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 Journalof Mathematical Education in Science and Tech- nology, 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 Jour- nal 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 Interna- tional 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): 10987654321

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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, Miklós 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 Undergrad- uate 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. Sánchez 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 Simula- tion, David Williamson, Marilou Mendel, Julie Tarr, and Deborah Yoklic

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Student Manual for Mathematics for Business Decisions Part 2: Calculus and Optimiza- tion, 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 xi 1 Two Classical Inequalities 1 1.1 AM-GMInequality ...... 1 1.2 Cauchy-SchwarzInequality ...... 8 Exercises ...... 12 References...... 15 2 A New Approach for Proving Inequalities 17 Exercises ...... 22 References...... 23 3 Means Generated by an Integral 25 Exercises ...... 30 References...... 32 4 The L’Hˆopital Monotone Rule 33 Exercises ...... 37 References...... 38 5 Trigonometric Identities via Complex Numbers 39 5.1 APrimerofcomplexnumbers ...... 39 5.2 FiniteProductIdentities ...... 41 5.3 FiniteSummationIdentities ...... 43 5.4 Euler’sInﬁniteProduct ...... 45 5.5 Sumsofinversetangents ...... 48 5.6 TwoApplications ...... 49 Exercises ...... 51 References...... 53 6 Special Numbers 55 6.1 GeneratingFunctions ...... 55 6.2 FibonacciNumbers ...... 56 6.3 Harmonicnumbers ...... 58 6.4 BernoulliNumbers ...... 61

vii

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viii Contents

Exercises ...... 69 References...... 72 7 On a Sum of Cosecants 73 7.1 Awell-knownsumanditsgeneralization ...... 73 7.2 Roughestimates...... 74 7.3 Tyinguptheloosebounds...... 76 7.4 FinalRemarks...... 79 Exercises ...... 79 References...... 81 8 The Gamma Products in Simple Closed Forms 83 Exercises ...... 89 References...... 91 9 On the Telescoping Sums 93 9.1 Thesumofproductsofarithmeticsequences ...... 94 9.2 The sum of products of reciprocals of arithmetic sequences ...... 95 9.3 Trigonometricsums ...... 97 9.4 Somemoretelescopingsums ...... 101 Exercises ...... 106 References...... 108 10 Summation of Subseries in Closed Form 109 Exercises ...... 117 References...... 119 11 Generating Functions for Powers of Fibonacci Numbers 121 Exercises ...... 128 References...... 130 12 Identities for the Fibonacci Powers 131 Exercises ...... 140 References...... 142 13 Bernoulli Numbers via Determinants 143 Exercises ...... 149 References...... 152 14 On Some Finite Trigonometric Power Sums 153 14.1 Sums involving sec2p .k=n/ ...... 154 14.2 Sums involving csc2p .k=n/ ...... 156 14.3 Sums involving tan2p .k=n/ ...... 157 14.4 Sums involving cot2p .k=n/ ...... 159 Exercises ...... 162 References...... 164 15 Power Series of .arcsin x/2 165 15.1FirstProofoftheSeries(15.1) ...... 165 15.2SecondProofoftheSeries(15.1) ...... 167 Exercises ...... 171

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Contents ix

References...... 173 16 Six Ways to Sum .2/ 175 16.1Euler’sProof...... 176 16.2ProofbyDoubleIntegrals ...... 177 16.3ProofbyTrigonometricIdentities ...... 181 16.4ProofbyPowerSeries...... 182 16.5ProofbyFourierSeries ...... 183 16.6ProofbyComplexVariables...... 183 Exercises ...... 184 References...... 187 17EvaluationsofSomeVariantEulerSums 189 Exercises ...... 197 References...... 199 18 Interesting Series Involving Binomial Coefﬁcients 201 18.1 An integralrepresentation and its applications ...... 202 18.2SomeExtensions ...... 208 18.3 Searching for new formulas for ...... 210 Exercises ...... 212 References...... 215 19ParametricDifferentiationandIntegration 217 Example1...... 217 Example2...... 218 Example3...... 219 Example4...... 220 Example5...... 220 Example6...... 222 Example7...... 223 Example8...... 224 Example9...... 225 Example10 ...... 226 Exercises ...... 228 References...... 230 20FourWaystoEvaluatethePoissonIntegral 231 20.1UsingRiemannSums ...... 232 20.2UsingAFunctionalEquation ...... 233 20.3UsingParametricDifferentiation ...... 234 20.4UsingInﬁniteSeries...... 235 Exercises ...... 235 References...... 239 21 Some Irresistible Integrals 241 21.1MonthlyProblem10611...... 241 21.2MonthlyProblem11206...... 243 21.3MonthlyProblem11275...... 244

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x Contents

21.4MonthlyProblem11277...... 245 21.5MonthlyProblem11322...... 246 21.6MonthlyProblem11329...... 247 21.7MonthlyProblem11331...... 249 21.8MonthlyProblem11418...... 250 Exercises ...... 252 References...... 254 Solutions to Selected Problems 255 Index 297 About the Author 301

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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 twoways.The first is toprovideacolorfultourof 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 em- ploying 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 al-

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xii 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 Pólya 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 assim- ilate 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 xiii

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 re- sponsibility. 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 thework 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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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 motivationnecessary to work throughall thesteps ofa rigorousproof. 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

n a1 a2 an pa1a2 an C CC ; (1.1) Ä n

with equality if and only if all ak .1 k n/ are equal. Ä Ä 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 2, we have D a1 a2 pa1a2 C : (1.2) Ä 2 2 This is equivalent to a1 a2 2pa1a2 0 or .pa1 pa2 / 0, which is always true. C Moreover, equality in (1.2) holds if and only if a1 a2. Next, we show that H.2/ and D H.n/ together imply H.n 1/. Let C a1 a2 an an 1 An 1 C CC C C : C D n 1 C 1

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2 Excursions in Classical Analysis

Then 1 An 1 Œ.n 1/An 1 .n 1/An 1 C D 2n C C C C 1 Œ.a1 an/ .an 1 An 1 An 1/ D 2n CC C C C C CC C n 1 terms 1 n n „ n 1 ƒ‚ … n pa1a2 an n an 1An1 2n C C C Â q Ã 2n n 1 a1a2 anan 1 An1; C C q where the ﬁrst inequality follows from H.n/, while the second inequality follows from H.2/ in the form a b pab C Ä 2 with n n n 1 a pa1a2 an; b an 1 An1: D D C C Now, H.n 1/ follows from q C 2n n 1 An 1 a1a2 an 1 An1: C C C

As before, equality holds if and only if a1 a2 an and an 1 An 1, or D D D C D C equivalently, if and only if a1 a2 an 1. D DD C 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 2 for n. H.2/ follows directly from .pa1 pa2 / 0 as noted in the ﬁrst proof. Moreover, H.2/ is self-generalizing: for example, H.2/ can be applied twice to get

1=2 1=2 1=4 .a1a2/ .a3a4/ a1 a2 a3 a4 .a1a2a3a4/ C C C C : Ä 2 Ä 4 This conﬁrms H.4/, and then H.4/ can be used again with H.2/ to establish

1=4 1=4 1=8 .a1 a4/ .a5 a8/ a1 a2 a8 .a1a2 a8/ C C CC : Ä 2 Ä 8

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1. Two Classical Inequalities 3

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 k .a1; a2;:::;an/ into a longer sequence .b1;b2;:::;b2k / to which we can apply H.2 /. To see this, take the case n 3 ﬁrst. Set bi ai .i 1;2;3/ and let b4 be the value that D D D yields the equality b1 b2 b3 b4 a1 a2 a3 C C C C C : 4 D 3 This leads to a1 a2 a3 b4 C C : D 3 Applying H.4/ gives

4 a1 a2 a3 a1 a2 a3 .a1 a2 a3/=3 a1 a2 a3 a1a2a3 C C C C C C C C C 3 Ä 4 D 3 s Â Ã which then simpliﬁes to the desired result H.3/:

3 a1 a2 a3 pa1a2a3 C C : Ä 3

In general, for n<2k, similar to the case n 3, set D

ai ; i 1;2;:::;n; bi a1 an D k D CC A; i n 1;:::;2 : ( n D D C

k Applying H.2 / to .b1;b2;:::;b2k /, we obtain

1=2k k k 2k n a1 an .2 n/A 2 A a1 anA CC C A; Ä 2k D 2k D Á which simpliﬁes to, after eliminating A on the left-hand side and then raising both sides to the power of 2k=n, 1=n a1 a2 an .a1a2 an/ C CC : Ä 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.n 1/, H) (II) H.n/ and H.2/ H.2n/. H) This leads to a short instructive proof of (1.1) as follows: To prove (I), set a1 a2 an 1 A C CC : D n 1

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4 Excursions in Classical Analysis

Then H.n/ yields

n 1 n 1 n n k 1 ak A .n 1/A A n ak A D C C A Ä n D n D k 1 ! P ! Â Ã YD and hence n 1 n 1 n 1 a1 a2 an 1 ak A C CCC : Ä D n 1 k 1 Â Ã YD For (II), we have

2n n 2n

ak ak ak D 0 1 k 1 k 1 ! k n 1 YD YD DYC n n n@ 2n A ak ak Ä n 0 n 1 k 1 ! k n 1 XD DXC 2n 2n@ A a k ; Ä 2n k 1 ! XD where in the ﬁrst inequality we use the induction hypothesis H.n/, and in the second inequality we use H.2/. Let us now turntoan alternate approach toprove (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 a A k ; .1 k n/ k n D pa1a2 an Ä Ä which are normalized in the sense that n

Ak 1: D k 1 YD Now, the AM-GM inequality says that if A1A2 An 1 then D

A1 A2 An n; (1.3) C CC

with equality if and only if A1 A2 An 1. This observation yields the third D DD D induction proof.

Third Proof. Let H.n/ stand for the hypothesis that, subject to A1A2 An 1, D inequality (1.3) is valid for n. For n 2, if A1A2 1, we have D D 2 A1 A2 2 A1 A2 0; C () p p Á which is always true. Assume that A1A2 An 1 1 and set C D

Bk Ak .1 k n 1/; Bn AnAn 1: D Ä Ä D C

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1. Two Classical Inequalities 5

Since B1B2 Bn 1, H.n/ implies that D

B1 B2 Bn A1 An 1 AnAn 1 n: C CC D CC C C

We have to show that A1 An An 1 n 1. To see this, it sufﬁces to prove that CC C C C

An An 1 AnAn 1 1; C C C C

which is equivalent to .An 1/.An 1 1/ 0. While this does not generally solely C Ä follow from the assumption that A1A2 An 1 1, due to the symmetry of the Ak ’s, we C D may order the numbers A1; A2;:::;An 1 in advance such that An Ak An 1 for all C Ä Ä C 1 k n 1. Thus, we have Ä Ä n 1 n 1 AnC A1A2 An 1 AnC1 Ä C Ä C

and so An 1 An 1, which implies that .An 1/.An 1 1/ 0. This establishes the Ä Ä C C Ä desired result H.n 1/. The equality case means that C

A1 A2 AnAn 1 1 and .An 1/.An 1 1/ 0; D DD C D C D

which implies A1 An An 1 1. DD D C D The additive inequality (1.3) yields a quite different proof of (1.1). Indeed, it also con- nects 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/ f ..a1; a2;:::;an//; D

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 max- imized when all variables are equal. Hardy’s following proof provides one of the best applications of this principle.

Fourth Proof. Let

n a1 a2 an G pa1a2 an; A C CC : (1.4) D D n

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

a1 min a1; a2;:::;an ; a2 max a1; a2;:::;an : D f g D f g Then it is clear that

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6 Excursions in Classical Analysis

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

A; a1 a2 A; a3;:::;an (1.5) f C g

and let G1 and A1 be the corresponding geometric and arithmetic means. Since

A.a1 a2 A/ a1a2 .A a1/.a2 A/ > 0; C D we have A.a1 a2 A/a3 an > a1a2 an; C from which it follows that G1 >G. On the other hand,

nA1 A .a1 a2 A/ a3 an nA; D C C C CC D

i.e., A1 A. Thus, we have shown that the transformationdeﬁned by (1.5) does not change D the arithmetic mean while the geometric mean is increased. In particular, if all numbers in (1.5) are the same then we have

A A1 G1 > G; D D 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. Ä Now, we have Gk Ak and D

A A1 Ak Gk >Gk 1 > >G1 > G: D DD D This ﬁnally proves (1.1) as desired. As an example of applying the smoothing transformation above, consider

.a1; a2; a3; a4/ .1; 7; 5; 3/: D The transformation successively yields the sets of positive numbers

.1; 7; 5; 3/ .4;4;5;3/ .4;4;4;4/: H) H)

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 Ä ÄÄ 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 Ä ÄÄ Ä Ä and so there exists one k with ak G ak 1. It follows that Ä Ä C k n G 1 1 ai 1 1 dt dt 0 (1.6) a t G C G G t i 1 Z i Â Ã i k 1 Z Â Ã XD DXC

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1. Two Classical Inequalities 7

or equivalently, n n ai 1 ai 1 dt dt: G t i 1 ZG i 1 ZG XD XD Here the left-hand side gives

n ai G A n 1 ; G D G i 1 Â Ã XD while the right-hand side becomes

n n .ln ai ln G/ ln ai n ln G 0: D D i 1 i 1 ! XD YD 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 a2 an G. D DD D 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 followingstronger inequality:

Weighted Arithmetic Mean and Geometric Mean Inequality Let a1; a2;:::;an and p1;p2;:::;pn be positive real numbers such that p1 p2 pn 1. Then C CC D p1 p2 pn a a a p1a1 p2a2 pnan; (1.7) 1 2 n Ä C CC

with equality if and only if all ak .1 k n/ are equal. Ä Ä P´olya’s Proof. It is well known that

x exp.x 1/ for all x R: Ä 2 For 1 k n, using this inequality repeatedly yields Ä Ä pk ak exp.ak 1/ and a exp.pk ak pk/ Ä k Ä and so n p1 p2 pn a a a exp pkak 1 : (1.8) 1 2 n Ä k 1 ! XD To obtain the desired arithmetic mean bound, the normalization process is used again. Let

ak bk with A p1a1 p2a2 pnan: D A D C CC

Replacing ak in (1.8) by bk yields

n a1 p1 a1 p2 an pn ak exp pk 1 1: A A A Ä A D k 1 ! Á Á Á XD This is equivalent to (1.7). In the case of equality, noticing that a a a k < exp. k 1/ unless k 1; A A A D

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8 Excursions in Classical Analysis

we see that (1.8) is strict unless ak A for all k 1;2;:::;n. Thus, one has equality in D D (1.7) if and only if all ak’s are equal. Similar to the AM-GM inequality, the Cauchy-Schwarz inequalityis 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 highlightthis, we present four proofs of the Cauchy-Schwarz inequality.

1.2 Cauchy-Schwarz Inequality

Let ak ;bk .1 k n/ be real numbers. Then Ä Ä n 2 n n 2 2 akbk a b ; (1.9) Ä k k k 1 ! k 1 ! k 1 ! XD XD XD

with equality if and only if there is a t R such that ak tbk .1 k n/. 2 D Ä Ä First, we demonstrate that the Cauchy-Schwarz inequality is a consequence of the AM- GM inequality.

First Proof. Let H.n/ represent the hypothesis that inequality (1.9) is valid for n. For n 2, by the AM-GM inequality, we have D 2 2 2 2 2 .a1b1 a2b2/ a b 2.a1b2/.a2 b1/ a b C D 1 1 C C 2 2 a2b2 a2b2 a2b2 a2b2 Ä 1 1 C 1 2 C 2 1 C 2 2 .a2 a2/.b2 b2/; D 1 C 2 1 C 2 which conﬁrms H.2/. For the case of equality, appealing to the equality condition in the AM-GM inequality, we have a1b2 a2b1, or equivalently, ai tbi .i 1;2/ with D D D t a1=a2 b1=b2. Now, we prove that H.2/ and H.n/ together imply H.n 1/. D D C Indeed, using the AM-GM inequality and H.n/ yields

n n n 2 2 2 2 2 ak bk an 1bn 1 2 an 1 bk bn 1 ak C C Ä v C v C k 1 ! u k 1 !u k 1 ! XD u XD u XD t n t n 2 2 2 2 an 1 bk bn 1 ak Ä C C C k 1 ! k 1 ! XD XD and hence

n 1 2 n 2 n C 2 2 akbk ak bk 2 akbk an 1bn 1 an 1bn 1 D C C C C C C k 1 ! k 1 ! k 1 ! XD XD XD n 2 n n 2 2 2 2 2 2 akbk an 1 bk bn 1 ak an 1bn 1 Ä C C C C C C C k 1 ! k 1 ! k 1 ! XD XD XD

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1. Two Classical Inequalities 9

n n n n 2 2 2 2 2 2 2 2 ak bk an 1 bk bn 1 ak an 1bn 1 Ä C C C C C C C k 1 ! k 1 ! k 1 ! k 1 ! XD XD XD XD n 1 n 1 C C a2 b2 : D k k k 1 ! k 1 ! XD XD For the case of equality, again, the equality condition in the AM-GM inequality implies

akbn 1 bkan 1 and ak tbk; for all k 1;2;:::;n: C D C D D

Therefore, ak tbk for all k 1;2;:::;n;n 1. D D C 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 ¤ Ä Ä ak bk Ak and Bk ; D n 2 1=2 D n 2 1=2 k 1 ak k 1 bk D D which are normalized inP the sense that P n n A2 1 and B2 1: (1.10) k D k D k 1 k 1 XD XD With these assignments, the inequality (1.9) is equivalent to

A1B1 A2B2 AnBn 1: (1.11) C CC Ä To see this, adding the well-known elementary inequalities

1 2 2 2 AkBk .A B / . .Ak Bk/ 0/ Ä 2 k C k () for k from 1 to n, in view of (1.10), we have

n n 1 2 2 A1B1 A2B2 AnBn A B 1: C CC Ä 2 k C k D k 1 k 1 ! XD XD Moreover, subject to (1.10), it is easy to see that (1.11) is equivalent to

2 2 2 .A1 B1/ .A2 B2/ .An Bn/ 0: C CC

Thus, the equality in (1.9) occurs if and only if Ak Bk for all 1 k n; that is, if and D Ä Ä only if n n 1=2 2 2 ak tbk for all 1 k n with t a = b : D Ä Ä D k k k 1 k 1 ! XD XD 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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10 Excursions in Classical Analysis

Third Proof. For n 2, the Cauchy-Schwarz inequality just says D 2 2 2 2 2 .a1b1 a2b2/ .a a /.b b /: C Ä 1 C 2 1 C 2 This is a simple assertion of the elementary identity

2 2 2 2 2 2 .a a /.b b / .a1b1 a2b2/ .a1b2 a2b1/ : 1 C 2 1 C 2 C D In general, let n n n 2 2 2 Dn a b akbk : D k k k 1 ! k 1 ! k 1 ! XD XD XD We have already seen that D2 is a perfect square. In view of the symmetry, we can rewrite Dn as n n n n 1 2 2 2 2 Dn .a b a b / ai bi aj bj : D 2 i j C j i i 1 j 1 i 1 j 1 XD XD XD XD This quickly yields the well-known Lagrange’s identity

n n 1 2 Dn .ai bj aj bi / : D 2 i 1 j 1 XD XD 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 .a1; a2;:::;an/; y .b1;b2;:::;bn/: D D Recall the inner product on Rn is deﬁned by

.x; y/ a1b1 a2b2 anbn D C CC and the norm is given by x .x; x/. In vector form, the Cauchy-Schwarz inequality k k D becomes p .x; y/2 x 2 y 2: Äk k k k Schwarz observed that the quadratic function

f .t/ xt y 2 x 2t 2 2.x; y/t y 2 Dk C k Dk k C Ck k is nonnegative for all t R and hence the discriminant 2 .x; y/2 x 2 y 2 0: k k k k Ä 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 11

n Application 1. Prove that sequence 1 1 is monotonic increasing and bounded. C n n o Proof. By the AM-GM inequality, we have

1 n 1 1 1 nC1 1 1 < n 1 1 1 : s C n n 1 C n C D C n 1 Â Ã C Ä Â Ã C This is equivalent to n n 1 1 1 C 1 < 1 ; C n C n 1 Â Ã Â C Ã n which shows that 1 1 is monotonic increasing as required. To ﬁnd an upper bound of C n the sequence, for n>5, using the AM-GM inequality once more, we have

6 nC1 5 1 5 n 1n 5 < 6 .n 5/ 1 ; s 6 n 1 6 C D n 1 Â Ã C Ä C and so .5=6/6 < .n=.n 1//n 1, or equivalently, .1 1=n/n 1 < .6=5/6. Therefore, C C C C n n 1 6 1 1 C 6 1 < 1 < <3; .for n > 5/: C n C n 5 Â Ã Â Ã Â Ã n In view of the monotonicity of the sequence, for all n 1, we obtain 1 1 < 3. It is C n well known that the sequence converges to the natural base e.

Application 2. (Monthly Problem 11430, 2009). For real x1;:::;xn, show that x x x 1 2 n < pn: 1 x2 C 1 x2 x2 CC 1 x2 x2 C 1 C 1 C 2 C 1 CC n Solution Notice that

x1 xn x1 xn j j j j : 1 x2 CC 1 x2 x2 Ä 1 x2 CC 1 x2 x2 C 1 C 1 CC n C 1 C 1 CC n

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

x x x 2 S 2 1 1 1 2 1 n D 1 x2 C 1 x2 x2 CC 1 x2 x2 Â C 1 C 1 C 2 C 1 CC n Ã x 2 x 2 x 2 n 1 2 n : Ä 1 x2 C 1 x2 x2 CC 1 x2 x2 "Â C 1 Ã Â C 1 C 2 Ã Â C 1 CC n Ã # For k 1, we have D 2 x x2 1 1 1 1 ; 1 x2 Ä 1 x2 D 1 x2 Â C 1 Ã C 1 C 1

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12 Excursions in Classical Analysis

and 2 2 xk xk 2 2 2 2 2 2 1 x1 xk ! Ä .1 x1 xk 1/.1 x1 xk/ C CC C CC C CC 1 1 2 2 2 2 ; D 1 x1 xk 1 1 x1 xk C CC C CC for 2 k n. Combining these inequalities results in Ä Ä x 2 x 2 1 1 n 1 < 1: 1 x2 CC 1 x2 x2 Ä 1 x2 x2 Â C 1 Ã Â C 1 CC n Ã C 1 CC n 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

an bn a0 a; an 1 C b0 b; bn 1 anbn; .n 1; 2; : : :/: D C D 2 I D C D D p 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 a b pab bn bn 1 an 1 an C ÄÄ Ä C Ä C Ä ÄÄ 2 and 1 an 1 bn 1 .an bn/: C C Ä 2 Hence the sequences an and bn both converge to the same limit, which is called the f g f g Gauss arithmetic-geometric mean and denoted by agm.a; b/. In 1799 Gauss successfully found agm.1; p2/ to great precision, but it wasn’t until 1818 that he ﬁnally established

1 2 =2 dx agm.a; b/ ; D 0 a2 sin2 x b2 cos2 x ! Z C which later was used to found the branchp 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 b/2 a2 b2 C : x y Ä x C y C Then use this self-generalizing inequality to prove the Cauchy-Schwarz inequality.

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1. Two Classical Inequalities 13

3. Let ak >0 for all 1 k n and a1a2 an 1. Prove the following inequalities. Ä Ä D n (a) (1) .1 a1/.1 a2/ .1 an/ 2 . C C C (b) (2) am am am 1 1 1 for any positive integer m n 1. 1 C 2 CC n a1 C a2 CC an n 4. Let ak >0 for all 1 k n and k 1 ak 1. Prove the following inequalities. Ä Ä D D a2 a2 P a2 1 (a) 1 2 n . a1 a2 C a2 a3 CC an a1 2 C C C n a n (b) k . 2 ak 2n 1 k 1 XD n am 1 k m>1 (c) m 2 , where . 1 ak .n 1/n k 1 XD n n ak 1 (d) pak . p1 a pn 1 k 1 k k 1 XD XD n ak (e) 1 , where a0 0. Ä p1 a0 a pa an Ä 2 D k 1 k 1 k XD C CC CC 5. For all nonnegative sequences a1; a2;:::; and positive integers n and k, use leap- forward fall-back induction to prove

n n n n a1 a2 ak a a a C CC 1 C 2 CC k : k Ä k Â Ã This is a special case of the power mean inequality, which we will explore more in Chapter 3.

6. Let ak .k 1;2;:::;n/ be positive real numbers. Let A and G be the arithmetic and D geometric means of ak ’s respectively, and H be the harmonic mean of ak ’s deﬁned by n H : D 1=a1 1=a2 1=an C CC Prove that nA H .n 1/G: C C In addition, if AH Gn; n 3, prove that n an nGn.n 1/: k k 1 XD n 7. Let a1 < a2 < < an and 1=ak 1: Prove that for any real number x and k 1 Ä n 2, D P 2 n 1 1 : 2 2 2 a x Ä 2.a1.a1 1/ x / k 1 k ! XD C C

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14 Excursions in Classical Analysis

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

k k 9. Let a1 a2 an >0. Let i 1 ai i 1 bi for all 1 k n. Show that D Ä D Ä Ä n 2 n 2 (a) i 1 ai i 1 bi , P P D Ä D n 3 n 2 (b) Pi 1 ai Pi 1 ai bi . D Ä D 10. The Schwab-SchoenbergP P Mean. Let a and b be positive numbers with a

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 n 1 1 K : a a a Ä a n 1 1 2 n n 1 n XD C CC XD 12. Monthly Problem 11145 [2005, 366; 2006, 766]. Find the least c such that if n 1 and a1;:::;an >0 then

n n k c ak: k Ä k 1 j 1 1=aj k 1 XD D XD Remark. The inequality above canP be generalized as: for any positive p,

p p 1 k p 1 1 C ap : k Ä p k k 1 j 1 1=aj ! Â Ã k 1 XD D XD P 13. Putnam Problem 1978-A5. Let 0

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

1=n 1=n 1=n .a1a2 an/ .b1b2 bn/ Œ.a1 b1/.a2 b2/ .an bn/ : C Ä C C C

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1. Two Classical Inequalities 15

15. Carleman’s Inequality. Let a1; a2;:::; be a positive sequence with 0 < n1 1 an < . Prove that D 1 P 1 1=n 1 .a1a2 an/ e an: Ä n 1 n 1 XD XD Remark. Pólya gave an elegant proof of this inequalitythat depends on littlemore than the AM-GM inequality. Several refinements of Carleman’s inequality were obtained recently. See “On an infinity series for .1 1=x/x and its application” (Int. J. Math. C Math. Anal., 29(2002) 675–680) for details.

References [1] H. Alzer, A proof of the arithmetic mean-geometric mean inequality, Amer. Math. Monthly, 103(1996) 585. [2] E. Beckenbach and R. Bellman, An Introduction to Inequalities, The Mathematical Association of America, 1961. [3] A. L. Cauchy, Analyse Algebrique (Note 2; Oeuves Completes), Ser. 2, Vol. 3, Paris, (1897) 375–377. [4] G. H. Hardy, J. E. Littlewood and G. P´olya, Inequalities, Cambridge University Press, Cam- bridge, 1952. [5] J. M. Steele, The Cauchy-SchwarzMaster Class, Cambridge University Press, Cambridge, 2004.

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A New Approach for Proving Inequalities

A beautiful approach for a specific problem frequently leads to wide applicability, thereby solving a host of new, related problems. We have seen in the preceding chapter some memorable approaches from a great variety of sources for proving the AM-GM inequality and the Cauchy-Schwarz inequality. There are many more different proofs recorded in the monograph [2] by Bullen et.al. In this regard, one may wonder whether there are any new approaches to re-prove these two inequalities? In this chapter, we introduce a unified elementary approach for proving inequalities [3, 4]. We will see that this new approach not only recovers these two classical inequalities, but also is strong enough to capture the famous Hölder, Triangle and Minkowski inequalities, as well as Ky Fan’s inequality. The unified elementary approach is rarely seen and depends on little more than the following well-known fact in analysis: inf f .x/ inf g.x/ inf .f .x/ g.x//; x E C x E Ä x E C 2 2 2 where E is a subset of R. In general, this suggests that n n

inf fk.x/ inf fk .x/: (2.1) x E Ä x E k 1 2 2 k 1 XD XD where fk E R R .k 1;2:::;n/. Moreover, if all fk.x/ are convex in E, the W ! D convexity allows us to conclude:

Lemma 1. Let all fk .x/ be convex in E. Then the equality holds in .2.1/ if and only if all n inﬁmums of the fk.x/ and the infumum of k 1 fk .x/ are achieved at the same point in E. D P We leave the proof of Lemma 1 to the reader (See Exercise 1). We ﬁrst prove the AM-GM inequality in terms of (2.1).

Theorem 2.1. Let a1; a2;:::;an be positive real numbers. Then

n a1 a2 an pa1a2 an C CC ; (2.2) Ä n

with equality if and only if all ak’s are equal.

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18 Excursions in Classical Analysis

Proof. Clearly, the key ingredient here is how to properly deﬁne the fk.x/, which will then enable us to employ (2.1). In order to accomplish this, consider the logarithm of both sides of (2.2), which becomes

n n k 1 ak ln ak n ln D : Ä n k 1 ÂP Ã XD Adding n to both sides yields

n n k 1 ak .1 ln ak/ n n ln : (2.3) C Ä C Dn k 1 ÂP Ã XD So to prove that (2.2) is true, it is enough to show (2.3) holds. With (2.1) in mind, we simply need fk.x/ that has a minimum value of 1 ln ak; for example, C x fk.x/ akx ln x or fk.x/ ln x: D D ak C

Once we determine the form of the fk .x/, the rest becomes a routine calculation. Indeed, let fk.x/ akx ln x;E .0; /. Since D D C1 1 1 f .x/ ak ; f .x/ ; k0 D x k00 D x2

fk has a unique critical point at xkc 1=ak, where it has minimum value D fk.xkc/ 1 ln ak; D C as desired. Similarly, the function

n n

f .x/ fk .x/ ak x n ln x D D k 1 k 1 ! XD XD n has a unique critical point xc n= k 1 ak and its minimum value is D D n P k 1 ak f .xc / n n ln : D C Dn ÂP Ã Finally, applying (2.1), we establish (2.3) and thus confirm the desired inequality (2.2). Furthermore, since all the fk are convex, by Lemma 1, we conclude that the equality in (2.2) holds if and only if all xkc’s are equal to xc ; that is, all ak ’s are equal. Before proceeding, let us take another look at the proof. It seems that the key insight in the foregoing proof is how to construct the functions fk.x/. Once fk.x/ have been defined, the remaining steps are a straightforward exercise in calculating infimums. Next, we turn our attention to proving the Cauchy-Schwarz inequality in terms of (2.1).

Theorem 2.2. Let ak ;bk .1 k n/ be real numbers. Then Ä Ä n 2 n n 2 2 akbk a b ; (2.4) Ä k k k 1 ! k 1 ! k 1 ! XD XD XD with equality if and only if there is a t R such that ak tbk .1 k n/. 2 D Ä Ä

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2. A New Approach for Proving Inequalities 19

Proof. Since ak bk ak bk , withoutloss of generality, we may assume that ak;bk > Äj jj j 0. In view of the algebraic identity

2 2 2 bk bk a x ak px 2akbk; k C x D px C Â Ã we deﬁne 2 2 bk fk.x/ a x ; E .0; /: D k C x D C1

Now, fk.x/ has minimum value of 2akbk, which occurs at xkc bk=ak . Furthermore, D we ﬁnd that n n n 2 2 k 1 bk f .x/ fk.x/ a x D D D k C x k 1 k 1 ! P XD XD n 2 n 2 1=2 has a unique critical number at xc . b = a / , and f .x/ has minimum D k 1 k k 1 k value D D P n nP 1=2 2 2 f .xc / 2 a b : D k k k 1 k 1 ! XD XD Applying (2.1) yields n n n 1=2 2 2 ak bk a b ; Ä k k k 1 k 1 k 1 ! XD XD XD which is equivalent to (2.4). By Lemma 1, we have that equality holds in (2.4) if and only if all xkc xc. Setting t xc yields ak tbk for all 1 k n. D D D Ä Ä 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 Ä Ä n 1=2 n 1=2 n 1=2 2 2 2 .ak bk/ a b ; (2.5) C Ä k C k k 1 ! k 1 ! k 1 ! XD XD XD

with equality if and only if there is a t R such that ak tbk .1 k n/. 2 D Ä Ä

Proof. Since ak bk ak bk , without loss of generality, we may assume that C Äj jCj j ak;bk >0. Modifying the functions used in the proof of (2.4), we deﬁne

2 2 ak fk.x/ .ak bk/ x ; E .0; /: D C C x D 1

By completing the square, we ﬁnd that minE fk.x/ f .xkc/ 2ak.ak bk/, where f g D D C xkc ak=.ak bk/. Similarly, we obtain that D C n n n 2 2 k 1 ak f .x/ fk.x/ .ak bk/ x D D D C C x k 1 k 1 ! P XD XD

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n 2 1=2 n 2 1=2 has a minimum value at xc . k 1 ak / =. k 1 .ak bk/ / . In particular, D D D C P n 1=2 Pn 1=2 2 2 f .xc/ 2 a .ak bk/ : D k C k 1 ! k 1 ! XD XD Inequality (2.1) thus ensures

n n 1=2 n 1=2 2 2 2 ak .ak bk/ 2 a .ak bk/ : (2.6) C Ä k C k 1 k 1 ! k 1 ! XD XD XD Since n n n n 2 2 2 2 ak.ak bk/ a .ak bk/ b ; C D k C C k k 1 k 1 k 1 k 1 XD XD XD XD (2.6) becomes

n n n 1=2 n 1=2 n 2 2 2 2 2 a .ak bk/ 2 a .ak bk/ b : k C C k C Ä k k 1 k 1 k 1 ! k 1 ! k 1 XD XD XD XD XD Completing the square on the left-hand side, we have

2 n 1=2 n 1=2 n 2 2 2 .ak bk/ a b : 8 C k 9 Ä k k 1 ! k 1 ! k 1 < XD XD = XD Taking the square: root on both sides, we ﬁnally obtain ;

n 1=2 n 1=2 n 1=2 2 2 2 .ak bk/ a b ; C k Ä k k 1 ! k 1 ! k 1 ! XD XD XD 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 xc . Set t xc. Then all ak tbk . D D D 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 n n n pk = iD1 pi k 1 pkak ak Dn : P Ä pk k 1 P k 1 YD D p q Moreover, for p;q > 1;1=p 1=q 1, replacingPfk.x/ by akx =p bkx =q in the C D C proof of Theorem 2.2 leads to H¨older’s inequality

n n 1=p n 1=q p q ak bk a b : Ä k k k 1 k 1 ! k 1 ! XD XD XD Observe that inequality (2.1) can be extended to

n n n

inf fk.x/ inf fk.x/ fk.y/ (2.7) x E Ä x E Ä k 1 2 2 k 1 k 1 XD XD XD

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2. A New Approach for Proving Inequalities 21

for all y E. The relaxed right-hand side bound becomes more convenient for both prov- 2 ing and discovering inequalities. To demonstrate this, we present a new elementary proof of the Ky Fan inequality. n Theorem 2.4. Let ak .0;1=2;pk > 0.1 k n/ with k 1 pk 1. Deﬁne 2 Ä Ä D D n n n 1 P pk A pkak;G a ; H pk=ak D D k D k 1 k 1 k 1 ! XD YD XD and

n n n 1 pk A0 pk.1 ak /; G0 .1 ak / ; H 0 pk=.1 ak/ : D D D k 1 k 1 k 1 ! XD YD XD Then H G A : (2.8) H 0 Ä G0 Ä A0 with equality if and only if all ak’s are equal.

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 x 1 x 1 x fk.x/ pk ln ; E .0; 1=2: D ak 1 ak C x D Â Ã Since 1 1 p .1 2x/ f .x/ p ; f .x/ k > 0; k0 k k00 2 2 D ak .1 ak / x.1 x/ D x .1 x/ Â Ã fk.x/ is strictly convex and has a unique critical point at x ak E. In lightof (2.7), we D 2 have

pk 1 ak y 1 y 1 y fk.ak / ln fk .y/ pk ln D ak Ä D ak 1 ak C y Â Ã Â Ã and

n pk n n 1 ak pk pk 1 y ln y .1 y/ ln : ak Ä ak 1 ak C y k 1 Â Ã k 1 ! k 1 ! YD XD XD Thus, n pk 1 ak y 1 y 1 y ln ln : ak Ä H H C y k 1 Â Ã 0 YD Taking y H=.H H / E leads to D C 0 2 G H ln 0 ln 0 ; G Ä 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 onlyif all ak’s are equal. This completes the proof.

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

n 1=p n 1=p n 1=p p p p .ak bk/ a b : C Ä k C k k 1 ! k 1 ! k 1 ! XD XD XD

3. Let ak >0 for all 1 k n. Use (2.1) to prove that Ä Ä n n kD1 ak n k 1 ak ak D P a : n Ä k ÂP Ã k 1 YD

4. Let ak >0 and bk 0 for all 1 k n. Use (2.1) to prove that Ä Ä n n bk n kD1 bk bk k 1 bk nD P : ak ak k 1 Â Ã Â Pk 1 Ã YD D P 5. Monthly Problem 11189 [2005, 929; 2007, 645]. Let a1; a2;:::;an be positive. Let an 1 a1 and p>1. Prove that C D n ap p n p 1 n k a .a a /1=.p 1/: k p=.p 1/ k k 1 ak ak 1 2 2 C C k 1 C k 1 k 1 XD C XD XD 6. Under the conditions stated in Theorem 2.4, prove that G A : G0 Ä A0 7. An Interpolation Inequality. Under the conditions stated in Theorem 2.4, prove that A 1 A 0 : G Ä G G Ä G 0 C 0

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2. A New Approach for Proving Inequalities 23

8. Let ak .0;1=2.1 k n/, which do not all coincide. Deﬁne 2 Ä Ä n n 1=n 1 1 1 f .x/ x 1 ; x .0; 1=n/: D ak C ai ak 2 k 1 i 1 Â Ã ! YD XD Show that f .x/ is continuous, strictly decreasing, and therefore H G f .1=n/ f .x/ f .0/ ; x .0; 1=n/: 1 H D Ä Ä D G 2 0

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. Sandorand V.Szabo,On an inequality for the suminﬁmums 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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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/ .a b/=2 D C 2. the geometric mean: G.a; b/ pab D 3. the harmonic mean: H.a; b/ 2ab=.a b/ D C 4. the logarithmic mean: L.a; b/ .b a/=.ln b ln a/ D 5. the Heronian mean: N.a; b/ .a pab b/=3 D C C 6. the centroidal mean: T .a; b/ 2.a2 ab b2/=3.a b/ D C C C These are all positively homogeneous, in the sense that

M.a; b/ M.a; b/ for all >0; D and symmetric, in the sense that M.a; b/ M.b; a/. Moreover, all of the named means D satisfy a third property, called intermediacy, as well:

min.a; b/ M.a; b/ max.a; b/: Ä Ä 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/. Ä Ä 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

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26 Excursions in Classical Analysis

B¢ O B b a Figure 3.1. Pappus’s construction: OA A.a;b/;AB G.a;b/;BB H.a;b/ D D 0 D 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] b xt 1 dx f .t/ a C : (3.1) D b t R a x dx This encompasses all the means at the beginningR 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

3 f . 3/ H.a; b/; f . / G.a; b/; f . 1/ L.a; b/; D 2 D D

1 f . / N.a; b/; f .0/ A.a; b/; f .1/ T .a; b/: 2 D D D 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) Ä Ä Ä Ä Ä with equality if and only if a b. D

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3. Means Generated by an Integral 27

To prove that f .t/ is strictly increasing for 00.By the quotient rule,

b t 1 b t b t 1 b t a x C ln xdx a x dx a x C dx a x ln xdx f 0.t/ : (3.3) D b t 2 R R . a x dx/R R Thus, it sufﬁces to prove that the numeratorR for this quotient is positive. Since the bounds of the deﬁnite integrals are constants, the numerator for this quotient can be written as

b b b b xt 1 ln xdx xt dx xt 1 dx xt ln xdx C C Za Za Za Za b b b b xt 1 ln xdx yt dy yt 1 dx xt ln xdx D C C Za Za Za Za b b xt yt .x y/ ln x dxdy: D Za Za Substitutingin a different manner, we can write the same numerator as

b b b b xt 1 ln xdx xt dx xt 1 dx xt ln xdx C C Za Za Za Za b b b b t 1 t t 1 t y C ln y dy x dx x C dx y ln y dy D Za Za Za Za b b xt yt .y x/ ln y dxdy: D Za Za Averaging the two equivalent expressions shows that the numerator equals

1 b b xt yt .x y/.ln x ln y/dxdy > 0: 2 Za Za 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 positivityof the numerator starts with u u u u F.u/ xt 1 ln xdx xt dx xt 1 dx xt ln xdx: D C C Za Za Za Za Clearly F.a/ 0 and for u a, D u F .u/ ut xt .u x/.ln u ln x/dx 0; 0 D Za 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 ab.ln b ln a/ G2.a; b/ f . 2/ ; D b a D L.a; b/

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28 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:

G2.a; b/ H.a; b/ < G.a; b/: Ä L.a; b/

Furthermore, deﬁning the power mean by

ap bp 1=p Mp.a; b/ C ; D 2 Â Ã we get the equalities

M1.a; b/ A.a; b/; M0.a; b/ lim Mp.a; b/ G.a; b/; M 1.a; b/ H.a; b/: D D p 0 D D ! Observing that

1 M1=2.a; b/ G.a; b/ A.a; b/ ; D 2 C 1 N.a; b/ G.a; b/ 2A.a; b/ ; D 3 C we challenge the reader to choose values of t to show that

1 L.a; b/ < M1=3.a; b/ < 2G.a; b/ A.a; b/ < M1=2.a;b/ < N.a;b/ mean value theorem to ln x on Œa;b, we have

ln b ln a 1 b a D c or b a c L.a; b/; D ln b ln a D where a

F .c/ F.b/ F.a/ 0 ; G .c/ D G.b/ G.a/ 0 where a ˇ. Then D D 1=.˛ ˇ/ ˛ ˛ 1=.˛ ˇ/ b ˛ 1 ˇ.b a / x dx c.a;b ˛;ˇ/ a : (3.4) I D ˛.bˇ aˇ / D b xˇ 1 dx Â Ã Ra ! R

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3. Means Generated by an Integral 29

Since a < c.a;b ˛;ˇ/< b, and positively homogeneous and symmetry are also obvious, I it follows that c constitutes a continuous mean. Thus,

f .t/ c.a;b t 2; t 1/; D I C C is just one member of the general mean family of (3.4). In particular, setting ˇ 1 in D (3.4), we obtain the Stolarsky mean

˛ ˛ 1=.˛ 1/ b a S˛.a; b/ ; .˛ 0;1/ D ˛.b a/ ¤ Â Ã which is monotonic increasing in ˛. Note that

S 1.a; b/ G.a;b/;S2.a; b/ A.a; b/: D D The limiting cases give

S0.a; b/ lim S˛.a; b/ L.a; b/ D ˛ 0 D ! and 1 b a 1=.b a/ S1.a; b/ lim S˛ .a; b/ e .b =a / I.a; b/: D ˛ 1 D D ! 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/: Ä Ä It is also interesting to see that

I.a; b/ A.a; b/ L.a; b/ ln : G.a; b/ D L.a; b/ Â Ã Finally, to gain a more general perspective on the topic of means, we state a set of axioms, which, ifsatisﬁed by a class of functions,entitlethose 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/ F.b; a/, D 4. F.ta;tb/ tF.a;b/ for all t>0, D 5. a

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30 Excursions in Classical Analysis

where f .s/ is positive, continuous and strictly increasing for 0

~~1 s x.x/dx f .s/ f .s/ ; D D 1 R s .x/dx then f satisﬁes these conditions and R~~

b x.x=b/dx F.a; b/ bf .a=b/ a D D b R a .x=b/dx deﬁnes a mean. Moreover, if is positive continuousR on .0; 1 and = is strictly increasing, then f < f on .0; 1/. The means generated by (3.1) craft continuousanalogs 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 ef- fective 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 de- ﬁned in 0 x 1 prove that Ä Ä 1 xf 2.x/dx 1 f 2.x/dx 0 0 : 1 Ä 1 R 0 xf.x/dx R 0 f.x/dx R R 3. Putnam Problem 1979-B2. Let 0

1 1=t lim Œbx a.1 x/t dx : t 0 C ! Z0 4. Slope mean: Deﬁne S.a; b/ as the slope of the line which bisects the angle formed by the lines y ax and y bx. Find an explicit form for S.a; b/ and then prove D D that S.a; b/ is a mean and satisﬁes the inequality

~~H.a; b/ S.a; b/ A.a; b/: Ä Ä~~

~~5. We see that M2=3.a; b/ I.a; b/. Is it possible to ﬁnd 2=3 < t < 1 such that Ä I.a;b/ < Mt .a; b/? 6. For 1<˛<1=2 or 2<˛, show that~~

~~S˛ .a; b/ M.˛ 1/=3 .a; b/: Ä C~~

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~~3. Means Generated by an Integral 31~~

7. Show that 1 2 2 .2A.a; b/ G.a; b// I.a; b/ A.a; b/ 1 G.a; b/: 3 C Ä Ä e C e Â Ã 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, ˛A ˇG I > C > .A Gı /1=. ı/ > pAG: ˛ ˇ C C 9. Let agm.a; b/ be the Gauss arithmetic-geometric mean. Prove that A.a; b/ G.a; b/ L.a; b/ agm.a; b/ C N.a; b/: Ä Ä 2 Ä 10. Let 0

~~Figure 3.3. Areas in the ﬁgure~~

12. (Some series involving means).Let x .b a/=.b a/. Prove that D C 1 2k (a) ln.A=G/ k1 1 2k x , D D 1 2k (b) ln.I=G/ k1 1 2k 1 x , D P D C (c) ln.A=I / 1 x2k. Pk1 1 2k.2k 1/ D D C 13. Open Problem.P Notice that

~~1 ln I.a; b/ lnŒta .1 t/bdt: D C Z0 Using various numerical methods to approximate the integral enables us to obtain some new inequalities. For example,~~

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~~32 Excursions in Classical Analysis~~

~~(a) For a b, show that ¤ 1=3 A2.a; b/G.a; b/ < I.a;b/;~~

(b) For a b, show that ¤ 2 1 I 2.a; b/ < A2.a; b/ G2.a; b/: 3 C 3 Along these lines, one may discover more new interesting inequalities. 14. Open Problem. The logarithmic mean, somewhat surprisingly, has applications to the economic index analysis. Further applications, however, require a natural generalization to a logarithmic mean of n variables. Try to deﬁne a logarithmic mean L of n variables which satisﬁes the inequality G L A. Since the form of L.a; b/ does Ä Ä not suggest a reasonable generalization, you may start with

~~1 1 dx~~

~~L.a; b/ D 0 xb .1 x/a Z C as Stolarsky suggested in [5]. Similarly, try to deﬁne an identric mean I of n variables such that G L I A. Ä Ä Ä~~

References [1] R. L. Burden and J. D. Faires, Numerical Analysis, 8th edition, Brooks/Cole, Belmont, 2005. [2] H. Chen, Means generated by an integral, Math. Magazine, 78 (2005) 397–399. [3] H. Eves, Means appearing in geometric ﬁgures, Math. Magazine, 76 (2003) 292–294. [4] D. S.Mitrinovic, J.E. PecaricandA. E.Fink, Classical and New Inequalities in Analysis,Kluwer Academic Publishers, Boston, 1993. [5] K. Stolarsky, Generalizations of the logarithmic means, Math. Magazine, 48 (1975) 87-92.

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~~The L’H^opital Monotone Rule~~

In elementary calculus, we learn that if f .x/ is continuous and has a positive derivative on .a; b/ then f .x/ is increasing on .a; b/. However, if we are trying to prove the monotonicity of xp 1 f .x/ ; .p>q>0;x 1/ (4.1) D xq 1 in this way, we see the numerator of the derivative is

.p q/xp q 1 pxp 1 qxq 1; C C 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 Œa;b R be continuous functions W ! that are differentiable on .a; b/ with g 0 on .a; b/. If f =g is increasing (decreasing) 0 ¤ 0 0 on .a; b/, then the functions

~~f .x/ f .b/ f .x/ f .a/ and g.x/ g.b/ g.x/ g.a/ are likewise increasing (decreasing) on .a; b/.~~

Proof. We may assume that g .x/ > 0 and f .x/=g .x/ is increasing for all x 0 0 0 2 .a; b/. By the mean value theorem, for a given x .a; b/ there exists c .a; x/ such that 2 2 f .x/ f .a/ f .c/ f .x/ 0 0 ; g.x/ g.a/ D g .c/ Ä g .x/ 0 0 and so f .x/ g.x/ g.a/ g .x/ f .x/ f .a/ 0: 0 0 33

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~~34 Excursions in Classical Analysis~~

~~Therefore,~~

f .x/ f .a/ 0 f .x/.g.x/ g.a// g .x/.f .x/ f .a// 0 0 0: g.x/ g.a/ D .g.x/ g.a//2 Â Ã 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/ xp 1 and g.x/ xq 1. Consider D D f .x/=g.x/; if x 1; G.x/ ¤ D p=q; if x 1: D We have f .x/=g .x/ .p=q/xp q , which is increasing on .1; / as long as p>q>0. 0 0 D 1 Hence, by LMR, G.x/ is increasing on .1; /. In particular, G.x/ > G.1/ p=q. This 1 D yields the following bonus inequality:

~~xp 1 xq 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 2 tan x >2; .0~~

2 sin x tan x ; if x 0; F.x/ x C x ¤ D 2; if x 0: ( D Let f .x/ sin2 x x tan x; g.x/ x2. Now D C D 2 f 0.x/ 2 sin x cos x x sec x tan x D C C and f .x/ 00 cos2 x sin2 x sec2 x x tan x sec2 x: g00.x/ D C C

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~~4. The L’H^opital Monotone Rule 35~~

~~Since~~

~~f .x/ 0 x 1 00 3 tan x sec2 x.1 cos4 x/ sin x cos x 2x tan2 x sec2 x g .x/ D C sin x cos4 x C Â 00 Ã Â Ã > 0;~~

f .x/=g .x/ is increasing on .0; =2/. Thus, using LMR twice and noticing that f .0/ 00 00 D f .0/ g.0/ g .0/ 0, we deduce that f .x/=g .x/ and so F.x/ f .x/=g.x/ is 0 D D 0 D 0 0 D increasing on .0; =2/. Our third example is related to the well-known Jordan’s inequality: 2 x sin x x; .0 x /: Ä Ä Ä Ä 2 Now, by using LMR, we improve Jordan’s inequality as follows. If x Œ0; =2, then 2 2 x 2 x. 2x/ sin x 2 x 2 x. 2x/; C 2 Ä Ä C 2 2 x 1 x. 2 4x2/ sin x 2 x 2 x. 2 4x2/; C 3 Ä Ä C 2 2 x 1 x. 4 16x4/ sin x 2 x 2 x. 4 16x4/; C 25 Ä Ä C 5 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

1 sin x 2 2 x : (4.3) 2 5 Ä 4 16x4 Ä 5 4 5 Let f1.x/ sin x=x; g1.x/ 16x ; f2.x/ sin x x cos x; g2.x/ x . Then D D D D f .x/ 1 sin x x cos x 1 f .x/ 10 2 ; 5 g10 .x/ D 64 x D 64 g2.x/ f .x/ 1 sin x 20 : 3 g20 .x/ D 5 x 3 Since sin x=x 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 f .x/ f .=2/ sin x 1 1 x 2 4 4 g1.x/ g1.=2/ D 16x decreasing on .0; =2/. Thus, the desired inequality (4.3) follows from sin x 2 sin x 1 lim x 2 ; lim x 2 : x 0 4 16x4 D 5 x =2 4 16x4 D 2 5 ! ! Our ﬁnal example is MonthlyProblem 11369 [2], which was recently posed by Donald Knuth: Prove that for all real t, and all ˛ 2, e˛t e ˛t 2 .et e t /˛ 2˛: (4.4) C Ä C

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~~36 Excursions in Classical Analysis~~

First, observe that the inequality is invariant when t is replaced by t. Hence, it is sufficient to show the inequality holds for t 0. Next, set x et . Hence (4.4) is equivalent to D .x2 1/˛ x2˛ 1 C 2˛ 2: (4.5) x˛ To apply LMR, let f .x/ .x2 1/˛ x2˛ 1 and g.x/ x˛ . Then D C D f .x/ 2x.x2 1/˛ 1 2x2˛ 1 0 : C ˛ 1 g0.x/ D x Define ˛ 1 f .x/ 1 F.x/ 0 2x x 2x˛ : D g .x/ D C x 0 Â Ã To show that f .x/=g .x/ is increasing for x>1, it suffices to prove that F .x/ 0 for 0 0 0 x>1. Indeed, ˛ 1 1 2.˛ 1/ F .x/ 2 x ˛ 2˛x˛ 1: 0 D C x x2 1 Â Ã Ä C Using Bernoulli’s inequality .1 t/p 1 pt for t 0;p 1; C C which can be proved via LMR by considering .1 t/p =.1 pt/ as well, we have C C ˛ 1 ˛ 1 1 1 x x˛ 1 1 x˛ 1 .˛ 1/x˛ 3: C x D C x2 C Â Ã Â Ã Therefore, for x>1 and ˛ 2, 2.˛ 1/.˛ 2/x˛ 3.x2 1/ F .x/ 0: 0 x2 1 C Now, LMR deduces that f .x/ f .1/ .x2 1/˛ x2˛ 2˛ 1 C C g.x/ g.1/ D x˛ 1 is increasing for x>1. In particular, applying l’Hôpital’s rule yields f .x/ f .1/ .x2 1/˛ x2˛ 2˛ 1 lim lim C C 2˛ 2: x 1 g.x/ g.1/ D x 1 x˛ 1 D ! ! Thus, f .x/ f .1/ .x2 1/˛ x2˛ 2˛ 1 C C 2˛ 2: g.x/ g.1/ D x˛ 1 This implies the desired (4.5) and therefore proves (4.4).

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 x f .x/ x .1 cos.1=t//dt; g.x/ x: D C D Z0 Then f .x/=g.x/ is increasing on .0; /, whereas f .x/=g .x/ is not monotonicon .0; /. 1 0 0 1 This also shows that the converse of LMR is false.

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~~4. The L’H^opital Monotone Rule 37~~

~~Exercises 1. For x > 1, prove that x ln.1 x/ F.x/ C D x ln.1 x/ C is increasing.~~

~~2. Prove that f .x/ .1 x/1=x is decreasing but g.x/ .1 x/1 1=x is increasing D C D C C on .0; /. 1 3. For 0~~

sin x 3 (a) x > cos x. (b) 2x x >3. sin x C tan x Remark. Using t 2 2t 1 with t x= sinx in the second inequality yields D x 2 x > 2: sin x C tan x Á 4. Monthly Problem 11009 [2003, 341; 2005, 92–93]. Let a>b c>d>0. Prove that ax bx f .x/ D cx d x 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 1, otherwise dividethe ratio by d x. Now, let y cx;˛ D D D ln b= ln c; ˇ ln a= ln c. Then ˇ>˛ 1. One can rewrite f .x/ as D yˇ y˛ F.y/ y .0; 1/ .1; /: WD y 1 2 [ 1 5. In Wilker’s inequality, prove that

~~2 sin x tan x 2 16 x C x ; .0~~

~~2 2 2 2 1 x x. 2x/ cos x 1 x x. 2x/: C 2 Ä Ä C 2~~

~~7. For 0~~

~~4 x x x < tan < : 1 x2 2 2 1 x2 Á~~

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~~38 Excursions in Classical Analysis~~

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; /. Ä Ä 1 Prove that n x 2ai n x ai i i > 2: sin xi C tan xi i 1 Â Ã i 1 Â Ã YD YD 9. For =n x =n, prove that Ä Ä sin nx 1 : n sin x 3

~~10. Prove that~~

~~n sin kx (a) k 1 k > 0; for all 0~~

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~~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 1 k n sin : (5.1) n D 2n 1 k 1 Â Ã YD However, the proofs for these identitiesare widely scattered throughoutbooks 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 identitiesin [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 1 k .n 1/.n 1/ csc2 C : (5.2) n D 3 k 1 Â Ã XD 5.1 A Primer of complex numbers Historically, complex numbers were ﬁrst introduced by Cardano in the course of solving quadratic and cubic equations. He noticed that if the “new numbers” 1 p 1 were ˙ treated as ordinary numbers with the added rule that p 1 p 1 1, they did solve the D equation x2 2x 2 0. However, for a long period of time, it was felt that no meaning C C D could actually be assigned to i p 1 and it was therefore termed “imaginary”. In 1748, D Euler discovered the amazing identity ei cos i sin : (5.3) D C 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.

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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 nine- teenth 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 x i y and the positive real axis, where 0 < 2. The angle is D C Ä called the argument and is denoted arg z: Thus, we have z in a polar coordinate D representation: z r.cos i sin /; (5.4) D C where r z x2 y2; tan y=x: Dj jD C D (5.4) gives us a geometric method forp performing complex multiplication:for any complex numbers z1 and z2, z1z2 z1 z2 j jDj jj j and arg.z1z2/ arg z1 arg z2 .mod 2/: (5.5) D C In general, this leads to the well-known De Moivre’s formula: for a positive integer n,

~~.cos i sin /n cos n i sin n: (5.6) C D C Now, appealing to Euler’s formula (5.3), we get the following two popular identities:~~

~~1 eix 1 .eix=2/2 2 cos2.x=2/ 2i sin.x=2/ cos.x=2/ 2 cos.x=2/eix=2 (5.7) C D C D C D and~~

1 eix 1 .eix=2/2 2 sin2.x=2/ 2i sin.x=2/ cos.x=2/ 2i sin.x=2/eix=2: (5.8) D D D 5.1.2 Roots of Unity The solutions of the equation zn 1 0 are called the nth roots of unity. De Moivre’s D formula can be reversed to ﬁnd all roots of unity, namely

1; !; !2; :::; !n 1 ; f g where ! ei.2=n/ is called the nth primitive root of unity. Geometrically, these numbers D are the vertices of a regular n-gon inscribed in the unit circle z 1, and have many nice j jD properties that interconnect algebra, geometry, combinatorics and number theory. Since

~~zn 1 .z 1/.zn 1 zn 2 1/; (5.9) D C CC~~

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clearly, the factor z 1 accounts for the root 1, and so the other roots satisfy zn 1 zn 2 1 0; (5.10) C CC D which yields zn 1 zn 2 1 .z !/ .z !n 1 /: (5.11) C CC D 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 1 and then taking modulus gives D n 1 n 1 !k : (5.12) D j j k 1 YD 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), k 1 !k 1 ei.2k=n/ 2i sin eik=n; D D n Â Ã we see that (5.1) follows from (5.12) immediately. Next, for z 1, (5.11) gives D n 1 1 Œ1 . 1/n . 1/n 1 .1 !k /: 2 D C k 1 YD By (5.7), we have k 1 !k 1 ei.2k=n/ 2 cos eik=n; C D C D n Â Ã and so n 1 n k 1 . 1/ cos : (5.13) n D 2n k 1 ˇ Â Ãˇ YD ˇ ˇ Since ˇ ˇ k .n ˇ k/ ˇ k .n k/ sin sin ; cos cos ; n D n n D n Â Ã Â Ã Â Ã Â Ã appealing to (5.12) and (5.13), we ﬁnd that

~~n 1 k pn sin ; 2n D 2n 1 k 1 Â Ã YD n k p2n 1 sin C ; 2n 1 D 2n k 1 Â Ã YD C n k 1 cos : 2n 1 D 2n k 1 Â Ã YD C~~

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~~Next, notice that~~

n 1 n 1 k 1 sin x ei.x k=n/ e i.x k=n/ C n D 2i C C k 0 Â Ã k 0 YD YD Á n 1 1 eik=n ix e2xi ei. 2k=n/ D .2i/n k 0 YD Á n 1 1 ei.n 1/=2 inx e2xi ei. 2k=n/ : D .2i/n k 0 YD Á Since ! k !n k , setting z e2xi in (5.9) we have D D n 1 e2xi ei. 2k=n/ e2nxi 1; D k 0 YD Á so that

n 1 k 1 1 sin x .ei=2/n 1.enxi e nxi / sin nx C n D .2i/n D 2n 1 k 0 Â Ã YD or equivalently n 1 k sin nx 2n 1 sin x : (5.14) D C n k 0 Â Ã YD Replacing x by x =2 in (5.14) yields C n 1 n=2 n 1 k sin nx . 1/ 2 cos x ; when n is even; D C n k 0 Â Ã YD n 1 k cos nx . 1/.n 1/=22n 1 cos x ; when n is odd: D C n k 0 Â Ã YD Finally, notice that the equation z2n 1 0 has 2n roots evenly spaced in angle around D the unit circle, with a separation of =n radians. Setting exp.i=n/, the factorization D of z2n 1 yields n 1 z2n 1 .z k/: D k n YD Combining the conjugate factors gives

~~n 1 k z2n 1 .z2 1/ z2 2z cos 1 : D n C k 1 Â Â Ã Ã YD Dividing by zn on both sides yields~~

~~n 1 k zn z n .z z 1/ z 2 cos z 1 : D n C k 1 Â Â Ã Ã YD~~

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Setting z eix and appealing to De Moivre’s formula, we obtain D n 1 sin nx k 2n 1 cos x cos : (5.15) sin x D n k 1 Â Â ÃÃ YD Similarly, using the factorizations of z2n 1 1 and z2n 1 respectively, we can establish C C other product identities:

n 2 2k 2n 1 .z 1/ z 2z cos 1 z C 1; 2n 1 C D k 1 Â Â Ã Ã YD C n 2k .z 1/ z2 2z cos 1 z2n 1 1; C C 2n 1 C D C k 1 Â Â Ã Ã YD C n 1 .2k 1/ z2 2z cos C 1 z2n 1: 2n C D C k 0 Â Â Ã Ã YD 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 !n k , we see that D 1; ! 1; ! 2;:::;! .n 1/ f g are also roots of zn 1. Thus, n 1 1 zn .1 z!k /; (5.16) D k 0 YD from which the partial fraction decomposition of 1 zn has the form of n 1 1 A k ; 1 zn D 1 z!k k 0 XD n where Ak are constants to be determined. Multiplying by 1 z on both sides yields n 1 n 1 z Ak 1: 1 z!k D k 0 XD Since 1 zn 0; if i k; lim i ¤ z !k 1 z! D n; if i k; ! D we obtain Ak 1=n and so D n 1 1 1 1 1 zn D n 1 z!k k 0 XD

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or equivalently n 1 n 1 : (5.17) 1 zn D 1 z!k k 0 XD In view of the identity 1 a ib a ib ; a ib D .a ib/.a ib/ D a2 b2 C C C setting z xei˛ in (5.17) and then extracting the real part gives D n n 1 n.1 x cos n˛/ 1 x cos.˛ 2k=n/ C : (5.18) 1 2xn cos n˛ x2n D 1 2x cos.˛ 2k=n/ x2 k 0 C XD C C Replacing x by 1=x in (5.18) yields

~~n n n 1 2 nx .x cos n˛/ x x cos.˛ 2k=n/ C : (5.19) 1 2xn cos n˛ x2n D 1 2x cos.˛ 2k=n/ x2 k 0 C XD C C Now, subtracting (5.19) from (5.18) we get~~

~~2n n 1 2 n.1 x / 1 x : (5.20) 1 2xn cos n˛ x2n D 1 2x cos.˛ 2k=n/ x2 k 0 C XD C C Alternately, adding (5.19) and (5.20) we have~~

n n 1 n.1 x cos n˛/ 1 x cos.˛ 2k=n/ C : (5.21) 1 2xn cos n˛ x2n D 1 2x cos.˛ 2k=n/ x2 k 0 C XD C C For ˛ 0, we may specify (5.20) and (5.21), respectively, as D n 1 n 1 n.1 x / C ; (5.22) 1 2x cos.2k=n/ x2 D .1 xn/.1 x2/ k 0 XD C and n 1 1 x cos.2k=n/ n : (5.23) 1 2x cos.2k=n/ x2 D 1 xn k 0 XD C 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

~~n 1 k lnsin nx .n 1/ ln 2 lnsin x ; D C C n k 0 Â Ã XD and then differentiate, the result is seen to be~~

~~n 1 k cot x n cot nx: (5.24) C n D k 0 Â Ã XD~~

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~~Replacing x by x =2 in (5.24) gives C n 1 k n 1 tan x n tan nx : (5.25) C n D C 2 k 0 Â Ã Â Ã XD Differentiating (5.24) yields~~

~~n 1 k csc2 x n2 csc2 nx: C n D k 0 Â Ã XD This establishes (5.2) as follows:~~

~~n 1 k .n 1/.n 1/ csc2 lim .n2 csc2 nx csc2 x/ C : n D x 0 D 3 k 1 Â Ã ! XD Similarly, differentiating (5.25) gives~~

~~n 1 k n 1 sec2 x n2 sec2 nx : C n D C 2 k 0 Â Ã Â Ã XD 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~~

2 1 x sin x x 1 : (5.26) D k2 2 k 1 Â Ã YD 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 Pn.sin x/; D where Pn.t/ is an nth degree polynomial. 3 For n 3, we see the required polynomial is just P3.t/ 3t 4t since sin 3x D D 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 i sin mx .cos x i sin x/m C D C and take its imaginary part, which is

~~m m 1 m 2 m 3 m 4 m 5 sin mx sin x cos x sin x cos x sin x cos x : D 1 3 C 5 C ÂÂ Ã Â Ã Â Ã Ã~~

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In particular, if m 2n 1, appealing to cos2 x 1 sin2 x, we arrive at D C D .2n 1/2 12 sin.2n 1/x .2n 1/ sin x 1 C sin2 x C D C 3Š Â Œ.2n 1/2 12Œ.2n 1/2 32 C C sin4 x : C 5Š C Ã That is, sin.2n 1/x .2n 1/ sin x P.sin2 x/; (5.27) C D C where P.t/ is an nth degree polynomialwith P.0/ 1.Note that if r1;r2;:::;rn are roots D of P.t/, then in factored form

n t P.t/ 1 : D rk k 1 Â Ã YD This is self evident, since substituting t 0 gives P.0/ 1, just as substituting t rk D D D yields P.rk/ 0 for k 1;2;:::;n. D D Next, due to (5.27), if x is a solution of sin.2n 1/x 0, but sin x 0, then sin2 x C D ¤ is a solution of P.t/ 0. Consider D k x ; for k 1;2;:::;n: D 2n 1 D C For each of these values we have .2n 1/x k, and thus sin.2n 1/x 0, while C D C D 0

~~2 2 2 r1 sin .=.2n 1//; r2 sin .2=.2n 1//;:::;rn sin .n=.2n 1//: D C D C D C Thus, in factored form,~~

n sin2 x sin.2n 1/x .2n 1/ sin x 1 : C D C sin2.k=.2n 1// k 1 ! YD C Replacing x by x=.2n 1/ yields C x n sin2.x=.2n 1// sin x .2n 1/ sin 1 C ; (5.28) D C 2n 1 sin2.k=.2n 1// Â Ã k 1 ! C YD C 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 . To avoid using such an advanced result, ! 1 we rewrite sin x as An Bn , where .K 1/ > x ;n>K and K K C j j x K sin2.x=.2n 1// An .2n 1/ sin 1 C ; K D C 2n 1 sin2.k=.2n 1// Â Ã k 1 ! C YD C n sin2.x=.2n 1// Bn 1 C : K D sin2.k=.2n 1// k K 1 ! DYC C

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For ﬁxed K, 1 k K, we have Ä Ä x lim .2n 1/ sin x n C 2n 1 D !1 Â C Ã and sin2.x=.2n 1// x2 lim C : n sin2.k=.2n 1// D k2 2 !1 C Thus, K 2 n x def lim A x 1 AK : n K D k2 2 D !1 k 1 Â Ã n n YD n Since BK sin x=AK , we see that limn BK BK exists. To estimate BK , using the D !1 D well-known inequality 2 < sin <; for 0< <=2; we have x x2 sin2 < 2n 1 .2n 1/2 C C and k 4 k2 2 sin2 > .k K 1; : : : ; n/: 2n 1 2 .2n 1/2 D C C C Thus, sin2.x=.2n 1// x2 1 C >1 sin2.k=.2n 1// 4k2 C and n x2 1>Bn > 1 : K 4k2 k K 1 Â Ã DYC Since x2=.4k2/ converges, we see that the inﬁnite product

~~2 P 1 x 1 4k2 k K 1 Â Ã DYC converges to 1 as K goes to inﬁnity and so~~

lim BK 1: K D !1 Thus, we finally establish Euler’s infinite product (5.26) by 2 1 x sin x lim AK BK x 1 : D K D k2 2 !1 k 1 Â Ã YD As a corollary, from cos x sin 2x=.2 sin x/, we find that D 2 1 4x cos x 1 : (5.29) D .2k 1/2 2 k 1 Â Ã YD Moreover, for x =2, (5.26) leads to the famous Wallis’s formula, namely D 2 1 1 1 3 3 5 5 7 7 1 : (5.30) D 4k2 D 2 2 4 4 6 6 8 k 1 Â Ã YD

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~~5.5 Sums of inverse tangents The evaluation of inverse tangent sums of the form~~

1 arctan f .k/ k 1 XD 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 principleof the argument for complex multiplication provides an equally simple method. n Let zk ak ibk .1 k n/ with ak;bk R. Consider Pn zk . Appealing D C Ä Ä 2 D k 1 to (5.5), we have D Q n n bk arg.Pn/ arg.zk / arctan .mod 2/: (5.31) D D ak k 1 k 1 Â Ã XD XD

Similarly, for a convergent inﬁnite product P k1 1 zk ; D D 1 bQ arg.P / arctan k .mod 2/: (5.32) D ak k 1 Â Ã XD 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 n n j .x rk/ aj x ; (5.33) C D k 1 j 0 YD XD show that n a1 a3 a5 arctan rk arctan C : D a0 a2 a4 k 1 XD C 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 1=i, (5.33) becomes D n n j n .1 irk/=i aj i ; C D k 1 j 0 YD XD which is equivalent to

~~n n j .1 irk/ aj i .a0 a2 a4 / i .a1 a3 a5 /: C D D C C C k 1 j 0 YD XD 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~~

~~2 1 z sin z z 1 : (5.34) D k2 k 1 Â Ã YD For z x i y, D C sin z sin x cosh y i cos x sinh y: D C Appealing to (5.32), we recover an elegant formula in [3] by Glasser and Klamkin, namely~~

1 2xy y tanh y arctan arctan tan 1 .mod 2/; k2 x2 y2 D x tan x k 1 Â Ã Â Ã XD C Á which, for x y 1=p2, becomes D D 1 1 tanh =p2 arctan arctan .mod 2/: (5.35) k2 D 4 tan =p2 k 1 Â Ã ! XD 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 2 d I ; . x < 1/ 2 D 0 1 2x cos x j j Z C via the limitof a Riemann sum. This integral is usuallyevaluated by using either the residue theorem or Fourier series. Since 1 2x cos x2 .1 x /2 >0 for x < 1; C j j j j the integrand is continuous and integrable. Partition the interval Œ0;2 into n equal subin- tervals by the partition points 2k k 1 k n : D n W Ä Ä By (5.22), we get

~~n 1 2 1 I lim D n n 0 2k 2 1 !1 k 0 1 2x cos n x XD C 2 @ n.1 xn/ Á A lim C D n n .1 xn/.1 x2/ !1 2 : D 1 x2~~

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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 n 1 n.1 x / 1 nC 2 2 ; 2k 2 D .1 x /.1 x / .1 x/ k 1 1 2x cos n x XD C Á and letting x 1, we re-prove (5.2) as ! n 1 n k n.1 x / 1 .n 1/.n 1/ csc2 4 lim C C ; n D x 1 .1 xn/.1 x2/ .1 x/2 D 3 k 1 Â Ã ! Â Ã XD where l’Höpital’s rule has been used to calculate the limit. Similarly, we assume that n is odd first. Setting ˛ in (5.20), we have D n 1 2n 1 n.1 x / : 2k 2 D .1 xn/2.1 x2/ k 0 1 2x cos n x XD C C C Á Letting x 1, we obtain ! n 1 2n k n.1 x / sec2 4 lim n2; (5.36) n D x 1 .1 xn/2.1 x2/ D k 0 Â Ã ! XD C where l’Höpital’s rule has been used again. If n is even, setting ˛ in (5.20) and then regrouping yields D n 1 2n 1 n.1 x / 1 : 2k 2 D .1 xn/2.1 x2/ .1 x/2 k 0;k n=2 1 2x cos n x D X¤ C C C Á Letting x 1, we have ! n 1 k .n 1/.n 1/ sec2 C : (5.37) n D 3 k 0;k n=2 Â Ã D X¤ Appealing to the trigonometric identities 1 tan2 x sec2 x and 1 cot2 x csc2 x, by C D C D (5.35)–(5.37), we find that

n 1 k .n 1/.n 2/ cot2 n D 3 k 1 Â Ã XD and n 1 2 k n.n 1/; if n is odd; tan .n 1/.n 2/ (5.38) n D if n is even, k n=2. k 0 Â Ã ( 3 XD ¤ If we replace the power 2 in these identitiesby 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 ! ei.2=n/ and n be a positive integer. Prove that D n 1 k 2k n (a) k 1 1 ! n . D j j D k n 1 k 1 n.n 1/=2 Q C n=2 (b) k 1 sin n 2 n . D D ÁÁk Qn 1 k 1 n.n 1/=2 C n n=2 (c) k 1 cos n 2 Œ1 . 1/ . D D ˇ Áˇ Q ˇ ˇ 2. Prove that theˇ sum of theˇ 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 n r k sin.kx/ and r k cos.kx/: k 1 k 0 XD XD Find their limits as n if r <1. ! 1 j j 4. Let n 3. Find n 1 .1 2 cos.k=n// : C k 1 YD~~

~~5. Deﬁne Pn.x/ by~~

~~n 2 n 2k 2.n k/ Pn.x/ x .1 x/ : D k k 0 Â Ã XD Show that for 0 x 1, Ä Ä 1 2n Pn.x/ : 22n n Â Ã 6. Prove that~~

~~n 1 .2k 1/ 1 n 1 (a) k 1 sin 2nC 2 . D D Qn 1 .3k 1/ Á p3 (b) k 1 sin 3nC 2n . D D Á 7. ProveQ that~~

~~sin x x (a) 1 cos : x k 1 2k D D sin x Q 2 Á 2 k 1 k m (b) x cos .x=2/ k1 1 sin .x=2 C / m 1 cos.x=2 /. D C D D 8. Putnam Problem 1949-A6.P For every x, proveQ that~~

~~n sin x 1 1 2 cos.2x=3 / C : x D 3 n 1 YD~~

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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,

~~sin x 1 1 x 3x 5x 2 cos 2 cos 2 cos : x D 6 6k C 6k C 6k k 1 Â Ã YD The details and further examples can be found in [7]. 9. Monthly Problem 11383 [2008, 757]. Show that~~

~~1 1 pn2 2npn2 4n 3 arccos C C C C : .n 1/.n 2/ D 3 n 1 ! XD C C Remark. The answer seems to be =6. 10. Monthly Problem 5486 [1967, 447; 1968, 421–422]. Prove that~~

~~n .2k 1/ 2n k 4 csc2 C n2 and csc2 n.n 1/: 2n D 2n 1 D 3 C k 0 Â Ã k 1 Â Ã XD XD C 11. Prove that~~

~~2 2 1 x 1 4x sinh x x 1 ; cosh x 1 : D C n2 2 D C .2n 1/2 2 n 1 Â Ã n 1 Â Ã YD YD Use the latter result to prove that~~

~~n 1 5 cosh./ lim n 1 : n k C 4k2 D !1 k 1 Â Ã YD 12. Prove that~~

1 2xy x y arctan arctan tan tanh .mod 2/: .2k 1/2 x2 y2 D 2 2 k 1 Â Ã XD C Á 13. Prove that 1 x x . 1/k arctan arctan tanh ; 2n 1 D 4 k 1 Â Ã XD C Á where it is assumed that =2 arctan t =2. Ä Ä 14. Putnam Problem 1986-A3. Evaluate

~~1 arccot.k2 k 1/; C C k 1 XD where arccotx for x 0 denotes the number in the interval 0< =2 with Ä cot x. D~~

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~~15. Prove that~~

n 1 k n 1 tan2 x n2 sec2 nx n: C n D C 2 k 0 Â Ã Â Ã XD In general, let n 1 m k Sm.x/ tan x : D C n k 0 Â Ã XD Prove that 1 Sm 1.x/ Sm0 .x/ Sm 1; for m 2;3;:::: C D m D 16. In view of (5.22), deﬁne

~~n 1 2 1 x Fp .x/ : D .1 2x cos.2k=n/ x2/p k 0 XD C If n 0 .mod 4/, show that ¤ n 1 p p 1 p sec .2k=n/ 2 . i/ .Fp .i//: (5.39) D k 0 XD Moreover, prove the following recursive relation~~

~~Fp .x/ x d Fp.x/ F1 p .x/ : C D 1 x2 C p dx 1 x2 Â Ã 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 .2k 1/ .2k 1/ 2n.n 1/ sec2 C 2n2; sec2 C C : 4n D 4n 2 D 3 k 0 Â Ã k 0 Â Ã XD XD C~~

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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~~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 functionsprovideone 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 chan- nels: the discrete and the continuous.” He further observes that “A generating function is a clotheslineon 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 a0; a1; a2;::: f g is the power series 2 G.x/ a0 a1x a2x : D C C C Now it is clear that the unknown numbers an are arranged neatly on the clothesline: n indexing from 0, an is the coefficient of x in the generating function. Therefore, once

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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 x <1, j j 1 1 x x2 x3 : C C C CD 1 x This indicates that the generating function of the sequence 1;1;1;::: is 1=.1 x/. On f g the other hand, if G.x/ ln.1 x/, we have D G.n/.0/ 1 : nŠ D n Therefore ln.1 x/ generates the sequence 1;1=2;1=3;::: . f g Sometimes a sequence an has a generating function whose properties are quite com- f g plicated, while the related sequence an=nŠ has a generating function that is quite simple. f g This leads to beneath definition:

~~Deﬁnition 2 The exponentialgenerating functionfor the inﬁnitesequence a0; a1; a2;::: f g is the power series~~

a1 a2 2 a3 3 G.x/ a0 x x x : D C 1Š C 2Š C 3Š C Thus the series for ex is the exponential generating function for 1;1;1;::: , and the f g series for cos x and sin x, for the periodic sequences 1;0; 1;0::: and 0;1;0; 1;::: , f g f g respectively. The magic of generating functions is that we can carry out all sorts of manipulationson 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 Fn 2 Fn Fn 1 for n 1; (6.1) C D C C with F1 F2 1 are revealed, where Fn denotes the nth Fibonacci number. D D

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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

~~1 n G.x/ Fn x : D n 1 XD In view of (5.1), then~~

2 3 4 5 G.x/ x x F3x F4x F5x D C C C C C 2 3 4 5 x x .F1 F2/x .F2 F3/x .F3 F4/x D C C C C C C C C 2 3 2 2 3 x x.F1x F2x F3x / x .F1x F2x F3x / D C C C C C C C C x xG.x/ x2G.x/ D C C and so x G.x/ : D 1 x x2 To determine the coefﬁcient of xn in G.x/, we factor

1 x x2 .1 ˛x/.1 ˇx/; D where ˛ .1 p5/=2; ˇ .1 p5/=2. Using the method of partial fractions and the D C D geometric series expansion yields x G.x/ D .1 ˛x/.1 ˇx/ 1 1 1 D p5 1 ˛x 1 ˇx Â Ã 1 1 .˛n ˇn/xn: D p 5 k 1 XD Thus, we obtain Binet’s formula

~~1 n n Fn .˛ ˇ / (6.2) D p5~~

~~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 Fn n Ge.x/ x : D nŠ k 1 XD Differentiating Ge.x/ successively gives~~

~~F3 2 F4 3 F5 4 F6 5 Ge .x/ F1 F2x x x x x ; 0 D C C 2Š C 3Š C 4Š C 5Š C~~

~~F4 2 F5 3 F6 4 F7 5 Ge 00.x/ F2 F3x x x x x : D C C 2Š C 3Š C 4Š C 7Š C~~

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

~~Ge .x/ Ge .x/ Ge.x/ 0: 00 0 D~~

This, together with Ge.0/ 0;Ge .0/ F1 1, yields D 0 D D 1 ˛x ˇx Ge.x/ .e e /: D p5 Finally, the formula (6.2) is an immediate consequence of the power series expansion of eax. In light of (6.2), Fibonacci numbers grow very fast. In fact, F45 1;134;903;170. That D 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:

Fn 1 1=n lim C lim .Fn/ ; (6.3) n F D n D !1 n !1 where .1 p5/=2 1:618034::: was called the golden ratio by the ancient D C D Greeks. Recently, Viswanath in [8] took a fresh look at Fibonacci numbers by studying the Fibonacci-like recurrence:

F n 2 Fn 1 Fn; F0 0; F1 1; C D˙ C ˙ D D where each sign is chosen randomly with probability 1=2. He unexpectedly found that ˙ 1=n lim . Fn / 1:13198824:::: n j j D !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 surprisinglymany 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 Hn are deﬁned by f g 1 1 1 Hn 1 : (6.4) D C 2 C 3 CC n~~

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~~These numbers satisfy the recurrence relation~~

~~1 Hn Hn 1 ; .for n>1/ D C n~~

~~and H1 1. Hence D~~

1 n 1 n Hnx x Hnx D C n 1 n 2 XD XD 1 1 n x Hn 1 x D C C n n 2 Â Ã XD n 1 n 1 x x Hnx : D C n n 1 ! n 1 XD XD Since it is valid to interchange the order of summation and integration for a convergent power series, for x <1, j j x x x 1 1 1 1 dt xn t n 1dt . t n 1/dt ln.1 x/: n D D D 1 t D n 1 n 1 Z0 Z0 n 1 Z0 XD XD XD

~~Thus, we ﬁnd that the generating function of Hn is f g~~

1 n ln.1 x/ Hnx : (6.5) D 1 x n 1 XD Unlike the Fibonacci numbers, a closed form of Hn for general n does not exist. More- over, even as limn 1=n 0, Hn itself is divergent! Modern calculus textbooks prove !1 D 1 this fact either by establishing the estimates H2n 1 n or by comparing Hn with C 2 11 dx=x. But the following argument is equally simple and insightful. Suppose that limn Hn S. Then R !1 D 1 1 1 1 1 1 1 1 S 2 C 4 CC 2n CD 2 C 2 CC n C D 2 Â Ã and so the sum of the odd numbered terms, 1 1 1 C 3 CC 2n 1 C C must be the other half of S. But this is impossible since

~~1 1 > 2n 1 2n for every positive integer n. On the other hand, deﬁne the alternate signed harmonic numbers by~~

~~n 1 1 1 . 1/ C H 0 1 : n D 2 C 3 CC n~~

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H2n0 exhibits the following particular beautiful form — Catalan’s identity: 1 1 1 1 H 0 1 2n D 2 C 3 C 2n 1 2n 1 1 1 H2n 2 D 2 C 4 CC 2n Â Ã 1 1 1 H2n Hn ; D D n 1 C n 2 CC 2n C C and 1 n 1 1 dx lim H2n0 lim ln 2: n D n n 1 k=n D 0 1 x D !1 !1 k 1 Z XD C C Now we study the asymptotic behavior of Hn as n . Deﬁne ! 1 1 n 1 an ln C : D n n Â Ã Since 1 1 1 1 xdx an dx ; D n n x D n.n x/ Z0 Â C Ã Z0 C a simple estimation on the integral yields 1 dx 1 0 < an < n2 D n2 Z0

~~and so n1 1 an converges by the series comparison test. Therefore, D n P n 1 1 lim .Hn ln n/ lim ak ln C an (6.6) n D n C n D !1 !1 k 1 Â Ã! n 1 XD XD exists. Euler denoted this limit as , namely~~

lim .Hn ln n/ 0:5772156649:::; (6.7) D n D !1 where is called Euler’s constant. A careful analysis of the series remainder in (6.6) will yield nice estimates on . Indeed, a closed form of the remainder can be established and various asymptotic estimates for the remainder, which imply the well-known Young and De Temple inequalities, are then derived. See Exercise 9. Although there is compelling evidence that is not rational, a proof has not yet been found. Next, due to the symmetry, we have

~~n n k Hk 1 k D i k k 1 k 1 i 1 XD XD XD 1 1 1 D 2 0 i k C i k 1 1 i k n i k ÄXÄ Ä XD 1 @ n 1 A H 2 : D 2 n C k2 k 1 ! XD~~

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This elegant identity is a special case of what today is known as an “Euler sum”, which we will explore more rigorously in Chapter 17. Finally, we turn tothe linkbetween the Riemann hypothesis and the harmonic numbers. It is well known that the distribution of prime numbers among all positive integers does not display any regular pattern. However, in 1859, Riemann observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function 1 1 .z/ 1 D C 2z C 3z C called the Riemann zeta function. In an eight page paper Riemann made his bold conjecture about the zeros of his Zeta function: except for negative even integers, all other zeros of .z/ lie on the vertical line Re.z/ 1=2. Using computers, mathematicians have man- D aged to show that Riemann’s hypothesis is true for the ﬁrst 1.5 billion zeros of the zeta function. A proof of the Riemann hypothesis will shed light on many of the mysteries surrounding the distribution of prime numbers, and satisfy the curiosity of many generations of mathematicians. It will also have major consequence for what has come to be a crucial component of modern life — internet security. In 2000, the Clay Mathematics Institute (claymath.org) offered a $1 million prize for proof of the Riemann hypothesis. Re- cently, Jeffrey Lagarias in [3] made a surprising observation: the Riemann hypothesis is equivalent to the following elementary statement: for all n 1, Hn .n/ Hn ln.Hn/e ; Ä C with equality only for n 1, where .n/ d denotes the sum of the divisors of n; D D d n for example, .6/ 12. A weaker version of thisj inequality appears in Exercise 10. D P 6.4 Bernoulli Numbers There are many methods for introducing the Bernoulli numbers. The most versatile way was conceived by Euler. Instead of using recurrences, as we did in the previous sections, he observed that the Bernoulli numbers, Bn, possess an exponential generating function

x 1 B n xn: (6.8) ex 1 D nŠ n 0 XD Rewrite the left-hand side in the alternative form x xe x=2 x cosh x sinh x x x x 2 2 coth : (6.9) ex 1 D ex=2 e x=2 D 2 sinh x D 2 2 2 2 Notice that .x=2/ coth.x=2/ is an even function. Therefore we ﬁnd that all but one of the Bernoulli numbers of odd terms are zero: 1 B1 ;B2k 1 0; k 1;2;3;:::: (6.10) D 2 C D D

~~In general, to obtain the Bn’s, we invoke the Cauchy product~~

~~B2 2 B3 3 1 2 1 3 B0 B1x x x x x x x: C C 2Š C 3Š C C 2Š C 3Š C D Â Ã Â Ã~~

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~~Recalling that the Cauchy product of two convergent power series is given by~~

~~1 i 1 j 1 k ai x bj x ai bj x ; (6.11) 0 1 D 0 1 i 0 ! j 0 k 0 i j k XD XD XD CXD @ A @ A we obtain B0 1 and the recursive formula D~~

B0 1 B1 1 Bn 1 1 0; n 2: 1Š nŠ C 1Š .n 1/Š CC .n 1/Š 1Š D Multiplying through by nŠ and then appealing to the binomial coefﬁcients yields a briefer form, n 1 n Bk 0; .for n 2/: (6.12) k D k 0 Â Ã XD k If B is interpreted as the Bernoulli number Bk, in view of the binomial theorem, (6.12) is abbreviated to .1 B/n Bn: C D By taking n 2;4;6;8 in (6.12), we ﬁnd that D 1 1 1 1 B2 ;B4 ;B6 ;B8 : D 6 D 30 D 42 D 30

Historically, Bernoulli numbers and polynomials were introduced in Jacob Bernoulli’s book Ars Conjectandi (The Art of Conjecturing).To ﬁnd the area under the curve y xk D from 0 to 1, it comes down to evaluating

k k k Sk.n/ 1 2 .n 1/ D C CC as Archimedes did in the case of k 2. Jacob Bernoulli gave an elegant description of D Sk.n/. In the course of doing this, he discovered these special numbers and polynomials. Let the generating function of Sk.n/ be f g

~~1 S .n/ G.n; x/ k xk: D kŠ k 0 XD Interchanging the summation order gives~~

~~n 1 k 1 i G.n; x/ xk D kŠ i 1 k 0 XD XD n 1 nx nx e 1 e 1 x eix ; D D ex 1 D x ex 1 i 1 XD~~

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~~where the second term is the expected Bernoulli numbers which were deﬁned in (6.8). On the other hand, using the power series expansion of enx and (6.11) yields~~

i 1 i 1 n x 1 B G.n; x/ C j xj D .i 1/Š 0 jŠ 1 i 0 ! j 0 XD C XD i@1 A 1 n B C j xk D 0 .i 1/ŠjŠ1 k 0 i j k XD CXD C @ k A k 1 1 k 1 k 1 j x C Bj n C : D 0k 1 j 1 kŠ k 0 j 0 Â Ã XD C XD @ A Deﬁne the kth degree Bernoulli polynomial by

~~k k k j Bk.x/ Bj x : (6.13) D j j 0 Â Ã XD~~

~~Equating the coefﬁcients in those two expressions of G.n; x/ gives~~

~~1 Sk.n/ .Bk 1.n/ Bk 1/: D k 1 C C C~~

~~In the following, we summarize some basic properties and formulas regarding Bk.x/ and Bk.~~

~~6.4.1 The generating function of Bk.x/~~

~~From (6.8) and (6.10), we have the generating function of Bk.x/,~~

~~k k k 1 t 1 k k j t Bk.x/ Bj x kŠ D 0 j 1 kŠ k 0 k 0 j 0 Â Ã XD XD XD @ k j k j A 1 t Bj .xt/ D 0 j Š.k j /Š 1 k 0 j 0 XD XD @ Ak 1 B 1 .xt/ k t k D kŠ kŠ k 0 ! i 0 ! XD XD text : D et 1~~

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~~In particular, in terms of the given Bernoulli numbers, the ﬁrst six Bernoulli polynomials are~~

B0.x/ 1; D 1 B1.x/ x ; D 2 2 1 B2.x/ x x ; D C 6 3 3 2 1 B3.x/ x x x; D 2 C 2 4 3 2 1 B4.x/ x 2x x ; D C 30 5 5 4 5 3 1 B5.x/ x x x x: D 2 C 3 6 6.4.2 Difference, Differentiation and Integration

~~Let GB .x; t/ denote the generating function of Bk.x/:~~

~~xt k te 1 t GB .x; t/ Bk.x/ ; D et 1 D kŠ k 0 XD~~

~~from which the following three properties of Bk.x/ are derived. 1. Note that~~

~~.x 1/t xt te C xt te xt GB .x 1;t/ te te GB .x; t/: C D et 1 D C et 1 D C Equating coefﬁcients of t k on both sides gives~~

~~k 1 B0.x 1/ B0.0/; Bk.x 1/ Bk.x/ kx .k 1/: (6.14) C D C D C ~~

~~2. Next, differentiating GB .x; t/ with respect to x yields~~

~~2 xt k @ t e 1 t GB .x; t/ B .x/ : @x D et 1 D k0 kŠ k 0 XD But clearly 2 xt k 1 t e 1 t C Bk.x/ ; et 1 D kŠ k 0 XD therefore equating coefﬁcients of t k =kŠ gives~~

~~Bk 0.x/ kBk 1.x/: (6.15) D Repeating this process leads to~~

~~.n/ kŠ Bk .x/ Bk n.x/: D .k n/Š~~

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~~3. Replacing k by k 1 in (6.15) and then integrating from a to x gives C x 1 Bk.x/dx .Bk 1.x/ Bk 1.a//: (6.16) D k 1 C C Za C This, appealing to (6.14), implies~~

~~t 1 C k Bk.x/dx t ; D Zt which, when t 0, gives D 1 Bk.x/dx 0 .k 1/: (6.17) D Z0 Moreover, (6.16) leads to a recursive relation~~

~~x Bk.x/ k Bk 1.t/dt Bk; k 1: (6.18) D C Z0~~

This equation and (6.17), together with B0 1, B0.x/ 1, provide another method for D D deﬁning the Bernoulli polynomials and Bernoulli numbers recursively. See Exercise 12. Indeed, Lehmer observed that, beginning with Bernoulli’s discovery, there have been six approaches for deﬁning the Bernoulli polynomials. See [4] for details.

6.4.3 Euler-Maclaurin summation formula As an applicationof (6.15), we derive a remarkable formula known as the Euler-Maclaurin summation formula, which precisely relates a discrete sum with a continuous integral. Along the way we also show how Bernoulli polynomials and Bernoulli numbers arise naturally. Start with 1 1 f.x/dx f .x/B0.x/ dx: D Z0 Z0 Appealing to B .x/ B0.x/, integration by parts gives 10 D 1 1 f.x/dx B10 .x/f .x/ dx 0 D 0 Z Z 1 1 B1.x/f .x/ B1.x/f .x/dx D j0 0 Z0 f .0/ f .1/ 1 C B1.x/f .x/ dx: D 2 0 Z0

In view of B1.x/ .1=2/B .x/, integration by parts again gives D 20 1 1 1 1 1 f.x/dx .f .0/ f .1// .B2.1/f .1/ B2.0/f .0// B2.x/f .x/dx D 2 C 2Š 0 0 C 2 00 Z0 Z0 1 1 1 1 .f .0/ f .1// B2.f .1/ f .0// B2.x/f .x/ dx: D 2 C 2Š 0 0 C 2 00 Z0

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~~Thus, by (6.10) and (6.12), repeating integration by parts yields~~

1 k 1 B2i .2i 1/ f.x/dx .f .0/ f .1// .f .1/ D 2 C .2i/Š Z0 i 1 XD 1 .2i 1/ 1 .2k/ f .0// B2k.x/f .x/dx: C .2k/Š Z0 Replacing f .x/ by f .x j / for any positive integer j gives C k 1 1 B f .x j/dx .f .j / f .j 1// 2i .f .2i 1/.j 1/ C D 2 C C .2i/Š C Z0 i 1 XD 1 .2i 1/ 1 .2k/ f .j // B2k.x/f .x j /dx: C .2k/Š C Z0 Next, for any positive integer N >1,

N N 1 j 1 N 1 1 f.x/dx C f.x/dx f .x j/dx; D D C Z0 j 0 Zj j 0 Z0 XD XD N 1 N 1 1 .f .j / f .j 1// f .j / .f .0/ f .N //; 2 C C D 2 C j 0 j 0 XD XD N 1 .f .2i 1/.j 1/ f .2i 1/.j // f .2i 1/.N / f .2i 1/.0/: C D j 0 XD Thus we establish the Euler-Maclaurin summation formula:

N N 1 k B f.x/dx f .j / .f .0/ f .N // 2i .f .2i 1/.N / D 2 C .2i/Š Z0 j 0 i 1 XD XD 1 N 1 .2i 1/ 1 .2k/ f .0// B2k.x/f .x j /dx: (6.19) C .2k/Š C Z0 j 0 XD Equivalently, we can write this as a relation between a ﬁnite sum and an integral

~~N N 1 f .j / f.x/dx .f .0/ f .N // D C 2 C j 0 Z0 XD k B2i .2i 1/ .2i 1/ .f .N / f .0// RN ; (6.20) C .2i/Š C i 1 XD where~~

~~1 N 1 1 .2k/ RN B2k.x/f .x j/dx D .2k/Š C Z0 j 0 XD N 1 .2k/ B2k.x x /f .x/dx: D .2k/Š b c Z0~~

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

n n 1 ln k ln xdx .ln 1 ln n/ D C 2 C i 1 Z0 XD 1 1 1 1 2 2 1 24 24 B2 B4 B6 : C 2Š n 1 C 4Š n3 13 C 6Š n5 15 C Â Ã Â Ã Â Ã By using the known Bernoulli numbers,we have 1 1 1 1 ln nŠ n ln n ln n Rn: D C 2 C 12n 360n3 C 1250n5 C Now exponentiating both sides yields

~~1 1 1 nŠ nne npneRn exp : D 12n 360n3 C 1250n5 C Â Ã Appealing to the power series of ex, we obtain~~

~~1 1 nŠ nne npneRn 1 : (6.21) D C 12n C 288n2 C Â Ã A careful analysis shows that (see Exercise 13)~~

~~eRn p2 as n : ! ! 1 This, together with (6.21), yields the famous Stirling’s formula~~

~~1 1 nŠ p2nnne n 1 : (6.22) D C 12n C 288n2 C Â Ã~~

~~6.4.4 Power series in terms of Bk Appealing to (6.10), we rewrite (6.9) as~~

x x 1 B coth 2n x2n: 2 2 D .2n/Š n 0 XD Substituting x by 2xi and noticing that coth.ix/ i cot x, we ﬁnd that D 2n 1 2 B2n 1 1 2 x cot x . 1/n x2n 1 x2 x4 x6 : (6.23) D .2n/Š D 3 45 945 n 0 XD Furthermore, by the trigonometric identities

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

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~~68 Excursions in Classical Analysis~~

~~we obtain~~

2n 2n 1 n 1 2 .2 1/B2n 2n 1 tan x . 1/ C x ; (6.24) D .2n/Š n 1 XD 2n 1 1 n 1 2.2 1/B2n 2n x csc x 1 . 1/ C x : (6.25) D C .2n/Š n 1 XD 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/ 2=6. Now, we conclude this chapter by deriving Euler’s amazing D formula:for any positive integer k,

~~k 1 2k 1 1 1 . 1/ 2 B2k .2k/ 2k: (6.26) D n2k D .2k/Š n 1 XD Recall Euler’s inﬁnite product (5.26), namely~~

~~2 sin x 1 x 1 : x D n2 2 n 1 Â Ã YD Taking the logarithm of both sides and then differentiating yields~~

~~1 1 2x cot x ; x D 2n2 x2 n 1 XD which results in Euler’s elegant partial fraction expansion:~~

~~1 1 2x 1 1 1 1 cot x : (6.27) D x 2n2 x2 D x n x n x n 1 n 1 Â Ã XD XD C With the previous established power series expansion (6.23), we now have proven that~~

~~2 k 2k 1 x 1 . 1/ 2 B2k 1 2 1 x2k: (6.28) 2n2 x2 D C .2k/Š n 1 k 1 XD XD In terms of a geometric series expansion,~~

~~2 2 2 2 2k x x =. n / 1 x : 2n2 x2 D 1 x2=. 2n2/ D 2kn2k k 1 XD Thus (6.28) becomes~~

~~2k k 2k 1 1 x 1 . 1/ 2 B2k 1 2 1 x2k: 2kn2k D C .2k/Š n 1 k 1 ! k 1 XD XD XD Interchanging the order of summation on the left yields~~

~~k 2k 1 1 1 1 . 1/ 2 B2k 1 2 x2k 1 x2k: C 2kn2k D C .2k/Š k 1 n 1 ! k 1 XD XD XD~~

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~~6. Special Numbers 69~~

Now (6.26) follows from equating the coefﬁcients of x2k on both sides. In particular, for k 2;3;4,we get D 4 6 8 1 1 1 1 1 1 ; ; : n4 D 90 n6 D 945 n8 D 9450 n 1 n 1 n 1 XD XD XD In lightof (6.26), wesee that .2k/ is a rational multipleof 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 1/ analogous to C (6.26). The question whether .2k 1/ is irrational has been asked since the time of Euler. C 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 1/ for k 2. C Exercises

~~1. Let Fn be the Fibonacci numbers and~~

~~an F0 F1 F2 Fn: D C C CC~~

~~Find a generating function for an and then prove that~~

~~an Fn 2 1: D C~~

2. Let a0 1. For n 1, deﬁne an by D n 1 an .n i/ai 1: D C i 1 XD Show that an F2n 1. D 3. Monthly Problem 11258 [2006, 939; 2008, 949]. Let Fn denote the nth Fibonacci number, and let i denote p 1. Prove that

~~1 F3k 2F1 3k 1 C i .1 p5/: F k i F k D C 2 k 0 3 2 3 XD C~~

~~4. Let Hn be the nth harmonic number. For all n 2, prove that n 1 Hk nHn n D k 1 XD and H H H H 2 2 2 3 n : n 2 C 3 CC n Â Ã 5. Prove that 1 H ln2.1 x/ 2 n xn D n 1 n 1 XD C~~

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and n 1 1 H ln3.1 x/ 6 k xn 2: D n 2 k 1 C n 1 k 1 ! XD C XD C 6. Let n k 1 1 1 . 1/ an 1 ln 2 : D 2 C 3 C k k 1 ! XD Does the sequence an converge? If so, to what value? f g 7. Monthly Problem 11382 [2008, 665]. Show that if p is prime and p>5, then

p 1 2 p 1 H H k k .mod p2/: k Á k2 k 1 k 1 XD XD 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.

~~n 8. Prove that if f .x/ n1 1 anx converges, then D D P 1 1 n f .x/ f .tx/ anHnx dt: D 1 t n 1 Z0 XD In particular, show that~~

~~1 1 1 1 Hn .2/; Hn 1 .3/: n2n 1 D n2 D n 1 n 1 XD XD~~

~~9. Let Rn Hn ln.n 1/ and be Euler’s constant. Prove that D C~~

~~1 ak Rn ; D .n 1/.n 2/ .n k/ n 1 XD C C C where 1 1 1 a1 ; ak x.1 x/ .k 1 x/dx .k > 1/: D 2 D k Z0 For the reader familiar with Stirling numbers of the ﬁrst kind s.k; i/, explicitly,~~

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

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~~6. Special Numbers 71~~

~~11. Prove that~~

1 (a) tan x n1 n 1=2 x . D D1 C . 1/n (b) sec x Pn1 n 1=2 x . D D1 C C 2 1 (c) 1 . sin2 x Pn .x n/2 D D1 C 12. Deﬁne b0 1;bP0.x/ 1 and D D x bn.x/ n bn 1.t/ dt bn; D C Z0 1 where the constant bn is chosen so that bn.t/ dt 0. Show that bn.x/ Bn.x/ 0 D D and bn Bn. D R 13. Show that nŠ lim p2: n nnpne n D !1 14. Show that n 1 1 m B 1 ln n 2k : D k 2n C 2k n2k C k 1 k 1 XD XD 15. Show that, for n 1, n 1 2n B2k 2 1 C : 2k .2k 2/ D .2n 1/.2n 2/ k 0 Â Ã XD C C C 16. Monthly Problem E 3237 [1987, 995; 1989, 364–365]. Prove that

~~m 1 n 1 1 k m m j Bm n Bk j ; D n.1 nm/ k k 0 Â Ã j 1 XD XD for any positive integer m;n>1.~~

~~17. Prove that~~

~~n 2k 1 1 . 1/ k 1 .2/ C . 1/ C B2k 1.1=4/: .2n 1/2k 1 D 2.2k 1/Š C n 0 C XD C C 18. Prove Raabe’s multiplication identity~~

~~m 1 1 k n Bn x m Bn.mx/: m C m D k 0 Â Ã XD~~

Comment. Lehmer in [4] proves that the nth Bernoullipolynomial 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 Re- source. 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 equivalent to the Riemann Hypothesis, Amer. Math. Monthly, 109 (2002) 534–543. [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, Jonathanand 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 sequencesand 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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~~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 n 1 k the form of k 1 csc n . D P Á 7.1 A well-known sum and its generalization The well-known sum we have in mind is

~~n 1 k sin cot : (7.1) n D 2n k 1 Â Ã XD Á 73~~

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~~74 Excursions in Classical Analysis~~

~~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 1 sinŒ.n 1/x=2 sin.nx=2/ sin.kx/ : (7.2) D sin.x=2/ k 1 XD To see this, appealing to r n 1 1 r r 2 r n 1 (7.3) C C CC D r 1 and De Moivre’s formula, we get~~

n 1 n 1 i.n 1/x e 1 sin.kx/ Im eikx Im eix D D eix 1 k 1 k 1 ! ! XD XD ei.n 1/x=2 e i.n 1/x=2 Im einx=2 D eix=2 e ix=2 ! sinŒ.n 1/x=2 Im.einx=2/ D sin.x=2/ sinŒ.n 1/x=2 sin.nx=2/ ; D sin.x=2/ as proposed. Now (7.1) follows from (7.2) by setting x =n. D In analogy to (7.1) we seek to compute in closed form

n 1 n 1 1 k In csc : (7.4) D sin.k=n/ D n k 1 k 1 Â Ã XD XD To begin with, based on the identity(5.2) and other favorable results [2] on the power sums n 1 2p k 1 csc .k=n/, it seemed to me that there ought to be a closed form for In. I spent manyD hours reading related journal papers, using Mathematica to explore and test potential P 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 wewillsee, thisproblemis hard enough that theresultswill be publishableyet 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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~~7. On a Sum of Cosecants 75~~

~~Jordan’s inequality: For 0~~

~~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 1 < Hn ln n < : (7.6) 2.n 1/ 2n C In addition, for convenience, we introduce the big O notation: given two sequences an and bn such that bn 0 for all n. We write f g f g ~~

~~an O.bn/ D~~

~~if there exists a constant M >0 such that an Mbn for all n. j j Ä Now we proceed to the estimates for In. Since~~

~~k .n k/ csc csc ; n D n Â Ã Â Ã the second inequality of (7.5) implies~~

n=2 n=2 b c k b c 1 In 2 csc n : Ä n Ä k k 1 Â Ã k 1 XD XD Notice that 1 1; if n 2m; n D 2 n=2 D 1 1 ; if n 2m 1: b c C 2m D C This, together with the second inequality of (7.6), yields that, for n 3, 1 In n.ln n ln 2/ 1 O : Ä C C C n Â Ã

Next, to estimate In from below, for n 3, we rewrite .n 1/=2 k 2 bk 1 c csc n ; if n is odd; In 8 D D P .n 1/=2 Á k ˆ1 2 bk 1 c csc n ; if n is even: < C D P Á By the ﬁrst inequality of:ˆ (7.5) we have

~~.n 1/=2 Œ.n 1/=2 b c k 2n 1 In 2 csc : (7.7) n k k 1 Â Ã k 1 XD XD~~

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~~76 Excursions in Classical Analysis~~

~~Clearly, if n 2m, D n 1 n ln ln.n 2/ ln 2 1; 2 D I 2. .n 1/=2 1/ D b cC while if n 2m 1, D C n 2 n 1 ln ln.n 1/ ln 2 1 : 2 D I 2. .n 1/=2 1/ D n 1 b cC C Recall that for 0~~

~~1 1 1 1 1 cot x ; D x C x i C x i i 1 Â Ã XD C from csc ˛ cot.˛=2/ cot ˛, we ﬁnd D 1 1 1 1 1 csc x . 1/i ; D x C x i C x i i 1 Â Ã XD C thereby implying that~~

~~k n 1 1 1 1 csc . 1/i : n D k C k ni C k ni Â Ã i 1 Â Ã! XD C~~

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~~7. On a Sum of Cosecants 77~~

Since i i 1 . 1/ 1 1 . 1/ ; k ni D n k C k ni i 1 i 2 XD XD shifting the index i to i 1 on the right-hand side gives C i i 1 . 1/ 1 1 . 1/ : k ni D n k C .n k/ ni i 1 i 1 XD XD C Noticing n 1 n 1 n 1 n 1 1 1 1 1 ; ; n k D k .n k/ ni D k ni k 1 k 1 k 1 k 1 XD XD XD C XD C and interchanging the summations yields

n 1 n 1 1 i 1 1 In . 1/ D k C k ni C k ni k 1 i 1 Â Ã! XD XD C n 1 n 1 2n 1 2n 1 1 . 1/i : (7.9) D k C k ni k 1 i 1 k 1 XD XD XD C Next, recall 1 1 Hn ln n O : D C C 2n C n2 Â Ã This leads to

~~n 1 1 1 Hn.i 1/ Hni k ni D C n.i 1/ k 1 XD C C 1 1 1 2 ln 1 O.i /; D C i n.i 1/ C n Â Ã C so that~~

~~n 1 i 2 1 1 1 1 1 . 1/ O. 1 i / . 1/i . 1/i ln.1 1=i/ i 1 : k ni D C n i 1 C Dn i 1 k 1 i 1 i 1 P XD XD C XD XD C Appealing to Wallis’ formula (5.30), which asserts that~~

~~1 2n 2n ; 2n 1 2n 1 D 2 n 1 YD C we obtain~~

~~1 1 2n 2n . 1/i ln.1 1=i/ ln ln.=2/: C D 2n 1 2n 1 D i 1 n 1 ! XD YD C Furthermore, in view of the well-known facts that~~

~~i 2 1 . 1/ 1 1 ln 2 1 and ; i 1 D i 2 D 6 i 1 i 1 XD C XD~~

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~~78 Excursions in Classical Analysis~~

we have n 1 1 1 1 . 1/i ln.=2/ O : k ni D C n i 1 k 1 Â Ã XD XD C This, in conjunction with (7.9), yields 2n In .ln n ln.=2// O.1/: (7.10) D C C To sharpen the preceding term O.1/, we use (6.25) in the form of

~~1 2i 1 csc x ai x (7.11) D i 0 XD with a0 1 and ai >0. Since D .ln.tan.x=2///0 csc x; (7.12) D from (7.11), integrating (7.12) provides~~

~~1 a ln.tan.x=2// ln.x=2/ i x2i : (7.13) D 2i i 1 XD On the other hand, by (7.11) we have~~

Œn=2 Œn=2 2n 1 1 2i 1 2i 1 In 2 ai .=n/ k : (7.14) Ä k C k 1 i 1 k 1 XD XD XD Since n=Œn=2 3, along the same lines we used in the rough estimate, we get Ä Œn=2 2n 1 2n 3 < .ln n ln 2/ : k C C C k 1 XD To estimate the double sum in (7.14), noticing

~~Œn=2 n=2 2i 1 2i 1 2i 1 k < t dt .n=2/ ; 0 C k 1 Z XD and using the equations (7.11) and (7.13), we have~~

Œn=2 1 2i 1 2i 1 2n 1 ai 2i 1 2i 1 2 ai .=n/ k < .=2/ 2 ai .=2/ 2i C i 1 k 1 i 1 i 1 XD XD XD XD 2n 2 .ln.tan.=4// ln.=4// 2 csc.=2/ D C Â Ã 2n 4 ln.=4/ 2 : D C Collecting all this together, ﬁnally, we have established 2n 1 In < .ln n ln.=2// 2 : (7.15) C C

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~~7. On a Sum of Cosecants 79~~

~~7.4 Final Remarks~~

In the course of searching for the optimal bounds for In, we see the process evolve in the following basic pattern: we start with a few familiar tools, which obtain relatively simple results; we wish to improve these results and ﬁnd that the familiar tools no longer serve for this purpose but reveal the direction of the improvement; then we appeal to either deeper mathematical tools or new ideas to reach a better solution of the problem at hand. As students begin to understand and work with this kind of methodology, they may develop their own internal monitors and become their own questioners. In this regard, In again provides a good example. Let 2n Jn .ln n ln.=2//: D C

~~Mathematica 6.0 tabulates some selected differences of Jn In:~~

n Jn In 3 0.0287351170881755128 4 0.0216641822966619335 5 0.0173744361336004932 6 0.0144985127666732041 7 0.0124376321763192829 8 0.0108888316916454100 9 0.0096825731521195440 10 0.0087166472057682210 100 0.0008726545781646500 1000 0.0000872664525524908 10000 0.0000087266462638730 100000 0.0000008726648900160

~~The numerical evidence now suggests that In is bounded by Jn along.We now have another new problem worthy of further investigation:~~

~~Open Problem: For n 3, prove that 2n In .ln n ln.=2//: Ä C~~

~~Exercises 1. Show that n x x csc cot cot x: 2k D 2n 1 k 0 C XD Á Á 2. Evaluate the limit n n 2k L lim : D n 2n k !1 k 1 XD~~

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~~3. Let n 2. Prove that n 1 n 1 k 1 .2k 1/ csc .2k 1/ cot C : n D n C 2n k 1 Â Ã k 0 Â Ã XD XD~~

k 1 4. Let An k1 n 1 . 1/ =k. Consider the inequalities D D C P 1 1 < An < : 2n a j j 2n b C C Kazarinoff (Amer. Math. Monthly, 62 (1955) 726–727) proved this estimate with a D 2 and b 2. Find the smallest a and the largest b such that this estimate holds D for every n 1 or for every n n0, some positive integer. If replacing An by k 1 An k1 n 1 . 1/ =.2k 1/, establish the corresponding estimates. D D C 5. DetermineP the best possible constants ˛ and ˇ such that inequalities

~~e 1 n e e 1 2n ˛ Ä C n Ä 2n ˇ C Â Ã C hold for every n 1. 6. Monthly Problem E 3432 [1991, 264; 1992, 684–685]. Prove that for every positive integer n we have~~

1 1 < Hn ln n < : 2n 2=5 2n 1=3 C C Show that 2=5 can be replaced by a slightly smaller number, but 1=3 cannot be replaced by a slightly large number. Remark. The best possible constant to replace 2=5 is 1=.1 / 2. 7. Prove (7.10) by applying the trapezoidal rule to the integral

~~1 1=2n 1 1 dx; sin.x/ x.1 x/ Z1=2n Â Ã This approach is due to Michael Renardy. See www.siam.org/journals/problems/downloadfiles/05-001s.pdf~~

~~8. Prove the complete asymptotic expansion~~

~~2k 1 2 k 2n 2n 1 2 1 2 In .ln n ln.=2// B ; C C k.2k/Š 2k n2 k 1 Â Ã XD~~

~~where B2k are Bernoulli numbers.~~

~~9. Open Problem. Can one establish (7.10) via the Euler-Maclaurin summation formula?~~

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~~7. On a Sum of Cosecants 81~~

~~10. Open Problem. In 1890 Mathieu deﬁned S.x/, now called Mathieu’s series, as~~

1 2k S.x/ : WD .k2 x2/2 k 1 XD C Various inequalities have been established on S.x/. For example, Alzer et al. proved that 1 1 < S.x/ < ; for all x R: x2 1=.2.3// x2 1=6 2 C C Although this estimate is asymptotically sharp, it leaves a big gap for x 0. Can one ﬁnd a best possible estimate of S.x/ applicable on the entire R? The following interesting observation,

~~1 1 1 t S.x/ 1 e kt t .xt/ dt 1 .xt/dt; sin t sin D x 0 D x 0 e 1 k 1 Z Z XD may be useful.~~

References [1] J. Borwein, personal communication. [2] H. Chen, On some trigonometric power sums, Internat. J. Math. Math. Sci., 30 (2002) 185–191. [3] ——, Bounds for a trigonometric sum — a casestudy of undergraduate math research, Internat. J. Math. Ed. Sci. Tech., 37 (2006) 215–223. [4] J. Gallian, The Duluth Undergraduate Research Program, available at www.d.umn.edu/ jgallian/progdesc.html [5] Wolfram, “The best-known properties and formulas for the cosecant function,” available at functions.wolfram.com/ElementaryFunctions/Csc/introductions/Csc/05/ [6] D. Zeilberger, The method of undetermined generalization and specialization, Amer. Math. Monthly, 103 (1996) 233–239.

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~~The Gamma Products in Simple Closed Forms~~

~~The art of doing mathematics consists in ﬁnding that special case which contains all the germs of generality. — D. Hilbert~~

To be a good problem solver, it is not just what you know, but how you use what you know. As we have seen in previous chapters, many formulas and theorems emerge from the analysis of specific calculations and special cases. Halmos once said “the source of all great mathematics is the special case, the concrete example. It is frequent in mathematics that every instance of a concept of seemingly great generality is in essence the same as a small and concrete special case.” Thus, to solve a difficult problem, one should initiallymake the problem concrete and physical whenever possible. Often, building a base case will give us understanding and insight. After recognizing the similarities between the base case and the original problem, one may eventually extract the idea or technique that obtains the more general result. This principle of “understanding special cases deeply” often gives one the feeling ‘oh, I see now’ and often offers a route to solving a problem. In this chapter, we demonstrate how to use this principle by studying products involving the gamma function. Of the so-called “special functions”, the gamma function is one of the most fundamental (see Chapter 1 of [1]). This function is closely related to factorials and crops up in many unexpected places in analysis and statistics, especially in systematically building integration formulas, summing infinite series, and defining probability distributions. In the wake of these processes, many remarkable identities appeared. These identities are simple enough to introduce to studentsin elementary analysis but deep enough to have called forth contributions from the finest mathematicians. As a first example we choose a particularly striking identity — Gauss’s multiplication formula. It states that for any real number x>0 and any positive integer n,

~~n 1 k x .2/.n 1/=2n1=2 nx .nx/; (8.1) C n D k 1 Â Ã YD where .x/ denotes the usual gamma function, deﬁned by~~

~~.x/ 1 t x 1e t dt: D Z0 83~~

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~~Integration by parts establishes the recursion~~

.x 1/ x.x/: (8.2) C D Letting x 1 in (8.1), using (8.2), and simplifying, we get D n 1 k .2/.n 1/=2n 1=2: (8.3) n D k 1 Â Ã YD The gamma products we have in mind are variations of (8.3). We are interested in ﬁnding simple values for productsof fewer factors. Let A denote a proper subset of 1;2;:::;n 1 f g and .k; n/ be the greatest common divisor of k and n. Consider the gamma product

~~k : (8.4) n k A;.k;n/ 1 Â Ã 2 Y D Emboldened by Gauss’s success, we may ask two things: I. Does (8.4) still retain a closed form?~~

II. Is it possible to identify those sets A such that (8.4) possesses a value analogous to (8.3)? There are countless interesting answers for Problem I from the well-known reﬂection formula .x/.1 x/ (8.5) D sin x to the more involved Chowla-Selberg formulas in [4] such as

~~2 1=2 13=4 1=2 .1=4/ 4pK.k1/; .1=8/.3=8/ .p2 1/ 2 K.k2/; D D~~

where K.k/ is a complete elliptic integral of the ﬁrst kind and ki is the ith elliptic integral singular values (see Chapter 1 of [2] and [3, 6, 7] for details). However, Problem II apparently has not received much attention. Most existing results are mere consequences of (8.3) such as

.1=4/.3=4/ p2; .1=8/.3=8/.5=8/.7=8/ 2p2 2: D D In the following, motivated by a recently posed Monthly problem, we ﬁrst investigate a base case and solve it by two different methods. The kernels of sophisticated ideas in both methods enable us to identity a large class of nontrivial gamma products that solve Prob- lem II. Our study here, solely using a special case of (8.1), (8.3) and (8.5), is particularly accessible to undergraduate analysis students. Surprisingly, in contrast to the usual derivations in [3, 6, 7], elliptical integrals, elliptic integral singular values and zeta functions do not manifest at all. Recently, Monthly Problem 11426 in [5], proposed by Glasser, asks the reader to ﬁnd

~~.1=14/ .9=14/ .11=14/ : .3=14/ .5=14/ .13=14/~~

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~~8. The Gamma Products in Simple Closed Forms 85~~

~~In the course of solving this problem, we actually ﬁnd~~

.1=14/ .9=14/ .11=14/ 4 3=2; (8.6) D .3=14/ .5=14/ .13=14/ 2 3=2: (8.7) D These will serve as our base case for investigating Problem II. We will give two proofs. Althoughone proof is enough to establish the truth of these identities, we will see the ideas used in each proof are insightfulfor determining the general setsof A that solve Problem II.

First Proof. Start with Legendre’s duplication formula, which is the special case of (8.1) with n 2, namely D .2x/ 1=222x 1 .x/.x 1=2/: (8.8) D C Setting x 1=14; 9=14 and 11=14 in (8.8) successively, we have D .1=7/ 1=22 6=7 .1=14/ .4=7/; (8.9) D 1=2 2=7 .9=7/ 2 .9=14/ .8=7/; (8.10) D .11=7/ 1=224=7 .11=14/ .9=7/: (8.11) D Applying (8.2) on both sides of (8.10) and (8.11) respectively yields

~~.2=7/ 1=22 5=7 .9=14/ .1=7/; (8.12) D .4=7/ 1=22 3=7 .11=14/ .2=7/: (8.13) D The product of the identities (8.9), (8.12) and (8.13) gives~~

.1=7/.2=7/.4=7/ 3=22 2 .1=14/.4=7/.9=14/.1=7/.11=14/.2=7/: D This proves (8.6) as desired after removal of the common factors. The identity(8.7) follows in a similar fashion. The above proof relies on the cancellation, but there is no apparent general principle to show us how to group the gamma products to make the cancellation work. In the following, we present another proof by iterating (8.8) with different selected values. As a result, the gamma product is nearly always ready for cancellation.

Second Proof. Setting x 1=14; 2=14 and 4=14 in (8.8) recursively yields D .2=14/ 1=22 6=7 .1=14/ .1=14 1=2/; (8.14) D C .4=14/ 1=22 5=7 .2=14/ .2=14 1=2/; (8.15) D C .8=14/ 1=22 3=7 .4=14/ .4=14 1=2/: (8.16) D C Since .8=14/ .1=14 1=2/; D C multiplying the identities (8.14)–(8.16) and then eliminating the common factors yields

~~.1=14/.2=14 1=2/.4=14 1=2/ 4 3=2; (8.17) C C D which is equivalent to (8.6).~~

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~~Next, we prove (8.7) using (8.6). Multiplyingboth sides of (8.6) by the corresponding reﬂection part of each argument yields~~

~~.1=14/.13=14/.9=14/.5=14/.11=14/.3=14/ 4./3=2.3=14/.5=14/.13=14/: (8.18) D By (8.5), the left-hand side of (8.18) is reduced to~~

~~ 3 : sin.=14/ sin.3=14/ sin.5=14/ So to show that (8.7) is true, it is enough to show~~

8 sin.=14/ sin.3=14/ sin.5=14/ 1: D Since sin ˛ cos.=2 ˛/, appealing to the double angle formula for sine, we get D 8 sin.=14/ sin.3=14/ sin.5=14/ 8 sin.=14/ cos.2=14/ cos.4=14/ D 4 sin.2=14/ cos.2=14/ cos.4=14/= cos.=14/ D sin.8=14/= cos.=14/ D sin.=2 =14/= cos.=14/ D C 1; D as required. It is interesting to notice that (8.7) can be rewritten as

.1=2 4=14/.1=2 2=14/.13=14/ 2 3=2: (8.19) D When we look back on this special case, the successive selection of the values 2k=14 offers us a convenient way to cancel common factors. Indeed, the final product on the left- hand side will leave only one gamma value. Thus, once this value on the left-hand side is removed, we obtain a gamma product which solves Problem II as expected. These are not isolated problems. It is remarkable that the ideas used in the special case are strong enough to extend to several entire families of gamma products. We first extend the special case to all even positive integers. Observe that, in the special case, the product of any two terms of (8.9)–(8.11) is not sufficient to cancel unnecessary terms thereby leading to a simple closed form; meanwhile, except the triples of (8.6) and (8.7), other combinations of three products do not have simple closed form either. In general, thisforces us to take thegamma product over all odd terms and results in Theorem 8.1. For any positive integer k,

~~2k 2i 1 2.2k 1/=2 k; (8.20) 2.2k/ D i 1 Â Ã YD 2k 1 C 2i 1 2.2k 1/=2 k: (8.21) 2.2k 1/ D C i 1 Â Ã YD C~~

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~~8. The Gamma Products in Simple Closed Forms 87~~

~~Proof. We prove (8.20) only. (8.21) can be treated similarly. By (8.3) we have~~

~~4k 1 j .2/.4k 1/=2.4k/ 1=2: 2.2k/ D j 1 Â Ã YD Splittingthe even and odd indices j yields~~

~~2k 2k 1 2j 1 2j .2/.4k 1/=2.4k/ 1=2: (8.22) 2.2k/ 2.2k/ D j 1 Â Ã j 1 Â Ã YD YD Using (8.3) once more gives~~

2k 1 2k 1 2j j .2/.2k 1/=2.2k/ 1=2: (8.23) 2.2k/ D 2k D j 1 Â Ã j 1 Â Ã YD YD (8.20) now follows from (8.22) and (8.23). The use of the recursive values of 2i =14 in (8.8) enables us to remove inductively the common factors. Thus, selection of the recursive values and cancelation help us reframe our problem so that an underlying product, whether it solves Problem II or not, could be found more efﬁciently. Indeed, setting x 1=n;2=n;:::;2m 1=n in (8.8) successively, D we ﬁnd

1=2 2=n 1 .2=n/ 2 .1=n/ .1=n 1=2/; D 2 C .22=n/ 1=222 =n 1 .2=n/ .2=n 1=2/; D C m 1 1=2 2m1=n 1 m 2 mk 2 .2 =n/ 2 .2 =n/ .2 =n 1=2/; D m C .2m=n/ 1=222 =n 1 .2m 1=n/ .2m 1 =n 1=2/: D C Multiplying these identities and then canceling the common factors we have

m m 1 k 2 m 1 2 1 m=222.2 1/=n m : (8.24) n D n n C 2 Â Ã Â Ã k 0 ! YD Thus, (8.24) will admit a simpleclosed form as long as one can eliminate .2m=n/. To this end, set 2m=n 1=n 1=2, which is equivalent to n 2.2m 1/. In this case, appealing D C D to (8.24) we actually have proved Theorem 8.2. Let m be a positive integer and n 2.2m 1/. Then D m 1 k 1 2 1 2m 1 m=2: (8.25) n n C 2 D Â Ã k 1 ! YD Next, we see (8.19) is the base case for the following generalization. Theorem 8.3. Let m be a positive integer and n 2.2m 1/. Then D m 1 k n 1 1 2 2 m=2: (8.26) n 2 n D Â Ã k 1 ! YD

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~~Proof. Let S denote the left-hand side of (8.26). Multiplying both sides of (8.25) by S yields~~

~~m 1 k k 1 n 1 2 1 1 2 m 1 m=2 2 S: n n n C 2 2 n D Â Ã Â Ã k 1 ! ! YD By (8.5) and its equivalent~~

1 1 x x ; 2 2 C D cos x Â Ã Â Ã we have 2 m=2 S : m m 1 k D 2 sin.=n/ k 1 cos.2 =n/ D Now, to establish (8.25), it remains to show that for all m 2, Q m 1 2m sin.=n/ cos.2k=n/ 1: D k 1 YD Repeatedly using the double angle formula and in view of 2m=n 1=n 1=2, we have D C m 1 k m 1 m 1 k 2 2 sin.2=n/ 2 2m sin cos cos n n D cos.=n/ n k 1 ! k 1 ! Á YD YD sin.2m=n/ 1; D cos.=n/ D as desired. Setting m 3, we recover (8.6)–(8.7) from (8.25)–(8.26), respectively. Moreover, we D have

.1=30/ .17=30/ .19=30/.23=30/ 8 2; D .7=30/ .11=30/ .13=30/.29=30/ 2 2; D .1=62/ .33=62/ .35=62/.39=62/.47=62/ 16 5=2; (8.27) D .15=62/ .23=62/ .27=62/.29=62/.61=62/ 2 5=2: (8.28) D It appears that .1=n/ has no simple closed form for n>3. So the three theorems above characterize some families of gamma products with simple closed forms. We do not yet know whether these are the only families possessing such simple closed forms. If we dispense with the restriction .k; n/ 1 in Problem II, (8.24) may yield more D simple closed forms. For example, let 2m=n 2=n 1=2, i.e., n 2.2m 2/. (8.24) D C D becomes

~~m 1 k 1 2 1 m=2 m 1 2=n 2 : n n C 2 D Â Ã k 0;k 1 ! DY¤~~

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~~8. The Gamma Products in Simple Closed Forms 89~~

~~In particular,~~

.1=12/.7=12/.5=6/ 211=6 3=2; D .1=28/.15=28/.9=14/.11=14/ 241=14 3=2; D .1=60/.31=60/.17=30/.19=30/.23=30/ 2119=30 5=2: D Along these lines, (8.24) will generate many more gamma products in simple closed forms. The details are left to the reader. To conclude this chapter, it is worth noticing that for a given problem there might be different ways to form a base case. For each base case, there may be more than one way to the solution. Some base cases might be more helpful than others in solving the problem at hand. Formulating good base questions may be just as important as posing general questions. In my personal experience, most of the Monthly problems could be good base problems due to the quality of the problems and richness in mathematical ideas. Indeed, many posed problems emerge from special cases in the proposer’s research. Trying to generalize a Monthly problem often leads to a good undergraduate research problem.

~~Exercises 1. Prove that 1 1 ln .x/dx ln.2/: D 2 Z0 2. Prove that x 1 C ln .t/ dt x ln x x C D C Zx and determine the value of C .~~

~~3. Recall that~~

~~1 2 =2 dx agm.a; b/ : D 0 a2 sin2 x b2 cos2 x ! Z C Prove that p 2 3=2 agm.1; 1=p2/ : D 2.1=4/~~

k 4. For all 0 < ai ;bi <1 with i 1 .ai bi / 0, prove that D D P k 1 .n a1/.n a2/ .n ak / .1 bi / : .n b1/.n b2/ .n bk/ D .1 ai / n 1 i 1 YD YD 5. Weierstrass’ product formula states that

~~1 1 x xe x 1 e x=n; .x/ D C n n 1 YD Á~~

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~~90 Excursions in Classical Analysis~~

~~where is Euler’s constant deﬁned by (5.8). Using this formula, prove~~

~~2 1 x sin x x 1 : D n2 n 1 Â Ã YD 6. For 0~~

n n 1 1 . 1/ 1 2x lim sin x D x C x2 n2 D n x k n 1 !1 k n XD XD directly from the reﬂection formula without using Euler’s sine product formula. 7. Putnam Problem 1984-A5. Let R be the region consisting of all triples .x;y;z/ of nonnegative real numbers satisfying x y z 1. Let w 1 x y z. Express C C Ä D the value of the triple integral

~~x1y9z8w4 dx dy dz •R in the form aŠbŠcŠdŠ=nŠ, where a;b;c;d and n are positive integers.~~

~~Note. This is the special case of Liouville’s integral: for all pi > 0; .1 i n/, Ä Ä~~

p1 1 p2 1 pn 1 L f .x1 x2 xn/x x x dx1dx2 dxn: D C CC 1 2 n xZ1;:::;xnZ 0 I x1 xn 1 CC Ä 8. Prove that .p /.p / .p / 1 1 2 n p1 p2 pn 1 L f .t/t C CC dt: D .p1 p2 pn/ C CC Z0 Use this result to derive the volume and surface area of the n-dimensional unit ball, where Putnam Problem 1951-B7 is a special case. 9. Using the above result, prove Gauss’ multiplication formula. 10. For each positive integer n, show that

n 1 2 2k 1 2n 2k 1 C n2 2: 2n 2n D k 0 Ä Â Ã Â Ã XD 11. Prove that .x/ 1 1 1 0 : .x/ D C x n 1 n n 1 Â Ã XD C 12. Open Problem. By (8.1) with n 3 we have D .3x/ .2/ 133x 1=2 .x/.x 1=3/.x 2=3/: D C C Find some products, as general as possible, in the simple closed forms.

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~~8. The Gamma Products in Simple Closed Forms 91~~

~~13. Open Problem. Similar to the ﬁrst proof, we have~~

.3=62/ .17=62/ .37=62/.43=62/.55=62/ 8 5=2; D .5=62/ .9=62/ .41=62/.49=62/.51=62/ 8 5=2; D .7=62/ .19=62/ .25=62/.45=62/.59=62/ 4 5=2; D .11=62/ .13=62/ .21=62/.53=62/.57=62/ 4 5=2: D (a) Using these results as the base cases, can you ﬁnd some general formulas similar to (8.25) and (8.26)? (b) Adding the identities (8.27) and (8.28) that derived in the text, we see the numerator set

.1;33;35;39;47/;.15;23;27;29;61/;.3;17;37;43;55/ f .5;9;41;49;51/;.7;19;25;45;59/;.11;13;21;53;57/ g is an exactly disjoint partition of all odd numbers less than 62, exclusive of 31, where .1=2/ p is well known. They are the only number sets that admit D the simple closed forms. Can you generalize this observation? Remark. Recently, Nijenhuis characterized the case n 2.mod 4/ via group Á theory approach. See arXiv:0907.1689v1. 14. Let p be prime and n pm. Show that D k .2/.n/=2 ; n D pp .k;n/ 1 Â Ã YD where .n/ is the Euler-phi function.

References [1] G. Andrews, R. Askey & R. Roy, Special Functions, Encyclopedia of Mathematics and Its Applications 71, Cambridge University Press, 1999. [2] J. Borwein and P. Borwein, Pi and the AGM, John Wiley, New York, 1987. [3] J. Borwein and J. Zucker, Fast evaluation of the Gamma function for small rational fractions using complete elliptic integrals of the ﬁrst kind, IMA J. Numerical Analysis, 12(1992) 519–526. [4] S. Chowla andA. Selberg, On Epstein’s zetafunction, J. Reine Agnew. Math., 227(1967)86–110. [5] M. L. Glasser, Problem 11426, Amer. Math. Monthly, 116(2009) 365. [6] Eric W. Weisstein, “Elliptic Integral Singular Value.” From MathWorld—A Wolfram Web Resource. mathworld.wolfram.com/EllipticIntegralSingularValue.html [7] ——, “Gamma Function.” From MathWorld—A Wolfram Web Resource. mathworld.wolfram.com/GammaFunction.html

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~~On the Telescoping Sums~~

In analysis, there are only a few series with partial sums that we can calculate in closed form and almost all of these come from telescoping sums. In this chapter, we apply this technique to a variety of problems, beginning with some that are arithmetic and trigonometric, moving to others that involve Apéry-like formulas and famous families of numbers, and finally ending with a class of problems that can be solved algorithmically. We begin with Gauss’s legendary method for summing arithmetic series. Consider an arithmetic series of n terms with first term a, last term b and common difference d. Write the sum twice in the form of

~~S a .a d/ .b d/ b; D C C CC C S b .b d/ .a d/ a: D C CC C C Summing the terms vertically in pairs yields 2S .a b/n, which has a solution D C .a b/n S C : D 2~~

k When we turn to a geometric series k1 1 ar , since the general terms are not additively symmetric, Gauss’s pairing trick can notD be applied anymore. However, rather than writing P the partial sum S twice, we look at S and rS:

~~S a ar ar 2 ar n 1; D C C CC rS ar ar 2 ar 3 ar n: D C C CC Instead of adding, we subtract the two equalities yielding~~

~~2 n 1 n S rS a ar ar ar ar ar : (9.1) D C CC Now, all terms on the right cancel out except for ﬁrst and last; solving for S yields~~

~~a.1 r n/ S : D 1 r 93~~

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Since the expression (9.1) on the right contracts the way a telescope does, it is called a telescoping sum. But this is not the end of the story. Following the idea above, in general, we call a sequence ak telescoping if there exists a sequence bk with the property that f g f g

~~ak bk 1 bk; for all k 1. D C In this case, we have~~

~~n n~~

~~ak .bk 1 bk/ D C k 1 k 1 XD XD .b2 b1/ .b3 b2/ .bn 1 bn/ D C CC C bn 1 b1: D C~~

~~In particular, if limk bk exists, we ﬁnd the inﬁnite series !1~~

~~1 ak lim bk b1: D k k 1 !1 XD~~

This technique is useful in two ways. First, given a sequence bk it is easy to generate n f g a sequence ak such that k 1 ak bn 1 b1. This enables us to construct many f g D C convergent and divergent inﬁniteD series. Second, it determines the sums of the convergent P series. A natural question arises, namely, how can we telescope a series? That is, how can we ﬁnd such a sequence bk from thegiven one? As usual there are no general methods to f g aid the search for the sequence. In elementary calculus, this technique is usually introduced in a few examples, and students regard it as a clever but unexpected trick. Now, we are ready to explore various approaches for constructing telescoping sums, and to show how they can be systematically used to evaluate various series. More importantly, we will see that this method is not as exotic as one might think.

~~9.1 The sum of products of arithmetic sequences~~

~~Let ak be an arithmetic sequence with a nonzero common difference d. Thus, an f g D a0 nd. For any nonnegative integer m, we have C~~

ak ak 1 ak m 1 ak 1ak ak m akak 1 ak m.ak m 1 ak 1/ C C C C D C C C C .m 2/d akak 1 ak m; D C C C and so 1 ak ak 1 ak m .akak 1 ak m 1 ak 1ak ak m/ : C C D .m 2/d C C C C C Therefore, the sequence akak 1 ak m is telescoping, and f C C g n 1 ak ak 1 ak m .anan 1 an m 1 a0a1 a1 m/ : (9.2) C C D .m 2/d C C C C k 1 XD C

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~~9. On the Telescoping Sums 95~~

In particular, setting ak k in (9.2) yields D n n.n 1/ .n m 1/ k.k 1/ .k m/ C C C ; (9.3) C C D m 2 k 1 XD C from which the closed formulas for the sums of powers of positive integers follow easily. Indeed, taking m 0 and m 1 in (9.3), respectively, yields D D n n.n 1/ k C (9.4) D 2 k 1 XD and n n.n 1/.n 2/ k.k 1/ C C : (9.5) C D 3 k 1 XD Therefore, n n n.n 1/.2n 1/ k2 Œk.k 1/ k C C : (9.6) D C D 6 k 1 k 1 XD XD Similarly, substituting m 2 in (9.3) gives D n n.n 1/.n 2/.n 3/ k.k 1/.k 2/ C C C : C C D 4 k 1 XD Combining this formula with (9.4) and (9.5) results in

n n n2.n 1/2 k3 Œk.k 1/.k 2/ 3k.k 1/ k C : (9.7) D C C C C D 4 k 1 k 1 XD XD In most beginning calculus textbooks, formulas (9.4), (9.6) and (9.7), which often are 1 k proved by mathematical induction, are useful in the evaluationof such integralsas 0 x dx by appealing to the definition of a Riemann integral. Students often ask how to find these R formulas in the first place. Using (9.3), we now have a method for their discovery.

~~9.2 The sum of products of reciprocals of arithmetic sequences~~

Under the same assumptions in the preceding on ak , for any positive integer m, since f g 1 1 ak ak m C ak 1ak 2 ak m akak 1 ak m 1 D akak 1 ak m C C C C C C C md ; D akak 1 ak m C C it follows that 1 1 1 1 : akak 1 ak m D md ak ak 1 ak m 1 ak 1ak 2 ak m C C Â C C C C C Ã

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~~Therefore, the sequence 1 is telescoping, and ak akC1 akCm n o n 1 1 1 1 : (9.8) akak 1 ak m D md a1a2 am an 1an 2 an m k 1 C C Â C C C Ã XD~~

~~For ak a kd and m 1;2, we may specify (9.8) respectively, as D C D n 1 1 1 1 (9.9) .a kd/Œa .k 1/d D d a d a .n 1/d k 1 Â Ã XD C C C C C C and~~

n 1 .a kd/Œa .k 1/d/Œa .k 2/d k 1 XD C C C C C 1 1 1 : (9.10) D 2d .a d/.a 2d/ Œa .n 1/dŒa .n 2/d Â C C C C C C Ã The well-known sums n 1 1 1 (9.11) k.k 1/ D n 1 k 1 XD C C and n 1 1 1 (9.12) k.k 1/.k 2/ D 4 2.n 1/.n 2/ k 1 XD C C C C follow immediately from setting a 0; d 1 in (9.9) and (9.10). Sometimes it is worth- D D while to manipulate a sequence into a slightly different form before applying (9.8). For example, since

~~1 k 1 1 1 C ; k.k 2/ D k.k 1/.k 2/ D .k 1/.k 2/ C k.k 1/.k 2/ C C C C C C C with (9.11) and (9.12), this leads to~~

~~n 1 3 2n 3 C : k.k 2/ D 4 2.n 1/.n 2/ k 1 XD C C C Generally, applying partial fractions, we get~~

~~n 1 n m 1 (9.13) k.k m/ D m k.k n/ k 1 k 1 XD C XD C and n m l 1 n 1 : (9.14) .k l/.k m/ D m l .k l/.k l n/ k 1 k 1 XD C C XD C C C~~

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~~9. On the Telescoping Sums 97~~

~~Furthermore, appealing to formulas (9.8), (9.13) and (9.14), we can establish the following inﬁnite sums:~~

1 1 1 1 1 1 ; k.k m/ D m C 2 CC m k 1 Â Ã XD C 1 1 1 1 1 ; .m>l/; .k l/.k m/ D m l l 1 CC m k 1 Â Ã XD C C C 1 1 1 : (9.15) .a kd/ Œa .k m/d D md.a d/.a 2d/ .a md/ k 1 XD C C C C C C 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 n n 1 sin x sin C S sin.x k/ C 2 2 : (9.16) WD C D k 0 sin 2 XD Write S twice in the form of Á S sin x sin.x / sin.x n/; D C C CC C S sin.x n/ sinŒx .n 1/ sin x: D C C C CC Summing the terms vertically in pairs and using the sum to product formula ˛ ˇ ˛ ˇ sin ˛ sin ˇ 2 sin C cos ; C D 2 2 we get n n n 2S 2 sin x cos k : D C 2 2 k 0 Á XD Á To eliminate the cosine sum, we multiply both sides by sin =2 and then apply the product to sum formula 1 sin ˛ cos ˇ Œsin.˛ ˇ/ sin.ˇ ˛/ (9.17) D 2 C to obtain n n n 1 n 1 2S sin =2 sin x sin .k 1/ C sin k C : D C 2 C 2 2 k 0 Â Â Ã Â ÃÃ Á XD Telescoping the sum yields n 1 n 1 2S sin =2 2 sin x C sin C ; D C 2 2 Â Ã Â Ã 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 n n 1 cos x sin C cos.x k/ C 2 2 : (9.18) C D k 0 sin 2 XD Á Furthermore, combining (9.16) and (9.18), we get another interesting formula, sin x sin.x / sin.x n/ n C C CC C tan x : cos x cos.x / cos.x n/ D C 2 C C CC C Á Along the same lines, one can establish the following alternating sums which are perhaps less well known:

n n 1 n sin C . / sin x . / . 1/k sin.x k/ 2 C C 2 C ; C D k 0 cos 2 XD n n 1 Á n sin C . / cos x . / . 1/k cos.x k/ 2 C C 2 C : C D k 0 cos 2 XD Á Finally, we derive a closed form for

~~n 1 k Ts r sin k WD k 1 XD in the same manner as we handled the sum in (9.1). Notice that~~

2 2 n 2 n 1 .1 2r cos r /Ts r sin r sin 2 r sin.n 2/ r sin.n 1/ C D C CC C 2r 2 sin cos 2r n 1 sin.n 2/ cos n 3 n 1 2r sin.n 1/ cos r sin r sin.n 3/ CC r n sin.n 2/ r n 1 sin.n 1/: C C C Let Œr k denote the coefﬁcient of r k. Combining the like terms yields

~~sin 2 2 sin cos ; k 2; D Œr k sin.k/ 2 sin.k 1/ cos sin.k 2/; 2~~

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~~9. On the Telescoping Sums 99~~

9.3.2 Angles in geometric sequences We begin with the trigonometric identity 3 1 sin3 x sin x sin.3x/: D 4 4 Replacing x by 3kx, multiplying both sides by 1=3k, and then summing over k from 0 to n, gives

n 3 k n sin .3 x/ 1 k 1 k 1 sin.3 x/ sin.3 C x/ 3k D 4 3k 1 4 3k k 0 k 0 Â Ã XD XD 3 sin.3n 1x/ sin x C : D 4 3n 4 Similarly, we ﬁnd n cos3.3kx/ 3 cos.3n 1x/ . 1/k cos x . 1/n C : 3k D 4 C 3n 4 k 0 XD Next, it is straightforward to verify that sin.2x/ 1 .cot x 3 cot.3x//: 1 2 cos.2x/ D 4 C Replacing 2x by x=3k and multiplyingboth sides by 1=3k yields k sin.x=3 / 1 1 k 1 k 1 cot.x=2 3 / cot.x=2 3 / : 3k.1 2 cos.x=3k// D 4 3k 3k 1 C Â Ã Summing over k from 0 to n, we now obtain n sin.x=3k/ 1 1 cot.x=2 3n/ cot.x=2/ : 3k.1 2 cos.x=3k// D 4 3n k 0 Â Ã XD C Moreover, using the identities tan x cot x 2 cot.2x/; D sec2 x 4 csc2.2x/ csc2 x; D and csc x cot.x=2/ cot x D respectively, the interested reader can correspondingly establish

~~n 1 2k tan.2k x/ cot x 2n cot.2nx/; D k 1 XD n 1 22k sec2.2k x/ 22n csc2.2nx/ csc2 x; D k 0 XD n csc.2k x/ cot.x=2/ cot.2nx/: D k 0 XD~~

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9.3.3 Sums related to the inverse tangent function 2 MonthlyProblem 10292, 1993 asks for the value of n1 1 arctan.2=n /. Appealing to the identity D P ˛ ˇ arctan ˛ arctan ˇ arctan ; (9.21) D 1 ˛ˇ Â C Ã we have 2 arctan arctan.n 1/ arctan.n 1/: n2 D C Â Ã The sum telescopes to

lim .arctan.n 1/ arctan.n/ arctan 1/ 3=4: n C C D !1 More often, the same idea can be extended to those situations in which the telescopic nature is hidden by a function. For k 1 and any nonnegative function f , (9.21) gives f .k/ f .k 1/ arctan f .k/ arctan f .k 1/ arctan C : C D 1 f .k/f .k 1/ Â C C Ã Thus,

n f .k/ f .k 1/ arctan C arctan f .1/ arctan f .n 1/: 1 f .k/f .k 1/ D C k 1 Â Ã XD C C For instance, with f .k/ 1=.2k 1/ we get D n 1 1 arctan arctan 2k2 D 4 2n 1 k 1 Â Ã Â Ã XD C and with f .k/ kxk;x 0, D n 1 kx k x arctan C xk arctan.nxn/ arctan.x/: 1 k.k 1/x2k 1 D k 1 Â C Ã XD C C In general, let m be a positive integer and let f be monotonic increasing with f .0/ 0 D and limx f .x/ . We have !1 D 1 m 1 1 f .k m/ f .k/ m f .i/ f .i 1/ arctan C arctan : (9.22) 1 f .k/f .k m/ D 2 1 f .i/f .i 1/ k 0 Â Ã i 1 Â Ã XD C C XD C To see this, ﬁrst notice that

~~f.k m/ m dx 1 dx m 1 C 2 2 2 D 0 1 x D f.k/ 1 x Z k 0 Z C XD C f.1/ f.2/ f .m 1/ dx dx dx .m 1/ 2 .m 2/ 2 2 : C f.0/ 1 x C f.1/ 1 x CC f .m 2/ 1 x Z C Z C Z C~~

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~~After carrying out all the integrations, we obtain~~

1 arctan f .k m/ arctan f .k/ C n 0 XD m .m 1/ arctan f .1/ arctan f .0/ .m 2/ arctan f .2/ arctan f .1/ D 2 arctanf .m 1/ arctan f .m 2/ m 1 m arctan f .i/ arctan f .i 1/ : D 2 i 1 XD Now, (9.22) follows from (9.21) immediately. When m 2, f .x/ x and m 1, D D D f .x/ 2x, (9.22) yields D 1 2 3 1 2 arctan and arctan : .k 1/2 D 4 .2k 1/2 D 2 k 0 Â Ã k 0 Â Ã XD C XD C After shifting the index in the ﬁrst identity, we have solved the Monthly Problem 10292 again. It is interesting to see the case where m 1; f .0/ 0; f .k/ 2k 1 a for a>0. D D D C Here (9.22) gives

~~1 2 arctan.a 1/ arctan ; C C .2k a/2 D 2 k 1 Â Ã XD C where the value is independent of a. Differentiating this with respect to a yields~~

1 4.2k a/ 1 C : 4 .2k a/4 D a2 2a 2 k 1 XD C C C C In particular, for a 0 and a 1, unexpectedly, we get D D 1 2k 1 1 4.2k 1/ 1 and C : 1 4k4 D 2 4 .2k 1/4 D 5 k 1 k 1 XD C XD C C 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 involvingthe Fibonacci numbers in which the telescoping technique can be utilized. Since Fk Fk 2 Fk 1; D C C the series Fk is telescoping and f g n

~~Fk Fn 2 1: D C k 1 XD~~

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~~Moreover, since 1 1 1 ; Fk 1Fk 1 D Fk 1Fk FkFk 1 C C we ﬁnd~~

~~n 1 1 1 ; Fk 1Fk 1 D FnFn 1 k 2 C C XD n F 1 1 k 2 : Fk 1Fk 1 D Fn Fn 1 k 2 C C XD~~

~~In general, for any sequence ak with f g~~

~~ak 1 ˛ak ak 1; .k 3/ C D C where ˛ is a nonzero constant, we have~~

~~n 1 ak .an 1 an a1 a2/: D ˛ C C k 2 XD 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 ak , let ¤ f g 1 a1a2 ak 1 b1 ; bk .k 2/: D x D x.x a1/ .x ak 1/ C C Then a1a2 ak 1 bk bk 1 : C D .x a1/ .x ak / C C The telescoping sum results in~~

n a1a2 ak 1 1 a1a2 an : (9.23) .x a1/ .x ak / D x x.x a1/ .x an/ k 1 XD C C C C This, along with k 1 1 1 5 1 . 1/ .3/ C ; D k3 D 2 2k k 1 k 1 k3 XD XD k Â Ã 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 1 1 .2/ 3 : (9.24) D k2 D 2k k 1 k 1 k2 XD XD k Â Ã~~

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~~To prove (9.24), we set x n and ak k in (9.23) thereby giving D D n .k 1/Š 1 nŠ : .n 1/.n 2/ .n k/ D n n.n 1/ .2n/ k 1 XD C C C C This leads to~~

~~1 n .k 1/Š .n 1/Š n2 D n.n 1/ .n k/ C n.n 1/ .2n/ k 1 XD C C C n 1 .k 1/Š .n 1/Š : D n.n 1/ .n k/ C n2.n 1/ .2n 1/ k 1 XD C C C Therefore,~~

~~n 1 1 1 .k 1/Š 1 .n 1/Š : (9.25) n2 D n.n 1/ .n k/ C n2.n 1/ .2n 1/ n 1 n 1 k 1 n 1 XD XD XD C C XD C For the double series in (9.20), we rearrange the series and sum in columns obtaining~~

n 1 .k 1/Š 1 1 1 .n 1/Š n.n 1/ .n k/ D .k 1/.k 2/ .n k 1/ n 1 k 1 n 1 "k n # XD XD C C XD XD C C C C 1 .n 1/Š ; D n.n 1/ .2n/ n 1 XD C where we have used the following telescoping sum from the identity (9.8):

~~1 1 .k 1/.k 2/ .n k 1/ k n XD C C C C 1 1 1 1 D n .k 1/.k 2/ .n k/ .k 2/.k 3/ .n k 1/ k n Â Ã XD C C C C C C C 1 : D n.n 1/ .2n/ C Hence the right-hand side of (9.25) becomes~~

~~1 1 1 1 .n 1/Š .n 1/Š 3 n.n 1/ .2n/ C n2.n 1/ .2n 1/ D n.n 1/ .2n/ n 1 Ä n 1 XD C C XD C 1 1 3 D 2n n 1 n2 XD n Â Ã 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 illustratethis approach by two examples.

~~Example 1. Monthly Problem E 3352 (1989). Show that~~

~~1 1 e : (9.26) kŠ.k4 k2 1/ D 2 k 0 XD C C Since .k2 k 1/.k2 k 1/ k4 k2 1; C C C D C C partial fractions gives~~

~~2 k 1 k 1 C : k4 k2 1 D k2 k 1 k2 k 1 C C C C C Thus,~~

1 1 k 1 k 1 C kŠ.k4 k2 1/ D 2 kŠ.k2 k 1/ kŠ.k2 k 1/ C C Â C C C Ã 1 k k 1 1 : D 2 .k 1/Š.k2 k 1/ kŠ.k2 k 1/ C .k 1/Š Â C C C C C Ã b k 1 Setting k 2ŒkŠ.k2 k 1/ yields D C 1 1 1 bk 1 bk : kŠ.k4 k2 1/ D C C 2 .k 1/Š C C C This transforms the left-hand side of (9.26) into

1 1 1 1 1 1 1 e : 2 C 2 .k 1/Š D 2 kŠ D 2 k 0 k 0 XD C XD Example 2. Transform .3/ into a faster converging series. We start with 1 1 S ; D .k 1/k.k 1/ k 2 XD C which telescopes and, by (9.12) sums to 1=4 after shifting the index of summation. There- fore,

~~5 1 1 1 5 1 1 .3/ : D 4 k3 k k3 D 4 k3.k2 1/ k 2 Â Ã k 2 XD XD 5 3 Here the nth term is approximately n whereas before it was n .~~

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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 ak whenever ak 1=ak f g C 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:

n If ak 1=ak is rational in k, is the sum k 1 ak telescoping? C D Gosper’s algorithm proceeds in two steps, each ofP which is fairly straightforward. Step 1 is to express the ratio in the special form

~~ak 1 p.k 1/ q.k/ C C ; ak D p.k/ r.k/ where p;q; and r are polynomials satisfying~~

~~gcd.q.k/; r.k h// 1; for all integers h 0: C D Step 2 is to ﬁnd a nonzero polynomial solution c.k/ of~~

~~q.k/c.k 1/ r.k 1/c.k/ p.k/; (9.27) C D whenever possible. If c.k/ exists, we ﬁnd~~

~~r.k 1/c.k/ bk ak; ak bk 1 bk: D p.k/ D C~~

~~Let us use Gosper’s algorithmto show that the sum~~

~~n .4k 1/ kŠ Sn C D .2k 1/Š k 0 XD C is telescoping. The term ratio~~

~~ak 1 4k 5 C C ak D 2.4k 1/.2k 3/ C C is rational in k as expected. The choices~~

~~p.k/ 4k 1; q.k/ 1; r.k/ 2.2k 3/ D C D D C clearly satisfy the conditions in Step 1. Thus, equation (9.27) becomes~~

~~c.k 1/ 2.2k 1/c.k/ 4k 1; C C D C and the constant polynomial c.k/ 1 is a solution. Hence D 2.2k 1/ .4k 1/ kŠ kŠ bk C C 2 D 4k 1 .2k 1/Š D .2k/Š C C~~

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satisfies ak bk 1 bk and D C nŠ Sn bn 1 b0 2 : D C D .2n 1/Š C (9.26) is also a good application of Gosper’s algorithm. Rewriting (9.26) in the equivalent form 1 1 1 0; kŠ.k4 k2 1/ 2kŠ D k 0 Â Ã XD C C k2 Gosper’s algorithm finds that bk 2 . This yields the sum as b b0 0 as D 2ŒkŠ.k k 1/ 1 D expected. C 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 2 example related to our discussion in 9.3.3 is to ﬁnd n1 1 arctan.1=n / in closed form. The answer turnsout tobe a curiousidentity(5.34), whichD cannot be derived by telescoping P methods.

~~Exercises 1. Find the followingsums in closed form:~~

12 22 n2 (a) 1 3 3 5 .2n 1/.2n 1/ . C CC C n (b) n k2 : k 1 k D Â Ã P 2. Evaluate n n cos kx sin.n k/x: k k 0 Â Ã XD 3. Prove that n tan nx sec kx sec.k 1/x : D sin x k 1 XD 4. Prove that 2 1 1 1 3 .2n 1/ 4 1 : C n 1 2 4 2n D n 1 Â Ã XD C

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5. Evaluate 2n 1 4n 2n . 1/k : 2k k k 0 Â Ã Â Ã XD 6. Let n be a positive integer. Prove that 1 1 1 1 <2 1 : 2p1 C 3p2 CC .n 1/pn pn 1 C Â C Ã 7. Find the sum of the series 1 ..k/ 1/: k 2 XD 8. Find thesum of 1 1 iŠjŠ : .i j 1/Š i 1 j 1 XD XD C C 9. Prove that 1 (a) k1 1 arctan k2 k 1 4 . D C C D P 2akÁ (b) k1 1 arctan a2k4 a2k2 1 2 . D C D P k 1 2 Á (c) k1 1 . 1/ C arctan k2 4 . D D Á 10. Let FPn be the nth Fibonacci number. Show that

~~1 Fn 2 Fn 3 arctan C : 1 FnFn 2 D 4 n 1 Â C Ã XD C 11. Putnam Problem, 1977-A4. For 0~~

~~2n 1 x : 2nC1 n 0 1 x XD 12. Putnam Problem, 1978-B2. Express~~

~~1 1 1 m2n mn2 2mn n 1 m 1 XD XD C C as a rational number. 13. Putnam Problem, 1977-B1. Evaluate the inﬁnite product~~

3 1 n 1 : n3 1 n 2 YD C Hint. Using a telescoping product. A telescoping product is the multiplicative analogue of a telescoping sum. Here each factor is written as a quotient in such a way that successive factors cancel out.

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~~14. Putnam Problem, 1981-B1. Find~~

~~1 n n lim .5h4 18h2k2 5k4/ : n n5 C !1 " h 1 k 1 # XD XD 15. Open Problem.Let P.x/=Q.x/ be a rational function with real coefﬁcients in which degQ 2 degP . Evaluate C~~

~~1 P.k/ arctan Q.k/ k 1 XD in closed form. When the leading coefﬁcient of Q.x/ is 1, show that~~

~~n 1 P.k/ arctan arg..1 aj // .mod 2/; Q.k/ D k 1 j 1 XD XD~~

~~where aj .1 j n/ are the roots of Q.x/ iP.x/. Ä Ä~~

References [1] R. Apéry, Irrationalitéde .2/ et .3/, Astérisque, 61 (1979) 11–13. [2] R. Gosper,Indefinite hypergeometric sums in MACSYMA, Proc. MACSYMA Users Conference, Berkeley, CA, (1977) 237–252 [3] M. Petkovsek, H. Wilf, and D. Zeilberger, A=B, A. K. Peters, Ltd., Wellesley, MA, 1996. [4] ——, How to do Monthly problems with your computer, Amer. Math. Monthly, 104 (1997) 505– 519.

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~~Summation of Subseries in Closed Form~~

~~One of the problems in the Second Putnam Mathematical Competition(1939) was to prove that if~~

3n 3n 1 3n 2 1 x 1 x 1 x u.x/ ; v.x/ C ; w.x/ C ; D .3n/Š D .3n 1/Š D .3n 2/Š n 0 n 0 n 0 XD XD C XD C then u3 v3 w3 3uvw 1. The suggested solution relied on C C D u v w ex; C C D u !v !2w e!x; C C D 2 u !2v !w e! x; C C D where ! e2i=3, a primitive cubic root of unity. Notice that all u, v and w are subseries D of ex. As a byproduct, solving the above system of equations yields

~~1 x x=2 p3x u.x/ e 2e cos ; D 3 C 2 !!~~

~~1 x x=2 p3x x=2 p3x v.x/ e e cos p3e sin ; D 3 2 ! C 2 !!~~

~~1 x x=2 p3x x=2 p3x w.x/ e e cos p3e sin : D 3 2 ! 2 !!~~

Later, for any positive integer k and m 0;1;:::;k 1, Rubel and Stolarsky in [12] D proved 2 k1 xkn m 1 e!x e! x e! x C ex ; (10.1) .kn m/Š D k C !m C !2m CC !.k 1/m n 0 ! XD C where ! e2i=k is a primitive kth root of unity. D 109

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~~These results motivate a question that does not seem to have received much attention: n let f .x/ n1 0 anx : Consider a subseries of f : D D P n anx ;~~

~~n N1 X2~~

~~where N1 is a subset of N with N1 and N1 N. It is interesting to see whether there ¤; ¤ is a closed-form expression for the function f1.x/ such that~~

n f1.x/ anx : D n N1 X2 In this chapter, we extend (10.1) to general analytical functions, and prove the following theorem. Theorem 10.1. Let k be a positive integer and let ! e2i=k be a primitive kth root of n D unity. If f .x/ n1 0 anx , then for any integer m, we have D D P k 1 n m m k 1 j m j anx amx am k x C ! f .! x/; (10.2) D C C CD k n m .mod k/ j 0 Á X XD

where aj 0 when j <0. D n Proof. Since f .x/ n1 0 anx , we have D D k 1 P k 1 j m j j m 1 j n ! f .! x/ ! an.! x/ D j 0 j 0 n 0 XD XD XD k 1 1 .n m/j n an ! x : (10.3) D 0 1 n 0 j 0 XD XD @ A If n m kl for some integer l, appealing to !k 1, we have !.n m/j !klj D D D D .!k /lj 1 and so D k 1 !.n m/j k: D j 0 XD On the other hand, if n m kl, ¤ k 1 .n m/k 1 ! !.n m/j 0: D 1 !n m D j 0 XD Thus, the summands in the right-hand side of (10.3) are for only those values of n for which n m are multiples of k, i.e., n m .mod k/. This proves (10.2) as desired. Á In view of (10.2), the left-hand side is a subseries of f that for each n N has a block n 2 of k terms of the form anx , a process known as multisection [8]; while the right-hand side is an appropriate linear combination of the original series. DeMorgan [5] called (10.2) a series multisection formula. It is easy to see that (10.1) is an immediate consequence

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of (10.2). Formula (10.2) somehow seems to be not widely known and remains under- appreciated. For example, a recent University Math Contest asked contestants to determine the closed form of u.x/. Instead of applying (10.1) or (10.2), the proposed solution requires solving the differential equation u .x/ u.x/. The derivation of (10.2) is within easy 000 D reach of undergraduate analysis students. To demonstrate its power and wide applicability, we illustrate it by a variety of examples. Our discussion will lead to the discovery of results some of which have been known for a long time, some of which were found only recently, as well as some which appear to be new. However, these results do not require the extensive background and advanced tools that the original discoveries required.

Example 1. In [7], Hardy and Williams proposed to determine the real function of x whose power series is x3 x9 x15 : 3Š C 9Š C 15Š C Clearly, this is a subseries of ex. Consider the primitive 6th root of unity,

1 p3 ! e2i=6 i: D D 2 C 2 Then ! 3 !3 1. Appealing to (10.2) with k 6; m 3, we have D D D D 3 9 15 6n 3 x x x 1 x C 3Š C 9Š C 15Š CD .6n 3/Š n 0 XD C 1 2 2 .ex e!x e! x e x e !x e ! x / D 6 C C 1 .sinh x sinh.!x/ sinh.!2 x// D 3 C 1 x p3x sinh x 2 sinh cos : D 3 " 2 2 !# Á Example 2. It is well known that

~~n 2 3 1 x x x ln.1 x/ x ; x < 1: (10.4) D n D C 2 C 3 C j j n 1 XD Therefore, (10.2) yields~~

kn m k 1 1 x 1 C ! j m ln.1 !j x/; x < 1: (10.5) kn m D k j j n 0 j 0 XD C XD In particular, for x <1, j j 2n 1 3 5 1 x x x 1 1 x C x ln C ; 2n 1 D C 3 C 5 CD 2 1 x n 0 Â Ã XD C 4n 3 3 7 11 1 x x x x 1 1 x 1 C ln C tan 1 x; 4n 3 D 3 C 7 C 11 CD 4 1 x 2 n 0 Â Ã XD C

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~~where in the second identity we have used that 1 1 ix tan 1 x ln C : D 2i 1 ix Â Ã As an application, (10.5) will provide uniform proofs of many formulas that appear in the classical table [6].~~

Example 3. In the following, we evaluate some combinatorial sums in closed form. When (10.2) is applied to n n f .x/ .1 x/n xi ; D C D i i 0 Â Ã XD it gives .n m/=k k 1 b c n kj m 1 j m j n x C ! .1 x! / ; kj m D k C j 0 Â C Ã j 0 XD XD where x denotes the greatest integer of x. This enables us to derive various identities b c listed in [4] such as

.n m/=k k 1 n b c n 1 j .n 2m/j 2 cos cos ; kj m D k k k j 0 Â C Ã j 0 Â Ã XD XD n=k k 1 n b c n 1 j nj 2 cos cos ; kj D k k k j 0 Â Ã j 0 Â Ã XD XD n=3 b c n 1 n 2n 2 cos : 3j D 3 C 3 j 0 Â Ã XD Á A more challenging problem is to evaluate the combinatorial expression

n kn 1 kn 1 Tn.k/ C C ; D ki 1 ki 2 i 1 Â Ã Â Ã XD where the case k 3 occurs in an oceanographic problem modeling the Gulf stream [9]. D When k 1, we have the telescoping sum D n n 1 n 1 n 1 1 Tn.1/ C C C n.n 1/: D i 1 i 2 D n 1 D 2 C i 1 Â Ã Â Ã Â Ã XD Similarly, by the binomial theorem,

~~n 2n 1 2n 1 Tn.2/ C C D 2i 1 2i 2 i 1 Â Ã Â Ã XD 2n 1 C i 1 2n 1 2n 1 . 1/ C C C 1 D i C 2n i 0 Â Ã Â Ã XD 2n 1 .1 x/ C x 1 .2n 1/ 1 2n: D j D C C D~~

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For k 3, we rewrite Tn.k/ as kn 1 kn 1 Tn.k/ C C ; D i i i 1 .mod k/ Â Ã i 2 .mod k/ Â Ã Á X Á X where i ranges from k 2 to kn 1. Applying (10.2) yields k 1 k 1 1 j j kn 1 1 2j j kn 1 Tn.k/ ! .1 ! / ! .1 ! / D k C C k C C j 0 j 0 XD XD k 1 1 !j .1 !2j /.1 !j /kn D k C j 1 XD k 1 1 2j j j knj=2 j=2 j=2 kn ! .! ! /! .! ! / D k C j 1 XD k 1 1 !2j . 2i sin 2j=k/. 1/nj .2 cos j=k/kn: D k j 1 XD But !2.k j/ ! 2j ; sin 2.k j /=k sin 2j=k, and cos.k j /=k cos j=k, D D D so the last sum can be cut in half. Thus

.k 1/=2 1 b c nj kn 2j 2j Tn.k/ . 1/ . 2i sin 2j=k/.2 cos j=k/ .! ! / D k j 1 XD .k 1/=2 1 b c . 1/nj .2 sin 2j=k/.2 sin 4j=k/.2 cos j=k/kn: D k j 1 XD Example 4. We now demonstrate how to use (10.2) to compute series involvingthe central binomial coefﬁcients. By the binomial series, we have the generating function for the central binomial coefﬁcients as

~~1 1 2n xn 1 2x 6x2 20x3 ; x < 1=4: p1 4x D n D C C C C j j n 0 Â Ã XD Applying (10.2) to this series gives~~

k 1 1 2.kn m/ kn m 1 1 C x C ; x < 1=4: kn m D k !j mp1 4!j x j j n 0 Â C Ã j 0 XD XD If we take k 2;4 and m 0, respectively, then D D 1 4n 1 1 1 x2n ; 2n D 2 p1 4x C p1 4x n 0 Â Ã Â Ã XD C 1 8n 1 1 1 2.1 p1 16x2/ x4n C C ; 4n D 4 0p1 4x C p1 4x C s 1 16x2 1 n 0 Â Ã XD C C @ A

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~~114 Excursions in Classical Analysis~~

~~where we have used~~

a pa2 b2 a pa2 b2 pa bi C C C C i: ˙ D s 2 ˙ s 2 Setting x 1=8, we recover Lehmer’s two sums that appeared in [10]: D 4n 1 2n 3p2 p6 Â Ã C ; 82n D 6 n 0 XD 8n 1 4n 15p2 5p6 6 5 p10 Â Ã C C C : 84n D 60 n 0 p XD Similarly, since (see (15.1))

~~1 1 1 2n 2.sin 1 x/2 .2x/2n ; D n2 n n 1 Â Ã XD applying x.d=dx/ twice on both sides gives~~

2 1 1 x x sin x 1 2n .2x/2n: 1 x2 C .1 x2/3=2 D n n 1 Â Ã XD By (10.2), letting k 4 and m 0, appealing to sin 1 z sinh 1.iz/=i, we ﬁnd that D D D 1 4 1 1 1 4n x x sin x x sinh x .2x/4n : 2n D 1 x4 C 2.1 x2/3=2 .1 x2/3=2 n 1 Â Ã XD C Setting x 1=2 yields D 1 1 4n 1 p3 2p5 ln ; 2n D 15 C 27 25 n 1 Â Ã XD where .1 p5/=2, the golden ratio as we have seen in Chapter 6. D C A comparable method studying the series involving the central binomial coefﬁcients proposed in [1, 2, 13] is mostly based on the identity

~~1 1 n .n 1/ t k.1 t/n k dt; k D C Â Ã Z0 a particular case of Euler’s beta function. Here (10.2) provides us an alternate method and maintains the exposition at an elementary level.~~

~~Example 5. We begin withan inﬁnitesum, which eventuallyleads to an elementary proof of the Gauss formula on the digamma function at rational values. For 0~~

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~~10. Summation of Subseries in Closed Form 115~~

~~in closed form. To use (10.2), we consider the power series~~

~~1 1 k m kn f1.x/ x ; x < 1: D n 1 m kn C j j n 0 Â Ã XD C C The second series in f1.x/ was already evaluated in (10.5). Using the previous result, we get~~

~~k 1 m k k j m j f1.x/ x ln.1 x / ! ln.1 ! x/ D C j 0 XD k k 1 1 x xm k ln .xm k 1/ ln.1 x/ ! j m ln.1 !j x/: D 1 x C j 1 XD~~

To see the connection between S.k; m/ and f1.x/, we turn to Abel’s continuity theorem, n which claims: If g.x/ 1 cnx converges for x < 1 and if 1 cn converges, D n 0 j j n 0 then D D P P 1 lim g.x/ cn: x 1 D ! n 0 XD Thus, by Abel’s continuity theorem, we have

k 1 j m j S.k; m/ lim f1.x/ ln k ! ln.1 ! /: D x 1 D C ! j 1 XD When the sum on the right-hand side is translated into a sum of real and imaginary terms, it is easy to show that the imaginary terms cancel out and that the real part of the j th summand is given by 1 2mj 2j 2mj sin.2j=k/ cos ln 2 2 cos sin arctan : 2 k k k 1 cos.2j=k/ Â Ã Â Ã Finally, since sin.2j=k/ j j arctan arctan cot ; 1 cos.2j=k/ D k D 2 k Â Ã Â Ã appealing to the well-known identities

n sin jx sin..n 1/x=2/ sin.nx=2/= sin.x=2/ D C j 1 XD and n 1 sin nx n cos..2n 1/x=2/ j sin jx ; D 4 2.x=2/ 2 sin.x=2/ j 1 sin XD we obtain k 1 2mj sin.2j=k/ m sin arctan cot k 1 cos.2j=k/ D 2 k j 1 Â Ã XD

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~~116 Excursions in Classical Analysis~~

~~and so~~

k 1 m 1 2mj 2j S.k; m/ ln k cot cos ln 2 2 cos : (10.6) D 2 k C 2 k k j 1 Â Ã XD It is remarkable that S.k; m/ is linked to the evaluation of the reciprocal of polygonal numbers in closed form.

~~Figure 10.1. The ﬁrst four polygons~~

A polygonal number isa number represented by dotsarrayed in the shape of a polygon. Figure 10.1 graphically illustrates the process by which the polygonal numbers are built up. Starting with the nth triangular number Tn, we have

~~n.n 1/ Tn 1 2 n C : D C CC D 2~~

~~Notice that~~

~~2 n 2Tn 1 n Sn C D D gives the nth square number,~~

~~1 n 3Tn 1 n.3n 1/ Pn C D 2 D~~

~~gives the nth pentagonal number, and so on. In general, let pr .n/ be the nth r-gon number. Then 1 1 pr .n/ nŒ.n 1/r 2.n 2/ nŒ.r 2/n .r 4/: D 2 D 2~~

Recently, Problem 1147 [11] in the Pi Mu Epsilon Journal asks one to ﬁnd the sum of the reciprocal of the pentagonal numbers. The published solution [3] to this problem used Euler’s integral. In terms of S.k; m/, we obtain the following intriguinggeneralization: for r 5,

~~1 1 2 1 1 r 2 2 S.r 2; 2/: p .n/ D r 4 n 1 2 .r 2/n D r 4 n 1 r n 0 Â Ã XD XD C C~~

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~~10. Summation of Subseries in Closed Form 117~~

~~In particular, appealing to (10.6), we have~~

1 1 1 .9 ln 3 p3/; p .n/ D 3 n 1 5 XD 1 1 1 .9 ln 3 p3/; p .n/ D 12 C n 1 8 XD 1 1 1 .6 ln 2 /; p .n/ D 6 C n 1 10 XD 1 1 1 .4 ln 2 3 ln3 p3 /: p .n/ D 10 C C n 1 14 XD For readers familiar with the digamma function, recall that d .x/ .x/ Œln .x/ 0 : D dx D .x/ Thus, 1 1 k .k=m/ .1/ S.k; m/: D n 1 m kn D n 0 Â Ã XD C C The proceeding calculation based on (10.2) indeed recaptured the famous Gauss formula: .m=k/ S.k; m/ D C k 1 m 1 2mj 2j ln k cot cos ln 2 2 cos ; D 2 k C 2 k k j 1 Â Ã XD where is Euler’s constant.This shows that .x/ can be evaluated by elementary functions when x is a rational number. The above examples are exploratory, interesting and well within grasp of analysis students. Thus, (10.2) opens up a new world of results that would not otherwise have been obtained except by more specialized techniques. Combining with the differential and integral operators, we extend (10.2) to the formulas

k 1 n 1 j m j nanx ! F.! x/; where F.x/ xf .x/; D k D 0 n m .mod k/ j 0 Á X XD k 1 a 1 n xn ! j m G.!j x/; where G.x/ x f.t/dt : n D k D 0 t n m .mod k/ j 0 Á X XD R These will enable us to discover many more identities. The details are left to the reader.

~~Exercises 1. Show that .n 1/=3 b c n 1 .n 2/ 2n 2 cos : 3j 1 D 3 C 3 j 0 Â C Ã Â Ã XD~~

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~~118 Excursions in Classical Analysis~~

~~2. Show that .n 1/=2 b c n .1 px/n .1 px/n xj C : 2j 1 D 2px j 0 Â C Ã XD 3. Find a closed form of n=2 b c n x2k : 2k 2k 1 k 0 Â Ã XD C 4. For any a>0, prove that~~

~~n 2n . a/j .1 a/n cos.2n tan 1 pa/: 2j D C j 0 Â Ã XD 5. Let the generating function of Fibonacci numbers be given by~~

~~1 n x Fnx : D 1 x x2 n 1 XD Find the generating functions for F2n and F2n 1 respectively via (10.2). C 6. Determine the real function of x whose power series is~~

4k 3 3 7 11 1 x C x x x .4k 3/Š D 6 C 5040 C 39916800 C k 0 XD C 7. Find 1 1 x4n 1 .4n 1/.4n 3/ C n 0 XD C C in closed form. 8. Monthly Problem 2991 [1922, 356; 1924, 51–52]. Sum the infinite series 3x2 4z4 6z6 S2.x/ 1 ; D C 2Š C 4Š C 6Š C where the numerators of the coefficients form a series of numbers whose third differences are all equal to 2. 9. It is well-knownthat ex is bounded for x<0. Determine all possible proper subseries of ex which remain bounded for x<0. 10. Open Problem on Ramanujan q-series. Define

~~1 A.q/ .1 qn/ 1 q q2 q5 q7 q12 q15 q22 : D D C C C C n 1 YD Now, we trisect the series into~~

~~12 15 51 57 117 126 A0 1 q q q q q q ; D C C C 7 22 40 70 100 145 A1 q q q q q q q ; D C C C C C 2 5 26 35 77 92 155 A2 q q q q q q q : D C C C C C~~

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~~10. Summation of Subseries in Closed Form 119~~

Look for algebraic relations among A0; A1; A2 by using Mathematica or Maple. For 2 2 2 example, Berndt and Hart recently found that A2A A0A A1A 0. For an 0 C 1 C 2 D 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, Mas- sachusetts, 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 undergraduatemath- ematics 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 the central binomial coefﬁcients, Amer. Math. Monthly, 92 (1985) 449-457. [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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~~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 Fn 2 Fn 1 Fn, we found that the generating function C D C C for Fibonacci numbers is 1 n x Fnx : (11.1) D 1 x x2 n 1 XD In general, for any positive integer k, let the generating function of F k be f n g

1 k n Gk.x/ F x : D n n 1 XD It is natural to ask whether there is an explicit formula for Gk.x/. When k 2, there is a D recursive relation 2 2 2 2 Fn 3 2Fn 2 2Fn 1 Fn : (11.2) C D C C C To prove this, using Binet’s formula

~~1 n n Fn .˛ ˇ /; (11.3) D p5~~

where ˛ .1 p5/=2; ˇ 1=˛ .1 p5/=2; appealing to ˛ˇ 1, we square D C D D D (11.3) to get 1 F 2 ..˛2/n 2. 1/n .ˇ2/n/: n D 5 C This shows that F 2 is a linear combination of the nth power of 1;˛2; and ˇ2. Thus, the n corresponding characteristic equation becomes

~~.t 1/.t ˛2/.t ˇ2/ t 3 2t 2 2t 1 0 C D C D from which (11.2) follows.~~

~~121~~

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~~122 Excursions in Classical Analysis~~

~~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 2 Fn 3 aFn 2 bFn 1 cFn ; C D C C C C where a;b and c are constants to be determined. Putting n 1;2;3, respectively, yields D the system of equations~~

~~32 22a 12b 12c; D C C 52 32a 32b 12c; D C C 82 52a 32b 22c: D C C Solving for a;b;c leads to (11.2). Once (11.2) has been proved, we have~~

2 2 2 2 3 2 4 G2.x/ F1 x F2 x F3 x F4 x ; D C 2 2 C 2 3 C 2 4 C 2 5 2xG2.x/ 2F1 x 2F2 x 2F3 x 2F4 x ; 2 D C 2 3 C 2 4 C 2 5 C 2 6 2x G2.x/ 2F1 x 2F2 x 2F3 x 2F4 x ; 3 D C 2 4 C 2 5 C 2 6 C2 7 x G2.x/ F x F x F x F x : D 1 C 2 C 3 C 4 C Summing the terms vertically and then appealing to (11.2), we obtain

2 3 2 .1 2x 2x x /G2.x/ x x ; C D and so 1 2 n x.1 x/ G2.x/ F x : (11.4) D n D .1 x/.1 3x x2/ n 1 XD C C Similarly, we have 3 3 3 3 3 Fn 4 3Fn 3 6Fn 2 3Fn 1 Fn (11.5) C D C C C C which results in

2 1 3 n x.1 2x x / G3.x/ F x : (11.6) D n D .1 x x2/.1 4x x2/ n 1 XD C 4 5 Theoretically, repeating this process, we can ﬁnd recurrence relations for Fn ; Fn ;:::, and thereby determine the corresponding generating functions. However, this process becomes tediousand not feasible for larger k.As amatter offact a proofof a general explicitformula in print has not been found. In 1962, Riordan [2], instead of giving an explicit form, found that Gk.x/ satisﬁes the recurrence relation

k=2 k 2 b c i i .1 akx . 1/ x /Gk.x/ 1 kx . 1/ aki Gk 2i .. 1/ x/=i; C D C i 1 XD 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 3;4. D 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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~~11. Generating Functions for Powers of Fibonacci Numbers 123~~

functions for Fkn and the Lucas numbers Lkn based on Binet’s formulas. Then, we f g f g k introduce the power-reduction formula, which enables us to show that Fn is a linear combination of either Lkn or Fkn , from which we establish an explicit formula for Gk.x/. f g f g Finally,we extend our results to more general sequences including Lucas and Pell numbers.

~~We start with establishing the generating functions for Fkn and Lkn , where the f g f g Lucas numbers Ln are deﬁned by f g~~

~~L1 1; L2 3; D D~~

~~Ln Ln 1 Ln 2 (for n 3). D C Similar to (11.3), we have Binet’s formula for Ln, namely~~

~~n n Ln ˛ ˇ : (11.7) D C~~

~~We now prove the generating functions for Fkn and Lkn are f g f g~~

~~1 n Fk x Fknx ; (11.8) D 1 L x . 1/k x2 n 0 k XD C~~

1 n 2 Lk x Lknx : (11.9) D 1 L x . 1/k x2 n 0 k XD C To see this, appealing to ˛ˇ 1, by (11.3) and (11.7), we have D kn kn 1 n 1 ˛ ˇ n Fknx x D p n 0 n 0 5 XD XD 1 1 1 ˛kn xn ˇkn xn D p 5 n 0 n 0 ! XD XD 1 1 1 D p5 1 ˛k x 1 ˇk x Â Ã 1 .˛k ˇk/x k 2 D p5 1 Lk x . 1/ x C F x k ; k 2 D 1 Lk x . 1/ x C 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 2 and k 3. This process not D D only reveals how to approach the general case but also leads to new proofs for G2.x/ and G3.x/. For k 2, by (11.9), D 1 n 2 3x L2nx : D 1 3x x2 n 0 XD C

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~~124 Excursions in Classical Analysis~~

~~In terms of L2n, we have~~

~~2 1 2 n n 2 n 1 n F ..˛ / 2. 1/ .ˇ / / .L2n 2. 1/ /; (11.10) n D 5 C D 5 and so~~

~~1 2 n 1 1 n 1 n n F x L2nx 2 . 1/ x n D 5 n 1 n 0 n 0 ! XD XD XD 1 2 3x 2 D 5 1 3x x2 1 x Â C C Ã x.1 x/ : D .1 x/.1 3x x2/ C C For k 3, by (11.8) and (11.1), we have D~~

~~1 n 2x 1 n 3x F3nx ; Fn. x/ : D 1 4x x2 D 1 x x2 n 0 n 0 XD XD C Again, appealing to~~

~~3n 2n n n 2n 3n 3 ˛ 3˛ ˇ 3˛ ˇ ˇ 1 n Fn C .F3n 3. 1/ Fn/; (11.11) D 5p5 D 5~~

~~we ﬁnd that~~

1 3 n 1 1 n 1 n F x F3nx 3 Fn. x/ n D 5 n 1 n 0 n 0 ! XD XD XD 1 2x 3x D 5 1 4x x2 C 1 x x2 Â C Ã x.1 2x x2/ : D .1 x x2/.1 4x x2/ C These cases are not isolated examples. In fact, they essentially reframe the problem as fol- k lows: once expressing Fn as a linear combination of Fnk and Lnk , we can ﬁnd an explicit formula of the underlying Gk.x/ by using the generating functionsof 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,~~

~~k 1 k 2k 2k i.n 1/ k.n 1/ 2k 5 Fn . 1/ C L2.k i/n . 1/ C ; (11.12) D i C k i 0 Â Ã Â Ã XD k k 2k 1 2k 1 i.n 1/ 5 Fn C C . 1/ C F.2.k i/ 1/n: (11.13) D i C i 0 Â Ã XD~~

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~~11. Generating Functions for Powers of Fibonacci Numbers 125~~

~~To prove (11.12), starting with the binomial theorem, we have~~

2k 2k 5kF 2k .˛n ˇn/2k . 1/i ˛inˇ.2k i/n n D D i i 0 Â Ã XD k 1 2k 2k . 1/i ˛inˇ.2k i/n . 1/k ˛knˇkn D i C k i 0 Â Ã Â Ã XD 2k i 2k in .2k i/n . 1/ ˛ ˇ : C i i k 1 Â Ã DXC Replacing i by 2k i in the last sum and appealing to 2k 2k ; i D 2k i Â Ã Â Ã we obtain

~~2k k 1 i 2k in .2k i/n i 2k .2k i/n in . 1/ ˛ ˇ . 1/ ˛ ˇ : i D i i k 1 Â Ã i 0 Â Ã DXC XD Hence~~

k 1 2k 2k 5kF 2k . 1/i .˛/in.ˇ/in .˛2.k i/n ˇ2.k i/n/ . 1/k .˛ˇ/kn: n D i C C k i 0 Â Ã Â Ã XD By ˛ˇ 1 and (11.7), this proves (11.12). Since L1 1, we may rewrite (11.12) as D D k 1 2k 1 2k i.n 1/ k.n 1/ 2k Fn . 1/ C L2.k i/n . 1/ C L1 ; D 5k i C k i 0 Â Ã Â Ã ! XD 2k which indicates that Fn is a linear combination of Lucas numbers. Similarly, we can prove the formula (11.13). Finally,we are ready to derive an explicitformula for Gk.x/. Using the power-reduction formulas (11.12) and (11.13) respectively, we obtain

k 1 k i 2k 1 i n k 2k 1 k n 5 G2k.x/ . 1/ L2.k i/n.. 1/ x/ . 1/ .. 1/ x/ D i C k i 0 Â Ã n 0 Â Ã n 0 XD XD XD k 1 i i 2k 2 L2.k i/. 1/ x k 2k 1 . 1/ . 1/ ; D i 1 L . 1/i x x2 C k 1 . 1/k x i 0 Â Ã 2.k i/ Â Ã XD C k k i 2k 1 1 i n 5 G2k 1.x/ . 1/ C F.2.k i/ 1/n.. 1/ x/ C D i C i 0 Â Ã n 0 XD XD k 2k 1 F2.k i/ 1x C C : D i 1 L . 1/i x x2 i 0 Â Ã 2.k i/ 1 XD C

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~~In particular, we ﬁnd that the generating functions for G4;G5;G6 are as follows:~~

1 2 7x 4.2 3x/ 6 G4.x/ C D 25 1 7x x2 1 3x x2 C 1 x Â C C C Ã x.1 4x 4x2 x3/ C ; D .1 x/.1 7x x2/.1 3x x2/ C C C 1 5x 10x 10x G5.x/ D 25 1 11x x2 C 1 x x2 C 1 4x x2 Â C Ã x.1 7x 16x2 7x3 x4/ C C ; D .1 4x x2/.1 x x2/.1 11x x2/ C 1 2 18x 15.2 3x/ 6.2 7x/ 20 G6.x/ C D 125 1 18x x2 C 1 3x x2 1 7x x2 1 x Â C C C C C Ã x.1 12x 53x2 53x3 12x4 x5/ C C : D .1 x/.1 18x x2/.1 3x x2/.1 7x x2/ C C C C When k 2;3, it is interesting to see that the homogeneous recursive relations (11.2) and D (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 4 4 4 Fn 5 5Fn 4 15Fn 3 15Fn 2 5Fn 1 Fn : C D C C C C C C 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 derivationsof 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:

~~k 1 2k 2k in kn 2k Ln . 1/ L2.k i/n . 1/ ; D i C k i 0 Â Ã Â Ã XD k 2k 1 2k 1 in Ln C C . 1/ L.2.k i/ 1/n: D i C i 0 Â Ã XD~~

~~k Let the generating function of L be Hk.x/. Then, f ng~~

k 1 i 2k 2 L2.k i/. 1/ x 2k 1 H .x/ ; 2k i 2 k D i 1 L2.k i/. 1/ x x C k 1 . 1/ x i 0 Â Ã Â Ã XD C k i 2k 1 2 L2.k i/ 1. 1/ x H2k 1.x/ C C : C D i 1 L . 1/i x x2 i 0 Â Ã 2.k i/ 1 XD C

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~~11. Generating Functions for Powers of Fibonacci Numbers 127~~

~~Explicitly,~~

1 n 2 x H1.x/ Lnx ; D D 1 x x2 n 1 XD 2 1 2 n 4 7x x H2.x/ L x ; D n D .1 x/.1 3x x2/ n 1 XD C C 2 3 1 3 n 8 13x 24x x H3.x/ L x C ; D n D .1 x x2/.1 4x x2/ n 1 XD C 2 3 4 1 4 n 16 79x 164x 76x x H4.x/ L x C C : D n D .1 x/.1 3x x2/.1 7x x2/ n 1 XD C C C Next, recall that the Pell numbers are deﬁned by Pn 2Pn 1 Pn 2 with P1 1; P2 2 D C D D and the Pell-Lucas numbers satisfy the same recursive relation but with Q1 2; Q2 6. D D Setting s 1 p2; t 1 p2; D C D we are able to ﬁnd:

~~Binet-type Formulas~~

~~1 n n n n Pn .s t /; Qn s t D 2p2 D C I~~

~~the generating functions of Pkn and Qkn : f g f g 1 n Pk x Pknx ; D 1 Q x . 1/k x2 n 0 k XD C 1 n 2 Qk x Qknx D 1 Q x . 1/k x2 I n 0 k XD C and the power-reduction formulas:~~

k 1 3k 2k 2k i.n 1/ k.n 1/ 2k 2 Pn . 1/ C Q2.k i/n . 1/ C ; D i C k i 0 Â Ã Â Ã XD k 3k 2k 1 2k 1 i.n 1/ 2 Pn C C . 1/ C P.2.k i/ 1/n: D i C i 0 Â Ã XD k n Finally, setting Gk.x/ n1 0 Pn x ; we have D D P k 1 i 3k i 2k 2 Q2.k i/. 1/ x 2 G2k.x/ . 1/ D i 1 Q . 1/i x x2 i 0 Â Ã 2.k i/ XD C 2k 1 . 1/k ; C k 1 . 1/k x Â Ã k 3k 2k 1 P2.k i/ 1x 2 G2k 1.x/ C C : C D i 1 Q . 1/i x x2 i 0 Â Ã 2.k i/ 1 XD C

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~~128 Excursions in Classical Analysis~~

~~In particular,~~

1 n x G1.x/ Pnx ; D D 1 2x x2 n 1 XD 1 2 n x.1 x/ G2.x/ P x ; D n D .1 x/.1 6x x2/ n 1 XD C C 2 1 3 n x.1 4x x / G3.x/ P x ; D n D .1 2x x2/.1 14x x2/ n 1 XD C 2 1 4 n x.1 x/.1 14x x / G4.x/ P x C C : D n D .1 x/.1 6x x2/.1 34x x2/ n 1 XD C C Exercises 1. Let be the golden ratio. For positive integer n 2, prove that n Fn Fn 1: D C 2. Deﬁne

~~u0 a; u1 b D D and~~

~~un pun 1 qun 2 D 2 with p 4q. Find the generating function for un . ¤ f g 3. Let~~

~~1 k n Gk.x/ u x : D n n 0 XD Prove that~~

~~1 n .Gm qGm 1/G1.x/ Œa .b pa/x Gm nx : D C C n 0 XD 4. Monthly Problem 3801 [1936, 580; 1938, 636–637]. Show that~~

cot 1 1 cot 1 2 cot 1 5 cot 1 13 cot 1 34 ; D C C C C where these integers constituteevery other term of the Fibonacci series and satisfy the recurrence un 1 3un un 1. C D 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 129~~

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

an 1 akan k : C D k 0 XD Use the generating function to show that an .2n/Š=.nŠ.n 1/Š/. D C 6. Find an explicit formula for an if an nan 1 .n 1/Š for n 1, and a0 0. D C C D 7. Let a1 a2 1, and let D D n 2 1 an 1 .n k 1/kak ; if n>2: D n k 1 XD Find an explicit formula for an. 8. For n 2, show that n 2 FnFn 1 Fk C Fn 2 1: Fn 2 Fk 1 C k 1 C XD 9. Prove that

~~F.k 1/nLn L.k 1/nFn Fkn C ; D 2 5F.k 1/nLn L.k 1/nFn Lkn C : D 2 10. Prove that~~

k k i k i Fkn Fn Fn 1 Fi ; D i i 0 Â Ã XD k k i k i Lkn Fn Fn 1 Li : D i i 0 Â Ã XD 11. Monthly Problem 10765 [1999, 864; 2001, 978–979]. Fix positive integers k and n with n>2k 1. Prove that C k k k Fn Fn k fn 2k b c b C c p p p 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 N by S0.x; y/ x;S1.x; y/ y, and the recurrence 2 D D Sn.x; y/ Sn 1.x; y/ Sn 2.x; y/ for all n N. Show that D C 2 x2 xy y2 inf Sn.x; y/ j C j; n N j j Ä 5 2 r and determine when equality holds.

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~~130 Excursions in Classical Analysis~~

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

~~1 1 n anx : 1 2x x3 D n 0 XD Prove that, for each integer n 0, there is an integer m such that 2 2 an an 1 am: C C D~~

~~14. Open Problem. Let fn satisfy that~~

~~fn afn 1 bfn 2 cfn 3 D C C~~

~~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 afn 1 bfn 2 cfn 3 dfn 4‹ D C C C~~

References [1] T. Mansour, A formula for the generating functions of powers of Horsdam’s sequence, Aus- tralasian 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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~~Identities for the Fibonacci Powers~~

~~In this chapter, using generating functions, we establish the following identities for any positive integers n and k,~~

k k k F ai .k/ F ; n k 1 D n k i C C i 0 C XD 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 Fn are defined by f g

F1 F2 1; D D Fn 2 Fn 1 Fn (for n 1). (12.1) C D C C In the previous chapter, we proved that for any n 1, 2 2 2 2 Fn 3 2Fn 2 2Fn 1 Fn ; (12.2) C D C C C 3 3 3 3 3 Fn 4 3Fn 3 6Fn 2 3Fn 1 Fn : (12.3) C D C C C C 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 identitiesin the form

~~k k k F ai .k/ F ; (12.4) n k 1 D n k i C C i 0 C XD 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 4 Fn 5 aFn 4 bFn 3 cFn 2 dFn 1 eFn : (12.5) C D C C C C C C C C 131~~

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~~132 Excursions in Classical Analysis~~

~~Start with~~

4 4 4 4 4 4 F5 F4 F3 F2 F1 a F6 4 4 4 4 4 4 F6 F5 F4 F3 F2 b F7 0 4 4 4 4 4 1 0 1 0 4 1 F7 F6 F5 F4 F3 c F8 B 4 4 4 4 4 C B C D B 4 C B F F F F F C BdC BF C B 8 7 6 5 4 C B C B 9 C B F 4 F 4 F 4 F 4 F 4 C BeC BF 4 C B 9 8 7 6 5 C B C B 10C @ A @ A @ A which results from setting n 1;2;3;4;5 in (12.5) respectively. Solving this system, for D 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 4 4 4 Fn 5 5Fn 4 15Fn 3 15Fn 2 5Fn 1 Fn : C D C C C C C C Along the same lines, we get~~

5 5 5 5 5 5 5 Fn 6 8Fn 5 40Fn 4 60Fn 3 40Fn 2 8Fn 1 Fn C D C C C C C C C C I 6 6 6 6 6 6 6 6 Fn 7 13Fn 6 104Fn 5 260Fn 4 260Fn 3 104Fn 2 13Fn 1 Fn C D C C C C C C C C C I F 7 21F 7 273F 7 1092F 7 1820F 7 1092F 7 273F 7 n 8 D n 7 C n 6 n 5 n 4 C n 3 C n 2 C C 6 6 C C C C C 21Fn 1 Fn C I F 8 34F 8 714F 8 4641F 8 12376F 8 12376F 8 4641F 8 n 9 D n 8 C n 7 n 6 n 5 C n 4 C n 3 C C 8 C 8 8 C C C C 714Fn 2 34Fn 1 Fn C C C I 9 9 9 9 9 9 Fn 10 55Fn 9 1870Fn 8 19635Fn 7 85085Fn 6 136136Fn 5 C D C C C C C C C 9 9 9 9 9 85085Fn 4 19635Fn 3 1870Fn 2 55Fn 1 Fn C C C C C C C I F 10 89F 10 4895F 10 83215F 10 582505F 10 1514513F 10 n 11 D n 10 C n 9 n 8 n 7 C n 6 C C 10 C 10 C 10 C 10 C10 10 1514513Fn 5 582505Fn 4 83215Fn 3 4895Fn 2 89Fn 1 Fn : C C C C C C C C The coefﬁcients of the above identities are listed below: 1 1 2 2 1 0 1 3 6 3 1 B 5 15 15 5 1 C B C B C B 8 40 60 40 8 1 C B C : B 13 104 260 260 104 13 1 C B C B 21 273 1092 1820 1092 273 21 1 C B C B 34 714 4641 12376 12376 4641 714 34 1 C B C B 55 1870 19635 85085 136136 85085 19635 1870 55 1 C B C B 89 4895 83125 582505 1514513 1514513 582505 83215 4895 89 1 C B C @ A

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~~12. Identities for the Fibonacci Powers 133~~

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/ Fk 1 Fibonomial.k 1; 1/; D C D C a1.k/ Fk 1Fk Fibonomial.k 1; 2/; D: C D C : i i.i 1/=2 ai .k/ . 1/ C C Fibonomial.k 1;i 1/ (12.6) D: C C I : k k.k 1/=2 ak.k/ . 1/ ; D C C where a Fibonomial coefﬁcient is deﬁned by

FnFn 1 Fn k 1 Fibonomial.n; k/ C : (12.7) D FkFk 1 F1 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

1 2 n x.1 x/ G2.x/ F x : (12.8) D n D .1 x/.1 3x x2/ n 0 XD C C 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

~~2 3 2 .1 2x 2x x /G2.x/ x x : C D 2 Now if G2.x/ is the generating function of Fn then xG2.x/ is the generating function 2 f g for the shifted sequence Fn 1 , and consequently we have f g~~

~~2 3 2 1 2 2 2 2 n .1 2x 2x x /G2.x/ x x .F 2F 2F F /x ; C D C n n 1 n 2 C n 3 n 3 XD and so 2 2 2 2 Fn 2Fn 1 2Fn 2 Fn 3 0 for all n 3; C Á which is equivalent to (12.2).~~

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~~134 Excursions in Classical Analysis~~

~~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 un satisﬁes the recursive formula f g~~

~~un a1un 1 a2un 2 akun k (12.9) D C CC if and only if its generating function, G.x/, has the form g.x/ G.x/ 2 k D 1 a1x a2x ak x for some polynomial g.x/.~~

n Proof. By the definition of a generating function, G.x/ n1 0 unx , we have D D P j 1 n aj x G.x/ aj un j x D n j XD for each j 1;2;:::;k. Hence D 2 k G.x/Œ1 a1x a2x akx 2 k G.x/ a1xG.x/ a2x G.x/ akx G.x/ D 1 n g.x/ .un a1un 1 a2un 2 akun k /x ; D C n k XD where g.x/ is a polynomial. Thus, un satisfies (12.9) if and only if G.x/ has the asserted f g 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 Lemma 12.1. Let ˛ .1 p5/=2; ˇ .1 p5/=2: If the generating function of F k D C D f n g is in the form of

~~1 k n Pk .x/ Gk.x/ F x ; (12.10) D n D Q .x/ n 0 k XD then k i k i Qk.x/ .1 ˛ ˇ x/ (12.11) D i 0 YD and the degree of Pk.x/ is less than the degree of Qk.x/.~~

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~~12. Identities for the Fibonacci Powers 135~~

~~Proof. By Binet’s formula and the binomial theorem, we have~~

k 1 k n 1 1 k k i ni n.k i/ n F x . 1/ ˛ ˇ x n D 5k=2 i n 0 n 0 i 0 Â Ã ! XD XD XD k 1 k k i 1 ni n.k i/ n . 1/ ˛ ˇ x D 5k=2 i i 0 Â Ã n 0 XD XD k 1 k . 1/k i : D 5k=2 i 1 ˛i ˇk i x i 0 Â Ã XD If k is odd, grouping the i-th term and the .k i/-th term together, we get .k 1/=2 k i i 1 1 k . 1/ . 1/ F k xn n D 5k=2 i 1 ˛i ˇk i x C 1 ˛k i ˇi x n 0 i 0 Â Ã ! XD XD .k 1/=2 1 k 1 1 . 1/i D 5k=2 i 1 ˛k i ˇi x 1 ˛i ˇk i x i 0 Â Ã Â Ã XD .k 1/=2 k i i i k i 1 k .˛ ˇ ˛ ˇ /x . 1/i : D 5k=2 i .1 ˛k i ˇi x/.1 ˛i ˇk i x/ i 0 Â Ã XD If k is even, in view of the fact that ˛ˇ 1, grouping the i-th term and the .k i/-th D term together, with the exception of the middle term, we get

1 k n Fn x n 0 XD k=2 1 1 k . 1/k i . 1/i k . 1/k=2 D 5k=2 i 1 ˛i ˇk i x C 1 ˛k i ˇi x C k=2 1 . 1/k=2x i 0 Â Ã ! Â Ã XD k=2 1 1 k 1 1 k . 1/k=2 . 1/i D 5k=2 i 1 ˛i ˇk i x C 1 ˛k i ˇi x C k=2 1 . 1/k=2x i 0 Â Ã Â Ã Â Ã XD k=2 1 1 k 2 .˛k i ˇi ˛i ˇk i /x k . 1/k=2 . 1/i C : D 5k=2 i .1 ˛k i ˇi x/.1 ˛i ˇk i x/ C k=2 1 . 1/k=2x i 0 Â Ã Â Ã XD 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 k 1 i C i .1 r x/ 1 bi x : (12.12) D C i 0 i 1 YD XD Then, for i 1;2;:::;k;k 1, D C k 1 k k i 2 i i.i 1/=2 .r C 1/.r 1/ .r C 1/ bi . 1/ r : (12.13) D .r 1/.r 2 1/ .r i 1/

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~~136 Excursions in Classical Analysis~~

~~Proof. Set k f .x/ .1 r i x/: D i 0 YD Then f satisﬁes the functional equation~~

~~.1 x/f .rx/ f .x/.1 r k 1x/; D C which is equivalent to~~

k 1 k 1 C i i C i k 1 .1 x/ 1 bi r x 1 bi x .1 r x/: C D C C i 1 ! i 1 ! XD XD Equating the coefﬁcients of xi .i 1;2;:::;k;k 1/ on both sides, we have D C i i 1 k 1 r bi r bi 1 bi r C bi 1; D and so k 1 i 1 i 1 k i 2 r C r r .r C 1/ bi bi 1 bi 1: D 1 r i D r i 1 Iterating this relationship to write bi 1 in terms of bi 2 and continuinguntil bi is expressed as a multiple of b1, we have

i 1 i 2 k i 2 k i 3 2 r r .r C 1/.r C 1/ bi . 1/ bi 2 D .r i 1/.r i 1 1/ : : i 1 i 2 k i 2 k i 3 k i 1 r r 1.r C 1/.r C 1/ .r 1/ . 1/ b1: D .r i 1/.r i 1 1/ .r 2 1/ Since k 1 2 k r C 1 b1 .1 r r r / ; D C C CC D r 1 we obtain

i 1 i 2 k 1 k k i 2 i r r 1.r C 1/.r 1/ .r C 1/ bi . 1/ ; D .r 1/.r 2 1/ .r i 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

~~k i 1 Qk.x/ 1 ai .k/x : (12.14) D C i 0 XD Multiplying (12.10) by (12.14), we have~~

~~k 1 k n 1 k n i 1 Pk.x/ F x ai .k/F x : D n n C C n 0 n 0 i 0 XD XD XD~~

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~~12. Identities for the Fibonacci Powers 137~~

~~n k 1 Equating the coefﬁcients of x C C on both sides gives~~

k k k 0 F ai .k/F : D n k 1 n k i C C i 0 C XD This proves (12.4) as desired. Thus, it remains to verify the formulas stated in (12.6). Replacing r by ˛=ˇ and x by ˇkx in (12.12), in view of the exponents of ˇ in (12.13), namely 1 1 1 1 k.i 1/ i.i 1/ .i 1/.i 2/ .i 1/.2k i 2/ i.i 1/ C 2 C 2 C C 2 C C D 2 C and ˛ˇ 1, we have D k.i 1/ ai .k/ bi 1ˇ C D C .˛k 1 ˇk 1/.˛k ˇk/ .˛k i 1 ˇk i 1/ . 1/i . 1/i.i 1/=2 C C C C : D C .˛ ˇ/.˛2 ˇ2/ .˛i 1 ˇi 1/ C C By the deﬁnition of Fibonomial coefﬁcients (12.7), we get

~~i i.i 1/=2 ai .k/ . 1/ . 1/ Fibonomial.k 1;i 1/: D C C C This proves (12.6) as required. To demonstrate the power of the both theorems, we highlighttwo applications.~~

~~2 1. Let the sequence Un satisfy Un 2 aUn 1 bUn with a 4b > 0 and the f g C D C C C generating function for the kth power be given by~~

~~1 p .x/ U kxn k : n D q .x/ n 0 k XD Based on the proof of Lemma 12.1, we ﬁnd that~~

~~k i k i qk.x/ .1 s t x/; D i 0 YD 2 where s and t are roots of x ax b 0. Furthermore, the degree of pk.x/ is less than D the degree of qk .x/. Thus, similar to Theorem 12.2, we have~~

k k k U ci .k/U ; n k 1 D n k i C C i 0 C XD where the coefﬁcients ci ;i 0;1;2;:::;k; are given by D k 1 k 1 k k k i 1 k i 1 i i.i 1/=2 .s C t C /.s t / .s C t C / ci . 1/ . b/ : D C .s t/.s2 t 2/ .si 1 t i 1/ C C In particular, for Lucas numbers Ln , since f g n n Ln 2 Ln 1 Ln and Ln ˛ ˇ ; C D C C D C

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~~138 Excursions in Classical Analysis~~

~~we observe that the denominator of the generating function of Lk is the same as that of f n g F k . Thus, Theorem 1 implies that identity (12.4) holds for Lucas numbers: f n g~~

~~k k k L ai .k/ L ; n k 1 D n k i C C i 0 C XD~~

~~where ai .k/ are given by (12.6). Explicitly, for any positive integer n,~~

~~2 2 2 2 Ln 3 2Ln 2 2Ln 1 Ln; C D C C C 3 3 3 3 3 Ln 4 3Ln 3 6Ln 2 3Ln 1 Ln; C D C C C C 4 4 4 4 4 4 Ln 5 5Ln 4 15Ln 3 15Ln 2 5Ln 1 Ln: C D C C C C C C~~

~~For the Pell numbers Pn , which are deﬁned by f g~~

Pn 2 2Pn 1 Pn; C D C C we have a 2;b 1;s 1 p2; t 1 p2. Thus, Theorem 1 implies that D D D C D k .sk 1 t k 1/.sk t k/ .sk i 1 t k i 1/ P k . 1/i i.i 1/=2 C C C C P k : n k 1 D C C .s t/.s2 t 2/ .si 1 t i 1/ n k i C C i 0 C C C XD Explicitly, for any positive integer n,

2 2 2 2 Pn 3 5Pn 2 5Pn 1 Pn ; C D C C C 3 3 3 3 3 Pn 4 12Pn 3 30Pn 2 12Pn 1 Pn ; C D C C C C 4 4 4 4 4 4 Pn 5 29Pn 4 174Pn 3 174Pn 2 29Pn 1 Pn : C D C C C C C C 2. Deﬁne k k k Fn Fn 1 Fn k C k k k C ˇ Fn 1 Fn 2 Fn k 1 ˇ Dk .Fn/ ˇ C C C C ˇ : D ˇ : : :: : ˇ ˇ : : : : ˇ ˇ ˇ ˇ F k F k F k ˇ ˇ n k n k 1 n 2k ˇ ˇ C C C C ˇ ˇ ˇ We show that Theorem 12.2 canˇ be used to simplify the computationˇ of Dk.Fn/ as

~~nk.k 1/=2 Dk.Fn 1 / . 1/ C Dk.F1/: C D Indeed, since~~

k k k k Fn 1 Fn 2 Fn k Fn k 1 C C k k k C kC C ˇ Fn 2 Fn 3 Fn k 1 Fn k 2 ˇ Dk.Fn 1 / ˇ :C :C C: C C: C ˇ ; C D ˇ : : :: : : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ F k F k F k F k ˇ ˇ n k 1 n k 2 n 2k n 2k 1 ˇ ˇ C C C C C C C ˇ ˇ ˇ ˇ ˇ

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~~12. Identities for the Fibonacci Powers 139~~

~~applying (12.4) on the last column and then repeatedly using the column operations, we have~~

k k k k Fn 1 Fn 2 Fn k ak.k/Fn C C k k k C k ˇ Fn 2 Fn 3 Fn k 1 ak.k/Fn ˇ Dk.Fn 1 / ˇ :C :C C: C : ˇ C D ˇ : : :: : : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ F k F k F k a .k/F k ˇ ˇ n k 1 n k 2 n 2k k n ˇ ˇ C C C C C ˇ ˇ k k k ˇ ˇ Fn Fn 1 Fn k ˇ C C F k F k F k k ˇ n 1 n 2 n k 1 ˇ . 1/ ak.k/ ˇ C C C C ˇ D ˇ : : :: : ˇ ˇ : : : : ˇ ˇ ˇ ˇ F k F k F k ˇ ˇ n k n k 1 n 2k ˇ ˇ C C C C ˇ k ˇ ˇ . 1/ ak.k/ Dk .Fn/: D ˇ ˇ Iterating this relationship, we get

~~2k 2 Dk.Fn 1 / . 1/ ak.k/ Dk .Fn 1/ C D nk nk . 1/ a .k/ Dk .F1/ DD k nk.k 1/=2 . 1/ Dk.F1/: D C In particular, for k 1, we have D~~

Fn Fn 1 D1.Fn/ C D Fn 1 Fn 2 ˇ C C ˇ ˇ n 1 ˇ .ˇ 1/ D1.F1/ˇ D ˇ ˇ n 1 F1 F2 . 1/ D F2 F3 ˇ ˇ ˇ ˇ n 1 ˇ 1 1 ˇ . 1/ ˇ ˇ D 1 2 ˇ ˇ ˇ ˇ . 1/n 1:ˇ ˇ D ˇ ˇ 2 n 1 That is, FnFn 2 Fn 1 . 1/ , which is the well-known Cassini’s identity. C C D 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 cos2 1. Thus, it is well worth C D knowing if the process of manipulatingFibonacci identitiesis also algorithmic.This should be a nice project for undergraduate research.

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~~140 Excursions in Classical Analysis~~

~~Exercises 1. For n 2, prove that 2 2 n 1 5.Fn 1 Fn 1 / 2. 1/ C .mod .Fn 1 Fn 1//: C C Á C C 2. Show that 4 Fn Fn 2Fn 1Fn 1Fn 2 1: C C D 3. Use a computer to show that~~

~~9 7 2 5 4 3 6 8 256F9n 625F 4500F L 3150F L 420F L 6FnL : D n C n n C n n C n n C n~~

~~Can you ﬁnd a similar formula for L9n? 4. Let p 1 i. Prove that D n 1 .n 1/=2 k b c 2k Fn 1 2i cos 3 2 cos D n D C n k 1 Â Ã k 1 Â Ã YD YD~~

~~2 n 1 ni 1 p5 2 n n 1 p5 i C sin ln C i sinh ln C ; D p5 2 1 p5 ! D p5 2 1 p5 ! n 1 .n 2/=2 .2k 1/ b c .2k 1/ Ln 1 2i cos C 3 2 cos C D 2n D C n k 1 Â Ã k 1 Â Ã YD YD~~

n n 1 ni 1 p5 n n 1 n 1 p5 2i i C cos ln C 2i i C cosh ln C : D 2 1 p5 ! D 2 1 p5 ! Comment: This should strengthen our conviction that manipulating Fibonacci identities is algorithmic. 5. Pascal’s triangle exhibits some interesting patterns of divisibilityby various numbers. Rewrite (12.4) as

~~k 1 C . 1/i.i 1/=2Fibonomial .k 1;i 1/ F k 0: C C C n k i D i 0 C XD The triangle of Fibonomial coefﬁcients begins with (see Sloane A010048 [5])~~

~~1 1 1 1 1 1 1 2 2 1 13 6 31 1 5 15 15 5 1~~

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

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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 n bi .1 x/ f .x/ 1 ai x : D C i 1 XD Find a simple expression (not involvingany sums) for f .1/ in terms of b1;b2;:::;bn and n ( but independent of a1; a2;:::;an.)

7. Putnam Problem 1993-A2. Let xn be a sequence of nonzero real numbers such f g that 2 xn xn 1xn 1 1 for n 1;2;3:::: C D D Prove there exists a real number a such that xn 1 axn xn 1 for all n 1. C D 8. Define u0 0; u1 1 D D and un aun 1 bun 2: D C Also, for any nonnegative integer m, define m 1; if j 0; um umj C1 D : j ; if j 1;:::m u D ( uj u1 Â Ã D Prove that k 1 C i.i 1/=2 i.i 1/=2 k 1 k . 1/ C b C un i 0: i D i 0 Â Ãu XD In particular, for k 2, D 2 2 2 2 2 3 2 un 3 .a b/un 2 b.a b/un 1 b un: C D C C C C C 9. Let i i j n n j aij a b ; 0 i;j n D n j C Ä Ä Â Ã and An 1 .aij / be a matrix of order n 1. Find all the eigenvalues of An 1 and C D C C prove that k uk.n 1/ tr.An 1 / C : C D uk 10. Define the q–Fibonacci polynomials as f.n;a;b/ af.n 1;a;b/ qn 2bf .n 2;a;b/ D C with initial values f.0;a;b/ 0;f.1;a;b/ 1: Prove that D D n 1 n 1 j j 2 n 1 2j j f.n;a;b/ q a b ; D j j 0 Ä XD where n .1 qn/.1 qn 1/ .1 qn k 1/ k D .1 q/.1 q2/ .1 qk/ Ä denotes a q–binomial coefficient.

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References [1] A. Brousseau,A sequenceof 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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~~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)):~~

~~n 1 n B0 1; Bk 0: (13.1) D k D k 0 Â Ã XD Let bk Bk=kŠ. Rewriting (13.1) explicitly yields the following system of linear equa- D tions:~~

1 b1 0; C 2Š D b1 1 8 2Š b2 3Š 0; ˆ C C D ˆ b1 b2 1 ˆ b3 0; ˆ 3Š C 2Š C C 4Š D ˆ ˆ <ˆ b1 b2 1 .n 1/Š .n 2/Š bn 1 nŠ 0; ˆ C CC C D ˆ b1 b2 1 ˆ nŠ .n 1/Š bn .n 1/Š 0; ˆ C CC C C D ˆ or equivalently :ˆ

1 b1 ; D 2Š b1 1 8 b2 ; 2Š C D 3Š ˆ b1 b2 1 ˆ b3 ; ˆ 3Š C 2Š C D 4Š ˆ ˆ <ˆ b1 b2 1 .n 1/Š .n 2/Š bn 1 nŠ ; ˆ C CC D ˆ b1 b2 1 ˆ bn : ˆ nŠ C .n 1/Š CC D .n 1/Š ˆ C ˆ : 143

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~~Solving for bn by means of Cramer’s rule, we ﬁnd~~

1 0 0 1 2Š ˇ 1 1 0 1 ˇ ˇ 2Š 3Š ˇ ˇ 1 1 1 ˇ ˇ 1 ˇ ˇ 3Š 2Š 4Š ˇ bn ˇ ˇ : D ˇ : : : :: : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ 1 1 1 1 ˇ ˇ .n 1/Š .n 2/Š .n 3/Š nŠ ˇ ˇ ˇ ˇ 1 1 1 1 ˇ ˇ ˇ ˇ nŠ .n 1/Š .n 2/Š .n 1/Š ˇ ˇ C ˇ Using the basic propertiesˇ of the determinants, factoring out the coefﬁcientˇ 1 and then ˇ ˇ permutating the columns, we obtain

1 1 0 0 2Š ˇ 1 1 1 0 ˇ ˇ 3Š 2Š ˇ ˇ 1 1 1 ˇ ˇ 4Š 3Š 2Š 0 ˇ n ˇ ˇ Bn nŠbn . 1/ nŠ ˇ ˇ ; for all n N: (13.2) D D ˇ : : : :: : ˇ 2 ˇ : : : : : ˇ ˇ ˇ ˇ 1 1 1 ˇ ˇ nŠ .n 1/Š .n 2/Š 1 ˇ ˇ ˇ ˇ 1 1 1 1 ˇ ˇ ˇ ˇ .n 1/Š nŠ .n 1/Š 2Š ˇ ˇ C ˇ ˇ ˇ In particular, appealing toˇ the well-known fact (see (6.10)) ˇ

~~B2n 1 0; n 1;2;3;:::: C D D we get an added bonus from (13.2), namely~~

1 1 0 0 2Š ˇ 1 1 1 0 ˇ ˇ 3Š 2Š ˇ ˇ 1 1 1 ˇ ˇ 0 ˇ ˇ 4Š 3Š 2Š ˇ ˇ ˇ 0; for n 3;5;7;:::: ˇ : : : :: : ˇ D D ˇ : : : : : ˇ ˇ ˇ ˇ 1 1 1 ˇ ˇ nŠ .n 1/Š .n 2/Š 1 ˇ ˇ ˇ ˇ 1 1 1 1 ˇ ˇ ˇ ˇ .n 1/Š nŠ .n 1/Š 2Š ˇ ˇ C ˇ ˇ ˇ This was theˇ proposed problem 3784 which appearedˇ in the American Mathematical Monthly, 1936. Furthermore, using B0 1;B1 1=2 and B2n 1 0 for n 1, we may rewrite D D C D x 1 1 B 1 x 2n x2n ex 1 D 2 C .2n/Š n 1 XD as 2 3 x x x 2 4 6 x x 1 b2x b4x b6x : (13.3) D C 2Š C 3Š C 2 C C C Â Ã Á

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~~Applying the Cauchy product of power series on the right hand, and then equating the coefﬁcients of x2k 1 for k 1;2;::: n in (13.3), we have C D~~

1 1 b2 0; C 3Š 2.2Š/ D b2 1 1 8 3Š b4 5Š 2.4Š/ 0; ˆ C C D ˆ b2 b4 1 1 ˆ b6 0; ˆ 5Š C 3Š C C 7Š 2.6Š/ D ˆ ˆ <ˆ b2 b4 b 1 1 0; ˆ .2n 3/Š .2n 5/Š 2.n 1/ .2n 1/Š 2Œ2.n 1/Š ˆ C CC C D ˆ b2 b4 b 1 1 0; ˆ .2n 1/Š .2n 3/Š 2n .2n 1/Š 2Œ.2n/Š ˆ C CC C C D ˆ :ˆ or equivalently

1 b2 ; D 2 .3Š/ b2 3 8 3Š b4 2 .5Š/ ; ˆ C D ˆ b2 b4 5 ˆ b6 ; ˆ 5Š C 3Š C D 2 .7Š/ ˆ (13.4) ˆ <ˆ b2 b4 b 2n 3 ; ˆ .2n 3/Š .2n 5/Š 2.n 1/ 2Œ.2n1/Š ˆ C CC D ˆ b2 b4 b 2n 1 ; ˆ .2n 1/Š .2n 3/Š 2n 2Œ.2n 1/Š ˆ C CC D C ˆ :ˆ where we have used the identity

~~1 1 2k 1 : 2 Œ.2k/Š .2k 1/Š D 2 Œ.2k 1/Š C C~~

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

1 1 0 0 3Š ˇ 3 1 1 0 ˇ ˇ 5Š 3Š ˇ ˇ 5 1 1 ˇ ˇ 7Š 5Š 3Š 0 ˇ n 1 .2n/Š ˇ ˇ B2n . 1/ ˇ ˇ : D 2 ˇ : : : :: : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ 2n 3 1 1 ˇ ˇ .2n 1/Š .2n 3/Š .2n 5/Š 1 ˇ ˇ ˇ ˇ 2n 1 1 1 1 ˇ ˇ ˇ ˇ .2n 1/Š .2n 1/Š .2n 3/Š 3Š ˇ ˇ C ˇ ˇ ˇ ˇ ˇ On the other hand, equating the coefﬁcients of x2k for k 2;3;:::;n 1 in (13.3), we D C

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~~have~~

b2 1 1 0; 2Š C 4Š 2 .3Š/ D 8 b2 b4 1 1 4Š 2Š 6Š 2 .5Š/ 0; ˆ C C D ˆ b2 b4 b6 1 1 ˆ 6Š 4Š 2Š 8Š 2 .7Š/ 0; ˆ C C C D ˆ ˆ <ˆ b2 b4 b2.n1/ 1 1 Œ2.n 1/Š Œ2.n 2/Š 2Š Œ2nŠ 2Œ.2n 1/Š 0; ˆ C CC C D ˆ b2 b4 b2n 1 1 ˆ 0; ˆ .2n/Š C Œ2.n 1/Š CC 2Š C Œ2.n 1/Š 2Œ.2n 1/Š D ˆ C C ˆ :ˆ or equivalently

b2 1 ; 2Š D 4Š 8 b2 b4 2 4Š 2Š 6Š ; ˆ C D ˆ b2 b4 b6 3 ˆ ; ˆ 6Š C 4Š C 2Š D 8Š ˆ (13.5) ˆ <ˆ b2 b4 b2.n1/ n 1 ; ˆ Œ2.n 1/Š C Œ2.n 2/Š CC 2Š D 2Œ.2n/Š ˆ ˆ b2 b4 b2n n ; ˆ .2n/Š 2.n 1/Š 2Š Œ2.n 1/Š ˆ C CC D C ˆ :ˆ where we have used the identity

~~1 1 k : 2 Œ.2k 1/Š Œ2.k 1/Š D Œ2.k 1/Š C C C~~

~~Therefore, applying Cramer’s rule to (13.5) and recognizing B2n .2n/Š b2n, we ﬁnd D another determinant expression for non-vanishing Bernoulli numbers, namely~~

1 1 0 0 4Š 2Š ˇ 2 1 1 0 ˇ ˇ 6Š 4Š 2Š ˇ ˇ 3 1 1 ˇ ˇ 8Š 6Š 4Š 0 ˇ n 1 n ˇ ˇ B2n . 1/ 2 .2n/Š ˇ ˇ : D ˇ : : : :: : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ n 1 1 1 1 ˇ ˇ .2n/Š Œ2.n 1/Š Œ2.n 2/Š 2Š ˇ ˇ ˇ ˇ n 1 1 1 ˇ ˇ ˇ ˇ Œ2.n 1/Š .2n/Š Œ2.n 1/Š 4Š ˇ ˇ C ˇ ˇ ˇ Now, we conclude this chapter withˇ three observations. ˇ

~~1. The determinant expressions for Bernoulli numbers provide us an alternate method to explore the nature of some determinants. For example, recall that B2n alternate in~~

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~~13. Bernoulli Numbers via Determinants 147~~

~~sign. Thus,~~

1 1 0 0 4Š 2Š ˇ 2 1 1 0 ˇ ˇ 6Š 4Š 2Š ˇ ˇ 3 1 1 ˇ ˇ 0 ˇ ˇ 8Š 6Š 4Š ˇ ˇ ˇ > 0; for all n 1. ˇ : : : :: : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ n 1 1 1 1 ˇ ˇ .2n/Š Œ2.n 1/Š Œ2.n 2/Š 2Š ˇ ˇ ˇ ˇ n 1 1 1 ˇ ˇ ˇ ˇ Œ2.n 1/Š .2n/Š Œ2.n 1/Š 4Š ˇ ˇ C ˇ ˇ ˇ The positivityˇ of determinants has caught our attentionˇ since it can not be conﬁrmed by the well-known Hadamard Theorem [4]. 2. The method used in this note can be extended to some other important numbers. For instance, let k k k Sk.n/ 1 2 n : D C CC For positive integer m, the binomial theorem gives

m 1 m .x 1/m xm xi : C D i i 0 Â Ã XD Adding up these identities for x 1;2;:::;n, we get D m 1 m m .n 1/ 1 Si .n/: (13.6) C D i i 0 Â Ã XD Taking m 1;2;:::;k in (13.6) respectively, we obtain a system of linear equations D for Si .n/; 0 i k 1. Cramer’s rulegives, after some simpliﬁcations, Ä Ä 10 0 0 n 1 C 2 ˇ 12 0 0 .n 1/ ˇ ˇ C ˇ ˇ 3 ˇ ˇ 13 3 0 .n 1/ ˇ ˇ : : : : : C: ˇ 1 ˇ : : : :: : : ˇ Sk 1.n/ ˇ : : : : : ˇ : D kŠ ˇ ˇ ˇ k 1 k 1 k 1 ˇ ˇ 1 .n 1/k 1 ˇ ˇ 1 2 k 3 C ˇ ˇ Â Ã Â Ã Â Ã ˇ ˇ ˇ ˇ k k k ˇ ˇ 1 .n 1/k ˇ ˇ 1 2 k 2 C ˇ ˇ Â Ã Â Ã Â Ã ˇ ˇ ˇ ˇ ˇ Appealing to theˇ deﬁnition of determinant, we see that Sk 1.n/ is a polynomial inˇ n with degree k. In particular, we have

~~n 1 0 n 1 2 1 C 2 n.n 1/.2n 1/ S2.n/ i ˇ 1 2 .n 1/ ˇ C C ; D D 3Š ˇ C ˇ D 6 i 1 ˇ 1 3 .n 1/3 ˇ XD ˇ C ˇ ˇ ˇ ˇ ˇ ˇ ˇ~~

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100 n 1 C n 1 2 0 .n 1/2 2 2 3 1 ˇ ˇ n .n 1/ S3.n/ i ˇ C 3 ˇ C : D D 4Š ˇ 1 3 3 .n 1/ ˇ D 4 i 1 ˇ C ˇ XD ˇ 1 4 6 .n 1/4 ˇ ˇ ˇ ˇ C ˇ ˇ ˇ The interested reader is encouragedˇ to ﬁnd the determinantˇ expressions for Fibonacci and other special numbers.

3. Our discussion above pivoted on the application of Cramer’s rule on recursive relations. There are many other uses of linear algebra in analysis. For example, we can use linear algebra to study sequences deﬁned by a linear recursive relation that is associated with a matrix. Let a b A D c d Â Ã be an arbitrary 2 2 matrix and let T a d tr.A/; D ad bc. Deﬁne D C D D

~~xn 1 Txn Dxn 1: C D It is not hard to show that~~

~~n=2 b c n i n 2i i xn T . D/ ; (13.7) D i i 0 Â Ã XD and~~

n axn Dxn 1 bxn A : (13.8) D cxn dxn Dxn 1 Â Ã As an illustration, applying (13.7) and (13.8) to the Fibonacci numbers Fn, where T D 1, we recover the famous identities D D n=2 b c n i Fn D i i 0 Â Ã XD and n 1 1 Fn 1 Fn C : 1 0 D Fn Fn 1 Â Ã Â Ã Finally, it is interesting to notice that to solve difﬁcult enumeration problems on plane partitions, George Andrews in [1] brought to the attention of the mathematical community the study of determinants involvingbinomialnumbers. In [2], Bressoud gives an interesting brief account on the historical facts surrounding the discovery of these determinants. As might be expected, there are still a large number of open problems. For a list of such determinants and the contexts in which they have appeared, you might enjoy the excellent survey paper by Krattenthaler [5].

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Exercises 1. Let b i j aij x dx: D C Za Show that n n 2 2 nŠ .n 1/2 2k det.aij / .b a/ : D .2n 1/Š C k k 1 Â Ã C YD 2. Deﬁne b1 1 0 0 ˇ b2 b1 2 0 ˇ ˇ ˇ ˇ : : : : : ˇ An.b1;b2;:::;bn/ ˇ : : : :: : ˇ : D ˇ ˇ ˇ ˇ ˇ bn 1 bn 2 bn 3 1 n ˇ ˇ ˇ ˇ bn bn 1 bn 2 b1 ˇ ˇ ˇ n ˇ ˇ Evaluate An when bn n =nŠ. ˇ ˇ D ˇ ˇ 3. Deﬁne a1 a2 a3 an ˇ 1 a1 a2 an 1 ˇ ˇ ˇ D .a ; a ;:::;a / ˇ 0 1 a1 an 2 ˇ : n 1 2 n ˇ ˇ D ˇ : : : : ˇ ˇ : : : :: : ˇ ˇ : : : : : ˇ ˇ ˇ ˇ 0 0 0 1 ˇ ˇ ˇ ˇ ˇ Prove the following: ˇ ˇ ˇ ˇ (a) If 1 1 1 An Dn ; ;:::; ; D 2Š 4Š .2n/Š Â Ã then 2 An 1 lim : n A D 4 !1 ˇ n ˇ ˇ ˇ (b) Let B be the Bernoulli numbers.Thenˇ ˇ n ˇ ˇ

~~1 1 1 Bn D2n ; ;:::; : 2Š 3Š .2n 1/Š D .2n/Š Â C Ã 4. Reciprocal of a power series. Let~~

~~1 n f .x/ anx : D n 0 XD~~

~~If a0 0, then the reciprocal of the power series exists and is determined by ¤~~

~~1 1 n Œf .x/ bnx ; D n 0 XD~~

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where a1 a0 0 0 n a2 a1 a0 0 . 1/ ˇ ˇ bn ˇ ˇ : D n 1 ˇ : : : :: : ˇ a0C ˇ : : : : : ˇ ˇ ˇ ˇ a a a a ˇ ˇ n n 1 n 2 1 ˇ ˇ ˇ ˇ ˇ 5. Let ˇ ˇ sinh x 1 a 2n x2n: sin x D .2n/Š n 0 XD Show that 11 0 0 ˇ 3 3 ˇ ˇ 1 0 ˇ n ˇ 3 1 ˇ . 1/ ˇ Â Ã Â Ã ˇ a2n ˇ : : : : : ˇ : D .2n 1/Š ˇ : : : :: : ˇ C ˇ ˇ ˇ ˇ ˇ 2n 1 2n 1 2n 1 ˇ ˇ 1 . 1/n C . 1/n 1 C C ˇ ˇ 2n 1 2n 3 3 ˇ ˇ Â Ã Â Ã Â Ã ˇ ˇ ˇ ˇ ˇ ˇ ˇ 6. Let 1=.2i 2j 3/Š; j i 1; Aij C Ä C : D 0; j>i 1 C Prove that n 1 2n . 1/ C .2 2/B2n det.Aij / : D .2n/Š 7. Prove the Vandermonde determinant evaluation

~~j 1 det .xi / .xj xi /: 1 i;j n D Ä Ä 1 i~~

~~det .pj .xi //; 1 i;j n Ä Ä j 1 where pj aj x lower terms: D C n k 8. Let Sk.n/ i 1 i and D D P n Ak.n/ i1i2 ik: D i1;i2;:::;ik 1; i1~~

~~Find a closed form for Ak.n/ in terms of Sk.n/. For example,~~

~~n n 2 .j 1/j 1 2 A2.n/ ij .S .n/ S2.n//: D D 2 D 2 1 i;j 1;i~~

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9. Putnam Problem 1978-A2. Let a;b;p1;p2;:::;pn be real numbers with a b. ¤ Deﬁne f .x/ .p1 x/.p2 x/ .pn x/. Show that D p1 a a a a ˇ ˇ ˇ b p2 a a a ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ b b p3 a a ˇ ˇ ˇ ˇ ˇ bf .a/ af.b/ det ˇ ˇ : ˇ : : : : : : ˇ D b a ˇ : : : :: : : ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ b b b p a ˇ ˇ n 1 ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ b b b b p1 ˇ ˇ ˇ ˇ ˇ 10. Putnam Problemˇ 1992-B5. Let Dn denote the valueˇ of the .n 1/ .n 1/ determinant 3 1 1 1 1 ˇ ˇ ˇ 1 4 1 1 1 ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ 1 1 5 1 1 ˇ ˇ ˇ : ˇ ˇ ˇ ˇ ˇ : : : : : : ˇ ˇ : : : : : : : ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ 1 1 1 1 n 1 ˇ ˇ C ˇ Is the set Dn=nŠ bounded?ˇ ˇ f g ˇ ˇ 11. More well-known numbers via determinants. Given a sequence an , deﬁne f g01

a1 a0 0 0 ˇ a2 a1 a0 0 ˇ ˇ ˇ ˇ : : : : : ˇ Dn.a0; a1; a2;:::;an/ ˇ : : : :: : ˇ : D ˇ ˇ ˇ ˇ ˇ an 1 an 2 an 3 a0 ˇ ˇ ˇ ˇ an an 1 an 2 a1 ˇ ˇ ˇ ˇ ˇ Problem 3 indicates ˇ ˇ ˇ ˇ n 1 . 1/ C .2n/Š B2n Dn.1;1=3Š;:::;1=.2n 1/Š/: D 22n 2 C Motivated by this result, we have the following interesting results: n n (a) Let f .x/ 1 anx ; g.x/ 1 bnx . If f .x/g.x/ 1, then a0 0 D n 0 D n 0 D ¤ and D D P nPDn.a0; a1;:::;an/ bn . 1/ n 1 : D a0C n (b) Let D.x/ n1 0 Dn.a0; a1; a2;:::;an/x . If an 1=.2n 1/Š, show that D D D C P x 2xex D.x2/ ; D. x2/ : D sin x D e2x 1

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~~(c) Let Hn be the harmonic numbers. Prove that~~

n 1 . 1/ n 1 Hn BnC1 : D .n 1/Š n (d) Let the Stirling numbers of the second kind be deﬁned by m x m n .e 1/ 1 n x : mŠ D m nŠ n m XD Deﬁne m n m n S .m/ C = C ; n D m m Â Ã and prove that

~~.n 1/ .n 1/ .n 1/ Hn Dn 1.S0 C ;S1 C =1Š;:::;Sn C1 =.n 1/Š/: D~~

~~12. Bernoulli polynomials via determinants. Let Bn.x/ be the Bernoulli polynomials. Prove that~~

1 x x2 x3 xn 1 xn 1 1=2 1=3 1=4 1=n 1=.n 1/ ˇ ˇ ˇ C ˇ ˇ 01 1 1 1 1 ˇ ˇ ˇ ˇ 00 2 3 n 1 n ˇ n ˇ ˇ . 1/ ˇ ˇ Bn.x/ ˇ 3 n 1 n ˇ : D .n 1/Š ˇ 0 0 0 ˇ ˇ 2 2 2 ˇ ˇ Â Ã Â Ã Â Ã ˇ ˇ : : : : : : ˇ ˇ : : : : :: : : ˇ ˇ : : : : : : : ˇ ˇ ˇ ˇ n 1 n ˇ ˇ 00 0 0 ˇ ˇ n 2 n 2 ˇ ˇ ˇ ˇ Â Ã Â Ã ˇ ˇ ˇ ˇ ˇ References [1] G. Andrews and D. Stanton, Determinants in plane partitions enumeration, Europe J. Combina- torics, 19 (1998) 273–282. [2] D. M. Bressoud, Proofs and Conﬁrmations:The Story of the Alternating Sign Matrix Conjecture, The Mathematical Association of America, 1999. [3] C. Dubbs and D. Siegel, Computing determinants, College Math. J., 18 (1987) 48–50. [4] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985. [5] C. Krattenthaler, Advanced determinant Calculus, available at www.mat.univie.ac.at/ slc/wpapers/s42kratt.pdf

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~~On Some Finite Trigonometric Power Sums~~

~~In Chapter 5, we established the trigonometric identity (see (5.22))~~

n 1 n 1 n.1 x / C ; 2k 2 D .1 xn/.1 x2/ k 0 1 2x cos n x XD C and reproduced the following elegant formulaÁ (see (5.2)) n 1 k .n 1/.n 1/ csc2 C n D 3 k 1 Â Ã XD along with n 1 k tan2 n.n 1/; if n is odd: n D k 1 Â Ã XD In contrast to Fourier series, these ﬁnite power sums are over angles equally dividing the upper half plane. Moreover, these beautiful and somewhat surprising sums often arise in both analysis and number theory. In this chapter, by using generating functions, we extend the above results to the power sums as shown in identities (14.9)–(14.20) and in the Appendix. We begin with establishing two auxiliary trigonometric identities. To this end, recall (5.15) n 1 sin n .cos cos.k=n// 21 n : (14.1) D sin k 1 YD Taking the logarithm of both sides gives n 1 ln.cos cos.k=n// .1 n/ ln 2 lnsin n lnsin : D C k 1 XD Differentiating with respect to yields n 1 sin n cot n cot : (14.2) k D C k 1 cos cos. n / XD 153

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~~Replacing by leads to n 1 sin n cot n cot : (14.3) k D C k 1 cos cos. n / XD C Adding (14.2) and (14.3) and dividing both sides by 2 we obtain~~

~~n 1 sin cos n cot n cot : 2 2 k D C k 1 cos cos . n / XD Now, dividing by sin and consolidating yields~~

n 1 cos n cot n csc : (14.4) 2 2 k D k 0 cos cos . n / XD Next, substituting cos2 x 1 sin2 x into (14.4), we have D n 1 cos n cot n csc : (14.5) 2 k 2 D k 0 sin . n / sin XD In the following, let n and p be positive integers. We show how to use the identities (14.4) and (14.5) to establish explicit formulas for various ﬁnite trigonometricpower sums.

~~14.1 Sums involving sec2p.k=n/ Let n 1 2p k Sp .n/ sec : D n k 0 Â Ã XD There are two cases, depending on whether n is even or odd.~~

~~Case 1: n odd. To construct a generating function for Sp.n/ of the form f g~~

~~1 2p G.n; t/ Sp.n/t ; (14.6) D p 1 XD we rewrite (14.4) as~~

~~n 1 2 cos n cot n cot : (14.7) D 2 2 k k 0 cos cos . n / XD By expanding the summand as a geometric series we have~~

~~2 cos n 1 k n 1 cos. n / 1 k n cot n cot Ä sec2p . / cos2p ; (14.8) D 2 D n k 0 cos p 1 k 0 ! XD 1 k XD XD cos. n / Ä ~~

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~~14. On Some Finite Trigonometric Power Sums 155~~

where the last summation order has been changed. Thus, comparing (14.8) and (14.6) with t cos , we ﬁnd the generating function for Sp.n/ , namely D f g G.n; t/ n cot n cot D nt nt cot.n arccos t/ tan.n arcsin t/; D p1 t 2 D p1 t 2 where we have used the fact that n is odd and the relation arccos =2 arcsin . D Therefore, n 1 2p k .2p/ Sp.n/ sec G .n; 0/=.2p/Š; D n D k 1 Â Ã XD where G.2p/ indicates the 2p-th derivative of G with respect to t. For p 1, computing the indicated derivative directly, we ﬁnd D n 1 00 k G .n; 0/ sec2 n2: (14.9) n D 2Š D k 0 Â Ã XD However, for p 2, the expression of G.2p/.n; t/ becomes very complicated and inconve- nient. So, we now demonstrate a derivationthat is carried out with the help of Mathematica rather than by hand. For example, in Mathematica,the identity (14.9) is reproduced with the following commands: G[t_] :=n*t*Tan[n*ArcSin[t]]/Sqrt[1-t^2]; D[G[t],{t,2}]/2!/.t->0 n^2 Similarly, we have D[G[t],{t,4}]/4!/.t->0; Simplify[%] 1/3 n^2(2 + n^2) Thus, n 1 2 k n sec4 .n2 2/: (14.10) n D 3 C k 0 Â Ã XD Further results obtained by Mathematica are listed at the end of the chapter.

Case 2: n even. Deleting the term for k n=2 from Sp.n/, we reset D n 1 2p k Sp .n/ sec : D n k 0;k n=2 Â Ã D X¤ Similarly, separating the term corresponding to k n=2 from (14.7), we have D n 1 2 cos 1 n cot n cot : C D 2 2 k k 0;k n=2 cos cos . n / D X¤

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As in (14.8), we ﬁnd the generating function of Sp .n/ has the form f g G.n; t/ 1 n cot n cot ; D C where t cos . Thus, explicitly, we have D nt G.n; t/ 1 cot.n arcsin t/; D p1 t 2 where that n is even and the identity arccos =2 arcsin have been used. Therefore, D

~~n 1 .2p/ 2p k G .n; t/ Sp.n/ sec lim : D n D t 0 .2p/Š k 0;k n=2 Â Ã ! D X¤ As in Case 1, we may use Mathematica.~~

~~g[t_] :=1 - n*t*Cot[n*ArcSin[t]]/Sqrt[1-t^2]; D[g[t],{t,2}]/2!; Limit[%,t->0] 1/3 (n^2 - 1)~~

~~Thus, n 1 k 1 sec2 .n2 1/ (14.11) n D 3 k 0;k n=2 Â Ã D X¤ and~~

~~g[t_] :=1 - n*t*Cot[n*ArcSin[t]]/Sqrt[1-t^2]; D[g[t],{t,4}]/4!; Limit[%,t->0] 1/45 (-11 + 10 n^2 + n^4)~~

~~giving n 1 k 1 sec4 .n2 1/.n2 11/: (14.12) n D 45 C k 0;k n=2 Â Ã D X¤ 14.2 Sums involving csc2p.k=n/ Let n 1 2p k Cp.n/ csc D n k 1 Â Ã XD and its generating function be~~

~~1 2p F.n; t/ Cp .n/t : D p 1 XD~~

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~~On separating off the term for k 0 from (14.5), and multiplying the remaining sums D by tan sin , we ﬁnd~~

~~n 1 2 sin 1 n cot n tan D 2 k 2 k 1 sin . n / sin XD 2 sin n 1 k n 1 sin. n / 1 k Ä csc2p . / sin2p : D 2 D n k 1 sin p 1 k 1 ! XD 1 k XD XD sin. n / Ä This implies~~

~~F.n; t/ 1 n cot n tan D with t sin . Explicitly, we have D nt F.n; t/ 1 cot.n arcsin t/: D p1 t 2~~

~~Since this coincides with the generating function of Sp,~~

~~.2p/ Cp.n/ F .n; 0/=.2p/Š Sp.n/: D D~~

~~Therefore, in view of (14.11) and (14.12), we ﬁnd~~

~~n 1 n 1 k k 1 csc2 sec2 .n2 1/; (14.13) n D n D 3 k 1 Â Ã k 0;k n=2 Â Ã XD D X¤ n 1 n 1 k k 1 csc4 sec4 .n2 1/.n2 11/: (14.14) n D n D 45 C k 1 Â Ã k 0;k n=2 Â Ã XD D X¤~~

~~14.3 Sums involving tan2p.k=n/~~

~~Let~~

~~n 1 2p k Tp.n/ tan D n k 0 Â Ã XD and its generating function be~~

~~1 2p T .n; t/ Tp.n/t : D p 0 XD~~

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~~Then~~

n 1 1 k T .n; t/ tan2p t 2p D n p 0 k 0 Â Ã! XD XD n 1 1 D 1 t 2 tan2.k=n/ k 0 XD n 1 2 cos .k=n/ D .1 t 2/ cos2.k=n/ t 2 k 0 XD C n 1 2 2 n 1 t =.1 t / C : D 1 t 2 C 1 t 2 cos2.k=n/ t 2=.1 t 2/ k 0 C C XD C Substituting t cot yields D n 1 2 2 n 1 2 t =.1 t / cos C : (14.15) cos2.k=n/ t 2=.1 t 2/ D cos2.k=n/ cos2 k 0 k 0 XD C XD If n is odd, then cot.n/ tan.n arctan t/. Thus, in view of (14.4), we arrive at D n T .n; t/ .1 t tan.n arctan t//: (14.16) D 1 t 2 C C If n is even, after deleting the term for k n=2 from Tp.n/, we deﬁne D n 1 2p k T .n/ tan p D n k 0;k n=2 Â Ã D X¤ and its generating function by

~~1 T .n; t/ T .n/t 2p : D p p 0 XD Then~~

n 1 1 k T .n; t/ tan2p t 2p D 0 n 1 p 0 k 0;k n=2 Â Ã XD D X¤ n@1 A 1 D 1 t 2 tan2.k=n/ k 0;k n=2 D X¤ n 1 2 cos .k=n/ D .1 t 2/ cos2.k=n/ t 2 k 0;k n=2 D X¤ C n 1 2 2 n 1 t =.1 t / C : D 1 t 2 C 1 t 2 cos2.k=n/ t 2=.1 t 2/ k 0 C C XD C

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Setting t cot and inview of(14.15), (14.4) andnoticingthat cot.n/ cot.n arctan t/, D D we arrive at n T .n; t/ .1 t cot.n arctan t//: (14.17) D 1 t 2 C Implementing (14.16) and (14.17) on Mathematica yields

~~T[t_] :=n*(1 + t*Tan[n*ArcTan[t]])/(1 + t^2); D[T[t],{t,2}]/2!/.t->0; Factor[%] (-1+n) n~~

~~n 1 k tan2 n.n 1/ (14.18) n D I k 1 Â Ã XD D[T[t],{t,4}]/4!/.t->0; Factor[%] 1/3 (-1 +n)n(-3+ n+n^2)~~

~~Thus n 1 k n.n 1/ tan4 .n2 n 3/ (14.19) n D 3 C I k 1 Â Ã XD 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 1 k 1 tan2 .n 1/.n 2/ (14.20) n D 3 I k 1;k n=2 Â Ã D X¤ D[Tc[t],{t,4}]/4!; Factor[ Limit[%, t->0]] 1/45 (-2 + n)(-1 + n)(-13 + 3n + n^2)~~

~~Thus n 1 k .n 1/.n 2/ tan4 .n2 3n 13/: (14.21) n D 45 C k 1;k n=2 Â Ã D X¤ 14.4 Sums involving cot2p.k=n/ Let n 1 2p k Cp.n/ cot D n k 1 Â Ã XD and its generating function be~~

~~1 2p A.n; t/ Cp.n/t : D p 0 XD~~

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Then n 1 1 k A.n; t/ cot2p t 2p D n p 0 k 1 Â Ã! XD XD n 1 1 D 1 t 2 cot2.k=n/ k 1 XD n 1 2 1 cos .k=n/ D 1 .1 t 2/ cos2.k=n/ k 1 XD C 2 n 1 2 n t 1=.1 t / C : D 1 t 2 1 t 2 cos2.k=n/ 1=.1 t 2/ k 0 C C XD C Substituting t tan yields D n 1 2 n 1 2 1=.1 t / cos C : cos2.k=n/ 1=.1 t 2/ D cos2.k=n/ cos2 k 0 k 0 XD C XD Therefore, n A.n; t/ .1 t cot.n arctan t// T .n; t/: D 1 t 2 D As an immediate consequence,C (14.20) and (14.21) deduce that

n 1 k .n 1/.n 2/ cot2 ; (14.22) n D 3 k 1 Â Ã XD n 1 k .n 1/.n 2/ cot4 .n2 3n 13/: (14.23) n D 45 C k 1 Â Ã XD We now have established a variety of finite trigonometric power sum identities. The reader shouldbe 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 Mathe- matica 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 1 2 k n sec6 .2n4 5n2 8/; n D 15 C C k 0 Â Ã XD n 1 2 k n sec8 .17n6 56n4 98n2 144/; n D 315 C C C k 0 Â Ã XD~~

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~~14. On Some Finite Trigonometric Power Sums 161~~

~~n 1 2 k n sec10 .62n8 255n6 546n4 820n2 1152/: n D 2835 C C C C k 0 Â Ã XD When n is even:~~

n 1 2 k .n 1/ sec6 .2n4 23n2 191/; n D 945 C C k 0;k n=2 Â Ã D X¤ n 1 2 k .n 1/ sec8 .3n6 43n4 337n2 2497/; n D 14175 C C C k 0;k n=2 Â Ã D X¤ n 1 2 k .n 1/ sec10 .2n8 35n6 321n4 2125n2 14797/: n D 93555 C C C C k 0;k n=2 Â Ã D X¤

~~2. Power sums of cosecant~~

n 1 2 k .n 1/ csc6 .2n4 23n2 191/; n D 945 C C k 1 Â Ã XD n 1 2 k .n 1/ csc8 .3n6 43n4 337n2 2497/; n D 14175 C C C k 1 Â Ã XD n 1 2 k .n 1/ csc10 .2n8 35n6 321n4 2125n2 14797/: n D 93555 C C C C k 1 Â Ã XD 3. Power sums of tangent When n is odd:

n 1 k n.n 1/ tan6 .2n4 2n3 8n2 8n 15/; n D 15 C C k 1 Â Ã XD n 1 k n.n 1/ tan8 .17n6 17n5 95n4 95n3 213n2 213n 315/: n D 315 C C C k 1 Â Ã XD n 1 k n.n 1/ tan10 .62n8 62n7 448n6 48n5 1358n4 1358n3 n D 2385 C C C k 1 Â Ã XD 2232n2 2232n 2835/: C When n is even:

~~n 1 k .n 1/.n 2/ tan6 .2n4 6n3 28n2 96n 251/; n D 945 C C k 1;k n=2 Â Ã D X¤~~

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n 1 k .n 1/.n 2/ tan8 .3n6 9n5 59n4 195n3 457n2 n D 14175 C C k 1;k n=2 Â Ã D X¤ 1761n 3551/; C n 1 k .n 1/.n 2/ tan10 .2n8 6n7 52n6 168n5 546n4 n D 93555 C C k 1;k n=2 Â Ã D X¤ 1974n3 3068n2 13152n 22417/: C C 4. Power sums of cotangent

n 1 k .n 1/.n 2/ cot6 .2n4 6n3 28n2 96n 251/; n D 945 C C k 1 Â Ã XD n 1 k .n 1/.n 2/ cot8 .3n6 9n5 59n4 195n3 475n2 1761n 3551/; n D 14175 C C C k 1 Â Ã XD n 1 k .n 1/.n 2/ cot10 .2n8 6n7 52n6 168n5 546n4 1974n3 n D 93555 C C C k 1 Â Ã XD 3068n2 13152n 22417/: C Exercises 1. Prove that n 1 sin x n sin nx : cos x cos. 2k=n/ D cos nx cos n k 0 XD C 2. Let ! be the nth primitive root of unity and 0 j n 1. Prove that Ä Ä 1 !i !j 2 n.n2 1/: j j D 12 i j X¤ 3. Let ! be the nth primitive root of unity and let

~~n 1 k N gN .n/ .! 1/ : D k 1 XD~~

Show that gN .n/ is a polynomial with rational coefﬁcients and degree at most N, and determine gN .n/ for N 1;2;3: D 4. Evaluate n cos N.2k 1/=n ; sin2.2k 1/=.2n/ k 1 XD in terms of the integers N and n, 0 N n. Ä Ä

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~~14. On Some Finite Trigonometric Power Sums 163~~

~~5. Let a and b be positive integers such that 0~~

~~N 2N m N mx NŠ cos cos : N N N D N x N x m 1 C2 Š 2 Š XD Á 7. Prove that (a) If n is even, then~~

~~n 1 k . 1/ cos n csc n csc : 2 2 k D k 0 cos cos . n / XD (b) If n is odd, then~~

~~n 1 k . 1/ cos.k=n/ n csc n csc : 2 2 k D k 0 cos cos . n / XD 8. For odd integer n, ﬁnd the generating function for the alternating sums of 2p 1 sec C .k=n/. 9. If n is odd, prove that~~

~~2p 1 2p 1 p k p 1 kn 2p Sp.n/ n . 1/ C C : D 2p 1 j 1 k 1 Â Ã j k Â C Ã XD XD 10. Let n;p be positive integers and let~~

~~n 1 2p 2p k Sn;p csc : D 2n 2n k 1 Â Ã Á XD~~

Find a closed form expression for Sn;p . 11. Evaluate N 1 N 1 N 1 sin.kj=N / sin.ij=N / S : D 1 cos.j=N / i 1 j 1 k 1 XD XD XD 12. Let i j .i j / aij sin sin cos ; i;j 1;2;:::;n 1: D n n n D

~~Find the eigenvalues of the matrix A .aij /. D~~

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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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~~Power Series of .arcsin x/2~~

~~The power series expansion~~

~~2n 1 1 .2x/ .arcsin x/2 ; x 1 (15.1) D 2 2n j j Ä n 1 n2 XD n Â Ã plays an important rolein the evaluation of series involving the central binomial coefﬁcient. Some classical examples [1] are~~

2 1 1 1 .2/; 2n D 18 D 3 n 1 n2 XD n Â Ã n 1 1 . 1/ 2 C .3/; 2n D 5 n 1 n3 XD n Â Ã 1 1 17 17 4 .4/; 2n D 3456 D 36 n 1 n4 XD n Â Ã 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/ .arcsin x/2. We begin with deriving a differential equation for y.x/. Differen- D tiating y.x/ gives p1 x2 y .x/ 2 arcsin x: (15.2) 0 D 165~~

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~~Differentiating (15.2) and then multiplying by p1 x2 shows that y.x/ satisﬁes the ordinary differential equation~~

~~.1 x2/y .x/ xy .x/ 2: (15.3) 00 0 D Since this is a differential equation with variable coefﬁcients, it is natural to seek a solution via the power series representation~~

~~1 n y.x/ anx : (15.4) D n 0 XD Termwise differentiation of (15.4) yields~~

~~2 1 n 1 1 n y0.x/ a1 2a2x 3a3x nanx .n 1/an 1x D C C CD D C C n 1 n 0 XD XD and~~

~~1 n 1 1 n y00.x/ 2a2 6a3x n.n 1/an 1 x .n 2/.n 1/an 2 x : D C CD C C D C C C n 1 n 0 XD XD Since a0 y.0/ 0; a1 y .0/ 0, substituting the series of y .x/ and y .x/ into D D D 0 D 0 00 (15.3) yields~~

1 2 n 2a2 6a3x .n 2/.n 1/an 2 n an x 2: C C C C C D n 2 XD n Equating coefﬁcients of x successively, we obtain a2 1; a3 0 and, in general, D D n2 an 2 an; for n 2: C D .n 2/.n 1/ C C For n 2, this determines

a2n 1 0; C D n 1 2 2n Œ2 .n 1/Š 1 2 a2n ; D .2n/.2n 1/ 4 3 D 2 2n n2 n Â Ã where we have used the fact that 2n .2n/Š : n D .nŠ/2 Â Ã Thus, we ﬁnd that 2n 1 1 .2x/ y.x/ ; D 2 2n n 1 n2 XD n Â Ã which converges for x 1 by the Ratio Test. This proves (15.1) as desired. j j Ä

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~~15.2 Second Proof of the Series (15.1) The second proof uses two different evaluations of the parametric integral~~

=2 d f .x/ ; for x < 1: (15.5) D 1 x sin j j Z0 For the ﬁrst evaluation, setting t tan.=2/, we have D =2 d 1 dt 2 1 x sin D .1 x2/ .t x/2 Z0 Z0 C 2 t x 1 arctan D p1 x2 p1 x2 0 Â Ã ˇ 2 1 x ˇ arctan C ;ˇ D p1 x2 1 x ! r where we have used the identity

~~˛ ˇ arctan ˛ arctan ˇ arctan C : C D 1 ˛ˇ Â Ã Next, appealing to the identity~~

s 1 2 arctan ps arcsin ; D s 1 C 2 Â C Ã which can be veriﬁed by showing that both sides have the same derivatives, and setting s .1 x/=.1 x/, our original function becomes D C arcsin x f .x/ : (15.6) D 2 p1 x2 C p1 x2 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

~~=2 1 f .x/ sinn d xn: D n 0 Z0 ! XD To proceed, for n 2, we integrate by parts to get 1 n 1 =2 sinn d sinn 1 cos sinn 2 d: D n C n Z Z0 Evaluating at 0 and =2 yields~~

~~=2 n 1 =2 sinn d sinn 2 d: D n Z0 Z0 Since =2 =2 sin0 d ; sin d 1; D 2 D Z0 Z0~~

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~~by induction, we obtain the following famous Wallis’s formulas:~~

~~=2 1 3 .2k 1/ 2k sin2k d (15.7) 2k 1 0 D 2 4 .2k/ 2 D 2 k Z C Â Ã and =2 2k 2k 1 2 4 .2k/ 2 sin C d : D 1 3 .2k 1/ D 2k Z0 C .2k 1/ C k Â Ã Therefore,~~

~~2n 1 1 2n 2n 1 2 2n 1 f .x/ x x C : (15.8) D 2 22n n C 2n n 0 Â Ã n 0 .2n 1/ XD XD C n Â Ã Recalling the binomial series expansion~~

~~1 1 1 2n x2n; (15.9) p 2 D 22n n 1 x n 0 Â Ã XD and comparing (15.6) and (15.8), we obtain~~

~~2n arcsin x 1 2 2n 1 x C : (15.10) p1 x2 D 2n n 0 .2n 1/ XD C n Â Ã Finally, integrating (15.10) yields~~

2n 1 2 1 2 2n 2 .arcsin x/ x C ; 2 D 2n n 0 .2n 1/.2n 2/ XD C C n Â Ã that is, after replacing n 1 by n, C 2n 1 1 2 .arcsin x/2 x2n; D 2.n 1/ n 1 .2n 1/.2n/ XD n 1 Â Ã 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/ 22n 1Œ.n 1/Š2; for all n 1: D The direct calculation is quite tedious.~~

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~~2. Integrating (15.9) gives~~

~~x dt 1 1 2n x x2n 1: arcsin 2n C D 0 p1 t 2 D 2 .2n 1/ n Z n 0 Â Ã XD C In view of the Cauchy product of arcsin x and itself, equating the coefﬁcients of x2n, we obtain the combinatorial identity~~

~~n 4n 3 1 2k 2.n k 1/ 2 : .2k 1/.2n 2k 1/ k n k 1 D 2n k 0 C Â Ã Â Ã n2 XD n Â Ã Furthermore, the Cauchy product of arcsin x and .arcsin x/2 leads to~~

n 1 1 3Š 2n 1 .arcsin x/3 x2n 1: D .2n 1/22n n .2k 1/2 C n 1 Â Ã k 0 ! XD C XD C In particular, for x 1, we get D n 1 3 1 3Š 2n 1 : .2n 1/22n n .2k 1/2 D 8 n 1 Â Ã k 0 ! XD C XD C 3. Letting x sin t in (15.1), we have D 2n 1 1 2 t 2 sin2n t; for t =2: D 2n j j Ä n 1 n2 XD n Â Ã Integrating with respect to t from 0 to =2 gives

~~3 2n 1 =2 1 2 sin2n t dt: 24 D 2n 0 n 1 n2 Z XD n Â Ã Applying Wallis’s formula (15.7), we get the well-known Euler formula~~

~~2 1 1 .2/: 6 D n2 D n 1 XD 4. Dividing (15.1) by x and integrating from 0 to x, we have~~

~~x 2 2n .arcsin t/ 1 1 .2x/ dt : (15.11) 0 t D 4 2n Z n 1 n3 XD n Â Ã 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,~~

~~2n 1 .2x/ 2 3 2 2x Hypergeometric F3Œ 1;1;1;1 ; ;2;2 ;x : 2n D 4 f g f2 g n 1 n3 XD n Â Ã Mathematica also deduces~~

2n 1 2 7 2 ln 2 .3/: (15.12) 2n D 2 n 1 n3 XD n Â Ã Moreover, replacing x by sin x and then setting t sin in (15.11), we have D x 2n 2 1 2 2 cot d sin2n x: (15.13) 0 D 2n Z n 1 n3 XD n Â Ã Setting x =2 and manipulating (15.12) and (15.13) yields an integral representa- D tion for .3/: 2 2 8 =2 .3/ ln 2 2 cot d: D 7 7 Z0 Integrating by parts converts this into 2 2 16 =2 .3/ ln 2 lnsin d: D 7 C 7 Z0 5. Using the power series expansion of arcsin x, Ewell in [3] rediscovered the Euler formula 2 1 .2n/ .3/ 1 4 : (15.14) D 7 .2n 1/.2n 2/22n ( n 1 ) XD C C In the following, we give a different proof of (15.14). Indeed, recall (see (6.23) and (6.26)) 2n 1 .2n/ cot 1 2 (15.15) D 2n I n 1 XD by (15.13), we get

2n 2 1 2 1 .2n/ 2.n 1/ 1 2 2n x x C sin x: (15.16) 2 .n 1/ 2n D 2n n 0 C n 1 n3 XD XD n Â Ã For x =2, appealing to (15.12), we obtain D 2 1 1 .2n/ 1 7 2 . 2 ln 2 .3//: 8 4 .n 1/22n D 4 2 n 1 XD C

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~~15. Power Series of .arcsin x/2 171~~

~~Thus, 2 2 1 1 .2n/ .3/ ln 2 : (15.17) D 7 2 C .n 1/22n n 1 ! XD C By the identity 1 1 1 2 ; n 1 D 2n 1 .2n 1/.2n 2/ C Â C C C Ã and appealing to the existing sum (see Exercise 10)~~

~~1 .2n/ 1 .1 ln 2/ (15.18) .2n 1/22n D 2 n 1 XD C we see that Euler’s formula (15.14) follows from (15.17) immediately.~~

~~Furthermore, integrating (15.16) we obtain~~

2n 2 x 1 3 1 .2n/ 2n 3 1 2 2n x 2n x C sin t dt: 6 .n 1/.2n 3/ D 2n 0 n 0 C C n 1 n3 Z XD XD n Â Ã For x =2, this gives another series representation for .3/: D 1 .2n/ .3/ 2 1 2 : (15.19) D .2n 2/.2n 3/22n ( n 1 ) XD C C 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 b b 1 f .x/ cot xdx 2 f .x/ sin 2nx dx: D Za n 1 Za XD Using this result veriﬁes (5.12) directly. 2. Setting x p2=2 in (15.10) yields D 2 n 1 1 .nŠ/ 2 C : D .2n 1/Š n 0 XD C Can you give another proof? 3. Using the power series expansion of (15.9) and 1 x arcsin t .arcsin x/2 dt; 2 D 0 p1 t 2 Z prove that 2 1 1 1 1 : 8 D C 32 C 52 CC .2k 1/2 C C

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~~4. Prove that~~

~~n 1 2n 3 1 1 .2x/ .arcsin x/4 ; D 2 i 2 2n n 1 (i 1 ) n2 XD XD n Â Ã n 1 i 1 2n 45 1 1 1 .2x/ .arcsin x/6 ; D 4 8 i 2 j 2 9 2n n 1~~

n 1 2 n 2 arcsin x 1 i 1 .1 .2i/ / 2n 1 i 1 .1 .2i 1/ / 2n 1 e D C x D C x : D .2n/Š C .2n 1/Š C n 0 Q n 0 Q XD XD C 6. Setting x 1=2 in (15.11) yields D 1=2 2 1 1 .arcsin x/ S 4 dx: WD 2n D x n 1 n3 Z0 XD n Â Ã Show that =3 1 1 S 2 x cos nx dx D n n 1 Z0 XD and n 1 sin 3 .3/ 3 S: D 2 n2 4 n 1 XD 7. Show that x=2 2 2 1 nŠ 1 xex =4 e t dt x2n: C D .2n/Š Z0 n 0 XD 8. Monthly Problem 4293 [1948, 254; 1950, 46–47]. Evaluate

~~n 1 .3 p5/ 2n n3 n 1 XD in closed form. 9. Monthly Problem 11400 [2008, 948]. Evaluate~~

~~1 .2n/ n.n 1/ n 1 XD C in closed form.~~

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~~15. Power Series of .arcsin x/2 173~~

~~10. Prove that~~

~~1 .2n/ 1 1 .2n/ 1 .ln ln 2/ and .1 ln 2/: .2n/22n D 2 .2n 1/22n D 2 n 1 n 1 XD XD C 11. Prove that~~

~~10 1 .n 1/.2n/ .2/ 1 12 ; D 3 C .n 1/.n 2/.n 3/2n ( n 2 ) XD C C C 2 4 1 .2n/ .3/ 1 6 : D 27 .2n 1/.2n 3/22n ( n 1 ) XD C C 12. Ramanujan Identity. Show that~~

~~3 7 1 1 .3/ 2 : D 180 k3.e2k 1/ k 1 XD This is a hyperbolic series approximation in which the “error” is~~

~~7 3 .3/ 0:003743: j 180 j~~

References [1] D. H. Lehmer, Interesting series involving the central binomial coefﬁcient, Amer. Math. Monthly, 92 (1985) 449–457. [2] Z. A. Melzak, Companion to Concrete Mathematics, John Wiley & Sons, New York, 1973. [3] J. A. Ewell, A new series representation for .3/, Amer. Math. Monthly, 97 (1990) 219–220.

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~~Six Ways to Sum .2/~~

~~Read Euler, read Euler, he is the master of us all. — P. Laplace~~

The history of mathematics is valuable because it shows the motivations underlying the developments of many branches of mathematics. Euler’s recorded work on infinite series provides a particularly instructive example. In the eighteenth century infinite series were, as they are today, considered an essential part of analysis. In comparison, at the dawn of the seventeenth century infinite series were littleunderstood and infrequently encountered. For example, Jakob Bernoulli handled the divergence of the harmonic series perfectly, but he could not find the exact value of the convergent series 1 1 1 .2/ 1 : WD C 22 C 32 CC n2 C Indeed, he included in his Tractatus [3] a plea for help in summing this series: If anyone finds and communicates to us that which thus far has eluded our efforts, great will be our gratitude. Before Euler, it appeared that all efforts at seeking an exact evaluation had proven futile, and even an accurate numerical evaluation was extremely difficult because of the slow decay of the terms. Indeed, since 1 1 1 1 1 1 1 < < n n 1 D n.n 1/ n2 n.n 1/ D n 1 n C C we have 1 1 1 1 < < ; N 1 n2 N n N 1 C DXC so that to compute directly the sum with an accuracy of six decimal places would require at least a million terms. In 1731, at the age of 24, Euler worked on this problem by calculating a numerical approximation. To this end, he transformed the series into the followingrapidly convergent one: 1 1 .2/ .ln 2/2 2 D C n2 2n n 1 XD 175

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~~with 1 1 ln 2 : D n 2n n 1 XD Using this he calculated .2/ accurately to six places and obtained the value~~

~~.2/ 1:644944:::: D Four years later, he ﬁnally arrived at the following brilliant result:~~

1 1 1 2 .2/ 1 : (16.1) D C 22 C 32 CC n2 CD 6 This truly remarkable formula yielded .2/, the ﬁrst nontrivial value of the Riemann zeta function. Since the ﬁnite sums of the series in (16.1) are always rational, no one expected the irrational number , the ratio of the circumference of a circle to the diameter, to appear in the value of the sum. One of the most memorable gems of 18th century mathematics, (16.1) has a prominent place in mathematical history. A number of extremely elegant and clever proofs have been discovered (and rediscovered). In this chapter, we present six of these proofs.

16.1 Euler’s Proof We begin with the heuristic reasoning that led Euler to his wonderful discovery. Euler’s idea was based on the extension of Newton’s symmetric functions of the polynomial roots to the case where the polynomial is replaced by a power series. Let

x2 x2 x2 1 1 1 1 ˛x2 ˇx4 . 1/n x2n: r 2 r 2 r 2 D C C Â 1 Ã Â 2 Ã Â n Ã We have 1 1 1 1 1 1 ˛ ; ˇ 2 2 2 2 2 2 2 2 2 D r1 C r2 CC rn D r1 r2 C r1 r3 CC rn 1rn and so on. In particular 1 1 1 2 4 4 4 ˛ 2ˇ: r1 C r2 CC rn D Euler’s idea was to apply these relations to the case when the polynomial is replaced by a power series sin x x2 x4 x6 1 : x D 3Š C 5Š 7Š C The function sin x=x has the roots ; 2; 3;:::; and so the above formulas give ˙ ˙ ˙ the following. First 1 1 1 1 ; 2 C 4 2 C 9 2 CD 3Š which leads at once to 1 1 1 2 .2/ 1 : D C 22 C 32 CC n2 CD 6

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~~Onecan go on and on,which is what Euler did,calculating .2k/ up to k 6. In particular, D 1 1 1 4 .4/ 1 : D C 24 C 34 CC n4 CD 90~~

Euler’s proof is not considered valid today since the power series is not polynomial and does not share all properties of polynomials. Indeed, when Euler communicated (16.1) to his friends, the following objection was brought up: if f .x/ is any function to which this method is applied, f .x/ and exf .x/ both have the same roots and yet they should lead to different formulas. It took him about ten years to ﬁnally succeed in obtaining Euler’s inﬁnite product for the sine (see (5.26))

2 sin x 1 x 1 : (16.2) x D n2 2 n 1 Â Ã YD Once this formula is established, all the objections disappear. Moreover, based on (16.2), Euler then succeeded in gettinga closed formula for all the .2k/, as we showed in Chapter 6. Euler returned to the evaluation of .n/ many times. In 1737, he discovered the fabu- lous product formula

.s/ .1 1=ps/; where p runs over all primes: D p Y Investigations on .n/ for odd n led him in 1749 to establish the functional equation of the zeta function s .1 s/ 2.2/ s cos .s/.s/; D 2 subsequently forgotten for over a century until resurrected by Riemann in 1859. Euler was one of the greatest masters of the theory of inﬁnite series, one whose height very few can hope to reach. Yet we can all learn from his impact on modern mathematics and inspire students to look at Euler’s work for new insights and focus.

16.2 Proof by Double Integrals The proof presented ﬁrst appeared as an exercise in William LeVeque’s number theory text- book from 1956, and was later rediscovered in the Mathematical Intelligencer by Apostol in 1983. The proof consists of two different evaluations of the double integral

~~1 1 1 I dxdy: WD 0 0 1 xy Z Z First, we expand the integrand as a geometric series, apply the monotone convergence~~

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~~theorem, and then integrate term by term. This results in~~

1 1 1 I xnyndxdy D Z0 Z0 n 0 XD 1 1 1 xndx yndy D n 0 ÂZ0 Ã ÂZ0 Ã XD 1 1 .2/: D .n 1/2 D n 0 XD C Next, we evaluate I by substitution x y y x u C ; v : WD 2 WD 2 This yields dudv I 2 D 1 u2 v2 ZZS C where S is the square with vertices .0; 0/; .1=2; 1=2/; .1; 0/ and .1=2; 1=2/ as shown in Figure 16.1.

~~y v 1~~

~~D 1/2~~

~~x u 1 S 1~~

~~1/2~~

~~Figure 16.1. Transformed region of integration~~

~~In view of the symmetry of the square we get~~

~~1=2 u dv 1 1 u dv I 4 du 4 du: D 1 u2 v2 C 1 u2 v2 Z0 ÂZ0 C Ã Z1=2 ÂZ0 C Ã Using dx 1 arctan.x=a/ C a2 x2 D a C Z C yields~~

~~1=2 1 u 1 1 1 u I 4 arctan du 4 arctan du: D 0 p1 u2 p1 u2 C 1=2 p1 u2 p1 u2 Z Â Ã Z Â Ã Set u 1 u g.u/ arctan ; h.u/ arctan : WD p1 u2 WD p1 u2 Â Ã Â Ã~~

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~~We have 1 1 1 g0.u/ ; h0.u/ : D p1 u2 D 2 p1 u2 Thus~~

1=2 1 I 4 g .u/g.u/du 4 2h .u/h.u/du D 0 C 0 Z0 Z1=2 1=2 1 2 g2.u/ 4 h2.u/ D 0 1=2 2 2.=6/2 0 0 4.=6/2 D C D 6 as expected. Note that 1 1 1 1 1 1 1 1 .2/; 22 C 42 C 62 CD 22 C 22 C 32 C D 4 Â Ã so the sum of odd terms

~~1 1 1 1 1 3 .2/: 12 C 32 C 52 CD .2n 1/2 D 4 n 0 XD C Thus Euler’s formula (16.1) is equivalent to~~

~~2 1 1 : (16.3) .2n 1/2 D 8 n 0 XD C In a similar fashion, we may express the series in (16.3) as a double integral, namely~~

~~1 1 1 1 1 J dxdy: WD .2n 1/2 D 1 x2y2 n 0 Z0 Z0 XD C To compute the double integral, we ignore the boundary of the domain and consider x; y in the interior of the unit square~~

~~D .x; y/ 0~~

~~2 2 1 1 y 1 1 x .u; v/ tan x ; tan y D 0 1 x2 s 1 y2 1 r @ A we have sin u sin v .x; y/ ; (16.4) D cos v cos u Â Ã and u; v lie in the triangle~~

~~S .u; v/ u>0;v>0;u v <=2 Df W C g whenever .x; y/ D. See Figure 16.2. 2~~

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~~y v p/2 1~~

~~D S~~

~~x u 1 p/2~~

~~Figure 16.2. Transformed region of integration~~

The Jacobian determinant is cos u sin u sin v cos v cos2 v sin2 u sin2 v det 0 1 1 1 x2y2: sin u sin v cos v D cos2 u cos2 v D B 2 C B cos u cos u C @ A Thus (16.4) is a bijective transformation between D and S. Hence

~~1 2 2 J 1 dudv Area of S D S D D 2 2 D 8 ZZ Á as required. Recently, Noam Elkies in [5] generalized (16.4) to the n-variable transformation~~

sin u1 sin u2 sin un 1 sin un x1 ; x2 ; :::;xn 1 ; xn (16.5) D cos u2 D cos u3 D cos un D cos u1 and proved the following two properties: 1. The Jacobian determinant of the transformation (16.5) is

~~@.x1;x2;:::;xn/ 2 1 .x1x2 xn/ ; @.u1; u2;:::;un/ D ˙~~

~~the sign and chosen according to whether n is even or odd. C 2. The transformation (16.5) maps the polytope~~

~~…n .u1; u2;:::;un/ ui > 0;ui ui 1 < =2 .1 i n/ WD f W C C Ä Ä g one-to-one into the open unit cube~~

~~Sn .x1;x2;:::;xn/ 0~~

~~In particular, he found that the volume of the polytope …n is~~

~~nk 1 . 1/ 1 1 S.n/ : D .2k 1/n D .4k 1/n k 0 k XD C D1X C~~

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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 sin2 x D 4 sin2.x=2/ cos2.x=2/ 1 1 1 D 4 sin2.x=2/ C cos2.x=2/ Â Ã 1 1 1 : D 4 sin2.x=2/ C sin2.. x/=2/ Â C Ã Letting x =2 and repeatedly applying this identity yields D 1 1 1 1 1 D sin2.=2/ D 4 sin2.=4/ C sin2.3/=4 Â Ã 1 1 1 1 1 D 16 sin2.=8/ C sin2.3/=8 C sin2.5=8/ C sin2.7/=8 Â Ã 2n 1 1 1 D 4n sin2..2k 1/=2n 1/ k 0 C XD C 2n1 1 2 1 : D 4n sin2..2k 1/=2n 1/ k 0 C XD C In the range of 0

~~1 1 1 > > cot2 t 1: sin2 t t 2 D sin2 t Thus, for N 2n and t .2k 1/=.2N /, we have D D C N=2 1 8 1 1 1 > >1 : 2 .2k 1/2 N k 0 XD C Taking the limit N yields ! 1~~

~~8 1 1 1 ; D 2 .2k 1/2 k 0 XD C which is equivalent to (16.3). The second one is based on the trigonometric identity~~

~~N k N.2N 1/ cot2 : (16.6) 2N 1 D 3 k 1 Â Ã XD C~~

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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 < < csc2 t; t 2 using (16.6) yields

N.2N 1/ N 2N 1 2 2N.N 1/ < C < C ; 3 k 3 k 1 Â Ã XD that is 2 2N 2N 1 N 1 2 2N 2N 2 < < C : 6 2N 1 2N 1 k2 6 2N 1 2N 1 k 1 C C XD C C Both the left-hand and the right-hand side converge to 2=6 for N , which ends the ! 1 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

x dt 1 1 2n x x2n 1; arcsin 2n C (16.7) D 0 p1 t 2 D 2 .2n 1/ n Z n 0 Â Ã XD C which is valid for x 1. Putting x sin t we get j j Ä D 1 1 2n t sin2n 1 t D 22n.2n 1/ n C n 0 Â Ã XD C for t =2. Integrating from 0 to =2 and using Wallis’s formula j j Ä =2 2 4 .2n/ sin2n 1 t dt C D 3 5 .2n 1/ Z0 C yields (16.3) again. Indeed, Choe’s proof can be traced back to the following original proof of Euler. Let

~~1 t k Ik dt: WD 0 p1 t 2 Z An integration by parts shows that k 1 Ik 2 C Ik C D k 2 C~~

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~~from which Ik can be calculated for all k. In particular, we have~~

2 4 .2n/ I2n 1 : (16.8) C D 3 5 .2n 1/ C In view of 1 x arcsin t .arcsin x/2 dt; 2 D 0 p1 t 2 Z setting x 1, then replacing arcsin t in the integrand by its power series expansion (16.7) D and making use of (16.8), yields

2 1 1 1 1 8 D C 32 C 52 CC .2n 1/2 C C 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 1 Œ.n 1/Š 1 1 .2/ 3 3 : (16.9) D .2n/Š D 2n n 1 n 1 n2 XD XD n Â Ã 16.5 Proof by Fourier Series Let f .x/ x.1 x/. Consider the Fourier cosine series of f .x/ on Œ0; 1. Notice that D 1 1 a0 2 f.x/dx ; D D 3 Z0 1 1 . 1/n an 2 f .x/ cos nxdx 2 C : D D n2 2 Z0 Therefore

a0 1 f .x/ an cos nx D 2 C n 1 XD 1 1 cos 2nx : D 6 2n2 n 1 XD Since f is continuous, of bounded variation on Œ0; 1 and f .0/ f .1/, the Fourier series D of f converges to f pointwise. Setting x 0 we again get .2/ 2=6. D D 16.6 Proof by Complex Variables The ﬁnal proof by Russell in [7] uses only Euler’s formula

~~eix cos x i sin x D C~~

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~~and the well-known logarithmic series. It begins with the deﬁnite integral~~

~~=2 I ln.2 cos x/dx: WD Z0 Now 2 cos x eix e ix eix.1 e 2ix/ implies D C D C 2ix ln.2 cos x/ ix ln.1 e /: D C C In terms of the logarithmic series expansion, we have~~

~~=2 n 1 . 1/ I ix e 2nix dx D n Z0 n 1 ! XD 2 1 1 i : D 8 .2k 1/2 k 0 ! XD C But I is real, forcing 2 1 1 8 D .2k 1/2 k 0 XD C and thereby yielding an added bonus~~

=2 ln.cos x/dx ln 2: D 2 Z0 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~~

~~1 1 .2/ .ln 2/2 2 ; D C n2 2n n 1 XD 3 8 .ln 2/ 1 1 ln 2 1 .3/ : D 7 3 C 2n n2 C n3 n 1 Â Ã! XD~~

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~~2. Another proof of Euler’s formula by double integral. Prove that~~

1 2k 2 2k 1 tanh x (a) t dt 2 dx, 2k 1 D 1 D cosh x C Z Z1 2k 2k 1 tanh x tanh y 1 (b) ; cosh2 x cosh2 y D cosh.x y/ cosh.x y/ k 0 XD C 2 1 1 1 1 (c) 1 1 dxdy : .2k 1/2 D 4 cosh.x y/ cosh.x y/ D 8 k 0 Z1Z1 XD C C 3. Recall that (for example, see (5.28))

~~n x sin2.x=.2n 1// sin x .2n 1/ sin 1 C : D C 2n 1 sin2.k=.2n 1// Â Ã k 1 ! C YD C Let sin2.x=.2n 1// uk C : D sin2.k=.2n 1// C Show that~~

~~n n k 1 uk sin x uk Dn 1 : 1 uk Ä .2n 1/ sin.x=.2n 1// Ä 1 uk P k 1 k 1 C D C C XD P Based on this inequality, give a proof of Euler’s formula.~~

~~4. Monthly Problem 3996 [1941, 341; 1942, 483–485]. Sum the series~~

~~1 Œ.n 1/kŠ ; .nk/Š n 1 XD where k is any integer greater than unity.~~

~~5. Monthly Problem 4318 [1948, 586; 1950, 345–346]. Show that~~

~~1 1 2 11 G .2/; .n sinh n/2 D 3 30 n 1 XD where G is Catalan’s constant, which is deﬁned by~~

~~n 1 . 1/ G : D .2n 1/2 n 0 XD C 6. Show that~~

~~n k 1 n k 1 1 1 n . 1/ 1 1 n . 1/ C 2 ln 2 and C .2/: 2n k k D 2n k k2 D n 1 k 1 Â Ã n 1 k 1 Â Ã XD XD XD XD~~

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Comment. Recall Euler’s Series Transformation, which states that if f .t/ n D n1 0 ant and n1 0 an converges, then D D P P n 1 t 1 n n f t ak : 1 t 1 t D k Â Ã n 0 k 0 Â Ã ! XD XD One may establish that

n k n 1 1 n x 1 x 2 : 2n k kp D np n 1 k 1 Â Ã n 1 XD XD XD 7. (H. Wilf) For every integer n 0, show that n 4 2 2 1 nŠ 1 k 3 : 2i C .2n/Š Œ.k n 1/Š2 D 6 i 1 i 2 k 0 C C XD i XD Â Ã Remark. This gives us a family of series which all converge to 2=6. When n 0 we D have the usual series for .2/, and when n we have the faster convergent series ! 1 (16.9). 8. Show that 1 1 1 1 1 1 ln.xy/ .3/ dxdydz dxdy D 0 0 0 1 xyz D 0 0 1 xy Z Z Z Z Z 1 ln.1 x/ ln x =2 lnsin x ln cos x dx 4 dx: D x D sin x cos x Z0 Z0 9. Show that 1 1 ln.1 x/ ln x .3/ dx: D 2 0 x.1 x/ Z Use this identity to show that

1 1 1 1 .3/ : D 2 i j 1 i 1 j 1 ij 2 C XD XD j Â Ã 10. Deﬁne 1 1 1 An : D i1i2 in.i1 i2 in/ i1 1 in 1 XD XD C CC Show that 1 n n ln x An . 1/ dx nŠ.n 1/: D 1 x D C Z0 11. Deﬁne i1 i2 in 1 1 . 1/ C CC Bn : D i1i2 in.i1 i2 in/ i1 1 in 1 XD XD C CC

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~~Show that~~

~~1 n n ln .1 x/ Bn . 1/ C dx D 0 x Z n n n n 1 1 n.n 1/ .n i 1/ n i . 1/ nŠ.n 1/ ln C 2 C ln 2 : D C n 1 2kki 1 k 1 i 1 C ! C XD XD 12. Open Problem. Motivated by~~

~~i 1 1 1 . 1/ ln.1 x/ ln.1 x/ 5 C dx .3/; ij.i j / D x D 8 i 1j 1 Z0 XD XD C deﬁne i1 i2 im 1 1 . 1/ C CC Cn : D i1i2 in.i1 i2 in/ i1 1 in 1 XD XD C CC For 0~~

~~1 m n m ln .1 x/ ln .1 x/ I.n; m/ C dx: D x Z0 13. Let 0<˛<1=2. Prove that~~

~~n n k 1 1 n . ˛/ .p 1/ : C D .1 ˛/n 1 k .k 1/p 1 n 0 C k 0 Â Ã C XD XD C In particular, n n k k 1 n . 1/ 2 .p 1/ 2 : C D k .k 1/p 1 n 0 k 0 Â Ã C XD XD C~~

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 2 2 [4] B. R. Choe, An elementary proof of k1 1 1=n =6, Amer. Math. Monthly, 94 (1987) 662–663. D D P n [5] N. D. Elkies, On the sums k1 .4k 1/ , Amer. Math. Monthly, 110 (2003) 561–573. D1 C [6] D. Kalman, Six ways to sumP 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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~~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

m 1 1 1 1 1 1 S.m; n/ 1 H m; (17.1) WD .k 1/n C 2 CC k D .k 1/n k k 1 Â Ã k 1 XD C XD C and was successful in obtaining several explicit values of these sums in terms of the Rie- mann zeta function .k/. Indeed, he established the beautiful formula

~~1 1 S.1; 2/ Hk .3/ D .1 k/2 D k 1 XD C as well as the more general relations: for all n>2,~~

~~n 2 n 1 S.1; n/ .n 1/ .k 1/.n k/: (17.2) D 2 C 2 C k 1 XD Two slight variant sums of Euler’s deﬁnition, which are known nowadays as Euler sums, are deﬁned as~~

~~1 1 S.m; n/ H .m/; with H .m/ 1 1=2m 1=km; (17.3) D kn k k D C CC k 1 XD and m 1 1 1 1 T .m; n/ 1 : (17.4) D kn C 2 CC k k 1 Â Ã XD It is easy to see that~~

~~T .2; n/ S.2; n/ 2S.1; n 1/ .n 2/: D C C C C 189~~

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~~190 Excursions in Classical Analysis~~

~~Over the centuries progress on evaluating S.m; n/ and T .m; n/ for m 2 has been mini- mal. In 1948, as Monthly Problem 4305, Sandham proposed to prove that~~

1 1 17 T .2; 2/ H 2 .4/; (17.5) D k2 k D 4 k 1 XD which apparently remained unnoticed until1993 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

~~2 1 1 Hk .3/ ln 2; 2kk2 D 12 k 1 XD 1 1 7 H .2/ .4/; k2 k D 4 k 1 XD 1 1 7 H 2 .5/ .2/.3/; k3 k D 2 k 1 XD and~~

k 1 1 . 1/ 5 Hk .3/: k2 D 8 k 1 XD In particular, Borwein and Bradley in [3] recently collected thirty-two beautiful proofs of S.1; 2/ .3/. D Motivated by these results, replacing Hk by 1 1 hk 1 ; (17.6) D C 3 CC 2k 1 we study the following new variant Euler sums

~~1 ak hk k 1 XD where the ak are relatively simple functions of k.~~

~~We begin by deriving some series involving hk . Since~~

~~x k dt 1 x ln.1 x/ ; D 0 1 t D k Z k 1 XD replacing x by x gives k 1 k 1 . 1/ x ln.1 x/ : C D k k 1 XD~~

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~~17. Evaluations of Some Variant Euler Sums 191~~

~~Averaging these two series yields~~

~~1 1 x 1 1 ln C x2k 1: (17.7) 2 1 x D 2k 1 Â Ã k 1 XD In terms of the Cauchy product and partial fractions, we have~~

1 1 x 1 1 1 1 ln2 C x2k 4 1 x D .2k 1/ 1 C .2k 3/ 3 CC 1 .2k 1/ Â Ã k 1 Â Ã XD 1 1 1 1 1 1 1 1 x2k D 2k 2k 1 C 1 C 2k 3 C 3 CC 1 C 2k 1 k 1 ÄÂ Ã Â Ã Â Ã XD 2k 1 1 1 x 1 : D C 3 CC 2k 1 k k 1 Â Ã XD Appealing to (17.6), we have

1 hk 1 1 x x2k ln2 C : (17.8) k D 4 1 x k 1 Â Ã XD This enables us to evaluate a wide variety of interesting series via specialization, differentiation and integration. First, setting x 1=2, we ﬁnd D

~~1 h 1 k ln2 3: (17.9) 22k k D 4 k 1 XD For x p2=2, D 1 h 1 k ln2.3 2p2/: (17.10) 2k k D 4 C k 1 XD Putting x .p5 1/=2 , the reciprocal of the golden ratio, we get D D N~~

~~1 h 1 k 2k ln2.2 p5/: (17.11) k N D 4 C k 1 XD Furthermore, for any ˛ 2, putting x .p5 1/=2˛ and x .p5 1/=2˛ in (17.8) D C D respectively, we get~~

~~2k 1 hk p5 1 1 .2˛ 1/ p5 C ln2 C C (17.12) ˛2k k 2 D 4 p k 1 ! .2˛ 1/ 5 ! XD and 2k 1 hk p5 1 1 .2˛ 1/ p5 ln2 C : (17.13) ˛2k k 2 D 4 p k 1 ! .2˛ 1/ 5! XD C~~

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~~Recalling the Fibonacci numbers, which are deﬁned by~~

~~F1 1; F2 1; Fk Fk 1 Fk 2 for k 2 D D D C and Binet’s formula~~

~~k k 1 p5 1 1 p5 Fk C ; D p5 0 2 ! 2 ! 1 @ A and combining (17.12) and (17.13), we ﬁnd~~

~~2 2 1 hk p5 ˛ ˛ 1 ˛ ˛p5 1 F2k ln C ln C C : (17.14) ˛2k k D 20 ˛2 ˛ 1 ˛2 ˛p5 1 k 1 Â Ã ! XD C In particular, for ˛ 2, D~~

1 hk p5 F2k ln 5 ln.9 4p5/: (17.15) 22k k D 4 C k 1 XD Another step along this path is changing variables. Setting x cos in (17.8) leads to D 1 h k cos2k ln2 .cot.x=2// : (17.16) k D k 1 XD Integrating both sides from 0 to , and using

~~ 2k cos2k d D 22k k Z0 Â Ã and 3 ln2 .cot.x=2// d ; D 4 Z0 we ﬁnd 2 1 h 2k k : (17.17) 22k k k D 4 k 1 Â Ã XD Next, for 0~~

~~1 2k x 1 x hkx ln C : (17.18) D 2.1 x2/ 1 x k 1 Â Ã XD Setting x 1=2,we get D 1 h 1 k ln 3: (17.19) 22k D 3 k 1 XD For x p2=2, D 1 h p2 k ln.3 2p2/: (17.20) 2k D 2 C k 1 XD~~

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~~17. Evaluations of Some Variant Euler Sums 193~~

~~We can now deduce a result similar to (17.15):~~

~~1 hk p5 F2k .10 ln.5 2p5/ 3p5 ln 5 5 ln 5/: (17.21) 22k D 50 C C k 1 XD Finally, for 0~~

~~1 1 ln u 1 u 1 1 ln du 2 u2k ln u du: .1 x/=.1 x/ u 1 u D 2k 1 .1 x/=.1 x/ Z C Â Ã k 0 Z C C XD C Since 1 1 u2k ln u du u2k 1 ln u u2k 1 C; D 2k 1 C .2k 1/2 C C Z C C we ﬁnd~~

2k 1 1 hk 1 1 x 1 x 1 1 1 x C x2k ln x ln2 2 ln k2 D 2 1 x C 1 x .2k 1/2 1 x k 1 Â Ã Â Ã k 0 Â Ã XD C C XD C C 2k 1 1 1 1 1 1 x C 2 2 : (17.23) C .2k 1/3 .2k 1/3 1 x k 0 k 0 Â Ã XD C XD C C In order to proceed comfortably, we employ the classical polylogarithm function

n 1 x Lin.x/ ; D kn k 1 XD which is the generating function of the sequence 1=kn . The special case n 2 is the f g D dilogarithm introduced by Euler. Clearly, .2/ Li2.1/ and D n 1 x 1 .Lin.x/ Lin. x//: .2k 1/n D 2 k 0 XD C Appealing to (17.23) and

~~1 1 1 1 1 1 7 .3/; .2k 1/3 D k3 .2k/3 D 8 k 0 k 0 k 0 XD C XD XD~~

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~~we ﬁnally obtain~~

1 hk 7 1 1 x x2k .3/ ln x ln2 k2 D 4 C 2 1 x k 1 Â Ã XD C 1 x 1 x x 1 ln Li2 Li2 C 1 x 1 x 1 x Â C Ã Â Â C Ã Â C ÃÃ 1 x x 1 Li3 Li3 : 1 x 1 x Â Â C Ã Â C ÃÃ Taking the limit as x approaches 1, we get

1 h 7 k .3/: (17.24) k2 D 4 k 1 XD For x 1=3, D 1 h 7 1 k .3/ ln 3 ln3 2 32k k2 D 8 2 k 1 XD 1 3 ln 2 ln 2 Li2. 1=2/ Li3. 1=2/; C 3 C C where we have used 2 1 2 Li2.1=2/ ln 2 D 12 2 I 2 7 1 3 Li3.1=2/ .3/ ln 2 ln 2: D 8 C 6 12 Moreover, manipulating (17.22) yields

~~1 1 hk 2k 1 1 2 1 t 2 .1 x / ln C dt: k D 2 x t 1 t k 1 Z Â Ã XD Appealing to~~

k 1 1 k 1 2k 2.i 1/ 2.i 1/ 1 x hk x dt x dt dx D D D 1 x2 i 1 Z0 Z0 i 1 ! Z0 XD XD we have 2 1 2k 1 hk 1 hk 1 x 2 2 2 dx k D k 0 1 x k 1 k 1 Z XD XD 1 1 1 1 1 1 t ln2 C dt dx: D 2 1 x2 t 1 t Z0 Â Zx Â Ã Ã Interchanging the order of the integration, we get

~~2 1 t 1 hk 1 1 2 1 t 1 2 ln C 2 dx dt: k D 2 0 t 1 t 0 1 x k 1 Z Â Â Ã Z Ã XD 1 1 1 1 t ln3 C dt: D 4 0 t 1 t Z Â Ã~~

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~~17. Evaluations of Some Variant Euler Sums 195~~

Using the substitution x .1 t/=.1 t/ and the well-known fact that D C 1 6 xk 3 xdx ; ln 3 0 D .k 1/ Z C we ﬁnd 2 1 3 1 hk 1 ln x 2 2 dx k D 2 0 1 x k 1 Z XD 1 1 1 x2k ln3 xdx D 2 0 k 0 Z XD 1 1 45 3 .4/: (17.25) D .2k 1/4 D 16 k 0 XD C Another path from (17.8) is via complex variables and the formula: 1 1 1 z tan 1.iz/ tanh 1 z ln C : i D D 2 1 z Â Ã Replacing x by ix in (17.8), we obtain

k 1 1 . 1/ hk x2k .tan 1 x/2: (17.26) k D k 1 XD This series may be evaluated explicitly at the values x 2 p3; p3=3; 1: D k 1 2k 2 1 . 1/ hk .2 p3/ 3 .2/; (17.27) k D 144 D 72 k 1 XD k 1 2 1 . 1/ hk 1 .2/; (17.28) 3k k D 36 D 6 k 1 XD k 1 2 1 . 1/ hk 3 .2/: (17.29) k D 16 D 8 k 1 XD Similarly, differentiating and integrating (17.26), we derive the corresponding formulas

~~1 k 1 2k x 1 . 1/ hk x tan x; (17.30) D 1 x2 k 1 XD C k 1 x 1 2 1 . 1/ hk 2k .tan t/ 2 x 2 dt: (17.31) k D 0 t k 1 Z XD In particular, we ﬁnd~~

~~k 1 1 . 1/ hk p3 ; (17.32) 3k D 24 k 1 XD k 1 1 . 1/ hk 7 G .3/; (17.33) k2 D 4 k 1 XD~~

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~~where G is Catalan’s constant, given by~~

~~k 1 . 1/ G : D .2k 1/2 k 0 XD C Finally, observing that 1 hk H2k Hk; (17.34) D 2 we ﬁnd from (17.24)~~

~~1 1 1 1 1 1 1 11 H2k hk Hk .3/: (17.35) k2 D k2 C 2 k2 D 4 k 1 k 1 k 1 XD XD XD Furthermore, using partial fraction yields~~

1 1 1 1 1 1 1 1 ij.i j / D i 2 j i j i 1 j 1 i 1 j 1 Â Ã XD XD C XD XD C 1 1 Hi 2.3/: D i 2 D i 1 XD Similarly, i j 1 1 . 1/ 1 C .3/: ij.i j / D 4 i 1 j 1 XD XD C The difference gives 1 7 .3/: ij.i j / D 8 i;j >1;i j odd C XC D Setting i j 2k 1 and using partial fractions, we have C D C 2k 1 1 1 ij.i j / D j.2k 1 j /.2k 1/ i;j >1;i j odd C k 1 j 1 C C XC D XD XD 2k 1 1 1 1 D .2k 1/2 j C 2k 1 j k 1 j 1 Â Ã XD C XD C 1 1 2H2k: D .2k 1/2 k 1 XD C Thus, 1 1 7 H2k .3/: (17.36) .2k 1/2 D 16 k 1 XD C Subsequently, we have

~~1 1 1 1 1 1 1 1 H2k H2k .2k 1/2 D .2k 1/2 C .2k 1/3 C 2k.2k 1/2 k 1 k 1 k 1 k 1 XD XD C XD XD 21 1 .3/ . 2 8 ln 2/: (17.37) D 16 C 8~~

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~~17. Evaluations of Some Variant Euler Sums 197~~

~~From this and the known result~~

1 1 1 2 2 Hk . ln 2 8 ln2 7.3//; (17.38) .2k 1/2 D 4 C k 1 XD we ﬁnally get 1 1 7 3 hk .3/ .2/ ln 2: (17.39) .2k 1/2 D 16 C 4 k 1 XD 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

~~1 1 S.m; n/ H .m/; D .2k 1/n k k 0 XD C 1 1 T.m; n/ h.m/; D .2k 1/n k k 0 XD C 1 1 U.m;n;r/ H .m/: D kn kr k 1 XD It follows that~~

~~1 1 1 1 S.m; n/ H .m/ H .m/ D .2k/n 2k C .2k 1/n 2k 1 k 1 k 0 C XD XD C 2 nU.m;n;2/ T.m; n/ 2 nS.m; n/: D C C One can compute a special case, a result similar to Euler’s formula (17.2):~~

S.1; 2p 1/ .2 2 .2p 1//.S.1; 2p 1/ .2 2p// C D C C C 2p 2.2p 1/ ln 2 21 k.k/.2p 2 k/; (17.40) C C C k 2 XD where 1 1 x .x/ .1 2 /.x/: D .2k 1/x D k 0 XD C 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~~

~~i 1 1 .i 1/Š.j 1/Š 1 j i Hj 1 .2/ and . 1/ C .2/: .i j /Š D j j 1 D i 1j 1 i 0 j 0 Â Ã XD XD C XD XD C~~

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~~2. Evaluate 1 1 : i 2 j 2 i;j 1;i j DX¤ 3. Show that, for x <1, j j~~

~~1 k 2k arctan x . 1/ hk 1x : C D x.1 x2/ k 0 XD C~~

4. For positive integers k, let gk 0 2n 1 k 1=.2n 1/. Show that D Ä C Ä C P k 1 k k 1 . 2/ x 1 gkx ex : kŠ k D kŠ k 1 k 1 XD XD Use this to prove: n k 1 n . 2/ gn : D k k k 1 Â Ã XD 5. Let S.m; n/ be deﬁned by (17.3). Show that

~~S.m; n/ S.n; m/ .m/.n/ .m n/: C D C C~~

~~6. Show that Li2.x/ satisﬁes the functional equation~~

Li2.x/ Li2.1 x/ .2/ ln x ln.1 x/: C D 7. Let G be Catalan’s constant. Prove that n 1 . 1/ (a) Hn G ln 2: 2n 1 D 2 n 0 XD C n 1 . 1/ 1 (b) hn ln 2 G: 2n 1 D 8 2 n 0 XD C n 1 1 . 1/ C 7 (c) hn G .3/: n2 D 4 n 1 XD n 1 2 (d) hn 1 2G: 2n C D n 0 .2n 1/ XD C n Â Ã 8. Show that 1 2 S.2; 2/ . t/2 ln2.2 sin.t=2//dt: D 4 Z0 9. Ramanujan’s Constant G.1/ (see [1] Entry 11. p 255). Ramanujan claimed that

~~k 1 1 1 1 . 1/ 1 1 G.1/ hk : WD 8 k3 D 4 .4k 1/3 p .2k 1/3 k 1 k 1 3 3 k 1 XD XD C XD C~~

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~~17. Evaluations of Some Variant Euler Sums 199~~

~~Disprove this identity. Show that 15 G.1/ S.2; 2/ .4/: D C 16~~

~~10. Prove (17.40). 11. For x <1, deﬁne j j 1 h f .x/ k x2k 1: D 2k 1 k 1 XD Show that x 1 2 1 f ln .1 x/ Li2.x/: 2 x D 8 C 2 Á 12. Prove that, for n>1,~~

~~n 1 1 h 1 1 1 k .2n 1/ 2 .2i/ .2n 1 2i/ : k2n D 4 C C 2 C C 2 k 1 i 1 Ä ! XD XD~~

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/ı 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. Avail- able 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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~~Interesting Series Involving Binomial Coefficients~~

~~In his wonderfulpaper [4], Lehmer deﬁnes a series to be interesting ifthe sum has a closed form in terms of well-known constants. Using the power series~~

~~2n 2x arcsin x 1 .2x/ ; (18.1) p1 x2 D 2n n 1 n XD n Â Ã he establishes a large class of interesting series, including~~

~~1 1 1 1 1 1 1 2p3 ; (18.2) D 2 C 6 C 20 C 70 CD 3 C 27 n 1 2n XD n Â Ã 1 1 1 1 1 1 p3 ; (18.3) 2n D 2 C 12 C 60 C 280 CD 9 n 1 n XD n Â Ã and~~

~~1 1 1 1 1 1 1 2: (18.4) 2n D 2 C 24 C 180 C 1120 CD 18 n 1 n2 XD n Â Ã However, he points out that the series~~

1=2 2 1 1 1 1 1 1 .arcsin y/ 4 dy 2n D 2 C 48 C 540 C 4480 CD 0 y n 1 n3 Z XD n Â Ã 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 attentionon 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~~

n 1 1 x x.1 t/ 2 dt for x <4. (18.5) 2n D 0 .1 xt.1 t// j j n 1 Z XD n Â Ã 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

~~1 ˇ.m; n/ t m 1.1 t/n 1 dt D Z0 and for all positive integers m and n,~~

.m 1/Š.n 1/Š ˇ.m; n/ : D .m n 1/Š C We have 1 1 2n nŠ nŠ n t n 1.1 t/n dt: (18.6) n D .2n/Š D Â Ã Z0 By the ratio test, we see that the series in (18.5) converges for x <4. Next, interchanging j j the order of the summation and the integration yields

~~n 1 1 x 1 n n 1 n nx t .1 t/ dt: (18.7) 2n D 0 n 1 Z n 1 XD n XD Â Ã Appealing to the well-known series~~

1 1 nzn 1 for z <1 D .1 z/2 j j n 1 XD and t.1 t/ 1=4 for 0 t 1, we establish (18.5) from (18.7). Ä Ä Ä We now demonstrate some examples of interesting series via its specialization, integration, and differentiation of (18.5). Starting by setting x 1 in (18.5), we recover (18.3) D

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~~18. Interesting Series Involving Binomial Coefficients 203~~

as 1 1 1 .1 t/dt 1 2p3 2 2 : 2n D 0 .t t 1/ D 3 C 27 n 1 Z C XD n Â Ã For x 1, letting .1 p5/=2 be the Golden-Ratio, .1 p5/=2, we obtain D D C N D n 1 1 1 . 1/ .1 t/dt 2 2 2n D 0 .t t 1/ n 1 Z XD n Â Ã 1 .1 t/dt 2 D 0 Œ.t /.t / Z N 1 4p5 ln : D 5 C 25 In general, noting that

~~1 x.1 t/ 1 1 d.1 xt.1 t// x 1 dt dt .1 xt.1 t//2 D 2 .1 xt.1 t//2 C 2 .1 xt.1 t//2 Z0 Z0 Z0 x 1 dt ; D 2 .1 xt.1 t//2 Z0 and taking into account that~~

~~dt 2at b 2a dt C ; .at 2 bt c/2 D .4ac b2/.at 2 bt c/ C 4ac b2 at 2 bt c Z C C C C Z C C we ﬁnd that, for x>0,~~

~~n 1 x x 4px x arctan : (18.8) 2n D 4 x C .4 x/3=2 4 x n 1 Âr Ã XD n Â Ã Next, dividing (18.5) by x and then integrating from 0 to x, we have~~

~~n 1 1 x x.1 t/ x x dt 2 arctan : (18.9) 2n D 0 1 xt.1 t/ D 4 x 4 x n 1 n Z r Âr Ã XD n Â Ã In the same manner, from (18.9), we ﬁnd that~~

~~n 1 2 1 x 1 x ln.1 xt.1 t// dt 2 arctan : (18.10) 2n D 0 t D 4 x n 1 n2 Z Â Âr ÃÃ XD n Â Ã In return, setting x 1 in (18.9) and (18.10), we arrive at (18.3) and (18.4) respectively. D~~

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~~If we divide both sides of (18.10) by x and integrate, we obtain~~

2 t n x 1 x arctan 4 t 2 dt: 2n D 0 qt ÁÁ n 1 n3 Z XD n Â Ã This fails to give a closed form in terms of simple constants even for x 1. However, D integrating (18.10) and then setting x 1, we have D

1 2 1 1 x p3 1 2 arctan dx 1 2: 2n D 0 4 x D C 3 18 n 1 n2.n 1/ Z Â Âr ÃÃ XD C n Â Ã We consider this to be one of the most interesting series deduced from (18.5). Differentiating (18.5) gives

~~n 1 1 1 2 1 nx .1 t/ 2xt.1 t/ 2 dt 3 dt 2n D 0 .1 xt.1 t// C 0 .1 xt.1 t// n 1 Z Z XD n Â Ã and setting x 1 respectively, we get D˙~~

1 n 2 2 p3; D 3 C 27 n 1 2n XD n Â Ã 1 n 6 4 . 1/n 1 p5 ln : D 25 C 125 n 1 2n XD n Â Ã In general, for any positive integer k, applying Leibniz’s rule for the k-th derivative of the product to (18.5), we have

~~d k .k 1/Š.t.1 t//k .1 xt.1 t// 2 C : dxk D .1 xt.1 t//k 2 C Appealing to~~

~~m m 1 t dt t .at 2 bt c/n D .2n m 1/a.at 2 bt c/n 1 Z C C C C .n m/b t m 1 dt .2n m 1/a .at 2 bt c/n Z C C .m 1/c t m 2dt ; C .2n m 1/a .at 2 bt c/n Z C C~~

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~~we ﬁnd that for x 1, respectively, D˙ k 1 n pk qk p3 D C I n 1 2n XD n Â Ã k 1 n 1 n . 1/ rk sk p5 ln D C n 1 2n XD n Â Ã~~

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 n 1 x 2n n 0 .2n 2k 1/ XD C C n Â Ã 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

~~1 1 2n .2n 1/ˇ.n 1; n 1/ .2n 1/ t n.1 t/n dt; n D C C C D C Â Ã Z0 we ﬁnd~~

n 1 1 1 x 1 dt xnt n.1 t/n dt : (18.11) 2n D 0 D 0 1 xt.1 t/ n 0 .2n 1/ Z n 0 Z XD C n XD Â Ã Setting x 1; 1; 1=2;2, respectively, we ﬁnd that D 1 1 1 dt 2 p 2 3; 2n D 0 t t 1 D 9 n 0 .2n 1/ Z C XD C n Â Ã n 1 1 . 1/ dt 4 p 2 5 ln ; 2n D 0 1 t t D 5 n 0 .2n 1/ Z C XD C n Â Ã n 1 1 . 1/ 2dt 4 2 ln 2; 2n D 0 2 t t D 3 n 0 .2n 1/2n Z C XD C n Â Ã n 1 1 2 dt 1 2 : 2n D 0 1 2t 2t D 2 n 0 .2n 1/ Z C XD C n Â Ã

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~~Next, we derive the closed form of~~

~~n 1 x 2n n 0 .2n 3/ XD C n Â Ã by transforming it into two known series. Indeed, since~~

~~2n n 1 2.n 1/ C C ; n D 2.2n 1/ n 1 Â Ã C Â C Ã it follows that~~

n 1 x 1 2.2n 1/ C : 2n D 2.n 1/ n 0 .2n 3/ n 0 .2n 3/.n 1/ C XD C n XD C C n 1 Â Ã Â C Ã Replacing n 1 by n gives C n n 1 1 x 1 2.2n 1/x : 2n D 2n n 0 .2n 3/ n 1 .2n 1/n XD C n XD C n Â Ã Â Ã By the partial fraction decomposition 2n 1 4 1 ; .2n 1/n D 2n 1 n C C we obtain from (18.9) and (18.11)

n 1 x 8 4.8 x/ x x arctan : (18.12) 2n D x C x2 4 x 4 x n 0 .2n 3/ r Âr Ã XD C n Â Ã Setting x 1; 1; 1=2;2 respectively yields D 1 1 14 8 p3; 2n D C 9 n 0 .2n 3/ XD C n Â Ã n 1 . 1/ 36 8 p5 ln ; 2n D 5 n 0 .2n 3/ XD C n Â Ã n 1 . 1/ 68 16 ln 2; 2n D 3 n 0 .2n 3/2n XD C n Â Ã n 1 2 3 4 : 2n D C 2 n 0 .2n 3/ XD C n Â Ã

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~~Similarly, we have~~

~~n 1 x 16 .x 24/ 2n D 9x2 C n 0 .2n 5/ XD C n (18.13) Â Ã 4 x x .128 16x x2/ arctan : C 3x3 4 x 4 x r Âr Ã In particular, we ﬁnd that~~

1 1 400 74 p3; 2n D 9 C 9 n 0 .2n 5/ XD C n Â Ã n 1 . 1/ 368 572 p5 ln ; 2n D 9 C 15 n 0 .2n 5/ XD C n Â Ã n 1 . 1/ 1504 724 ln 2; 2n D 9 C 3 n 0 .2n 5/2n XD C n Â Ã n 1 2 104 23 : 2n D 9 C 6 n 0 .2n 5/ XD C n Â Ã In general, for any integer k 0, we successively ﬁnd that 1 1 r1k s1k p3; 2n D C n 0 .2n 2k 1/ XD C C n Â Ã n 1 . 1/ r2k s2k p5 ln ; 2n D C n 0 .2n 2k 1/ XD C C n Â Ã n 1 . 1/ r3k s3k ln 2; 2n D C n 0 .2n 2k 1/2n XD C C n Â Ã n 1 2 r4k s4k 2n D C n 0 .2n 2k 1/ XD C C n Â Ã for appropriate rational numbers rik;sik.1 i 4/. Ä Ä 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 sufﬁ- ciently 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 n 1 x (18.14) n 1 kn XD ln Â Ã where k and l are positive integers with k>l. In view of (18.6), we have

1 1 kn ln 1 .k l/n ln t .1 t/ dt: ln D Â Ã Z0 Therefore, n 1 l 1 k l 1 x xt .1 t/ l dt: l k l 2 (18.15) kn D 0 .1 xt .1 t/ / n 1 Z XD ln Â Ã Observing that 1 dt 1 dt ln 2; ; 2 t D 1 t 2 D 4 Z0 Z0 C 1 dt 2 1 dt 4 p3 and p5 ln ; 1 t t 2 D 9 1 t t 2 D 5 Z0 C Z0 C we see that there exists a closed form evaluation for (18.15) in terms of ; ln2; p3; p5 ln , as long as the denominator of the integrand 1 xt l .1 t/k l has factors of the ˚ form « t 2;1 t 2;1 t t 2;1 t t 2 : f C C C 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 3;l 1, (18.15) becomes D D n 1 2 1 x x.1 t/ dt: (18.16) 3n D .1 xt.1 t/2/2 n 1 Z0 XD n Â Ã When x 1=2;1 xt.1 t/2 .t 2 1/.t 2/=2. Since D D C 1 t 2 4 8 8.3t 4/ 4.2t 9/ C ; .1 t.1 t/2=2/2 D 25.t 2/2 C 125.t 2/ 25.1 t 2/2 125.1 t 2/ C C we obtain 1 1 2 6 11 ln 2 : (18.17) 3n D 25 125 C 250 n 1 2n XD n Â Ã

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~~d Applying x dx to (18.16) shows that~~

n 1 2 4 2 1 nx 2x t.1 t/ x.1 t/ 2 3 2 2 dt 3n D 0 .1 xt.1 t/ / C .1 xt.1 t/ / n 1 Z Ä XD n Â Ã from which we get, for x 1=2, D 1 n 81 18 79 ln 2 : (18.18) 3n D 625 3125 C 3125 n 1 2n XD n Â Ã Similarly, we have

~~2 1 n 561 42 673 ln 2 : 3n D 3125 C 15625 C 31250 n 1 2n XD n Â Ã In particular, by solving (18.17) and (18.18) for , we recover Gosper’s identity:~~

1 50n 6 : (18.19) 3n D n 0 2n XD n Â Ã Example 2. When k 4;l 2, (18.15) becomes D D n 1 2 1 x xt.1 t/ 2 dt: 2 2 2 (18.20) 4n D 0 .1 xt .1 t/ / n 1 Z XD 2n Â Ã Since t 2.1 t/2 1 .t 2 t 1/.t 2 t 1/ D C and t.1 t 2/ t 1 1 t ; .1 t 2.1 t/2/2 D 4.t 2 t 1/2 C 4.t 2 t 1/ C substituting x 1 into (18.20) yields D 1 1 1 1 2 p3 p5 ln : D 15 C 27 25 n 1 4n XD 2n Â Ã Similarly,

~~1 3 4 2 1 n 2t .1 t/ t.1 t/ 2 dt 2 2 3 2 2 2 4n D 0 .1 t .1 t/ / C .1 t .1 t/ / n 1 Z Ä XD 2n Â Ã 8 1 1 p3 p5 ln D 75 C 54 125 I~~

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2 1 5 6 3 4 2 1 n 6t .1 t/ 6t .1 t/ t.1 t/ 2 2 2 4 2 2 3 2 2 2 dt 4n D 0 .1 t .1 t/ / C .1 t .1 t/ / C .1 t .1 t/ / n 1 Z Ä XD 2n Â Ã 11 5 1 p3 p5 ln : D 75 C 324 250 18.3 Searching for new formulas for Based on Gosper’s identity(18.19), an algorithmwas 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

~~1 an b C : D 3n n 0 2n XD n Â Ã Rewriting 1 1 3n .3n 1/ t 2n.1 t/n dt; n D C Â Ã Z0 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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~~18. Interesting Series Involving Binomial Coefficients 211~~

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 p .n/ m ; (18.21) D kn n 0 bn XD ln Â Ã~~

~~where pm.n/ is an mth degree polynomial in n and b is an integer. Using the same idea as before, and appealing to~~

1 1 kn .kl 1/ t ln.1 t/.k l/n dt; ln D C Â Ã Z0 we have n 1 l k l 1 x .1 x/ RHS of (18.21) .kl 1/pm.n/ dx: D C b Z0 n 0 ! XD Since 1 n q.x/ .kl 1/pm.n/x ; C D .1 x/k 2 n 0 C XD where q.x/ has degree m 1. Substituting x t l .1 t/k l =b, we obtain C D 1 P.t/dt ; RHS of (18.21) l k l k 2 (18.22) D 0 .t .1 t/ b/ Z C where P.t/ is a polynomial of degree k.m 1/. C 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 1 or .t 1/2 1; that is, it must have a zero at i or 1 i, thereby C C C 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~~

2 1 150n 230n 36 C ; D 3n n 1 2n XD n Â Ã 2 27 1 10n 15n 2 p3 C : D 4 n 1 4n XD 2n Â Ã 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 func- n ˛ tion P.x/=Q.x/ with Q.x/ qi .x/ i , where qi .x/ are either linear or irreducible D i 1 quadratic factors, then D Q P.x/ P .x/ P .x/ dx 1 2 dx; Q.x/ D Q .x/ C Q .x/ Z 1 Z 2 where n n ˛i 1 Q1.x/ qi .x/ Q2.x/ qi .x/: D I D i 1 i 1 YD YD 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 repeti- tion give us the transcendental part. This enables us to perform the experiment on (18.22): if 1 P.x/ R.x/ dx r.x/; Q.x/k 2 D Qk 1.x/ C Z0 C C 1 we would expect that R.0/ R.1/ 0 and r.x/ . D D 0 D In the last two decades, computer algebra systems have become an active tool in math- ˇ ematical research. They are often used to experimentˇ 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 computer- aided research will play a more signiﬁcant role.

~~Exercises 1. Prove that n n k n 2n 1 k 1 n 1 2 1 2n 1 C 2 C ; C : n D 2n k 1 2n D 22n 1 k 1 k 0 k 0 C k 0 C k 0 C XD k XD XD 2k XD Â Ã Â Ã 2. Let m be a positive integer. Show that~~

~~1 m m 1 1 m lnŒ1 t .1 t/ dt ; 2mn D 2 t.1 t/ n 1 n2 Z0 XD mn Â Ã n 1 m m 1 . 1/ m lnŒ1 t .1 t/ dt C : 2mn D 2 t.1 t/ n 1 n2 Z0 XD mn Â Ã~~

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~~18. Interesting Series Involving Binomial Coefficients 213~~

~~n G G . 1/ 3. Let be Catalan’s constant which is deﬁned by n1 0 .2n 1/2 . Let D D C P 2 1 1 2k a.n/ : D .2k n/24k k k 0 Â Ã XD C Prove that a.n/ satisﬁes the recurrence~~

~~4G 4G a.0/ 2 ln 2 ; a.1/ ; D D 2 .n 1/2a.n/ .n 2/2a.n 2/ ; for n 2: D C 4. Let G be Catalan’s constant. Prove that~~

1 1 8G ln.2 p3/ (a) C : 2n D 3 3 n 0 .2n 1/2 XD C n Â Ã n 2 1 . 1/ (b) 3 ln2./: 2n D 6 n 0 .2n 1/2 XD C n Â Ã 2n 1 2 (c) 2G: 2n D n 0 .2n 1/2 XD C n Â Ã n 2n 2 1 . 1/ 2 1 (d) ln2.1 p2/: 2n D 8 2 C n 0 .2n 1/2 XD C n Â Ã 5. Prove that

~~4n 1 2 (a) 2 8G 14.3/: 2n D n 1 n3 XD n Â Ã 4n 1 2 7 (b) .3/ G: 2 D 2 n 0 3 2n XD .2n 1/ C n Â Ã 6. Ramanujan Problem. Given~~

~~j 1 2j x 1 1 x <1: j 4j D p1 x Ä j 0 Â Ã XD Determine 1 2j 1 : j .2j 1/24j j 0 Â Ã XD C~~

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~~7. Prove that~~

~~1 1 1 1 ln 2; 3n D 10 5 n 1 n2n XD n Â Ã 1 1 1 2 ln 2: 3n D 24 n 1 n2 2n XD n Â Ã 8. Show that~~

2 1 1 575n 965n 273 ln 2 C ; D 6 3n n 1 2n XD n Â Ã n 37 1 1 . 1/ . 728n 17/ ln 2 C ; D 24 C 24 3n n 1 4n XD n Â Ã 2 3 1 882n 1295n 296 ln 2 C : D 16 4n n 1 4n XD 2n Â Ã 9. Show that 1 1 p .n/ 4 ; D 325272 8n n 0 . 4/n XD 4n Â Ã where

~~2 3 4 p4.n/ 89286 3875948n 34970134n 110202472n 115193600n : D C C 10. Monthly Problem 11356 [2008, 365]. Prove that, for any positive integer n,~~

~~2 n n k 24n.nŠ/4 Â Ã : 2n D .2n/Š.2n 1/Š k 0 .2k 1/ C XD C 2k Â Ã 11. It is known that~~

~~1 1 Hn ln x ln.1 x/ 2.3/ 2 dx: n2 D D 1 x n 1 Z0 XD Prove that~~

~~p 1 p 1 1 Hn . 1/ ln x ln.1 x/ . 1/p .p 1/.p/ dx: np D C C .p/ 1 x n 1 Z0 XD~~

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~~18. Interesting Series Involving Binomial Coefficients 215~~

~~12. Open Problems: Determine the exact sum of~~

~~n n n 1 1 1 . 1/ 1 . 1/ 2 ; ; : 2n 2n 2n n 0 .2n 1/3 n 0 .2n 1/3 n 0 .2n 1/3 XD C n XD C n XD C n Â Ã Â Ã Â Ã~~

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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~~Parametric Differentiation and Integration~~

~~x2 A mathematician is a person to whom 1 e dx p=2 is as 0 D obvious as 1 1 2. — W. Thomson C D R~~

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], emi- nent 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 sin x 1 dx : (19.1) x D 2 Z0 This integral is generally evaluated by using contour integration and thus requires the theory of complex functions. Here we consider the parametric integral

~~sin x I.p/ 1 e px dx; .p > 0/: D x Z0 Differentiating under the integral sign yields~~

~~1 I .p/ 1 e px sin xdx ; 0 D D 1 p2 Z0 C 217~~

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which has solutions I.p/ arctan p C; D C where C is an arbitrary constant. Since limp I.p/ 0, it follows that C =2 and !1 D D I.p/ arctan p; D 2 from which (19.1) follows by setting p 0. D Example 2 Show that the probability integral

~~2 p G 1 e x dx : (19.2) D D 2 Z0 One clever way of evaluating this integral resides in the introduction of polar coordinates~~

2 2 G2 1 1 e .x y / dxdy: D C Z0 Z0 Here, we offer a modiﬁcation that uses the double integrals but applies parametric integration. Notice that if p>0, the change of variable x pt yields D 2 2 G p 1 e p t dt: D Z0 2 Multiplying both sides by e p and then integrating p from 0 to leads to 1 1 p2 2 1 p2 1 p2t 2 G e dp G pe dp e dt: D D Z0 Z0 Z0 Exchanging the order of the right-hand side integrations gives

2 2 1 dt G2 1 1 pe .1 t /p dpdt 1 ; D C D 2 1 t 2 D 4 Z0 Z0 Z0 C 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 1 1 xp dx ; .p 1/; D p 1 ¤ Z0 C k times with respect to p gives 1 . 1/k kŠ xp . x/k dx ln k 1 0 D .p 1/ Z C C 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 didin Example 2, or by specialization, where thegiven 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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~~19. Parametric Differentiation and Integration 219~~

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

b I.p/ f.x;p/dx: D Za Leibniz’s Rule. If f and its partial derivative fp are continuous on the rectangle R D .x; p/ x Œa;b;p Œp1;p2 , then I.p/ is differentiable in .p1;p2/ and f W 2 2 g b I .p/ fp .x;p/dx: 0 D Za

Fubini’s Theorem. If f is continuous on the rectangle R .x; p/ x Œa;b;p D f W 2 2 Œp1;p2 , then for any t Œp1;p2, g 2 b b lim f.x;p/dx lim f.x;p/dx p t D p t ! Za Za ! and t t b b t I.p/dp dp f.x;p/dx dx f .x; p/ dp: D D Zp1 Zp1 Za Za Zp1 Now, we single out two integrals and demonstrate how to apply these two theorems.

Example 3 To evaluate 1 arctan x dx; 0 xp1 x2 Z we introduce a parametric integral 1 arctan.px/ T .p/ dx: D 0 xp1 x2 Z Differentiating with respect to p by Leibniz’s rule gives 1 dx T 0.p/ : D 0 .1 p2x2/p1 x2 Z C The substitution x sin yields D =2 d 1 2 =2 1 T 0.p/ arctan. 1 p tan / : D 1 p2 sin2 D 2 C 0 D 2 2 Z0 1 p 1 p C C p ˇ C p ˇ p

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Appealing to T .0/ 0, we have D T .p/ ln.p 1 p2/: D 2 C C In particular, p 1 arctan x dx T .1/ ln.1 p2/: 0 xp1 x2 D D 2 C Z Example 4 To compute =2 a b sin x dx I ln C ; .a>b>0/; D a b sin x sin x Z0 Â Ã observing that 1 a b sin x 1 dy ln C 2ab ; 2 2 2 2 sin x a b sin x D 0 a b y sin x Â Ã Z we get =2 1 dy I 2ab dx : 2 2 2 2 D 0 0 a b y sin x Z Z For a >b >0;1=.a2 b2y2 sin2 x/ is continuous on Œ0; =2 Œ0; 1.By Fubini’stheorem, interchanging the order of integrations, we obtain

1 =2 dx I 2ab dy 2 2 2 2 D 0 0 a b y sin x Z Z 1 dy 2ab D 0 2a a2 b2y2 Z bp arcsin : D a Â Ã The following example warns us that switching the order of integrations without justi- ﬁcation may change the answer.

Example 5 Consider y2 x2 f .x; y/ on Œ0; 1 Œ0; 1: D .x2 y2/2 C We have 1 x 1 1 f.x;y/dx D x2 y2 0 D 1 y2 Z0 C ˇ C and so ˇ 1 1 dy f.x;y/dx : D 4 Z0 Z0 However, 1 1 dx f .x; y/ dy : D 4 Z0 Z0

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~~19. Parametric Differentiation and Integration 221~~

~~Here the discontinuity of f .x; y/ at .0; 0/ causes the discrepancy. For an improper integral~~

I.p/ 1 f.x;p/dx; D Za 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 Œp1;p2 there is a constant A a, independent of p, such that x A 2 implies A 1 f.x;p/dx f.x;p/dx 1 f.x;p/dx < : ˇ a a ˇ D A ˇZ Z ˇ ˇZ ˇ ˇ ˇ ˇ ˇ There is a veryˇ convenient test for uniform convergence,ˇ ˇ due to Weierstrass.ˇ ˇ ˇ ˇ ˇ M-Test. If f .x; p/ is integrable on every ﬁnite interval and there is an integrable function .x/ on Œ0; / such that 1 f .x; p/ .x/; for all x a;p Œp1;p2; j j Ä 2 then I.p/ converges uniformly on Œp1;p2. When the integrand is in product form, the following criterion is sometimes useful.

Dini Test. If A f.x;p/dx ˇ a ˇ ˇZ ˇ is uniformly bounded in A and p, g.x;ˇ p/ is monotonicˇ in x and g.x; p/ 0 uniformly ˇ ˇ ! in p as x , then ˇ ˇ ! 1 1 f.x;p/g.x;p/dx Za converges uniformly on Œp1;p2.

~~As an immediate consequence of the Dini test, if a1 f.x/dx converges, then~~

1 xp R1 x2p e f.x/dx and e f.x/dx Za Za 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

~~1 f.x;p/dx and 1 f.x;p/dp; Za Zb Fubini’s theorem further requires that one of the double integrals~~

~~1 1 f .x; p/ dxdp and 1 1 f .x; p/ dpdx j j j j Zb Za Za Zb 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; / .b; /. Suppose 1 1 the improper integrals~~

~~1 f .x; y/ dy and 1 f.x;y/dx Zb Za exist and converge uniformly for x and y restricted to every ﬁnite interval, respectively. In addition, suppose that, for s;t >a,~~

~~t f.x;y/dx M.y/ Ä ˇZs ˇ ˇ ˇ ˇ ˇ and b1 M.y/ dy exists. Then ˇ ˇ R 1 dx 1 f .x; y/ dy 1 dy 1 f.x;y/dx: D Za Zb Zb Za~~

~~As a nice application of this theorem, the integral 1 1 e xy dy x D Z0 enables us to evaluate (19.1) in one line:~~

1 1 xy 1 1 xy 1 dy sin xdx e dy dy e sin xdx : D D 1 y2 D 2 Z0 Z0 Z0 Z0 Z0 C 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 2 1 ˛2x2 ˇ e x2 dx: (19.3) Z0 Let y ˛x;p ˛ˇ. Then D D 2 2 2 ˇ2 2 p 1 ˛ x 1 1 y 2 e x2 dx e y dy: D ˛ Z0 Z0 If p2 1 y2 J.p/ e y2 dy; D Z0 then p2 1 p y2 J .p/ 2 e y2 dy; 0 D y2 Z0 and the substitution z p=y yields D 2 1 z2 p J .p/ 2 e z2 dz 2 J.p/: 0 D D Z0

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~~19. Parametric Differentiation and Integration 223~~

~~Appealing to (19.2), we ﬁnd~~

~~2 p J.p/ J.0/e 2p 1 e y dy e 2p e 2p D D D 2 ÂZ0 Ã and so ˇ2 1 ˛2x2 p 2˛ˇ e x2 dx e : D 2˛ Z0 Remark. In general, by substitution, one can establish that~~

~~B 2 1 1 f Ax dx 1 f .y2/dy; .A>0;B>0/: x D A Z0 "Â Ã # Z0 A related Putnam problem is 1968-B4, which asks us to prove~~

~~1 1 f x dx 1 f.x/dx: x D Z1 Â Ã Z1 Example 7 Let ˛;ˇ>0. Show that~~

~~1 ˛x2 1 ˇ 2=4˛ H.ˇ/ e cos ˇxdx e : (19.4) D D 2 ˛ Z0 r Differentiating with respect to ˇ and then integrating by parts gives~~

~~2 H .ˇ/ 1 xe ˛x sin ˇx dx 0 D Z0 1 2 1 sin ˇx d.e ˛x / D 2˛ Z0 1 ˛x2 ˇ 1 ˛x2 e sin ˇx 1 e cos ˇxdx D 2˛ 0 2˛ Z0 ˇ ˇ ˇ H.ˇ/: D 2˛ Moreover, by (19.3),~~

~~2 1 2 1 H.0/ 1 e ˛x dx 1 e x dx : D D p˛ D 2 ˛ Z0 Z0 r Consequently, the desired integral follows by solving the initial value problem. It is interesting to note that the similar integral~~

~~1 ˛x2 e sin ˇx dx Z0 has no elementary closed form.~~

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~~Example 8 Let ˛;ˇ>0. Evaluate the Laplace integrals~~

~~cos ˇx x sin ˇx L .ˇ/ 1 dx L .ˇ/ 1 dx: 1 2 2 and 2 2 2 D 0 ˛ x D 0 ˛ x Z C Z C Applying Leibniz’s rule to L1.ˇ/ gives~~

~~x sin ˇx L .ˇ/ 1 dx: 10 D ˛2 x2 Z0 C~~

We caution that here L10 .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 x2 cos ˇx 1 dx; ˛2 x2 Z0 C which can be rewritten as

~~=2ˇ 2 .2n 1/=2ˇ 2 x cos ˇx 1 x cos ˇx dx . 1/n 1 C j j dx: ˛2 x2 C C ˛2 x2 Z0 n 1 Z.2n 1/=2ˇ C XD C Since~~

.2n 1/=2ˇ x2 cos ˇx C dx j2 2 j .2n 1/=2ˇ ˛ x Z C 2 .2n 1/=2ˇ x C min 2 2 cos ˇx dx .2n 1/=2ˇ x .2n 1/=2ˇ ˛ x .2n 1/=2ˇ j j Â Ä Ä C C Ã Z 2..2n 1/=2ˇ/2 2 ; as n , D .˛2 ..2n 1/=2ˇ/2/ˇ ! ˇ ! 1 C the series, hence the corresponding integral, is divergent. On the other hand, by observing that ˛2 sin ˇx sin ˇx x sin ˇx x.˛2 x2/ D x ˛2 x2 C C and sin ˇx sin x 1 dx 1 dx ; x D x D 2 Z0 Z0 we have ˛2 sin ˇx L .ˇ/ 1 dx: 10 D 2 C x.˛2 x2/ Z0 C Hence 2 1 ˛ cos ˇx 2 L00.ˇ/ dx ˛ L1.ˇ/; 1 D ˛2 x2 D Z0 C which leads to ˛ˇ ˛ˇ L1.ˇ/ c1e c2e ; D C

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~~19. Parametric Differentiation and Integration 225~~

~~where c1 and c2 are arbitrary constants. Since dx L .ˇ/ 1 ; 1 2 2 Ä 0 ˛ x D 2˛ Z C L1.ˇ/ is uniformly bounded. Therefore, c1 0; c2 L1.0/ D D D 2˛ and consequently~~

~~ ˛ˇ ˛ˇ L1.ˇ/ e and L2.ˇ/ L .ˇ/ e : D 2˛ D 10 D 2~~

The value of L1 also can be captured by using the parametric integration. Indeed, noticing 1 2 2 1 e t.˛ x / dt; 2 2 C ˛ x D 0 C Z we have 1 1 t.˛2 x2 / L1 cos ˇx dx e dt: D C Z0 Z0 Interchanging the integral order yields

~~1 ˛2t 1 tx2 L1 e dt e cos ˇx dx: D Z0 Z0 (19.4) implies the inner integral~~

~~2 1 2 1 e tx cos ˇx dx e ˇ =4t : D 2 t Z0 r Finally, appealing to (19.3), we arrive at~~

~~p 1 ˛2 t ˇ 2=4t dt L1 e D 2 pt Z0 1 ˛2 s2 ˇ 2=4s2 2 p e ds .let t s / D D Z0 e ˛ˇ : D 2˛ Example 9 Evaluate the Fresnel integrals~~

~~1 2 1 2 Fc cos.x /dx and Fs sin.x /dx: D D Z0 Z0 It is standard to substitute x for x2, so~~

~~1 1 cos x 1 1 sin x Fc dx and Fs dx: D 2 px D 2 px Z0 Z0~~

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~~Recall that 1 2 1 xy2 e dy: px D p Z0 Then~~

1 1 1 xy2 Fc cos xdx e dy D p Z0 Z0 1 1 1 xy2 dy e cos xdx D p Z0 Z0 1 dy 1 1 1 : 4 D p 0 1 y D p 2p2 D 2 2 Z C r Similarly, 1 Fs : D 2 2 r Example 10 Evaluate e ax e bx F 1 dx; .a;b>0/: D x Z0 Recall that b e ax e bx e yx dy : D x Za This yields

b b b 1 b F 1 dx e yx dy dy 1 e yx dx dy ln : D D D y D a Z0 Za Za Z0 Za Remarks. Here F belongs to the family of Frullani integrals [2, pp. 406–407], which are deﬁned by f .ax/ f .bx/ 1 dx .a;b>0/: x Z0 In general, if f .x/ is continuous on Œ0; / and 1 f . / lim f .x/ x C1 D !1 exists, then f .ax/ f .bx/ b 1 dx .f .0/ f . // ln : x D C1 a Z0 If limx f .x/ has no ﬁnite limit, but !1 f .x/ 1 dx x ZA exists, then f .ax/ f .bx/ b 1 dx f .0/ ln : x D a Z0

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~~19. Parametric Differentiation and Integration 227~~

Similarly, if f is not continuous at x 0, but D A f .x/ dx x Z0 exists, then f .ax/ f .bx/ b 1 dx f . / ln : x D 1 a Z0 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

~~e px cos qx e ax cos bx 1 dx; .p;a>0/; x Z0 it is instructive to check that differentiating under the integral sign with one parameter or using parametric integration via~~

~~1 1 e xt dt x D Z0 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~~

@I p 1 e px cos qxdx ; @p D D p2 q2 Z0 C @I q 1 e px qxdx ; sin 2 2 @q D 0 D p q Z C which have solutions 1 I.p; q/ ln.p2 q2/ C; D 2 C C where constant C is independent of p and q. Since I.a; b/ 0, we obtain that D 1 C ln.a2 b2/ D 2 C and so 1 a2 b2 I.p; q/ ln C : D 2 p2 q2 C 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 1 p2 p2 1 sin.x2/ cos.px/dx cos sin : D 2 2 4 4 Z0 r Ä Â Ã Â Ã He then differentiated under the integral sign with respect to p yielding

~~p p2 p2 1 x sin.x2/ sin.px/dx sin cos : (19.5) D 4 2 4 C 4 Z0 r Ä Â Ã Â Ã~~

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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 y2 x2 y2 x2 1 dy 1 dx 1 dx 1 dy : 2 2 2 and 2 2 2 1 1 .x y / D 4 1 1 .x y / D 4 Z Z C Z Z C Since both y2 x2 y2 x2 1 dy 1 dx 1 dx 1 dy j 2 2 j2 and j 2 2 j2 1 1 .x y / 1 1 .x y / Z Z C Z Z C diverge, Fubini’s theorem does not apply to the given function.

Exercises 1. Show that 1 2xy2 1 2xy2 dx: lim 2 2 2 lim 2 2 2 y 0 0 .x y / ¤ 0 y 0 .x y / ! Z C Z Â ! C Ã 2. Show that sin px 1 dx x Z0 is uniformly convergent for p p0 >0, but fails to converge uniformly for p 0. 3. Evaluate 1 x 1 dx: ln x Z0 1 xp 1 Hint: Consider 0 lnx dx: 4. Putnam ProblemR 2005-A5. Evaluate 1 ln.x 1/ C dx: x2 1 Z0 C There are many ways to solve thisproblem. Try to introduce a parametric integral and then differentiate it.

5. Let ˛;ˇ;k>0. Evaluate 1 cos ˛x (a) 1 e kx dx. x Z0 sin ˛x sin ˇx (b) 1 e kx dx. x x Z0 6. Let a;b >0. Prove that ln.1 a2x2/ (a) 1 C dx ln.1 ab/: b2 x2 D b C Z0 C arctan ax (b) 1 dx ln.1 a/: x.1 x2/ D 2 C Z0 C

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~~19. Parametric Differentiation and Integration 229~~

arctan ax arctan bx .a b/a b (c) 1 dx ln C : 2 Ca b 0 x D 2 a b Z 7. For a;b >0, show that cos ax cos bx 1 dx .b a/: x2 D 2 Z0 8. Evaluate 1 1=2 ax px1 x e dx: Z0 Remark. For a p 1985, this is the Putnam Problem 1985-B5. In contrast to D D 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 epx eqx 1 dx: px qx 0 x.e 1/.e 1/ Z C C 10. Evaluate

1 x2 2 2 1 x2 2 2 e cos.a =x /dx and e sin.a =x /dx: Z0 Z0 11. Prove that x2 x2 1 xe dx a 1 e dx 2 2 .a > 0/: 0 px2 a2 D p 0 x a Z C Z C 12. Evaluate 2 e x dx 1 : .x2 a2/2 Z0 C When a2 1=2, this is Monthly Problem 4212 [1946, 397; 1947, 601–603]. D 13. Monthly Problem 3766 [1936, 50; 1938, 56–58]. Evaluate

~~1 x 2 e ln xdx: Z0~~

~~14. Let J0.x/ be the zero-order Bessel function. Evaluate~~

~~1 sx e J0.x/ dx .s > 0/: Z0 15. For p>0 and 0~~

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~~16. Monthly Problem 11101 [2004, 626; 2006, 270–271]. Evaluate~~

~~b dx 1 a arctan 0 pa2 x2 pa2 x2 Z Â C Ã C in closed form for a;b;> 0. 17. Monthly Problem 11113 [2004, 822; 2006, 573–574]. Evaluate~~

~~kpx2 y2 1 1 e C sin.ax/ sin.bx/ Ik .a; b/ dxdy D 0 0 xy x2 y2 Z Z C in closed form for a;b;k >0. p 18. Monthly Problem 11225 [2006, 459; 2007, 750]. Find~~

~~1 n x ln.1 x=n/ lim C dx: n n 0 1 x !1 Z C~~

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, Ban- tam 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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~~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 I.x/ ln.1 2x cos x2/d D C Z0 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 x <1; I.x/ j j D 2 ln x ; if x >1. j j j j 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/ 0. D 2. I. x/ I.x/. D 3. I.x/ 2 ln x I.1=x/; .x 0/. D j jC ¤ To verify these results, first note that for x <1, j j .1 x /2 1 2x cos x2 .1 x /2: j j Ä C Ä Cj j 231

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~~Thus, taking the logarithm and integrating with respect to from 0 to , we ﬁnd~~

2 ln.1 x / I.x/ 2 ln.1 x /: j j Ä Ä Cj j Letting x 0, we have that I.0/ 0. Next, applying the substitution ˛ in I.x/, ! D D we have I.x/ ln.1 2x cos ˛ x2/d˛ I. x/: D C C D Z0 Finally, if x 0, we have ¤ 2 1 I.x/ ln x2 1 cos d D x C x2 Z0 Ä Â Ã ln x2 d I.1=x/ 2 ln x I.1=x/: D C D j jC Z0 In light of the third fact, the main formula follows easily once we show that I.x/ 0 for D x <1. This will be the goal of the next four sections. j j 20.1 Using Riemann Sums Since 1 2x cos x2 .1 x /2; for x < 1; C j j j j the integrand is continuous and integrable. Partition the interval Œ0; into n equal subin- tervals by the partition points k xk 1 k n : D n W Ä Ä We have the corresponding Riemann sum for I.x/, namely

~~n k 2 Rn ln 1 2x cos x D n n C k 1 Â Â Ã Ã XD n 1 k ln .1 x/2 1 2x cos x2 : (20.1) D n C n C " k 1 Â Â Ã Ã# YD Recall that (see (5.2))~~

~~n 1 2n k x 1 1 2x cos x2 : (20.2) n C D x2 1 k 1 Â Â Ã Ã YD Substituting the identity (20.2) into (20.1), we have~~

~~ x 1 2n Rn ln C .x 1/ : D n x 1 Â Ã Since x < 1;x2n 0 as n . Hence, we obtain j j ! ! 1 x 1 2n I.x/ lim Rn lim ln C .x 1/ 0: D n D n n x 1 D !1 !1 Â Ã~~

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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 1 I.x/ I. x/ I.x2 /: (20.3) D D 2 Applying the identity~~

.1 2x cos x2/.1 2x cos x2/ 1 2x2 cos.2/ x4; C C C D C we obtain I.x/ I. x/ ln .1 2x cos x2/.1 2x cos x2/ d C D C C C Z0 ln.1 2x2 cos 2 x4/d: D C Z0 Setting ˛ 2, we have D 1 2 I.x/ I. x/ ln.1 2x2 cos ˛ x4/d˛ C D 2 C Z0 1 1 2 I.x2 / ln.1 2x2 cos ˛ x4/d˛: D 2 C 2 C Z Using the substitution˛ 2 t in the last integral shows that it is exactly thesame as the D ﬁrst integral. Since the two terms on the left are the same (recalling that I.x/ I. x/), D we obtain (20.3) as desired. Applying equation (20.3) repeatedly, we ﬁnd that

1 1 1 n I.x/ I.x2/ I.x4/ I.x2 /: D 2 D 22 DD 2n n Again we assume that x <1, so that x2 0 as n and consequently j j ! ! 1 1 n I.x/ lim I.x2 / 0: D n 2n D !1 Remark. Equation (20.3) holds for any x. In particular, we have that I.0/ 0 and D I. 1/ 0. The latter equation leads to an added bonus: ˙ D =2 =2 ln.sin /d ln.cos /d ln 2; D D 2 Z0 Z0 since

~~ =2 I.1/ ln.2 2 cos /d 2 ln 2 4 ln.sin /d D D C I Z0 Z0 =2 I. 1/ ln.2 2 cos /d 2 ln 2 4 ln.cos /d: D C D C Z0 Z0~~

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~~These two integrals are improper. To show convergence, for example, using integration by parts, we have~~

=2 =2 ln.sin /d lim ln.sin /d D 0 Z0 ! Z =2 cos lim ln.sin / lim d D 0 0 sin ! ! Z =2 cot d: D Z0 =2 Since cot is Riemann integrable on Œ0; =2, 0 ln.sin /d converges. R 20.3 Using Parametric Differentiation Since I.x/ is differentiable for x <1, we apply Leibniz’s rule to I.x/ to ﬁnd j j 2 cos 2x I 0.x/ C d: D 1 2x cos x2 Z0 C Clearly, I .0/ 0. We now show that I .x/ 0 for x 0. First, we prove that 0 D 0 D ¤ 1 x2 d : (20.4) 1 2x cos x2 D Z0 C The integrand in (20.4) is called Poisson’s kernel. It is used to derive solutionsto the two- dimensional Laplace’s equation on the unit circle, and it also plays an important role in summation of Fourier series. Computingthe 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 tan.=2/, we have D 1 x2 dt d 2.1 x2/ 1 2x cos x2 D .1 x/2 .1 x/2t 2 Z C Z C C 1 x 1 x 2 arctan C t C 2 arctan C tan.=2/ C: D 1 x C D 1 x C Â Ã Â Ã The fundamental theorem of calculus gives

1 x2 1 x 2 d lim 2 arctan C tan.=2/ : 0 1 2x cos x D 1 x D Z C ! Â Ã Using (20.4) and x 0, we get ¤ 1 1 x2 I .x/ 1 d 0: 0 2 D x 0 1 2x cos x D Z Â C Ã Thus, we have I .x/ 0 for x < 1 and so I.x/ is a constant. Since I.0/ 0, we have 0 D j j D shown that I.x/ 0 for all x <1. Á j j

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~~20.4 Using Infinite Series We ﬁrst show that~~

n 1 x ln.1 2x cos x2/ 2 cos n; (20.5) C D n n 1 XD where the series converges uniformly for x <1. Once we establish this, integrating (20.5) j j with respect to from 0 to , will show that I.x/ 0 once again. D To prove (20.5) and to keep the evaluation at an elementary level, instead of using the Fourier series, we start with 1 x2 1 x2 1 x2 ; 1 2x cos x2 D 1 x.ei e i / x2 D .1 xei /.1 xe i / C C C where we have used the relation 2 cos ei e i . Now, decomposing into partial D C fractions yields 1 x2 1 1 1 : 1 2x cos x2 D C 1 xei C 1 xe i C Thus, the geometric series expansion leads to

2 1 x 1 1 2 xn cos n: (20.6) 1 2x cos x2 D C n 1 C XD n The series (20.6) converges uniformly since 1 x converges for x <1. Removing n 1 j j j j 1 from the right-hand side of (20.6) and then dividingbyD x, we have P 2 cos 2x 1 2 xn 1 cos n: (20.7) 1 2x cos x2 D n 1 C XD 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 n 1 z ln.1 z/ ; . z < 1/ D n j j n 1 XD to prove (20.5) and~~

~~x sin 1 sin n arctan xn; . x < 1/: 1 x cos D n j j Â Ã n 1 XD~~

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~~2. Show that 1 1 x x dx n n: D Z0 n 1 XD 3. Let f be a continuous function. Prove that~~

2 f .a cos b sin /d 2 f . a2 b2 cos /d: 0 C D 0 C Z Z p 4. Show that .z/.z 1=3/.z 2=3/ 33z C C .3z/ is periodic of period 1. Hence it is a constant. Determine the constant. 5. Show that 1 ln.1 2x cos x2/ I./ C dx D x Z0 satisﬁes the functional equation

~~ 1 I I I./; 2 C 2 D 2 Â Ã Â Ã 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~~

~~f .x/ f .y/ f . x2 y2/; for all x and y; D C show that p 2 f .x/ 0 or f .x/ e˛x ; D D where ˛ is an arbitrary constant. Use this to prove that~~

2 p 2 1 e x cos.px/dx e p =4: D 2 Z0 7. Let 2 2 F.p/ 1 e pt cos x2 dx and G.p/ 1 e pt sin x2 dx: D D Z0 Z0 Show that F and G satisfy p F 2.p/ G2.p/ and 2F.p/G.p/ ; D 4.1 p2/ D 4.1 p2/ C C 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,

~~x f 2.x/ Œf 2.t/ .f .t//2 dt 1990: D C 0 C Z0~~

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9. For all p, show that =2 dx p : 0 1 tan x D 4 Z C Remark. For p p2, this is Putnam Problem 1980-A3. D 10. Deﬁne the parametric integral n In.z/ cos.zt/ cos t dt: D Z0 2 2 (a) Establish the reduction formula .n z /In .z/ n.n 1/In 2 .z/. D (b) Show that I .z/ lim n 1: n I .0/ D !1 n (c) Deduce the Euler sine product formula. 11. The theta function is deﬁned by

~~1 2 .s/ e n s 1 2e s 2e 4s 2e 9s ; s>0: D D C C C C X1 Prove that .1=s/ ps .s/; that is, D~~

~~1 2 1 2 e n =s ps e n s : D X1 X1 12. (P. Bracken) Let ˛ and ˇ be positive real numbers with ˛ˇ , and let y be a real D number. Prove that~~

1 1 1 1 1 e ˛k cos.˛yk/ : 2 C D ˛ 1 .y 2ˇj /2 k 1 j XD D1X C C 13. Monthly Problem 11036 [2003, 743; 2005, 569-572]. For 0 a p3, evaluate Ä Ä 1 ln.1 x2 xpa2 x2/ I.a/ C C dx D 1 p1 x2 Z in closed form. 14. Monthly Problem 11072 [2004, 259; 2005, 845-846]. Evaluate the integral sin ax sin ay I.a; k/ 1 1 exp. k x2 y2/ dydx: D 0 0 C x y Z Z p 15. Using 1 1 1 1 1 . 1/k ; sin t D t C t k C t k k 0 Ä XD C prove that sin x 1 dx : x D 2 Z0

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~~16. Evaluate ln cos x 1 j j dx: x2 Z0 17. Let Bn.t/ be the Bernoulli polynomials. Show that~~

2n 1 1=2 n 1 .2/ C .2n 1/ . 1/ C B2n 1.t/ cot.t/dt: C D .2n 1/Š 0 C C Z 18. Let =2 n In .lnsin t/ dt: D Z0 (a) Deﬁne 1 .2k 1/ŠŠ 1 D.s/ : D .2n/ŠŠ .2k 1/s k 0 XD C Show that . 1/n =2 D.n 1/ .lnsin t/n dt: C D nŠ Z0 (b) Find I.n/ explicitly for n 2;3;4. D (c) Show that

1 1 0 0 0 2 1 2 0 0 . 1/n ˇ ˇ In ˇ 3 2 1 3 0 ˇ ; D 2 ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ n n 1 n 2 n 4 1 ˇ ˇ ˇ ˇ ˇ where ˇ ˇ 1 1 1 1 1 k k .1 2 /.k/: D 1k 2k C 3k 4k CD (d) Generalize these results to

~~=2 p q n In.p; q/ .lnsin t ln cos t/ dt: D Z0 19. Monthly Problem 11041 [2003, 843; 2005, 655–657]. Let~~

n 1 1 1 . 1/ ˛.k/ ; ˇ.k/ : D .2n 1/k D .2n 1/k n 0 n 0 XD C XD C For x <1, deﬁne j j 1 1 f .x/ ˛.2k 1/x2k; g.x/ ˇ.2k/x2k 1: D C D k 1 k 1 XD XD Show that 1 1 x 1 x 1 f .x/ ln 2 f C f C 2 D 2 2 C 2 Â Â Ã Â ÃÃ and 1 x 1 x 1 g.x/ f C f : D 2 2 2 Â Â Ã Â ÃÃ

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~~20. Four Ways to Evaluate the Poisson Integral 239~~

~~20. For positive integer p 2, let ~~

~~1 ˛.2k/ A.p/ 2 p 1 ; D .2k/.2k 1/ .2k p 1/ k 1 XD C C p 1 1 ˇ.2k 1/ B.p/ C ; D 2p 1 .2k 1/.2k 2/ .2k p/ k 0 XD C C C where ˛.k/ and ˇ.k/ are deﬁned as in Exercise 19 above. Prove that~~

~~2p 1 =2 1 =2 xp 1 A.p/ xp 1 cot xdx; B.p/ dx: D .p 1/Š D 2 .p 1/Š sin x Z0 Z0 In particular,~~

~~=2 1 ˇ.2k 1/ 1 x B.2/ C dx G: D 2 .2k 1/.2k 2/ D 2 0 sin x D k 0 Z XD C C~~

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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~~Some Irresistible Integrals~~

~~I could never resist an integral. — G. H. Hardy~~

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 includingan 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 Monthlyhad 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 ax2 x p bx2 1 1 e 1 y2=2 1 1 e C < e dy < C 2 p2 2 Z1 hold for all x>0. The inequalities give tight bounds for the normal distribution. In the following, we show that a 1=2 and b 2= are the best possible constants for which the stated D D inequalities hold. First, by (19.2) we have

~~0 y2=2 1 y2=2 e dy e dy D 0 D 2 Z1 Z r and so the stated inequalities are equivalent to~~

~~p1 e ax2 p1 e bx2 < f .x/ < ; (21.1) 2 2~~

~~241~~

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~~242 Excursions in Classical Analysis~~

where x 1 y2=2 f .x/ e dy: D p2 0 Z Next, if the second inequality of (21.1) holds for all x>0, then the series expansions give p1 e bx2 pb 1 0 < f .x/ x O.x3/ 2 D 2 p2 ! C as x 0, which implies that b 2=. Similarly, if the ﬁrst inequality of (21.1) holds for ! all x>0, appealing to

~~ax2 p1 e 1 1 2 1 2 2 e ax e 2ax O.e 3ax / 2 D 2 4 16 C~~

1 1 1 y2=2 f .x/ e dy D 2 p2 x Z 1 1 1 1 y2 =2 d.e / D 2 C p2 x y Z x2=2 1 1 2 e e x =2 O ; 2 D 2 p2x C x ! we have p ax2 x2=2 1 e 1 ax2 1 x2=2 3ax2 e 0 < f .x/ e e O.e / O 2 2 D 4 p2x C C x !

2 as x . Dividing each side by e x =2 yields a 1=2. ! 1 Ä Finally, we show that (21.1) holds for all x>0 when a 1=2 and b 2=. To this D D end, we consider x x 1 2 2 f 2.x/ e .y z /=2 dydz: D 2 C Z0 Z0 Let

~~2 2 2 D Œ0;x Œ0;x; D1 .y; z/ 0 y;0 z; y z x D Df W Ä Ä C Ä g 2 2 2 D2 .y; z/ 0 y;0 z; y z .4=/x : Df W Ä Ä C Ä g See Figure 21.1. We have~~

1 .y2 z2/=2 1 .y2 z2 /=2 e C dydz e C dydz 2 D1 Ä 2 D Z Z Z Z (21.2) 1 .y2 z2 /=2 e C dydz; Ä 2 Z ZD2 where the ﬁrst inequality holds because D1 D; the second because D and D2 have .y2 z2/=2 .2=/x2 .y2 z2 /=2 the same area and e C e for .y; z/ D D2 while e C .2=/x2 Ä 2 e for .y; z/ D2 D. Evaluating the outer integrals in (21.2) via polar coordi- 2 nates, 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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~~21. Some Irresistible Integrals 243~~

~~y O x~~

Figure 21.1. Square D and quarter circles D1 and D2 21.2 Monthly Problem 11206 [2006, 180] Evaluate n 1 n 2 lim n n k !1 k 1 XD n o where x denotes x x , the fractional part of x. f g b c The problem posed originally in the February, 2006, Monthly was to evaluate n n 2 lim : n k !1 k 1 XD n o The crucial factor 1=n before the summation was missing. Since 1=x 2 is bounded and f g continuous on .0; 1/ except at each reciprocal of a positive integer, 1=x 2 is Riemann f g integrable. Therefore, n 1 n 2 1 1 2 lim dx: n n k D 0 x !1 k 1 Z XD n o Now, we show that 1 1 2 dx ln.2/ 1: (21.3) x D Z0 where is Euler’s constant. To see this, for x .1=.n 1/; 1=n/, we have 2 C 1 1 n<1=x

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~~244 Excursions in Classical Analysis~~

~~Let SN be the partial sum, namely~~

~~N n 1 n SN 1 2n ln C : D n C n 1 n 1 Â Ã XD C Since~~

N N N n 1 1 1 2 2N ; C n 1 D n 1 D n 1 n 1 Â Ã n 1 Â Ã n 1 XD C XD C XD C N n 1 n ln C .ln 2 ln 1/ 2.ln 3 ln 2/ N.ln.N 1/ ln N / n D C CC C n 1 XD N ln.N 1/ ln.N Š/; D C we ﬁnd N 1 SN 2N 2N ln.N 1/ 2 ln.N Š/: D n 1 C C n 1 XD C Recall the asymptotic formula of the harmonic numbers

Hk ln k O.1=k/ D C C and Stirling’s formula k k kŠ p2k : e Â Ã We obtain 1 N SN ln.2/ 1 .2N 1/ ln C O.1=N/; D C C N C from which (21.3) follows by letting N . ! 1 Using similar paths, one can evaluate more challenge integrals like

~~1 1 2 1 2 dx 4 ln.2/ 4 5; 0 x 1 x D Z which appeared in Pi Mu Epsilon Journal as Problem 1151, 2006.~~

21.3 Monthly Problem 11275 [2007, 165] Find 2 1 1 .x y/ ln..x y/=.x y// I1 C dxdy: WD y 0 x y xy sinh.x y/ Z D Z D C Substitution of x uy gives D 2 1 1 y.u 1/ ln..u 1/=.u 1// I1 C dudy: D y 0 1 u sinh.y.u 1// Z D Z C

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~~21. Some Irresistible Integrals 245~~

~~Interchanging the order of the integration yields~~

2 1 .u 1/ ln..u 1/=.u 1// 1 y I1 C dy du: D u sinh.y.u 1// Z1 ÂZ0 C Ã Substitution of t e .1 u/y leads to D C y 2 1 ln t 1 dy dt sinh.y.u 1// D .1 u/2 1 t 2 Z0 C C Z0 1 2 1 2k 2 t ln t dt D .1 u/ 0 Z k 0 C XD 2 1 1 D .1 u/2 .2k 1/2 k 0 C XD C 2 ; D 4.1 u/2 C where we have used the facts that

1 2 1 1 1 3 1 1 t 2k t dt : ln 2 and 2 2 0 D .2k 1/ .2k 1/ D 4 k D 8 Z k 0 k 1 C XD C XD Therefore 2 2 1 .u 1/ ln..u 1/=.u 1// I1 C du: D 4 u.1 u/2 Z1 C Finally, substituting s .u 1/=.u 1/, we get D C 2 1 s2 ln s 2 1 ln s 2. 2 8/ I1 ds ln s ds : D 2 1 s2 D 2 1 s2 D 16 Z0 Z0 Â Ã 21.4 Monthly Problem 11277 [2007, 259; 2008, 758–759] Evaluate the double integral

=2 =2 log.2 sin cos / sin I dd: 2 2 2 WD 0 0 2 2 sin cos sin cos Z Z C 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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~~246 Excursions in Classical Analysis~~

Lemma 21.1. If f is integrable on Œ0; 1, then =2 =2 1 f .sin cos / sin dd f.x/dx: (21.4) D 2 Z0 Z0 Z0 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 sin cos ; y sin sin ; z cos and D D D T T sin : k kDj j This yields =2 =2 f .sin cos / sin dd f.x/dS: D Z0 Z0 Z ZS On the other hand, viewing the x-axis as the polar axis and as the angle, we have =2 1 1 f.x/dS f.x/dxd f.x/dx; D D 2 Z ZS Z0 Z0 Z0 which proves (21.4). Clearly, replacing by =2 in (21.4), we also ﬁnd =2 =2 1 f .cos cos / cos dd f.x/dx: D 2 Z0 Z0 Z0 Now we are ready to use the lemma to evaluate I2. Setting f .x/ ln.2 x/= 2 D .2 2x x /, by (21.4), I2 becomes C 1 ln.2 x/ I dx: 2 2 D 2 0 2 2x x Z C Substituting x 1 tan ˛, we obtain D =4 I2 ln.1 tan ˛/d˛ D 2 C Z0 =4 p2 cos.=4 ˛/ ln d˛ D 2 cos ˛ Z0 ! =4 =4 ln 2 ln cos.=4 ˛/d˛ ln cos ˛d˛ D 2 8 C Z0 Z0 ! 2 ln 2; D 16 where the last two integrals are equal by symmetry about ˛ =8. D 21.5 Monthly Problem 11322 [2007, 835] Let N be a positive integer. Prove that

~~1 1 N 2 .x.1 x/y.1 y// 1 NŠ dydx 1 dt: .1 xy/. ln.xy// D t.t 1/ .t N / Z0 Z0 n N 1 Zn Â C C Ã DXC (21.5)~~

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~~21. Some Irresistible Integrals 247~~

~~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~~

~~1 1 k N 1 .xy/ .x.1 x/y.1 y// I3 dydx D 0 0 . ln.xy// Z Z k 0 XD 1 1 k N 1 .xy/ .x.1 x/y.1 y// dydx: D 0 0 . ln.xy// k 0 Z Z XD In view of the fact that .xy/k N 1 .xy/t dt C ; for 0~~

1 1 1 1 t N N I3 .xy/ .1 x/ .1 y/ dydx dt: D k N 0 0 k 0 Z C ÂZ Z Ã XD Using Euler’s beta function, we have 1 .t 1/.N 1/ xt .1 x/N dx B.t 1;N 1/ C C : 0 D C C D .t N 2/ Z C C Repeatedly using .x 1/ x.x/ yields C D .t 1/ 1 C : .t N 2/ D .t N 1/.t N / .t 1/ C C C C C C Appealing to .N 1/ NŠ, we have C D 2 1 1 NŠ I3 dt: D k N .t N 1/.t N / .t 1/ k 0 Z C Â Ã XD C C C C Replacing t 1 by t and then setting n k N 1, we arrive at C D C C 2 1 1 NŠ I3 dt D t.t 1/ .t N / n N 1 Zn Â Ã DXC C C as desired.

21.6 Monthly Problem 11329 [2007, 925] Let f .t/ 2 t ln .t/, and let be Euler’s constant. Derive the following integral iden- D tities: 1 lnln 2 1 f .t/ dt 2 f .t/ dt C ; D ln 2 Z0 Z0 1 . lnln 2/.1 2 ln 2/ 1 1 t f.t/dt 2 .t 1/f .t/ dt C C : D C ln2 2 Z0 Z0

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~~Using that .t 1/ t.t/, we see that f .t/ satisﬁes C D 2 f .t 1/ f .t/ 2 t ln t: (21.6) C D C Hence~~

1 1 f .t/ dt f .t/ dt 1 f .t 1/dt D C C Z0 Z0 Z0 1 1 1 1 t f .t/ dt f .t/ dt 2 ln t dt ; D C 2 C Z0 ÂZ0 Z0 Ã and so 1 1 f .t/ dt 2 f .t/ dt 1 2 t ln t dt: D C Z0 Z0 Z0 Recall the classical formula (for example, see [3, Formula 4.331.1])

~~ln ˛ 1 e ˛t ln t dt C : (21.7) D ˛ Z0 Setting ˛ ln 2, we obtain the ﬁrst desired integral identity. D Differentiating (21.7) with respect to the parameter ˛ yields~~

~~1 . ln ˛/ 1 te ˛t ln t dt C : D ˛2 Z0 Thus, using (21.6) and the ﬁrst integral identity, we obtain~~

~~1 1 t f.t/dt 2 .t 1/f .t/ dt 1 .t 2/2 t ln t dt D C C C Z0 Z0 Z0 1 . lnln 2/.1 2 ln 2/ 1 2 .t 1/f .t/ dt C C : D C ln2 2 Z0 Remark: This problem is related to the Laplace transform~~

~~L.˛/ 1 e ˛t .t 1/dt; D C Z0 where .t/ d ln .t/ is the digamma function. Integrating by parts yields D dt~~

~~L.˛/ ln ˛ ˛ 1 e ˛t ln .t/dt: D C Z0 In particular, L.ln 2/ ln.ln 2/ ln 2 1 f .t/dt: D C Z0 Refer to [4] for details.~~

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~~21. Some Irresistible Integrals 249~~

~~21.7 Monthly Problem 11331 [2007, 926] Show that if k is a positive integer, then~~

k 1 k 1 ln.1 t/ C C dt .k 1/ aj .j 1/; (21.8) t D C C Z0 Â Ã j 1 XD j where denotes the Riemann zeta function and aj is the coefﬁcient of x in k 1 x n1 .1 nx/. D QHere we have made a slight modiﬁcation: originally,the factor .k 1/ before the sum- C mation in (21.8) was missing. Let I4 denote the desired integral in (21.8). The substitution t .1 x/=x yields D 1 k 1 1 ln C xk 1 I x dx: 4 k 1 D 0 .1 x/ Z C Using 1 1 n k C xn; .1 x/k 1 D k C n 0 Â Ã XD we deduce 1 k 1 1 n k 1 C n k 1 I4 C ln x dx: D k x C n 0 Â Ã Z0 Â Â ÃÃ XD Setting s ln 1 , we have D x 1 k 1 1 C .k 1/Š ln xn k 1 dx 1 sk 1e .n k/s ds ; C C C C k 2 0 x D 0 D .n k/ Z Â Â ÃÃ Z C C and so

1 n k .k 1/Š 1 .n 1/.n 2/ .n k 1/.n k/ I4 C C .k 1/ C C C C : D k .n k/k 2 D C .n k/k 2 n 0 Â Ã C n 0 C XD C XD C Let k 1 k j t .1 jt/ aj t : (21.9) D j 1 j 1 YD XD k 1 Replacing t by 1=x and then multiplying by x C on both sides gives

~~k 1 k k 1 j x .x j / aj x : D C j 1 j 1 YD XD Thus, we have~~

~~k 1 k k 1 j .n 1/.n 2/ .n k 1/.n k/ .n k/ .n k j / aj .n k/ C C C C D C C D C C j 0 j 1 YD XD~~

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~~and~~

k 1 aj I4 .k 1/ D C .n k/j 1 n 0 j 1 C XD XD C k 1 1 .k 1/ aj D C .n k/j 1 j 1 n 0 C XD XD C k k 1 1 1 1 .k 1/ aj D C nj 1 i j 1 j 1 n 1 C i 1 C ! XD XD XD k k 1 k 1 aj .k 1/ aj .j 1/ .k 1/ : D C C C i 0 i j 1 j 1 i 1 j 1 XD XD XD @ A Noticing that 1=i .i 1;2;:::;k 1/ are solutions of (21.9) implies D k a j 0; for i 1;2;:::;k 1: i j D D j 1 XD Finally, we ﬁnd that k I4 .k 1/ aj .j 1/ D C C j 1 XD as desired.

21.8 Monthly Problem 11418 [2009, 276] Find 2 2 1 t sech t I5 dt WD a tanh t Z1 for complex a with a >1. j j Substitution of x tanh t gives D 1 arctanh2x I5 dx: D 1 a x Z Recall that 1 1 x arctanh x ln C : D 2 1 x Â Ã Substitution of s .1 x/=.1 x/ gives D C 2 1 1 ln sds I5 : D 2 0 .s 1/Œ.a 1/s a 1 Z C C C Rewrite I5 in the form of

~~1 2 2 1 ln sds 1 ln sds I5 : D 2 0 .s 1/Œ.a 1/s a 1 C 1 .s 1/Œ.a 1/s a 1 Z C C C Z C C C !~~

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~~21. Some Irresistible Integrals 251~~

~~Substituting s by 1=s in the second integral yields~~

1 1 ln2 s 1 ln2 s I5 : D 2 .s 1/Œ.a 1/s a 1 C .s 1/Œ.a 1/s a 1 Z0 C C C Z0 C C C ! The partial fraction decompositions 1 1 1 a 1 and .s 1/Œ.a 1/s a 1 D 2 s 1 .a 1/s a 1 C C C Â C C C Ã 1 1 1 a 1 C .s 1/Œ.a 1/s a 1 D 2 s 1 C .a 1/s a 1 C C C Â C C C Ã lead to 1 1 a 1 a 1 2 I5 C ln sds: D 4 .a 1/s a 1 .a 1/s a 1 Z0 Â C C C C Ã To proceed further, we recall the trilogarithm function deﬁned by

k 1 x Li3.x/ : WD n3 n 1 XD Let ˛ .a 1/=.a 1/. Appealing to D C 1 kŠ sn k sds . 1/k ; ln k 1 0 D .n 1/ Z C C we ﬁnd that 1 .a 1/ ln2 s 1 ln2 sds C ds ˛ .a 1/s a 1 D ˛s 1 Z0 C C Z0 C 1 1 ˛ . ˛/n sn ln2 sds D n 0 Z0 XD n 1 . ˛/ 2˛ 2 Li3. ˛/: D .n 1/3 D n 0 XD C Similarly, 1 .a 1/ ln2 s 1 ln2 s 1 ds 2 Li3 : .a 1/s a 1 D s ˛ D ˛ Z0 C C Z0 C Â Ã Therefore 1 1 I5 Li3. ˛/ Li3 : D 2 ˛ Â Â ÃÃ Finally, by the trilogarithmic identity (see mathworld.wolfram.com/Trilogarithm.html)

1 1 3 2 Li3. ˛/ Li3 .ln ˛ ln ˛/; ˛ D 6 C Â Ã we obtain 1 3 a 1 2 a 1 I5 ln C ln C : (21.10) D 12 a 1 C a 1 Â Â Ã Â ÃÃ Symbolically, it is interesting to see that Mathematica 6.0 displays

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~~252 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 1 dx 2 ˛ 0 .1 x /.1 x / Z C C is independent of ˛. 2. Let a be a positive number. Prove that a ln xdx ln a a dx : x2 ax a D 2 x2 ax a Z1 C C Z1 C C 3. Putnam Problem 1983-B5. Let a be the fractional part of a. Determine f g 1 n n lim dx: n n 1 x !1 Z n o 4. Let a denote the greatest integer a. Determine b c Ä n 1 2pn pn lim 2 : n n pk pk !1 k 1 Â Ã XD 5. Let n 1;3;4. Evaluate D 1 1 n dx: x Z0 6. Prove that 1 ln x ln.1 x/ 2 1 dx ln2 2: .1 x/2 D 24 2 Z0 C 7. Prove that 1 1 x 1 dxdy ; .1 xy/ ln.xy/ D Z0 Z0 1 1 x 1 dxdy ln.4=/: (J. Sondow) 0 0 .1 xy/ ln.xy/ D Z Z C 8. Monthly Problem 11148 [2005, 366; 2007, 80]. Show that .x8 4x6 9x4 5x2 1/dx 1 : 12 10 C8 6 C 4 2 0 x 10x 37x 42x 26x 8x 1 D 2 Z C C C 9. Monthly Problem 11152 [2005, 567; 2006, 945]. Evaluate

~~1 ln.cos.x=2// dx: x.1 x/ Z0 C~~

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~~21. Some Irresistible Integrals 253~~

~~10. Monthly Problem 11159 [2005; 567; 2007, 167]. For a < =2, evaluate j j =2 =2 cos dd : cos.a cos cos / Z0 Z0~~

~~11. Monthly Problem 11234 [2006, 568; 2008, 169–170]. Let a1;:::;an and b1;:::; bn 1 be real numbers with a1~~

.x a1/ .x an/ 1 h dx 1 h.x/ dx: .x b1/ .x bn 1/ D Z1 Â Ã Z1 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).

n n 12. Hadamard Product. Let f .z/ n1 0 anz and g.z/ n1 0 bnz be analytical D n D D D functions. Let h./ n1 0 anbn : Then D D P P P 1 2 h./ f.zei /g.e i /d; D 2 Z0 where z. D 13. Sequence of Deﬁnite Integrals. It is knownthat nŠ . 1/n2n 1 1 x lnn xdx. Let D C 0 1 1 R I dx: n 2 n D 0 .x x 1/ Z C C Show that In an bnp3 and determine the closed forms of an and bn. D C 14. Putnam Problem 1985-A5. Let 2 Im cos.x/ cos.2x/ cos.mx/ dx: D Z0 For which integers m;1 m 10; is Im 0? Ä Ä ¤ 15. Open Problem. Determine the maximum of

2 Ik cos.n1x/ cos.n2x/ cos.nk x/dx D Z0 where n1; n2;:::;nk are positive integers. 16. Open Problem. Let n N;a>1. Find 2 lnn xdx I.a/ 1 D 0 .x 1/.x a/ Z C C in closed form. Along the way you may use the functional equation

~~n n .2i/ 1 ln z Lin. z/ . 1/ Lin. 1=z/ Bn ; C D nŠ 2 C 2i Â Ã where Bn.x/ is the Bernoulli polynomial.~~

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~~254 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/ı 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. Glasserand D. Manna, On the Laplace transform of the psi-function, Contemporary Mathe- matics. 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, Solu- tions 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 .a b/ 4a; a b C C C which implies a2 1 4a2 1 .3a b/: a b D 4 a b 4 C C Repeatedly using this inequality gives~~

~~n 2 ak 1 1 1 .3a1 a2/ .3a2 a3/ .3an a1/ a a 4 C 4 CC 4 k 1 k C k 1 XD C 1 1 .a1 a2 an/ : D 2 C CC D 2~~

~~255~~

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Second Proof: Let an 1 a1. The Cauchy-Schwarz inequality yields C D 2 2 n n ak ak ak ak 1 0 1 D 0 pak ak 1 C C 1 k 1 k 1 C C XD XD p @ A @n 2 n A ak .ak ak 1/ Ä a a C C k 1 k C k 1 k 1 XD C XD n 2 n ak 2 ak; D ak ak 1 k 1 C C k 1 XD XD n from which and k 1 ak 1 the desired inequality follows. D D Third Proof: Let an 1 a1. The key observation is P C D n 2 n a2 a2 ak 1 k k 1 C C ; a a D 2 a a k 1 k C k 1 k 1 k C k 1 XD C XD C which is equivalent to n 2 n a2 ak k 1 C : ak ak 1 D ak ak 1 k 1 C C k 1 C C XD XD This immediately follows from n 2 2 n n ak ak 1 .ak ak 1/.ak ak 1/ C C C C .ak ak 1/ 0: a a D a a D C D k 1 k C k 1 k 1 k C k 1 k 1 XD C XD C XD Now, the original inequality is equivalent to n 2 2 ak ak 1 C C 1: a a k 1 k C k 1 XD C To prove this, notice that 2 2 ak ak 1 ak ak 1 C C C C ; ak ak 1 2 C C which follows directly after clearing out denominators and rearranging in the form 2 .ak ak 1/ 0. Therefore, C n 2 2 n n ak ak 1 ak ak 1 C C C C a 1: a a 2 D k D k 1 k C k 1 k 1 k 1 XD C XD XD n Remark. One may use a similar argument to prove that, if k 1 ak 1 and an 1 a1, then for any positive integer m, D D C D P n am 1 k : am 1 am 2a am 3a2 am 1 m k 1 k k k 1 k k 1 k 1 XD C C C C CC C 4.(e) By the AM-GM inequality, 1 1 a0 ak 1 ak an .1 a0 ak 1 ak an/ 1: C CC CC Ä 2 C CC C CC D Hence,p for all 1 k np, Ä Ä ak bk ak; WD 1 a0 ak 1pak an C CC CC p

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~~Solutions to Selected Exercises 257~~

and therefore n n b a 1: k k D k 1 k 1 XD XD This proves the ﬁrst inequality. To prove the second inequality, for 0 k n, deﬁne Ä Ä arcsin.a0 a1 a /: k D C CC k Clearly, 0 0 <1 < <n arcsin 1 : D D D 2 Notice that

~~ak sin k sin k 1 D k k 1 k k 1 2 cos C sin ; D 2 2 and~~

2 cos k 1 1 sin k 1 D q 2 1 .a0 a1 ak 1/ D C CC q 1 a0 ak 1 ak an: D C CC CC By the well-known inequality p p sin <; .0;=2/; 2 we have k k 1 ak < cos C .k k 1/ cos k 1.k k 1/; 2 Ä and therefore n n n ak bk .k k 1/ : D cos k 1 Ä D 2 kX1 kX1 kX1 Remark. Observe that D D D

ak .a0 ak/ .a0 ak 1/ b CC CC : k 2 2 D 1 .a0 a1 ak 1/ D 1 .a0 a1 ak 1/ C CC C CC The proposedsump can be viewed as a Riemann sumforp the function 1=p1 x2 on the interval Œ0;1 when the partition is

0 a0; a0 a1; a0 a1 a2;:::;a0 a1 an 1 f D C C C C CC D g It follows at once that n 1 dx 1 .1 0/ bk : Ä Ä 0 p1 x2 D 2 k 1 Z XD 7. There are two distinct cases. First, if x2 a .a 1/, since n 1 1, we have 1 1 k 1 ak D Ä 2 2 n n P 1 1 0 a2 x2 1 Ä 0 2a x 1 k 1 k k 1 k j j XD C XD @ A @ A 2 n 1 1 1 2 2 D 4x 0 ak 1 Ä 4x k 1 XD 1 @ 1 A 2 : Ä 2 a1.a1 1/ x C

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2 Next, if x < a1.a1 1/, applying the Cauchy-Schwarz inequality, 2 n n n 1 1 ak 0 a2 x2 1 Ä 0 a 1 0 .a2 x2/2 1 k 1 k k 1 k k 1 k XD C XD XD C @ A @n A @ A a k : Ä .a2 x2/2 k 1 k XD C Notice that ak 1 ak 1 for k 1;2;:::;n 1. We have C C D 2ak 2ak .a2 x2/2 Ä .a2 x2 1=4/2 a2 k C k C C k 1 1 D .a 1=2/2 x2 .a 1=2/2 x2 k C k C C 1 1 2 2 2 2 : Ä .ak 1=2/ x .ak 1 1=2/ x C C C Similarly, we have

2an 1 1 1 2 2 2 2 2 2 2 : .a x2/2 Ä .an 1=2/ x .an 1=2/ x Ä .an 1=2/ x n C C C C C Thus, telescoping yields n n 1 ak 1 1 1 1 1 2 2 2 2 2 2 2 .a x2/2 Ä 2 .a 1=2/ x .a 1=2/ x C 2 .an 1=2/ x k 1 k k 1 Ä k C k 1 C C XD C XD C 1 1 1 1 2 2 2 D 2 .a1 1=2/ x Ä 2 a1.a1 1/ x C C as expected. 8. Let s0 0; s a1 a2 a ; .k 1;2;:::;n/: D k D C CC k D Then 1 1 sk Œ.a1 ak/ .a2 ak 1/ .ak a1/ kak 1: D 2 C C C CC C 2 C Appealing to Abel’s summation formula n n n 1 k ak bk an bk .ak 1 ak/ bi ; D C k 1 k 1 k 1 i 1 XD XD XD XD we have n n ak sk sk 1 k D k k 1 k 1 XD XD n 1 1 1 1 sn s D n C k k 1 k k 1 Â C Ã XD n 1 1 1 1 1 1 sn s1 kak 1 n C 2 C 2 k k 1 C k 1 Â C Ã XD n 1 1 1 1 ak 1 sn s1 C D n C 2 C 2 k 1 k 1 C XD n 1 1 ak sn ; D n C 2 k k 1 XD

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~~Solutions to Selected Exercises 259~~

~~and so~~

n ak 2 2 sn .sn 1 an/ k n D n C k 1 XD 2 n 1 n 1 an an C an > an: n 2 C D n Â Ã Chapter 2 4 Deﬁne f .x/ a x b ln x; E .0; /. Since f .x/ a b =x; f .x/ b =x2, k D k k D 1 k0 D k k k00 D k fk.x/ has a unique critical point at xkc bk=ak, where it has minimum value fk.xkc / D n D bk bk ln.bk=ak/: Similarly, the function f.x/ k 1 fk.x/ has a unique critical point n n D D xc . k 1 ak/=. k 1 bk/ and its minimum value is D D D P P P n n k 1 bk f .xc/ b 1 ln : D k nD a k 1 Â Pk 1 k Ã XD D P By using (2.1) we obtain

n n b n b a ln k b ln k 1 k ; k a 0 k1 nD a k 1 k k 1 Pk 1 k XD XD D @ A P which is equivalent to the required inequality. 6. Deﬁne a 1 a 1 x f .x/ p k k ln ;E .0;1=2: k D k x 1 x x D Â Ã Notice that f .x/ p .a x/.2x 1/=.x2.1 x/2/. We ﬁnd that f has a minimum at k0 D k k k x a and its value is f .x / lnŒa =.1 a /pk . On the other hand, we have kc D k k kc D k k n A 1 A 1 x f.x/ f .x/ ln ; D k D x 1 x x k 1 Â Ã XD

which has a minimum at xc A and its value is f.A/ ln.A=A /. Now, applying (2.1) yields D D 0 n a pk A ln k ln ; 1 a Ä A k 1 Â k Ã 0 YD which is equivalent to the desired inequality. 7. Let f.x/ 1=.1 ex /. Since D C ex .ex 1/ >0; x .0; /; f .x/ 2 1 00 D .1 ex /3 D <0; x . ;0/; C 2 1 we see that f is convex in .0; / and concave in . ;0/. Therefore, for xk ln.1 a /=a .1 k n/, by Jensen’s1 inequality, 1 D ˙ k k Ä Ä n n k 1 pk f .xk/; xk .0; /; f pk xk Ä nD 2 1 0 1 D pk f .xk/; xk . ;0/: k 1 Pk 1 2 1 XD D @ A P 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 .t 2 1/.t 2 4t 1/ ln t C C : .1/ Ä 4t.t 2 t 1/ C C Proof. Let .t 2 1/.t 2 4t 1/ f.t/ C C ln t: D 4t.t 2 t 1/ C C Note that f.1/ 0, so it suffices to show that f 0.t/> 0 for t>1. This follows from the fact that D .t 1/4.t 1/2 f .t/ C > 0; for t >1: 0 D 4t 2.t 2 t 1/2 C C Lemma 2. For t 1 we have ln t t pt 1 ln C C 1: .2/ t 1 3t ! Proof. Note that left-hand side becomes continuous if we define it to have the value 1 at t 1. Thus, it suffices to show that it has a nonnegative derivative for t>1, in other words, D ln t 1 1 2pt C 0: .t 1/2 C .t 1/ 2pt.t 1/ 2t C C 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 b=a in (2). We find that D a.ln a ln b/ a pab b ln b 1 > ln C C ; a b C 3 ! and the first inequality follows upon exponentiation. Also note that the strict inequality holds unless a b. D 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/ .1 s/G.a; b/; for 0 < s < 1: C This is equivalent to a.ln a ln b/ ln b 1 > ln Œ2s.a b/ .1 s/pab: a b C C C The analogue of f 0.t/ in Lemma 1 becomes .t 1/2.1 t/2. s s2 4t 8st 2s2t st 2 s2t 2/ C C C C 0: t 2.s 2t 2st st 2/2 C C Letting t 1 yields s 2=3. Similarly, one can show that if ! Ä I.a;b/ < ˇA.a;b/ .1 ˇ/G.a;b/ C then ˇ 2=e.

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~~10. First, since S˛ .a;b/ is strictly increasing in ˛, for 0~~

~~ax bx 1=x a b C C 2 2 Â Ã for x 1 with equality if and only if x 1. Finally, we obtain D bx y ax y x y C C C A.a;b/y bx ax x for x;y 1 with equality if and only if x y 1. Remark.When 0~~

1 .b a/ 4 2A2=3 G2=3 1 .b a/ 4 exp < C < exp : 360 max.a;b/ I 2 360 min.a;b/ Â Ã ! Â Ã ! This yields the required inequality. Remark. Stimulated by this result, for p 2, one may prove that the double inequality ˛Ap .a;b/ .1 ˛/Gp .a;b/< I p.a;b/ < ˇAp.a;b/ .1 ˇ/Gp .a;b/; C C holds for all positive numbers a b if and only if ˛ .2=e/p and ˇ 2=3. ¤ Ä

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~~Chapter 4 1. Let f.x/ x ln.1 x/;g.x/ x ln.1 x/ and let D C D C~~

~~f1.x/ f .x/ .1 x/ ln.1 x/ x 0 C C C : g1.x/ WD g0.x/ D x~~

~~Since f .x/ 10 2 ln.1 x/ g10 .x/ D C C~~

is increasing,by the LMR, f1.x/=g1.x/ is increasing. An appealto the LMR again establishes the assertion. 6. Apply the LMR to 1 cos x 2 x : x 2 Â Ã . Á 7. Rewrite the inequality as 2 1 x2 < cot.x=2/ < : x ! 4 . Then apply the LMR. n 10. (a) We prove a stronger result: Let k 1 ak sin.2k 1/x 0;00; 0

sin kx =2 sin.x=2/ 2k sin.2k 1/t 2 dt: k D sin t sin t Zx=2 Â Ã Thus n =2 n 2n ak sin kx sin.x=2/ sin.2n 1/t 2 an dt: k D x=2 sin t sin t k 1 Z k 1 Â Ã XD XD n 2k 1 The proposed inequality follows from k 1 ak r sin.2k 1/t > 0 for 0 0: D sin x k 1 XD

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~~Solutions to Selected Exercises 263~~

Chapter 5 2. Let P ei0 . Recall that the vertices of a regular n-gon are given by !k .1 k n/. Thus, D Ä Ä n n P !k 2 .P !k /.P !/ j j D k 1 k 1 XD XD n . P 2 .P ! P!/ ! 2/ D j j C Cj j k 1 XD n 2k 2 1 cos 0 D C n k 1 Â Â ÃÃ XD 2n: D 7. Repeatedly using the double angle formula sin 2 2 sin cos , we obtain D x x sin x 2 sin cos D 2 2 x x x 22 sin cos cos D 22 2 22 x x x x 23 sin cos cos cos D 23 2 22 23 : : x x x x 2n sin cos cos cos : D 2n 2 22 2n

~~But from elementary calculus we have that limt 0 sin t=t 1, and therefore ! D x lim 2n sin x: n 2n D !1 This proves the identity (a). Next, let~~

n n x 2 x 2 x Pn cos and Qn x cos x sin Pk : D 2k D 2 C 2k 1 k 1 k 1 C YD XD Since x x x x x sin x cos x cos x sin2 x cos2 2 sin ; D 2 2 C 4 4 C 2 Á we have x x sin x Q1 xP2 cos 2P1 sin : D 4 C 2 By induction, x n x sin x Qn xPn 1 cos n 1 2 Pn sin n : D C 2 C C 2 Appealing to that Pn 1 and j j Ä x n x 2 x n x lim xPn 1 cos 2 Pn sin lim Pn x cos 2 sin 0; n C 2n 1 C 2n D n 2n 1 C 2n D !1 C Á !1 C Á we establish that sin x lim Qn D n !1 which is equivalent to the desired identity (b).

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~~12. Replacing x by .x iy/=2 in the inﬁnite product of cosine (5.29) yields C 2 2 .x iy/ 1 x y 2xyi cos C 1 C : 2 D .2k 1/2 k 1 ! YD On the other hand,~~

~~.x iy/ x y x y cos C cos cosh i sin sinh : 2 D 2 2 2 2 It follows from (5.32) that~~

1 2xy x y arctan arctan tan tanh .mod 2/: .2k 1/2 x2 y2 D 2 2 k 1 Â C Ã XD Á Chapter 6 4. For n 2, we have n 1 n 1 k 1 H .interchange the order of sums/ k D i k 1 k 1 i 1 XD XD XD 1 1 1 .n 1/ 1 .n 2/ Œn .n 2/ Œn .n 1/ D C 2 CC n 2 C n 1 nHn 1 .n 1/ nHn n: D D This identity can also be established by using Abel’s summation formula. Next, we have

~~1 1 2 1 1 2 H 2 H 2 1 1 k k 1 D C 2 CC k C 2 CC k 1 Â Ã Â Ã 1 1 1 1 2 1 D k2 C k C 2 CC k 1 Â Ã 1 2 1 H D k2 C k k k Â Ã H 1 2 k : D k k2 Telescoping yields~~

~~H H H 1 1 1 H 2 H 2 2 2 3 n : n 1 D 2 C 3 CC n 22 C 32 CC n2 Â Ã Â Ã Thus, we get~~

H H H 1 1 1 H 2 2 2 3 n 1 n D 2 C 3 CC n C 22 32 C n2 Â Ã Â Ã H2 H3 Hn 1 1 1 >2 1 2 C 3 CC n C 1 2 2 3 C .n 1/ n Â Ã Â Ã H H H 1 2 2 3 n D 2 C 3 CC n C n Â Ã as desired. Note that telescoping is used in the last summation again.

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~~Solutions to Selected Exercises 265~~

~~8. Appealing to 1 1 n n 1 1 t Hn .1 t t /dt dt; D 0 C CC D 0 1 t Z Z we have~~

1 n 1 n 1 n 1 t anHnx anx dt D 1 t n 1 n 1 Z0 XD XD 1 n n 1 anx an.tx/ dt D 1 t n 1 Z0 XD 1 f.x/ f.tx/ dt: D 0 1 t Z 9. Rewrite Rn as n k 1 n 1 1 C 1 t Rn dx dt: D k k x D 0 k.k t/ k 1 Z ! k 1 Z C XD XD Thus

1 1 t Rn dt D 0 k.k t/ k n 1 Z C DXC 1 1 1 1 1 1 1 t dt t dt D 0 k.k t/ k.k 1/ C 0 k.k 1/ k n 1 Z Â C C Ã Z k n 1 C DXC DXC 1 1 1 t.1 t/ 1 dt tdt: D 0 k.k 1/.k t/ C n 1 0 k n 1 Z C C C Z DXC Deﬁne 1 1 1 t.1 t/ R1.n/ dt; a1 t dt: D 0 k.k 1/.k t/ D 0 k n 1 Z C C Z DXC Then a1 Rn R1.n/ : D C n 1 C Repeating this process, we have

1 1 1 t.1 t/.2 t/ 1 R1.n/ dt t.1 t/dt: D 0 k.k 1/.k 2/.k t/ C .n 1/.n 2/ 0 k n 1 Z C C C C C Z DXC 1 Let the sum be R2.n/ and a2 1=2 t.1 t/dt. Then D 0 R a1 a2 Rn R2.n/ : D C n 1 C .n 1/.n 2/ C C C By induction, for N 2, we have N ak Rn RN .n/; D .n 1/.n 2/ .n k/ C k 1 C C C XD where 1 1 t.1 t/ .N t/ RN .n/ dt D 0 k.k 1/ .k N/.k t/ I k 1 Z C C C XC

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1 1 a t.1 t/ .k 1 t/dt: k D k Z0 The required answer follows from two estimates: 1 1 .k 2/Š a .k 1/Š 6k Ä k Ä 6k I 1 n k 1 C R .n/ : 6.n 2/k.k 1/ Ä k k Ä 6n.k 1/ C C Â Ã C n x 13. Note that nŠ 1 x e dx. The substitution of x pnt n yields D 0 D C n R n n 1 t pnt nŠ n pne 1 e dt: D pn C pn Z Â Ã Let the integrand be fn.t/. Note that n t pnt pnt n ln.1 t=pn/ t 2=2 O.1=pn/ fn.t/ 1 e e e : D C pn D C C D C Â Ã Hence, the expected limit follows from the dominated convergence theorem

1 1 1 t 2 =2 lim fn.t/dt lim fn.t/dt e dt p2: n pn D n D D !1 Z Z1 !1 Z1 To justify this, we show that R supn fn dx is bounded.Indeed, for t 0, fn.t/ is decreasing t j j in n and so supn fn .1 t/e , and that is integrable. For t 0, fn.t/ is increasing in n, j j D CR Ä t 2=2 which shows that in this case the sup occurs as n , and so has the value e . This is also integrable. ! 1 .k/ k .k 1/ 14. Let f.x/ 1=x. Then f .x/ . 1/ kŠx C . The Euler-Maclaurim summation formula givesD D m 1 B2k 2k 1 .2k 1/Š 2k 1 Hn ln n .1 1=n/ .. 1/ . 1/ .2k 1/Š/ Rn.m/ D C 2 C C .2k/Š n2k C k 1 XD m 1 B2k 1 ln n .1 1=n/ 1 Rn.m/: D C 2 C C 2k n2k C kX1 Â Ã Thus D m 1 B2k lim .Hn ln n/ R .m/; D n D 2 C 2k C 1 !1 kX1 and so D m 1 B2k 1 Hn ln n Rn.m/ R .m/: D C C 2n 2k n2k C 1 kX1 This implies the desired formula. D

Chapter 7 2. The limit is 2. For n 3 we have n k n 1 n k n 2 n 2 2n k D 2n n k k 1 k 0 XD XD n 1 1 n n k using 1 D 2k n k n k D C n k k 0 Â Ã XD n 1 n 1 1 k : D 2k C 2k.n k/ k 0 k 1 XD XD

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~~Solutions to Selected Exercises 267~~

~~Since~~

n 1 n 1 k 1 k 2k.n k/ Ä 2.n 1/ C .k2 k/.n k/ k 1 k 2 XD XD n 2 1 2 1 D 2.n 1/ C n 1 k k 1 XD 1 2 .ln n O.1=n// 0; as n Ä 2.n 1/ C n 1 C C ! ! 1 we obtain n 1 1 L lim 2: D n 2k D !1 k 0 XD 4. Consider the sequence an deﬁned by f g 1 1 1 1 An ; 2n an Dj j D n 1 n 2 C n 3 C C C C 1 or an An 2n. For every n 1, we have Dj j 1 1 1. 1 < an <1 ; C p.n 1/2 1 n 1 C pn2 1 n C C C C C C 2. an is strictly decreasing and converges to 1; 3. The best constantsare a a1 1=.1 ln 2/ 2 1:258891:: : and b 1. D D D D 5. The best possible constants are ˛ 11=6 and ˇ .4 e/=.e 2/. D D Chapter 8 1. By using Gauss’s multiplication formula and .1/ 1, we have D n n 1 1 i 1 i ln ln n n D n n i 1 Â Ã i 1 Â Ã! XD YD 1 ln .2/.n 1/=2n 1=2 D n 1 n 1 1 Á ln.2/ ln n : D n 2 2 Â Ã Letting n yields the desired answer. ! 1=2 Remark. If 0 ln sin x dx ln 2=2 has been proved, a short proof can be derived by applying the reﬂection formula.D R 2. Using .x 1/ x.x/ and Leibniz’s rule, we have C D d x 1 C ln .t/dt ln .x 1/ ln .x/ ln x: dx D C D Zx It is clear that .x ln x x/ ln x. Thus 0 D x 1 C ln .t/dt x ln x x C; D C Zx where C is an arbitrary constant. Now setting x 0 yields C 1 ln.2/. D D 2

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3. Notice that =2 =2 d : agm.1;1=p2/ D 0 2 Z 1 sin =2 Substitution of x sin2 =.2 sin2 / yields q D =2 1 dx p2 : agm.1;1=p2/ D 0 p1 x4 Z Next, substitution of u x4 gives D 1 =2 p2 1=4 1 1=2 1 u .1 u/ du: agm.1;1=p2/ D 4 0 Z Appealing to the beta function, we have

=2 p2 B.1=4;1=2/: agm.1;1=p2/ D 4 Finally, since .1=2/ p and D .1=4/.1=2/ 2.1=4/.1=2/ 2.1=4/p2 B.1=4;1=2/ ; D .3=4/ D = sin.=4/ D 2p we obtain the proposed answer. 6. By the reﬂection formula, we have

t x 1 1 t x 1 t x 1 1 dt dt 1 dt sin x D 0 1 t D 0 1 t C 1 1 t Z C Z C Z C 1 x 1 x 1 n n 1 n 1 t t x 1 x k k . 1/ C t C C dt .t t / . 1/ t dt D 0 1 t D 0 C 2 C 1 t 3 Z C Z k 0 C XD n 4 5 . 1/k Rn; D x k C k n XD where 1 n x n x 1 1 1 Rn .t C t C /dt : j j Ä ˇ 0 C ˇ Ä n x 1 C n x 2 ˇZ ˇ C C C ˇ ˇ 11. Take the logarithmic derivativeˇ of the Weierstrass productˇ formula. ˇ ˇ Chapter 9 2. Let the required sum be S. Appealing to Gauss’s pairing trick, we have

n n n n 2S cos kx sin.n k/x cos.n k/x sin kx D k C n k k 0 Â Ã k 0 Â Ã XD XD n n .cos kx sin.n k/x cos.n k/x sin kx/ D k C k 0 Â Ã XD n n sin nx 2n sin nx: D k D k 0 Â Ã XD Therefore S 2n 1 sin nx: D

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~~Solutions to Selected Exercises 269~~

3. Using tan ˛ tan ˇ sin.˛ ˇ/=.cos˛ cos ˇ/: D 1 3 .2n 1/ 2 4. Let an 4n 24 2n : Then D Á 1 1 3 .2n 1/ 2 an 1 an : C D n 1 2 4 2n C Â Ã Appealing to Wallis’s formula, telescoping yields that limn an 4=. !1 D 7. Inverting the order of summation, we ﬁnd that

1 1 1 1 1 1 1 ..k/ 1/ D nk D 0 nk 1 k 2 k 2 n 2 ! n 2 k 2 XD XD XD XD XD 2 @ A 1 1=n .the inner sum is a geometric series/ D 1 1=n n 2 XD 1 1 1 1 1 1: D n.n 1/ D n 1 n D n 2 n 2 Â Ã XD XD 9. (b). Let f.x/ ax.x 1/. The identity follows from (9.22) directly. D (c). Notice that

~~1 2 1 2 1 2 . 1/k 1 arctan arctan arctan : C k2 D .2k 1/2 .2k/2 k 1 Â Ã k 0 Â C Ã k 1 Â Ã XD XD XD We have~~

1 2 1 arctan .arctan.2k 2/ arctan.2k// ; .2k 1/2 D C D 2 k 0 Â C Ã k 0 XD XD 1 2 1 arctan .arctan.2k 1/ arctan.2k 1// ; .2k/2 D C D 4 k 1 Â Ã k 1 XD XD from which the required identity follows. 15. If the leading coefﬁcient of Q.x/ is 1, by using the identity 1 1 ix arctan x ln C ; D 2i 1 ix we have

1 P.k/ 1 1 Q.k/ iP.k/ 1 1 Q.k/ iP.k/ arctan ln C ln C : Q.k/ D 2i Q.k/ iP.k/ D 2i Q.k/ iP.k/ k 1 k 1 k 1 XD XD YD Let aj .1 j n/ be the roots of Q.x/ iP.x/. Since P.x/ and Q.x/ have real coefﬁcients, Ä Ä the roots of Q.x/ iP.x/ are complex conjugatesof the aj . Appealingto Exercise3 of Chapter 8, we obtain C

1 P.k/ 1 1 .n a1/.n a2/ .n an/ arctan ln N N N Q.k/ D 2i .n a1/.n a2/ .n an/ k 1 k 1 XD YD 1 1 .1 a1/.1 a2/ .1 an/ ln D 2i .1 a1/.1 a2/ .1 an/ k 1 N N N YD n arg..1 aj // .mod 2/: D j 1 XD

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2 2 For example, let P.x/ 2 and Q.x/ x . Thus aj are the roots of x 2i, which are 1 i and 1 i. Therefore D D C 1 2 arctan arg.. i// arg..2 i// .using .1 x/ x.x// k2 D C C C D k 1 XD arg.. i// arg.1 i/ arg.i/ arg..i// D C C C C 3 arg.1 i/ arg.i/ : D C C 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 n xk .1 x/n k D C k 0 Â Ã XD and then manipulating it yields~~

~~n n xk .x 1/n 1 1 C C : k k 1 D .n 1/x k 0 Â Ã C C XD By the multisection formula we have~~

~~n=2 2k n 1 n 1 b c n x .1 x/ .1 x/ C C C : 2k 2k 1 D 2.n 1/x k 0 Â Ã C C XD 4. By the multisection formula we have~~

n 2n 1 x2j .1 x/2n .1 x/2n : 2j D 2 C C j 0 Â Ã XD h i Taking x pai yields D n 2n 1 . a/j .1 pai/2n .1 pai/2n/ : 2j D 2 C C j 0 Â Ã XD h i Let arctan.pa/. Rewrite D 1 pai p1 ae i : ˙ D C ˙ We have

n 2n 1 . a/j .1 a/n.ei2n e i2n / .1 a/n cos.2n/: 2j D 2 C C D C j 0 Â Ã XD 5. By the multisection formula with k 2 and m 0, D D 2 1 2n 1 x x x F2nx : D 2 1 x x2 1 x x2 D 1 3x2 x4 n 1 Â Ã XD C C

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~~Replacing x2 by x yields~~

~~1 n x F2n x ; D 1 3x x2 n 1 XD C which is the generating function of F2n . Similarly, with k 2 and m 1, the multisection formula gives f g D D~~

~~2 1 2n 1 1 x x x.1 x / F2n 1 x C : C D 2 1 x x2 C 1 x x2 D 1 3x2 x4 n 1 Â Ã XD C C 2 Dividing by x and then replacing x by x yields the generating function of F2n 1 as f C g~~

~~1 n 1 x F2n 1x : C D 1 3x x2 n 1 XD C 7. For x <1, start with j j n 1 x ln.1 x/: n D n 1 XD Multiplying both sides by x and then integrating gives~~

~~n 2 1 x 1 C .2x x2 2 ln.1 x/ 2x2 ln.1 x// n.n 2/ D 4 C C n 1 XD C or n 2 2 1 x 2x x 2 ln.1 x/ 2x ln.1 x/ C C : n.n 2/ D 4x2 n 1 XD C Now, using the multisection formula yields~~

4n 1 1 x 1 1 1 x 1 C 1 ln 2 1 arctan x : .4n 1/.4n 3/ D 8 x2 1 x C C x2 n 1 ÄÂ Ã Â Ã XD C C C Letting x 1, we get a byproduct D 1 1 : .4n 1/.4n 3/ D 8 n 1 XD C C Chapter 11 n 2. Let G.x/ n1 0 unx . Then D D P 1 n G.x/ a bx unx D C C n 2 XD 1 n 1 n a bx p un 1x q un 2x D C C n 2 n 2 XD XD 1 n 2 1 n a bx px unx qx unx D C C n 1 n 0 XD XD a bx px.G.x/ a/ qx2G.x/: D C C Solving for G.x/ yields a .b ap/x G.x/ C : D 1 px qx2 C

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7. We begin to derive a closed form for bn nan. By the assumptions,we have WD n 1 b1 1; b2 2; and bn n .n k 1/b for n>2: D D D k k 1 XD n Let G.x/ n1 1 bnx . Then D D P n 1 1 G.x/ n .n k 1/b xn D 0 k1 n 1 k 1 XD XD @ n 1 A 1 1 nxn .n k 1/b xn D 0 k1 n 1 n 1 k 1 XD XD XD x x@2 A G.x/; D .1 x/2 .1 x/2 where the Cauchy product has been used in the second sum. Therefore, x G.x/ : D x2 .1 x/2 C To determine the coefﬁcient of xn in G.x/, we factor

~~x2 .1 x/2 2.x ˛/.x ˇ/; C D where ˛ .1 i/=2;ˇ .1 i/=2. Appealing to partial fractions and geometric series expansion,D we haveC D~~

1 ˛ ˇ 1 1 G.x/ Œ.1 i/n .1 i/nxn: D 2i x ˛ x ˇ D 2i C Â Ã n 1 XD n n i=4 Thus bn Œ.1 i/ .1 i/ =2i. Now, by using 1 i p2e , we obtain D C ˙ D ˙ n=2 ni=4 ni=4 n=2 bn 2 e e 2 sin.n=4/ an : D n D n 2i D n Remark. It is interesting to see that

1 a 1 sin.n=4/ n : 2n D n=2 D 4 n 1 n 1 2 n XD XD 8. Let n n 2 S1 F and S2 F : D k D k k 1 k 1 XD XD Appealing to n n 2 Fk Fn 2 1 and Fk FnFn 1; D C D C k 1 k 1 XD XD we have n 2 n 2 FnFn 1 Fk S2 Fk C : Fn 2 Fk 1 D S1 Fk k 1 C k 1 XD XD Using the power mean inequality

~~2 .a1 a2 am/ a2 a2 a2 C CC 1 C 2 CC m m~~

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~~yields 2 2 .S1 Fk/ S2 F : k n 1 Therefore n 2 n S2 Fk S1 Fk S1 Fn 2 1: S1 F n 1 D D C k 1 k k 1 XD XD Chapter 12 4. Recall that Fibonacci polynomials are deﬁned by~~

fn.x/ xfn 1.x/ fn 2.x/ .n 2/ D C with f1.x/ 1 and f2.x/ x. Similar to Binet’s formula, we have D D ˛n .x/ ˇn.x/ fn.x/ D ˛.x/ ˇ.x/ 2 iz iz where ˛.x/; ˇ.x/ .x px 4=2. Let x 2i cos z. Then ˛.x/ ie ;ˇ ie . Therefore, D ˙ C D D D inz inz n 1 e e n 1 sin nz fn.x/ i iz iz i : . / D e e D sin z Appealing to (5.15), we obtain

n 1 n 1 k fn.x/ .2i/ cos z cos : D n k 1 Â Ã YD Since cos z x=.2i/, we get D n 1 k fn.x/ x 2i cos ; D n k 1 Â Ã YD and so n 1 k Fn fn.1/ 1 2i cos : D D n k 1 Â Ã YD Regrouping the factors above yields the proposed factorization without the complex number i. The third equality follows from by setting x 1. D 9. Let a pa2 4b ˛;ˇ ˙ C : D 2 First, by using induction on k, we prove that i n i .ukx buk 1/ .uk 1x buk/ C C C

n i n i1 n ik 1 kn i 2i1 2ik1 ik i1 ik n ik a b CC x : D i1 i2 ik i1;iX2;:::;ik Â Ã Â Ã Â Ã Notice that n i xi .ax b/n i an i j bj xn j ; C D j 0 j n i Â Ã ÄXÄ which implies the proposed result is true for k 1. Now suppose the equation is true for some 1 D n positive integer k. Substituting a bx for x and multiplying by x , we see the left side of the equation becomes C i n i .auk buk 1x buk/ .auk 1 bukx buk 1/ ; C C C C C C

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~~which is equivalent to~~

~~i n i .uk 1x buk/ .uk 2x buk 1/ : C C C C C Expanding the right side of the equation and simplifying, we obtain~~

n i n i1 n ik .k 1/n i 2i1 2ik ikC1 i1 ikC1 n ikC1 a C b CC x i1 i2 ik 1 i1;i2X;:::;ikC1Â ÃÂ Ã Â C Ã as desired. Next, multiplying both sides of the equation by xi and summing over i, we have

~~n i n i i .ukx buk 1/ .uk 1x buk/ x C C C i 0 XD~~

n i n i1 n ik 1 kn i 2i1 2ik1 ik i1 ik n i ik a b CC x C : D i1 i2 ik i;i1;iX2 ;:::;ik Â ÃÂ Ã Â Ã n k n The coefﬁcient of x on the right side of the equation is tr.An 1/, while the coefﬁcient of x on the left side of the equation is C

~~i n i s i s t n i t uk.buk 1/ uk 1.buk/ s t i s t n Â Ã Â Ã C CXC D i n i i s s n i s s .buk 1/ ukuk 1 .buk/ : D s s i s n Â Ã Â Ã C CXÄ Let an be the right side above. Then~~

1 n 1 i i i s 2s i s 1 n i n i s anx b uk 1uk x C .uk 1x/ D s s C n 0 i;s 0 Â Ã i s n Â Ã XD XD CXD 1 i i i s 2s i s s 1 b uk 1uk x C .1 uk 1x/ D s C i;s 0 Â Ã XD 1 s 2s 2s s 1 i i s b uk x .1 uk 1x/ .buk 1x/ D C s s 0 i s Â Ã XD X 1 bsu2s x2s .1 u x/ s 1.1 bu x/ s 1 D k k 1 k 1 s 0 C XD 1 : D 2 2 1 .uk 1 buk 1/x .buk 1uk 1 buk/x C C C C Appealing to

~~k k 2 k k uk 1 buk 1 ˛ ˇ ; buk 1uk 1 buk ˛ ˇ ; C C D C C D we ﬁnd that~~

~~1 1 ˛k ˇk : 2 2 D k k k k 1 .uk 1 buk 1/x .buk 1uk 1 buk/x ˛ ˇ 1 ˛ x 1 ˇ x ! C C C C Thus, k tr.Ak 1/ an ukn k =uk: C D D C~~

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~~Finally, let the eigenvalues of An 1 be 0; 1;:::;n and let the characteristic polynomial be C~~

P.x/ det.xI An 1/: D C Then n n P 0.x/ 1 1 k 1 k x j P.x/ D x j D j 0 k 0 j 0 XD XD XD nk k nk k 1 k 1 k 1 k 1 ˛ C ˇ C x tr.A / x D n 1 D ˛k ˇk k 0 C k 0 XD XD n n 1 1 x k 1 ˛jk ˇ.n j/k : D D x ˛j ˇn j k 0 j 0 j 0 XD XD XD Therefore n P.x/ .x ˛j ˇn j /; D j 0 YD which implies that the eigenvalues of An 1 are C n n 1 n 1 n ˛ ; ˛ ˇ;:::;˛ˇ ;ˇ :

~~Remark. This solution contains kernels of sophisticated ideas used to study the sequencesdeﬁned by the second-order recursive relations.~~

~~Chapter 13 7. Let j 1 D.x1; x2;:::;xn/ det .x /: D 1 i;j n i Ä Ä If xi xi with i1 i2, then D 0. Thus, 1 D 2 ¤ D~~

~~.xi xi / D: 1 2 ˇ Repeating this process yields ˇ .xj xi / D: 1 i~~

~~D c .xj xi / D 1 i~~

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~~Along the same lines, we have~~

~~det .pj .xi // a1a2 an .xj xi /: 1 i;j n D Ä Ä 1 i~~

det .aij / .xi xj / .bi aj /: 1 i;j n D Ä Ä 1 i0: k 0 XD This can be viewed as a system of linear equations in n 1 unknowns b0;b1;:::;bn. By n nC1 Cramer’s Rule, we have bn . 1/ D.a0; a1;:::;an/=a0C . (b). Notice that D n 2n n 2n 1 1 . 1/ x 1 1 . 1/ x sin x C : .2n 1/Š D x .2n 1/Š D x n 0 n 0 XD C XD C Appealing to the result of (a), we have

~~2 1 n 2 n x D.x / . 1/ Dn.a0; a1;:::;an/. x / : D D sin x n 0 XD Moreover, since 2 x x 1 x n sinh x e e ; .2n 1/Š D x D 2x n 0 XD C we get~~

x 2 1 n 2n 2x 2xe D. x / . 1/ Dn.a0; a1;:::;an/x : D D ex e x D e2x 1 n 0 XD Chapter 14 1. Setting z bei =a in (5.16) and then multiplying by an on both sides yields D n 1 an bnein a bei. 2k=n/ : D C k 0 YD Á Taking modulus on both sides gives

~~n 1 a2n 2anbn cos n b2n .a2 2ab cos. 2k=n/ b2/: C D C C k 0 YD Dividing this equation by anbn and regrouping gives~~

~~2n 2n n 1 2 2 a b a b C 2 cos n C 2 cos. 2k=n/ : anbn D ab C k 0 ! YD~~

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Let a=b eix . Then b=a e ix, D D a2 b2 a2n b2n C 2 cos x and C 2 cos nx: ab D anbn D Therefore, n 1 cos nx cos n 2n 1 .cos x cos. 2k=n//: D C k 0 YD Now, the proposed identity follows by taking the logarithmic derivatives with respect to x. 5. Let ! e2i=n. Then D n 1 n 1 bk 1 sin2a x .!bk=2eix ! bk=2e ix/2a n C D .2i/2a k 0 Â Ã k 0 XD XD a n 1 2a . 1/ 2a !.2a j/bk=2e.2a j/ix. 1/j ! jbk=2e j ix D 22a j k 0 j 0 Â Ã XD XD a 2a n 1 . 1/ 2a e2ix.a j/ !bk.a j/: D 22a j j 0 Â Ã k 0 XD XD Recall that n 1 n; if n b.a j /; !bk.a j/ j D 0; otherwise k 0 XD Since 0

k 1 cos cos k . 1/ C ; k 1;2;:::;n 1; ak n lim k 1 D D k sin sin n D . 1/ =2; k 0;n: ! ( C D If n is even, replacing by in (1) gives n n a k : .2/ sin sin n D cos cos x k 0 C XD

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Now, adding (1) and (2) we arrive at the desired answer (a). If n is odd, replacing by in (1) gives n n a k : .3/ sin sin n D cos cos k 0 C k XD The desired answer (b) follows from adding (1) and (3). 8. Let n 1 k 2p 1 k S2p 1 . 1/ sec C C D n k 0 Â Ã XD and its generating function be

~~1 2p 1 G.x/ S2p 1 x C : D C p 0 XD Appealing to Exercise 6(b),~~

n 1 1 2p n cos 1 cos . 1/k C ; sin sin n D cos1 2p .k=n/ p 0 k 0 C XD XD With x cos , this shows that D nx sin.n=2/ G.x/ sec.n arcsin x/: D p1 x2 Remark. If n is even, by using Exercise 6(b), we can ﬁnd the generating function of n 1 k 2p k k 0;k n=2 . 1/ sec n as D ¤ P Á nx . 1/n=2 1 csc.n arcsin x/ : p1 x2 Â Ã 9. Recall that the generating function in this case is nx G.x/ tan.n arcsin x/: D p1 x2 Since

~~1 e2ni .1 e2ni /=2 1 e2ni D 1 .1 e2ni /=2 C 2 m.1 e2ni /m D m>0 X m m 2 m . 1/k e2kni ; D k m>0 k 0 Â Ã X XD we obtain~~

~~nxi 1 e2ni G.x/ D p1 x2 1 e2ni m C m 2kn n 2 m . 1/k . 1/j x2j .1 x2/kn j : D k 2j 1 m>0 k 0 Â Ã j >0 Â Ã X XD X~~

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Appealing to the binomial theorem, the last summation can be expressed as p 2kn kn j . 1/p x2p : 2j 1 p j p>0 j 0 Â Ã Â Ã X XD The desired answer follows from application of the Vandermonde convolution formula n 1 2x r x k r 2n r x n C 2 C : r 2k n k D C 2n r k 0 Â C Ã Â Ã Â C Ã XD 12. For n 2, clearly the eigenvalue of A is 1. If n 3, let the characteristic polynomial of A be D n 1 P./ det.I A/ n 1 . 1/kc n k 1; D D C k k 1 XD where c is the sum of all the principal minors of A of order k. Now, for k 3 and n 4, we k show that c 0: To see this, deﬁne B .bij / where k D D bij sin xi sin yj cos.xi yj /; .i;j 1;2;3/: D D It is easy to verify that det.B/ 0 for all xi and yj . This implies that any 3-rowed minor of A vanishes, which clearly provesD that c 0. Thus, k D P./ n 3 2 trA Œ.trA/2 trA2=2 : D C By using the trigonometric summationn formulas, we have o n 10n2 trA ; trA2 ; D 2 D 64 and so the eigenvaluesof A are 0 (n 3 multiple), n=8 and 3n=8. Remark. A related problem is Putnam Problem 1999-B5: Let 2=n and aij D D cos.i j/. Determine the eigenvalues of A .aij /. C D Chapter 15 2. Appealing to Wallis’s formula for an odd power, we have =2 2 2n 2n 1 .nŠ/ 2 cos C x dx ; 0 D .2n 1/Š Z C and so 2 n 1 =2 n 1 .nŠ/ 2 1 1 C 2 cos x cos2 x dx .2n 1/Š D 2 n 0 Z0 n 0 Â Ã XD C XD =2 cos x 2 dx : D 0 1 cos x=2 D Z One can also prove this identity via the gamma and beta functions: 2 n 1 1 .nŠ/ 2 1 C 2n 1.n 1/.n 1/=.2n 2/ .2n 1/Š D C C C C n 0 n 0 XD C XD 1 1 1 2n 1B.n 1;n 1/ 2n 1 xn.1 x/n dx D C C C D C n 0 n 0 Z0 XD XD 1 1 1 2 .2x/n.1 x/n dx 2 Œ1 2x.1 x/ 1 dx D D Z0 n 0 Z0 XD 2 arctan 2.x 1=2/ 1 : D 0 D ˇ ˇ

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~~Remark. The reader may recognize that~~

~~2 n 1 1 .nŠ/ 2 C n x Pn.x/dx; .2n 1/Š D 1 C Z where Pn.x/ is the Legendre polynomial of degree n. Appealing to its generating function~~

~~1 n 2 1=2 Pn.x/t .1 2xt t / ; D C n 0 XD one obtains~~

2 n 1 1 1 1 .nŠ/ 2 C 1 n 2 1=2 x Pn.x/dx .1 x / dx ; .2n 1/Š D D D n 0 Z 1 n 0 Z 1 XD C XD 5. The desired result is obtained by setting a 1 in the following generalization. Let y ea arcsin x n p 2D D D n1 0 anx : Then y0 ay= 1 x , or D D P .1 x2/.y /2 a2y2: 0 D

~~Differentiating both sides with respect to x and then dividing both sides by 2y0 yields~~

~~.1 x2/y xy a2y 0: 00 0 D Substituting the power series into this differential equation and equating coefﬁcients of like powers of x, we have~~

~~n2 a2 an 2 C an; for n 0: C D .n 1/.n 2/ C C Appealing to a0 1 and a1 a, for n 1, by a simple inductive argument, we obtain D D a2.a2 22/.a2 42/ .a2 .2n 2/2/ a2n C C C ; D .2n/Š~~

~~and a2.a2 12/.a2 32/ .a2 .2n 1/2/ a2n 1 C C C : C D .2n 1/Š C Setting a 1 yields the proposed power series. Remark. ConsiderD~~

~~arcsin2 x arcsin3 x ea arcsin x 1 .arcsin x/a a2 a3 : D C C 2Š C 3Š C~~

~~Equating coefﬁcients of ak on both sides gives the power series for arcsink x;k 1. For example, for k 4, we obtain D~~

~~1 1 b arcsin4 x 2n x2n; 4Š D .2n/Š n 2 XD where n 1 2 2 2 n 1 2 4 .2n 2/ 2.n 1/ 2 1 b2n 2 Œ.n 1/Š : D .2k/2 D .2k/2 k 1 k 1 XD XD~~

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~~6. Substituting x sin.t=2/ yields D 1 =3 S t 2 cot.t=2/dt: D 2 Z0 Integrating by parts gives~~

~~=3 =3 S t 2d.ln.2 sin.t=2//dt 2 t ln.2 sin.t=2//dt: D D Z0 Z0 Recall that n 1 z ln.1 z/ ; z < 1: D n j j n 1 XD Setting z eit and exporting the real part gives D 1 cos nt ln.2 sin.t=2// ; 0~~

2 1 sin.n=3/ 1 cos.n=3/ S 2 2.3/: D 3 n2 C n3 n 1 n 1 XD XD Appealing to 1 cos.n=3/ 1 .3/; n3 D 3 n 1 XD we obtain 1 sin.n=3/ 3 .3/ S: D 2 n2 4 n 1 XD 7. Rewrite the identity as x 2 2 1 nŠ ex e t dt .2x/2n 1: D .2n/Š Z0 n 1 XD The function on the left is the unique solution of the initial value problem y 2xy 1 y.0/ 0: 0 D D n Set y.x/ n1 0 anx . Substituting the series into the differential equation and equating coefﬁcientsD of likeD powers of x yields P 2 an 1 an 1 for all n 1: C D n 1 C Appealing to a0 0 and a1 1, we have a2n 0 and D D D nŠ 2n 1 a2n 1 2 : D .2n/Š Remark. You may also show that each side of the proposed identity is a solution of the initial value problem 2y xy 2y 0; y.0/ 1; y .0/ 0: 00 0 D D 0 D 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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~~10. Dividing (15.15) by gives~~

~~2n 1 1 .2n/ 1 1 cot : 2n D 2 n 1 Â Ã XD Integrating this identity from 0 to =2 yields~~

~~=2 1 .2n/ 1 1 1 =2 1 2n cot d ln ln.=2/: .2n/2 D 2 0 D 2 sin 0 D 2 n 1 Z Â Ã Â Ã ˇ XD ˇ ˇ Next, integrating (15.15) from 0 to =2 yields~~

~~=2 1 .2n/ 1 cot d : .2n 1/22n 1 D 2 2 n 1 C Z0 ! XD C Further, integrating by parts, we have~~

=2 =2 1 cot d ln sin =2 ln sin d ln 2; D 0 D 2 Z0 Z0 ˇ and so ˇ 1 .2n/ 1 .1 ln 2/: .2n 1/22n D 2 n 1 XD C Remark. In general, one can sum the series .2n/ for any odd p via n1 1 .2n p/22n D C P =2 p .p 1/ 1 1 .2n/ xp 1 csc2 x dx C 2 : C D 2p p .2n p/22n Z0 n 1 ! XD C A more challenging problem is to evaluate the series involving .2n 1/ such as C

~~1 .2n 1/ C .2n 1/22n n 1 XD C 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 n 1. Setting x 1=2 in the power series of ln.1 x/ yields ln 2 n1 1 1=.2 n/. Appealing to the CauchyD product, we have D D P~~

~~1 1 1 1 ln2 2 D 2n n 2m m n 1 ! m 1 ! XD XD n 1 1 1 1 (using the partial fraction) D .n i/i 2n n 1 i 1 ! XD XD n 1 1 1 1 1 Hn 1 ; D 2n 1 n i D 2n 1 n n 1 i 1 n 1 XD XD XD~~

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~~where Hn are the harmonic numbers. Therefore,~~

2 1 1 1 Hn 1 1 1 ln 2 2 C 2n n2 D 2n 1 n C 2n 1 n2 n 1 n 1 n 1 XD XD XD 1 1 1 Hn 1 D 2n 1 n C n n 1 Â Ã XD 1 H n : D 2n 1 n n 1 XD n 1 Now we show that this last series is indeed .2/. In Exercise 6.8, set an 1=n2 , f.x/ 2 ln.1 x=2/. Then D D 1 1 1 1 2 ln 2 2 ln.1 t=2/ ln.1 t/ Hn C dt 2 C dt: n2n 1 D 1 t D t n 1 Z0 Z0 XD But

~~1 1 n 1 n 1 ln.1 t/ 1 . 1/ 1 . 1/ 1 C dt C t n 1 dt C .2/: t D n D n2 D 2 Z0 Z0 n 1 n 1 XD XD Similarly, the second identity follows from~~

n 1 1 H ln3 2 6 k : D 0 .n 2/2n 2 k 11 n 1 C C k 1 C XD XD @ A 6. Set t 1=2 in Euler’s series transformation or show that D n k 1 n 1 n . 1/ C n k 1 . x/ dx k k D k 0 k 1 Â Ã k 1 Â Ã Z XD XD n 1 .1 x/n 1 1 t n 1 1 dx Hn: D 0 x D 0 t 1 D k D Z Z k 1 XD

~~9. Substituting x by 1 x yields 1 ln.1 x/ ln x 1 ln.1 x/ ln x .3/ dx dx: D 0 x D 0 1 x Z Z Averaging these two integrals gives~~

~~1 1 ln.1 x/ ln x .3/ dx: D 2 0 x 1 x Z By expanding the two factors in the integrand into the power series respectively, we have~~

~~1 i 1 j 1 1 1 x 1 . x/ .3/ dx D 2 i 0 j 1 Z0 i 1 ! j 1 XD XD 1 @ A 1 1 1 1 xi 1.1 x/j 1 dx: D 2 ij i 1j 1 Z0 XD XD~~

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~~Appealing to the beta function, we ﬁnd that~~

~~1 i 1 j 1 .i 1/Š.j 1/Š x .1 x/ dx B.i;j/ ; 0 D D .i j 1/Š Z C and so 1 1 1 1 .3/ : D 2 i j 1 i 1j 1 ij 2 C XD XD j Â Ã 10. First, we rewrite An as an integral:~~

1 1 1 1 i1 i2 in 1 An x C CC dx: D i1i2 in 0 i1 1 in 1 Z XD XD Since all summands are positive, the monotone convergencetheorem permits us to interchange i the order of summation and integration. Appealing to i1 1 x =i ln.1 x/, we obtain D D 1 i1 in P 1 n 1 1 x 1 x n ln .1 x/ An dx . 1/ dx: D 0 x i1 in D 0 x Z i1 1 in 1 Z XD XD Next, in view of the well-known fact that

1 k n x ln t n 1 n .k 1/x n nŠ t ln t dt D . 1/ x e C dx . 1/ n 1 ; 0 D 0 D .k 1/ Z Z C C we have 1 n 1 ln t 1 dt t k lnn t dt . 1/nnŠ.n 1/: 0 1 t D 0 D C Z k 0 Z XD Therefore, 1 n n ln x An . 1/ dx nŠ.n 1/: D 0 1 x D C Z Chapter 17 2. By using the partial fraction decomposition

~~1 1 1 1 ; i 2 j 2 D 2i j i j i Â C Ã for positive integers i and N with N >i, we have~~

1 1 1 S.i;N/ D 2i j i j i j 1;j i Â C Ã DX¤ 1 1 1 1 1 D 2i 1 i CC .i 1/ i C .i 1/ i CC N i Â C Ã 1 1 1 1 1 C 2i 1 i CC .i 1/ i C .i 1/ i CC N i Â C C C C C Ã 1 3 .HN i HN i / 2 ; D 2i C 4i

~~where Hk are the harmonic numbers. Appealing to~~

~~Hn ln n O.1=n/; D C C~~

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~~we ﬁnd that N i lim .HN i HN i / lim ln C 0; N C D N N i D !1 !1 and so 3 lim S.i;N/ : N D 4i 2 !1 Therefore, the proposed sum is~~

~~K K 3 1 2 lim lim S.i;N/ lim : K N D 4 K i 2 D 8 !1 i 1 Â !1 Ã !1 i 1 XD XD Remark. Two related problems are to evaluate~~

~~1 1 1 1 and : i 2j 2 ij.i j/ i;j 1;.i;j / 1 i;j 1;.i;j / 1 C D X D D X D~~

~~7. (a) Rewriting Hn as an integral of a ﬁnite geometric series and then appealing to Leibniz’s formula and the power series expansionsof arctan x, we have~~

n n 1 n 1 . 1/ 1 . 1/ 1 t Hn dt 2n 1 D 2n 1 1 t n 0 n 0 Z0 XD C XD C n 1 2n 1 1 . 1/ x x 2 C dx . set t x2/ D 2n 1 1 x2 D n 0 Z0 XD C 1 x 1 arctan x 2 dx 2 2 dx D 2 0 1 x 0 1 x Z Z 1 1 x 1 dx 2 dx 2 arctan x 2 D 2 0 1 x C 0 4 1 x Z Z Á ln 2 G; D 2 C where we have used the fact that 1 dx 1 1 x dx 2 arctan x 2 2 arctan 2 0 4 1 x D 0 1 x 1 x Z Z Â C Ã Á 1 1 n arctan udu 1 . 1/ u2ndu D u D 2n 1 Z0 Z0 n 0 XD C n 1 . 1/ G: D .2n 1/2 D n 0 XD C (b). Replacing x by ix in (17.18) yields

~~1 x . 1/kh x2k arctan x: k D 1 x2 n 1 XD C Integrating this identity from 0 to 1 leads to~~

~~k 1 1 . 1/ x h arctan x dx 2k 1 k D 1 x2 n 1 Z0 XD C C 1 1 ln.1 x2/ ln 2 C 2 dx .using integration by parts/: D 8 C 2 0 1 x Z C~~

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~~Now, the proposed answer follows from~~

1 ln.1 x2/ C 2 dx ln 2 G: 0 1 x D 2 Z C 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 eix in D n 1 z x ln.1 z/ ln 2 sin.x=2/ i n D D j j C 2 n 1 XD and then exporting the real and imaginary parts yields

~~1 sin.nx/ x 1 cos.nx/ and ln 2 sin.x=2/ : n D 2 n D j j n 1 n 1 XD XD Next, multiplying these series together and manipulating gives~~

x 1 sin.mx/ cos.nx/ 1 1 sin.m n/x ln 2 sin.x=2/ C 2 j j D mn D 2 mn m;n 1 m;n 1 XD XD n 1 1 1 sin.nx/ 1 1 sin.nx/ 1 1 D 2 k.n k/ D 2 n k C n k n>k>0 n 1 k 1 Â Ã X XD XD n 1 1 sin.nx/ 1 ; D n k n 1 k 1 XD XD from which the proposed identity follows by applying Parseval’s identity. Remark. If one uses arctanh instead of ln, one is led to

~~ 3 ln2.tan.x=4//dx ; D 4 Z0 and hence to another proof of (17.25):~~

~~2 4 1 h 45 k .4/: k2 D 32 D 16 k 1 XD 11. Appealing to (17.18), we have~~

~~1 1 1 x f .x/ h x2k 2 ln C : 0 D k D 2x.1 x2/ 1 x k 1 Â Ã XD Hence,~~

~~2 x x 2 1 ln.1 x/ f ln.1 x/ ln.1 x/ : .2 x/2 0 2 x D 4x.1 x/ D 2x 4.1 x/ Á Integrating the equation above from 0 to x yields the desired identity.~~

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~~Chapter 18 2. Rewrite~~

~~1 1 m 1 1 2mn D 2 2mn 2 n 1 n2 n 1 n.2mn 1/ XD mn XD mn 1 Â Ã Â Ã m 1 1 : D 2 2mn 2m 2 n 0 .n 1/.2mn 2m 1/ C XD C C mn 1 Â Ã By using the beta function, we have~~

~~1 1 1 m 1 1 mn m 1 mn m 1 t C .1 t/ C dt: 2mn D 2 n 1 0 n 1 n2 n 0 C Z XD mn XD Â Ã Now, the proposed identity follows from~~

mn m mn m 1 t .1 t/ C C lnŒ1 t m.1 t/m: n 1 D n 0 XD C 1 4(c). Recall that G 0 .arctan x=x/dx. Set x tan.t=2/. Appealing to the double angle formula of sine, we haveD D R 1 =2 tdt 1 =2 t G dt: D 4 tan.t=2/ cos2.t=2/ D 2 sin t Z0 Z0 By using (18.1) we obtain

=2 t 1 2u arcsin u du 2G dt .u sin t/ D sin t D p 2 2u2 D Z0 Z0 1 u 1 2n 2n 1 .2u/ du 1 2 D 2n 2u2 D 2n Z0 n 1 n n 1 2n.2n 1/ XD n XD n Â Ã Â Ã 2n 1 2 : D 2n n 0 .2n 1/2 XD C n Â Ã Remark. If 1 =2 1 sin x G ln C dx D 4 0 1 sin x Z Â Ã has beenproved, applying Wallis’s formula of odd power yields the following alternative proof:

~~=2 1 1 2G .sin x/2n 1 dx D 2n 1 C Z0 n 0 XD C 1 1 2 4 6 .2n/ D 2n 1 3 5 7 .2n 1/ n 0 C C XD 2n 1 2 : D 2n n 0 .2n 1/2 XD C n Â Ã~~

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~~8 We derive the ﬁrst identity by using Mathematica. First we set~~

~~2 1 an bn c ln 2 C C : D 3n n 1 2n XD n Â Ã Feeding the general sum to Mathematica and integrating gives Sum[(a n^2 + b n + c) (3 n + 1) x^n, {n, 1, Infinity}]~~

~~(4ax+4bx+4cx+12ax^2-2bx^2-9cx^2+2ax^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/ =2 arctan x D 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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~~Chapter 19 5(b). Let I denote the required integral. Differentiating I with respect to ˛ yields~~

~~sin ˇx cos ˛x I .˛/ 1 e kx dx: 0 D x Z0 By using the product to sum formula we have~~

~~1 sin.˛ ˇ/x sin.˛ ˇ/x I .˛/ 1 C e kx dx 1 e kx dx : 0 D 2 x x ÂZ0 Z0 Ã Appealing to Example 1, we ﬁnd that~~

~~1 ˛ ˇ ˛ ˇ I .˛/ arctan C arctan C ; 0 D 2 k k Â Ã and so~~

˛ ˇ ˛ ˇ ˛ ˇ ˛ ˇ k k2 .˛ ˇ/2 I C arctan C arctan ln C : D 2 k 2 k C 4 k2 .˛ ˇ/2 C C Remark. Letting k 0, we get ! 1 sin ˛x sin ˇx 2 ˇ; if ˛ ˇ; dx : x x D ˛; if ˛ ˇ Z0 2 Ä n In general, if ˛ > i 1 ˛i and all ˛i >0, then D P 1 sin ˛x sin ˛1x sin ˛nx dx ˛1˛2 ˛n: x x x D 2 Z0 7. Appealing to cos ax cos bx b sin yx dy x D Za and sin yx 1 dx ; x D 2 Z0 we have

cos ax cos bx b sin yx 1 dx 1 dydx x2 D x Z0 Z0 Za b sin yx 1 dxdy .b a/: D x D 2 Za Z0 9. This is a Frullani integral with f.x/ ex =.1 ex / and so the answer is .1=2/ln.p=q/. D C 11. Let f.a/ denote the right side of the equation. Substituting x at gives D a2t 2 1 1 e dt f.a/ 2 : D p 0 1 t Z C By Leibniz’s rule, we have

~~a2t 2 2a 1 a2t 2 1 e dt f 0.a/ e dt 2 : D p 0 0 1 t Z Z C !~~

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~~Appealing to 2 2 p 1 e a t dt ; D 2a Z0 we ﬁnd that f.a/ satisﬁes an initial value problem~~

p f .a/ 2af.a/ 1; f.0/ 0 D D 2 and so 2 2 f.a/ ea 1 e t dt; D Za which is identical with the left side of the equation after substituting t 2 x2 a2. D C 14. Recall that 2n 1 1=2 x J0.x/ : D n .2n/Š n 0 Â Ã XD Appealing to the binomial theorem, we obtain

1 sx 1 1=2 1 1 sx 2n e J0.x/dx e x dx D n .2n/Š Z0 n 0 Â Ã Z0 XD 1 1=2 1 1 e t t 2n dt D n .2n/Šs2n 1 n 0 Â Ã C Z0 XD 1 1=2 .2n 1/ C D n .2n/Šs2n 1 n 0 Â Ã C XD 2 1 1 1=2 1 1 1 1 : D s n s2 D s 2 D p 2 n 0 Â Ã Â Ã 1 .1=s/ 1 s XD C C p 2 Remark. This result indicates that the Laplace transform of J0.x/ is 1=p1 s . C Chapter 20 2. Since all summands are positive, the monotone convergence theorem permits us to interchange the order of summation and integration. Substituting x e t yields D 1 t x x dx 1 ete e t dt D Z0 Z0 1 1 1 t ne nt e t dt D nŠ Z0 n 0 ! XD 1 1 1 t ne .n 1/t dt D nŠ C n 0 Z0 XD 1 1 1 t ne nt dt .shifting the index n by one/ D .n 1/Š n 1 Z0 XD n 1 n 1 n 1 s s e ds .set s nt/ D .n 1/Š 0 D n 1 Z XD n 1 n 1 .n/ n n: D .n 1/Š D n 1 n 1 XD XD

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~~11. Deﬁne 1 2 f.x/ e .n x/ s : D C 1X Clearly, f .x 1/ f.x/; i.e., f is periodic and has period 1. Expanding f as a Fourier series C D~~

~~1 2nxi ane ;~~

~~1X we have~~

1 sx2 2nxi an e e dx D Z1 2 2 1 e sx cos.2nx/dx .see (19.4)/ D Z0 1 n2 =s e ; D r s and so 1 .n x/2 s 1 1 n2 =s 2nxi e C e e : D r s 1X 1X The required identity follows from this equation by setting x 0. D 12. First Proof. Recall the well-known Poisson summation formula, for r <1, j j 2 1 1 r 1 2 rk cos.kx/ : C D 1 2r cos x r2 k 1 C XD Taking r e ˛ ; x ˛y yields D D 2˛ 1 1 1 1 e e ˛k cos.˛yk/ 2 C D 2 1 2e ˛ cos.˛y/ e 2˛ k 1 C XD 1 e2˛ 1 : D 2 1 2e˛ cos.˛y/ e2˛ C On the other hand, the partial fraction gives

~~1 1 1 1 ; 1 .y 2ˇj /2 D 2i y i 2ˇj y i 2ˇj C C Â C C C Ã where i p 1. Thus, the Euler formula D 1 1 cot.x/ D x j j C D1X leads to~~

1 1 1 1 1 1 1 .y 2ˇj /2 D 2i y i 2ˇj y i 2ˇj j j Â C C C Ã D1X C C D1X y i y i cot cot C D 4ˇi 2ˇ 2ˇ Â Â Ã Â Ã Ã ˛ ˛.y i/ ˛.y i/ cot cot C ; .using ˛ˇ /: D 4i 2 2 D Â Â Ã Â ÃÃ

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In view of the trigonometric identity sin.t s/ 2 sin.t s/ cot s cot t D sin s sin t D cos.t s/ cos.t s/ C and sin.ix/ i sinh x; cos.ix/ coshx, we have D D ˛.y i/ ˛.y i/ 2i sinh.˛/ e2˛ 1 cot cot C 2i ; 2 2 D cosh.˛/ cos.˛y/ D 1 2e˛ cos.˛y/ e2˛ Â Ã Â Ã C and so 2˛ 1 1 1 1 e 1 1 1 e ˛k cos.˛yk/: ˛ 1 .y 2ˇj /2 D 2 1 2e˛ cos.˛y/ e2˛ D 2 C j C C C k 1 D1X XD Second Proof. This proof is basedon the Poisson summation formula, which asserts that

1 1 T f.kT/ f.j=T/; .1/ D O k j D1X D1X where the Fourier transform fO of f is deﬁned by f ./ 1 f.x/e 2xi dx: O D Z1 Let f.x/ e x cos.xy/. Then D j j f ./ 1 e x cos.xy/e 2xi dx O D j j Z1 2 1 e x cos.xy/ cos.2x/dx D Z0 1 e x Œcos.y 2/x cos.y 2/xdx D C C Z0 1 1 : D 1 .y 2/2 C 1 .y 2/2 C C C Now, the required identity follows from (1) by choosing T ˛ and some simpliﬁcations. D 16. To eliminate the absolute value, we divide the interval Œ0; / into 1 n ;.n 1/ ; .n 0;1;2;:::/: 2 C 2 D h i When n 2k, D k =2 =2 C ln cos x ln cos t j 2 j dx 2 dt k x D 0 .t k/ I Z Z C and when n 2k 1, D k ln cos x =2 lncos t j 2 j dx 2 dt: k =2 x D 0 .t k/ Z Z Therefore,

=2 1 ln cos x 1 1 1 1 j 2 j dx ln cos t 2 2 2 dt: 0 x D 0 8 t C .t k/ C .t k/ 9 Z Z k 1 Ä C < XD = 2 Noticing the expression in the bracket is: indeed 1= sin t (see Exercise 6.11), integrating; by parts, we ﬁnd that ln cos x =2 ln cos t 1 j j dx dt : x2 D sin2 t D 2 Z0 Z0

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~~17. First, since 1 2n 1 lim B2n 1.t/ cot.t/ C B2n; t 0 C D 2n ! Â Ã the integral converges. Next, appealing to the trigonometric series~~

n 1 2.2n 1/Š 1 sin.2kt/ B2n 1.t/ . 1/ C C .0 t 1 n 1;2;:::/; C D .2/2n 1 k2n 1 Ä Ä I D C k 1 C XD we have 2n 1 1=2 1=2 n 1 .2/ C 1 1 . 1/ C B2n 1.t/ cot.t/dt 2 2n 1 sin.2kt/ cot.t/dt: .2n 1/Š 0 C D k 0 C Z k 1 C Z XD Recall that k sin.2k 1/t C 1 2 cos.2kt/: sin t D C i 1 XD For any positive integer k, we have

=2 sin.2k 1/t C dt ; . / sin t D 2 Z0 and so 1=2 1 =2 sin.2k 1/t =2 sin.2k 1/t 1 sin.2kt/ cot.t/dt C dt dt : D 2 sin t C sin t D 2 Z0 Z0 Z0 ! Thus, we prove the formula as proposed. Remark. Appealing to n 1 cos.2nt/ sin2.nt/ sin.2k 1/t ; D 2 sin t D sin t k 1 XD repeatedly using . / yields =2 sin.nt/ 2 n dt : sin t D 2 Z0 Â Ã Chapter 21 1. Let I denote the required integral. Rewrite I as 1 dx 1 dx I 2 ˛ 2 ˛ : D 0 .1 x /.1 x / C 1 .1 x /.1 x / Z C C Z C C The substitution t 1=x in the second integral yields D 1 dx 1 t ˛ dt I 2 ˛ 2 ˛ : D 0 .1 x /.1 x / C 0 .1 t /.1 t / Z C C Z C C Thus 1 .1 x˛ /dx 1 dx I C2 ˛ 2 ; D 0 .1 x /.1 x / D 0 1 x D 4 Z C C Z C which is independent of ˛. Remark. Let f.x/ be a rational function and let F.x/ f.x/ f .1=x/=x2. Then D C 1 1 f.x/dx F.x/dx: D Z0 Z0

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4. For any real number x, we have 0; if 2x is even; 2x 2 x b c b c b c D 1; if 2x is odd: b c Thus, n 2pn pn Sn 2 WD pk pk k 1 Â Ã XD n 1 .where 2 n=k is odd/ D b c k 1 XD p f .n/ n 1 .where 2 n=k 2j 1/; D b c D C j 1 k 1 XD XD p where f.n/ . 2pn 1/=2 . Next, we see that 2pn 2j 1 if and only if Db b c c b pk c D C 4n 4n

1 1 1 1 lim Sn 4 n n D .2j 1/2 .2j 2/2 !1 j 1 Â C C Ã XD 2 2 1 2 4 1 3: D " 8 ! 24 4 !# D 3 7. Let the proposed integral be I . We use the geometric series, an elementary integral .xy/n 1 .xy/t dt; .0

1 1 n 1 1 1 .xy/ 1 I .1 x/ dxdy 1 .1 x/.xy/t dxdy dt D ln.xy/ D Z0 Z0 n 0 n 0 Zn Z0 Z0 ! XD XD 1 1 1 1 1 n 2 1 dt ln C D .t 1/2 .t 1/.t 2/ D n 1 n 1 n 0 Zn Â Ã n 0 Â Ã XD C C C XD C C lim .HN ln.N 1// : D N C D !1

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~~13. First, integrating by parts yields~~

~~1 1 2.2n 1/ In 1 1 In: D 3n 3n 1 C 3n C Â Ã~~

Appealing to I1 p3=9, we see that D In an bnp3; D C where an and bn are rational numbers. By bn 1 .2.2n 1/=3n/bn, we have C D 1 2n bn 1 : D 3n 2 n C C Â Ã Next, an satisﬁes a recursive relation 1 1 2.2n 1/ an 1 1 an: D 3n 3n 1 C 3n C Â Ã A telescoping process gives

n 2 1 1 k 1 n k .2n 1/.2n 3/ .2n 2k 3/ an 1 1 2 .3 3 / C : C D 3n 3n 1 C n.n 1/ .n k 1/ Â Ã k 2 C XD 15. One possible formulation: by repeated use of the product to sum identity, we rewrite the integral as 2 n 2 cos.a1n1 a2n2 a n /x dx C CC k k Z0 X n where ai 1 for 1 i n and the summation is taken over all the 2 combinations of the D ˙ Ä Ä values of the ai . Appealing to the fact that

~~2 2; if n 0; cos.nx/dx D D 0; if n is any nonzero integer; Z0 the integral is a maximum if the number of solutions .a1; a2;:::;ak/ to the equation~~

~~a1n1 a2n2 a n 0 C CC k k D is a maximum.~~

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~~Index~~

Abel’s continuity theorem, 115 bounds for normal distribution, 241 Abel’s summation formula, 258, 264 absolute integrability, 221 Carleman’s inequality, 15 AM-GM inequality, 1, 2, 4–8, 11, 12, 17, 255, Cassini’s identity, 139 256 Catalan’s constant, 185, 196, 198, 213, 286 Alzer’s proof, 6 Catalan’s identity, 60 Cauchy’s leap-forward fall-back induction, Cauchy product, 62, 145, 169, 272, 276, 282 2 Cauchy-Schwarz inequality, 1, 8–12, 17–19, Hardy’s proof via smoothing transforma- 256, 258 tion, 5 proof by AM-GM inequality, 8 Pólya’s proof, 7 proof by Lagrange’s identity, 10 proof by normalization and orderings, 4 proof by normalization, 9 proof by regular induction, 1 proof via infimums, 18 proof via infimums, 17 Schwarz’s proof via quadratic functions, Apery’s constant, 69, 108 10 Apery-like formula, 102, 183 central binomial coefficients, 113, 114, 119, asymptotic formula, 60, 67, 70, 78, 80, 244 165, 173 for a sum of cosecants,78 characteristic equation, 121 for harmonic numbers, 60, 70, 244 complete elliptic integral of the first kind, 84 Stirling approximation, 67 Cramer’s rule, 143–148, 276 Bernoulli, 55, 61–65, 67, 71, 143, 146, 149, 152, 160, 238 De Moivre’s formula, 40, 43, 45, 74 inequality, 36 digamma function, 114, 117, 248 numbers, 55, 61–65, 67, 143, 146, 149, Dini test, 221 160 numbers via determinants, 143 elliptic integral singular values, 84, 91 polynomials, 63–65, 71, 238, 253 Epstein’s zeta function, 91 polynomials via determinants, 152 Euler, 39, 45, 47, 52, 53, 60, 65, 68, 70, 71, Raabe’s multiplication identity, 71 76, 77, 117, 170, 177, 193, 237, beta function, 114, 202, 247, 284 247, 252, 253, 264 big O notation, 75 constant, 60, 70, 90, 117, 243, 247, 252 Binet’s formula, 57, 121, 123, 127, 135, 192 dilogarithm, 193 binomial coefficients, 62, 106, 160, 163, 201, formula for , 60, 77, 244 202, 208 formula for .2k/, 68 binomial series, 113, 168 formula for .3/, 170 binomial theorem, 112, 125, 135, 270, 290 formula for ei , 39

~~297~~

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~~298 Index~~

functional equation of the zeta function, 134, 137, 154, 156–158, 163, 193, 177 271, 278 infinite product for cosh x, 52 exponential, 56, 57 infinite product for sinh x, 52 for Bernoulli numbers, 61 infinite product for cosine, 47, 264 for Bernoulli polynomials, 63 infinite product for sine, 45, 177, 237 for Fibonacci cubes, 122 partial fraction for cotangent, 68, 76 for Fibonacci numbers, 57 partial fraction for secant, 71 for Fibonacci squares, 122 partial fraction for tangent, 71 for harmonic numbers, 59 Euler formula for .2/, 169, 175–177, 181– for powers of Fibonacci numbers, 125 183, 185, 186 ordinary, 55 proof by complex variables, 183 golden ratio, 58, 128, 203 proof by double integrals, 177, 185 Gosper’s algorithm, 105, 106 proof by Euler, 176 Gosper’s identity, 209, 210 proof by Fourier series, 183 proof by power series, 182 Hölder’s inequality, 20 proof by trigonometric identities, 181 Hadamard product, 253 Wilf’s family of series, 186 Hadamard theorem, 147 Euler sums, 61, 189, 190, 194, 195, 197, 199 harmonic numbers, 55, 58, 59, 61, 69, 152, Euler’s series transformation, 186, 283 283, 284 Euler-Maclaurin summation formula, 65, 66, Hermite-Ostrogradski formula, 212 80, 266 higher transcendent, 201 Eves’s means via a trapezoid, 26 Identities for Fibonacci powers, 131 Fibonaccinumbers,55–59, 69, 101, 107, 118, Identities for Fibonaccisquaresand cubes,131 121, 131, 148, 192 Inverse Symbolic Calculator, 245 Fibonomial coefficients, 131, 133, 137, 140 inverse tangent sums, 100 finite product identities for cosine, 42 Jensen’s inequality, 259 finite product identities for sine, 41, 42 Jordan’s inequality, 35, 75 finite summation identities involving cosine, 44 Knuth’s inequality, 35 Fourier series, 183, 291 Kober’s inequality, 37 Fresnel integrals, 225 Ky Fan’s inequality, 21 Frullani integrals, 226, 289 Fubini’s theorem, 219–222, 228 L’Hôpital monotone rule, 33, 262 L’Hôpital’s monotone rule, 34–36 Gallian’s REU model, 73 L’Hôpital’s rule, 33, 36, 50 gamma function, 83–86, 90, 91, 267 Lagrange’s identity, 9, 10 Chowla-Selberg formula, 84 Laplace integrals, 224 Gauss multiplication formula, 83, 90, 267 Laplace transform, 290 Legendre’s duplication formula, 85 leap-forward fall-back induction, 13 reflection formula, 84, 86, 90, 267, 268 Legendre polynomial, 280 gamma product, 83, 85, 86, 88, 89 Leibniz’s rule, 204, 219, 222, 224, 234, 267, Gauss, 40, 93, 97, 114, 117, 218, 261, 268 289 formula of function, 117 Liouville’s integral, 90 formula of digamma function, 114 Lucas numbers, 123, 125, 126, 138 fundamental theorem of algebra, 40 pairing trick, 93, 97, 268 Mathematica, 53, 74, 79, 119, 130, 132, 139, probability integral, 218 155, 156, 159, 169, 208, 210, 211, quadrature formula, 261 251, 288 Gauss arithmetic-geometric mean, see means Mathieu’s inequality, 81 generating functions, 55–57, 59, 61, 63, 64, mean value theorem, 28 72, 118, 121–126, 128, 129, 133, means, 25–32

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~~Index 299~~

arithmetic mean, 25 power-reduction formula, 124–127 centroidal mean, 25 primitive root of unity, 40 Gauss arithmetic-geometric mean, 12, 31 principle of the argument, 48 general definition, 29 Putnam problems, 14, 30, 51, 52, 90, 107– geometric mean, 25 109, 130, 141, 151, 223, 228, 229, harmonic mean, 13, 25, 26 236, 237, 252, 253, 255, 279 Heronian mean, 25 identric mean, 29, 32 q–binomial coefficient, 141 logarithmic mean, 25, 32 q–Fibonacci polynomials, 141 power mean, 28 Schwab-Schoenberg mean, 14 Ramanujan q-series, 118 slope mean, 30 Ramanujan identity for .3/, 173 Stolarsky mean, 29 Ramanujan problem, 213 Minkowski’s inequality, 22 Ramanujan’s constant G.1/, 198 Monthly problems, 11, 14, 22, 31, 35, 37, 38, random Fibonacci numbers, 58, 72 48, 49, 52, 54, 70, 71, 80, 84, 101, Riemann hypothesis, 61, 72 104, 106, 118, 128, 129, 144, 163, Riemann sums, 49, 231, 232 172, 185, 190, 214, 229, 230, 237, Riemann zeta function, 61, 68, 165, 176, 189, 238, 241, 243–247, 249, 250, 252, 249, 281 253 roots of unity, 40, 109 Rubel and Stolarsky formula, 109 Open problems, 31, 32, 79–81, 90, 91, 108, 118, 130, 187, 215, 253 sampling theorem, 53, 54 Pappus’s construction on means, 25 series multisection formula, 110, 270, 271 parametric differentiation, 217, 222–224, 227– Simpson’s 3=8 rule, 31, 261 229, 231, 234 Sloane’s integer sequences,72, 133, 140, 142 parametric integration, 167, 217–220, 225, 226, smoothing transformation, 5 228 special numbers via determinants, 151 Parseval’s identity, 286 Stirling numbers of the first kind, 70 partial fraction decomposition of 1 zn, 43 Stirling numbers of the second kind, 152 Pell numbers, 126, 127, 138 Stirling’s formula, 67, 244 Pell-Lucas numbers, 127 pentagonal numbers, 116 telescoping sums, 93, 94, 96, 97, 101, 103– Poisson integral, 231, 235 106, 264, 269, 295 Poisson summation formula, 291, 292 theta function, 237 polygonal numbers, 116 trapezoidal rule, 80 polylogarithm function, 193 Triangle inequality, 19 power mean inequality, 13, 272 trigonometric power sums, 45, 153, 154, 156, power series, 67–69, 150, 165, 169, 172, 182, 157, 159, 160, 163, 164 191 alternating sums, 163 for arcsin x, 169, 182 cosecant, 45, 156 2 for arcsin x, 165 cotangent, 159 3 for arcsin x, 169 secant, 45, 154 2n for arcsin x when n 2;3;4, 172 tangent, 157 D for cot x, 67 various, 160 for csc x, 68 trilogarithm function, 251 1 2 1 x for 4 ln 1Cx , 191 trilogarithmic identity, 251 for ln2.1 x/, 69Á for ln3.1 x/, 69 uniform convergence,221 for sinh x=sin x, 150 for tan x, 68 Vandermonde convolution formula, 279 for earcsin x, 172 Vandermonde determinant, 150

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~~300 Index~~

Wallis’s formula, 47, 77, 168, 169, 182, 245, 269, 279, 287 Weierstrass M-test, 221 Weierstrass product formula, 89, 268 weighted arithmetic-geometric mean inequality, 7, 20 Wilker’s inequality, 34, 37 WZ method, 106

~~Young’s inequality, 75~~

~~zero-order Bessel function, 229~~

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~~About the Author~~

Hongwei Chen was born in China, and received his PhD from North Carolina State University in 1991. He is currently a professor of mathematics at Christopher Newport University. He has published more than ﬁfty research articles in classical analysis and partial differential equations.

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