

(17.25)

Another path from (17.8) is via complex variables and the formula:   1 1Cz 1 1 1 tan .iz/ D tanh z D ln : i 2 1 z Replacing x by ix in (17.8), we obtain 1 X . 1/k 1 hk 2k x D .tan k

1

x/2 :

(17.26)

kD1

p p This series may be evaluated explicitly at the values x D 2 3; 3=3; 1: p 2k 1 2 k 1 X . 1/ hk .2  3 3/ D D .2/; k 144 72

(17.27)

kD1

1 X . 1/k 1 hk 1 2 D .2/; D k 36 6 3 k

(17.28)

1 X . 1/k 1 hk 2 3 D D .2/: k 16 8

(17.29)

kD1

kD1

Similarly, differentiating and integrating (17.26), we derive the corresponding formulas 1 X

. 1/k

kD1

1

hk x 2k D

x tan 1 C x2

1

x;

Z x 1 X . 1/k 1 hk 2k .tan 1 t/2 x D 2 dt: 2 k t 0

(17.30)

(17.31)

kD1

In particular, we find p 1 X . 1/k 1 hk 3 D ; k 24 3

(17.32)

kD1

1 X . 1/k 1 hk DG k2

kD1

7 .3/; 4

(17.33)

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where G is Catalan’s constant, given by GD

1 X

kD0

. 1/k : .2k C 1/2

Finally, observing that hk D H2k

1 Hk ; 2

(17.34)

we find from (17.24) 1 1 1 X X 1 1 1 X 1 11 H D h C Hk D .3/: 2k k k2 k2 2 k2 4

kD1

kD1

(17.35)

kD1

Furthermore, using partial fraction yields 1 X 1 X i D1 j D1

1

1

XX 1 1 D ij.i C j / i2 i D1 j D1



1 j

1 i Cj



1 X 1 D Hi D 2.3/: i2 i D1

Similarly,

1 X 1 X . 1/i Cj 1 D .3/: ij.i C j / 4 i D1 j D1

The difference gives

1 7 D .3/: ij.i C j / 8

X

i;j >1;i Cj Dodd Setting i C j D 2k C 1 and using partial fractions, we have

X

i;j >1;i Cj Dodd

1 X 2k X 1 1 D ij.i C j / j.2k C 1 j /.2k C 1/ kD1 j D1

D D Thus,

1 X

kD1

Subsequently, we have 1 X

kD1

1 .2k

1/2

H2k D

1 X

kD1

1 X

2k  X 1 1 1 C 2 .2k C 1/ j 2k C 1

1 X

1 2H2k : .2k C 1/2

kD1

kD1

j D1

1 7 H2k D .3/: .2k C 1/2 16

j



(17.36)

1

1

kD1

kD1

X X 1 1 1 H C C 2k .2k C 1/2 .2k 1/3 2k.2k 1/2

21 1 D .3/ C . 2 16 8

8 ln 2/:

(17.37)

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17. Evaluations of Some Variant Euler Sums

From this and the known result 1 X

kD1

1 1/2

.2k

Hk D

1 2 . 4

 2 ln 2

8 ln 2 C 7.3//;

(17.38)

we finally get 1 X

1

kD1

1/2

.2k

7 3 .3/ C .2/ ln 2: 16 4

hk D

(17.39)

There are many avenues for generalizing Euler sums. We close this chapter by posing some other new variant Euler sums that are generalizations of (17.38), (17.39) and (17.35) respectively. Define S.m; n/ D T .m; n/ D

1 X

kD0

1 H .m/ ; .2k C 1/n k

1 X

kD0

1 .m/ h ; .2k C 1/n k

1 X 1 .m/ U.m; n; r / D H : k n kr kD1

It follows that S.m; n/ D

1 X

kD1

D2

n

1

X 1 1 .m/ .m/ H2k C H n .2k/ .2k C 1/n 2kC1 kD0

U.m; n; 2/ C T .m; n/ C 2

n

S.m; n/:

One can compute a special case, a result similar to Euler’s formula (17.2): S.1; 2p C 1/ D .2

2

.2pC1/

/.S.1; 2p C 1/

2.2p C 1/ ln 2 C where .x/ D

1 X

kD0

2p X

21

.2 C 2p// k

kD2

1 D .1 .2k C 1/x

.k/.2p C 2

2

x

k/;

(17.40)

/.x/:

But, we have been unable to determine S.m; n/ in closed form for m  2. Future exploration is left to the interested reader.

Exercises 1. Show that 1 X 1 X .i i D1j D1

1/Š.j 1/Š D .2/ .i C j /Š

and

1 X i X i D0 j D0

  i Hj C1 . 1/ D .2/: j j C1 j

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2. Evaluate

1 X

i;j D1;i ¤j

1 i2

j2

:

3. Show that, for jxj < 1, 1 X

kD0

. 1/k hkC1 x 2k D P

4. For positive integers k, let gk D

02nC1k

arctan x : x.1 C x 2 /

1=.2n C 1/. Show that

1 1 X X . 2/k 1 x k gk x k e D : kŠ k kŠ x

kD1

kD1

Use this to prove: gn D

 n  X . 2/k n k k

1

:

kD1

5. Let S.m; n/ be defined by (17.3). Show that S.m; n/ C S.n; m/ D .m/.n/ C .m C n/: 6. Show that Li2 .x/ satisfies the functional equation Li2 .x/ C Li2 .1

x/ D .2/

ln x ln.1

x/:

7. Let G be Catalan’s constant. Prove that 1 X . 1/n  (a) Hn D G ln 2: 2n C 1 2 nD0

(b)

1 X  . 1/n hn D ln 2 2n C 1 8

1 G: 2

nD0

1 X . 1/nC1 (c) hn D  G n2 nD1

(d)

1 X

nD0

2n

.2n C 1/

7 .3/: 4

  hnC1 D 2G: 2n n

8. Show that 1 S.2; 2/ D 4

Z

2

.

t/2 ln2 .2 sin.t=2//dt:

0

9. Ramanujan’s Constant G.1/ (see [1] Entry 11. p 255). Ramanujan claimed that G.1/ WD

1 1 1 X 1  X . 1/k h D k 8 k3 4 .4k C 1/3 kD1

kD1

1  X 1 p : 3 3 kD1 .2k C 1/3

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Disprove this identity. Show that G.1/ D S.2; 2/ C

15 .4/: 16

10. Prove (17.40). 11. For jxj < 1, define f .x/ D Show that f 12. Prove that, for n > 1, 1 X hk 1 D k 2n 4

kD1

1 X

kD1

hk x 2k 2k 1

 x  1 D ln2 .1 2 x 8

1 .2n C 1/ C 2

2

n X1 i D1

1

:

1 x/ C Li2 .x/: 2



.2i / .2n C 1

1 2i / C 2

!

:

References [1] B. Berndt, Ramanujan’s Notebooks, Part I, New York, Springer-Verlag, 1985. [2] D. Borwein and J. Borwein, On some intriguing sums involving .4/, Proc. Amer. Math. Soc., 123 (1995) 1191–1198. [3] J. Borwein and D. Bailey, Thirty-Two Goldbach Variations, Available at users.cs.dal.ca/~ jborwein/32goldbach.pdf [4] J. Borwein, D. Bailey and R. Girgensohn, Experimentation in Mathematics, A. K. Peters, MA, 2004. [5] H. Chen, Evaluations of some variant Euler sums, J. Integer Seq., 9 (2006), Article 06.2.3. Available at www.cs.uwaterloo.ca/journals/JIS/VOL9/Chen/chen78.pdf [6] V. S. Varadarajan, Euler Through Time: A New Look at Old Themes, AMS, 2006. [7] E. Weisstein, Euler Sum, From MathWorld—A Wolfram Web Resource. Available at mathworld.wolfram.com/EulerSum.html

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18 Interesting Series Involving Binomial Coefficients In his wonderful paper [4], Lehmer defines a series to be interesting if the sum has a closed form in terms of well-known constants. Using the power series 1 X 2x arcsin x .2x/2n p D  ; 2n 1 x2 nD1 n n

(18.1)

he establishes a large class of interesting series, including p 1 1 1 1 2 3 1 C C C C


D C ; D   2 6 20 70 3 27 2n nD1 n p 1 X 1 1 1 1 1 3 C C C


D ;  D C 2 12 60 280 9 2n nD1 n n 1 X

1

(18.2)

(18.3)

and 1 X

nD1

1 1 1 1 1 2 1 C C C


D  :  D C 2 24 180 1120 18 2n 2 n n

(18.4)

However, he points out that the series 1 X

nD1

1 1 1 1 1 C C C


D 4  D C 2 48 540 4480 2n 3 n n

Z

1=2 0

.arcsin y/2 dy y

is a “higher transcendent” and does not admit a simple closed form. Motivated by such results, it is natural for us to ask when a given series involving the central binomial coefficients is interesting. In this chapter, we focus our attention on a class 201

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of series that possess integral representations. First we replace the power series expansion (18.1) by an integral representation, namely 1 X

nD1

xn  D 2n n

Z

1

0

x.1 t/ dt for jxj < 4. .1 xt.1 t//2

(18.5)

This, along with its integration, differentiation and specialization, will enable us to recover and extend many of Lehmer’s related results by a different approach. Moreover, we will see that whether a series is interesting or not is closely linked to the partial fraction decomposition of the integrand. Next, we extend this technique to more general binomial series. Finally, in a spirit of experimental mathematics, we show how to search for rapidly converging series for  .

18.1

An integral representation and its applications

We begin with deriving the proposed integral representation (18.5). Instead of using the series (18.1), we turn to the binomial coefficients directly (see [2] and [3]). Recall that the ˇ-functionis defined by Z

ˇ.m; n/ D

1

tm

1

t/n

.1

1

dt

0

and for all positive integers m and n, .m 1/Š.n 1/Š : .m C n 1/Š

ˇ.m; n/ D We have

  2n n

1

D

nŠ nŠ Dn .2n/Š

Z

1

tn

1

.1

t/n dt:

(18.6)

0

By the ratio test, we see that the series in (18.5) converges for jxj < 4. Next, interchanging the order of the summation and the integration yields 1 X

nD1

xn  D 2n n

Z

1

0

1 X

nx n t n

1

.1

t/n dt:

(18.7)

nD1

Appealing to the well-known series 1 X

nD1

nz n

1

D

1 .1

z/2

for jzj < 1

and t.1 t/  1=4 for 0  t  1, we establish (18.5) from (18.7). We now demonstrate some examples of interesting series via its specialization, integration, and differentiation of (18.5). Starting by setting x D 1 in (18.5), we recover (18.3)

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as 1 X

Z 1 .1 t/dt 1 D C D   2 2 .t t C 1/ 3 2n 0 nD1 n p For x D 1, letting  D .1 C 5/=2 be the Golden-Ratio, N 1

p 2 3 : 27 p 5/=2, we obtain

D .1

Z 1 1 X .1 t/dt . 1/n 1   D 2 t 1/2 2n 0 .t nD1 n Z 1 .1 t/dt D N 2 /.t / 0 Œ.t p 4 5 1 D C ln : 5 25 In general, noting that Z

1 0

x.1 t/ dt D .1 xt.1 t//2 D

x 2

1 2 Z

Z

0 1

0

1

d.1 xt.1 t// x C .1 xt.1 t//2 2 dt ; .1 xt.1 t//2

Z

1

dt xt.1

.1

0

t//2

and taking into account that Z

dt D 2 .at C bt C c/2 .4ac

2at C b 2a C 2 2 b /.at C bt C c/ 4ac b 2

Z

at 2

dt ; C bt C c

we find that, for x > 0, 1 X

nD1

p r  xn x 4 x x arctan C :   D 4 x 4 x .4 x/3=2 2n n

(18.8)

Next, dividing (18.5) by x and then integrating from 0 to x, we have 1 X

nD1

xn   D 2n n n

Z

1 0

x.1 t/ dt D 2 1 xt.1 t/

r

x 4

x

arctan

r

x 4

x



:

(18.9)

In the same manner, from (18.9), we find that 1 X

nD1

xn  D 2n n2 n

Z

1 0

1 ln.1 t

xt.1

 r 2 x t// dt D 2 arctan : (18.10) 4 x

In return, setting x D 1 in (18.9) and (18.10), we arrive at (18.3) and (18.4) respectively.

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If we divide both sides of (18.10) by x and integrate, we obtain 1 X

nD1

n

x   D2 2n n3 n

Z

x 0



arctan

q

t 4 t

t

2

dt:

This fails to give a closed form in terms of simple constants even for x D 1. However, integrating (18.10) and then setting x D 1, we have 1 X

nD1

1

 D 2n n2 .n C 1/ n

Z

1 0

p 2  r x 3 dx D 1 C  2 arctan 4 x 3

1 2  : 18

We consider this to be one of the most interesting series deduced from (18.5). Differentiating (18.5) gives Z 1 1 X nx n 1   D 2n 0 .1 nD1 n

.1 t/ dt C xt.1 t//2

Z

1 0

2xt.1 t/2 dt .1 xt.1 t//3

and setting x D ˙1 respectively, we get 1 X

nD1

1 X

. 1/n

1

nD1

2 2 p n  3;   D C 3 27 2n n n 6 4 p 5 ln : C   D 25 125 2n n

In general, for any positive integer k, applying Leibniz’s rule for the k-th derivative of the product to (18.5), we have dk .1 dx k

xt.1

t//

2

D

.k C 1/Š.t.1 t//k : .1 xt.1 t//kC2

Appealing to Z

t m dt D .at 2 C bt C c/n

tm 1 .2n m 1/a.at 2 C bt C c/n 1 Z .n m/b t m 1 dt .2n m 1/a .at 2 C bt C c/n Z .m 1/c t m 2 dt C ; .2n m 1/a .at 2 C bt C c/n

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we find that for x D ˙1, respectively, 1 X

nD1

1 X

. 1/n

1

nD1

p nk   D pk C qk  3I 2n n p nk   D rk C sk 5 ln  2n n

for explicitly given rational numbers pk ; qk ; rk and sk . Finally, we show how the method of integral representation can be extended to the series 1 X xn   2n nD0 .2n C 2k C 1/ n for all integers k  0. These series are not covered in [4] and appear to be new. As in the derivation of (18.5), using   2n n

1

D .2n C 1/ˇ.n C 1; n C 1/ D .2n C 1/

Z

1

t n .1

t/n dt;

0

we find 1 X

nD0

xn

  D 2n .2n C 1/ n

Z

1 0

1 X

n n

n

x t .1

t/ dt D

nD0

Z

1 0

1

dt xt.1

t/

:

(18.11)

Setting x D 1; 1; 1=2; 2, respectively, we find that 1 X

nD0

1 X

nD0

1 X

nD0

1

  D 2n .2n C 1/ n . 1/n   D 2n .2n C 1/ n . 1/n

  D 2n n .2n C 1/ 2 n

1 X

nD0

2n

  D 2n .2n C 1/ n

Z Z Z Z

1

t2

0

1

dt 4p D 5 ln ; 2 1Ct t 5

1

2dt 4 D ln 2; 2 C t t2 3

0

0

1 0

2 p dt D  3; t C1 9

1

dt 1 D : 2t C 2t 2 2

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Next, we derive the closed form of 1 X

xn

nD0

  2n .2n C 3/ n

by transforming it into two known series. Indeed, since     nC1 2n 2.n C 1/ D ; n nC1 2.2n C 1/ it follows that 1 X

nD0

1 X 2.2n C 1/ D    : 2n 2.n C 1/ nD0 .2n C 3/ .2n C 3/.n C 1/ n nC1

xn

Replacing n C 1 by n gives 1 X

nD0

1 X 2.2n 1/x n 1   D  : 2n 2n nD1 .2n C 1/n .2n C 3/ n n

xn

By the partial fraction decomposition 2n 1 4 D .2n C 1/n 2n C 1

1 ; n

we obtain from (18.9) and (18.11) 1 X

nD0

xn

 D 2n .2n C 3/ n

8 4.8 x/ C x x2

r

x 4

x

arctan

r

x 4

x



:

(18.12)

Setting x D 1; 1; 1=2; 2 respectively yields 1 X

nD0

1 X

nD0

1 X

nD0

1

 D 2n .2n C 3/ n

8C

36 p 5 ln ; 5

. 1/n  D8 2n .2n C 3/ n . 1/n

  D 16 2n n .2n C 3/ 2 n

1 X

nD0

2n

 D 2n .2n C 3/ n

14 p  3; 9

4C

68 ln 2; 3

3 : 2

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Similarly, we have 1 X

nD0

xn

16 .x C 24/ 9x 2

 D 2n .2n C 5/ n

(18.13) C

4 .128 3x 3

16x

In particular, we find that 1 X

nD0 1 X

nD0 1 X

nD0

r r  x x 2 x / arctan : 4 x 4 x

1

  D 2n .2n C 5/ n . 1/n   D 2n .2n C 5/ n . 1/n

  D 2n n .2n C 5/ 2 n

1 X

nD0

2n

  D 2n .2n C 5/ n

400 74 p C  3; 9 9 368 572 p C 5 ln ; 9 15 1504 724 C ln 2; 9 3 104 23 C : 9 6

In general, for any integer k  0, we successively find that 1 X

nD0

1 X

nD0

1 X

nD0

1

p 3;

. 1/n

p 5 ln ;

  D r1k C s1k  2n .2n C 2k C 1/ n   D r2k C s2k 2n .2n C 2k C 1/ n . 1/n

.2n C 2k C

1 X

nD0

1/ 2n 2n

.2n C 2k C 1/

  D r3k C s3k ln 2; 2n n   D r4k C s4k  2n n

for appropriate rational numbers ri k ; si k .1  i  4/. In summary, these derivations indicate that  A given series has a closed form if it yields an integral representation;  The closed form is in terms of well-known constants if the integrand displays a sufficiently simple factorization.

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18.2

Some Extensions

In this section, we extend the integral representation of (18.5) to the more general series involving binomial coefficients 1 X xn (18.14)   kn nD1 ln where k and l are positive integers with k > l . In view of (18.6), we have   kn ln Therefore,

1 X

nD1

Observing that

Z

1

Z

1

D ln

xn   Dl kn ln 1 0

dt 2

t

Z

Z

1

t ln

1

.1

t/.k

l/n

dt:

0

1 0

xt l 1 .1 .1 xt l .1

D ln 2;

dt 2 p D  3 and 2 t Ct 9

Z

1 0

Z

t/k l dt: t/k l /2

(18.15)

dt  D ; 2 1Ct 4

1

4p dt D 5 ln ; 2 t 5 0 1 0 1Ct p ˚ we there exists a closed form evaluation for (18.15) in terms of ; ln 2;  3; p see that 5 ln  , as long as the denominator of the integrand 1 xt l .1 t/k l has factors of the form ft 2; 1 C t 2 ; 1 t C t 2 ; 1 C t t 2 g: With this as a guideline, as with the help of Mathematica, we can search for interesting series. In the following we display two examples of partial fraction decompositions and the corresponding closed forms. We use Mathematica to carry out the partial fraction decompositions and integrations. Example 1.

For k D 3; l D 1, (18.15) becomes Z 1 1 X xn x.1 t/2 D dt:   xt.1 t/2 /2 3n 0 .1 nD1 n

When x D 1=2; 1 .1 we obtain

xt.1

t/2 D .t 2 C 1/.t

2/=2. Since

1 t2 4 8 D C 2 2 2 t.1 t/ =2/ 25.t 2/ 125.t 1 X

nD1

1 2  D 25 3n 2n n

(18.16)

2/

8.3t 4/ 25.1 C t 2 /2

6 11 ln 2 C : 125 250

4.2t C 9/ ; 125.1 C t 2 / (18.17)

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18. Interesting Series Involving Binomial Coefficients d Applying x dx to (18.16) shows that

Z 1 1 X 2x 2t.1 nx n   D .1 xt.1 3n 0 nD1 n

t/4 x.1 t/2 C t/2 /3 .1 xt.1 t/2 /2



dt

from which we get, for x D 1=2, 1 X

nD1

n 81   D 625 3n 2n n

18 79 ln 2 C : 3125 3125

(18.18)

Similarly, we have 1 X

nD1

n2 561 42 673 C ln 2 C :  D 3125 15625 31250 3n n 2 n

In particular, by solving (18.17) and (18.18) for  , we recover Gosper’s identity: 1 X 50n 6   D : 3n nD0 2n n

Example 2.

(18.19)

When k D 4; l D 2, (18.15) becomes 1 X

xn   D2 4n 2n

nD1

Z

1 0

xt.1 t/2 dt: .1 xt 2 .1 t/2 /2

(18.20)

Since t 2 .1 and

t/2

1 D .t 2

t

1/.t 2

t C 1/

t 1 1 t t.1 t 2 / D C ; 2 2 2 2 2 2 .1 t .1 t/ / 4.t t 1/ 4.t t C 1/

substituting x D 1 into (18.20) yields 1 X

nD1

1 1 1 p C  3   D 15 27 4n 2n

2 p 5 ln : 25

Similarly, 1 X

nD1

n

  D2 4n 2n D

Z

1 0



2t 3.1 t/4 t.1 t/2 C .1 t 2 .1 t/2 /3 .1 t 2 .1 t/2 /2

8 1 p C  3 75 54



dt

1 p 5 ln I 125

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nD1

Excursions in Classical Analysis

n2   D2 4n 2n D

18.3

Z

1 0



6t 5 .1 t/6 6t 3 .1 t/4 t.1 t/2 C C .1 t 2 .1 t/2 /4 .1 t 2 .1 t/2 /3 .1 t 2 .1 t/2 /2

p 11 5 C  3 75 324



dt

1 p 5 ln : 250

Searching for new formulas for 

Based on Gosper’s identity (18.19), an algorithm was developed for computing the nth decimal of  without computing the earlier ones (see fabrice.bellard.free.fr/pi). This is a radical idea, since all previous algorithms for the nth digit of  required the computation of all previous digits. Now, we present a proof from the perspective of experimental mathematics: suppose we wish to see if  can be expressed as D

1 X an C b  : 3n nD0 2n n

Rewriting   3n n

1

D .3n C 1/

Z

1

t 2n .1

t/n dt;

0

we feed the general sum to Mathematica Sum[(a n + b)(3 n + 1)x^n, {n, 0, Infinity}] (-b - 4ax - bx - 2ax^2 + 2bx^2)/(-1 + x)^3 % /. x -> t^2(1 - t)/2 (-b - 2a(1 - t)t^2 - 1/2b(1 - t)t^2 1/2a(1 - t)^2t^4 + 1/2b(1 - t)^2t^4)/(-1 + 1/2(1 - t)t^2)^3 Simplify[%] (4(ax^2(4 - 4x + x^2 - 2x^3 + x^4) b(-2 - x^2 + x^3 + x^4 - 2x^5 + x^6)))/(2 - x^2 + x^3)^3 Integrate[%, {t, 0, 1}] (2a(405 + 79\[Pi] - 18Log[2]) + 25 b (270 + 11\[Pi] - 12Log[2]))/6250 We regroup the coefficients of  and ln 2 Collect[%, {Pi, Log[2]}] (810 a + 6750b)/6250 + ((158 a + 275b)\[Pi])/6250 + ((-36a - 300b)Log[2])/6250

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Now we simply search for values of a and b that cause all but the second summand to vanish and the second to equal  . This can be done by Solve[{810a + 6750b == 0, 158a + 275b == 6250, 36a + 300b == 0}, {a, b}] {{a -> 50, b -> -6}} This recaptures (18.19). We then ask if there are other such formulas in the form of 1 X

D

pm .n/  ; kn n b ln

nD0

(18.21)

where pm .n/ is an mth degree polynomial in n and b is an integer. Using the same idea as before, and appealing to   kn ln

1

D .kl C 1/

Z

1

t ln .1

t/.k

l/n

dt;

0

we have RHS of (18.21) D Since

1 X

nD0

Z

1 0

1 X

nD0

.kl C 1/pm .n/

.kl C 1/pm .n/x n D

x l .1

x/k b

l

!n

dx:

q.x/ ; .1 x/kC2

where q.x/ has degree m C 1. Substituting x D t l .1 t/k l =b, we obtain Z 1 P .t/dt RHS of (18.21) D ; l .1 .t t/k l b/kC2 0

(18.22)

where P .t/ is a polynomial of degree k.m C 1/. Now, we want the integral in (18.22) to equal  . A decent way to generate  is to have either arctan x or arctan.1 x/ after integration. This would imply that t l .1

t/k

l

b

must have the factor t 2 C 1 or .t 1/2 C 1; that is, it must have a zero at i or 1 C i , thereby restricting k; l and b. Experimenting with Mathematica, we can find formulas for  in the following cases (there could be many more, see [1]) k

l

b

m

3

1

2

1 or 2

4

2

1

2

8

4

4

4

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For example, we have D

D

1 X

nD1

150n2 C 230n   3n 2n n

1 27 p X 3 4 nD1

36

10n2 C 15n   4n 2n

;

2

:

It is interesting to see that these results discovered experimentally indeed are based on the rigorous mathematics of the Hermite-Ostrogradski Formula in [5]: Given a rational funcQ tion P .x/=Q.x/ with Q.x/ D niD1 qi .x/˛i , where qi .x/ are either linear or irreducible quadratic factors, then Z Z P1 .x/ P2 .x/ P .x/ dx D C dx; Q.x/ Q1 .x/ Q2 .x/ where

Q1 .x/ D

n Y

qi .x/˛i

1

i D1

I Q2 .x/ D

n Y

qi .x/:

i D1

This formula says what is intuitively clear — in a partial fraction decomposition, the repeated factors of the decomposition give us the rational part and the factors without repetition give us the transcendental part. This enables us to perform the experiment on (18.22): if Z 1 P .x/ R.x/ dx D kC1 C r .x/; kC2 Q.x/ Q .x/ 0 ˇ1 we would expect that R.0/ D R.1/ D 0 and r .x/ˇ D  . 0

In the last two decades, computer algebra systems have become an active tool in mathematical research. They are often used to experiment with various examples, to formulate credible conjectures and to decide if a potential result leans in the desired direction. With the continued advance of computing power and accessibility, it is expected that computeraided research will play a more significant role.

Exercises 1. Prove that n X

kD0

n 1 n C 1 X 2k  D ; 2n kC1 n

n X

kD0

k

kD0

2nC1 1 2n C 1 X 2k   D 2nC1 : 2 kC1 2n

2k

kD0

2. Let m be a positive integer. Show that 1 X

nD1 1 X

nD1

1 n2



2mn mn

D

. 1/n  D

n2

2mn mn

m 2

Z

m 2

Z

1

lnŒ1

1

lnŒ1 C t m .1 t/m dt : t.1 t/

0

0

t m .1 t/m dt ; t.1 t/

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18. Interesting Series Involving Binomial Coefficients

3. Let G be Catalan’s constant which is defined by G D a.n/ D

1 X

 2

1 .2k C n/24k

kD0

. 1/n nD0 .2nC1/2 .

P1

2k k

Let

:

Prove that a.n/ satisfies the recurrence a.0/ D 2 ln 2 1/2 a.n/ D .n

.n

4G ; 

2/2 a.n

a.1/ D 2/ C

4G ; 

2 ; 

for n  2:

4. Let G be Catalan’s constant. Prove that (a)

1 X

nD0

(b)

1 X

nD0

(c)

(d)

1 X

1

8G   D 3 2n .2n C 1/2

p 3/

:

n

. 1/n 2   D 6 2n .2n C 1/2

3 ln2 ./:

n

22n 1/2

  D 2G: 2n n

nD0

.2n C

1 X

. 1/n 22n 2   D 8 2n .2n C 1/2

nD0

 ln.2 C 3

p 1 2 ln .1 C 2/: 2

n

5. Prove that (a)

1 X

nD1

(b)

1 X

nD0

n3

24n  2 D 8 G

14.3/:

2n n

24n .2n C

1/3

 2 D 2n n

7 .3/ 2

 G:

6. Ramanujan Problem. Given  1  X 1 2j x j Dp j 4j 1 x j D0 Determine

1  x < 1:

 1  X 1 2j : j .2j C 1/2 4j

j D0

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7. Prove that 1 X

nD1 1 X

n2

n2

nD1

2

1 1   D  10 3n n

1 ln 2; 5

n

1 1 2   D  24 3n n

ln 2:

n

8. Show that ln 2 D

ln 2 D

ln 2 D

1 1 X 6

575n2 C 965n   3n 2n

nD1

1 X

D

1 X

nD1

. 1/n . 728n C 17/   ; 3n 4n n

882n2 C 1295n   4n n 4

nD1

9. Show that

;

n

37 1 C 24 24 3 16

273

296

:

2n

1 X 1 32 52 72

nD0

p4 .n/  ; 8n n . 4/ 4n

where p4 .n/ D

34970134n2 C 110202472n3

89286 C 3875948n

115193600n4:

10. Monthly Problem 11356 [2008, 365]. Prove that, for any positive integer n, n X

kD0

 2 n k

.2k C 1/

  D 2n 2k

24n .nŠ/4 : .2n/Š.2n C 1/Š

11. It is known that Z 1 1 X Hn ln x ln.1 x/ D 2.3/ D 2 dx: n2 1 x 0

nD1

Prove that Z 1 p 1 X Hn . 1/p ln p D . 1/ .p C 1/€.p/ C np €.p/ 0

nD1

1

x ln.1 1 x

x/

dx:

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12. Open Problems: Determine the exact sum of 1 X

nD0

1 .2n C 1/3

 ; 2n n

1 X

nD0

1 X . 1/n . 1/n 2n  ;  : 2n 2n nD0 .2n C 1/3 .2n C 1/3

n

n

References [1] G. Almkvist, C. Krattenthaler and J. Petersson, Some new formulas for , Experimental Math. 12 (2000) 441–456. [2] A. Chen, Accelerated series for  and ln 2, Pi Mu Epsilon J., 12 (2008) 529–534. [3] J. Borwein, D. Bailey and R. Girgensohn, Experimentation in Mathematics, A. K. Peters, MA, 2004. [4] D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92 (1985) 449–457. [5] T. N. Subramaniam and Donald E. G. Malm, How to integrate rational functions, Amer. Math. Monthly, 99 (1992) 762–772.

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19 Parametric Differentiation and Integration A mathematician is a person to whom obvious as 1 C 1 D 2.

R1 0

e

x2

p dx D =2 is as — W. Thomson

In this chapter, we present an integration method that evaluates integrals via differentiation and integration with respect to a parameter. This approach has been a favorite tool of applied mathematicians and theoretical physicists. In his autobiography [3], eminent physicist Richard Feynman mentioned how he frequently used this approach when confronted with difficult integrations associated with mathematics and physics problems. He referred to this approach as “a different box of tools”. However, most modern texts either ignore this subject or provide only few examples. In the following, we illustrate this method with the help of some selected examples, most of them being improper integrals. These examples will show that the parametric differentiation and integration technique requires only the mathematical maturity of calculus, and often provides a straightforward method to evaluate difficult integrals which conventionally require the more sophisticated method of contour integration. To illustrate the basic idea, we begin with two examples.

Example 1 Show that

Z

1 0

 sin x dx D : x 2

(19.1)

This integral is generally evaluated by using contour integration and thus requires the theory of complex functions. Here we consider the parametric integral Z 1 sin x I.p/ D e px dx; .p > 0/: x 0 Differentiating under the integral sign yields Z 1 0 I .p/ D e px sin x dx D 0

1 ; 1 C p2 217

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which has solutions arctan p C C;

I.p/ D

where C is an arbitrary constant. Since limp!1 I.p/ D 0, it follows that C D =2 and I.p/ D

 2

arctan p;

from which (19.1) follows by setting p D 0.

Example 2 Show that the probability integral GD

Z

1

e

x2

0

dx D

p  : 2

(19.2)

One clever way of evaluating this integral resides in the introduction of polar coordinates Z 1Z 1 2 2 G2 D e .x Cy / dxdy: 0

0

Here, we offer a modification that uses the double integrals but applies parametric integration. Notice that if p > 0, the change of variable x D pt yields Z 1 2 2 GDp e p t dt: 0

p2

Multiplying both sides by e and then integrating p from 0 to 1 leads to Z 1 Z 1 Z 1 2 2 2 2 e p t dt: pe p dp G
e p dp D G 2 D 0

0

0

Exchanging the order of the right-hand side integrations gives Z 1Z 1 Z 1 1 dt  2 2 pe .1Ct /p dpdt D G2 D D ; 2 2 1 C t 4 0 0 0 from which (19.2) follows easily. These two examples suggest a natural question: how should one introduce a parameter within the integrand? In many integrals, especially in the formulas of integrals, parameters are already present. For example, differentiating the formula Z 1 1 x p dx D ; .p ¤ 1/; p C 1 0 k times with respect to p gives Z 1 0

x p .ln x/k dx D

. 1/k kŠ .p C 1/kC1

as a new integral formula. However, there also exist integrals containing no parameter, like (19.1) and (19.2). In such cases, a parameter is generally introduced either by substitution, as we did in Example 2, or by specialization, where the given integral is viewed as a special case of a parametric integral. See Example 3 below. The foregoing examples also require some justification of the following three operations:

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1. Differentiation with respect to a parameter under the integral sign; 2. Integration with respect to a parameter under the integral sign; 3. Exchanging the limit and integral operations. Many theorems have been devoted to determining when the order of these processes may be interchanged (see [1] and [5]). For proper integrals, our discussion and illustrations will be based on the following Leibniz’s rule for differentiation and Fubini’s theorem for interchanging orders of integration. Let I be an integral whose integrand f .x; p/ contains a parameter p, namely Z b I.p/ D f .x; p/ dx: a

Leibniz’s Rule. If f and its partial derivative fp are continuous on the rectangle R D f.x; p/ W x 2 Œa; b; p 2 Œp1 ; p2g, then I.p/ is differentiable in .p1 ; p2 / and Z b I 0 .p/ D fp .x; p/ dx: a

Fubini’s Theorem. If f is continuous on the rectangle R D f.x; p/ W x 2 Œa; b; p 2 Œp1 ; p2g, then for any t 2 Œp1 ; p2, Z b Z b lim f .x; p/dx D lim f .x; p/dx p!t

and

Z

t

p1

I.p/dp D

a

Z

a

t

dp

p1

Z

p!t

b

f .x; p/ dx D

a

Z

b

dx a

Z

t

f .x; p/ dp:

p1

Now, we single out two integrals and demonstrate how to apply these two theorems.

Example 3 To evaluate

we introduce a parametric integral

Z

1 0

T .p/ D

arctan x p dx; x 1 x2 Z

1 0

arctan.px/ p dx: x 1 x2

Differentiating with respect to p by Leibniz’s rule gives Z 1 dx 0 T .p/ D p 2 0 .1 C p x 2 / 1

x2

:

The substitution x D sin  yields Z =2 p ˇ=2 d 1  1 T 0 .p/ D p Dp arctan. 1 C p 2 tan /ˇ0 D : 2 2 2 2 1 C p2 1 C p sin  0 1Cp

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Appealing to T .0/ D 0, we have

p  ln.p C 1 C p 2 /: 2

T .p/ D In particular, Z

1 0

p  arctan x p dx D T .1/ D ln.1 C 2/: 2 x 1 x2

Example 4 To compute ID observing that

Z

=2

ln 0



a C b sin x a b sin x






dx ; sin x

.a > b > 0/;

  Z 1 1 a C b sin x ln D 2ab 2 sin x a b sin x 0 a

dy ; b 2 y 2 sin2 x

we get I D 2ab

Z

=2

dx 0

Z

1

dy a2

0

b2y2

sin2 x

:

For a > b > 0; 1=.a2 b 2 y 2 sin2 x/ is continuous on Œ0; =2Œ0; 1. By Fubini’s theorem, interchanging the order of integrations, we obtain Z 1 Z =2 dx I D 2ab dy 2 2 a b y 2 sin2 x 0 0 Z 1 dy D 2ab p b2y2 0 2a a2   b D  arcsin : a The following example warns us that switching the order of integrations without justification may change the answer.

Example 5 Consider f .x; y/ D We have

and so

However,

Z

y2 x2 .x 2 C y 2 /2

1 0

f .x; y/dx D Z Z

1

dy 0

1

0

1

dx 0

Z

Z

0

x2

on Œ0; 1  Œ0; 1:

ˇ1 x 1 ˇ D 0 2 Cy 1 C y2

f .x; y/ dx D

1

f .x; y/ dy D

 : 4

 : 4

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Here the discontinuity of f .x; y/ at .0; 0/ causes the discrepancy. For an improper integral I.p/ D

Z

1

f .x; p/ dx;

a

to avoid the absurdity that appeared in Example 5, uniform convergence and absolute integrability are usual requirements. We say that I.p/ converges uniformly on Œp1 ; p2  if for every  > 0 and p 2 Œp1 ; p2  there is a constant A  a, independent of p, such that x  A implies ˇZ ˇ ˇZ ˇ Z A ˇ 1 ˇ ˇ 1 ˇ ˇ ˇ ˇ f .x; p/ dx f .x; p/ dx ˇ D ˇ f .x; p/ dx ˇˇ < : ˇ ˇ a ˇ a A There is a very convenient test for uniform convergence, due to Weierstrass.

M-Test. If f .x; p/ is integrable on every finite interval and there is an integrable function .x/ on Œ0; 1/ such that jf .x; p/j  .x/;

for all x  a; p 2 Œp1 ; p2;

then I.p/ converges uniformly on Œp1 ; p2 . When the integrand is in product form, the following criterion is sometimes useful. Dini Test.

If

ˇZ ˇ ˇ A ˇ ˇ ˇ f .x; p/ dx ˇ ˇ ˇ a ˇ

is uniformly bounded in A and p, g.x; p/ is monotonic in x and g.x; p/ ! 0 uniformly in p as x ! 1, then Z 1

f .x; p/g.x; p/ dx

a

converges uniformly on Œp1 ; p2.

As an immediate consequence of the Dini test, if Z 1 Z e xp f .x/ dx and a

R1 a

1

f .x/ dx converges, then e

x2 p

f .x/ dx

a

are uniformly convergent for p > 0. When the range of parameter p also becomes infinite, in order to switch the order of integration, besides the uniform convergence of the integrals Z 1 Z 1 f .x; p/ dx and f .x; p/ dp; a

b

Fubini’s theorem further requires that one of the double integrals Z 1Z 1 Z 1Z 1 jf .x; p/j dxdp and jf .x; p/j dpdx b

a

a

b

exists. Combining with the M-test, we have the following theorem to justify the desired steps in our subsequent examples.

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Extended Fubini’s Theorem. Let f .x; y/ be continuous on .a; 1/  .b; 1/. Suppose the improper integrals Z 1 Z 1 f .x; y/ dy and f .x; y/ dx b

a

exist and converge uniformly for x and y restricted to every finite interval, respectively. In addition, suppose that, for s; t > a, ˇZ t ˇ ˇ ˇ ˇ ˇ  M.y/ f .x; y/ dx ˇ ˇ s

and

R1 b

M.y/ dy exists. Then Z 1 Z 1 Z dx f .x; y/ dy D a

b

1

dy

b

Z

1

f .x; y/ dx:

a

As a nice application of this theorem, the integral Z 1 1 D e xy dy x 0 enables us to evaluate (19.1) in one line: Z 1 Z 1 Z 1 Z xy sin x dx e dy D dy 0

0

0

1

xy

e

sin x dx D

0

Z

0

1

dy  D : 2 1Cy 2

Now, we turn to more selected examples, in which most of the justifications of applications of Leibniz’s rule or Fubini’s theorem are straightforward — the verifications are left to the reader.

Example 6 Let ˛; ˇ > 0. Evaluate

Let y D ˛x; p D ˛ˇ. Then Z 1 e

Z ˛2 x 2

1

ˇ2 x2

dx D

If J.p/ D

Z

J 0 .p/ D 2

Z

then

and the substitution z D p=y yields J .p/ D 2

ˇ2 x2

(19.3)

dx:

0

0

0

˛2 x 2

e

Z

1 0

1

1 ˛

e

Z

1

e

y2

p2 y2

dy:

0

y2

p2 y2

dy;

0 1 0

e

p e y2

z2

p2 z2

y2

p2 y2

dy;

dz D 2 J.p/:

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19. Parametric Differentiation and Integration

Appealing to (19.2), we find 2p

J.p/ D J.0/e and so

Z

1

D

Z

˛2 x 2

e

1



y2

e

dy e

0

ˇ2 x2

0

2p

p  e 2˛

dx D

p

 e 2

D

2˛ˇ

2p

:

Remark. In general, by substitution, one can establish that  # Z 1 " Z 1 B 2 1 dx D f .y 2 / dy; f Ax x A 0 0

.A > 0; B > 0/:

A related Putnam problem is 1968-B4, which asks us to prove  Z 1  Z 1 1 f x dx D f .x/ dx: x 1 1

Example 7 Let ˛; ˇ > 0. Show that H.ˇ/ D

Z

1

˛x 2

e

0

1 cos ˇx dx D 2

r

 e ˛

ˇ 2 =4˛

(19.4)

:

Differentiating with respect to ˇ and then integrating by parts gives Z 1 2 0 H .ˇ/ D xe ˛x sin ˇx dx 0

1 D 2˛

Z

1

sin ˇx d.e

˛x 2

/

0

1 D e 2˛

˛x 2

ˇ1 sin ˇx ˇ 0

ˇ 2˛

Z

1

e

˛x 2

cos ˇx dx

0

ˇ H.ˇ/: 2˛

D Moreover, by (19.3), H.0/ D

Z

1

e

˛x 2

0

1 dx D p ˛

Z

1 0

e

x2

1 dx D 2

r

 : ˛

Consequently, the desired integral follows by solving the initial value problem. It is interesting to note that the similar integral Z 1 2 e ˛x sin ˇx dx 0

has no elementary closed form.

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Example 8 Let ˛; ˇ > 0. Evaluate the Laplace integrals Z 1 Z 1 cos ˇx x sin ˇx dx and L2 .ˇ/ D dx: L1 .ˇ/ D 2 2 ˛ Cx ˛2 C x2 0 0 Applying Leibniz’s rule to L1 .ˇ/ gives L01 .ˇ/

Z

D

1

x sin ˇx dx: ˛2 C x2

0

We caution that here L01 .ˇ/ can not be directly differentiated under the integral sign anymore, because the resulting integral is divergent. Indeed, differentiating L0 .ˇ/ under the integral sign yields Z 1 2 x cos ˇx dx; ˛2 C x2 0 which can be rewritten as Z

=2ˇ 0

Since Z .2nC1/=2ˇ .2n 1/=2ˇ

1

X x 2 cos ˇx dx C . 1/nC1 2 2 ˛ Cx nD1

x 2 j cos ˇxj dx ˛2 C x2   min

.2nC1/=2ˇ .2n 1/=2ˇ

x2 ˛2 C x2

.2n 1/=2ˇx.2nC1/=2ˇ

D

Z

2 2..2n 1/=2ˇ/2 ! ; .˛ 2 C ..2n 1/=2ˇ/2 /ˇ ˇ

Z

x 2 j cos ˇxj dx: ˛2 C x2

.2nC1/=2ˇ .2n 1/=2ˇ

j cos ˇxj dx

as n ! 1,

the series, hence the corresponding integral, is divergent. On the other hand, by observing that ˛ 2 sin ˇx sin ˇx x sin ˇx D 2 2 x.˛ C x / x ˛2 C x2 and

Z

1 0

sin ˇx dx D x

we have L01 .ˇ/

 C 2

D

Hence L001 .ˇ/ D

Z

1 0

Z

Z

1 0

1 0

sin x  dx D ; x 2

˛ 2 sin ˇx dx: x.˛ 2 C x 2 /

˛ 2 cos ˇx dx D ˛ 2 L1 .ˇ/; ˛2 C x2

which leads to L1 .ˇ/ D c1 e ˛ˇ C c2e

˛ˇ

;

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19. Parametric Differentiation and Integration

where c1 and c2 are arbitrary constants. Since Z 1 dx  D ; L1 .ˇ/  2 2 ˛ Cx 2˛ 0 L1 .ˇ/ is uniformly bounded. Therefore, c1 D 0; c2 D L1 .0/ D

 2˛

and consequently  e 2˛

L1 .ˇ/ D

˛ˇ

L2 .ˇ/ D L01 .ˇ/ D

and

 e 2

˛ˇ

:

The value of L1 also can be captured by using the parametric integration. Indeed, noticing Z 1 1 2 2 D e t .˛ Cx / dt; ˛2 C x2 0 we have

L1 D

Z

1

cos ˇx dx

0

Interchanging the integral order yields Z 1 Z 2 L1 D e ˛ t dt 0

Z

1

e

t .˛2 Cx 2 /

dt:

0

1

tx 2

e

cos ˇx dx:

0

(19.4) implies the inner integral Z

1

e

tx 2

0

cos ˇx dx D

Finally, appealing to (19.3), we arrive at p Z 1  2 L1 D e ˛ t 2 0 Z 1 p 2 2 D  e ˛ s

r

1 2

ˇ 2 =4t

 e t

 e 2˛

˛ˇ

:

dt p t

ˇ 2 =4s2

ds

.let t D s 2 /

Fs D

Z

0

D

ˇ 2 =4t

:

Example 9 Evaluate the Fresnel integrals Z 1 Fc D cos.x 2 / dx 0

and

1

sin.x 2 / dx:

0

It is standard to substitute x for x 2 , so Z Z 1 1 cos x 1 1 sin x Fc D p dx and Fs D p dx: 2 0 2 0 x x

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Recall that

1 2 p Dp x 

Z

1

xy 2

e

dy:

0

Then Z

1 Fc D p 

1

cos x dx

0

Z

1 Dp 

1

Z

1

e

xy 2

dy

0

dy

0

1 Dp 

Z

1

Z

1

xy 2

e

cos x dx

0

1 dy 1  Dp p D 4 1Cy 2  2 2

0

Similarly, 1 Fs D 2

r

r

 : 2

 : 2

Example 10 Evaluate F D Recall that

Z

Z

1

dx

0

Z

ax

e

e

yx

a

yx

a

dx;

b

b

e

bx

e x

0

Z

This yields F D

1

dy D

dy D

Z

b

dy a

ax

e

.a; b > 0/: bx

e

:

x

Z

1

e

yx

0

dx D

Z

b a

1 b dy D ln : y a

Remarks. Here F belongs to the family of Frullani integrals [2, pp. 406–407], which are defined by Z 1 f .ax/ f .bx/ dx .a; b > 0/: x 0 In general, if f .x/ is continuous on Œ0; 1/ and f .C1/ D lim f .x/ x!1

exists, then

Z

1

f .ax/

f .bx/ x

0

dx D .f .0/

If limx!1 f .x/ has no finite limit, but Z 1 A

exists, then

Z

1 0

f .ax/

f .x/ dx x

f .bx/ x

b f .C1// ln : a

b dx D f .0/ ln : a

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227

Similarly, if f is not continuous at x D 0, but Z exists, then

Z

1 0

f .ax/

A 0

f .x/ dx x

f .bx/ x

b dx D f .1/ ln : a

Finally, we conclude this chapter with two comments. First, sometimes one may need to deal with integrals with more than one parameter. For example, to evaluate Z 1 e px cos qx e ax cos bx dx; .p; a > 0/; x 0 it is instructive to check that differentiating under the integral sign with one parameter or using parametric integration via Z 1 1 D e xt dt x 0 fails to yield an explicit value. However, let I.p; q/ denote the integral to be computed. Differentiating I.p; q/ with respect to p and q respectively gives Z 1 @I p D e px cos qx dx D ; 2 @p p C q2 0 Z 1 @I q D e px sin qx dx D ; 2 C q2 @q p 0 which have solutions

1 ln.p 2 C q 2 / C C; 2 where constant C is independent of p and q. Since I.a; b/ D 0, we obtain that I.p; q/ D

C D

1 ln.a2 C b 2 / 2

and so I.p; q/ D

1 a2 C b 2 ln 2 : 2 p C q2

Second, we present two examples of what can go wrong when differentiating under the integral sign or interchanging the order of integrations is not valid. In 1851 Cauchy obtained the result r   2  2  Z 1 1  p p sin.x 2 / cos.px/ dx D cos sin : 2 2 4 4 0 He then differentiated under the integral sign with respect to p yielding r   2  2  Z 1 p  p p x sin.x 2 / sin.px/ dx D sin C cos : 4 2 4 4 0

(19.5)

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This formula has subsequently been reproduced and still appears in standard tables today. However, Talvila in [4] recently proved that the integral in (19.5) is divergent! Next, direct calculation shows Z 1 Z 1 Z 1 Z 1 y2 x2  y2 x2  dy dx D and dx dy D : 2 C y 2 /2 2 C y 2 /2 .x 4 .x 4 1 1 1 1 Since both Z 1

dy

1

Z

1 1

jy 2 x 2 j dx .x 2 C y 2 /2

Z

and

1

dx

1

Z

1 1

jy 2 x 2 j dy .x 2 C y 2 /2

diverge, Fubini’s theorem does not apply to the given function.

Exercises 1. Show that lim

y!0

Z

1 0

2xy 2 ¤ .x 2 C y 2 /2

2. Show that

Z

1 0

Z

1 0



2xy 2 lim y!0 .x 2 C y 2 /2



dx:

sin px dx x

is uniformly convergent for p  p0 > 0, but fails to converge uniformly for p  0. 3. Evaluate

Hint: Consider

R1

xp 1 0 ln x

Z

1 0

x 1 dx: ln x

dx:

4. Putnam Problem 2005-A5. Evaluate Z 1 0

ln.x C 1/ dx: x2 C 1

There are many ways to solve this problem. Try to introduce a parametric integral and then differentiate it. 5. Let ˛; ˇ; k > 0. Evaluate Z 1 1 cos ˛x kx (a) e dx. x 0 Z 1 sin ˛x sin ˇx kx (b)
e dx. x x 0 6. Let a; b > 0. Prove that Z 1 ln.1 C a2 x 2 /  (a) dx D ln.1 C ab/: 2 2 b Cx b 0 Z 1 arctan ax  (b) dx D ln.1 C a/: 2/ x.1 C x 2 0

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19. Parametric Differentiation and Integration

(c)

Z

1

arctan ax
arctan bx  .a C b/aCb : dx D ln x2 2 aa
b b

0

7. For a; b > 0, show that Z

1

cos ax

cos bx x2

0

8. Evaluate

Z

1

1=2

x

ax px

e

 .b 2

dx D 1

a/:

dx:

0

Remark. For a D p D 1985, this is the Putnam Problem 1985-B5. In contrast to Bernau’s answer for the Putnam problem, here differentiating under the integral sign with respect to p yields an alternative solution. 9. For p > q > 0, evaluate the integral Z 1

e px e qx dx: x.e px C 1/.e qx C 1/

0

10. Evaluate Z 11. Prove that

1

e

x2

2

2

cos.a =x / dx

Z

and

0

Z

1 0

2

xe x dx a p Dp  x 2 C a2

12. Evaluate

Z

1 0

Z

1 0

1

e

x2

sin.a2 =x 2/ dx:

0

2

e x dx x 2 C a2

.a > 0/:

2

e x dx : .x 2 C a2 /2

When a2 D 1=2, this is Monthly Problem 4212 [1946, 397; 1947, 601–603]. 13. Monthly Problem 3766 [1936, 50; 1938, 56–58]. Evaluate Z 1 e x ln2 xdx: 0

14. Let J0 .x/ be the zero-order Bessel function. Evaluate Z 1 e sx J0 .x/ dx .s > 0/: 0

15. For p > 0 and 0 < s < 1, evaluate Z Hint: use

1 0

1 1 D xs €.s/

cos px dx: xs Z

1

ys

1

e

xy

dy:

0

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16. Monthly Problem 11101 [2004, 626; 2006, 270–271]. Evaluate   Z 1 b dx a arctan p p a2 C x 2 a2 C x 2 0 in closed form for a; b; > 0. 17. Monthly Problem 11113 [2004, 822; 2006, 573–574]. Evaluate p Z 1Z 1 2 2 e k x Cy sin.ax/ sin.bx/ Ik .a; b/ D p dxdy 0 0 xy x 2 C y 2 in closed form for a; b; k > 0.

18. Monthly Problem 11225 [2006, 459; 2007, 750]. Find Z 1 n x ln.1 C x=n/ dx: lim n!1 n 0 1Cx

References [1] T. M. Apostol, Mathematical Analysis, 2nd edition, Addison-Wesley, 1974. [2] H. Jeffreys and B. S. Jeffreys, Methods of Mathematical Physics, 3rd ed, Cambridge University Press, Cambridge, England, 1988. [3] R. P. Feynman, “Surely You’re Joking, Mr. Feynman!”: Adventures of a curious character, Bantam Books, New York, 1985. [4] E. Talvila, Some divergent trigonometric integrals, Amer. Math. Monthly, 108 (2001) 432–436. [5] ——, Necessary and sufficient conditions for differentiating under the integral sign, Amer. Math. Monthly, 108 (2001) 544–548.

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20 Four Ways to Evaluate the Poisson Integral In general, it is difficult to decide whether or not a given function can be integrated via elementary methods. In light of this, it is quite surprising that the value of the Poisson integral Z 

I.x/ D

ln.1

0

2x cos  C x 2 / d

can be determined precisely. Even more surprising is that we can do so for every value of the parameter x. In this chapter, using four different methods, we show that  0; if jxj < 1; I.x/ D 2 ln jxj; if jxj > 1. Our integral is one of several known as the Poisson integral. All are related in some way to Poisson’s integral formula, which recovers an analytic function on the disk from its boundary values, a relationship we mention below. However, none of our methods involves complex analysis at all. The first one uses Riemann sums and relies on a trigonometric identity. The second method is based on a functional equation and involves a sequence of integral substitutions. The third method uses parametric differentiation and the half angle substitution. Finally, we finish with an approach based on infinite series. It is interesting to see how wide a range of mathematical topics this chapter exploits. These evaluations are suitable for an advanced calculus class and provide a very nice application of Riemann sums, functional equations, parametric differentiation and infinite series. We begin with three elementary observations: 1. I.0/ D 0. 2. I. x/ D I.x/. 3. I.x/ D 2 ln jxj C I.1=x/;

.x ¤ 0/.

To verify these results, first note that for jxj < 1, .1

jxj/2  1

2x cos  C x 2  .1 C jxj/2 : 231

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Thus, taking the logarithm and integrating with respect to  from 0 to  , we find 2 ln.1

jxj/  I.x/  2 ln.1 C jxj/:

Letting x ! 0, we have that I.0/ D 0. Next, applying the substitution  D  we have Z  I.x/ D

Finally, if x ¤ 0, we have Z I.x/ D D

Z

0

 0  0

˛ in I.x/,

ln.1 C 2x cos ˛ C x 2 / d˛ D I. x/:



 ln x 1

2 1 cos  C 2 x x

2



d

ln x 2 d C I.1=x/ D 2 ln jxj C I.1=x/:

In light of the third fact, the main formula follows easily once we show that I.x/ D 0 for jxj < 1. This will be the goal of the next four sections.

20.1

Using Riemann Sums

Since 1

2x cos  C x 2  .1

jxj/2 ;

for jxj < 1;

the integrand is continuous and integrable. Partition the interval Œ0;   into n equal subintervals by the partition points   k xk D W 1kn : n We have the corresponding Riemann sum for I.x/, namely     n k  X ln 1 2x cos C x2 Rn D n n kD1 "   # n Y1   k 2 2 D ln .1 C x/ 1 2x cos Cx : n n

(20.1)

kD1

Recall that (see (5.2)) n Y1

kD1

 1

2x cos



k n



C x2



Substituting the identity (20.2) into (20.1), we have   x C 1 2n Rn D ln .x n x 1

D

x 2n 1 : x2 1

(20.2)

 1/ :

Since jxj < 1; x 2n ! 0 as n ! 1. Hence, we obtain   x C 1 2n I.x/ D lim Rn D lim ln .x n!1 n!1 n x 1

 1/ D 0:

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Remark. The above method relies on the trigonometric identity (20.2), which is of interest in its own right.

20.2

Using A Functional Equation

The functional equation we have in mind is I.x/ D I. x/ D

1 I.x 2 /: 2

(20.3)

Applying the identity .1

2x cos  C x 2 /.1 C 2x cos  C x 2 / D 1

2x 2 cos.2/ C x 4 ;

we obtain I.x/ C I. x/ D D

Z

Z

 0 

 ln .1 ln.1

0

 2x cos  C x 2 /.1 C 2x cos  C x 2 / d

2x 2 cos 2 C x 4 / d:

Setting ˛ D 2, we have I.x/ C I. x/ D D

1 2

Z

2

2x 2 cos ˛ C x 4 / d˛

ln.1 0

1 1 I.x 2 / C 2 2

Z

2

ln.1 

2x 2 cos ˛ C x 4 / d˛:

Using the substitution ˛ D 2 t in the last integral shows that it is exactly the same as the first integral. Since the two terms on the left are the same (recalling that I.x/ D I. x/), we obtain (20.3) as desired. Applying equation (20.3) repeatedly, we find that I.x/ D

1 1 1 n I.x 2 / D 2 I.x 4 / D


D n I.x 2 /: 2 2 2 n

Again we assume that jxj < 1, so that x 2 ! 0 as n ! 1 and consequently I.x/ D lim

n!1

1 n I.x 2 / D 0: n 2

Remark. Equation (20.3) holds for any x. In particular, we have that I.0/ D 0 and I.˙1/ D 0. The latter equation leads to an added bonus: Z =2 Z =2  ln.sin / d D ln.cos / d D ln 2; 2 0 0 since I.1/ D

Z

I. 1/ D

Z

Z

=2

2 cos / d D 2 ln 2 C 4

=2

ln.2 C 2 cos / d D 2 ln 2 C 4

Z



ln.2 0  0

ln.sin / dI

0

ln.cos / d: 0

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These two integrals are improper. To show convergence, for example, using integration by parts, we have Z

=2 0

ln.sin / d D lim

!0

Z

=2

ln.sin / d 

D lim  ln.sin / !0

D

Z

!0

Z

=2 

 cos  d sin 

=2

 cot  d:

0

Since  cot  is Riemann integrable on Œ0; =2,

20.3

lim

R =2 0

ln.sin / d converges.

Using Parametric Differentiation

Since I.x/ is differentiable for jxj < 1, we apply Leibniz’s rule to I.x/ to find Z  2 cos  C 2x 0 I .x/ D d: 1 2x cos  C x 2 0 Clearly, I 0 .0/ D 0. We now show that I 0 .x/ D 0 for x ¤ 0. First, we prove that Z



1

0

1 x2 d D : 2x cos  C x 2

(20.4)

The integrand in (20.4) is called Poisson’s kernel. It is used to derive solutions to the twodimensional Laplace’s equation on the unit circle, and it also plays an important role in summation of Fourier series. Computing the value of this integral is often used to show the usefulness of the residue theorem – a relatively advanced tool. We give a more straightforward method using the half angle substitution. Setting t D tan.=2/, we have Z

1

Z dt x2/ 2 .1 x/ C .1 C x/2 t 2     1Cx 1Cx D 2 arctan t C C D 2 arctan tan.=2/ C C: 1 x 1 x

1 x2 d D 2.1 2x cos  C x 2

The fundamental theorem of calculus gives Z

 0

1

  1 x2 1Cx d D lim 2 arctan tan.=2/ D : 2x cos  C x 2 ! 1 x

Using (20.4) and x ¤ 0, we get I 0 .x/ D

1 x

Z

 0

 1

1

1 x2 2x cos  C x 2



d D 0:

Thus, we have I 0 .x/ D 0 for jxj < 1 and so I.x/ is a constant. Since I.0/ D 0, we have shown that I.x/  0 for all jxj < 1.

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20.4

Using Infinite Series

We first show that 2x cos  C x 2 / D 2

ln.1

1 X xn cos n; n nD1

(20.5)

where the series converges uniformly for jxj < 1. Once we establish this, integrating (20.5) with respect to  from 0 to  , will show that I.x/ D 0 once again. To prove (20.5) and to keep the evaluation at an elementary level, instead of using the Fourier series, we start with 1

1 x2 D 2x cos  C x 2 1

1 x.e i

x2 D C e i / C x 2 .1

where we have used the relation 2 cos  D e i C e fractions yields 1

1 x2 D 1C 2x cos  C x 2 1

i

x2 xe i /.1 xe 1

i /

;

. Now, decomposing into partial

1 C xe i 1

1 xe

i

:

Thus, the geometric series expansion leads to

1

1 X 1 x2 D1C2 x n cos n: 2x cos  C x 2 nD1

(20.6)

P n The series (20.6) converges uniformly since 1 nD1 jxj converges for jxj < 1. Removing 1 from the right-hand side of (20.6) and then dividing by x, we have 1 X 2 cos  2x D 2 xn 1 2x cos  C x 2 nD1

1

cos n:

(20.7)

Since the series (20.7) converges uniformly, integrating from 0 to x term by term, we have established (20.5) as desired. Remark. As a bonus, we have another proof of (20.4) that follows from integrating the series (20.6) from 0 to  term by term. We have seen a variety of evaluations of the Poisson integral. The interested reader is encouraged to investigate additional approaches.

Exercises 1. Use ln.1

z/ D

to prove (20.5) and arctan



x sin  1 x cos 



1 X zn ; n nD1

D

.jzj < 1/

1 X sin n n x ; n

.jxj < 1/:

nD1

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2. Show that

Z

1

x 0

x

dx D

1 X

n

n

:

nD1

3. Let f be a continuous function. Prove that Z

2 0

f .a cos  C b sin / d D 2

4. Show that 33z


Z

 0

p f . a2 C b 2 cos / d:

€.z/€.z C 1=3/€.z C 2=3/ €.3z/

is periodic of period 1. Hence it is a constant. Determine the constant. 5. Show that I./ D

Z

1

ln.1

0

satisfies the functional equation     I CI  2

2x cos  C x 2 / dx x

 2



D

1 I./; 2

and then use this equation to evaluate the integral. 6. If f is a real-valued continuous function on R and satisfies the functional equation p f .x/
f .y/ D f . x 2 C y 2 /; for all x and y; show that

f .x/ D 0

2

f .x/ D e ˛x ;

or

where ˛ is an arbitrary constant. Use this to prove that p Z 1  x2 e cos.px/ dx D e 2 0 7. Let F .p/ D

Z

1

e

pt 2

0

2

cos x dx and G.p/ D

Show that F and G satisfy F 2 .p/

G 2 .p/ D

p 4.1 C p 2 /

and

Z

p2 =4

1

:

e

pt 2

sin x 2 dx:

0

2F .p/G.p/ D

 ; 4.1 C p 2 /

and solve the simultaneous quadratic equations for F .p/ and G.p/. 8. Putnam Problem 1990-B1. Find all continuously differentiable functions f on the real line such that, for all x, Z x f 2 .x/ D Œf 2 .t/ C .f 0 .t//2  dt C 1990: 0

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9. For all p, show that Z

Remark.

=2

dx  D : p 1 C tan x 4

0

p For p D 2, this is Putnam Problem 1980-A3.

10. Define the parametric integral In .z/ D

Z



cos.zt/ cosn t dt: 0

(a) Establish the reduction formula .n2 (b) Show that lim

n!1

z 2 /In .z/ D n.n

1/In

2 .z/.

In .z/ D 1: In .0/

(c) Deduce the Euler sine product formula. 11. The theta function is defined by .s/ D

1 X

n2 s

e

1

Prove that .1=s/ D

D 1 C 2e

s

C 2e

4s

C 2e

9s

C


;

s > 0:

p s .s/; that is, 1 X

n2 =s

e

1

D

1 p X s e

n2 s

:

1

12. (P. Bracken) Let ˛ and ˇ be positive real numbers with ˛ˇ D  , and let y be a real number. Prove that 1

1 X C e 2 kD1

˛k

cos.˛yk/ D

1 1 X 1 : ˛ 1 C .y C 2ˇj /2 jD 1

13. Monthly Problem 11036 [2003, 743; 2005, 569-572]. For 0  a  p Z 1 ln.1 C x 2 x a2 C x 2 / I.a/ D p dx 1 1 x2

p 3, evaluate

in closed form. 14. Monthly Problem 11072 [2004, 259; 2005, 845-846]. Evaluate the integral Z 1Z 1 p sin ax sin ay I.a; k/ D exp. k x 2 C y 2 / dydx: x y 0 0 15. Using   1 1 1 X 1 1 D C . 1/k C ; sin t t t k t C k kD0

prove that

Z

1 0

sin x  dx D : x 2

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16. Evaluate

Z

1

ln j cos xj dx: x2

0

17. Let Bn .t/ be the Bernoulli polynomials. Show that 2nC1 Z 1=2 nC1 .2 / .2n C 1/ D . 1/ B2nC1 .t/ cot. t/ dt: .2n C 1/Š 0 18. Let In D (a) Define D.s/ D Show that

Z

=2

.ln sin t/n dt: 0

1 X .2k 1/ŠŠ 1
: .2n/ŠŠ .2k C 1/s

kD0

. 1/n nŠ (b) Find I.n/ explicitly for n D 2; 3; 4. (c) Show that ˇ ˇ 1 1 ˇ ˇ   2 1 . 1/n ˇˇ In D  ˇ 3 2 2 ˇ





ˇ ˇ   D.n C 1/ D

n

Z

=2

.ln sin t/n dt: 0

0 2 1


n 2

n 1

0 0 3


n 4


















ˇ ˇ ˇ ˇ ˇ ˇ; ˇ ˇ ˇ ˇ

0 0 0


1

where

1 1 1 1 C k C


D .1 21 k /.k/: k k 1 2 3 4k (d) Generalize these results to Z =2 In .p; q/ D .ln sinp t ln cosq t/n dt: k D

0

19. Monthly Problem 11041 [2003, 843; 2005, 655–657]. Let ˛.k/ D For jxj < 1, define f .x/ D

nD0

1 X

kD1

Show that f .x/ C and

1 X

1 ; .2n C 1/k

˛.2k C 1/x 2k ;

ˇ.k/ D

1 X

nD0

g.x/ D

1 X

. 1/n : .2n C 1/k

ˇ.2k/x 2k

1

:

kD1

     1 1 xC1 x 1 ln 2 D f Cf 2 2 2 2

g.x/ D

   1 xC1 f 2 2

f



x

1 2



:

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20. For positive integer p  2, let A.p/ D 2 p B.p/ D

p 2p

1 X

1

kD1 1 1

1 X

kD0

˛.2k/ .2k/.2k C 1/


.2k C p

1/

;

ˇ.2k C 1/ ; .2k C 1/.2k C 2/


.2k C p/

where ˛.k/ and ˇ.k/ are defined as in Exercise 19 above. Prove that 2p 1 A.p/ D .p 1/Š

Z

=2

x 0

p 1

cot x dx;

1 B.p/ D 2
.p 1/Š

Z

=2 0

xp 1 dx: sin x

In particular, Z 1 ˇ.2k C 1/ 1 =2 x  X D dx D G: B.2/ D 2 .2k C 1/.2k C 2/ 2 0 sin x kD0

References [1] T. M. Apostol, Mathematical Analysis, 2nd edition, Addison-Wesley, 1974. [2] E. W. Weisstein, “Definite integral.” From Mathworld — A Wolfram Web Resource. mathworld.wolfram.com/about/author.html

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21 Some Irresistible Integrals — G. H. Hardy

I could never resist an integral.

Over the years, we have seen a great many interesting integral problems and clever solutions published in the Monthly. Most of them contain mathematical ingenuities. In this chapter, we record eight of these intriguing integrals. They are so striking and so elegant that we can not resist including an account of them in this book. I still remember how exciting I found them on first encounter. In presenting the combination of approaches required to evaluate these integrals, I have tried to follow the most interesting route to the results and endeavored to highlight connections to other problems and to more advanced topics. At the time this chapter was written, the Monthly had not yet published solutions of about half of the problems. I have referenced the solutions published up to that time and look forward to those forthcoming.

21.1

Monthly Problem 10611 [1997, 665; 1999, 75]

Find the largest value of a and the smallest value b for which the inequalities p p Z x 2 2 1 C 1 e ax 1 1 C 1 e bx y 2 =2 < p e dy < 2 2 2 1 hold for all x > 0. The inequalities give tight bounds for the normal distribution. In the following, we show that a D 1=2 and b D 2= are the best possible constants for which the stated inequalities hold. First, by (19.2) we have r Z 0 Z 1  2 2 e y =2 dy D e y =2 dy D 2 1 0 and so the stated inequalities are equivalent to p p 2 1 e ax 1 e < f .x/ < 2 2

bx 2

;

(21.1) 241

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where

Z x 1 2 f .x/ D p e y =2dy: 2 0 Next, if the second inequality of (21.1) holds for all x > 0, then the series expansions give ! p p 1 e bx2 b 1 0< f .x/ D x C O.x 3 / p 2 2 2 as x ! 0, which implies that b  2= . Similarly, if the first inequality of (21.1) holds for all x > 0, appealing to p 1 e ax2 1 1 ax2 1 2ax2 2 D e e C O.e 3ax / 2 2 4 16 Z 1 1 1 2 f .x/ D p e y =2dy 2 2 x Z 1 1 1 1 2 D Cp d.e y =2/ 2 2 x y ! 2 1 1 e x =2 x 2 =2 D p ; e CO 2 x2 2x we have 0 < f .x/

p 1

e 2

ax 2

1 D e 4

ax 2

1

p e 2x

x 2 =2

C O.e

3ax 2

/CO

e

x 2 =2

x2

!

2

as x ! 1. Dividing each side by e x =2 yields a  1=2. Finally, we show that (21.1) holds for all x > 0 when a D 1=2 and b D 2= . To this end, we consider Z xZ x 1 2 2 f 2 .x/ D e .y Cz /=2 dydz: 2 0 0 Let D D Œ0; x  Œ0; x; D1 D f.y; z/ W 0  y; 0  z; y 2 C z 2  x 2 g

D2 D f.y; z/ W 0  y; 0  z; y 2 C z 2  .4= /x 2 g:

See Figure 21.1. We have Z Z 1 e 2 D1

.y 2 Cz 2 /=2

1 dydz  2 

1 2

Z Z

e

.y 2 Cz 2 /=2

dydz

D

Z Z

(21.2) e

.y 2 Cz 2 /=2

dydz;

D2

where the first inequality holds because D1  D; the second because D and D2 have 2 2 2 2 2 the same area and e .y Cz /=2  e .2=/x for .y; z/ 2 D D2 while e .y Cz /=2  2 e .2=/x for .y; z/ 2 D2 D. Evaluating the outer integrals in (21.2) via polar coordinates, we obtain 2 2 1 e x =2 1 e 2x = < f 2 .x/ < ; 4 4 which is equivalent to (21.1).

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z

x

y

O

x

Figure 21.1. Square D and quarter circles D1 and D2

21.2

Monthly Problem 11206 [2006, 180]

Evaluate lim

n!1

n 1 X n n o2 n k kD1

where fxg denotes x

bxc, the fractional part of x.

The problem posed originally in the February, 2006, Monthly was to evaluate n n o X n 2 lim : n!1 k kD1

The crucial factor 1=n before the summation was missing. Since f1=xg2 is bounded and continuous on .0; 1/ except at each reciprocal of a positive integer, f1=xg2 is Riemann integrable. Therefore, Z 1  2 n 1 X n n o2 1 lim D dx: n!1 n k x 0 kD1

Now, we show that

 2 1 dx D ln.2 / 1: x 0 where is Euler’s constant. To see this, for x 2 .1=.n C 1/; 1=n/, we have   1 1 n < 1=x < n C 1; D n; x x Z

1

(21.3)

and so

Z

1 0

  2 1 Z 1=n X 1 1 dx D x x nD1 1=.nC1/ 1  X D 1 nD1

2 n dx

 nC1 n 2n ln C : n nC1

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Let SN be the partial sum, namely SN D

N  X 1

2n ln

 nC1 n C : n nC1



N X

nD1

Since N  X 1C

nD1

N X

nD1

n nC1

n ln



N  X D 2

1 nC1

nD1

nC1 D .ln 2 n

D 2N

ln 1/ C 2.ln 3

D N ln.N C 1/

nD1

1 ; nC1

ln 2/ C


C N.ln.N C 1/

ln N /

ln.N Š/;

we find N X

SN D 2N

nD1

1 nC1

2N ln.N C 1/ C 2 ln.N Š/:

Recall the asymptotic formula of the harmonic numbers Hk D ln k C C O.1= k/ and Stirling’s formula kŠ 

 k p k 2 k : e

We obtain SN D ln.2 /

C1

.2N C 1/ ln

1CN C O.1=N /; N

from which (21.3) follows by letting N ! 1. Using similar paths, one can evaluate more challenge integrals like Z

 2  2 1 1 dx D 4 ln.2 / x 1 x

1 0

4

5;

which appeared in Pi Mu Epsilon Journal as Problem 1151, 2006.

21.3

Monthly Problem 11275 [2007, 165]

Find I1 WD

Z

1

Z

1

.x

y/2 ln..x C y/=.x xy sinh.x C y/

y//

y.u

1/2 ln..u C 1/=.u u sinh.y.u C 1//

1//

yD0 xDy

dxdy:

Substitution of x D uy gives I1 D

Z

1

Z

yD0 1

1

dudy:

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Interchanging the order of the integration yields Z 1  Z 1 .u 1/2 ln..u C 1/=.u 1// y I1 D dy du: u sinh.y.u C 1// 1 0 Substitution of t D e Z 1 0

.1Cu/y

leads to

y dy D sinh.y.u C 1//

2 .1 C u/2

D

2 .1 C u/2

D

Z Z

1

ln t dt 1 t2

1

1 X

0

0

t 2k ln t dt

kD0

1 X 2 1 .1 C u/2 .2k C 1/2 kD0

2

D

 ; 4.1 C u/2

where we have used the facts that Z 1 1 1 X 1 1 3 X 1 2 t 2k ln t dt D and D D : 2 2 2 .2k C 1/ .2k C 1/ 4 k 8 0 kD0

Therefore I1 D

2 4

Z

1 1

.u

kD1

1/2 ln..u C 1/=.u u.1 C u/2

Finally, substituting s D .u 1/=.u C 1/, we get Z Z   2 1 s 2 ln s 2 1 ln s I1 D ds D 2 2 0 1 s 2 0 1 s2

ln s

1//



du:

ds D

 2. 2 8/ : 16

21.4 Monthly Problem 11277 [2007, 259; 2008, 758–759] Evaluate the double integral I2 WD

Z

0

=2 Z =2 0

log.2 sin  cos / sin  dd: 2 2 sin  cos  C sin2  cos2 

Bailey and Borwein [2] recently evaluated its numerical value to exceedingly high precision and found the exact value via the Inverse Symbolic Calculator, an online numerical constant recognition tool available at oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html They then established an analytical evaluation via double series expansions and Wallis’s formula. We now calculate this integral based on a lemma stated below. The lemma enables us to convert a class of double integrals into single integrals, which leads to a straightforward evaluation of I2 . We begin with the lemma.

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Lemma 21.1. If f is integrable on Œ0; 1, then Z =2 Z =2 Z 1  f .sin  cos / sin  dd D f .x/ dx: 2 0 0 0

(21.4)

Proof. Let S be the unit sphere in the first octant. The double integral can then be viewed as a surface integral over S with the angle  measured from the z-axis. Thus, x D sin  cos ; y D sin  sin ; z D cos  and kT  T k D j sin j: This yields Z

0

=2 Z =2 0

f .sin  cos / sin  dd D

Z Z

f .x/ dS: S

On the other hand, viewing the x-axis as the polar axis and  as the angle, we have Z Z Z =2 Z 1 Z 1  f .x/ dS D f .x/dxd D f .x/ dx; 2 0 S 0 0 which proves (21.4). Clearly, replacing  by =2  in (21.4), we also find Z =2Z =2 Z 1  f .cos  cos / cos  dd D f .x/ dx: 2 0 0 0

Now we are ready to use the lemma to evaluate I2 . Setting f .x/ D ln.2 .2 2x C x 2 /, by (21.4), I2 becomes Z 1  ln.2 x/ I2 D dx: 2 0 2 2x C x 2

Substituting x D 1

tan ˛, we obtain

 2

Z

 D 2

Z

I2 D

 D 2

x/=

=4

ln.1 C tan ˛/ d˛

0 =4 0

p 2 cos.=4 ln cos ˛

 ln 2 C 8

Z

˛/

!

d˛ Z

=4

ln cos.=4

˛/ d˛

0

=4

ln cos ˛ d˛ 0

!

2 ln 2; 16 where the last two integrals are equal by symmetry about ˛ D =8. D

21.5

Monthly Problem 11322 [2007, 835]

Let N be a positive integer. Prove that Z 1Z 1 1 X .x.1 x/y.1 y//N dydx D .1 xy/. ln.xy// 0 0

nDN C1

Z

1 n



NŠ t.t C 1/


.t C N /

2

dt: (21.5)

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Let I3 denote the left-hand integral in (21.5). Expanding 1=.1 xy/ in a geometric series and using the monotone convergence theorem yields Z 1Z 1 X 1 .xy/k .x.1 x/y.1 y//N dydx I3 D . ln.xy// 0 0 kD0 1 Z 1Z 1 X .xy/k .x.1 x/y.1 y//N dydx: D . ln.xy// 0 0 kD0

In view of the fact that Z 1

kCN

we have I3 D

1 Z X

kD0

.xy/t dt D 1

0

Z

1 0

Z

.xy/kCN ; . ln.xy//

1

.xy/t .1

kCN

for 0 < xy < 1;

x/N .1

y/N dt



dydx:

Since the integrand is positive, exchanging the order of integration gives  1 Z 1 Z 1 Z 1 X I3 D .xy/t .1 x/N .1 y/N dydx dt: kD0

kCN

0

0

Using Euler’s beta function, we have Z 1 €.t C 1/ €.N C 1/ x t .1 x/N dx D B.t C 1; N C 1/ D : €.t C N C 2/ 0 Repeatedly using €.x C 1/ D x€.x/ yields

1 €.t C 1/ D : €.t C N C 2/ .t C N C 1/.t C N /


.t C 1/

Appealing to €.N C 1/ D N Š, we have 2 1 Z 1  X NŠ I3 D dt: .t C N C 1/.t C N /


.t C 1/ kCN kD0

Replacing t C 1 by t and then setting n D k C N C 1, we arrive at 2 Z 1 1 X NŠ I3 D dt t.t C 1/


.t C N / n nDN C1

as desired.

21.6

Monthly Problem 11329 [2007, 925]

Let f .t/ D 2 t ln €.t/, and let be Euler’s constant. Derive the following integral identities: Z 1 Z 1

C ln ln 2 f .t/ dt D 2 f .t/ dt ; ln 2 0 0 Z 1 Z 1 . C ln ln 2/.1 C 2 ln 2/ 1 t f .t/ dt D 2 .t C 1/f .t/ dt : ln2 2 0 0

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Using that €.t C 1/ D t€.t/, we see that f .t/ satisfies t

2 f .t C 1/ D f .t/ C 2

ln t:

(21.6)

Hence Z

1 0

f .t/ dt D D

Z

Z

1

f .t/ dt C

0 1

Z

1

1 f .t/ dt C 2

0

f .t C 1/ dt

0

Z

1 0

f .t/ dt C

Z

1

2

t



ln t dt ;

0

and so Z

1 0

f .t/ dt D 2

Z

1

0

f .t/ dt C

Z

1

2

t

ln t dt:

0

Recall the classical formula (for example, see [3, Formula 4.331.1]) Z

1

e

˛t

0

C ln ˛ : ˛

ln t dt D

(21.7)

Setting ˛ D ln 2, we obtain the first desired integral identity. Differentiating (21.7) with respect to the parameter ˛ yields Z

1

te

˛t

0

. C ln ˛/ : ˛2

1

ln t dt D

Thus, using (21.6) and the first integral identity, we obtain Z

1 0

t f .t/ dt D 2 D2

Z

0

Z

0

.t C 1/f .t/ dt C

Z

.t C 1/f .t/ dt

. C ln ln 2/.1 C 2 ln 2/ ln2 2

1

1

1 0

.t C 2/2

t

ln t dt 1

:

Remark: This problem is related to the Laplace transform L.˛/ D where

.t/ D

d dt

Z

1

e

˛t

.t C 1/ dt;

0

ln €.t/ is the digamma function. Integrating by parts yields L.˛/ D

ln ˛ C ˛

Z

1

e

˛t

ln €.t/ dt:

Z

1

0

In particular, L.ln 2/ D

ln.ln 2/ C ln 2

f .t/dt:

0

Refer to [4] for details.

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21. Some Irresistible Integrals

21.7

Monthly Problem 11331 [2007, 926]

Show that if k is a positive integer, then Z

1 0



ln.1 C t/ t

kC1

dt D .k C 1/

k X

j D1

(21.8)

aj .j C 1/;

where  denotes the Riemann zeta function and aj is the coefficient of x j in Qk 1 x nD1 .1 nx/.

Here we have made a slight modification: originally, the factor .k C 1/ before the summation in (21.8) was missing. Let I4 denote the desired integral in (21.8). The substitution t D .1 x/=x yields

kC1 k 1 Z 1 ln x1 x I4 D dx: kC1 .1 x/ 0 Using  1  X 1 nCk D xn; k x/kC1 nD0

.1 we deduce

 Z 1   kC1 1  X 1 nCk x nCk I4 D ln k x 0

1

dx:

nD0

Setting s D ln Z

1 0

1 x


, we have

  kC1 1 ln x nCk x

1

dx D

Z

1

s kC1 e

.nCk/s

0

ds D

.k C 1/Š ; .n C k/kC2

and so  1  1 X X .n C 1/.n C 2/


.n C k .k C 1/Š nCk I4 D D .k C 1/ kC2 k .n C k/ .n C k/kC2 nD0 nD0

1/.n C k/

:

Let t

k Y1

j D1

.1

j t/ D

k X

aj t j :

(21.9)

j D1

Replacing t by 1=x and then multiplying by x kC1 on both sides gives x

k Y1

j D1

.x

j/ D

k X

aj x kC1

j

:

j D1

Thus, we have .n C 1/.n C 2/


.n C k

1/.n C k/ D .n C k/

kY 1

j D0

.n C k

j/ D

k X

j D1

aj .n C k/kC1

j

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and I4 D .k C 1/ D .k C 1/ D .k C 1/

D .k C 1/

1 X k X

aj .n C k/j C1

nD0 j D1 k X

1 X

aj

nD0

j D1

1 .n C k/j C1

k X

1 X

aj

k X

aj .j C 1/

nD1

j D1

j D1

Noticing that 1= i .i D 1; 2; : : : ; k

1 nj C1

k X1 i D1

1 i j C1

!

0 1 k 1 k X 1 @X aj A .k C 1/ : i ij i D1

j D1

1/ are solutions of (21.9) implies

k X aj D 0; ij

for i D 1; 2; : : : ; k

j D1

1:

Finally, we find that I4 D .k C 1/ as desired.

21.8

k X

j D1

aj .j C 1/

Monthly Problem 11418 [2009, 276]

Find I5 WD for complex a with jaj > 1.

Z

1

Z

1

1

t 2 sech2 t dt a tanh t

Substitution of x D tanh t gives I5 D

1

Recall that arctanh x D Substitution of s D .1 C x/=.1 I5 D

1 2

x/ gives Z

1 0

arctanh2 x dx: a x   1 1Cx ln : 2 1 x

ln2 s ds : .s C 1/Œ.a 1/s C a C 1

Rewrite I5 in the form of ! Z 1 Z 1 1 ln2 s ds ln2 s ds I5 D C : 2 1/s C a C 1 .s C 1/Œ.a 1/s C a C 1 0 .s C 1/Œ.a 1

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21. Some Irresistible Integrals

Substituting s by 1=s in the second integral yields Z 1 Z 1 1 ln2 s ln2 s C I5 D 2 1/s C a C 1 0 .s C 1/Œ.a 0 .s C 1/Œ.a C 1/s C a

1

!

:

The partial fraction decompositions 1 1 D .s C 1/Œ.a 1/s C a C 1 2 1 .s C 1/Œ.a C 1/s C a

1 D 1 2

 

1 sC1

.a

a 1 1/s C a C 1



1 aC1 C sC1 .a C 1/s C a

1

and 

lead to

 Z  1 1 aC1 a 1 ln2 s ds: I5 D 4 0 .a C 1/s C a 1 .a 1/s C a C 1 To proceed further, we recall the trilogarithm function defined by Li3 .x/ WD Let ˛ D .a C 1/=.a

1 X xk : n3 nD1

1/. Appealing to Z 1 s n lnk s ds D . 1/k 0

kŠ ; .n C 1/kC1

we find that Z

1 0

.a C 1/ ln2 s ds D ˛ .a C 1/s C a 1

Z

ln2 s ds ˛s C 1 0 Z 1 X n D˛ . ˛/ 1

nD0

D 2˛

1

s n ln2 s ds 0

1 X . ˛/n D 2 Li3 . ˛/: .n C 1/3 nD0

Similarly,

Therefore

Z

1 0

.a .a

1/ ln2 s ds D 1/s C a C 1

Z

1 0

ln2 s D sC˛

2 Li3



1 ˛



:

   1 1 I5 D Li3 . ˛/ Li3 : 2 ˛ Finally, by the trilogarithmic identity (see mathworld.wolfram.com/Trilogarithm.html)   1 1 3 Li3 . ˛/ Li3 D .ln ˛ C  2 ln ˛/; ˛ 6 we obtain

     1 aC1 aC1 ln3 C  2 ln : 12 a 1 a 1 Symbolically, it is interesting to see that Mathematica 6.0 displays I5 D

(21.10)

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Excursions in Classical Analysis

Assuming[a > 1, {Integrate[t^2*Sech[t]^2/(a - Tanh[t]), {t, -Infinity, Infinity}]}] {-(1/12) Log[(-1 + a)/(1 + a)] (\[Pi]^2 + Log[(-1 + a)/(1 + a)]^2)} which is equivalent to (21.10).

Exercises 1. Prove that

Z

is independent of ˛.

1 0

dx .1 C x 2 /.1 C x ˛ /

2. Let a be a positive number. Prove that Z Z a ln x dx ln a a dx D : 2 C ax C a 2 C ax C a x 2 x 1 1 3. Putnam Problem 1983-B5. Let fag be the fractional part of a. Determine Z 1 n nno lim dx: n!1 n 1 x 4. Let bac denote the greatest integer  a. Determine  p  n  p  2 n 1 X n p 2 p : lim n!1 n k k kD1 5. Let n D 1; 3; 4. Evaluate

6. Prove that

Z

Z 1

0

1 0

 n 1 dx: x

2 ln x ln.1 x/ dx D 2 .1 C x/ 24

1 2 ln 2: 2

7. Prove that Z 1Z 0

Z 1Z 0

1 0

0

1

.1

x 1 dxdy D ; xy/ ln.xy/

x 1 dxdy D ln.4= /: .1 C xy/ ln.xy/

(J. Sondow)

8. Monthly Problem 11148 [2005, 366; 2007, 80]. Show that Z 1 .x 8 4x 6 C 9x 4 5x 2 C 1/ dx  D : 12 10 C 37x 8 6 C 26x 4 2C1 x 10x 42x 8x 2 0 9. Monthly Problem 11152 [2005, 567; 2006, 945]. Evaluate Z 1 ln.cos. x=2// dx: x.1 C x/ 0

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21. Some Irresistible Integrals

10. Monthly Problem 11159 [2005; 567; 2007, 167]. For jaj < =2, evaluate Z =2 Z =2 cos  dd : cos.a cos  cos / 0 0 11. Monthly Problem 11234 [2006, 568; 2008, 169–170]. Let a1 ; : : : ; an and b1 ; : : : ; bn 1 be real numbers with a1 < b1 < a2 <


< an 1 < bn 1 < an , and let h be an integrable function from R to R. Show that  Z 1  Z 1 .x a1 /


.x an / h dx D h.x/ dx: .x b1 /


.x bn 1 / 1 1 Comment. A more detailed analysis of this type of integrals appeared in Glasser’s “A remarkable property of definite integrals” (Math. Comp., 60 (1983) 561–563). P P1 n n 12. Hadamard Product. Let f .z/ D 1 nD0 an z and g.z/ D nD0 bn z be analytical P1 functions. Let h./ D nD0 an bn  n : Then Z 2 1 h./ D f .ze i /g.e i / d; 2 0 where  D z.

R1 13. Sequence of Definite Integrals. It is known that nŠ D . 1/n 2nC1 0 x lnn x dx. Let Z 1 1 In D dx: 2 C x C 1/n .x 0 p Show that In D an C bn 3 and determine the closed forms of an and bn . 14. Putnam Problem 1985-A5. Let Z 2 Im D cos.x/ cos.2x/


cos.mx/ dx: 0

For which integers m; 1  m  10; is Im ¤ 0? 15. Open Problem. Determine the maximum of Z 2 Ik D cos.n1 x/ cos.n2 x/


cos.nk x/ dx 0

where n1 ; n2 ; : : : ; nk are positive integers. 16. Open Problem. Let n 2 N; a > 1. Find Z 1 I.a/ D 0

lnn x dx .x C 1/.x C a/

in closed form. Along the way you may use the functional equation   .2 i /n 1 ln z Lin . z/ C . 1/n Lin . 1=z/ D Bn C ; nŠ 2 2 i where Bn .x/ is the Bernoulli polynomial.

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Excursions in Classical Analysis

References [1] T. M. Apostol, Mathematical Analysis, 2nd edition, Addison-Wesley, 1974. [2] D. H. Bailey and J. M. Borwein, “Solution to Monthly Problem 11410”. Available at crd.lbl.gov/~ dhbailey/dhbpapers/amm-11275.pdf [3] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, 6th edition, edited by A. Jeffrey and D. Zwillinger. Academic Press, New York, 2000. [4] M. Glasser and D. Manna, On the Laplace transform of the psi-function, Contemporary Mathematics. Available at locutus.cs.dal.ca:8088/archive/00000375/01/ContempMathFinal.pdf [5] E. W. Weisstein, “Definite integral.” From Mathworld — A Wolfram Web Resource. Available at mathworld.wolfram.com/about/author.html [6] Wikipedia, Polylogarithm. Available at en.wikipedia.org/wiki/Polylogarithm

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Solutions to Selected Problems Here the solutions do not include any Monthly and Putnam problems. For Monthly problems, please refer to the journal. The second pair of numbers indicates when and where the solution has been published. For Putnam problems and solutions before 2000, see 1. The William Lowell Putnam Mathematical Competition. Problems and Solutions: 1938–1964, by A. M. Gleason, R. E. Greenwood and L. M. Kelly. 2. The William Lowell Putnam Mathematical Competition. Problems and Solutions: 1965–1984, by G. L. Alexanderson, L. F. Klosinski and L. C. Larson. 3. The William Lowell Putnam Mathematical Competition 1985–2000: Problems, Solutions and Commentary, by K. S. Kedlaya, B. Poonen and R. Vakil. All three are currently in print, and should be available for purchase through the MAA online bookstore (www.maa.org/EBUSPPRO/). For the problems and solutions after 2000, see www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml.

Chapter 1 4. (a) We present three distinct proofs as follows: First Proof: For any positive numbers a and b, the AM-GM inequality yields 4a2 C .a C b/  4a; aCb which implies a2 1 4a2 1 D  .3a aCb 4 aCb 4 Repeatedly using this inequality gives n X

kD1

ak2 ak C akC1



1 .3a1 4

D

1 1 .a1 C a2 C


C an / D : 2 2

a2 / C

1 .3a2 4

b/:

a3 / C


C

1 .3an 4

a1 /

255

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Excursions in Classical Analysis Second Proof: Let anC1 D a1 . The Cauchy-Schwarz inequality yields 0 12 0 12 n n X X p ak @ ak A D @
ak C akC1 A p ak C akC1 kD1

kD1



n X

ak2

a kD1 k n X

D2

C akC1

n X




ak2




ak C akC1

kD1

.ak C akC1 /

kD1 n X

ak ;

kD1

P from which and nkD1 ak D 1 the desired inequality follows. Third Proof: Let anC1 D a1 . The key observation is n X

kD1

ak2 ak C akC1

D

n 2 2 1 X ak C akC1 ; 2 ak C akC1 kD1

which is equivalent to n X

kD1

ak2 ak C akC1

D

This immediately follows from n a2 X k

kD1

2 akC1

kD1

2 akC1

ak C akC1

:

n n X X .ak C akC1 /.ak akC1 / D .ak ak C akC1

D

ak C akC1

n X

kD1

akC1 / D 0:

kD1

Now, the original inequality is equivalent to n 2 X ak2 C akC1

kD1

ak C akC1

 1:

To prove this, notice that 2 ak2 C akC1

a C akC1  k ; ak C akC1 2 which follows directly after clearing out denominators and rearranging in the form .ak akC1 /2  0. Therefore, n 2 X ak2 C akC1

kD1

ak C akC1



n n X X ak C akC1 D ak D 1: 2

kD1

kD1

Remark. One may use a similar argument to prove that, if for any positive integer m, n X

Pn

kD1

ak D 1 and anC1 D a1 , then

akm

kD1

akm

1

C akm

2a kC1

C akm

3 a2 kC1

m 1 C


C akC1



1 : m

4.(e) By the AM-GM inequality, p

1 C a0 C


C ak

1

p

ak C


C an 

1 .1 C a0 C


C ak 2

1

C ak C


C an / D 1:

Hence, for all 1  k  n,

ak bk WD p  ak ; p 1 C a0 C


C ak 1 ak C


C an

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Solutions to Selected Exercises and therefore

n X

kD1

bk 

n X

kD1

ak D 1:

This proves the first inequality. To prove the second inequality, for 0  k  n, define k D arcsin.a0 C a1 C


C ak /: Clearly, 0 D 0 < 1 <


< n D arcsin 1 D

 : 2

Notice that ak D sin k

sin k k C k D 2 cos 2

1 1

sin

k

k 2

1

;

and cos k

1

By the well-known inequality

q 1 sin2 k 1 q D 1 .a0 C a1 C


C ak 1 /2 p p D 1 C a0 C


C ak 1 ak C


C an :

D

 2 .0; =2/;

sin  < ; we have ak < cos and therefore

n X

kD1

k C k 2

bk D

Remark. Observe that

n X

kD1

1

.k

k

ak cos k

1

1/



 cos k

n X

.k

1 .k

k

1/

kD1

k

D

1 /;

 : 2

.a0 C


C ak / .a0 C


C ak 1 / p : 1 .a0 C a1 C


C ak 1 1 .a0 C a1 C


C ak 1 /2 p The proposed sum can be viewed as a Riemann sum for the function 1= 1 x 2 on the interval Œ0; 1 when the partition is bk D p

ak

/2

D

f0 D a0 ; a0 C a1 ; a0 C a1 C a2 ; : : : ; a0 C a1 C


C an D 1g It follows at once that 1
.1

0/ 

n X

kD1

bk 

Z

1

0

dx  p D : 2 2 1 x

P 7. There are two distinct cases. First, if x 2  a1 .a1 1/, since nkD1 a1  1, we have k 0 12 0 12 n n X X 1 1 A @ A @ 2ak jxj ak2 C x 2 kD1 kD1 0 12 n X 1 1 1 @ A  D 4x 2 ak 4x 2 kD1

1 1 
: 2 a1 .a1 1/ C x 2

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Excursions in Classical Analysis Next, if x 2 < a1 .a1 0

1/, applying the Cauchy-Schwarz inequality, 12 0 10 1 n n n X X X 1 1 a k @ A @ A@ A ak ak2 C x 2 .ak2 C x 2 /2 kD1

kD1



kD1

n X

ak : C x 2 /2

.ak2 kD1

Notice that akC1  ak C 1 for k D 1; 2; : : : ; n

1. We have

2ak 2ak  2 2 2 2 2 .ak C x / .ak C x C 1=4/2

Similarly, we have 2an  .an .an2 C x 2 /2

Thus, telescoping yields n 1 n X 1 X ak  2 .ak .ak2 C x 2 /2 1 D 2 .a1

as expected. 8. Let

.ak



.ak

1 1=2/2 C x 2 1 1=2/2 C x 2

1 1=2/2 C x 2

kD1

kD1

D

1 .ak C 1=2/2 C x 2 1 : .akC1 1=2/2 C x 2

1  .an C 1=2/2 C x 2 .an

1 1=2/2 C x 2

.akC1

1 Œ.a1 C ak / C .a2 C ak 2 Appealing to Abel’s summation formula sk D

n X

kD1

1 : 1=2/2 C x 2

 1 1 C 1=2/2 C x 2 2 .an

1 1=2/2 C x 2

1 1 1  2 a1 .a1 1/ C x 2 1=2/2 C x 2

s0 D 0; sk D a1 C a2 C


C ak ;

Then

we have

ak2

ak bk D an

n X

bk

kD1

n n X X ak sk sk D k k

kD1

1/ C

.k D 1; 2; : : : ; n/:




C .ak C a1 / 

n X1

.akC1

ak /

k X

1 ka : 2 kC1

bi ;

i D1

kD1

1

kD1

n X1  1 1 D sn C n k

 1 s k C1 k

kD1

 D

n 1  1 1 1 X 1 sn C s1 C n 2 2 k kD1

 1 kakC1 kC1

n 1 1 1 1 X akC1 sn C s1 C n 2 2 k C1 kD1

n 1 1 X ak D sn C ; n 2 k kD1

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Solutions to Selected Exercises and so n X ak 2 2  sn D .sn 1 C an / k n n kD1   2 n 1 nC1  an C an D an > an : n 2 n

Chapter 2 4 Define fk .x/ D ak x bk ln x; E D .0; 1/. Since fk0 .x/ D ak bk =x; fk00 .x/ D bk =x 2 , fk .x/ has a unique critical point at xkc D bk =ak , where Pn it has minimum value fk .xkc / D bk bP ln.b =a /: Similarly, the function f .x/ D k k k P kD1 fk .x/ has a unique critical point xc D . nkD1 ak /=. nkD1 bk / and its minimum value is f .xc / D

n X

kD1

 bk 1

By using (2.1) we obtain

Pn  bk ln PnkD1 : kD1 ak

0 1 Pn n X bk bk @ bk A ln PkD1 ; ak ln n ak kD1 ak

n X

kD1

kD1

which is equivalent to the required inequality. 6. Define



fk .x/ D pk

ak x

1 ak 1 x

ln

1

x x



; E D .0; 1=2:

Notice that fk0 .x/ D pk .ak x/.2x 1/=.x 2 .1 x/2 /. We find that fk has a minimum at xkc D ak and its value is fk .xkc / D lnŒak =.1 ak /pk . On the other hand, we have f .x/ D

n X

kD1

fk .x/ D



A x

1 1

A x

ln

1

x x



;

which has a minimum at xc D A and its value is f .A/ D ln.A=A0 /. Now, applying (2.1) yields ln

pk n  Y A ak  ln 0 ; 1 ak A

kD1

which is equivalent to the desired inequality. 7. Let f .x/ D 1=.1 C ex /. Since f 00 .x/ D

ex .ex 1/ D .1 C ex /3



> 0; < 0;

x 2 .0; 1/; x 2 . 1; 0/;

we see that f is convex in .0; 1/ and concave in . 1; 0/. Therefore, for xk D ˙ ln.1 ak /=ak .1  k  n/, by Jensen’s inequality, 0 1  P n X  nkD1 pk f .xk /; xk 2 .0; 1/; @ A P f pk xk D  nkD1 pk f .xk /; xk 2 . 1; 0/: kD1

This proves the proposed inequalities.

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Chapter 3 7. We show the first inequality based on the following two lemmas. The first lemma yields a rational bound for ln t . Lemma 1. For t  1 we have ln t 

.t 2

1/.t 2 C 4t C 1/ : 4t .t 2 C t C 1/

.1/

Proof. Let 1/.t 2 C 4t C 1/ ln t: 4t .t 2 C t C 1/ Note that f .1/ D 0, so it suffices to show that f 0 .t / > 0 for t > 1. This follows from the fact that .t 1/4.t C 1/2 f 0 .t / D 2 2 > 0; for t > 1: 4t .t C t C 1/2 f .t / D

.t 2

Lemma 2. For t  1 we have

ln t t 1

tC

ln

! p t C1  1: 3t

.2/

Proof. Note that left-hand side becomes continuous if we define it to have the value 1 at t D 1. Thus, it suffices to show that it has a nonnegative derivative for t > 1, in other words, p ln t 1 1C2 t C p  0: .t 1/2 .t 1/ 2 t .t C 1/ C 2t In fact, it suffices to show this with t replaced by t 2, and this reduces to (1). To prove the first inequality, we may assume b > a and set t D b=a in (2). We find that ! p a C ab C b a.ln a ln b/ C ln b 1 > ln ; a b 3 and the first inequality follows upon exponentiation. Also note that the strict inequality holds unless a D b. Remark. The constants 2=3 and 2=e in the double inequality indeed are the best possible. In fact, since G.a; b/ < I.a; b/ < A.a; b/, one may assume that I.a; b/ > sA.a; b/ C .1

s/G.a; b/;

for 0 < s < 1:

This is equivalent to a.ln a ln b/ C ln b a b

1 > ln Œ2s.a C b/ C .1

p s/ ab:

The analogue of f 0 .t / in Lemma 1 becomes .t

1/2 .1 C t /2. s C s 2 4t C 8st 2s 2 t t 2 .s C 2t 2st C st 2/2

st 2 C s 2 t 2 /

 0:

Letting t ! 1 yields s  2=3. Similarly, one can show that if I.a; b/ < ˇA.a; b/ C .1

ˇ/G.a; b/

then ˇ  2=e.

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Solutions to Selected Exercises 10. First, since S˛ .a; b/ is strictly increasing in ˛, for 0 < a < b; ˛  2, we have   b ˛ a˛ aCb ˛ 1 ˛ b a 2

with equality if and only if ˛ D 2. Next, let A D ax ; B D b x . Then axCy D A.xCy/=x ; b xCy D B .xCy/=x . Applying the inequality above implies   x C y A C B y=x B .xCy/=x A.xCy/=x  : B A x 2 Recall that Mp .a; b/ is strictly increasing in p. Hence  x  a C b x 1=x aCb  2 2 for x  1 with equality if and only if x D 1. Finally, we obtain b xCy bx

xCy axCy  A.a; b/y ax x

for x; y  1 with equality if and only if x D y D 1. Remark. When 0 < x; y < 1, the inequality is reversed, namely b xCy bx

x Cy axCy  A.a; b/y : ax x

But for x; y  0, one can prove that b xCy bx

x Cy axCy  G.a; b/y : ax x

11. We show that L.a; b/  M1=3 .a; b/ by using Simpson’s 3/8 rule:       Z d d c .d c/5 .4/ 2c C d c C 2d f .x/ dx D f .c/ C 3f C 3f C f .d / f ./; 8 3 3 6480 c where  2 .c; d /. Let f .x/ D ex . Replacing c and d with ln a and ln b, respectively yields the proposed inequality. 13. (b) Recall the Gauss quadrature formula with two knots, namely,     Z 1 1 1 1 1 1 1 1 f .t / dt D f C p C f p C f .4/ ./; 2 2 2 2 4320 2 3 2 3 0

0 <  < 1:

Let f .t / D lnŒt a C .1

t /b. Then   1 2 2 1 ln I.a; b/ D ln A .a; b/ C G 2 .a; b/ 2 3 3

.b a/4 ; 720Œa C .1 /b4

which implies 1 exp 360



.b a/ max.a; b/

4 !

1 2A2 =3 C G 2 =3 < < exp 360 I2



.b a/ min.a; b/

4 !

:

This yields the required inequality. Remark. Stimulated by this result, for p  2, one may prove that the double inequality ˛Ap .a; b/ C .1

˛/G p .a; b/ < I p .a; b/ < ˇAp .a; b/ C .1

ˇ/G p .a; b/;

holds for all positive numbers a ¤ b if and only if ˛  .2=e/p and ˇ  2=3.

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Chapter 4 1. Let f .x/ D x ln.1 C x/; g.x/ D x

ln.1 C x/ and let

f1 .x/ f 0 .x/ .1 C x/ ln.1 C x/ C x WD 0 D : g1 .x/ g .x/ x Since f10 .x/ D 2 C ln.1 C x/ g10 .x/ is increasing, by the LMR, f1 .x/=g1.x/ is increasing. An appeal to the LMR again establishes the assertion. 6. Apply the LMR to 

cos x x

1

2 

.

 2

 x :

7. Rewrite the inequality as . 1 x2 2 < cot.x=2/  x

!


0; k

0 < x < :

kD1

Notice that d dt



sin.mt / m sinm t



sin.m

D

1/t

sinmC1 t

:

For m D 2k, integrating this equation from x to =2 yields sin.2kx/ 2k sin

2k

D

x

Z

=2

sin.2k

x

1/t

2kC1

sin

t

dt:

Replacing x by x=2 yields sin kx D2 k

Z

=2 x=2



sin.x=2/ sin t

2k

sin.2k 1/t dt: sin t

Thus   Z =2 X n n X ak sin kx sin.x=2/ 2n sin.2n 1/t D2 an dt: k sin t sin t x=2

kD1

kD1

Pn

2k 1 sin.2k The Pn proposed inequality follows from kD1 ak r a sin.2k 1/t  0. When all a D 1, we have k kD1 k n X

kD1

sin.2k

1/x D

1/t > 0 for 0 < r < 1 since

sin2 .nx/ > 0: sin x

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Solutions to Selected Exercises

Chapter 5 2. Let P D ei0 . Recall that the vertices of a regular n-gon are given by ! k .1  k  n/. Thus, n X

kD1

! k j2 D

jP

D

n X

kD1 n X

.jP j2

kD1 n X

D2

! k /.P

.P

kD1

!/

.P ! C P !/ C j!j2 /

 1

  2k cos 0 C n

D 2n:

7. Repeatedly using the double angle formula sin 2 D 2 sin  cos  , we obtain x x cos 2 2 x x x D 22 sin 2 cos cos 2 2 2 2 x x x x D 23 sin 3 cos cos 2 cos 3 2 2 2 2 :: : x x x x D 2n sin n cos cos 2


cos n : 2 2 2 2

sin x D 2 sin

But from elementary calculus we have that lim t !0 sin t =t D 1, and therefore lim 2n sin

n!1

x D x: 2n

This proves the identity (a). Next, let Pn D

n Y

cos

kD1

kD1

Since sin x D cos we have

n X x x x sin2 and Qn D x cos2 C x Pk : k kC1 2 2 2

x x x x cos C x sin2 2 2 4

sin x D Q1

xP2 cos

x cos2

x x C 2 sin ; 4 2

x x C 2P1 sin : 4 2

By induction, sin x D Qn

xPnC1 cos

x x C 2n Pn sin n : 2nC1 2

Appealing to that jPn j  1 and lim

n!1



xPnC1 cos

x 2nC1

C 2n Pn sin

 x x x 2 n D lim P x cos C 2 sin D 0; n n!1 2n 2nC1 2n

we establish that sin x D lim Qn n!1

which is equivalent to the desired identity (b).

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12. Replacing x by .x C iy/=2 in the infinite product of cosine (5.29) yields 1 Y .x C iy/ cos D 2

1

x2

kD1

y 2 C 2xyi .2k 1/2

!

:

On the other hand, cos

.x C iy/ x y D cos cosh 2 2 2

i sin

x y sinh : 2 2

It follows from (5.32) that 1 X



arctan

kD1

2xy 1/2

.2k

x2 C y 2



 y  x tanh D arctan tan 2 2

.mod 2/:

Chapter 6 4. For n  2, we have n X1

kD1

Hk D

n X1

k X 1 i

.interchange the order of sums/

kD1 i D1

D .n

1 C


C Œn 2 1/ D nHn n:

1/
1 C .n

D nHn

1

.n

2/


.n

2/


1 n

2

C Œn

.n

1/


1 n

1

This identity can also be established by using Abel’s summation formula. Next, we have Hk2

Hk2 1

   2 1 2 1 1 1 1 C C


C D 1 C C


C 2 k 2 k 1   1 1 1 1 D 2 C2
1 C C


C k 2 k 1 k   2 1 1 D 2 C Hk k k k 1 Hk : D2
k k2

Telescoping yields Hn2

H12 D 2



H2 H3 Hn C C


C 2 3 n





1 1 1 C 2 C


C 2 22 3 n



:

Thus, we get 

H2 H3 C C


C 2 3  H2 H3 C C


C >2 2 3  H2 H3 D2 C C


C 2 3

Hn2 D 2

  Hn C 1 n   Hn C 1 n  Hn 1 C n n

 1 1 1 C


22 32 n2  1 1 1 C


1
2 2
3 .n 1/
n

as desired. Note that telescoping is used in the last summation again.

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Solutions to Selected Exercises 8. Appealing to Z

Hn D we have

1 X

1 0

.1 C t C


C t n

an Hn x n D

nD1

D D 9. Rewrite Rn as Rn D

n X

kD1

Thus

Rn D D D

1 X

Z

kDnC1 1 X

Z

kDnC1 1 X

Z

kDnC1

1 0

t 1 0

Define R1 .n/ D

an x n

nD1 1 X

nD1 Z 1 0

kC1

k

Z

1

/ dt D Z



0

kDnC1

1

0

1 X

kDnC1

Z

1

an .tx/n dt 1 t

!

D

n Z X

1

0

kD1

t dt: k.k C t /

 Z 1 1 X 1 dt C t dt k.k C 1/ 0

kDnC1

Z

Rn D R1 .n/ C

t dt: 0

Z

1

t dt: 0

a1 : nC1

t .1 t /.2 t / 1 dt C k.k C 1/.k C 2/.k C t / .n C 1/.n C 2/ R1 0

t .1

Rn D R2 .n/ C

1 k.k C 1/

1

t .1 t / dt; a1 D k.k C 1/.k C t /

0

Let the sum be R2 .n/ and a2 D 1=2

0

1 tn dt; 1 t

1 tn dt 1 t

t .1 t / 1 dt C k.k C 1/.k C t / nC1

Repeating this process, we have Z

1

f .x/ f .tx/ dt: 1 t

1 dx x

1 k.k C t /

1 X

0

an x n

Then

R1 .n/ D

1

Z

t dt k.k C t /

1 0

Z

1 k

1 X

1

Z

1

t .1

t / dt:

0

t /dt . Then a1 a2 C : nC1 .n C 1/.n C 2/

By induction, for N  2, we have

Rn D

where RN .n/ D

N X

kD1

ak C RN .n/; .n C 1/.n C 2/


.n C k/

1 Z X

kC1

0

1

t .1 t /


.N t / dt I k.k C 1/


.k C N /.k C t /

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Excursions in Classical Analysis Z 1 1 t .1 t /


.k k 0 The required answer follows from two estimates: ak D

1

t / dt:

1 2/Š  ak  .k 1/ŠI 6k   1 1 nCk  Rk .n/  : k 6.n C 2/k.k C 1/ 6n.k C 1/ p x n e x dx. The substitution of x D nt C n yields  Z 1  p t n pnt nŠ D nn ne n 1 C p e dt: p n n 1 .k 6k

13. Note that nŠ D

R1 0

Let the integrand be fn .t /. Note that   t n pnt e fn .t / D 1 C p De n

p

p nt Cn ln.1Ct = n/

De

p t 2 =2CO.1= n/

:

Hence, the expected limit follows from the dominated convergence theorem Z 1 Z 1 Z 1 p 2 f .t / dt D lim f .t / dt D e t =2 dt D 2: lim n n p n!1

1 n!1

n

R

1

To justify this, we show that R supn jfn j dx is bounded. Indeed, for t  0, fn .t / is decreasing in n and so supn jfn j D .1 C t /e t , and that is integrable. For t  0, fn .t / is increasing in n, 2 which shows that in this case the sup occurs as n ! 1, and so has the value e t =2 . This is also integrable. 14. Let f .x/ D 1=x. Then f .k/ .x/ D . 1/k kŠx .kC1/ . The Euler-Maclaurim summation formula gives m X 1 B2k .. 1/2k Hn D ln n C .1 C 1=n/ C 2 .2k/Š kD1

D ln n C Thus

1 .2k

1/Š

 m X B2k 1 .1 C 1=n/ C 1 2 2k

1 n2k

kD1

D lim .Hn n!1

and so Hn D ln n C C

1 2n

This implies the desired formula.

ln n/ D

. 1/2k

n2k



1

1/Š/ C Rn .m/

.2k

C Rn .m/:

m X 1 B2k C C R1 .m/; 2 2k kD1

m X B2k 1 C Rn .m/ 2k n2k

kD1

R1 .m/:

Chapter 7 2. The limit is 2. For n  3 we have

n n 1 n X 2k n X 2n k D 2n k 2n n k kD1

kD0

D D

n X1

kD0 n X1

kD0

1 n 2k n k

 using

n n

k

D1C

k n

k



n X1 1 k C : 2k 2k .n k/ kD1

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Solutions to Selected Exercises Since n X1

kD1

n X1 1 k  C 2.n 1/ .k 2 2k .n k/

k k/.n

kD2

D 

1 2.n

C

1/ 1

2.n

C

1/

2 n

1 2

n

1

n X2

k/

1 k

kD1

.ln n C C O.1=n// ! 0; as n ! 1

we obtain L D lim

n!1

n X1

1 D 2: 2k

kD0

4. Consider the sequence fan g defined by 1 1 D jAn j D 2n C an nC1 or an D jAn j 1. 1 C

1

1 1 C nC2 nC3




;

2n. For every n  1, we have

1 p .nC1/2 C1CnC1

< an < 1 C

p

1 ; n2 C1Cn

2. an is strictly decreasing and converges to 1; 3. The best constants are a D a1 D 1=.1

ln 2/

5. The best possible constants are ˛ D 11=6 and ˇ D .4

2 D 1:258891 : : : and b D 1. e/=.e

2/.

Chapter 8 1. By using Gauss’s multiplication formula and €.1/ D 1, we have    ! n nY1 1 X i i 1 ln € D ln € n n n n i D1 i D1  1  D ln .2/.n 1/=2 n 1=2 n   1 n 1 1 D ln.2/ ln n : n 2 2 Letting n ! 1 yields the desired answer. R =2 Remark. If 0 ln sin x dx D  ln 2=2 has been proved, a short proof can be derived by applying the reflection formula. 2. Using €.x C 1/ D x€.x/ and Leibniz’s rule, we have d dx It is clear that .x ln x

Z

xC1

ln €.t /dt D ln €.x C 1/

x

ln €.x/ D ln x:

x/0 D ln x. Thus Z

xC1 x

ln €.t /dt D x ln x

x C C;

where C is an arbitrary constant. Now setting x D 0 yields C D

1 2

ln.2/.

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Excursions in Classical Analysis

3. Notice that =2

p D agm.1; 1= 2/ Substitution of x D sin2 =.2

Z

p D agm.1; 1= 2/ Next, substitution of u D

0

sin2  / yields =2

x4

=2

d q 1

p Z 2

1

0

:

sin2 =2

dx p : 1 x4

gives

p Z 1 2 u1=4 p D 4 0 agm.1; 1= 2/ =2

1

u/1=2

.1

1

du:

Appealing to the beta function, we have p 2 p D B.1=4; 1=2/: 4 agm.1; 1= 2/ =2

Finally, since €.1=2/ D

p

 and

B.1=4; 1=2/ D

p €.1=4/€.1=2/ € 2 .1=4/€.1=2/ € 2 .1=4/ 2 D D p ; €.3=4/ = sin.=4/ 2 

we obtain the proposed answer. 6. By the reflection formula, we have Z 1 x 1 Z 1 x 1 Z 1 x 1  t t t D dt D dt C dt sin x 1 C t 1 C t 1 Ct 0 0 1 2 3 Z 1 Z 1 x 1 n X . 1/nC1 t nC1 5 t Ct x x 1 x 4 k k dt D .t Ct / . 1/ t C dt D 1Ct 1Ct 0 0 kD0

D where

n X

kD n

. x

1/k k

C Rn ;

ˇZ ˇ 1 ˇ jRn j  ˇ .t nCx C t n ˇ 0

xC1

ˇ ˇ 1 ˇ C / dt ˇ  ˇ nCx C1 n

1 : x C2

11. Take the logarithmic derivative of the Weierstrass product formula.

Chapter 9 2. Let the required sum be S. Appealing to Gauss’s pairing trick, we have 2S D D D

n   X n cos kx sin.n k

kD0 n X

kD0 n X

kD0

Therefore S D 2n

  n .cos kx sin.n k

k/x C

 n  X n cos.n n k

k/x sin kx

kD0

k/x C cos.n

k/x sin kx/

  n sin nx D 2n sin nx: k

1 sin nx:

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Solutions to Selected Exercises 3. Using tan ˛ 4. Let an D 4n

tan ˇ D sin.˛  2 1
3


.2n 1/ 2
4


2n

ˇ/=.cos ˛ cos ˇ/: : Then 

1 nC1

an D

anC1

1
3


.2n 1/ 2
4


2n

2

:

Appealing to Wallis’s formula, telescoping yields that limn!1 an D 4=.

7. Inverting the order of summation, we find that 1 X

1/ D

..k/

kD2

D D 9. (b). Let f .x/ D ax.x

1 X

kD2 1 X nD2 1 X

nD2

1 X 1 k n nD2

!

D

1=n2 1

n.n

1/

1 1 X 1 @ A nk

nD2

kD2

.the inner sum is a geometric series/

1=n 1

0

1 X

D

1  X

nD2

1 n

1 n

1



D 1:

1/. The identity follows from (9.22) directly.

(c). Notice that 1 X

. 1/kC1 arctan

kD1



2 k2



D

1 X

arctan

kD0



2 .2k C 1/2



1 X

arctan

kD1



2 .2k/2



:

We have 1 X

 1 X 2 D .arctan.2k C 2/ .2k C 1/2 kD0 kD0   1 1 X X 2 arctan D .arctan.2k C 1/ .2k/2 arctan



kD1

arctan.2k// D

 ; 2

1// D

arctan.2k

kD1

 ; 4

from which the required identity follows. 15. If the leading coefficient of Q.x/ is 1, by using the identity arctan x D

1 1 C ix ln ; 2i 1 ix

we have 1 X

arctan

1 1 Y 1 X Q.k/ C iP.k/ 1 Q.k/ C iP.k/ P.k/ D ln D ln : Q.k/ 2i Q.k/ iP.k/ 2i Q.k/ iP.k/ kD1

kD1

kD1

Let aj .1  j  n/ be the roots of Q.x/ iP.x/. Since P.x/ and Q.x/ have real coefficients, the roots of Q.x/CiP.x/ are complex conjugates of the aj . Appealing to Exercise 3 of Chapter 8, we obtain 1 X

kD1

arctan

1 Y P.k/ 1 .n D ln Q.k/ 2i .n

D

1 ln 2i

D

n X

kD1 1 Y kD1

€.1 €.1

arg.€.1

aN1 /.n a1 /.n

aN2 /


.n a2 /


.n

a1 /€.1 aN1 /€.1 aj //

aNn / an /

a2 /


€.1 aN2 /


€.1

an / aNn /

.mod 2/:

j D1

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Excursions in Classical Analysis For example, let P.x/ D 2 and Q.x/ D x 2 . Thus aj are the roots of x 2 and 1 i . Therefore 1 X

arctan

kD1

2 D arg.€. i // C arg.€.2 C i // k2

2i , which are 1 C i

.using €.1 C x/ D x€.x//

D arg.€. i // C arg.1 C i / C arg.i / C arg.€.i // 3 : D arg.1 C i / C arg.i / D 4

The sum of the first few terms shows that no modification by a multiple of 2 is needed.

Chapter 10 3. Integrating the binomial theorem n   X n x k D .1 C x/n k

kD0

and then manipulating it yields n   X xk .x C 1/nC1 1 n D : k k C1 .n C 1/x

kD0

By the multisection formula we have bn=2c X kD0



n 2k



.1 C x/nC1 .1 x/nC1 x 2k D : 2k C 1 2.n C 1/x

4. By the multisection formula we have n   X 1h 2n 2j x D .1 C x/2n C .1 2j 2

j D0

i x/2n :

p Taking x D ai yields

n   X p 1h 2n . a/j D .1 C ai /2n C .1 2j 2

p

j D0

p Let  D arctan. a/. Rewrite 1˙

i ai /2n / :

p p ai D 1 C a e˙i :

We have n   X 1 2n . a/j D .1 C a/n .ei 2n C e 2j 2

j D0

i 2n

/ D .1 C a/n cos.2n /:

5. By the multisection formula with k D 2 and m D 0, 1 X

nD1

F2n x 2n D

1 2



1

x x x2

x 1 C x x2



D

1

x2 : 3x 2 C x 4

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Solutions to Selected Exercises Replacing x 2 by x yields 1 X

nD1

F2n x n D

x ; 3x C x 2

1

which is the generating function of fF2n g. Similarly, with k D 2 and m D 1, the multisection formula gives   1 X x x x.1 x 2 / 1 C D : F2nC1 x 2nC1 D 2 2 2 1 x x 1Cx x 1 3x 2 C x 4 nD1

Dividing by x and then replacing x 2 by x yields the generating function of fF2nC1 g as 1 X

F2nC1 x n D

nD1

7. For jxj < 1, start with

1

1 X xn D n

1 x : 3x C x 2

ln.1

x/:

nD1

Multiplying both sides by x and then integrating gives 1 X

x nC2 1 D .2x C x 2 C 2 ln.1 n.n C 2/ 4

nD1

or

1 X

nD1

2x C x 2 C 2 ln.1 x/ xn D n.n C 2/ 4x 2

Now, using the multisection formula yields  1 X x 4nC1 1 1 D .4n C 1/.4n C 3/ 8 x2

2x 2 ln.1

x/

x//

2x 2 ln.1

x/

:

    1 x 1 1 ln C 2 1 C 2 arctan x : 1Cx x

nD1

Letting x D 1, we get a byproduct 1 X

nD1

1  D : .4n C 1/.4n C 3/ 8

Chapter 11 2. Let G.x/ D

P1

nD0

un x n . Then G.x/ D a C bx C

1 X

un x n

nD2 1 X

D a C bx C p

un

nD2 1 X

D a C bx C px

1x

n

un x n

q

qx

nD1

D a C bx C px.G.x/

1 X

un

nD2 1 X 2

2x

n

un x n

nD0

a/

qx 2 G.x/:

Solving for G.x/ yields G.x/ D

a C .b ap/x : 1 px C qx 2

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7. We begin to derive a closed form for bn WD nan . By the assumptions, we have n X1

b1 D 1; b2 D 2; and bn D n Let G.x/ D

P1

nD1

.n

k

for n > 2:

1/bk

kD1

bn x n . Then G.x/ D D D

0

1 X

n X1

@n

nD1 1 X

.n

1/bk A x n

k

kD1 1 X

nx n

nD1

nD1

0 @

n X1

.n

k

kD1

x2 G.x/; .1 x/2

x x/2

.1

1

1

1/bk A x n

where the Cauchy product has been used in the second sum. Therefore, x x 2 C .1

G.x/ D

x/2

:

To determine the coefficient of x n in G.x/, we factor x 2 C .1 where ˛ D .1 C i /=2; ˇ D .1 expansion, we have  1 ˛ G.x/ D 2i x ˛ Thus bn D

Œ.1 C i /n

.1

an D

x/2 D 2.x

˛/.x

ˇ/;

i /=2. Appealing to partial fractions and geometric series ˇ x

ˇ



D

1 1 X Œ.1 C i /n 2i

i /n x n :

nD1

i /n =2i . Now, by using

bn 2n=2 en i=4 e D n n 2i

.1

1˙i D

n i=4

D

p

2e˙ i=4 , we obtain

2n=2 sin.n=4/ : n

Remark. It is interesting to see that 1 1 X X an sin.n=4/  D D : n=2 n 2n 4 2 nD1 nD1

8. Let S1 D Appealing to n X

kD1

we have

n X

kD1

Fk D FnC2

1 and

FnC2

n X

kD1

n X Fn FnC1

kD1

n X

Fk and S2 D

Fk2

Fk

Using the power mean inequality

2 a12 C a22 C


C am 

1

D

Fk2 :

kD1

Fk2 D Fn FnC1 ;

n X S2

kD1

S1

Fk2 Fk

:

.a1 C a2 C


C am /2 m

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Solutions to Selected Exercises yields Fk2 

S2 Therefore

n X S2

kD1

Fk2

S1

Fk



.S1 Fk /2 : n 1

n X S1 n

Fk D S1 D FnC2 1

kD1

1:

Chapter 12 4. Recall that Fibonacci polynomials are defined by fn .x/ D xfn

1 .x/

C fn

.n  2/

2 .x/

with f1 .x/ D 1 and f2 .x/ D x. Similar to Binet’s formula, we have ˛n .x/ ˛.x/

fn .x/ D where ˛.x/; ˇ.x/ D .x ˙ Therefore,

p

ˇ n .x/ ˇ.x/

x 2 C 4=2. Let x D 2i cos z. Then ˛.x/ D i e

fn .x/ D i n

1e

i nz

i nz

e e

eiz

iz

D in

1

iz ; ˇ

D i eiz .

sin nz : sin z

./

Appealing to (5.15), we obtain fn .x/ D .2i /n

nY1

1

kD1

 cos z

cos

k n



:

Since cos z D x=.2i /, we get fn .x/ D

nY1

kD1

and so Fn D fn .1/ D

 x

nY1

kD1

2i cos

 1

k n

2i cos



;

k n



:

Regrouping the factors above yields the proposed factorization without the complex number i . The third equality follows from  by setting x D 1.

9. Let

˛; ˇ D

a˙

p

a2 C 4b : 2

First, by using induction on k, we prove that

D

X

i1 ;i2 ;:::;ik

.uk x C buk 1 /i .ukC1 x C buk /n      n i n i1 n ik 1 k n i 2i1





a i1 i2 ik

Notice that

x i .ax C b/n

i

D

X

0j n i

 n

i j



an

i 2ik

1

ik i1 C


Cik n ik

i j j n j

b x

b

x

:

;

which implies the proposed result is true for k D 1. Now suppose the equation is true for some positive integer k. Substituting a C bx 1 for x and multiplying by x n , we see the left side of the equation becomes .auk C buk

1x

C buk /i .aukC1 C buk x C bukC1 /n

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;

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Excursions in Classical Analysis which is equivalent to .ukC1 x C buk /i .ukC2 x C bukC1 /n

i

:

Expanding the right side of the equation and simplifying, we obtain      X n i n i1 n ik .kC1/n i 2i1


2ik


a i1 i2 ikC1

ikC1 i1 C


CikC1 n ikC1

b

x

i1 ;i2 ;:::;ikC1

as desired. Next, multiplying both sides of the equation by x i and summing over i , we have n X

i D0

D

.uk x C buk X

i;i1 ;i2 ;:::;ik



1/

n i1

i

.ukC1 x C buk /n

i

 n

i1 i2










i i

x

n

ik ik

1



ak n

i 2i1


2ik

1

ik i1 C


Cik nCi ik

b

x

:

The coefficient of x n on the right side of the equation is tr.AknC1 /, while the coefficient of x n on the left side of the equation is X i  n i  usk .buk 1/i s utkC1 .buk /n i t s t i CsCt Dn

X i  n i  D .buk s s

1/

i s s n i s uk ukC1 .buk /s :

i Csn

Let an be the right side above. Then 1 X

nD0

an x n D D D D D

1   X i b i uik s

i;sD0 1 X

 1 X n

i CsDn

  i b i uik s

s 2s i Cs .1 1 uk x

2s b s u2s k x .1

ukC1 x/

i;sD0 1 X

s 2s i Cs 1 uk x

i s

ukC1 x/

s 1



.ukC1 x/n

s 1

X i  .buk s

sD0 1 X

2s b s u2s k x .1

ukC1 x/

1

.ukC1 C buk

1 /x C .bukC1 uk 1

i s

1 x/

i s

i s s 1

.1

buk

1 x/

s 1

sD0

1

bu2k /x 2

:

Appealing to ukC1 C buk

1

D ˛k C ˇ k ; bukC1 uk

bu2k D ˛k ˇ k ;

1

we find that

1

.ukC1 C buk

1 1 /x C .bukC1 uk

1

bu2k /x 2

D

1 ˛k

ˇk

˛k 1 ˛k x

ˇk 1 ˇk x

!

:

Thus, tr.AkkC1 / D an D uk nCk =uk :

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Finally, let the eigenvalues of AnC1 be 0 ; 1 ; : : : ; n and let the characteristic polynomial be P.x/ D det.xI

AnC1 /:

Then n 1 X X P 0 .x/ 1 D D x P.x/ x j j D0

D D

1 X

kD0 1 X

kD0

k 1

x

k 1

k 1

tr.AknC1 / D n X

1 X

j /k

j D0

P.x/ D

n Y

k 1

x

kD0

˛j k ˇ .n

Therefore

jk

j D0

kD0

x

n X

D

j D0

˛j ˇ n

.x

n X

j

˛nkCk ˛k

ˇ nkCk ˇk

1 ˛j ˇ n

x

j

:

/;

j D0

which implies that the eigenvalues of AnC1 are ˛n ; ˛n

1

ˇ; : : : ; ˛ˇ n

1

; ˇn:

Remark. This solution contains kernels of sophisticated ideas used to study the sequences defined by the second-order recursive relations.

Chapter 13 7. Let D.x1 ; x2 ; : : : ; xn / D

det

j 1

1i;j n

.xi

/:

If xi1 D xi2 with i1 ¤ i2, then D D 0. Thus, ˇ xi2 / ˇ D:

.xi1 Repeating this process yields

Y

.xj

1i 0 kD0

x 2 /k n

j

:

j >0

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Solutions to Selected Exercises Appealing to the binomial theorem, the last summation can be expressed as   p  X X 2k n kn j . 1/p x 2p : 2j 1 p j p>0

j D0

The desired answer follows from application of the Vandermonde convolution formula     n  X 1 C 2x r x k r x Cn D 22nCr : r C 2k n k 2n C r kD0

12. For n D 2, clearly the eigenvalue of A is 1. If n  3, let the characteristic polynomial of A be P./ D det.I

A/ D n

1

C

n X1

. 1/k ck n

k 1

;

kD1

where ck is the sum of all the principal minors of A of order k. Now, for k  3 and n  4, we show that ck D 0: To see this, define B D .bij / where bij D sin xi sin yj cos.xi

.i; j D 1; 2; 3/:

yj /;

It is easy to verify that det.B/ D 0 for all xi and yj . This implies that any 3-rowed minor of A vanishes, which clearly proves that ck D 0. Thus, n o P./ D n 3 2 trA C Œ.trA/2 trA2 =2 : By using the trigonometric summation formulas, we have

n 10n2 ; trA2 D ; 2 64 and so the eigenvalues of A are 0 (n 3 multiple), n=8 and 3n=8. Remark. A related problem is Putnam Problem 1999-B5: Let  D 2=n and aij D cos.i C j /. Determine the eigenvalues of A D .aij /. trA D

Chapter 15 2. Appealing to Wallis’s formula for an odd power, we have Z =2 .nŠ/2 22n cos2nC1 x dx D ; .2n C 1/Š 0 and so

n Z =2 1  1 X X .nŠ/2 2nC1 1 D2 cos x cos2 x dx .2n C 1/Š 2 0 nD0 nD0 Z =2 cos x D dx D : 1 cos2 x=2 0

One can also prove this identity via the gamma and beta functions:

1 1 X X .nŠ/2 2nC1 D 2nC1 €.n C 1/€.n C 1/= €.2n C 2/ .2n C 1/Š nD0 nD0 Z 1 1 X X D 2nC1 B.n C 1; n C 1/ D 2nC1 nD0

D2

Z

0

.2x/n .1

nD0

D 2 arctan 2.x

x/n dx D 2

Z

x n .1

x/n dx

0

nD0

1 1 X

1

1

Œ1

2x.1

x/

1

dx

0

ˇ1 1=2/ˇ0 D :

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Excursions in Classical Analysis Remark. The reader may recognize that .nŠ/2 2nC1 D .2n C 1/Š

Z

1 1

x n Pn .x/ dx;

where Pn .x/ is the Legendre polynomial of degree n. Appealing to its generating function 1 X

nD0

Pn .x/t n D .1

2xt C t 2 /

1=2

;

one obtains Z 1 X .nŠ/2 2nC1 D .2n C 1/Š

nD0

1

1 X

1 nD0

x n Pn .x/ dx D

Z

1

.1

x2/

1=2

1

dx D ;

5. The result is obtained by setting a D 1 in the following generalization. Let y D ea arcsin x D p Pdesired 1 n : Then y 0 D ay= 1 x 2 , or a x n nD0 .1

x 2 /.y 0 /2 D a2 y 2 :

Differentiating both sides with respect to x and then dividing both sides by 2y 0 yields .1

x 2 /y 00

xy 0

a2 y D 0:

Substituting the power series into this differential equation and equating coefficients of like powers of x, we have anC2 D

n2 C a 2 an ; .n C 1/.n C 2/

for n  0:

Appealing to a0 D 1 and a1 D a, for n  1, by a simple inductive argument, we obtain a2n D

a2 .a2 C 22 /.a2 C 42 /


.a2 C .2n .2n/Š

and a2nC1 D

a2 .a2 C 12 /.a2 C 32/


.a2 C .2n .2n C 1/Š

2/2 /

;

1/2 /

:

Setting a D 1 yields the proposed power series. Remark. Consider ea arcsin x D 1 C .arcsin x/a C

arcsin2 x 2 arcsin3 x 3 a C a C


: 2Š 3Š

Equating coefficients of ak on both sides gives the power series for arcsink x; k  1. For example, for k D 4, we obtain 1 X 1 b2n 2 arcsin4 x D x n; 4Š .2n/Š nD2

where b2n D

n X1

kD1

22 42


.2n .2k/2

2/2

D 22.n

1/

Œ.n

1/Š2

n X1

kD1

1 : .2k/2

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Solutions to Selected Exercises 6. Substituting x D sin.t =2/ yields 1 2

SD

Z

=3

t 2 cot.t =2/ dt:

0

Integrating by parts gives Z =3 SD t 2 d.ln.2 sin.t =2// dt D

2

0

Recall that

=3

t ln.2 sin.t =2// dt: 0

1 X zn ; n

z/ D

ln.1

Z

jzj < 1:

nD1

Setting z D ei t and exporting the real part gives

1 X cos nt ; n

ln.2 sin.t =2// D and so

Z 1 X 1 =2 t cos nt dt: n 0

S D2

nD1

Integrating by parts twice leads to SD

0 < t < 2;

nD1

1 1 X 2 X sin.n=3/ cos.n=3/ C 2 3 n2 n3 nD1

2 .3/:

nD1

Appealing to 1 X 1 cos.n=3/ D .3/; 3 n3

nD1

we obtain

.3/ D 7. Rewrite the identity as ex

2

Z

x

e

1  X sin.n=3/ 2 n2

3 S: 4

nD1

t2

0

dt D

1 X

nD1

nŠ .2x/2n .2n/Š

1

:

The function on the left is the unique solution of the initial value problem y0

2xy D 1

P1

y.0/ D 0:

Set y.x/ D nD0 an x n . Substituting the series into the differential equation and equating coefficients of like powers of x yields anC1 D

2 an nC1

1

for all n  1:

Appealing to a0 D 0 and a1 D 1, we have a2n D 0 and a2n

1

D

nŠ 2n 2 .2n/Š

1

:

Remark. You may also show that each side of the proposed identity is a solution of the initial value problem 2y 00 xy 0 2y D 0; y.0/ D 1; y 0 .0/ D 0:

Since the coefficients of the differential equation are continuous on R, the proposed identity follows from the existence-uniqueness theorem of ode.

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Excursions in Classical Analysis

10. Dividing (15.15) by  gives 1 X .2n/ 2n  2n

1

D

nD1

1 2



 cot  :

1 

Integrating this identity from 0 to =2 yields 1 X

nD1

.2n/ 1 D 2 .2n/22n

Z



=2

0

1 

cot 



 ˇ 1  1 ˇ=2 ln D ln.=2/: ˇ 2 sin  0 2

d D

Next, integrating (15.15) from 0 to =2 yields 1 X

.2n/ 1 D .2n C 1/22nC1 2

nD1

Z

 2

!

=2

 cot  d : 0

Further, integrating by parts, we have Z

=2 0

and so

1 X

nD1

=2

x

pC1

0

=2

ln sin  d D

0

.2n/ 1 D .1 .2n C 1/22n 2

.2n/ nD1 .2nCp/22n

.p C 1/ p csc x dx D 2p 2

1 p

1  ln 2; 2

ln 2/:

P1

Remark. In general, one can sum the series Z

Z

ˇ=2  cot  d D  ln sin  ˇ0

for any odd p via

1 X

2

nD1

.2n/ .2n C p/22n

!

:

A more challenging problem is to evaluate the series involving .2n C 1/ such as 1 X

nD1

.2n C 1/ .2n C 1/22n

in closed form. A systematic treatment of these series can be found in Srivastava and Choi’s “Series Associated with the Zeta and Related Functions.”

Chapter 16 1. Setting x D 1=2 in the power series of ln.1 Cauchy product, we have 2

ln 2 D D D

1 X

nD1 1 X

nD1 1 X

nD1

1 2n n n X1 i D1

!

1

1 X




mD1

1 .n

1 2n

x/ yields ln 2 D

n

i /i

n X1 i D1

!

1 2m m 1 2n

P1

nD1

1=.2n n/. Appealing to the

! (using the partial fraction)

1 X 1 Hn 1 D ; i 2n 1 n nD1

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Solutions to Selected Exercises where Hn are the harmonic numbers. Therefore, ln2 2 C 2

1 1 X X Hn 1 1 1 D C n 2 n 1 n 2 n 2 n 2 1 n2 nD1 nD1   1 X 1 1 D H C n 1 2n 1 n n

1 X

nD1

D

nD1 1 X

Hn : n 2 1n

nD1

Now we show that this last series is indeed .2/. In Exercise 6.8, set an D 1=n2n 2 ln.1 x=2/. Then 1 X

nD1

1 n2n 1

Hn D

Z

1

2 ln 2 C 2 ln.1 1 t

0

t =2/

dt D 2

Z

1 0

1,

f .x/ D

ln.1 C t / dt: t

But Z

1 0

Z

ln.1 C t / dt D t

1 1 X

0

nD1

. 1/nC1 n t n

1

dt D

1 X . 1/nC1 1 D .2/: 2 n2

nD1

Similarly, the second identity follows from 1 0 1 n X X 1 H k A @ ln3 2 D 6 : kC1 .n C 2/2nC2 nD1

kD1

6. Set t D 1=2 in Euler’s series transformation or show that n   n   Z 1 X X n . 1/kC1 n D . x/k k k k 0

kD1

9. Substituting x by 1

1

dx

kD1

D

Z

1

ln.1

1

x/n x

.1

0

1

dx D

Z

1

0

n X tn 1 1 D D Hn : t 1 k kD1

x yields .3/ D

Z

x/ ln x dx D x

0

Z

1 0

ln.1 x/ ln x dx: 1 x

Averaging these two integrals gives .3/ D

1 2

Z

1

0

ln.1 x/ ln x
dx: x 1 x

By expanding the two factors in the integrand into the power series respectively, we have 1 ! 0 1 Z 1 X X . x/j 1 1 1 xi 1 A dx .3/ D
@ 2 0 i j i D1

Z 1 1 1 1 XX 1 D xi 2 ij 0

j D1

1

.1

x/j

1

dx:

i D1j D1

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Excursions in Classical Analysis Appealing to the beta function, we find that Z

1

xi

1

x/j

.1

1

0

and so .3/ D

dx D B.i; j / D

1 1 1 XX 2

i D1 j D1

.i

1/Š .j 1/Š ; .i C j 1/Š

1  i Cj ij 2 j

1

:

10. First, we rewrite An as an integral: An D

1 X

i1 D1






1 X

in D1

1 i1 i2


in

Z

1 0

x i1 Ci2 C


Cin

1

dx:

Since all summands are positive, the monotone convergence theorem permits us to interchange P i the order of summation and integration. Appealing to 1 x/, we obtain i D1 x = i D ln.1 An D

Z

1

0

Z 1 n 1 1 X x in ln .1 x/ 1 X x i1 n


dx D . 1/ dx: x i1 in x 0 i1 D1

in D1

Next, in view of the well-known fact that Z 1 Z xD ln t t k lnn t dt D . 1/n 0

we have

Therefore,

Z

1 0

1

xn e

.kC1/x

0

dx D . 1/n

nŠ ; .k C 1/nC1

1 Z 1 X lnn t dt D t k ln n t dt D . 1/n nŠ.n C 1/: 1 t 0 kD0

An D . 1/n

Z

1 0

lnn x dx D nŠ.n C 1/: 1 x

Chapter 17 2. By using the partial fraction decomposition 1 i2

j2

D

1 2i



1 j

i

1 j Ci



;

for positive integers i and N with N > i , we have   X 1 1 1 S.i; N / D 2i j i j Ci j D1;j ¤i   1 1 1 1 1 D C


C C C


C 2i 1 i .i 1/ i .i C 1/ i N i   1 1 1 1 1 C C


C C C


C 2i 1 C i .i 1/ C i .i C 1/ C i N Ci 1 3 D .HN Ci HN i / ; 2i 4i 2 where Hk are the harmonic numbers. Appealing to Hn D ln n C C O.1=n/;

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Solutions to Selected Exercises we find that lim .HN Ci

HN

N !1

and so

D lim ln

i/

N !1

lim S.i; N / D

N !1

Therefore, the proposed sum is lim

K!1

K  X

 lim S.i; N / D

N !1

i D1

3 4

N Ci D 0; N i

3 : 4i 2

lim

K!1

K X 1 D i2

i D1

2 : 8

Remark. Two related problems are to evaluate 1 X

i;j D1;.i;j /D1

1 2 i j2

1 X

and

i;j D1;.i;j /D1

1 : ij.i C j /

7. (a) Rewriting Hn as an integral of a finite geometric series and then appealing to Leibniz’s formula and the power series expansions of arctan x, we have Z 1 1 1 X X 1 tn . 1/n . 1/n Hn D dt 2n C 1 2n C 1 0 1 t nD0 nD0 Z 1 1 X . 1/n x x 2nC1 D2 dx . set t D x 2 / 2n C 1 0 1 x2 nD0 Z 1 Z 1  arctan x x dx 2 dx D 2 0 1 x2 1 x2 0 Z 1 Z 1   dx  1 x  D dx C 2 arctan x 2 2 0 1 x 4 1 x2 0  D ln 2 C G; 2 where we have used the fact that   Z 1  Z 1  dx 1 x  dx D 2 arctan 2 arctan x 4 1 C x 1 x2 1 x2 0 0 Z 1 Z 1 X 1 arctan u du . 1/n 2n D D u du u 2n C 1 0 0 nD0

D

1 X

nD0

1/n

. D G: .2n C 1/2

(b). Replacing x by ix in (17.18) yields 1 X

nD1

. 1/k hk x 2k D

x arctan x: 1 C x2

Integrating this identity from 0 to 1 leads to 1 X . 1/k h D 2k C 1 k

nD1

D

Z

1

x arctan x dx 1 C x2 Z  1 1 ln.1 C x 2 / ln 2 C dx 8 2 0 1 C x2 0

.using integration by parts/:

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Excursions in Classical Analysis Now, the proposed answer follows from Z

1

ln.1 C x 2 /  dx D ln 2 2 1 C x2

0

G:

Remark. Bradley assembles a fairly exhaustive list of formulas involving Catalan’s constant, see citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.26.1879 8. Setting z D eix in 1 X zn D n

z/ D

ln.1

nD1

ln j2 sin.x=2/j C



x 2

i

and then exporting the real and imaginary parts yields 1 X sin.nx/  x D n 2

1 X cos.nx/ D n

and

ln j2 sin.x=2/j:

nD1

nD1

Next, multiplying these series together and manipulating gives 

x 2

ln j2 sin.x=2/j D D

1 X

m;nD1

1 2

1 X

1 sin.mx/ cos.nx/ D mn 2

1 X

n>k>0

1 sin.nx/ D k.n k/ 2

nD1

1 n 1 X sin.nx/ X 1 ; D n k nD1

m;nD1

1 X

sin.m C n/x mn

 n 1 1 sin.nx/ X 1 C n k n k kD1

kD1

from which the proposed identity follows by applying Parseval’s identity. Remark. If one uses arctanh instead of ln, one is led to Z



0

ln2 .tan.x=4// dx D

3 ; 4

and hence to another proof of (17.25): 1 X h2k

D

hk x 2k

2

kD1

k2

4 45 D .4/: 32 16

11. Appealing to (17.18), we have f 0 .x/ D

1 X

kD1

D

1 2x.1

x2 /

ln



1Cx 1 x



:

Hence, 2 .2

x/2

f0

 x  x 2 D ln.1 2 x 4x.1 x/

x/ D

1 ln.1 2x

x/

ln.1 4.1

x/ : x/

Integrating the equation above from 0 to x yields the desired identity.

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Solutions to Selected Exercises

Chapter 18 2. Rewrite 1 X

nD1

n2



1 1 m X D 2 2mn nD1 n.2mn mn 1 m X 2

D

1   2mn 2 1/ mn 1 1

.n C 1/.2mn C 2m

nD0

1/

: 2

 2mn C 2m mn 1

By using the beta function, we have 1 X

nD1

Z 1 1 m X 1 1  D t mnCm 2 nC1 0 2mn 2 nD0 n mn

1

t /mnCm

.1

1

dt:

Now, the proposed identity follows from 1 X t mnCm .1 t /mnCm D nC1

lnŒ1

t m .1

t /m :

nD0

4(c). Recall that G D of sine, we have

R1 0

.arctan x=x/ dx. Set x D tan.t =2/. Appealing to the double angle formula Z

1 GD 4

=2 0

t dt 1 D tan.t =2/ cos2 .t =2/ 2

Z

=2 0

t dt: sin t

By using (18.1) we obtain Z =2 2G D 0

D

Z

D

1 X

1

0

nD0

Z 1 2u arcsin u du t
dt D p .u D sin t / sin t 1 u2 2u2 0 1 1 X X 22n .2u/2n du  
2 D   2u 2n 2n nD1 n nD1 2n.2n 1/ n n 22n

.2n C 1/2



2n n

:

Remark. If

  Z 1 =2 1 C sin x ln dx 4 0 1 sin x has been proved, applying Wallis’s formula of odd power yields the following alternative proof: GD

2G D D D

Z

1 =2 X

0

1 X

nD0 1 X

nD0

nD0

1 .sin x/2nC1 dx 2n C 1

1 2
4
6


.2n/
2n C 1 3
5
7


.2n C 1/ 22n .2n C 1/2

 : 2n n

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Excursions in Classical Analysis

8 We derive the first identity by using Mathematica. First we set ln 2 D

1 X an2 C bn C c   : 3n nD1 2n n

Feeding the general sum to Mathematica and integrating gives Sum[(a n^2 + b n + c) (3 n + 1) x^n, {n, 1, Infinity}] (4 a x + 4 b x + 4 c x + 12 a x^2 - 2 b x^2 - 9 c x^2 + 2 a x^3 2 b x^3 + 6 c x^3 - c x^4)/(-1 + x)^4 % /. x -> t^2 (1 - t)/2 1/(-1 + 1/2 (1 - t) t^2)^4 (2 a (1 - t) t^2 + 2 b (1 - t) t^2 + 2 c (1 - t) t^2 + 3 a (1 - t)^2 t^4 - 1/2 b (1 - t)^2 t^4 9/4 c (1 - t)^2 t^4 + 1/4 a (1 - t)^3 t^6 - 1/4 b (1 - t)^3 t^6 + 3/4 c (1 - t)^3 t^6 - 1/16 c (1 - t)^4 t^8) Simplify[%] -1/(2 - t^2 + t^3)^4 (-1 + t) t^2 (4 a (8 + 12 t^2 - 12 t^3 + t^4 - 2 t^5 + t^6) + (2 t^2 + t^3) (-4 b (-4 - t^2 + t^3) + c (16 - 10 t^2 + 10 t^3 + t^4 - 2 t^5 + t^6))) Integrate[%, {t, 0, 1}] 1/15625 (a (2805 - (673 + 14 I) ArcTan[1/3] - (673 - 14 I) ArcTan[1/2] + (673 + 14 I) ArcTan[2] + (673 - 14 I) ArcTan[3] + 42 Log[2]) + 5 (b (405 - (158 - 6 I) ArcTan[1/3] - (158 + 6 I) ArcTan[1/2] + (158 - 6 I) ArcTan[2] + (158 + 6 I) ArcTan[3] - 3 Log[64]) + 25 c (10 - (11 - 2 I) ArcTan[1/3] - (11 + 2 I) ArcTan[1/2] + (11 - 2 I) ArcTan[2] + (11 + 2 I) ArcTan[3] - Log[64])))

Next we make some simplifications based on arctan.1=x/ D =2

arctan x

FullSimplify[% /. ArcTan[1/3] -> (Pi/2 - ArcTan[3]) /. ArcTan[1/2] -> (Pi/2 - ArcTan[2])] 1/31250 (a(5610 + 673\[Pi] + 84Log[2]) + 5(2b (405 + 79\[Pi] - 3Log[64]) + 25c(20 + 11\[Pi] - 2Log[64]))) Expand[% /. Log[64] -> 6 Log[2]] (561a)/3125 + (81b)/625 + (2c)/25 + (673a\[Pi])/31250 + (79b\[Pi])/3125 + (11c\[Pi])/250 + (42aLog[2])/15625 - (18bLog[2])/3125 - 6/125cLog[2] Collect[%, {Log[2], Pi}] (561a)/3125 + (81b)/625 + (2c)/25 + ((673a)/31250 + (79b)/3125 + (11c)/250)\[Pi] + ((42a)/15625 - (18b)/3125 - (6c)/125)Log[2] Finally, we solve the system to establish the desired identity. Solve[{(561a)/3125 + (81b)/625 + (2c)/25 == 0, (673a)/31250 + (79b)/3125 + (11c)/250 == 0, (42a)/15625 - (18b)/3125 - (6c)/125 == 1}, {a, b, c}] {{a -> -(575/6), b -> 965/6, c -> -(91/2)}}

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Solutions to Selected Exercises

Chapter 19 5(b). Let I denote the required integral. Differentiating I with respect to ˛ yields Z 1 sin ˇx cos ˛x kx I 0 .˛/ D e dx: x 0 By using the product to sum formula we have Z 1 1 sin.˛ C ˇ/x kx I 0 .˛/ D e dx 2 x 0

Z

Appealing to Example 1, we find that  ˛Cˇ 1 I 0 .˛/ D arctan 2 k

1

sin.˛

ˇ/x x

0

˛Cˇ k

arctan



kx

e

 dx :

;

and so I D

˛ Cˇ ˛Cˇ arctan 2 k

˛

ˇ

˛

arctan

2

ˇ k

k k 2 C .˛ ˇ/2 ln : 4 k 2 C .˛ C ˇ/2

C

Remark. Letting k ! 0, we get Z In general, if ˛ >

Pn

1 0

i D1

Z

1

0

sin ˛x sin ˇx
dx D x x



 2 ˇ;  2 ˛;

if ˛  ˇ; : if ˛  ˇ

˛i and all ˛i > 0, then sin ˛x sin ˛1 x sin ˛n x 



dx D ˛1 ˛2


˛n : x x x 2

7. Appealing to cos ax

cos bx x

and

Z

we have Z

1

cos ax

cos bx x2

0

1 0

D

Z

b

sin yx dy a

 sin yx dx D ; x 2

dx D D

Z

1Z b

0 a Z bZ 1 a

0

sin yx dydx x  sin yx dxdy D .b x 2

a/:

9. This is a Frullani integral with f .x/ D ex =.1 C ex / and so the answer is .1=2/ ln.p=q/. 11. Let f .a/ denote the right side of the equation. Substituting x D at gives 1 f .a/ D p 

Z

1

e

0

a2 t 2

dt : 1 C t2

By Leibniz’s rule, we have 0

f .a/ D

2a p 

Z

0

1

e

a2 t 2

dt

Z

1

0

e

a2 t 2

dt

1 C t2

!

:

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Excursions in Classical Analysis Appealing to Z

1

e

a2 t 2

0

dt D

p

 ; 2a

we find that f .a/ satisfies an initial value problem f 0 .a/

2af .a/ D 1;

and so f .a/ D ea

Z

2

1

f .0/ D e

t2

p  2

dt;

a

which is identical with the left side of the equation after substituting t 2 D x 2 C a2 . 14. Recall that J0 .x/ D

 1  X 1=2 x 2n : n .2n/Š

nD0

Appealing to the binomial theorem, we obtain Z

1

e

sx

0

 Z 1 1  X 1 1=2 e sx x 2n dx n .2n/Š 0 nD0  Z 1 1  X 1 1=2 D e t t 2n dt n .2n/Šs 2nC1 0 nD0  1  X 1=2 €.2n C 1/ D n .2n/Šs 2nC1 nD0   2 1  1 1 X 1 1 1 1=2 D p D p D : 2 n 2 s s 1 C .1=s/ s 1 C s2 nD0

J0 .x/ dx D

p Remark. This result indicates that the Laplace transform of J0 .x/ is 1= 1 C s 2 .

Chapter 20 2. Since all summands are positive, the monotone convergence theorem permits us to interchange the order of summation and integration. Substituting x D e t yields Z

1

x 0

x

dx D

Z

1

et e

t

e

t

dt

0 Z 1

! 1 X 1 n nt D t e e t dt nŠ 0 nD0 Z 1 1 X 1 D t n e .nC1/t dt nŠ 0 nD0 Z 1 1 X 1 D t n e nt dt .shifting the index n by one/ .n 1/Š 0 nD1 Z 1 1 X n n D s n 1 e s ds .set s D nt / .n 1/Š 0 D

nD1 1 X

nD1

1 X n n €.n/ D n .n 1/Š

n

:

nD1

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Solutions to Selected Exercises 11. Define f .x/ D

1 X

.nCx/2 s

e

:

1

Clearly, f .x C 1/ D f .x/; i.e., f is periodic and has period 1. Expanding f as a Fourier series 1 X

an e2nxi ;

1

we have an D

Z

1

1 Z 1

D2 r D

and so

1 X

sx 2

e

e

2nxi

e

sx 2

dx

cos.2nx/ dx

.see (19.4)/

0

1 e s

n2 =s

;

.nCx/2 s

e

1

D

r

1 1 X e s 1

n2 =s 2nxi

e

:

The required identity follows from this equation by setting x D 0.

12. First Proof. Recall the well-known Poisson summation formula, for jrj < 1, 1 X

1C2 Taking r D e

˛; x

kD1

r k cos.kx/ D

1 r2 : 2r cos x C r 2

1

D ˛y yields

1 X 1 C e 2

˛k

kD1

cos.˛yk/ D

1 2 1

D

1 2 1

On the other hand, the partial fraction gives  1 1 D 1 C .y C 2ˇj /2 2i y p where i D 1. Thus, the Euler formula  cot.x/ D

e 2˛ 2e ˛ cos.˛y/ C e 1

e2˛ 1 : cos.˛y/ C e2˛

2e˛

1 i C 2ˇj 1 X

2˛

jD 1

1 y C i C 2ˇj



;

1 xCj

leads to 1 X

jD 1

 1 X

 1 1 y i C 2ˇj y C i C 2ˇj jD 1        y i yCi D cot  cot  4ˇi 2ˇ 2ˇ      ˛ ˛.y i / ˛.y C i / D cot cot ; 4i 2 2

1 1 D 2i 1 C .y C 2ˇj /2

.using ˛ˇ D /:

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Excursions in Classical Analysis In view of the trigonometric identity cot s

cot t D

2 sin.t s/ sin.t s/ D sin s sin t cos.t s/ cos.t C s/

and sin.ix/ D i sinh x; cos.ix/ D cosh x, we have     ˛.y i / ˛.y C i / 2i sinh.˛/ cot cot D D 2i 2 2 cosh.˛/ cos.˛y/ 1

e2˛ 1 ; cos.˛y/ C e2˛

2e˛

and so 1 ˛

1 X

jD 1

1 1 D 1 C .y C 2ˇj /2 2 1

1 X e2˛ 1 1 D C e 2e˛ cos.˛y/ C e2˛ 2

˛k

cos.˛yk/:

kD1

Second Proof. This proof is based on the Poisson summation formula, which asserts that 1 X

T

kD 1

f .kT / D

1 X

jD 1

where the Fourier transform fO of f is defined by Z 1 fO./ D f .x/e Let f .x/ D e

jxj

fO.j=T /;

2x i

.1/

dx:

1

cos.xy/. Then Z 1 fO./ D e jxj cos.xy/e 2x i dx 1 Z 1 D2 e x cos.xy/ cos.2x/dx 0 Z 1 D e x Œcos.y C 2/x C cos.y

2/xdx

0

D

1 1 C : 1 C .y C 2/2 1 C .y 2/2

Now, the required identity follows from (1) by choosing T D ˛ and some simplifications. 16. To eliminate the absolute value, we divide the interval Œ0; 1/ into h  i ; .n D 0; 1; 2; : : :/: n ; .n C 1/ 2 2 When n D 2k, Z kC=2 Z =2 ln j cos xj ln cos t dx D dt I 2 x .t C k/2 k 0 and when n D 2k 1, Z k Z =2 ln cos t ln j cos xj dx D dt: 2 x .t k/2 0 k =2 Therefore, Z

1 0

ln j cos xj dx D x2

Z

=2

0

8 1