AMS / MAA CLASSROOM RESOURCE MATERIALS VOL 35 i i “master” — 2011/5/9 — 10:51 — page i — #1 i i
10.1090/clrm/035
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 ver- sion 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, Mikl´os Laczkovich Counterexamples in Calculus, Sergiy Klymchuk Creative Mathematics, H. S. Wall Environmental Mathematics in the Classroom, edited by B. A. Fusaro and P. C. Kenschaft Excursions in Classical Analysis: Pathways to Advanced Problem Solving and 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. S´anchez Oval Track and Other Permutation Puzzles, John O. Kiltinen A Primer of Abstract Mathematics, Robert B. Ash Proofs Without Words, Roger B. Nelsen Proofs Without Words II, Roger B. Nelsen She Does Math!, edited by Marla Parker Solve This: Math Activities for Students and Clubs, James S. Tanton Student Manual for Mathematics for Business Decisions Part 1: Probability and 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
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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 ad- vanced 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 expe- rience 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 mathemat- ics by applying induction, recursion, and combinatorics to traditional problems in classical analysis. Advanced problem solving and research involve not only deduction but also experi- mentation, 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 prob- lems in which computing plays a significant role. These experimentally discovered results are indeed based on rigorous mathematics. Two interesting examples are the WZ-method for telescoping in Chapter 9 and the Hermite-Ostrogradski Formula for searching for a new formula for in Chapter 18. In his classic “How to Solve It”, George P´olya provides a list of guidelines for solving mathematical problems: learn to understand the problem; devise a plan to solve the prob- lem; 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 exist- ing 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 in- cludes 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 mathemat- ics 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 prob- lem 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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Solutions to Selected Problems
Here the solutions do not include any Monthly and Putnam problems. For Monthly prob- lems, 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
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256 Excursions in Classical Analysis
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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258 Excursions in Classical Analysis
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  C à XD n 1 1 1 1 1 1 sn s1 kak 1 n C 2 C 2 k k 1 C k 1  C à 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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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  à 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  Pk 1 k à 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  k à 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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260 Excursions in Classical Analysis
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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Solutions to Selected Exercises 261
10. First, since S˛ .a;b/ is strictly increasing in ˛, for 0 ax bx 1=x a b C C 2 2  à 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/  à !  à ! 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 262 — #276 i i 262 Excursions in Classical Analysis 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  à . 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;0 sin kx =2 sin.x=2/ 2k sin.2k 1/t 2 dt: k D sin t sin t Zx=2  à 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  à XD XD n 2k 1 The proposed inequality follows from k 1 ak r sin.2k 1/t > 0 for 0 i i i i i i “master” — 2011/5/9 — 10:51 — page 263 — #277 i i 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   Ãà 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). i i i i i i “master” — 2011/5/9 — 10:51 — page 264 — #278 i i 264 Excursions in Classical Analysis 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  C à 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  à  à 1 1 1 1 2 1 D k2 C k C 2 CC k 1  à 1 2 1 H D k2 C k k k  à 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  à  à 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  à  à H2 H3 Hn 1 1 1 >2 1 2 C 3 CC n C 1 2 2 3 C .n 1/ n  à  à H H H 1 2 2 3 n D 2 C 3 CC n C n  à as desired. Note that telescoping is used in the last summation again. i i i i i i “master” — 2011/5/9 — 10:51 — page 265 — #279 i i 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  C C à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 266 — #280 i i 266 Excursions in Classical Analysis 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  à 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  à 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  à 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 Z 1 !1 Z 1 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 for- mula 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  à 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  à XD n 1 n 1 1 k : D 2k C 2k.n k/ k 0 k 1 XD XD i i i i i i “master” — 2011/5/9 — 10:51 — page 267 — #281 i i 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  à i 1  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  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 268 — #282 i i 268 Excursions in Classical Analysis 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  à k 0  à XD XD n n .cos kx sin.n k/x cos.n k/x sin kx/ D k C k 0  à XD n n sin nx 2n sin nx: D k D k 0  à XD Therefore S 2n 1 sin nx: D i i i i i i “master” — 2011/5/9 — 10:51 — page 269 — #283 i i 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  à 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  à 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  à k 0  C à k 1  à XD XD XD We have 1 2 1 arctan .arctan.2k 2/ arctan.2k// ; .2k 1/2 D C D 2 k 0  C à k 0 XD XD 1 2 1 arctan .arctan.2k 1/ arctan.2k 1// ; .2k/2 D C D 4 k 1  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 270 — #284 i i 270 Excursions in Classical Analysis 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  à 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  à 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  à 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  à 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  à 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  à 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  à XD C C i i i i i i “master” — 2011/5/9 — 10:51 — page 271 — #285 i i Solutions to Selected Exercises 271 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  à 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 Ä à  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 272 — #286 i i 272 Excursions in Classical Analysis 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  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 273 — #287 i i Solutions to Selected Exercises 273 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  à YD Since cos z x=.2i/, we get D n 1 k fn.x/ x 2i cos ; D n k 1  à YD and so n 1 k Fn fn.1/ 1 2i cos : D D n k 1  à 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 2ik 1 ik i1 ik n ik a b CC x : D i1 i2 ik i1;iX2;:::;ik  à  à  à Notice that n i xi .ax b/n i an i j bj xn j ; C D j 0 j n i  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 274 — #288 i i 274 Excursions in Classical Analysis 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;:::;ikC1 Ã à  C à 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 2ik 1 ik i1 ik n i ik a b CC x C : D i1 i2 ik i;i1;iX2 ;:::;ik  Ã à  à 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  à  à C CXC D i n i i s s n i s s .buk 1/ ukuk 1 .buk/ : D s s i s n  à  à 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  à i s n  à 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  à 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  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 275 — #289 i i Solutions to Selected Exercises 275 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 276 — #290 i i 276 Excursions in Classical Analysis 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 i 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 277 — #291 i i Solutions to Selected Exercises 277 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  à 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  à XD XD a 2a n 1 . 1/ 2a e2ix.a j/ !bk.a j/: D 22a j j 0  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 278 — #292 i i 278 Excursions in Classical Analysis 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  à 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  à 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  à 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  à j >0  à X XD X i i i i i i “master” — 2011/5/9 — 10:51 — page 279 — #293 i i Solutions to Selected Exercises 279 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  à  à 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  C à  à  C à 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  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 280 — #294 i i 280 Excursions in Classical Analysis 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 281 — #295 i i Solutions to Selected Exercises 281 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. i i i i i i “master” — 2011/5/9 — 10:51 — page 282 — #296 i i 282 Excursions in Classical Analysis 10. Dividing (15.15) by gives 2n 1 1 .2n/ 1 1 cot : 2n D 2 n 1  à 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  à  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 283 — #297 i i Solutions to Selected Exercises 283 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  à 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  à k 1  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 284 — #298 i i 284 Excursions in Classical Analysis 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  à 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  C à 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  C à DX¤ 1 1 1 1 1 D 2i 1 i CC .i 1/ i C .i 1/ i CC N i  C à 1 1 1 1 1 C 2i 1 i CC .i 1/ i C .i 1/ i CC N i  C C C C C à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 285 — #299 i i Solutions to Selected Exercises 285 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  !1 à !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 for- mula 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  C à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 286 — #300 i i 286 Excursions in Classical Analysis 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  à 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  à 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. i i i i i i “master” — 2011/5/9 — 10:51 — page 287 — #301 i i Solutions to Selected Exercises 287 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  à  à m 1 1 : D 2 2mn 2m 2 n 0 .n 1/.2mn 2m 1/ C XD C C mn 1  à 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  à 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  à  à 2n 1 2 : D 2n n 0 .2n 1/2 XD C n  à Remark. If 1 =2 1 sin x G ln C dx D 4 0 1 sin x Z  à 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  à i i i i i i “master” — 2011/5/9 — 10:51 — page 288 — #302 i i 288 Excursions in Classical Analysis 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  à 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)}} i i i i i i “master” — 2011/5/9 — 10:51 — page 289 — #303 i i Solutions to Selected Exercises 289 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 à Appealing to Example 1, we find that 1 ˛ ˇ ˛ ˇ I .˛/ arctan C arctan C ; 0 D 2 k k  à 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 ! i i i i i i “master” — 2011/5/9 — 10:51 — page 290 — #304 i i 290 Excursions in Classical Analysis 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  à 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  à Z0 XD 1 1=2 1 1 e t t 2n dt D n .2n/Šs2n 1 n 0  à C Z0 XD 1 1=2 .2n 1/ C D n .2n/Šs2n 1 n 0  à C XD 2 1 1 1=2 1 1 1 1 : D s n s2 D s 2 D p 2 n 0  à  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 291 — #305 i i Solutions to Selected Exercises 291 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 Z 1 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  C C C à where i p 1. Thus, the Euler formula D 1 1 cot.x/ D x j j C D 1X leads to 1 1 1 1 1 1 1 .y 2ˇj /2 D 2i y i 2ˇj y i 2ˇj j j  C C C à D 1X C C D 1X y i y i cot cot C D 4ˇi 2ˇ 2ˇ   à  à à ˛ ˛.y i/ ˛.y i/ cot cot C ; .using ˛ˇ /: D 4i 2 2 D   à  Ãà i i i i i i “master” — 2011/5/9 — 10:51 — page 292 — #306 i i 292 Excursions in Classical Analysis 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˛  à  à 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 D 1X 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 D 1X D 1X where the Fourier transform fO of f is defined by f ./ 1 f.x/e 2xi dx: O D Z 1 Let f.x/ e x cos.xy/. Then D j j f ./ 1 e x cos.xy/e 2xi dx O D j j Z 1 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 293 — #307 i i Solutions to Selected Exercises 293 17. First, since 1 2n 1 lim B2n 1.t/ cot.t/ C B2n; t 0 C D 2n !  à 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  à 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 294 — #308 i i 294 Excursions in Classical Analysis 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  à 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  C C à 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  à n 0  à XD C C C XD C C lim .HN ln.N 1// : D N C D !1 i i i i i i “master” — 2011/5/9 — 10:51 — page 295 — #309 i i Solutions to Selected Exercises 295 13. First, integrating by parts yields 1 1 2.2n 1/ In 1 1 In: D 3n 3n 1 C 3n C  à 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  à Next, an satisfies a recursive relation 1 1 2.2n 1/ an 1 1 an: D 3n 3n 1 C 3n C  à 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/  à 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. i i i i i i “master” — 2011/5/9 — 10:51 — page 296 — #310 i i i i i i i i “master” — 2011/5/9 — 10:51 — page 297 — #311 i i 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 P´olya’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 i i i i i i “master” — 2011/5/9 — 10:51 — page 298 — #312 i i 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 H¨older’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’Hˆopital monotone rule, 33, 262 L’Hˆopital’s monotone rule, 34–36 Gallian’s REU model, 73 L’Hˆopital’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 i i i i i i “master” — 2011/5/9 — 10:51 — page 299 — #313 i i 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 i i i i i i “master” — 2011/5/9 — 10:51 — page 300 — #314 i i 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 inequal- ity, 7, 20 Wilker’s inequality, 34, 37 WZ method, 106 Young’s inequality, 75 zero-order Bessel function, 229 i i i i i i “master” — 2011/5/9 — 10:51 — page 301 — #315 i i 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 pub- lished more than fifty research articles in classical analysis and partial differential equations. 301 i i i i AMS / MAA CLASSROOM RESOURCE MATERIALS Excursions in Classical Analysis will introduce students to advanced problem solving and undergraduate research in two ways: it will provide a tour of classical analysis, showcasing a wide variety of problems that are placed in historical context, and it will help students gain mastery of mathematical discovery and proof. The author presents a variety of solutions for the problems in the book. Some solutions reach back to the work of mathematicians like Leonhard Euler while others connect to other beautiful parts of mathematics. Readers will frequently see problems solved by using an idea that might at first glance, might not even seem to apply to that problem. Other solutions employ a specific technique that can be used to solve many different kinds of problems. Excursions empha- sizes the rich and elegant interplay between continuous and discrete math- ematics by applying induction, recursion, and combinatorics to traditional problems in classical analysis. The book will be useful in students' preparations for mathematics competi- tions, in undergraduate reading courses and seminars, and in analysis courses as a supplement. The book is also ideal for self study, since the chapters are independent of one another and may be read in any order. … Based on his obviously very rich and far-reaching experience in this didactic realm, the author offers a colorful panorama of various topics in calculus, both elementary and advanced, as well as a wide variety of typical problems placed in their respec- tive historical contexts. … No doubt, this fine book will be of great use and value for students preparing for mathematics competitions, participating in undergraduate analysis courses, seminars, and research projects, or conducting any kind of self-study in the field. —Zentrallblatt Math Chen (Christopher Newport Univ) offers many enjoyable trips through classical anal- ysis, providing wonderful insights and remarkable connections. The focus is problem solving, while modeling various aspects of discovery, proof, and multiple solutions. … —CHOICE Magazine Chen's book is a wonderful tour of classical analysis and would serve as an excellent source of undergraduate enrichment/research problems. It recalls the type of gems in classical analysis, number theory, and combinatorics I first encountered in the books of Polya and Szego˝ as an undergraduate many years ago. Peruse the table of contents and see if some of the topics and subtopics don't grab you. —Henry Ricardo, MAA Reviews