Applied and Computational Complex Analysis - Volume 2: Special Functions, Integral Transforms, Asymptotics, Continued Fractions [1 ed.] 0471015253, 9780471015253

Presents applications as well as the basic theory of analytic functions of one or several complex variables. The first v

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Table of contents :
8 Infinite Products
8.1. Definition and Elementary Properties
8.2. Some Infinite Products Relevant to Number Theory
8.3. Product Representations of Entire Functions
8.4. The Gamma Function
8.5. Stirling's Formula
8.6. Some Special Series and Products
8.7. The Beta Function
8.8. Integrals of the Mellin-Barnes Type
Seminar Assignments
Notes
9 Ordinary Differential Equations
9.1. The Existence Theorem
9.2. Power Series Method
9.3. Linear Systems
9.4. Linear Systems with Isolated Singularities
9.5. Singularities of the First Kind: Formal Solutions
9.6. Scalar Equations of Higher Order: Method of Frobenius
9.7. Two Examples: The Equations of Kummer and Bessel
9.8. The Infinite Point: Equations of Fuchsian Type
9.9. The Hypergeometric Differential Equation
9.10. Quadratic Transforms: Legendre Functions
9.11. Singularities of the Second Kind: Formal Solutions
9.12. Singularities of the Second Kind of Special Second-Order Equations
Seminar Assignments
Notes
10 Integral Transforms
10.1. Definition and Basic Properties of the:?? Transformation
10.2. Operational Rules: Basic Correspondences
10.3. Ordinary Differential Equations: Systems
10.4. Convolution
10.5. Some Nonelementary Correspondences
10.6. The Fourier Integral
10.7. The Laplace Transform as a Fourier Transform
10.8. Dirichlet Series: Prime Number Theorem
10.9. Functions of Exponential Type
10.10. The Discrete Laplace Transform
10.11. Some Integral Transforms Related to the ?? Transform
10.12. Some Applications to Partial Differential Equations
Seminar Assignments
Notes
11 Asymptotic Methods
11.1. An Example: Asymptotic Power Series
11.2. The Algebra of Asymptotic Power Series
11.3. Analytic Properties of Asymptotic Power Series
11.4. Asymptotic Solutions of Differential Equations
11.5. The Watson-Doetsch Lemma
11.6. Extension of the Lemma
11.7. Asymptotic Formulas: Laplace's Method
11.8. The Method of Steepest Descent
11.9. General Asymptotic Expansions: Asymptotic Factorial Series
11.10. Generating Functions: Subtracted Singularities
11.11. The Euler-Maclaurin Summation Formula
11.12. The Numerical Evaluation of Limits: Romberg's Algorithm
Seminar Assignments
Notes
12 Continued Fractions
12.1. Definition and Basic Properties
12.2. Continued Fractions in Number Theory
12.3. Convergence of Continued Fractions with Complex Elements
12.4. RITZ Fractions (Formal Theory): Pade Table
12.5. The Convergence of RITZ Fractions: Examples
12.6. Tne Division Algorithm: Rational RITZ Fractions
12.7. SITZ Fractions: Approximants, Stable Polynomials
12.8. S Fractions: Generalized Value Functions, Convergence
12.9. S Fractions: The Representation of Tneir Generalized Value Functions by Stieltjes Transforms
12.10. Positive Symmetric Functions and Their Representation as Stieltjes Transforms
12.11. Existence and Convergence of the S Fraction Corresponding to a Stieltjes Transform
12.12. S Fractions: Expansions of Stieltjes Transforms
12.13. S Fractions: Expansions of Iterated Laplace Transforms
12.14. Moment Problems
Seminar Assignments
Notes
Bibliography
Appendix: Some additional problems on vol. I
Index
Recommend Papers

Applied and Computational Complex Analysis - Volume 2: Special Functions, Integral Transforms, Asymptotics, Continued Fractions [1 ed.]
 0471015253, 9780471015253

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PURE AND APPLIED MAmEMATICS A WUey-Interscience Series of Texts, Monographs, and Tracts Founded by RICHARD COURANT Editors: LIPMAN BERS, PETER HILTON, HARRY HOCHSTADT

ASH-Information Theory AURIN-Approximation of Elliptic Boundary-Value Problems BELLMAN and WING-An Introduction to Invariant Imbedding BEN-ISRAEL and GREVILLE-Generalized lnvenes: Theory and Applications BUCY and JOSEPH-Filtering for Stochastic Processes with Applications to Guidance CARTER-Simple Groups of Lie Type CLARK-Mathematical Bioeconomics COHN-Dillerence Algebra CORDUNEANU-Aimost Periodic Functions COURANT and FRIEDRICHS-Supersonic Flow and Shock Waves CURTIS and REINER-Representation Theory af Finite Groups and Associative Algebras DAVIS-Applied Nonstandard Analysis DUNFORD and SCHWARTZ--Linear Operaton Part 1-General Theory Part 2-Spectral Theory, Self Adjoint Operators in Hilbert Space Part 3-Spectral Operators EHRENPREIS-Fourier Analysis in Several Complex Variables FRIEDMAN-Dillerential Games HALE-Ordinary Dillerential Equations HARRIS-Mathematical Structures of Language HENRICI-Applied and Computational Complex Analysis, Volume 1 HENRIO-Applied and Computational Complex Analysis, Volume 2 HESTENES-Optimization Theory: The Finite Dimensional Cue RILLE-Ordinary Dillerential Equations in the Complex Domain HILTON and WU-A Coune in Modem Algebra HOCHSTADT-The Functions of Mathematical Physics HOCHSTADT-Integral Equations JACOBSON-Lie Algebra KOBAYASHI and NOMIZU-Foundations of Dillerential Geometry. In Two Volumes

APPLIED AND

COMPUTATIONAL COMPLEX ANALYSIS VOLUME2

Special Functions-Integral Transforms -Asymptotics-Continued Fractions

PETER HENRICI Professor of Mathematics Eidgenissische Tedmisdle Hochschule, Ziirich

A WILEY-INTERSCIENCE PUBUCATION

JOHN WILEY & SONS,

New York o Londono Sydney o Toronto

To MARIE-LOUISE

Copyright© 1977 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. No part of this book may be reproduced by any means, nor transmitted, nor translated into a machine language without the written permission of the publisher. Libnry of ConiiP"ess Cataloging in PubUcation Data: (R.msed)

Henrici, Peter, 1923Applied and computational complex analysis. (Pure and applied mathematics) "A Wiley-Interscience publication." Includes bibliographies. CONTENTS: v. 1. Power series-integration-conformal mapping--location of zeros.-v. 2. Special functions--integral transforms--asymptotics-continued fractions. 1. Analytic functions. 2. Functions of complex variables. 3. Mathematical analysis. I. Title. QA33l.H453 1974 515'.9 73-19723 ISBN 0-471-01525-3 (v.2)

r>REFACE L'. the present Volume II of our three-volume work we continue to discuss cLtgorithmic techniques that can be used to construct either exact or approxirnate solutions to problems in complex analysis. A focal point for these f;pplications is the evaluation and manipulation of solutions of analytic differential equations. Successive chapters deal with the representation of ~,.olutions by (convergent or divergent) series expansions, with the method of integral transforms, with asymptotic analysis, and with the representation ·;>f special solutions by continued fractions. The gamma function is dealt with in the opening chapter in the context of product expansions of analytic functions. Together with its companions, this volume provides a fair amount of information on some of the more important special functions of mathematical physics. However, our treatment of these functions is unconventional in its organization. Whereas the conventional treatment proceeds function by function, giving to each function its due share of series and integral representations, and of asymptotic analysis, our treatment proceeds by general methods and problems rather than by individual functions. Special results thus appear mainly as applications of general principles. The same methodology will be followed in Volume III; for instance, addition theorems will be considered in the context of partial differential equations. Although I hope that my program has enabled me to illuminate the basic properties of special functions such as the gamma function, the hypergeometric function, the confluent hypergeometric function, and the Bessel functions, it must be pointed out that a full in-depth treatment of any class of special functions was neither intended nor possible. For more detailed information the reader should turn either to specialized treatises or to the monumental Bateman manuscript project (Erdelyi [1953], [1955]), which provides an essentially complete collection of results known up to the early 1950s. To call this treatment of complex analysis computational is not meant to imply that I deal exhaustively with the problem of obtaining numerical values of a given special function for all possible values of the variable and of the parameters. This topic has grown into a far too specialized and refined iii

iv

PREFACE

science to be treated thoroughly in a book that also must deal with many other topics. Tne reader is referred to Gautschi [1975] for an excellent survey of the methods that are currently employed. Questions of computational efficiency, including the manipulation of power series, will be dealt with in Chapter 20 (Volume III) of the present work. The contents of individual chapters are, briefly, as follows. Chapter 8, on infinite products, features, after the necessary preliminaries, some products of importance in number theory, including Jacobi's celebrated triple product identity. Tne striking combinatorial implications of this identity seem appropriate as an eye-opener to the joys of classical analysis. We then proceed to a standard treatment of the gamma function, proving the equivalence of the definitions by Weierstrass, Gauss, Euler, and Hankel. Stirling's formula is obtained via the Weierstrass definition; derivations from the other three definitions are contained in Chapter 11. The chapter concludes with a discussion of integrals of the Mellin-Barnes type and their application to hypergeometric functions. The next chapter, on ordinary differential equations, begins with a standard presentation of the analytic theory from the matrix point of view. Here we can apply some of the material given in Chapter 2 on analytic functions with values in a Banach algebra. The treatment of the confluent and of the standard hypergeometric equations is more detailed than is customary in more theoretically oriented texts. In particular we present, on the basis of Riemann's epochal paper [1857], a complete theory of the linear and the quadratic transforms of the hypergeometric series. Because Legendre functions are merely hypergeometric functions permitting quadratic transforms, written in a different notation, we can dispose of these functions very quickly. Chapter 10, on integral transforms, begins with a broad discussion of the Laplace transform from an elementary point of view, avoiding advanced real variable theory. To present a clean solution of the inversion problem, we provide a self-contained discussion of the Fourier integral theorem (for piece-wise continuous L 1 functions). We next apply the Laplace transform to Dirichlet series and use this opportunity to give a short account of the Riemann zeta function and its connection with the prime number theorem. A presentation of Polya's theory of Laplace transforms of entire functions of exponential type, with its fascinating link between the growth of the original function in a given direction and the location of the singularities of the image function, follows. The next section, on discrete Laplace transforms, contains some generalizations of Polya's theory suggested by the late H. Rutishauser. The chapter concludes with a discussion of the Mellin transform, and of some simple applications of the integral transform idea to problems in mathematical physics.

PREFACE

v

In Chapter 11, on asymptotics, we have tried, first of all, to give a clear definition of asymptotic series, a concept that is notoriously difficult to absorb for the beginning student. We then prove the important result that the (generally diverging) formal series solutions to differential equations with irregular singular points are asymptotic to appropriate actual solutions. In addition to standard topics, such as Watson's lemma, Laplace's method, the method of steepest descent, Darboux's method, and the EulerMaclaurin sum formula, we then present some less orthodox subjects such as general asymptotic series (in particular, asymptotic factorial series, for which a useful analog of Watson's lemma is given), and the numerical evaluation of limits by the Romberg algorithm. The last chapter of this volume, on continued fractions, presented a special challenge to the expositor because the analytical theory of continued fractions is seldom presented in a larger context in a textbook. A novel feature here is the prominence given to Moebius transformations, and with them to the geometric point of view. This not only enables us to deal efficiently with the formal aspects of continued fractions, but also permits us to treat questions of convergence in an intuitively appealing manner. Once again, the qd algorithm makes its appearance; here it is used to establish some classical continued fractions representing hypergeometric functions. We then discuss the division algorithm and use it to give an alternate solution of the stability problem for polynomials. The second half of the chapter is devoted to continued fractions of the Stieltjes type. Contrary to other presentations, in which such fractions are merely incidental to a discussion of the moment problem, continued fractions and the functions represented by them here are at the center of interest. Our approach enables us to encompass in a very natural way topics of general interest such as the Stieltjes integral, normal families, Vitali's theorem, and the representation formulas of Herglotz, Hamburger, and Nevanlinna for functions with values in a circular region. Some of these topics will be required again in Volume III. We then proceed to the Carleman convergence criterion and more generally to various estimates for the truncation error, valid also when the corresponding power series has radius of convergence zero. Numerous applications, some of them new, should demonstrate the usefulness of the theory. As in Volume I, I have restrained myself from using an excess of specialized mathematical notation and terminology to make my subject matter accessible to readers with a variety of backgrounds. Although power series are still favored, this volume can be read without knowing in detail the formal power series approach to complex analysis presented in Volume I. Even within this volume, the chapters are reasonably self-contained to make our text useful also to the casual peruser. Courses of varying length on

vi

PREFACE

aspects of applied and computational analysis could be based on almost any combination of chapters; in fact, most of the material was presented in the form of such courses at the ETHZ. By exposing the student to a variety of techniques and applications, we are trying to educate applied mathematicians who are able to contribute to the progress of science by their general expertise as well as by specialized research. Once more, it is my pleasure to express my thanks to the many individuals who have helped me along in my expository endeavors. In addition to the teachers and collegues mentioned in the preface to Volume I, I wish to record my indebtedness to J.-P. Berrut, P. Geiger, M. Gutknecht, E. Hane, M.-L. Henrici, and J. Waldvogel, who have read parts of the manuscript, corrected errors, and suggested numerous improvements. M. Gutknecht, in addition, wrote the programs for drawing the graphs of the gamma function that appear in Chapter 8. R. Askey and J. F. Kaiser supplied valuable information. R. P. Boas provided not only encouragement but also some important references. During my stay at the Bell Laboratories in 1975, D. D. Warner substantially deepened my understanding of continued fraction theory. I also wish to express my appreciation to the staff of John Wiley & Sons, who once more handled all problems that arose in the production of a manuscript of mine in the most expert and professional manner. I dedicate this volume to my wife, who by her optimism and good judgment has been of invaluable help in making the many decisions that were necessary to shape my manuscript into its final form. PETER HENRIO Zurich, Switzerland September 1976

CONTENTS 8

Infinite Products

1

8.1. Definition and Elementary Properties 1 8.2. Some Infinite Products Relevant to Number Theory 7 8.3. Product Representations of Entire Functions 17 8.4. The Gamma Function 24 8.5. Stirling's Formula 37 8.6. Some Special Series and Products 46 8.7. The Beta Function 55 8.8. Integrals of the Mellin-Barnes Type 62 Seminar Assignments 73 Notes 73

9

Ordinary Dift'erential Equations

9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 9.8. 9.9. 9.10. 9.11. 9.12.

75

The Existence Theorem 75 Power Series Method 82 Linear Systems 91 Linear Systems with Isolated Singularities 105 Singularities of the First Kind: Formal Solutions 114 Scalar Equations of Higher Order: Method of Frobenius 124 Two Examples: The Equations of Kummer and Bessel 132 The Infinite Point: Equations of Fuchsian Type 143 The Hypergeometric Differential Equation 150 Quadratic Transforms: Legendre Functions 165 Singularities of the Second Kind: Formal Solutions 180 Singularities of the Second Kind of Special Second-Order Equations 187 Seminar Assignments 193 Notes 194 vii

viii

CONTENTS

10 Integral Transforms

10.1. 10.2. 10.3. 10.4. 10.5. 10.6. 10.7. 10.8. 10.9. 10.10. 10.11. 10.12.

195

Definition and Basic Properties of the:£ Transformation 199 Operational Rules: Basic Correspondences 209 Ordinary Differential Equations: Systems 220 Convolution 237 Some Nonelementary Correspondences 247 The Fourier Integral 261 The Laplace Transform as a Fourier Transform 276 Dirichlet Series: Prime Number Theorem 293 Functions of Exponential Type 305 The Discrete Laplace Transform 321 Some Integral Transforms Related to the:£ Transform 336 Some Applications to Partial Differential Equations 340 Seminar Assignments 351 Notes 351

11 Asymptotic Methods

. . . . . 353

11.1. 11.2. 11.3. 11.4. 11.5. 11.6. 11.7. 11.8. 11.9.

An Example: Asymptotic Power Series 353 The Algebra of Asymptotic Power Series 364 Analytic Properties of Asymptotic Power Series 370 Asymptotic Solutions of Differential Equations 380 The Watson-Doetsch Lemma 391 Extension of the Lemma 401 Asymptotic Formulas: Laplace's Method 409 The Method of Steepest Descent 416 General Asymptotic Expansions: Asymptotic Factorial Series 432 11.10. Generating Functions: Subtracted Singularities 442 11.11. The Euler-Maclaurin Summation Formula 450 11.12. The Numerical Evaluation of Limits: Romberg's Algorithm 461 Seminar Assignments 471 Notes 471

12 Continued Fractions . . . . . . . .

12.1. Definition and Basic Properties 473 12.2. Continued Fractions in Number Theory

. . . . . . . . 473 487

ix

CONTENTS

12.3. 12.4. 12.5. 12.6. 12.7. 12.8. 12.9. 12.10. 12.11. 12.12. 12.13. 12.14.

Convergence of Continued Fractions with Complex Elements 501 RITZ Fractions (Formal Theory): Pade Table 515 The Convergence of RITZ Fractions: Examples 529 Tne Division Algorithm: Rational RITZ Fractions 539 SITZ Fractions: Approximants, Stable Polynomials 551 S Fractions: Generalized Value Functions, Convergence 561 S Fractions: The Representation of Tneir Generalized Value Functions by Stieltjes Transforms 568 Positive Symmetric Functions and Their Representation as Stieltjes Transforms 586 Existence and Convergence of the S Fraction Corresponding to a Stieltjes Transform 596 S Fractions: Expansions of Stieltjes Transforms 616 S Fractions: Expansions of Iterated Laplace Transforms 62 7 Moment Problems 635 Seminar Assignments 639 Notes 640

Bibliography

642

Appendix: Some additional problems on vol. I

650

Index

653

8 INFINITE PRODUCTS

§8.1. DEFINmON AND ELEMENTARY PROPERTIES Let {an} be a sequence of complex numbers. It is intuitively clear what is to be understood by the infinite product: 00

n an= ala2a3

0

0

0

0

(8.1-1)

n=l

We are to form the sequence of partial products {pn}. where Pt :=a~. P2 := a1a2, P3 := a1a2a3, .... This sequence is somehow to be identified with the infinite product a 1a 2a 3 ... . It is clear, however, that to write down the factors of the product conveys more information than to write down merely the sequence of partial products. If one factor, say am is zero, then all partial products Pm are zero for m ;;:;?= n, and it is impossible to recover the values of the factors am from the sequence of partial products for m > n. (Contrary to this, the terms of an infinite series can always be recovered from the sequence of its partial sums.) For this reason we shall adopt the following formal definition (see Buck [1965], p. 158): An infinite product is an ordered pair [{an}~, {pn}~] of sequences, where at. a 2, ... are complex numbers, and where Pn := a 1a 2 · · · am n = 1, 2, .... The numbers an and Pn are, respectively, called the nth factor and the nth partial product of the infinite product [{an}, {pn}]. Once this definition is understood, it is completely acceptable to denote an infinite product by a symbol such as (8.1-1), which exhibits only the factors. Some difficulties also arise if we try to define the concepts of convergence and of value for infinite products. Proceeding as in the case of infinite series, it would be tempting to call the product (8.1-1) convergent if the limit lim Pn =: p

(8.1-2)

n->OO

exists, and to define p as the value of the product. In the interest of formulating simple necessary and sufficient conditions for convergence, it 1

2

INFINITE PRODUCI'S

is advantageous, however, to call a product of nonzero factors convergent only if the limit (8.1-2) exists and is different from zero. If a product has zero factors, the limit of its partial products always exists and has the value zero. Convergence would thus not depend on the whole sequence of factors. To avoid this exceptional situation, we call a product_ with zero factors convergent if the product of the nonzero factors converges in the foregoing sense. Thus, in summary, we adopt the following DEFINIDON

The product (8.1-1) is said to converge if and only if at most a finite number of its factors are zero and if the sequence of partial products formed with the nonzero factors has a limit which is different from zero.

Let fl~=l an be a convergent infinite product, and let Pn := a1a2 · · ·an, possible zero factors excluded. Then we have, for n sufficiently large, an = Pnf Pn -I· Because Pn -+ p :¢ 0 there follows (8.1-3) n-+00

Thus in a convergent infinite product the factors must tend to one. In view of this it is customary to write infinite products in the form n=l

so that an -+ 0 now is a necessary condition for convergence. It is easy to see that this condition is not sufficient by considering the example an := 1/ n, n = 1, 2, .... Here Pn = ( 1

+i) (1 +~) ···(1 +;) > 1 +(1+~ +•••+;),

and the product is divergent, because the harmonic series is divergent. The logarithm of a finite product equals the sum of the logarithms of the factors. We thus may expect to derive convergence criteria for products from convergence criteria for sums by taking logarithms. We are led to consider the infinite series 00 (8.1-4) L Log(1 +an), n=l

where, for any z :¢ 0, Log z denotes the principal value of the logarithm, here defined by the condition -'IT' < Im Log z ~ 1T. Let sn be the nth partial sum of (8.1-4). Then Pn = e•", and if sn -+s, it follows from the continuity of the exponential function that Pn -+ p := e• -:F 0. Thus the convergence of (8.1-4) is a sufficient condition for the convergence of the infinite product. We now shall show that this condition is also necessary. Suppose that

3

DEFINI'DON AND ELEMENTARY PROPERTIES

Pn -+ p ~ 0. We let t/J := Im Log p and define a single-valued branch log* z of log z by the condition t/J- '1T' < Im log* z:;;;; t/J +'IT'. Then log* z is continuous in the vicinity of z := p, and there follows ..

log*pn-+ Log p (n-+ oo).

(8.1-5)

We cannot be sure that sn = log*pn [because the branches of the logarithms in (8.1-5) have already been chosen] but it is certainly true that (8.1-6) where hn is some well-determined integer. We wish to show that limn .... oo sn exists, and in view of (8.1-5) this amounts to showing that hn = hn- 1 for all suffi.ciently large n. Taking the difference of two consecutive terms in (8.1-6), w~ find which we write in the form

(hn- hn-t)2'1T'i = Log(l +an)+ [log*Pn-t- Log p] -[log*Pn -Log P]. For sufficiently large values of n, lim Log(1 + an)l < 27T'/3 in view of an-+ 0, and IIm(log*pn- 1- Logp)l< 27T'/3 in view of (8.1-5). Thus ultimately

lhn- hn-1127T' < 27T', which implies that hn = const and Sn -+ s, as desired.



Altogether we have proved: THEOREM 8.la

n:=

An infinite product I (1 +an) with nonzero factors converges if and only if the series ~:=t bn converges, where bn := Log(1 +an) (principal value). A necessary condition for the convergence of the product fl (1 +an) or of the series I. Log(1 +an) is that an -+0. Now if an -+0, Log(1 +an) asymptotically behaves like an. In fact, from Log(1+z)=

r1~tdt= r

(1-1:t)dt

we have for lz I:;;;;!, integrating along the straight line segment, Log(l + z) = z(1 + wz), where lwl:;;; 1. Thus if lanl,.;;;f, then

!!ani,.;;; ILog(1 +an)! ,.;;;~lanl·

(8.1-7)

INFINITE PRODUCI'S

4

Hence I.jLog( 1 +an )j converges and diverges simultaneously with I. ian j. An infinite product for which the series I. Log(l +an) converges absolutely will be called absolutely convergent. In this terminology, we have obtained: THEOREM 8.1b

A necessary and sufficient condition for the absolute convergence of the product (1 +an) is the absolute convergence of the series ~:=1 an.

rc=l

The emphasis here is on absolute convergence. Simple examples (see problems 9 and 10) show that the theorem is not true if the words "absolute convergence" are replaced by "convergence." These definitions and theorems also apply to the pointwise convergence of infinite products whose factors depend on a variable. A difficulty arises if we wish to define uniform convergence, because of the vanishing of factors. For definiteness, assume that the functions an (z) are analytic on a region S, that none of the functions 1 +an (z) vanishes identically on S, and that at most finitely many of these functions assume the value zero on S. The product 00

n (1 +an(z))

n=l

is said to converge uniformly on S if the sequence of partial products formed with those factors that do not vanish on S converges to a limit ~0 uniformly for all z E S. With this convention, the following analog of Theorem 8.1b holds and is proved similarly: THEOREM 8.lc

A necessary and sufficient condition for the absolute and uniform convergence of the product II (1 +an(z)) is the absolute and uniform convergence of the series I. an(z).

The fundamental theorem on uniformly convergent sequences of analytic functions (Theorem 3.4b) shows that the values of a product of analytic factors that is uniformly convergent on a set·S define an analytic function on S, even if the factors with zeros are included. EXAMPLE

Let a"(z) := -z 2 /n 2 • We shall show that the product

n co

n=l

(

z2)

1-2 ·

n

(8.1-8)

converges uniformly on every bounded setS. Indeed, let k be an integer such that Sis

5

PROBLEMS

contained in the disk

lz I,;;; k. Omitting the first k factors, we obtain the product z 2) ( 1--

fl"'

n2

n=k+l

'

whose factors do not vanish on S. The series z z:"' ( -n 2

)

2

n=k+1

converges uniformly and absolutely on S, because it is majorized by the converging I. (1/n 2 ). Hence the uniform convergence follows by Theorem 8.1c. We series conclude that (8.1-8) represents an entire analytic function. Because the zeros of this function are located at z = ± 1, ±2, ... , we may expect it to be closely related to ( 1rz )- 1 sin(1rz ), which has the same zeros and the same value at z = 0. It is shown in §8.3 that the two functions are, in fact, identical.

e

PROBLEMS 1.

Show that

il (1-(-n1)") =_!.2

n~2

2.

In calculus it is shown that

2·2·4·4·6·6···2n·2n 2 n-oo 1 · 3 · 3 · 5 · 5 · 7 · · · (2n -1)(2n + 1)

1T

-=lim-------------(Wallis' formula). Show that this may be written 2= 'TT

3.

fl"'

n~t

( 1 -1-)

(2n) 2



Prove that 1 "' (1 +z 2 ") = - f1 1- Z

n~o

4.

uniformly on every compact set contained in Show that

Il (1 + .:_) n

lz I< 1.

e-z/n

n=l

5.

represents an entire analytic function with zeros at the negative integers. Let the real number~ be given in decimal representation,

Show that e< =

fl"' k=-m

e w-•a•.

6

INFINITE PRODUCI'S 6.

Show that for any z ~ 0,

n cos 2 co

(

-k

k-o

7.

8.

) sin 2z z =-2-. z

[Using trigonometric identities, express the partial products as sums of cosines. Then use the definition of the Riemann integral.] Find the value of Vieta's product,

Following D. H. Lehmer (Amer. Math. Monthly, 1935), show tliat the value of the infinite product

in which the successive denominators satisfy d,. = 6d,._ 1 - d,._ 2 , is purely imaginary. [Solve the recurrence relation for the d,.. The resulting formula has a meaning for nonintegral n, and there follows d,.+l

{;:

lim-= 3+v8. d,.

,. ... co

Letting t,. := tan(argp!14), where p,. is the nth partial product, show that· l = d,./2+1-dn/2 ] 11

2dn/2

.

"9. Show that the product

n 00

,_2

(

(-1)") 1+-.;;,

is divergent, although the series l: (-1)"/..r, is convergent. 10. Let 1 1 a2,.-t := -173, a2,. := ~1 , n = 2, 3, ....

n

n

-

Show that the product

is convergent (and, in fact, has the value 1), although the series l: a,. is divergent.

SOME INFINITE PRODUCI'S RELEVANT TO NUMBER THEORY

7

§8.2. SOME INFINITE PRODUCI'S RELEVANT TO NUMBER THEORY The main purpose here is the study of certain classical infinite products with variable factors. Although the functions defined by these products do not lie in the mainstream of general complex analysis, some of the identities that exist between them have striking combinatorial and numbertheoretical applications. The products in question are 00

p(z):=

00

n (1+zn), n=l

q(z) :=

n (1-zn). n=l

(8.2-1)

By the criterion of Theorem 8.1c, both products are uniformly and absolutely convergent in any disk lz I::;; p where p < 1. Hence they represent analytic functions that can be expanded in Taylor series for lz I< 1: 00

00

p(z) =: ~ anZn, n=O

q(z) =: ~ bnzn. n=O

(8.2-2)

Because the products contain no zero factors, they are (by the definition of convergence!) different from zero for lzl < 1. Thus their reciprocals are likewise analytic for lz I< 1; we put, in particular, . ~

1

n

- ()=. t... CnZ. q Z n=O

The coefficients am bm Cn can be evaluated very easily. Consider, for example, the nth partial product of p(z ), n

ll

Pn(Z) :=

(1 +zk).

k=1

This is a polynomial of degree !n(n + 1), which we write as Pn(Z)

• (oo)

=.

(n)

k

~ ak z .

k=O

Because Pn (z)-+ p (z) locally uniformly in lz I< 1, the basic theorem on convergence of sequences of analytic functions (Theorem 3.4b) implies that for each k = 0, 1, ... ,

n-+oo

By comparing coefficients of 1, z, . .. , zn in the relation Pn+t(Z) = (1 + Zn+l)pn(z)

8

INFINITE PRODUCfS

we see, however, that a~n+l) = a~n)

k

= 0, 1, ... 'n.

Thus for any fixed k the coefficients a~n> no longer change once n has reached the value k, and we find k =0, 1, ....

(8.2-3)

Thus we have, for instance, (1 +z)(l +z 2)(1 +z 3) = a 0+a1z +a 2 z 2 +a 3 z 3 +a~>z 4 + · · ·. As in the discussion of the "drawer problem" in §7 .3 we see that ak fork > 0 equals the number of ways (without regard for the order) in which the integer k can be written as a sum of distinct positive integers. For example, 1 = 1,

hence a1 = 1

2=2,

hence

a2= 1

3=3 =1+2,

hence a3=2

6=6 =1+5 =2+4 = 1+2+3, hence

a6 =4.

Putting

we find in an analogous manner that k =0, 1, 2, ....

The combinatorial interpretation analogous to the above is as follows: For k > 0, bk represents the excess of the number of ways in which k can be written as a sum of an even number of distinct positive integers over the number of ways it can be written as a sum of an odd number of distinct positive integers. For instance, from the data given we see that b6 = 0. A combinatorial interpretation also exists for the coefficients ck. Formally, this can be seen by expanding each term of the product 1I q (z) in a geometric series: 1 (1+z+z 2 + · · · )(1+z 2 +z 4 + · · ·) (1-z)(1-z 2 } • • • x(1+z 3 +z 6 + ·· ·}·· ·.

SOME INFINITE PRODUCTS RELEVANT TO NUMBER THEORY

9

Multiplying the series on the right, we see that only the first k - 1 factors can contribute to the coefficient of zk. This is confirmed by letting 1 _ 1 _. --n ~-· qn (z) 1n

k=1

Z

00

(n)

~ ck z

k

k =O

and noting the recurrence relation

1

1

1

1 n+1 --(1+z + · · ·)

qn+1(z) = qn(Z) 1-zn+l

qn(Z)

which implies, as before, that (n+1) _ (n)

ck

- ck ,

k =0, 1, ... , n.

Each contribution to a coefficient ck where k > 0 stems from a product of the form (8.2-4) where the n; are nonnegative integers such that n1 •

1 + n2 • 2 + · · · + nk · k = k

(n; ;;;?::0),

(8.2-5)

and each such product adds 1 to ck. A representation of a positive integer k of the form (8.2-5) is calied a partition of k. Because there is a one-to-one correspondence between products (8.2-4) equalling zk and partitions of k, the coefficient ck is equal to the total number of partitions of k. In the number-theoretical literature, the function 1r(k) := ck is called the partition function. We have established 1/q(z) as the generating function of the partition function. EXAMPLE

1

The number 6 has the partitions 6=6 =5+1=4+2=3+3 =4+1+1=3+2+1=2+2+2 =3+1+1+1=2+2+1+1 =2+1+1+1+1 = 1+1+1+1+1+1. There are 11 partitions; we conclude that c6 = 11.

The partition function is of considerable interest in number theory. To calculate the function from its definition is extremely laborious. For very large k, asymptotic formulas discovered by Rademacher are sufficiently

10

INFINITE PRODUCfS

accurate to permit the numerical determination of 1r(k) without error. Here we shall derive certain functional relations between the functions p and q that can be used to compute the partition function rapidly by recursion. As a first result we prove THEOREM 8.1a (Euler):

q(z2) p(z}= q(z),

lzl= h~L and it suffices to determine these coefficients fork= 0, 1, ... , n. Evidently, 1+3+5+ +(2n-1) n2 h (n) (8.2-9) n =z =z . o

o

o

From the functional relation

gn(Z, z2t) =

n

n (1 + z2k+l t)(1 + Z2k-3t-1)

k=l

1+z 2"+ 1t 1+z- 1t- 1 1+zt 1+z2n It tgn(Z,t)

=

1 +z2n+tt 2n gn(Z, t)

zt+z

SOME INF1NITE PRODUCfS RELEVANT TO NUMBER THEORY

13

we find n

(zt+ z2n)

L

n

hin)z2kl

k=-n

= (1 + z2n+lt) L

hin>l,

k=-n

and hence, comparing coefficients of l, z

2k+2nh(n)+ 2k-lh(n) _ h(n)+ 2n+lh(n) k z k-1- k z k-t. 2k+2n 1 h(n) -z (n) k-1- z2k 1(1- z2n 2k+2) hk ,

k = 1, 2, ... , and thus, using (8.2-9}, (n)- (1- z2k+2n+2)(1- z2k+2n+4) ... (1- z 4n) k2 hk( 1 -z2)( 1 -z4}···( 1 -z2n 2k) z .

In terms of the partial products n

qn(Z) :=

n (1- Zk) k=l

of q(z), hin> may be expressed as follows: h (n) k =

q2n (z 2 } k2 2 2 z qn+k(Z )qn-k(z )

If k is fixed and n ~oo. each of the partial products tends to q(z\ and on cancellation we find

as asserted.



Let us now consider some consequences of Theorem 8.2b. Multiplying (8.2-8) by q(z 2) yields n=l

n=-oo

Let O=s;;x < 1, and set z := x 312 , t := -x 112 • On the left there results

n (1- x3n-1)(1- x3n-2)(1- x3n). CX)

n=l

All positive integral powers of x occur here; hence this equals q(x). We thus have proved q(z) =

co

L (-1rz 0, let ek and ok denote the number of its partitions into an even and into an odd number of distinct positive integers. If k is not of the form !(3n 2+ n ), then ek = ok. If k = !(3n 2+ n) for some integer n, then ek - ok = (-1)". EXAMPLE

3

The partitions of k = 7 into unequal terms are 7=7

=1+6

=1+2+4

=2+5 =3+4

There are three partitions into an even number of terms and one less partition into an odd number of terms, which must be so because 7 = ~(3 · 22 + 2). We next establish a recurrence relation for the partition function ck. Because (co+clz +c 2z 2+ · · · )(b 0 +b1z +b2z 2+ · · ·) = 1, we find in view of b0 = 1 n

Cn

=- L k=l

Cn-kbk>

n= 1,2, ... ,

SOME INFINITE PRODUCfS RELEVANT TO NUMBER THEORY

15

where c0 = 1, and thus, using the values of bk found above and writing ck =: c(k) for clarity, c(n)=-

~

(-l)m{c(n-3m:-m)+c(n 3m:+m)},

!C3m 2 +m) 1, we define

1

f (z) := IIp 1-p

(8.2-11)

z.

The product here comprises the prime numbers only; p -z := e -z Logp. To prove the convergence of the product, it suffices (by our definition) to establish the convergence of the reciprocal product

II (1-p-z). p

+ iy, we shall show that this product converges, in fact, uniformly for x ~ x 0 , where x 0 > 1. This follows from Theorem 8.1c, because

If z = x

00

~~P-zl=~p-x< p

p

I

n=l

00

n-x:s;;

I

n-xo.

n=l

We now shall derive an alternate representation for f. Taking the partial product up to the greatest prime :s;;k, we get 1 ( 1 +2 -z +2 -2z + ... ) II -----=z= pO, Zn -+00, is it possible to find polynomials Pn(Z) such that the product (8.3-5) converges uniformly in every compact set of the plane? The product converges uniformly if and only if this is the case for the series l: rn(z), where rn(Z) :=Log{ ( 1- z:) eP"} =log( 1- ~) +pn(Z),

(8.3-6)

where the branch of log is chosen such that the imaginary part of rn (z) lies between -7r and +1r. We wish to choose the Pn such that the series (8.3-6) converges uniformly on every set lz I:;;;;. p. For lz I< lzn I we have

We choose for Pn the negative of a partial sum of this series, Pn(z)

z 1 ( z ) + ''' +1-( z )m" , :=-+-2

Zn

2 Zn

mn Zn

INFINITE PRODUCI'S

20

whose degree mn is yet to be determined. The function r!(z) :=Log( 1- z:) +pn(z) [equal to rn(z) up to a multiple of 21Ti] then has the representation r!(z)= _ _l_(_:_)m"+l{l+ mn+l _:__+ ... }. mn + 1 Zn mn + 2 Zn Replacing { } by a geometric series, we easily obtain the following estimate, valid for lzlO.

Proof. Let d := Log g- Log r. Then, by (B), d(x + 1) = Log[xg(x)]- Log[xr(x)] = d(x),

x > 0.

Thus dis periodic with period 1, and so are d'(x) and d"(x). From (8.4-17) we have (integral test) for x > 1

1

00

L (

(Log r(x))"=

n=O X

1

+n

)2 0. For 8 > 0 and TT/2 < cp < TT, now let Aa. denote the improper curve sketched in Fig. 8.4c consisting of pieces of the two rays arg s = ±cp extending to infinity, and of the circular arcs= &;T' -cp::::;; -r::::;; cp. Now let z be an arbitrary complex number, and for iarg sl < TT, lets -z be defined by its principal value. The integral I(z) :=

J

e•s-z ds

As,,.

then clearly exists for every choice of 8 >0 and cp E (TT/2, TT). In fact, the integral converges uniformly with respect to z in every bounded set of the z plane and thus, because the integrand depends analytically on z, is itself an entire analytic function of z. By Cauchy's theorem tlJ.e integral clearly does not depend on 8, and by estimating the integrand on large circles we see that the integral is also independent of cp in the permitted range of cp. To evaluate the integral, we suppose that Re z < 1. Then if 8 -+ 0, the integral along the circular part of Aa. tends to zero. Substituting t := -s we see that the

INFINITE PRODUCfS

34 lm.s

Res

Fig. 8.4c.

integral along the lower leg of Ao,q, by (8.4-22) equals

and the integral along the upper leg -1-e'"'roe -r (e;"t)-z dt = -e -;,.zr(l- z ).

We thus find I(z) = (e;1TZ- e-;,.z)f(l- z) = 2i sin 'TTZ r(l- z ).

By (8.4-12) there follows 27Ti I(z) = r(z)'

Thus we have obtained the following result due to Hankel:

35

PROBLEMS THEOREM 8.4b

If A 8.4> denotes the path depicted in Fig. 8.4c, and s -z has its principal value,

f

1

s -z d

27Ti A••.,

es

1 s = r(z)

(III')

for all complex z, independently of the choice of 8 > 0 and cf> E (7r/2, 1r). One may even choose cf> = 7T, provided that arg s = 7T on the upper edge of the cut and arg s = -7r on the lower edge. If z is an integer, the integrals along the upper and lower edges cancel each other, and we have for

n =0, ±1, ±2, ... -1- =1-

I

2m IsH

r(n)

s -nds es

·

This makes evident that 1 /r(n) = 0 for n = 0, -1, -2, ... and, by virtue of the residue theo.rem, that r(n) = (n -1)! for n = 1, 2, ... . PROBLEMS

1. Deduce the basic properties of r(z) from the Gauss definition (II). 2. Establish the Legendre duplication formula by means of (1). ~3. Show that for real y

lf 0, f(z + t) as a function of t satisfies the hypotheses of the Plana summation formula (Theorem 4.9c). Summing the series on the right of (8.5-1) by the summation formula we thus get

r< ), =t/(z)+ Jor f(z+~)d~+i Jor [f(z+i71)-f(z-i71)]e2'"'_ 1 (f(;)) 1 d71 00

00

or (8.5-3) We assert that the integral may be integrated under the integral sign with respect to the parameter z, that is, that for Re z > 0

1

00 r'(z) 1 2., 1 +Log z- 2 - - 2 2...., d.,, r( z ) =a -2z o 71 +z e - 1

(8.5-4)

where a is a constant of integration. All we have to show to justify this step is that the derivative of (8.5-4) once again yields (8.5-3). Evidently, the improper integral on the right of (8.5-4) converges uniformly with respect to z in every compact subset of Re z > 0. According to Theorem 4.1a, it thus represents an analytic function whose derivative can be calculated by differentiating under the integral sign. This process, in fact, yields (8.5-3). Our goal is an integral representation for log f(z ), and to this end we wish to integrate (8.5-4) once more. To avoid the multivalued function arctan under the integral, we transform the integral with respect to 71 by integration by parts. In view of

J

1 _ e2....,_ 1 d71-

J1 _e e- 2'"'

2....,

_ 1 ( -2....,) d71- 21TLog 1-e ,

this process, first applied to the proper integral between the limits 8 > 0 and

39

STIRLING'S FORMULA p., yields

1,. 8

e

z.,

[

z., ],.,.

1 1 ( -211'11) 21n1 1 ~+ d71= -2 Log 1-e ~+ - 71 z '11' 71 z

+- l Log( ,.,.

1

'11' 8

-\

2

1-e

-211'11)

8

2

( z 2 -.,2)2 d71 71 +z

The integrated part vanishes as 8-+ 0 and p.-+ oo, and the last integral exists as an improper integral. Hence we have

r"

1 1 z 2 -71 2 -211'11 r(z) =a+Logz- 2z- '11'Jo (7~ 2 +z2)2Log(1-e )d71. (8.5-5)

r'(z)

Now we can integrate and obtain, first operating formally, log r(z) = az + b + (z -

!) Log z -

z + J(z ),

(8.5-6)

where b is another constant of integration and J denotes the Binet function, 2 J(z) := -1 '11'

1 00

z ~Log

o 71

+z

1 1 -e 2 11'11 d71.

(8.5-7)

The justification is again by noting that the integral (8.5- 7) converges uniformly for Re z ;;;. 8 > 0 and hence may be differentiated under the integral sign. Thus the expression on the right of (8.5-6) indeed represents an analytic branch of log r(z ). For a proper choice of b we obtain the branch that is real for z > 0. Before determining the constants a and b, we collect some facts concerning the Binet function. THEOREM 8.Sa

TheBinetfunctionJ(z) (defined by (8.5-7)for Re z >0) can be continued to a function that is analytic in the cut plane larg z I< 'TJ', For every () such that 0 :s;; () < '11' there exists K ( ()) such that IJ(z >I :s;; K~~~)

(8.5-8)

for all z in the sector SB : larg z I:s;; 6. As a corollary, if z e S8 , lim J(z) = 0.

(8.5-9)

z-+OO

2 lt seems desirable to name this function, although no name appears to be generally used. J.P. M. Binet, a contemporary of Cauchy, obtained various representations of J.

40

INFINITE PRODUCfS

If()::$; Tr/4, (8.5-8) holds with K(fJ) = 1/12; in particular, if z = x > 0, 1 O 0, it follows that J(z) is continued analytically by J(z) :=log f(z)-{az +b +(z -t) Log z- z}. To obtain an explicit representation of the continued function similar to (8.5-7), we use the familiar device of rotating the path of integration. Let a be any number such that-( Tr/2) 0. lms

Res

Fig. 8.Sa.

For any fixed z > 0, the function

z s +z

s ~ """""2:'"""2 Log

1 2.... 1-e

is analytic for Re s > 0; hence its integral along Aa,a,,. is zero. Simple estimates show that the integrals along the circular arcs of radius 8 and p. vanish as 8 ~ 0 and p. ~co. Hence it follows that for z > 0, 1

J(z) =7r

ielaao o

Z

---z:---! Log

s +z

1

1 -e 2.... ds,

41

STIRLING'S FORMULA

where the integral is along the ray arg s =a. Parametrizing the ray by s = wu, O:s;;u 0, and hence represents a function that is analytic for Re wz > 0, that is, for larg z- a I< 71'/2. For z >0 this function agrees with J(z). It thus equals the analytic continuation of J(z) into the set larg z- a I< 71'/2. To establish (8.5-8), let w = eia in (8.5-11) be chosen such that a= t8. If O:s;; arg z:::;; 8, then

hence if u;;.;. 0, lu 2+ w2z 21;a. sin( max(~.

9) )1zl 2.

Straightforward estimation of the integral now yields (8.5-8) where 1

K(8)= 7T

fo

00

7T

1Log 1 _e\,.wuldu 0, and substituting (8.5-6) into the relation Log f(x + 1) = Log x +Log f(x) derived from f(x + 1) = x f(x) yields (x + 1)a +b +(x +4) Log(x + 1)-x -1 +J(x+ 1) =xa +b +(x +f) Logx -x +J(x),

which reduces to a= 1-(x +!) Log(1 +~) +J(x)-J(x + 1).

Letting x-+ oo, the expression on the right tends to zero by virtue of (8.5-10), and we conclude that a = 0. We next app~ (8.5-6) to the relation Logf(x)+Logf(x+4)= Log f(2x) +Log .J?T- (2x -1) Log 2, which is the logarithmic form of the duplication formula (8.4-18). Using a = 0 there results, after simplification, b =!Log 2+ Log.J;.+e(x),

where e (x)-+ 0 for x-+ oo by virtue of (8.5-9) and some calculus properties of the logarithm. We conclude that b =Log fu. We thus have proved: THEOREM 8.5b (Stir6ng's formula for r(z)).

For all z such that iarg z I< 1T, the analytic branch of log f(z) that is real for z>O is

log f(z) = Logfu+(z -4) Log z -z +J(z),

(8.5-12)

where J(z) denotes the Binet function defined by (8.5-7) and enjoying the properties described in Theorem 8.5a. Equivalently,

r::r( z ) =vL?TZ 2 z-1/2 e -z e J(z)

where the power of z has its principal value.

(IV)

43

STIRLING'S FORMULA

In (IV), e1 should be regarded as a correction factor by which the elementary function J2;.zz- 112 e -z, called Stirling's approximation to r(z ), should be multiplied to furnish the exact value of r(z). We know from (8.5-9) that the correction factor tends to 1 as z ~ oo in any sector larg zl :os;;; 8 where (J < 7T. Thus the relative error committed by omitting the correction factor tends to zero as z ~ oo. In fact, if larg z I:os;;; 7r/4, the precise estimate

,:::: 1 e 1/(12lzl> """"121z1 is available for the rel.ative error of Stirling's approximation. This makes it possible to compute f(z) to arbitrary accuracy for any complex z not 0 or a negative integer by using the formula r(z) = r(z +n) (z)n

[following from (8.4-9)] and choosing n such that (i) larg (z + n >I :os;;; 7r/ 4 and (ii) Stirling's approximation to r(z + n) yields the relative accuracy desired. Figure 8.5b shows a graph of lf(z >I for complex z which was calculated by the foregoing method. More precise asymptotic information on Binet's function J(z) will be obtained in §11.1, §11.11, and §12.12. Several computational applications of Stirling's formula are discussed in subsequent paragraphs. Here we use it to give a third proof of Euler's representation (III). Once more, let g be defined by g(z) :=

i

0

CXl

e -rz-1 d-r, -T

(8.5-13)

where -rz- 1 has its principal value. We have seen in example 3 of §4.1 that g is analytic in Re z > 0, and an integration by parts establishes the functional relation g(z + 1) = zg(z). As was shown in example 15 of §3.2, this relation enables us to continue g as a meromorphic function into the whole plane, with simple poles at z = 0, -1, -2, .... Let now r be defined by (I). From the above it follows, as before, that p(z) := g(z)/f(z) is an entire function, and that it is periodic with period 1. We wish to show that p is constant. Using Liouville's theorem and the periodicity, it suffices to show that p is bounded in a period strip, for instance in the strip 1 :os;;; Re z :os;;; 2. By (8.5-13), if z =x+iy,

INFINITE PRODUCI'S

44

tlr(zJ I

30

20

10

-5

Fig. 8.Sb.

thus g(z) is bounded in the strip. The required lower bound for jf(z )j is furnished by Stirling's formula. Taking real parts in (8.5-12), we find Log lrI= Log .J2;- x + (x- !) Log lz 1- y Arg z + Re J(z ). For 1::::; x::::; 2, only the term - y Arg z can become negatively infinite. In view of IArg zl < n/2, -y Arg z > -1TIYI/2. Thus IPI is bounded by p. exp(1TIYI/2), where p. is a constant. For an arbitrary entire function this would not suffice to conclude that it is constant. Here, however, we can use the periodicity of p to define, for w ;CO,f(w) := p((l/21Ti) log w). Becausep is periodic with period 1, it does not matter which branch of the logarithm is chosen, and because log w can be defined as an analytic function in the neighborhood of any w 0 ,C 0, it follows that f as a composition of analytic functions is analytic, with the sole possible exception of w = 0. Thus f may be

45

STIRLING'S FORMULA

expanded in a Laurent series, 00

L

f(w)=

(8.5-14)

n=-oo

where an=2 1 . 'TTl

r

w-n-lf(w)dw,

Jlwl=p

(8.5-15)

p being arbitrary. To obtain an estimate for ano we estimate f on the circle lwl =p. The values off on that circle are given by the values of p for z=(l/27ri)Iogw where lwl=p; that is, on the line Imz=(1/27r)Logp. The estimate obtained from Stirling's formula shows that for p -+ oo

and for p -+0

Using these estimates in (8.4-23) we immediately find that an = 0 for n :¢: 0, and hence that f, and thus p, is constant. PROBLEMS 1. Use Stirling's formula to compute f(0.8) to six decimal places. 2. If a and b are any two complex numbers, show that

f(z +a) . . f(z +b)- za-b (prmctpal value) as z -+ oo uniformly in every sector S8 where 8 < 1r. 3. Under what conditions on a does the binomial series

r (a)

n-o

n

n

X

converge (a) for x = 1, (b) for x = -1? 4. Show that the series

converges to lxl uniformly in the interval -1:s;; x :s;; 1. (This result is used in Lebesgue's proof of the Weierstrass approximation theorem.)

46

INFINITE PRODUCfS

5. Prove that if a is not an integer, and if Rea< 1,.then

i: q + 1. If p = q + 1, the ratio tests shows that the radius of convergence of the series is 1. To study the convergence for lzl = 1 if p = q + 1, we note that the nth term of the series equals n r(a1 +n) ... f(aq+1 +n) r(b1+n) · · · r(bq+n)r(1+n)z' up to a factor independent of n. By Stirling's formula (see, in particular, problem 2 of §8.5) the quotient of r functions is asymptotic to

48

INFINITE PRODUCfS

Thus for the absolute convergence of the series if lzl = 1 it is necessary and sufficient that (8.6-5)

We now evaluate several instances of the series where p = q + 1 and z = 1. If p = 1 and q = 0, then by the binomial theorem, if lzl < 1, 1F 0

~ -(a)n z n = t.. ~ (-a)( -z )n = ( 1 -z )-a [a; z ] = t.. 1 n=O

n.

n=O

n

(principal value), and hence if Rea < 0, using Abel's theorem (Theorem 2.2e) 1Fo[a; 1]= lim (1-z)-a =0. z-tol-

We next evaluate the hypergeometric series of Gauss,

(

)

F a, b; c; z := zF1

[a, b;c z]

for z = 1. The result is THEOREM 8.6b (Gauss formula)

For Re(a +b--c) 0). Hence its sum is analytic in D. On the subset of D described by (8.6-10) this function agrees with the expression on the right of (8.6-9), which is likewise analytic in D. Hence 4 the agreement persists throughout D. • Finally we shall prove Dougall's formula: TIIEOREM 8.6e

If a and b are not integers, and if 1 + Re(a +b)< Re(c +d), then

I

f(a+n)f(b+n) n=-oof(c+n)f(d+n)

Tr 2 f(c+d-a-b-1) sinTra sinTrb f(c-a)f(d-a)f(c-b)f(d-b)" (8.6-11)

Proof. Stirling's formula in combination with (8.4-12) shows that the doubly infinite series on the left converges properly (and not merely as a principal value). Because Tr cot TrZ is a summatory function with residues + 1, the sum of the series equals the sum of the residues of f(a+z)f(b+z) f(z) := Tr cot TrZ f(c +z)f(d+z)

at the poles z

= 0, ±1, ±2, ... of 1r·cot Trz.

For lzllarge, Re z ~o.

f(a + z)f(b + z) a+b-c-d -z f(c+z)f(d+z)

if Re z ~o. f(z +a)f(z +b) f(c + z)f(d + z)

)"+b-e-d sin

(z

Tr(z +c) sin Tr(z +d) sin Tr(z +a) sin Tr(z +b)"

Let ~ be a real number not congruent mod 1 to any of the poles of the function f. We integrate f along the square with vertical sides Re z = ~ ± n and horizontal sides Im z = ±n, where n is an integer destined to tend to oo. 4

Here we use the basic facts on analytic continuation of functions of several complex variables, which are analogous to the corresponding facts about functions of one variable given in Chapter 3. See also Chapter 17.

53

SOME SPECIAL SERIES AND PRODUCfS

On this square the absolute value of 1r cot

sin 1r(z +c) sin 1r(z +d) ( ) • ( b) SID 1T Z + a SID 1T Z +

1TZ •

is bounded, and the bound is independent of n. The integral

taken along the square tends to zero because of the condition Re(a + b- cd)< -1. It follows that the total sum of the residues off is zero, and hence that the value of the sum (8.6-11) equals the negative of the sum of the residues off arising from the poles of r(z +a) and r(z +b). The residue off at the pole z = -a-m (m = 0, 1, 2, ... ) is - 1r

cot 1ra

(-l)m --

m!

f(b -a-m) • f(c-a-m)r(d-a-m)

By (8.4-10) we have, for instance, f(b-a) f(b-a-m)= (b-a-m)m ( -1)mr(b- a) (a-b+l)m and consequently the sum of all residues by Gauss' theorem (applicable here because the conditions of convergence are satisfied) equals f(b-a) [a-c+1,a-d+l; -1r cot 1raf(c- a)f(d- a) 2 F 1 a -b + 1

1]

f(a-b+ l)r(c+d-a-b-1) = -1r cot ?Ta _ _f(b-a) .:......__::....__ r(c -a)f(d -a)

2

cot 1ra

r(c -b)f(d -b)

f(c +d-a-b -1)

= 1T sin 1r(a -b) f(c -a)f(d -a)f(c -b)r(d -b)" The residues arising from the poles at z = -b - m can be obtained from the foregoing formula by interchanging a and b. The quotient of r functions is unchanged, and in view of cot 1ra cot 1rb sin 1r(a-b) +sin 1r(b-a)

1

sin 1ra sin 1rb

S4

INFINITE PRODUCI'S

the negative of the sum of residues is just f(c +d-a-b -1) sin 'Tf'a sin 'Tf'h f(c -a)f(c -b)f(d -a)f(d -b)' 71' 2

as stated. • PROBLEMS 1.

Express the following series in closed form, subject to convergence conditions:

n~O

(b)

)"(:r

1

n~O

(d)

2.

(-

>"(:r

(- 1

Using the identity (1.6-11),

( 1 _ z )-a2Fl

ra, b; - 1 ~ z] _ F [a, c- b; z] -2

c

I

c

'

prove the formulas (a)

(b)

3.

2

F1

a,b;!] r(!}f((l+a+b)/2) [a+~+1 =r((1+a)/2)f((1+b)/2)'

F1

l

fa, 1- a;!]

2

c

f(c/2)f((c + 1)/2) =r((c+a)/2)f((1+c-a)/2)

Prove that

oo f(a+n) J~oo f(b +n) 4.

(a) as a special case of Dougall's formula; (b) directly by the calculus of residues. Prove: 4 (15. 15) ( ~) 1 . 5 13 17 °

5.

0 (Rea 1:

§8.7. THE BETA FUNCfiON The integral

L I

B(p, q) :=

Tp- 1(1-T)q-i dT,

(8.7-1)

where Rep> 0, Re q > 0, and where Tp-i and (1- T)q-i have their principal values, is known as the beta function of Euler and occurs frequently in applications. The beta integral can easily be evaluated in terms of r functions. We assume temporarily that Rep> 1, Re q > 1. The binomial series ~ (1-q)n n (1 - T )q-1 = t... T n=O n! then converges uniformly for 0 :s;; T :s;; 1. Substituting into the integral and interchanging summation and integration, we find

B(p,q)= ~ (1-q)n ( 1 7 p+n-i dT= ~ (1-q)n_1_ n=O n1 Jo n=O n! p+n

=.!.

I

Pn=O

(1-q)n (p)n =.!. 2p 1[1-q,p; 1] n! (p+1)n p p+1

1 r(p + 1)f(q > r(p )f(q) = {; r(p +q)r(l) r(p +q)' by Gauss' theorem 8.6b. Like B(p, q), the function on the right is, in each of the variables p and q, analytic in the half plane of positive real parts. We thus have

56

INFINITE PRODUCfS

tHEOREM 8.7a.

For Rep>O, Req>O, B(p, q):=

r

~- 1 (1-r)q-t dr = ~~~~;.

(8.7-2)

Some other forms of the beta function can be obtained by simple substitutions.

r. .

B(p,q)=2 Jo

B(p,q)=

/2

(sint/>) 2p- 1(cost/>) 2q-I d4> (r=sin24>)

f"' e- (1+e)-p-q 1

( e-l

de(r=

+eq-l ( B(p, q) = Jo (1 +~t+q de ~

1!e)

(8.7-3)

(8.7-4)

1)

-+"i .

(8.7-5)

As an application of the beta integral we now derive an integral representation for the hypergeometric function. If c ,c 0, -1, -2, ... , and if lzl < 1, 2Ft

~ (a)n(b)n n (a, b ; c; z ) = .t... ( ) 1 Z

n=O

C

nn.

r(c)

~ f(a+n)f(b+n)

f(a)f(b)n=O

r(c+n)n!

n

z ·

If Re c >Reb >0, then

f(b+n)f(c-b)=B(b+n c-b) r(c +n) '

and thus 2F 1

If

(

f(c) ~ f(a+n) nf 1 b+n-1( )c-b-ld ) a,b;c;z =f(a)f(b)f(c-b)n~o n! z JoT 1-r T.

lzl < 1, the series ~ f(a+n) n n .t... --'---'-z T

n=O

n!

converges uniformly to f(a)(1- zr)-a on the range of integration. We thus may interchange integration and summation to obtain 2

f(c) F 1 (a, b ; c; z ) = f(b )f(c _b)

Jor T b-1( 1- T )c-b-1( 1- rz )-adT, 1

57

THE BETA FUNCDON

ra

where (1- TZ has its principal value. The result SO far has been proved for lz I< 1. But the integral on the right has a meaning for any z in the complex plane cut along the real axis from+ 1 to +oo, and it converges uniformly with respect to z in any compact subset of the plane thus cut. It therefore represents an analytic function in the cut plane. Because that function coincides with 2F 1(a, b; c; z) for lzl < 1, it represents the analytic continuation of that function into the cut plane. We call that continuation the hypergeometric function (as opposed to the hypergeometric series, which is defined only for lzl < 1) and denote it by the symbol F(a, b; c; z). We then have THEOREM 8.7b.

For Re c > Re b > 0, the hypergeometric function possesses the representation

r

1 b-1(1 r(c) )-ad )c-b-1(1 ( b ; c; z ) = r(b)r(c Fa, -b) Jo T -11 -Tz T,

(8.7-6) valid for all z that are not real and ;;::.1.

It is a drawback of both representations given in the foregoing theorems that they are valid only under severe restrictions on the parameters. For the beta integral this drawback can be removed by a device due to Pochhammer (1890). In the complex t plane we consider a closed (but not simple) path A that originates, say, at t = t and encircles each of the points t = 1 and t = 0 twice, first in the positive, then in the negative sense (see Fig. 8.7a). 5 It is customary to denote such a path symbolically by ( 1+, 0+, 1-, 0-). The winding number (see §4.6) of A with respect to both t = 0 and t = 1 is zero. It follows that a continuous argument exists on A, and hence that log t can be defined as a locally analytic function on A, for instance by assignin~ it the principal value for t > 1. With this determination of log t, tp- 1 := e - 1)logr and.(1-t)q- 1 =e can likewise be defined as locally analytic functions. Thus the integral

I(p, q) :=

t

tp- 1(1- t)q- 1 dt

has a meaning for arbitrary (complex) values of p and q, and by Cauchy's theorem is independent of A as long as A stays clear of the points 0 and 1. To evaluate I(p, q), we temporarily assume p > 1, q > 1 and contract A to a path 5

Such a path is given, for instance, by t = {cosh[ sin( q,

+~) +it/1]} o~q, ~21r. 2

,

58

INFINITE PRODUCTS lm t

Ret

Fig. 8.7a.

consisting of four straight-line segments joining 0 and 1 and two pairs of small circles, one pair surrounding 0, the other surrounding 1. Schematically this path is indicated in Fig. 8.7b. lm t

Ret

Fig. 8.7b.

As the radii of the circles tend to zero, the contributions of the circles vanish, and the integral reduces to the sum of the four integrals along the straight line segments. Each ofthese integrals equals B(p, q) up to a factor of absolute value 1 because of the direction of integration and the momentary value of arg {tp-i (1- t)q -I}. Following the changes of the argument, we find I(p, q) = {1- e2'1Tiq + e2'1Ti(p+ql -e2'1Tip}B(p, q)

= (1- ez'IT;p)(l- e2'1Tiq)B(p, q).

(8.7-7)

59

THE BETA FUNCDON

If neither p nor q is an integer, we can solve for B(p, q ). Because I clearly is an entire analytic function of p and of q, we obtain Pocbhammer's integral, THEOREM 8.7c.

For all complex p and q that are not integers,

I

hr(p+q)

B(p,q)=-

.e

.

4 SID 1Tp SID 1Tq

tp- 1(1-t)q- 1 dt.

(8.7-8)

(1+,0+,1-,0-)

Pochhammer's integral may be cast into different forms by transformations of the variable. If, for instance, we set

s 1+s'

1 dt = (1 +s) 2 ds,

t=--

we get l(p, q) =

J

sp- 1(1 +s)-p-q ds,

As

where the closed curve As winds once positively and once negatively around each of the points s = 0 and s = -1. Setting further ds = 2i e 2 iu du, we obtain

I(p, q) = 2 1-p-q

f

ei(p-q)u(cos u)-p-q du,

Au

where Au is a figure-eight-shaped loop having winding number +1 with respect to 11'/2 and -1 with respect to -(11'/2), see Fig. 8.7c. If p +q < 1, the lm u

Fig. 8.7c.

INFINITE PRODUCI'S

60

lm u

Re u _!!. 2

2"

Fig. 8.7d.

loop may be contracted to a curve consisting of two straight-line segments joining -(71/2) and 7r/2, and two small circles around 7r/2 and -(7r/2), as shown in Fig. 8.7d. The contributions of the circles tend to zero with the radii, and the integrals along the line segments together yield

J

Tr/2

2l-p-qi(l-e- 2 "'i(p+q))

ei(p-q)u(cos u)-p-q du.

-(Tr/2)

The integral of the imaginary part is zero. Expressing I in terms of B(p, q), we thus find f"'/2

Jo

cos[(p-q)u](cos u)

p+q

7T r(l+p+q) du- 2p+q+l r(l +p)r(l +q)"

(8.7-9

)

Loop integrals of the Pochhammer type, valid for less-restricted values of the parameters than (8.7-6), can also be obtained for the hypergeometric function; see Problem 6. Other representations of the hypergeometric function are obtained in §8.8.

PROBLEMS

1.

Compute the volume of then-dimensional unit ball,

61

PROBLEMS introducing n-dimensional polar coordinates (p, l/J, 8 17 ••• , (Jn_ 2) where

Xn = P COS (Jn -2• Xn-1 = p sin (Jn-2 COS (Jn-3• Xn-2 = p sin Bn-2 sin (Jn-3 cos (Jn-4•

x3 = p sin (Jn- 2 •

• •

sin 82 cos 8 17

x2 = p sin (Jn-2 .. sin (Jl cos l/J, x 1 = p sin Bn- 2

• • •

sin 8 1 sin 1 the expression on the right tends to zero if k-+ co, and we thus have proved: THEOREM 8.8a

If none of a, b, cis zero or a negative integer, and if a - b is not an integer, then for larg(- z >I < 11' F(a,b;c;z)=(-z)

1)

-af(b-a)f(c) ( f(c-a)f(b)F a, 1+a-c; 1+a-b;;

(8.8-8)

+(-z)

-bf 1, but holds without that restriction by virtue of analytic continuation. The analytic continuation of any series of type pl'p-1 beyond lzl < 1 can be obtained in exactly the same manner. Because there now are p descending chains of poles, p series of the form pFp- 1[ ••• ; 1/z] will appear in the continuation formula. We next apply the method of Mellin-Barnes integrals to the confluent hypergeometric function 1F 1 (a; c; z ). All computations could be carried out in an analogous fashion for the function pl'p(at. ... , ap; Ct. ... , cp; z) for any integer p. By analogy it is easy to conclude that we now should integrate

f(s) := f(a +s)f(-s)(-zY f(c +s) along suitable closed curves. This function has one ascending and one descending chain of poles-at the points -a-n and n (n = 0, 1, 2, ... ), respectively. We assume that neither a nor care negative integers or zero, and begin by integrating along the rectangle Am introduced previously, with the usual convention that the two chains should be separated. To determine

69

INTEGRALS OF THE MELLIN-BARNES TYPE

the behavior of f(s) for Is I large, Re s ;;;:,: 0, we again use the formula f(a+s)l-lla -{3args l f(c+s) s e ' where, now, a - c =: a + i{:J, and

71'

1

lf(-s)l = lsllsin 7rsllf(s)l

We thus find

lf(s)l- J'#,lsla-Res-l/21z1Res • e[-arg(-z)-.,.+args] Ims+Res.

(8.8-9)

On the right side of Am, lsrRes ~ m -m. All other factors tend to oo at most like econst·m, and it thus is clear that the integral tends to zero form-+ oo. On the horizontal sides of Am the integrand is bounded by

where u := Res, and the integral with respect to u between the limits 0 and m +hends to zero form -+OO only if -arg(-z)-7r+arg s = -€, where € >0. This can be guaranteed only if larg(- z >I~ 71'/2- €, and it thus becomes necessary to confine z to a more restricted region than in the case of the ordinary hypergeometric function. If the foregoing conditions are met, then all integrals tend to zero except the integral along -Aco>, in the above notation. The integral equals 27T'i ~ (resns=n· Now, evidently, f(a +n) (-1t n (resf)s=n = -r( ) - 1-(-z).

c+n

n.

Summing all these yields 1F 1(a; b; z) times a r factor, and we have obtained: f(a) 1 J;oo f(a +s)f(-s) s tFt(a;c;z)=f( ) (-z) ds, . . r() 2 c 7T'l -•oo c +s

(8.8-10)

with the usual separating convention about the path of integration. We now stipulate, as before, that the path of integration is chosen such that by moving it to the left by one unit it crosses the first pole of the descending chain. Let A 0 may be arbitrarily small. The solution v satisfies an identity similar to (9 .1-7), and on subtracting the two identities, we find v(z) -w(z) =

r

[f(t, v(t)) -f(t, w(t))] dt.

Applying (9.1-5), this implies llv(z)-w(z)II:S:K

(9.1-8)

r

llv(t)-w(t)llldtl.

z

8(T) := sup llv(z)-w(z)ll· lz-ijo;;T

Then 8(T);:a=:O, and by the above 8(7') :S:

\~~~T

r

llv(t) -w(t)llldtl :S: KT8(T),

which is contradictory for any T such that KT < 1 and 8(T) ¥= 0. It follows that 8 (T) = 0 for 0 :s: T < K -I, which is sufficient to establish uniqueness. •

THE EXISTENCE THEOREM

81

EXAMPLES

We show by means of examples that the assertion of the theorem concerning y, the radius of the disk of analyticity of the solution, cannot be improved unless further assumptions on f are made. For a..:; PIp., this is illustrated by the scalar case (n = 1) where f is independent of w and has a singularity on the circle lz - I= a. Any solution

z

w(z) =

w+

r

/(t) dt

z

will then likewise have a singularity at the same point, or else /(t) = w'(t) would have to be analytic. For a >PIp. consider the scalar problem

w)]

1 ( 1+ji w'=f(w):= p. [ 2

1

1m

,

w(O)=O,

where p >0, p. >0, and m is a positive integer> 1. Here a= oo. Clearly,fis analytic for lwl

n 0 and n > n 0 • The number s, if it exists, is uniquely determined, and is called the value or the sum of the double series. To assert that the double series (9.2-1) is convergent is not the same as to assert that the iterated series co

co

L L

(9.2-2)

amn

m=On=O

is convergent. Consider, for example, the double series with the partial sums (-1)"

Smn

:=

m+ 1.

It is convergent with the value 0. However, the series co

I aon n=O has the nth partial sum so,n = (-1)" and is therefore divergent. Thus the iterated series (9.2-2) cannot even be formed in this case. (c) Cauchy criterion. The series (9.2-1) is convergent if and only if for every E > 0 there exists n0 = n 0 ( E) such that

whenever m~o n~o m2, n2>no. (d) Series with positive terms. Let am";;::.: 0, m, n = 0, 1, .... The series (9.2-1) is convergent if and only if there exists a constant u such that SmnII-+ o

(9.4-12)

for lz 1-+ 0, where p. := K + IIRII. Because P has an isolated singularity at z = 0, Theorem 4.4e now shows that this singularity is at most a pole. • Theorem 9.4c implies that in the representation (9.4-8) the matrix P, although not uniquely determined, either always has at most a pole or always has an essential singularity for a given differential equation. The question may be asked whether it is possible to determine the character of a singular point directly from the differential equation, without first constructing a fundamental solution. There indeed exists a simple sufficient condition for a singular point to be regular. DIEOREM 9.4d

Let the matrix A(z) have a pole of order 1 at z = 0. Then z = 0 is a regular singular point of (LHo).

Let w be a nonzero, log-holomorphic vector solution of (LH0 ), and let h(p, 8) = w(p e;6 ). Then, using the differential equation,

Proof.

ah (p, 8 ) = ddw (p eiB) eiB = A(p eiB)w(p e;B) e;B

ap

z

and hence

Let Oll-+ 0

asp-+ 0, 181 os;;, 8 1 • Because w is an arbitrary solution, the differential equation thus satisfies the criterion of Theorem 9.4c, and it follows that the singular point z = 0 is regular. • The converse of Theorem 9 .4d is not true for n ;;::. 2, as can be seen from Problem 3. PROBLEMS 1. The system w' = A(z )w, where . -1

Q_ 3

A(z) :=

-IZ

1

0

0

-(z +z- 3 )+-z- 1 4i 2i

0

3 3 --(z-z-)

0

0

0

0

-;z-1

-~(z -z-3 )

0

3 3 1 I --(z+z)+-z4i 2i

0

4

4

occurs in celestial mechanics in the theory of perturbed Keplerian motion. (a) Show that the system has the four linearly independent solutions

1 I 2; 1 Pi involves elements of q~, q2, ... , ~-I only. This circumstance enables us to determine the columns q~, ... , qn recursively as follows. For j = 1, (9.5-16) is [(m +At)I-R]qt =ct. In view of hypothesis 1e, m +A 1is not an eigenvalue of R, and thus the matrix in brackets is nonsingular. Thus the equation can be solved for q 1 • Assuming that q~, ... , 'li-t have been determined, Pi is known, and (9.5-16) may be written Again by virtue of hypothesis 1e this can be solved for q;. Thus the complete matrix Om can be determined under the assumption that 0 0 , ••• , Om -I have been determined, and because Oo has been found, the construction of a formal fundamental solution has thus been completed for differential equations satisfying hypothesis 1e. • It remains to deal with equations that do not satisfy hypothesis 1e, which include many equations of practical importance. (Among others, the differential equations satisfied by the Bessel and the Legendre functions are in this category for special values of the parameters.) This is accomplished by repeated applications of the following fact:

ORDINARY DIFFERENTIAL EQUATIONS

122

THEOREM 9.5d Let the distinct eigenvalues of R beAt. ... , Am (m ~ n ). There exists a matrix V(z ), analytic at z = 0 and nonsingular for z ~ 0, such that the function defined by w = Vw satisfies a differential equation of the form (9.5-1),

w

A' = ( z -IRA+ I... ~ z mAA)Aw, w m=O

where

R has the eigenvalues A1 -1, A2 , ••• , Am.

Proof. LetT be any nonsingular constant matrix. If w satisfies (9.5-1), the function := Tw then will satisfy

w

w' = Tw' = TA(z)w = TA(z )T- 1w. It thus satisfies a differential equation of the same form, where A is replaced by the similar matrix TAT- 1 . Now let the multiplicity of the eigenvalue A 1 of

R be p, and let T be so chosen that

R := TRT-1 = (RI 0

0)

Rz'

where R 1 is a p x p upper triangular matrix with the p-fold eigenyalue At. and where R 2 is a square matrix of order n- p with the eigenvalues A2 , ••• , Am. By the fact just mentioned, the function := Tw then satisfies

w

w'= (z- 1R+H(z))w, where H is analytic at z

= 0. Now let U :=

(z-0 I 1

0) I '

where the partitioning is similar to that of i. (The two unit matrices I appearing here do not necessarily have the same dimension.) Clearly, U is nonsingular for z ~ 0, and

is analytic at z

=

0. Now let

w:= Uw. For z ~ 0 this satisfies

w'=U'w+Uw = U'w+U(z- 1R+H(z))w = (U'U- 1+ z -lu:Ru- 1+ UH(z)U- 1)w.

SINGULARmES OF THE FIRST KIND: FORMAL SOLUTIONS

123

Now

Furthermore, if H(z) = ~:=oHmzm and

then UH( z )u-1 = z -1(0 0

"12) . . 0 +analytic matnx.

Thus w satisfies a differential equation of the form (9.5-1), where R is replaced by the matrix

i

:=

(Rt-1 H12) 0

Rz

which has the eigenvalues A1 -1, A2 , ••• , Am. Composing the foregoing transformations, the statement of the theorem is proved for v:= y-1u-1. • By repeated applications of the constructiop of Theorem 9.5d, a differential equation with a singularity of the first kind at z = 0 can be transformed into a differential equation satisfying hypothesis :le by a substitution of the form w = Vw, where V is nonsingular for z -:/= 0. For the transformed system a fundamental matrix W can be constructed by the algorithm given in Theorem 9.5c. A fundamental matrix of the original system is then given by W := vW. It has the form W(z) = P(z )zii where Pis a power series and i is a matrix whose eigenvalues coincide with those of R modulo one. More precisely, if the eigenvalues of Rare divided into sets of eigenvalues differing by an integer, the eigenvalues of Rconsist of the lowest representatives of each set. The multiplicity of each such eigenvalue equals the number of eigenvalues in the set to which it belongs. The construction of a fundamental system is simple enough for differential equations satisfying hypothesis :le, but it can be laborious for equations that do not satisfy this hypothesis, especially if two eigenvalues differ by a large integer. For systems that arise from a single nth order equation there exists a somewhat simpler method for constructing a fundamental system in the exceptional cases. This method, which is due to Frobenius, is discussed in

§9.6.

ORDINARY DIFFERENTIAL EQUATIONS

124 PROBLEMS

1.

Use the heuristic method sketched after the proof of Theorem 9 .Sa to construct a fundamental system for the equation w' = (z- 1ai+A)w,

2.

where a is a complex number and A a matrix of constants. Interpret the result. Transformation of variable at singular point of first kind. In the system

w' = (z- 1R+H(z))w, where H is analytic at z = 0, a new variable t is introduced by setting z = g(t), where g is analytic at t = t0 , g(t0 ) = 0, g'(t0 ) '¢ 0. Show that u(t) := w(g(t)) satisfies a linear system of differential equations with t0 as a singular point of the first kind, and that the residue of its coefficient matrix at t0 is again R. 3. Let W be a fundamental matrix near z = 0 of the system w' = (z -Ia + H(z ))w,

where H is analytic at z = 0, and let the matrix C (as in the proof of Theorem 9.4b) be defined by W(z e 2 ,.;) = W(z )C. If R satisfies hypothesis lie, show that the matrices C and V := e 2 ";a are similar. 4. (Continuation). By considering the example

H(z) :=

(~ ~)

show that C and V need not be similar if R does not satisfy hypothesis lie (Gantmacher).

§9.6. SCALAR EQUATIONS OF IDGHER ORDER: MEmOD OF FROBENIUS Here we are concerned with the scalar equation of the nth order, u 1. We differentiate the formal identity (9.6-10) with respect to the parameter A. In view of L' = 0, observation 9.6b yields L(U') = [1r'(A) + 1r(A) log z ]z".

If m 1 = 2, then 1r(A 1) = 1r'(A 1) = 0, and the foregoing relation shows that U' evaluated at A1 is another formal solution of L ( U) = 0, and hence another

ORDINARY DIFFERENTIAL EQUATIONS

130

actual solution of (9.6-1). Evidently, lF(A 1) = U1 log z + Uz,

(9.6-12)

where 00

Uz := zA'

L

c~(A 1 )zm.

m=O

The presence of a logarithmic term shows that lF(A 1) is linearly independent from U 1 • If m 1 > 2, it is clear that m 1 -1 differentiations of (9.6-10) would yield the required m 1 -1 additional linearly independent solutions. We next consider the situation in which the indicial polynomial1r has two simple zeros A1 and A2 such that A1 - A2 =: k is a positive integer. We assume that 1r(A 2 + j) ¥= 0 for integral j such that 0 < j < k and j > k. A solution U 1 belonging to the zero A1 can be found as indicated above. Again, the problem is to find a second, linearly independent solution. The method outlined above cannot be applied because 1r(A 2 + k) = 0. However, the method can be modified as follows. Instead of choosing c 0 := 1 as our initial coefficient, we now choose c0 :=A- A2 • The coefficients c., c 2, ... , ck-t then likewise will have the factor A- A2. Therefore in the equation for ck> 1r(A + k )ck = gk, not only the left side vanishes for A = A2 of order 1 but also the right. Thus ck is determined as a rational function of A which does not have Az as a pole. The same holds for the coefficients Cm where m > k. The series thus determined formally satisfies L(U)=11'(A)(A-A 2)zA

(9.6-13)

identically in A. If we now would set A = A2 , the first k terms in the series U would vanish, because they have the factor A- A2 • Thus we would obtain a series with leading term ZA 2 +k = zA', and because the series is uniquely determined by its leading coefficient, this series would merely be a multiple of the series U 1 found above. To get a linearly independent solution it is necessary to differentiate the identity (9.6-13) formally with respect to A before putting A = A2 • Doing so we obtain L(lF) = 1r(A)zA +(A- A2 )[1r'(A) +1r(A) log z ]zA.

Letting A = A2 now yields L(Uz)=O,

where Uz := lF(Az).

SCALAR EQUATIONS OF IHGHER ORDER

131

Tnus U 2 is a formal solution of L(U)=O, and hence an actual solution of (9.6-1). The leading term of u2 is z'\ hence u2 is linearly independent from the solution ul found before. The first k coefficients of u2 are what one would obtain by straightforward application of the method of undetermined coefficients; logarithmic terms enter from the term in z"• onward. If A1 has multiplicity m 1 and ,\ 2 has multiplicity m2 , it is readily seen that the same process carried out with c0 := (,\ - ,\ 2 )m' will yield a formal series U whose first m 1 -1 ,\-derivatives evaluated at ,\ 2 yield m1 formal solutions associated with A1o and whose next m 2 derivatives yield m 2 formal solutions associated with ,\ 2 • The case of several zeros differing by integers and having arbitrary multiplicities can be dealt with in a similar manner. Because the differential equations that are most studied are of the second order, the two special instances of the method of Frobenius given above suffice for most cases of interest. Some applications of the method are discussed in §9.7. PROBLEMS 1. Let z = 0 be a singular point of the first kind of equation (9.6-1). Show that for each characteristic exponent A there exists at most one solution u of the differential equation that can be represented in the form u(z) = zAh(z),

where h is analytic at z of such a solution. 2. The equation

= 0, h (O) = 1. State a sufficient condition for the existence

u+z- 1q 2 (z)umm!

L

nz

oo

(-z2)m

V=-- L ---'------'11'

m=o(2m+1)!

(9.7-26a)

12.

=v=smz, 11'Z

and

(9.7-26b)

=~cosz. 11'Z

'IWO EXAMPLES: THE EQUATIONS OF KUMMER AND BESSEL 141 Thus, the functions 1 112 and 1_ 112 are elementary in the sense that they can be represented as a composition of a finite number of rational or exponential functions or inverse functions thereof. We now show that, more generally, all functions 1. where"-! is an integer are elementary. This is a consequence of the relations, valid for arbitrary "•

(9.7-27)

which are easily proved by term-by-term differentiation of the series expansions. By iterating the relations (9.7-27), we have in an obvious symbolic notation

C~J" {z"J.(z)}= z"-"J._,(z), C~J" {z-•J.(z)}= (-1)"z-•-"J.+,.(z), n = 0, 1, 2, .... For"=±! this yields in view of (9.7-26) - ( d )"(sin:\

1,,2+,.(z)= (-z)"·hz/Tr

zdz

- ( zdz d 1_1/z-,.(z) = z".J2z/Tr

-z--J·

)"(cosz) -z- ,

(9.7-28)

n = 0, 1, 2, .... Because the derivatives of iin elementary function are elementary, these relations indeed establish the elementary character of the Bessel functions involved. It can be shown-but this is much more difficult-that the cases where "-! is an integer are the only ones where J. is an elementary function in the sense defined above.

PROBLEMS 1.

If 'Y is not an integer, show that a fundamental system for the equation

zu"+-yu'-ku = 0 is given by the functions u1(z) := oF1(-y; kz), Uz(z) := z'-yoF,(2--y; kz).

Use the method of Frobenius to obtain a fundamental system for 'Y = 1.

142

ORDINARY DIFFERENTIAL EQUATIONS

2,. Let a, b, c, d, e be arbitrary complex numbers, and let k be an integer. All

solutions of the differential equation 1 1 u"+-(a +bzk)u'+-(c+dzk +ez 2 k)u =0 z z2 can be represented in the form u(z) = za et~z•v(yzk),

3.

where a, {3, 'Y are suitable constants, and where v is a solution of Kummer's equation. Let A be a complex number. The differential equation z 2 u" + zu'- u = Az 2 (1- z 2)u

possesses a solution analytic at z

=

0 which is given by

u(z) = z e-zt~z\F1 (1- {3; 2; 4{3z 2 ),

where {3 satisfies -16{3 2 =A. Confirm by means of Kummer's first transformation that u is unchanged if {3 is replaced by - {3. 4. Let q be analytic in some region S. If u 1 and u 2 both are solutions of the differential equation u"+q(z)u =0, then the function v := u 1 u 2 satisfies vm+4q(z)v' +2q'(z)v = 0.

Use this result to obtain representations of the products

of', (I +y; z)oF,(l-y; z) in terms of single hypergeometric series. 5. Let v and z :;f:. 0 be two complex numbers. The difference equation in the variable n, 2(v+n) Yn+l Yn + Yn-1 = 0 z possesses the linearly independent solutions Y~l) := Jo+n{Z),

6.

Bessel's equation may be written z(zu')' = (v 2 - z 2)u.

7.

Conclude that for v > 0 and z real, 0 < z < v, the function J. is increasing and hence different from zero. By expanding the integral in series and interchanging summation and integration, establish Poisson's integral representation

( )( . 8)2• dfJ, (z/2)" f.,. ( J. z) = fJo cos z cos fJ sm valid for Rev>-!.

THE INFINITE POINT: EQUATIONS OF FUCHSIAN TYPE

143

§9.8. THE INFINITE POINT: EQUATIONS OF FUCHSIAN TYPE Let the function f have an isolated singularity at z = oo (see §4.5). We recall that f is said to be anai}'tic at z = oo if and only if the function /defined by /(z) := f(1/ z) has a removable singularity at z = 0. Furthermore, z = oo is said to be a zero or pole of I if z = 0 is a zero or pole off The order of the zero or pole of I at oo is defined to be the order of the zero or pole of/ at 0. Let the matrix A and the coefficients Pt. ... , Pn have isolated singularities at z = oo. To study the behavior of the solutions of the system w'=A(z)w

(9.8-1)

or of the nth order scalar equation u+p 1(z)uOwe now also assume thatRe(1--y)>O. (The set of triples (o:, {:J, -y) inC satisfying both conditions is open and connected.) Then both series on the right of (9.9-25) converge for z = 0, and letting z-+ 0 from the right we obtain 1 = as(o:, /3, 1 +o: + f3- -y) +bs(y -o:, 'Y -{:J, 1 +'Y-o:- /3) which permits the determination of b. By the Gauss formula (Theorem 8.6b) f(-y)f(-y-o:

-13)

a =s(o:,{:J, -y)=f(-y-o:)f(-y-/3)'

and using f( 'Y )f( 1 - 'Y) = 1r(sin 1T'Y) -t we get

b =sin 1r('Y-o:) sin 1r('Y- /3)- sin 1T'Y sin 1r('Y-o: - /3) f( 'Y )f(o: + f3- 'Y) sin ?TO: sin 1r/3 f(o:)f(/3) ·

162

ORDINARY DIFFERENTIAL EQUATIONS

It is easily seen thatthe value ofthe first fraction is 1. We thus have found:

F(a,

p; y; z) f(-y -a -{J)f(-y)

= f(-y-a)f(-y-p)F(a, {J; 1 +a +{J --y; 1- z)

(9.9-26)

+f(-y)f(a+!J--y\1- v-a-{3 f(a)f(p) z · F( y- a, y- P; 1 + y- a -

p; 1 - z)

At the moment, this formula is established only in a connected open subset of the parameter space. However, because for any fixed zESt n s3 both sides of (9.9-26) are meromorphic functions in each of the parameters, the formula remains valid for any set of parameters for which both sides are meaningful. The formulas (9.9-23) and (9.9-26) enable us to express any hypergeometric series of argument z in terms of series with argument 13 (z) or t4(z ). In view of the relations

the transformations 13 and t4 can be used to generate the full set of transformations 12 , . . . , 16 • We thus can express the function u 1 in terms of any ofthe pairs (u2k-1> u 2k ). For instance, to obtain a representation in terms of series of argument t6 , we apply (9.9-24) to each of the functions on the right of (9.9-26). Similarly, to find a representation in terms of series of argument 15 , we apply (9.9-26) to the function u7 appearing in the relation u 1 = u7 • A further application of (9.9-24) will then yield a representation in terms of 14 o 13 o 14 = 12 . This last representation is especially interesting, because it connects functions with the nonintersecting original domains of definition S 1 and S2. In Table 9.9a we have listed the explicit representations of u 1 in terms of series of argument 1k(z), k = 2, 3, ... , 6. Because u1 is analytic at z = 0, these representations are valid in the plane cut from 1 to oo. The powers of 1- z and -z have their principal values. A further set of representations is obtained by applying Euler's first transformation to each of the functions on the right.

THE HYPERGEOMETRIC DIFFERENTIAL EQUATION

163

Table 9.9a. Linear transforms and analytic continuation of the hypergeometric series

F(a, {3; y; z) f(p-a)f(y) -a ( 1) =f(y-a)f({3)(-z) Fa, 1+a-y; 1+a-{3;z f(a-{3)f(y) _13 ( 1) + f(y-{3)f(a) (-z) F {3, 1 +{3 -y; 1 +{3 -a;:;

f(y-a -{3)f(y) = f(y-a)f(y-{3)F(«r {3; 1 +a +{3 ~y; 1-z) +

f(a+{3-y)f(y)( )y-a-{3 ( ) f(a)f(p) 1-z F y-a, y-{3; 1 +y-a -{3·; 1-z

=(1-z)-aF(a, y-{3; y;

z~ 1 )

f({3 -a)f(y) -ad 1 ) =f(y-a)f({3)(1-z) r\a,y-{3;1+a-{3; 1 _z f(a -{3)f(y) _13 ( 1 ) +f(y-{3)f(a) (1-z) F {3; y-a; 1 +{3 -a; 1 _z

r(y-a -{3)r(y) -a ( 1) =f(y-a)r(y-{3)z Fa, 1-y+a; 1+a+{3-y; 1-:;

+

f(a +{3 -y)f(y) a-y( 1 )y-a-{3 f(a)f({3) z - z

· F(y-a, 1-a; 1 +y-a -{3;

1-;)

PROBLEMS 1. The Jacobi polynomials P~'"· 11 >(z) can be defined by

(l+a)n ( -n, l+a+/3+n; l+a;1-z) 2- .

p~a,IJ>(z):=-n-!-F

164

ORDINARY DIFFERENTIAL EQUATIONS

Establish the following alternate representations in terms of hypergeometric series: {J)

p~a, (z)=

(l+a+{3+n)n(z-1)n 2 ) n! - 2- F ( -n,-a-n;-a-{3-2n; 1 _z

(1 +f3)n ( l +z) =(-1) n-n-! -p -n, l+a+{3+n; 1+{3;-2-

z-1) =(1+a)n(1+z)n - - - - - F ( -n -a-n·1+a·-n! 2 ' ,.. ' 'z+1 =(-1t

(1+a+{3+n)(z+1)n ( 2 ) n! - 2F -n,-{3-n;-a-{3-2n; 1 +z

Z + 1) =(1-+~)n(Z - - --1)n - F ( -n-a-n·1+a·--

n!

2

'

'

,.., z-1 ·

2.

Find, by Frobenius' method, a second solution of the hypergeometric equation when y = 1. 3. The complete eUiptic integrals can for 1e1 < 1 be expressed in terms of hypergeometric series as follows:

f

w/2

K(k):=

0

E(k):=

.J

1

1T'

1 - e(sin 4> )

2

dtf>=-F(!.~; 1;e),

2

r·'"'2

Jo .J1-e(sint/>) 2 dt/>=~F(-!J; 1;e).

Show that for 11- ej < 1 K(k)=

~ (~)~(~~n{hn(l)-hn(~) +2Log2-~Log(l-e)}(l-et. n-O

n.n.

-4 Log 2- Log(l- k 2)}(1- er. 4. 5.

Find the analytic continuation of u 1 into the region S 2 by means of the relation t2 = t3 ° t4 ° t3. Show that a second-order Fuchsian equation with the four singular points a, b, c, oo and corresponding exponent pairs (a, 0), (~. 0), (y, 0), (15, 15') is of the form 1-a 1-{3 1-y) u"+ ( - - + - - + - - u'+

z -a

z -b z -c

88'z+p u =0 (z -a)(z -b)(z -c) '

where p may have any value. (Equation of Heun. The constant p is called the accessory parameter.)

QUADRATIC TRANSFORMS: LEGENDRE FUNCDONS

165

6. Prove Elliot's formula F(l+a, -~-y; 1+a+~;z)F(l-a,~+y; 1+~+y; 1-z) +F(!+a,l-y; 1 +a+~; z)F(-!-a, !+y; 1 +~ +y; 1-z) -F(!+a,!-y; 1 +a+~; z)F(!-a,!+-y; 1 +~+-y; 1-z) _ f(1 +a +~)f(l +~ +y) - f(~+a +~+y)f(l+~)'

and obtain as a special case Legendre's relation E(k)K(k')+ E(k')K(k)- K(k)K(k') = ~·

where E and K denote complete elliptic integrals, and k' 2 := 1-e.

§9.10. QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS In this section we continue our investigation of the hypergeometric differential equation by studying certain of its special cases whose solutions admit transformations where the variables are linked not by Moebius transformations, as in §9.9, but by certain rational functions whose numerators and denominators are of degree two. Let us consider a linear differential equation of order two that at z = 0 has a singular point of the first kind with exponents 0 and!. Of the two functions making up the fundamental system at this point, one can be chosen analytic, and the other as z 112 times an analytic function. By subjecting the independent variable to the transformations := z 112 , both functions will appear as analytic functions of s. Thus the singularity at z = 0 can be removed. On the other hand, new singularities will be introduced; generally, if z = z 0 were a singular point, both points s = ±z~ 12 will now be singular. If there is only one finite singular point besides z = 0, the total number of singular points remains the same, and if the given equation was hypergeometric, we may expect it to transform into another hypergeometric equation by the process. Consequently, let us consider the hypergeometric differential equation with singular points at 0, oo, 1 and corresponding exponents 0, !; a, {3; y, 8, where a +{3 +y +8 =!.By (9.8-15b), this equation is of the form

u"+(_!_+ 1 -y-8\ u' +(af3 +__I!_) 2z

1 u = 0. z-1 z(z-1)

z-1 -,

If u(z) = v(s), s= z 112 , then v·is readily seen to satisfy

2s s-

(

y8 )

4 s-

v"+(1-y-8)~ 1 v'+ a{3+~1 ~1 v=O.

s-

(9.10-1)

166

ORDINARY DIFFERENTIAL EQUATIONS

This may be written

-)v'

v" + (1- 'Y- c5) ( -1- + -1

s-1 s+1

+ ( yc5 _ yc5 + 4ap) s-1 s+1

2

2 v= 0 (s-1)(s+1) '

and a comparison with (9.8-15b) shows that this equation is hypergeometric with the singular points -1, oo, 1 and the corresponding exponents y, c5; 2a, 2p; y, c5 whose sum is 1 as required. In terms of Riemann's P symbol we thus have proved:

{0 1 }= {-1

P 0

00

a

y

t P

c5

z2

P

y

2a

1}

c5

2P

c5

00

y

z ,

where for consistency we have written z in place of s. The singular points -1, oo, 1 can be brought back to 0, oo, 1 in two different ways. Using the simplified version of the P symbol, we thus get

P{! ; ; z}=P{; 2

~;

; 1;z} (9.10-2)

As an application we derive a quadratic transformation due to Goursat. If = 0 and hence c5 = t- a - p, the set on the left by Theorem 9 .9d has the fundamental system 'Y

P; t; z 2) v2(z) = zF(a +t, P +t; i; z\ v1(z) = F(a,

(9.10-3)

On the other hand, the two functions

v3, 4 (z) =F( 2a, 2p; a

1±z)

+P +2;1 - 2-

are both members of the set on the right. They are linearly independent, and hence likewise form a fundamental system. Hence it is possible, for instance, to find constants a and b such that v 1 = av 3 + bv4 • Because v 1 is unchanged if z is replaced by -z, b =a. To determine a, we set z = 0 to find 1 = 2aF(2a, 2P; a

+P +!; !).

QUADRATIC TRANSFORMS: LEGENDRE FUNCfiONS

167

The hypergeometric series has the sum

f(!)f(a + {:J +!) f(a +t)r

' as shown ln Problem 2, §8.6. We thus may calculate a to find

=

r(

(9.10-4) 11+z) +F( 2a 2f:J·a+{:J+-·-11-z) 2a 2[:J·a+{:J+-·--

'

'

2' 2

,

'

2' 2

In a similar manner we find

(9.10-5) 1 1+z) 1 1-z) =F( 2a 2[:J· a +{:J +-· - -F( 2a' 2[:J·' a +{:J +-· , ' 2' 2 2' 2

A number of transforms can be derived from the following general principle. Suppose we have an identity of the form 0

P {*

{0 * 0 } * 0 } * * z = P * * * g(z)

(9.10-6)

where the corresponding exponents on either side are not all equal in pairs, and where g is analytic at 0 and satisfies g(O)=O.

(9.10-7)

Each of the sets (9.10-6) contains precisely one function that is analytic and equal to 1 at z = 0. This function can be represented by a hypergeometric series as indicated in Theorem 9.9d. Because the sets (9.10-6) are identical, the two representations are identical. Tne method will generally yield identities of the form F(*,*;*;z)=h(z)F(*,*;*;g(z)), where his an elementary factor. A first set of transforms is obtained by writing (9.10-2) as follows:

p{'>~'

2a

'Y

2{:J~z

} = P{O

a 'Y !f:J~

(2 _ 1)2 } z .

(9.10-8)

168

ORDINARY DIFFEREN11AL EQUATIONS

Here we may replace z by any of the functions tdz); the resulting P set on the left can then be written asaP set with argument z by suitably permuting the exponents. The arguments on the right become qdz) := (2tk(z)-1) 2.

It is readily seen that q1 = q3, q4 = q 5, q2 = q6. Because the P symbol on the

left has the same form for each pair of transformations (th t3), (t4, t5 ), (t2, t6 ), because two pairs of exponents are equal, nothing is lost by considering the cases k = 1, 4, 6 only. The functions qk do not satisfy (9 .1 0-7), because their values at z = 0 are 1 fork = 1, 4 and oo fork = 6. Therefore, we must subject the P set on the right to one more Moebius transformation, which brings qk (O) to 0. Fork = 1, 4 this can be done by either t3 or t6, fork = 6 by t2 or ts. We thus arrive at a total of six transformations of type (9.10-6), where g(z) = ti(qk(z)). for suitable values of j and k. Tne six resulting identities are presented in Table 9.10a. Table 9.10a

k

j

3 1

P{~ 0 a 4z(z -1)}

!

{3

P{~ :

0

u

u

4

{1'

p 8

1'

8

2a 2{3

z

}

1-1

(2z -1) 2

4z 2} ! - (1-z)

4z } 'Y

p{2a 2{3

1'

1'

8

8

3

= (1 +z)2

6

6

z} =

8

? 4(zz2-1)} 2

6 5

2

QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS

169

To obtain the desired identities between hypergeometric series, we now must in each case choose either a or 'Y such that one of the exponents at 0 becomes 0, and identify the analytic solutions by Theorem 9.9d. We thus obtain the six identities given in Table 9.10b, where the parameters a and p are not necessarily equal to those in Table 9.1 Oa.

Table 9.10b

Another set of six identities is obtained by interchanging the roles of z and g(z ). This leads to expressions involving J 1- z, Fz, and M. The root ~must be so defined as to become + 1 for z = 0. The roots Fz and M can be chosen arbitrarily, but the same determination must be chosen in all places where it appears in a formula. The resulting formulas are listed in Table 9.10c, p. 170. All these identities are based on relations between two different kinds of P sets, namely (i) P sets in which one set of exponents'has difference!, and (ii) P sets in which two sets of exponents are identical. A nontrivial identity between two P sets of type (ii) is obtained as follows. As shown, any P set of type (ii), where('}', c5) is the recurring pair of exponents, is equal to a set of type (i) of the form

0 a

P{t

{3

'Y

c5

} qk (z) ,

""" = -.I

Table 9.10c

F(a, {3; a +{3 +4; z) = ~

F(a, a +4; {3; z) = ~

k

j

1 1 - v' 1 - z) ( F 2a, 2(3; a +{3 +2; 2 ·

1

3

1 v'1-z-1) 1 c+v'1-zr 2a ( F 2a,a-{3+2;a+{3+-;~ 2 1-z+1 2

4

3

z.J-;~ ) (v'1-z+../-;)- 2aF(2a,a+{3; 2a+2{3;~ 1-z+ -z

6

5

~ (1-z)-aF ( 2a,2{3-2a-1; {3; y'1-z-1) 2 1-z

1

6

4

6

6

2

c

+J1- zr2aF ( 2a,2a-{3+1; {3; 1-Jl=-;) ~ 1+ 1-z 2

2.!; 1 2{3 -1; -) (1 +.f;r 2aF ( 2a, {3 --; 2

1 +.;-;

QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS

171

where k is 1, 4, or 6. We interchange the second and third pair of exponents and obtain 'Y a qk(z) } ! ~ f? qk (z)- 1 ·

P{O

Applying (9 .1 0-2) and thereby going back to a set of type (ii), this becomes

p{: tJ

2y a 2 ~ {3

( rk z

>} ,

where

either value of the square root being acceptable. The resulting identity is of the form (9.10-6) only if rk(O) = 0, which is the case only if qk(O) = oo, that is, for k = 6, if the root is chosen appropriately. In the remaining cases k = 1, 4 we have qk (0) = 1 and consequently rk (0) = oo, and we must apply tz or t 5 to bring the infinite point to zero. In this manner we obtain identities where a P set of type (ii) of argument z is expressed as another P set of type (ii) with argument where j and k are chosen as indicated previously. The results are listed in Table 9.10d. Table 9.10d k

j a {3 {3

p{2'Y a 1

p{;

2~

2a

'}'

2{3

~

2~

'}'

2a

~

2{3

a

{3

{3

P{2'}' a {3

a {3

p{2'Y a

a

2~

P{;

5

-4hJt-z} (J1-z-h) 2

2

z}= P{2'}' a

4

4h·h-z} (J1- z+Jz) 2

2Jz } (1-./;)2

5

z}= 2~

{3

{3

2Jz } (1 +Jz)2

2

172

ORDINARY DIFFERENTIAL EQUATIONS

Table 9.l0d-continued

k

6

j

P{2a 2{3

'Y

8

'Y 8

P{;

2-y a 28 {3

P{;

2-y

a

28

{3

(1-J1- z) 2 } 2J1-z

1

z}= (1-v'1- z) 2 } (1 +Jt- z) 2

4

To obtain identities between hypergeometric series, we again choose either a or 'Y such that one of the exponents at 0 becomes 0. Theorem 9.9d then enables us to identify the ana~ic solutions. (In the cases k = 1, 4, where the functions g have a factor.../ z at 0, it is necessary to replace z by z 2 before identifying analytic solutions.) In this manner we obtain the six identities shown in Table 9.10e. Because these transforms are based on identities between P sets of type (ii), the transforms obtained by interchanging the roles of z and g(z) are already contained in Table 9.10e. The 18 transforms listed in Tables 9.10b, 9.10c, and 9.10e are the basic quadratic transforms of Gauss and Kummer. From each formula given, three others can be obtained by subjecting one or both hypergeometric series to Euler's first identity. Thus a total number of 72 quadratic transforms can be obtained. Legendre's equation

The theory of quadratic transforms is intimately linked with the theory of a differential equation that is of considerable importance in mathematical physics, called the Legendre differential equation. This equation depends on two complex parameters, commonly denoted by IL and 11, and is given by 2

(1- z 2)u"- 2zu' + [ 11(11 + 1)- 1~ zz Ju = 0.

(9.10-9)

Table 9.10e

a

f3

4-r-;_~

(J1- z+J:::;)-2a F(a, 2+2; a +/3; (J1-z+.r-;_)2 a f3 1 z ) = Fa a.-+-+-· ~ ,,..,,2 2 2'

F(a,/3; a -13 + 1; z) =

a f3 -4.r-;J1- z) (J1-z-h)-2ap(a,2+2;a +/3; {~-..C\2

1 -4Jz ) (1-.JZ)-2a F(a, a -/3 + 2 ; 2a -2/3 + 1; ( 1 -Jz)2 1 4Jz ) (1 +v'Z)-2a F(a, a -{3 +2; 2a -2/3 + 1; (1 +.fZ)2 I (I-~)2) (I- z)-a 12 F ( a, 2/3 -a; /3 +-;~

2

F(a,/3; 2/3; z)=

.... c;!

)

A, 1 - ~

2 ( 1 1 (1-~) 2 ) ( 1+J1-z)- a Fa a- 12 +-· a+-· 2 ' ,.... 2',.... 2'o+J1-z) 2

174

ORDINARY DIFFERENTIAL EQUA110NS

The equation occurs, for instance, when Laplace's equation is separated in spherical, elliptical, or toroidal coordinates. A comparison with (9.10-1) shows that the set of solutions is

-1

p

1

00

t!:. z

t!:. v+1 2

IL

2

_I!:..

-v

2

'

2

and hence is identical with

l-~

p

t!:.

v+1

2 JL

-v

1-zj 2 '

2

Here (1-z)/2 can be replaced by (1 +z)/2 because the exponent sets belonging to 0 and to 1 are identical. Reducing one of the exponents at 0 and 1 to zero to express the solutions by hypergeometric series, the set of solutions of (9.10-9) can also be written in the form 12 ( z+1),.. P{o z-1 JL

v+1

0

-v

-JL

1-z} 2

= (z + 1)-,..12 p{ 0

v+1 0 1+z} (9.10-10) 2 . -v -JL z-1 JL In these expressions it is permissible to replace JL by -JL an.d v by -v -1 because this does not change the original set. Numerous solutions of the Legendre equation now can be written down by specializing parameters in the solutions uh .•• , u 24 obtained in §9.9. Since Hobson [1931] it has become traditional to employ as standard solutions the two functions 1

(z+1),..12 (

P~(z):=r(l-JL) z- 1

1-z)

F -v,v+1;1-JL;2-

(9.10-11)

and

QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS

175

called Legendre function of the first and second kind, respectively. It is assumed that larg zl < 71', larg (z ± 1)1 < 'TT'; consequently, these functions are undefined for real z :,;:;; 1. (The convention to define them as the average of the limits of the values above and below the cut is frequently used.) The function P~ is proportional to u 1, and the function Q~ is derived from u 3 by means of the last quadratic transformation in Table 9.10b. Evidently, P~ has been selected so as to have a simple behavior at the points z = ±1, and 0': has been selected so as to have a simple behavior at z = oo. The normalizing factors have been chosen to give a simple appearance to some formulas occurring in the theory. The factors f(l- 1.1.) and f(v +~) prevent the definitions from losing their meaning for 1.1. = 1, 2, ... and P = - ~. -1, ... , respectively. We note that for 1.1. = 0, v = n = 0, 1, 2, ... the function P~ reduces to a polynomial of degree n. This is called the Legendre polynomial. It is commonly denoted by Pn and has the simple representation

1-z)

(

Pn(z)=F -n,n+1; 1;2- .

(9.10-13)

It is obvious that the theory of linear and of quadratic transforms implies a large number of relations between Legendre functions. In fact, because the set of solutions of the Legendre equation is elementarily related to a P set of type (ii), every relation between hypergeometric series admitting quadratic transforms can be written as a relation between Legendre functions. We refer to Chapter III (written by F. Oberhettinger) of Erdelyi [1953] for a complete listing of these results. PROBLEMS 1. Obtain the following expansions either directly or as special cases of quadratic

transforms: (a)

(b)

a 1 a+ 1· z) ( 1+~~-.. = F (a - -+-· 2 2' 2 2' '

a 1 a ) F ( -+-,-+1;a+1;z 2 2 2

(c)

M(1+z)"'+(1-z)"'}=F( -~.~-~;~;z 2 )

(d)

2!z {(1 +z)"' -(1- z)"'}= F(~-~·

1-~; ~; z

2)

176

ORDINARY DIFFEREN11AL EQUATIONS

1 a 1 a 3 2) 2+iz)a -(./1-z 2-iz)a}=F(1----{(./1-z -+-· -· z 2aiz 2 2'2 2'2'

(f)

2.

Show that for -Tr/2 < c/J < Tr/2 and arbitrary a, cos a"' = '+'

F(~2' -~!. (sin '+'"')2 ) ' 2'2'

(1 1

sin ac/J a a 3 . 2) asinc/J =F 2+2'2-2;2;(smc/J) · 3.

The Chebyshev polynomials of the first kind are for n = 0, 1, 2, ... defined by

T" (x) := cos(n arccos x ). Show that

T"(x)=F( n, -n; ~; 1 ;x). 4. If P" denotes the Legendre polynomial, show that _ n (2n)! 1 1 2 P2n(x)- (-1) 22"(n!) 2F(.-n, n +;z; ;z; x ),

-( )"(2n+1)! ( 3 3. 2) P2n+l (X ) - -1 22"(n !) 2 xF -n, n +;z; :z, X , n =0, 1, 2, .... 5.

Cubic transforms. If a+ {3

=~and w :=

e;'"13 , show that

{0 1 } {1

P 0

i

6.

00

0

a z3

i

{3

tJ

tJ

{3

z}=nf:

tJ

0 0 3iJ3z(l-z)} (z-w) 3

"l.tJ i !

By identifying analytic members of the Riemann sets of problem 6, show that

( 1 2 F(3a, 3+a;3+2a;z)= 8.

a

Deduce from problem 5 that

p{: : : 7.

=P

3

1 2 3iJ3z(1-z)) 1-;z) F ( a,3+a;3+2a; (z-w)3 ·

By subjecting the Riemann sets of problem 7 to appropriate quadratic transformations, establish Goursat's cubic transforms:

(a)

1 9z)- 2a ( 5 27z 2(z -1)) F(3a,3a+!;4a+~;z)= ( 1-gF a,a+2;a+6; (B-gz)2

177

QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS (b)

F(3a, 3a +!; 2a +~; z) = (1-9z)- 2aF(a a+!. 2a +~· ' 2' 6'

27z(l-z) 2) (1-9z) 2

(c)

2 ) z)-3aF ( a a+-· 1 2a +-· 5 27zF(3a a +1. 4a +~· z)= ( 1-6 3 ' 4 ' ' ' 3' 6' (4-z) 3

(d)

S ) 1 5 27 Z ) ) I ( F3a,3-a;2a+6;z =(1-4z) -3aF ( a,a+-;2a+-;( 3 3 6 4z-1



Problems 9+16 deal with applications of the algorithm of the arithmetic-

geometric mean, due to Gauss. 9. The complete elliptic integrals of the first and of the second kind, conventionally denoted by K(k) and E(k), were expressed in terms of hypergeometric series in problem 3, §9.9. (a) If O.;; k < 1, k' := .(1- k 2 , show by means of a quadratic transform that 2 (1-k') K(k)=1+k'K 1+k'.

(9.10-14)

(b) For O {3 0 > 0, show that the sequence {anl decreases monotonically and the sequence {pn} increases monotonically, and that both sequences have a common limit f.L(a 0 , {30 ), called the arithmetic-geometric mean of a 0 and {3 0 • (b) If a 0 := 1, {3 0 := k', show that, in the notation of problem 9,

f3n=k~, an+l= 1+2 k~.

an

an

hence

·an=

1 +k' n. 2 m,n=1,2, ....

n-1

m=O

ORDINARY DIFFERENTIAL EQUATIONS

178 Conclude that

(9.10-17)

K(k)= 2#-£(1, k')"

(c) Letting

n =0, 1, 2, ...

(9.10-18)

show that under the initial conditions (b) kn

= 1'n. n = 1, 2, ...• an

and establish the inequality 1

2

kn+l,.; 4 k~ kn•

Conclude that the sequences {an} and Wn} tend to their common limit with quadratic convergence. 11. If 0 < k < 1, use quadratic and linear transforms to establish the identity E(k) = (1 +k')EG

:~:) -k'K(k).

(9.10-19)

12. Letting

Q(k) := K(k)-E(k) K(k)

'

show that in the notation of problem 9

Q(k)= (1 +2k') 2 Q(kl)+ik 2 , hence, using problem toe, that

Q(k)=! ~ 2m mrr [1+k;]\;.. 2 m-o ;-o 2 13.

(9.10-20)

Using the notation of problem 10, show thatthe mth term in the sum (9.10-20) equals 2m')I;., m = 1, 2, .... Hence, putting 'Yo := k0 = k,

and (9.10-21) [Note: The formulas (9.10-17) and (9.10-21) are very useful for the efficient numerical evaluation of K(k) and E(k).]

QUADRATIC TRANSFORMS: LEGENDRE FUNCTIONS

179

14. For-1p.

Verify that this equation possesses the solution w(z) = z 6 eA•p(z),

where p is analytic at oo, p(z) := exp(- a 2 -~-~-· · ·).

z

2z 3 3z 3

and that the series P constructed in the proof of Theorem 9.11a in this case represents p. 2. The first-order system obtained from the scalar second-order equation

u"-.!.u'-.!.u=O z

z

by letting wT := (u, u') does not satisfy the hypotheses of Theorem 9.11a, because the matrix A 0 has a nonlinear elementary divisor. Show that by introducing the new variable t := z 112 one can obtain a system satisfying the hypotheses, and establish the existence of formal solutions of the form u = e±2iz'l2z-l/4p(z''2), where p denotes a formal series in z- 1 •

187

SPECIAL SECOND-ORDER EQUATIONS 3. Show that the system of rank r > 1, w' = z' A(z)w,

where the matrix Ao has n distinct eigenvalues, possesses a formal solution of the form W=Pz"'e~",

where Pis a formal series in z- 1, A is a constant diagonal matrix, and z2

z•+1

r:=zrr +-r 2 r-1 +···+-r r+1 0 with diagonal matrices ro.... , r,. Formulate an algorithm for determining the ri and the coefficients of P. [It is now necessary to carry along r undetermined diagonal matrices.]

§9.12. SINGULARITIES OF TilE SECOND KIND OF SPECIAL SECOND-ORDER EQUATIONS In this section we adapt the construction of the preceding section to obtain formal solutions of the special scalar second-order equation u"-r(z)u =0,

where, for

(9.12-1)

lzl sufficiently large, wo implies 'f(s)-

r•

e-s.-F(T) dT'

0 be given. By (10.1-4) there exists w 0 such that T>w 0 implies IG(T)I w0 and Re(s -s0 )>0, the first term on the right is bounded by 2e. The second term is bounded by I (s

-

) IX -Re(s-so)T d I< Is- sol s0 E e T R ( )E. "' e s-so

If larg(s- so) I:,.,;; {3, where {3 < 'TT'/2, then Re(s- so) ;;;.Is- sol cos {3. Hence the last bound may be replaced byE/cos {3, and it follows that for x ;a.w > w 0 and SESp,

Because the bound on the right does not depend on s and tends to zero with E, the uniformity of the convergence is proved. • Lemma 10.1a implies, in particular, that if s 0 E C(F), then any s such that Re s > Re s0 also belongs to C(F), for any such s is contained in an angular domain larg(s- so) I:,.,;; {3 where {3 < 7r/2. The infimum aF of all real numbers a such that C(F) contains an s with Re s = a is called the abscissa of simple convergence of the Laplace integral (10.1-1). (If C(F) is empty, we set aF := oo.) If aF is finite, then C(F) contains all s such that Res> aF, and nos such that Re s 0, le -s,TF(T)I:,.,;; le-soTF(T)I, and since (10.1-5) holds for s = s 0 it also holds for s = s 1 • There follows THEOREM lO.lc

The set of absolute convergence of a Laplacian integral, if not empty, is either the full plane or an open or closed right half-plane.

BASIC PROPERTIES OF THE !£ TRANSFORMATION

203

The infimum of all numbers Res such that s satisfies (10.1-5) is called the abscissa of absolute convergence of the integral (10.1-1) and is denoted by {Jp. We put fJF= oo if (10.1-5) holds for nos. If 'YF denotes the growth indicator of F, then evidently ap.;;;; fJF.;;;; 'YF· The literature on the !£ transformation attaches some weight to the fact that in any of the statements -oo.;;;; ap, ap.;;;; {Jp, fJF.;;;; 'YH 'YF.;;;; oo either equality or inequality can actually occur. Nevertheless, as we shall see later, the numbers ap, {Jp, 'YF coincide in most cases of practical interest. They always coincide ifF is defined by a power series. U. !I! Transforms are Analytic TIIEOREM lO.ld

Let FE .0, ap < oo. Then f := :I!F is analytic in the interior of C(F), that is, at all points s such that Re s > ap. This is one of the basic facts about the !£ transformation. It makes it possible to apply the powerful tools of complex analysis to the solution of real variable problems. Proof of Theorem 10.1d. For Res> aF,

f(s) =lim fn(s), n-+oo

(10.1-6)

where n=0,1,2, .... By Theorem 4.1a, each In is analytic in Res> ap (in fact, entire). By Theorem 3.4b, the limit function f is analytic if the convergence of the limit (10.1-6) is uniform on every compact subset T of Res> aF. This, however, follows at once from Lemma 10.1a, for every such compact subset is contained in some angular domain larg(s- aF )I.;;;; {J, where {J < Tr/2, and the convergence has been shown to be uniform on such angular domains. • COROLLARY lO.le

Let FEn, aF < 00, I := :I!F. Then for n = 1' 2, ... the functions TnF(T) have the Laplace transforms (-1ffn>(s); their abscissa of convergence is o;;;;ap. Proof. It suffices to establish the Corollary for n = 1. We first show that the Laplacian integral

204

INTEGRAL TRANSFORMS

exists for Res >a.F. Let lJ > 0 be such that Re (s -8) >a.F. Then the function G defined by

G(T) := -

fXl e -(s-B)uF(u) du T

tends to zero forT-+ oo. Using integration by parts,

I Xe "'

-ST

T

F( T) d T=

IX Te -8T e -(s-8)TF(T) d T "'

= [Te -BTG(T)]! + Jx (1-lJT) e -BTG(T) dT. "' Given E > 0, this can be made To because fo' II -lJTie -aT dT exists. Thus h (s) exists. Using the notation of the preceding proof, the fact that fn -+ f uniformly on T by the Weierstrass double series theorem implies that f~-+ f'. By Theorem 4.1a,

Because the limit on the left exists as n -+ oo and equals f'(s ), the limit on the right exists and has the same value. But the limit on the right has been shown to equal -h (s ). •

III. Behavior of !£Fon the Boundary of C(F) Let Fen, and let a.F < oo. It is clear that f == I£F, as an analytic function, is continuous at all interior points of the set of convergence C(F). This means that lim f(s) =/(so) S--+So

holds at all interior points of C(F). It is less trivial that a similar result also holds at the boundary points of C(F). THEOREM lO.lf (Abelian 'Theorem)

Let FEn, f == I£F. If C(F) contains the boundary point So, then for every 13 such that 0..;, {3 < 7r/2,

lim f(s) =/(so),

s-+so

BASIC PROPERTIES OF THE !L TRANSFORMATION

205

where the approach of s to s 0 is restricted to the angular domain SfJ: larg(s -so)l :!iiO{J. Proof.

Let E > 0 be given. By Lemma 10.1a, w can be selected such that

If''

e-•TF(T) dTI

0 except at most on a set of points with no finite point of acCIImulation. By this equivalence relation the set .0 is divided up into equivalence dasses. Two functions are in the same equivalence class if and only if they are equivalent to each other. The class of all F equivalent to a given F 0 E .0 is denoted by [F0 ]. By the foregoing, all functions in one and the same equivalence class have the same:£ transform. We now show that, conversely, functions in different equivalence classes have different!£ transforms. THEOREM lO.lb

Let Fk E .0, {k := :£Fk (k = 1, 2). If f1(s) = fz(s) for all sin some right halfplane, then F 1 and F 2 belong to the same equivalence class. Actually a stronger result will be proved; see Corollary 10.1j. The proof is based on the following LEMMA lO.li

Let t/J be a complex valued, continuous function on the interval [0, 1] such that

1 1

t/J(x)xndx=O,

n=0,1,2,....

(10.1-7)

Then t/J vanishes identically. Proof of the Lemma. Without loss of generality we may assume t/J real. For t/J real the proof is indirect. Suppose t/J does not vanish identically. Then by the continuity of t/J, a:=

r

[t/J(x)] 2 dx>O,p :=

r

it!J(x)idx>O.

Let E := a/ p. Then E > 0. By the Weierstrass approximation theorem, there exists a polynomial 1r such that E

lt/J(x)-7r(x)i Re s0 , where G(T) :=

e -"'G(w)-e-""'uo

I"'

eO"o"G(-r)d-r.

Because u 0 E C(F'), lim..., ... oo G(w) exists; thus there exists JL such that IG(-r)l O. Hence lim e-c•-uo>"'G(w)=O,

....... oo

and -sw Ie u 0

J,"' e u

0

"G( T ) dT 1,:::: """'IL e -Re(s-u0 )w -+ 0 •

0

Thus (10.2-9) follows, implying (10.2-8) and rule III. • Rule III provides one of the links between the operational calculus of Heaviside and the .f£ transformation. If F(O+) = 0, rule III simply reads F'(-r) o-----e sf(s).

OPERATIONAL RULES; BASIC CORRESPONDENCES

213

The operation "differentiation with respect to T" thus is translated into the operation "multiplication by s" .It also becomes now clear why Heaviside's calculus generally breaks down for functions F such that F(O+)-¥- 0. Even in this more general case, however, rule III transforms differentiation, a transcendental operation requiring a limit process, into a simple algebraic operation. Repeated application of rule III under suitable conditions yields: RULE ID': Repeated ditlerentiation

Let Fe 0 be such that the derivatives F, ... , F" -I) exist for limits

p(o+) := lim p( 'T ),

T

> 0, let the

k=O, 1, ... ,n-1

T-+0

T>O

exist, and let Fpossess a (possibly generalized) nth derivative F"> e .0. Then if u 0 e C(F"\ u 0 ;;;. 0, then for all s such that Res> uo

Rule III, read backwards, translates the operation "multiplication by s" from the image space to the original space. We already know from Corollary 10.1e how to translate the operation "multiplication by T" from the original space to the image space. For completeness, we state the result as: RULE IV: Multip6cation by T of an original function

Let Fe.O, aFaF>

n= 1,2, .... EXAMPLE

uo.2-10)

3

From (10.2-2) there follows T

" o-----e --;;-:;t, n! s

n =0, 1, 2, ....

(10.2-11)

More generally (10.2-1) implies n

T

GT

e

n! o-----e (s-a )"+I'

n =0, 1, 2, ....

(10.2-12)

214

INTEGRAL TRANSFORMS

RULE V: Integration of an original function

Let FEn, Uo E C(F), Uo;;;. 0. Then if Res> Uo,

rF(u)duo--ef~)·

(10.2-13)

This readily follows by applying rule III to the original function G(T) := rF(u) du,

whose generalized derivative is F. From rule IV we can obtain a result concerning integration in the image space. Let F be a function in fi such that 1 G(T) := -F(T) T

is likewise inn, aa 0, J;;' F( T) dr exists. Theorem 10.1 f thus yields the result (unobtainable by elementary.methods)

I" o

cos a-r- cos {3T 7'

{3 ..~ .. r =Log- (a, {3 a

> 0).

216

INTEGRAL TRANSFORMS

RULE VD: Tnnslation of an original function

Let A > 0. Then F(T-A) o - e e-s"f(s). Proof.

Because F( T- A) = 0 for T 0 only-the solution of the linear scalar nth order equation x'( ) Cd ,\p. 2 1' o--e--8 82S 2(s 2+ a 2) s 1

1_}

=~-~{_!. _ _ s

8 1 8 2 a 2 s 2 s2+a 2 ·

Consulting a table of correspondences, there follows

ORDINARY DIFFEREN11AL EQUATIONS. SYSTEMS

227

III. Systems Engineering In engineering science the word system has come to describe any device that effectively realizes a transformation T from one function space E 1 to another function space E2. If xl EEl and x2 := TXh then XI is called the input or driving function of the system, and X 2 is the output or response caused by X1 (see Fig. 10.3a). Input driving function

Output response System

)

x,

)

Fig.IO.Ja.

Frequently in applications, input and output are functions of time, and the spaces E 1 and E 2 may both be taken as n, the space of original functions. It is then often possible to describe the system by a linear aifferential or integra-differential equation with constant coefficients. In this setting the.:£ transformation proves to be a tool whose usefulness goes beyond that of a mere computational device. By shifting the analysis from the original space to the image space, the theory of systems attains a transparency which would not be available otherwise. For the sake of illustration, consider the electric circuit depicted in Fig. 10.3b.

p

Fig.10.3b.

The meaning of the symbols is as follows: Ohmian resistance capacity of condenser A inductivity of coil U(-r) imposed voltage 2 1(-T) currene

p 'Y

2 Use of capital letters for voltage and current does not imply (as it does in engineering literature) that we are dealing with direct current.

INTEGRAL TRANSFORMS

228

We conceive the circuit as a system with input U and output I. Tne working of the system is described by Kirchhoff's law stating that the sum of all voltage drops equals the imposed voltage or electromotive force. The various voltage drops are as follows: Resistance: pi (Ohm's law)

char~e

Condenser:

.!. r.,. I(u) du 'Y Jo

capacity

Coil: A di (Faraday's law) d-r Kirchhoff's law thus yields

1 f..,. di pi+I(u) du+A-d = U(-r). 'Y

0

(10.3-8)

'T

We assume the system to be at rest initially: /(0) = 0. Letting U(-r)0----4 u(s), 1(-r) o----e i(s), we then have dl -d o------esi(s),

1. 0

'T

1

l(u)duo------e-i(s). s

Applying I£ to (10.3-8), there follows

.

1 .

.

P' +-r +sAr s-y

=u

(10.3-9)

or

Thus

i(s)=g(s)u(s),

(10.3-10)

where

1

g(s):=--1 p+-+sA s-y

The function g is called the transfer function of the system. Thus in the image space the working of our system is described in a very simple manner, as follows: The image function of the output is the product of

229

ORDINARY DIFFERENTIAL EQUATIONS. SYSTEMS

the image function of the input and of the transfer function of the system (see Fig. 10.3c). i(s)

u(s) • g(s)

)

)

Fig.10.3c.

We call a linear system any engineering system that maps n into n in such a manner that the images of inpbt and output are connected by a homogeneous relation of the form (10.3-10). In addition to electric circuits such as the one considered above, typical examples of linear systems are mechanical vibrating systems subject to outside excitation. Simple rules hold if several linear systems are connected. Consider, for instance, two linear systems with transfer functions g 1 and g 2 . Let F 0 be the input of the first sytem, and let its output F 1 be used as input for the second system, which then in turn produces the output F 2 (see Fig. 10.3d).

,----------------l I I

~1 I

ISy~em I ~2 I 1

System

I I

211----+i---;~._3 -

I I I I L ________________ j Fig. 10.3cl.

If the respective image functions are denoted by f 0 , ft, [2, we have

hence

We conclude: If several systems are connected serially, the transfer function of the combined system is the product of the transfer functions of the individual systems (see Fig. 10.3e). fo ):

Fig.10.3e.

r, = gJo

"

f2 = K2f1 = K2K,{o )

230

INTEGRAL TRANSFORMS

An important application of the foregoing arises in the case of control systems. Here the output of a system is used to drive a control unit. The output of the control unit is then subtracted algebraically from the input of the system and thereby modifies this input, as shown in Fig. 10.3/. If g 1 and

Control unit

Fig.l0.3f.

g 2 , respectively, denote the transfer functions of the system and of the control unit, then the following relations hold in the image space:

lz = gtift- /J), 1 /J=-g2f2 K

(K =gain factor). There follows

hence

The transfer function of the system including the control unit thus is (10.3-11) Stability. Generally, a system is called stable if a bounded input always produces a bounded output. It is easy to find a necessary and sufficient condition for the stability of a system whose transfer function g is rational, g(oo) = 0. For simplicity, we restrict ourselves to bounded input functions of the form F 1(T) := e;.,.. ; in principle, any bounded input function may be synthesized from input functions-of this particular form in view of Fourier's integral formula (see §l0.6). Because ei"" o___. (s- iw )-\ the image

ORDINARY DIFFERENTIAL EQUATIONS. SYSTEMS

231

function of the output F 2 is

1 fz(s) := -.-g(s). s-lw

Let the poles of g be St. . . . , sm, and let the order of sk be A. Then if sk :F iw for all k,lz has a partial fraction decomposition fz(s)

= g(i~) + I s -lw

rk(s),

k=l

where rk is a principal part of degree jk. We have rk(s) .---oPk(T)e•k", where Pk is a polynomial of degree ik -1. Thus the response of the system is Fz(T) = g(iw) e;"'" +

m

L Pk(T) e•k".

(10.3-12)

k=l

Clearly, this is bounded if and only if all Re sk.;;; 0, and if the poles such that Re sk = 0 have order one. However, if g has a purely imaginary pole of order one, say sk = iw 0 , then if w = w0 the partial fraction expansion of {2 contains a quadratic polynomial in (s- iw 0 ) - \ which again leads to an unbounded output. Thus we find that a necessary and sufficient condition for g to be the transfer function of a stable system is that all poles of g have negative real parts. ·Assume now that the system described by g is stable. Then in view of Re sk < 0, k = 1, ... , m, all terms in the sum (10.3-12) except the first tend to zero as T-+ oo, and after a sufficiently long time (which in practice may be a matter of microseconds) all that remains is the steady-state response F 2(T) = g(iw) e;"'".

This simply is the input, multiplied by a constant factor that depends on w. The function w-+ g(iw) is called the frequency admittance of the system. It is customary to write g(iw) = a(w) e;"' O.lf the

F(t)

Fig.10.3g.

u

gravitational constant is denoted by ')1, the differential equation of the motion is ,.,.A cit'+ P.'Y sin«= F( 1'), which for small 1«~>1 is approximated by (10.3-16)

Letting o - e t/J, Fo---e f and taking transforms, we get p.As 2 t/J + P.'Yt/J = f(s), 1 t/J(s)= p.A(s2+1'/A/(s).

(10.3-17)

234

INTEGRAL TRANSFORMS

Thus 1

1

g(s) := JLA s 2 +-y/A

is the transfer function of the system. Thus the shock response of the pendulum, that is, the response to a force of short duration at T = 0 and integrated strength 1, is G(T) :=

~sin fiT.

IL v-yA

A

(10.3-18)

This result can be confirmed as follows: Let the impulse have duration E. Because force equals mass times acceleration,

Thus as E-+0, 1

'(0+)=-. JLA

(10.3-19)

This indeed is the initial condition satisfied by (10.3-18). If the homogeneous pendulum equation is solved by means of Laplace transforms under the initial conditions (0) = 0 and (1 0.3-19), there results JLA(s 2 4p 2 'Y. Determine the shock response of the circuit shown in Fig. 10.3i.

6(r)

G(T)

Fig.l0.3i.

INTEGRAL TRANSFORMS 4.

A weakly damped (p 2 0 at which the current in the circuit passes through zero. /l(t)

'Y

Fig.10.3j.

5. The system

X"-X+Y+Z=O, X+Y"-Y+Z=O, X+Y+Z"-Z=O is to be solved under the initial conditions X(O) = 1, Y(O) = Z(O) = X'(O) = Y'(O) = Z'(O) = 0.

6.

Two identical pendula (mass p., length A) are coupled by means of a spring of strength K. Assuming the angular displacements ~o 2 to be small, their motion is governed by the differential equations p.A r = -p.yt- KA (t-2), p.A ~ = - JL y2 + KA ( I - 2)

7.

( y := gravitational constant). Determine the motion of the pendula when the second pendulum is at rest initially and the first is released from the amplitude a with initial velocity zero. (For compact notation, let (J) 2 := yf A, /3 2 := K/ A.) A number of n -1 beads of equal mass are attached to a taut elastic string of length 1 at equidistant points. The beads are initially at rest. At time T = 0 the first bead is given an impulse perpendicular to the string. Determine the subsequent motion of all beads. [The differential equations of motion are

XZ +A (2Xk- Xk-t- Xk+t) = 0, k = 2, 3, ... , n - 2,

x:-1 +A (2X,_t- X,_2) = 0, where A is proportional to the tension of the string and inversely proportional to the mass of a bead.]

CONVOLUTION

237

8. Let F be a continuous bounded function on the real line. It is well known that the solution of the initial value problem for the heat equation in an unbounded one-dimensional medium, (T>O, -OOF(p )G(u) dp du,

where Rw denotes the rectangle 0 < p, u < (J). Because the double integral converges absolutely, we may integrate over the triangle Sw : p > 0, u > 0, p + u < (J) instead and let (J)-+ oo. Using the variables T := p + u and p in place of p and u, we have

H

e-(p+u)sF(p)G(u)dpdu

s~

=

L"' e-s,-{f F(p)G(T-p) dp} dT

=

1"' e -s'"X(T) dT,

and the limit as (J)-+ oo (which is already known to exist) by definition of the Laplace transform equals x(s). • A modification of this theorem, which is somewhat more difficult to prove, states that if f(s) converges absolutely and g(s) converges simply for s = s0 , then x(s) converges at least simply for s = s 0 , and (10.4-3) holds for s = s0 and for Res> Re s0 • The convolution theorem must be regarded as yet another operational rule for the I£ transformation, which should be added to the eight rules given in § 10.2. The dual rule, which concerns the image of the ordinary product FG in the image space, may be established as a by-product of results on the Fourier integral theorem, see §10.6.

242

INTEGRAL TRANSFORMS

It has already been mentioned that the convolution of two functions in .{} obeys the commutative law. It is trivial that the distributive law holds, and it is easy to see that the associative law is likewise valid. That is, ifF, G, He 0, then F

* (G * H) = (F * G) * H.

This can be proved elementarily by an appropriate change of variables. If the transforms f e---o F, g e---o G, h e---o H have a finite abscissa of absolute convergence, then the following short proof based on the foregoing theorems is available: Both 'functions F * (G * H) .and (F * G) * H have the Laplace transform fgh. By Theorem 10.4a they are both continuous for T>O, thus by Theorem lO.lh they are identical. We turn to applications of the convolution theorem. The first application expresses the theorem in the language of system theory. A system with transfer function g and input Fo-e f produces an output that in the image space is represented by f(s)g(s). If g is an :£ transform, then the corresponding original function is the shock ·response of the system. Theorem 10.4b thus yields: THEOREM 10.4c

The output of a linear system is the convolution of the input with the shock response. EXAMPLE

1

The pendulum considered in Example 4 of §10.3 has the shock response

G(T):=

~sin.j!.T. A

IJ.Y-yA

Consequently, the motion of the pendulum caused by an arbitrary forcing function FeO is given by

(T) =

~

,.,_ v -yA

r 0

F(u) sin·fi (T-u) du. A

This result agrees with what one obtains classically by the method of variation of constants.

The following illustrations of the convolution theorem are more special.

CONVOLUTION EXAMPLE

2.43

2

Which original function corresponds to f(s} := s- 112 (principal value}? Because -n

s

1 n-1 (n-1}!T

...---a

for positive integers n, it seems reasonable to try F(T) := CT- 112 , where the constant c is yet to be determined. Under this assumption the following chain of correspondences holds: l s- 112 · s- 112 e--o F * F(T} = c 2 1 1 o--------e-= du = c 21r.

JT

s

0

../u(T-u)

Because c < 0 would yield a negative Laplace transform there follows c =

1r -

1 o----e ..../; . r ;pnnc1pa , ( " " va l lue }, J;.

112 ; thus

(044} 1 . -

an important correspondence that is generalized in §10.5. By Rule V (Integration) we have for n = 1, 2, ... (10.4-5) EXAMPLEJ

Which original function corresponds to 1 f(s) := ../s2+ 1?

(The square root for whichf(s} -s- 1 for

lsi large is to be taken.} We have

f(s} = (s- i}-1/2(s +i}-1/2,

where the roots have their principal values. By (1 0.4-4) and Rule VIII (Damping), (s-i)-112- 0

);e;T7 -112,

(s+i}-112- 0

);e-;T7 -112.

Thus by the convolution theorem the original function corresponding to f is F(T} := _!_ 1T

r

e-;.

0

vu

~du. T-u

Letting u := [(1 +~}/2]T we obtain F(T} =1 1T

J1 -1

-*_ d~. _e_

.J1-e

INTEGRAL TRANSFORMS

244

The integral on the right is recognized as Poisson's (or Parseval's) integral for the Bessel function ] 0 (problem 3, §4.5). We thus have found the important correspondence (10.4-6) The images of Bessel functions of arbitrary order, and of various combinations thereof, are computed in §10.5. EXAMPLE4

Many curious integrals can be evaluated by means of the convolution theorem. From (10.4-6) and 1 )2 1 ( -fs2+ 1 = s2+ 1 ~sin T

it follows that the convolution of ] 0 with itself equals sin T,

r

lo(u)lo('T -u) do-= sin 7".

EXAMPLE

5

(10.4-7)

Abel's integral equation

Let F be continuous for 'T;;;. 0 and differentiable for 'T > 0, F(O) = 0, and let F' En. We wish to solve Abel's integral equation for an unknown function G, 'T ;a.O.

(10.4-8)

Assuming the existence of a solution G En, let F c--. f, G o--e g, and apply G( T)

~to (1 0.4-8}. The expression on the left may be regarded as the convolution of with T - 112 • Thus the image of (10.4-8) is

X

.J;

g(s) .Js = f(s),

which is readily solved for g: 1 1 ..r;,. g(s) = - .f"st(s) = - - sf(s).

J;

'TT.Js

Because sf(s) e--o F'(T) + F(O) = F'(T), transforming back yields (10.4-9)

245

CONVOLUTION

This formula was derived under the hypothesis that a solution exists in n. It thus remains to be verified that the function G defined by (10.4-8) actually is a solution. The verification is simple: Being the convolution of two functions in 0, G is in 0. We thus may take Laplace transforms. Retracing the foregoing steps, we find that G actually solves (10.4-7). PROBLEMS

1. For nonnegative integers m and n, let I(m, n) :=

2.

r

Tm(l-T)" dT.

The integral /(m, n) may be regarded as a convolution. Use this fact to evaluate I(m,n). Let FEn. A form of Taylor's theorem implies that the n -fold indefinite integral ofF, p. The integral along the circular arc is bounded by

and thus tends to zero for p-+ oo. The integral along the real axis tends to Euler's integral for the f function. We thus have obtained THEOREM 10.5a

If Re "

> -1, the correspondence , T

f(v+l) C>-.

s

"+1

holds for Res> 0, where s"+ 1 has its principal value.

(10.5-1)

248

INTEGRAL TRANSFORMS

If v=n, n=O, 1,2, ... , then in view of f(n+1)=n! (10.5-1) is in accordance with the elementary correspondence (10.2-ll). For v = n +!, f(n + !) = J;@no and we obtain (1 0.4-5). We now shall make an algorithmically potent application of Theorem 10.5a. Still assuming Re "> -1, suppose a function FE 0 is for T > 0 represented by the infinite series 00

F(r) = r"

L

anTn

(10.5-2)

n=O

with complex coefficients a"' and suppose we wish to calculate f := 9!F. Applying the !£ transformation term by term and using Theorem 10.5a formally yields 00

g(s)=

L

anf(v+n+1)s-"-n-l.

(10.5-3)

n=O

For what s (if any) does g(s) representf(s)? The answer is not clear, because (1) the Laplace integral formed with F is improper, both at its upper and (if Re v < 0) at its lower limit, and (2) the convergence of the series in the integrand (unless the series terminates) is not uniform over (0, oo). However, the following result grants everything that can be hoped for. THEOREM 10.5b (Hardy's theorem)

Let the series g(s) be convergent for somes= s0 >0. Then the series (10.5-2) converges for all T > 0, and g(s) = f(s) := 9!F(s) for s = s0 and for all s such that Re s >so. Proof.

It is convenient to write bn := anf(1 +v+n)

throughout the proof. The series g(s), apart from the factors_,_,, is a power series in s _,. Because it converges for s = s0 , it has a positive radius of convergence, lim suplbnl 11n 0. Applying 2 termwise we get co

F(T) = ~

(-1)"(4a)"'Tn-l/2 (2 )I

n .

n-0

o----e

=

eo L

n-o

(-1)"(4a)" f{n + 1/2) (2n)! sn+l/ 2

(-a)"

1

.J;

n!

sn+l/2

..;;

..!;-- - - = - e -a/s

'

where we have used f(n + 1/2) = J;{~)n and (2n)! = 4nn !(~)n. We thus have (10.5-7)

EXAMPLE2

To compute the 2 transform of where lv is the Bessel function. The series Tp.+v

T"fv(T)=2v

00

(-T 2/4)"

L If{ ) n-on. v+ 1 +n

shows that FEn for Re (/1- +II)> -1. Hardy's theorem yields the transform f(s)=..!._ ~ (-1/4)"r(P-+11+2n+1)

2v n-O n!f(11+1+n)s"+v+l+ 2n.

We use the relations f-l.

(10.5-10)

2

-1. By the same quadratic transformation, there now results T

-1 Jp(T)o---e

1 11(s +.Jt +s 2)"

, Re11>0.

(10.5-11)

EXAMPLEJ

Let a, 'Y· 11 be complex numbers, Re 11 > -1, 'Y ~ 0, -1, -2, .... Applying Hardy's theorem to the confluent hypergeometric function

we obtain the following representation for f := I£F:

f(s)= n-o('Y)n ~ (a)n f(ll+1+n) n!SP+I+n

=

f~P~Il)

2F1

(a, 11+1; 'Y; ~).

(10.5-12)

The series converges for Is I> 1 and (unless it terminates) is singular at s= 1. Thus f, in general, cannot be continued into any open right half plane containing Res = 1 and is represented by the series in the whole set C(F).

254

INTEGRAL TRANSFORMS

Using Rule VIII (Damping), we find for Whittaker's function (9.12-14)

the correspondence p-

T

1

M ..,. (T)

o-------e f(v + #£ + 1/2) (s + 1/2t+,..+t/2

·2F 1(1£ -K + 1/2; v + #£ + 1/2; 21£ + 1; s + ~

12).

(10.5-13)

The function

(10.5-14) likewise a solution of the differential equation (9.12-13), is known as the Whittaker function of the second kind. If Re(v±,.)>-!, then ,p-I W,.,.(T) is inn and has the image r(-2,.)f(#£ + v + 1/2) (s + 112 )-,..-p- 112 2r(-,. -I(+ 1/2) · 2F 1(1£ -K +1/2,1£ +v+1/2; 21£ +1; s+~/2) +a similar term with#£ replaced by-,.. This simplifies by the linear transformation (9.9-26), and there results T

P-tw ( )o---er(p.+v+1/2)r(-,.+v+1/2)( + 1/ 2)-,..-p- 112 "ol
-1 ("'

j_ e o

-sT], ( ) ,T

d

T

= (v"i+7S )" ~ v1+s-

SOME NONELEMENTARY CORRESPONDENCES

255

for all s >0. From the known asymptotic behavior of J,(T) (see §11.5) we know that the integral exists for s = 0. By the Abelian theorem lO.lf we conclude that its value equals its limit as s -+ 0 + ; that is, we have

f"

J.,(T) dT = 1,

Rev> -1.

(10.5-16)

From (10.5-8) we deduce in a similar manner, using the Gauss formula (8.6-6) to evaluate the hypergeometric series at s = 0, the more general formula

2"r(v+~ +1)

oo

I

T"J.,(T)dT=

o

r(v-~+

1 ,

(10.5-17)

)

valid for Re #£ -1. The so-called discontinuous integrals of Weber and Scbafheitlin,

ioo T"J,.(aT)J.,(/3T) dT, although more difficult, can be evaluated in a similar manner; see Problems

10-13. The convolution theorem (Theorem 10.4b) is a further source of integral relations. For instance, let v F.,(T) := -J.,(T) (Rev> 0). T

Then by (10.5-11) the transformf.,.-..oF., satisfies

f,(s )/,. (s) =f.,+,. (s ). There follows for all T > 0 V#£

I., 0

J.,(u) J,.(T-u) ..~ U

(

uU = V + P,

--

T-U

)J.,+,.(T) .....;.__=..;....;.

(10.5-18)

T

(Re #£, Re 11 >0). From (10.5-10) there follows in a similar manner

r

J,(u)J,.(T-u) du

(Re#£, Rev,

r

Re(v+~£)>-1),

=

r

Jo(u)J,.+.,(T-u) du

(10.5-19)

and from (10.5-9)

u"J.,(u)(T-utJ,.(T-U)du

(10.5-20) _ r(v + 1/2)r(#L + 1/2) .,+,.J, ( ) - r(v + #£ + 1/2)r(l/2) T ll+l£ T •

256

INTEGRAL TRANSFORMS

Hardy's theorem permits the calculation not only of Laplace transforms, but also of inverse Laplace transforms. Whenever a Laplace transform can be extended to a function of the form s _., g(s ), where Re "> 0 and g is analytic at infinity, the following result, which is a mere reformulation of Theorem 10.5b, permits to find the original function. COROLLARY lO.Sc

Let Re v > 0 and let 00

f(s)=s-"

L

(10.5-21)

CnS-n,

n=O

where the series converges for lsi> p, and where (if v is not an integer) larg s I< 'TT. Then f is an analytic continuation of .!£F, where F is the original function defined for all T > 0 by

F(T) := ~ n=O

EXAMPLE

Cn

(10.5-22)

-r"+n-1

f(v+n)

4

Let a > 0, and let a)"

-1 -afs_ -1 ! (s)•.- s e -s /~o 7 s . co

~

Then f(s)

(

-n

•--O F(T) where

More generally, if Re v > -1, (10.5-23)

The method of inversion implied by Corollary 10.5c requiresf(s) to have a very special behavior at infinity. We therefore mention another method of inversion by series expansion that does not require any special behavior at oo and which is suggested by certain results of this section. Assume the Laplace transform f is analytic in Res > 0 (this can always be achieved by a translation s ~ s +a). The Moebius map s-1/2

t:s~z := s+1/2

257

SOME NONELEMENTARY CORRESPONDENCES

maps Res> 0 onto

lzl < 1. Therefore the function g(z) := /(t[-IJ(z)) =

is an:lytic in

t(! 1 +z) 21-z

lz I< 1 and can be expanded in powers of z: 00

g(z) =

I

anz".

n=O

There follows oo

f(s) = g(t(s)) = n~O a,

(s +-1/2)" 1/2 ' S

Res>O.

If original functions Fn such that F,(T)

s -1/2)" o--• ( --

s+1/2

were known, and if (as in the case of Hardy's theorem) term-by-term application of .ft were permissible, the function 00

F(T) := ~ a,F,(T) n=O

would have the Laplace transform f, and thus would be the desired original function. Clearly, the functions F, do not exist, because their presumed image functions do not vanish at infinity. However, by specializing parameters in (10.5-13) we find 'T

-1/2

Mn+I/2.o('T)

1

(

1

o--e s + 1/2 2FI -n, 1; 1 ; s + 1/2

)

(s-1/2)"

= (s + 1/2)"+!· The function on the left is related to the Laguerre polynomial L, (see §2.5 for definition and other properties); more precisely, 'T

-!/2M ( ) n+I/2,o 'T

= e -.,./2Ln ( 'T ) •

We thushave found: If

(s+1/2)f(s)=

oo L a, (s- -1/2)" 112 ,

n=O

S+

( 10.5-24)

INfEGRAL TRANSFORMS

258

then f

= 2F, at least formally,

where 00

F(r) := e-"12

L

anLn(r).

(10.5-25)

n=O

The following result concerning the validity of the foregoing operations is proved in Doetsch [1950], p. 301: THEOREM lO.Sd

Let f be analytic in Res >0 and be represented there by (10.5-24), where L.ian 12 < oo. Then f is the Laplace transform of a Lebesgue -integrable function F that is represented (in the L2(0, oo) metric) by the series (10.5-25). EXAMPLES

Let us pretend to be ignorant of the original function corresponding to

f(s) := s + 11/2 exp(- s +~/2) (a> 0). To expand (s + 1/2)/(s) in powers of

s-1/2 z:=s+1/2' we note that (s + 1/2)- 1 = 1- z. Consequently, ~ (az)". (s + 1/2}/(s) = e -a(l-zl = e-a t.... n-O n!

Clearly, the coefficients in this expansion satisfy the conditions of Theorem 10.5d. Thus f = :£F where

On the other hand, by (10.5-23) and Rule VIII (Damping} f(s) corresponds to e-T121 0 (2.J;;,). We thus have proved HiDe's expansion

(10.5-26)

PROBLEMS 1. Let Re 11- > 0, Re v > 0. Evaluate

r

7'"- 1 (1-rt-l

dr

by convolution and thus obtain a new proof of Theorem 8.7a.

259

SOME NONELEMENTARY CORRESPONDENCES 2.

For Re v>O and T>O, let

I

T

y(v, T) := 0 u .,-] e -a du, an incomplete r -function. Prove

y(v, T) o----e f(v) 3.

(1 +s)-•

s

.

Let FE 0 possess a generalized derivative F' E 0 (see§ 10.2). Then the integral equation

r

fo(T-u)X(u) du = F(T)

is solved by

4.

Let F(T) o---e f(s). Show that

1

1

T

G (T) :=

F(u)J0 (2.Ju(T-u)) du

1

o---e;t(s +;)·

[Interchange order of integrations and use (10.5-23).]

5. Establish the correspondence J.(2.J;;;.)J.(2J{i;) o----e; e-I•J. (2~) -s- ,

1

where J.(z) denotes the modified Bessel function of the first kind, I.(z) := e-i""121.(iz ).

6.

For T>O andRe p. >

-!,

1) -T T~'-K1'- (T)-2~'-f( 1J. + e

~

i., ( n=O

1/ 2 -p.),. L (2 ) 112 +p. )n+l ,. T · (

[K~'- (z) = modified Bessel function of the second kind, K (z) :=:!: L~'-(.~)-J~'-(z).] ~'2 SID#J.'TT 7.

If L,. denotes the Laguerre polynomial, prove by means of Laplace transforms

n = 0, 1, 2, ....

260 8.

INTEGRAL TRANSFORMS

Establish the generating function of the Laguerre polynomials, exp( -

...!!.__) 1-z

1-z

co

L

z"Ln(T),

n=O

lzl {3 > 0, and let Re(#£ + 11 + 1)> Re ,\ > -1. Show that

1co T-AJ,.(aT)Jv({3T) dT Wr((#£ + 11-,\ + 1)/2)

11.

(Continuation) If a> 0 and Re(#£ + 11 + 1) >ReX> 0, show that

1co T-AJ,.(aT)fv({3T) dT (a/2)A-Ir(,\)r((#£ + 11-,\ + 1)/2) 2f((,\ + 11- #£ + 1)/2)f((,\ +II+#£+ 1)/2)f((,\ +#£-II+ 1)/2) 12.

(Continuation) As a special case, show that for a >0, {3 >0, Re #£ >0

I

co1

0

_1

(f!.)"'.

{3 :so; a,

1 2#£

(a),. {i '

/3 ";;!!=a.

2#£ a

-J,.(aT)J,.({3T) dT = T

261

THE FOURIER INTEGRAL

13. (Continuation) Deduce Weber's formulae

0,

a >{3>0

oo,

a={J>O

.JBC;;l' {3 a 1 Ja2-p2'

f"'

l 0 (a-r) cos {3rd-r =

{3>a>O

a>{3>0

oo,

a={3>0

0,

{3>a>O

as special cases of the Weber-Schafheitlin integral.

§10.6. THE FOURIER INTEGRAL Of the two general methods for inverting the .:£ transformation we have discussed, the first (Corollary 10.5c) is valid only under the stringent hypothesis that .:t:F (apart from a factor s -") is analytic at infinity. The second method (Theorem 10 .Sd) is subject to conditions that are not easily checked; also, it is too specialized to allow much flexibility. In §10.7 we derive an inversion formula that is valid under the sole assumption that the Laplace integral converges absolutely for some s, that is, that fJF < oo. This includes all cases of practical interest. The inversion formula mentioned is based on a form of the Fourier integral theorem, the discussion of which is the main goal of the present section. We begin by a brief informal discussion of the theorem. After a short discourse on Fourier series, we then prove a version of the integral theorem that is sufficient for inverting all absolutely convergent.:£ transforms of functions in .0. I. Fourier theory: Informal discussion Let F be a complex-valued function of period 2A defined on the real line. Assuming that IFI is (possibly im!" 'lperly) Riemann integrable on [-A, A], the numbers

n = 0, ±1, ±2, ... ,

(10.6-1)

INTEGRAL TRANSFORMS

262

called Fourier coeftidents of F, can be defined, and the series (10.6-2) called the Fourier series ofF, can be formed. The theory of Fourier series is concerned with the question under what conditions the Fourier series of a function F converges, and to what sum. It is known that the convergence of the series at a point T depends only on the behavior of F in the immediate vicinity of the point T. IfF meets a certain condition (C) at T (see II below), then the series converges, and its value is F(T): F(T) =

oo n];_oo

an exp

(in'TTT) A .

(10.6-3)

We now drop the condition that F be periodic. Instead, we demand

f_:

IF(T)l dT 0 and for real w·we define

J A

GA(w) :=

e-;_.F(u) du,

(10.6-5)

-A

so that n =0, ±1,[*=2, ....

If -A< T 0 such that IF(T+u)-F(T)I :o;;;;p.lul 11

for

lui :o;;;;a.

We mention without proof that (C) also holds ifF is of bounded variation 3 3 A function F is said to be of bounded variation on an interval [a, fJ] if there exists a constant p. such that, whatever subdivision a =To0 be given. We

write (10.6-15) and choose 71 such that

268

INTEGRAL TRANSFORMS

This makes the first and the third integral in (10.6-15) less than E for all values of w. The middle integral becomes less than E for all sufficiently large lwl by the Riemann-Lebesgue lemma. This shows that jG(w)I 0, then F( T) = -21 lim

f''

e ;..,TG (w) dw.

1T 71_,."" j_71

(10.6-16)

The theorem is not perfect because the roles played by F and G are not symmetric. Whereas G exists as an ordinary improper integral,

(where 71 1 and 71 2 approach their limits independently), the integral in (10.6-16) exists only as a principal value integral, where the upper and the lower limit tend to ±oo symmetrically. One also writes (see §4.8)

l~~J-7171

71

=: PV

J"" . -00

It can be shown by means of examples that under the conditions stated the

ordinary integral

need not exist. Proof of Theorem 10.6d.

Let

F 71 (T) := -21 1T

f 71

ei=G(w) dw.

L71

In view of (10.6-12) the improper integral defining G converges uniformly with respect to w. Thus we may interchange integrations to obtain [ 71 e i={ Lao f"" e -i6JO'F(u) du}dw F71 (T) = 217T j_ 71

269

THE FOURIER INTEGRAL

f'"'

=1-

21r L"'

F(u){

f"'

t.,

e;,(.--ul

dw} du

= _!_ f"" F(u) sin[71(T-u)] du '7T

=1 -

T-u

-oo

J,"" {F(T+u)+.F(T-u)}--du. sin 71U u

0

'7T

N-om this we subtract the relation 1 F(T)='7T

J. "" 2F(T)--du, sin 71U u

0

which is a consequence of (10.2-15). This yields 1 F.,(T)-F(T)='7T

J."" S(T, u)--du, sin 71u u

0

where Sis the integrable function defined by (10.6-10). To show that F.,(T)-+F(T) for 71-+00, let E >0 be given. We write

L"" =

r r f". +

+

(10.6-17)

and choose x > 1 such that

J."" IF(T+u)+F(T-u)l du 1). This makes

II""

S(T, u)sinu11u dul .;;; 11"" {F(r+u)+F(r-u)}sinu71U dul+ 12F(r) 1""sinu71U dul ;:;;2E.

The middle integral in (10.6-17) tends to zero for 71-+ oo by the RiemannLebesgue Lemma, and the first tends to zero by hypothesis (C). This shows that the sum of all three integrals is 0 be two parameters, and let F(T) := exp(-'1TAT 2 +2iZT).

272

INTEGRAL TRANSFORMS

This clearly satisfies (10.6-12) and condition (a) of Theorem 10.6e. For the Fourier transform G of F we find by completing the square in the exponent G(w) =

L:

exp( -'ITAT 2 + 2izT- iwi) dT

=exp[ _ _!_(z-~) 2 ] 'TI'>< 2

J""

-ao

exp[ -'ITA(T-.!.:._+~) 2 ] dT. 'ITA 27TA

By Cauchy's theorem, the path of integration may be shifted to lm T = Re(z/ 'ITA - w/2'1TA. ), and we get G(w) = exp[- :A. (z- ~Y]

L:

exp(- 'ITA.s 2 ) ds

We apply the summation formula (10.6-21) where 71 = 1 and series on the left,

T

= 0. The resulting

ao

O(z,

A) :=

L exp(-7TA.k 2 +2ikz),

(10.6-22)

k=-00

is known as a (J function. (The parameter A. is usually denoted by -iT.) The series on the right can likewise be expressed by the above (J function, and we obtain Jacobi's identity ()(z, A.) =

For z

..!.. exp (- ~)o (!.).). Ji.

'7TA

(10.6-23)

iA. A.

= 0 there results in particular 8(0, A)= (1/Ji.)8(0, 1/A) or (10.6-24)

Setting q := e-..-", this relation expresses the value of the series ao

f(q) :=

L

q",

(10.6-25)

n=-oo

(where jqj < 1) in terms of f(q 1) where log q log q 1 = '17" 2 • Thus if q is real, the series f(q) (which is of importance in the theory of elliptic functions, and in the theory of partitions, see §8.2) can always be evaluated numerically with a value of q such that jq I< e _, = 0.04321 ... , where the convergence is very rapid indeed. PROBLEMS 1.

Find the Fourier transforms of the functions (a)

THE FOURIER INTEGRAL

(b)

F(r) :=

2.73

{a(l-l·rl) f3 ' 0,

2. The relation (10.6-13) holds for 1 a +r

F(r) :=

-2--2,

G(w) := ~ e -I.. Ia a

where a >0 (see example 4, §4.8). Use this result to express co

1

k~co a 2 +(kT/ +rV by a rapidly converging series when 11 > 0 is small. (Perform numerical tests for a= 1, T = 0, T/ =2-m, m = 0, 1, 2, ... ). 3. For real a, p > 0 show that if P/211' is not an integer, PV

~

~

n=-co

sin[(a+n)p] a +n

sin[(2m+1)11'a] SID 'TT'a

11'--~-.--~~

where m := [p/211' ]. Deduce that for p < 211'

I

PV

n--co

sin[(a+n)p]=Jco sinpr dr='TT'. a +n -co T

[Compare problem 1.] 4. Show that for 11 > 0 co (sin k 71) 2 =11' { 1+2m--m(m+1) 11' }, L -k.,

k=-CO

T/

T/

where m := [ 11/ 11' ]. Deduce that for 11 < 1r

I k--co

(sin k11) 2 = fco (sin r) 2 dr. k71 j_co T

Also check the cases 71 = 11', 71 = 11'/2. [Compare problem 1.]

5. Poisson's integral (9.7-29) shows that G(w) := 10 (w)

is the Fourier transform of

1 1 { F(r) := -;; ../1-r 2 '

0,

274

INTEGRAL TRANSFORMS

Conclude that for a > 0, 1 +2

T ;;.: 0

ao

1

10 (an) cos(anT) = 2 L .J , n-1 k a 2 -(27Tk +aT) 2

L

where the sum on the right involves those integers k for which the square root is real. 6. For some functions F satisfying the hypotheses of Theorem 10.6e it may happen that for certain ., > 0 T/

kfao F(k71)=

t:

F(T)dT,

compare the problems above. Explain, and state a general result to this effect. 7. Let a > 0. From the fact that ao 1 .

J-ao

~e--dT=2K0 (aiCdl> a T

(K0 =modified Bessel function of second kind), deduce that for y real, y ±271', ±471', ...

¢

0,

The following problems are concerned with the approximation of the integral

[:=I:

F(-r) d-r

(-oou0 , the integral around the curve is zero. By Theorem 10.7a, the contributions of the horizontal pieces tend to zero for 11-+ oo, and it thus follows that

Remark 2. By virtue of its derivation, the inversion formula also holds for r < 0 and then must yield the value zero. This also can be seen directly, as follows. By Cauchy's theorem,

i u~ip e•"t(s) ds = i where

fp

e•"t(s) ds,

rp

u-•p

denoteS the circular arcs =u+pei, -1Tj2:s;.tjJ :!S;.TT/2. Let P,p

:=max 1/(s)j. serp

If T < 0, the integral on the right as in Jordan's lemma is estimated by

By virtue of Theorem 10. 7a, f.Lp-+ 0 as p-+ oo. Thus for T < 0,

i

u+iOO

PV

u-ioo

e•"f(s) ds

= 0.

The complex inversion formula features, in a sense, an explicit representation of the original function Fin terms of its image function f. No undue confidence should be placed, however, in the immediate usefulness of this formula for the purpose of determining the original function by means of numerical integration, because (i) a separate integration would be required for each value of r; (ii) for large r, the integrand oscillates rapidly, thus the integral is difficult to evaluate numerically; (iii) the convergence of the improper integral is determined by the speed with which f(u ± iw) tends to zero for w-+ oo, which may be low. In the following subsections the inversion formula is applied to deriveunder suitable additional conditions on f-several algorithmically more useful representations of F in terms of its image. For the first time in this chapter, use is made of the fact that because!£ transforms are analytic, their analytic continuation into any given region, if it exists, is unique. Some additional inversion formulas based on this fact are derived in §10.9.

280

INTEGRAL TRANSFORMS

ID. The Heaviside Expansion Theorem Let FE 0, f3F < oo, and let I := 2F satisfy the following conditions: (I) I can be continued analytically into the whole complex plane, with the exception of finitely or infinitely many isolated singularities, say at the points s~o s2, .... (II) There exists a sequence of circles r n : Is I= Pm where Pn -+ oo, such that the quantities /Ln :=

sup ll(s)j

(10.7-7)

ser"

tend to zero for n -+ oo. (The fact that I is a Laplace transform merely guarantees that l(s)-+ 0 if s tends to infinity in some right half plane; we now postulate that l(s)-+ 0 on a certain sequence of whole circles.) w

0

Fig.l0.7a.

Let u > f3F, and let f n intersect the straight line Res = u in the points u ± iwn. We apply the residue theorem to the closed curve consisting of the straight line segment r~ joining u ± iwll and of the part

r: of r

n

lying to the

THE LAPLACE TRANSFORM AS A FOURIER TRANSFORM

281

left of Re s = u. This yields

-21 . 11'1

f.

r~+ r::

e'"f(s) ds

= L

l•k l 0, F(T):=Isinc.JTI,

T;;;.:O,

This satisfies (C) at all -r. We know (cf. Problem 9, §10.2) that 1 + eB.nrlw

(A)

F(T) 0--•---z----+21 -S 0. Thus the hypotheses of the expansion theorem are satisfied. In view of res {

2

S

c.J 1 +e-....'"'} 1 2 2 -nr/w =- - - - 2 +c.J 1-e s=2iwk '1T 1-4k

(10.7-9) yields after combining conjugate terms 2 4 1 lsin c.JTI = - - - L - 2 00

'1T

'1T k-1

4k -1

cos 2c.JkT.

IV. A Real Inversion Formula Here we assume that the .!£ transform I= .!l!F of an original function F(f3FO. We integrate e'.,.f(s ), where -r > 0, along the simple closed curve f shown in Fig. 10.7b. Because the integrand is analytic insider, the value of the w

r

a

Fig.l0.7b.

284

INTEGRAL TRANSFORMS

integral is zero by Cauchy's theorem for ailS > 0 and all p > 0. Letting 8-+ 0, the integral along the small semicircle tends to zero by (10.7-11), and the integrals along the horizontal parts by hypothesis (iii) tend to

r

e-={f(ei'"u)-f(e-i'"u)} du.

As p-+ oo, the integrals along the outer circular arcs tend to zero by (10.7-12), using Jordan's lemma. The integral along the vertical part of r tends to 21TiF(r) at every point T where Fsatisfies condition (C). Thus in the limit we get THEOREM 10.7d

Let FED., {3p 0 where F satisfies condition (C),

F(r)= 2 1 . 1Tl

(l() e-={f(e-;'"u)-f(e;'"u)}du.

Jo

(10.7-13)

If f(s) is real for real s > 0, then by the symmetry principle 1/ (21Ti) {f(e-;'"u)-f(e;'"u)} is real for u>O. Thus the above formula may be regarded as a "real" inversion formula. EXAMPLE

4

For Rea >0, the function F(T)

:=Ta-l

is in O.lts image

f(s)= f(a)

s"

satisfies the foregoing conditions for 0 u 0 ; moreover, the asymptotic behavior ofF is determined by the rightmost singularities of its Laplace transform. Do such statements also hold for original functions with nonrational transforms? The answer is, only under additional hypotheses on I or on F. To show that additional hypotheses are necessary, let u be any real number, and let G be any positive, monotonically increasing function defined for 'T > 0. G may tend to infinity with arbitrary rapidity. We shall construct a function Fen such that I := .!l!F is analytic for Re s > u, although the statement F(T) = o(G(T)) is not true (nor is it true for any

THE LAPLACE TRANSFORM AS A FOURIER TRANSFORM

287

function equivalent to Fin the sense of §10.1). This shows that there is, in general, no connection between the growth of an original function and the abscissa of analyticity of its !£ transform. The trick of our construction consists in letting F be of the order of G merely on a sequence of short intervals. We may assume u < 0 without weakening our assertion. Let 'Yn := G(n ), . {1 ,')In-12-n e '"'}' n = 1,2, ... , 8n:= mm and define F(T)

n=1,2, ... ,

n ~T 0 be given. By the hypothesis of uniform convergence, there exists 71 such that

288

INTEGRAL TRANSFORMS

for all -r > -r0 • By the Riemann-Lebesgue lemma there exists -r 1 (depending on E) such that

i"'71 e ;,.G(w) dw < 2

1

IE"

12 17"

for all -r>-rt. This implies that IF(-r)Imax(-ro, Tt).



It should be noted that the hypothesis of uniform convergence is certainly satisfied if, as hypothesized in Theorem 10.6c,

t:

IG(w )I dw < oo.

This condition is violated in many cases of practical importance, for instance, if 0 < ,\ :,;;;; 1 for G(w) := (a + iw) -", where a > 0 and the power has its principal value. On the other hand, this function satisfies the conditions of Haar's lemma, for by integration by parts we have

1 '7

_71

1 [ e iwT( a +.zw )-A]71-,., +,\ e iwT( a +.zw )-Adw =-:lT

J."'

T

-,.,

• )-A-1 dw, e iwT( a + zw

and the limitS of the tWO termS On the right as 7l -+ 00 exist uniformly forT> 1, say. The following applications of Lemma 10.7e to f£ transforms yield two results of Abelian type. THEOREM 10.7f

Let FE n satisfy condition (C) at all large -r, and let

t:

le-UTF(-r)l d-r u 0 where u 0 < u, with the exception of a pole at s = s0 (u0 < Re s 0 < u) where f has the principal part a1 a2 ak --+ +···+ k s-so (s-so) 2 (s-so) · If f(s)-+ 0 for s-+ oo, Res> u 0 , and if there exists u., u 0 < u 1< Re s 0 , such that PV

L:

e;"'"f(u1+ iw) dw

converges uniformly for T ;;a: To, then F(T) = P(T) e 50"

+ o(eu•"),

(10.7-22)

where 1 1 k-1 P(T) := a 1 + l!a2T+ · · · + (k _ 1)!akT

Proof.

By the complex inversion formula, if (C) is satisfied at T, 1

F(T)=27Tl.PV

Iu+ioo

e 5"f(s)ds.

u-ioo

Because f(s)-+ 0 for s-+ oo, Res> u 0 , we have by the residue theorem u 1 +ioo 1 F(T) = .PV es"f(s) ds +res[eS'Tf(s)]s=so· 2 7Tl o -ioo

1 1

290

INTEGRAL TRANSFORMS

The residue equals P(T) e'o", and we thus obtain e-u•"F(T)-P(T) eut. A difficulty in the application of the foregoing theorems in concrete situations arises because there are conditions imposed on the behavior of f(u + iw) for w-+ ±oo even for values of u < Re s0 • The following theorem of Tauberian type, which apparently concerns a much more special situation, avoids this difficulty. The theorem is of a more recondite character than Theorem 10.7g; we quote it without proof from Doetsch [1950], p. 524. TIIEOREM 10.7h (Ikehllra-Wiener theorem)

Let Fen be nonnegative and monotonically (but not necessarily strictly) increasing, and let C(F) contain the half plane Res> u 0 > 0. Iff := !£F can be extended to a function analytic for Re s ;;;. u 0 save for a pole of order one at s = u 0 with residue a > 0, then T-+00.

(10.7-23)

The theorems of the present section draw conclusions about the asymptotic behavior of F from certain assumptions on f := !£F. This is the situation that occurs when problems concerning functions in n are attacked by means of the I£ transformation; see the examples given in §10.9 and §10.10. The converse problem of describing the asymptotic behavior off= IEF for s -+ oo in terms of properties of F is also of frequent interest. This may be solved by means of the Watson-Doetsch lemma discussed in §11.5. PROBLEMS

1.

When is an I an :£F? Let I be analytic in a half plane H: Res> u 0 , and let (i)

(ii)

lim l(s) = 0;

... oo seH

r:

ll(u + iw >I dw < 00 for all u > Uo.

THE LAPLACE TRANSFORM AS A FOURIER TRANSFORM

Then for all real

T

and for u > u 0 the integral 1 F(T) := -2 . 1Tl

2.

4.

f.u+iao

enf(s) ds

(10.7-24)

u-i«>

exists and is independent of u, FE 0, and f(s) = !£F(s) for s E H. Prove: If FEn is a piecewise smooth function with discontinuities, and if f := !£F, then

t:

3.

291

if(u + iw >I dw = oo

for all u. [Thus condition (ii) of problem 1 is not necessary in order that f be a Laplace transform.] Find the Fourier series expansions of the following nonanalytic periodic functions by expanding their image functions in partial fractions: (a)

F(T):=(-1)[TJ

(b)

F(T) := B:(T) (Bernoulli function)

[See Problems 8 and 10, §10.2, for image functions.] Let O 0 is given and n 0 is chosen such that lrn I< E for n > n 0 , then if k>n 0 +1

II a~ I~Eisl I n=kn

u

n=k

{~n

1

(n+l)

u}+ k2! = 3€lsl. u

If iargsi~P. then isl/o-=1/cosf:J, and it follows that the series (10.8-1)

satisfies Cauchy's test for uniform convergence. • If the series (10.8-1) converges for somes, let u be the infimum of the real parts of all such s. If the series diverges for all s, set u := oo. The symbol u is called the abscissa of convergence of the series. Lemma 10.8a implies: THEOREM 10.8b

Either the series (10.8-1) converges for nos, or it converges for all s, or there exists a real number u such that the series converges for all s such that Re s > u and diverges for all s such that Re s < u. The question of convergence for s such that Res= u remains open; various cases are possible. If the series converges absolutely for some s = s0 , then it converges absolutely for all s such that Re s = Re s0 • Thus we have the following analog of Theorem 10.1c: THEOREM 10.8c

The set of all s for which the series (10.8-1) converges absolutely, if not empty, is either the full plane or an open or closed half plane. Let u < oo, and let T denote any compact subset of Res> u. Because T can be enclosed in an angular domain of the type described in Lemma 10.8a,

DIRICHLET SERIES: PRIME NUMBER 111EOREM

295

the series converges uniformly on T. Because each term of the series is an analytic function of s, there follows: THEOREM 10.8d

The sum of a Dirichlet series represents an analytic function in the interior of its domain of convergence. A uniformly convergent series of analytic functions may be differentiated term by term. We thus have: COROLLARY 10.8e

Let (10.8-1) have an abscissa of convergence u < oo, and let co a a(s) := ~ --7. Res >u. n=l n Then

'( ) _ ~ a,. Log n• as--.t... s n=l n In a similar manner one could deduce from Lemma 10.8a the analog for Dirichlet series of the Abelian theorem 10.1f and of Theorem 10.1g concerning the behavior for s ~ oo. EXAMPLE

1

The best-known example of a Dirichlet series is the (function of Riemann defined by co 1 ((s) := L --;. (10.8-2) n=l n The abscissa of convergence is clearly 1. Thus ((s) is analytic for Res> 1. EXAMPLE

2

Consider the Dirichlet series co

f(s) :=

(-1)"-1

L -.-. n-1

n

(10.8-3)

For real s > 0 this is an alternating series with terms whose absolute values decrease monotonically to zero. Hence by the Leibniz criterion the series converges. For Re s < 0 the series clearly diverges. Hence the abscissa of convergence is 0, and the sum f(s) is analytic for Res> 0. The abscissa of absolute convergence, on the other hand, is 1. We now show that the function (10.8-3) is closely related to the Riemann ( function. Suppose Res> 1, so that the series converges absolutely, and hence its

INTEGRAL TRANSFORMS

296

terms can be arranged in arbitrary order. From 1 1 r(s)= ( 1+-+-+· .. )

~

~

~

1 1. ) + (1 -+-+-+· T ~ ~ .. ,

there follows by subtraction

) 1 1 = 2 1-•(1 +-+-+··· 2'

3'

= 21-s((s).

Thus if 2 1-• f. 1, 1 ((s) = 1-2~-.f(s).

(10.8-4)

This formula continues ((s) analytically into the half plane Re s > 0, with the possible exception of those values of s for which 2 1 -• = 1. At s = 1,/(1) =Log 2. On the other hand, 1-2 1-• = 1-e-u this tends

to zero for 71 ~ oo. At the same time,

a.,(s)~s

f"

and the identity (10.8-8) follows.

e-•"A(T) dT,



D. The Prime Number 'Theorem For T>O, let L(T) :=

~

Logn,..;-r

A(n).

(10.8-10)

By virtue of (10.8-5) and Theorem 10.8f, the logarithmic derivative of the l function then admits the representation

Res> 1.

(10.8-11)

THEOREM 10.8g ForT~oo.

(10.8-12)

INTEGRAL TRANSFORMS

300

Proof. The proof is an application of the Wiener-Ikehara Tauberian theorem 10.7h. Clearly, the function Lis nonnegative and nondecreasing. The domain of convergence of the Laplace integral formed with L includes the half plane Res> 1. The Laplace transform I :=.f£L by (10.8-11) is l(s) =- ('(s}. s((s)

We already know that ( at s = 1 has a pole of order 1. It follows that ('I ( at the same point has a pole of order 1 with residue -1, and that l(s) has apple of order 1 with residue +1. Thus (10.8-12) follows from (10.7-23) if we can show that with the exception of the points= 1, (' / (is analytic on Res= 1. This is an immediate consequence of the following two lemmas: LEMMA 10.8h

The function

1

((s}--

s-1

is entire. Proof.

For every s such that Res> 1, the function

f(z) := (1 + z )-s = e -s Log(l+z) satisfies the hypotheses of the Plana summation formula (Theorem 4.9c). Because

there follows 1

1

((s) =-+-+Z(s},

(10.8-13)

2 s-1

where Z(s}:=i('[(1+iy)-s-(1-iy)-s]

~

e

21r!

dy. - 1

Because

· }-s - (1 - 'Y · }-s ._ -s Log(l+iy) (1 + 'Y .- e - e -s Log(l-iy) = (1 + /)-s12 (-2i) sin(s Arctan y),

DIRICHLET SERIES: PRIME NUMBER THEOREM

301

the integral representing Z may be written

Jo

1

oo

Z(s)=2

(l+y 2)-'12 sin(s Arctany)

e

Z1ry

dy. - 1

Here the integrand for each y ;;a: 0 is an entire analytic function of s, and because lsin(s Arctan y >I::::;; el*12 the integral converges uniformly with respect to s in every compact set of the s plane. Thus Z(s) is an entire function, tantamount to the assertion of Lemma 10.8h. • LEMMA 10.8i

For Res= 1, s ¢ 1, ((s) ¢0.

(10.8-14)

Proof. Although somewhat artificial, the following proof, due to E. Landau, has the advantage of being very short. Let

We already know that ((s) has a pole of order 1 at s = 1. Thus the residue of the logarithmic derivative at s = 1 is -1, which implies lim EA(1 +E)= -1. E->0

(10.8-15)

If ( has a zero of order m at s0 , then

lim (s - s 0 )A (s) = m; s-~>so

this also holds form = 0 (((s 0 ) ¢ 0). Assume now, contrary to the assertionof Lemma 10.8i, that (has a zero of order m 1 ;;a: 1 at a points= 1 + iw such that w ¢ 0. Then by the above lim EA (1 + iw +E) = m 1 •

E->0

(10.8-16)

The point 1 +2iw isnopoleof (. Itmayormaynot be a zero of(; in any case, lim EA(1 +2iw+E) =: m 2 ;;a:O. E->0

(10.8-17)

Confining E to real values and taking real parts, equations (10.8-15), (10.8-16), and (10.8-17) imply lim E{3A(1 +E) +4 ReA (1 + iw +E)+ Re A(1 + 2iw +E)}

E->0

(10.8-18)

302

INTEGRAL TRANSFORMS

On the other hand, the expansion (10.8-5) yields for e > 0 3A(1 +e) +4 Re A(1 +iw +e)+ Re A(1 +2iw +e) _

~ A(n)[ 3 + 4R e n-iw + R en -2iw] . n

- - .... --n:-; n=l

But 3+4 Re n -iw +Re n- 2 iw

= 2 + 4 cos(w Log n) + 1 + cos(2w Log n) = 2[1 + 2 cos(w Log n) + cos2 (w Log n )]

= 2[1 +cos(w Log n)] 2 ~ 0; hence e[3A(1 +e) +4 ReA (1 +iw +e)+ Re A(l +2iw +e)]:::; 0, contradicting (10.8-18). Thus the assumption that ((1 + iw) = 0 for a real w ;6 0 cannot be maintained, proving Lemma 10.8i, and Theorem 10.8g. • The interest of Theorem 10.8g lies in the fact that the function L is elementarily related to the distribution of primes. For T > 0, let 1r(T) denote the number of primes not exceeding -r. Examples: 1T(2) = 1, 1T(3.5) = 2, 1T(12) = 5. Although the distribution of the primes themselves appears to be fairly irregular, the function 1r(-r) has a smooth asymptotic behavior. The following theorem was conjectured by Gauss and proved simultaneously by Hadamard and De Ia Vallee-Poussin in 1894: THEOREM 10.8j (lbe prime number theorem)

For-r-+OO, T

1T(-r)---. Log-r

(10.8-19)

Proof. We estimate 1r(-r) from below and above in terms of the function L(-r). In the sum L(-r)=

~

A(n)

(10.8-20)

only the terms for which n = pm (p frime) yield a nonzero contribution. For a fixed p these are the powers p, p , ... , p', where r is the greatest integer such that r Log p ::;:;; T or r :=

[Lo~p].

303

DIRICHLET SERIES; PRIME NUMBER THEOREM

Thus L(-r) =

L

LogpE;T

[LoT

gp

JLogp,

and by omitting the brackets there follows

We thus have (10.8-21) On the other hand, if 1 < u < -r,

or, because obviously 1r(eu) 'Y· For simplicity in what follows, we call "Laplace transform ofF" and denote by !£F what is really the Laplace transform of the function G defined by (10.9-4). Hardy's theorem 10.5b taught us how to compute the Laplace transform of certain entire functions by term-by-term integration. In the present section we are generally concerned with the connection between the growth properties of functions FE n'Y and the location of the singularities off= !£F. To state our first result we denote, for all u ;;a: 0, by wu the class of all· functions f(s) that vanish at oo and are analytic outside the circle is I= u, and outside no smaller circle. By the Cauchy-Hadamard formula for the radius of convergence of a power series, a function 00

f(s)=

L

bns-n-l

n=O

belongs to wu if and only if

THEOREM 10.9a

For every 'Y ;;a: 0, the!£ transform defines a one-to-one mapping of the set fi-y onto the set w'Y.

In other words, every FE n'Y possesses a Laplace transform that can be extended to a function/ E w'Y. Conversely, every f E w'Y is an analytic continuation of the Laplace transform of a certain FE n'Y. In the terminology that is used in the general theory of entire functions, the class n.,. for 'Y > 0 is identical with the entire functions of order 1 and type y, and for 'Y = 0 with the entire functions that are either of order 1 and of minimal type or of order < 1.

5

307

FUNCfiONS OF EXPONENTIAL TYPE

Proof. (a) Let the entire function F, represented by the power series (10.9-1), belong to fly. This means that given any E >0, there exists 1< >0 such that the function p.(p) defined by (10.9-2) satisfies p.(p) 0 for which the series converges. By the Cauchy-Hadamard formula the series converges for s > u, where u ==lim sup In !anll'n. n-+OO

By (10.9-5) and Stirling's formula, In !an I~ KJ21Tn(y +Ef(1 +0(1/n)), and it is obvious that u ~ 'Y +E. Because E > 0 was arbitrary, we have u Thus I:= .ftF e Wu where u ~ 'Y· (b) Let IE Wu,

~ 'Y·

00

f(s)

= L bns -n-\ lsi >u.

(10.9-7)

n=O

By Hardy's theorem, the restriction of I toRe s > u is the Laplace transform (in the manner of speech used here) of the entire function G(z) :=

We shall show that GenT where

I

b~zn.

n=on.

T~u.

Because

lim sup Ibn ll/n = u, n-+00

308

INTEGRAL TRANSFORMS

there exists, for every E > 0, n 0 such that n > n 0 implies lbnlno

n.

~ lb~llzln+e(u+E)Izl:s:;e 0 was arbitrary, T :s: u. (c) Let FE .O.'Y for some 'Y ~ 0. By (a), .;t maps this onto some f E wu where u :s: 'Y· By (b), f is the image of some G E .O.T where T :s: u. Because f can be the image of only one analytic original function, G = F, hence 'Y = T = u. • Let FE .0.-y. Then the Laplace transform ofF converges absolutely at least for Res> 'Y· and thus the complex inversion formula (10.7-5) holds for any vertical path of integration to the right of Re s = 'Y· However, for functions of exponential type there holds yet another inversion formula which is more convenient because the path of integration has finite length. Let p > 'Y· and let f denote the circle s = pei, 0 :s: q, :s: 2?T. For arbitrary complex z we consider the integral

f.

I(z) := -2 1 . f(s) e sz ds, ?Tl r

where f := .:tF. On f, f(s) is represented by the series (10.9-6) which converges uniformly. Substituting the series and integrating term by term, we get J(z)= 2 1 . 1Tl

~

n=O

anf.

r

S

~~ 1 e'zds= ~

n=O

anzn=F(z),

because the residue of n !s -n-t e•z at s = 0 is zn. By Cauchy's theorem, f may be replaced by any closed curve having winding number + 1 with respect to the set lsi :s: 'Y· We thus have: THEOREM 10.9b (Pincherle's Theorem)

Let FE .0.-y ('Y ~ 0),/ := .:tF, and let r be any simple closed curve encircling the set lsi :s: 'Yin the positive sense. Then for all complex z, F(z) = 2 1 . ?Tl

f.r f(s) e•z ds.

(10.9-8)

It should be noted that the inversion formula (10.9-8), in contrast to (10.7-5), does not yield the value zero for z = T, let y(c/>) denote the infimum of all real numbers a such that for all sufficiently large T > 0,

IF(eiT)I ) thus defined is called the indicator function of F. If the growth parameter ofF is y, then clearly y( c/>) :s;; y for all cJ>.It can happen that y(c/>) < y, and even that y(c/>) < 0, for certain values of c/>. EXAMPLE

2

If F(z) = e•, then forT;;;. 0, IF(e'"'T)I = eTco•, thus y(r/J) =cos r/J.

The growth indicator 'YF defined in §10.1 for arbitrary functions FEn in the case of a function of exponential type clearly equals y(O). Because we know already from Theorem lO.ld that f := !tF is analytic for Res> 'YF> Theorem 10.9a can be modified as follows: THEOREM 10.9c

Let Fen.,. where y~O. and let y(c/>) be the indicator function of F. Then f := !tF can be extended to a function which is analytic in the union of the two sets lsi> 'Y (including s = oo) and Res> y(O). lm s

'Y (0) Res

Fig.10.9a.

INTEGRAL TRANSFORMS

310

Here we recall the following question on Laplace transforms. Let F be any function inn whose!£ transform has an abscissa of convergence ap ap, and in all examples that have been encountered this half plane was the maximal half-plane of analyticity; that is, it was not possible to continue f analytically into a half plane Res> a 0 where a 0 < ap. However, we have not proved this impossibility. In fact, examples have been constructed (see Doetsch [1958], p. 38) in which the continuation is possible. On the other hand, it will now be shown that ifF is of exponential type, f := !£F cannot be continued into a half plane Res> a 0 where a 0 < 'YF = y(O). Because f is analytic at co, this means that f has a singular point somewhere on the line Res= y(O). Let F satisfy the hypotheses of Theorem 10.9c, and let f :=!£F. We express F(T), where T > 0, in terms of f by means of the new inversion formula (10.9-8). Suppose f can be continued into a half plane Res> a 0 where a 0 < y(O). Then for every e > 0, the path of integration r could be deformed into a path of finite length A lying entirely in Res :s;; a 0 +e. The maximum of le.-sl on the path of integration would then not exceed e.-. Denoting the maximum of if(s)i on r by p., (10.9-8) would then for every e > 0 yield the estimate

Because e > 0 was arbitrary, it would follow that y(O) = a 0 , contrary to our assumption that a 0 < y(O). THEOREM 10.9d

Under the hypotheses of Theorem 10. 9c, f := !£Fcannot be continued analytically into any half plane Res> a 0 where a 0 < y(O); that is, f has a singular point on the straight line segment Res= y(O), lsi :s;; 'Y·

Thus far we have been concerned only with the Laplace transform of the restriction ofF to the positive real axis. F is defined, however, for arbitrary complex values. We now apply the foregoing results to the function T;;:.O, F.,(T) := F(e;"'T), where y, Res> y(

-it/> L an e li'TI= e f(e s).

n=O

S

[Thus rule II (similarity) in the case of functions of exponential type holds for arbitrary complex p.] We conclude from this thatf(e-i"'s) can be continued analytically into the half plane Res> 'Y(cfJ ), and not into any larger right half plane. This is the same as to say that f(s) can be continued analytically into the half plane H.,: Re(ei"'s) > 'Y(cfJ ), but not into any half plane Re(ei"'s) > a 0 where a 0 < 'Y(cfJ ). Any points on the boundary of H.p such that f cannot be continued analytically into s is henceforth called a singularity of f. The foregoing holds for every value of cfJ, 0:;;;; cfJ :;;;; 211". We conclude that f can be continued into the set

S:=

U

H.,,

Ooe;;.poe;;27T

the union of all half planes H.p. The continuation is not possible into any larger set which is a union of half planes. Let C denote the complement of S. The set C is contained in the disk I:; ; 'Y. thus bounded. It is the intersection of the closed half planes H.p: Re(ei"'s):;;;; 'Y(cfJ ), the complements of the open half planes H.p. Each H.p contains all singular points of f. By Theorem 10.9d, the boundary of each H.p contains at least one singular point. Thus for fixed cfJ, H.p is the intersection of all closed half planes Re(ei"'s):s;; a containing all singular points off. The set C thus may also be described as the intersection of all closed half planes containing all singular points of f. The intersection of all closed half planes containing a given set Tis commonly described as the convex huH of T (see § 6.5 for another example). Thus Cis the convex huH of the singularities of f. We next show that the function 'Y( cfJ) is completely determined by C. Let cfJ be fixed. Then because Cis the intersection of all closed half-planes fi"', 0:;;;;"':;;;; 211",

is

sup Re(ei"'s) = seC

inf

sup Re(ei"'s).

Ooe;;l/loe;;27T sefi.,.

The last supremum is +oo for all 1/1 except for 1/1 = cfJ, where it has the value 'Y(cfJ ). There follows 'Y(cfJ) =sup Re(ei"'s ).

(10.9-9)

seC

Now, if K is any bounded closed convex set in the plane, the function K(cfJ) defined for all real cfJ by K(cfJ) := supRe(e-i"'s) seK

(10.9-10)

INTEGRAL TRANSFORMS

312

is called the support function of K (see Fig. 10.9b). We thus see that the support function of C is just y( -

) is identically equal to 1? If such a function exists, then the convex hull of the singularities of its Laplace transform must be the closed unit disk lsi""' I. This requires f to be a noncontinuable power series with radius of convergence unity. According to the Weierstrassian example 16 of §3.2, ao

L

f(s) :=

s-n!-l

n=l

is such a function. Thus F(z) := .;J; 1{(z)=

1

2

co

z"'

L-

n=I(n!)!

1

1

6

24

=z+2z +720z +620448'401733'239439'360000z +··· has the required property.

Functions of Semie:xponential Type. Some of these results can be extended to functions F that are analytic and of exponential growth not in the whole plane but merely in some angular domain jarg z I< a where a > 0. For simplicity we consider only the case a= 1T/2. Let F be analytic in Re z > 0 and continuous in Re z ;a. 0, except possibly at z = 0. For p > 0 we now define f.L(p) := max jF(z)j. lzl=p Rez;;.O

The function F is said to be of semie:xponential type if f.L (p) is integrable at = 0, and if there exists a real number a such that

p

f.L(p) < eap for all sufficiently large p.

(10.9-11)

315

FUNCTIONS OF EXPONENTIAL TYPE

The infimum 'Y of all a such that (10.9-11) holds is again called the growth parameter of F. Although 'Y measures the overall growth of F, the growth along the ray arg z = c/J is measured as before by the indicator function '}'(c/J) which is the infimum of all a such that !F(e;"' T )I< ear for all sufficiently large T > 0.

The function '}'(c/J) is now defined only for -n/2.;:; c/J.;:; TT/2. ByTneorem 10.1d,/ := 5/?Fis analytic for Res> '}'(0). Similarly, if we let F(r) := F(e;"'r),

T

>0,

then f := 5/?F is analytic for Res> '}'(c/J ), -n/2.;:; c/J.;:; TT/2. It will be shown that for all these c/J, if Res> 'Y(O) and Re(e;"'s) > '}'(c/J ), (10.9-12)

f(s) = e;"'f(e;"'s).

When dealing with functions of exponential type, this relation could be read off the Laurent series off at oo. In the present case, there is no Laurent series, and a different approach is necessary. By Cauchy's theorem, the integral

t

e -szF(z) dz

has the same value whether extended along the straight line segment from z = 0 to z = p > 0 or along the straight line segment from z = 0 to z = e;"'p and then along the circular arc centered at 0 to z = p. The integral along the circular arc is bounded by 7Tp 2

I.L (p) - e

-lslpv
). This contradicts the definition of y(cf> ). Thus the assumption that f can be continued into a half plane Re(e;"'s)>8 where 8) is untenable. • As before we conclude that each line Re(e;"'s) = y(cf>) (-71/2 < cf> < 71/2) actually contains a point into which{ cannot be continued analytically. The totality of these points is again called the set of singular points of f. If cf> = ±7r/2, the lines Re(e;"'s) = y(cf>) do not necessarily contain singular

319

FUNCfiONS OF EXPONENTIAL TYPE

points. To state a result analogous to Theorem 10.9e, we enlarge the set of singular points by two "virtual" singular points -oo =F i'Y(±(17'/2)). A virtual point -oo+i~ by definition is contained in a half plane Re(e;s)os;;6 if and only if either -?T/2 < c/J < 7T/2 or c/J = -?T/2 and 6;;;.: ~or c/J = 7r/2 and -6.;;;; ~­ Then the set C may again be described as the intersection of all closed half planes containing all singularities of f (including the virtual singularities), and thus as the convex hull of these singularities. Defining the support function of a not necessarily bounded convex set again by ( 10.9-1 0), we have THEOREM 10.9h

Let F be a function of semiexponential type with indicator function 'Y(c/J ), let f := !£F, and let K(c/J) be the support function of the convex hull of the singularities off, including the virtual singularities. Then K ( cfJ) = 'Y( -c/J) holds for -1rj2.;;;; c/J.;;;; 7T/2. EXAMPLE

7

The function F(z) := (1 + z)- 1 is of semiexponential type with indicator function y(c/J) = 0. The virtual singularities are at -oo + iO. Thus C consists of the negative real axis. Indeed s = 0 is known to be a singular point of

EXAMPLE

8

Let

F(z) := e·-·. F, though entire, is not of exponential type, because for z real, z -+ -oo the exponent grows much more rapidly than linearly. But

LogiF(e'.,T)i =

e-TcosO>

cos(T sin cfJ ),

thus for -Tr/2.;;; cfJ .;;; Tr/2 .

'Y (cfJ ) =hmsup ... ~oo

LogiF(e'4>T)i T

O .

Thus F is of semiexponential type, and the set C again consists of the closed negative real axis. That this indeed is the convex hull of the singularities of the L_aplace transform is confirmed by the explicit representation (obtained by substituting ~ := e-T in the Laplacian integral) co 1 f(s) = L --:-n=on!(s+n) exhibiting its poles at s = 0, -1, -2, ....

320

INTEGRAL TRANSFORMS

EXAMPLE

9

Here we consider

1 F(z) := f(1 + z)' We know from §8.4thatFisentire. By Stirling's formula, if z -+OO, larg zl:>;;a
-y, and vanishing and analytic at s = oo. More precisely, f is analytic outside a certain compact convex set C, the hull of singularities off, contained in the disk

lsi,.;,; "Y· Because we do not wish to change the definition of F(O) if F(O) :fi 0, Polya's formula must now be modified to read

71

Jo F(n71) e-ns., =¥F(O)+PV n=~oof(s+ 2;n).

(10.10-4)

323

THE DISCRETE LAPLACE TRANSFORMATION

The variation of G( T) := e -s.-F(T) on (0, oo) is bounded if

i

00

(10.10-5)

jG'(T)j dT 1'· This range of validity will now be extended. Notice, however, that the expression on the right of (10.10-4) even ceases to have a meaning if s belongs to the set C., of all s such that 2win C s+--e 71

for some integer n. (The set C., consists of C and of all sets obtained by translating C by an integral multiple of 27ri/71.) We denote by Sn the complement of C., with respect to the unextended plane. Two cases are possible: (a) The set S., is connected. Because the diameter of C does not exceed 2y, this certainly happens if 711' < 7T. (b) The setS., is not connected. Then S., has a unique decomposition into components. Precisely one component contains the set Res> 1'· This is called the right component of S., and is denoted by In case (a) above, the right component of S., equals S.,.

s:.

THEOREM 10.10b

Let FE .n'Y, and let f denote the analytic continuation of .'£F into the exterior of C, the hull of singularities of f. Then f., := .'£.,F can be extended analytically the right component of the setS.,, and the analytic continuation is into represented by

s:,

00 71 ( 2Trin ) f.,(s) = 2F(O) + PV n=Y;_oo f s + 71 - ,

(10.10-6)

where the series on the right converges uniformly on every compact subset of

s:.

Proof. We begin by proving the last statement. Let s0 E C. Then f can be expanded in a series of powers of (s- s 0 )- 1 with no constant term and thus may be represented in the form ao 1 f(s)=-+( )2g(s), s-so s-so

324

INTEGRAL TRANSFORMS

where g is analyticin Sand bounded for lsi~ 'Y + 1, say. Let T be a compact subset of We know from a similar discussion in §7.10 that

s;.

PV n

I

.

1 =-coS+ 271m/ 71 -So

converges uniformly in T and thus represents an analytic function there. The same is true of the series

I

1 _{ s + 21rin) n=-oo(s+21Tin/71-so) 2 g\ 71 (even without the PV) because leaving aside the finitely many terms for which Is + 271'in/ 711 < 'Y + 1, its terms are dominated by const · n - 2 • We conclude that the series (10.10-6) converges uniformly on every compact subset T of and thus represents an analytic function in Because this function agrees with 71 for Re s sufficiently large, it represents the analytic continuation of 71 into

s;

t

t

s;.

s;. •

The series on the right of (10.10-6) converges merely like Inn -k where k ~ 2, whereas the series defining/71 for Res> 'Y converges like Ln qn where lq I< 1. Thus the result can sometimes be used to convert slowly converging series into rapidly converging ones. EXAMPLE

1

Letf(s) :=s-k-I (k a nonnegative integer). We know thatfis the image of (1/k!)Tk. Thus for 71 = 1 and k = 0 there follows 1

Q()

PV

and for k

=

L

n--=s-27Tin

1

Q()

2

n-1

-+ L e-•n

(10.10-7) '

1, 2, ... Q()

I

(

n--co S -

1 . )k+l 271"ln

(10.10-8)

In both formulas the series on the right for Re s > 0 converge like a geometric series. EXAMPLE

2

Let a > 0. Then from the correspondence 1

~e---o]0 (aT)

vs-+a-

we obtain by (10.10-6) where 71 = 1 for Res >0

THE DISCRETE LAPLACE TRANSFORMATION

325

If a is not an integral multiple of 21T, the series on the right still converges for s = 0. If

the series is considered a series of powers of e-•, it follows from Abel's theorem on power series (Theorem 2.2e) that its value for s = 0 equals the limit for s = 0+ of the expansion on the left. When s = 0, the terms of the series on the left where n = ±k, 21Tk >a are purely imaginary and cancel each other, and there remains only 1 1 , 1 -+ L lo(an)=-+2 L .J , 2 2 n~J a n~J a -4?T 2 n 2 Q()

(10.10-9)

where r := [a/21T]. From this point onward we assume that the set S., is connected, which means that all translates of C, the hull of singularities, are disjoint. In this situation there is yet another possibility to express {., in terms of f. For E > 0, let C denote the set of all points s + d where s E C and ld I,;; E, and let c; be the set of all s such that s + 27Tik/ 71 E C for some integer k (see Fig. 10.10a). If E is sufficiently small, the translates of Care disjoint, and S~. the complement of ~. is still connected. Let r E denote the boundary curve of the convex set C, oriented in the positive sense. For s e S~ we consider the integral h(s) :=

_!_.J. 27Tl r.

f(u)

1 -e .,(u

s)

d U.

The singularities of the integrand (considered as a function of s) occur only where

. 27Tik I.e., for s = u +-71

e"'(u-s) = 1,

s lies on fE or on one of its translates. Thus as a function of s the integrand is analytic in S~, and because the integrand also depends analytically on u, it is clear that h(s) is analytic in S~. We now evaluate the integral. Let Re TIS;;;!: 'Y + 2E. Then le .,(u-s)',;; e-.,E < 1 on rE. Expanding into a geometric series and integrating term by term, we then get (k integer), that is, when

r I

h = 2 1 . 7Tl Jr.

e -n(s-u)"'t du

n=O

~ -RST/_1 . ~...e n=O

f.

RUT/~( ) d 2 7Tl r. e 1 uu.

But, by Theorem 10.9b,

/.f.

m r.

f(u) enu., du =F(n71);

INTEGRAL TRANSFORMS

326 lm s

s,

Res

Fig. lO.lOa.

therefore 00

h(s)=

~ n=O

and we have obtained:

e-sn.,F(nT/)=f.,(s),

THE DISCRETE LAPLACE TRANSFORMATION

327

THEOREM lO.lOc

Let Fe fi,forsome y;;;.O, and letT/ be such thatS., is connected. Then for any s E S., the continuation f., of 2.,F is represented by 1 f.,(s) = 2 -.

i

f(u) .,(s r. 1 -e

7Tl

u)

du,

(10.10-10)

Where fE runs parallel tO the boundary of Cat distance E, and E iS Chosen SO small that S is exterior tO r E and tO all (27Ti/ T/)- translates of r E"

Now lets be such that Re T/S < -y- 2E. Then le .,l < e -.,E at all points of rE. Expanding the kernel ofthe integral (10.10-10) in powers of e"'(s-u) (in place of e-., 1, then the analytic continuation of p into lz I< 1 is given by p(z)=-L P(-n)z". n=l

EXAMPLE

4

The function F(r) := Cosh~a >0) belongs to the class 0 0 • Thus C again consists of the sole point s = 0. The function p(z) :=

I

Cosh& z-"

n=O

(lz I> 1) can be continued into the set lz I< 1 and there has the representation 00

I

p(z) = -

cos.Jna z".

n=l

We now turn to the problem of inverting the discrete Laplace transform. If by inversion we merely mean recovering the numbers F(n'Y/) from the function!., (s ), then this is merely a matter of determining the coefficients of the Taylor series of an analytic function and need not be discussed any further. If by inversion we mean determining F( T) from[., (s) also at points T that are not of the form n'Y/, then the inversion pro~lem clearly has no unique solution, because for r-# n'Y/ the definition ofF( r) may be changed arbitrarily without influencing/., (s ). The following theorem shows that matters are different if F is an entire function of exponential type. THEOREM lO.lOe Let F be an entire function of exponential type, let C be the hull of singularities off := 5£F, and let 'Y/ > 0 be so small that the set S., is connected. Then for arbitrary complex t, if r is any positively oriented simple closed curve in S,1 encircling C but none of its translates, F(t) = 2 1 . 7Tl

Proof.

J.r

e 15[ 71 (s) ds.

Using Polya's formula (10.10-6), we write /.,(s) = /(s) + g.,(s),

(10.10-13)

329

THE DISCRETE LAPLACE TRANSFORMATION

where

71

oo

(

211'in)

g71 (s):= ZF(O)+PV n=~oof s+--;;n .. o

Because g71 is analytic on and in the interior off,

L

e 15g71 (s) ds

= 0,

and by Theorem 10.9b we have

11,.
-y, and that it is periodic with period 27ri/ 71· It thus suffices to study the analytic behavior of{., in a horizontal strip of width >211"/71. By Theorem 10.9f, the function/ can be continued to a function analytic in the complement of the unbounded convex set

C := { s: Re(eis) ~ -y(cf> ),

-~~ cf> ~~}.

the convex hull of the singularities off (see Fig. 10.9d). It thus remains to continue g.,. The integrand in the integral (10.10-15) representing g., is an analytic function of s. If s is complex, s = u + iw, then the integrand is of the order of the larger of the two functions

The integral thus converges whenever the exponents are both negative, which is the case if

331

THE DISCRETE LAPLACE TRANSFORMATION

(Because y(±7r/2) ~ 'Y < 27T/ 71 by hypothesis, the interval / 71 has positive length.) It is obvious that the integral converges uniformly with respect tow whenever w is restricted to a closed subinterval of I.,. By the Theorems 4.1a and 3.4b it follows as usual that g., is analytic in the strip

T., := -oods}dT

=I"" F(T) dT. o z+T

By analytic continuation, this representation evidently holds for all z such that larg z I< '11'. The functional transformation

1

F(T) -dT o z+T 00

g(z)=ffF(z):=

(10.11-9)

is called the Stieltjes transformation. We write g ~ F in order to indicate that g = ffF. The following operational rules for the Stieltjes transformation are easily established: If F(T) o-------e g(z), then TF(T) o-----e

i""

F(T) dT-zg(z),

F(.f;) o-------e g(i.Jz) + g( -i.Jz)

(10.11-10) (10.11-11)

(principal values). Moreover, if p > 0, F(pT) o-----e g(pz),

(10.11-12)

; F(;) o-------e ~ g(~),

(10.11-13)

340

INTEGRAL TRANSFORMS

and for any complex a such that jarg a I< 71', F(T)

g(z)- g(a)

T+a

z-a

- - o-------e

(10.11-14)

If F exists and FEn, then

1 z

F(T) o-------e --F(O)- g'(z).

(10.11-15)

An inversion formula for Stieltjes transforms of functions in fi can be obtained by way of the theory of Cauchy integrals; see §13.5. In the later part of chapter 12 we encounter Stieltjes transforms defined in terms of Stieltjes integrals. The domain of definition of the Stieltjes transform as considered in Chapter 12 contains the nonnegative functions in n as a proper subset. It is seen there that an intimate connection exists between Stieltjes transforms and certain continued fractions. This forms the basis for numerically efficient algorithms for the evaluation of Laplace transforms.

§10.12. SOME APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS In some treatments of integral transform techniques great emphasis is placed on applications of integral transforms to initial or boundary value problems involving partial differential operators. The method consists in applying the transformation to one selected variable and treating the other variables as parameters. The result in the image space is a boundary value problem whose dimension is reduced by one. The desired solution then is obtained by translating back into the original space. To apply the method rigorously, it is usually necessary to make a number of assumptions, such as the assumption that differentiation with respect to the paramete's may be interchanged with the application of the integral transformation. These assumptions can usually be verified only a posteriori. This means that after the solution has been found by formal methods, it must be verified that the expression found actually is a solution of the posed problem. This drawback is shared by the method of integral transforms with other methods for solving partial differential equations. Another, more serious, criticism may be voiced. What is obtained by applying the integral transformation to a partial differential equation frequently is nothing more than what could have been found. by the completely elementary method of separating variables and integrating with respect to the separation parameter. Indeed, the decision to use a certain integral transformation already prejudices the values (real or complex) that the separation parameter is permitted to assume. In our view greater

PARTIAL DIFFERENTIAL EQUATIONS

341

freedom of action is obtained if variables are separated first, the method of integral transforms being used only afterwards to adjust the solution to the boundary conditions. We illustrate the method by three simple examples.

I. Heat Conduction We wish to find a solution U of the heat equation in one spatial dimension,

au rlu dT

(10.12-1)

= iJ~ 2 '

defined and continuous forT;;;;= 0, -oo < ~ < oo and satisfying the conditions lim

U(~, T)

T-+00

= 0 for all ~.

(10.12-2)

UU, 0) = F(~).

(10.12-3)

where F is a given function such that, say,

L: jF(~)I d~

This solution of (10.12-13} does not yet have the desired form e!""'v(p), because the integral (through the limits of integration) still depends on c/J. We can force it to be independent of c/J, however, by letting the path originate and terminate in an area where a parallel displacement of the path does not change the value of the integral. This is accomplished by letting the path extend to infinity in areas where the integrand is small. If Jl is real, p > 0, and f3 = + i71. the real part of the exponent of the exponential function is Re{ip cos f3 + ivf3} = p sinh 71 sine- JI'TI. For fixed this tends to -oo exponentially if 2mr < < (2n + 1}1r, for 71 ~ -00 for 71 ~ +oo if (2n + 1)'7T < < (2n + 2)1T (n =0, ±1, ±2, ... ), that is, in the shaded areas of Fig. 10.12a.

e

e.

Fig. 10.12a.

e

e

346

INTEGRAL TRANSFORMS

We now select a path f 1 from ~0 +ioo to ~ 1 -ioo where ~0 e(-1r,O), and a path f2 from ~~-iOO to ~2+ioo where ~2E(1T,21T), the precise choices of ~0 , ~t. ~2 being irrelevant, and define the functions ~tE(0,1T),

H~o{p) :=

cf.

eipcosf3+ivf3

d{3,

eipcos(3+iv(3

d{J,

f't

Jt}>(p) :=

cf

Jr2

(10.12-15)

where 1 -ivTr/2 . C ..=-e

(10.12-16)

1T

The functions H~1 > and li,2 > are called the first and second Hankel function of order v, respectively. A number of elementary properties of the Hankel functions can be derived directly from these integrals. Making the substitution {3 -+ 1T- {3 in the integral (10.12-17) yields

But 1T- r I is a path equivalent to - r 2' where the bar denotes complex conjugation. We thus have H~0 {p)=[Ji}>(p)]-.

(10.12-18)

Substituting {3-+ -{3 in

Jt.!~(p) =_!_f.

eipcos(3-iv(J3-.,.f2)

d{3

1T f't

we find

hence (10.12-19a) and by (10.12-18) (10.12-19b)

347

PARTIAL DIFFERENTIAL EQUATIONS

By virtue of their construction, the Hankel functions are solutions of the Bessel differential equation (10.12-14). It thus must be possible to express them linearly in terms of the functions 1., and Y.,, the Bessel functions of the first and second kind defined in §9. 7. THEOREM lO.lla

For all real ., and for all p > 0, H.,l)(p) = J.,(p) + iY.,(p ), Jt}>(p) = J.,(p)- iY.,(p ). Proof.

(10.12-20)

We define (for the purpose of the present proof only) .

S., := t{lt,'> + H.,2 >},

D ·= _!_{Jt.'>- Jt.2>} "0

2i

"

" 0

We then have to show that S., =1.,, D.,= Y.,. By (10.12-19), -i"'"~,.(2J} S -v = 2'{e ;.,'"~,.(IJ N ~ +e N ~ =COS

S

·

D .,.

P'm ., -SID P'TT

Hence if., is not an integer, D =cos .,7r5.,- s_, "

0

0

SID P'TT

In view of (9.7-20), Theorem 10.12a will be proved if we can show that S.,=J.,. We have S.,(p)

=_!_f.

eip cosfJ+iv Pn· Actually, we may assume that all Pn are the same, because (11.1-9) for n = 0 implies that lf(z)- aolo;; Yolz l- 1 for

lz I> Po,

and thus in particular that f is bounded for lz I> p 0 , z such that Pn >Po zn+V(z)- Fn(z))

S,

z

E

E

S. Hence for any n

is bounded, say by Ym on the set z E S, Po< lz j.;; Pn· Replacing 'Yn by max(ym Yn ), it follows that (11, 1-9} holds for lz I> p 0 , z E S.

356

ASYMPTOTIC METHODS

Relation (11.1-9) clearly implies

lzn[J(z)- Fn(z)]l ._;;;; 'Ynlzr' for lz I> Pm z E S, and hence that f and F satisfy PROPERTY (B)

For n = 0, 1, 2, ... , lim {zn[/(z)-Fn(z)]}=O.

(11.1-10)

z-+OO

zeS

This may be written yet differently. Since

znFn(z}=zn(ao+a,z- 1 + · · · +an_,z-n+l)+an

= znFn_,(z)+an (where F_ 1 := 0), and because lim an= am property (B) clearly implies PROPERTY (C)

For n = 0, 1, 2, .... lim {zn[f(z)- Fn_ 1(z)]} exists and equals an.

(11.1-11)

z-+00

zeS

THEOREM ll.la

Properties A, B and Care equivalent. Proof. We have already seen that (A)~(B)~(C); it thus remains to sho\\ that (C)~(A). Let n be any nonnegative Integer. Then (C) implies

an+!= lim {zn+ 1[/(z)-Fn(z)]}. Z-+00

zeS

Thus for lzl sufficiently large, lzi>Pn say, lzn+l[J(z)- Fn(Z)]I ._;;;; 1 + lan+ll

or

lan+!l If(z)-Fn(Z)I._;;;; 1 +lzln+l . But this is precisely (11.1-9} where 'Yn

= 1 +!an+ II· •

AN EXAMPLE; ASYMPTOTIC POWER SERIES

357

By virtue of Theorem ll.la, we may henceforth use the Properties (A), (B) and (C) as equivalent conditions for f to be represented asymptotically by F as z ~ oo, z E S. It is customary to use the notation

f=F,

z -+00,

zeS

to indicate that these conditions are satisfied. If S is a full neighborhood of oo, or if there is no ambiguity about the manner in which z is allowed to tend to oo, the qualifier "z E S" may be omitted. An important conclusion may be drawn from property (C). TIIEOREM 11.1b

A function f can in a given unbounded set S admit at most one asymptotic power series as z ~ oo, z E S.

Proof by contradiction. Let F=ao+a1z- 1+a 2 z- 2 + · · ·, G = b0 +b1z- 1+b2z- 2+ · · · be two distinct asymptotic power series representations off as z ~ oo, z E S. Let n be the smallest integer such that a,.-:1: bn. Then Fn- 1 = Gn- 1• Hence (11.1-9) implies that a,.= b,., which contradicts the definition of n. • EXAMPLES

of asymptotic series:

1 The computations carried out near the beginning of this section show that for every a such that Q.;;;a we expand the logarithm in powers of e- 2 '"' to obtain for k =0, 1, 2, ...

Although the convergence is not uniform near 71 = 0, summation and integration may be interchanged by Beppo Levi's theorem (see §8.4). Using Euler's integral in the form (10.5-1), this yields

359

AN EXAMPLE; ASYMPTOTIC POWER SERIES The series can be summed by (4.9-5) or (7.10-13), and we find (-1)k {3k = (2k

+ 1)(2 k + 2 ) B2k+2•

(11.1-15)

k = 0, 1, 2, ... ,

where B2k+ 2 is a Bernoulli number. Thus the asymptotic expansion for J(z) takes the final form oo B2k 1 J(z) ""k~l 2k(2k -1) z 2 k-l'

z-+ CXl,

zeS..

(11.1-16)

where a < 71'. Other ways of identifying the {3k are discussed in § 11.2 and § 11.11.

The definition of an asymptotic series given is applicable when z approaches oo, which is the situation that occurs most frequently. However, we also may consider the approach to a finite value. Let f be defined on a set with point of accumulation z 0 , and let F= a 0 +a1(z- z 0 )+a 2 (z- z 0 ) 2 + · · ·

be a formal power series in z- z 0 . We write Fn(z) := ao+al(z-zo)+ · · · +an(z-zot.

n =0,1, 2, ... ,

F_ 1(z ):=0. We say that F is an asymptotic power series for f as z-+ z 0 , z if F has any of the following three equivalent properties:

(A') For n = 0, 1, 2, ... , f(z)- Fn (z) = O((z- zot+ 1), which is to say that for every n such that

z-+ Zo,

for lz- Zol sufficiently small, z E S. (B')

For n = 0, 1, 2, ... , lim {(z- Zo)-n[/(z)- Fn(z)]} = 0.

z-+zo

zeS

(C') For n = 0, 1, 2, ... , z-+zo

zeS

We shall use the notation

f=F,

z-+ z 0 ,

zeS

S,

zeS,

= 0, 1, 2, ... there exists a

lf(z)- Fn(z)l,;;; 'Ynlz- zoln+l

E

real number 'Yn

ASYMPTOTIC MEDIODS

360

to indicate that f meets these conditions. It is clear that the analog of Theorem 11.1 b holds in this case. We finally note some variation in the use of the symbol =. For simplicity we assume that S is unbounded, and that z approaches oo. Analogous conventions may be made for z -+ z 0 • If h is a function defined on S and "F 0 for lz I sufficiently large, we write

z -+00,

ZES

z-+ 00,

ZES.

f=hF, to mean Furthermore, if g has the range S,

lim g(z) = oo,

z-+oo

and if the inverse function g[-ll exists, we shall write f(z) ""'ao + al[g(z

>r' + a2[g(z )r2+ ...

to indicate that z-+ oo,

PROBLEMS 1. The following rules for manipulating the 0 symbol hold for arbitrary integers m and n: (a) O(z-m) + O(z-") = O(z-min(m,n>), z-+ oo; (b) O(z-m)O(z-") = O(z-m-"), z-+ 00. 2. Letfbedefinedon an unbounded setS, letz 0 e C, and let, for z e T:=z 0 +S- 1, g(z) :=

1- ) . t(z-zo

Then f(z)=a 0+a 1z- 1+a 2z- 2+ · · ·,

z eS

z-+OO,

if and only if g(z)=ao+a1(z- Zo)+a 2(z- Z 0) 2 +· · ·, 3.

z-+ Z 0,

z

Let /(z)=a 0 +a 1z- 1 +a 2 z- 2 + · · ·,

z -+00,

zeS.

Then for n = 1, 2, z"[f(z)-(a 0+a 1z- 1+ · · · +a,.z-")] z -+00,

zeS.

E

T.

361

AN EXAMPLE; ASYMPTOTIC POWER SERIES 4.

Let h be a real, continuous function on [0, oo) such that all integrals n =0, 1, 2, ... ,

C,. := f"'.h(T)T" dT,

are absolutely convergent (example: h(T) = e-T). Show that the function f(z) :=

1oo h(T) dT,

zeS..,

z+T

admits for every a < '11' the asymptotic power series l(z) ==cot'- 1-c1z- 2 +c 2 z- 3 -

5.

Let 0 < 'Y < 1. Then the function

• • ·,

oo

z -+00,

zes...

k

f(z ):= L _'Y_ k=1

z+k

is defined for z E S.,.. Show that for every a < '11' it admits the asymptotic power series

z -+00, where

a,. :=

oo L

k"- 1 'Y"·

zeS,.,

n= 1,2, ....

k-1

6.

Let {3 >0, 'Y>O. The function / (z) :=

oo e_T,. 1 (1 +TZ 1)'Y

d

T,

zeS..,

possesses, for every a < '11', the asymptotic power series

i;

f(z)==.!. (-1)" ('Y)S((n + 1)/{3) z-", {3 n=O n!

z -+00,

zes...

7.

In the definition of asymptotic power series it was required that one of the three properties (A), (B), (C) holds for all nonnegative integers n. Show that it actually suffices that any of these properties holds on some unbounded set of integers. 8. The formal series F = La,.x -k is said to envelop the function I for x > 0, if for each nand all x >0

f(x) = F,.(x)+ 8,.(x)a,.+ 1x_"_\

9.

where 0,.. 8,. (x),.. 1. (The remainder is a positive fraction of the first neglected term.) Show that ifF envelops f, it represents I asymptotically as x -+ oo. Also show that the converse is not true. Prove: If, in problem 4, h has constant sign, then the series L( -1)"c,.x_"_ 1 envelopsl(x) for x >0. In particular, the series (11.1-1) envelops the exponential integral (11.1-4).

362

ASYMPTOTIC METHODS

Prove: The series (11.1-14) (considered as z- 1 times a series in z- 2 ) envelops the Binet function J(z) for z = x > 0. 11. The Taylor series at 0 of the functions e-x, Log(l + x ), (1 + x )-a where d > 0 (considered as series in x rather than x- 1) envelop these functions for x >0. State a general theorem that contains these examples as special cases. 10.

§11.2 THE ALGEBRA OF ASYMPTOTIC POWER SERIES We recall some conventions on formal power series. If F =a 0 +a 1z -1 a 2z -2 +· · ·,

G=bo+b1z- 1+b2z- 2+ · · · are two formal power series, and if a and {3 are any two complex numbers, we denote by aF + {3G the formal power series with nth coefficient aa, + {3b,, by FG the formal Cauchy product of the two series, and, if a 0 ;1:. 0, by r 1 the solution of the formal equation FX = 1. It is evident from§ 1.2 that with these definitions the formal power series form an algebra .st1 (see §2.1). The units of d, that is, the elements that possess reciprocals, are precisely the series F with a 0 ;1:. 0. We have seen in Chapter 2 that the formal power series with positive radius of convergence form a subalgebra of d. It will now be shown that the formal series that are asymptotic to some function in a given set S form another (larger) subalgebra. THEOREM 11.2a

LetS be an unbounded set in C, let the functions f and g be defined on S, and let F and G be two formal power series such that f(z)=F, g(z)=G, z ~oo, z E S. Then for arbitrary complex a and {3, (i) af(z) +{3g(z) = aF + {3G, furthermore (ii)

f(z)g(z)=FG

and, if ao ;t:. 0, (iii) as z -+ oo, z e S. Proof. As in §11.1 we use subscripts to denote partial sums. Since (aF + {:JG), = aF, + {:JG,, we have, using property (A), (af +{:Jg)(z)- (aF +{:JG),(z) = a[/(z)- F,(z)]+{:J[g(z)- G,(z)]

= O(z-n-1)+ O(z-n-1) =O(z-n-1)

THE ALGEBRA OF ASYMPTOTIC POWER SERIES

363

for n = 0, 1, 2, .... Thus the series aF + {3G has property (A) with regard to the function af + (3g, proving (i). Although it is not true in general that (FG)n = FnGm we have (FG)n(Z) = (FnGn)(z)+ O(z - n - 1 ).

Hence f(z )g(z)- (FG)n (z)

= f(z)g(z)- Fn(z)Gn(z)+ O(z -n- 1) = [f(z)- Fn(z)]g(z) +[g(z)- Gn(z)]Fn(Z) + O(z -n- 1 ) = O(z -n- 1)g(z)+ O(z -n- 1)Fn(z)+ O(z -n- 1), and this is O(z -n- 1), in view of the fact that the limits of g and Fn exist as z -+ oo, z E S. This establishes (ii). To prove (iii), we first note that, since by property (C) lim f(z)

=

ao

Z-+00

zeS

exists and is ¢ O,f- 1 is defined for lzl sufficiently large, z E S. To simplify the notation, let 1 a1 -1 G :=F-1 =--2z + ·· · ao

ao

Then _1__ 0 ( )=1-f(z)Gn(z) f(z) n z f(z) = f:z) {1-[Fn(z)+ O(z -n- 1)]Gn(z)}.

By the definition of G, Fn(z)Gn(Z) = 1 + O(z -n- 1 );

hence the last expression equals

/(~) {O(z -n- 1)+Gn(z)O(z -n- 1)}= O(z -n- 1) for z-+ oo, z E S. Thus the series G has property (A) with regard to 1/f, proving (iii). •

364

ASYMPTOTIC METIIODS

Formal power series, as was pointed out in Chapter 1, in addition to forming an algebra admit in certain cases the operation of composition. If

F=a1z- 1+a2z- 2+ · · · is a non-unit in d (leading coefficient zero) and if

G=bo+b1w+b2w 2+ · · · is any formal power series, we can substitute F into G. The result of this operation is known as the composition of G with F and denoted by Go F. The coefficients of are calculated as follows: Let, for k = 1, 2, ... ,

pk =:

a~k>z-k +a~kltz-k-1 +

... '

then c0 = b0 , and for n > 0, Cn is a finite sum, Cn ·.- b 1an(1) + b 2an(2 ) + • • • + bnan(n) ,

n = 1, 2, .... The question naturally arises whether the composition of functions possessing asymptotic power series mirrors itself in the composition of their asymptotic power series. The answer is affirmative. THEOREM ll.lb LetS be an unbounded set in C, let f be defined on S and f(z)=F=a~z- 1 +a 2 z- 2 +·

· ·,

z~oo,

ZES,

so that, in particular,

lim /(z) = 0.

z ... oo zeS

Let g be defined on T := f(S), and let

g(w)=G = ho+h1w +b2w 2+ · · ·,

w~o.

WET.

Then

g o f(z) = G o F,

z~oo,

zES.

Proof. We continue to denote partial sums of formal series by subscripts. Let n be a fixed nonnegative integer. Theorem 11.2a implies that for k = 0, 1, 2, ... , n

365

THE ALGEBRA OF ASYMPTOTIC POWER SERIES

Multiplying these relations by bk and summing, we get

Gn ° /(z) = Gn ° Fn(z)+O(z -n- 1). But

g 0 f(z)

= Gn ° f(z)+O((f(zt+ 1));

hence by the above and because /(z) = O(z - 1),

g 0 /(z) = Gn ° Fn(z)+ O(z -n- 1). The definition of composition implies

Gn °Fn(Z) = (GoF)n(z)+ O(z-n-l); thus

g 0 f(z)

= (G

0

F)n(z)+ O(z-n- 1),

and G oF is seen to possess property (A) with regard to g

of. •

EXAMPLE

We use Theorem 11.2b to identify once more the coefficients flk occurring in the asymptotic expansion (11.1-14} of Binet's function J(z}. These coefficients were calculated in § 11.1 using nontrivial tools, both from real analysis (Beppo Levi's theorem} and from complex analysis (the residue theorem}. Such tools are avoided in the purely formal calculation that follows, which requires merely that some asymptotic expansion is known to exist, plus one of the functional relations satisfied by the gamma function. The Binet function is connected with r(z} by Log r(z} =~Log 21T+(z -~}Log z- z +J(z},

(11.2-1}

where we suppose z > 0 to avoid any ambiguity in the choices of the logarithms. Taking logarithms in f(z + 1} = z f(z} we have Log r(z + 1} =Log z +Log r(z}; thus (11.2-1} after simplification implies J(z + 1}-J(z} = 1-(z +!}Log ( 1 +~).

(11.2-2}

The method now consists in computing the asymptotic power series of the functions on either side and comparing coefficients. The function J(z + 1} may be viewed as the composition of J(w- 1 } with w := (1 +z}- 1 • We have W""'Z- 1 -z- 2 +z- 3 -

z-+ CXl

• • ·,

and more generally for m = 1, 2, ... ,

wm

""'Z

-m

~

'-

n-o

(

-

(m}n 1}" - z -n

n!

'

z-+ CXl

ASYMPTOTIC MEmODS

366

(these expansions are, in fact, convergent). Assuming that J(z) admits an asymptotic expansion,

L

J(z)=

n=l

-n

z>O,

c,.z '

z -+00,

we thus have co

L d,.z-",

J(z+l)=

z>O,

z -+00,

n=l

where

d,.

- (n)o n+l (1),._1 -me" ---1)1 1-,-c.. -1+ .. · +(-1) (n- 1)!cl, (n

n = 1, 2, .... The coefficient of z-" in the asymptotic power series for J(z + 1) -J(z)

thus is n-1 {k}.. -~c ~ ( )"-k t... - 1 - - ck k=l (n-k)!

_ ( - -

n-1 { + 1) k-1 ~ - n 1),._ 1 t... k=o (k-1)!

ck.

On the other hand, by Theorem 11.2a,

=1-(z+~)(z- 1 -~z- 2 +lz- 3 -

r (_

1)"_ 1

=

n=2

n -1

2n(n + 1)

• • ·)

z-",

z-+ 00.

Comparing coefficients there follows "~ 1

t...

k=l

(- n + l)k-1 (k-1)!

ck =

n -1 2n(n+1)'

(11.2-3)

n = 1, 2, .... To identify the ck in terms of the Bernoulli numbers, it is only necessary to put

k =1,2, .... This yields n~l

k~l

that is,

(-n-l)k+l

(k + 1)!

(-1)k+lb k+l

=n-1 2 '

TilE ALGEBRA OF ASYMPTOTIC POWER SERIES

367

(n = 1, 2, ... ) or, replacing n by n -1, 1

n-t (

1-n · 2\~2

-n)k( -1)k

k!

bk =0,

n = 2, 3, .... If we define b 0 := 1, b 1 :=!, then this may be written n-t

(n)

k~O k bk =0, n = 2, 3, ... , which is identical with the recurrence relation of the Bernoulli numbers B" following directly from their generating function

In view of b0 = B 0 we thus have bk = Bk (k = 1, 2, ... ); hence ( -l)k+l ck

= k(k

+ 1) Bk+~>

k = 1,2, ... '

in agreement with (11.1-16).

PROBLEMS 1.

Let S be an unbounded set, and let f(z) =F= ao+atz- 1 +a 2 z- 2 + · · ·,

z -+00,

zeS,

where larg a 0 l < '7T. Show that larg f(z )j < '7T for Iz I sufficiently large, and that for any complex number a the function == ea Log/ (principal value) satisfies

r

r(z)=F'=a~{1+ :: Z-

1

+ ..

·r

z-+ oo,

2.

Let S be an unbounded set, k a positive integer, and

z -+00, Show that a branch of

t1

k

zeS.

can be defined such that

z-+oo, 3.

zeS.

zeS.

Let the function f be one-to-one on a set S having 0 as a point of accumulation, and let

z-+0,

zeS.

where a 1 ¢0. If T := f{S), and if pl-tl denotes the formal reversion ofF, show that W-+0,

weT.

ASYMPTOTIC METHODS

368

4.

Deduce the recurrence relation (11.2-3) of the coefficients of the asymptotic expansion of Binet's function from the functional relation 7T

f(z)f(l- z) = -.- .

sm TTZ

5.

If A is a matrix of functions having asymptotic expansions, we write zES.

z~oo,

6.

Show that if det Ao ?" 0, then the matrix A - I has an asymptotic expansion with leading term A{) 1 • If 1 denotes the Binet function, show that the asymptotic series for f(z) := 2J(z)-1(2z) envelops f for z = x >0 (see Problem 8, §11.1). Conclude that for all integers n > 0 On -) 1 - - -1- + 1 +r 2n& ( 2n) =ex ( - 7 5 3

7.

8n p n where 0< on< 1. Show that for all positive integers n,

r

2

192n

640n

840n

'

n.J;;,(:n) ,.,;exp(- 2 ~Arctau 2 ~J.

and that the error in the upper bound is O(n -s). Verify that already for n error is less than 0.1%.

=

1 the

§11.3. ANALY11C PROPERTIES OF ASYMPTOTIC POWER SERIES Let p;;,: 0, and let f be analytic for p < lz I< oo. We recall from §4.4 that f is represented by its Laurent series,

f(z)

= L

(11.3-1)

n=-oo

which converges for lzl > p and whose coefficients are uniquely determined by f. It is possible that only finitely many or no coefficients Cn with negative indices (corresponding to positive powers of z) are different from zero. The vanishing of such Cn depends on the existence of limz ... ro/(z). Three possibilities exist: (i) limz ... ro f(z) exists as a proper (finite) limit. In this case Cn = 0 for all n < 0; the isolated singularity at oo is removable in the sense that by defining f{oo) := c 0 , f becomes analytic at oo. (ii) limz ... ro f(z) = oo. Here there exists a positive integer m such that en = 0 for n < - m, C-m ¥0. The function f is now said to have a pole at oo, and m is called the order of the pole.

ANALY11C PROPERTIES OF ASYMPTOTIC POWER SERIES

369

(iii) limz-+oo /(z) does not exist either as a proper or as an improper limit. In this case Cn :F 0 for infinitely many negative indices n. The function f is now said to have an essential singularity at oo. Under what circumstances does a function/that is analytic for p < lzl < oo possess an asymptotic power series for z ~ oo if the approach is unrestricted? THEOREM 11.3a

Let f be analytic for p < lz I< oo. Then f admits an asymptotic power series for z ~ oo (approach unrestricted) if and only if the singularity at oo is removable, in which case the asymptotic series is identical with the Laurent series off at oo, and hence converges at least for lz I> p.

Proof. (a) Iff is analytic at oo, let its Laurent series be 00

L

/(z)=

(11.3-2)

CnZ-n,

n=O

and let

F.n (Z ) := Co+CtZ -1 + """

+CnZ

-n

,

n = 0, 1, 2, ....

Then for every nonnegative integer n zn[f(z)- Fn-l(z)]

= Cn +cn+lz-l +· .. '

lz I> p. There follows lim zn[/(z)- Fn-t(z)] = Cn

where the series on the right converges for Z-+00

for n = 0, 1, 2, .... Hence the formal series F := co+CtZ

-1

+c2z

-2

+ ···

with regard to f possesses property (C) of §11.1, and thus represents f asymptotically. (b) If/ admits an asymptotic power series F= co+c1z -I +c 2z - 2+ ···as z ~ oo, where the approach is unrestricted, then by (C) lim f(z) Z-+00

= c0 (approach unrestricted)

exists. By Theorem 4.4d (Riemann's theorem) z = oo is a removable singularity of f. The Laurent series thus involves no powers with positive exponents, and by (a) also is an asymptotic series. By the uniqueness of asymptotic series (Theorem 11.1b) F agrees with the Laurent series, and thus is convergent. •

370

ASYMPTOTIC METHODS

It is clear that a convergent power series is a more effective computational tool than a mere asymptotic series. A series such as (11.3-2) permits us (in principle) to compute the values off at any point z such that lz I> p with an arbitrarily small error. A nonconvergent asymptotic series (as the first example in § 11.1 shows) merely may enable us to compute fat any point z with a certain error, or to achieve an arbitrarily small error at some points

z ¥-00. Theorem 11.3a implies the following: If a function f that is analytic for p < lz I< oo admits an asymptotic power series for z-+ oo that is not identical

with its Laurent series, then z = oo cannot be a removable singularity of f. Neither can it be a pole off, because if it were a pole, then limz .... oo f(z) = oo, no matter whether or how the approach is restricted. Thus z must be an essential singularity. The following example shows that a function with an essential singularity at oo may indeed admit an asymptotic series in powers of z -I (which even may be convergent) if the approach to oo is suitably restricted. Let 0 =s:: a
0. An important conclusion to be drawn from our example is the fact that two different functions defined on the same unbounded set S may be admit the same asymptotic series. In fact, the functions f(z) := 0 and f(z) := e -z (z E Sa) both possess the asymptotic power series representation 0. The last remark is of interest in connection with the pr:oblem of summing an asymptotic series, which we state as follows: Given a formal power series F=a 0 +a 1z- 1 +a 2 z- 2 + · · · (11.3-3) and a sector S :a

< arg z < {3

({3 -a :s:: 27r), does there exist a function f, analytic for lz I sufficiently large, z E S, such that F is asymptotic to f as z-+ oo, z E S? Our example shows that

ANALYTIC PROPERTIES OF ASYMPTOTIC POWER SERIES

371

if the problem has any solution at all, this solution cannot be unique. The question of the existence of a solution, however, is answered positively without restrictions by the following theorem due to J. F. Ritt, which even gives an explicit formula for the "sum" f. For notational simplicity we assume, clearly without loss of generality, that Sis bisected by the positive real axis. THEOREM 11.3b (Ritt's theorem)

Let 0 0. Thus (11.3-6) is applicable with t:=z/Jlanr 1 and yields (11.3-7) Thus the series is majorized by I:=olztJ-nl, which converges uniformly on every compact subset of Sa n{lzl> 1}. It follows that{ is analytic. To prove the asymptotic relation (11.3-5), we establish property (B) of § 11.1. Let n be any nonnegative integer, and let Fn denote the nth partial sum of F. Then zn(f(z)-Fn(z)]= -zn

I, k=O

ak

exp(-,z/JI)z-k ak

ASYMPTOTIC MEmODS

372

If z E Sa, then Re z 13 ~ lz 113 cos(/3a ). In view of cos(/3a) > 0, every term in the first sum on the right is thus bounded by lz In-k lak Jexp( -lz 1'\,k), where 'Yk > 0, and thus tends to zero for z -+ oo, z E Sa. Using (11.3-7), the second sum is majorized by

I

00

1/3-1

" I 1-k- -'-z~-. z -l-lzl 1 L..., k=n+1 which again tends to zero for z-+ oo in view of {3 < 1. Thus property (B) is lzln+fJ

established, which completes the proof of Theorem 11.3b.



We conclude with some remarks concerning integration and differentiation of asymptotic power series. Beginning with integration, we first consider a function that possesses an asymptotic power series at some finite point z 0 that without loss of generality we assume to be 0. For simplicity we assume where a;;.: 0. We thus letfbe defined on Sa, that Sis an angular set of type and

t

f(z)=ao+a1z+a 2 z 2 +···,

z-+0,

zEt.

We call formal integral of the series on the right the series a, z az 3

aoz +2z +3z + · · ·.

The question is whether this formal integral is an asymptotic series for the indefinite integral J~/(t) dt. Before attacking the problem, it is necessary to note that the hypothesis of f possessing an asymptotic expansion does not guarantee the integrability of f. This is already seen by the function defined for x > 0 by f(x) := {

e -I! X. x rational,

0, x irrational. Tnis function has the asymptotic expansion 0 for x -+ 0, x > 0, but, being discontinuous at every x > 0, it is clearly not integrable. It is thus necessary to add integrability to our hypotheses. mEOREM 11.3c

Let f be defined on some set Sa (a~ 0), let

f(z)=ao+a 1 z+a 2 z 2 + · · ·, and let g(z) := rf(t) dt

z -+0,

(11.3-8)

ANALYTIC PROPERDES OF ASYMPfODC POWER SERIES

373

(the integral taken along the straight line segment from 0 to z) exist for every zESa. Then z-+0,

It is not necessary to assume that obviously is the only case of interest.

f

is analytic, although if a > 0 this

Proof. We denote the nth partial sum of the series on the right of (11.3-8) by Fn (z ). By property (B') there exists, for every E > 0 and for every positive integer n, ~ > 0 such that V(z)- Fn-1(z)l :s::;Eizl"- 1

if lz I< 8, z E Sa. Letting

i

z

Gn(z) :=

0

a1 2 Fn-1(t) dt = aoz +-zz +

an-1

n

· · · +---;;-z .

it follows that for the same values of z

I

jg(z)-Gn(z)j= f)f(t)-Fn-1(t)]dtl :s;; rl/(t)- Fn-1(t)lidtl 0

This implies lim z -n[g(z)- Gn(z)] = 0. Z-+00

zeSa

for n = 1, 2, .... Clearly, lim g(z) = Q.

z-+0

zeSa

Thus the series a1 2 a2 3 G := a 0 z+-z +-z +

2

3

.. ·

has property (B') with regard to the function g, establishing the assertion of the theorem. • The case in which f(z)=ao+atz - 1+a2z - 2 + · · ·,

z -+00,

Z E

Sa

(11.3-9)

ASYMPTOTIC METHODS

374

is more complicated because even for f continuous the improper integral

flO f(t) dt does not exist unless a 0 =a 1 = 0. However, the following can be established: THEOREM 11.3d

Let f be continuous on Sa, and let (11.3-9) hold. Then the improper integral h(z) :=

f"

~~] dt

[t(t)-a 0 -

(taken along the ray arg t = arg z) exists for z E Sa and possesses the asymptotic series z-+ 00, (11.3-10) Proof.

FortE Sa let /(t) := f(t- 1). By hypothesis, t-+0,

hence /(t)-a 0 -a 1t t2

t-+0,

We conclude that

. hA( z ) ._ .- 1lm

iz /(t)-ao-altdt

IJ-+0+ llz

t

2

exists. By the preceding theorem,

z -+0, or, what is the same, a3 -2 a4 -3 hA( z -1) =a 2 z -1 +2z +3z · + · · ·,

z -+00,

However, by substituting t-+ t- 1 in the integral,

J

IJ-lz

h(z-

1)=

lim

/J-+0+ z

[t(t)-a 0 - a 1] dt. t

ANALYTIC PROPERTIES OF ASYMPTOTIC POWER SERIES

375

Thus the improper integral defining h(z) exists and equals h(z - 1). Because h(z - 1) admits the asymptotic power series on the right of (11.3-10), the same holds for h(z). • We next discuss differentiation. Here we must first appreciate the fact that iff possesses an asymptotic power series and is differentiable, then f' need not possess an asymptotic power series. For example, let x>O.

Clearly f is differentiable, and f""' 0, x-+ oo, x > 0. But f'(x) = -e-x sin(e") + cos(e"), and this does not possess an asymptotic power series, because limx ...oof'(x) does not exist. To obtain results, the real and the complex cases must be discussed separately. In the real case, only a weak result may be proved inasmuch as the existence of an asymptotic series for f' must be postulated. THEOREM 11.3e

Let f be defined for x >0, let it possess the asymptotic power series (11.3-9) where a = 0, and let f' exist, be continuous, and possess an asymptotic power series. Then x-+oo, Proof.

By hypothesis, there exist b0 , bt. ... such that f'(x)""'bo+b1z- 1+b2z- 2+ · · ·, x-+oo,

itisto be shown thatb 0 = b1 = 0, bk+ 1 = -kak, k f(x) = f(xo) +

r xo

= f(xo)+

x>O.

x>O;

= 1, 2, .... Letx0 >0. Then

/'(~) d~

J" (bo+b1~- 1 ) d~+J" (f'(~)-bo-b 1 ~- 1 ] d~ xo

xo

where c is a constant. The integral has, by Theorem 11.3d, the asymptotic power series -

b 2x - i

b3 -2 b4 -3 -zx -Jx - · · ·,

X-+00,

x>O.

Because f admits an asymptotic power series, limx ...oof(x) exists, and it follows that b0 = b1 = 0. Because an asymptotic power series is unique, we

ASYMPTOTIC METHODS

376

may compare coefficients and find bk = -(k -l)ak-h



establishing the result.

f'

k =2, 3, ... '

Iff is analytic in a sector S" where a > 0, it is not necessary to assume that possesses an asymptotic power series.

THEOREM 11.3f

Let f be analytic inS" where a > 0, and let it possess the asymptotic power series (11.3-9). Then for every f3 such that 0 < f3 0 and Pn > 0 such that for t E Sa, ltl ~ Pm lf(t)-ao-a1t- · · · -ant-"I~"Ynltr"- 1 •

If jzj ~ (1- t:)- 1p"' this estimate holds for all t E f. Hence {11.3-11) implies for n = 0, 1, 2, ... that

z E S/3; Thus the formally differentiated series F'= -a 1z - 2 -2a 2 z -J -3a 3 z - 4 -

• · •

possesses property (A) with regard to f', establishing the result.



ANALYTIC PROPERTIES OF ASYMPTOTIC POWER SERIES

377

EXAMPLE

In § 11.1 we have obtained an asymptotic power series, valid for z -+ oo inS,. for every a < TT, for Binet's function J(z) appearing in Stirling's formula, f(z) = &exp[(z -~)Log z- z]eJ. We now convert this into an asymptotic expansion for the Theorem 11.2b,

r function

z -+00,

zES,.,

itself. By

(11.3-12)

where H is the composition of the exponential series with the asymptotic series for J(z). Lacking a simple explicit formula for the coefficients h", a recurrence relation can be established as follows. Theorem 11.3f permits us to differentiate (11.3-12) formally to obtain

z -+00,

z ES,..

Applying Theorem 11.3f to the asymptotic series for J(z ), H' by Theorem 11.2a equals

Comparing coefficients yields n = 1, 2, ....

The sum on the right terminates when the index of h becomes negative. From h0 = 1 and from the known values of the Bernoulli numbers the h" are now easily calculated. We obtain for a E (0, TT) 1 1 139 eJ=1+-z- 1 +-z- 2 - - - z - 3 +· · · 12 288 51840 '

z-+ oo,

PROBLEMS

1.

Let I be defined for lz I> p, and 00

f(z)::::=

L c"z-",

z -+00,

n=O

2.

where the approach to oo is unrestricted. Is I analytic at oo? Let a < TT/2. Show that z -+00,

zES,..

zES,..

(11.3-13)

378

ASYMPTOTIC METHODS

3. Let a < 1r/ 4, and for z E S,. let erfeo(z):= J; 2 1r

J"" e-'

2

dt,

erfcn+ 1(z):=

%

J"" erfc"(t) dt, %

n = 0, 1, ... , the integrations being performed along the ray arg t = arg z. Using the known result (see § 11.5) 1

erfeo(z)=

1

V'Tr

e

-z2 2F 0

-

(1:z, 1; -z -2 ),

z -+00,

z

ES,.,

Z

verify that

2 e -• 2 F. (" + 1 n +2 erfc"(z)=--- , - - ; -z- 2) , ..f;,.(2z)"+ 12 0 2 2

z -+00,

zeS,..

§11.4. ASYMPTOTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS Here we apply the theory of asymptotic power series to the study of singular points of ordinary differential equations. The notations and concepts introduced in Chapter 9 are used. Our discussion concerns the general system of order n, w'=A(z)w, (11.4-1) where A(z) is analytic at z series in powers of z -I,

= oo, and hence representable by a convergent A(z) =

00

~

Amz-m,

m=O

for lz Isufficiently large. We shall assume that A 0 ~ 0. The point z = oo then is an i"egular singular point of rank one of the system (11.4-1). As in §9 .11 we assume that the matrix Ao has n distinct eigenvalues. Without further loss of generality it then may be taken to be in diagonal form, A 0 = diag(A 1o A2 , . . . , An). Theorem 9.11a then stated the existence of a formal solution of ( 11.4-1) of the form (11.4-2) where P is a formal power series,

P=Po+P1z- 1 +P2z- 2+· · ·, and where~= diag(8 1, 8 2 , ••• , 8n )is a certain diagonal matrix depending on Ao and A1. The main objective here is to show that each column of the formal solution W is, in suitable sectors of the complex plane, an asymptotic series to a

ASYMPTOTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS

379

certain actual solution of the system (11.4-1). More specifically, we shall prove: THEOREM 11.4a

Let the columns of the formal solution matrix W be denoted by w 1 , w2 , ••• , wn· Then for every i, i = 1, 2, ... , n, the following holds: If S is a closed sector of the z plane containing none of the straight lines Re(Ai - Ai )z = 0 (j = 1, 2, ... , n; j ~ i), then there exists in S for lz I sufficiently large an actual solution v of the system (11.4-1) such that v(z)=wi,

zES.

z~oo,

(11.4-3)

Let i be one of the indices 1, 2, ... , n. This index will be kept fixed throughout the proof. We begin with some normalizations designed to simplify the notations of the proof. By a substitution of the form w ~ exp(Aiz )z..,•w in (11.4-1) we can achieve that Ai = 5i = 0 in the transformed differential equation. The formal solution wi then is a power series in z - 1 ,

Proof.

wi=piO+pilz

-1

+pi2z

-2

+···,

(11.4-4)

whose coefficients are complex n vectors. The condition on S now requires that it contains none of the straight lines Re Aiz = 0 (j ~ i). By a substitution of the form w(z)~w(ez) (lei= 1) we can achieve that Sis a sector Sa with a suitable a, 0 a. By Ritt's Theorem 11.3b there exists a vector f of functions analytic in Sp such that f(z)=wj,

(11.4-5)

z~oo,

(Here and subsequently, asymptotic relations between vector-valued functions are to be understood componentwise.) By virtue of Theorem 11.3f, f'(z)=w:,

(11.4-6)

z~oo,

Because wi is a formal solution, w; = Awi, where the product on the right is to be interpreted formally. Because the convergent series A(z) is asymptotic to its sum, it follows from (11.4-5) and from Theorem 11.2a that A(z )f(z) = Awi and hence that f'(z)- A(z)f(z) =w;- Awi = 0,

z~oo,

380

ASYMPTOTIC METHODS

A vector v is a solution of w' = A(z )w satisfying relation (11.4-3) as required by the theorem if and only if the function h := v-f

has the following properties: (i) h(z) =W;- W; and hence is asymptotic to the zero series, z~oo,

h(z)=O+Oz- 1 +0z- 2 +· · ·,

zESa;

(11.4-Sa)

(ii) h satisfies the differential equation

h'=v'-f' =A(z)v-f' = A(z)h-(f'- A(z)f) or

h' = A(z)h+b(z),

(11.4-Bb)

b(z) := A(z)f(z)-f'(z).

(11.4-Sc)

where By ( 11.4-7), b is asymptotic to zero,

b(z)=O+Oz - 1 +0z- 2 +· · ·,

z~oo,

To prove the existence of an h satisfying (11.4-8), we now set up an integral equation that is equivalent to (11.4-Sb). The integral equation is solved by the method of successive approximations, beginning with the initial approximation ho := 0. The solution is then shown to be asymptotic to zero. Some preliminaries are required to set up the integral equation. The formal solution matrix W = Pz 4 el\z satisfies

W'=A(z)W

(11.4-9)

as an identity between formal logarithmic series. Since det P 0 ~ 0, P- 1, and hence W- 1 := e -1\z Z -Ap- 1 exist formally, and (11.4-9) implies

w·w-

1

=A,

(11.4-10)

which is the same as (P'+z- 1 P~+PA)P- 1 =A.

Let m

P(m)

:= ~ k=O

Pkz-k,

m =0, 1, 2, ....

(11.4-11)

ASYMPTOTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS

381

We define

W (m) ·P .= (m)Z4Az e . Clearly, W!ml may be regarded as a function of z. Since det P 0 ~ 0, P~> is defined as a function of z for lzl sufficiently large, lzl ~p. say. It follows that for such lzl also -1 _ -Az -4.p-1 W (m)-e Z (m)

is defined. We set

Evidently, A(m) is analytic at oo; moreover, it follows by comparison with (11.4-11) that A(m)(Z) = A(z) + O(z -m- 1).

(11.4-13)

By construction, the matrix W!ml is a fundamental system of the differential equation (izl~p)

(11.4-14)

near z = oo. By (11.4-13), this system can be made arbitrarily "close" to the given system w' = A(z )w by making m sufficiently large. This is used later on. By the general theory of linear analytic systems of differential equations (cf. (9.3-7)), a solution of the nonhomogeneous equation (11.4-15)

w' = A(m)(z)w+a(z)

is given by the variation of constants formula, w(z) =

r

K(z, t)a(t) dt,

where (11.4-16) More generally, if the kernel K is decomposed as follows, K(z, t) = Ko(z, t) + K1 (z, t),

(11.4-17a)

where each K; is a solution of the homogeneous system,

a az

-K;(z, t) = A(ml(z)K;(z, t),

i =0, 1,

(11.4-17b)

382

ASYMPTOTIC METHODS

then the function w(z) :=

r

Ko(z, t)a(t) dt-

r

K 1(z, t)a(t) dt,

where lal>p, lbl>p, again is a solution of (11.4-15), as is immediately verified by differentiation. This statement also holds forb = oo provided that the improper integral converges locally uniformly with respect to z. How does the foregoing relate to our task of finding a solution of (11.4-8b) that satisfies (11.4-8a)? Write (11.4-8b) in the form h' = A(m)(Z )h + (B(m)(Z )h + b(z )),

(11.4-18)

where B(m)(Z) := A(z)- A(m)(Z). Considering this as a differential equation of the form (11.4-15) with nonhomogeneous term a(z) := B(z )h(z) + b(z ), it follows that if h is a solution of the integral equation h(z) =

r

Ko(z, t)[B(t)h(t) + b(t)] dt (11.4-19)

- ioo K,(z, t)[B(mJCt)h(t)+b(t)] dt, then h is a solution of (11.4-18), and hence of (11.4-8b). We now choose the integer m, the decomposition (11.4-17) of the kernel, and a, the lower limit of integration, in such a way that the process of successive approximations applied to (11.4-19) converges and furnishes a solution with the correct asymptotic behavior. Let 10 be the set of all integers j E { 1, 2, ... , n} such that z-+ oo,

and let h be the complementary set. We denote by Do the diagonal matrix that has ones in the jth position where j E / 0 , and zeros elsewhere, and by D 1 the matrix I- 0 0 • With this we define K;(z, t) := W(t) ( ) = P(m) (z ) z A e AzDi e -Att -Ap-1 (m) t '

(11.4-20)

i = 0, 1. The effect of these definitions is to make the real parts of the exponents in the exponential factors negative in Ko for Re(z- t) > 0 and negative in K 1 for Re(z- t) < 0.

ASYMPI'OTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS

383

Bounds on the kernels K; are required to prove existence and convergence of the successive approximations. In Ko, we may assume lz I~ It I; in KI> lz I~ It I. Furthermore, larg z I~ a and larg tl ~a. Because Pcm>(z) and P~>(z) are analytic for p ~ lz I~ oo, the contributions of these factors to the norms IlK; II are bounded. Because Sa contains none of the rays Re Aiz = 0, there exist constants 'Y and JL > 0 such that for z, t e Sa, writing ( := lz I, T := ltl, jexp(Ai(z -

t))l ~ 'Y e-,..u-..>,

jelo,

and (because i e 11 and A;= 0) jexp(Ai(z- t))l ~ y,

(~T,

Letting ~ := max IRe ~il, it follows that for a suitable constant

K,

IIKo(z, t)ll ~ K( 8T -IJ e-,..u-.. >, IIK1(z, t)II~KC 8 T 8 ,

(~T.

(11.4-21)

We recall that as a consequence of (11.4-13), there exists for each m a constant /3m such that T~p.

(11.4-22)

We now fix m to satisfy m > 2~, m > 1. With this choice of m we attempt to construct a sequence of successive approximations {hd to a solution of (11.4-19) in the following obvious way: ho(z) := 0; hk+I(z) :=

r

Ko(z, t)[Bcmlhk(t)+b(t)] dt

-fXl K1(z, t)[Bcm)hk(t)+b(t)] dt, k =0, 1, 2, ... ;

(11.4-23)

lzl~lal.

These approximations still depend on a, which is going to be chosen conveniently later. However, we assume a to lie on the axis of symmetry of Sa, and hence to be real. The paths of integration in (11.4-23) are chosen as shown in Fig. 11.4a. In the first integral we first integrate from t = a to t = az := z iaz -II along the circular segment ltl =a, then from az to z along the ray arg t = arg z. The integration from z to oo is again along the ray arg t = arg z.

384

ASYMYfOTIC METHODS lm

e

Fig.U.4a.

LEMMA 11.4b

The successive approximations hk exist and are analytic for there exist constants Xk such that .,;;;::a.

lz I;;::: a, z e Sa, and (11.4-24)

Proof. The assertion is clearly true for k = 0. Assume the assertion to be true for some k ;;::: 0. Let gk := B(m~k +b. Because b is asymptotic to zero, there exists by (11.4-22) a constant "Yk such that

llgk(t)ll.s:;; "Yd5 -m-\

T;;;::a.

Using the estimates (11.4-21), we now find

llf' II[ and, since m >

2~,

Ko(Z, t)gk (t) dtll.s:;;

"YkK~('5a-m,

K 0 (z, t)gk(t) d41.s:;; "YkKm- 1( 8a-m,

ASYMPTOTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS

385

Thus hk+l exists and satisfies (11.4-24) where . XkK('TT' 1 1 ) Xk+t .=am 2+ m + m-28 ·

Because the integrands in (11.4-23) depend analytically on z and because the improper integral converges uniformly with respect to z on compact subsets of Sa, hk+ 1 is analytic. The assertion of the lemma thus holds for k + 1, and hertce for all k. • LEMMA 11.4c

If a is chosen sufficiently large, the sequence {hk} converges uniformly on every compact subset of lz I~a, z E Sa. Proof.

r

Let dk := hk -hk- 1. It follows from (11.4-23) that dk+1(z)=

Ko(z, t)B(t)dk(t) dt

-1

00

K1(z, t)B(m)(t)ddt) dt.

Now suppose that (11.4-25)

-r~a.

Proceeding much as in the proof of the preceding lemma, we then have, letting f3 := f3m,

II



Ill:

Ko(Z, t)B(m)(t)dk(t)

Ko(Z, t)B(m)(t)dk (t)

dtii~Kf3lhfC 8 a-m,

d41 ~ K/3lAm - 1(

8

a-m,

and

111

00

K1(z, t)B(t)dk(t)

d11 ~ K/38k(m -28)- 1(

8 Cm.

Let now a be so chosen that

am

~2Kf3

('T-+-+--. T' 1 1 ) 2

m

m-28

Then the foregoing estimates show that lldk+1 (z >II~ ok+1(8,

(~a,

(11.4-26)

ASYMPTOTIC METHODS

386

In view of d 1 = ht. (11.4-25) is true for k = 1 by virtue of Lemma 11.4b. There follows lldk(z)ll:o:;;2 1-k8t(8, (~a. k = 1, 2, ... , which implies the uniform convergence of the series 00

L k=l

00

dk(z) =

L

(hdz)-hk-t(z))

k=l

on every set a.:;;;; lzi.:;;;Pt. and hence of the sequence {hk}. Our argument also shows that the limit function h satisfies (~a.

(11.4-27)

By virtue of the uniform convergence, h is a solution of the integral equation (11.4-19), and hence of the differential equation (11.4-Sb). It remains to prove: LEMMA 11.4d

The limit function h is asymptotic to zero. Proof. It suffices to show that for p = 0, 1, 2, ... , there exist constants Tip such that (11.4-28) (~a. By (11.4-27) this is true for p = 0. Let g =: Bcm>h +b. If (11.4-28) is true for some p ~ 0, we have, since b = 0, llg(z)ll:o:;;PTI,£8-p-m-t, (~a.



Much as in the preceding lemmas we now estimate II

Ko(z, t)g(t) d11.:;;;;Kp71p'ia8-m-p'8 e -,..u-a>.

The integral from az to z is split in two by first integrating to the halfway point z 1 := !(az +z). This yields

II



Ko(z, t)g(t)

d11.:;;; K/371p(m +p)- 1a -m-p(8 exp( -i((-a) ).

a.

II( Finally,

Ko(z, t)g(t) dtii.:;;;KpTip(m +p)- 1 ((;a) -m-p(8

ASYMPTOTIC SOLUTIONS OF DIFFERENTIAL EQUATIONS

387

By virtue of m > 1, each of these estimates is O{(a-p-t) for (-+ oo. This shows that (11.4-28) is true, with p increased by one. The induction step is thus complete, and with it the proof of Lemma 11.4d and of Theorem 11.4a. • The next result now follows easily. COROLLARY 11.4e

Let S be a closed sector of the complex plane containing none of the critical lines Re{Ai- Ai )z = 0, i, j = 1, 2, ... , n, i -:1: j. Then equation (11.4-1) possesses in Sa fundamental system of solutions Vt. v2 , ••• , vn such that v;{z)=wi,

z-+ oo,

zeS,

i = 1, 2, ... , n.

(11.4-29)

Proof. The sectorS satisfies the hypotheses of Theorem 11.4a for every i. It thus follows that n solutions vi satisfying (11.4-29) exist, and it remains only to be shown that they form a fundamental system. This follows from the fact that for z -+ oo det(Vt. ... , Vn) = det{{P0 + O(z - 1))z 4 eAz}

= z1:a, ez l:"'{det Po+ O(z -t)} -:1:0 • •

By construction, the solutions v; mentioned in the corollary depend on the sector S. By the general theory of analytic differential equations, each of these solutions can, of course, be continued analytically across the boundaries of S, and across the critical lines Re(A; - Ai )z = 0. However, it must not be assumed that the asymptotic relation {11.4-29) remains valid for the continued function. Typically, the function v; will be represented asymptotically by a different linear combination of the formal solutions wi in each sector bounded by the lines Re(A; -Ai)z = 0. After its discoverer, the discontinuous change in the asymptotic representations is called the Stokes phenomenon. To illustrate the results of this section, we consider their implications for the special second-order equation u"-r(z)u =0, (11.4-30) where r(z) =

00

L

TnZ -n

n=O

for lzl sufficiently large, 4"o -:1:0. As shown in §9.12, the usual reduction to a first-order system here yields a system (11.4-1) with n = 2 where Ao has the eigenvalues ±r~12 • There is only one critical line, namely, the line

ASYMYf011C METHODS

388

Re r~12 z = 0. Thus in each of the two half planes separated by this line there exist two actual solutions that are asymptotic to the two formal solutions constructed in §9.12. EXAMPLE

To study a specific case, we consider the equation

u"+ ( 1- 112 ~~ 14 )u = 0 satisfied by z 112 v where vis a solution of Bessel's equation (9.7-9). Since r0 = -1, the critical line is given by Re iz = 0, and thus coincides with the real axis. The two formal solutions found by the algorithm of §9.12 were v(l)

= ei• F.

vn) =

2

0

(!-v !+ v· - 1-) 2 '2 '2iz '

e-iz F. (!-v !+ v· __1_) 2 0 2 '2 ' 2iz ·

The actual solutions having these formal solutions as asymptotic expansions are in §11.5 identified as certain Hankel functions. In fact, there it is shown that

In this case the sectors in which the asymptotic expansions hold are larger than what could be inferred from the general theory. Nevertheless, the Stokes phenomenon is present. To observe it, consider the Bessel function z 1121v(z). In view of lv = !(H~1 ) + H~2)) we have z 112J(z)=H 0 1)+ 0 11 )},

z -+00,

z

E_S.,.-..

(11.4-31)

If arg z is increased by 2'17", the expression on the left is multiplied by exp((v +!)2'1Ti), whereas the expression on the right is unchanged. Thus (11.4-31) cannot hold in any larger sector. PROBLEM

1.

Show that Theorem 11.4a remains true even if A(z) is not represented by a convergent expansion in z-•, but merely possesses the asymptotic expansion A(z)=

L m=O

in a suitable sector Sa.

Amz-m,

Z-+00

389

THE WATSON-DOETSCH LEMMA

§11.5. THE WATSON-DOETSCH LEMMA This deals with the asymptotic behavior as z ~ oo, z E Sa of functions defined by a Laplacian integral, f(z) := (..Pc/J)(z) :=

I'Xl e

-zrcfJ( T)

dT.

(11.5-1)

The main point is that this behavior is fully determined by the behavior of cfJ near T = 0. It will be shown, for instance, that if cfJ is an original function of bounded exponential growth that possesses an asymptotic power series for T ~ 0, T > 0, then its Laplace transform I for every e > 0 possesses an asymptotic power series as z ~ oo, z E S,.;2 -E whose coefficients are very easily calculated from the asymptotic series for cfJ. If cfJ is analytic in a sector Sa, then the asymptotic representation for f even holds for z E S,.; 2 +a-E· Because a large number of important special functions, notably solutions of ordinary differential equations, may be regarded as Laplace transforms, the method has numerous applications. We recall from Chapter 10 the following facts about the Laplace transform: (i) Let cfJ be an original function with growth indicator -y (see §10.1 for definitions). Then f:=..PcfJ is analytic at least for Re z > -y. (ii) Let cfJ = cfJ (t) be analytic for jarg tl..;; a where 0 O ltl=r

jargtl"'a

be integrable at T = 0 and satisfy . Log/L(T) -y :=hmsup

-y, where -a ..;; 1/1..;; a. If a > 7T/2, this means that f can be continued across parts of the negative real axis. The continuations from the top down and from the bottom up will, in general, be different. We use the symbol I to denote the continued function, with the understanding that it must be thought of as a Iog-holomorphic function as defined in §9.4. The proof of these assertions is given explicitly in§ 10.9 for a = 7T/2; the general case can be dealt with similarly. The precise formulation of the result alluded to initially is as follows: THEOREM 11.5 (Watson-Doetsch lemma)

Let cfJ satisfy either condition (i) or condition (ii), and let cfJ(t)= := l{ao+at{ +a 2 t 2c +· · ·}

(11.5-2)

ASYMPTOTIC METHODS

390

as t-+ 0 and t > 0 [case (i)] or t-+ 0 and t e Sa [case (ii)], where band care complex numbers such that Re b > -1, Re c > 0, and where the powers have their principal values. Then for every 8 > 0 the function f := !£q, satisfies

~

f(z)=F:=--br Z

f(b+~+mc)

am

m=O

{11.5-3)

Z

as Z-+ 00 and Z E S"'/2-8 in case (i) and Z -+ 00 and Z E S"'/2+a-8 in case (ii). Proof. Before proceeding to the proof proper, it is instructive to see how the result {11.5-3) comes about in a purely formal way. Substituting the series (11.5-2) into (11.5-1) and integrating term by term as if the interval of integration were finite and the series uniformly convergent, we obtain

!( Z ) = 100 e 0

-ZT

T

b

00

~

t..

amT

me

d

T

m=O

=

00

~

t..

100 am

m=O

e

-ZT

T

b+mc

d

T

0

and this is formally identical with F in view of Euler's integral in the form (10.5-1). It is obvious that the foregoing formal procedure is invalid, but it suggests how the result was found and how it might be proved. We now carry out the proof for case (i); the proof for the other case is obtained by rotating the path of integration exactly as done in §10.9. Let n be any nonnegative integer, and let 0 be given, and let (sin al+n-y+l Et

:= 2r{P +ny+ 1) E,

where {3 := Re b, y := Re c. By property (B') of § 11.1 there exists 71 > 0 such that for OO.

Here we consider an incomplete f fnnc:tion defined by r(a, z) :=

J"". e-•sa-l ds,

(11.5-7)

where larg z I< 1r, larg a I< 1r, and the path of integration runs parallel to the real axis. We shall determine a series asymptotic to r(a, z) for z-+ oo, z E S3 .,12 _8 (a> 0), where a is fixed. (The behavior for jajlarge and z fixed, or when a and z tend to oo simultaneously, is more complicated.) We assume temporarily that jarg zl < 11'/2. Rotating the path of integration, we then have r(a, z)=

t

e-•sa-l ds,

where the path A runs from z to oo in the direction arg z. Letting s = z(l + r), where 0,.. r < oo, we obtain

all powers having their principal values. The integral now has precisely the form considered in Theorem 11.5 where

Clearly, cfJ satisfies the conditions (ii) for every a< 1r. Because cfJ is analytic at t = 0, it is asymptotic to its Taylor series:

cfJ=

I: "". n.

n-o

t-+o.

393

THE WATSON-DOETSCH LEMMA Watson's lemma thus applies with b = 0, c = 1, and

·= (-1)" (1-n!a)" '

n=0,1,2, ....

an.

In view of f(n + 1) = n! we thus find (1-a)l (1-a)2 r( a,z ) =z a-1 e - · { 1---z-+-z2- - ·

• •

}

z-+ oo,

,

(11.5-8) EXAMPLE

2 The complex error integral.

Let

erf(z):=

2

1 V1T

J,r· e -•

dx.

2

(11.5-9)

o

It is well known that

Ioo

e-•2 ds =

J;. 2 '

0

hence by Cauchy's theorem 2 foo erf(z) = 1- J; z e-• 2 ds, where the path of integration is parallel to the real axis or, if jarg zl < 1T/4, along the ray arg s = arg z. To reduce the integral to a form where the Watson-Doetsch lemma is applicable, set dT

ds=z--. 2v'1 +T

This yields erf(z) = 1 -

ze-z2loo e -z

~

V1T

0

1

2 T

.....,=:==dT. v1+T

Again, the function

satisfies the conditions (ii) of Theorem 11.5 for every a < 1r. Replacing z by z 2 we infer

e-·• {

1/2 (1/2) 2

}

erf(z)=l-- 1 - - + - - - · · · , J;z z2 z4

z -+00,

zES3 ,..14 _ 11•

(11.5-10)

394

ASYMPTO'DC METHODS

EXAMPLE

3 Bessel Functions.

In the first place, we consider the Hankel functions of the first and second kind and of unrestricted order 11 that were defined in § 10.12 and shown to be related to the Bessel functions as follows: Jr~1 > = J~ + iY~,

(11.5-11)

J1 = J~- iY~.

Watson's lemma enables us to study the asymptotic behavior of these functions, and thereby of the Bessel functions, as z -+ oo in suitable sectors of the complex plane. Our first task is to represent these functions by Laplacian integrals. Among the means at our disposal, the simplest way is to express the Hankel functions by Whittaker functions and to make use of the key formula (10.7-15). Using!9.7-20), Jl can be expressed in terms of J~ and L~. Expressing the Bessel functions by confluent hypergeometric series of argument -2iz as in (9.7-11) and using (9 .12-14), we find that 11!,1>is a linear combination of two Whittaker functions of the first kind that is proportional to a Whittaker function of the second kind. There results (11.5-12) Similarly, we find

Jl(z) =exp[ (~~+;r;J~ wo.~C2iz).

(11.5-13)

Now the integral (10.7-15) can be used. Replacing z by iz in (11.5-12), we

fin~

Jl(iz) = -i exp (- iii1T) ~ W0.~(2z) 2

"'(z)~ ~

If 1T)] F: (1 -exp [ (z -111T --1TZ 2 4 2 i

z -+00,

2

0

1 1 ) -+II--~~·' 2

-1r +8 .;;arg z .;;21r-8.

' 2iz '

(11.5-14)

395

THE WATSON-DOETSCH LEMMA

In a completely analogous manner one obtains from (11.5-13) 1111" 11")] F. (1 1 1 ) H:21(z)= ~-exp [ -i ( z---2 0 -+" --~~· - -

~

2

1rZ

z-+ oo,

4

2

'2

'

2iz ' (11.5-15)

-21r+8 ,..arg z ,..11"-8.

As suggested in § 11.4, these asymptotic series agree up to constant factors with the formal solutions (9.12-18) of Bessel's equation. In view of the relations y = _!_{Jt.ll- Jt.2l}

.,

2i

...

... ,

it is now an easy matter to obtain asymptotic expansions for 1~ and Y~ valid as z -+ oo, z E -11· In view of the exponential factors, only Jt,.21 is relevant if 0 < arg z ... 11"- 8, and only It,.11 if -1r + 8 ,.. arg z < 0. However, the asymptotic behavior for z-+ oo, z = x > 0 is of special interest. Here both It,.11 and It,.21 have the same order of magnitude. It is customary to write the asymptotic formulas in a manner that renders this fact more explicit. Let

s..

P~(z) := ~#;{ exp[ -i(z -"; -~) JJt..ll(z)+exp[i( z -"; -~) ]Jt..2>(z) }. (11.5-16)

O~(z) := ~#;{ exp [ -i(z- "; -~) JJt..1>(z) -exp[ ;(z- "; -~) JJt..2>(z)}. It follows from (11.5-14) and (11.5-15) that

p ( )= ~ (- 1)m(l/2-11hm(l/2+11)2m ~ Z m=O (2m)!(2z) 2m ' Q ( )= ~ (- 1)m(l/2-11hm+l(1/2+11)2m+l

~ z

m-0

(2m+ 1)!(2z)2m+l

z-+ oo,

'

(11.5-17)

zeS.,._11•

It is merely a matter of some algebra to verify that

The expansions (11.5-17) can now be used to calculate numerical values of the Bessel functions for sufficiently large positive values of z, and also to determine the approximate location of the zeros. For many purposes it is sufficient to consider the

ASYMPTOTIC METHODS

396

leading terms of the asymptotic expansions only. For x-+ oo, x > 0 we have

(11.5-19)

Y~(x) =/!;{sin ( x- .,; -~) + O(x-

1 ) }.

From an orthodox numerical point of view, all the foregoing expansions are useless unless strict numerical bounds for the remainder terms can be given. Although the construction of such bounds is implicitly contained in the proof of Theorem 11.5, this construction is probably difficult to carry through in many concrete cases. We return to this question in §12.l:f-13, where expressions in terms of convergent continued fractions are given for the remainders in many of these expansions. In the three applications of Theorem 11.5 given thus far the asymptotic series for c/J was convergent for sufficiently small values of ltl. In the following example this series is divergent for all t ~ 0. EXAMPLE

4

Products of Hankel Functions.

An integral equivalent to

H~0 (z)lt},(z) = i exp~?Ti/ 2 ) {"' e-• ·-r-• e 112·H~·,(;J 2

d-r

(11.5-20)

is established in Watson's treatise ([1944], p. 439). The integral is of the form (11.5-1) where

(!...)

A.( ) ·= i exp(iv?T/2) t_e 1 112,H0 , .,t. ~ 2t 1T

.

By (11.5-14), t-+0,

hence it follows from Theorem 11.5 that H(l)( )ft:2l( )=~ ~ (- 1)" (1/2-v)n(l/2+v)n(l/2)n ,

Z

,

Z

i...

1Tn~o

=

I 2n+l n.z

:z3Fo(i-v, ~+v, ~;- :

(11.5-21)

2)

for Z -+ 00, Z 2 E S2..-B, that is, Z E S.. _8• The same result could have been obtained by direct multiplication of the asymptotic series for H~1 , and ~2,, using Dixon's formula (see §1.6, problem 12). It should be noted that the divergence of the series for f/1

397

DIE WATSON-DOETSCH LEMMA causes the series for considered.

5tO,

)"

and describe the curve z := C(u) + iS(u), -oo < u < oo (Cornu spiral). Deduce from (10.7-15) the asymptotic expansion WKJ.' (z)=e -z/2z

K

1 1 1) F.0 ,.... u-K+2' -u-K+-· ,.... 2' -z ' (

2

(11.5-22) z -+00,

4. The function f'(z) 1/J(z) := f(z)

can be shown to have the integral representation 1/!(z)=

[co(e:·- 1 ~-;~.)dT,

Rez>O.

Establish that

z

5.

-+00,

zES,_ 8 ,

where B~o B 2 , ••• , are the Bernoulli numbers. Let the function erfc,. be defined as in Problem 3, § 11.3, and obtain its asymptotic expansion from the integral representation erfc,.(z)=

2 1

fco (t-z)"e-•' dt

v1rn! •

6.

obtained from a well-known formula for repeated integration. Equation (11.5-19) suggests that the Bessel function 1. (x) has, for large values of n, zeros close to the points

n= 1,2, ....

ASYMPTOTIC METHODS

398 Show how to construct a formal series

such that, with the definitions of Example 3, Xn-+ 00.

In particular establish the leading terms

4v 2 -1 a1=---8-,

(4v 2 -1)(28v 2 -31) 384

[That the series X indeed is an asymptotic expansion for the large zeros is shown by Watson [1944], p. 506.] 7. For x>O and n =0, 1, 2, ... , let

Show that for each fixed n, X-+ 00.

8.

r

The Sievert integral is for x > 0 and 0 ~ (J ~ TT/2 defined by S(x, 8) :=

e-xtcos dl/J.

Show that

s(X,~)= e-x L~ e-xTl/l(T) dT, where l/J(T) := (1 + T)- 1(2T + T2 )- 112 , and deduce that

( ID 2

Sx-= ,

9.

#;2

5 -1 129 -2 -xe -x {1--x +-x -···}, 8 128

X-+ 00.

Show further that for any fixed 8 > 0 this also is the ·asymptotic expansion of S(x, 8) for x-+ oo. As a special case of (10.7-15), obtain the representation

(11.5-23) and obtain (11.5-22) in a different way. Also, obtain an estimate for the remainder by applying Taylor's formula (with remainder term) to (1 + T ),.+K- 112 •

EXTENSION OF TilE LEMMA

399

§11.6. EXTENSION OF THE LEMMA In many applications the function whose asymptotic behavior for z ~ oo is to be investigated appears in the form f(z)

r r =

e -zo/l(u) du

a

or, more generally, f(z)

=

e-zl/l(u)lJ(u) du.

(11.6-1)

a

Here 1/1 is a real function that increases monotonically on (a, {J), 1/J(u) ~ oo for u ~ {J, and 6 is a possibly complex valued function whose growth is suitably restricted so as to make the integral convergent at least for Re z sufficiently large. Without loss of generality, we may assume that 1/J(a) = 0, for if this is not the case, we can divide the integral bye -zi/ICa>. Furthermore, by a mere shift of the independent variable the lower limit of integration can be assumed to be zero .. By substituting a new variable of integration, integrals such as the foregoing are easily reduced to a form in which the Watson-Doetsch lemma is applicable. We assume that 1/J' exists and is positive on (0, {J). Then by letting 1/J(u) := -r in (11.6-1) we get /(z) =

1'"' e -ZT6(1/1[- 1l(-r))l/l[- 1l' d-r,

(11.6-2)

where 1/J[- 1) denotes the inverse function of 1/J. If the required conditions are satisfied, Theorem 11.5 thus becomes applicable with cfJ := ()

0 " ' [ -1)

• "'[ -1],'

which by virtue of 1/1[- 11' = 1/1'- 1 o 1/1[- 11 and by the associative law of composition is the same as (11.6-3) The Lagrange-Biirmann formula affords an easy way to determine the coefficients of the required asymptotic power series for cfJ as -r ~ 0 if asymptotic power series for 1/J and () are known. We assume u~o.

u>O,

where k is chosen so that ak ¥- 0. Because 1/J grows monotonically, we have ak > 0. Furthermore, we let

O(u)::::8=bo+b1u+bzu 2 +· · ·,

u~o.

u>O.

400

ASYMPTOTIC MEmODS

We then have mEOREM 11.6a

If, in addition to the foregoing hypotheses, 1/1' possesses an asymptotic power series as u-+ 0, u > 0, then r-+0,

r>O. (11.6-4)

It is understood that w-< 1 +mJ/k := ('1' 11 k)- 1 -m, where .rr1/k._a1/k Y

k

.-

u(

- - u + ·· · )

l+ak+1

1/k

(11.6-5)

,

ak

al1k>O. If Z

is a formal Laurent series in x, say, res Z denotes, as usual, the coefficient of x - 1 • Proof of (11.6-4). Let c/J be defined by (11.6-3). The assertion of the theorem is equivalent to the assertion that the function x(v) := kvk-1 c/J(vk)

admits the asymptotic expansion

L

x(v)=

res(8w-O.

m=O

By the distributive law of composition and by the formula for the derivative of the inverse function, c/J = (6 o 1/J(-1])(1/J -1 o 1/J[-1]) = (6 o 1/J(-1])1/J[-I]r. Thus if 1r denotes the kth power function, 1r(u) := uk (u ;;a.O). X= (c/J o 1r)1r' = (0 o 1/1[- 11o 1T)(I/I[-tlt o 1r)1r'.

Because the range of 1/1 is [O. oo ), the function (J)

:= 1/11/k ='IT[ -1] 0

1/J

is well defined, and

yields (11.6-6)

EXTENSION OF THE LEMMA

401

Let 0 := '11 11 k as defined in (11.6-5). By Problem 1, §11.2, and by Theorem 11.3e, w(u)=.O,

u-+0,

w'(u)=.O',

u>O

and by Problem 2, c..}-1l(v) =nHl,

v-+0,

v>O.

By Theorem 11.2b (or its analog, in which v is substituted for z - 1) x(v) = (8l.tl'- 1) 0 l.l.l[-l](v) = (8.0'- 1)

n[-1].

0

According to one version of the Lagrange-Biirmann theorem (Corollary 1.9c), (8.0'- 1) o .nHl =

00

I

res(en-m- 1)vm,

m=O

establishing (11.6-4). • Theorem 11.5 now yields immediately: THEOREM 11.6b

Let the functions 1/1 and fJ be such that cfJ := (81/1'- 1) o r/J[-ll satisfies the conditions (i) of § 11.5. Then the function f defined by ( 11.6-1) admits for every S > 0 the asymptotic power series ~ (llO.

By the basic functional relation we also have f(z)=z- 1f(z+1)=z- 1

f"'

e-•r•dr.

Here we make the substitution T = zu. Because the integrand vanishes exponentially at co, we may after a rotation of the path of integration again integrate along the real axis. This yields

(principal values), and /(z) := z-• e•r(z) =

1"" (ue -"Y du. 1

Clearly, this is of the form

where 1/J(u) := u-1-Logu.

The function 1/1 does not satisfy the hypotheses of Theorem 11.6b, because it does not increase monotonically. However, because lim 1/J(u) =co,

a-+0+

lim 1/J(u) =co,

u-+

1/J{l) = 0,

and 1/J'(u) = 1-.!.., u

EXTENSION OF THE LEMMA

403

we find that 1/1 decreases monotonically from oo to 0 on (0, 1) and increases monotonically from 0 to oo on (1, oo). Thus we may write /(z)=

r

exp[-z,1(u)]du+

I"'

exp[-z,2(u)]du,

where

1/11(u) := 1/J{l-u)=-u-Log(l-u),

r/12(u) := r/l(l+u)=u-Log(1+u). Clearly, these functions satisfy the required conditions. We have

u-+ 0, u

> 0; hence k = 2; furthermore,

(J

= 1 == 1. Because '\f1 2 = '\f1 1o (-X),

res('l'2(1+m)/2) = (-1)m res('l'l(l+m)/2). The odd terms in the asymptotic expansion thus cancel, and putting m = 2q, r(~ + q) = J;{~)9, we find 00

f(z)==z' e-•..r:;,. ~ res('IJ1- 112 -9 )(~)9 z- 112 -9 ,

z -+00,

z

q=O

es. .

-8

(11.6-8) where

This, at last, is a "closed formula" for the asymptotic expansion of r. The residues are computed easily by means of the J. C. P. Miller formula (Theorem 1.6c), and we find f(z) ,.,.,fi;,. z'- 112 e-•{1 +-hz- 1+ 2 ~8 z- 2 + · · ·}, in agreement with ( 11.3-13).

EXAMPLE

3 1be Legendre polynomials P.(x) for x > 1.

To examine the asymptotic behavior of these polynomials for n -+ oo, when x > 1 is fixed, we start from the integral representation (see Chapter 18 or Szego [1959], p. 88)

Pn(x)=1

1"' (x+x' cos a)" da,

1T 0

n =0, 1, 2, ... ,

(11.6-9)

ASYMPTOTIC MEmODS

404

where x' := ../x 2 - 1 for brevity. To apply Theorem 11.6b, we put x' g:=x+x and can write (x+x')"f" [1-{(1-cos a)]" da. P"(x) = - 7T

0

The integral is of the form (11.6-1), with z replaced by the discrete variable n, (} = 1, and l{l(a) := -Log[1-{(1-cosa)], From

1-g(~ -;4 + 7~ 0 -· · ·) 2

1/l(a)=-Log [

4

6

]

we find by expanding the logarithm and rearranging l/l(a)='ft=p2a 2+p4a 4 +· · ·,

a~o,

where p 2 k is a polynomial of degree k in g, P2({) =!{,

p4({) = ke-ohg, P6(g) =-he-

-1se + 1iog, ....

Evidently, k = 2. Because 'I' contains only even powers, the powers appearing in w-o+ml/ 2 have the same parity as 1 + m, and the residue is therefore different from zero only for even values of m, m = 2q. By virtue of f(~+q) = ../;(!)q there follows n~oo.

(11.6-10)

Straightforward computation yields

thus the leading terms of the expansion are given by P"(x)

(x+x')"+ 112 { c:---;

v2Trnx·

_2 } x-2x' 1 +--+ O(n ) , 8x'n

n

~

oo.

This result was derived for x real, x > 1. It can be shown, however, that (11.6-10) also holds when xis replaced by any complex number outside the real interval [-1, 1]; see (11. 9-14 ). For real values of x such that -1 ,; x ,; 1, the asymptotic behavior of P" (x) for n large is best studied by the method of Darboux; see § 11.10.

405

EXTENSION OF THE LEMMA PROBLEMS 1.

Let Re(b -a) >0. Use Euler's beta integral in the form f(z+a) f(z+b)

1

1 Tz+a-1(1--r)b-a-ldT

1

f(b-a) o

to obtain an asymptotic series f(z+a) f(z +b)

2.

(1-a-b)(b-a)

a-b{

2z

1+

z

}

z -+00,

+... '

z

e s.. -a·

Let En(x) be defined as in Problem 7, §11.5. Show that there exist polynomials bm (x) of degree m, m = 0, 1, 2, ... , such that for fixed x > 0 n -+OO,

where, in particular, b 1(x) = 1-x,

b0 (x)= 1,

3. The coefficients in Adams' formula of numerical integration are given by

·=

'Yn.

I

I (u)n _,_ uu, o n!

n =0, 1, 2, ....

Show that 1 ogn

n -+00.

'Yn =-L-+O((Logn)- 2 ),

4.

(Sidney Spital.) By considering leading terms of appropriate asymptotic expansions, show that



1

!!.'!! f(Ax)

foo x

Ax- I

-T

e

d-r =

T

{?' 2.

O 1. 5.

(John Whittlesey.) Determine the asymptotic behavior of the modified Bessel function In (z) := ;-nIn (iz ), n = 0, 1' 2, as z -+ oo, z E /2-B from the integral representation 0

6.

0

0

'

s..

Let 8" be defined by n n2 nn-l nn 1 1 +-+-+ ... +---+-8 =- e" 1! 2! (n-1)! n! " 2 '

n= 1,2, ....

406

ASYMPTODC METHODS Show that (J" =

1 +~{fe-n.,,(.,., du-

L"' e-n.,(u) du}.

where 1/1 1 and 1/12 are defined as in Example 2, and deduce that QO

L

6" = 1 +~n

res('l'-m-l)m!n-m

m=O

n 7.

-+00.

(Ramanujan.) In oqe form of chess match, 2n games are played. Wins count 1 point each, draws t losses are worth 0. To win the match, the defender needs to score at least n, whereas the challenger must achieve at least n +~. We suppose that the two players are of equal strength and that the probability of a draw is a constant 8. The probability of the defender's keeping his title then is ~(1 +a")' where

._ ~ (2n)! 2n-2 k(1- ~ 2 k a"( 8).- k:;-o(k!) 2 (2n-2k)! 8 -2--} . (a) Study the asymptotic behavior of "IT" (8) as n-+ oo, paying particular attention to small values of 8. (b) Show that "IT" is not a monotonic function of 8, that the unique point 8" where "1Tn(8) assumes its minimum satisfies 2 Log Bn-(Log ( _ 2) 8= - - 8n) - +On n 4n 4n '

n -+OO,

and that an(8n) an(O)

1 2'

n -+00.

---+-

(c) Show that the unique point

3

8! > 0 such that "IT" (8!) ="IT" (0) satisfies

3

8 * =---+O(n-) " 4 32n ' 2

n -+00.

[Use the integral representation

8.

Let n and m be integers, 0 < n ,.; m. The numbers

l(n-1 2m

n-1 n-2 n-1 n-2 n-3 m m m m m

8(m, n) :=- - - + - - - - + - - - - - - + · · ·

)

are of interest in the combinatorics of computer science, especially in the theory of hashing (see Knuth [1974]). Study the behavior of 8(m, am) as m -+00. In

407

ASYMPTOTIC FORMULAS; LAPLACE'S METHOD particular, show that for 0 0 be given, and choose 5 such that the numbers ± [ det H J112 6 (5 ) det(H=FTJ(5)1)

lie between 1 =F E/2. It is clear that for this 5 there exists x 0 such that x > x 0 implies n/2

.JdetH(...:_)

Ke-,.(B)x

27T

O.

(Polya and Szego). Form= 1, 2, ... and p = 1, 2, ... , let

(m)P .

m

T(p, m) := k~o (-l)k k

It is clear that T(p, m) = 0 when m is odd. Show that

2

{-1)mT{p, 2m)--( J2 cos~)

~

p-1

(2?Tm)

1-p12 (

2p

2 cos~)

2mp

,

m-+oo.

2p

Compare the result with the closed formula (-1)mT{p 2 ) = (pm )! ' m (m!)P'

m=1,2; ...

known to hold for p = 2, 3 (see Problems 11 and 14, §1.6), and show that this formula cannot hold for any p > 3 (De Bruijn).

§11.8. mE MEmOD OF STEEPEST DESCENT Here we consider integrals of the form

t ==

L

ezg(()h(() d(.

(11.8-1)

where the functions g and h are analytic in a region S of the ( plane and where A is an appropriate path in S. Again we wish to determine the asymptotic behavior off as a function of the parameter z, which to begin with we assume real and positive.

THE METHOD OF STEEPEST DESCENT

417

The method of steepest descent (also called saddle point method for reasons to become apparent subsequently) is a pattern of thought that is frequently helpful in solving the foregoing problem. It makes essential use of the freedom in selecting the exact shape of A, which is at our disposal by virtue of Cauchy's theorem. We first describe the method in intuitive terms. From our discussion of Laplace's method it should be clear that, for any choice of A, the main contribution to the integral will arise from those portions of A where lezg(')l = ez Reg({) is largest. Since for z > 0, ez Reg(') is a monotonic function of Reg({), these will also be the portions where Re g({) is largest. The contributions of these maxima will be most pronounced if the path is selected in such a way that Re g({) changes as rapidly as possible along it. It is easy to determine the general equation of any path of most rapid change of Reg({). Let l = ~ + i71, w(~. 71) := Reg({). By calculus, the direction of strongest increase (or steepest ascent) of w is given by the vector grad w or, in complex notation, grd w := iJw/ a~+ i(iJw/ iJ71 ). By the CauchyRiemann equations, grd w = g'((). We call steepest path for g({) any path that at every point where g'(l)-:/:- 0 runs in the direction of g'((). Let ( = (( T ), a :;;;. T:;;;. {:J, be a piecewise regular steepest path. Then arg ('( r) = -arg g'((( r)) for all T E [a, {:J] except at isolated points. Thus d drg(((r)) = g'(((r))('( r)

(11.8-2)

is real wherever defined, which implies that Im g(((r)) = const.,

a:;;;. T

:s;;.p.

We thus have established that along any steepest path the function g has constant imaginary part. Along a steepest path A, where do the local maxima of Re g(((r)) occur? By the rules of calculus, such maxima can occur at the end-points of the interval [a, {:J ], that is, (i) at the endpoints of A. Further maxima can occur where (d/dr)g(((r))=O, which by (11.8-2) means that g'(((r))('(r) = 0. Because the parametrization of a piecewise regular steepest path may be assumed to be such that(' exists and is different from zero at all points where g'(()-:/:- 0, this leaves only (ii) the points where g'(() = 0. By the maximum principle, unless at the same time g({) = 0, these are the points where the surface (-+ lg(()l has a saddle point-hence the name saddle-point method for the method of steepest descent. It is now clear that any steepest path A is a sum of arcs Ai with the following properties: Any two consecutive arcs Ai-l and Ai meet at a point (i

ASYMPT011C ME1HODS

418

where g'((i) = 0, and the function Reg(() is monotone along any Ai. If A is a steepest path, the integral (11.8-1) can be represented ~s

f(z)

= "[.fi(z), i

where /i(z):=

f ezg(()h(() d(.

(11.8-3)

AI

The asymptotic evaluation of the functions [j is easily accomplished by the Watson-Doetsch lemma. Without loss of generality we may assume that the integration ·is in the direction of decreasing values of Re g, for if the orientation of Ai is otherwise it suffices to change the sign of the integral. If Ai begins at (i> we parametrize Ai by setting T

:= g((j)- g(().

This parameter is legitimate, for since Im g(() = Im g((j) on Ai> Tis real, and because Reg(() is decreasing, Tis increasing. The function ( = (( T) decreasing Ai thus is well defined. If T increases to p, we have /i(z) = ezg((/)

r

e -ZTh(((T))('(T) dT,

(11.8-4)

and to apply the Watson-Doetsch lemma it only remains to determine the asymptotic series of h(((T))('(T) as T~O, T>O. Let w := (- (i, and put (11.8-5) g((i)-g(()=akwk+ak+lwk+ 1 +· · · := Gi, h(() = bo+ b 1w + b2 w 2 + · · the curves Im g(() = Im g((;). The new path should be composed of such curves. (C) Evaluate the integrals along the differentiable sub-arcs of A by the Watson-Doetsch lemma. Step (A), unless the function g is unduly complicated, is easy. The step that is likely to give the most trouble is (B). It may be necessary to introduce auxiliary subarcs along which the condition lm g(() = const. is violated. If so, the contribution of these subarcs must be estimated separately and shown to be negligible compared to the other contributions, or they must be evaluated accurately. Due regard must also be paid to singularities, if any, of the functions g and h. Step (C) is completely mechanized by formula (11.8-9) once the series G; and~ and the directions exp(i 0. Let t =: z 112C, x := z 312, then we obtain Ai(x) = (2TT)- 1x 112f(x 312 ), where f(x) :=

t:

exp(ix(ll" 3 +()] d(.

(11.8-10)

420

ASYMPTOTIC METHODS

This is of the form (11.8-1) where h(() := 1, g(() := iHC+().

The integral converges very slowly-it is not even absolutely convergent-and thus at first sight seems to present a difficult case for asymptotic evaluation. However, it can be evaluated easily by the saddle-point method. Following the procedure outlined, we first determine the saddle points. These are given by g'(() = i((2 + 1) = 0, and hence are

We next determine the curves Im g(() = lm g((k) (k = 1, 2) to find a suitable path of integration. Setting ( = ~ + i71, we have Im g(() = Im i(l( 3 +() = l~ 0. For .simplicity, it is assumed that q is real, q > 0. The saddle points of the integrand are located where g'(() = 0 or

qcos(=l.

(11.8-18)

The location of the saddle points in the strip -11' < Re ( < 11' depends on the size of q. If q > 1, there are two real saddle points; if 0 < q < 1, there are two complex saddle points that are purely imaginary, and if q = 1, there is exactly one saddle point, at ( = 0. Accordingly, we distinguish three cases. (i) q > 1. Let a be defined by cos a = q -l, 0 O.

425

THE METHOD OF STEEPEST DESCENT

The same result could have been deduced from the.relation H.,2 >(qv)= fiO.

Because A2 runs opposite to A1 on the relevant part of the path of integration, the expression on the right of (11.8-23) also gives the asymptotic expansion of -Jt;,2 >(v/Cos a), and hence of iY.,(v/Cos a).

427

'I1IE METHOD OF STEEPEST DESCENT

If we tried to obtain the asymptotic expansion of J,(v/Cos a) from these results, we would obtain the zero series, indicating that this function is exponentially small compared to the leading term in (11.8-23). The explanation lies in the fact that we now integrate along At+ A2; hence the contributions of the segments CDE and EDC cancel each other, and the higher saddle point does not come into play. The appropriate path of integration is the curve ACB, with the sole saddle point at ( = ia. Here we have, putting

( = ia +w, g(ia)-g(() =Tga(l-cos w)+i(sin w -w), and hence, expanding in powers of w,

G

Tg a 2{ 1 = --w 2

2i Ctg a

2

w --w

4!

{I)

2 2i Ctg a

+

5!

w

3

}

+· · · .

Because

0 and x >0 let f(x) :=

Show that when

a

r

eiz'• d(.

is fixed and x--+ oo,

I

f(x)=exp(i7T)r(!+1)x- 11 "'+ ~iz 2a a rax

m-0

[ 1 -~ 1 /:)Jm (rx)

3. Let A denote a curve as indicated in Fig. ll.Be. Obtain the asymptotic expansion of 1/r(z) for z--+ oo from Hankel's integral

_1r(z)___ 1_f - 27Ti e ( C

A

zd

'-

431

111E METHOD OF STEEPEST DESCENT lm

~

Re ~

Fig.11.8e.

4. For s > 0, fJ > 0 let

G(s, {J) := s e -~

2

2

~I

J

e 4, . 2'[1- (1/

A

llr d(, '

where the (plane is cut between ( = 0 and ( = 1 and where A is a path encircling the cut counterclockwise. Show that for s = 1 there holds the asymptotic expansion G(1, fJ)..,

3~ J bm cos (11~+{J~m) sin~~ r(;)C~) 1

3

m/ ,

fJ-+ oo, where b 1 = 1, b3 = -(3/5), b5 = 3/35, b2 = b4 = · · · = 0. Also show that for fixed 8, 0 < 8 < ~/2, G(cos8,fJ)= 5.

2

J

1 ~11

sin 8

sin[fJI< 'IT', let F(a, b, z, c) denote the value of the hypergeometric function obtained by analytic continuation of the hypergeometric series along the ray from 0 to c. Then the asymptotic behavior ofF for z-+ oo, z E Sa and fixed a, b, c is described by the hypergeometric series (even though it may be divergent); that is, there holds the asymptotic expansion in 1/(z)"' 00

F(a,b;z;c)= ~

n=O

1

(a)n(b)ncn n.1

-(), Z n

(11.9-12)

z-+ oo,

Proof. We first assume Reb > 0. For lz I sufficiently large we also have Re z >Reb since z E Sa. Under these circumstances there holds Riemann's integral representation (8. 7-8), f(z) f ( F (a, b; Z; C ) = f(b )f{z _b) Jo 1 1

)-a b-1(

CT

T

1-

)z-b-1

T

d

T.

Apart from the r factor, this is precisely of the form (11.9-8) where z is replaced by z - b and

For T
0. We use the decomposition k-l

F(a,b;z;c)= ~ n=O

(a)n(b)ncn 1

n.

1

(a)t(b)kck kl() F(a+k,b+k;z+k;c), n .Z k

-()+ Z

and on applying the foregoing result to the hypergeometric function on the right the desired formula again follows. •

438

ASYMPT011C METHODS

We utilize Theorem 11. 9c to determine the asymptotic behavior as n -+ oo of the Legendre polynomials Pn (z) for a fixed complex value of z that is not real and outside the interval (-1, 1). (For the excepted case the asymptotic behavior was discussed by means of Watson's Lemma in§ 11.5.) We have

,j

1-z)

Pn(z) =~·\ -n, n + 1; 1; - 2- ,

n=O, 1,2, ....

The hypergeometric function appearing here is not of the form to which Theorem 11. 9c is directly applicable, because the large parameter appears in the numerator and not in the denominator. However, because of the large number of transformations of the hypergeometric series at our disposal, we may hope to find another hypergeometric representation of Pn (z) in which n occurs in the denominator only. It is convenient to consider separately the cases in which n is even and where n is odd. If n is even, n = 2m, then Pn is an even polynomial. By Goursat's formula (9.10-4) we then have

1+z) ( 1-z)} P 2m(z)=21{ F (-2m,2m+1; 1;22 - +F -2m,2m+1; 1;m {1/2)m ( 1 1 =(-1) ~F -m,m+2•2;z

2) .

By Euler's first transformation this becomes

1 1. 1. Z 2 ) ( 1)m{l/2)m -2m-l p{ ---;;;-!w \m+2,m+2'2'z 2 -1' where w := (1 + z) 112(1- z) 112 with larg(l + z )I< 1r, larg(l- z )I< 1r. Applying Goursat now in the opposite direction, we obtain

( - 1)m

(2m)! -2m-l 22m(3/2}zm W

1 iz )} 3 1 iz) +F( ····-+( · {F2m+1 2m+1·2m+-·--' ' 2' 2 2w '2 2w '

where the parameters are the same in both cases. We can get rid of the large numerator parameters by Euler's second transformation. This yields

~- .!_-~) ( - 1)m J2 (2m)! -1/2. {( +. )·-2m-1/2p(.! .!. 2 1T (3/2hm w w lZ 2'2' m+2'2 2w . )-2m-1/2 (1 1 2 3 1 iz ) } + ( w-zz F 2'2; m+2;2+2w ·

GENERAL ASYMPTOTIC EXPANSIONS

439

An analogous computation can be carried through for odd values of n using (9.10-5) twice. The two results may be combined into the single formula P. ( ) -

J2 ~ -1/2. {e -i.,./4( z+lw · )n+1/2p(! !. ~ . !-~) 2'2'2+n,2 2w

n z - 71' (3/2)n w

· 14 (z -iw)"+ 11 2 F (1 1 -+n3 1 iz ) } +e'.,. - -· -+2'2'2 '2 2w ·

(11.9-13)

This is a representation of the required type, in which the third parameter of the hypergeometric series is large and positive and all other quantities are fixed. To discuss the leading terms of the asymptotic expansion, we discuss two cases. (i) If z is real, -1 < z < 1, let z = cos 8, where 0 < (J < 71'. Then z ± iw = e±iB' and the foregoing reads

23 / 2 n 1 f.'"""""":Re 1 {exp [ in+( 1) 8-i71'] Pn(cos8)=---·71' (3/2)n vsin (J 2 4 · F(!.!. ~+n· !_cot 8)} 2'2'2 '2 2 ° In view of n! (3/2)n

1 r; 2 V~'

n~oo,

the leading term yields the asymptotic formula Pn(cos 8)- [ 71'n

r

~in 8

12

{ cos( ( n

+~)8 -~) + O(n -

1

) },

n~oo.

(11.9-14) (ii) If z is not real, we first assume Im z >0. Then z +iw = z +~, where iw=~=.Jz+1.Jz-1,0 lz- iwj. Hence the second term in (11.9-13) is exponentially small compared to the first, and the asymptotic behavior is given by

n~oo.

(11.9-15) If Im z < 0, then iw = -~, Im iw < 0; hence lz + iwl < lz- iwj. Only the second term in (11. 9-13) is now significant. However, in view of the

440

ASYMPTOTIC METHODS

reversal of sign the appearance of the asymptotic expansion is as in (11.9-15). PROBLEMS

1.

Let v be a complex number, but not an integer. Show: The sequence g0 (n), gt(n), ... , where

is asymptotic. 2.

For k = 0, 1, 2, ... , let ao

:E

- (1- z)" Log(1- z) =

5~c(n)z",

lzllt1l· For lt-t1lmax(O, Rev). We have r(t) = bo+b1(t-t1)+ ... +bm-1(t-t1)m- 1+(t-tdmrm(t),

where rm is analytic in It I 1 and the series is divergent.

450

ASYMYfOTIC METHODS

PROBLEMS 1.

Let q,. denote the probability that in n tosses of an ideal coin no run of three consecutive heads appears. Clearly, q0 = q 1 = q 2 = 1, and in probability theory it is established that q,.=!q,._ 1 +!q,._ 2 +kq,._ 3 • Show that the q,. possess the generating function

and deduce the asymptotic formula 1.2368398446 q,.- (1.0873780254)"+ 1 ' 2.

n-+OO

(Feller [1957], p. 260.) For n = 1, 2, 3, ... , let 1-n (2-n) 2 (-1)"- 1 a == 1+--+---+···+---. " 1! 2! (n -1)!

3.

Find the generating function of the sequence {a,.} and show that a,.-+ 0, n-+ oo. What is the asymptotic behavior of the numbers p,. defined by J~ ---=

Sin Jz

4.

00

L p,.z"? n-O

Let x > 1. From the generating function of the Jacobi polynomials, 00

L

P:·11 (x)t" = 2a+IIR- 1(1- t +R)-a(l +t+ R)- 11,

n=O

where R := (1-2tx +t 2) 112 , deduce by Darboux' method that for n -+00,

.J

. (x2 -l)-1/4( 27Tn )-112(x + x2 _ 1]"+1/2.

5.

Develop an analog of Darboux' Theorem for sequences with generating functions of the form p(t) =Log( 1-t)r(t),

where r is analytic in It! :s;;jtll·

§11.11. THE EULER-MACLAURIN SUMMATION FORMULA The objective in this section is the derivation of a classical formula, due to Euler and Maclaurin, that concerns functions f defined on the positive real

THE EULER-MACLAURIN SUMMATION FORMULA

451

line and, roughly speaking, provides a link between L~ =o f(n) and J~ f. The formula as such does not have the form of an asymptotic expansion, but it is frequently useful for obtaining asymptotic expansions of sums of the above focm as n ~oo. The starting point is a problem in numerical integration. Let f be a continuous complex-valued function on [0, 1] that has continuous derivatives of all orders. How can the integral

r

(11.11-1)

f(x) dx

be expressed in terms of the values of f and of the derivatives of f at the endpoints of the interval of integration? The problem can be solved in many different ways by the following general procedure. Let p 1 be any polynomial such that p~ (x) = 1. Applying integration by parts to ( 11.11-1) we then gee

r

f(x) dx

=[pt(x)f(x)]~- rPt(x)f'(x) dx.

The procedure may be repeated. If p 2 is any polynomial such that p;(x) = Pt (x"}, we have

r

Pt(x)f'(x) dx

= [p2(x)f'(x)]~- rp 2 (x)f"(x) dx.

Continuing in this manner, we get the following preliminary result: Let {pd denote any sequence of polynomials such that

k

(i)

Then there holds for m function f:

= 0, 1, 2, ....

= 0, 1, 2, ... and for any infinitely differentiable

(11.11-2)

1 I

+(-l)m+!

Pm+l(x)fm+l)(x) dx.

This formula is now simplified by subjecting the sequence {pd to the additional condition that (ii)

r

Pk(x) dx = 0,

k

=

1, 2, ....

0

3

Here and in the following we use the notation [g(x )]: := g(b)- g(a ).

452

ASYMYfOTIC METHODS

By (i), each Pk may be determined from Pk- 1 up to an additive constant. Condition (ii) now fixes that constant. Thus there exists one and only one sequence of polynomials satisfying both conditions (i) and (ii). In view of (i) it follows from (ii) that (11.11-3)

k=2,3, ....

Even more can be said. Fork= 1, (ii) in view of p;(x) = 1 clearly implies (11.11-4)

P1(x)=x-i.

Thus p 1(x +i) is an odd function. It follows that p 2 (x +i) is even and [using (ii)] that p 3 (x +!)again is odd. Generally, Pk (x +!)has the same parity ask. For odd k ;;;.: 3 this means that Pk (O) = - Pk (1), which is compatible with (11.11-3) only if Pk (O) = pk(1) = 0. Thus (11.11-3) may be replaced by the more precise statement P2dO) = P2k(1), P2k+1(0) = P2k+1(1) = 0,

k = 1, 2, ....

(11.11-5)

With this choice of the polynomials pk, the integration formula ( 11.11-2) now reads as follows: ft(x) dx =W(0)+/(1)]

-r m

P2k(O)(f2k-1l(l)-f2k-ll(O)]

(11.11-6)

k=1

-1

1

P2m+1(x)f 2m+ 1)(x) dx,

m=1,2, ....

The first term on the right is the result of evaluating the integral by one application of the trapezoidal formula. The remaining terms express the error of that formula in terms of the derivatives of odd orders of the integrand at the two endpoints, and in terms of an integral involving a derivative of arbitrarily high order. Let us now identify the polynomials Pk and the numbers Pk (0). We introduce the generating power series co

F(x) :=

r

k=O

Pk(x)l.

1HE EULER-MACLAURIN SUMMATION FORMULA

453

IfF has a positive radius of convergence for all x E [0, 1] and thus represents a function f(x, t), termwise differentiation with respect to x in view of (i) yields

f,.(x, t) =

00

00

00

k=O

k=I

k=O

L p~(x)l = L Pk-t(x)tk = t L Pk(x)l

= tf(x, t). Solving the ordinary differential equation in x, there follows f(x, t) = g(t) e'"

(11.11-7)

where g is yet to be determined. The condition (ii), interpreted in terms off, yields

i

l

f(x,t)dx=

0

00

L

k=O

11 Pk(x)dx·tk=1 0

for all t. On the other hand, from (11.11-7), if t ¢ 0, l t 1 f(x, t) dx = g(t) ~. 0 t

i

Together with the previous result this yields t

g(t)=-,-1,

e-

hence tetJc f(x, t)=-,-1. e-

The singularity at t = 0 is obviously removable, and on settingf(x, 0) := 1 the function f becomes analytic at t = 0. Its Taylor series has radius of convergence 27T for all x, and its Taylor coefficients are seen to satisfy (i) and (ii). Because these properties determine the Pk uniquely, there follows k

t etJc ltl (x) dx. To obtain the desired summation formula, let f be defined and have continuous derivatives up to order 2m + 1 for all x;;::. 0. By a shift of variable we then obtain from (11.11-11) for I= 0, 1, 2, ... 1+1

i

f(x)dx=i[f(l)+/(1+1)]

(11.11-12) 1

-(2m+ 1)!

I,

1+1 B2m+t(X -[x])/2m+l)(x)

dx.

It is convenient to define the Bernoulli function of order k by BZ(x):=Bk(x-[x]),

k=0,1,2,....

(11.11-13)

The Bernoulli function Bt is not a polynomial; rather, it is the periodic continuation with period 1 of the restriction of the polynomial Bk (x) to the interval [0, 1]. 4 The Bernoulli polynomials and the Bernoulli numbers are traditionally denoted by the same symbol, Bn. To avoid confusion, an argument is always used to denote the polynomial.

THE EULER-MACLAURIN SUMMATION FORMULA

455

Summing the expressions (11.11-12) from I = 0 to I= n -1, where n is an arbitrary positive integer, we find that the integrals can be combined into integrals between the limits 0 and n, and that the values of the derivatives at intermediate points cancel. We thus find (Euler-Maclaurin summation formula)

THEOREM 11.11

Let f be a complex valued function defined for x ;;;.: 0 having 2m + 1 continuous derivatives. Then the following identity holds for n = 1, 2, ... : ~f(O) + f(l) + f(2) +

· · · + f(n -1) +!f(n)

=i"f(x)dx+

I

r k=l

0

+( 2 m~ 1)!

Bzk [/2 k-t)(n)-f 2 k-t)(O)]

(2k)!

(11.11-14)

B!m+t(x)f 2 m+t)(x) dx.

It is obvious that this formula may convey information concerning the behavior of L.7=o f(l) as n ~ oo (the factors ! are, of course, of trivial consequence), and several examples to this effect follow. The formula in itself, however, is not asymptotic. It will become so only if the functions / 2k-O(n), k = 1, 2, ... , form an asymptotic sequence, and if the remainder integrals are suitably bounded. EXAMPLE

1

Let f(x) := (1 + x)- 1 • We have fzk-!)(x)= -(2k-1)!(1+xrzk,

k=1,2, ... , and therefore, replacing n by n-1 in (11.11-14) and transposing terms, 1 1 1 1 1 1 +-+-+ · · · +--+---Log n n-1 2 n 2 3 m B 1 k~! 2~k[1-n-2k]=2+

f" I

B!m+l(x)x-2m-2 dx.

Here we let n __,. oo while keeping m fixed. We know from (8.4-6) that the limit of the expression on the left equals Euler's constant y. The integral on the right converges in view of the boundedness of Bim+ 1(x). We thus obtain 11 1 1 1 1+-+-+ · · · +--+---Logn n-1 2n 2 3 (11.11-15)

456

ASYMPTOTIC METHODS

The integral is bounded by max

0"'%"'1

1 2m-l =const · n- 2m-l IB*2m+! (x)l--n2m+ 1 '

thus the series obtained by letting m -+ oo has property (A) of §11.1 where z = n. We thus get "" B2k 2k 1 1 1 1 1+-+-+ · · · +--+--Logn=='Y- L - n 2 3 n -1 2n k-1 2k '

n-+ oo.

(11.11-16)

This series is useful for the numerical calculation of Euler's constant; see § 11.12. EXAMPLE

2 Once again: Binet's function.

Let z > O,f(x) := Log(z + x ). We have f2k-ll(x)= (2k-2)! (z+x)2k 1•

k=1,2, ....

r

In view of

/(x) dx = (z +n) Log(z +n)-z Logz -n,

the sum formula yields form= 0, 1, 2, ... n

L Log(z +I)= (z +n +!) Log(z +n)-(z -!) Logz -n 1=0

B2k

m

+k~t 2k(2k -1) + -1 2m+ 1

1

[

(z +n)2 k

1

z2!-~J

f." (z B~m+! d +x) 2m+t · X

0

Subtracting from this the same formula where z = 1, m = 0, and n is diminished by 1, and also subtracting n Log z, we get (z)n+l ( z+n+-1) Log--z+n ( z--1) Logz-1- f."- 1 B~(x) Log--= -dx n !n• 2 n 2 1+ x 0

Here we let n -+ oo, keeping z and m fixed. By (II), §8.4, the limit on the left exists and equals -Log f(z ). Thus the limit must exist on the right, which in particular implies the existence of ~:=

l

ao

o

Bf(x) -1-dx.

+x

THE EULER-MACLAURIN SUMMATION FORMULA

457

We thus find Logf(z) = (z -!)Log z- z + 1 +{3 m B2k 1 1 f"" Bfm+l(x) + k~l 2k(2k -1) z 2 k-t +2m+ 1 J0 (z + x) 2 m+t dx. Recalling (IV), §8.5, this may be recast as a formula for Binet's function J(z). Because we know that J(z)-+0 for z -+OO, there follows 1 +{3 = Logh; and thus ~ B2k 1 1 J."" B~m+ 1 (X) J(z)= k::t2k(2k-1) z 2 k-t+2m+1 a (z+x) 2 m+tdx,

(11.1 1- 17)

which may be regarded as a quantitative version (with remainder term) of the asymptotic expansion of Binet's function. EXAMPLE

3 Numerical integration.

Let g be 2m + 1 times continuously differentiable on [0, 1]. We apply the summation formula to f(x) := g(hx), where h := 1/n. Writing x1 := lh, l = 0, 1, ... , n, there results

r

h[ig(xo) + g(x1) + · · · + g(Xn-1) +ig(xn)]

=

g(x) dx +

h2m+l +(2m+ 1)!

k~1 (~~)![ga if 1 1 Log n >a --y- 2n + 8n2' where 1' denotes Euler's constant, and s,. < a if 1 Logn 0 there follows (-1)kcm,k >0,

k = 1, 2, ... , m.

(11.12-9)

= 0, 1, 2, ...) be defined by

Let now the functions rm (m

q(h) = ao+a1h +a2h 2+ · · · +am-1hm- 1+rm(h).

(11.12-10)

By (11.12-2) we have for h >0 lrm(h)I~ILmhm,

m =0, 1, 2,....

(11.12-11)

Evaluating the numbers qn-k,o=q(pn-k) by (11.12-10), substituting in (11.12-5) and rearranging the summation we get

_

-

_

-

m-1 ~

1...

p=O

m

ap

~

1...

k=O

m-1 ~

1... Qp{J

p=O

Cm,kfJ

(n-k)p

nptm (p-p) +

m

+

m

~

1...

k=O

~

1...

k=O

Cm,kTm

(pn-k)

Cm,kTm (pn-k) .

By ( 11.12-8), all terms of the first sum except the first term vanish, and we are left with m

qn,m =ao+

L

k=O

Cm,krm(pn-k).

(11.12-12)

Using the estimate (11.12-11), this yields

,:::: Iqn,m -aoI""'Jl.m

m

~ I...

k=O

ICm,k IP (n-k)m,

which by virtue of (11.12-9) may be written lqn,m -aol,;;; ILmPnmtm(-p -m).

(11.12-13)

The representation ( 11.12-7) yields ( -m) (1+p1-m)(1+p2-m) ... ( 1 +pm-m) tm -p = (1-p)(l-p2) ... (1-pm) _ -[m(m-1)]/2 (pm- 1+ 1)(pm-2+ 1) · · · (1 + 1) 2 (1-p)(1-p ) • • • (1-pm)

-p

-[m(m-1)]/2 =p

2 (1 +p)(1 +p 2) • • • (1 +pm) 1+pm (1-p)(1-p 2) • • • (1-pm)"

466

ASYMPTOTIC METHODS

Defining the convergent infinite product

1+

oo

m

t(p) := mQI 1 _;m•

(11.12-14)

we thus have tm(-p-m).s;;2t(p)p-[m(m-i)]f 2,

m = 1, 2, ....

Using this in (11.12-13), we obtain THEOREM 11.12c

Let the constants /Lm satisfy (11.12-2), and let t(p) be defined by (11.12-14). Then the following estimate holds for the elements qn,m of the Romberg array defined by (11.12-3): lqn,m -

aol ~ 2t(p )~t,P m[n-(m-1)/ 21, n=0,1,2, ... ;

m = 1, 2, ... , n.

(11.12-15)

By taking logarithms it is easily shown that 1 1/(1-p) t(p)< ( ;

+p)

1-p

more accurate values of t(p) may be computed as shown in Problem 6. We thus find, for instance, t(!) = 8.256 ... '

t(;}) = 1.969 ... . Thus if bounds for the constants ILm are known, explicit error estimates for the Romberg scheme are easily obtained from (11.12-15). Moreover the following is a direct consequence:

COROLLARY ll.lld Let l'!{mpm 12 ~0. m~oo. Then each diagonal of the Romberg scheme converges to a 0 faster than linearly. That is, for any A > 0 (no matter how large) and fork= 0, 1, 2, ... , there holds

lim Amlqm+k,m -aoi = 0.

m-+oo

Proof.

Letting n

= m + k in (11.12-15) we get

Am jqm +k,m - aol.;;;; 2t(p )JLmA mp m.

467

THE NUMERICAL EVALUATION OF LIMITS

But /Lm

,\ m

P

m(m/2+k+i/2) = [(

1/m m/2)(,\

/Lm P

P

k+i/2)]m

'

and the conclusion follows from the fact that, by virtue of the hypothesis, the expression in brackets is < 1 for sufficiently large values of m. • The hypothesis made on the growth of /Lm is a very mild one. Stirling's formula shows that it is satisfied, for any p < 1, for such strongly diverging series where /Lm = O{(pm)!) or /Lm = O((m !)P), where pis an arbitrary fixed integer. We now present several applications of the Romberg algorithm.

EXAMPLE

1

Numerical integration over a finite interval

Let f satisfy the conditions of Example 3, § 11.11, and let q(h) denote the trapezoidal value of J~f(x) dx calculated with step h. Then by the result of the example q(h) admits an asymptotic expansion of the form (11.12-1) where his to be replaced by h 2 • Successive halving of the integration step thus corresponds to the value p =!in the Romberg scheme. By Problem 16, § 11.11, the constants f.Lm are no larger than const.(2m)! (27Tp 1

r

2 m.

Thus at most

and the hypothesis of the corollary is clearly fulfilled. The resulting method of integration is known as Romberg integration and is widely used in numerical practice. For algorithmic details the reader is referred to numerical analysis texts. EXAMPLE

2

Calculation of natural logarithms

Let x > 1, and let it be desired to calculate Log x. By calculus, Log x =lim q(h), h~o

where

xh-x-h q(h) :=

2h

,

h>O.

Here q admits the convergent expansion

Although the coefficients of this expansion depend on Log x, the quantity to be computed, the Romberg algorithm is applicable, because no knowledge of these

ASYMPTOTIC METHODS

468

coefficients is required. Because the series proceeds in powers of h 2 , a successive halving of h is consistent with p = 1. The elements in the first column of the array are ln := qn,o=q(T")=2" Sin(z-n Logx). (11.12-16) The evaluation of hyperbolic functions is avoided by using the recurrence relation x 2-1 x-1 lo=h, /1 = .[;_, (11.12-17)

f2L

ln+l=lnV~/ +/ , n

n=1,2, ... ,

n-1

which is advantageous also from the point of view of numerical stability. By way of illustration, we show in Table 11.12a the scheme resulting for x = 10. Table 11.12a.

4.950000000 2.846049894 2.431876170 2.334508891 2.310541291

Romberg Scheme for Log 10

2.144733192 2.293818262 2.302053132 2.302552090

2.303757266 2.302602123 2.302585354

2.302583788 2.302585088 2.302585093

The constants /Ln are easily estimated as follows: If h.;;; 1, then (11.12-2) holds with h 2 in place of h and ao (Logx)2m+l (Logx)2n+l (Logx)2n+l ILn

:=

:E

m-n (2m+ 1)!

.;;;

(2n + 1)!

el.o•% =x

(2n + 1)!

.

Using t(l} < 2, Theorem 11.12c thus yields the following estimate for the diagonal elements, simplified by approximating the factorials by Stirling's formula: e Log X ] 2m+! lqm,m-Logxjo;;;4x [ zm12(2m+1) 0

For m = 5 (which requires the evaluation of four square roots) and x .;;; 10 the error thus can be guaranteed to be less than 10-9 • EXAMPLE

3

Calculation of Arcsin X

Let 0 .;;; x .;;; 1, and let a := Arcsin x. Clearly, a=limq(h), h .. o

where q(h) :=

In view of

sin~ha)_

THE NUMERICAL EVALUATION OF LIMITS

469

the limit a of q(h) can be computed from the Romberg scheme with left column s,. :=q,.,0 = q(T") = 2" sin(2-"a ),

where p = 1/4. The values s,. are obtained most conveniently from the recursive scheme SoFX,

n =0, 1, 2, ... ,

(11.12-18a) (11.12-18b)

which differs from (11.12-17) only in the initial conditions. For the error of the diagonal elements we find as in example 3 ]2m+l ea m =0, 1,2, .... • jqm,m -aj :s;;4 [ 2m/2( 2m + 1) EXAMPLE

4 Euler's constant

For n = 1, 2, ... , let 1 1 1 1 y(n):= 1 +-+-+···+--+--Log n. 2 3 n-1 2n The definition of Euler's constant shows that lim,. ..oo y(n) = 'Y· More precisely, by (11.11-16 ), y(n)""'y-

B2k 1 L -----v;-, k-t 2k n DO

n -+00.

Thus the function 1 1 1 h= 1·4'9'16····· satisfies (11.12-1). To evaluate 'Y by the Romberg algorithm, we use the initial values 1 -+.!._!__n Log2 q 0 :=...,(2")= 1+·!+!.+. · · + "· 2 3 2" -1 2 2" ' , which means that we are taking p = 1. Part of the resulting scheme is shown in Table 11.12b. The last entries in the last row are correct to all places.

Table ll.llb. Romberg Scheme for Euler's Constant 0.5000000000 0.5568528194 0.5720389722 0.5759156012 0.5768902710 0.5771342926

0.5758037592 0.5771010231 0.5772078108 0.5772151609 0.5772156332

0.5771875074 0.5772149300 0.5772153653 0.5772156509 0.5772156624 0.5772156635 0.5772156647 0.5772156649 0.5772156649

ASYMPTOTIC METHODS

470

By Problem 3, §11.11, we have JL. :s;;JB 2 .Jn- 1 • Using an obvious bound for the Bernoulli numbers, Theorem 11.12c thus yields the following estimate for the errors of the diagonal entries of the array:

n =I, 2, .... For instance (neglecting rounding errors) no more than 187 steps of the algorithm are necessary to compute y to 10,000 decimal places, without using explicit values of Bernoulli numbers.

PROBLEMS 1. 2.

Calculate TT/2 by choosing x = 1 in Example 3. Let x be a given real number. Show how to compute x) Arctan . _tan(h a _ ______:_ ___:__ :=Arctan x = 11m ,., h h-o

by means of the Romberg algorithm, using the starting values

Also show that t 0 =x,

3.

n

= 1, 2, ....

Let r be a rational function with a zero of order 2 at co and with no poles at the positive integers. Using the fact that the partial sums of

L

r(n)

n=l

4.

admit an asymptotic expansion in powers of n _,as n ~co, show how to speed up the convergence of the series by means of the Romberg algorithm. Invent a speed-up algorithm for computing a 0 := lim q., if q. is known to admit the generalized asymptotic expansion

n ~co,

5.

6.

where 1 = y 0 > y 1 > · · · > 0, and where the "Yk are known and the ak are unknown. In the notations of the preceding problem, speed up the convergence of q. if both the .ak and the "Yk are unknown. In the notation of §8.2, ( ) =p(p) q(p)" tp

471

NOTES

Show that

and use the known power series for the infinite product q to calculate accurate numerical values of t(!) and t(!).

SEMINAR ASSIGNMENTS

1. Study the numerical effectiveness of an asymptotic expansion such as ( 11.1-1) or ( 11.1-16) by plotting, in the complex plane, the curves where an accuracy of 1o-p (p = 1, 2, ... ) can be attained by breaking off the expansion at the most favorable term. 2.5 Study the probability density Fn (r) of a sum 71 = L~ = 1 ~k of independent random variables, each having the probability density O:s;;r:s;; 1, otherwise. Start from the recurrence relation

n = 1, 2, ... , to obtain the correspondence

(a) Obtain a piecewise polynomial representation of Fn. (b) Use the complex inversion formula to obtain

F(r) =

2~

L: [sin~7~ )r 2

cos wrdw

and study the asymptotic behavior of Fn (T) as n ~ oo. In addition to the main saddle point at w = 0 the subsidiary saddle points at the solutions of w/2 = tg w/2 should be considered. Graph some of the results and compare with the explicit values obtained under (a).

NOTES The following are some specialized texts dealing with asymptotic methods: Erdelyi [1956], De Bruijn [1961], Berg [1968], Sirovich [1971], Nayfeh [1973], Dingle [1973], Olver [1974]. The last work and Wasow [1965] are 5

Suggested by Dr. J. Waldvogel.

472

ASYMPrO'DC METHODS

oriented toward differential equations. A major portion of Doetsch [1955] is devoted to asymptotic methods. §11.1. The enveloping property defined in Problem 8 is discussed by Polya and Szeg() [1925], Vol. I, p. 26. In some older texts this is taken as the defining property of an asymptotic expansion. §11.2. Problems 6 and 7 are suggested by, but not identical with, Shafer [1975]. §11.3. For Theorem 11.3b see Ritt [1916]. The remaining results are proved as in Erdelyi [1956]. §11.4. The proof of Theorem 11.4a is an amalgamation of ideas of Coddington and Levinson [1955], Chapter 4, and Wasow [1965], Chapter 4. §U.S. For an even more general version of the Watson-Doetsch lemma (Theorem 11.5) see Doetsch[1955], p.45. For the incomplete ffunction both the definition by (11.5-7) and the definition given in Problem 2, §10.5, are in use. §11.6. See Buckholtz [1963] for an interesting generalization of Problem 6. For Problem 7 see Henrici and Hoffmann [1975]. Concerning the origin of Problem 8 see Knuth [1974]. §11.7. Theorem 11.7 is from De Bruijn [1961], and Example 2 is a simplified version of Problem 4, which is taken from the same source. §11.8. For a generalization of the method of steepest descent see Chester, Friedman, and Ursell [1957]. Many further applications of the method occur in Szego [1959] and in most books on special functions. §11.9. General asymptotic series are discussed in Erdelyi [1956]. §11.10. For Problem 1 see Feller [1957], p. 260. §11.11. For the history of the Euler-Maclaurin sum formula see Whittaker and Watson [1927], p. 127. Some interesting examples are in the appendix of Hardy [1910]; Problem 8 is taken from there. For Problem 6 see Wilkinson [1961], p. 284. Problem 14 was used for numerical calculations in the ETH dissertation of W. Gander (unpublished). Boas and Stutz [1971] compare several methods of estimating sums by integrals. §11.12. Extrapolation to the limit was used informally by Richardson and Gaunt [ 1927]. A systematic algorithm was proposed by Romberg [1955]. A rigorous analysis of the Romberg process, as applied to integration, was given by Bauer, Rutishauser, and Stiefel [1962]. Rutishauser [1963] discusses applications to numerical differentiation, Filippi [1966] to the computation of elementary functions. The polynomials t(p) are expressible in terms of the "Gauss binomial coefficients" defined by 1--y":• ... 1--yn-:+1 [n]r := 1--y" 1--y 1--y 1--y (Zurich lecture by G. Polya on January 29, 1970). For a variant of Romberg quadrature see Laurie [1975].

12 CONTINUED FRACTIONS Like infinite series and infinite products, continued fractions can be considered as arising through the composition of certain types of Moebius transformations. Again like infinite series and products, continued fractions can be used to represent certain types of analytic functions. Contrary to representations by power series, continued fraction representations may converge in regions that contain isolated singularities of the function to be represented, and contrary to representations by infinite products, continued fraction representations in many cases converge very rapidly. In addition to the representation of analytic functions, continued fractions also have interesting applications in stability theory, in asymptotics, and in number theory. Wherever possible, the treatment of continued fractions given here is based on Moebius transformations. Indeed, the very definition of continued fractions makes use of that concept. In comparison to some earlier presentations of the theory, the explicit use of Moebius transformations, in addition to simplifying some concepts and some proofs, injects a geometric note into the theory, which is deemed desirable.

§12.1. DEFINITION AND BASIC PROPERTIES A continued fraction is often defined as "an expression of the form" a1

az

bl+---------------bz + -------------

(12.1-1)

which it has become customary to write in a typographically more convenient form as follows: ~+...!!.U+~+.... (12.1-2) lbl ~ ~ 473

474

CONTINUED FRACI'IONS

Such an "expression," however, is at best a prescription to perform certain algebraic operations. As a formal definition, it stands on the same level as the definition of an infinite series as "an expression of the form" a1 +a2+a3+· · ·. To obtain an operative definition of a continued fraction, we recall the definitions of an infinite series and of an infinite product given at the beginning of §8.1. To prepare for the analogy to continued fractions, the definition of an infinite series given there might also be phrased as follows: An infinite series is a pair of sequences [{an}';", {sn}';"], where the an are complex, and where, denoting by tk the Moebius translation tk : u-+ ak + u, k = 1, 2, ... ,

and

= amTm-l + bmPm-h qm = amSm-l + bmqm-h

Pm

m = 2, 3, .... In view of (12.1-7) the latter relations may be written (in an obvious abbreviated form)

(Pm-2)

+ bm (Pm-1). ( Pm) =am qm qm-2 qm-l

(12.1-Sa)

Together with the initial conditions Po= 0, Pt = a1 and qo = 1, q1 = b1 following from (12.1-7) and from the fact that M1,1 = Mt. these relations permit the recursive computation of the sequences {pm} and {qm}· Under the simpler initial conditions P-1 = 1, Po=O (12.1-Sb) • q_l =0, qo= 1 the relations (12.1-Sa) are seen to hold already from m = 1 onward. The numbers Pm and qm are called the mth numerator and the mth denominator of the continued fraction (12.1-5). Both the numerators and the denominators satisfy the same recurrence relation; only the initial conditions are different. For reference we note THEOREM 12.1a

Let tk be defined by (12.1-3). Thenforn tion

= 1, 2, ... , the Moebius transforma-

is associated with the matrix

M 1,n

= (Pn-1 qn-l

Pn) ' qn

(12.1-9)

478

CONTINUED FRACDONS

where the Pn and the qn are defined by the recurrence relations (12.1-8). Because the nth approximant is given by Wn

= tl,n(O),

we immediately obtain COROLLARY 12.lb

For n

= 1, 2, ... , the nth approximant of the continued fraction Pn qn

(12.1-5) is (12.1-10)

Wn=-.

The evaluation of Wn by the descending method, that is, by the evaluation of the recurrence relations (12.1-8) and the quotient (12.1-10), normally requires one division, 2n - 3 additions, and 4n -6 multiplications. The number of multiplications is cut in half if either all partial numerators or all partial denominators are equal to 1. This can always be achieved by an equivalence transformation (see below). EXAMPLE

2

To evaluate the c.f. of example 1 by the descending method, we form the following scheme: m

-1

am bm Pm qm

0 0

2

3

4

5

6

2 -1

7 6

-1 5

3 4

-2 3

-5 2

2 -1

12 1

58 6

268 27

688 69

36 3

0

1

We again see that the continued fraction has the value w6 = 36/3 = 12. At the same time, the algorithm yields the values of all approximants, 2 w. = -1 =-2,

w2 =r=f-J+rfJ= 12 =12 etc. -1 6 I '

Equivalent Continued Fractions. Two continued fractions are called equivalent if they have the same sequence of approximants. The equivalence of two continued fractions C and C* is denoted by the symbol

C=C*. It is dear that if one of two equivalent continued fractions is convergent, the other is likewise convergent, and has the same value. It will now be shown

DEFINmON AND BASIC PROPERTIES

479

that any continued fraction is equivalent to a continued fraction whose partial numerators are 1, and any continued fraction whose partial denominators are different from zero is equivalent to a continued fraction with partial denominators 1. Let {em}~ be any sequence of nonzero complex numbers. The Moebius transformation 1m

am

:u~--­

bm+U

evidently can be represented in the form tm(u)=

c,.am Cmbm+CmU

and this may be regarded as the composition of two transformations, 1m= Sm o rm, where

Thus we have

Here the transforms

may be regarded as a single RIT transform. Because rm(O) = 0 we have, if c 0 := 1,

Thus we obtain the important relation

cP am= cP Cm-!Cmam m~l bm- m~l Cmbm '

(12.1-11)

valid for an arbitrary sequence of nonzero complex numbers c 0 = 1, c., c 2, •••• It should be noted that although equivalent continued fractions have the same approximants, their numerators and denominators are, in general, not the same. If we denote by Pm• qm the numerators and denominators of the

480

CONTINUED FRACDONS

fraction on the left of (12.1-11) and by p!, q! the corresponding quantities of the fraction on the right, then it follows from (12.1-8) that

p!, = C1C2 · · · CmPm•

(12.1-12)

q!,=c1c2 · · · Cmqm,

m=1,2, .... To find for the c. f. on the left of (12.1-11) an equivalent c. f. with partial denominators 1, it evidently suffices to pick the Cm such that Cmbm = 1, which is always possible if bm ¢- 0. We thus find b-i cP am= cP b-i m-1 m am (12.1-13) m=l bm m=i 1

(b 0 := 1). To find an equivalent c. f. with partial numerators 1, the Cm have to be chosen such that m=2,3, .... This implies

1 hence (12.1-14) We thus have obtained

cP m=l

am_ cP 1 bm = m=l bmcm'

(12.1-15)

where the em are defined by (12.1-14). The fraction on the right is called the reduced form of the fraction on the left.

Improper Continued Fractions. In some contexts it is advantageous to consider continued fractions of the form ho+

+ ~ +· · · 2

hi

b2

(12.1-16)

where b0 is some complex number. Such fractions might be called improper continued fractions. Their formal definition can be given along the same lines as that of the "proper" c. f. considered above, with the following differences: The sequences {bn} and {Wn} now begin with the elements b0 and w 0 , and we have

481

DEFINfi10N AND BASIC PROPERTIES

where tt. t2 , ••• , are defined as before, but where t 0 : u-+ b0 + u. All the remaining definitions and terminology remain unchanged. We leave it to the reader to verify that

n = 0, 1, 2, ... , where the sequences{pn}and {qn} again satisfy the difference equation (12.1-8a), but now obey the less symmetric initial conditions

= 1, q_l = 0,

P-1

Po=bo

(12.1-17)

qo= 1

in place of (12.1-8b). Thus the nth approximant Wn fraction (12.1-16) is again given by n

= to.n(O) of the improper

= 0, 1, 2, ....

Even and Odd Part of a Continued Fraction. By the even [odd] part of a proper or improper continued fraction C is meant a continued fraction whose sequence of approximants consists of the approximants of even [odd] order of C. Thus if {wn} denotes the sequence of approximants of C, the sequence of approximants of an even part of Cis {w 2 n}, and that of an odd part {w 2 n+ 1}. Thus if C converges, both its even and its odd part converge and have the same value; on the other hand, if both an even and an odd part of C converge, then C itself converges if and only if both parts have the same value. In later applications we mainly require the even and odd parts of the continued fraction (12.1-18)

If ak tdu) :=-, 1 +u

then ak tkotk+I(u)=1+ak+d(1+u)

ak(1+u) 1+ak+1+u

CONTINUED FRACfiONS

482

where

* (u ) ..-_ __a:.:. ka.. .:.k:.. .;+.. .:.1_ tk+1 1 +ak+l +u

sdu) := ak- u, Thus because sdO) = tdO), if n equals W2n

= t1 ° t2 ° · · ·

0

l2n-1

= t1 ° (s2 ° t!> (s4 ° 0

= (t1 ° s2)

0

(t~ 0 s4) 0

= 1, 2, ... , the 2nth approximant of C ° l2n(O)

rn ° ... ••• 0

0

(s2n-2 ° t!n-1) 0 S2n(O)

(t!n-1 ° S2n)(O)

where

Thus an even part of C is

c· ._ ~.-11 +a2

a2a3

1

(12.1-19)

l1 +a3+a4

A similar computation yields the odd part (12.1-20)

The Difference between Consecutive Approximants. It was shown in Theorem 12.la that the matrix M 1,m ·= · (Pm-1 qm-1

Mk :=

(01

ak)

bk '

Pm)

qm

k = 1, 2, ....

From the fact that the determinant of a product of matrices equals the product of the determinants of the factors, we immediately obtain the important relation (12.1-21)

483

DEFINITION AND BASIC PROPERTIES

m = 1, 2, .... If qm- 1qm

~0.

we find on dividing that

Wm- Wm-1 =Pm _Pm-1 = (-l)m+lala2 ... am. qm qm-1 qm-lqm

(12.1-22)

Obviously, Wm = (wm- Wm-l)+(wm-1- Wm-2)+ · · · +(wl- Wo)+ Wo. Thus if no denominator vanishes, we find that the approximants of a proper continued fraction can by virtue of w 0 = 0 be written in the form al ala2 ( 1)m+l ala2 . .. am Wm =-----+· • ·+• qoql qlq2 qm-lqm

A continued fraction Cis called equivalent to an infinite series S if for n = 1, 2, ... , the nth approximant of C equals the nth partial sum of S. We again indicate the equivalence of C and S by the symbol C == S. This result permits us to state that if the denominators qh q 2 , ••• , of the continued fraction on the left are all different from zero, then

$

am==

m=l bm

r

(-l)m+l ala2 .. . am.

qm-lqm

m=l

(12.1-23)

There follows, in particular, that the fraction on the left is convergent if and only if the series on the right is convergent. The Convergence of a Continued Fraction with Positive Elements. We now discuss a general criterion for the convergence of continued fractions with positive elements. Because any such fraction is equivalent to a reduced fraction (partial numerators one), we begin by discussing such fractions. The simple main result is as follows. THEOREM 12.1c

Let f3m > 0, m

= 1, 2, ....

Then the continued fraction

c := $

__!__

m=l f3m

is convergent if and only if the infinite series 00

S :=

L

m=l

f3m

is divergent. Proof.

The denominators qm of C satisfy the recurrence relation qo= 1,

(12.1-24)

m = 2, 3, .... It is evident that qm > 0, m = 0, 1, 2, ... ; thus as a special case

484

CON11NUED FRACI'IONS

of (12.1-24),

The fraction C converges if and only if the alternating series on the right converges. From (12.1-24) therefollowsqm+ 1>qm-t. m = 1, 2, ... ; hence

m = 1, 2, ... ; hence the absolute values ofthe terms ofthe alternating series on the right are monotonically decreasing. Thus by the Leibniz criterion the series is convergent if and only if p2.1-25) Now again by (12.1-24), because qm;;:;. 'Y := min(l, 13 1), m

qm-1qm

= 1, 2, ... ,

= qm-2qm-1 + 13mq 2m-1 ;;.qm-2qm-1 +13m'Y 2 ;;:;. (131 + 132 + · · • + 13mh 2·

Thus if S diverges, (12.1-25) holds, and Cis convergent. On the other hand, once more from (12.1-24),

= qm-2 + (1 + 13m)qm-1.;;; (1 + 13m)(qm-l +qm-2); hence by induction, because q0+q 1= 1 +13t. qm-1 +qm

qm -1 + qm ... (1 + 131)(1 + 132) ... (1 +13m) < etl' +{3 2 +···+{3"'. Hence if S converges, and if u denotes the sum of S, m=1,2, ... ; thus by the inequality of the geometric and arithmetic mean,

qm-1qm .;;;!(qm-1 +qm) 2o;;;!e 2u for all m = 1, 2, .... Consequently, (12.1-25) does not hold, and C fails to converge. • We leave it to the reader to deduce from Theorem 12.1c convergence criteria for continued fractions

;p

am m=113m where am >0, 13m >0, by using the equivalence relations (12.1-13) and (12.1-15).

485

DEFINmON AND BASIC PROPERTIES PROBLEMS 1.

Evaluate the continued fraction

2.

by both the ascending and the descending method. Let am >0, 13m >0, m = 1, 2, ... , and let

I

Jf3mf3m+l =

m=l

am+l

00.

Show that the continued fraction

3.

4.

is convergent. Discuss the convergence of the continued fraction ao a cl>m-1 b' where a and b are complex numbers, a ¥- 0, by explicitly solving the difference equation for the numerators and the denominators. Check the result of Problem 2 when a and b are positive. Show that the numerators Pn and the denominators qn of the improper continued fraction ao

bo+ cl> m-1

a .....!!!

bm

can be represented by determinants, as follows:

Pn =

ho 1 a1 bl 1 a2 b2 1

0

qn =

an-I bn-1 1 an bn

bl 1 a2 b2 1 a3 b3 I

0

0

0

an-I bn-1 1 an bn

(Elements not shown are zero.) [Expand the determinants in terms of the elements of the last row and thus show that they satisfy the recurrence relations (12.1-Sa).]

486

CONTINUED FRACTIONS

5.

Prove that no two of any three consecutive approximants of an infinite continued fraction with nonzero elements can have the same value. The numerators q. of the infinite continued fraction

6.

1 1+ 1J+ 1 1+· ..

~ ~

fb;

are on occasion denoted by [b~> b 2 , ••• , b.] and are called Gauss brackets. The empty bracket is given the value 1. Prove the following relations:

7.

[b., b.-~> . .. , btl;

(a)

[b~> b2, ... , b.]=

(b)

[bh b 2, ... , b.]= b 1[b 2, b3 ,

(c)

l[bh····b.] [b 2, ... , b.]

Let c, -:/- 0, i

=

••• ,

b.]+[b 3 , b4 ,

••• ,

b.];

[b., ... ,b.-1]1=(-1)" ' [b 2, ... , b.- 1]

0, 1, .... Prove Euler's relations

I

c3/c2

ll+c 3 /c 2

8.

Show that for arbitrary z -:/- 0 ez

=

1 +~-~

;~; 1_4:~; 1_ 1:~;I_ ....

9. Prove: 1

m

oo

ci> - = - . e-1

m~lm

10.

Showthatfor-l~x~l

9·x ·x GJ1 + ~ +1 5-3x 3-x

2 2 1 ~5·x 2 +· · · 2 + 7-5x

2

Arctan x =

2

and, consequently,

~= _!j+_!j+~+~+~+ ....

11.

4 11 12 12 12 12 Let am> 0, 13m> 0, m = 1, 2, .... Show that the even and the odd part of the fraction

;; am m=l

12.

/3m

always converge. Let /3 > 0. Show that the continued fraction oo

mk

-

m~l

/3

converges for k ~ 2 and diverges for k > 2.

CONTINUED FRACI10NS IN NUMBER THEORY

13.

487

Let the elements of the fraction cf>(am/bm) be positive. Show that there exist constants c~> c2 , ••• , such that the denominators of the equivalent fraction (12.1-11) all are 1. [Solution:

§12.2. CONTINUED FRACilONS IN NUMBER THEORY We expect the reader to be familiar with the Eudidean algorithm. Let k = k 0 and I= k 1 be two positive integers. The Euclidean algorithm consists in determining integers k 2 > k 3 > · · · ;;;;. 0 and b0 , b~o ... (b 0 ;;;;. 0, b; ;;;;.1, i = 1, 2, ... ) such that

ko= hok1 +k2, kl

= blk2+kJ,

(12.2-1)

Because a decreasing sequence of nonnegative integers necessarily reaches the value zero in a finite number of steps, the Euclidean algorithm necessarily terminates; that is, there exists an index n such that kn+ 2 = 0, the last relation (12.2-1) having the form (12.2-2) It follows from the elements of number theory that form = 0, 1, ... , n -1 every divisor of krn and krn+l is also a divisor of krn+ 2· Thus every divisor of k 0 and k 1 is a divisor of kn+l· Conversely, every divisor of kn+l is seen to be a divisor of k 0 and k 1 • Tnus kn+l is the greatest common divisor of k 0 and k~o and the Euclidean algorithm is commonly regarded as a convenient tool (as opposed to the decomposition in prime factors) to determine this greatest common divisor. EXAMPLE

1

k 0 = 63, k 1 = 24 yields

63=2. 24+15 24= 1. 15+9 15=1·9+6 9=1·6+3 6=2·3+0

Thus the greatest common divisor of 63 and 24 is 3.

CONTINUED FRACDONS

488 EXAMPLE2

ko = 64, k 1 = 25 yields 64=2 ° 25+14 25 = 1 ° 14+ 11 14=1·11+3 11=3·3+2 3=1·2+1 2=2·1+0 Thus the greatest common divisor of 64 and 25 is 1.

In a slightly different vein, we here look at the Euclidean algorithm as a tool to obtain a certain standardized representation for rational numbers. For m = 0, 1, ... , n let

Then the relations (12.2-1) may be written

1 Po=bo+-, PI

1 P2

PI =bl+-, (12.2-3) 1

Pm=bm+--, Pm+l Pn =bn. Starting with any rational number p := k/ I, the numbers p; and b; resulting in the foregoing manner through the Euclidean algorithm may also be characterized by the conditions p 0 = p,

m=O, 1, ...

(12.2-4)

Here, for any real number ~ the symbol [~] denotes the greatest integer not exceeding ~Let us now define the Moebius transformations

t0 (u) := bo+u,

1 tm(U) := -b- , m+u

m = 1, 2, ... , n.

CONTINUED FRACDONS IN NUMBER THEORY

489

Then the relations (12.2-3) show that

1

-=

m= 1,2, ... ,

tn(O),

n-1,

Pn

Thus evidently P = too t I

o

tz

o ••• o

tn (0),

which we also write p

= bo+_!j+ __!j+ ... +_!j. ~ ~

lh:-

(12.2-5)

Thus the rational number p has been represented as the value of an improper, terminating, reduced continued fraction ~hose partial denominators are integers, bi > 0 (i = 1, 2, ... , n -1), b" > 1. Such a continued fraction is called simple. The foregoing illustrations of the Euclidean algorithm yield the following examples of representations of rational numbers by terminating simple continued fractions: 63

24 = 2

_!j _!j _!j _!j +11+11+11+12,

_!j 1 1 _!j _!j _!j 64 25 = 2 +11+1t+IJ+Il+rz. Because the value of any terminating simple continued fraction obviqusly is a rational number, we have THEOREM 12.2a

A real number is rational if and only if it is the value of a terminating simple continued fraction. It is easy to see that a rational number can be represented in one way only by a simple terminating continued fraction. Diophantine Equations. Let k and I be two positive integers that are mutually prime, that is, whose greatest common divisor is 1. Here we consider the problem of finding all pairs of integers (x, y) solving the equation kx -ly = 1,

(12.2-6)

490

CONTINUED FRACfiONS

called a Diophantine equation. It follows from simple facts in the theory of ideals that this equation always has ·a solution. The theory of continued fractions provides an algorithm for actually constructing the solution. Let the number k/ l be expanded in a terminating simple continued fraction:

~ = ho + .-!.J + ____!_j + · · · + ____!_j. 1

lbl

~

~

Let Pt. qk denote the numerators and denominators of this fraction. Then by (12.1-21), because all a;=:= 1, m = 1, 2, ... , n. (12.2-7) Naturally, Pn ~ qn [' the value of the continued fraction. Because (12.2-7) form = n shows that Pn and qn have no common divisors, there follows Pn = k, qn = [. To find a special solution of (12.2-6), we distinguish two cases according to the partity of n. If n is odd, let x := qn-h y := Pn-I· Then by (12.2-7) kx -[y = PnX -qny = (-l)n-l = 1, and the equation is solved. If n is even, let x := 1- qn _1, y := k - Pn -I· Then kx -ly = Pn(qn -qn-1) -qn(Pn- Pn-1) = -(-1r-l = 1, and the equation is again solved. The general solution is now easily found by observing that if (x 0 , y0 ) and (xh y 1) are distinct special solutions, then k(x 1 - x0 ) -l(y 1 - y0 ) = 0, hence Y1- Yo k --=-

xi-xo l' and thus, because k and 1 are mutually prime, Y1 = Yo+km, X1 =xo+lm, where m is some integer. Conversely, any such (x~> y 1) is a solution, hence the general solution is given by

(x, y) = (xo, Yo)+ m(l, k), where m is an arbitrary integer and (x 0 , y0 ) is any special solution, for instance the solution found above. EXAMPLEJ

61x -48y = 1. From the expansion in a simple continued fraction 61 1 l_] _!j _!j _!j 48 = +13+11+12+14

491

CONTINUED FRACDONS IN NUMBER THEORY we find the numerators and denominators 1 3 1 1

4 3

1

2

4

5 4

14 11

61 48

Because n = 4iseven, a special solution is (x 0 , y 0 ) = (48-11, 61-14) = (37, 47), and the general solution is (x, y) = (37, 47)+m(48, 61).

Representation of I"ational Numbers. Let the number~ now be irrational. Then it still can be subjected to the algorithm (12.2-4); that is, we can still determine a sequence of integers b; (b; > 0 for i > 0) and of real numbers ~i (~; > 0 for i > 0) by the conditions ~o=~,

~m

1

(12.2-8)

= bm +-1:-, ~m+1

m = 0, 1, 2, .... The process cannot terminate, for if it would, then~ would be value of a terminating simple continued fraction, and thus rational, which would be contrary to our hypothesis. By analogy with the rational case, we expect that

~ = ho+___!j+___!j+_!j+ · · ·. r-b-;- ~ I b3

(12.2-9)

By Theorem 12.1c, the continued fraction on the right certainly converges. The problem is to show that its value is ~- By the definitions of the numbers b;, if the t; denote the same Moebius transformations as before, we have form= 1, 2, ...

~ = to ot 1 ot2 o... otm(-1-). ~m+1

Thus by Theorem 12.1a, if the numerators and denominators of the fraction are denoted by Pm and qm, ~ = Pm-1 + Pm~m+1 qm-1 +qm~m+1'

m= 1,2, ....

There follows ~-Pm

qm

= (Pm-1 +pm~m+1)qm -pm(qm-1 +qm~m+1) qm(qm-1 +qm~m+1) Pm-1qm -pmqm-1 qm(qm-1 +qm~m+1)'

CONTINUED FRACDONS

492

and thus by (12.1-21), because all q; >0 and

'

~_Pml,.:;:

all~;>

1,

1

qm ~ qm(qm-l +qm)"

Because qm -+00 (see the proof of Theorem 12.1c), there follows ~:

~=

. Pm, 1Im qm

m-+CO

which is the same as (12.2-9). We call a continued fraction such as the one appearing on the right of (12.2-5) a nontenninating simple continued fraction. We have just shown that every irrational real number can be represented as a nonterminating simple continued fraction. It is easy to see that this representation is unique. Let

be two such representations. Because the value of a proper simple continued fraction always lies between 0 and 1, there follows immediately

bo=co=[e]. Having proved that b; = c; for i = 0, 1, ... , m -1, we may similarly conclude from

that bm =em. Every nonterminating simple continued fraction converges by Theorem 12.1c. Its value must be irrational, by Theorem 12.2a. In summary, we thus have THEOREM 12.2b

Every irrational number can be represented by a nonterminating simple continued fraction, and the value of every nonterminating simple continued fraction is irrational. The representation of a given irrational~ is (12.2-9), where the b; are determined by the algorithm (12.2-8).

493

CONTINUED FRACfiONS IN NUMBER THEORY EXAMPLE

For { =

4

J6, the algorithm begins as follows: ho=[J6]=2,

Because { 3 = { 1, there follows {4 = {2 ; hence b1 = b3 = b5 = · · · = 2, b2 = b4 = b6 = · · · = 4. Hence the desired representation is

J6=2+ 1 I+ 1 1+_2_1+ 1 I+· ..

12141214

'

the partial denominators periodically repeated. EXAMPLE

5

In §12.6 it will be shown accidentally that

rfoJ rN rfs-J

--= + + + + +··. ee-1 +1 ~ 2 ~ 6 10 14 18 ' showing that e is not rational.

Ultimately Periodic Continued Fractions. A nonterminating simple continued fraction bo +__!_j + ___!j+_!Jb1 + ...

~ ~ lb3

(12.2-10)

is called ultimately periodic if there exist integers k ;;;= 0 and m ;;;= 1 such that bn+m

= bn for all n ;;;= k.

(12.2-11)

For instance, the continued fraction for J6 given above is ultimately periodic with k = 1 and m = 2. We now investigate the nature of the irrational numbers that are represented by ultimately periodic simple continued fractions. Let ~ be the value of (12.2-10). This means that the sequence of approximants

494

CONTINUED FRACI'IONS

converges to f Thus a fortiori the sequence wk> wk+m• wk+ 2m, ... converges to f Putting

we have wk+nm = r o inl(O), where s[nl =so so· · · o s (n times), the n-fold iterate of s. Thus r o sfnl(O)-+ ~ for n -+co, which is to say that T'Jn := sfnl(O)-+ r[-tJ(~}, properly or improperly. From the fact that T'Jn = S(TJn-t), it follows by letting n-+ co that TJ = s( TJ ), and thus that TJ is a fixed point of s (see §6.12). The fixed points of s are easily determined. Let p~ and q~ denote the numerators and denominators of the continued fraction

c := _!_j+

1

I+

1

I+· ...

~~~

Then the p~ and

q~

are positive integers, and by Theorem 12.1a ( ) _P!-tu +p!

s u - qm-tU+qm * *. Thus ., is a solution of the quadratic equation P'!-tTJ +p! = TJ(q'!-tTJ +q!},

or (12.2-12) which has integer coefficients. Both solutions of this equation are real and ;CO, and precisely one solution is positive. Because all approximants of Care positive, TJ equals the positive solution of (12.2-12). Once more, by Theorem 12.1a, r(u)=lo 0 1t

0 • •

·otk-t(U)=

Pk-2u +pk-t , qk-2u +qk-t

where the p; and q; denote the numerators and denominators of the full fraction (12.2-10), and thus are again integers. The value of the full fraction thus equals ~:.

( )

~=rTJ=

Pk-2TJ+Pk-t ' qk-271 +qk-1

(12.2-13)

where TJ denotes the positive solution of (12.2-12). A solution of an algebraic equation of degree 2 whose coefficients are integers is called an algebraic number of degree 2 or a quadratic irrationality. Clearly, the number r(TJ) defined by (12.2-13) shares with TJ the property

CONTINUED FRACI10NS IN NUMBER THEORY

495

of being a quadratic irrationality. We thus have proved one-half of the following celebrated theorem due to Lagrange: THEOREM 12.2c

Every simple continued fraction that is ultimately periodic represents a quadratic i"ationality. Conversely, the representation of any real quadratic i"ationality by a simple continued fraction is ultimately periodic. Proof. It only remains to prove the second assertion. Let the quadratic equation satisfied by~ be

h~ 2 + k~ + 1= 0,

(12.2-14)

where h, k, 1 are integers, k 2 - 4hl > 0, and let

~=bo+__!_j+__!_j+ 1 I+· .. ~ fb;- ~

(12.2-15)

be the simple continued fraction representation of~. We wish to show that the fraction (12.2-15) is ultimately periodic. For n = 0, 1, 2, ... , let

so that, by Theorem 12.1a, ~=too

tl o ... o t ( -1-) = :.Pn-1 . .:. :. . . .:. .+ _. . Pn~n+l :. . .:.::.:::.:. .;. .:. n ~n+l qn-l +qn~n+l'

Pn and qn denoting, as usual, the (integral) numerators and denominators of (12.2-15). From (12.2-14) we have 2

h[Pn-l + Pn~n+l] + kPn-1 + Pn~n+l + 1= O qn-1 +qn~n+l qn-1 +qn~n+l

or, after removing denominators,

hn~~+l + kn~n+l + ln

=

0,

where hn := hp~+kpnqn +lq~, kn:= 2hPn-IPn +k(pn-lqn +pnqn-1)+2lqn-lqm ln := hp~-1 + kPn-lqn-l + lq~-Jo

These numbers are integers for every n.

(12.2-16)

CONTINUED FRACfiONS

496

Equation (12.2-16) can be simplified by using the fact that for n = 0, 1, 2, 0

0

0'

as follows from the proof of Theorem 12.1c, and hence n =1,2, ... ,

which by virtue of (12.2-14) simplifies. Using the fact that the qn increase monotonically and q 0 = 1, we thus find for hn the bound In a similar manner one may establish that lknj=s;:;4jhjj~j+2jkj+2jpj,

llnl,.;:; jhj(2j~j + 1) + jkj. Thus for each of the coefficients hno kno ln of (12.2-16) there are only finitely many possibilities, which implies that eventually a triple of coefficients must repeat itself. In fact, there exists a triple (h *, k *, I*) that occurs infinitely many times among the triples (hm km ln). In particular, there exist three distinct indices n; such that (hn,, kn,, ln.)= (h*, k*, 1*). At least two of the three solutions of the corresponding quadratics (12.2-16) must be identical. If, say, ~n 1 +1 = ~n2 +t. then by the algorithm (12.2-8) ~n,+m = ~n 2 +m for all positive integers m, showing that the simple continued fraction representing ~ is ultimately periodic. • Some examples of such ultimately periodic representations of quadratic irrationals, in addition to 4, are

5

1+J5=1+ 11+ 11+_!j+· 2

6

111111

00

'

497

CONTINUED FRACDONS IN NUMBER THEORY

7 Some of these fractions are purely periodic, that is, there exists an integer m such that bn+m = bn for all n;;;,: 0; another example is

F7+2=4+~+~+~+l: I+~+~+· ... Which algebraic property. characterizes quadratic irrationals with purely periodic continued fraction representations? Let

~ = bo+___!j+___!_j+ · · · +_!_j+_!j+ 1 J+· ·· lho Wt ibm ~ ~ be such a quadratic irrational. By the above,

~

(12.2-17)

satisfies the equation

~ = Pm-1 +pm~

qm-1 +qmf that is, qme+(qm-1 +pm)~-Pm-1 = 0.

(12.2-18)

Now consider the fraction 'T/ :=bm + ____!___j+___!_j+' '

~

~-

'+_!_j+_!_j+~+' ' ' , bm-1 I bo I bm

with the period reversed. This likewise is a simple purely periodic continued fraction, and its value is a fixed point of the Moebius transformation S := Sm o Sm -I

o ••• o

So,

where

k =0, 1, ... , m. Let us compute the matrix associated with s. If M

._

k .-

(bk1 01)

is the matrix associated with sk> then the matrices Mk,l := (;:

::) =MkMk-1 · · · Mo

498

CONTINUED FRACTIONS

satisfy the recurrence relation

that is,

which on comparing elements yields

A= bJk-1 +hk-h hk = fk-1· Thus M k,l_ (ek

A )' A-1

e_ 1 = 1,

fo= 1,

ek-1

where

It thus follows that the ek and fk satisfy exactly the same recurrence relations and initial conditions as the numerators and denominators pk and qk of the fraction for ~· Hence ek = Pk> A = qk, k = 0, 1, ... , and s(u) =

The fixed point relation 71

PmU +qm . Pm-lu+qm-1

= s(71) thus becomes

Pm-171 2 +(qm-l-pm}71-qm

= 0.

(12.2-19)

The last equation is related to (12.2-18), the equation satisfied by~. in the following obvious manner: If 71 is a solution of (12.2-19), then -1/71 is a solution of (12.2-18). We conclude that 71 = -1/f, where f is one of the solutions of (12.2-18). Now f cannot be the solution represented by the fraction (12.2-17), because both~ and 71 > 0. It follows that~· is the second solution of (12.2-18). This is known as the (algebraic) conjugate of the quadratic irrational ~. Because 71 > 1, we have -1 < f < 0. We thus have obtained: THEOREM 12.2d

Any purely periodic simple continued fraction represents a quadratic irrati~n­ ality ~ > 1 whose algebraic conjugate satisfies -1 < ~· < 0. There holds

CON'IlNUED FRACDONS IN NUMBER THEORY

f = -1/71, where

71

499

is the value of the continued fraction with the period

reversed. As an illustration, recall the fraction for~= (1 +.f5)/2 considered in S. The equation satisfied by ~ is ~- 1 = 0. The algebraic conjugate is ~· = (1-./5)/2. It so happens that f~ = -1, and thus 71 =~.which obviously must be the case if the period is 1. The algebraic conjugate of the number 2 + J7 considered after example 7 is 2-J7. By virtue of Theorem 12.2d,

e-

- _1- = J7 + 2 = 1 + _!__j + _!__j +__!_j +_!__j+_!__j+_!__j +_!__j + .... 2-../7 3

It 11 14 It It 11 14

Remarkably, the converse of Theorem 12.2d is also true. Thus every quadratic irrationality~> 1 whose algebraic conjugate f satisfies -1 < f < 0 is represented by a purely periodic simple continued fraction. The proof of this result, which was first given by Galois, is too long to be included here. Assuming the truth of Galois' theorem, we briefly determine the nature of the continued fraction representation of the numbers Ji, where I> 1 is not a perfect square. The algebraic conjugate f = -Ji does not satisfy the condition of Theorem 12.2d. However, if we let b 0 := [~1]. and if~ := ..fi+ b 0 , then f = -..fi+ b 0 satisfies -1 < f < 0, by the definition of b0 , and thus the expansion of ..fi+ b 0 is purely periodic. It thus follows that

Ji=b 0 +_2j+_2j+· · ·+____!_j+~+_2j+· · ·, ~lb;

~~~

(12.2-20)

the period consisting of the integers b~o b2 , .•• , 2b0 • It is seen that the expansion in example 7 is of this special form.

Pelt's Equation. This is the name given to the equation (12.2-21) where I is an integer, not a perfect square. The equation is to be solved in integers x and y. References to this equation are scattered throughout the history of number theory; among the mathematicians who contributed to the theory of Pell's equation we find Archimedes, Fermat, Euler (but no Pell). Here we wish to record an algorithm for solving Pell's equation, which is based on continued fractions. Let the expansion of .Jl be given by (12.2-20), and let Pi> q; denote the numerators and denominators of this fraction. Then by Theorem 12.1a, r.

Pm-1

+ Pm~m+1

qm-1

+qm~m+1'

vl=-----

500

CONTINUED FRACI10NS

where

Hence

or Ji{qm-1 +qmbo-Pm} = Pm-1 +pmbo-qml,

implying that both sides of this equation are zero. Hence qm-I = Pm -qmbo, Pm- 1 =qml-pmb0 , which on substitution into the basic relation (12.1-21) yields

or

p!-lq! = (-l)m-1. Hence x := Pm• y := qm is a solution of Pell's equation if m is odd. If m is even, then replacing m by 2m + 1 in the foregoing computation yields 2 lq22m+1 = 1• P2m+1-

hence x := p 2 m+h y := q 2 m+ 1 is a solution in this case. EXAMPLES

To solve x 2 -7y 2 = 1. From the expansion given in 7 we see that m = 3, and the approximant p 3 / q 3 is required. We find b;

2

1

1

1

2 1

3 1

5 2

8 3

This yields x = 8, y = 3, and indeed 82 -7 · 32 = 1. PROBLEMS

The reader will probably want to test the algorithms of this section in some examples of his own invention. In addition, we recommend: 1. Rediscover Euler's simple continued fraction expansion for e = 2.71828 18284 .... 2. Find the simple continued fraction expansions for JS, Ji5, .J24, ../35.

CONVERGENCE OF CONTINUED FRACDONS

501

3. Show that for m = 2, 3, ...

Jm 2 -1 = m -1 +L2__ +L____2__J2___ +L____2__ +· · · .

11 Tm-=tTl 11 Tm-=tTl

4.

Solve Pell's equation

form =2, 3, .... §12.3. CONVERGENCE OF CONTINUED FRACDONS WITH COMPLEX ELEMENTS

In mathematics much can be learned from a careful study of special examples. To form an idea about the convergence of continued fractions with complex elements, we consider the special fraction C :=

.£ZI+ a 1+-~j+· .. ~ ~ ~ '

(12.3-1)

where a is any complex number. Here the recurrence relations (12.1-8) satisfied by the numerators Pn and the denominators qn are (12.3-2) qn

= aqn-2 +qn-1•

The standard method for solving such difference equations with constant coefficients is to assume a solution in the form Pn = (-s t -the minus sign is inserted for later convenience-that on substituting into (12.3-2) yields (12.3-3) If this equation has two distinct solutions s 1 and s 2-which happens if and only if a¢- -*-then the first equation (12.3-2) has the general solution Pn

= c1(-s1t +cz(-s2t·

Determining the constants c 1 and c2 from the initial conditions (12.1-Sb), we get n=-1,0,1,···. In a similar way one finds n =-1, 0, 1, · · ·.

S02

Thus if a

CONTINUED FRACI10NS ;C -!,

the nth approximant of C is Wn

Pn (-sd" -(-s2)" =-=sls2( )n+l ( )n+h qn -s2 - -s!

and the sequence {wn} is seen to converge if and only if ls2i > lstl.

is 1 ;C is 2 i. If, say, 1

(12.3-4)

then by writing

we see that the value of C is w := lim Wn =s,. n-+CO

To see what the condition (12.3-4} means in terms of a, we note that the two solutions of the characteristic equation (12.3-2) are

They have distinct moduli if and only if the square root is not pure imaginary, which is the case if and only if a is not real and~-!. The solution of smaller modulus is then given by St

=

-!+.J!+a,

where the root symbol denotes the principal value of the square root. It remains to discuss the case a = -! where the characteristic equation Then the general solution of the (12.3-2) has the sole solution s 1 = difference equation for Pn is

-t

Pn

= (cl +c2n)(!)",

and the initial conditions (12.18b} yield Pn = -n (2l}n+l . Similarly, we find

Thus

n -=w =Pn n qn 2(n + 1)'

(12.3-5)

CONVERGENCE OF CONTINUED FRAcriONS

503

and w =lim Wn again exists and equals-!. In summary, we have obtained: THEOREM 12.3a

If a is complex, the continued fraction (12.3-1) converges if and only if a is not real and < - !. The value of the fraction, if convergent, is

w := -!+J~+a (principal value). This method settles the question of convergence of C in a rather pedestrian way. By availing ourselves of the theory of fixed points (§6.12) and of circular arithmetic (§6.6), we now treat the same problem by a method that will lead to an important generalization. For the fraction C, the R/Ttransformations (12.1-3) defining the approximants all are identical, tk = t : u ~

Hence the approximants

Wn

a

1 + u.

of C satisfy n =0, 1, 2, ... ;

wo=O,

(12.3-6)

they are obtained by iterating the function t. Thus if the sequence {wn} converges, its limit w satisfies w = t(w), that is, w is a fixed point of t. Because t(w) = a(l + w)- 1 , the fixed points oft are precisely the solutions s 1 and s 2 of (12.3-3). To decide whether the sequence {wn} converges, and to which one of the two fixed points it converges if it converges, we apply Theorem 6.12a. According to this result, convergence will occur if there exists a simply connected region S with closure S' such that t(S') c S and w0 = 0 E S'. To determine a suitable S, assume a~-!, so that s 1 ~ s 2 , and let the S; be numbered Slich that (12.3-7) For O::s;;p. 0 be ~iven, and let m be chosen such that 5m < E for all C E ;Ji, Let D(m := t!,m(D). In view of OED, Wm =t!,m(O)Et!,m(D)=D(m). If n>m, then by (12.3-11) we also have W,. =t!

ot2o • • •

= t 1,m o tm+I

ot,.(O)Et! o · • •o

ot2o • • •

ot,.(D)

t,. (D) c t1,m (D)= D

for all CE ;Ji, Thus both Wm and w,. belongtoD,hence iw.. - wml.:;;;5m m, showing that the sequence{w,.}is a Cauchy sequence. Because Dis closed and Wm ED for all m, it follows that w =lim Wm lies in D. • A closed set having the two properties (12.3-11) for a family ;Ji of continued fractions is called an eigendomain of ;Ji. In all applications that follow, the eigendomains are circular regions, and the estimation of the diameters 5,. is facilitated by circular arithmetic (see §6.6). As a first application of the contraction principle we prove the following classical result. THEOREM 12.3c (Worpitzky's theorem) The family

~of continued fractions

~

___E__U

a3 I

C=ll+lt+lt+· .. such that

k

= 1, 2, ... '

(12.3-13)

507

CONVERGENCE OF CONTINUED FRACDONS

converges uniformly. The value w of any CE :Ji satisfies approximants Wn there holds

lwl:o;;;i, and for the

n =0, 1, ....

(12.3-14}

Proof.

Here we have k

Let D := [0;

=

1,2, ....

n Evidently 0 ED. By the rules of circular arithmetic, ak 1 -1 4 2! I tk(D)= 1 +D=ak[1;2] =[3ak;3ak].

In view of lakl:o;;;;}, tk(D)cD. It follows that Dis an eigendomain of the familv :Ji. Let m be any positive integer, and let D := D, ·- tm-k (D,

Pk

k = 0, 1, ... , m. Lemma 12.3e shows that v c D = [s1- v 2s2. vls1- s2l] "

1-

,2 '

1-

,2 '

k >I.

Hence Pk.;;;u,

k>1,

where (12.3-24) By circular arithmetic,

lam-k IPk ldk+ll Pk+i.;;; 11 +d"l2-p~ = 11 +dkiPk· Using

the

fact

that

1 +s 1= -s 2, we

have

for

k >I

11 +d"l;;:::

l1+s~l-ld" -s~l;a:.ls21-u, hence

where

._lsd+u

A---~~-.

s2 -u

(12.3-25)

The desired relation Pm ~ 0 (m ~ oo) uniformly in ~is assured if the a in (12.3-23) is chosen such that the u in (12.3-24) becomes so small that A < 1. This requires 2u < ls21-1sll• which in turn holds if

2v 1-v2

1-0 1+0'

-- 0, let ~a denote the family of continued fractions

where E; > 0, {3; > 0, {3; the parabolic region

E; ;;;.: a,

i = 1, 2, ... , and where the parameter z lies in

Pa:Rez+2a2:1zl. Show that all approximants of all C E ~a are finite. [Decompose the constituents I; of C to show that the nth approximant of Cis

where S;:

u-+ z(1-___2__). {3;+U

Letting H:Re u >-a, show that s;(H)cH, i = 1, 2, .... Because oo~H and z E H, there follows w,. + z E H.]

515

RITZ FRACfiONS: FORMAL THEORY; PADE TABLE

7.

Let z, ~;, Ei satisfy the conditions of Problem 6 for some a > 0, and let the sequence of polynomials {pd be aefined by P-t := 0, Po:= 1, Pk = (pk

8.

+ z >Pk-1- ZEkPk-2,

k = 1, 2, ....

(12.3-28)

Show that no Pk vanishes in the set Pa. Let F be a formal power series, co

where all

ak

> 0 and

(a_ 1 := 0). Show that for z EPa all partial sums ofF are different from zero. [Let sk(z) := L:~oamzm. The polynomials Pk := aj; 1sk satisfy (12.3-28)

where

9. No partial sum of the exponential series has a zero in the set Pt. 10. Determine values of a, ~. 1' such that the confluent hypergeometric series 1F 1 (~; 1'; z) is different from zero in the set Pa. [Use the theorem of Hurwitz (Theorem 4.10f) in connection with the preceding problems.] 11. Let C be a continued fraction belonging to the family~" defined in Problem 6. If ~k -+ oo(k -+ oo ), C converges for all z EPa, uniformly on compact subsets.

§12.4. RITZ FRACfiONS: FORMAL mEORY; PADE TABLE It was shown at the beginning of §12.1 that infinite series and infinite products can be defined by forming compositions of Moebius transformations of the types (T) and (R), respectively. By letting the transformations tm occurring in the definitions depend on a parameter z, basic results on the representation of analytic functions are obtained. For instance, if

m =0, 1,2, ... , then the value of the composite Moebius transformation t0 o t 1 o u = 0 is a polynomial in z, too t1

o • • • o

tn(O)= ao+atZ +·

• • • o

tn at

· · +anzn.

[The backward evaluation of this expression, beginning with tn (0), yields the familiar Horner rule for the evaluation of a polynomial (see §6.1).] Letting n ~ oo we are led to power series, and thus to the very foundations of analytic

516

CON'11NUED FRACI10NS

functions. Similarly by the composition of rotations of the special form tm: U-+ (1 +amz)u

we can obtain the infinite product representations of the kind considered in §8.3. In a similar spirit we now consider RIT transformations zam tm:u-+-1-, +u

m=1,2, ... ,

(12.4-1)

where ah a 2 , ••• , are arbitrary nonzero complex constants, and where z is a complex parameter. By composing such transformations we obtain continued fractions whose approximants depend on the parameter z. It is customary to omit the factor z from the transformation t 1 • The resulting RIT fraction

(12.4-2)

=.!.

~ zam

Z m=l

1

will be called a RITZ fraction. The sequences of the numerators {pn} and of the denominators {qn} of C satisfy the initial conditions Po=O,

(12.4-3a)

qo= 1,

and the recurrence relations Pn

= ZQnPn-2 + Pn-1>

(12.4-3b)

where n = 2, 3, .... These relations show that both Pn and qn now are polynomials in z. An easy induction argument shows that Pn has degree

[n~1] and qn has degree

[~].

RITZ FRACDONS: FORMAL THEORY~ PADE TABLE

517

From the initial conditions and from the recurrence relations it is also clear that

n =0, 1, 2, ....

(12.4-4)

We shall require: THEOREM 12.4a

For n Proof.

= 0, 1, 2, ... , the polynomials p,. and q,. have no common zeros. The relation (12.1-21) shows that for n = 1, 2, ... , and for all z, Pn-l(z)q,.(z)-p,.(z)q,._l(z)=(-1)"a 1 a 2 · · · a,.z"- 1 •

(12.4-5)

If z #: 0 were a common zero of p,. and q,., the expression on the left would have to vanish. This is impossible, because the expression on the right is #: 0. It follows from (12.4-4) that z = 0 is not a common zero of p,. and q,., because it is not a zero of q,.. •

Being the ratio of two polynomials, the nth approximant of C(z) is a rational function of z. By virtue of (12.4-4) this rational function is analytic at z = 0 and thus can be expanded in a Taylor series, p,.(z) = c~"> +c~">z +c~">z 2 + · · · , q,.(z)

(12.4-6)

that converges for lz Isufficiently small. At first sight it would appear that the coefficients c~"> depend on n as well as on k. This, however, is true only up to a point, as the following result shows. THEOREM 12.4b

For all positive integers k, the coefficients c~"> have the same values for all

n>k. Proof. yields

In (12.4-5) we replace n by n + 1 and divide by q,.(z)q,.+ 1(z). This Pn+l(z) p,.(z) = (- 1),.a1a2 · · · a,.+1z" q,.+l(z) q,.(z) q,.(z)q,.+l(z) ·

(12.4-7)

Here we expand both sides in powers of z. Because q,.(O)q,.+ 1(0)#:0, the expansion on the right begins with the term in z". Thus the same must be true for the expansion of the expression on the left. From (12.4-6), the coefficient of zk in the expansion on the left is c~"+ 1 >-c~o

forn =0, 1 andk = 1, 2, ....

(12.4-10)

RTIZ FRACI10NS: FORMAL THEORY; PADE TABLE

519

If q~">, e~"> denote the elements of the quotient-difference scheme associated with P, the RITZ fraction correspo.nding to Pis given by

C=l ~o 1_1qrz 1_ 1e~>z 1_ 1q~:>z I_ I e~:>z 1_....

(1Z.4-ll)

EXAMPLE 1

From example 1, §7 .6, we conclude that the RITZ fraction corresponding to the series is

C=

z ~z r?z r+J+rf1 +rf1 +rTl 1 + 1 + 1 +· .. .

Quite generally, Theorem 12.4c enables us to construct the RITZ fraction corresponding to a given series P easily by means of the qd algorithm.

Proof of Theorem 12.4c. We first show that condition (12.4-10) is sufficient for the existence of a corresponding RITZ fraction. Let us call a rational function r = p/ q of type (k, l) if deg p :!S; k and deg q :!S; l. We then have: LEMMA 12.4d

If the series P satisfies (12.4-10), thenforn one rational function Wn = Pnl qn of type

=

0, 1, 2, ... , there exists precisely

([ n; 1]. [i]) such that

(12.4-12) Proof. (a) Let n be even, n =2m. Because wn must be analytic at 0, the required polynomials Pn and qn may be assumed in the form qn(Z) = 1 +d1z +d2z 2+· · · +dmzm, Pn(z) = eo+e1z +e2z 2+ · · · +em-lZm-I.

Condition (12.4-12) is equivalent to Pqn- Pn

= O(z 2 m)

or (12.4-13a)

CONTINUED FRACfiONS

520

Comparing the coefficients of zm, zm+I, ... , zzm-I yields for the d; the system of linear equations (12.4-14a)

= 0, 1, ... , m -1, whose determinant He:;.>¥- 0 by hypothesis. Thus the d; are uniquely determined. Comparing coefficients of z 0 , z 1 , ••• , zm-I then uniquely determines the e;. (b) If n is odd, n =2m+ 1, we must assume i

qn(z) = 1 +d1z +dzz 2 +· · · +dmzm, Pn(Z) = eo+e1z +ezz 2+· · · +emzm.

The required identity now is (co+cl!z + ... +Czm+IZZm+I + .. ·)(1 +dtZ + ... +dmzm) -(eo+etZ +· · · +emzm) = O(z 2 m+I).

(12.4-13b)

. We now fi rst compare t h e coeffi c1ents o f z m+l , z m+Z , ... , z 2m to get t h e system (12.4-14b) where i = 1,2, ... , m. The determinant now is It~>¥- 0, thus again a unique solution exists, which in turn determines thee; by comparing the coefficients of z 0 , z 1, ••• , zm in (12.4-13b). • For the following we require the leading coefficients of the 0 terms on the right of the relation (12.4-12). We write

n

= 1, 2, ....

(12.4-15)

For n =2m we get from (12.4-13a)

Together with (12.4-14a) this may be considered a system of m + 1 equations for them+ 1 unknowns d1, d 2, · · ·, dm; Szm· The determinant is -If:// and thus ¥- 0 by hypothesis. Solving for Szm by Cramer's rule yields Szm

He;;,~~ =

ft:Ol '

m=O, 1, ....

(12.4-16a)

m

For n =2m+ 1 we get from (12.4-13b)

Together with (12.4-14b) this forms a system of m + 1 linear equations for the unknowns· d 1 , d 2 , ••• , dm; Szm + 1 • The determinant now is -It~>¥- 0.

521

RITZ FRACDONS: FORMAL THEORY; PADE TABLE

Solving as before by Cramer's rule now yields w ,.{1)

n;,+1 S2m+1 = J-t.1l •

m =0, 1, 2, ....

(12.4-16b)

m

We note that under the hypothesis (12.4-10) all sn ¥-0, n we may define

=

0, 1, .... Thus

a1 := so=co,

(12.4-17)

n = 1,2, ... , which implies n =0, 1, ....

(12.4-18)

We now assert: LEMMA 12.4e The rational functions Wn := Pnl qn are precisely the approximants of the continued fraction

We show that the polynomials Pn and qn constructed in the proof of Lemma 12.4d satisfy the appropriate recurrence relations (12.4-3). Clearly, the correct initial conditions (12.4-3a) are satisfied. By the relations (12.415) and (12.4-18) we have for n = 0, 1, 2, ...

Proof.

Pqn -pn

= (-1ra1a2 · · · an+1zn + · · ·,

thus for n > 1 P(qn -qn-1- zanqn-2)-(pn -pn-1- zanPn-2) = (-1ra1a2 · · · anzn- 1{an+1z -1 + 1}+ O(zn+ 1)

= (-1ra1a2 · · · an+1zn +O(zn+ 1). The expression qn -qn- 1- zanqn- 2 is a polynomial of degree [n/2] with constant term zero. It thus may be assumed in the form d 1z +· · · +dmzm where m := [n/2]. Similarly Pn- Pn- 1 - zanPn- 2 is a polynomial of degree [(n-1)/2] and thus may be written e0 +e 1z+···+em•Zm' where m' := [(n -1)/2]. In the relation P(qn -qn-1- zanqn-2)-(pn _Pn-1- zanPn-2) = O(zn) (12.4-19) . • lds f or we now compare t he coeffi c1ents o f z m'+1 , z m'+2 , ... , z m'+m . ·Th"1s y1e

522

CONTINUED FRACTIONS

the d; the m homogeneous equations C;dm

+ci+ldm-1 + · · · +ci+m-ldl = 0,

i=m'-m+1, ... ,m'. Because the determinant Jt,:'-m+l>~o bd hypothesis, it follows that all d; = 0. Comparing coefficients of z , z 1, ..• , zm' shows that also all e; = 0. Thus the two expressions in parentheses in (12.4-19) vanish identically for all n > 1, which is to say that the recurrence relations (12.4-3b) hold. •

From (12.4-17) and (12.4-16) there now follows immediately rriO)

rrll)

.l"lm+l .l"lm-1

S2m

a2m+l

= - - - = ----;-;;m'r

.H.;;,·

S2m-1

(0)

~=-em'

.H.;;,'

rril) rriO) _ S2m+l _ .Hm+l .H.;,. _ a2m+2- - - - - rriO) S2m m n;,.+l

Jt.O

(0)

-qm •

by the basic relations (7 .6-5) and (7 .6-6). Thus the RITZ fraction (12.4-11) certainly is a fraction corresponding toP. To show that it is the only fraction corresponding toP under the hypothesis (12.4-10), let C * -_1 af 1

I+I ah I+1 afz I+··· 1

1

be any fraction corresponding to P. Then its approximants of Theorem 12.4b satisfy

w! by the proof

P- w! = (-1rafa~ · · · a!+ 1 z" + O(z"+ 1)

(12.4-20)

for n = 0, 1, 2, .... Because

w! is of type

([n; 1]. [~]). there follows w! = Pnfqm where Pn and qn are the unique polynomials constructed in the proof of Lemma 12.4d. There furthermore follows (12.4-21)

n = 0, 1, 2, ... ; hence, because af = a 1 = c0 and all

an~ 0,

a!= an for all

n;;ii!':l.

We finally show that the condition (12.4-10) is necessary for the existence of a corresponding RITZ fraction. Thus let the condition be violated, and in the sequence rriO)

rrll)

rrlO)

rril)

rriO)

rrll)

.Ni ,ni ,n2 ,n2 ,nj ,nj , ...

let the kth element be the first that is zero. We assume that k is even, k =2m;

Rnz FRACI10NS:

523

FORMAL THEORY; PADE TABLE

the proof for k odd is similar. Thus we have

JI.Ol ¥: 0, 11.1) ¥: 0,

0

0

0

'

It:!.> ¥: 0, It~>= 0,

and the existence of a unique rational function

Wn

(12.4-22)

of type

([n; 1]. [~]) satisfying (12.4-12) can be proved for n = 0, 1, ... , 2m, as in the proof of Lemma 12.4d. Assume now that a corresponding RITZ fraction

C* = I a! I+ I afz I+ I atz I+··· 1 1 1 exists, which implies that i= 1, 2,

af¥0,

0

0

0

(12.4-23)

0

Then the approximants wf of C* satisfy (12.4-20), and by the modified Lemma 12.4d wf = p;/q; follows fori= 0, 1, ... , 2m. However, in view of (12.4-22) the coefficient s 2 m-I in Pq2m-1- P2m-l = Szm-IZ 2 m-l +

•••

now equals zero. There follows -a!af · · · afm = s2m-I = 0,

contradicting (12.4-23). • The following is a trivial extension of Theorem 12.4c that is frequently used. COROLLARY U.4f

Let the series P be normal, that is, let ~t,.n> ¥:0 for all n = 0, 1, 2, ... and all k = 1, 2, .... Then for n = 0, 1, 2, ... the formal power series

possesses the corresponding RITZ fraction

C := n

c,.l_l q~n>z 1_1 e~n)z 1_1 q~n>z 1_1 e~n)z I_ ...

It

1

1

1

1



This simply follows from Theorem 12.4c by considering the series P := c~+c!z+· ··where ct=cn+k· •

524

CONTINUED FRACTIONS

The Pade Table. Let the series P be normal, as defined in Corollary 12.4f. Then the method used in the proof of Lemma 12.4d more generally yields the following: THEOREM 12.4g

If Pis normal, then for every pair (k, I) of nonnegative integers there exists precisely one rational function wk,l of type (k, I) such that

(12.4-24) Thus with every normal series P we may associate the two-dimensional array of rational functions wk,l· The basic formal properties of this array were established by Frobenius [1881]. Today the array is commonly known as the Pade table associated with P (Pade [1892]). Among the more recent contributors to the formal theory of the Pade table we mention Wynn [1966]. See also the excellent survey article by Gragg [1972]. A considerable number of applications of aspects of the Pade table to problems in computational physics are discussed by Baker and Gammel [1970]. From the point of view of the Pade table, the study of RITZ fractions merely amounts to a study of the two diagonals wn-l,n and Wn,n of the Pade table. In fact, many of the results about the Pade table that appear in the literature are merely results about RITZ fractions in disguise. Because the geometric point of view introduced by the Moebius transformations tm is not applicable to the Pade table, we do not pursue this subject here. Expansions in z -I. On occasion it is convenient to denote the parameter in the transformation tm by z -I in place of z. This yields a continued fraction

C -~~ +I a2z-l I+I aJz-1 I+· .. 1

1

00

=

z amz m=l

1

-1

1

(12.4-25)

which corresponds to the formal series

in the following sense: If the nth approximant of Cis expanded in powers of z-\ the expansion agrees with P through the term cn_ 1z-n+l. The nth numerator and the nth denominator of C now are polynomials (of degree [(n -1)/2] and [n/2], respectively) in z- 1• However, if we subject the fraction C to an equivalence transformation of type (12.1-11), where

RITZ FRACfiONS: FORMAL THEORY; PADE TABLE

525

c 1 = c 3 = c5 = · · · = z, c0 = c 2 = c4 = · · · = 1, we obtain the equivalent fraction . r.1z + ~2 + ~3 + ~4 + ... C .= z 1 z 1

'

(12.4-26)

whose approximants are unchanged, but whose numerators and denominators by (12.1-12) now are

(1)

z [(n+1)/2]p,. z

an

dz

[(n+1)/2]

(1)

q,. - ,

z

respectively, and thus are polynomials (of degree [(n + 1)/2]) in z. It is customary to drop the factor z in the fraction (12.4-26). This yields the continued fraction (to be called RITZ- 1 fraction) (12.4-27) It corresponds to the formal series

P := c 0 z

-1

+c 1z

-2

-3

+czz +· · ·

in the following sense: For n = 1, 2, ... , the expansion of the nth approximant in powers of z - 1 agrees with P through the term c,.- 1z - ... We now denote the numerators and denominators of the RITZ- 1 fraction (12.4-27) by p,. and q,., respectively. They are polynomials of the respective degrees [(n -1)/1] and [(n + 1)/2] that satisfy the recurrence relations

fio=O, 4o= 1,

(l2.4-28a)

and

p,. = a,.fi,.-z+b,.p,_.,

q,. =a,.4,.-z+b,.q,_.,

(12.4-28b)

where b,. = z for n 9dd and b,. = 1· for n even. It is of interest to consider the even and the odd parts of the RITZ- 1 fraction (12.4-27). Because the formulas of §12.1 apply only to the case in which the partial denominators are 1, we first carry through the computation for the fraction (12.4-25). Its even and odd parts are by (12.1.,19} and (12.1-20)

526

CONTINUED FRACfiONS

and

Treating these fractions to an equivalence transformation (12.1-11) where Cm = z, m = 1, 2, ... , and dividing by z, we obtain the even and the odd parts of the RITZ- 1 fraction (12.4-27):

C

·-~

.-, z +az

C

:=

a1{ 1 z

aza3

a 4a5

1

I z +a3+a4 I z +as+a6 azaJ

I

a4as

I z+az+a3 I z+a4+as

1

I

...

(even part),

} (odd part).

The foregoing relations are of interest if viewed in connection with the quotient-difference scheme associated with a normal formal power series 2

P=co+c1z+c2z +· · ·. By Corollary 12.4f, the even part of the RITZ- 1 fraction corresponding to

(12.4-29)

e1(n)q2(n)

.--z---q....,~"-:-_-e....,~....,,.>,..._ ·

}

··

(12.4-30)

Thus for m = 1, 2, ... , the mth approximant of the fraction O,.(z), if expanded in powers of z-\ agrees with P,. through the term c,.+ 2 mz-zm- 1 • This implies that the mth approximant of (

)

_

Cn+1

zO,. z -c,. -, z -q1 -e1

I

e~")q~") z -q2_e2

agrees with

zP,.- c,. = Cn+1Z - 1 + Cn+2z -z + ... = Pn+1 through the term c,.+ 2mz- 2m. The same is true of the fraction (n+1) (n+l)

qt e1 (n+l) (n+1 z-et -qz

Rl7Z FRACfiONS: FORMAL THEORY; PADE TABLE

527

Making the plausible assumption that the even part (12.4-29) of the RITZ- 1 fraction corresponding to a given series P is uniquely determined-we have not proved this-we are thus led to conjecture that (n)+

qm

(n)

em

(n)_

(n+l)+

em -em-1

(n) _ qm+l-

(n+l)

qm

'

(n+l) (n+l)

qm

(12.4-31)

em

for n = 0, 1, ... and m = 1, 2, .... These relations are indeed true; they are identical with the so-called rhombus rules that were proved by an entirely different method in §7.6. It is likely, however, that the rhombus rules were discovered by Rutishauser in the foregoing manner. Another connection with the qd scheme is as follows. The mth denominator of the fraction En (z) is the polynomial 4zm (z) defined by (12.4-28). The basic formulas (12.1-8) thus yield

tlzm+z(z) = (z -e~>-q~~~)4zm(z)-e~>q~>4zm-z(z),

m = 1, 2, .... Exactly the same recurrence relations are also satisfied by the Hadamard polynomials p~>(z) associated with the qd scheme (see §7.7). Because 4o(Z) = 1 = P~n>(z ),

tlz(Z) = Z -q~n) = P~n)(z), there follows m=O, 1,2, ....

.. ( ) (n)( ) qzm Z -Pm Z,

(12.4-32)

If the identity (12.1-23) is applied to the fraction En(z), we finally obtain the important formula

(12.4-33) where

or by (7.7-14) r,.(n)

(n)

-

S m -I -

n;,.

-;-;r.ar-·

H;;;:....l

(12.4-34)

We recall that the equivalence symbol used in (12.4-33) means that for m = 1, 2, ... , the mth approximant of the fraction on the left equals the mth partial sum of the series on the right. Convergence neither of the series nor of the fraction is implied.

528

CONTINUED FRACDONS

PROBLEMS 1. Show that the continued fraction

corresponds to the power series

p := ~ (-4)"(1/2)" z". n-o

2.

(n + 1)!

Show that to the power series 1 +(1-a)z +(1-a)(1-aq)z 2 +(1-a)(l-aq)(l-aq 2 )z 3 +· · ·

there corresponds the continued fraction 11_ (1-a)zl 1 1

It 3.

a(l-q)zl

I

1

I

q(1-aq)zl

I

1

aq(l-q 2 )zl 1

...

Show that the formal power series ao

p :=

L

1"2z"

n=O

gives rise to a qd scheme where

and that it therefore corresponds to Eisenstein's continued fraction 1

1_ _E_l

1(1 2 -1>z 1

ltltl 4.

h 1

13 (1 4 -1>z 1

Ill

1

1

Show that the qd scheme associated with the series

P:= 1 +

ao {n-1 1-la+k}

L

n-1

n ------:;:;:k 1-1

z"

k-0

is given by (n) _

k-l

qk - - 1 (n)

a+n+k-1

ek =-1

5.

(1- 1a+n+k-1)(1- (y+n+k-2) (1-ly+n+2k-3)(1-ly+n+2k-2)'

Let the formal power series

( 1 - I k)(1 - I y-a+k-1) (1-ly+n+2k 2)(1-l"+n+2k I)"

THE CONVERGENCE OF RITZ FRACI10NS: EXAMPLES

529

possess a corresponding RITZ- 1 fraction. Show that the equivalent reduced fraction is given by

C=1 b~z l_l ; 2 l_l b~z l_l ; 4 I_,.·' where

6. A continued fraction of the form

is called a J fraction (J for Jacobi). Show that with every J fraction we can associate a formal power series

with the following property: For m = 1, 2, 3, ... the expansion of the mth approximant of the fraction in powers of z- 1 agrees with P through the term involving z - 2 m. 7. Prove that given a formal series P= c0 z- 1 +c 1z- 2 + · · ·,there exists at most one J fraction to which it is associated. There exists precisely one such fraction if and only if It!;,> "F- 0, m = 1, 2, .... In the affirmative case the associated J fraction is equivalent to the even part of the RITZ- 1 fraction corresponding toP.

§12.5. THE CONVERGENCE OF RI1Z FRACfiONS: EXAMPLES In§ 12.4 we have dealt exclusively with the formal aspects of the correspondence between RITZ fractions and power series. At no point did we assume the convergence either of the fraction C or of the corresponding series P. The reader now undoubtedly expects to learn about the mutual implications of convergence of either C or P. The conjecture would seem hopeful, for instance, that C (as a function of z) converges in some region containing the origin if and only if P has a positive radius of convergence. Disappointingly, an example constructed by Perron, ([1957], II, p. 158) proves the "if" part of this conjecture to be false. Thus we either must assume the convergence of C, or we can assert the convergence of C under additional hypotheses on the elements of C (in addition, possibly, to assumptions on P). All results given subsequently are of this kind. Fortunately, they are still strong enough to permit us to deal very completely with a number of concrete special cases. In §12.9 et seq. we see that C may

530

CONTINUED FRACfiONS

converge in large regions (not containing 0) even if P has radius of convergence 0. A connection between C and P then exists via the theory of asymptotic expansions. Throughout this section we assume that (12.5-1) is a power series satisfying the determinant condition (12.4-10), and that C(z) =I

~1 l+l a:z l+l a;z I+ I a;z 1+ ...

(12.5-2)

is its corresponding RITZ fraction. THEOREM ll.Sa

LetS be a region containing 0, and let C(z) converge locally uniformly on S. Thenifw(z) denotes the valueofCatz, thefunctionf:z ~ w(z) is analytic on S; moreover, the corresponding series Pis the Taylor series off at 0. Proof. The hypothesis implies that for every compact subset T c: S the approximants w,(z) of C(z) are defined for all z e T and for all sufficiently large n, and that w,(z)~ w(z) uniformly for z e T. Because the approximants are rational, they are analytic wherever defined; hence by the general form of the Weierstrass double series theorem (Theorem 3.4b) the limit function f is analytic. By the same theorem, if as in ( 12.4-6) c ~~~> denotes the kth Taylor coefficient of w,(z) at 0, then for each k the limit Iim, .... oo c~"> exists and equals the kth Taylor coefficient of f at 0. But, by the very definition of the corresponding series, lim, .... oo c~"> = ck [in fact, c~"> = ck for n > k ], hence the Taylor series off at 0 equals P. • The following theorem asserts the convergence of C(z) under special hypotheses on the a;. THEOREM ll.Sb

Let la;j=s;;p., i = 1, 2, .... Then C(z) converges uniformly for lzl=s;; 1/4p.. Proof. If lzl=s;; 1/4p. and la;j:s;;p., then ja;zj=s;;1. The uniform convergence (uniform with respect to z, for fixed a;) follows from Worpitzki's theorem 12.3c. • By the preceding theorem, the function defined by C(z) is analytic in the disk lz I< 1I 4p., and the corresponding series P thus has radius of convergence ;?:!: 1/4p..

mE CONVERGENCE OF R11Z FRACOONS: EXAMPLES

531

mEOREM ll.Sc

Let the sequence {a;} tend to zero. Then C(z) converges, properly or improperly, for all z, and the function f represented by it is analytic at 0 and meromorphic in the whole compleK plane. Proof. Let p.>O be arbitrary. Let m be such that lanim. Denote by wf(z), wf(z), ... the approximants of the continued fraction m+2Z

+~ +··· . 1 By the preceding theorem, wj(z)-+ w*(z), an analytic function, uniformly for all z in D,.. : lz I.:;;; 1/4p.. The nth approximant of C(z) for n > m is Wn(z) =It 0

t2 ° · ·

· lm{w!-m(z)) 0

or, by virtue of Theorem 12.1a, (12.5-3) The limits

exist and are analytic in D,... Because q(O) = 1, q is not identically zero. Thus for all zeD,.. such that q(z)-¥- 0, . p(z) hm Wn(z)=-( n-+oo q z) exists. The limit function is analytic at 0 and, being the ratio of two analytic functions in D,.., it is meromorphic in D,... Because this holds for every JL > 0, the assertion of Theorem 12.5c follows. • mEOREM ll.Sd

Let a -¥- 0, and let S denote the region arg (z +

4~) ¥-arg(- L)·

(12.5-4)

If {a;} is any sequence of nonzero complex numbers such that lim; ....oo a; =a, then C(z) converges, properly or improperly, for all z E S, and represents a function f meromorphic in S whose Taylor series at 0 is P.

532

CONTINUED FRACfiONS

Proof. Let z 0 E S. Then az 0 is not real and :s;;; -*·Thus by Perron's theorem 12.3d, there exists Eo= E0 (z 0 ), 0 0 be the minimum of the radii of the covering disks, and let m be such that in the given fraction (12.5-2) la,-al m. We assert that the continued fraction C*(z) :=I am;1Z l+l am;2Z I+ ... converges uniformly for z Let z E T. Then

E

T.

Eo(zo) lz -zol< 1 +lal+lzol

(12.5-7)

for a suitable center z 0 of a covering disk; hence if n > m

la,z -azol:s;;;la, -allzol+lz -zollal+la, -a liz -zol :s;;; {

lzol lal 1 + + I 2 } Eo(zo) 1+lal+lzol 1+lal+lzol (l+lal+ zol>

:s;;; Eo(zo).

Thus by the Perron theorem C*(z) converges uniformly m D(z0 ), and, because finitely many such disks cover T, uniformly in T. The remainder of the proof is similar to the proof of Theorem 12.5c. Let w*(z) be the analytic function defined by C*, and let p, and q, denote the numerators and the denominators of C. Then if {wt(z )} is the sequence of approximants of C*, the relation (12.5-3) again holds. Because the limits of the numerator and of the denominator exist, C( )=Pm-l(z)w*(z)+pm(Z)

z

qm_,(z)w*(z)+qm(z)'

a meromorphic function analytic at 0.



THE CONVERGENCE OF RITZ FRACTIONS: EXAMPLES

533

These results possess interesting illustrations. A first group of examples may be drawn from the series P(z) := 2F1(a, 1; c; bz) =

~

((a))"(bzt,

n=O C n

where a, b, c are complex, b -¥- 0, c -¥- 0, -1, -2, . . . . The qd scheme associated with this series can be constructed explicitly. The initial conditions (7 .6-3) readily yield (n) a+n b q 1 = b+n '

n =0, 1, 2, ....

(12.5-8)

The continuation rules (7.6-4) then furnish, as may be verified by a somewhat tedious calculation, ~n>=

q

(a +n +k -1)(c +n +k -2) b (c+n +2k -3)(c +n +2k -2) '

k=2,3, .. . , (12.5-9a)

(n)_

ek -

k(c-a+k-1) "b (c +n +2k -2)(c +n +2k -1) '

k= 1,2,. 0 0' (12.5-9b)

n = 0, 1, 2, .... Thus the corresponding continued fraction where a 2 k = -q~O). a 2 k+t = -e~O) can be written down explicitly. We see that b 4

0

hm ak

=--

k-+oo

exists. We thus may apply Theorem 12.5d. The regionS here is the complex plane cut along the ray arg(z -1/b) = arg(l/b). We know from §9.9 that the function defined by 2F 1(a, 1; c; bz) for lz I< 1 can be continued to a function f that is analytic in all of S. It follows that C(z) converges (properly) and equals f(z) for all z e S. We consider some illustrative special cases. EXAMPLE

1

For lzl< 1, Log(l+z)=z 2F 1(1, 1;2;-z). For a= 1, c = 2, b = -1 we find from (12.5-9) --

a2k-

(0)_

k

qk - 2(2k -1)'

a

2k+l--

k _ e _ _ _ k

-

2(2k + o·

534

CON'I1NUED FRACI10NS

Hence for all z that are not real and ,.. -1,

-rf ~/2 ~/6 I

Log(1 +z ) -

1

+

+

1

+ 2z/6l + l2z/10 I+ l3z/10 I+· .. 1 1 1 '

1

which by an equivalence transformation may be thrown into the form (12.5-10) To check the numerical performance of this fraction, we list in Table 12.5 some numerators, denominators, and approximants resulting for z = 1.

Table 12.5. Approximants of (12.5-10) for z = 1

a,. b,.

0 0

1 1

1 2

1 3

4 4

4 5

9 6

9 7

p,. q,.

0 1

1 1

2 3

7 10

36 52

208 300

1572 2268

12876 18576

w,.

0

1.00000 0.66667 0.70000 0.69231 0.69333 0.69312 0.69315

Because the elements of the continued fraction are positive, we may deduce from the proof of Theorem 12.1c that the approximants alternately furnish upper and lower bounds for the value of the fraction. (Truncation error estimates for complex z ar.e given in §12.11.) We thus see that already the seventh approximant yields an accuracy that would require as many as 50,000 terms of the series Log 2 = Log{1 + 1)= 1-~+l-!+· · ·. EXAMPLE

2

For lzl< 1, Arctan z =z-lz 3 +!z 5 -~z 7 +· · · =2F1(t 1J; -z 2). Here the partial numerators of the corresponding continued fraction are (2k -1) 2 a2k = (4k,.... 3)(4k -1)'

a 2k+t = (4k -1)(4k + 1)"

The fraction converges for all z such that z 2 is not real and ,.. -1, that is, for all z not of the form z = iy where y is real andy~ ;;:.1. After an equivalence transformation we get z I 1z 2 I 4z 2 I 9z 2 I 16z 2 1 Arctan z =rt+l3+l5+~+~+ · · · .

(12.5-11)

THE CONVERGENCE OF

Rnz FRACDONS:

EXAMPLES

535

EXAMPLE 3

In a similar manner one finds Log 1 +z = 2zj_ ~~

1-z

j_ iz~- ~z~j_l 16z 2 1_ ...

I t 13 15 17

9

'

(12.5-12)

which holds for all z that are not real and ,.;:; -1 or ;;;.: 1. EXAMPLE

4

The general binomial series is

Here (12.5-9) yields a2 =a,

k-a azk+l

= 2(2k -1)'

a

-

Zk+Z-

k+a 2(2k + 1)'

k = 1, 2, .... After an equivalence transformation to remove denominators, we thus find

(1

+z

)-"-_!_j ~

-lt+IT+I

(1-a)z I I (1+a)z 1 I (2-a)z I I (2+a)z 2 + 3 + 2 + 5

1

+ (3-a)z l+l (3+a)z I+· ..

I

2

1

'

which converges for all z not real and ,.;:; -1. For a =!, z = 1 we can recover the fraction for J2 given in §12.2.

Another group of examples may be deduced from the series P(z) :=

1F1 (1;

c; bz) =

00

(bz)"

n=O

C n

I - )(,

where b and c are again complex, c oF 0, -1, -2, .... Again the qd scheme is known explicitly here; we find (n)

q1

b

= c+n' kb (c +n +2k -2)(c +n +2k -1)'

k= 1,2, ... ,

(12.5-13) (c+n +k -2)b qk = (c+n+2k-3)(c+n+2k-2)' (nJ

k=2,3, ....

536

CONTINUED FRACfiONS

Thus again the corresponding continued fraction C(z) can be written down explicitly with kb

b

a2=--,

a 2k+l = (c

c

k

+2k -2)(c +2k -1)'

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

+2k -l)(c +2k)'

= 1, 2, .... It is evident that

Thus Theorem 12.5c is applicable. Because 1F 1 (1; c; bz) defines an entire function, equality between function and corresponding series holds for all complex z. EXAMPLE

5

The most familiar special case is

There follows

for all complex z. After an equivalence transformation this becomes

e

._!j~~~~~~~

=~-~+r:z--1:3+12:-f5;+[2 -~+···.

(12.5-14)

In conclusion we present applications of some of the foregoing expansions to two problems in numerical analysis. EXAMPLE

6 Difference Operators of Maximum Order.

A multistep method for the numerical solution of ordinary differential equations is defined by two polynomials without common zeros, p(() = ak(k +ak-l(k-l + · · · +ao,

u(() = f:Jk(k + f:Jk-l(k-l + · · · + f:Jo (ak t:- 0), which for maximum accuracy should be chosen such that the order p of the resulting method is as large as possible. Here p is the largest integer such that for all sufficiently differentiable functions y(T), ash-+ 0, aky(T+ kh)+ak_,y(T+(k -l)h)+ · · · +a 0y(T) - h{{:Jky'(T + kh)+ · · · + {:J 0y'(T)} = O(hp+ 1 ).

(12.5-15)

THE CONVERGENCE OF RITZ FRACTIONS: EXAMPLES

537

The relation (12.5-15) must hold, in particular, for y (-r) = e•, which after cancelling e• yields

On letting z := eh -1, we see that this holds if and only if (12.5-16) as z-+ 0, where p 1(z) := p(l + z), u 1(z) := u(l + z). If p >0, then p 1 (0) = 0. Because p 1 and u 1 have no common zeros, u 1(0) ~ 0. Thus (12.5-16) is equivalent to Log(l +z)-PI(z) = O(zP+I), U1(z)

where p should be as large as possible. It follows from Lemma 12.4d that to achieve order p, z- 1p 1(z) and u 1 (z) should be chosen as the numerator and the denominator of the pth approximant of the continued fraction corresponding to 1 ...::_ Log(l + Z) = 1 z

I

2Z

I 2 + 3Z -

· · · ,

given by (12.5-10). In view of the factor z this produces polynomialsp 1 and u 1 ofthe respective degrees [(p + 1)/2] and [p/2]. Thus if k is given, the order p = 2k can be achieved. Unfortunately, the operators thus obtained cannot normally be used, because they are subject to numerical instability (see Henrici [1962]). EXAMPLE

7 Rational Approximations to e"A.

The linear initial value problem for time-independent systems in both ordinary and partial differential equations has the form dx

-=Ax d-r '

x(O) =s.

(12.5-17)

Here xis either a vector, of dimension m say, and accordingly A is an m x m matrix, or x is a function defined on a certain region of the space variables, and A is a linear operator acting in the space of these functions. In either case the solution of (12.5-17) may be written symbolically as (12.5-18) To compute e•A by means of the Jordan canonical form as described in §2.6 is possible only if A is a matrix, and may not always be feasible even then. On the other hand, computation by power series may not be practical because it may be too expensive to compute high powers of A. One thus is led to approximate e•A by a rational expression,

538

CONTINUED FRACTIONS

where p,. and q,. are chosen such as to produce maximum agreement with the exponential series. By Lemma 12.4d, choosing for p,. and q,. the numerators and denominators of the continued fraction corresponding to e• produces agreement up to O(z"}, and this is the only choice featuring this property. Table 12.5 gives the polynomials p,. and q,.. Table U.S.

Approximants of RITZ Fraction Corresponding to e•

n

p,.(z)

q,.(z)

1 2

1 1 1+!z 1+!z 1 +!z +tzz 2

1 1-z 1-!z 1-~z +kz 2 1-!z +tzz 2

3

4 5 6

1+~z+roz 2

1-~z +roz 2 -~z 3

7

1 +4z +toz 2 + tioz 3

1-4z +toz 2-tioz 3

To compute the value of the approximate solution at

T,

p,.(TA) y := q,.(TA) 5 ' it is not necessary to compute the inverse of the operator q,. (T A), because y may be found as the solution of the linear equation q,.(TA)y = p,.(TA)s.

(12.5-19)

In certain applications to partial differential equations A is a sparse matrix of very high order that is not stored explicitly. To solve (12.5-19) by elimination may then be impractical, and an iterative method may be preferred. In this case it would be wasteful to compute the vectors y,. :=q,. (TA)y and s,. := p,. (TA)s by first forming the polynomials q,.(TA) and p,.(TA) using matrix-matrix multiplications. It is more economical to use the recurrence relation Pk(TA) = Pk-t(TA) +akTAPk-2(TA) (k = 2, 3, ... ) which on multiplying from the right by s and taking into account the

initial conditions (12.4-3a) yields 5o=O, k = 2, 3, .... Similarly for they,. we find

Yo=y,

Yt=y,

k = 2, 3, .... This method has the further advantage that it is not necessary to specify n in advance.

THE DIVISION ALGORITHM. RATIONAL

Rnz FRACI'IONS

539

PROBLEMS 1.

If z z2 zn-l z" e• = 1+-+-+· · · + - - + - r (z) 1! 2! (n-1)! n!" '

show that z I z I 11 r"(z)=lt-ln+l+ln+2

(n+1)zl+

I n+3

2zJ

Jn+4 I

(n+2)zl+···

n+5

for all complex z. Obtain (12.5-14) as a special case. 2. Discuss the convergence of Eisenstein's fraction given in Problem 3, §12.4, and show, for instance, that for ItI< 1 ~ t"'= 1 I_ t I_ n=O

It fT

,3_, I_ ,s I_

~

t'-t31_ ....

It 1 1

3. Show that for n = 1, 2, ...

i

z

0

rf

1 (n + Wz" --dt= + ~2Zn + ~2Zn +.---___..;. _ __._1 1 + t" 1 n+ 1 2n + 1 I 3n + 1 (2n) 2 z" I (2n + Vz" I I (3n) 2 z" I (3n + 1) 2 z" I +~+1 5n+1 + 6n+1 +1 7n+1 +· ..

[Expand the integral in powers of z.]

§12.6. THE DIVISION ALGORITHM. RATIONAL Rl7Z FRACDONS It has been shown in §12.4 how to construct the RITZ fraction corresponding to a normal formal power series in a purely arithmetic manner by means of the qd algorithm. Here we show that the RITZ fraction corresponding to a convergent formal power series P can also be obtained directly from the function represented by P. This construction is based on the following fact. THEOREM 12.6a

Let ai ~ 0, i = 1, 2, ... , and let the RITZ fraction C(z)

:=I~~

1+1

a:z 1+1 a;z I+ ...

be uniformly convergent in a neighborhood N of 0. For z

k= 1,2, ... ,

~ 0,

let

540

CONTINUED FRACDONS

so that zak

[-Il

k

:v-+--1, v

tk

=

1, 2, ....

If f(z) denotes the value of C(z) for z e N, define for z e N, z ¥= 0

/k(z) := ~~-ll(h-I(z)),

fo(z) := zf(z),

k = 1, 2,...

(12.6-1)

Then fork= 0, 1, 2, ...

(12.6-2) Note: We have changed the earlier definition of t 1 (see § 12.4) to avoid an exceptional situation fork= 1. The idea of Theorem 12.6a may be simply stated thus: Letting z-+ 0 in a convergent RITZ fraction evidently yields its first partial numerator. From

fo(z) = t 1 ° t2 ° · · · 0 tn ° · · · (0) there follows I' (0) [-1] [-I]( I' ( )) 0 (1 Jk =tk[-1] otk-1°""" JQZ

=

=

tk+l

1

0

tk+2 ° tk+n ° • • • (0)

+ I zak+2 I+ ... 1

'

whence (12.6-2) follows immediately. The justification of these steps is somewhat more elaborate than expected because the transformations tk as well as fo depend on z. Proof of Theorem 12.6a.

Let

I ak+2z 1 I ak+Jz 1 Ck (z ) ._ .- ~ ~+ 1 + 1 + ...

and suppose that.Ck (z) converges uniformly in a neighborhood Nk of 0, and that for z e Nk> z ¥= 0 (12.6-3) By hypothesis, these assumptions are true for k = 0. Assuming that they are true for some integer k ;;J!: 0, we shall show that they are true for k + 1. Let {w!(z)} denote the sequence of approximants of Ck(z), (12.6-4)

THE DIVISION ALGORITHM. RATIONAL

Rnz FRACOONS

541

n = 1, 2, .... The uniform limit of this sequence by Theorem 12.5a is an analytic function whose value at 0 is ak+t· Hence by (12.6-3), (12.6-2) follows for the index k. Furthermore, because ak+t "# 0, there existp >0 and n 0 such that

for 0 < lz I:::;; p and n > n 0 • It follows that the sequence of functions (12.6-5) converges, uniformly for 0 < lzl :o;;;p, to

~~-}t(fdz)) = fk+t(z). The functions (12.6-5) all have removable singularities at z Hence as n ~ oo, 1

*

[-1]

= 0 with value 0.

1

-tk+t(Zwn(Z))~- ft+t(Z)

z

z

likewise holds uniformly for 0 < lz I:::;; p. (This follows from the fact that the absolute value of the function 1 [-1] * 1 -tk+l (zwn(z))--[k+t(Z),

z

z

which has a removable singularity with value 0 at 0, takes its maximum on lzl = p.) However, by (12.6-4) the functions 1

[-1]

*

-

1

-tk+t(Zwn(z)) --tk+2 ° tk+J 0

z

z

• • • 0

tk+n(O)

are the approximants of Ck+ 1(z). We thus have verified the induction hypothesis with k increased by 1, and in doing so have proved (12.6-2). • Theorem 12.6a suggests the following division algorithm for computing the RITZ fraction corresponding to a series P representing a function f analytic at 0. Let fo(z) := zf(z ),

(12.6-6 a)

at := lim .!./o(z)

(12.6-6b)

Z-+0

Z

CONTINUED FRACTIONS

542

(which of course means a 1 := /(0)). If/k- 1 and ak have been determined and if ak :f; 0, put (-11

fk(z) := tk

akz Jk-1 z

(fk-1(z))=~()-1,

(12.6-7a) (12.6-7b)

and let k + 1 -+ k. The division algorithm is said to terminate after k steps if ak+1 =0. THEOREM U.6b Let f be analytic at 0, and let the division algorithm described above never terminate. Then the RITZ fraction

1 + I a2z I+ ~3Z +··· C( z= ) ~ 1 1 1

(12.6-8)

co"esponds to the power series representing fat 0. (Convergence of Cis not implied.) Proof. by

Denoting by Pn and qn the numerators and denominators of C and

Wn (z)

Pn(z)

= -(-)' qn Z

n =0, 1, 2, ...

its approximants, it is to be shown that for n = 0, 1, 2, ...

f(z)- Wn(Z) = ( -l)"a1a2 · · · an+1z" +0(z"+ 1).

(12.6-9)

Evidently, this is true for n = 0. For an induction proof we assume the truth of (12.6-9) with n replaced by an integer n -1 ;;J!: 0. Clearly by (12.6-7a) fn (z) = t~-11 o t~-_1} o

••• o

~~-11(/o(z)).

By the fundamental Theorem 12.1 a, taking into account that t 1 (z) contains a factor z,

hence

THE DIVISION ALGORITIIM. RATIONAL RTIZ FRACDONS

543

In view of (12.6-6a) there follows fn(z)= _

qn(z)f(z)-pn(z) .

qn-1 (z )f(z)- Pn-1 (z) Because thepolynomialsqm(O) = 1, m = 0, 1, ... , wemaywritefor z ¢0, lzl sufficiently small, fn(z) =

f(z)- Wn(Z) qn-1(z) f(z)- Wn-1(z)" qn(z)

(12.6-10)

In view of the induction hypothesis, f(z)- Wn-1 (z) = ( -1)"- 1a1a2 · · · anz"- 1+ O(z").

This shows thatf,. is analytic for z ¢0, lzl sufficiently small. By (12.6-7b), the singularity is removable, and fn(Z) = an+1Z + O(z 2).

Because qn(z)/qn-l(z) = 1 + O(z), (12.6-10) now implies that (12.6-9) holds, and the induction step is complete. • Classical examples for the division algorithm arise in the theory of hypergeometric series. Perhaps the simplest of these is the following. For s ¢ 0, -1, -2, ... let (12.6-11) By comparing coefficients in the power series expansion, the entire function Fs is easily seen to satisfy the recurrence relation

z

F 8 (z)=Fs+l(z)+ s(s+ 1 )Fs+2(z).

On dividing by Fs+ 1(z) there follows

z

F.+2(z) s(s + 1)Fs+l(z)

1 Fs+l(z) Fs(z)

1.

(12.6-12)

If for some fixed s ¢0, -1, -2, ... and fork= 0, 1, 2, ... we define ~ ( ) ._

Jk

z .a

Z Fs+k+2(z) (s+k)(s+k+1) Fs+k+1(z)'

1 ·-...,--------

k+t·-

(s+k)(s+k+1)'

(12.6-13) (12.6-14)

544

CONTINUED FRACDONS

then replacing s by s + k in (12.6-12) yields akz

fk(z)=~()-1,

k

= 1, 2, ... ,

Jk-l z

and from (12.6-13) there follows lim .!-ft(z) = ak+l· z-+0 Z

Thus applying the division algorithm to the function 0 F 1(s+2;z) f(z):=!fo(z)= 1 z s(s+1) 0 F 1(s+1;z)

(12.6-15)

yields the constants ak defined by (12.6-14 ). It follows that the power series representing f at 0 has the corresponding RITZ fraction

z 1

z

+ (s + l)(s + 2) + (s + 2)(s + 3) + ... 1 1 .

Because the ak tend to zero, Theorem 12.5c is applicable, and the continued fraction converges, to f(z ), for all complex z. The convergence is proper for all z at which f(z) is finite. After an equivalence transformation we thus obtain _1_ 0 F 1(s+2;z)_ 1 I+ z I+ z 1+ z I+··. s + 1 0 F 1 (s + 1; z) -ls+'l r-.;+2 ls+3 1s"+4

(12.6-16)

(s ¥:--1, -2, ... ), the convergence being proper for all z such that oFt(s+l; z)¢0. The following special cases are worth mentioning. EXAMPLE

1

In view of sin z = zoFt

(i; - ~}

cos z =oFt

(i; - ~)

we get from (12.6-16) for s =-!after a further equivalence transformation ~__£j__£J__£J

tan z =[1 - 13 - 15- 17-. · · or, replacing z by iz,

e' -e-• e' +e •

rf ~ rfJ2

+ 32 + 5 + ~2 + ... . 7

(12.6-17)

(12.6-18)

THE DIVISION ALGORITHM. RATIONAL RTIZ FRACTIONS EXAMPLE

545

2

For arbitrary 11 the Bessel function of order 11 is defined (see §9.7) by (z/2)" ( lv(z):=r( 11 + 1)oF1 11+ 1;

2

z -4 · )

Thus lv+l(z) Z J.(z) = 211+2

of'1(11+2; -z 2 /4) of'1(11+ 1; -z 2 /4)'

and (12.6-16) yields

I

I

I

I

lv+l(z) z Z2 Z2 Z2 lv(z) =~-[2;+4-[Tv+6_rz;;+8"_

···.

(12.6-19)

Setting 11 =-!in view of (9.7-26) once again yields (12.6-17).

Rational RITZ fractions. A terminating RITZ fraction C( z ) =

1 + ~ 1

r¥2Z+···+I 1

anz 1

I

(12.6-20)

where a; ;t. 0, i = 1, 2, ... , n, will be called an n-tenninating RITZ fraction. It is clear that every n-terminating RITZ fraction represents a rational function r; in fact, ( ) _Pn(z)

r z - qn (Z )' where Pn and qn denote the nth numerator and the nth denominator of C. We call a rational function r = p/ q of precise type (m, n) if the polynomials p and q have no factors in common, and if the degrees of p and of q are precisely m and n, respectively. It then follows from Theorem 12.4a and from the material immediately preceding it that the n-terminating RITZ fraction (12.6-20) is of precise type ([(n -1)/2], [n/2]). The question naturally arises, to what extent rational functions of type ([(n -1)/2], [n/2]) can be represented by ann-terminating RITZ fraction. The proof of Theorem 12.6a shows that if ~~

~

r(z)=ll +1 1 +···+11' where a; ;t. 0, i = 1, 2, ... , n, then k=0,1, ... ,n-1,

(12.6-21)

546

CONTINUED FRACDONS

where the functions rk are defined recursively by r 0 (z) := zr(z ),

k

(12.6-22)

= 1, 2, ... , n.

This means that the division algorithm (12.6-7) can be carried out for the function r for at least n steps. Trivially, there exist rational functions for which the division algorithm breaks down earlier; for instance, a 1 = 0 if r(O) = 0. However, the following holds: THEOREM 11.6c

Let n > 0, and let r be a rational function of precise type ([(n -1)/2], [n/2]). Then r can be represented by an n -terminating RITZ fraction if and only if the limits (12.6-21) (with the rk defined by (12.6-22)) all exist and are different from zero. In the affirmative case,

___E_Jj ~ r(z)=ll+ll+ ...

anZ

I

+~.

(12.6-23)

Proof. It remains to establish the sufficiency of the existence of the nonzero limits (12.6-21). We assert: Fork= 0, 1, ... , n,

rdz) is of precise type

([ n-k2 + 1] , [n- -k]) 2- .

Because r0 (z) = zr(z) and 0 is not a pole of r, this is true for k the truth of (12.6-24) for some k ;;::O. Then

(12.6-24)

= 0. Assume

ak+l. . type ([n-k] rk (z) IS of prectse - 2 - , [n-k+1]) . 2

In view of the existence of (12.6-21), rk (0) = 0, thus ak+tZ. . rk (z) ts of prectse type

([nk] [n- k -1]) - 2- , , 2

and the same holds for

establishing (12.6-24) with k increased by 1. Thus the assertion holds up to k = n. It follows that rn-l is of precise type (1, 0), that is, rn_ 1(z) = anz, and rn(z) = 0. Defining the transforms tk as in Theorem 12.6a, we have rk (z) = t~- 11 (rk-1 (z )),

THE DIVISION ALGORITHM. RATIONAL RITZ FRACDONS

547

and thus evidently ro(z) = t1 ° t2 ° · · · 0 tn(O), which on interpreting the composition as a continued fraction and dividing by z yields (12.6-23). • It remains to discuss the details of the division algorithm defined by (12.6-6) and (12.6-7). We assume that{ in a neighborhood of 0 is defined by 2

c0 +c 1z +c2z + · · ·

f(z)=

2

d 0 +d 1 z +d2z

do¢0.

,

+ ···

Evidently,

and

We now set 't"oo

fdz) =

(k)

n+l

L..n=O Cn Z 't"oo d(k) n • L..n=O n Z

k =0, 1, 2, ...

Then (12.6-7b) yields (k)

Co ak+l = d~kl•

(12.6-25)

and from (12.6-7a) we get 't"

(k+l)

't" d(k)

n+l

L.. Cn Z 't"d(k+l) n L.. n Z

=

ak+l L..

n Z ~ (k) n L.. Cn Z

n

1,

which shows that d~k+ll = c~kl and thus (k+l)

Cn

(k-1) (k) = ak+lCn+l -cn+h

n=O, 1,2, ....

(12.6-26)

The relation (12.6-25) now becomes c~k> ak+l

= -=-vc=r>· Co

k

= 0, 1, 2, ....

(12.6-27)

The algorithm of determining the ak+l in this manner is a variant of the so-called Routh algorithm. The computation starts from the first two rows of

548

CONTINUED FRACI'IONS

coefficients c~0 l = c,.,

(-l)_d

Cn

-

n'

n =0, 1, 2, ... ,

(12.6-28)

and then proceeds as indicated in Table 12.6a. Table 12.6a The Routh Algorithm (Arrows indicate flow of computation)

EXAMPLE

J

The rational function r () z =

1+z+z 2 1+2z+3z 2 +4z 3

is of precise type ([(n -1)/2], [n/2]) for n = 6. We seek its representation by a 6-terminating RITZ fraction. The Routh algorithm yields the following scheme: 2

3

4

1 1 1 1

2

-1

-3

4

-1

-1

-4 7 7

-4

-28 Thus the desired representation is ____!_j

~ ~ ~ ~ ~

r(z) =11+11-11-11+11-lt' as can be verified directly.

THE DIVISION ALGORITHM. RATIONAL RITZ FRACI'IONS

549

The coefficients c ~k >, and thus the elements ak of the corresponding RITZ fraction, can be expressed in terms of determinants involving only the coefficients en and dn. much like the elements of the qd scheme can be expressed in terms of Hankel determinants. It readily follows from ( 12.6-26) and (12.6-28) that for n = 0, 1, 2, ... (I)

I

1 Co

Cn =do do

(2) Cn

Cn+ll dn+l'

(3) Cn

1 doco

Co

CJ

Cn+2

do

dl

dn+2

0

Co

Cn+l

co

CJ

c2

Cn+3

do

dl

d2

dn+3

0

co

c1

Cn+2

1 0

do

dl

dn+2

do

I

Co do

c11 dl

To formulate a general result, let hin> denote the k-dimensional column vector whose elements are the first k elements of the sequence en. dn, en-~> dn-J. ... , coefficients with negative indices being zero. Let D~n> := det(h~O>, h~0 , ... , h~k- 2 >, h~k-l+n>),

D&n> := 0,

(12.6-29) where k

= 1, 2, ... ; n = 0,

1, 2, ... , and set

Dk :=

Dk >, 0

k = 0, 1, 2, ....

(12.6-30)

Then, if I possesses a corresponding RITZ fraction, Dk ¥- 0 for k 0, 1, 2, ... , and

=

1 D k+l ---n do Dk-1,

k

= 1,2, ... '

(12.6-31)

_ 1 Dk+l co - - - do Dk-1'

k

= 1, 2, ....

(12.6-32)

(n)

c

(k)_

thus

With the ad hoc definition D_ 1 := d 01 , D_ 2 := d 02 (12.6-32) remains true even for k = - 1 and k = 0. Then generally 0 and Im z := arg z can be chosen in the interval (0, 11"). Because all a; > 0, there follows arg tn (0) = arg anZ = f/>, hence arg(1 +tn(O))= arg(l +anz)E (0, 1, at least one of these arcs initially lies in U. This is a contradiction, since by (iii) the image of U under z-+ 1/r(z) lies in U. There follows 1 r(z) = c 1(z -~){1 + O(z -~)}. In view of 1/r(U) c U, c 1 >0. Thus all poles of r are real, negative, and simple, and have positive residues. It thus follows that the partial fraction expansion of a positive symmetric rational function has the form

r(z)=ao+

m

ak

k=l

z +~k

L --,

(12.7-3)

where ao;;a.O, ak >0 (k = 1, 2, ... , m), and 00, i = 1, ... , m. (These a; necessarily the same as in (12.7-3).) Evidently, ak+1 =lim z- 1rdz) = a 0 + z-+0

and~;

are not

~ ~;>0.

i= 1 ~i

By (12.6-7a),

z

-1

1 { ak+1 rk+l(z) =1 ( ) z z rk z

(12.7-4)

By the proof of Theorem 12.6c, the function z- 1rk+I(z) isoftype (m -1, m). Let us find its partial fraction expansion. Its poles are the zeros of z- 1rk(z). [z = 0 is not a pole, in view of the definition of ak+I·] The real function x- 1rk(x) jumps from -oo to oo at each pole-~;, decreases monotonically in between, and tends to the limit a 0 >0 for x~ ±oo. Them zeros - . , h . . . , -Tim of z- 1rdz) thus are all real; moreover they satisfy -oo< -Tim< -~m < -Tim-1 < -~m-1 < · · · < -~2 0. The factor representation of p then shows

hence r(z)

1 1-[p( -Fz)/p(Fz)]

=-

:;e 0.

Fz 1 +[p( -Fz)/p(Fz)]

Because z = 0 is not a pole of r, the poles of r thus can lie only on the negative where ~ > 0 be a pole of r. This requires real axis. Let z = -

e

p(i~) + p( -i~) = 0.

(12.7-8)

If the pole had multiplicity > 1, this would require that also p'(i~)- p'(- i~)

= 0.

Combining this with (12.7-8), we would find p'(ig) +p'(- ig) = 2 Re p'(ig) = 0 p(ig) p(- i~) p(i~)

which contradicts that in view of p being stable,

Rep'(i~)=Re ~

k~t

p(i~)

- 1 ->0. ig-zk

Thus the poles of r are simple, and the residue at the pole z = be 2

res r(-g ) =

-e is found to

[12Re p'(i~)J-l > 0. p(ig)

Because r is analytic at oo and obviously r(x) > 0 for x ;:,: 0, r is a rational Stieltjes transform. (b) Let r, the Hurwitz alternant of the real polynomial p, be a rational Stieltjes transform. Denoting by sand t the numerator and the denominator of r, we evidently have (12.7-9) It is to be shown that p is stable, that is, that all zeros of p have negative real parts. Now evidently z = 0 is not a zero of p, since 0 is not a pole of r. Let z :;e 0 be a zero of p. Then from (12.7-9) (

rz

2) _

s(z 2 ) _ _ !. , 2) z t z

-(

SITZ FRACTIONS: APPROXIMANTS, STABLE POLYNOMIALS

557

on the other hand, by virtue of the hypotheses on r, 2



m

~

L

r(z )=ao+

i=l

where a 0 ;;;.0, a; >0,

~;

z

+~;

>0, i = 1, ... , m. We thus have

1 m --=ao+ L

z

i=l

a; -2--.

z

+~;

(12.7-10)

If z>O, (12.7-10) is evidently contradictory. If O