Special Functions 9780521789882, 9780521623216

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Special Functions Special functions, which include the trigonometric functions, have been used for centuries. Their role in the solution of differential equations was exploited by Newton and Leibniz, and the subject of special functions has been in continuous development ever since. In just the past thirty years several new special functions and applications have been discovered. This treatise presents an overview of the area of special functions, focusing primarily on the hypergeometric functions and the associated hypergeometric series. It includes both important historical results and recent developments and shows how these arise from several areas of mathematics and mathematical physics. Particular emphasis is placed on formulas that can be used in computation. The book begins with a thorough treatment of the gamma and beta functions, which are essential to understanding hypergeometric functions. Later chapters discuss Bessel functions, orthogonal polynomials and transformations, the Selberg integral and its applications, spherical harmonics, g-series, partitions, and Bailey chains. This clear, authoritative work will be a lasting reference for students and researchers in number theory, algebra, combinatorics, differential equations, mathematical computing, and mathematical physics. George E. Andrews is Evan Pugh Professor of Mathematics at The Pennsylvania State University. Richard Askey is Professor of Mathematics at the University of WisconsinMadison. Ranjan Roy is Professor of Mathematics at Beloit College in Wisconsin.

ENCYCLOPEDIA OF MATHEMATICS AND ITS APPLICATIONS

EDITED BY G.-C. ROTA Editorial Board B. Doran, M. Ismail, T.-Y. Lam, E. Lutwak Volume 71

Special Functions 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 46 47 48 49 50 51 52 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

N. H. Bingham, C. M. Goldie, and J. L. Teugels Regular Variation P. P. Petrushev and V. A. Popov Rational Approximation of Real Functions N. White (ed.) Combinatorial Geometries M. Pohst and H. Zassenhaus Algorithmic Algebraic Number Theory J. Aczel and J. Dhombres Functional Equations in Several Variables M. Kuczma, B. Choczewski, and R. Ger Iterative Functional Equations R. V. Ambartzumian Factorization Calculus and Geometric Probability G. Gripenberg, S .-O. Londen, and O. Staffans Volterra Integral and Functional Equations G. Gasper and M. Rahman Basic Hypergeometric Series E. Torgersen Comparison of Statistical Experiments A. Neumaier Interval Methods for Systems of Equations N. Korneichuk Exact Constants in Approximation Theory R. Brualdi and H. Ryser Combinatorial Matrix Theory N. White (ed.) Matroid Applications S. Sakai Operator Algebras in Dynamical Systems W. Hodges Basic Model Theory H. Stahl and V. Totik General Orthogonal Polynomials G. Da Prato and J. Zabczyk Stochastic Equations in Infinite Dimensions A. Bjorner et al. Oriented Matroids G. Edgar and L. Sucheston Stopping Times and Directed Processes C. Sims Computation with Finitely Presented Groups T. Palmer Banach Algebras and the General Theory of *-Algebras F. Borceux Handbook of Categorical Algebra I F. Borceux Handbook of Categorical Algebra II F. Borceux Handbook of Categorical Algebra III A. Katok and B. Hasselblatt Introduction to the Modern Theory of Dynamical Systems V. N. Sachkov Combinatorial Methods in Discrete Mathematics V. N. Sachkov Probabilistic Methods in Discrete Mathematics P. M. Cohn Skew Fields R. Gardner Geometric Topography G. A. Baker Jr. and P. Graves-Morris Pade Approximants J. Krajicek Bounded Arithmetic, Propositional Logic, and Complexity Theory H. Groemer Geometric Applications of Fourier Series and Spherical Harmonics H. O. Fattorini Infinite Dimensional Optimization and Control Theory A. C. Thompson Minkowski Geometry R. B. Bapat and T. E. S. Raghavan Nonnegative Matrices with Applications K. Engel Sperner Theory D. Cverkovic, P. Rowlinson, S. Simic Eigenspaces of Graphs F. Bergeron, G. Labelle, and P. Leroux Combinatorial Species and Tree-Like Structures R. Goodman and N. Wallach Representations and Invariants of the Classical Groups

ENCYCLOPEDIA OF MATHEMATICS AND ITS APPLICATIONS

Special Functions GEORGE E. ANDREWS

RICHARD ASKEY

CAMBRIDGE UNIVERSITY PRESS

RANJAN ROY

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi, Dubai, Tokyo, Mexico City Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521789882 © George E. Andrews, Richard Askey, and Ranjan Roy 1999 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1999 First paperback edition 2000 A catalogue record for this publication is available from the British Library ISBN 978-0-521-62321-6 Hardback ISBN 978-0-521-78988-2 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work are correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter.

To Leonard Carlitz, Om Prakash Juneja, and Irwin Kra

Contents

Preface

page xiii

1

The Gamma and Beta Functions 1.1 The Gamma and Beta Integrals and Functions 1.2 The Euler Reflection Formula 1.3 The Hurwitz and Riemann Zeta Functions 1.4 Stirling's Asymptotic Formula 1.5 Gauss's Multiplication Formula for Y(mx) 1.6 Integral Representations for Log F(x) and x//(x) 1.7 Kummer's Fourier Expansion of Log V (x) 1.8 Integrals of Dirichlet and Volumes of Ellipsoids 1.9 The Bohr-Mollerup Theorem 1.10 Gauss and Jacobi Sums 1.11 A Probabilistic Evaluation of the Beta Function 1.12 The /7-adic Gamma Function Exercises

2

The Hypergeometric Functions 2.1 The Hypergeometric Series 2.2 Euler's Integral Representation 2.3 The Hypergeometric Equation 2.4 The Barnes Integral for the Hypergeometric Function 2.5 Contiguous Relations 2.6 Dilogarithms 2.7 Binomial Sums 2.8 Dougall's Bilateral Sum 2.9 Fractional Integration by Parts and Hypergeometric Integrals Exercises

Vll

1 2 9 15 18 22 26 29 32 34 36 43 44 46 61 61 65 73 85 94 102 107 109 111 114

viii

Contents

3

Hypergeometric Transformations and Identities 3.1 Quadratic Transformations 3.2 The Arithmetic-Geometric Mean and Elliptic Integrals 3.3 Transformations of Balanced Series 3.4 Whipple's Transformation 3.5 Dougall's Formula and Hypergeometric Identities 3.6 Integral Analogs of Hypergeometric Sums 3.7 Contiguous Relations 3.8 The Wilson Polynomials 3.9 Quadratic Transformations - Riemann's View 3.10 Indefinite Hypergeometric Summation 3.11 The W-Z Method 3.12 Contiguous Relations and Summation Methods Exercises

124 125 132 140 143 147 150 154 157 160 163 166 174 176

4

Bessel Functions and Confluent Hypergeometric Functions 4.1 The Confluent Hypergeometric Equation 4.2 Barnes's Integral for XFX 4.3 Whittaker Functions 4.4 Examples of i F\ and Whittaker Functions 4.5 Bessel's Equation and Bessel Functions 4.6 Recurrence Relations 4.7 Integral Representations of Bessel Functions 4.8 Asymptotic Expansions 4.9 Fourier Transforms and Bessel Functions 4.10 Addition Theorems 4.11 Integrals of Bessel Functions 4.12 The Modified Bessel Functions 4.13 Nicholson's Integral 4.14 Zeros of Bessel Functions 4.15 Monotonicity Properties of Bessel Functions 4.16 Zero-Free Regions for i F\ Functions Exercises

187 188 192 195 196 199 202 203 209 210 213 216 222 223 225 229 231 234

5

Orthogonal Polynomials 5.1 Chebyshev Polynomials 5.2 Recurrence 5.3 Gauss Quadrature 5.4 Zeros of Orthogonal Polynomials 5.5 Continued Fractions

240 240 244 248 253 256

Contents

5.6 5.7 5.8

Kernel Polynomials Parseval's Formula The Moment-Generating Function Exercises

ix

259 263 266 269

6

Special Orthogonal Polynomials 6.1 Hermite Polynomials 6.2 Laguerre Polynomials 6.3 Jacobi Polynomials and Gram Determinants 6.4 Generating Functions for Jacobi Polynomials 6.5 Completeness of Orthogonal Polynomials 6.6 Asymptotic Behavior of P^(x) for Large n 6.7 Integral Representations of Jacobi Polynomials 6.8 Linearization of Products of Orthogonal Polynomials 6.9 Matching Polynomials 6.10 The Hypergeometric Orthogonal Polynomials 6.11 An Extension of the Ultraspherical Polynomials Exercises

277 278 282 293 297 306 310 313 316 323 3 30 3 34 339

7

Topics in Orthogonal Polynomials 7.1 Connection Coefficients 7.2 Rational Functions with Positive Power Series Coefficients 7.3 Positive Polynomial Sums from Quadrature and Vietoris 's Inequality 7.4 Positive Polynomial Sums and the Bieberback Conjecture 7.5 A Theorem of Turan 7.6 Positive Summability of Ultraspherical Polynomials 7.7 The Irrationality of ? (3) Exercises

355 356 363

8

The Selberg Integral and Its Applications 8.1 Selberg's and Aomoto's Integrals 8.2 Aomoto's Proof of Selberg's Formula 8.3 Extensions of Aomoto's Integral Formula 8.4 Anderson's Proof of Selberg's Formula 8.5 A Problem of Stieltjes and the Discriminant of a Jacobi Polynomial 8.6 Siegel's Inequality 8.7 The Stieltjes Problem on the Unit Circle 8.8 Constant-Term Identities 8.9 Nearly Poised 3 F2 Identities 8.10 The Hasse-Davenport Relation

371 381 384 388 391 395 401 402 402 407 411 415 419 425 426 428 430

Contents 8.11

A Finite-Field Analog of Selberg s Integral Exercises

Spherical Harmonics 9.1 Harmonic Polynomials 9.2 The Laplace Equation in Three Dimensions 9.3 Dimension of the Space of Harmonic Polynomials of Degree k 9.4 Orthogonality of Harmonic Polynomials 9.5 Action of an Orthogonal Matrix 9.6 The Addition Theorem 9.7 The Funk-Hecke Formula 9.8 The Addition Theorem for Ultraspherical Polynomials 9.9 The Poisson Kernel and Dirichlet Problem 9.10 Fourier Transforms 9.11 Finite-Dimensional Representations of Compact Groups 9.12 The Group SU(2) 9.13 Representations of SU(2) 9.14 Jacobi Polynomials as Matrix Entries 9.15 An Addition Theorem 9.16 Relation of SU(2) to the Rotation Group SO(3) Exercises

434 439 445 445 447 449 451 452 454 458 459 463 464 466 469 471 473 474 476 478

10 Introduction to ^r-Series 10.1 The q -Integral 10.2 The g-Binomial Theorem 10.3 The q -Gamma Function 10.4 The Triple Product Identity 10.5 Ramanujan's Summation Formula 10.6 Representations of Numbers as Sums of Squares 10.7 Elliptic and Theta Functions 10.8 g-Beta Integrals 10.9 Basic Hypergeometric Series 10.10 Basic Hypergeometric Identities 10.11 g-Ultraspherical Polynomials 10.12 Mellin Transforms Exercises

481 485 487 493 496 501 506 508 513 520 523 527 532 542

11 Partitions 11.1 Background on Partitions 11.2 Partition Analysis 11.3 A Library for the Partition Analysis Algorithm

553 553 555 557

Contents 11.4 11.5 11.6 11.7

Generating Functions Some Results on Partitions Graphical Methods Congruence Properties of Partitions Exercises

12 Bailey Chains 12.1 Rogers's Second Proof of the Rogers-Ramanujan Identities 12.2 Bailey's Lemma 12.3 Watson's Transformation Formula 12.4 Other Applications Exercises

xi 559 563 565 569 573 577 577 582

586 589 590

Infinite Products A.I Infinite Products Exercises

595 595 597

Summability and Fractional Integration B.I Abel and Cesaro Means B.2 The Cesaro Means (C, a) B.3 Fractional Integrals B.4 Historical Remarks Exercises

599 602 604 605 607

C

Asymptotic Expansions C. 1 Asymptotic Expansion C.2 Properties of Asymptotic Expansions C.3 Watson's Lemma C .4 The Ratio of Two Gamma Functions Exercises

611 611 612 614 615 616

D

Euler-Maclaurin Summation Formula D.I Introduction D.2 The Euler-Maclaurin Formula D.3 Applications D.4 The Poisson Summation Formula Exercises

617 617 619 621 623 627

E

Lagrange Inversion Formula E. 1 Reversion of Series E.2 A Basic Lemma E.3 Lambert's Identity E.4 Whipple's Transformation Exercises

629 629 630 631 632 634

B

599

xii

F

Contents

Series Solutions of Differential Equations F.I Ordinary Points F.2 Singular Points F.3 Regular Singular Points

Bibliography Index Subject Index Symbol Index

637 637 638 639 641 655 659 661

Preface

Paul Turan once remarked that special functions would be more appropriately labeled "useful functions." Because of their remarkable properties, special functions have been used for centuries. For example, since they have numerous applications in astronomy, trigonometric functions have been studied for over a thousand years. Even the series expansions for sine and cosine (and probably the arc tangent) were known to Madhava in the fourteenth century. These series were rediscovered by Newton and Leibniz in the seventeenth century. Since then, the subject of special functions has been continuously developed, with contributions by a host of mathematicians, including Euler, Legendre, Laplace, Gauss, Kummer, Eisenstein, Riemann, and Ramanujan. In the past thirty years, the discoveries of new special functions and of applications of special functions to new areas of mathematics have initiated a resurgence of interest in this field. These discoveries include work in combinatorics initiated by Schiitzenberger and Foata. Moreover, in recent years, particular cases of long familiar special functions have been clearly defined and applied as orthogonal polynomials. As a result of this prolific activity and long history one is pulled different directions when writing a book on special functions. First, there are important results from the past that must be included because they are so useful. Second, there are recent developments that should be brought to the attention of those who could use them. One also would wish to help educate the new generation of mathematicians and scientists so that they can further develop and apply this subject. We have tried to do all this, and to include some of the older results that seem to us to have been overlooked. However, we have slighted some of the very important recent developments because a book that did them justice would have to be much longer. Fortunately, specialized books dealing with some of these developments have recently appeared: Petkovsek, Wilf, and Zeilberger [1996], Macdonald [1995], Heckman and Schlicktkrull [1994], and Vilenkin and Klimyk

xiv

Preface

[1992]. Additionally, I. G. Macdonald is writing a new book on his polynomials in several variables and A. N. Kirillov is writing on /^-matrix theory. It is clear that the amount of knowledge about special functions is so great that only a small fraction of it can be included in one book. We have decided to focus primarily on the best understood class of functions, hypergeometric functions, and the associated hypergeometric series. A hypergeometric series is a series Ea n with an+i/an a rational function of n. Unfortunately, knowledge of these functions is not as widespread as is warranted by their importance and usefulness. Most of the power series treated in calculus are hypergeometric, so some facts about them are well known. However, many mathematicians and scientists who encounter such functions in their work are unaware of the general case that could simplify their work. To them a Bessel function and a parabolic cylinder function are types of functions different from the 3 — j or 6 — j symbols that arise in quantum angular momentum theory. In fact these are all hypergeometric functions and many of their elementary properties are best understood when considered as such. Several important facts about hypergeometric series were first found by Euler and an important identity was discovered by Pfaff, one of Gauss's teachers. However, it was Gauss himself who fully recognized their significance and gave a systematic account in two important papers, one of which was published posthumously. One reason for his interest in these functions was that the elementary functions and several other important functions in mathematics are expressible in terms of hypergeometric functions. A half century after Gauss, Riemann developed hypergeometric functions from a different point of view, which made available the basic formulas with a minimum of computation. Another approach to hypergeometric functions using contour integrals was presented by the English mathematician E. W. Barnes in the first decade of this century. Each of these different approaches has its advantages. Hypergeometric functions have two very significant properties that add to their usefulness: They satisfy certain identities for special values of the function and they have transformation formulas. We present many applications of these properties. For example, in combinatorial analysis hypergeometric identities classify single sums of products of binomial coefficients. Further, quadratic transformations of hypergeometric functions give insight into the relationship (known to Gauss) of elliptic integrals to the arithmetic-geometric mean. The arithmetic-geometric mean has recently been used to compute n to several million decimal places, and earlier it played a pivotal role in Gauss's theory of elliptic functions. The gamma function and beta integrals dealt with in thefirstchapter are essential to understanding hypergeometric functions. The gamma function was introduced into mathematics by Euler when he solved the problem of extending the factorial function to all real or complex numbers. He could not have foreseen the extent of its importance in mathematics. There are extensions of gamma and beta functions

Preface

xv

that are also very important. The text contains a short treatment of Gauss and Jacobi sums, which are finite field analogs of gamma and beta functions. Gauss sums were encountered by Gauss in his work on the constmctibility of regular polygons where they arose as "Lagrange resolvents " a concept used by Lagrange to study algebraic equations. Gauss understood the tremendous value of these sums for number theory. We discuss the derivation of Fermat's theorem on primes of the form An + 1 from a formula connecting Gauss and Jacobi sums, which is analogous to Euler's famous formula relating beta integrals with gamma functions. There are also multidimensional gamma and beta integrals. The first of these was introduced by Dirichlet, though it is really an iterated version of the onedimensional integral. Genuine multidimensional gamma and beta functions were introduced in the 1930s, by both statisticians and number theorists. In the early 1940s, Atle Selberg found a very important multidimensional beta integral in the course of research in entire functions. However, owing to the Second World War and the fact that thefirststatement and also the proof appeared in journals that were not widely circulated, knowledge of this integral before the 1980s was restricted to a few people around the world. We present two different evaluations of Selberg's integral as well as some of its uses. In addition to the above mentioned extensions, there are g-extensions of the gamma function and beta integrals that are very fundamental because they lead to basic hypergeometric functions and series. These are series Ec n where cn+\/cn is a rational function of qn for a fixed parameter q. Here the sum may run over all integers, instead of only nonnegative ones. One important example is the theta function Yl°?oo ^w ^ n - This and other similar series were used by Gauss and Jacobi to study elliptic and elliptic modular functions. Series of this sort are very useful in many areas of combinatorial analysis, a fact already glimpsed by Euler and Legendre, and they also arise in some branches of physics. For example, the work of the physicist R. J. Baxter on the Yang-Baxter equation led a group in St. Petersburgh to the notion of a quantum group. Independently, M. Jimbo in Japan was led by a study of Baxter's work to a related structure. Many basic hypergeometric series (or g-hypergeometric series), both polynomials and infinite series, can be studied using Hopf algebras, which make up quantum groups. Unfortunately, we could not include this very important new approach to basic series. It was also not possible to include results on the multidimensional U(n) generalizations of theorems on basic series, which have been studied extensively in recent years. For some of this work, the reader may refer to Milne [1988] and Milne and Lilly [1995]. We briefly discuss the qgamma function and some important g-beta integrals; we show that series and products that arise in this theory have applications in number theory, combinatorics, and partition theory. We highlight the method of partition analysis.

xvi

Preface

P. A. MacMahon, who developed this powerful technique, devoted several chapters to it in his monumental Combinatory Analysis, but its significance was not realized until recently. The theory of special functions with its numerous beautiful formulas is very well suited to an algorithmic approach to mathematics. In the nineteenth century, it was the ideal of Eisenstein and Kronecker to express and develop mathematical results by means of formulas. Before them, this attitude was common and best exemplified in the works of Euler, Jacobi, and sometimes Gauss. In the twentieth century, mathematics moved from this approach toward a more abstract and existential method. In fact, agreeing with Hardy that Ramanujan came 100 years too late, Littlewood once wrote that "the great day of formulae seem to be over" (see Littlewood [1986, p. 95]). However, with the advent of computers and the consequent reemergence of computational mathematics, formulas are now once again playing a larger role in mathematics. We present this book against this background, pointing out that beautiful, interesting, and important formulas have been discovered since Ramanujan's time. These formulas are proving fertile and fruitful; we suggest that the day of formulas may be experiencing a new dawn. Finally, we hope that the reader finds as much pleasure studying the formulas in this book as we have found in explaining them. We thank the following people for reading and commenting on various chapters during the writing of the book: Bruce Berndt, David and Gregory Chudnovsky, George Gasper, Warren Johnson, and Mizan Rahman. Special thanks to Mourad Ismail for encouragement and many detailed suggestions for the improvement of this book. We are also grateful to Dee Frana and Diane Reppert for preparing the manuscript with precision, humor, and patience.

The Gamma and Beta Functions

Euler discovered the gamma function, F(x), when he extended the domain of the factorial function. Thus F(x) is a meromorphic function equal to (x — 1)! when x is a positive integer. The gamma function has several representations, but the two most important, found by Euler, represent it as an infinite integral and as a limit of a finite product. We take the second as the definition. Instead of viewing the beta function as a function, it is more illuminating to think of it as a class of integrals - integrals that can be evaluated in terms of gamma functions. We therefore often refer to beta functions as beta integrals. In this chapter, we develop some elementary properties of the beta and gamma functions. We give more than one proof for some results. Often, one proof generalizes and others do not. We briefly discuss the finite field analogs of the gamma and beta functions. These are called Gauss and Jacobi sums and are important in number theory. We show how they can be used to prove Fermat's theorem that a prime of the form An + 1 is expressible as a sum of two squares. We also treat a simple multidimensional extension of a beta integral, due to Dirichlet, from which the volume of an n-dimensional ellipsoid can be deduced. We present an elementary derivation of Stirling's asymptotic formula for n! but give a complex analytic proof of Euler's beautiful reflection formula. However, two real analytic proofs due to Dedekind and Herglotz are included in the exercises. The reflection formula serves to connect the gamma function with the trigonometric functions. The gamma function has simple poles at zero and at the negative integers, whereas esc nx has poles at all the integers. The partial fraction expansions of the logarithmic derivatives of F(JC) motivate us to consider the Hurwitz and Riemann zeta functions. The latter function is of fundamental importance in the theory of distribution of primes. We have included a short discussion of the functional equation satisfied by the Riemann zeta function since it involves the gamma function. In this chapter we also present Kummer's proof of his result on the Fourier expansion of log T(x). This formula is useful in number theory. The proof given

2

1 The Gamma and Beta Functions

uses Dirichlet's integral representations of log F(x) and its derivative. Thus, we have included these results of Dirichlet and the related theorems of Gauss.

1.1 The Gamma and Beta Integrals and Functions The problem of finding a function of a continuous variable x that equals n! when x = n, an integer, was investigated by Euler in the late 1720s. This problem was apparently suggested by Daniel Bernoulli and Goldbach. Its solution is contained in Euler's letter of October 13,1729, to Goldbach. See Fuss [1843, pp. 1-18]. To arrive at Euler's generalization of the factorial, suppose that x > 0 and n > 0 are integers. Write

^ ± ^ (* + !)»

(1.1.1)

where (a)n denotes the shifted factorial defined by (fl) B =a(a + l)---(a + i i - l )

forn > 0, (a) 0 = 1,

(1.1.2)

and a is any real or complex number. Rewrite (1.1.1) as

Since hm

= 1,

x

we conclude that nln

*

jc!=Jm^

.

(1.1.3)

Observe that, as long as x is a complex number not equal to a negative integer, the limit in (1.1.3) exists, for n\nx

(

n

V

TT-

and ^

x(x -

1.1 The Gamma and Beta Integrals and Functions Therefore, the infinite product

n( 1+ 7) 0+7 converges and the limit (1.1.3) exists. (Readers who are unfamiliar with infinite products should consult Appendix A.) Thus we have a function I I ( J C ) = lim

'

(1.1.4)

fc^oo (x + 1)* defined for all complex x ^ — 1, —2, — 3 , . . . and Ti(n) = n\. Definition 1.1.1 For all complex numbers x ^= 0, — 1, — 2 , . . . , the gamma function T(x) is defined by k\kx~l lim ——-.

T(x)=

k^oc

(1.1.5)

(x)k

An immediate consequence of Definition 1.1.1 is F(JC + 1) =xT(x).

(1.1.6)

Also, l)=/i!

(1.1.7)

follows immediately from the above argument or from iteration of (1.1.6) and use of F(l) = l.

(1.1.8)

From (1.1.5) it follows that the gamma function has poles at zero and the negative integers, but 1/ F(x) is an entire function with zeros at these points. Every entire function has a product representation; the product representation of 1/F(JC) is particularly nice. Theorem 1.1.2 00

r(x)

r /

)e-x/n\,

(1.1.9)

where y is Euler's constant given by

y= lim ( V l - l o g n ) .

(1.1.10)

4

1 The Gamma and Beta Functions

Proof. 1 = lim

*(* + ! ) . . . ( * + „ _ i) :

- V*7* n=\

The infinite product in (1.1.9) exists because

and the factor e~*/n was introduced to make this possible. The limit in (1.1.10) exists because the other limits exist, or its existence can be shown directly. One way to do this is to show that the difference between adjacent expressions under the limit sign decay in a way similar to l/n2. • One may take (1.1.9) as a definition of T(x) as Weierstrass did, though the formula had been found earlier by Schlomilch and Newman. See Nielsen [1906, p. 10]. Over seventy years before Euler, Wallis [ 1656] attempted to compute the integral / J y/1 - x2dx = ±f^(l-x)l/2(l+x)l/2dx. Since this integral gives thearea of a quarter circle, Wallis's aim was to obtain an expression for n. The only integral he could actually evaluate was Jo xp{\ — x)qdx, where p and q are integers or q = 0 and p is rational. He used the value of this integral and some audacious guesswork to suggest that

4 A (l.l.H) Of course, he did not write it as a limit or use the gamma function. Still, this result may have led Euler to consider the relation between the gamma function and integrals of the form Jo xp(l — x)qdx where p and q are not necessarily integers. Definition 1.1.3

The beta integral is defined for Rex > 0, Re y > 0 by B(x,y)=

I tx~\\-t)y~ldt. Jo

(1.1.12)

1.1 The Gamma and Beta Integrals and Functions

5

One may also speak of the beta function B(x,y), which is obtained from the integral by analytic continuation. The integral (1.1.12) is symmetric in x and y as may be seen by the change of variables u = 1 — t. Theorem 1.1.4

Remark 1.1.1 The essential idea of the proof given below goes back to Euler [1730, 1739] and consists of first setting up a functional relation for the beta function and then iterating the relation. An integral representation for T(x) is obtained as a byproduct. The functional equation technique is useful for evaluating certain integrals and infinite series; we shall see some of its power in subsequent chapters. Proof. The functional relation we need is B{x, y) = ^-^B(x, y First note that for Re x > 0 and Re y > 0, l

y + 1).

(1.1.14)

r

B(x,y + 1)= / tx-l(l-t)(l-t)y-ldt Jo = B(x,y)-B(x + l,y). However, integration by parts gives B(x,y + 1)= \-tx(l-ty] L*

(1.1.15)

tx(l-ty~ldt

+- / Jo x Jo

= -B(x + l,y). (1.1.16) x Combine (1.1.15) and (1.1.16) to get the functional relation (1.1.14). Other proofs of (1.1.14) are given in problems at the end of this chapter. Now iterate (1.1.14) to obtain + y x + y+ y(y + 1 ) Rewrite this relation as B(x, y) =

x

B(X

+ 2) =

...

x =

y n

B(x, y + n). (y)n

)nJo v —1

n rn

/

J. \ n-\-y—1

(,+,)„„!„y />.(!_£-, [ \ \ ( L \

n\n*+y-^ (y)nn Jo Jo

\

n

du

6

1 The Gamma and Beta Functions

As n -> oo, the integral tends to /0°° tx~xe~ldt. This may be justified by the Lebesgue dominated convergence theorem. Thus B(x, y) =

F(y) ^

f°° 1 , / tx-le-'dt.

(1.1.17)

Set y = 1 in (1.1.12) and (1.1.17) to get

I = f f-Ut = B(x, 1) = -P^x

(

r(x +1) Jo

Jo

Then (1.1.6) and (1.1.8) imply that /0°° tx-le-'dt = V(x) for R e x > 0. Now use this in (1.1.17) to prove the theorem for Rejc > 0 and Re v > 0. The analytic continuation is immediate from the value of this integral, since the gamma function can be analytically continued. • Remark 1.1.2 Euler's argument in [1739] for (1.1.13) used a recurrence relation in x rather than in y. This leads to divergent infinite products and an integral that is zero. He took two such integrals, with y and y = m, divided them, and argued that the resulting "vanishing" integrals were the same. These canceled each other when he took the quotient of the two integrals with y and y = m. The result was an infinite product that converges and gives the correct answer. Euler's extraordinary intuition guided him to correct results, even when his arguments were as bold as this one. Earlier, in 1730, Euler had evaluated (1.1.13) by a different method. He expanded (1 — t)y~l in a series and integrated term by term. When y = n + 1, he stated the value of this sum in product form. An important consequence of the proof is the following corollary: Corollary 1.1.5 For Re x > 0 r(jc) = /

tx-xe-fdt.

(1.1.18)

Jo The above integral for T(x) is sometimes called the Eulerian integral of the second kind. It is often taken as the definition of V(x) for Rex > 0. The Eulerian integral of the first kind is (1.1.12). Legendre introduced this notation. Legendre's F(x) is preferred over Gauss's function U(x) given by (1.1.4), because Theorem 1.1.4 does not have as nice a form in terms of n (x). For another reason, see Section 1.10. The gamma function has poles at zero and at the negative integers. It is easy to use the integral representation (1.1.18) to explicitly represent the poles and the

1.1 The Gamma and Beta Integrals and Functions

7

analytic continuation of T(x): pOO

p\

F(x) = I tx-xe-tdt+ \ Jo J\

tx-le-ldt tX le dt

(1U9)

~ " -

The second function on the right-hand side is an entire function, and the first shows that the poles are as claimed, with (— l)n/n\ being the residue at x = —n, n = 0,1,.... The beta integral has several useful forms that can be obtained by a change of variables. For example, set t = s/(s + 1) in (1.1.12) to obtain the beta integral on a half line, (U.2O)

sY+y Then again, take t = sin2 6 to get

T / 2 sin 2 - 1 0 cos2^"1 Odd = r ( * ) r ° ° . Jo 2T(x + y)

(1.1.21)

Put x = y = 1/2. The result is

_

2r(i)

2'

or

(1.1.22)

Since this implies [F(|)] 2 = 7r/4, we have a proof of Wallis's formula (1.1.11). We also have the value of the normal integral pOQ

/

e~x2dx = 2

pOO

pOO

e~x2dx = /

r ^ V d f = T(l/2) = Jn.

(1.1.23)

J-oo Jo Jo Finally, the substitution t = (u — a)/(b — a) in (1.1.12) gives

rb / (b-u)x~l(u-a)y~ldu Ja

= (b-a)x+y-lB(x,

y) = (b-a)x+

r(x+y) (1.1.24)

The special case a = — l,b = 1 is worth noting as it is often used:

-\

y)

(1.1.25)

8

1 The Gamma and Beta Functions A useful representation of the analytically continued beta function is

j)(

(

a

This follows immediately from Theorem 1.1.2. Observe that B(x, y) has poles at x and y equal to zero or negative integers, and it is analytic elsewhere. As mentioned before, the integral formula for Y(x) is often taken as the definition of the gamma function. One reason is that the gamma function very frequently appears in this form. Moreover, the basic properties of the function can be developed easily from the integral. We have the powerful tools of integration by parts and change of variables that can be applied to integrals. As an example, we give another derivation of Theorem 1.1.4. This proof is also important because it can be applied to obtain the finite field analog of Theorem 1.1.4. In that situation one works with a finite sum instead of an integral. Poisson [1823] and independently Jacobi [1834] had the idea of starting with an appropriate double integral and evaluating it in two different ways. Thus, since the integrals involved are absolutely convergent, oo

/

x x y x

/

ppoo y l s

/»oo

poo {s+t)

t - s - e-

dsdt

= /

/ x l

t - e-'dt

s - e' ds

= T(x)V(y).

/

Jo — v) to the double integral, Apply Jo the change of variables s =Jouv and t = u(l and observe that 0 < u < oo and 0 < v < 1 when 0 < s, t < oo. This change of variables is suggested by first setting s + t = u. Computation of the Jacobian gives dsdt = ududv and the double integral is transformed to rl

OO

/

e~uux+y~ldu / vx~\l - v)y~ldv = T(x + y)B(x, y). Jo

A comparison of two evaluations of the double integral gives the necessary result. This is Jacobi's proof. Poisson's proof is similar except that he applies the change of variables t = r and s = ur to the double integral. In this case the beta integral obtained is on the interval (0, oo) as in (1.1.20). See Exercise 1. To complete this section we show how the limit formula for T (x) can be derived from an integral representation of F(x). We first prove that when n is an integer > 0 and Rex > 0,

/' Jo

-t)'dt=

n

x(x

-

This is actually a special case of Theorem 1.1.4 but we give a direct proof by induction, in order to avoid circularity in reasoning. Clearly (1.1.27) is true for

1.2 The Euler Reflection Formula n = 0, and

f tx-\\-t)n+ldt= Jo

tx-\\-t)(\-t)ndt

[ Jo

n\

n\

This proves (1.1.27) inductively. Now sett = u/n and let« -> oo. By the Lebesgue dominated convergence theorem it follows that

f

tx~le~fdt = lim —

forRejc > 0.

Jo

Thus, if we begin with the integral definition for F(JC) then the above formula can be used to extend it to other values of x (i.e., those not equal to 0 , - 1 , — 2 , . . . ) . Remark 1.1.3 It is traditional to call the integral (1.1.12) the beta function. A better terminology might call this Euler's first beta integral and call (1.1.20) the second beta integral. We call the integral in Exercise 13 Cauchy's beta integral. We shall study other beta integrals in later chapters, but the common form of these three is Jc[li(t)]p[l2(t)]qdt, where l\(t) and l2if) are linear functions of t, and C is an appropriate curve. For Euler's first beta integral, the curve consists of a line segment connecting the two zeros; for the second beta integral, it is a half line joining one zero with infinity such that the other zero is not on this line; and for Cauchy's beta integral, it is a line with zeros on opposite sides. See Whittaker and Watson [1940, §12.43] for some examples of beta integrals that contain curves of integration different from those mentioned above. An important one is given in Exercise 54. 1.2 The Euler Reflection Formula Among the many beautiful formulas involving the gamma function, the Euler reflection formula is particularly significant, as it connects the gamma function with the sine function. In this section, we derive this formula and briefly describe how product and partial fraction expansions for the trigonometric functions can be obtained from it. Euler's formula given in Theorem 1.2.1 shows that, in a sense, the function 1/ F(JC) is half of the sine function. Theorem 1.2.1 Euler's reflection formula:

*) = -r^—. Sin7TJt

(1.2.1)

10

1 The Gamma and Beta Functions

Remark The proof given here uses contour integration. Since the gamma function is a real variable function in the sense that many of its important characterizations occur within that theory, three real variable proofs are outlined in the Exercises. See Exercises 15, 16, and 26-27. Since we shall show how some of the theory of trigonometric functions can be derived from (1.2.1), we now state that sin x is here defined by the series

^

x

+

.

The cosine function is defined similarly. It is easy to show from this definition that sine and cosine have period In and that eni = -l.SeeRudin[1976,pp. 182-184]. Proof. Set y = 1 - JC, 0 < x < 1 in (1.1.20) to obtain f00 tx~l r(*)r(l-*)= / r—dt. Jo 1 + *

J

(1.2.2)

I+

To compute the integral in (1.2.2), consider the integral

where C consists of two circles about the origin of radii R and e respectively, which are joined along the negative real axis from —R to —e. Move along the outer circle in the counterclockwise direction, and along the inner circle in the clockwise direction. By the residue theorem

L

zx dz = -2TTI,

(1.2.3)

c 1-z when zx~l has its principal value. Thus -2jti=

f

iRe iRxeix®

rxdO+

ff€ ttx~le eixn —

ff~K iee iexeix® ffR ttx~*e~ e i dt+ I ^>d0+ — dt.

Let R —• oo and 6 -^ 0 so that the first and third integrals tend to zero and the second and fourth combine to give (1.2.1) for 0 < JC < 1. The full result follows by analytic continuation. One could also argue as follows: Equality of (1.2.1) for 0 < x < 1 implies equality in 0 < Rex < 1 by analyticity; for Rex = 0, x ^ 0 by continuity; and then for x shifted by integers using Y(x + 1) = xT(x) and sin(x + n) = — sinjc. • The next theorem is an immediate consequence of Theorem 1.2.1. Theorem 1.2.2

n=\

1.2 The Euler Reflection Formula OO

/

- + 5Z

71 COt TTJt —

It

!

sin7rjc

X

n

1

(•

1 ™

v

11

\1

( v

!•l l A

^

,

^l.Z.DJ

k=—n

. (-1)" n=l

' x2 - n2

Y A

n

\r A-

n

lim

7i tan n x

/

(1.2.7)

h

J

:+±-Jc'

k=—n

(-1)*

it sec 7i x = lim

(1.2.8)

+ x + -'

2

1

2

sin 7t x

n=—oo

( • •")

(X +

Formula (1.2.4) follows from the product formula

proved in the previous section and from Theorem 1.2.1 intheformr(jc)r(l—x) = —jcF(jc)r(—JC) = 7r/sin7rjc.

Formula (1.2.5) is the logarithmic derivative of (1.2.4), and (1.2.6) follows from (1.2.5) since esc x = cot | — cot JC. The two formulas (1.2.7) and (1.2.8) are merely variations of (1.2.5) and (1.2.6). Formula (1.2.9) is the derivative of (1.2.5). • It is worth noting that (1.2.6) follows directly from (1.2.1) without the product formula. We have poo JCCSC7TX

tx-\

P\

= /

Jo p\

= / Jo

tx-\

dt = /

l + t

POO

Jo l + t

tx-l_^_t-x

dt -

tx-\

dt + /

dt

Jx l + t

~l+rx)

dt

l+t

n+l

where (tn+x 0

+tn~x+l)dt

n +x + 1

/i — x +2

Thus (1.2.6) has been derived from (1.2.1). Before going back to the study of the gamma function we note an important consequence of (1.2.5).

12

1 The Gamma and Beta Functions

Definition 1.2.3 pansion

The Bernoulli numbers Bn are defined by the power series exr2k

^

J

It is easy to check that -^—^ + | is an even function. The first few Bernoulli numbers are Bx = - 1 / 2 , B2 = 1/6, B4 = -1/30, B6 = 1/42. Theorem 1.2.4 For each positive integer k,

n=\

Proof. By (1.2.10) eix + e-ix jccotx = ix-fx z^

=

2ix *x + ~2~i^c

=

^ .,, 1 ~ / ,(~ 1) B2k

and (1.2.5) gives the expansion X2

°°

oo

oo

x2k

xcotx = i + 2 y ^ -r—T—r = i — 2 y ^ y ^ —~——. ^

x2 - n2n2

^

^

«2fc7r2/:

Now equate the coefficients of xlk in the two series for x cot* to complete the proof. • Eisenstein [1847] showed that a theory of trigonometric functions could be systematically developed from the partial fractions expansion of cot x, taking (1.2.5) as a starting point. According to Weil [1976, p. 6] this method provides the simplest proofs of a series of important results on trigonometric functions orginally due to Euler. Eisenstein's actual aim was to provide a theory of elliptic functions along similar lines. A very accessible account of this work and its relation to modern number theory is contained in Weil's book. Weil refers to lim^oo X]-« ak a s Eisenstein summation. Theorem 1.2.2 shows that series of the form 1 -

^(x+ n)k' where k is an integer, are related to trigonometric functions. As we shall see next, the "half series" oo

y

l

1

1.2 The Euler Reflection Formula

13

bears a similar relationship to the gamma function. In fact, one may start the study of the gamma function with these half series. Theorem 1.2.5 r'(l) = - y ,

(1.2.12)

r(x)

Proof. Take the logarithmic derivative of the product for 1 / F (x). This gives

-r(x)

=

The case x = 1 gives (1.2.12). The other two formulas follow immediately.

•

Corollary 1.2.6 Log F(x) is a convex function of x for x > 0. Proof

The right side of (1.2.14) is obviously positive.

•

Remark The functional equation (1.1.6) and logarithmic convexity can be used to derive the basic results about the gamma function. See Section 1.9. We denote Fr (JC)/ F (x) by i/r (x). This is sometimes called the digamma function. Gauss proved that i/r(x) can be evaluated by elementary functions when x is a rational number. This result is contained in the next theorem. Theorem 1.2.7 1

1

1

xls(x+n) =X - +Jt +—1 - + , p\

it

X+ft— 1 up

n = 1,2, 3 , . . .

cos

2nnp i

q

,

(1.2.15)

. *n\ ogU sin I, i) (1.2. 16)

where 0 < p < q;^ means that when q is even the term with index n = q/2 is divided by 2. Here lq/2j denotes the greatest integer in q/2. Proof

The first formula is the logarithmic derivative of T(x + n) = (x + n - 1)(% + n - 2) • • • xT{x).

14

1 The Gamma and Beta Functions

We derive Gauss's formula (1.2.16) by an argument of Jensen [1915-1916] using roots of unity. Begin with Simpson's dissection [1759]: If fix) = EZo"nX\then oo

kn m

i k—1

-J2™-Jmf(uJx),

"}Takn+mx + =

7=0

n=0

where w = e2ni^k is a primitive &th root of unity. This is a consequence of £ * l j wJm = 0, m # O(modfc). Now by (1.2.13) 00

1

+ 1

p + nq - — ) t p + n q = : lim s(t)

= lim V ( —

by Abel's continuity theorem for power series. From the series — log(l — t) Y1T=\ tn/n>an cos log 2 sin — .

q

frf

q

V

•

q )

1.3 The Hurwitz and Riemann Zeta Functions The half series OO

j

forx>0,

(1.3.1)

called the Hurwitz zeta function, is of great interest. We have seen its connection with the gamma function for positive integer values of s in the previous section. Here we view the series essentially as a function of s and give a very brief discussion of how the gamma function comes into the picture. The case x = 1 is called the Riemann zeta function and is denoted by f (s). It plays a very important role in the theory of the distribution of primes. The series converges for Res > 1 and defines an analytic function in that region. It has a continuation to the whole complex plane with a simple pole at s = 1. The analytic continuation of f (s) up to Res > 0 is not difficult to obtain. Write the series for f (s) as a Stieltjes integral involving [*J. Thus for Res > 1 [xjdx

The last integral converges absolutely for Re s > 0 and we have the required

16

1 The Gamma and Beta Functions

continuation. The pole at s = 1 has residue 1 and, moreover,

= lim f,

.4H n

/

I r I— r

\

= lim ( V - - l o g n ) =y. m=\

(1.3.2)

/

The best way to obtain analytic continuation to the rest of the plane is from the functional relation for the zeta function. We state the result here, since the gamma function is also involved. There are several different proofs of this result and we give a nice one due to Hardy [1922], as well as some others, in the exercises. In Chapter 10 we give yet another proof. Theorem 1.3.1 For all complex s, ((1)/2)

s).

(1.3.3)

If 5 < 0, then 1 — s > 1 and the right side provides the value of t;(s). This relation was demonstrated by Euler for integer values of s as well as for s = 1 /2 and s = 3/2. He had proofs for integer values of s, using Abel means. An interesting historical discussion is contained in Hardy [1949, pp. 23-26]. The importance of t;(s) as a function of a complex variable in studying the distribution of primes was first recognized by Riemann [1859]. The last section contained the result B2k7r2k.

t(2k) = — S

(2*)!

The following corollary is then easy to prove. Corollary 1.3.2

?(1 - 2k) = —B2k, ?(0) = —

and ((-2k) =0 for k = 1, 2, 3 , . . . . (1.3.4)

Corollary 1.3.3 , v ,

Proof

(1.3.5)

2

From the functional equation and the fact that l-s

V 2

1.3 The Hurwitz and Riemann Zeta Functions

17

we have

Now (1.3.2) implies that (s - l)f(s) = 1 + y (s - 1) + A(s - I) 2 H the logarithmic derivative of (1.3.6) to get

.So take

Set 51 = 1 and use Gauss's result in Theorem 1.2.7 with p = 1 and q = 2. This proves the corollary. • There is a generalization of the last corollary to the Hurwitz zeta function £ (JC , s). A functional equation for this function exists, which would define it for all complex s, but we need only the continuation up to some point to the left of Re s = 0. This can be done by using the function f 0 ) . Start with the identity 00

s

?(*, s) - (Us) ~ sx$(s + D) = x~ + ]TV 5 [(1 + x/ny5 - (1 - sx/n)]. n=\

The sum on the right converges for Re s > — 1, and because £ (s) is defined for all s, we have the continuation of f (*, s) to Res > — 1. The following theorem is due to Lerch. Theorem 1.3.4

s=0

Proof. The derivative of the equation t;(x + 1, s) = t;(x, s) — x to s at s = 0 gives

s

with respect

(1.3.8) s=0

\

u

*

/

5=0

For Re 5 > 1,

9JC 2

so

(139)

18

1 The Gamma and Beta Functions

Now (1.3.8) and (1.3.9) together with (1.2.14) of Theorem 1.2.5 imply that = C + logF(x). ds

/ s=0

To determine that the constant C = — \ log lit, set x = 1 and use Corollary 1.3.3. This completes the proof of Lerch's theorem. • For a reference to Lerch's paper and also for a slightly different proof of Theorem 1.3.4, see Weil [1976, p. 60]. 1.4 Stirling's Asymptotic Formula De Moivre [1730] found that n\ behaves like Cnn+l^2e~n for large n, where C is some constant. Stirling [1730] determined C to be V27r; de Moivre then used a result of Stirling to give a proof of this claim. See Tweddle [1988, pp. 9-19]. This formula is extremely useful and it is very likely that the reader has seen applications of it. In this section we give an asymptotic formula for T{x) for Rex large, when Imjc is fixed. First note that log T(x + n + 1) = Yl=\ l°g(k + x) + l o £ r ( * + 1). We then employ the idea that an integral often gives the dominant part of the sum of a series so that if the integral is subtracted from the series the resulting quantity is of a lower order of magnitude than the original series. (We have already used this idea in Equation (1.3.2) of the preceding section.) In Appendix D we prove the Euler-Maclaurin summation formula, a very precise form of this idea when the function being integrated is smooth. Two fuller accounts of the Euler-Maclaurin summation formula are given by Hardy [1949, pp. 318-348] and by Olver [1974, pp. 279-289]. Theorem 1.4.1 T{x) ~ \/7jzxx~xl1e~x

as Rex —> oo.

Proof. Denote the right side of the equation n-\

log r(jc + n) = ^2 l°S(k + *) + loS r(jc + 1) by cn, so that cn+i -cn =log(x + n). By the analogy between the derivative and the finite difference we consider cn to be approximately the integral of log(jc + ri) and set

cn = (n + x) \og(n +x)-(n+x)

+ dn.

1.4 Stirling's Asymptotic Formula

19

Substitute this in the previous equation to obtain log(jc + n) = (TI + 1 + x)log(n + 1 + x) - (n + x) login + JC) + dn+l

-dn-\.

Thus 4+1 - dn = 1 - in + x + 1) log 1 +

1

n+x

V

Proceeding as before, take

and substitute in the previous equation to get

1

/ 2

I2(n+x)

1

\(n+x)3J'

Now

therefore, limn_>oo(^n — eo) = K\ix) exists. Set 1 12(/I+JC)

\ ( * + * ) :2

) J'

where Kix) = K\ix) + eo. The term in + JC)"1 comes from completing the sum in (1.4.1) to infinity and approximating the added sum by an integral. So we can write cn = in + x) login + x) - in + x) - - login + x)

where K(x) = log C(x). This implies that 12(n+x) (1.4.2)

20

1 The Gamma and Beta Functions

We claim C{x) is independent of x. By the definition of the gamma function hm

r(n + x) r( + )

ny

=

r(x)

(x)n

?zJ

lim

r()^(y)n

=

r(x) roo •

r(y) T(x)

= 1.

(1.4.3)

Now, from (1.4.2) and (1.4.3) we can conclude that

_V(n+x) 1 = hm n n^oo

=

r(n)

C(x) . / x\n _ C(x) hm 1 + - \ e x = . C(0)n^oo\

nJ

C(0)

Thus C(x) is a constant and F(JC) ~ Cx x ~ 1/2 ^~ x

as Rex -> oo.

To find C, use Wallis's formula: r

= hm

2(n\)

1

(2)!

p

C

This gives C = \/2n and proves the theorem. Observe that the proof gives the first term of an error estimate. • We next state a more general result and deduce some interesting consequences. A proof is given in Appendix D. For this we need a definition. The Bernoulli polynomials Bn(x) are defined by tn ef — 1

^—'

(1A4)

n\

n=0

The Bernoulli numbers are given by Bn (0) = Bn for n > 1. Theorem 1.4.2 number,

For a complex number x not equal to zero or a negative real

log z

\

l r° 2m Jo

/, /

7=1

(x + 0 2 m

•dt.

(1.4.5)

T/i^ v^/w^ of log x is the branch with log JC real when x is real and positive.

The expansion of log V(x) in (1.4.5) is an asymptotic series since the integral is easily seen to be O(x~2m+l) for |argx\ < it - 8, 8 > 0.

1.4 Stirling's Asymptotic Formula

21

From this theorem the following corollary is immediately obtained. Corollary 1.4.3 For 8 > 0 and |argx\ < n — 8, F(x) ~ V2nxx~l/2e~x Corollary 1.4.4

as

\x\ -> oo.

When x = a + ib, a\ < a < a2 and \b\ - • oo, then

\T{a + ib)\ = sFhx\b\a-ll2e-n^l2\\

+ 0(l/|fc|)],

where the constant implied by O depends only on a\ and a2. Proof. Take \b\ > I, a > 0. It is easy to check that the Bernoulli polynomial B2 - B2(t) = t -t2. Thus \\B2 - B2(t)\ < \\t{\ - t)\ < \ for 0 < t < 1. So (1.4.5) with m = 1 is log T(a + ib)= (a + ib

j log(a + ib) - (a + ib) + - log 2TT + R(X),

and dt _ I i r°° dt _ I r°° ~ 8 Jo k + ^ l 2 8^0 (a + t)2 + b2 S\b\

_j \b\ a'

Now R e f a + ife - 2- ) l o g ( a + i b ) \ = (a--) 2

IA

/

J V

/

\og(a2 + b 2 ) 1 ' 2 - barctan

Also,

log(a2 +b2)l/2 = Uogb2 + ^log^l + ^)=

log |*|

Moreover,

b a f^, arctan - + arctan - = < 2 n a b [-f,

iffe>0, ifb 0 is removed by a finite number of uses of the functional equation (1.1.6) and the corollary follows. Observe that the proof only uses a = o(\b\) rather than a bounded. • Corollary 1.4.5 For |argjt| < n — 8, 8 > 0,

Corollary 1.4.4 shows that T{a + ib) decays exponentially in the imaginary direction. This can be anticipated from the reflection formula, for cosh**'

or e7ib

_i_

e

—Tib

• 27te~nlbl

as b -> ±oo,

or as b - • ±oo.

n-+n Similarly,

4

-ib sin 7rZ?/

and as

±oo.

Since T(x) increases rapidly on the positive real axis and decreases rapidly in the imaginary direction, there should be curves going to infinity on which a normalized version of T(x) has a nondegenerate limit. Indeed, there are. See Exercise 18. 1.5 Gauss's Multiplication Formula for T(mx) The factorization

together with the definition of the gamma function leads immediately to Legendre's duplication formula contained in the next theorem. Theorem 1.5.1

1.5 Gauss's Multiplication Formula for F(mx)

23

This proof suggests that one should consider the more general case: the factorization of (a)mn, where m is a positive integer. This gives Guass's formula. Theorem 1.5.2

T{ma){2nim-X)'2 = m ^ ^ F ^ F fa + -J-Y • • F (a + ^ - = - ^ . (1.5.2) Proof. The same argument almost gives (1.5.2). What it gives is (1.5.2) but with (27r)"r i m-5

replacedby

F( - ] . . . F ( — - ) =: P. \mj \ m J To show that (1.5.3) is true, we show that

(1.5.3)

By the reflection formula mj

\

mj

sin —

So it is enough to prove m_,

JZ . 2n . (m — l)n sin — sin — • • • sin mm m Start with the factorization

2

x — —

= m.

= ]J(x - exp(2kni/m)).

X

Let x —> 1 to obtain m-\

m = TT(1 —Qxp(2k7ti/m)) k=\

i

= 2

n

sin — sin mm

(m — 1)TT

2it

sin

. m

This proves (1.5.3). • Remark 1.5.1 A different proof of (1.5.1) or (1.5.2) that uses the asymptotic formula for F(JC) and the elementary property T(x + 1) = xF(x) is also possible. In fact it is easily verifed that *

F(l/2)F(2x)

satisfies the relation g(x + 1) = g(x). Stirling's formula implies that g(x) ~ 1 as x -> cxDsothatlim^^oo^^+w) = 1 when n is an integer. Since g(x-\-n) — g{x) we can conclude that g (x) = 1. A similar proof may be given for Gauss's formula. This is left to the reader.

24

1 The Gamma and Beta Functions

An elegant proof of the multiplication formula using the integral definition of the gamma function is due to Liouville [1855]. We reproduce it here. The product of the gamma functions on the right side of (1.5.2) is pOO

pOO

/ e-t'x^dxj Jo

Jo

pOO

e-x*x$+il/m)-ldx2---

poo poo poo / ... /

=

Jo

e

e-U^2+-+xm)xa-lxa+a/m)

Jo Jo Jo Introduce a change of variables:

z'H

_ X\ —

_

_

, X2 — X2, - • • » Xm — Xm.

The Jacobian is easily seen to be mzm~l x2 x3

and the integral can be written

/

•••/

JO

Jo

exp - ( x2 + x3 + • • • + xm + I V

)

X2X3'"XmJj

KX2-"Xm)

X2X3'"Xm m

Set t = x2 + x3 -\ poo

poo

/

1- xm + z /(x2x3 • - • xm), and rewrite the integral as poo

••• /

Jo Jo Jo

e-'zma-Xx{Vm)-Xxflm)-1

• • • x«n-l)lm)-ldzdx2

...dxM. (1.5.4)

First compute poo poo

I

in L poo m—\ poo ~

Jo ,._-,

Jo

Clearly, pOO

dl

m-\

=

—YYIZ1'1~V

/

I

Jn Jo

rn — L

pOO

'''

/

-t TT

I e~l

Jc\ Jo

,

,

{j/m)-\dX2"'dXm

_T J

J =

X

2 ' ' ' Xm

Now introduce a change of variables, -Xm),X3

=X3,...,Xm

=Xm,

1.5 Gauss's Multiplication Formula for F(rax)

25

and ' 1 — *^3

• *^4

I *** I

in

• Jv\

I 1. 2 4k Now use integral (1.6.2) for log T{x) in (1.7.3) so that i

i

i

sin2knxdudx logw

But /

sin 2knxdx = 0,

/ x sin 2knxdx = — Jo

and

rl ux

sin 2knxdx =

The first two integrals are easy to solve and the third is the imaginary part of 1 fl u Jo

1 M-1 u logu + 2kni'

1.7 Kummer's Fourier Expansion of Log F (x)

31

Therefore,

-i

—2kn 2 u((\ogu) +4k2n2) K

1 \ du 2kn ) logw'

2knt

or, with u = e~

, -Iknt

Take £ = 1 and we have

Moreover, x = 1 in Dirichlet's formula (Theorem 1.6.1) gives 2n Jo

2TT

V

l+

tjt

where y is Euler's constant. Therefore, ~

Y

1

By (1.6.1), thefirstintegral is log 2n and a change of variables from t to 1/f shows that the second integral is 0. Thus

To find Dk, observe that 1

2nt _ n~2nt e-2knt f°° e~ — e~

kDk-Dx = — In Jo

1t

1

dt = — log/:, "" 2ll In

where the integral is once again evaluated by (1.6.1). Thus Dk =

(y + log2A:7r),

k = 1, 2, 3 , . . . .

The Fourier expansion is then logF(x) = -log27T + ^ C ° S o 7 7 r k=\ k=\

X

+-(y+

LK

n

Iog27r)"

}2

k=\

, 1 v^log/: .

Apply (1.7.1) and (1.7.2) to get the result.

•

Kummer's expansion for log (F (X)/V2TT) and Theorem 1.3.4 have applications in number theory. Usually they give different ways of deriving the same result. This suggests that the Hurwitz zeta function itself has a Fourier expansion from

32

1 The Gamma and Beta Functions

which Kummer's result can be obtained. Such a result exists and is simply the functional equation for the Hurwitz function: r(x,s) =

2F(1— s) ) . 1 ^coslmnx — < sm -us >

1 ^ sinlmixx | hcos -ns > >. (1.7.4)

m=l

I

m=l

J

The functional equation for the Riemann zeta function is a particular case of this when x = 1. See Exercises 24 and 25 for a proof of (1.7.4) and another derivation of Kummer's formula. 1.8 Integrals of Dirichlet and Volumes of Ellipsoids Dirichlet found a multidimensional extension of the beta integral which is useful in computing volumes. We follow Liouville's exposition of Dirichlet's work. Liouville's [1839] presentation was inspired by the double integral evaluation of the beta function by Jacobi and Poisson. Theorem 1.8.1 If V is a region defined by xt > 0, / = 1, 2 , . . . ,n,and^Xi then for Re, at > 0,

< 1,

Proof The proof is by induction. The formula is clearly true for n = 1. Assume it is true for n = k. Then for a (k + 1)-dimensional V

f'.

• dxk+l l-xi-Jt2

= f f~xx••• f' Jo Jo 1 =

—

xk

*

\rlxrl---xakk+T1dxk+i---dXl

Jo r\

/

pl-Xy—Xk-x i-1 x\

• • • /

(*k+\ Jo

• ( 1 - JCI

Jo

xk)ak+ldxkdxk-x'

Now set xk = (1 — xx — • • • — xk-x)t

&k+\ Jo Jo

Ofjfc-1

• • dxx .

to get

Jo

Jo

ff-\..f-'-"-

Jo Jo

Xakk_T\\

-xx

Jo

xk.x)ak+ak+ldxk.x

• • • dxx

1.8 Integrals of Dirichlet and Volumes of Ellipsoids

33

Compare this with the integral to which the change of variables was applied and use induction to get

r(ak)T(aM + 1)

(«* +a*+i)(n?=i !>,-))I"(a*

This reduces to the expression in the theorem.

•

Corollary 1.8.2 IfV is the region enclosed by xt > 0 and ]T(JC;/ 0, J2"=\ xi = h then

n

[••• J x ? - l

This is a surface integral rather than a volume integral, but it can be evaluated directly by induction or from Corollary 1.8.2. It is also a special case of Theorem 1.8.5 when f(u) is taken to be the delta function at u = 1. This function is not continuous, but it can be approximated by continuous functions. 1.9 The Bohr-Mollerup Theorem The problem posed by Euler was to find a continuous function of x > 0 that equaled n\ at x = n, an integer. Clearly, the gamma function is not the unique solution to this problem. The condition of convexity (defined below) is not enough, but the fact that the gamma function occurs so frequently gives some indication that it must be unique in some sense. The correct conditions for uniqueness were found by Bohr and Mollerup [1922]. In fact, the notion of logarithmic convexity was extracted from their work by Artin [1964] (the original German edition appeared in 1931) whose treatment we follow here. Definition 1.9.1 A real valued function f on (a, b) is convex if f(kx + (1 - X)y) < Xf(x) + (1 -

X)f(y)

for x, y e (a, b) andO < X < 1. Definition 1.9.2 A positive function f on(a, b) is logarithmically convex if log / is convex on (a,b). It is easy to verify that if / is convex in (a, b) and a < x < y < z < b, then f(y) ~ fM y-x

^ f{z) - fix) ^ f{z) - f(y) ~ z-x ~ z-y

With these definitions we can state the Bohr-Mollerup theorem:

1.9 The Bohr-Mollerup Theorem

35

Theorem 1.9.3 If f is a positive function on x > 0 and (i) / ( I ) = 1, (ii) f(x + 1) = xf(x), and (Hi) f is logarithmically convex, then f(x) = T(x)for x > 0. Proof Suppose n is a positive integer and 0 < x < 1. By conditions (i) and (ii) it is sufficient to prove the theorem for such x. Consider the intervals [n, n + 1], [n + 1, n + 1 + JC], and [n + 1, n + 2]. Apply (1.9.1) to see that the difference quotient of log /(JC) on these intervals is increasing. Thus

1 /(* + !+*) ^ / ( / i + 2) < - log < log . /(") * fin + 1) /(w + 1) Simplify this by conditions (i) and (ii) to get log

/(n + 1)

x logrc < log

[fr+W)(jc+wl)

L n\ Rearrange the inequalities as follows:

*/(*)[ ^ < x log(rc + 1).

J

Therefore, /(JC)

n\i

= lim

and the theorem is proved.

•

This theorem can be made the basis for the development of the theory of the gamma and beta functions. As examples, we show how to derive the formulas poo POO

T(JC) = /

fx l e-e-t t - dt,

J C > 0,

Jo and

I

c > 0 and y > 0. (1.9.2) o L \x -t y) We require Holder's inequality, a proof of which is sketched in Exercise 6. We state the inequality here for the reader's convenience. If / and g are measurable nonnegative functions on (a, b), so that the integrals on the right in (1.9.3) are finite, and p and q positive real numbers such that \/p + \/q = 1, then < I / fpdx 1 / gqdx I . (1.9.3) \Ja ) \Ja ) It is clear that we need to check only condition (iii) for log F(JC). This condition can be written as >0

and a + )8 = l.

(1.9.4)

36

1 The Gamma and Beta Functions

Now observe that /»O

=

/

Jo

and apply Holder's inequality with a = \/p and ft = l/q to get (1.9.4). To prove (1.9.2) consider the function

r(x + y)B(x,y) T{y) Once again we require the functional relation (1.1.14) for B(x, y). This is needed to prove that / (x +1) = xf (x). It is evident that / (1) = 1 and we need only check the convexity of log fix). The proof again uses Holder's inequality in exactly the same way as for the gamma function. We state another uniqueness theorem, the proof of which is left to the reader.

Theorem 1.9.4 If f(x) is defined for x > 0 and satisfies (i) / ( I ) = 1, (ii) f(x + 1) = xfix), and (iii'jlinwoo fix + n)/[nxf(n)] = 1, then fix) = For other uniqueness theorems the reader may consult Artin [1964] or Anastassiadis [1964]. See Exercises 26-30 at the end of the chapter. Finally, we note that Ahern and Rudin [1996] have shown that log |F(JC + iy)\ is a convex function of x in Rex > 1/2. See Exercise 55. 1.10 Gauss and Jacobi Sums The integral representation of the gamma function is

C

=Jor e-,-d± h

t

Here dt/t should be regarded as the invariant measure on the multiplicative group (0, oo), since diet) _ dt ct ~~ t ' To find the finite field analog one should, therefore, look at the integrand e~cttx. The functions e~ct and tx can be viewed as solutions of certain functional relations. This point of view suggests the following analogs. Theorem 1.10.1 Suppose f is a homomorphism from the additive group of real numbers R to the multiplicative group of nonzero complex numbers C*, that is,

/ : / ? - > C*

1.10 Gauss and Jacobi Sums

37

and

f(x + y) = f(x)f(y).

(1.10.1) cx

/ / / is differentiable with /'(0) = c # 0, then f(x) = e . Remark 1.10.1 We have assumed that f(x) ^ 0 for any x but, in fact, the relation g(x + y) = g(x)g(y), where g : R —> C, implies that if g is zero at one point it vanishes everywhere. Proof. First observe that / ( 0 + 0) = / ( 0 ) 2 by (1.10.1). So /(0) = 1, since /(0) cannot be 0. Now, by the definition of the derivative, J w

r-+o

t

t-+0 = f (JC) lim t^o

t t

= cf(x). Sof(x)=ecx.

m

Remark 1.10.2 In the above theorem it is enough to assume that / is continuous or just integrable. To see this, choose a y e / ? such that f* f(t)dt / 0. Then

fix) jy f(t)dt = jj f(X + odt = jxx+y

f{t)dt. so

This equation implies that if / is integrable, then it must be continuous and hence differentiable. Corollary 1.10.2 Suppose g is a homomorphism from the multiplicative group ofpositive reals R+ to C*, that is, g(xy) = g(x)g(y).

(1.10.2)

c

Then g(x) = x for some c. Proof Consider the map f = g o exp : R —• C*, where exp(jc) = ex. Then / satisfies (1.10.1) and g(ex) = ecx. This implies the result. • A finite field has pn elements, where p is prime and n is a positive integer. For simplicity we take n = 1, so the field is isomorphic to Z(/?), the integers modulo p. The analog of / in (1.10.1) is a homomorphism

Since Z(/?) is a cyclic group of order p generated by 1 we need only specify Also, tff(l)p = if(0) = 1 and we can choose any of the pth roots of unity as the

38

1 The Gamma and Beta Functions

value of V^(l). We therefore have p different homomorphisms ^.(JC)

=

2nijx/p e

,

j = 0 , 1 , . . . , / ? - 1.

(1.10.3)

These are called the additive characters of the field. In a similar way the multiplicative characters are the p — 1 characters defined by the homomorphisms from Z(p)* to C*. Here Z(p)* = Z(p) - {0}. Since Z(p)* is a cyclic group of order p — 1, we have an isomorphism Z(/?)* = Z(p — 1). The p — 1 characters on Z(/?)* can be defined by means of this isomorphism and (1.10.3). We denote a multiplicative character by either x or 77, unless otherwise stated. It is now clear how to define the "gamma" function for a finite field. Definition 1.10.3 For an additive character ^ and multiplicative character Xi we define the Gauss sums gj(Xi), j = 0 , 1,...,/? — 1 by the formula P-\

x=0

where we extend the domain of Xi by setting X/(0) = 0. It is sufficient to consider g(x) •= gi(x)> f° r when j ^ 0

gj(x) = X)xW^;W

= XU)8(X).

d-10.5)

This formula corresponds to /0°° e~jxxs~ldx = T(s)/js, where j is a nonzero complex number with positive real part. When j = 0 in (1.10.4) the sum is )i which can be shown to be zero when x (x) ^ 1 for at least one value of x. Theorem 1.10.4

For a character x>

, Remark 1.10.3 point in Z(/?)*.

,

:

d)

V p-l ifX=id. The identity character is the one that takes the value 1 at each

The result is obvious for x = id. If x ¥" id, there is a y e Z(p)* such l.Then

1.10 Gauss and Jacobi Sums

39

which implies the theorem. There is a dual to (1.10.6) given by the following theorem: • Theorem 1.10.5

For the sum over all characters we have

Proof. It is sufficient to observe that if x / 1, then there is a character x such that x (x) 7^ 1. The theorem may now be proved as before. • We now define the analog of the beta function. Definition 1.10.6 defined by

For two multiplicative characters x

Y,

an

d r\ the Jacobi sum is

(1.10.8)

x+y=l

The following theorem gives some elementary properties of the Jacobi sum. We denote the trivial or identity character by e. The reader should notice that the last result is the analog of the formula B{x, y) = r(x)T(y)/[r(x + y)]. Theorem 1.10.7 hold:

For nontrivial characters x

an

d ??, the following properties

J(e,X) = 0.

(1.10.9)

J(e,e) = p-2.

(1.10.10)

1

(l.io.ii) (1.10.12) then J ( x , ly) = * ( x ) * ( > y ) . g(x*i) Remark 1.10.4 From the definition of characters it is clear that the product of two characters is itself a character and so the set of characters forms a group. The additive characters form a cyclic group of order p and the multiplicative characters a cyclic group of order p — 1. Also, x~l(x) — x(x~!) — VxOO and since IxOOl = 1 it follows that x - 1 ( ^ ) Proof The first part of the theorem is a restatement of Theorem 1.10.3 and the second part is obvious. To prove (1.10.11), begin with the definition x

JC/0,1

Now note that as x runs through 2 , . . . , / ? — 1, then x(l — x) runs through 1 , . . . , p — 2. The value y = p — \ = — 1 (mod p) is not assumed because x = y ( l + y ) ~ 1 . Therefore,

J(x,x~l)=

J]

40

1 The Gamma and Beta Functions

by Theorem 1.10.4. This proves the third part. The proof of the fourth part is very similar to Poisson's or Jacobi's proofs of the analogous formula for the beta function. Here one multiplies two Gauss sums and by a change of variables arrives at a product of a Jacobi sum and a Gauss sum. Thus, for x V / e, liziylp

X(x)r](t-x)e27tit/p. x+y=0

= 0 since XV ¥" id. The

The first sum is J2X X(*)*7(—*) second sum with x = st is

"

This proves the fourth part of the theorem.

s)

•

We were able to evaluate F (s) in a nice form for positive integer values and halfinteger values of s. Evaluations of special cases of Gauss sums are also possible and important, but in any case the magnitude of the Gauss sum can always be found. Theorem 1.10.8 For nontrivial multiplicative and additive characters x and \jf,

Proof. By (1.10.5) it is enough to prove that |gi(x)| 2 = \g(x)\2 = P

\g(x)\2 =

Setx = ty. Then

ty^O

E

^O or 1

1.10 Gauss and Jacobi Sums

41

The first sum is p — 1 and the inner sum in the second term is — 1. Thus

and the result is proved.

•

Corollary 1.10.9 If x, r\ and x *7 are nontrivial characters, then \J{X,ri)\ = Jp. Proof. This follows from Theorems 1.10.7 and 1.10.8.

(1.10.13) •

As an interesting consequence we have: Corollary 1.10.10 If p = An + 1 is a prime, then there exist integers a and b such that p = a2 + b2. Proof The group Z(/?)* is of order p — 1 = An, which is also isomorphic to the group of multiplicative characters on Z(/?)*. Since the latter group is cyclic there exists a character x of order 4 that takes the value ± 1 , =b". It follows that J(X, X) = a + bi for integers a and b. Since x 2 7^ id, apply Corollary 1.10.9 to obtain the desired result. • Corollary 1.10.10 is a theorem of Fermat, though Euler was the first to publish a proof. See Weil [1983, pp. 66-69]. Later we shall prove a more refined result that gives the number of representations of a positive integer as a sum of two squares. This will come from a formula that involves yet another analog of the beta integral. We have seen that characters can be defined for cyclic groups. Since any abelian group is a direct product of cyclic groups, it is not difficult to find all the characters of an abelian group and their structure. The following observation may be sufficient here: If xi is a character of a abelian group G\, and xi °jGi> then we can define a character x : G\ x G2 -» C* by / ( * , y) = X\MX2(y)We thus obtain n additive characters of Z(n) and (n) multiplicative characters of Z(n)*. The Gauss and Jacobi sums for these more general characters can be defined in the same way as before. Gauss [1808] found one derivation of the law of quadratic reciprocity by evaluating the Gauss sum arising from the quadratic character. (A character x 7^ id is a quadratic character when x 2 = id.) Details of this connection are in Exercise 37 at the end of the chapter. One problem that arises here, and which Gauss dealt with, is evaluating the sum G = Ylx=o el7Tlx2/N. As in Theorem 1.10.8 one can show that G2 — ±N depending on whether N = I (A) or 3 (4). The problem is to determine the appropriate square root for obtaining G. According to Gauss, it took him four years to settle this question. Dirichlet's evaluation of Ylx=o e2nix2/N by means of Fourier series is given in Exercise 32.

42

1 The Gamma and Beta Functions Jacob! and Eisenstein also considered the more general Jacobi sum J(Xi,X2, • • •, Xt) =

Yl

Xi(h)X2(t2) • • • xiift).

(1.10.14)

This is the analog of the general beta integral in Theorem 1.8.6. Eisenstein's result, corresponding to the formula in Theorem 1.8.6, follows. Theorem 1.10.11 If xi, X2> • • • > Xi are nontrivial characters and X\Xi" ' Xt ^ nontrivialy then J(Xu X2, • • •, Xi) =

; :—• (1.10.15) g(XiX2'-Xi) The proof of this is similar to that of Theorem 1.10.7, and the reader shouldfillin the details. In Section 1.8 the volume of n-dimensional objects of the form a\xs^ + a^x^ + hfljfcjCfc* < b was determined by means of the gamma function. In the same way, for finite fields, the number of points satisfying a\x\x +a2Xs22 H \-akXskk = b can be found in terms of Gauss sums. Gauss himself first found the number of points on such (but simpler) hypersurfaces and used this to evaluate some specific Gauss sums. Weil [1949] observed that it is easier to reverse the process and obtain the number of points in terms of Gauss sums. For an account of this the reader should see Weil [1974]. It may be mentioned that Weil's famous conjectures concerning the zeta function of algebraic varieties over finite fields are contained in his 1949 paper. It also contains the references to Gauss's works. One may also consult Ireland and Rosen [1991] for more on Jacobi and Gauss sums and for references to the papers of Jacobi and Eisenstein. The form of the Gauss sums also suggests that they are connected with Fourier transforms. Let T denote the vector space of all complex valued functions on Z(N), the integers modulo N. Let F be the Fourier transform on T defined by N-l

(Ff)(n) = -y=Y, fWe2ninx/N**N

(1.10.16)

x=0

It can be shown that the trace of this Fourier transform with respect to the basis {So, S\, ...,SN], where &x(y)

fO, 11,

x^y, x = y,

is the quadratic Gauss sum Ylx=o e2ltlx2/N. Schur [1921] gave another evaluation of this sum from this fact. The details are given in Exercise 47. One first proves that the fourth power of F is the identity so that the eigenvalues are ± 1 , ± / and the essential problem is to find the multiplicity of these eigenvalues.

1.11 A Probabilistic Evaluation of the Beta Function

43

Discrete or finite Fourier analysis was not applied extensively before 1965 because of the difficulty of numerical computation. This changed when Cooley and Tukey [1965] introduced an algorithm they called the Fast Fourier Transform (FFT) to reduce the computation by several orders of magnitude. The reader may wish to consult the paper of Auslander and Tolimieri [1979] for an introduction to FFT, which emphasizes the connection with group theory. Some of the earlier instances of an FFT algorithm are mentioned here. Computational aspects are also interesting. See de Boor [1980] and Van Loan [1992, §1.3].

1.11 A Probabilistic Evaluation of the Beta Function When a and ft are positive integers, l

dx=

/

(a

It seems that it should be possible to arrive at this result by a combinatorial argument. But working with only a finite number of objects could not give an integral. Here is a combinatorial-cum-probabilistic argument that evaluates the integral. Choose points at random from the unit interval [0, 1]. Assume that the probability that a point lies in a subinterval (a, b) is b — a. Fix an integer n and let P(xk < t) denote the probability that, of n points chosen at random, exactly k of them have values less than t. The probability density function for P(xk < t) is

pit) = lim ^ v } A?^O

P(xk{x)

and TOO

/ Jo

Jo

,*-l

dt =:{x)SX-\

t+s t-\-s

(b) Deduce that

(c) Use the second formula in (a) to get /•oo

[Hx)f= Jo

i

/

poo

t

4 T ( / 7+ s s + 1 V Jo

t

50

1 The Gamma and Beta Functions then change the order of integration to obtain 2

_

poo +x—x l

r°°t - \ogt

~ Jo

t-\

(d) Deduce y

_

[°° ty-\

[(/)(x)]2dx = / Jo / \-y-y

t-y

—dt.

t

— A

(e) Integrate (b) with respect to s over (0, oo) and use (d) to derive

000 / [4>(t)]2dt = 2 [ Ji-x Jo

l

\ lOgtdt = 2(j)Xx). 1 +1

(f) Show that 0(JC) = 0(1 — x) implies (j)f{\) = O and x

/ -x

px

[c/)(t)]2dt = 2 [c/)(t)]2dt. Jl/2

(g) Deduce that Jl/2 (h) Show that 0 satisfies the differential equation 00 r/ — (0') 2 = 0 4 (i) Solve the differential equation with initial condition 0 ( | ) = TT and 0 r (l) z= Otoget0(;c) = n csc7rjc. 17. Show that -, Rex > 0,Rey > 0,« > 0.

Jo Vat + 18. Show that

19. Prove that for a > 0, sinaut and /•OO f°° cos ax

/

1

. ! stc(nb/2)

:—JJC = -na

Jo

*b

2

,

0 < ReZ? < 1.

T(b)

20. For A > 0, JC > 0 and — n/2 < a < TT/2, prove that Jo

Exercises

51

and /•OO

/

Jo

fX-ig-ktcosa s m ( ^ sina)dt = X~xT(x) sin ax.

21. Prove that n-s/2r(s/2)$(s) (a) Observe that 00

E

= 7r- ( 1 - 5 ) / 2 r((l - s)/2)S(l - s) as follows:

sin(2/2 + l);c = (—1) 7T/4 for run < x < (ra + l)7T,ra= 0, 1, 2n + l

n—\

(b) Multiply the equation by xs~l(0 < s < 1) and integrate over (0, oo). Show that the left side is F(s) sin(>7r/2)(l - 2" 5 " 1 )f (s + 1) and that the right represents an analytic function for Res < 1 and is equal to (c) Deduce the functional equation for the zeta function. (Hardy) 22. Let C be a contour that starts at infinity on the negative real axis, encircles the origin once in the positive direction, and returns to negative infinity. Prove that

This formula holds for all complex s. (a) Note that the integral represents an analytic function of s. (b) C may be taken to be a line from — oo to — 1 an integer, and returns to positive infinity. Hint: First prove that

and then apply the ideas of the previous exercise. Note also that £ 0 , s) is now

52

1 The Gamma and Beta Functions

defined as a meromorphic function by the contour integral with a simple pole at ^ = 1. 24. Prove the functional equation 2F(1 — s) f ^ coslmnx ^ K ^sin(7r5/2) V V + C o s ( s 7 r / 2 ) V r yl s s;; ^sin(7r5/2) ?(*, s) = (2ixy~ *-^ m[~s +Cos(s7r/2) *-^ m~ K

m=\

m=\

}. )

Hint: Let Cn denote the line along the positive real axis from oo to (In + l)n, then a square with corners (2n + 1)TT (± 1 d= /), and then the line from (2n + 1) to oo. Show that r

ts-i^-xt

-dt — the sum of the residues at d= Imni, m = 1 , . . . , n, where C is the curve in the previous exercise. Note that the sum of the residues at ±2mni is -2(2m7t)s-leins

sin(2m7rx + its/2).

Now let n —> oo and show that \n —>• 0. 25. Show that the functional equation for f (JC, 5) easily implies (a) the functional equation for f (5), (b) Kummer's Fourier expansion for log T{x)/\/2jt. The next five problems are taken from Artin [1964]. 26. For 0 < x < 00, let 0(JC) be positive and continuously twice differentiable satisfying (a) 0(JC + 1) = 0(JC), (b) 0(f)0(*±±) = d0(jc), where J is a constant. Prove that 0 is a constant. Hint:Letg(x)

^(^))

= £2 log 0(JC). Observe that g(jc + l) = g(jc)and|(g(f) +

= gW.

27. Showthat0(jc) = F(X)F(1—JC) sin 7TJC satisfies the conditions of the previous problem. Deduce Euler's reflection formula. 28. Prove that a twice continuously differentiable function / that is positive in 0 < x < 00 and satisfies (a) f(x +1) = xf(x) and (b) 22x~xf{x)f{x + \) = ^/nf(2x) is identical to F(JC). 29. It is enough to assume that / is continuously differentiable in the previous problem. This is implied by the following: If g is continuously differentiable, g(x + 1) = g(x), and £(f) + g(*±!) = g(jc), then g = 0. Hint: Observe that

A:=0

The left side tends to J* g'{x)dx = g(l) - g(0) = 0 as n -> 00.

Exercises

53

30. Prove that the example g(x) = Y1T=\ h sin(2n7rx) shows that just continuity is insufficient in the previous problem. 31. Suppose/ and g are differentiable functions such that fix+y) = f(x)f(y) — g(x)g(y)mdg(x+y) = f (x)g (y)+g(x)f(y). Prove that /(*) = eaxcosbx and g(x) = eax sin bx, unless f(x) = g(x) = 0. 32. Prove that ^=o e2nixl/N = l-^VN, where i = ° < * < I. Note that /(0) = / ( l ) and extend fit) as a periodic function to the whole real line. (b) Note that f(t) = E-oo ane2**"', where an = f* fit)e-2*intdt. Conclude that /(0) = E t " o e27Tix2/N = E-oo *»• (c) Show that an = e~2«iNn2/4 ]%n2/2) e (d) Show that (

r

"odd

(e) Use Exercise 19 to evaulate the integral. Another way is to take N = I in id). (Dirichlet) 33. If p is an odd prime, then there is exactly one character x2 that maps Zip)* onto {±1}. Recall that Zip)* is the integers modulo p without 0. Prove that X2ia) = 1 if and only if x2 = a mod/7 is solvable, that is, a is a square in Zip)*. Usually one writes x2 (a) = (~)»which is called the Legendre symbol. 34. Prove that if a is a positive integer prime to p, then ap~1/2 = ( £ ) (mod p). Here p is an odd prime. (Use the fact that Zip)* is a cyclic group.) 35. For pan odd prime, use the previous problem to prove that (—) = (— l)^" 1 )/ 2

and ip = (-1)^ 22- 1} / 8 . (Use 2^ 2 = ieni/4 + (mod/?). Consider the two cases /? = ±1 (mod8) and p = ±3 (mod8) separately.) 36. Prove the law of quadratic reciprocity: For odd primes p and q, ( - ) ( - ) = s h o w t h a t s2 = (-^P- ( T h e P r o o f i s similar (a) For S = J2x=l(pe2nix/p^ to that of Theorem 1.10.8.) (b) Use (a) and Exercise 34 to prove that Sq~l = ( - l ) V ^ (£) (modg).

(c) Show that ^ = E ^ i ( f V 2 7 r ^ / / ? = ( f ) ^ (mod^). (d) Deduce the reciprocity theorem from (b) and (c). (Gauss) 37. For integers a and N with N > 0, define G(a, A0 = E^To ^27r/flx2/iV. (a) For p prime, show that G(l, /?) = E ? = i ( f )^ I > J : / p (b) For p prime show that G( 0and> oo, and suppose that 4>(n) —> 0, (p(n ± /y) —• 0, to get d>(iy) - (b(-i Jo (ii) Deduce Hermite's formula (for reference, see Whittaker and Watson, [1940, p. 269]) x~s

f (*, j ) = — - +

xx~s

f00 (x2 + t2)~s/2 sin(s arctan^/jc)

+ 2 /

——

^

-

1 1 f™ Axt dt (iii) Conclude that f (JC, 2) = ^-^ 2JC2 H x h Jo/ (x2 + t2)2(e2*t Jo 43. (a) For x//(x) = T\x)/ T(JC), note that f'(x) = $(x, 2). (b) Deduce that f00

1

dt.

-1)'

ltdt (x2 + t2)(e27Tt - 1)'

(Use part (iii) of the previous exercise.) (c) Deduce Binet's second formula

in roc) = ([xx - I )) hue in, --J Cx++ -ln(27r) + 2 /

2J

V

2

Jo

1

where x is complex and Re x > 0. (d) Use Hermite's formula in the previous problem to obtain Lerch's formula (1.3.7) for ( £ ? ( * , s)) 5 = 0 . 44. Prove the following properties of Bernoulli polynomials:

U ~p 1

n=M

(c)

Bn(x) =

^ k=0

(d) Bn{\ — x) = (— \)nBn(x). (e)

Bn(lx) =

45. Prove that \dj

£>2q — 1 v-^

L-^J/ — ^ v

56

1 The Gamma and Beta Functions and 00

B2q(x-[x]) = 2(-ir

^ (Inn)2*

(b) Deduce

and

n=0

46. Prove that B2n = (p-l)\2n

P

where G^n is some integer and p is a prime such that p — 1 divides In. (Clausen-von Staudt) Hint: Define Y^=o ft*" = Y^T=o fa" ( m o d k)if ^ divides an - Z?n for all n > 0. Show that /z3 z 5 z7 \

(a) (ez - I) 3 = 2f - + - + - + • • • J (mod4). (b) For prime /?,

(c) For composite m > 4 (^ _ i)^-i = o (modm).

Deduce the result on Bernoulli numbers. (See Polya and Szego [1972, Vol. II, p. 339]. 47. Let C(Z(n)), where Z(n) is the integers modulo n, be the set of all complex functions on Z(w), where n is an odd positive integer. Define F: C(Z(n)) -> C(Z(/i))by n-l

= — V f{k)elnikx'n

for x G Z(/i).

(a) Show that Trace F = ^ ^ ^ I Q e2nik^n. Hint: Use the functions 8x,x e Z(n), where 1.

Define the Dirichlet L-function by

The series converges for Re s > 0, when x is a nontrivial character, that is, X (n) T^ 1 for at least one w e Z(m)*. 51. (a) Prove that when x is nontrivial ^ m—1

L(x,D = — (b) Show that if x is primitive

in— + ++ V xW logg sin— m m *—J V m m ) (c) Prove that when x is even, ^ x (^)^ = 0> a n d also, when x is odd, £x(*)log (d) Prove that

{

^^ £

( ) X (k) log sin ^ ,

^ E m ,

52. Prove that (a) 1 — | + i — ^H = 5(b) 1 + 1 - 1 - 1 + 1 + ^ - . . . = ^ (c) 1 - 1 + 1 - 1 + 1 - 1 + . . . = ^

when x is even,

when x is odd. (Madhava-Leibniz) (Newton) (Euler)

Exercises

59

(d) 1 + 1 - 1 + 1 - 1 - 1 + 1 + . . . = ^ . (e)

i _ I _ I +I+I _ I _ i +I+± _ . . .

(Euler) 2 j 1^5

=

The series for n/4, usually called Leibniz's formula, was known to Madhava in the fourteenth century. See Roy [1990]. Newton [1960, p. 156] produced his series in response to Leibniz's formula by evaluating the integral

I

l+x2

'

Adx 0 1+*4 in two different ways. Series (c) and (d) are attributed to Euler by Scharlau and Opolka [1985, pp. 30 and 83]. Define the generalized Bernoulli numbers by the formula 00

V^Xfa z—4 emx __ Y a=\

„

Xn

n=0

53. (a) Prove the following functional equation for L(x, s), X primitive:

, 8(x)f2nY L(x,l-s) ) ~ 2i* \m) r ( S ) c o s ^ ' where 8 = 0 or 1 according as x is even or odd. Hint: Consider the integral

I

Jc

emi - 1

-dt,

where C is as in problems 23 and 24. Follow the procedure given in those problems. (b) For any integer n > 1, show that

L(XA-n)

= - ^ ^ . n (c) For n > 1 and n = 8 (mod 2) (8 as defined in (a)), prove that

54. Let P be any point between 0 and 1. Show that

r

The notation implies that the integration is over a contour that starts at P, encircles the point 1 in the positive (counterclockwise) direction, returns to P, then encircles the origin in the positive direction, and returns to P. The 1—, 0— indicates that now the path of integration is in the clockwise direction, first around 1 and then 0. See Whittaker and Watson [1940, pp. 256-257].

1 The Gamma and Beta Functions

60 55. Let (a) (b) (c)

G(z) = log F(z). Show that If x > 1/2, then Re G"(x + iy) > 0 for all real y. If x < 1/2, then Re G"(x + iy) < 0 for all sufficiently large y. If 1/2 < a < fe, then

is an increasing function of y on (—oo, oo). (d) The conclusion in (c) also holds if 0 < a < 1/2 and b > I — a. (Ahern and Rudin) 56. Show that

This problem was given without the value by Amend [1996]. FOXTROT © 1996 Bill Amend. Reprinted with permission of Universal Press Syndicate. All rights reserved.

The Hypergeometric Functions

Almost all of the elementary functions of mathematics are either hypergeometric or ratios of hypergeometric functions. A series Ec n is hypergeometric if the ratio cn+\/cn is a rational function of n. Many of the nonelementary functions that arise in mathematics and physics also have representations as hypergeometric series. In this chapter, we introduce three important approaches to hypergeometric functions. First, Euler's fractional integral representation leads easily to the derivation of essential identities and transformations of hypergeometric functions. A secondorder linear differential equation satisfied by a hypergeometric function provides a second method. This equation was also found by Euler and then studied by Gauss. Still later, Riemann observed that a characterization of second-order equations with three regular singularities gives a powerful technique, involving minimal calculation, for obtaining formulas for hypergeometric functions. Third, Barnes expressed a hypergeometric function as a contour integral, which can be seen as a Mellin inversion formula. Some integrals that arise here are really extensions of beta integrals. They also appear in the orthogonality relations for some special orthogonal polynomials. Perceiving their significance, Gauss gave a complete list of contiguous relations for 2^1 functions. These have numerous applications. We show how they imply some continued fraction expansions for hypergeometric functions and also contain three-term recurrence relations for hypergeometric orthogonal polynomials. We discuss one case of the latter in this chapter, namely, Jacobi polynomials. 2.1 The Hypergeometric Series A hypergeometric series is a series J2 cn s u c n m a t Cn+\/cn is a rational function of n. On factorizing the polynomials in n, we obtain cn+i _ "

in + ai)(n + a2) • - (n + ap)x + b2)--(n + bq)(n + 1)' 61

62

2 The Hypergeometric Functions

The x occurs because the polynomial may not be monic. The factor (n + 1) may result from the factorization, or it may not. If not, add it along with the compensating factor (n + 1) in the numerator. At present, a reason for inserting this factor is to introduce n! in the hypergeometric series Y^ cn • This is a convenient factor to have in a hypergeometric series, since it often occurs naturally for many cases that are significant enough to have been given names. Later in this chapter we shall give a more intrinsic reason. From (2.1.1) we have

^

Cn

= c0 y -r-

"(ap)nxn — :='cOpFq[

au...,ap ;

Here the b[ are not negative integers or zero, as that would make the denominator zero. For typographical reasons, we shall sometimes denote the sum on the right side of (2.1.2) by pFq(a\,..., ap\ b\,..., bq\ x) or by pFq. It is natural to apply the ratio test to determine the convergence of the series (2.1.2). Thus,

An immediate consequence of this is the following: Theorem 2.1.1 The series pFq(a\,..., ap\ b\,..., bq\ x) converges absolutely for all x if p < q and for \x\ < 1 if p = q + 1, and it diverges for all x ^ 0 if p > q + 1 and the series does not terminate. Proof It is clear that \cn+\/cn\ —• 0 as n -+ oo if p < q. For p = q + 1, lim^oo \cn+\/cn\ = |JC|, and for p > q + 1, \cn+i/cn\ -> oo as n -> oo. This proves the theorem. • The case |x | = 1 when p = q + 1 is of great interest. The next result gives the conditions for convergence in this case. Theorem 2.1.2 The series q+\ Fq {a\,..., aq+\; b\,..., bq; x) with \x \ = 1 converges absolutely ifRe(Y^bi — J^at) > 0. The series converges conditionally if x = el° ^ 1 and 0 > Re(^fr; — J^a/) > — 1 and the series diverges if

Proof

The coefficient of nth term in q+\ Fq is (a\)n • • • (aq+i)n

2.1 The Hypergeometric Series

63

and the definition of the gamma function implies that this term is

as rc -» oo. Usually one invokes Stirling's formula to obtain this, but that is not necessary. See Formula (1.4.3). The statements about absolute convergence and divergence follow immediately. The part of the theorem concerning conditional convergence can be proved by summation by parts. • This chapter will focus on a study of the special case 2F\(a, b; c; JC), though more general series will be considered in a few places. The 2F\ series was studied extensively by Euler, Pfaff, Gauss, Kummer, and Riemann and most of the present chapter and the next one is a discussion of their fundamental ideas. We saw that 2F1 (a, b; c; JC) diverges in general for x = 1 and Re(c — a — b) < 0. The next theorem due to Gauss describes the behavior of the series as x ->• 1". A proof is given later in the text, where it arises naturally.

Theorem 2.1.3 7f Re(c - a - b) < 0, then 2Fx(a,b;c;x) c a b

* ™- (1 - x) - ~

T{c)T(a + b-c) ~

r(a)T(b)

'

and for c = a + b, . 2F\ {a, b;a + b;x) _ F(a + b) x ™- l o g ( l / ( l - * ) ) = r(a)F(b)' The next result about partial sums of 2F\ (a, b; c; 1) is due to Hill [1908]. It can be stated more generally for p+\ Fp. The proof is left as an exercise. Theorem 2.1.4 Let sn denote the nth partial sum of 2F\(a,b; c; 1). For Re(c - a - b) < 0, r(c)na+b~c and for c = a + b,

^ r(c)iogrc Sn

~

r(a)T(b)'

The theorem is easily believable when we note that the nth term is T(C)

~T(a)T(b)

na+b-c-X

n

The necessary result would now follow if we replace the sum with an integral.

64

2 The Hypergeometric Functions

Many of the elementary functions have representations as hypergeometric series. Here are some examples: (2.1.3) n"11 X = 2XFJ^ F J ^ tan" X=X

l

2 ; -x2x)

;

(2.1.4)

(2.1.5) (2.1.6) This last relation is merely the binomial theorem. We also have sinx = x0F{ f~;

-x2/4\

;

^ )

(2.1.7)

(2.1-8)

()

(2.1.9)

The next set of examples uses limits:

(hbl)

(2.1.10)

a 1 2Fl( ' ';-?-);

(2.1.11)

b)

coshx= lim b

\ l / 2 4a^y (2.1.12)

;x)= /

a,b

x

lim zFif" 1 ' ; 4 ) a,b^oo

\

c

(2-1.13)

ab)

The example of log (1 — JC) = — JC 2^1(1, 1;2;JC) shows that though the series converges for I x I < 1, it has a continuation as a single-valued function in the complex plane from which a line joining 1 to 00 is deleted. This describes the general situation; a 2^1 function has a continuation to the complex plane with branch points at 1 and 00. Definition 2.1.5 series

The hypergeometric function 2^\(a, b\ c\ x) is defined by the

(«x for \x\ < 1, and by continuation elsewhere.

2.2 Euler's Integral Representation

65

When the words "hypergeometric function" are used, they usually refer to the function 2F\(a,b; c\ x). We will usually follow this tradition, but when referring to a hypergeometric series it will not necessarily mean just 2F\. Hypergeometric series will be the series defined in (2.1.2). 2.2 Euler's Integral Representation Contained in the following theorem is an important integral representation of the 2Fi function due to Euler [1769, Vol. 12, pp. 221-230]. This integral also has an interpretation as a fractional integral as discussed in Section 2.9. Theorem 2.2.1

If Re c> Re b > 0, then

in the x plane cut along the real axis from 1 to oo. Here it is understood that arg t = arg(l — t) = 0 and (1 — xt)~a has its principal value. Proof. Suppose at first that \x\ < 1. Expand (1 — xt)~a by the binomial theorem given in (2.1.6) so that the right side of the formula becomes {a)n n

x

C tn+b~l

This is a beta integral, which in terms of the gamma function is T(n + b)r(c-b) Substitute this in the last expression to get T(c)

^

( a ) n r ( n + b) n_

r(fe) 4^ n\T(n + c)

fa,b 2

~

\ c '

This proves the result for |JC| < 1. Since the integral is analytic in the cut plane, the theorem holds for x in this region as well. • The integral in Theorem 2.2.1 may be viewed as the analytic continuation of the 2F1 series, but only when Rec > Reb > 0. The function (1 — xt)~a in the integrand is in general multivalued and one may study the multivalued nature of 2F\(a, b; c\ x) using this integral. To discuss analytic continuation more deeply would require some ideas from the theory of Riemann surfaces, which goes beyond the scope of this book. See Klein [1894]. It is also important to note that we view 2F\(a, b; c; x) as a function of four complex variables a,b,c, and x instead of just x. It is easy to see that ^ 2F\(a,b; c; x) is an entire function of a, b, c if x is fixed and \x\ < 1, for in this case the

66

2 The Hypergeometric Functions

series converges uniformly in every compact domain of the a, b, c space. Analytic continuation may be applied to the parameters a,b,c. The results may at first be obtained under some restrictions, and then extended. For example: Theorem 2.2.2 (Gauss [1812]) (a)n(b)n

n\(c)n

=

For Re(c - a - b) > 0, we have

2L

fa,b

\

Y(c)Y(c-a-b)

=

\ c'J J V

T(c-a)T(c-b)

Proof. Let x ->• 1 ~ in Euler's integral for 2 F\. The result is, by Abel's continuity theorem, /n h

\

r(r\

rl

, .

b

-t)c~a~b-xdt

~\\

_ T(c)T(c-a-b) " T(c-a)T{c-bY when Rec > Reb > 0 and Re(c — a — b) > 0. The condition Rec > Re& > 0 may be removed by continuation. It is, however, instructive to give a proof that does not appeal to the principle of analytic continuation. Our first goal is to prove the relationship ^ ) ( c(c — a — b)

»

'

)

(2.2.1)

If An = — - — n\(c)n

and Bn = f

then

and

c(nAn - (n + l)A n + i) = ———

\n

n!(c+l) ! ( + l ) n _iLL

c+ n

So, since the right sides in the last two expresions are equal,

c(c-a-

b)An = (c - a)(c - b)Bn + cnAn - c(n + \)An+\

and N

N

c(c — a — b) ^^ An = (c — a)(c — b) ^ 0

0

Bn — c(N +

l)AN+\.

2.2 Euler's Integral Representation

67

Now let AT -> oo and observe that (N + l)AN+\ ~ \/Nc~a~b -> 0, because Re(c — a — b) > 0. This proves (2.2.1). Iterate this relation n times to get

r(c-a)T(c-b)

(a,b *

\

T(c + n - a)T(c + n - b)

;1

„,

.

.„,

.

v

772*1

It is an easy verification that the right side -> 1 as n -+ oo. This proves the theorem for Re(c — a — b) > 0. The theorem is called Gauss's summation formula. • The case where one of the upper parameters is a negative integer, thereby making the 2^ia finite sum, is worthy of note. This result was essentially known to the thirteenth century Chinese mathematician Chu and rediscovered later. See Askey [1975, Chapter 7]. Corollary 2.2.3 (Chu-Vandermonde)

Euler's integral for 2^1 can be generalized to pFq. Rewrite it as

Thus, integrating a 1 F o with respect to the beta distribution th~x (1 — t)c~b~l gives a 2 / 7 i, that is, a parameter b is added in the numerator and c in the denominator of the original \Fo(a',t). More generally, we have

, ap+l

\ ' )

=

r(ap+l)r(bq+l-a-a ) )J0 p+lp+l fl . A ^

b\,...,bq

(22 2)

J

when R e ^ + i > Rea^+i > 0. This condition is needed for the convergence of the integral. By a change of variables the expression on the right of (2.2.2) also equals

(2.2.3) Note also that (2.2.2) can be used to change the value of a denominator or numerator parameter in pFq{a\,

...,ap;b\,...,bq;x).

For example, take ap+\ = bq in

68

2 The Hypergeometric Functions

(2.2.2) to get /

p q

i * i , . . . , up

\

\bu ..., bq-u V i ' ~J

T(bq)Y{bq+l - bq)

. / tbi-\l-t)bi+l-bi-lpFq( Jo

u '"\p;xt)dt. \bi,- — < > b q /

(2.2.4)

It should be remarked that when x is a complex variable in (2.2.2) to (2.2.4), then the pFq is in general a multivalued function. Thus, the variable x has to be restricted to a domain where the pFq in the integrand is single valued. One must take care to state the conditions for single-valuedness. We note a special case of (2.2.4). Theorem 2.2.4

For Rec > Red > 0, x ^ 1, and |arg(l — x)\ < it,

One pecularity of Euler's integral for 2^1 is that the 2^1 is obviously symmetric in the upper parameters a and b, whereas it is not evident that the integral remains the same when a and b are interchanged. Erdelyi [1937] has presented a double integral from which the two representations can be obtained:

[r(c -a)T(c

-b)J0 Jo

• (1 - s)c~a~l (1 - tsxYcdtds.

(2.2.5)

The next theorem gives an important application of Euler's integral to the derivation of two transformation formulas of hypergeometric functions. Theorem 2.2.5

oo.

Jo This expression equals 1 and we have the result. The reader should try tofindthe beta integral that corresponds to Gauss's formula for 2F\ at x = 1. 2.3 The Hypergeometric Equation The hypergeometric function satisfies a second-order differential equation with three regular singular points. This equation was found by Euler [1769] and was extensively studied by Gauss [1812] and Kummer [1836]. Riemann [1857] introduced a more abstract approach, which is very important. Our treatment will basically follow Riemann, in a more explicit form given by Papperitz [1889]. The reader who has never seen series solutions of differential equations with regular singular points might find it helpful to read Appendix F first. Let p(x) and q(x) be meromorphic functions. Suppose that the equation =0

(2.3.1)

has regular singularities at the finite points a, p, y and that the indicial equations at these points have solutions a\,a2\b\,b2\ and c\,c2 respectively. Assume that a\ — a2,b\ — b2, and c\ — c2 are not integers. Set x = \/t so that the differential equation is transformed to -q(l/t)y

= O.

(2.3.2)

Since oo is an ordinary point, 2x — x2p(x) and x4q (JC) are analytic at oo. Moreover, since a, p, y are regular singular points, A pyx) — x — a

B

C

x-p

\-ui(x)

' x-y

and

(x- -a)(x

- ] 8 ) ( J C - -y)q(x)

D x —a

where u\(x) and u2(x) are analytic functions.

E

F

x — ft x — y

74

2 The Hypergeometric Functions

The last two relations together with the analyticity of 2x — x2p(x) and x4q (x) at infinity imply that A + B + C = 2 and u\ (JC) = U2(x) = 0. Suppose a solution has the form X^^Lo an(x ~ ° 0 n + \ where the exponent X satisfies the indicial equation X(X

- 1) + XA +

-£ (a-

= 0.

P)(a-y)

Since a\ and a2 are roots of this equation, ax + a2 = \ — A and D Therefore, A = 1 — a\ — ci2 and D = (a — /3)(a — y)a\a2Similarly, B = \-bi-b2

and

E = (0 - oo in the above differential equation to obtain x2(x - \fd-\ 1 + {(1 -axax

a2)x(x - I) 2 + (1 - bx - b2)x2(x - 1 ) } ^ dx

+ {axa2(\ - JC) + bxb2x + cxc2x(x - \)}y = 0.

(2.3.4)

The hypergeometric equation is obtained from this one by another simplification. Write this equation in the form (2.3.1). If y satisfies (2.3.1) and y = xxf, then / satisfies d2f dx2

( V

2X\df x J dx

( \

Xp(x) x

X(X-l)\ x1 J

This equation also has 0, 1, oo as singular points, but the exponents are different. Equation (2.3.4) has exponents a\ and a2 at 0; the new equation has exponents a\ — X and a2 — X at zero. The exponents at oo, however, are c\ + X and c2 + X. By this procedure we can arrange that one exponent at 0 and one exponent at 1 be equal to 0. (At x = 1, we set y = (1— x)kf(x).) Thus the new equation has exponents 0, a2 — a\\ 0, b2 — b\\ c\ + a\ + b\, c2 + a\ + b\. This brings considerable simplification in (2.3.4) since the terms a\a2 and b\b2 vanish. It is traditional to write a = c\ + a\ + b\, b = c2 + a\ + b\, and c = 1 + a\ — a2. After simplification, the equation becomes JC(1 - x)—^ + [c - (a +ft+ 1)*]— - flftv = 0. dx dx1

(2.3.5)

This is Euler's hypergeometric differential equation. It has regular singularities at 0, 1, and oo with exponents 0,1 — c; 0, c — a —ft;and a,ftrespectively. Unless specifically stated, we assume that c, a —ft,and c — a —ftare not integers. Riemann [1857] denoted the set of all solutions of the equation in Theorem 2.3.1 by a

y C\

b2 In particular, the set of solutions of (2.3.5) is denoted by

(2.3.6)

76

2 The Hypergeometric Functions

Our earlier discussion implies that

{

0 ax a2

oo 1 ci bi c2 b2

( 0 = P 0) T(c)T(c-a-b) T(c-a)T(c-b)

+

T{2-c)T(c-a-b) T(\-a)T(\-b)

2.3 The Hypergeometric Equation

79

After some tedious trigonometric calculation, which comes in after applying Euler's reflection formula to the second term on the right, we arrive at the value of B required by the theorem. This proves (2.3.11). Suppose ReZ? > Rta. The right side of (2.3.12) as x -> oo is ~C(-x)~a. To see the behavior of the left side, apply Pfaff's transformation. Then

' x- 1

r(c-a)roo' The assumption that Reb > Re a was used in the last step to evaluate the 2F\ by Gauss's formula. It follows that

The value of D follows from the symmetry in a and b. Corollary 2.3.3 /

r(c)T(c-a-b)

r(c)T(a + b — c)

a,b

Y-a-b

F ( c~a'c~^

> \ (2.3.13)

(c)n

\b+\-n-c

)

Proof. In (2.3.11), replace x by 1 - x and c by 0 + 6 + 1 - c. Then (2.3.14) follows from (2.3.13). Just take a = —n and recall that j ^ — ^ = 0 when n is a nonnegative integer. • The first part of Theorem 2.1.3 also follows from (2.3.13) above. It should also be noted that, since Pfaff's formula 2 Fi(fl,

b; c; x) = (1 - *)- fl 2 Fi(a, c - b; c; x/(x - 1))

gives a continuation of 2^1 from |JC| < 1 to Rex < ^, then (2.3.13) gives the continuation to Rex > 1/2 cut along the real axis from x = 1 to x = 00. The cut comes from the branch points of (1 — x)c~a~b, and once this function is defined on a Riemann surface, 2^1 (a, b\ c; x) is also defined there.

80

2 The Hypergeometric Functions Now consider the function S(x)

dt

-f

We show how Theorem 2.3.2 can Jo be employed to find the asymptotic expansion of the function given above. Wong [ 1989, p. 18] used this function to demonstrate that a certain amount of care should be taken when finding the asymptotic expansion of a function. In this instance, when the method of integration by parts is applied, one gets the expansion 3n(n — IV / ~ "

(23 15)

-

which is obviously incorrect, since the integral is positive and every term of the expansion is negative. However, for t > 1 we have

n=0

If this series is substituted in the integral, then term-by-term integration produces the divergent integrals poo

f Jo

f—n—1/3

x+t

-dt.

JO

If these are interpreted in a "distributional" sense the value of the above integral can be set equal to 2TT

(-1)*

With this interpretation, S(x) has the expansion (after termwise integration) S(X)

7= V ±2liLx-n-W

^T^o

nl

as x -> oo.

(2.3.16)

The correct result is, however, the sum of the two expansions in (2.3.15) and (2.3.16). We obtain this from Theorem 2.3.2. First note that for Re( 0 and Re b > 0,

r

Jo

•xt)~adt

a,b

\ ; 1— x )

.

(2.3.17)

This follows from Euler's integral representation of a 2^1 given in Theorem 2.2.1. To reduce this integral to Euler's form, seU = u/{\ — u). From (2.3.17) it follows

2.3 The Hypergeometric Equation

81

that 1

pOO

s(X) = -

(i + tyl/\i + t/xyldt

x Jo 1

3

-

•

-

,

.

-

-

!

1

.

Apply (2.3.11) to get 3 fr(4/3)F(-2/3) fr(4/3)r(-2/3)

/ I , 1 1\

r(2/3)r(4/3) 2/3 r(i)r(i) x 2

l

/1/3,1/3 A 1 \ 1/3 ' j c j /

1/3 3, 1/3 1/3 ; This is equivalent to the sum of the series in (2.3.15) and (2.3.16). Remark 2.3.1 We have seen that 2F\ (a, b; c; x) is one solution of the hypergeometric equation. This fact was used to show that another independent solution is

xl~c2Fi(a + 1 - c, b + 1 - c\ 2 - c\ x). Here we show how the other solutions can be obtained formally from the series for 2F\. We should write the hypergeometric series as a bilateral series

Since F(l + JC) has poles at x = —1, —2,... , the series has no negative powers of x. It is clear that a change of variables n ->rc+ m, where m is an integer, does not change the bilateral series. Consider the transformation n —> n + a, where a is a noninteger. Now the series takes the form

n=-oo

Wn+aWn+a

(c)«(l)«

~~

nf^

~ xn. (c + Of)n(l + £*)„

(A)

82

2 The Hypergeometric Functions

In the latter expression, the terms with negative values of n vanish if we set c + Qf = l o r l + Q f = l. The last condition gives back the original 2^1 series. The first case, where a = 1 — c, gives X

i _ c p ( f l + l - c)n(b + l - c ) / ^-J (l)n(2-c)n

n

= X

i_c „ ( a + l - c , b + l - c 2*1 I \ X V 2-c

which is the second independent solution. The solutions at 00 are obtained in a similar manner by changing n to —n. In this case (A) becomes

where k is a constant. Since (a + a)-n = (—l) n /(l — a — a)n, write the last series as

Once again, to eliminate that portion of the sum involving negative values of n, take either a = — a ora = — b. In the first case we get ex~a2F1 (a + \—c,a\ a + \—b\ 1/JC) and in the second case cx~b2F\(b -\-l—c,b;b-\-l—a; l/x). Remark 2.3.2 In Section 2.1 we proved that the series 2^1 (0, b\ c; x) has radius of convergence 1. We now reverse the point of view taken in Remark 2.3.1 and see how to obtain convergence from the theory of differential equations. Since the singularities of the equation are at 0, 1, and 00, the radius of convergence is at least 1. If it is more than 1, then the series is an entire function. Moreover, by (2.3.12) it is a linear combination of x~a f\(x) and x~bf2(x), which are solutions at 00. This is possible only if either a or b is an integer. Otherwise, both solutions at 00 are multivalued. (Both a and b cannot be integers since a — b is not an integer.) Liouville's theorem shows us that the integer must be negative and that the 2^1 is a polynomial. Hence, if the 2^1 is an infinite series then the radius of convergence must be 1. We now consider the case where c is an integer. Suppose c is a positive integer; then r(a)T(b)

(a,b

\

^F(a+k)r(b

+ k)

k

and _ r ( a + i - c ) r ( f e + i - c ) !_c

/fl + c - i , f e + i - c 2-c

r ( a + 1 - c + k)T(b + l k\T(2-c

c

+ k)

k X

2.3 The Hypergeometric Equation

83

are equal. Tofindthe second solution in this case, suppose a and b are not negative integers. Consider the limit FI-F2

a

lini = — (Fi - F2) \c=n c^n c — n ac r(a + k)r(b + k)T'(n + k)

k

k\T(n+k)T(n

i_» v ^

{ + ~ n + t)r(fr + 1 - w

j r(a + 1 - « + ife)

r(fe + 1 - n + k) \

— hmx r ' ( 2 - c + Jk) V(2-c

k

The second series is

and the first n — 1 terms in the third series are zero because

l rn

,,. = 0

1 \Z n-\-K)

fork = 0, 1 , . . . , n — 2. The same is not true in the fourth series, because r r (2— c + k)/ F(2 — c + k) has poles at these points. By Euler's reflection formula,

Set x = n — k — 1 to get s*

I 1r\

—c The fourth series can now be written as n-\

_x(k-\)\T(a-k)Y(b-k) _k

r(a + l-n + k)r(b + l-n + k) kwin + k)

k

84

2 The Hypergeometric Functions

So, when a and b are not negative integers, the second solution is

'

\x\\ogx (

(^i)!r(.w^) ^

1

}

(n-Jk-l)!

*

'

(23

*18)

where ^(JC) = r r (x)/r(jc). If a is a negative integer, say —m, then ^ (a + £) is undefined for some values of k. Consequently, the solution given above does not work. To resolve this difficulty observe that

lim {f{a +k)-

ir (a)} = if(l+m-k)-

^(1 + m) for k < m. (2.3.19)

a—t—m

Now, if \jf{a)2F\(a, b; c; x) is subtracted from the second term in (2.3.18), then the resulting series is again a solution of the hypergeometric equation except that now we may let a tend to the negative integer —m. The reader should verify (2.3.19) and also that in this case the second solution is

-m,b

^

\, ;xjlogx

-k

T{b) The case where both a and b are negative integers may be treated in the same way. When c = 0, — 1, — 2 , . . . , then the indicial equation shows that the first solution is

i_c „ / u - l T U - c + l SFli 2 c The second solution in this case can be obtained from (2.3.18) by replacing a and b with a — c + 1 and b — c + 1 respectively. Theorem 2.3.2 and its corollary must be modified when c,a — b, or c — a — b is an integer. The reader should work out the necessary changes. An interesting history of the hypergeometric equation is contained in Gray [1986].

2.4 The Barnes Integral for the Hypergeometric Function

85

2.4 The Barnes Integral for the Hypergeometric Function In a sequence of papers published in the period 1904-1910, Barnes developed an alternative method of treating the hypergeometric function 2F1. A cornerstone of this structure is a contour integral representation of 2F\(a,b; c; x). An understanding of this representation can be obtained through the concept of a Mellin transform. We begin with a simple and familiar example: F(s) = Jo°° xs~le~xdx. It turns out that it is possible to recover the integrated function e~x in terms of a complex integral involving T(s). This inversion formula is given by e~x = — : /

x~sr(s)ds,

c> 0.

(2.4.1)

2TH Jc-i00

This can be proved by Cauchy's residue theorem. Take a rectangular contour L with vertices c ±iR, c — (N + ^) zb iR, where N is a positive integer. The poles of T(s) inside this contour are atO, — 1,...,— N and the residues are (— iy /j! at j = 0, 1 , . . . , N. Cauchy's theorem gives ^7 JL x'sF(s)ds = ^=0 (-l)jxj/jl. Now let R and N tend to infinity and use Theorem 1.4.1 and Corollary 1.4.4 to show that the integral on L minus the line joining c — iR to c + iR tends to zero. This proves (2.4.1). The Mellin transform of a function f(x) is defined by the integral F(s) = Jo°° xs~l f(x)dx. We have studied other examples of Mellin transforms in Chapter 1. The integral / 0 x^~1(l — x)f~ldx is the transform of 0, and

r

-dx Jo ( 1 + * ) ' " Jo is the transform of f(x) = 1/(1 + x ) f . Once again, one can prove that rF(t) l _ —^1 (1 - xy~t-i = 2ni + 2ni

fc+ioc T(s) /f x~s -—ds, T( + t) JJc_i,

Re t > 0 and c> 0, (2.4.2)

and 1 (1+*)'

1 2JTIT(0

7 c _ i0

jc~ J r(,s)r(f - s)ds,

0 < c < Re t.

(2.4.3)

The phenomenon exhibited by (2.4.1) through (2.4.3) continues to hold for a fairly large class of functions f(x). Thus, if F(x) = /0°° xs~lf(x)dx then f(x) = x s IJTJ fc-i00 ~ F(s)ds is true for a class of functions. We do not develop this theory as a whole but prove a few interesting cases. The Mellin transform is further discussed in Chapter 10 with a different motivation.

86

2 The Hypergeometric Functions

The above discussion shows that if we want a complex integral representation for the hypergeometric function we should find its Mellin transform. Now,

••»; -A* c

J -b) Jo

0

tb-Hl-t)c-b-\l+xt)-°dtdx

\ fl V(c) ^su,/1-^ / tb-\\-ty-b-x T(b){c-b)Jo r(s)T(a-s)r(c) fl b_s_x n

' - b) Jo t

/*00

S—1

/ -^—-djcA Jo (l+xty y-b-\d

F(s)F(a - s)F(c) F(b - s)F(c - b) T{a)T(b)T(c-b) T{c-s) T(c)

r(s)F(a-s)r(b-s)

r(a)V(b)

(2.4.4)

V(c-s)

These formal steps can be justified by assuming min(Re a, Re b) > Re s > 0. The auxiliary condition Re c > Reb can be removed by analytic continuation or via contiguous relations, which are treated in Section 2.5. Note that we integrated 2^1 at —x because, in general, 2F\(a, b\ c\ x) has branch points at x = 1 and x = 00 and in (2.4.4) the integral is over the positive real axis. There is another proof of (2.4.4) in Exercise 35. We expect, by inversion, that r(a)F(b)

(a,b

\ _

Pk+io° r(s)F(a - s)F(b - s)

1

(2.4.5)

where min(Re k > 0 and c ^ O , - 1 , —2, This is Barnes's formula and it is the basis for an alternative development of the theory of hypergeometric functions. It should be clear that we can represent a pFq by a similar integral. The precise form of Barnes's [1908] theorem is given next. Theorem 2.4.1 F(a)F(b) ^ (a,b

\

1

fio° F(a + s)F(b + s ) r ( - s )

/

xV f

|arg(—JC)| < TV. The path of integration is curved, if necessary, to separate the poles s = —a — n, s = —b — n, from the poles s = n, where n is an integer > 0. (Such a contour can always be drawn if a and b are not negative integers.) Proof Let L be the closed contour formed by a part of the curve used in the theorem from —{N + \)i\o(N + \)i together with the semicircle of radius Af + \

2.4 The Barnes Integral for the Hypergeometric Function

87

drawn to the right with 0 as center. We first show that the above integral defines an analytic function in |arg(—JC)| < n — 8, 8 > 0. By Euler's reflection formula, write the integrand as

By Corollary 1.4.3, this expression is asymptotic to

Set s = it to get eit(\og\x\+iaig(-x)) p~Ht

pTlt

for |arg(—x) \ < n — 8. This estimate shows that the integral represents an analytic function in |arg(—JC)| < n — 8 for every 8 > 0, and hence it is analytic in |arg(—x)\ < n. We now show that the integral represents the series —xn

for U| < 1.

This will prove the theorem by continuation. (Note that, if we start with the series for 2F\ (a, b; c; JC), then Barnes's integral gives the continuation to the cut region |arg(-*)| < 7T.) On the semicircular part of the contour L the integrand is

smsn for large N. For s = (N + \)ew and |JC| < 1,

sin sit

_

Q\

Since — n + 8 < arg(—x) < n — 8, the last expression is

In 0 < |0| < | , c o s O > -^ and in | < \0\ < | , |sin0| > ^ . So, since log|jt| < 0, the integrand is 0{Na+b-c-xeT2{N+^l°^) for 0 < \0\ < f and l 0(Na+b-c-ie-%W+ 2)) f o r | < |6>| < | . This implies that the integral on the semicircle -> 0 as TV -> oo. Since the pole s = n of the integrand has residue )

n\F(c the theorem is proved.

•

88

2 The Hypergeometric Functions

We can recover the asymptotic expansion contained in (2.3.12) from Barnes's integral quite easily. Suppose a —bis not an integer. Move the line of integration to the left by m units and collect the residues at s = — a — n and s = —b — n. The residue at s = —a — n is

T(b-a-n)T(a+n){-x)-a-n n\Y{c-a-n) 1

'

T(a)T{b-a) (q T(c-a) n\(l+a-b)n

after a little simplification. Thus,

r(a)T(b)

i ds

)

+ ( X))

r(a)r(b - a) y («)„(! +a-c)n _„ T(c-a) ^ n\(l+a-b)n * „r(b)r(a -b)1^ (b)n(l +b-c)n r(c-b) ^^ n\(\+b-a)n

_n X X

'

(2.4.6) where m (a) is the largest integer n such that a +n < m. We define m(b) similarly. The integral is equal to

r(a-m+s)r(b-m 2ni

J_i

+ s)7t

— m +^)r(l — m

_f / -(-x)sds.

For |arg(—x)\ < n — 0, the last integral is a bounded function of m and x. This implies that the expression is O(l/xm), so that we have an asymptotic expansion for 2F\ (a, b; c\ x) in (2.4.6). If a — b is an integer, then some of the poles of T(a + s)F(b + 5) are double poles and logarithmic terms are involved. The reader should work out this case as an exercise. To gain insight into the next result of Barnes [1910], suppose F(s) and G(s) are the Mellin transforms of f(x) and g(x) respectively. The problem is to determine how the Mellin transform of f(x)g(x) is related to F(s) and G(s). Formally, it is easily seen that poo

/ Jo

1

/*oo

rc+ioo

xs-lf(x)g(x)dx = — / xs~lg(x) / Z7Tl

1

JO pc+ioo

= —: / llti Jc-ioo

= ^ - I

F(t)x-ldtdx

Jc-ioo poo

F{t) I

xs~t~lg(x)dxdt

JO

F(t)G(s-t)dt.

(2.4.7)

2.4 The Barnes Integral for the Hypergeometric Function

89

The case of interest to us is where s = 1. Then oo

i

pc+ioo

/ f(x)g(x)dx = — / Apply this to the Mellin pairs xb /(*> = ( f T T ^

F

F(t)G(l - t)dt.

(2.4.8)

^ =

and xd (i +x)-'

G(s) =

r(c)

to obtain 1 - . , 2ni —too

-I

_

.

-ds

oo

for a suitable &. By renaming the parameters, this can be written as / r(a+s)T(b + s)r(c-s)r(d-s)ds 2ni J_ioc Tin +rW(n +dW(h + rW(b + d) ^-—A

(2.4.9)

This formula is due to Barnes and the above proof is due to Titchmarsh [1937]. It is correct when Re(a, b,c,d) > 0. We give another proof, because we have not developed the general theory of Mellin transforms in a rigorous way here. But first note that, if we take f(x) = xa+(\ - x)b-a~l

and g(x) = xc+~l(l -

x)d+~c~l

in (2.4.8), we get S

2TTI A-KX) r(b + s)T(d-s)

-ds =

r(b-a)r(d-c)r(b

+

d-iy

max(-a, -b) < k < min(c, d). (2.4.10) Theorem 2.4.2 If the path of integration is curved to separate the poles of T {a + s)F(b + s)from the poles ofT(c-s)Y(ds), then I := J L /

Y{a + s)F(b + s)T(c - s)V(d - s)ds

T(a + c)T(a + d)T(b + c)V(b + d) Note that a + c, a-\-d,b + c, b + d cannot be 0 or a negative integer.

90

2 The Hypergeometric Functions

Proof. As in the proof of the previous theorem, use Euler's reflection formula to write the integrand as F(l — c + s)(l — d + s)

sin7r(c — s) sinTt(d — s)

Also, let L be a closed contour formed by a part of the curve in the theorem together with a semicircle of radius R to the right of the imaginary axis. By Stirling's theorem (Corollary 1.4.3), the integrand is O(sa+b+c+d-2e-27T]lmsl) as |si -> oo on L. So the integral in Theorem 2.4.2 converges, but for JL we see that Im s can be arbitrarily small when \s\ is large. Thus, we have to assume that Re(a + b + c + d — 1) < 0 to ensure that j L on the semicircle tends to 0 as R —> oo. By Cauchy's residue theorem, _ y . T(a + c + n)Y(b + c + n)T(d - c -

n)(-l)n

n=0

^

T{a + d + n)T(b + d

T(a + d)T(b 4- d)F(c - d)2Fx V

(^^ \+d-c

The 2^is can be summed by Gauss's formula. After some simplification using Euler's reflection formula and trigonometry, the right side of the theorem is obtained. This is under the condition Rc(a + b + c + d — 1) < 0 . The complete result follows by analytic continuation of the parameters a,b,c,d. • Theorem 2.4.2 is the integral analog of Gauss's summation of the 2F\ at JC = 1. Moreover, if we let b = e — it, d = f — it, and s = itx in the theorem and let t ->• oo, we get, after some reduction employing Stirling's formula,

r y0

F(a + c-\-e + f)

Thus, Barnes's integral formula is an extension of the beta integral on (0, 1) and so will be called Barnes's beta integral. It is also called Barnes's first lemma. Theorem 2.4.2 can also be used to prove that a,b \ = T{c)T{c-a-b) c ' y ~ r(c-a)T(c-b)2

p

T(c)T{a + b-c)^

f l

a,b \a+b-c+l'

^c_a_b

(c-a,c-b

2.4 The Barnes Integral for the Hypergeometric Function

91

c — a — b / integer. The proof is an exercise. This result was derived from the hypergeometric differential equation in the previous section. The next theorem gives an integral analog of the Pfaff-Saalschiitz identity. Theorem 2.4.3 For a suitably curved line of integration, so that the decreasing sequences of poles lie to the left and the increasing sequence of poles lies to the right of the contour: 'io° T(a + s)T{b + s)T(c + s)V(l - d - s)T(-s)

J

ds -d + a)T(l-d T{e - a)T{e - b)T(e - c)

~ where d + e — a Proof

Start with the following special case of Theorem 2.4.2:

F(c - a)T(c - b)Y{a + n)T(b + n) fi0°

1 =

/

Y(a + s)T(b + s)T{n - s)T(c - a - b - s)ds.

Multiply both sides by (d)n/[n\(e)n] and sum with respect to n. The result is

r(c-a)F(c-b)r(a)r(b) T(c)

V c,e

1

-a-b-

2Ki J-ioo^n\{e)n 1 ri0° ~2ni J_ioo

a

s

e

a

s

s)ds

f-s,d i^ ^

i r°° r(^) r(a + s)r(b + s)r(e-d + s)r(c-a-b-s)r(-s) ds •i J-ioo T{e-d) T(e + s) (2.4.11) Take c — d, so that the 3F2 on the left becomes 2^1. Thus we have r ( f l ) r ( £ ) I » r ( e - a - b)V(c - a)T(c - b) T(c)T{e-a)T(e-b) T(^) 1 fiocr(a T(e-c)2niJ_ioo I J-io

+ s)F(b + s)F(e - c + s)T(c - a - b T(e + s)

-ds.

This is the required result after renaming the parameters. Note that some of the operations carried out in the proof can be done only after appropriate restrictions on

92

2 The Hypergeometric Functions

the parameters. These restrictions can be removed later by analytic continuation. The reader can check the details. • A corollary of the proof of the previous theorem is the following interesting formula: Theorem 2.4.4 fa,b,c 3 /

\

H d,e

;l

)

Y(d)Y(d-a-b) =

(

Y(d-a)Y(d-b)3F2{e,l+a Y(d)Y(e)Y(d +

a,b,e-c

+

b-d'1

+ e - a - b - c)Y(a + b - d)

Y(a)Y(b)Y(d

+ e - a - b)Y(e - c)

fd — a,d — b,d + e — a — b — c \

Proof

d + e — a — b,d + I — a — b '

\ /

As a consequence of Cauchy's theorem, (2.4.11) is equal to

Y(e-d)

Y(a + n)Y(b + n)Y(e -d + n)Y(c -a-bn\Y(e + n) ±e-a-b-d

n)(-\)n

+ n)Y(a + b-c-

n)(-l)n

Set this equal to the 3F2 on the left of (2.4.11) and the theorem is obtained after reduction. • Corollary 2.4.5

Ifd + e = a + b + c+l, then

a,b,c

\ ; l d,e ' J

T(d)T(e)T(d-a-b)Y(e-a-b)

Y(d-a)Y(d-b)Y(e-a)Y(e-b) 1 a + b-d

r(d)Y(e) Y(a)Y(b)Y(d + e - a - b) d-a,d-b,\

Proof. When d + e = tf + £ + c + l , t h e first 3 F2 on the right in Theorem 2.4.4 becomes 2^1. Evaluate the 2^1 by Gauss's formula and get the result. • Note that when a or b is a negative integer the second expression on the right vanishes because of the factor l/[Y(a)Y(b)] and we recover the Pfaff-Saalschiitz formula. Thus this result is the nonterminating form of the Pfaff-Saalschutz identity.

2.4 The Barnes Integral for the Hypergeometric Function

93

The second term on the right in Theorem 2.4.4 vanishes if we take c = e+n — 1, where n > 1 is an integer. The formula obtained is 3 2

U,b,e \

+ n-\\_Y(d)Y{d-a-b) d,e ' ) ~ Y(d-a)F(d-b)3

( a,b,l-n \ \a + b-d + l,e' ) ' (2.4.12)

2

This leads to an interesting result about the partial sums of 2F\(a, b; e; 1). Set d = a + b + n + € and let 6 -> 0 to get 'a, b, e + n — 1

[ton terms] =

2F

(2.4.13) A particular case, where a = b = ^ and e = 1, was given by Ramanujan in the following striking form: 1

/ 1 \

2

1

/1.3\2

1

2AJ n + 2 2

Bailey [1931, 1932] also proved the next, more general, theorem. Theorem 2.4.6 • m) (x, y, u + m — 1 3F2 ;1

to n terms ;I )

to m

terms.

There is a simple proof from Theorem 3.3.3. Remark The Mellin transform can be seen as the Fourier transform carried over to the multiplicative group (0, oc) by means of the exponential function. In F(s) = /0°° xs~lf(x)dx, write s = a + it and x = e2nu to get F(a + it) = 2TT / ^ ( / ( e 2 ™ ) • e2*™^2*1"*du =: 2;r / ^ g(u)e2nitudu. A new feature in the theory of Mellin transforms is that F(s) is analytic in a vertical strip. Just as the gamma function, a special case of a Mellin transform, has a finite field analog, so does the more general case. Let Fq denote a finite field with q elements and F* its multiplicative part. Let / be a complex-valued function on F*. Its Mellin transform is defined on the group of characters of F* as F(x) = ^2aeF* x(a)f(a)The reader may verify that this has an inversion given by f(s) = ~zi J2X x( f l )^(x)- There is also an analog of Barnes's formula (Theorem 2.4.2) due to Helversen-Pasotto [1978]. A proof based on Mellin transforms is given in Helversen-Pasotto and Sole [1993].

94

2 The Hypergeometric Functions

2.5 Contiguous Relations Gauss defined two hypergeometric functions to be contiguous if they have the same power-series variable, if two of the parameters are pairwise equal, and if the third pair differ by 1. We use F(a±) to denote 2F\ (a ± 1, b; c\ x) respectively. F(b±) and F(c±) are defined similarly. Gauss [1812] showed that a hypergeometric function and any two others contiguous to it are linearly related. Since there are six functions contiguous to a given 2F\, we get (^) = 15 relations. In fact, there are only nine different relations, if the symmetry in a and b is taken into account. These relations can be iterated, so any three hypergeometric functions whose parameters differ by integers are linearly related. These relations are called contiguous relations. In this section we show how Gauss's fifteen relations are derived. Then we briefly point out connections with continued fractions and orthogonal polynomials. Contiguous relations can be iterated and we use the word contiguous in the more general sense when the parameters differ by integers. It is easily verified that —2^1 dx

\x) =—2Fi( \

c

)

c

;x). \

c+l

2.5.1)

)

Since this 2F\ satisfies the equation JC(1 — x)y" + [c - (a + b + l)x]y' - aby = 0, we get the contiguous relation

c(c ic-ia + b+Dx)

^

+

1,»+1. \ c+l )

/«,». \ c

\ )

Q

By means of transformation formulas, this can be changed into other contiguous relations. Apply Pfaff's transformation c

)

\

c

x — 1

to each term in the above equation. After a little simplification where we set u = x/(x — 1) and replace c — b by b, the result is c

) -2Fi

(2.5.3)

2.5 Contiguous Relations

95

This is a contiguous relation due to Euler, who derived it in a different way. If we apply Euler's transformation

to (2.5.2), then we get another contiguous relation:

c-(2c-a-b

+ l)x

F

( a,b c+1 \ c +2

c(c+l)

This is one of the relations Gauss obtained. Euler's method for obtaining (2.5.3) was to use his integral representation of 2F\. By direct integration he found a formula of which the following is a particular case: 1

o

ta-\\-t)c-a-\\-tx)-bdt

rl

= (c + (a + 1 - b)x) I ta{\- t)c~a~\\

r1

Jo

+ l)x / ta+\\ - t)c-a-\\ Jo

-(c-b

- tx)~bdt

- txYbdt.

(2.5.30

This is identical with (2.5.3). See Exercise 23 for a simple way of proving this identity. As another example of how the integral can be used, observe that

(1 - xt)~a = (1 - xt)-a~\\

- xt) = (1 - xtya-x[\ - x + (1 - t)x}.

Substitute the right side in

to obtain 2

r fa'b /y f

I

U

\ * V 1 , A I

i

n | 1 \ X

, j?(a

+

1 b

'

y \ . L A lyi I

\

, (c-^x

* V 1 —L« A | ~T~

v c

;

, n I Z •*

A c+i

The above examples show how contiguous relations arise. Now we give a derivation of Gauss's basic contiguous relations. It is enough to obtain a set of six relations from which Gauss's fifteen are obtained by equating the (!)) pairs of

96

2 The Hypergeometric Functions

them. The first three in this set are

dF JC—=fl(F(fl+)-F),

(2.5.5)

dx dF x— = b(F(b+) - F), dx

f

(2.5.6)

{c\){F(c)-F),

(2.5.7)

dx where F:=2F, I •"";*)

and

and so on. The nth term of F( 0,

n = 0, 1 , . . . .

Conversely, any set of polynomials that satisfies this recurrence relation is orthogonal with respect to a positive measure, which may not be unique. The three-term recurrence relation for Jacobi polynomials comes from the contiguous relation 2b(c - a)(b - a - l ) 2 F i [~1'b \ - [(1 - 2 x ) { b - a - l )

3

+ c

l

;x

+ ( b - a)(b + a - l)(2c - b - a -

1 ) ] 2 F J "' " ; x

Vc -2a(b-c)(b-a

+ l)2Fi[

;^)=0,

(2.5.15)

after proper identification. In particular we require that a = —n, where n is a positive integer. But (2.5.15) continues to hold when the series does not terminate. Another contiguous relation that gives a set of orthogonal polynomials is +

^ ' b;x)+[c-2a-(b-

-(c-a)2Fl(a~~*'b;x\

=0.

a)x]2Fl (2.5.16)

We shall study orthogonal polynomials in detail in later chapters. That will provide the natural setting for some of the contiguous relations in the sense implied by the above remarks.

2.5 Contiguous Relations

101

Kummer [1836] considered the problem of extending the contiguous relations to pFq, but he stopped with the remark that for 3 F 2 (a, b, c; d, e; x) the formulas are more complicated. In particular, the linear contiguous relations require four functions, a 3F2 and three contiguous to it. Kummer also noted that only when x = 1 did the formulas simplify. That is the key to three-term relations for some higher p and q. These will be discussed in the next chapter. Remark 2.5.2 It should be noted that the continued fraction from Gauss's relation (2.5.11) contains a continued fraction for arctanx as a special case. The Taylor series for arctanx converges very slowly when x = 1, but the convergence of the continued fraction is extremely rapid and was once useful in computing approximations of n. See Exercise 25. Remark 2.5.3 The orthogonality relation (2.5.14) is a generalization of the wellknown fact from trigonometry:

I

cosmO cos nO dO =0

form^n.

(2.5.17)

This becomes clear by setting x = cos (9 and cos nO = Tn(x). Then Tn(x) is a polynomial of degree n and (2.5.17) becomes 1

Tm(x)Tn(x)(l -xyl/2(l+xyl/2dx

= 0 form / n.

It is not difficult to show that Tn(x) = CP^l/2^l/2)(x), Another set of polynomials is

where C = (2rc)!/[22w(rc!)2].

sin(n + 1)0 Un(cos0) = Tn(x) and Un(x) are called Chebyshev polynomials of the first and second kind respectively. The three-term recurrence relation for Tn(x) is xTn(x) = -Tn+i(x) +

-Tn-\(x).

This is the trigonometric identity 2 cos 0 cos nO = cos(n + 1)0 + cos(n - 1)0. Several properties of the Chebyshev polynomials translate to elementary trigonometric identities and they form the starting point for generalizations to Jacobi and other sets of orthogonal polynomials. For this reason the reader should keep them in mind when studying the "classical" orthogonal polynomials. A number of exercises at the end of this chapter deal with Chebyshev polynomials. We also

102

2 The Hypergeometric Functions

refer to Vn (cos 0) =

sin{(2n + 1)0/2} —— sin 0/2

and

Wn (cos 0) =

cos{(2rc + 1)0/2} — cos 0/2

as Chebyshev polynomials of the third and fourth kinds respectively. These polynomials are of lesser importance. 2.6 Dilogarithms All the examples given in Section 2.1 of special functions expressible as hypergeometric functions were either 2 Fi or of lower level. The dilogarithm function is an example of a 3F2. This function was first discussed by Euler and later by many other mathematicians including Abel and Kummer. But it is only in the past two decades that it has begun to appear in several different mathematical contexts. Its growing importance is reflected in the two books devoted to it and its generalization, the polylogarithm. See Lewin [1981, 1981a]. Here we give a few elementary properties of dilogarithms. The reader may also wish to see Kirillov [1994] and Zagier [1989]. The latter paper gives a number of interesting applications in number theory and geometry. The dilogarithm is defined by the series 00

xn

U2(x) := Y^ — > 71=1

for

(2- 6 - 1 )

1*1 < 1-

U

From the Taylor expansion of log(l — t) it follows that

-I

X

U2(x) = - I

^-^dt.

(2.6.2)

The integral is defined as a single-valued function in the cut plane C — [1, 00); so we have an analytic continuation of Li2Qc) to this region. The multivaluedness of Li2(jt) can also be studied easily. There are branch points at 1 and 00. If Li2(x) is continued along a loop that winds around x = 1 once, then the value of Li2 changes to Li2(jc) — 2ni log x. This is easily seen from the integral definition. We now obtain the hypergeometric representation from the integral I 1 \

fl

/I 1

\

/I

1 1

by formula (2.2.2). Though it is possible to develop the properties of the dilogarithm without any reference to the theory of hypergeometric series, we note one example where Pfaff's transformation is applicable.

2.6 Dilogarithms

103

Theorem 2.6.1 Li2(jc) + Li2(jc/(jc - 1)) = -±[log(l - x)]2 (Landen's transformation). Proof. By Pfaff's transformation (Theorem 2.2.5)

Set u = t/(t — 1) in the last integral to get rx/{x-\)

; u) 2 ; u )\-u

Jo rx/(X-l)

,

rx/{ -l) rx/{XX-l)

X

X l

,

u

h l

= - / ;u)du-2FA 2F1[ V2 / ^o u-l \ 2 Jo The first integral is Li2(jc/(jc — 1)) and the second integral is

\u\du. J

2

- /

2

Jo

This proves the result.

X

[log(l - x)Y.

•

We give another proof since it involves a different expression for the dilogarithm as a hypergeometric function. This expression is given by )

(2.6.3)

Let x be in the region {JC||JC| < 8 < 1} n {x\\x/(x - 1)| < 8 < 1} = S8, where 5 > 0. Apply Pfaff's transformation to (2.6.3) to get

\ l ; -?—) - l}

Li2(x) = l i m - U d o e2[

= lim 1 J (1 - € log(l - x) + ^[log(l - x)]2 +

Now, oo

i n=\

x

7

n=l

x

'

n=

104

2 The Hypergeometric Functions

Thus,

e2

if/ U2{x)

= lim - { (

1 - e l o g ( l -x)

+ - [ l o g ( l - x)f

3

+

O(e )

n=l

The limit operation may be justified by the fact that for x e S$ and \e\ < 1/2, the relevant series represent analytic functions of x and e. This proves the theorem again. There is another proof in Exercise 38. Theorem 2.6.2 n-\

1

If of = 1,

then

-Li 2 (jc n ) = V U2(cokx). n *-^

^Li 2 (x 2 ) = Li 2 (x) + Li 2 (-x).

(2.6.4) (2.6.5)

2

+ Li 2 (l - x) = ^ - - logJC log(l - x). (2.6.6) 6 Prao/ To prove (2.6.4) start with the factorization (1 - tn) = (1 - t)(l - cot) • • • (1 — oon~lt). Take the logarithm and integrate to get U 2 (JC)

_

rx iog(i - cot)dt

Jo

t

rx iog(i - «r~

" J o

A change of variables shows that the integral on the left is ^Li 2 (x n ). This proves (2.6.4), and (2.6.5) follows by taking n = 2. To derive (2.6.6) integrate by parts: U2w

= - r M I ^ = - log* logo - x) - r JO

t

The last integral, after a change of variables u = 1 — t, is

JO

^Ldt. A — *

2.6 Dilogarithms

105

But

by Theorem 1.2.4. The proof of the theorem is complete.

•

Apparently, the only values JC at which Li2(jc) can be computed in terms of more elementary functions are the eight values x = 0, ± 1 , \, ~l±2: , ^ ^ , ^ y ^ Theorem 2.6.3 Li2(0) = 0,

(2.6.7) 2

Li 2 (D - ^ - ,

(2.6.8)

Li 2 (-1) - - ^ - ,

(2.6.9)

6

it2

1\

1

(2.6.10)

" T2 ~ 2 Li2

(2.6.11)

2

V5-1

(2.6.12)

Li2 Li2

(2.6.13) -it2

-

1

(2.6.14)

10

Proo/ Relation (2.6.7) is obvious and (2.6.8) was done in the proof of Theorem 2.6.2. For (2.6.9), observe that 1 1 ^ 2 2

IT

2

+

1

1

1

1

^ 2

+

Tt"\

~ 6 ~ ~ 2 * '~6~

Set x = \ in (2.6.6) to get (2.6.10).

/ 1 ^ 2

+

-7T

V22

1 +

4 ^

+

"

106

2 The Hypergeometric Functions

The identities (2.6.11) and (2.6.12) can be derived as follows: Landen's transformation and (2.6.5) combine to give

Li2 (-?-rr) + Iu2(x2)

- Li 2 (-x) = ~[log(l - x)f.

(2.6.15)

\x — \ J 2 2 Set the variables in the first two dilogarithmic functions equal to each other. Then x/(x - 1) = x2 and x2 - x - 1 = 0. A solution of this is x = (1 - \/5)/2. Substitute this x in (2.6.15) to obtain

To find another equation involving Li 2 ((3 — V5)/2) and Li 2 ((l — \/5)/2) take JC = (3 — V5)/2 in (2.6.6) to arrive at it2

Now solve these equations to obtain the necessary result. The proofs of the formulas (2.6.13) and (2.6.14) are left as exercises. • There are also two variable equations for the dilogarithm. The following is usually attributed to Abel though it was published earlier by Spence. See Lewin [1981a] for references. The formula is

Li2 \-±— • - ^ 1 = Li2 [-^_1 + Li2 [ - ^ 1 - U2(x) - Li2(v) - log(l-x)log(l-y).

(2.6.16)

This is easily verified by partial differentiation with respect to x or y and is left to the reader. More generally we can define the polylogarithm by the series Limjc := V — *-^ nm

for |JC| < 1, m = 2, 3 , . . . .

(2.6.17)

n=\

The relation d 1 —Li m (x) = - L i ax x is easy to show and one can use this to define the analytic continuation of Lim (;t). The polylogarithm can be expressed as a hypergeometric function as well. The formula is Li w (z) =xm+iFm(

^ '""

\

2.7 Binomial Sums

107

We do not go into the properties of this function any further but instead refer the reader to Zagier's [1989] article and the books mentioned earlier. 2.7 Binomial Sums One area where hypergeometric identities are very useful is in the evaluation of single sums of products of binomial coefficients. The essential character of such sums is revealed by writing them as hypergeometric series. Sums of binomial coefficients that appear to be very different from one another turn out to be examples of the same hypergeometric series. One reason for this is that binomial coefficients can be taken apart and then rearranged to take many different forms. A few examples given below will explain these points. Consider the sum

j+l

U

To write this as a hypergeometric series, look at the ratio CJ+\/CJ as we did when we defined these series. A simple calculation shows that cj+i =

(J-k)(J-n) h- I 1 \ ( j

( j

C

I / \

So and Co = Now, as explained in Section 2.1, we could introduce j \ = (1)7 in the numerator and denominator to get k-\ n We have learned to sum two 3F2 series: the balanced and the well-poised series (see Section 2.2). This is neither though it is actually "nearly" poised. However, we have not yet considered this type of series. But there is another way out. Note that the denominator has (2)7, which can be written as (l) ; +i. Then

s=:k-

(n k-l\

(-k)

^

(-k - lU-n

) (n + l)(k + I) f^

k-l

ff f n l (n + l)(k + 1 )

- \)t

(l)d-k)t

k l

;

2 The Hypergeometric Functions

108

The 2F1 can be evaluated by the Chu-Vandermonde formula and after simplification we get S =

1

ifc + 1

As another example take the sum 2k\(-l

Here ck+i _ (k + n + \)(k + \)(k - rn + n) *±i)(lfc +2) ~^T after simplification. Thus S=

\ _i

'-l-n+m,n,-± 3^2

m —1

m

'

AW

~^~'2~

I

/

The 3F2 is balanced and we can apply the Pfaff-Saalschiitz identity. We get S=

(m~l — n\ (UL\ \ 2 )n+\-m V 2 Jn+l-m

m

_ 1

/n+l-m

which simplifies to (^_j )• The reader may verify that m-r k

+ s\/n + r - s\ / r +k\ ) \ n - k ) { )

also reduces to an example of a Pfaff-Saalschiitz series. As the final example take the series 2n

= S,

where we are assuming that I = min(^, m, n). This reduces to the series (-\Y(2m)\(2n)\ (m - l)\(m + l)\(n - l)\(n +

(-21, -m-l,-n;i • m - I + 1, n - I + l

This is a well-poised series (see Section 2.2) that can be summed by Dixon's formula. However, this result cannot be applied directly because we get a term

2.8 Dougall's Bilateral Sum

109

F(l — I)/ F(l — 21) that is undefined. A way around this is to use the following case of Dixon's formula: -21 - 2e, -m - I - e, -n - I - e m-l-€ + l,n-l-€ +l ' ~

T(l-i€)V(l +m-l r ( l -212e)r(l + m)T{\ + n ) r ( l + m + n)

Apply Euler's reflection formula to the right side to get

T(l+m-e-

€)F(1 +n-l-

e)r(l

In the limit as e - • 0, this expression is t

(21 - 1)! (m - £)!(/! - £)! (£-1)! m!n!(

Thus

Another example that gives a well-poised 3 F 2 is YH=i ^(k+ )(k+n)' Examples can be multiplied but the discussion above is sufficient to explain how hypergeometric identities apply to the evaluation of binomial sums. See Exercise 29 in Chapter 3. 2.8 Dougall's Bilateral Sum The bilateral series

is the subject of this section. In fact, the hypergeometric series 2 ^2 («, b\ c; 1) should be regarded as a special case of the bilateral series where d — 1, since 1/ r(n) = 0 for nonpositive integers n. This explains why we introduced 1/n! in the nth term of the hypergeometric series. The above remark gives us a way of evaluating the above bilateral series. For d = 1, 2 , . . . , the sum reduces to a series that can be evaluated by Gauss's summation of 2F1 (a, b\ c; 1). The following theorem of Carlson allows one to evaluate the bilateral series from its values at d = 1, 2,

110

2 The Hypergeometric Functions

Theorem 2.8.1 Iff(z) is analytic and bounded for Re z > 0 and iff(z) = Ofor z = 0, 1, 2 , . . . , then f(z) is identically zero. Remark The boundedness condition can be relaxed. We need only assume that f(z) = O(ek^), where k < n. The simple proof given below of the particular case is due to Selberg [1944]. Proof. As a consequence of Cauchy's residue theorem, 1

2ni

J

{ z

a ) (

z

\ ) ( z n )

Z

for n > a > 0. Then, for a > 1, 1/(0)1
1. This implies the theorem. Theorem 2.8.2 (Dougall) For 1 + Re(a + b) < Re(c + d), ^ F(a+n)F(b + n) n ^ n

T(c + d-a-b-l) it1 sinTra sinTTZ? T(C - a)r(d - a)T(c - b)T(d - b)'

Proof For Red > Re (a + b — c) + 1, the functions on both sides are bounded analytic functions of d. Let m be an integer in this half plane. For d = m the series on the left is Y(a+n)Y(b + n)

_ r(c n=—m-\-l

=

T(a-m

+ l)T(b -m + l)^(a-m

T(c-m

j^0

+ l)t(b - m

This series can be summed by Gauss's 2^1 formula. Thus Dougall's result can be verified for d equal to an integer in the half plane. Carlson's theorem now implies Theorem 2.8.2. • Gauss's 2^1 sum is itself a consequence of Theorem 2.8.1. We have to prove that 2

F

H c

2.9 Fractional Integration by Parts and Hypergeometric Integrals

111

This relation is true for b = 1, 2 , . . . by the Chu-Vandermonde identity, and then by the above argument the general case follows. For a different example where Carlson's theorem applies, consider the formula

f It is easy to prove this by induction when a is a positive integer. We saw this in Chapter 1. The integral is bounded and analytic in Re a > 8 > 0 and so is the right side of the formula. This proves the result. 2.9 Fractional Integration by Parts and Hypergeometric Integrals Theorem 2.2.1 gives Euler's integral representation for a hypergeometric function. One drawback of this representation is that the symmetry in the parameters a and b of the function is not obvious in the integral. We observed that Erdelyi's double integral (2.2.5) gives the two different representations by changing the order of integration. In this section, we show how fractional integration by parts can be used to transform the integral for 2F\ (a, b\ c\ x) to another integral in which a and b have been interchanged. In fact, fractional integration by parts is a powerful tool and we also use it to prove a formula of Erdelyi. This contains some of the integral formulas considered in this chapter as special cases. It also has implications in the theory of orthogonal polynomials, which will be discussed in Chapter 6. Let (//)(*) := /

f{t)dt

Ja

and

(hf)(x):= f I f(h)dtldt= f Ja

Ja

(x-t)f(t)dt.

Ja

Inductively, it follows that, for a positive integer n,

(Inf)(x) := f

f ... f"

Ja

Ja

= T—!"f77 in-

f(tn^)dtn^

• -dhdt

Ja

[\x-t)n-lf(t)dt.

1)! Ja

A fractional integral Ia, for Re a > 0, is then defined by

:= - ^ - f\x - ty~lf(t)dt. r(a) Ja

(2.9.1)

112

2 The Hypergeometric Functions

The restriction Re a > 0 can be removed by using contour integrals. An interpretation of Euler's integral for iF\(a, b; c; x) as a fractional integral is now evident. The fractional derivatives can also be defined by the formal relation

Red > 0, x ^ 1, |arg(l - JC)| < n. Use (2.9.2) to see that (2.9.4) is equal to

r(c)

(i-tyAlso, ? !

" (1 ~ xt)~a dtb~d

=

db~d ^ T(a + r)tb+r~xxr dtb~d f^0 F(a)r\ + r)r(b + r)td+r~lxr r(a)F(d + r)r\ (a, b 4

T(d)

Substitute this in the last integral to complete the proof of (2.9.6). We now state and prove the formula of Erdelyi [1939] mentioned earlier. Theorem 2.9.1

For Rec > Re/x > 0, x / 1, |arg(l — JC)| < n, we have

r(/x)r(c-/x) "k — a , \ — b fJL

\ )

[ a + b - X , X - \ i (1 - 0 * V

C-fJL

l-Xt

dt.

114

2 The Hypergeometric Functions

Proof. Apply Euler's transformation in Theorem 2.2.5 to the 2^1 inside the integral in (2.9.6). The result is c-X-l

By (2.9.2) and the series representation of 2F1, we see that l l

X-a,X-b

d»-1 f t^1

\

(X-a,X-b

\xt

Substitute this in the last integral and use the fractional-integration-by-parts formula (2.9.3) to get „,

M

d(i - ty-^

f\-a,k-b 2^1 (

—

',xt

"

(\-xty

Write the expression in curly braces as (1 -

T(c-A) r\T{c-X)

Take the (/x — A)th derivative of this expression to obtain

x-a-b t 1 ~ ty-*-1 r ( c — /x)

(a + b-k,c-\

(1-0*

V

' x— i

c — /x

By Pfaff's transformation (Theorem 2.2.5), this is equal to

„-

l-Xt

C-/JL

Substitute this in the last integral. The result is proved.

•

Exercises 1. Complete the proof of the second part of Theorem 2.1.2 concerning conditional convergence.

Exercises

115

2. Suppose that | YTk=iak | and YTk=\ \ak I tend to infinity in the same way, that is, n

£

n

\ak\ < K

k=\

k=\

for all n and K is independent of n. Prove that

lim p ^ = lim **, n^oo } _ ^ k = l ak

«->oo an

provided that the right-hand limit exists. 3. Use the result in Exercise 2 to prove Theorem 2.1.4. 4. (a) Show that |((1 + x)n + (1 - x)n) = 2 * I ( - T I / 2 , -{n + l)/2; 1/2; Find a similar expression for ^((1 + x)n — (1 — x)n). (b) Show that (1+*) 1 1 = 1 + wx 2 Fi(l - /i, 1; 2; -JC).

JC2).

5. Derive the Chu-Vandermonde identity by equating the coefficient of xn on each side of (1 - x)~a{\ - x)~b = (1 - x)~{a+b\ 6. Suppose that log(l + JC) is defined by the series (2.1.3). Use Pfaff's transformation (Theorem 2.2.5) to show that log(l + x) = -log(l + JC)" 1 . 7. Show that Pfaff's transformation is equivalent to -n,c-b C

\ 5 1 )

(b)n = 7—, (c)n

n = 0, 1 , . . . .

This is one way of removing the restriction Re c > Re b > 0 used in the proof of Theorem 2.2.5. 8. Prove the identities (2.1.10) to (2.1.13). 9. Show that 1*1(0; c; JC) = exiFi(c - a; c\ -JC). 10. Suppose x is a complex number not equal to zero or a negative integer. Show that

_n\(n+x)

Jx

Note that the i F\ series exhibits the poles and residues of 11. Show that

F(JC).

...,bq

b\,...,bq when p < q, Res > 0, Re a > 0, and term-by-term integration is permitted.

116

2 The Hypergeometric Functions

12. Show that 2

2

\

;sin2;

/ 1+m

sinmx = msinxiFi I

2

1-m

'

V

2

\

; sin2x I.

2

)

13. Prove that the functions a,b and 2a, 2b, a both satisfy the differential equation x2(x - \)ym -3x(a+b

+

+ {[2(a2 + b2 + 4aZ?) + 3(fl + b) + 1)]JC - (a + ft)(2fl + 2Z? + 1 ) } / + Z?)j = 0. Thus prove Clausen's identity (see Clausen [1828]), 2

2a,2b,a+b

Also prove that a,b

\

(\-a,\-b

\X

14. Show that the Pfaff-Saalschtitz identity (2.2.8) can be written in a completely symmetric form as 3

(a,b,c \ 7r 2 r(d)r(e)[cos7rdcos7re + cos7r«cos7rZ?cos7rc] 2 V d, e ' ) ~ T(d - a)T(d - b)T(d - c)T(e - a)T(e - b)V(e - c)

when the series terminates naturally and d+e = a+b-\-c-\-\. This observation was made by R. William Gosper.

Exercises

117

15. Prove that

k=0

where Tn(x) is the Chebyshev polynomial of the first kind. Find a similar expression for Un(x), the Chebyshev polynomial of the second kind. 16. Prove the following analog of Fermat's little theorem: If JC is a positive integer and p an odd prime then 7^(JC) = TI(JC) (mod p). 17. Pell's equation is JC2 — Dy2 = 1, where D is a square free positive integer. Let S be the set of all positive solutions (JC, y) of Pell's equation. Let (jq, y\) be the solution with least JC in S. Show that (Tn(x\), y\Un-i(x\)), n > 1, is a solution and thus that Pell's equation has infinitely many solutions, if it has one. 18. Let Sn(x) = Un-i(x) with £/_I(JC) = 0. Prove that gcd(SnQc), Sm(x)) = Sgcd(m,n)(x), where JC, m, and n are positive integers. 19. (a) Show that ( 1 - J C 2 ) V J C = 0,

where p{x) is any polynomial of degree < In — 1. Deduce that D(«>«)/V\

^

'

"•"

'

p(a,-1/2)

T^ M

/r*

2

i\

\^X

— 1).

(b) Show that P (a ' a Vx1

F

r(2w+Q!

(X)

(c) What do (a) and (b) mean for the Chebyshev polynomials of the first and second kind respectively? 20. Prove that -n,n+a

+p+l

^(-11,-n-p

l+x\

x-l -n,-n-a

\

2

">\-a-B-2n'l-x

118

2 The Hypergeometric Functions

21. Let x = cos#. Prove that dn~l sin 2 "" 1 0 (-I)"" 1 (2*)! . ^ = sm nO. dxn~l n 2nn\ 22. Prove that the Jacobi polynomials P^a^\x) relation

„ ^ (Jacobi)

satisfy the three-term recurrence

p +a = {In + a + 0 + l){(2w + a + £ + 2)(2n x(2n+a + p + 2)P^\x) =0, n = 1, 2, 3, (Compare this with (2.5.15).) 23. Prove Euler's contiguous relation expressed as the integral formula (2.5.30 by observing that 0=/ -(/ f l (l-O c " f l (lReoReoO. Jo dt 24. Prove the following contiguous relations: (a) (c-2a-(ba)x)F + a (I - x)F(a+) - (c - a)F(a-) = 0. (b) (c - a - 1)F + aF(tf+) - (c - l)F(c-) = 0. (c) (fe - fl)(l - x)F -(c- a)F(a-) + (c - b)F(b-) = 0. (d) c(fe - (c - fl)jc)F - bc{\ - x)F(b+) - (c - b)F{b~) = 0. 25. Prove (a)

*(

f l > 1

^

- i *

l-

2

2

i ^ 2 x 22x

x 1 V 2 V 32*2 arctanx = -— —— —— — 1+3+ 5+ 7+

(c)

w

.

* = _Lii ££.... 4

1 + 3 + 5 + 7+

The 9th approximate gives JT/4 correctly up to seven decimal places,

(e)

i o g i±i = A _^_ _L_ _J_... —1

r— £v— 2 r _ Z v _

A

arcsinjc

^r?

JC 1 • 2 x

i_

3_

2A

2

3A

1 • 2JC

5_

2

4A

3• 4X2 3-
0, as a Fourier series £ Ame2nima. Express Am as an integral over (—00, 00). Use the result in (a) to deduce Hankel's formula for 1/T(s). See Exercise 1.22. 38. (a) Verify Theorem 2.6.1 by differentiating both sides of the equation. (b) Verify the Abel-Spence identity (2.6.16). (c) Prove the identities (2.6.13) and (2.6.14). 39. (a) Prove that Li2(x) - Li2(y) + Li2 (y- ) + Li2 I — — I - Li

x)

Li-yJ

*W-y)

= £--logxlog[i^ (b) Prove that Prove that Li2 ,

x(l-y)

xd-y)2]

. F

l-yl , . F n - y

= TLi2 -x +TLi2 — U 1-xJ *L yl-x

y(l-x)>\ + Li2

+ -\og2y. 1

X \

(Kummer)

L

See Lewin [1981a]. (c) Prove that Li 2 (x)+Li 2 (y)-Li 2 (xy) f x(\-y) Li2 1 - xy ) \ 1 - ^ / 1-x \ ( 1-y \ lQ g 1 — lQg 1 17 •

(Rogers)

(d) Show that if 0 < x < 1 and f(x) e C 2 ((0, 1)) and satisfies (2.6.6) and the functional relation in part (c), then f(x) = Li 2 (x). See Rogers [1907].

122

2 The Hypergeometric Functions

40. Suppose x is a primitive Dirichlet character mod N. Show that

E x(n)u2( n

1

1

-

where g(x) is the Gauss sum defined in Exercise 31. 41. Suppose n is a positive integer. Define .flog/?, .v.& ^ , n., = pk, " 1 0^,

(a) Show £(s) = Y\p{\ — p s) (b) Show that

—

a power of a prime, otherwise.

for Res > 1.

fA(«) = >

for Re s > 1.

(c) Let f(x) = ^2n• 1 on both sides gives 2 f l

'2\;l|=l.

This implies, by Gauss's summation, that

r( a\ > a2 > • • • > an > - - • > bn > bn-\ > • • • > bo = b. Also, , ^ , an+bn an+\ — bn+\ < an+\ — bn = —

an-bn bn = — - — .

Thus an-bn
i.

This requires some computation, and (3.2.9) follows. We started with an elliptic integral and then introduced the arithmetic-geometric mean. Clearly Gauss was approaching the problem from the opposite direction, as presented above. Let us now consider another way of arriving at the elliptic integral from the arithmetic-geometric mean. Gauss also studied the function

/w-

'

Now 2t

2

, l - t

2

)

=(l+t2)f(t2).

(3.2.10)

136

3 Hypergeometric Transformations and Identities

Assume that / is analytic about t = 0. We want the analytic function that satisfies / ( 0 ) = 1 and the functional relation in (3.2.10). Since / is clearly even, f(x) = g(x2) for some function g and At2 Replace t2 with x to get

Write g(x) = J2T=o CnXn, and use this functional equation to get ai=ao/4,

I2 • 3 2 02 = ^ 7 ^ 0 ,

ai=22

I 2 • 32 • 52 42 62a0

This suggests that

n Formula (3.1.11), 'a, b suggests the same identity. It is possible to show directly that f(x) = g(x2) is analytic. See the first few pages of Borwein and Borwein [1987]. However, it is easier to do the argument in the other direction, as done above. Definition 3.2.4

The complete elliptic integral of the second kind is defined as rn/2

E := E(k) := / (1 - k2 sin2 0)1/2d0. (3.2.11) Jo A theorem of Legendre connects the complete elliptic integral of the first kind with that of the second kind. Before proving it we state a lemma about the Wronskian of an hypergeometric equation. (For the reference to Legendre's book, where the result appears, see Whittaker and Watson [1940, p. 520].) Definition 3.2.5 If y\ and j2 are two solutions of a second-order differential equation, then their Wronskian is W(y\, y^) := y\y'2 — Lemma 3.2.6 Ify\ and y2 are two independent solutions of the hypergeometric equation y" + (c — {a + b + \)x)y' — aby = 0, then

where A is a constant.

3.2 The Arithmetic-Geometric Mean and Elliptic Integrals Proof.

137

Multiply the equation

JC(1 - x)y% + (c-(a + b + l)x)yf2 - aby2 = 0 by y\ and subtract from it the equation obtained by interchanging y\ and y2. The result is x(l - x)(yxy'2' - m") +(c-(a

+ b + l)x)(yiy'2 - y2y[) = 0

or JC(1 - x)W'(yu y2) + (c-(a + b + l)x)W = 0. Now solve this equation to verify the result in the lemma.

•

We shall need particular cases of the following two independent solutions of the general hypergeometric equation: (3.2.12) y2 = xl-c(l

- x)c-a-b2Fi(

^~ a^~

b

; 1 — JCV

(3.2.13)

Observe that from (3.2.11) E(k) = -2Fi(i'~2;k2Y Theorem 3.2.7

(3.2.14)

EK' + E'K - KK' = f, where K' := K(k'), E' := E{k'), and

Proof. Set x = k2 so that 1 — x = k'2. The contiguous relation (2.5.9) gives us x(l - J C ) — = -E - - ( 1 - x)K. dx 2 2

(3.2.15)

Similarly, -x{\ - x)

= -E' - -K'. dx 2 2 Multiply (3.2.15) by K' and (3.2.16) by K and add to get

(3.2.16)

EK' + E'K - KK' = 2JC(1 - x)W(K\ K). With a = b = \ and c = 1, Lemma 3.2.6 gives W(/T, AT) = A/JC(1 - JC). SO £ATr + E'K — KK' is a constant. An examination of the asymptotic behavior of K as x —> 1 shows that the constant must be (TC/2). This is left to the reader. •

138

3 Hypergeometric Transformations and Identities

It is possible to prove the following more general result of Elliot [1904] in exactly the same way. The proof is left as an exercise. The formula in Theorem 3.2.7 is called Legendre's relation. Theorem 3.2.8 ;

a+bll

a+b+1

)

\

b+c

Salamin [1976] and Brent [1976] independently combined Legendre's relation with the arithmetic-geometric mean to find an algorithm for approximating it. We conclude this section with a brief sketch of this application. Some of the details are left for the reader to work out. Lemma 3.2.9

If{an} and [bn] are sequences in Definition 3.2.2, then

where n/2 rpn/2

Jn := / Jo

(al cos2 0+b\ sin2 0)

1/2

d6,

and In has —^ as the power in the integrand. Lemma 3.2.10

E(k)= ( l - ^ 2 n - 1 c 2 where c2n=a2n-

b\.

Proof. From (3.2.9) we know that /„ = I {a, b) =: I. By Lemma 3.2.9,

2(jn+l-a2n+ll)-(jn-a2nl) = (anbn-2a2n+l+a2n)l

3.2 The Arithmetic-Geometric Mean and Elliptic Integrals

139

Rewrite this equation as

2"+l(jn+l-a2n+1l)-2"(jn-a2nl)

= 2^c\l

and sum it from n = 0 to n = m to get /

7-2

\

m

(yTO+i-am+1/J = I a ~ 2 ^

2

^ j

7

'

(3.2.17)

where J := Jo. Now, Jm+l -am+llm+l)

l2

=2?m+

2

shl2 Odd

[^

cm+l

Since c^ +1 tends to zero quadratically, the last term tends to zero. Let m -> oo in (3.2.17) and then take a = 1 and b = kr. The lemma is proved. • Theorem 3.2.11 ix =

^ _ y^ 00

2nc2'

where cl = a2 — bl with CLQ = 1 and bo = 4=.

Proof. Take & = -4=. Then k' = 4 j and Legendre's relation becomes — = {IE - K)K,

where £ = £ ( -r= } and K = , V > 2/

This implies oo

"I Cn

n=0 Since K = n/(2M(l,

~ ~2' J

- ^ ) ) , the result follows.

•

An algorithm based on this theorem has been used to compute millions of digits of n. Define 7tm : =

Then nm increases monotonically to n. Note that cn = y/a% — b2n = c^_x/Aan. The an and bn are computed by the arithmetic-geometric mean algorithm. For more information on the computation of 7T, see Berggren, Borwein, and Borwein [1997].

140

3 Hypergeometric Transformations and Identities 3.3 Transformations of Balanced Series

In the previous chapter we saw how a general 2F\ transforms under a fractional linear transformation and how to evaluate the sum of the series when x = 1. In the case of quadratic transformations, there were restrictions on the parameters. For higher p+\Fp, transformations and summation formulas do not exist in general. There are, however, two classes of hypergeometric series for which some results can be obtained. Definition 3.3.1 A hypergeometric series

is called k-balanced where k is a positive integer, if x = 1, if one of the ais is a negative integer, and if

i=0

i=l

The condition that the series terminates may seem artificial, but without it many results do not hold. The case k = 1 is very important, and then the series is called balanced or Saalschutzian. Definition 3.3.2 tions

If the parameters in the hypergeometric series satisfy the relaa0 + 1 = a\ + b\ = • • • = ap + bp

the series is called well poised. It is nearly poised if all but one of the pairs of parameters have the same sum. In Section 3.4 we shall give a connection between the two kinds of series considered in Definitions 3.3.1 and 3.3.2. We begin with a study of balanced series. The main theorem of this section is the following result of Whipple, which transforms a balanced 4F3 to another balanced 4F3. Theorem 3.3.3 (e - a)n(f - a)n *

F

X4

where

( \

—n,a,d — b,d —

3.3 Transformations of Balanced Series

141

Proof. Start with Euler's transformation:

Rewrite this with different parameters:

Suppose c — a — b = f — d — e and multiply the two identities to get

c ••')'«{' 7

••')->'•{ 7

-j^/^-

The coefficient of x" on the left side is

A (a)k(b)k(f ~ d)n-k(f - e)n-k

f^Q

(c) t *!(/) B _ t (n - *)!

This expression can be rewritten as

(f-d)n(f-e)n «!(/)„

( \c,d-f-n

a,b,l-f-n,-n + l,e-f-n

+

Equating this to the coefficient of xn on the right side, we obtain 4^31

,

r'

'

'. 1

£

i 1' 1

_ Wn(g)/i f -n,c-a,c-b,\f ~n \ 4 ~ (f - d)n(f - e)n \ cA-d-nA-e-n ' / This is equivalent to the statement of the theorem. This result is due to Whipple [ 1926]. For a different proof see Remark 3.4.2. • The next result was given by Sheppard [1912]. Corollary 33.4

'-n,a,b d,e

;

\ {d-a)n{e-a)n 3 V ~ (d)n(e)n

Proof. Let / —• oo but keep f — c fixed in Theorem 3.3.3 so that

142

3 Hypergeometric Transformations and Identities

A similar change takes place on the right side and the end result is

n,a,b

\

(e-a)n

(

-n,a,d-b

Sheppard's transformation is obtained by applying this transformation to itself. The corollary is proved. • The formula in Corollary 3.3.4 has some interesting special cases. For example, suppose the left side is ^-balanced, that is, d + e = k — n + a + b. Then the right side is a sum of k terms. In particular, k = 1 gives back the PfaffSaalschiitz identity. Corollary 3.3.5 \ Y(e)Y(d + e — a— b — c) ^(a,b,c 2 \ A ; l I = FT TFZTi Z ^H \ a, e J V(e — a)i(d + e — b — c) when the two series converge. 3f

( a,d — b,d — c . ;1 AA. u \a, a -\- e — b — c

Proof. Let n -> oo and keep f + n fixed. Since the number of terms of the series tends to infinity, Tannery's theorem may be used to justify the calculation. The left side in Theorem 3.3.3 becomes a, b, c d,e ' To find the right side, write

(e - a)n(f - a)n (e)n(f)n

T(e - a + n)T(f -a+n) r(e)T(f) T(e - a)T(f - a) T(e + n)T(f + n) Y(e) r(fl-/ + l)r(-w-/ + l) Y(e-a T(e - a) F(a - f - n + \)T(-f + 1) T(e + n) '

where Euler's reflection formula was used to derive the second equality. Recall that

l-f-n

= d + e-a-b-c.

(e - a)n(f - a)n (e)n(f)n

So

T(e)T(d + e - a - b - c) / T ( a - / + 1)I> - a + n)\ r(e-a)r(d + e-b-c) \ T(-f + 1)I> + n) ) '

Asn —> oo and n + / i s fixed, we have— / —> oo and the expression in parentheses equals 1 in the limit. The corollary follows. • Corollary 3.3.5 was given by Kummer [1836]. If we apply Kummer's transformation to itself we get a theorem of Thomae [1879]:

3.4 Whipple's Transformation

143

Corollary 3.3.6 2,b,c

A ; 11 =

r(d)T(e)r(s)

_(d-a,e-a,s

3F2(

A ; 1 1,

where s = d + e — a — b — c.

3.4 Whipple's Transformation The main result of this section is an important formula of Whipple [1926] that connects a terminating well-poised 1F^) with a balanced 4F3. We prove it by a method of Bailey, which requires that we know the value of a general well-poised 3F2 at x = 1. In Chapter 2, we showed that the latter result, known as Dixon's theorem, is a consequence of Dougall's theorem. See Remark 2.2.2 in Chapter 2. Because Dougall's theorem is itself a corollary of Whipple's transformation, it would be nice if we had a direct proof of Dixon's formula, that is, one that does not use Dougall's formula. Several such proofs are known. We give one that follows from a quadratic transformation given in Section 3.1. Theorem 3.4.1 a, -b, -c r((q/2) + l)r(g + b + l)r(a + c + l)r((a/2) + b + c + 1) T(a + l)r((a/2) + b + l)r((a/2) + c + \)T(a + b + c + 1)' Proof. If a = —n, a negative integer, in the quadratic transformation (3.1.15), then both sides of the equation are polynomials in x. Take x = 1. If a is an even negative integer, then we get -2n,b,c \ ;A -2n-b, 1- 2 n - c ' /

(2n)\(b + c + n)n (fe

If a is an odd negative integer, then /

-2n-l,b,c

\

Thus Theorem 3.4.1 is verified when a is a negative integer. Now suppose that c is a positive integer and a is arbitrary. In this case both sides are rational functions of a and the identity is true for an infinite number of values of a. Thus, we have shown that the identity holds if c is an integer and a and b are arbitrary. In the general case, for Re c > Re(—a/2 — b—l), both sides of the identity are bounded analytic functions of c and equal for c = 1, 2, 3, By Carlson's theorem the result is proved. •

144

3 Hypergeometric Transformations and Identities

Kummer's identity, which gives the value of a well-poised 2F\ at x = - 1 , is a corollary of Dixon's theorem. To see this, let c -> oo. Remark 3.4.1 We noted earlier that the balanced identities in their simplest form come from the factorization

In a similar sense, the well-poised series comes from

(i-xrb(i+xyb = a-xV Equate the coefficient of x2n from both sides to get (b)k(b) -k £"

t

i

Ln-

This is Kummer's identity for 2

^a-b+l

when a is a negative even integer. As in the proof of Theorem 3.4.1, we can now obtain the general result of Kummer. The 2F\ result was so special that Kummer failed to realize that series of a similar nature could be studied at the 3 F2 and higher levels. This is not surprising for well-poised series are not on the surface. We also need the following lemma to prove Whipple's theorem. It is proved by Bailey's [1935] method mentioned at the beginning of this section. Lemma 3.4.2 / 5

a,b,c,d,-m

^

\ a - b-Y \,a - c + l , a - d+ \,a + m + 1 ' (a/2), a -b-c+l,d,-m a — b + \,a — c + l,d — m — a/2'

Proof. By the Pfaff-Saalschiitz identity for a balanced 3 F2 we have

(-n)r(a-b-c+l)r(a + n)r r\(a - b + l)r(a - c + l)r

(b)n(c)n (a - b + \)n{a - c + !)„ '

3.4 Whipple's Transformation

145

So 5

a, b, c, d, —m I,a - cc + l,a- l,a-d-\ d+

\ a -b+

l,a+m

(a)n(d)n(-m)n , A (-n)r(a - b - c + l)r(a + n)r r a i\(a — d + l)n(a + m + \)n _ K — b + l)r(a — c + l ) r

(n - r)\r\(a -b + \)r(a - c + l) r (a - d + \)n{a + m + l) n (set t = n — r), (a)t+2r(d)t+r(-m)t+r(a

-b-c+

l)r(-l)r

r=0 ?=0

{a)2r(d)r(—m)r(a — b — c + l) r (—l) r b + l) r (a — c + l) r (a — J + l) r (a + m + l ) r

m

^

r!(a - d + r + 1),(a + m + r + 1),

The inner sum can be computed by Dixon's identity (Theorem 3.4.1). An easy calculation then gives the required relation. • Lemma 3.4.2 transforms a terminating well-poised 5F4 to a balanced 4F3. Corollary 3.4.3 a,(a/2)

+ l,c,d,-m

a/2, a-c+l,a-d '

+ l,a+m

\ _ (0 + l)m(a - c - d + l) m + l, ' 1JI — (a - c + \)m(a - d + l) m

Take Z? = (a/2) + 1 in Lemma 3.4.2. The 4 F 3 reduces to a balanced

Theorem 3.4.4 p 7

a(a/2)

+

\a/2, a-b+ l,a-c+l,a-d (a + l)m(a-d-e + l)m — J + l) m (a — e + l)m

l,b,c,d,e,-m d+ l + + l,a-e + ( a-b \a — b +\,a

l + + l' l, c — c +\,d + e — a — m'

146

3 Hypergeometric Transformations and Identities

Proof. The proof of this theorem is exactly the same as that of Lemma 3.4.2 except that one uses Corollary 3.4.3 instead of Dixon's theorem. Thus a/2, a -b + l,a-c+l,a-d

+

(a)n((a/2) +

^

n\((a/2))n(a -d+

l)n(d)n(e)n(-m)n

l)n(a - e + l)n(a + m + \)n

-n)r(a-b-c+l)r(a r\(a-b+

l,a-e

+ n)r

l)r(a-c+

l) r

After a calculation similar to the one in Lemma 3.4.2, this sum equals

£

(a)2r((a/2) + r\(a -b + \)r(a - c + \)r(a -d+

l)r(d)r(e)r(-m)r l)r(a/2)r(a - e + \)r(a + m + \)r

(a + 2r)t((a/2) + r + \)t (d + r),0 + r)t(-m + r)? t\((a/2) + r),(a - d + r + l),(a - ^ + r + l) f (« + m + r + 1 ) /

t=Q

The inner sum can be evaluated by Corollary 3.4.3 and the result follows after a straightforward calculation. • Since the two elementary identities in Remark 3.4.1 are not related, we see the very surprising nature of Whipple's identity. In a later chapter we give a more natural proof of Whipple's theorem as a consequence of some properties of Jacobi polynomials. We refer to the above 7 F6 as a very well poised 7 F6. The word "very" refers to the factor

((a/2) + 1)^ = a + 2k (a/2)k " a The 5 F4 in Corollary 3.4.3 is also very well poised. Remark 3.4,2 Theorem 3.3.2 is a particular case of Theorem 3.4.4 because of the symmetry in the parameters b, c, d, e in the 1F^. Whipple also stated a more general form of Theorem 3.4.4. Theorem 3.4.5 .

7

6

")

V«*/-/

I

^-5 " ? ^ » ' • J ^ ? ./

>/2,a-Z?+l,«-c+l,a-J+l,a-^

*

+ l,a-/

+ l'

T(a - d + l ) r ( « - e + l ) r ( a - / + l ) r ( a - d - e - / + l)

- e- f

a-b-c

+

l,d,e,f

3.5 Dougall's Formula and Hypergeometric Identities

147

provided the series on the right side terminates and the one on the left converges. Proof. This is a consequence of Carlson's theorem and Theorem 3.4.4. The reader should work out the details or see Bailey [1935, p. 40]. • Theorem 3.4.6 f

6 5

I*,

yi*/

**y

|

J.,

I/,

1^,

U-,

W

^

>/2, a - f r + l , f l _ c + l , 00 to see this. Part (i) of the last theorem is due to Watson [1925], who proved it for the case where a = —n, a negative integer. Watson's theorem can also be obtained by equating the coefficient of xn on each side of the quadratic transformation (3.1.11). The general case then follows by Carlson's theorem. Another way is to multiply Equation (3.1.3) by (JC - x2)c~l and integrate over (0, 1). This only works in the terminating case. Remark We end this section with the following comments on well-poised series. Let /./ x

r- (

—n9a\,a2,...

f(x)=q+iFJ H

\ 1 — n — a\, ...,

,aq q

1— n — aq

\-x

150

3 Hypergeometric Transformations and Identities

Then the polynomial f(x) satisfies the relation

A polynomial g (x) that satisfies g (x) = xn g (1 /x) is called a reciprocal polynomial of degree n. These polynomials have the form g(x) =ao + axx + a2x2 + • • • + a2xn~2 + axxn~x + aoxn. Note that /(JC) is reciprocal if either q or n is even. It is easy to check that if g is the reciprocal polynomial given above then g{x) = ao(l + x)n + axx{\ + x)n~2 + • • • + a v x v (l + x) w " 2v for some ao, « i , . . . , dv which can be defined in terms of ao,a\, [n/2]. It can be shown that

(3.5.1)

Here v =

0 deg pk - df + 1.

The last value of d is used only if it is an integer greater than deg pk — d' + 1. This completes Gosper's algorithm. It decides whether a partial sum of a hypergeometric series can be expressed as a hypergeometric term and gives its value if it does. Zeilberger [1982] extended the scope of this algorithm by taking ck as a function of two variables n and k rather than just k. We discuss the Wilf-Zeilberger method here and in the next section. This method is very powerful in proving hypergeometric identities. Suppose the identity to be proved can be written as

k

where A(n) ^ 0 and n > 0. Divide both sides by A(n) and write the identity as ]TF0X,£) =

1.

(3.10.11)

k

This implies that

Earlier we were trying to express F(n,k), or rather T(n,k), as the difference 0

2n ukf \f

2a

\f

2b

m+nj

184

3 Hypergeometric Transformations and Identities

v

k=0

*

'

*

*

U

'

30. Prove that a, 1 + (3/2, J/2, (J + l)/2, (3 - J, 1 + 2(3 - J + m, -m 7

6

' a / 2 , 1 + a - d/2,a + (1 - d ) / 2 , 1 + a", d - a -

OT,

l+a

+ m'

(l+a)m(l+2a-2d)m 2a-d)m

See Bailey [1935, p. 98]. 31. Prove that ab

'

_ ~3

2a,2b,a + b 2

\2a + 2b-

l,a

See Bailey [1935, p. 86]. 32. Prove Theorem 3.6.1. 33. Prove that tow terms

3F2

n)

F

(x,y,v

+n - l

y + n) 3 2 y v,x + y + n

;1

\

J

to m terms.

34. Prove formula (3.8.2). 35. Prove that Wilson's polynomial in Definition 8.1 is symmetric in a, b, c, and d.

Exercises

185

36. Show that 2)/2,b,c,d,e,-n + l-c,a + l-d,a

(a/2), a + l-b,a

+ l-e,a

+

(a + \)n(a - b - c)n(a - b - d)n(a - c - d)n (a + l - b)n(a + 1 - c)n(a + 1 - d)n(a - b - c - d)n n(n + 2a - b - c - d)(a -b-c-d) c)(a - b - d)(a - c - d)

1 + (a-b-

when e = 2a -\- n — b — c — d. The 7 F$ is 4-balanced and very well poised. 37. Prove formulas (3.5.2), (3.5.3), and (3.5.4) connected with reciprocal polynomials. 38. Prove Bailey's cubic transformations: a, 2b — a — 1, a + 2 — 2b x

(a)

= (l-xy«3F2\

(b)

a a+l a±2 3' 3 ' 3

^

-

a

b

-21X

+ l 4(1-.

' a,b-\,a + l-b 2b, 2a + 2 - 2b ; x X\

"

/

a a+\ a+2 3' 3 ' 3

3^2

27X2 '

For comments on these cubic transformations and for the reference to Bailey, see Askey [1994]. 39. Show that 2a

,a

2F\

and a, a + 1/2

2a 40. Define

#

\ _

1

' X ) ~ JT^x~ n-\

0, which is necessary for convergence in (4.1.6), consider the integral r(o+)

or r(0+)

/

f

\

c-a-l

dt.

(4.1.10)

These integrals are also solutions of the confluent equation, but without the restrictions on x and a needed in (4.1.6). From (4.1.10) and (4.1.8), we can once again obtain the 2^0 asymptotic expansion for large |JC|, when |arg x\ < n — 8 < n. Relations among solutions of the hypergeometric equation suggest corresponding relations among solutions of the confluent equation. These can then be proved rigorously. Similarly, transformations of hypergeometric functions imply transformations of the 1 F\ function. The following are a few examples. In Pfaff's transformation,

change x to x/b and let b -+ 00 to get Kummer's first transformation, (4.1.11) A similar procedure applied to the quadratic transformation, /

1

\

Ax leads to Kummer's second transformation, ~~1/2>X)'

xj^e^oFif

(4.1.12)

Finally, the three-term relation a+1 —b (a + l-c)F(l-b)2

r(a)F(c-b)

l

a,b \ c '

{ X)

~

lFl

{

2-c

192

4 Bessel Functions and Confluent Hypergeometric Functions

suggests that

i_c j , (a + \ -c 2-c 9 \ (4.1.13) Formulas (4.1.11) and (4.1.12) can be proved directly. Thus, the coefficient of xn on the right side of (4.1.11) is

-»,c-a «'/ 'V c

(c)kk\(n-k)\

(a)n which is the coefficient of xn on the left side. There is a similar proof of (4.1.12). We give a proof of (4.1.13) in the next section where we approach this topic from a different point of view. 4.2 Barnes's Integral for \F\ We can find the contour integral representation for \F\{a\ c\ x) by computing its Mellin transform. This is similar to finding such a representation for the hypergeometric function. Let )dx. By Kummer's first transformation (4.1.11), /•CO

/•OO

5 _i

_x _ ( c — a

^ > (c — a)n _x

\

s+n_\j

Jo n= C n \ c -a,s z

\

c

\ T{c) r(s)r(a-s) ) r(a) r(c-s)

tl

By Mellin inversion, we should have x

ds

or (4.2.2)

4.2 Barnes's Integral for {F{

193

Of course, once we have seen Barnes's integral for a 2^1, this can be written by analogy. In (4.2.2) we have — x > 0, but this can be extended. The next theorem gives the extension and is due to Barnes. Theorem 4.2.1

For |arg(—x) \ < n/2 and a not a negative integer or zero,

where the path of integration is curved, if necessary, to separate the negative poles from the positive ones. The proof follows the same lines as that of Theorem 2.4.1. The reader should work out the details. Again as in Chapter 2, this representation of 1F1 can be used to obtain an asymptotic expansion by moving the line of integration to the left. The residues come from the poles of F(a + s) at s = —a — n. The result is contained in the next theorem. Theorem 4.2.2 x

For Re x < 0,

i( 0, iFi(a;c;x)~————-2F0I T(a)xc~a Proof

; \

—

x/

This follows from Theorem 4.2.2 after an application of (4.1.11).

•

Now note that the 2^0 in Theorem 4.2.2 suggests the integral J = —: /

T(-s)r(l - c - s)Y(a + s)xsds.

(4.2.3)

2TH J_ioo

Again the line of integration is suitably curved. By moving the line of integration to the left and picking up the residues at s = — k — a, where k > 0 is an integer, we get

1

27Zi

r—a—n+i

J-a-n-io

- c - s)T(a + s)xsds.

(4.2.4)

To ensure the validity of this formula, we need an estimate of the integrand on s = o + iT, where T is large and — a — n < a < 0. By Stirling's formula (see

194

4 Bessel Functions and Confluent Hypergeometric Functions

Corollary 1.4.4), \T(-s)r(l-c-s)r(a+s)xs\ — M The expression on the right-hand side dies out exponentially when |argx| < 3n/2 — 8 < 37r/2. We assume this condition and (4.2.4) is then true. The last integral in (4.2.4) is equal to n

rioo

— /j-too r(a+n- s)T(l

2TT/

+a-c + n- s)T(s - n)xsds = 0(x~a~n)

when |JC | is large. Thus J ~T{a)T(l+a-c)x-a2F0[

"'* ' "

";--),

(4.2.5)

the asymptotic expansion being valid for |argx\ < ?>TT/2. However, when the line of integration is moved to the right, it can be seen that J = I » r ( l - c) 1*1 (*; xj + I > + 1 - c)T(c - l)xl~c\Fx(a ^]_~ °'; (4.2.6) when |argx| < 37r/2. This proves the next theorem. Theorem 4.2.4

For |argx \ < ?>n/2, (4.2.5) and (4.2.6) hold and

r ( a ) r ( l - c ) i F i ( " ; * ) + T(a + 1 - c)T(c - l)xv-cxFx ( " ^ Z

0

>x

Observe that this is the same as relation (4.1.13). This theorem gives the linear combination of the two independent i F\ that produce the recessive solution of the confluent equation. This is of special interest in numerical work. To clarify this point, consider the simpler equation y" — y = 0, which has independent solutions sinhx and coshx as well as ex and e~x. This equation has an essential singularity at oo and, in the neighborhood of this point for Rex > 0, e~x is the recessive solution. Any other solution independent of e~x is a dominant solution. Thus the combination Aex + Be~x can be computed very accurately from values of ex and e~x. However, A cosh x — B sinh x creates problems, especially when A ^ B and x has a large positive real part.

4.3 Whittaker Functions

195

4.3 Whittaker Functions Whittaker [1904] gave another important form of the confluent equation. This is obtained from Kummer's equation (4.1.3) by a transformation that eliminates the first derivative from the equation. Set y = ex/2x~c/2co(x) in (4.1.3). The equation satisfied by co is

Two independent solutions of this equation can have a more symmetric form if we set

c = 1 + 2m,

c

a= k

or

m=

c-\ , 2

1 a = - +m — k. 2

The result is Whittaker's equation:

w + U + i + lzi!' {

4

x

x2

From the solutions (4.1.4) of (4.1.3), it is clear that when 2m is not an integer, two independent solutions of (4.3.1) are

( j

2 m

)

(4.3.2)

and

'

2

( j ~ ™ ^

k

\

(4.3.3)

The solutions M^±m(;c) are called Whittaker functions. Because of the factors X2 ±m , the functions are not single valued in the complex plane. Usually one restricts x to |argx| < it. Formulas for i Fi obviously carry over to the Whittaker functions. Kummer's first formula, for example, takes the form x-*-mMktm(x) = (-xyi-mM-k,m(-x).

(4.3.4)

A drawback of the functions M&,±m (JC) is that one of them is not defined when 2m is an integer. Moreover, the asymptotic behavior of the solution of Whittaker's equation is not easily obtained from these functions. So we use the integral in

196

4 Bessel Functions and Confluent Hypergeometric Functions

(4.1.10) to derive another Whittaker function, Wk,m(x). This is defined by

(0+)

/

l

(-t)-k—2+m

t\k-±+m

1+ e~fdt, (4.3.5) o V XJ where arg JC takes its principal value and the contour does not contain the point t = — JC. Moreover, |arg(—01 < n, and when t approaches 0 along the contour, arg(l + t/x) —> 0. This makes the integrand single valued. It is easily verified that Wk,m(x) is also a solution of (4.3.1). Note that Whittaker's equation (4.3.1) is unchanged when x and k change sign. Thus W-k,m(—x) is also a solution and is independent of Wk,m (x)> This is clear when one considers the asymptotic expansion for Wk,m(x). The reader should verify that the remarks after (4.1.10) imply that

x '

|JC| -> oo, (4.3.6)

when | arg x | < n — 8 < n. Consequently, W±jkfm(=bc) = ^ ± x / 2 ( ± ^ ) ± * | l + C This shows that Wk,m(x) and W-k,m(—x) are linearly independent. 4.4 Examples of \F\ and Whittaker Functions This section contains some important examples of i Fi and Whittaker functions that occur frequently enough in mathematics, statistics, and physics to be given names. (a) The simplest example is given by ex = xFi(a;a;x).

(4.4.1)

(b) The error function is defined by erf JC = —= e~f2dt = 1 - erfcjc n V Jo where

2

f°°

erfex = —7= /

2

e~l dt.

V* Jx It is easy to see that erf JC = -^=iFi(l/2; 3/2; x).

(JC real),

(4.4.2)

4.4 Examples of i F\ and Whittaker Functions

197

To express the error function in terms of Wk,m(x) we need to write (4.3.5) as an integral over (0, oo). Assume that Re (k — \ — m) < 0; then (4.3.5) can be written as

Wk.m(x) = e-x/2xkT

(k+l--m\-

/>00

• /

e~'r^+m(i •

Jo = —^r- / r v2 -k + m) Jo

e-'r*-5 +m (l+f/jt)*-5 +m 0. Note the relation of (4.4.3) with the integral in (4.1.6). Now set t = u2 — s2, where u is the new variable. Then e-x/lvkr)*2

roo

/

t

v

,

7 _ _2\ k-\+m

e-*udu.

Js Set k = - 1 / 4 , m = 1/4, and x = s2 to get /•OO

W.1/4A/4(s2)

= 2es2/2^

/

e-"2du.

JS

Thus ^ ^ " x 2 / 2 x " 1 / 2 W _ i / 4 i/ 4 (x 2 ).

erf x = 1

(4.4.4)

An asymptotic expansion for erf x can be derived from this formula and (4.3.6). (c) The incomplete gamma function is defined by

y(a, x) = / e^f-Ut

= V(a) - /

J0

e'U^dt = F(a) - F(a, x).

Jx

(4.4.5) After expanding e~f as a series in t and term-by-term integration, it is clear that xa y(a,x) = —\F\(a; a + 1; x). (4.4.6) a The reader may also verify that T(a,x) = e-x/2xa-^Wa_^4(x). (d) The logarithmic integral li(jc) is defined by X dt

• _ [ ~ Jo l o g * '

(4.4.7)

198

4 Bessel Functions and Confluent Hypergeometric Functions Check that

If JC is complex, take |arg(—log;c)| < n. Additional examples of Whittaker functions such as the integral sine and cosine and Fresnel integrals are given in Exercise 4. (e) The parabolic cylinder functions are also particular cases of the Whittaker functions. To see how these functions arise, consider the Laplace equation d2u

d 2u

—2 H

d 2u

j" H

2

=

°*

(4.4.8)

T h e coordinates of the parabolic cylinder %,rj,z are defined by x = - ( £ 2 -r)2),

y = %rj,

z = z.

(4.4.9)

Apply the change of variables (4.4.9) to the Laplace equation. The result after some calculation is 1

/d2u

d2u\

fu__

This equation has particular solutions of the form U(t-)V(ri)W(z), which can be obtained by separation of variables. The equation satisfied by U, for example, has the form

where o and X are constants. After a slight change in variables, this equation can be written as =0

-

(4A10)

Equation (4.4.10) is called Weber's equation. It can be checked that Dn(x)

= 2n2+i*x-i W f + i 5 _i (x2/2)

(|argjc| < 3TT/4)

(4.4.11)

is a solution of (4.4.10). The constant factor is chosen to make the coefficient of the first term in the asymptotic expansion of Dn{x) equal to one. Dn(x) is called the parabolic cylinder function. When n is a positive integer, Dn(x) is e~4*2 times a polynomial, which, except for a constant factor, is Hn(x/y/2), where Hn(x) is the Hermite polynomial of degree n. These polynomials will be studied in Chapter 6.

4.5 Bessel's Equation and Bessel Functions

199

(f) In the study of scattering of charged particles by spherically symmetric potentials, we can take (see Schiff [1947, Chapter V]) the solution of the Schrodinger equation

I 2 V w + Vu = 2

22/z

Eu

to be of the form 00

u(r,9) = r

i=0

where Pi is the Legendre polynomial of degree I and yt satisfies the equation

k

2

h = Ti By a change of variables we can take k = 1. The Coulomb potential is given by U(r) = 2rj/r, so the equation for y is rl

r

Comparison of this equation with Whittaker's equation (4.3.1) shows that yt

= ri+xe-irxFx(l

+ 1 - irj; 21 + 2; 2i>).

The function 0>€(?7, r) := e~iriFi(i + 1 - irj\ 21 + 2; 2ir) is called the Coulomb wave function. 4.5 Bessel's Equation and Bessel Functions Bessel functions are important in mathematical physics because they are solutions of the Bessel equation, which is obtained from Laplace's equation when there is cylindrical symmetry. The rest of this chapter gives an account of some elementary properties of Bessel functions. When k = 0 and m = a in Whittaker's equation (4.3.1), we get

4

t-1

If we set y{x) — «fx~W{2ix), then y satisfies the equation (4.5.1)

200

4 Bessel Functions and Confluent Hypergeometric Functions

This equation is called Bessel's equation of order a. It is easily verified that

is a solution of (4.5.1). Ja(x) is the Bessel function of the first kind of order a. From (4.1.12), we have another representation of Ja{x). This is

Equation (4.5.1) is unchanged when a is replaced by —a. This means that J-a (x) is also a solution of (4.5.1). One can check directly that when a is not an integer, Ja(x) and J-a(x) are linearly independent solutions. When a is an integer, say a = n, then J.n(x) = (-l)nJn(x).

(4.5.4)

Therefore /_„ (x) is linearly dependent onJn(x).A second linearly independent solution can be found as follows. Since (—1)" = cos nix, we see that Ja (x) cos not — J-a(x) is a solution of (4.5.1), which vanishes when a is an integer. Define

When a = n is an integer, Ya(x) is defined as a limit. By L'Hopital's rule,

Yn(x) = lim Ya(x) = - { ^ - (-i jr

[da

Note that

This implies that Ja(x) is an entire function in a. Thus the functions ^ and ^ ^ in (4.5.6) are meaningful. Moreover, as functions of JC, Ja (x) are analytic functions of x in a cut plane. Thus we can verify that Yn (JC) is a solution of Bessel's equation (4.5.1) when a, = n is an integer. We can conclude that (4.5.5) is a solution of (4.5.1) in all cases. Ya(x) is called a Bessel function of the second kind. Substitution of the series for Ja(x) in (4.5.6) gives, after simplification, X Yn(x) =72-J - -71^ I g n{x) in T 2 n X /

1 °°

(—\)k

k\

k=0

l)](x/2)zlc+n.

Here « is a nonnegative integer, |arg JC| < n, and i/(x) — T'(x)/ T{x).

(4.5.7)

4.5 Bessel's Equation and Bessel Functions

201

Note that Bessel's equation can be written

Suppose a is not an integer. It is easy to deduce from (4.5.8) that

d ( dJ (x)\ d ( dJ. {x)\ J-a(x)—- x— a - Ja(x) — x— a =0 dx \ dx J dx \ dx ) or x[j-a(x)J'a(x)

- Ja(x)J!_a(xj\

= C = constant.

To find C, let x ->• 0 and use the series (4.5.2) and Euler's reflection formula. The result is C = 2sina7T /jr. Thus the Wronskian W(Ja(x), /_«(*)) = Ja(x)J'_a(x) by W(Ja(x),

J-a(x))

- J-a(x)Jfa{x)

is given

= — 2 sin aTT/TTX,

for a ^ integer, and

W(Ja(x),Ya(x))=2/7tX not only when a ^ integer, but also for a = n by continuity. Many differential equations can be reduced to the Bessel equation (4.5.1). For example,

u=xaJa(bxc) satisfies

" V^

[^f

f l ^ ] „ = 0.

(4.5.9)

When x = 1/2, b = 2/3, c — 3/2, and a2 = I/a, this equation reduces to u" + xu = 0.

(4.5.10)

This is the Airy equation and it has a turning point at x = 0, so solutions oscillate for x > 0 and are eventually monotonic when x < 0. As such, solutions of the Airy equation can be used to approximate solutions to many other more complicated differential equations that have a turning point. For example, the differential equation after (6.1.12) has x = \/2n + 1 as a turning point. Airy functions can be used to uniformly approximate Hermite polynomials in a two-sided neighborhood of the turning point. See Erdelyi [I960].

202

4 Bessel Functions and Confluent Hypergeometric Functions

4.6 Recurrence Relations There are two important differentiation formulas for Bessel functions: —xaJa(x)= dx

> ^

v

yv

F(n+a-

00

n=0

/

i\n v 2n+2a-l

v

and, similarly, -^x~aJa(x) dx From the series for cosine and sine,

= -x~aJa+l(x).

(4.6.2)

(4.6.3) itx

and J-i/2(x) = \—cosx.

(4.6.4)

V 7tX

Rewrite (4.6.1) and (4.6.2) as aJa(x)+xJ'a(x)

=xJa-i(x)

and -aJa(x) +xJ'a(x) = -xJa+\(x). Elimination of the derivative J'a gives r

^

J

,

^

20i

Ja-\{X) + /a+i(jc) =

r ,

^

,

Ja(x).

A

*

^

(4.6.5)

X

Elimination of Ja(x) gives Ja-\(x) — Ja+\(x) = 2J'a{x).

(4.6.6)

It follows from (4.6.1) and (4.6.2) that 1 dV x dx

„

and / 1

rl \

n

(4.6.8) When these are applied to (4.6.3) and (4.6.4), we obtain

nx

\x dx J

\ x J

4.7 Integral Representations of Bessel Functions

203

and

dx The following two formulas can now be proved by induction (the details are left to the reader): — < sin(x — nn/2) = \ nx

\

k=0

'^(2k)\{n-2k)\{2x)*

V

(4.6.11)

\{n ~ 2k)\(2x)2k

(-l)*(/i +2ik+l)! (4.6.12) 4.7 Integral Representations of Bessel Functions a

Set y = x u in Bessel's equation

/' + V + (l-a2/*2)y = 0. Then w satisfies the equation xu" + (2a + l)wr + xu = 0.

(4.7.1)

Since equations with linear coefficients have Laplace integrals as solutions, let

[ extf(t)dt,

c where A is a constant. Substitute this in (4.7.1). Then

0 = / f(t)(xt2 + {la + \)t + *)£?*'* Jc

= J fit) \(t2 + 1)1 + (2« + l

204

4 Bessel Functions and Confluent Hypergeometric Functions

after integration by parts. This equation is satisfied when [ex'(t2 + l)f(t)]c=0

(4.7.2)

and |-[(f2 + I) fit)] = (2a + I) f(t). (4.7.3) ot Equation (4.7.3) holds when f(t) = (t2 + l)a~l/2. Replace t with V 1 1 ^ so that (4.7.2) holds when C is the line joining —1 and 1 and Re a > —1/2. The above calculations imply that a ixt 2 a l/2 y = Ax / e (l - t ) - dt

(4.7.4)

J—l

is a solution of Bessel's equation. We assume that arg(l — t2) — 0. To see that (4.7.4) gives an integral representation of Ja(x) for Re a > —1/2, expand the exponential function in the integrand as a series and integrate. The result, after an easy calculation involving beta integrals, is

An application of Legendre's duplication formula (Theorem 1.5.1), 22kT{k + \)T(k + 1/2) = V5r(2ifc)!, gives y(x) = A^T{a

+ \/2)2aJa(x).

Therefore, *JTZ\ (a + 1/2)

when Re a > —1/2. Put t = cos 0 to get the Poisson integral representation Ja(x) = —= (x/2)a / eixcosesin2a OdO ^7tF(a + 1/2) 70 1 Z*71 (x/2)a / cos(xcos^) sin OdO, (4.7.6) = _ _

V^r(af +1/2)

y0

for Re a > —1/2. An important consequence of (4.7.5) is Gegenbauer's formula, which gives a Bessel function as an integral of an ultraspherical polynomial. The formula is

f (4.7.7) for Rev > - 1 / 2 . When v ->• 0, we get Bessel's integral (4.9.11) for Jn(x).

4.7 Integral Representations of Bessel Functions

205

To prove this, take a = v + n in (4.7.5) and integrate by parts n times to get (i)n(x/2Y

T

fl ixt dn 2 / e (1 — t )

Jv+n{x) = By Rodrigues's formula (2.5.13),

dn{\ - t2)v+n-1'2

_ (-2)nn\T(v + n + \/2)T{2v)

_

2

Use of this in the previous integral gives Gegenbauer's formula (4.7.7). Condition (4.7.2) is also valid when C is a closed contour on which elxt{t2 — iy*+i/2 r e t u r n s to its initial value after t moves around the curve once. We take C as shown in Figure 4.1, and write an integral on C as

f(t)dt. JA

Here 1+ means that 1 is circled in the positive direction, and —1— means —1 is circled in the negative direction. We are interested in the integral

= xa f

'

eixt(t2-l)a-1/2dt,

JA

which is defined for all a, since C does not pass through any singularity of the integrand. When Re a > —1/2, we can deform C into a pair of lines from — 1 to 1 and back. We choose arg(£2 — 1) = 0 at A; then

y(x) = / [ f

eix'[(l - t2)e-nif-ll2dt + f

= xa2i sin (--a\n

eix'[(l - ,

f eix'(l - t2)"-l/2dt

Ja(x).

Figure 4.1

206

4 Bessel Functions and Confluent Hypergeometric Functions

This gives Hankel's formula: {2

r- }{x/2V^/ V^r In I JA

Ja(x) =

eixt(t2 -

\Y-Xlldt,

(4.7.8)

, n = 0, 1, 2 , . . . , and arg(r 2 - 1) = 0 at A.

when a

We now prove another formula of Hankel, J-a(x) =

r(l/2-ofK / Q f (x/2) a f

- niI J

eixt(t2-l)a-l/2dt,

(4.7.9)

i(X

when Rex > 0, — 3n < arg(^2 — 1) < n, and a + ^ 7^ 1,2, The contour in (4.7.9) is shown in Figure 4.2. It is assumed that the contour lies outside the unit circle. To prove (4.7.9) we need the following formula of Hankel (see Exercise 1.22): 1

1

F(Z)

1 r(0+)

H0+)

-i- = — /

z e'r dt

=— /

27tl J.oo

e'rhlt,

(4.7.10)

lit I J-ooeie

for\0\ 0, we have |argx\ = |0| < \n. Set u = ei7t/2xt. Then (0+) Jioc

=

(-l)ke-a7Tix2k-2a

2ni

Y(\-2a + 2k)

Substitute this in (4.7.11) and apply Legendre's duplication formula. This proves (4.7.9). Now modify the paths in (4.7.7) and (4.7.8) to those shown in Figures 4.3 and 4.4 respectively. Take Rex > 0 in (4.7.7) and (4.7.8). When the horizontal parts of these curves are made to go to infinity we get 1

Ja(x) =

f J\+ioO

+

e't'tfJ-l+i0C

(4.7.12)

208

4 Bessel Functions and Confluent Hypergeometric Functions

and

J-a(x) =

Y(\-a)(x/2)a

ea7li

/(1+)+ / (

2ni

J l+ioo

l+)

2 a e» 0, a + \ ^ 1,2, Moreover, arg(^2 — 1) = —n at 1 + /oo and arg(r2 — 1) = n at — 1 -Moo. Note also that H%2(x) = y ^ ( c o s x + i sinx) = y - ^ ^ = H^ix)

(4.7.21)

4.8 Asymptotic Expansions

209

and (4.7.22)

References to the work of Bessel, Poisson, Gegenbauer, and Hankel can be found in Watson [1944, Chapters 2, 3, and 6]. 4.8 Asymptotic Expansions Set t = 1 -+- iu/x in (4.7.19). The integral becomes H0+) /

fV'2du.

(4.8.1)

Joo

Compare this with (4.3.5) and (4.3.6) to obtain the asymptotic expansion (4.8.2) Hankel introduced the notation

(4a 2 - I 2 )(4a 2 - 32) • • • (4a2 - (2k - I) 2 ) Then (4.8.2) can be written as k-i

(4.8.3) A similar argument gives k-\

(4.8.4) lj=o

Since and

Ya(x) = ——

we have from (4.8.3) and (4.8.4) cos x —

an — sin

x —

2

4 7 ^

ix \ ^ A {—V

(2x)2J (4.8.5)

210

4 Bessel Functions and Confluent Hypergeometric Functions

and

Ya(x)

an

sin x

~2~

7=0

y^(-iy

\.

t

(4.8.6)

7=0

when |argx| < it. Note that (4.6.11) and (4.6.12) are special cases of (4.8.5).

4.9 Fourier Transforms and Bessel Functions Many special functions arise in the study of Fourier transforms. In Chapter 6, we shall see a connection between Hermite polynomials, which were mentioned in Section 4.4, and Fourier transforms in one variable. Here we consider a connection with Bessel functions and as a byproduct obtain a generating function for Jnix). Start with the Fourier transform in two dimensions. We have 1

/*OO

F(M, u) = — /

f*OQ

/

fix, y)ei(xu+yv)dxdy.

(4.9.1)

2TT J _ O O J _ O O

Introduce polar coordinates in both (x, y) and (w, v) by y — rsinO;

x = rcos0,

u = Rcos(f>,

v^Rsincf).

Then p2j[

rOG

i

F(u, v) = —

fircos0,rsin0)eirRcos(0-(p)rd0dr.

/

2iz Jo Jo Expand / as a Fourier series in 0, 00

/ ( r cos(9, r sin(9)= V

fn(r)eine,

to get °°

, u) = V

poo

/

r

i

/n(r)r —

The relation with Bessel functions comes from the inner integral. Since the integrand is periodic (of period 2TT) it is sufficient to consider the integral Fnix) = — / " eixcoseein0d6. 2TT JO

(4.9.3)

4.9 Fourier Transforms and Bessel Functions

211

Expand the exponential and integrate term by term to get i

F

°°

-k k

p

k in »M = T~ E l-TT / cos 0e °d0.

(4.9 A)

Now 2k cosk 6 = (eiG + e~w)k

= eik

,

So, writing k = n + 2m we have

'n + 2m' m = inJn(x)

(4.9.5)

This relation is interesting as it gives the Fourier expansion of oo

eixcos0 =

J2

lxcos6 e

:

i"Jn(x)eine

n=—oo oo

injn(x) cos n6.

= J0(x) +2^2

(4.9.6)

n=\

The last equation follows from J-n (x) = (— \)n Jn (x). Equating the real and imaginary parts gives 00

cos(xcos6>) = Jo(x) + 2^2(-l)nJ2n(x)cos2nO

(4.9.7)

n=\

and oo

sin(jt cos/9) = 2^(-l)V2*+i(*)cos(2/i + 1)0.

(4.9.8)

n=0

For an interesting special case, take 6 = n/2 in (4.9.7) to get ^

(4.9.9)

71 = 1

It is worth mentioning Miller's algorithm at this point. The series (4.9.9) shows that for any given x = xo and sufficiently large n, J2n (*o) is small. So take J^n C*o) to be 0 and /2(n-i) Uo) = c, which is to be determined. Use the recurrence relation (4.6.5) to compute /2(/i-2)(*o) ami so on down to /2U0) and Jo(xo) as multiples of c. Since (4.9.9) can be approximated by

k=\

212

4 Bessel Functions and Confluent Hypergeometric Functions

we obtain an approximate value of c and hence also of the values of the Bessel functions hkixo)- This is an example of Miller's algorithm. See Gautschi [1967, p. 46]. There is another way of looking at (4.9.6). Put t = iew'. Then oo

exp(x(f-l/f)/2)= Y, -A. (*)'"•

—1/2. By Carlson's theorem, we can conclude that (4.9.12) is true for these values of n. We end this section with a proof of the inequalities: For JC real, |70(*)l < 1,

and

\Jm{x)\ < 1/V2

form = 1, 2, 3, ....

(4.9.13) Thefirstinequality follows from (4.9.12), but we have another proof which verifies them all at once. Change t to — t in the generating function (4.9.10) to get oo

exp(-x(f - 1/0/2) = YJ {-\)n Jn{x)t".

(4.9.14)

4.10 Addition Theorems

213

Multiply (4.9.10) by (4.9.14) to get OO

00

1=£

E

n=—oo

n=—oo

00

OO

" Y, (-\TJm(x)Jn-m(x). n=—oo

m=—oc

Equate the coefficients of the powers of t and use J-n {x) — ( - l)n Jn (x) to obtain OO

OO

] T Jt(x) = 702(x) + 2 ] T J*(x) = 1

(4.9.15)

and OO

(-l)mJm(x)Jn-m(x) = 0 forra^O.

(4.9.16)

>O

The inequalities in (4.9.13) follow from (4.9.15). Remark 4.9.1 Observe that, in Bessel's formula (4.9.11), n has to be an integer, for when n = a, a noninteger, the integral on the right side of (4.9.11) is no longer a solution of Bessel's equation (4.5.1). However, Poisson's integral formula (4.9.12) holds for all n as long as Re n > —1/2. We also remark that Jacobi obtained the direct transformation of (4.9.12) to (4.9.11) by the argument given here in reverse. For references and details of Jacobi's proof, see Watson [1944, §§2.3-2.32]. Note also that Jacobi's formula mentioned after (4.9.12) is really a consequence of Rodrigues's formula (2.5.130 when applied to Chebyshev polynomials of the second kind. 4.10 Addition Theorems In this section, we prove a useful addition theorem of Gegenbauer. First we show that OO

Jn(x+y)=

Y

Jm{x)Jn-m(y).

(4.10.1)

m=-oo

This follows immediately from the fact that exp((x + 3O(f - 1/0/2) = exp(x(f - 1/0/2) exp(y(f - 1/0/2), for this implies OO

OO

Y, Jn(x + y)t"= Y,

OO

J

m(x)tm Y •

214

4 Bessel Functions and Confluent Hypergeometric Functions

The result in (4.10.1) is obtained by equating the coefficients of tn. Observe that (4.9.16) follows from this addition formula. To state the second addition theorem, suppose a, b, and c are lengths of sides of a triangle and c2 = a2 + b2 — lab cos 6. Then oo

Mc)=

Y,

Jm(a)Jm(b)eim0.

(4.10.2)

m=—oo

Set d = aew - b. Then c2 = dd, so c and d have the same absolute value. Thus there is a real \fr such that c = (aei0

-b)eif.

A short calculation shows that the last relation implies c sin0 = a sin((9 + yfr + 0) — b sin(i/f + 0). By (4.9.11),

1

/-27T

Since x// is independent of 0 and the integrand is periodic, by (4.9.10),

/O(C) = — ff ""eiiatW Jm(a)eime—

= V

/

Jm(a)eime—

00

= V The last equation comes from (4.9.11). Since 7_m (x) = (— 1 )m Jm (x), we can write the addition formula in the following form: 00

Me) = Ma) Mb) + 2^2 Jm(a)Jm(b) cos me.

(4.10.3)

4.10 Addition Theorems

215

Observe that

Id c dc

Id ab sin 6 dO'

(4.10.4)

then apply this operator to (4.10.3), and use (4.6.2) to get mJm(a)Jm(b) — c

"—'

Rewrite this as

22^(m + 1) ra=0

Apply (4.10.4) to the last formula; use (4.6.2) again to get Cm ra=0

In general, we have the following result for derivatives of ultraspherical polynomials: — C^(cos6>) = -2X sin OC^l (cos 0). dO Now apply induction to see that ^a+m(^) Ja+m(b)

(4.10.5)

m=0

when a = 0, 1, 2, By (6.4.11), C^(cos#) is a polynomial in a; hence by the remarks we made after formula (4.9.12) the two sides of (4.10.5) are bounded analytic functions in a right half plane. Carlson's theorem now implies the truth of (4.10.5) for values of a in this half plane. By analytic continuation, (4.10.5) is then true for all a except a = 0, — 1, —2,.... Equation (4.10.5) is called Gegenbauer's addition formula. We state without proof the following result of Graf: no-)

= Yl

a — bew ) '

Ja+m(a)Jm(b)eim0

(4.10.6)

*-^ AW = — 0 0

when b < a. When a, b, and 6 are complex, we require that \be±l°/a\ < 1 and c• —• a as b -> 0. Graf's formula contains (4.10.1) and (4.10.2) as special cases. See Watson [1944, §11.3]. Exercise 29 gives a proof of (4.10.6) when a is an integer.

216

4 Bessel Functions and Confluent Hypergeometric Functions

4.11 Integrals of Bessel Functions Expand the function F{u, v) in (4.9.2) as a Fourier series: oo

F(u,v) = n——oo pOQ

=

p2jT

1

-L oo

/

elRra*

-j

p2n

°

eiRrcose

— / -

27t

poo i

J

0

n'

pin

Jo 2nJ0

°°

m f^

[by (4.9.6)] 00

in\ I

V

Jn(Rr)fn{r)rdr\ein(t).

(4.11.1)

n=—oo

Hence /•OO

(-/) w F n (/?)= / fn{r)Jn{Rr)rdr. Jo The inverse Fourier transform of (4.9.1) is 1

pOO

f(x, y) = —

(4.11.2)

pOQ

/

F(«,

v)e-i{xu+yv)dudv.

2?T J_oo J-oo

If a calculation similar to (4.11.1) is performed, we obtain poo

fn(r) = (~i)n / Fn{R)Jn{Rr)RdR. (4.11.3) Jo The integral in (4.11.2) is called the Hankel transform of order n of the function fnir). Then (4.11.3) is called the inverse Hankel transform. For a function f(x) that is smooth enough and vanishes sufficiently fast at infinity, we have more generally the Hankel pair of order a: pOQ

F(y)= / Jo

f(x)Ja(yx)xdx

(4.11.4)

and pOQ

/ ( * ) = / F(y)Ja(xy)ydy. Jo

(4.11.5)

4.11 Integrals of Bessel Functions

217

To obtain an interesting integral, multiply Gegenbauer's formula (4.10.5) by C"(cos 0) and use the orthogonality relation which follows from (2.5.14), namely

The result is

Jo

C

C "

cosO)sinZa

-2a)

OdO =

Ja+n(a)Ja+n(b) aaba (4.11.6)

n\T(a)

where a, b, and c are sides of a triangle, that is, c2 = a2 + b2 — 2ab cosO. Rescale a, b, and c and take n = 0 to arrive at sin 2« 0 d 0

/

Jo

=

c*

Z r ( 2 a ) J —1/2, /

l

Ja (ax)Ja(bx)Ja(cx)x

a

dx =

JO

(4.11.8) iov \a — b\ < c < a + b. The value of the integral is 0 otherwise. If the formula for the area of a triangle (denoted by A) in terms of its sides is used, then the right side of (4.11.8) can be written oa—1

A

2a—1

There are important generalizations of integral (4.7.6) due to Sonine. These are contained in the next theorem. Theorem 4.11.1

For Re /x > — 1 and Re v > — 1, xv+l

M+y+1 X

~ 2*T(v + l ) Jo

rn/2

^ X Sm

^in^+16>cosy+16>(i6>. (4.11.10)

218

4 Bessel Functions and Confluent Hypergeometric Functions

The integrals (4.11.9) and (4.11.10) are referred to as Sonine's first and second integrals. Proof, (i) The proof is simple. Expand J^ (x sin 6) as a power series and integrate term by term. Thus i

2 v r(v + l ) 7 0

Mxsiad)sin'l+lecos2v+1ede

oo

-m + l)r(y + l;

JO

The last integral is a beta integral equal to

Substitution of this in the above series gives the result. (ii) In this case expand both J^{x sin#) and Jv(y cos 6) in power series and integrate term by term. The details are left to the reader. Observe that (4.11.9) is a special case of (4.11.10). Divide both sides of (4.11.10) by yv and let y -+ 0. The result is (4.11.9). • Corollary 4.11.2

For Re a > -1 /2,

—f

^—* — f cos(jccos6>)sin2a 0dO. r( + 1/2)VT J 2)V7T Jo Take /x = - 1 / 2 , v + 1/2 = a and recall that Ja(x) =

(4.7.6)

J-\/i(x) = W — cos*. V

7TJC

Sonine's first integral (4.11.9) can also be written as y M+v+1 (x) z i (^y

By Hankel inversion, 2T(v + 1) r

J

^

x )

Mxt)xdx = r"(l - r2);

(4.11.12)

for Re/x > — 1, Re y > — 1. We now turn to the computation of the Laplace transform of Bessel functions. Hankel evaluated the transform of t^~lJa(yt) in terms of a 2^1 function. For special values of a and /x, the 2^1 reduces to more elementary functions. We

4.11 Integrals of Bessel Functions

219

consider this class of integrals next. The simplest integral of this kind was found by Lipschitz. His result is the following: For Re (JC ± iy) > 0, > e~xtUyt)dt

1 =

.

.

(4.11.13)

From the asymptotic expansion for Bessel functions (4.8.5), it is clear that Re(jc ± iy) > 0 is sufficient for the convergence of the integral. Use (4.9.11) to see that /•OO

1

/ e-xtJ0(yt)dt Jo ft

/»OO

Pit

= - / e~xt / Jo Jo dO

eiytcosed0dt

7t Jo x — iy cos 0 1 The more general result, which gives the Laplace transform of tfl~lJa(yt), toHankel[1875]. Theorem 4.11.3

is due

For Re (a + /x) > 0 and Re (JC ± iy) > 0,

A« + M)/2.(«+M+l)/2 \

Qf + l

f\ X2J

Proof. First assume that \y/x | < 1. Substitute the series for Jv (yt) in the integral to get

(-l) m (y/2) g + 2 w r ( q + /x + 2m)

Since \y/x\ < 1, the final series is absolutely convergent. This justifies the termby-term integration. So we have (4.11.14) under the restriction \y/x\ < 1. The complete result follows upon analytic continuation, since both sides of (4.11.14) are analytic functions of y when Re (JC ± iy) > 0. This proves the theorem. • When /z = a + lom-|-2, the 2^1 in (4.11.14) reduces to a 1F0, which can be summed by the binomial theorem for \y/x\ < 1. The results are in the next corollary.

220

4 Bessel Functions and Confluent Hypergeometric Functions

Corollary 4.11.4 For Re (x ± iy) > 0,

(4.11.15) and /•OO

/

Jo

e~xtJa

,

when Re a > — 1. (4.11.16)

Corollary 4.11.5 For Re (JC ± iy) > 0, f°° , \(xz + v 2 ) 1/2 -xT / r^/aW^^-—^—-, Jo ay01 and / Jo

e-xtJa (yt)dt=

\(x2 4- v 2 ) 1/2 - JCF ] , a ( \ \ 2^/2

Apply Exercise 3.39.

w^Rea>0,

wte/iRea>-l.

(4.11.17)

(4.11.18)

•

The formulas in the two corollaries are limits of some formulas for Jacobi polynomials introduced in Chapter 2. Recall that these are defined by

a+1 Theorem 4.11.6 For real a and ft, lim n-aP^( cos-) = lim n~a P^P) (I - ^ ] = (x/2yaJa(x). n^oo \ nj n^oo \ 2nZ) Proof. This follows easily from Tannery's theorem (or the Lebesgue dominated convergence theorem). Suppose a is not a negative integer. Termwise convergence is easily checked. Moreover, domination by a convergent series is seen from the fact that, for large ft, n~a(n + a + 0 + l)*(a + l) n (2w+ot+ £)*(« +l) w ft- a C k 2k k !(n-Jfc)!(a + l)k2 n ~ k\(a + l)k(2n) n\ ~ k\(a where C is a constant that holds for all k, 0 < k < n. When a = — £ is a negative integer, use the fact that

to obtain the desired result.

4.11 Integrals of Bessel Functions

221

The integral formulas (4.11.15) to (4.11.18) are limits of generating-function formulas for Jacobi polynomials, three of which will be proved in Chapter 6. The corresponding formulas are

(4.11.19) (2w + a + /? + l)F(w + a

CXJ

V —^— P^~l\x)rn

= 2a(l - r + R)~a,

(4.11.21)

n=0

where R = (1 - 2xr + r 2 ) 1 / 2 , and (l - r + /?)~ a (l + r + /?)"^.

(4.11.22)

n=0

A proof of a result more general than (4.11.21) is sketched in Exercise 7.31. The other generating-function formulas are proved in Chapter 6 (Section 6.4). The next theorem gives another infinite integral of a Bessel function due to Hankel. Theorem 4.11.7

For Re (/x + v) > 0,

(4.11.23) Proof. The condition Re (/x + v) > 0 is necessary for convergence at zero. The asymptotic behavior of Jv(x) given by (4.8.5) shows that the integral converges absolutely. Thus the integral can be evaluated using term by term integration. This gives /•oo

/

°°

Jv{at) e-p2t2t^~ldt = Y^

/

i \m / „ /o\v+2m

m!rfy

D

222

4 Bessel Functions and Confluent Hypergeometric Functions

Since the integral on the right-hand side equals

the result follows. Corollary 4.11.8

• For Re (/x + v) > 0,

Jo (4.11.24) Proof. Apply Kummer's first i F\ transformation (4.1.11) to Hankel's formula in Theorem 4.11.7. • An important particular case we use later is Mat) tv+1e-"2'2dt =

a2/4p \ 2)v+1e-

Rev > - 1 .

(4.11.25)

See Watson [1944, Chapters 12 and 13] for references. 4.12 The Modified Bessel Functions The differential equation

x dx

\

x2 I

where x is real, arises frequently in mathematical physics. It is easily seen that Ja(ix) is a solution of this equation. Moreover, for x, real e~oe7r^2Ja(xeni^2) is a real function. We then define the modified Bessel function of the first kind as Ia(x) = e-"*i/2Ja(xe*i/2) = e3a7Ti/2Ja(xe-3ni/2)

(-7T < arg;c < TT/2) f-7t < argjc < nj i ix.

(4.12.2)

When a is not an integer, Ia(x) and I-a(x) are two independent solutions of (4.12.1). When a = n is an integer, then

4.13 Nicholson's Integral

223

To deal with this situation, define the modified Bessel function of the second kind: Ka(x) := — [/_„(*) - Ia(x)l 2 sin an It is immediately verified that

rr

l\n(x) = \ —sinhjc V nx

rr

and I-\n(x) = \ —coshx. V nx

(4.12.3)

(4.12.4)

Thus 'n

(4.12.5)

We see that Ja(x) corresponds to the sine and cosine functions whereas Ia(x) corresponds to the exponential function. Perhaps this is why the nineteenth century British mathematician, George Stokes, took Ia (JC), rather than the Bessel function, as the fundamental function. The asymptotic expansions for Ia(x) and Ka(x) can be obtained in the same way as those for Ja(x) and Ya(x). Thus (|argx| < 3TT/2),

(-TT/2

< argjc < 3;r/2),

(4.12.6)

(4.12.7)

and

(-37T/2 < argx < n/2). (4.12.8)

Here (a, n) = (-\)n(a + l/2) n (-a + l/2)n/nl 4.13 Nicholson's Integral Integral representations for modified Bessel functions can be obtained from those for Bessel functions. Similarly, there are formulas for integrals of modified Bessel functions. As one example, take y = i and Rex > 1 in (4.11.18) to get / e~xtIa(t)dt = [x~v(* Jo

-^L.

(4.13.1)

224

4 Bessel Functions and Confluent Hypergeometric Functions

Set x = cosh/3; then (4.13.1) can be written as ro /

(4.13.2)

sinh£

Jo Now, since Ka(t) ~ J—e-'

as t -> oo,

(4.13.3)

we see from (4.13.2), on replacing a with —a, that

I

tcoshfi Ka(t)dt e-

= _ZL_sinhQ^ sinof7r sinh/3

when Re(cosh^) > - 1 . (4.13.4)

Let a -^ 0 to get o

poo PO

(4.13.5)

/

Jo /o When )S = /7r/2, we have (4.13.6)

Jo Jo

2

Nicholson's formula, 8 f°° N(x) = —z K0(2x sinhO cosh(2ar) dt = J*(x) + Y*(x), ^ Jo

Rex > 0,

(4.13.7) generalizes the trigonometric identity sin2 x + cos2 x = 1 as can be seen by taking a = 1/2 and applying (4.13.6). We present Wilkins's [1948] verification of (4.13.7). This is done by showing that both sides of (4.13.7) satisfy the same differential equation and then analyzing their asymptotic behavior. We show first that 2 N(x) as x -+ oo, (4.13.8) 7tX

where N(x) denotes the left side of (4.13.7). It is sufficient to prove that 8

f°°

lim jtN(jc) = lim — /

2

x^ 0 (2jcsinhr)cosh^ = - .

(4.13.

The second equation follows from (4.13.6). For the first equation, we show that F(x,t) = xK0(2x sinhO(cosh2ar — coshr)

4.14 Zeros of Bessel Functions

225

converges boundedly to 0. The dominated convergence theorem then implies (4.13.9). From (4.13.3) we have \F(x,t)\

< A0(x csch01/2e~2-*sinh'|cosh2Gtf - c o s h r | < A0(x/t)l/2e-2xs[nht(2\a\

+ I)r(sinh2|or|f + sinhO-

The second inequality follows upon an application of the mean value theorem to cosh 2at — cosh t, recalling the fact that csch t < \/t. Let x > 1. Since sinh t > t and (xt)lt2e~xt is bounded, we have |F(JC,

ih

01 < A(sinh2|a|f

This proves (4.13.9). For the next step, check that the product of any two solutions of y "+p y'+qy = 0 satisfies the equation y'" + 3py" + (2p2 + p' + 4q)yf + (4pq + 2q')y = 0. Apply this to Bessel's equation to see that {H^l\x)}2, {#v(2)(*)}2 and J2(x) + Y2(x) are independent solutions of 4

Using differentiation under the integral sign, the reader should verify that N(x) satisfies this differential equation. The differential equation

satisfied by KQ(X) is also required in the calculation. Thus we have

N(x) = A{J2(X) + Y2(x)} + B{H^l\x)}2 + C{//f} (x)}\ Let x -+ oo and use (4.8.3) to (4.8.6) to obtain 1 = A + £?2l'(*-5«*-i*)fl + e-2i(x-]2aTt-^)c

+ o(l).

Hence B = C = 0, A = 1, and Nicholson's formula is proved. 4.14 Zeros of Bessel Functions It is easily seen that all nontrivial solutions of the Bessel equation (4.5.1) have simple zeros except possibly at zero. The first derivatives of such solutions also have simple zeros except possibly at zero and ±a. From (4.8.5) we can conclude that for real a, Ja (x) changes sign infinitely often as x - • oo. This implies that Ja(x) and J'a{x) have infinitely many positive zeros. The conclusion for J'a(x) follows from the mean value theorem.

226

4 Bessel Functions and Confluent Hypergeometric Functions

Suppose that y'^i, 7^,2, ••• are the positive zeros of Ja(x) in ascending order. Then, for a > — 1, 0 < 7«,l < ja+l,\ < ja,2 < ja+\,l < ja,3
x2n > • • • be the zeros ofP^P)(x) xkn = cos Okn, 0 < Okn < 7t • Then for a fixed k,

in [ - 1 , 1] and let

lim nOkn = ja,kIn particular, Ja(x) has an infinite number of positive zeros. In the next chapter, we prove that all zeros of P^a^\x) lie in (—1, 1) when of, j3 > —1. This, combined with Hurwitz's theorem, shows that x~a Ja(x) has only real zeros for a > — 1. For Hurwitz's theorem one may consult Hille [1962, p. 180]. Another method of obtaining the reality of the zeros of Ja (x) for a > — 1 is to establish the formula (b2 -a2)

f tJa(at)Ja(bt)dt Jo

= x[ja(bx)J^(ax)

- Ja(ax)J'a(bx)].

(4.14.2)

To prove this, note that Ja(ax) satisfies the differential equation

Multiply this equation by Ja(bx) and multiply the corresponding equation for Ja(bx) by Ja(ax); subtract to get d ( dJa(ax)\ d ( dJa(bx)\ 2 2 Ja(bx)— x— - Ja(ax)—[ x— ) = (b2 - al)x Ja(ax)Ja(bx) dx\ dx J dx\ dx )

4.14 Zeros of Bessel Functions

227

or — L\xJa(bx)J'a(ax) ax

- xJa(ax)J^(bx)]

J

= (b2 -

a2)xJa(ax)Ja(bx).

Formula (4.14.2) is simply an integrated form of this. Now if a is a complex zero of Ja(x), then so is a. Take x = 1, b = a in (4.14.2) and note that the integrand tJa(at)Ja(dt) > 0. Hence the left side of (4.14.2) is nonzero but the right side is zero. This contradiction implies that Ja(x) has no complex zeros. An argument using differential equations can also be given to show that Ja(x) has an infinity of positive real solutions for real a. This technique goes back to Sturm. The version of Sturm's comparison theorem given below is due to Watson [1944, p. 518]. Theorem 4.14.2

Let u\(x) and u2(x) be the solutions of the equations

^

= o,

—^ + 02(*)w2 = o

such that when x = a ui(a) = u2(a),

u\(a) = u'2{a).

Let 0i(x) and 02(x) be continuous in the interval a < x < b, and u[(x) and u'2(x) be continuous in the same interval. Then, if(p\(x) > 02(x) throughout the interval, \u2(x)\ exceeds |MI(JC)| as long as x lies between a and the first zero of u\(x) in the interval. Thus the first zero ofu\{x) in the interval is on the left of the first zero of u2(x). Proof Without loss of generality, we assume that u\(x) and u2(x) are both positive immediately to the right of x =a. Subtract u2 times the first equation from u\ times the second to get d2ui

d2U2 1

dx

Ul

:(0l(jt)- -

dx1

(fo(x))uiu2 > 0.

Integration gives

\

du2

L

dx

du\ dx

-M2—-

I***

Since the expression in the brackets vanishes iJitaa, we have ui

du2

—

2

dx

du\ > 0. dx

Hence

d(u2/ui) ^ dx

0

228

4

Bessel Functions and Confluent Hypergeometric Functions

or

]

x

>o, a

which implies that U2(x) > u\(x). This proves the theorem.

•

Suppose \a\ > \,a real, and take 0I(JC) = 1 — (a2 - 1/4)/x2 and 02(x) = 1 — ( a 2 - 1/4) /c2. Then for x > c, wehave0i(jc) > 0 2 (JC). Note that wi = xl/2Ja(x) is a solution of u'[ + 0I(JC)MI = 0. Denote the general solution by xl/2Ca(x). It is clear that w2 = Acos&uc + 5 sin&>;c, where co2 = 1 — (a2 — l/4)/c 2 . It follows from Theorem 4.14.2 that if c is any zero of Ca(x) then the next larger zero is at most c + 7t/co. When |a| < 1/2, take 0 2 00 = co2 < 1. Thus for real a, Ja(x) has an infinite number of real zeros. Essentially, Sturm's theorem says that the greater the value of 0, the more rapid are the oscillations of the solutions of the equation as x increases. Theorem 4.14.2 can also be used to prove that the forward differences of the positive zeros of Ja(x) are decreasing for |a| > 1/2 and increasing for \a\ < 1/2. Suppose \a | > l/2andlety a?w _i < j a n < ja,n+1 be three successive positive zeros of Ja(x). Now set 0i(JC) = 1 — (a2 — \/4)/x2 and 02(JC) = 0I(JC — k), where k = j a n — ja,n-\ • Now 0i (JC) is an increasing function, so 0i (x) > 0 2 (JC). Consider the interval [jan, ja,n+\]- At x = j a n , ux = Ja(x) = 0 and u2 = Ja(x - k) = 0. By Sturm's theorem, u\ oscillates more rapidly and hence ja,n+\ — ja,n < ja,n < ja,n-\' A similar argument applies to the case \a\ < 1/2. It should be clear that the same argument works for the general solution of Bessel's equation. We end this section with an infinite product formula for Ja(x), a real. For large x, the asymptotic formula (4.8.5) for Ja(x) suggests that the asymptotic behavior of the zeros is given by J C - (m + (2a + 1)/4)TT.

(4.14.3)

Since the zeros of Ja (x) are real and simple, one expects that the number of zeros of x~aJa(x) between the imaginary axis and the line Re x = (m + (a + l)/4)7r, for large x, is m. This is true. See Watson [1944, §15.4]. It follows that the entire function x~a Ja (x) has the product formula

1 - */Ja,n) exp^/y^)} JJ{(1 + x/ja,n) exp(-x/ 7a , n )}. n=\

(4.14.4)

n=\

This result continues to hold when a is not real. See Watson [1944, §15.41].

4.15 Monotonicity Properties of Bessel Functions

229

4.15 Monotonicity Properties of Bessel Functions Sturm's comparison theorem for differential equations stated in the previous section gave information about the zeros of the solutions of the differential equation 2 _

l / 4

^ >

0 forn = 0, 1,..., AT, where the Nth derivative exists in the open interval /, and the lower-order derivatives are continuous in I. Then {-l)nAnMk>0 In particular

230

4 Bessel Functions and Confluent Hypergeometric Functions

Moreover, ify(x) denotes another solution of (4.15.2) with zeros at x\, JC2,... and ifx\ > x\, then {-\)nkn{xk-xk)>Q

for n = 0,...,N;

fc=l,2,....

This theorem yields results on Bessel functions when applied to equation (4.15.1). Two independent solutions of this equation are *JxJv(x) and 11111 I J I c l l l l l I L/LJ&ll JL V i Jrj Z

8 Z*00 / J \ + —- / KQ{2X sinh r) cosh2v^ (tanh t cosh2vf) dt. Ttl Jo V "^ / The first term on the right is zero, for by definition (4.12.3), it follows that Ko(x) behaves like log x as x —• 0, while (4.12.6) gives the behavior of K0(x) as x -> oo. Thus O

POO

p'{x) = —- / KQ{2X sinh t) tanh t cosh 2vr[tanh t - 2v tanh2vt]dt. nl Jo It is easy to check that the expression in brackets is negative for |v| > 1/2 and the rest of the integrand is positive. So p'(x) < 0. Similarly, it can be shown that /•OO 8 P°°

^n\x) = — /

^"1)(2

• {tanh/ -

2vtmh2vt}dt.

It is clear from (4.15.5) that (-l)nK^n\x) > 0 for x > 0, n = 0, 1, 2 , . . . . Thus, the conditions of Theorem 4.15.1 hold for Equation (4.15.1) when |v| > 1/2. So we have the following corollary:

4.16 Zero-Free Regions for \ F\ Functions

231

Corollary 4.15.2 Let cvk, cvk denote the kth positive zeros in ascending order of any pair of nontrivial solutions of Bessel's equation (4.15.1) with |v|>l/2. Suppose X > — 1 and set

Mk= jVk+X

xx/2\Cv(x)\xdx.

Then, for k = 1,2,..., (-l)nAnMk

> 0 forn =

(-\)n-lAncvk>0 (-l)nAn(cv,m+k

-cvk)

0,1,...,

forn = 1 , 2 , . . . , > 0 forn =

0,1,...

with m a fixed nonnegative integer, provided that cv?m+i > cv\. In particular, (-l)n-lAnjvk>0,

(-l)n-lAnyvk>0

forn = 1,2,...

and (-l)nAn(jvk-yvk)>0

forn = 0, 1 , . . . ,

where j v k , yvk denote the kth positive zeros of Jv(x) and Yv(x) respectively. Remark 4.15.1 Lorch, Muldoon, and Szego [1970] have extended Theorem 4.15.1 to a study of higher monotonicity properties of

Mk= [ ^ W(x)\y(x)\xdx, Jxk where W(x) is a function subject to some restrictions. As an example, it is possible to take W(x) = x~x/2 when y is a solution of (4.15.1). This implies the monotonicity of PVk+l

\Cv(x)\dx,

where the integral is the area contained by an arch of a general Bessel function instead of xl/2 times a Bessel function. For further extensions of the results see also Lorch, Muldoon, and Szego [1972]. 4.16 Zero-Free Regions for \F\ Functions We end this chapter with some results of Saff and Varga [1976] on zero-free regions for sequences of polynomials satisfying three-term recurrence relations. These polynomials can be partial sums of i Fi functions, so we may invoke a theorem of Hurwitz on zeros of an analytic function which is the limit of a sequence of analytic functions to obtain zero-free regions for i F\ functions. Saff and Varga's basic theorem is the following:

232

4 Bessel Functions and Confluent Hypergeometric Functions

Theorem 4.16.1

Let {pk(z)}Q be a finite sequence of polynomials satisfying

Pk(z) = [ — + 1 pk-dz) - -pk-2(z), \bk J ck where p-i(z) Let

k = 1, 2 , . . . , n,

(4.16.1)

:= 0, poU) = Po ¥" ®> and the bk andck are positive real numbers.

a := min{fc*(l - bk-i/ck)

: k = 1,2, ...,n],

bo = O.

(4.16.2)

TTien, j/a? > 0, the parabolic region Pa = {z=x

contains no zeros of pk(z),

+ iy :y2 < Aa(x + a ) , * > - a }

(4.16.3)

k = 1, 2 , . . . , n.

Proof. Suppose z 6 /*« is a fixed complex number that is not a zero of any pk(z), k — 1 , . . . ,72. Set A** := M*(z) : = zPk-\(z)/bkpk{z)

for & = 1 , . . . , n.

The proof depends on the following two facts: 1. The polynomials pk(z) and pk-\(z) k= l , . . . , / i . 2. Re/x* < 1 forifc = l , . . . , n .

have no zeros in common for each k,

Assume these results for now and suppose that for some k, pk(z) is zero at a point zo e Pa. Observe that k ^ 1, for, /?i(z) = /?oU + ^i)/^i has a zero at — b\, which by (4.16.2) is < —a and hence cannot be in Pa. Suppose 2 < k < n. Since pk(zo) = 0, it follows from (4.16.1) that

(zo/h + l)pk-i(zo) = (zo/ck)Pk-2(zo). Fact 1 implies that pk-i(zo) j^ 0 because pk and p^-i have no common zeros. So we can divide by pk-\(zo) to get Ck

/

, / \

ZoPk-2(Zo)

,

,

— = w-i(zo)h-\bk (zo + **) = bk-iPk-i(zo)

,.

1/:

A,

(4.16.4)

The second fact and the continuity of /x give Re fik-i(zo) < 1. Then by (4.16.4)

Rez0 < -h(l

- bk-i/ck) < -a,

which contradicts the assumption that zo ^ Pa- This means that pk(z) (k = 1,...,«) have no zeros in Pa. It remains only to prove our two assumptions. It is evident from (4.16.1) that none of the polynomials pk{z), k = 0, 1 , . . . , n vanish at 0. So suppose pk(zo) = pk-i(zo) = 0, where zo / 0, for some k > 1. By a repeated application of (4.16.1) we get po(zo) = 0, which contradicts the assumption that po / 0. Thus pk(z) and pk-\(z) have no common zeros.

4.16 Zero-Free Regions for i F\ Functions

233

We now prove that Re fik(z) < 1 by induction. Clearly, z

z + bi Since z e Pa, Rez > — a > —b\. Thus Re/xi < 1. Now, it follows easily from (4.16.1) that z + bk

-bkckXbk-\lik-\

or

where Tkiw) is a fractional linear transformation defined by £ = Tk(w) =

Z

—r

z + bk -bkck

.

bk-\w

The function 7^ has a pole at the point Wk = (z + bk)/{bkC^xbk-\) whose real part, by (4.16.2) and (4.16.3), is seen to be >1. So 7^ maps Re u; < 1 into a bounded disk with center §* '= 7\(2 — Wk), where 2 — ibk is the point symmetric to the pole Wk with respect to the line Re w = 1. By the definition of Tk

Moreover, the point 0 = r^(oo) lies on the boundary of this disk so its radius is I& |. Thus the real part of any point in the disk does not exceed

where the first inequality follows from (4.16.2) and the second from (4.16.3). Now, by the induction hypothesis, Re/z^-i < 1; thus Re/x^ = Re Tk(iik-\) < 1. This proves the theorem. • Corollary 4.16.2 For an infinite sequence ofpolynomials {pk (z) }Q° satisfying the three-term relation (4.16.1), suppose that a = inf \bk (1 ~~ bk—\Ck J } > 0.

Then the region Pa defined in the theorem is zero-free for the polynomials Pk(z). Moreover, if Pk(z) —> f(z) ^ 0 uniformly on compact subsets of Pa, then f(z) is also zero-free in Pa. Proof

The first part is obvious. For the second part use Hurwitz's theorem.

•

The next corollary applies to the polynomial squence obtained from the partial sums of a power series.

234

4 Bessel Functions and Confluent Hypergeometric Functions

Corollary 4.16.3 and

Suppose Sk(z) '.= Ylj=o aiz^

K

ak-i

cik-2\\

> > 0,

))JJ

ak

Then the polynomials Sk(z),k = 1,2,...

nave

strictly positive coefficients /A

where a^i/ao = 0.

,n have no zeros in Pa.

Proof. First observe that sk(z) = (—

+ 1 )sk-dz) - —sk-2(z),

k = 1, 2,..., /i,

when s-i — 0. Then apply Theorem 4.16.1 to obtain the required result.

•

Note that a consequence of the above results is that the partial sums sn (z) = Y^o zn /n\ of the exponential function have the parabolic region, y2 < 4(x+l),x > — 1, as a zero-free region. This region is sharp, both because of the zero at z = — 1 for s\(z) = 1 + z and asymptotically as n —> oo in sn(z). The next corollary concerns the more general i F\ confluent hypergeometric function. The proof is left to the reader. Corollary 4.16.4 Suppose sn(z) is the nth partial sum of \F\{c\ d; z). Then sn(z), n = 0, 1, 2 , . . . have no zeros in the region (i) Pd/C, ifO - 1 / 2 , „ r72

2 T(a + l/2)Ja(x) and

= —T=Z(X/2) 71

V "

.

/

la

cos(xsm^)(cos^) J^

Jo 1

ri

r ( a + 1/2)/«(JC) = -4=(JC/2)« /

Deduce that I/«(*)! < | x / 2 r ^ | u | / r ( a + l),

e~xt(l -

t2)a~l/2dt.

where JC = W + /I;.

236

4 Bessel Functions and Confluent Hypergeometric Functions

10. Use (4.9.11) to obtain Neumann's formulas (Watson [1944, p. 32]): J2(x) = — / J2n(2x sin0)d0 = — / J0(2x sin0) cos 2n0d0. n Jo x Jo 11. Show that, for |argx\ < TT/2, 2;nT(l/2) Deduce that, when Re a > —1/2,

Tia

e-xt(\-t2)a-l/2dt\.

) /

J-\

Hence

=

J

r(« + 1/2) Jo

e

12. Prove that for x > Oanda > -1/2 cosxt 13. Show that for a > — 1 and c > 0

14. Prove the following result of Sonine and Schafheitlin: S := / - ^ Jo

^ tv-«-fi

dt ^ 2 ^

fa,p

and 5 = 1 ^

provided the integral is convergent.

b 2\

;-x

forO 0 and Re (y — a — P) > 0. 16. Show that

y-> (-l) B 2fi(-w, - a - w; )8 + 1; b2/a2)(ax/2)

n\{a + X)n

'h Deduce that

17. Show that for a, b > 0 and - 1 < Re a < 2Re £ + 3/2

xa+lJa(bx)

a'-W dX

K

(Note that /0°° e-(x2+a2)ttpdt = J2+$+i forRe0 > -1.) 18. Show that (x/2)a f°° _f _x2/44t a ^ a t o = —z— / e~ ~ / t~ ~ dt, |argx| < n/ 1 Jo 19. Prove that for a > 0, b > 0, y > 0, and Re 0 > - 1

la> +

»*a-'-1 Ka.^(yVa2

Consider the case a = 1/2, ft = 0. 20. Prove the followmg formula for Airy's integral: 3 := I f cos(f +xt)dt = ^ * Jo 3n

See Watson [1944, §6.4].

+ b2).

238

4 Bessel Functions and Confluent Hypergeometric Functions

21. Let (f)(x) be a positive monotonic function in Cl(a,b) and let y(x) be any solution of the differential equation

Show that the relative maxima of |v|, as x increases from a to b, form an increasing or decreasing sequence accordingly as 0 (x) decreases or increases. [Hint: For f(x) = {y(x)}2 + {/(x)} 2 /0(JC) show that sgn/'(*) = —sgn^C*).] (Sonine) f 22. Suppose that &(JC) and (p(x) are positive and belong to C (a, b). If y(x) is a solution of the equation

then show that the relative maxima of \y\ form an increasing or decreasing sequence accordingly as k(x)(/)(x) is decreasing or increasing. (Butlewski) 23. Show that u = xa Ja (bxc) satisfies the differential equation

24. Take (x) = 1 + J-.J,,,

JC^O

lJ L

^ . Use (4.8.5) and Exercise 21 to prove that f VVTT

i f — 1 / 2 < or < 1/2,

[ finite and > ^/2/n

if a > 1/2 .

For Exercises 21, 22, and 24 and the references to Sonine and Butlewski, see Szego[1975,pp. 166-167]. 25. Let a = X — 1/2, 0 < A. < 1. Denote the positive zeros of Ja(x) by j \ < h < J3 < " ' a n d the zeros of the ultraspherical polynomial C^(cos#) by 0i < 02 < • • • < 0n. Use Theorem 4.14.2 to show that 0k [Note that u = (sin 6)kC^(cos 6) satisfies the equation

and compare this with the equation satisfied by \f6Ja{(n + 26. Suppose - 1 / 2 < a < 1/2 and mil < x < (m + l/2)7r, m = 0, 1, 2 , . . . . Show that Jv(x) is positive for even m and negative for odd m. [Note that when x = (m + 9/2)7t with 0 < 0 < 1, 2(7T/4) g aU)

~ r ( + 1/2)V5F(2 + 0)« i

^

COS(7T^/2)

{(2m + 0)2

2}i/2-«

U

Exercises

239

and show that sgn Ja(mic + 6n/2) = sgn [(-l) m {i^ + (vm - vm-{)

= sgn(-l)m, where r2r

cos(7tu/2) 2 2 r _ 2 {(2m + 0) - u J2rand

m ' _ f22 Jim Jim

-du

^

27. Show that

= X-\x{j'a{x)}2

- Ja{x)j-x

Deduce that f(x) = AJa(x) + BxJ'a(x) ^ 0 has no repeated zeros other than x = 0. (This result is due to Dixon. See Watson [1944, p. 480].) 28. Let f(x) = AJa(x) + BxJ'a{x) and g(x) = CJa(x) + DxJ'a{x) with AD BC ^ 0. Prove that the positive zeros of f(x) are interlaced with those of g(x). (Show that 0(JC) = f(x)/g(x) is monotonic.) 29. Prove Graf's formula (4.10.6) when a is an integer by using the identity ie ie ea(t-\/t)/2e-b(te- -l/(te- ))/2

where u = (a — be~ld)/c.

=

ec{tu-\/{tu))/2 ^

Orthogonal Polynomials

Although Murphy [ 1835]firstdefined orthogonal functions (which he called reciprocal functions), Chebyshev must be given credit for recognizing their importance. His work, done from 1855 on, was motivated by the analogy with Fourier series and by the theory of continued fractions and approximation theory. We start this chapter with a discussion of the Chebyshev polynomials of the first and second kinds. Some of their elementary properties suggest areas of study in the general situation. The rest of this chapter is devoted to the study of the properties of general orthogonal polynomials. Orthogonal polynomials satisfy three-term recurrence relations; this illustrates their connection with continued fractions. We present some consequences of the three-term relations, such as the Christoffel-Darboux formula and its implications for the zeros of orthogonal polynomials. We also give Stieltjes's integral representation for continued fractions which arise from orthogonal polynomials. In his theory on approximate quadrature, Gauss used polynomials that arise from the successive convergents of the continued fraction expansion of log(l + x)/ (1 — x). Later, Jacobi [1826] observed that these polynomials are Legendre polynomials and that their orthogonality played a fundamental role. We devote a section of this chapter to the Gauss quadrature formula and some of its consequences, especially for zeros of orthogonal polynomials. We also prove the Markov-Stieltjes inequalities for the constants that appear in Gauss's formula. Finally, we employ a little elementary graph theory to find a continued fraction expansion for the moment-generating function. In the past two decades, combinatorial methods have been used quite successfully to study orthogonal polynomials. 5.1 Chebyshev Polynomials We noted earlier that the example of the Chebyshev polynomials should be kept in mind when studying orthogonal polynomials. The Chebyshev polynomials of the first and second kinds, denoted respectively by Tn(x) and Un(x), are defined 240

5.1 Chebyshev Polynomials

241

by the formulas ^

^

(5.1.1)

and IT

, .

(2n

=

(5.1.2)

where x = cos 6. The orthogonality relation satisfied by Tn(x) is given by r+l

/

Tn(x)Tm(x)(l - x2)~l/2dx

= 0,

when

m ^ n.

For x = cos 0, this is the elementary result:

r

I cos mO cos nO dO = 0, Jo

when

m ^ n.

Similarly, the orthogonality for (5.1.2) is contained in

r I sinin + 1)0 sin(m + 1)0 dO = 0,

when

m ^ n.

Jo To motivate our later discussion of orthogonal polynomials, we note a few results about Chebyshev polynomials. The three-term recurrence relation, for example, is given by 2xTm(x) = Tm+l{x) + Tm_i(x),

(5.1.3)

2cos0cosm0 =cos(rn + l)0+cos(m - 1)0.

(5.1.4)

which is equivalent to

The last relation is contained in the linearization formula 2 cos mO cos nO = cos(m +n)0 +cos(m — n)0

(5.1.5)

or Tm(x)Tn(x) = -(Tm+n(x) + Tm.n(x)). (5.1.6) In a more general context, one is interested in the problem of determining the coefficients a(k, m, n) in m+n

(5.1.7)

242

5 Orthogonal Polynomials

where {pn(x)} is a sequence of polynomials with pn{x) of degree n exactly. A simple but important special case of this is

It is usually difficult to say very much about the coefficients a(k,m,n). Later we shall see an important example which generalizes (5.1.5) and the next formula about Chebyshev polynomials of the second kind Un(x): sin(m + 1)0 sm(n + 1)0 sinO

^ sin(m + n + 1 - 2k)0 = > , :

sinO

^—f k=0

(5.1.8)

sin (9

where m A n = min(m, n). This formula is easily verified by noting that sin(ra + n + 1 - 2k)0 sin 0 = \[cos(m + n - 2k)0 - cos(m -f n - 2k + 2)0]. The dual of (5.1.8) is given by sin(7i + 1)0 sin(n + 1)0 =

sin(n + \)\l/dtff,

/ 2

(5.1.9)

j0—(j)

whereas the dual of (5.1.5) is essentially the same formula, that is, cosnOcosncj) = - (cos n(0 + 0) + cos n(0 — 0)). In Fourier analysis, one represents a periodic function f(x) by a series of sines and cosines. This involves analyzing partial sums of the form n

1

-ao + ^ ( a m cos mO + bm siwmO), m=\

where i

p2n

i

am = — /(0)cosm0J0 x Jo

r2n

and bm = — / /(0)sinm0 J0. n Jo

Therefore, 2l 20

^ncosr nO

+

+ bm sinmO)

m=\ n

i r2n x Jo i r2n "l 2 x Jo

V^(cos mO cos ra0 + sin mO sin m0) f(„*(*) = y ,

Pn*(xk)

(Note that the degree of P* (x) is less than the degree of Pn (x).) By Theorem 5.5.4, it follows that P^iXk)

Ja

P^Xk)(t-Xk)

5.6 Kernel Polynomials

259

The last equality follows from the Gauss quadrature formula in Theorem 5.3.2. The result is proved. • Theorem 5.5.5 is due to Stieltjes [1993, paper LXXXI]. The next theorem was presented by Markov [1895, p. 89]. Theorem 5.5.6

Let [a, b] be a finite interval For any x g[a,b]

n^oc Pn(X)

Proof. For any x g [a,b], the function - ^ is a continuous function of t in [a, b]. This observation taken together with Theorems 5.3.3 and 5.5.5 imply the result. • Remark 5.5.1 Since A& > 0, it is an immediate consequence of (5.5.8) that the zeros of P* and Pn alternate. Remark 5.5.2 In Theorem 5.5.6, we may take x to be a complex number that does not lie in [a,b]. If we denote the right side of (5.5.9) by F(x), then the inversion formula of Stieltjes is given by a(c) - a(d) =

1

fd lim / lm{F(u + iv)}du.

IX w->0+ Jc

Thus, the distribution can be recovered from F.

5.6 Kernel Polynomials In Section 5.1 we saw that the partial sum of the Fourier series of a function when expressed as an integral gave us the Chebyshev polynomials of the third kind: Vn(x) = sin(ft + 1/2)6/sin (0/2), where x = cosO. More generally, we get the kernel polynomials when we study partial sums involving orthogonal polynomials. Let {pn(x)} be a sequence of polynomials orthogonal with respect to the distribution da(t) on an interval [a,b]. As before, — oo < a < b < oo. Let / be a function such that Ja f(t)pn(t)da(t) exists for all n. The series corresponding to the Fourier series is given by aoPoM + a\p\(x) H

h anpn(x) + • • •,

(5.6.1)

where an = f f(t)pn(t)da(t)/ f {pn(t)}2da(t). Ja I Ja

(5.6.2)

260

5 Orthogonal Polynomials

In this section we assume that the denominator of an is one, that is, the sequence {pn(x)} is orthonormal. Then the nth partial sum Sn(x) is given by

Sn(x) = J2Pk(x)

Ja

k=o

f(t)pk(t)da(t)

= / f(t)Kn(t,x)da(t),

(5.6.3)

Ja where n

^

.

(5.6.4)

k=0

Definition 5.6.1 For a sequence of orthonormal polynomials {pn(x)}, the sequence {Kn(xo, x)}, where Kn(x0, x) = 2_^ Pk(xo)Pk(x), k=0

is called the kernel polynomial sequence. Lemma 5.6.2 If Q(x) is a polynomial of degree < n, then

Q(x)= f

Kn(t,x)Q(t)da(t).

Ja

Proof

Clearly, n

Q(x) = Yak for some constants ak. Multiply both sides by pj(x) and integrate. Orthogonality gives us

r"

/ Q(t)Pj(t)da(t) Ja

The lemma follows immediately.

= aj.

•

Theorem 5.6.3 Suppose XQ < a are both finite. The sequence {Kn(xo,x)} is orthogonal with respect to the distribution (t — xo)da(t). Proof. In Lemma 5.6.2, let Q(t) = (t — xo)Qn-i(t), where Qn-\ is an arbitrary polynomial of degree n — \. The theorem follows. • Remark 5.6.1

A similar result is obtained when b < XQ are both finite.

5.6 Kernel Polynomials

261

Remark 5.6.2 In the case of the Chebyshev polynomials Tn (x) with xo = a = — 1, we see that for x = cos 6, Kn(-l,cos0)

(-l)ncosn0

= - - cosO+ cos20- ••• + cos(n +1/2)0 cos(• xk in (5.6.6). Thus,

K(xk,xk) = — This proves (5.6.8). The kernel polynomials also have a maximum property, as contained in the next theorem. Theorem 5.6.5 Let xo be any real number and Q(x) an arbitrary polynomial of degree < n, normalized by the condition

f

(Q(t))2da(t) = l.

Ja

The maximum value of(Q(xo))2 is given by the polynomial

and the maximum itself is Kn(xo, xo). Proof

Since Q(x) is of degree < n, we have

Q(x) = aopo(x) + a\p\(x) H

+ anpn(x).

5.7 Parseval's Formula

263

The normalization condition gives

By the Cauchy-Schwartz inequality

k=0

Equality holds when a^ = Apk(xo), where A is determined by

This proves the theorem.

• 5.7 ParsevaFs Formula

Let L,P(a, b) denote the class of functions / such that

fb / \f\Pda(x) 0. In this section, we are interested in the space L%(a, b). By the Cauchy-Schwartz inequality, we infer the existence of

rb / Ja

f(x)xnda(x),

for n > 0. Theorem 5.7.1 such that

Suppose f e L^(a,b). Let Q(x) be a polynomial of degree n,

k=0

where {pn(x)} is the orthonormal sequence of polynomials for da. The integral / [/(*) - Q(x)]2da(x) Ja

(5.7.1)

becomes a minimum when a k=

f(x)pk(x)da(x). Ja

(5.7.2)

264

5 Orthogonal Polynomials

Moreover, with ak as in (5.7.2), »b

n

a

J2 t-J

U(x)fda(x).

Proof. Let ck = Ja f(x)pk(x)da(x).

(5.7.3)

By the orthonormality of [pn{x)}, we get

0 < I [fix) - Q(x)fda(x) = f \f(x) - J2akPk(x)} da(x) Ja

Ja

L

k=0

-•

b

/

n

n

[f(x)fda(x) -2^2akck + ^ ^ k=0 k=0 [f(x)]2da(x) - Y, c\ + 5Z(fl* " Ck)2'

/

k=0

k=0

The last expression assumes its least value when ak — Q . This proves both parts of the theorem. • Corollary 5.7.2

For f e L^(a,b) and ak as in (5.7.3), we have pb

oo a

Y n-

/ UM]2da(x).

(5.7.4)

Proof. The sequence of partial sums sn = YH=o al ^s increasing and bounded.

•

The inequality (5.7.4) is called Bessel's inequality. We now seek the situation where equality holds. Assume that [a, b] is a finite interval. We shall use the following result from the theory of integration. Lemma 5.7.3 For f e L^(a, b), and a given e > 0, there exists a continuous function g such that rb

[f(x)-g(x)]2da(x)0,

(5.7.7)

Since ak = 0 for all k, it follows from Parseval's formula (5.7.5) that rb

[ [f(x)]2da(x)=0. Ja

This implies the result. Exercises 24-28 give results similar to Corollary 5.7.5 for infinite intervals. Stone [1962] contains proofs of these results. Remark 5.7.1 We could also argue as follows: With Qn (x) as in (5.7.6) it follows from (5.7.7) and the Cauchy-Schwartz inequality that b \ [f(x)]2da(x))

2

/ rb =(

f(x)[f(x)-Qn(x)]da(x)

< / [f(x)fda(x) f Ja

[f(x)-Qn(x)]2da(x).

Ja

So rb

/ [f(x)fda{x) < 46 Ja

and the result follows. Corollary 5.7.6 With f as in Corollary 5.7.5 andsn(x) = YH=o ^kPk{x)} where ak is given by (5.7.2), it follows that rb

\\sn(x)-f(x)\\2=

/ Ja

[sn(x)-f(x)]2da(x)^0

as

n —• oo.

266

5 Orthogonal Polynomials

Proof. This result is contained in the proof of Theorem 5.7.4.

•

The results of Theorem 5.7.4 and its corollaries are in general false when [a, b] is not finite. As an example (see Exercise 1.20), one may take da(x) = exp(—JCMCOS/X7T)^JC, f(x) = sin(xM sin/x7r), 0 < /x < 1/2. There are, however, important examples of orthogonal polynomials over infinite intervals, such as the Laguerre polynomials on (0, oo) and Hermite polynomials on (—oo, oo). We prove in the next chapter that Theorem 5.7.4 continues to hold in these cases.

5.8 The Moment-Generating Function In this section we obtain a continued fraction expansion for the moment-generating function X^«>o lJLnXn, where !JLn = (\,tn)

= / tnda(t). Ja

(5.8.1)

The treatment here follows Godsil [1993] and the reader should consult this book for further information on the methods of algebraic combinatorics in the theory of orthogonal polynomials. We assume a minimal knowledge of graph theoretical terminology. Let G be a graph with n vertices. The adjacency matrix A = A(G) is the n x n matrix defined as follows: If the /th vertex is adjacent to the 7th vertex, Atj = 1; otherwise it is zero. An edge {/, 7} in G is considered to be composed of two arcs, (/, 7) and (7, /). A walk in a graph is an alternating sequence of vertices and arcs where each arc joins the vertices before and after it in the sequence. If the first vertex is the same as the last one in the sequence, then it is a closed walk. The number of arcs in a walk is called the length of the walk. The following result is easily checked by induction: The number of walks in G from vertex / to vertex j of length m is given by (Am)/7, that is, the entry in the /th row and 7th column of the matrix Am. We shall have to consider graphs with weighted arcs so that the entry (A)// is the weight of the arc (/, 7). We continue to denote this matrix by A = A(G). Let 0(G,JC) = d e t ( * / -

and

n>0

A(G))

5.8 The Moment-Generating Function

267

Thus Wtj (G, x) is the generating function for the set of all walks in G from vertex i to vertex j . Let W(G, JC) be the matrix whose entries are W(j(G, x). Then

n>0

From the fact that A adj(A) = det(A)/, where adj (A) is the adjoint of A, it follows that

W(G, x) = JC" X 0(G, x~lylad](x"lI

- A).

(5.8.2)

This implies that

Va(G,x) =x~l

(5.8.3)

where G\i is the graph obtained from G by removing vertex i. The connection of the above discussion with orthogonal polynomials is obtained as follows: Suppose {pn(x)} is an orthogonal polynomial sequence satisfying the three-term recurrence pn+i(x)

= (x - an)pn(x)

- bnpn-i(x),

n>\.

(5.8.4)

It is assumed that the polynomials are monic. Let A denote the matrix 1

at

b2

1

\ where the rows and columns of the matrix are indexed by the nonnegative integers. Let An denote the square matrix obtained from A but taking the first n rows and columns. Observe that when det(x/ — An) is expanded about the last row, we get

det(*/ - An) = (x - an-i Thus pn(x) is the characteristic polynomial of An. Observe that matrix A is the adjacency matrix of a particular weighted directed graph G whose vertex set is indexed by the nonnegative integers. If only the first n vertices of G are taken, then the adjacency matrix of the subgraph is An. Denote this subgraph by Gn. We need the next lemma to derive the continued-fraction expression for the moment-generating function X)n>o^» tn)xn, which is understood in the sense of a formal power series. We assume that /XQ = (1, 1) = 1.

268

5 Orthogonal Polynomials

Lemma 5.8.1

For nonnegative integers n,

Proof. First note that (Ak)oO = (A^)Oo for k < In + 1, because no closed walk starting at 0 and of length < In + 1 can include a vertex beyond therathvertex. This implies (pn(A))oo = (Pn(An))oo. We have already noted that for n > 1, ^(x)isthe characteristic polynomial of An. By the Cayley-Hamilton theorem, pn(An) = 0, and hence (I, Pn) = (Pn(A))00 for n > 1 and for n = 0 by definition. Now xn is a linear combination of po, P\,. • •, /?„, so the result follows. • Theorem 5.8.2

With an and bn as in (5.8.4),

1 — xao— 1 — xa\— 1 — xa2— Proof. Let An^ be the matrix obtained from An by removing the first k rows and columns. Set qn-k(x) = det(/

-xAn,k).

Observe that 4>(Gn, x) = det(x/ - An) = xn det(/ - x~l An,0) =

xnqn(x-x)

and 0(G n \O, JC) = det(jc/ — Aw>i) = JC""1 det(/ — x ' U ^ i ) =

xn~lqn-\(x~l).

By (5.8.3), we can conclude that x lWw(Gn,x

l

7-=x

) = ——

_t

.

(5.8.5)

Expansion of det(/ — x An) about the first row gives

- x2biqn-2(x)

qn(x) = (1 -xao)qn-i(x) or qn-\(x) ^ n (x)

l

= 2

1 - X(2o - X ^l^-2(^)/^n-l(^) '

8

Exercises

269

By Lemma 5.8.1, = lim This combined with (5.8.6) proves the theorem.

Exercises 1. (a) Prove the positivity of the Poisson kernel for Tn(x) in the interval — 1 < r < 1 by showing that °°

m

1 -r2 1 - 2r cos 0 + r 2 '

(b) Compute ^ s i n ( m + l)(9sin(m + sin 0

^—'

sin

1x

" -r ,

m=0

which is the Poisson kernel for Un(x). Observe that it is positive in the interval — 1 < r < 1. (c) Show that the Poisson kernel for sin (ft + 1/2)0 is ^ rn sin(n + 1/2)0 sin(n + 1/2)0 - cos(6>+-0)/2] l -2r cos(6> -0)/2+r 2 ] ' 2. Suppose / has continuous derivatives up to order n in [a, b] and x\ < X2 < • • • < xn are points in this interval. Prove the following Lagrange interpolation formula with remainder:

/ (n) (£) f(x) = Ln(x) H

—(x - xi)(x -x2)'"(xn\

xn),

where a < min(x, JCI,JC2, . . . ,xn) < § < max(x, JCI, . . . , xn) < b. Here Ln (x) is the Lagrange interpolation polynomial (defined by (5.3.1)) that takes the value f(xt) at JC,-, / = 1,2,..., w. A discussion of the results in Exercises 3-10 can be found in Natanson [1965]. This book also contains the references to the works of Hermite and Fejer mentioned in the exercises. 3. With the notation of Exercise 2, suppose that n LnW

=

^

k=0

A

k(X

~ *l)(* - X 2 ) ' - ( X -

Xk).

270

5 Orthogonal Polynomials Show that j^

x

(Xj -Xi)'"

(Xj - Xj-i)(Xj

- Xj+i) • • • (Xj -

Xk)

Now let Xjr = a + (j - l)h for j = 1, 2 , . . . , n. Show that

where and Alf(Xj) = 4. Let tj{x) be defined by (5.3.2) with P(x) = (JC - JCI)(JC - JC2) • • • (JC - jcn). Check that €}(JC7-) = P"(xj)/2Pf(xj). Now show that the function //(JC) defined by

*(*> = E ^ f» " TTTTT^ " ^)1J fyx) + ^. j;-(x - XJ)12J(X) L ^ KXj) j=l

; =1

r

(where j i , yi,..., yn, y[,..., y n is a given set of 2n real numbers) satisfies H(XJ) = yj and Hf(xj) = y'j, where /T denotes the derivative of //. Prove also that if / is as in the previous problem with derivatives of order 2n, then

where yj = f(xj) and )/ = f'(xj) in the definition of //. Again £ lies in the same interval as before. 5. Apply Gauss quadrature to the formula for f(x) in the previous problem to obtain f(x)da(x)

= YJ^kf{xk) k=\

+

J

—^

\^n)'

/

Pt(x)da(x),

a sin / cos 2/2 + 1 f—' 2/1 + 1 \ 2n + 1

where - 1 < | < 1. Obviously the various f are not necessarily the same.

5 Orthogonal Polynomials

272 9. Prove that

Tn(x)dx

where xk = cos

(2k l)7T 2n

>o,

.

Hint: Write the integral as

n

"*"* Jo cos 0 — cos 0k

where 0k = (2^w1)7r • Apply the Christoffel-Darboux formula and integrate term by term. (Fejer) 10. Prove that

Unix) -xk)

dx > 0,

(Fejer)

where Un(xk) = 0. 11. Suppose {Pn(x)} is an orthogonal polynomial sequence. Let JC&, n, denote the zeros of Pn(x). Suppose that

k=l,2,

Pn-\(x)

is a partial fraction decomposition of Pn-\(x)/Pn(x). 12. With the notation used in Lemma 9.5.2, show that

Prove that «^ > 0.

provided bt ^ 0, Bt ^ 0 (1 < i < n). 13. Show that if {Pn(x)} satisfies rn\X)

==

for

\/\n—\X T~ t>n-

n

and P_i(x) = 0 , then Aox + 5 0 1 Ci A\x + . 0 C2

0 0 1 An-2x

0 0 0

273

Exercises

14. Suppose An = 1 for all n in Exercise 13 and let Cn = \dn\2 = dndn. Then the zeros of Pn(x) are the eigenvalues of the matrix: Bo di -Bi

k 0

0 ••• d2 •••

0 0 0

0 0

0 0

d2 -B2d3 0

0 0

0

dn-2 -Bn-2

0

dn-l -Bn~l

15. Prove the following recurrence relations for Laguerre and Hermite polynomials respectively: \)Lan{ x ) - (n +a)Lan_1(x),

(a) (n + DLan+l(x) = (-

n = 0,1,2,3,... (b)

Hn+l(x) = 2xHn(x) - 2nHn_x(x),

n = 0, 1, 2,

16. Let {/?nOt)}o° be an orthonormal sequence of polynomials with respect to the distribution da(x). Let

fb

HInn= = fI *xnnda(x), C

n = 0,1,2,....

Ja

Show that

Pn(x)=Cn Mn-1

1

x

x2

•••

xn

where Cn is a constant given by Cn = (Dn_i Dn)~l/2, when Dw is the positivevalued determinant [fik+m]k,m=o,i,...,«• 17. With the notation of Exercise 16, prove that \1\X - fA2

- /X2

= Cn

274

5 Orthogonal Polynomials

18. With the notation of Exercise 16, prove that pb

/~i

rb K J_

a

i=o xi - Xj)2da(x0)da(xi)

• • • da(xn-\)

and i

rb

pb

xt - Xj)2da(xo)da(xi)

• • • da(xn).

(For the reference to Heine, see Szego [1975, p. 27].) (Heine) 19. Let 1 > x\(a, P) > Jt2(a?, /?)••• > xn(a, P) > —1 be the roots of the Jacobi polynomial Pn(Qt/3)(jc). Show that, for a > - 1 and p > - 1 , da

'dp

Proceed as follows: (a) Take a = - 1 , b = 1, and da(x) = (1 - x)a(l + x)? in the Gauss quadrature formula (Theorem 5.3.2). Take the derivative with respect to a to get

j f(x)(l-xr(l+xf\og(l-x)dx n

n

7=1

(b) Take f(x) = {P^\x)}2/(x /

7=1

- xk) to show that (P^'^ixW2

r-l -*)-log(l-

x

J-

-xk

2

da V dx

Now observe that the expression in curly braces and x — xk have opposite signs. Prove | j > 0 in a similar way. (Stieltjes) 20. This result generalizes Exercise 19. Let a)(x, x) be a weight function dependent on a parameter r such that co(x, r) is positive and continuous for a < x < b,T\ < r < T2. Assume that the continuity of the partial derivative | ^ for a < x < b, X\ < x < x^ and the convergence of the integrals

i:

dx

occur uniformly in every closed subinterval x' < x < x" of (ri, r 2 ). If the

Exercises

275

zeros of Pn(x)=Pn(x, r) (the polynomials orthogonal with respect to co(x, r)) are JCI(T)>JC2(T)> • • • >xn(r), then the kth zero Xk(r) is an increasing function of r provided that §f/&> is an increasing function of x, a 1 (or / is a bounded measurable function), and (for a given a > 0) {X n

x dx = 0 for n = 0, 1, 2 , . . . , o then f(x) = 0 a.e. 26. Suppose / : (—oo, oo) -> R is continuous and lim^ioo /(JC) = 0. Prove that / can be uniformly approximated by functions of the form e~a2xlp(x) where p{x) is a polynomial. 27. Show that if / e Lp(-oo, oo), /? > 1, and r

2 2

°°

f(x)e~a then/(*) =0a.e.

x

xndx = 0,

n = 0, 1, 2 , . . . ,

276

5 Orthogonal Polynomials

28. Show that

k=0

(n + a + l)T(n x-y 29. (a) Show that the Legendre polynomial Pn(x) is a solution of (1 - x2)y" - 2xy' + n(n + l)y = 0. (b) Show that Qn(x) — ^ J_x ^j-dt, x $ [— 1, 1] is another solution of the differential equation in (a). (c) Show that Qn(x) = Pn(x)Q0(x) - Wn-i(x), where Wn.x(x) is a polynomial of degree n — 1 given by r1 Pn(x) - Pn(t) 1 x -t For - 1 < x < 1, define Qn(x) = Pn(x)Qo(x) - Wn-i(x) with QQ(x) = (d) Prove the following recurrence relations: (2n + l)xPn(x)

= (n + l)Pn+i(x)

+nP n _i(jc),

(2n + l ) i Q n W = (/i + l)Gn+i W + nQn-i(x),

n = 0, 1 , . . . ,

n = 1, 2 , . . . .

(e) Prove that (2* + l)Qk(x)Qk(y) =

+

and

x

k=o

(f) Show that G n (x) has « + 1 zeros in — 1 < x < 1. See Frobenius [1871].

x—y

Special Orthogonal Polynomials

Special orthogonal polynomials began appearing in mathematics before the significance of such a concept became clear. Thus, Laplace used Hermite polynomials in his studies in probability while Legendre and Laplace utilized Legendre polynomials in celestial mechanics. We devote most of this chapter to Hermite, Laguerre, and Jacobi polynomials because these are the most extensively studied and have the longest history. We reproduce Wilson's amazing derivation of the hypergeometric representation of Jacobi polynomials from the Gram determinant. This chapter also contains the derivation of the generating function of Jacobi polynomials, by two distinct methods. One method, due to Jacobi, uses Lagrange inversion. The other employs Hermite's beautiful idea on the form of the integral of the product of the generating function and a polynomial. This generating function is then used to obtain the behavior of the Jacobi polynomial P^a^\x) for large n. We quote a theorem of Nevai to show how the asymptotic behavior of P^a?/9) (x) gives its weight function. It is important to remember that the classical orthogonal polynomials are hypergeometric. We apply Bateman's fractional integral formula for hypergeometric functions, developed in Chapter 2, to derive integral representations of Jacobi polynomials. These are useful in proving positivity results about sums of Jacobi polynomials. We then use Whipple's transformation to obtain the linearization formula for the product of two ultraspherical polynomials. This clever idea is due to Bailey. One of the simplest examples of linearization is the formula cos mO cos nO = -[cos(ra + n)6 + cos(m — n)0]. We observe that a linearization formula for a set of orthogonal polynomials is equivalent to the formula for an integral of the product of three of these polynomials. We briefly discuss the connection between combinatorics and orthogonal polynomials. In recent years, this topic has been studied extensively by Viennot, Godsil, 277

278

6 Special Orthogonal Polynomials

and many others. We content ourselves with two combinatorial evaluations of an integral of a product of three Hermite polynomials. This chapter concludes with a brief introduction to g-ultraspherical polynomials. This discussion is motivated by a question raised and also answered by Feldheim and Lanzewizky: Suppose f(z) is analytic and \f(rel0)\2 is a generating function for a sequence of polynomials /?n(cos#). Do the /?n(cos#) produce an orthogonal polynomial sequence other than the ultraspherical polynomials? The answer involves an interesting nonlinear difference equation that can be neatly solved. 6.1 Hermite Polynomials The normal integral J^ e~x dx, which plays an important role in probability theory and other areas of mathematics, was computed in Chapter 1. The integrand e~%1 has several interesting properties. For instance, it is essentially its own Fourier transform. In fact,

e~xl = A= I™ e~t2e2ixtdt.

(6.1.1)

V ft J—oo

This can be proved in several ways. (See Exercise 1.) The integral is uniformly convergent in any disk |JC | < r and is majorized in that region by the convergent integral 1

r

e- 1.

(6.2.6)

This arises from the identity dF — OX

which implies

dhl(x) dx

dl

Lan_l(x)=0,

dx

forw>l.

(6.2.7)

Eliminate Lan_x{x) from (6.2.5) and (6.2.7) to get

dx dx a + (2n + 2 + a - x)L n(x) - (n + l)Lan+l(x) = 0, for n > 0. Replace n by n — 1 in this equation and eliminate (d/dx)L^_l(x) means of (6.2.7) to get (6.2.6), the required result. Now differentiate (6.2.6) and then apply (6.2.6) and (6.2.7) to arrive at dx1

dx

=0,

fovn>0,

by

(6.2.8)

Thus u = L°^(x) satisfies the second-order linear differential equation xu" + (a + 1 - x)u' + nu = 0.

(6.2.9)

Because the normal integral is a particular case of the gamma integral, it should be possible to express Hermite polynomials in terms of Laguerre polynomials. Such a relationship exists and is given by H2m(x) = {-\)m22mm\L^I\x2)

(6.2.10)

H2m+i(x) = (-l)m22m+lrn\xLlJ2(x2).

(6.2.11)

and

6.2 Laguerre Polynomials

285

To prove that H2m(x) = CL~ 1/2 (x 2 ) for some constant C, it is sufficient to show that, for any polynomial q(x) of degree < 2m — 1,

L-l/2(x2)q(x)e-x2dx=0. f —oo

A general polynomial is the sum of an even and an odd polynomial. When q is odd, the integral is obviously zero. When q is even, it can be written as q(x) = r(x2), where r is a polynomial of degree < m — 1. Then for y = x2, the above integral becomes />O

/»OO

// Jo

by the orthogonality of L~ 1/2 (y). The value of C can be found by setting x = 0. Relation (6.2.11) can be proved in the same way, or by differentiating (6.2.10). There is another way in which the normal integral is related to the gamma integral. The normal integral is a limit of the gamma integral. This gives another connection between Laguerre and Hermite polynomials. By Stirling's formula,

Jo \ a. We give a proof below that uses (6.2.21) and the orthogonality and the completeness of L® (x). A proof of completeness is in Section 6.5. Observe that / Jo

x^L^n(x)Li(x)e-xdx

„

i (6.2.23)

288

6 Special Orthogonal Polynomials

The orthogonality relation (6.2.3) applied to (6.2.23) implies /•o

Lan(t)t

(6.2.24) By (6.2.3), (6.2.23), and (6.2.24)

L (

" °[r(i8-a)i ^

for w = 0, 1, Now the completeness of L£(f) gives (6.2.22). The formula for the Poisson kernel for Laguerre polynomials is given by

(6.2.25) when \r\ < 1, a > —1, and Ia is the modified Bessel function of order a. A simple proof of (6.2.25) is obtained by using the generating function (6.2.4) and the derivative formula (6.2.19). Write the left side of (6.2.25) as oo n \ - ^ rot, \ n \~^ JL^\X)r 7

1

1

* —

(6 2 26)

- -

To find a closed form for the inner sum, start with the generating-function formula oo

T ^ La+k(x)rn

= e~xr^x~r) l{\ — r)a+k+l

n=0

Multiply both sides by xk+ae~x, result is

take the £th derivative, and apply (6.2.19). The

00

n=0

dxk

n=0

_ ry+k+i L^,

(\-r)"n\

(a + I)** Fxik + a + 1; a + 1; —JC/(1 - r)).

(6.2.27)

6.2 Laguerre Polynomials

289

Apply Kummer's transformation (4.1.11) to the 1F1 to see that the expression (6.2.27) is equal to

Xa

k\Lak(x/(\-

Use this for the inner sum in (6.2.26) to get

The sum in the last expression can be written as

l-r

j=0

y* (xyr/(l - r)2y y . (-yr/(l - r))k

y

=

+ Djif^ e-yr/d-r) (±^LYIa(2^7/(l

k\ _ r))m

\VxJ The last equation follows from the series expansion for the modified Bessel function Ia(x). This proves (6.2.25). We can obtain the Hankel transform and its inverse from (6.2.25). The argument given here can be made rigorous. See Wiener [1933, pp. 64-70] where the Fourier inversion formula is derived from the Poisson kernel for Hermite polynomials. Let

Then (6.2.25) can be written as

H(x,y,t):=

—

1 -t .

n=0

(6.2.29)

290

6 Special Orthogonal Polynomials

Let f(x) be a sufficiently smooth function that dies away at infinity. Then f{x) has the Fourier-Laguerre expansion

/o Multiply (6.2.29) by f(y) and integrate to get f(y)fn(y)dy Let t -> e

1ZI

t".

-70

»=o to arrive at °°

/•OO

/»OO

f(y)fn(y)dy =: gW.

irn(x) (6.2.30)

Thus, /»OO

*(*) = / f(y)Ja(yfiy)dy. (6.2.31) Jo Nowg(i) has a Fourier-Laguerre expansion that, by the definition of g(x), implies that POO

POO

/ g{x)fn{x)dx = (-1)" / f{x)fn{x)dx. Jo Jo By a derivation similar to that of (6.2.31), we have POO

/ Jo

POO

OO^

g(x)Ja(\/xy)dx

= y,(— l ) " / Jo n=0

g(y)ifn(y)dy

ifn(x)

'00

f(y)fn(y)dy = fix).

fn{x) (6.2.32)

We can write (6.2.31) and (6.2.32) as the Hankel pair POO

fiy2)Ja(xy)ydy

/

= g(x2),

0

(6.2.33)

P \

g(x2)Ja(xy)xdx =

Jo This may be the place to point out that (6.2.21) contains Sonine's first integral (4.11.11) as a limiting case. This follows from the fact that lim n-aLan(x/n) = x-a/2Ja(2^).

(6.2.34)

6.2 Laguerre Polynomials

291

If in the Hankel inverse of Sonine's first integral (4.11.12), we change x to x/t and, in the formula thus obtained, change t to 1/s, we get 1

Jo

when X > a > — 1. The Hankel inverse of this is

for /z < A < 2/z + 3/2. Write it as

for — 1 < ^ < X < 2/X + 3/2. This is the analog of (6.2.22). Two analogs of Sonine's second integral are n

Lak(x)Ltk(y)

(6.2.35)

and

Jo

L£(0)

Ln (6.2.36)

Formula (6.2.36) is due to Feldheim [1943]. Formula (6.2.35) is an immediate consequence of the generating function (6.2.4), and (6.2.36) is proven by using the series representations of Laguerre polynomials, the value of the beta integral, and the Chu-Vandermonde sum. When y = 0, (6.2.35) is equivalent to

This is an easy consequence of the generating-function formula (6.2.4). The details are given in Section 7.1. This formula is equivalent to DO

/

n

x

x

k

x x e~ dx

^

"

^

(6.2.38)

292

6 Special Orthogonal Polynomials

This can be used to show that the Fourier-Laguerre expansion of xa~^L^(x) in terms of L@ (x) is given by the formula xae~xLak{x)

(6.2.39) for a > (ft — 1) /2. To understand this condition on a andft,needed for convergence, see Theorem 6.5.3 and the remark after that. Note that (6.2.39) is the inversion of (6.2.37). Just as the Sonine integral and its inversion can be used to solve dual integral equations, (6.2.37) and (6.2.39) help to solve a dual sequence equation involving Laguerre polynomials. Theorem 6.2.1 Let a, A, C be given, such that c > (k — 2a — l)/2, a, A > — 1. Then if an, bn are given (and are small enough) and if xcf(x)Lan(x)xae-xdx,

an=

n = 0, 1 , . . . , N,

Jo

f{x)Lkn(x)xxe-xdx,

bn= [

Jo

n = N + 1, W + 2 , . . . ,

and if fi = a + c, then

(6-2.40) Proof

By (6.2.37), we have

-k + p-a)

~ak = / )

f(x)Lfe)xfie-xdx.

Jo

However, by (6.2.39) we have oo

-

W + A •

n + \)T(k + A

bk

/»0

=

/

Jo Now use the Fourier-Laguerre expansion of f(x) to get (6.2.40). This makes more than merely formal sense if the bn s are small enough. This proves the theorem. •

6.3 Jacobi Polynomials and Gram Determinants

293

Similarly, (6.2.21) and (6.2.22) can be used to solve dual-series equations involving Laguerre polynomials. Theorem 6.2,2 Let a > 8 > — 1, and let a < min(• 0 in the formulas involving C^(JC). The polynomials C^(x) are called ultraspherical polynomials or Gegenbauer polynomials. They are orthogonal with respect to the weight function (1 - x 2 ) A ~ (1/2) when X > -(1/2). An important expression for C^(JC) follows immediately from the generating function (6.4.10). Let x = cos 6, factor the left-hand side using 1 - 2r cos(9 + r 2 = (1 - rei6){\ -

re~w),

expand by the binomial theorem, and equate the coefficients of rn. The result is

The second equation is true because C^(cos 0) is real and the real part of elin~lk)0 is cos(rc - 2k)0. When X > 0, (6.4.11) implies

Another hypergeometric representation for C^(x) is obtained by taking a different factorization:

(X)k k\

(2xr)k (l+r2)k+x

6.4 Generating Functions for Jacobi Polynomials

? k\

303

n\

n,k

n,k

l-n-X

' '

n=\)

Thus, C^(x) = —-^-(2x)n2Fi( ' ; — J. (6.4.12) n\ \ 1 —n—X x1 J Even though ultraspherical polynomials are special cases of Jacobi polynomials, their great importance compels us to note and prove some of their properties. When X ->• 0, we get the Chebyshev polynomials of the first kind:

A _o

n + X

,

X

nK

f 1,

« = 0,

\2Tn(x),

n = l,2,...,

= Tn(x).

(6.4.13)

(6.4.130

When X —>• oo, we have

The Rodrigues formula (2.5.130 takes the form (1-I2)X-1/2CJW

=

(

~2)"(^" - ^ ( l - x 2 ) ^ " - ' / 2 ,

(6.4.14)

n\(n + 2X)n dxn and the formulas for the derivative and the three-term recurrence relation are ;)

(6.4.15)

ax and nCxn(x) = 2(n + X- \)xCkn_x(x) -(n + 2X- 2)&n_2(x) for n > 2 and C^(x) = 1, Cf(jc) = 2Xx. It should also be noted that u(0) = (sin0)kC^(cos0) equation

(6.4.16)

satisfies the differential

304

6 Special Orthogonal Polynomials

A proof of (6.4.15) from the generating function for C^(JC) is as follows: The derivative of (6.4.10) with respect to x gives

n=0

n=0

and hence (6.4.15). To get (6.4.16) take the derivative of (6.4.10) with respect to r. The result is 00

2\{x - r)(l - 2xr + r2)~x = (1 - 2xr + r 2 ) ^nC x n (x)r n - 1 . The left side is 2X(x - r) J2Z=o C$(x)rn. Relation (6.4.16) follows on equating the coefficients of rn on each side. We now state a property of the relative extrema of ultraspherical polynomials. Let y ^ \ k = 1 , . . . , n - 1 denote the zeros of the derivative of P^a)(x). Order the zeros so that yk,n(a) < yk-\,n(a) a n d set yo,n(&) = 1. J«,n(^) = — 1. Define ta)(l),

* = 0, l , . . . , / i .

(6.4.18)

These numbers satisfy the inequality fikin(a) < R « - i W ,

a > -1/2, fc = 1, 2, . . . , n - 1,

n = 1, 2 (6.4.19)

but the inequality is reversed for —1 < a < —1/2. For a = —1/2, /x^,w(of) = 1. For a = 0, (6.4.19) was observed by Todd [ 1950] after studying graphs of Legendre polynomials. Todd's conjecture was proved by Szego [1950]. We prove (6.4.19) with a = 0 by an argument that generalizes to give (6.4.19). It is left to the reader to prove the general case. We begin by stating some necessary results about Jacobi polynomials. The following two identities follow directly from Remark 5.6.3: (n + a + \)P^\x)

- (n + l)Fn(+f \x) {x),

(6A20)

^ \ x ) .

(6.4.21)

For the reader's convenience, we again note that ^+l\)

(6.4.22)

6.4 Generating Functions for Jacobi Polynomials

305

and

P^H-x) = (-\)nP^a\x).

(6.4.23)

Following convention, we denote the Legendre polynomial Pn(0'0) (x) by Pn (x). By (6.4.20), (6.4.23), and then (6.4.21), we have [Pn(x)f - [Pn+i(x)]2 = [/»„(*) - Pn+l(x)][Pn(x) + Pn+l(x)] P 1, then subtracting a term like g does not make f — g continuous. More terms have to be subtracted. These can be determined by expanding (1 — rew)~x in powers of (1 — re~w) and vice versa. We have, about r = el6, -x

e2i6

1 fir) = (1 - re'-W\-k ")

pLlO

(i

1

e2w

- emrk

Vi

e-i6)k

m _ 1/ If for an integer n, n — X > 0, we can take n

g=

{l-re-

w

2i0 x

T\\-e r Ys

Theorem 6.6.1 Let f(z) = YA anZn be analytic in \z\ < r, r < oo, and have a finite number of singularities on \z\ = r. Assume that g(z) = Y™ ^nZn is also analytic in \z\ < r and that f — g is continuous on\z\ = r. Then an—bn = o(r~n) as n —> oo. Proof

By Cauchy's theorem and the hypothesis on / — g,

\z\=r

zn+x

Z

~ 2nrn Jo

The Riemann-Lebesgue lemma for Fourier series implies that the last integral tends to zero as n —> oo. This proves the theorem. •

312

6 Special Orthogonal Polynomials

In fact, in the above theorem it is not necessary to assume continuity off — g on \z\ = r. The same conclusion can be obtained by assuming that / — g has a finite number of singularities on |z| = r and at each singularity Zj, say,

where Oj is a positive constant. For these refinements and further examples, see Olver [1974, §8.9] or Szego [1975, §8.4]. The generating function for P^a^\x) is -

r

+ yj\-2rx+r2ya(\

+ r + y/l - 2rx + r 2 ) ^ ( l - 2r;c + r 2 ) - 1 / 2 .

Take JC = cos 6. The above function has singularities at r = e±l6. In the neighborhood of r = el°, the generating function behaves as

This implies that P^(cos 0 forrc = 1,2,....

6.7 Integral Representations of Jacobi Polynomials

313

Theorem 6.6.2 If the series 00

I

R

+

n=0

(

l/i+l

\

_°_

\AnAn+iJ

2

converges,

£/*erc J ^ caw &£ expressed in the form

Here i/sf(x) is continuous and positive in (—8, 8), supp (\/rf) = (—8, 8), and ifj(x) is a step function constant in (—8,8). Furthermore, the limiting relation 2 nlii^sup{V/(x)Vs

- x2Pfe)/hn}

= 1

holds almost everywhere in supp(d^). The singularity in the generating function for Laguerre polynomials is more complex. This makes it difficult to apply Darboux's method. Fejer, however, has shown how this can be done. See Szego [1975, §§8.2-8.3]. 6.7 Integral Representations of Jacobi Polynomials Integral representations for hypergeometric functions imply the existence of such representations for Jacobi polynomials. A few important and useful integral representations are given in this section. Recall Bateman's [1909] fractional integral formula: a,b

\ _ Yjc + i t i x 1 - ^ 1 *

(rx . c - \ ,

,w,.-i

^ fq, (a*b b

Jo

\ \ c'J '

where Re c > 0, Re /x > 0, and |JC| < 1, if the series is infinite. This formula is a particular case of (2.2.4) and we use it to prove the next theorem, which is called the Dirichlet-Mehler formula. Theorem 6.7.1 For 0 < 0 < TC, the nth Legendrepolynomial is given by JZ Jo (2COS0-2COS0) 1 / 2

-2-f Proof

sin(w + | ) 0 (2cos6>-2cos0) 1 / 2 ^ '

n Je

Takea = -n,b = /i + l,c = /x = l/2,x = sin2 (0/2), and t = sin 2 (0/2)

in Bateman's formula. This gives (-/i, n + 1; 1/2; sin2(0/2))cos(0/2)J0 x Jo

(sin2(0/2) - sm2((l)/2))1/2

314

6 Special Orthogonal Polynomials

The 2F1 in the integral is the hypergeometric form of the fourth Chebyshev polynomial given by

cos(0/2) To get the other form of the integral change 0 ton — 0 and 0 to n — 0. Then use

the fact that Pn(-x) = (-l)nPn(x).

•

One way to use this theorem is to show, as Fejer did, that the sum of Legendre polynomials, YTk=o PkM, is greater than or equal to 0 for 0 < x < 1, since y , P/1/2'"1/2)(cos6>) _ y , sinjk + \)0 _ / s in(/i + l/2)0\2 > Z . ^(-l/2,l/2) (1) " I . sin(0/2) ~[ ) ± sin(e/2) For the reference to Fejer and other related results, see Askey [1975, Lecture 3]. Theorem 6.7.2 For //, > 0, - 1 < x < 1, (a)

(I - x)a+^—

p(a+ti,p-ii)(x\

—

(1 — y)

, ^ — ( y — xr

dy, a > — 1;

p(a-fi,P+fi)( (

(c)

x (y — JC)M dy, a > — 1;

x

(^ — y)M "y 5 ^ > —i-

To obtain (a) use the hypergeometric representation of the Jacobi polynomials and apply Bateman's formula with an appropriate change of variables, (b) now follows from (a). Apply P^\-x) = (-l)nP^a)(x).

6.7 Integral Representations of Jacobi Polynomials

315

Finally, (c) and (d) are derived from (a) and (b) respectively by an application of the Pfaff transformation:

Note that when this is applied to Bateman's formula we get

\ c JJ

Jo

The theorem is proved. We also note that Bateman's formula and the results in the theorem are all particular cases of Theorem 2.9.1. • An important integral of Feldheim [1963] and Vilenkin [1958] can be obtained from (c) by using the quadratic transformation

P£a\x)

Pf'-Wjlx2 - 1)

and

and taking 0 = ±1/2. The result is Cnv(cos X> -111, 0 < 0 < n

-sin 2 (9cos 2 0)- 1 / 2 ) d(p.

Cfcos0)

y>A> —.

2

316

6 Special Orthogonal Polynomials

Here C°n(cos0) ^ =

C£(cos0) lim —2— = cos n9.

^(D

The final theorem of this section is the Laplace integral representation for ultraspherical polynomials. It is due to Gegenbauer [1875].

Theorem 6.7.4 ForX>0,

lx2 o 2 A rnlyi^ {A)) z

Proof. Recall that

x

k=0

Rewrite this using the beta integral. Then

; fl yM-l(l

(n\ei(n-2Wdy

- yf^-k-l

Set y = sin2 ^ to get '

' /

[cos^ + i sin 0(cos2 ^r - sin2

x sin2'1"1 ^ cos2*"1 fdx//. Now let 0 = 2x// to get the result in the theorem.

•

6.8 Linearization of Products of Orthogonal Polynomials The addition theorem for cosines implies the formula cos mO cos nO = | cos(w + m)0 + ^ cos(>z — m)0. In the previous chapter, we noted that this result pertains to Chebyshev polynomials of the first kind, p ^ A - i ^ ) ^ ^ where JC = cos 0. This is called a linearization formula because it gives a product of two polynomials as a linear combination of

6.8 Linearization of Products of Orthogonal Polynomials

317

other polynomials of the same kind. More generally, given a sequence of polynomials {pn(x)} one would like to know something about the coefficients a(k,m, n) in m+n Pm{x)pn(x) = ^fl(*,lfl,/!)/*(*).

(6.8.1)

If the pn (x) are orthogonal with respect to a distribution da(x), then a{k,m,n) = — / pm(x)pn(x)pk(x)da(x).

(6.8.2)

Thus the problem of the evaluation of the integral of the product of three orthogonal polynomials of the same kind is equivalent to the linearization problem. As another example of a linerization formula, recall the identity sin(n + 1)0 sin(m + 1)0 _ sin 6 sin 6

mi

^ w ) sin(n + m + 1 - 2k)0 f-* sin 6

k=0

This comes from the addition formula for sines. The addition formula is contained in an important special case of (6.8.1), that is, xmxn

=xm+n

when x = ew. One way of obtaining linearization formulas would be to look for those polynomials for which the integral (6.8.2) can be computed. A simpler integral would involve only the product of two polynomials; but this would yield the orthogonality relation. As we have seen, using the generating function is one way of obtaining orthogonality in some cases. The simplest generating function is for a Hermite polynomial, since it involves only the exponential function, which can be multiplied by itself without resulting in something very complicated. For example, to get orthogonality, note that oo^

m\n\

m,n _ /

e2xr-r

00

2

+2xs-s2-x2^x

J —OO /»O

=

/

Therefore, rOO

/ J —OO

Hm(x)Hn(x)e-xldx

=

318

6 Special Orthogonal Polynomials

Similarly, to find the integral of the product of three Hermite polynomials, consider °° y .

Ht(x)Hm(x)Hn{x)r^mtn^dx

00 poo

-I '""" 2

-s2+2xt-t2-r2

J

J — oo

oo

POO

= /

r -(x-r-s-t)2 e-0

f^ k\(l -X-

n)k(X - s)k(2X + m - s)k

(l — A ) 5 ( 1 — 2A — m)s ( —n,X,—s,m-\-X —s 4F3 (l-X-m)s \l-X-n,X-s,2X + m-s' (6.8.8)

320

6 Special Orthogonal Polynomials

This 4F3 is balanced. Recall Whipple's formula (Theorem 2.4.5), which transforms a balanced 4F3 to a very well poised 7 F 6 :

a + l-b-c,d,e,-s a + l-b,a

+ l-c,d

+ e-a-s*

\ ;1 J

(a + 1 - d)s(a + 1 - e)s (a + l)s(a + 1 - d - e)

( a,l+a/2,b,c,d,e,-s \ \a/2, a -\- I — b, a -\- I — c, a -\- I — a, a -\- i — e, a -\- L -\- s

J

when s = 0, 1, 2, Take a = —A — m —n,b = — m, c = 1— 2A — m — n -f d = A, and e = —n. Then (6.8.8) is transformed to n)se~2sW

1 - A) 5 (l - 2A - m -

— X — m — n)s

k

(-A - m - n)k(l - (A + m + n)/2)k(-m)k k\(-(X + m + w)/2)ik(l - A - n)* yi

^_.A

m

n ~r~ iJ/jtvAy^v^

r^)k\

")k

(A — 5")jt(l — 2X — m — n)k(l — A — m ) ^ ( l — A — m — n + s)k Reverse the order of summation, set s = k + £,, and simplify. Put this in (6.8.7) and use (6.8.6) for the inner sum to get an identity that reduces to Dougall's identity (6.8.4). • The limit X -> 0 gives the identity for cos mO cos nO and X = 1 is the identity for sin(m + 1)0 sin(rc + 1)0. When X ->• 00, Dougall's identity reduces to

The next corollary was given by Ferrers [1877] and Adams [1877]. Corollary 6.8.3 For Legendre polynomials Pn (x),

J2 iv 2k + 2nM^ + 1— *—^ 02m ^1 (l/2)k(l/2)m-k(l/2)n-k(m+n-k)l 7777777

^

(

^

)

(0.5.9)

k\(m - k)\(n Proof. Take X = 1/2 in Dougall's identity. • Remark 6.8.2 The coefficients a(k,m,n) in (6.8.4) are positive for A > 0. Moreover, Theorem 6.8.2 implies a terminating form of Clausen's formula. See Exercise 3.17(d) for the statement of Clausen's formula.

6.8 Linearization of Products of Orthogonal Polynomials

321

Corollary 6.8.4 For X > - 1 / 2 and A ^ 0

f f

C^x)C^(x)Cxn(x)(l-x2)k-^2dx (k)s-t(k)s-m(k)s-ns\ (s - e)l(s - m)\(s - n)\(k)s

2l-2xnr(s + 2k) ' [r(A)] 2 s!(s + A ) '

when I + m + n = 2s is even and the sum of any two ofl,m,n the third. The integral is zero in all other cases.

is not less than

This is straightforward from Dougall's identity. It contains (6.8.3) as a limiting case. Integrals involving products of some orthogonal polynomials also have combinatorial interpretations. In Section 6.9, we show how (6.8.3) can be computed combinatorially. The coefficients a(k, m, n) in (6.8.1) can also be computed in terms of gamma functions when pn(x) = P^a^\x) and a, ft differ by one. This once again covers the cases of the third and fourth Chebyshev polynomials. Hsu [1938] showed how to use the result in Theorem 5.11.6 to go from (6.8.10) to the corresponding integral of Bessel functions: poo

f Jo

Ja(at)Ja(bt)Ja{ct)tl-adt 0

if a, b, c are not sides of a triangle, ; of a triangle of area

This integral was evaluated by a different method in Section 4.11. It is possible to define a second solution of the differential equation for ultraspherical polynomials that converges to the second solution Ya(x) of the Bessel equation. In fact, there is an analog of the second solution for general orthogonal polynomials {pn(x)}. For simplicity, suppose that the orthogonality measure (or distribution) da(t) has support in a finite interval [a, b] and that, on [a, b],

da(t) = co(t)dt, where co (t) is continuously differentiate and square integrable. Define the function of the second kind qn outside [a, b] by qn(z)= f ^-co(t)dt, Ja Z - t and on the cut (a, b) by

z€C,z^[fl,i»],

qn{x) = ^lirn l-{qn{x + iy) + qn(x - iy)) = j - f^-co(t)dt,

(6.8.11)

(6.8.12)

322

6 Special Orthogonal Polynomials

a < x < b. Note that, on the cut, qn is the finite Hilbert transform on {a, b) of the function copn. The ultraspherical function of the second kind, D^(x), is defined by x) = -qn(x) 71

^ ^ ( 1 - t2)x~l/2dt.

= - l

IX J-\X

(6.8.13)

—t

It can be shown that lim nx~2xDxn (l - - ^ ) = - - ^ ^ - y / 4 - A / 2 F A _ 1 / 2 ( V ) 0 -

(6.8.14)

Askey, Koornwinder, and Rahman [1986] have considered the integral rl

DA / n

\/^. /

\/^»A/

\/i

2\2A.— 1 J

^X j C m ^XJC^ ^XJ^l — I j

CiX.

/ / : o i r\

^O.O.lJJ

1

It vanishes if the parity of I + m + n is even. It also vanishes when l + m +nis odd and there is a triangle with sides £, m, n. In the other cases its value is

m - I - l)/2)l(-A)( n - w - €+ i) /2 ((/i -m-£

+ \)/2)

They also evaluate a more general integral in which D^{x) is replaced by a function of a different order, D%(x). The proof of these results used Whipple's 7 F 6 transformation. Corresponding integrals for Bessel functions are also derived. For these results and for references, the reader should see their paper. This work arose from a special case studied by Din [1981]. For general Jacobi polynomials, P^a^\x), the linearization coefficients can not be found as products. Hylleraas [1962] found a three-term recurrence relation for these coefficients from a differential equation satisfied by the product PJla^){x)P^){x). In addition to the case a = p, Hylleraas showed that when a = P + l, the linearization coefficients are products. For many problems, only the nonnegativity of these coefficients is necessary. Gasper used Hylleraas's recurrence relation to determine the values of (a, P) with all of the linearization coefficients nonnegative. For many years, the best representation of these coefficients was as a double sum. Finally, Rahman [1981] showed that these coefficients can be written as a very well poised 2-balanced 9/% These series satisfy three-term contiguous relations and comprise the most general class of hypergeometric series that satisfy a three-term recurrence relation. Although Rahman's result was unexpected, in retrospect it would have been natural to expect such a result. If Jacobi polynomials are normalized to be positive at x = 1, as they are when a > — 1, then the linearization coefficients are nonnegative when ct>P> — \, when a~\- p>0, —\ [m/2] in the first sum, then p(Km, [(m + l)/2]) = 0. So the first sum is [m/2]

Jk=O

In the second sum, change k to k + 1. It becomes

k=0

This proves the theorem.

•

Our next objective is to give a combinatorial evaluation of the integral

Hnx(x)Hni(x)--Hnk(x)e~x2dx.

= / J — oo

For the casefc= 2, its evaluation will give the orthogonality of the Hermite polynomials. A related integral is /•OO

J(nun2,...,nk)=

/

HeHl (x)Hen2(x)

• • • Henk{x)e~x

/2

dx,

J — OO

where Hem(x) = 2~m/2Hm(x/y/2).

A change of variables shows that

/(ni, / i 2 , . . . , nO = 2 ( f l l + - + n *- 1 ) / 2 /(/ii, / i 2 , . . . , n*). It is convenient to shorten the notation by writing n = (n\, ni,..., ftfc)- Also, let 4 ° = /Oil, . . . , 7!;_i, flj - 1, ni+U • • • , /Ijfe). Here the i\h parameter nt is reduced by 1. Similarly, /i*' ; ) will mean that the ith and jth parameters are reduced by 1. The next lemma gives a recurrence relation for/a-

326

6 Special Orthogonal Polynomials

Lemma 6.9.2 h = £* =2 ni 4U)

and J

o= ^ -

Proof. First observe that the Rodrigues-type formula (6.1.3) for Hermite polynomials gives

Also, (6.1.11) implies

H'em = — Hem{x) = mHem-\(x). dx Applying integration by parts, we get poo

h =

JJI\

\ (-1)" 1 -—e-x2/2Hen2(x) J-oo dX l

J

• • • Hem

(x)dx

-

i=2

Since /^ is the normal integral,

J-c

the lemma is proved. A combinatorial object that satisfies the same functional relation as in Lemma 6.9.2 is obtained as follows: Let V\, V^» • • •» V* be a disjoint set of vertices. Let V = Vi U V2 U • • • U Vk. Let |Vt\ = m so that \V\ = ^ = 1 n,-. Construct a graph G from V by putting an edge between every pair of vertices that does not belong to the same V,. G is called the completefc-partitegraph on V\ U V2 U • • • U V^. Let P(n\,ri2, •.., «&) = -P« denote the number of complete matches on G, that is, the number of matches that use all the vertices of G. We set P$ = 1, in accordance with the earlier convention. It is clear that if Xl/=i nt ^s a n °dd number, then P% = 0. We also define P$iJ) similarly to / i u ) .

6.9 Matching Polynomials

Lemma 6.9.3

327

Pn = £*L 2 nt P$U) and Pg = 1.

Proof. Choose a specific vertex in V\. This vertex can be matched with any of the rit vertices in Vt,i ^ 1. Once one such match is made, the rest can be completed in pi ways. This implies that

1=2

and the proof of the lemma is done.

•

Since J^ and PH satisfy the same recurrence relation we have the following: Theorem 6.9.4 h = V2nPn, Theorem 6.9.5

h = (2ni+n2+-+"*7r)1/2PH.

P(m, n) = m\Smn.

Proof. In this case V = V\ U V2, \V\\ = m, and \V2\ = n. If m ^ n, then the vertices of V\ cannot be matched with the vertices of V2 to give a complete matching. So P(m, n) = 0 for m ^ n. If m = n, then the number of complete matchings is m! and the theorem is proved. • This theorem implies the orthogonality of the Hermite polynomials, for we have the well-known result oo

Hm(x)Hn(x)e~x2dx

= 2mm\^/7t8mn.

-oo

Now suppose V = V\ U V2 U V3, where V\,V2, V3 have l,m,n elements respectively. If l+m+n is odd, or if I > ra+w, then it is easy to see that P(£, m, n) = 0. The next theorem considers the other situations.

Theorem 6.9.6 Suppose l + m+n is even and s = (l-\-m-\-n)/2. Suppose also that the sum of any two of£,m,n is greater than or equal to the third. Then P(i,m,n) =

t\m\n\ (s - l)\(s - m)\(s - n)\

Proof. Without loss of generality, we assume m > n. After all vertices in V\ are matched with vertices in V2 and V3, the same number of vertices must be left over in V2 and V3 for a complete matching to be possible. This means that there are m — n more matchings of V\ into V2 than V\ into V3 in a given complete match. So if x denotes the number of V\, V2 pairs and y the Vi, V3 pairs, then x + y = I and x — y = m — n. Therefore, x = s — n and y = s — m. This implies that there are (s — £) V2, V3 pairs. There are ( 5 l n ) ways of choosing elements in V\ to pair with elements in V2. The remaining elements in Vi then pair with elements in V3. Moreover, for any given 2(5 — n) elements, taking s — n from V\ and s — n from

328

6 Special Orthogonal Polynomials

V2, there are (s — n)\ ways of doing the pairing. All this means that

P(l9m,n)=(

l )( m \( U \s-n)\(s-l)\(s-m)\ \s — nj\s — I) \s - m)

ilmlnl (s - l)\(s - m)\(s - n)\ The theorem is proved.

•

The theorem implies that

V

2

) ' \

) m \

2

2 )

•

when t + m + n is even and the sum of any two of €, m, n is not smaller than the third. Otherwise the above integral is zero. It is possible to compute P(k,l,m,n) as well, but the result is a single series rather than a product. The reader should read the paper of Azor, Gillis, and Victor [1982] for this and other results. For further results on matching polynomials see Godsil [1981]. We now give a different approach to the theorem that 7^ = V^tPn. First observe that the matching polynomial of G, a(G, x), can be written as a(G) = a(G, x) = ^ ( - l ) | Q f | x m " 2 | o f | , a

where a runs through all the matchings of G and |a| = the number of edges in the matching a. Let the disjoint union of two graphs G\ and G2 be denoted by Gi UG 2 . See Figure 6.3.

Lemma 6.9.7 a(Gx U G2) = a(Gi)a(G 2 ). Proof. Suppose G\ has m vertices and G2 has n vertices. Then m-2\p\

6.9 Matching Polynomials

329

The last relation follows because every matching y breaks up uniquely into a matching a of G\ and /3 of G2. The lemma is proved. • Let 0 be a linear operator on polynomials defined by a(Knk))

•

The expression ^^{—Y)^pm(Kn-2\a\)

can also be written as

where «/ is a matching in Ky.. Finally, we can rewrite this as

Ct\,...,OLk,y

330

6 Special Orthogonal Polynomials

where y runs through all the complete matchings of Kn-2\a\ with |a | = \ct\\-\ h \ak\. The matchings a\, 012, •. •, oik, y, taken together, give a complete matching ofKv. To complete the final step, we need one more lemma that uses the concept of a colored complete matching of Ky. For each matching a\, a 2 ,.. •, oik, y, color the edges in each a, red and the edges in y blue. Thus all the red edges are homogeneous and the blue edges are either homogeneous or inhomogeneous. The set of all matchings a\,..., oik, y in the summation is the set X of all matchings of Ky in which only the homogeneous edges are red. Let K I , where Y is the set of all matchings in which there are no red edges and all the blue edges are inhomogeneous. These are the complete matchings in the A:-partite subgraph in Ky. If r(a) denotes the number of red edges in a, then by Lemma 6.9.8 we have shown that

aeX

Also, by definition

aeY

The next lemma will complete the proof of Pa = La. Lemma 6.9.9 \ ^ /_i\r(a) _ Q / J aeX-Y

Proof. First define an involution 0 on X — Y. Number the edges of Ky arbitrarily. For any a e X — Y, consider the set of all homogeneous edges of a. This set is nonempty. Consider the smallest edge in this set and change its color from red to blue or from blue to red. This gives a new matching a' = 6(a) in X — Y. Clearly 0(0(a)) — a. It is also clear that ( - l ) r ^ + (-i) r ^(«)) _ Q. This proves the lemma and the theorem. • The above proof follows DeSainte-Catherine and Viennot [1983]. Also see Viennot [1983]. 6.10 The Hypergeometric Orthogonal Polynomials The hypergeometric representations of the Jacobi, Laguerre, and Hermite polynomials, which we have extensively studied in this chapter, are respectively

6.10 The Hypergeometric Orthogonal Polynomials

331

given by

n\

" \

a+1

2 y

(a + \)n ^ ( ~n n!

and

In Chapter 3, the Wilson polynomials were introduced. These polynomials can be represented as 4F3 hypergeometric functions: Wn(x2;a,b, c, d) (a + b)n(a + c)n(a + d)n n,n+a+b

+ ix,a-ix + c + d-l,a a + b, a + c, a + d

\ )

We saw that Jacobi polynomials are limiting cases of Wilson polynomials and in turn the Laguerre and Hermite polynomials are limits of Jacobi polynomials. A question arises as to whether there are hypergeometric orthogonal polynomials at the 3F2 level. In fact there are such polynomials. A few are treated in this section and others are given in the exercises. For a more complete treatment the reader should see Koekoek and Swarttouw [1998]. It is easily seen that lim

Wn(x2; a, b, c, d) n\ \ ' ' ' (a + d)n

( —n,a + ix,a — ix (a + + b)(a b)n(a + + c))F[ n3F2[ (a ' , ' ; \ a + b,a + c =: Sn(x2;a,b,c) (6.10.2)

and Wn((x + 0 2 ; a ~ it, b - it, c + it, d + it) t^So (-2)nnl r

_ ,n(a + c)n(a +d)n n\ =: pn(x;a,b,c,d).

( —n,n + a + b + c + d — \,a + ix \ a + c,a + d (6.10.3)

The polynomials Sn(x2; a, b, c) and pn(x\ a, b, c, d) are called the continuous dual Hahn and continuous Hahn polynomials respectively. Their orthogonality and recurrence relations can be obtained from those of Wilson polynomials, which were derived in Chapter 3. We restate them here for convenience. When Re(a, b, c, d) > 0 and the nonreal parameters occur in conjugate pairs, the orthogonality is

332

6

Special Orthogonal Polynomials

given by 1

r

T(a + ix)T(b + ix)T(c + ix)T(d + ix)

2^ Jo

r(2ix)

•Wm(x2; a, b, c, d)Wn(x2; a, b, c, d)dx (n+a + b + c + d— \)nn! T(n+a+b)Y(n + a + c)Y(n+a + d)Y{n+b+c)Y(n+b+d)Y(n+c+d)

(

(6.10.4) The recurrence relation is -{a2 +x2)Wn(x2)

= AnWn+l(x2)

- {An + Cn)Wn(x2) (6.10.5)

where (a + b)n(a + c)n(a + d)n' • d - \)(n + a + b)(n +a+ c)(n + a + d)

(n+a

-b + c + d- l)(2n +a + b + c + d] and Cn =

n(n + b + c- \)(n + b + d- \){n + c + d - 1) - 2) (2/i +tf + b +

These polynomials also satisfy a difference equation that is a dual of the recurrence relation. This is given by = B(x)y(x + i) - [B(x) + D(x)]y(x) + D(x)y(x-i),

(6.10.6)

where =

Wn(x2;a,b,c,d),

B(x) =

(a — ix)(b — ix)(c — ix)(d — ix) 2ix(2ix - 1)

D(x) =

(a + ix)(b + ix)(c + ix)(d + ix) 2ix(2ix + 1)

and

The corresponding results for the continuous dual Hahn polynomials are 2n

T(a + ix)T(b + ix)T(c + IJ T(2ix)

Sm(x2)Sn(jt2)dx Vc)n\8mn.

(6.10.7)

6.10 The Hypergeometric Orthogonal Polynomials

333

Here Sn(x2) — Sn(x2\ a, b, c) and a, b, c are either all positive or one is positive and the other two are complex conjugates with positive real parts, -(a2 + x2)~Sn{x2) = AnSn+l(x2)

- (An + Cn)Sn(x2) + CnSn-i(x2),

(6.10.8)

where Sn(x2) = Sn(x2)/[(a + b)n(a + c)n], An = (n + a -f b)(n + a + c), and Cn = n(n + b + c - 1); ny(x) = B(x)y(x + i) - [B(x) + D(x)]y(x) + D(x)y(x - i),

(6.10.9)

where yW =

Sn(x2), (a — ix)(b — ix)(c — ix)

"

2/JC(2/JC - 1)

'

and (a + ix)(b D(X)

=

+ /JC)(C +

ix)

2ix(2ix + 1)

•

In the case of the continuous Hahn polynomials, the results are 1 f°° — / T(a + ijc)r(fe + /x)r(c - ix)T(d 2n J_oo r(n + a + c)r(n + a

+ d)r(n

ix)pm(x)pn(x)dx

+ b + c)r(n + b

+ d)

=

,,inim dmn,

(o.lu.lU)

when Re(a, Z?, c, d) > 0, c = a, and d = b\ (a + i'x)/?„(*) = A n ^ n + i(x) — (Art + Cn)pn(x) -f C n p n _i(x), where pn(x) =

n

Pn(x'i a, b, c, d),

i (a + c)n(a + d)n

(n+a-\-b-\-c + d — l)(n + a + c)(n + « (2n + a + b + c-\-d — l)(2n + a + b + c

cn =

n(n + b + c — \)(n + Z? + d — 1) (2ft + tf + Z? + c + d - 2){2n + a + / ? + c + J - l ) '

(6.10.11)

334

6 Special Orthogonal Polynomials

and

n(n+a+b

+ c + d-

l)y(x)

= B(x)y(x + /) - [B(x) + D(x)]y(x) + D(x)y(x - i),

(6.10.12)

where

y(x) =

pn(x\a,b,c,d),

B(x) = {c- ix)(d - ix), and D(x) = (a + ix)(b + ix). We observed earlier that Jacobi polynomials are limits of Wilson polynomials. In this case, however, the difference equation for the Wilson polynomials becomes the differential equation for Jacobi polynomials. See the exercises for other examples of hypergeometric orthogonal polynomials. For recent developments on some polynomials considered here and their extensions and applications, see Nevai [1990]. 6.11 An Extension of the Ultraspherical Polynomials The generating function of the ultraspherical polynomials is given by the product of (1 — rew)~x and its conjugate. More generally, Fejer [1925] studied a sequence of polynomials defined as follows. Let f(z) = YlT=o an^n be a function that is analytic in a neighborhood of z = 0, with real coefficients. The generalized Legendre polynomials or the LegendreFejer polynomials are defined by

n=0

k=0

oo

n

k=0

pn(cos0)rn.

(6.11.1)

n=0

Feldheim [1941a] and Lanzewizky [1941] independently asked whether the pn(cos0) give rise to orthogonal polynomials other than the Gegenbauer polynomials. We know that if the pn(x) are orthogonal with respect to some positive measure, then they must satisfy xpn(x) = AnPn+\{x) + BnPn(x) + Cn/?n_i(x),

u = 0, 1, 2 , . . .

(6.11.2)

6.11 An Extension of the Ultraspherical Polynomials

335

with AnCn+\ > 0 and A, Bn, Cn+\ real. We can normalize to take p-\(x) = 0, po(x) = 1. The converse is also true, although we did not prove it. So, to find polynomials pn(x) that are orthogonal, it is enough to derive those that satisfy the three-term recurrence relation. Note that if pn(cos0) = ^2akan-kcos(n

— 2k)9,

k=0

then by 9 - • 6 + n we obtain

(-l)npn(cos0).

pn(-cos0) =

Therefore, if pn(x) satisfies (6.11.2), it must, in fact, satisfy 2xpn(x) = Anpn+\{x) +

Cnpn-\(x),

with An, Cn real, AnCn+\ > 0, n = 0, 1, 2 , . . . . This implies that n

n+\

2 cos 0 2^ ak^n-ic cos(rc — 2k)0 = An Y ^ akan+\-k cos(n + 1 — 2k)0 k=0

k=0 n-\

+ Cn ^ 2 akan-\-k cos(n - 1 - 2k)6. (6.11.3) Now use the trigonometric identity 2cos# cos(rc - 2k)0 = cos(n + 1 - 2k)0 + cos(n - 1 - 2k)0 to write the left side of (6.11.3) as n

dn-k cos(n + 1 — 2k)0 + ^2 ak^n-k cos(n — 1 — 2k) 0. &=0

k=0

Substitute this in (6.11.3) and equate the coefficient of cos(n + 1)0 to get Anaoan+i = aoan or A -

a

»

The coefficient of cos(« — 1 — 2k)0 gives h

u an-k-\

an

6

336

Special Orthogonal Polynomials

Take k = 0 and 1 to obtain an equation for the variables a. To simplify this equation, set sn = an/an-\. We obtain the nonlinear difference equation S\+Sn

= S2 + Sn-\

or Sn+\(sn - Sn-i

+ S\ - S2) = S\Sn - 52*11-1 •

For further simplification, set sn = tn + s\. The equation becomes

Wife - tn-l ~ t2) = -t2tn-Uh = 0. Write rn = t2Un\ then

un+i(un - un-X - 1) = -un-u u\ = 0. For linear difference equations we get polynomial solutions ]T) Anqn. Such a solution is not possible here and as there is no general method for solving nonlinear equations, we try the simplest rational expression as a possible solution, keeping in mind that u\ = 0. Set

withal < l,foiun(q,A,B) = un(q~\ A/(Bq), B~l). Then - qn~x) - (\ - Bqn)

A(l-qn) Bq n+l 1 - Bq

\-Bqn

A{\ - qn) - (1 - Bqn n+1

1 - Bq

A(\-qn-z) \-Bqn~l '

For this to be true, we must have B = 1 and (1 - qn~l)(A - 1 - (A - q)qn~l) = (1 - qn~2)(A - 1 - (A This is identically true for A — 1 = q. Thus

- qn7, So u/l

which shows that for some a and /? Sn

=

(6.11.4)

6.11 An Extension of the Ultraspherical Polynomials

337

This gives An =

q

~

Q

n^

(6.11.5)

and after some simplification Cn = —

^-~—-.

(6.11.6)

The recurrence relation for pn(x) is given by 2xa(l - Pqn)pn = (1 - qn+X)pn+\ + c*2(l -

p2qn-l)pn-U

with (1 - qn+l)(l - P2qn) We have taken \q\ < 1. If q = 1, then the value of un in (6.11.4) is defined by the limit as q -> 1. This case also gives rise to orthogonal polynomials. For example, if 0 = qx in (6.11.5) and (6.11.6), then An = (n + l)/(n + A), Cn = (n + 2k — l)/(n + X) (with a = 1), and the recurrence relation for ultraspherical polynomials is obtained. Clearly, there are other cases where division by zero may be involved in (6.11.5) and (6.11.6). These do not lead to orthogonal polynomials of all degrees unless q is a root of unity. Consider the situation where this problem does not arise and let us see what polynomials we get. We need an expression for an. We have

A B _iA B _2-"A 0

a0

(1

Therefore, ^

_

q2}...

(l

_

a qn)

which suggests that we take ao = « = 1- In this case, the polynomial is


h

;

. :

(l-q)---(l-qk)d-q)---(l-qn-k)

COS(« -

.... 2k)0.

This expression may appear a little strange at this point. As pointed out before, taking P = qx and letting q -> 1 gives the ultraspherical polynomials. One should keep this procedure in mind. In Chapter 10, we give an introduction to objects of

6 Special Orthogonal Polynomials

338

this kind. By the methods developed there, it can be shown that the generating function is given by

= 1fr1 n=0

(\-rei0qn)(\-re-i0qn)

This may appear like a much more complicated expression than the generating function for the ultraspherical polynomials that it is supposed to extend. But there is one sense in which it is simpler. Recall that the singularities of the generating function can be used to get information about the asymptotic behavior of the polynomials and the weight function (see Theorem 6.6.2). The generating function for the ultraspherical polynomials has algebraic singularities, whereas the singularities here are simple poles. These are easier to deal with. The poles closest to the origin are at r = el° and r — e~l°. Near r = el°, the generating function behaves like

°°r(l-^ 2 V)(1[ (1 - e2ie2ieqqnn)(\)(\ - qnn+x)

E

(1 - re~w)'

n=\J

So,

n=0

Write the infinite product as Re'*. Then 1/2

- fig")

l\^ (1 _ qn+X) l^n

2 cos(n6

V)

a s n —>• o c .

By Theorem 6.6.2, we expect the weight function to be

^

J

- 1JL

(i-^V)(i-g-^V) (l_pe2Wqn)(1_pe-2Wqn)

(1 - 2 cos 26qn+q2n) 2 2n -2pcos26qn +fi2qq2n)') These infinite products are well known in the theory of elliptic functions. In Chapter 10, we hope to convince the reader that they are quite natural and tractable, unwieldy though they may appear now.

Exercises

339

Exercises 1. Evaluate the integral rOO

/*OO

I =

e-

x2 2ixt

e dx

e~x2 cos2xtdx

=2 JO

J-OQ

(a) by contour integration, (b) by expanding cos 2x t in powers of JC and integrating term by term, and (c) by showing that the integral satisfies the differential equation

*L = -2,1. dt 2. Prove that F(JC, r) := e2xr~r = Y^Lo ^ ^ r W by s n o w m g that \d"Fn ] [ dr J r=0 3. Prove that / ?\

n/1

.

H (x}

a->oo

Hint: You can use generating functions, recurrence relations, or Rodrigues's formula. 4. Prove that un = e~x /2Hn(x) satisfies the equation

uf; + (2n +

l-x2)un=0.

Deduce that — (ufnum - u'mun) + 2{n - m)umun = 0. Hence prove the orthogonality of Hermite polynomials, that is, /»OO

/

umundx = 0 for m ^ n.

J—oo

5. Use the generating function for Hermite polynomials in Exercise 2 to prove that . v , v ^ Hk(x)Hn-k(y) u , Hn(x cos u + y sinu) = n\ 2_^ —7 6. Let n be a nonnegative integer. Show that x2n = and 2n+1

= 22n+1

. n _k k cos u sin u.

340

6 Special Orthogonal Polynomials

7. Define sgnx

_ f 1, ~\-l,

x > 0, xn(x) = e~x2/2Hn(x)/y/2nn\, n = 0, 1, 2, Denote the relative maxima of \(pn(x)\, as x decreases from +oo to 0, by Mo,n> /xi,n, /X2,w, Prove that Mr,* > MIMI+1,

w > r > 0.

Deduce that \(/)n(x)\ < max |0O(*)I = 1. (See Szasz [1951].) 11. Show that the Fourier transform of un (x) = e~x2/2Hn (x) is inun (x) by filling in and completing the following steps: poo

/>oo

\ un(x)eixydx J-oo

jn

= / e~x n eixy+x J-oo dx poo

= (-i)"ey/2 J-oo

/2

dx

JH

e-'—eW^dx dy"

/

dyn J12. Let V'ViOO — (2"-v2whi/2Hni^/lnx)e nx . Suppose / is square integrable on (—00, 00) and g is its Fourier transform. Let 00

fix) ~ ^2an\l/nix),

gix) '

n=0

n=0

00

00

^nix),

Xgix)

(a) Show that an — inbn. (b) Use the recurrence relation for Hermite polynomials to obtain V47ZX \jfnix) = Vrc + 1

Exercises

341

(c) Use (a) and (b) to show that \f\7icn = \Jn + \an + *fnan-\, *fkndn = i~n~l[ — / ^—oo

^

\f(x)\2dx

J—oo

with equality only if f(x) is almost everywhere equal to a constant multiple Of exp(—7TJC2). (e) Rescale to show that (d) implies that (for p > 0) pOO

p2

pOO

x2\f(x)\2dx + p-2

1

pOQ

x2\g{x)\2dx > — /

J—OO —oo

JJ—oo — OO

^

\f(x)\2dx.

J — OO

(f) Show how that (e) implies Heiseni Heisenberg's inequality: r

AmUx

poo oo

\L

-| "I 1/2 r

poo p

\ \ \l\J

2

2

1 1/2

x \g(x)\ dx\

-^

/.oo

>— j

(See de Bruin [1967].)

13. Show that

14. Show that

15. Prove that 2 , 2x _ ( - ! ) " ^ p H2k(x)H2n-2k(y) y

\f(x)\2dx.

22n

k^

16. Prove that

17. For Re(of + 1, 0) > 0, prove that

and

r(a + l)r(a + ^ + 2) f[ nLa(xt) ' *a n -(i-ty

RLfL(x(l-t)) m dt-. R

342

6 Special Orthogonal Polynomials

18. Prove

(a)

Y V ( V )Hk(x)(xy)n-k

(b)

= Hn(x + y).

19. Prove the identity

20. Show that for a > - 1 , r > 0, and JC > 0,

21. Prove that for Legendre polynomials Pn(x), r = e /o(v 1 — xzr). n=0

More generally,

X > -1/2. 22. Suppose Pn (JC) is the Legendre polynomial of degree n. Then Turan's inequality states that

[Pn{x)f - Pn.l(x)Pn+l(x)

> 0,

n > 1,

- 1 < x < 1.

This exercise sketches a proof of Turan's inequality. See Szego [1948] on which this is based. (a) Show that if the polynomial

has all real roots, then

(b) The following is a result from entire function theory: Suppose oo

f(y) = lim 5n(y//i) = V

-^/

Exercises

343

is an entire function with the factorization

n=0

where a > 0, ft and fin are real, and YlT=o Pn *s convergent. Then Sn(y) has all real roots. To obtain Turan's inequality, use Exercise 21 and (4.14.3). 23. (a) Use Exercise 19 to show that Sn(y) in Exercise 22(a) with uk = Pk(x) has all real roots and thus obtain another proof of Turan's inequality. (b) Extend Turan's inequality to the polynomials Hn(x), L®(x), and C£(JC). Prove these inequalities by two different methods. 24. For the Legendre polynomial Pn(x), prove the following results:

Pn(x)e-itxdx = rn^

(a)

/-oo

— Jn

10

ifx>lorjc — 1. Show that -i d v

dy

(1 - x2) - \1 + [fi - a - (a + 0 + 2)JC] / + Aj = 0 ax ax has a nontrivial polynomial solution if and only if X has the form n(n + a + ^ + 1), where n is a nonnegative integer. This solution is CP^a^\x), where C is a constant. 26. Prove the following results for ultraspherical polynomials: (a)

limx-"C^(x) = 2 " ^ .

f; ( \t* /2) "

(b) where

^1/2

344

6 Special Orthogonal Polynomials

27. Use Rodrigues's formula to prove

when x / ± 1 , and when the contour of integration is a simple closed curve, around t = x in the positive direction, that does not contain t = ± 1.

2n (l yr(l

J~

28. Prove that (a) F

i

+1/2

;1

"X

(b) =

(2A)2m+i

/ - m , ra + A, + 1

(2^TT)!X2Jpli

X l/2

;1

"X

2

The following problems define some important hypergeometric orthogonal polynomials. For a given nonnegative integer N appearing in the definition of a discrete orthogonal polynomial, we use the notation p

Fq(au

. . . , av\ bu .. •, bq\ x) := ^

-JTTT-

j—-x

29. The Racah polynomials are defined by

:=4

3

; V a + l,j8 + « + l , y + l for n = 0, 1, 2 , . . . , N, and where A.(*) = x(x + Y + S + 1) and one of the bottom parameters is—N. Show that the orthogonality relation is given by

(y + s + i)x((y + s + 3)/2),(q + i)x(P + s + i)x(y + i), xl((y + 8 + l)/2)AY +S-a

=

(w + a + j6 + l)H(j6 + !),(« - 3 + !)„(« + /3 - y + l)Bw! (a + fi + 2) 2n (a + !)„(/} + 5 + l) n (y + !)„

345

Exercises

where ifa + 1 = -N, (y+8 + (y+8-a

M=

2)N(8-a)N

ifp + 8

= -N,

Show that the recurrence relation is given by

X(x)Rn(X(x)) = AnRn+l(X(x)) - (An + Cn)Rn(X(x)) + Q where

An =

(n + a + p + l)(/i + a + l)(n + P + 8 + l)(/i + y + 1) (2n + a + P + l)(2n + a + p + 2)

and

n(n + P)(n + a + p — y)(n + a — 8) Show that the difference equation satisfied by y(x) = Rn(X(x)) is

n(n + a + l)y(x) = B(x)y(x + 1) - [B(x) + D(x)]y(x) + D(x)y(x - 1), with

B(x) =

(x + a + l)(x + P + 8 + 1)(JC + y + l)(x + y + 8 + 1) (2x + y + 8 + 1)(2JC + y + 8 + 2)

and

+ 8)(x - p + y)(x -a + y+8) (2x + y+8)(2x + y+8 + l)

D(x) =

30. The Hahn polynomials are limits of Racah polynomials defined by Bin Rn(X(x); a, p, -N - 1 , 5 ) = Qn(x; a, p, N) or

lim^ Rn(Hx); a, p,y,-p-N-i)

= Qn(x\ a, 0, N).

Show that

(a) Qn(x\a,p,N) = 3P N

(b)

Y^ —

-^-

-^^-Qmix; of, p, N)Qn(x\ a, j8,

x=0

+a

8mn-

346

6 Special Orthogonal Polynomials (c) -xQn(x)

= AnQn+x(x) n

- {An + Cn)Qn{x) + CnQn-i(x),

where

_ (n + of + p + l)(n + a + !)(# - n) ~ (2

and _ (d) Qw(*) satisfies the difference equation 7i (n + a + £ + l)v(x) = B(x)y(x + 1) - [B(x) + D(jc)]y(x) +

D(JC)};(X

- 1),

where

D(x)

=x(x-p-N-l).

(e) Use (b) to show that Qn(x; a, p, N) = CnQn(N - x; p, a, N), where Cn = ( - l ) n ( a + l)n/(P + l) n . Deduce Corollary 3.3.2. 31. Define the dual Hahn polynomials by Rn(X(x); y, 8, N) := lim Rn(X(x); -N - 1, p, y, 8) and deduce properties corresponding to (a) through (d) in the previous problem. Observe that dual Hahn is obtained from the Hahn by interchanging n and x. 32. Show that, for the Hahn polynomials Qn(x) defined in Exercise 31, the following limit formula holds: n

lim Qn(Nx;a,p,N)=

'

'-.

33. The Meixner polynomials can be defined by l i m Qn ( x \ b-l, N-+OQ

Show that (a)

0

(c) (c-l)xMn(x)

= c(n + b)Mn+l(x)-[n +

^J^^Srnn, and 0 < c < 1.

(n+b)c]Mn(x)+nMn-l(x).

(Note that an application of the Pfaff transformation (Theorem 2.2.5) shows that Meixner polynomials satisfy a three-term recurrence, which,

Exercises

347

by Favard's theorem, implies that they are also orthogonal with respect to a positive measure when c > 1. The reader is encouraged to find this orthogonality relation, which is obtainable from (b).) (d)

n(c - \)Mn{x) = c(x + b)Mn(x + 1) - [x + (x + b)c]Mn(x) + xMn(x - 1).

Observe the duality in n and x exhibited by relations (c) and (d). 34. A way of defining Krawtchouk polynomials is given by Kn(x\ p, N) := lim Qn{x\ pt, (1 - p)t, N). Prove the following relations: (a) Kn(x',p,N)

= 2Fi(-n,-x',-N',l/p), / AT\

N

(b)

w = 0, 1 , . . . , # .

) ^ ( 1 " P)N~xKm(x\ p, N)Kn(x; p, N)

Yl (

{-\)nn\f\-p\n Omn-

(c)

(-N)n V P -xKn(x) = p(N - n)Kn+i(x) - [p(N-n)+n(l +

- p)]Kn(x)

n(l-p)Kn-l(x).

(d) -nKn(x) = p(N - x)Kn(x + 1) - [p(N - x) + x(l - p)]Kn(x) +

x(l-p)Kn(x-l).

Note the relationship of (c) and (d) as in the case of the Meixner polynomials given in the previous exercise. In fact Kn(x; p, N) = Mn(x; -tf, p/(p - 1)). 35. Define the Charlier polynomials by Cn(x; a) := lim Mn(x; b, a/(a + b)) or Cn(x;a)=

lim

Kn(x;a/N,N).

N->oo

Deduce that (a)

Cn(x; a) = 2 F 0 ( - n , -JC; - ; oo

(b)

V —Cm{x; a)Cn(x; a) = n\a-neaSmn, a > 0. x=0

(c)

-I/a).

x

X

-xCn(x)

-

= aCn+l(x) - (n + a)Cn(x) + nCn-i(x).

348

6 Special Orthogonal Polynomials (d)

-nCn(x) = aCn(x + 1) - (x + a)Cn(x) + xCn(x - 1).

Compare (c) and (d) as in the previous two exercises. 36. Prove that lim (2a)n/2Cn(^x + a; a) = (-l)nHn(x). 37. The hypergeometric representation of the Meixner-Pollaczek polynomials is

These polynomials can be obtained as limits of continuous dual Hahn (or continuous Hahn) polynomials. Show that Sn((x — t)2; X + it, X — it, tcotc/)) i P»(x; 0) = lim . t-^oo n\(t/ sm(j))n

(a)

po /

(b)

J—o

(2 sm 4>)2Xn! "" (c)

(n + 1)P* + ,(JC) -

(d)

2[JC sin0

X>0

'

+ (n

e'*(X. - ix)P^(x + i) + 2i[xcos