This is the first primary introductory textbook on complex variables and analytic functions to make extensive use of fun
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Table of contents :
Contents
Preface
1 Complex Numbers
2 Functions of a Complex Variable
3 Analytic Continuation
4 Introduction to Complex Integration
5 Residue Calculus
6 Gamma, Zeta, and Related Functions
7 Elliptic Functions
8 Conformal Mappings
9 Transforms
10 Wiener–Hopf and Riemann–Hilbert Methods
11 Special Functions Defined by ODEs
12 Steepest Descent for Approximating Integrals
Bibliography
Index
Complex Variables and Analytic Functions
Complex Variables and Analytic Functions An Illustrated Introduction Bengt Fornberg University of Colorado Boulder, Colorado
Cécile Piret
Michigan Technological University Houghton, Michigan
Society for Industrial and Applied Mathematics Philadelphia
Copyright © 2020 by the Society for Industrial and Applied Mathematics 10 9 8 7 6 5 4 3 2 1 All rights reserved. Printed in the United States of America. No part of this book may be reproduced, stored, or transmitted in any manner without the written permission of the publisher. For information, write to the Society for Industrial and Applied Mathematics, 3600 Market Street, 6th Floor, Philadelphia, PA 191042688 USA. Trademarked names may be used in this book without the inclusion of a trademark symbol. These names are used in an editorial context only; no infringement of trademark is intended. MATLAB is a registered trademark of The MathWorks, Inc. For MATLAB product information, please contact The MathWorks, Inc., 3 Apple Hill Drive, Natick, MA 017602098 USA, 5086477000, Fax: 5086477001, [email protected], www.mathworks.com. Mathematica is a registered trademark of Wolfram Research, Inc. Publications Director Executive Editor Developmental Editor Managing Editor Production Editor Copy Editor Production Manager Production Coordinator Compositor Graphic Designer
Kivmars H. Bowling Elizabeth Greenspan Mellisa Pascale Kelly Thomas David Riegelhaupt Claudine Dugan Donna Witzleben Cally A. Shrader Cheryl Hufnagle Doug Smock
Library of Congress CataloginginPublication Data Names: Fornberg, Bengt, author.  Piret, Cécile, author. Title: Complex variables and analytic functions : an illustrated introduction / Bengt Fornberg (University of Colorado, Boulder, Colorado), Cécile Piret (Michigan Technological University, Houghton, Michigan). Description: Philadelphia : Society for Industrial and Applied Mathematics, [2020]  Series: Other titles in applied mathematics ; 165  Includes bibliographical references and index.  Summary: “This book is the first primary introductory textbook on complex variables and analytic functions to use predominantly functional illustrations” Provided by publisher. Identifiers: LCCN 2019030487 (print)  LCCN 2019030488 (ebook)  ISBN 9781611975970 (paperback)  ISBN 9781611975987 (ebook) Subjects: LCSH: Functions of complex variablesTextbooks.  Analytic functionsTextbooks. Classification: LCC QA331.7 .F67 2020 (print)  LCC QA331.7 (ebook)  DDC 515/.942dc23 LC record available at https://lccn.loc.gov/2019030487 LC ebook record available at https://lccn.loc.gov/2019030488
is a registered trademark.
Contents Preface 1
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Complex Numbers 1.1 How to think about different types of numbers . . . . . . 1.2 Definition of complex numbers . . . . . . . . . . . . . . 1.3 The complex number plane as a tool for planar geometry 1.4 Stereographic projection . . . . . . . . . . . . . . . . . . 1.5 Supplementary materials . . . . . . . . . . . . . . . . . 1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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Functions of a Complex Variable 2.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Some elementary functions generalized to complex argument by means of their Taylor expansion . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Additional observations on Taylor expansions of analytic functions . . 2.4 Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Multivalued functions—Branch cuts and Riemann sheets . . . . . . . 2.6 Sequences of analytic functions . . . . . . . . . . . . . . . . . . . . . 2.7 Functions defined by integrals . . . . . . . . . . . . . . . . . . . . . . 2.8 Supplementary materials . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Analytic Continuation 71 3.1 Introductory examples . . . . . . . . . . . . . . . . . . . . . . . . . . 71 3.2 Some methods for analytic continuation . . . . . . . . . . . . . . . . 73 3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
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Introduction to Complex Integration 4.1 Integration when a primitive function F (z) is available 4.2 Contour integration . . . . . . . . . . . . . . . . . . . 4.3 Laurent series . . . . . . . . . . . . . . . . . . . . . . 4.4 Supplementary materials . . . . . . . . . . . . . . . . 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
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Residue Calculus 119 5.1 Residue calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 5.2 Infinite sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 v
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Analytic continuation with use of contour integration Weierstrass products and Mittag–Leffler expansions . Supplementary materials . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . .
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Elliptic Functions 7.1 Some introductory remarks on simply periodic functions . 7.2 Some basic properties of doubly periodic functions . . . . 7.3 The Weierstrass ℘function . . . . . . . . . . . . . . . . 7.4 The Jacobi elliptic functions . . . . . . . . . . . . . . . . 7.5 Supplementary materials . . . . . . . . . . . . . . . . . 7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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Conformal Mappings 8.1 Relations between conformal mappings and analytic functions . 8.2 Mappings provided by bilinear functions . . . . . . . . . . . . 8.3 Riemann’s mapping theorem . . . . . . . . . . . . . . . . . . 8.4 Mappings of polygonal regions . . . . . . . . . . . . . . . . . 8.5 Some applications of conformal mappings . . . . . . . . . . . 8.6 Revisiting the Jacobi elliptic function sn(z, k) . . . . . . . . . . 8.7 Supplementary materials . . . . . . . . . . . . . . . . . . . . 8.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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209 211 212 214 215 219 222 226 227
Transforms 9.1 Fourier transform . . . . . . . . . . . . . . . 9.2 Laplace transform . . . . . . . . . . . . . . . 9.3 Mellin transform . . . . . . . . . . . . . . . . 9.4 Hilbert transform . . . . . . . . . . . . . . . 9.5 ztransform . . . . . . . . . . . . . . . . . . . 9.6 Three additional transforms related to rotations 9.7 Supplementary materials . . . . . . . . . . . 9.8 Exercises . . . . . . . . . . . . . . . . . . . .
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Wiener–Hopf and Riemann–Hilbert Methods 10.1 The Wiener–Hopf method . . . . . . . . . . 10.2 A brief primer on Riemann–Hilbert methods 10.3 Supplementary materials . . . . . . . . . . 10.4 Exercises . . . . . . . . . . . . . . . . . . .
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Gamma, Zeta, and Related Functions 6.1 The gamma function . . . . . . 6.2 The zeta function . . . . . . . 6.3 The Lambert Wfunction . . . 6.4 Supplementary materials . . . 6.5 Exercises . . . . . . . . . . . .
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Special Functions Defined by ODEs 291 11.1 Airy’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 11.2 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
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Hypergeometric functions . . . . . Converting linear ODEs to integrals The Painlevé equations . . . . . . Exercises . . . . . . . . . . . . . .
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Steepest Descent for Approximating Integrals 12.1 Asymptotic vs. convergent expansions . 12.2 Euler–Maclaurin formula . . . . . . . . 12.3 Laplace integrals . . . . . . . . . . . . . 12.4 Steepest descent . . . . . . . . . . . . . 12.5 Supplementary materials . . . . . . . . 12.6 Exercises . . . . . . . . . . . . . . . . .
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Bibliography
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Index
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Preface The topics of Complex Variables and Analytic Functions are of fundamental importance not only in pure mathematics, but also throughout applied mathematics, physics, and engineering. Their theorems and formulas also simplify many results from calculus and for functions of real variables. There is already a wide choice of textbooks available, ranging from timeless classics such as by Whittaker and Watson [41], Copson [8] and Ahlfors [2],1 to many later ones, raising the question of why anyone would want to see yet another one. Our main motivation lies in the evolution that has occurred in other fields. For the last half century or so, it has been unthinkable to use introductory text books for calcu2 lus that do √ not graphically illustrate the basic elementary functions, such as f (x) = x , f (x) = x, f (x) = sin x, f (x) = log x, etc. When we first taught a course on complex variables and analytic functions, we became puzzled about why complex variables texts should not also visually build on the real cases, familiar to all students, and then liberally illustrate how these same functions extend away from the real axis. While formulas alone for some students might provide a feasible alternative to visual impressions, we are aiming this text at students that find the latter to be helpful for gaining at least their initial intuitive feeling for the subject. Although we have included an abundance of illustrations (and give brief code templates for displaying analytic functions with MATLAB and Mathematica), this book is an introduction to the classical theory of complex variables and analytic functions. It contains enough materials to support a twosemester course, but has been structured to make it easy to omit chapters or sections as needed for a onesemester course (offering a lot of flexibility in course emphasis). SIAM’s website for this book, available from www.siam.org/books/ot165, contains “Notes to Instructors,” with suggestions for different one and twosemester syllabi, ideas for supplementary student projects, etc. Once a solution manual for all the exercises in the text has been developed, it will also provide information for how instructors can get access to it. Textbooks often differ with regard to the order in which topics are covered. One strategy is to make sure each step follows rigorously from previous steps. While that can have some appeal, it might not necessarily be the best order for developing an initial conceptual understanding, and it also does not correspond to how mathematical problem solving and research is carried out.2 Introducing key ideas early on might require certain issues to be revisited later, once additional tools have fallen in place. In either case, the end knowledge 1 Supplementing these texts, Jahnke and Emde’s “Tables of Functions” [28] has excellent (precomputer era) illustrations, but lacks textbooktype materials. 2 The eminent mathematician Paul Halmos writes in his autobiography [24, page 321]: “Mathematics is not a deductive science  that’s a cliché. When you try to prove a theorem, you don’t just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork. You want to find out what the facts are, and what you do is in that respect similar to what a laboratory technician does.”
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Preface
will be similar, but it is our belief that the latter approach is better suited for an introductory text. A case in point concerns singularities, where we here bring these up early on, before having developed all the tools of contour integration that are needed for a certain (Laurentexpansion based) singularity characterization. We have in this book never sacrificed correctness for simplicity. However, there inevitably arise situations where it becomes unavoidable to choose a compromise path between strictest mathematical rigor and more heuristic arguments. Believing that the former is better suited for finalizing proofs than for gaining insights and for solving problems, we have not shied away from the latter when that is more appropriate (but have then made this choice clear in the text). The emphasis that is given here to separate topics also differs somewhat from many other texts. For example, we view analytic continuation as fundamental to a good understanding of the nature of analytic functions. Instead of just listing one approach for this task (the circlechain theorem, which incidentally is quite impractical in many contexts), we have included numerous approaches. We also discuss and illustrate multivalued functions and their Riemann sheets more extensively than what is customary in introductory texts. The main applications of analytic functions have changed quite significantly during the last decades. For example, conformal mapping was more central to applied mathematics before recent advances in scientific computing (which have vastly extended the range of equations that can be addressed effectively). On the other hand, analytic function techniques are gaining in significance in other areas (such as when used in conjunction with numerical methods). Rather than discussing different applications (say, why scientists and engineers routinely use Fourier or Laplace transforms, complex exponentials, etc.), our focus has been on providing the analytic function based tools that are needed across a wide range of fields. The present material originated from lecture notes first developed by Bengt Fornberg in 2006 and then used jointly by us in 2008 (at the University of Colorado, Boulder). After that, they were mostly put aside until Cécile Piret revitalized them when resuming teaching the topic in 2015 at Michigan Technological University. As noted above, a main purpose of extending from the real axis to the complex plane is to greatly simplify a vast number of tasks in applied mathematics, engineering, etc. The renowned mathematician Jacques Hadamard expressed this succinctly (paraphrasing an earlier statement from 1900 by Paul Painlevé): “The shortest path between two truths in the real domain passes through the complex domain.” However, complex numbers go far beyond being just a mathematical trick that, once having done its magic, ought to quickly and gracefully disappear. It is our hope that this book will make its readers not only appreciate their utility, but also come to regard them as equally “natural” as, say, negative integers. Acknowledgments: While developing the present book, we received extensive and thorough comments from numerous experts, both on educational and on technical aspects. We want to extend special thanks to Dr. Tom Bogdan and Professors Willy Hereman, Paul Martin, Nick Trefethen, and Grady Wright for many suggestions, as well as much encouragement. From the first concept suggestion to the final product, it has been a delight to work with Elizabeth Greenspan at SIAM. Personally, for love and support, B.F. thanks Natasha and C.P. thanks Erin, Maximilien, and Juliette.
Chapter 1
Complex Numbers
1.1 How to think about different types of numbers 1.1.1 Integers, rational numbers, and real numbers The origins of counting are lost in prehistory. For a very long time, the only numbers used were positive integers, in modern notation 1,2,3, . . . . Such numbers can be used to count apples, oranges, days, frying pans, etc. The earliest number system extension was to allow for rational numbers, such as 3/7 or 11/8. These are clearly useful in measuring things like distances and weights. They can still be associated with counting (like pieces of a whole pie), and they simplify division by allowing all cases (apart from division by zero). An important point to note is that it is usually best not to think of a rational number as a pair of two integers. Even when represented as a ratio of two integers, the combination is usually best viewed as a single number (and we use it as such in, say, representing a single position on the real axis), not as some sort of a 2component vector. Zero and negative numbers originally made little sense. However, much algebra—such as subtraction—became much easier if these were allowed. Else there would need to be extra rules about when subtraction is permitted. Also, there are many applications for which both the input data and the final answer are positive, but when nevertheless going via negative numbers during intermediate steps makes it easier to reach the final answer. We have now given up on the idea that numbers need to directly correspond to counting objects. The Greeks, over 2000 years ago, noted that even the rational numbers were insufficient to measure all distances, such as the length of the diagonal of a square with side length one. This discovery was deeply agonizing, since it caused fear that the Gods would avenge this human exposure of their imperfection in creation. The discovery of the irrational numbers which, together with rational numbers, form the set of real numbers was kept secret for quite some time.
1.1.2 Complex numbers The historical perspective above may be useful as a background when we now extend the real number system to complex numbers. There are no 3+5i apples, just as there are no −7 apples, so we are already used to numbers being generalized past being merely counters for 1
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Chapter 1. Complex Numbers
objects. Much like for rational numbers, a complex number can be represented by means of a pair of numbers (and we graphically use these two real numbers as coordinates in picturing a complex number’s location in the complex plane). Nevertheless, it is often best to think about z = 3 + 5i as a single number, just as 3/7 is a single number. This book is largely about what happens ´ to our usual functions, such as y = sin x, and to our calculus tools, such as df (x)/dx, f (x)dx, etc., when we use z complex instead of x real. Any other way of thinking of z than as a single number would for such work be very awkward.3 It is natural to wonder whether one next should continue generalizing complex numbers to some kind of hypercomplex ones, with more than two components. It can be shown that such attempts (of which there have been many) will require very severe sacrifices—rules such as a·b = b·a and/or others that we want to take for granted will have to be abandoned, typically causing much greater losses than gains. Complex numbers in many ways is THE most natural number system possible. Making excursions into the complex plane often provides the simplest solution strategy also for problems where all ingredients as well as the final answers are all real. Also, analytic functions (meaning functions w = f (z) for which df /dz exists, the primary topic of this book) are in many ways much simpler than real functions f (x). A few aspects which we will encounter early on in this book are the following: 1. If a function f (z) can be differentiated once, it can be differentiated infinitely many times. 2. If an analytic function f (z) is defined uniquely on any interval, no matter how short, it is automatically defined uniquely away from that line segment as well. 3. If the magnitude of an analytic function is everywhere bounded by some constant, the only possibility is that the function is identically constant. 4. A polynomial equation of degree n will always have exactly n roots. The term complex is unfortunate, since generalizing to complex numbers simplifies much of calculus and algebra. The complex number system is the most natural system in which to do most mathematics. Other cases, such as integers, rational numbers, real numbers, etc., are just restrictive subclasses, with often more difficult rules.
1.2 Definition of complex numbers We recall that negative numbers were introduced in order to always make subtraction possible, with rather immediate practical applications. In contrast, when complex numbers were first conceived (to have some formal notation for all square roots and for solutions to all quadratic equations, such as x2 + 1 = 0), their practical utility was at first very unclear. A key step in advancing these complex numbers from being mostly meaningless notational abstractions occurred when Girolamo Cardano (1501–1576) described a for3 mula for solving a general cubic equation. One √ case he considered was z − 15z − 4 = 0, with the three √ roots z1 = 4, z2,3√= −2 ± 3. However, his procedure gave one root as z1 = (2 + −121)1/3 + (2 − −121)1/3 . On observing that, in our modern notation, 3 Roger Penrose, in his bestseller The Road to Reality [37] writes: “When we get used to playing with these complex numbers, we cease to think of a + i b as a pair of things, namely the two real numbers a and b, but think instead of a + i b as an entire thing on its own, and we could use a single letter, say z, to denote the whole complex number z = a + i b.”
1.2. Definition of complex numbers
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(2+i)3 = 2+11i and (2−i)3 = 2−11i, z1 can be simplified to z1 = (2+i)+(2−i) = 4. For a purely realvalued problem, a temporary excursion into the world of complex numbers had produced a realvalued solution that no previously available systematic approach had been able to reach. This is likely the first known example of the (much later formulated) quote in the Preface: “The shortest path between two truths in the real domain passes through the complex domain.” √ In 1777, Euler assigned the symbol i to the imaginary units satisfying i = −1. Not until in the early 19th century did the complex number plane become recognized as a natural extension of the number line. The real axis becomes then just a special case.
1.2.1 Complex number system: Algebraic introduction Let i have the property that i2 = −1. Assume that all the standard rules of algebra continue to hold: 1. z1 + z2 = z2 + z1 and z1 z2 = z2 z1 (commutativity in addition and multiplication), 2. (z1 + z2 ) + z3 = z1 + (z2 + z3 ) and (z1 z2 ) z3 = z1 (z2 z3 ) (associativity in addition and multiplication), 3. (z1 + z2 ) z3 = z1 z3 + z2 z3 and z1 (z2 + z3 ) = z1 z2 + z1 z3 (distributivity). Next, let z1 = a + ib and z2 = c + id, where a, b, c, d are real. Then the following hold: 1. z1 + z2 = (a + c) + i(b + d) and z1 − z2 = (a − c) + i(b − d), 2. z1 z2 = (a + ib)(c + id) = (ac − bd) + i(ad + bc), z1 a + ib ac + bd a + ib c − id (ac + bd) + i(bc − ad) 3. = = 2 + = = z2 c + id c + id c − id c2 + d2 c + d2 bc − ad . i 2 c + d2 The results are in these cases complex numbers. While the steps in cases 1 and 2 are straightforward, the idea in Case 3 (eliminating a complex factor c+id from a denominator by multiplying by c−id c−id ) is also very often applicable.
1.2.2 Complex number system: Graphical representation A complex number z = x + iy is made up of a real part x, denoted by Re z, and of an imaginary part,4 y = Im z. The complex zplane was originally devised by Caspar Wessel, a DanishNorwegian cartographer, in a paper published in 1797, but which would receive little initial attention outside of Scandinavia, maybe because it was written in Danish.5 This plane can be viewed as a twodimensional Cartesian coordinate system whose horizontal axis corresponds to the real part of the numbers and whose vertical axis corresponds to the imaginary part of the numbers. We can thus display z on the complex plane as a point with abscissa x and ordinate y, as in Figure 1.1. We can also display operations on the numbers in the complex plane. Addition and subtraction are illustrated in Figure 1.2. One can add and subtract by forming parallelograms, 4 Note
that y = Im z in itself is a realvalued number. a result, the complex plane is sometimes referred to as the Argand plane, after an independent description of it by J.R. Argand in 1813. 5 As
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Chapter 1. Complex Numbers
Im(z) z = x +iy
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1.2. Definition of complex numbers
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in the same way that one would perform addition or subtraction of two vectors. In order to display multiplication and division of two complex numbers, we need first to represent z1 and z2 in polar form. A complex number has a magnitude and an √argument (or angle); cf. Figure 1.3. The point z1 = a + ib has a magnitude r1 = z1  = a2 + b2 and an argument θ1 . The point z2 has a magnitude r2 and an argument θ2 . Then, z1 = r1 (cos(θ1 )+i sin(θ1 )) and z2 = r2 (cos(θ2 ) + i sin(θ2 )). We obtain z1 z2 = r1 (cos(θ1 ) + i sin(θ1 )) · r2 (cos(θ2 ) + i sin(θ2 )) = r1 r2 ((cos(θ1 ) cos(θ2 ) − sin(θ1 ) sin(θ2 )) + i(cos(θ1 ) sin(θ2 ) + sin(θ1 ) cos(θ2 ))) = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) and similarly z1 r1 = (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )). z2 r2 The magnitude of a product is r1 r2 , the product of the magnitudes, and the argument of a product is θ1 + θ2 , the sum of the arguments. Similarly, the magnitude of a quotient is r1 /r2 and the argument of a quotient is θ1 − θ2 . The product rule can be expressed as follows:6 z1 · z2  = z1  · z2  and arg(z1 · z2 ) = arg(z1 ) + arg(z2 ), z1 /z2  = z1 /z2  and arg(z1 /z2 ) = arg(z1 ) − arg(z2 ). The complex conjugate of a complex number z = x + iy is z¯ = x + iy = x − iy (i.e., swapping the sign of the imaginary part). A useful formula for finding the magnitude of a complex number is r2 = z2 = z¯z. One can easily check that • z1 ± z2 = z1 ± z2 , • z1 · z2 = z1 · z2 , • z1 /z2 = z1 /z2 . If p(z) is a polynomial with realvalued coefficients, it follows from these rules that p(z) = p(z). This is a special case of the Schwarz reflection principle described in Section 3.2.2.
1.2.3 Examples of polar rules 1. i2 = −1. This makes sense, since arg(i) = π/2 and i = 1. If we square i, we square the magnitude and double the argument; see the illustration in Figure 1.3. This had better be true, since it is the property that initiated the whole present topic of complex variables. 2. We can graphically find solutions to z k + 1 = 0, k = 1, 2, 3, by alternatively considering z k = −1. We find all the points on the unit circle (the circle of radius 1 centered at the origin) for which k arg(z) = π + 2nπ, where n is an integer. For example, if k = 2, arg(z) = π/2 + nπ when n = 0, 1. The k = 3 case is displayed in Figure 1.4(a). 6 The
dot “·” indicating multiplication will usually later be omitted unless it is helpful for clarity.
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Chapter 1. Complex Numbers
Im(z) z1
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Im(z)
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1 Re(z)
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Figure 1.3. If we square i, we square the magnitude (here equal to one) and double the argument (here π/2), i.e., the result ends up at −1.
3. We can similarly find solutions to z k − 1 = 0 by considering that z k = 1. These k solutions are called the kth roots of unity. They are also equispaced around the unit circle. The third roots of unity are shown in Figure 1.4(b). 4. The product rule implies de Moivre’s formula n
(cos θ + i sin θ) = cos nθ + i sin nθ
(1.1)
in the special case of n integer. After we, in the next chapter, have introduced functions of complex variables, such as f (z) = ez , we can simplify (and generalize) many polar rule results. One such case, noted at the end of Section 2.2.2, removes the restriction on n in (1.1). Example 1.1. Use the polar rules to simplify z = √ 1+i√ . 1+i 3
1.2. Definition of complex numbers
7
(a) z 3 + 1 = 0.
(b) z 3 − 1 = 0. Figure 1.4. Illustrations of the solutions to two single cubic equations.
√ For the numerator, we get 1 + i = 2 and √ arg(1 + i) = arctan 1= √ √ π/4. Regarding 3, i.e., 1 + i 3 = 1 + 3 = 2, and the denominator, we start by considering 1 + i √ √ √ √ π arg(1 + i 3) = arctan 3 = π/3. Thus, z = 2/ 2 = 1 and arg z = π4 − 12 · π3 = 12 . π π 7 Hence z = cos 12 + i sin 12 . 7 Since both (+z)2 and (−z)2 evaluate to z 2 , the reverse process of taking a square root will give two answers, √ of the opposite sign. It is common to let z denote the value in the right halfplane, but to include both values when instead writing z 1/2 . Multivalued functions are discussed further in Section 2.5.
8
Chapter 1. Complex Numbers
We should note that the function arctan x (with x real) always returns a value in the range [− π2 , π2 ] while the argument θ of a complex number has no corresponding limitation. If z is in the 1st or 4th quadrants (counting these counterclockwise starting from x > 0, Im(z) , but we need to either add π (if in 2nd quadrant) or y ≥ 0), we can use θ = arctan Re(z) subtract π (if in 3rd quadrant) to this result. Example 1.2. Express z = −1 − i in polar form. p √ We obtain r = Re(z)2 + Im(z)2 = 2 and, since z is in the 3rd quadrant, θ = −1 8 (arctan −1 ) − π = π4 − π = − 3π 4 .
1.3 The complex number plane as a tool for planar geometry Most geometric problems in planar geometry can be solved by means of analytic geometry (which has little or nothing to do with analytic functions). It is a branch of algebra that is used to model geometric objects by representing points, lines, etc., by formulas in Cartesian coordinates x, y in 2D (and x, y, z in 3D). There are certain cases where the complex plane z = x + iy provides convenient algebraic shortcuts over more general formulations in x and y. We give here just a single classical example, with several further ones in the Exercises. George Gamow’s delightful book One, Two, Three . . . Infinity [21] describes the following treasure hunt problem: Example 1.3. A young adventurer found among his greatgrandfather’s belongings a piece of parchment with instructions to find a hidden treasure. It described the location of the island, with the following further instructions (here translated to contemporary English): “Find the meadow with a single oak tree and a single pine tree. Start from the gallows, walk to the oak, counting your steps. Turn at the oak right and walk the same distance again. Put there a stake in the ground. Return to the gallows, walk to the pine (counting your steps), turn this time left, walk the same distance, and place another stake. Dig halfway between the stakes, and the treasure is there.” The adventurer found the island, the meadow, and the two trees, but all traces of the gallows were lost. He dug desperately in many places, and finally returned home, ruined. Could he have done better? Figure 1.5(a) shows the adventurer just having landed on the island. Had he known the basics of complex arithmetic, he would have placed a complex plane over the island, in such way that the oak was located at −1 and the pine at +1, and denoted the unknown location of the gallows by Γ. Part (b) of the figure shows this same complex plane, reduced to its basic mathematical essentials, and also two arbitrarily chosen guesses Γ and Γ0 for the location of the gallows. The instructions for the two walks starting from Γ give stake positions A and B satisfying A − (−1) = i (Γ − (−1)), B − 1 = i (1 − Γ), 8 It
is common to use the range (−π, π] for θ, but z = r (cos θ + i sin θ) still holds if we to a value for θ add any integer multiple of 2π.
1.3. The complex number plane as a tool for planar geometry
9
(a) Sketch of island
A' 2
A
1.5 1 treasure
B 0.5
B'
oak 2
1.5
1
0.5 pine1
0.5
1.5
2
0.5
' 1
(b) Superposed complex plane Figure 1.5. Illustration of the treasure hunt in Example 1.3: (a) Sketch of island (reproduced from [21], with permission from Dover, Inc.), (b) Simplified illustration showing only the mathematical essentials (in case of two different locations Γ and Γ0 of the gallows; A, B and A0 , B 0 are the stake locations).
= i .9 To find the treasure, it was therefore completely from which it follows that A+B 2 unnecessary to know the starting point of the walks. He could have found the treasure immediately at the location +i. 9 We utilized here that adding a constant to a complex number amounts to a translation in the complex plane, and multiplication by i a rotation of 90◦ around the origin, leaving magnitudes intact.
10
Chapter 1. Complex Numbers
Figure 1.6. The stereographic projection of the lines Re z = 0 and Im z = 0 (in bold) and two arbitrary circles (viewed from a location above the complex plane, with Im z large and positive). The North Pole (denoted by a black dot) maps to infinity, and the South Pole maps to zero.
1.4 Stereographic projection Although addition, subtraction, multiplication, and division are all well defined in the numbering system defined so far, there still persists a notable limitation. We cannot divide by zero. In terms of complex numbers, this becomes naturally handled if we project the complex number plane to a sphere of radius one, placed on top of the origin in the plane, in the way shown in Figure 1.6. We clearly get a onetoone map between the plane and the sphere. However, ∞ (infinity, in all directions) in the plane maps to the North Pole, a point much like any other point on the sphere. There is a tradeoff, however, in defining complex numbers as points on the sphere instead of on the plane. By doing so, we lose the nice immediate illustrations of the arithmetic operations +, −, ∗, /, but instead gain a better way to think about divisions by zero. Maybe more importantly, the sphere helps in thinking about singularities (such as poles and branch points, introduced later); these may be located in the finite part of the complex plane and/or at infinity.
Examples 1 Consider the function f1 (z) = z−(1+i) . It diverges to ∞ at z = 1 + i, a point on the plane with a matching point on the sphere, where it is said to have a pole singularity. Next, consider the function f2 (z) = z. This function goes to ∞ in the limit of z → ∞ following any direction in the complex plane. The limits in all of these directions map on the sphere to its top (North Pole). Thus, on the sphere, the two functions f1 (z) and f2 (z) do not differ in “character,” but only in where the singularity is located. This stereographic projection has a number of nice properties, such as the following:
• Locally, angles between intersecting curves are preserved between the plane and the sphere. • A circle in the complex plane maps to a circle on the sphere. • A circle on the sphere which contains the North Pole maps to a straight line in the plane.
1.5. Supplementary materials
11
These properties are all discussed further in the Exercises in Section 1.6.
1.4.1 Formulas It is easy to find the correspondence between points z = x + iy in the plane and points (X, Y, Z) on the sphere (satisfying X 2 + Y 2 + (Z − 1)2 = 1): Plane → Sphere X=
4x x2 +y 2 +4
Y =
x2 +y 2 +4
Z=
2(x2 +y 2 ) x2 +y 2 +4
Sphere → Plane
4y
x=
2X 2−Z
y=
2Y 2−Z
1.5 Supplementary materials 1.5.1 Constructing regular polygons with ruler and compass Geometric constructions using just a ruler and a compass are possible only when the result can be expressed algebraically as solving a sequence of linear and quadratic equations. Constructing regular polygons is closely related to properties of roots of unity. Example 1.4. Show that the regular pentagon can be constructed using ruler and com2π pass.10 Derive algebraic expressions for cos 2π 5 and sin 5 . The task is equivalent to finding the nontrivial roots to z 5 − 1 = (z − 1)(z 4 + z 3 + z 2 + z + 1) = 0. Writing these roots in turn as z1 , z2 , z3 , z4 , their sum is −1 (by the relation between roots and coefficients of a polynomial; −1 is the negative of the coefficient for z 3 in this quartic polynomial). Noting that z4 = z1 and z3 = z2 , we next group them in pairs z + z + z2 + z3 = −1. 1 {z }4  {z } y1 y2
(1.2)
Then y1 + y2 = −1 (directly from (1.2)) and also y1 · y2 = (z1 + z4 )(z2 + z3 ) = z1 z2 + z1 z3 + z4 z2 + z4 z3 = z3 + z4 + z1 + z2 = −1. Again using the relations between roots and√coefficients of a polynomial, y1 and y2 are the roots of y 2 + y − 1 = 0, i.e., √ 1 1 y1 = 2 ( 5 − 1) and y2 = 2 (− 5 − 1). Next, y1 = z1 + z4 = z1 + 1/z1 gives the q √ √ quadratic z12 − y1 z1 + 1 = 0, with solution z1 = 14 ( 5 − 1 ± 2i 52 + 25 ), from which q √ √ 1 2π we read off cos 2π = ( 5 − 1) and sin = (5 + 5)/8. 5 4 5 10 In one of the most famous mathematical discoveries of all times, C.F. Gauss (in 1796) used a related argument (involving successive stages of grouping roots) to find that a regular polygon with N corners is possible to construct with ruler and compass if and only if N = 2k p1 p2 · · · · · ps , where k = 0, 1, 2, . . . and pi , n i = 1, 2, . . . , s, are distinct prime numbers of the form p = 2(2 ) + 1, n = 0, 1, 2, . . .. At present, it is unknown whether there are any further such “Fermat primes” than what are obtained by n = 0, 1, 2, 3, 4. The cases n = 5, 6, . . . , 32 are all known to give composite numbers; some evidence suggests that the Fermat number sequence will never produce anyq further prime beyond n = 4. The n = 2 case gives p p √ √ √ √ 2π 1 cos 17 = 16 (−1 + 17 + 34 − 2 17 + 2 17 + 3 17 − 170 + 38 17), with a similar formula . This case is discussed in some detail in [11]. for sin 2π 17
12
Chapter 1. Complex Numbers
1.6 Exercises Exercise 1.6.1. Find the magnitude and the argument of the following complex numbers: (a) 1 − i.√ (b) 1 + i 3. 4 1+i √ (c) . √ 2 (d) 3 + 4i. (e) The√roots of z 7 + 128 = 0. 3 π (f) 1+i 1+i . Compute from this cos 12 . 1 (g) (1−i)8 . (h) √ 1+i√ . 1+i 3 4 1+i (i) . 1−i (j) ii/2 . Exercise 1.6.2. Express each of the following numbers in the form a + bi, where a and b are real: 1 . (a) 6+2i (b) (c)
(2+i)(3+2i) . 1−i 2 3 1 + 1+i
.
√ (d) The roots of z 2 + 32i z − 6i = 0. (e) The roots of z 5 = 2. √ (f) The roots of z 2 − i 3z − 1 = 0. 4 1−i . (g) 1+i (h) i1/2 . √ 2 1−i 3 (i) . 2+2i Exercise 1.6.3. Show that if z0 is a root to the polynomial equation z n +a1 z n−1 +a2 z n−2 + · · · + an = 0 with real coefficients, then so is z0 . Exercise 1.6.4. Show that the only prime number of the form n4 + 4 with an n integer is 5, obtained for n = ±1. Hint: Find the roots of z 4 + 4 = 0 and thus split z 4 + 4 as a product of four linear factors. Rearrange these four factors into two quadratic ones, with no imaginary parts remaining. Exercise 1.6.5. Show that the product of two positive integers that each can be written as the sum of two squares is itself the sum of two squares. Hint: Solution options include to (i) follow the idea of Exercise 1.6.4 or to (ii) utilize the relation z1  z2  = z1 z2 . Exercise 1.6.6. Let z and w be any two complex numbers. Prove the following relations: (a) z + z¯ = 2 Re(z). (b) Re(z) ≤ z. (c) z − w ≤ z + w.
1.6. Exercises
13
(d) zw = zw. (e) wz + wz ≤ 2wz. (f) z − z¯ = 2 i Im(z). Exercise 1.6.7. Draw the set of points that satisfy the following: (a) Im(z + 2) = 3. (d) z − 1 + z + 1 = 3. (b) z − i < 2. (e) z − 1 − z + 1 = 2. (c) z − i + 2 = z + 2i − 1. Exercise 1.6.8. Show that the lines Re(a z +b) = 0 and Re(c z +d) = 0 are perpendicular if and only if Re(a c) = 0. Exercise 1.6.9. Determine which of the following sets of points form a triangle ∆αβγ with a 90o corner: α=1−i α=1+i α=1−i β = 5 + 5i , β = 5 + 5i , β = 4 − 2i . (b) (c) (a) γ = −2 + i γ = −4 − 2i γ = −1 − i Exercise 1.6.10. Let z, w be complex numbers, and define E = zw + zw + 2zw. Show that E is real and nonnegative. Exercise 1.6.11. h One can i associate a complex number a + ib with a real antisymmetric a −b 2 × 2 matrix b a . Show that all the complex arithmetic operations (+ − ×/) then match the equivalent realvalued matrix operations (with “/” corresponding to a multiplication with the matrix inverse). h i a + bi c − di Comment: Considering regular matrix algebra for 2×2 matrices −c forms − di a − ib one entry point to the subject of quaternions (historically the bestknown attempt to generalize complex numbers). Exercise 1.6.12. Carry out the algebra Cardano used for finding all the roots to the (reduced) cubic equation z 3 − 15z − 4 = 0.11 Hint: The classic approach for solving a general cubic equation x3 + bx2 + cx + d = 0 proceeds as follows: Set x = y −b/3 to obtain a reduced cubic of the form y 3 +py +q = 0. 3 Next set y = z − p/(3z) to obtain z 3 − (p/3z) 0, which is a quadratic equation q + q =√ 3 3 in z , with two (out of six) solutions z1,2 = −q/2 ± R with R = (p/3)3 + (q/2)2 . Straightforward algebra will then show that z1 z2 = −p/3, from which in turn follows that y = z1 + z2 satisfies the reduced cubic. With one root known, it then suffices to solve a quadratic equation for the remaining roots. Exercise 1.6.13. Let α, β, γ be complex numbers representing the corners of a triangle. Show that this triangle is equilateral (all sides the same length) if and only if α2 + β 2 + γ 2 − αβ − αγ − βγ = 0.
(1.3)
11 G. Cardano (1501–1576) attributes in his 1540 publication the solution method to N.F. Tartaglia (1500–1557),
who likely was not the first one either to find it. Cardano’s student L. Ferrari (1522–1565) had by then already found a way to reduce a general quartic equation to a cubic one. N.H. Abel showed in 1823 that no closed form solution (using radicals, e.g., square root, cube root, etc.) is possible for the general quintic and higher degree polynomial equations (sometimes known as the Abel–Ruffini theorem).
14
Chapter 1. Complex Numbers
Hints: The following are two solution options: (i) If the triangle is equilateral, then γ − β = (β − α) C, α − γ = (γ − β) C, β − α = (α − γ) C, √
√
where C = − 21 + i 23 if α, β, γ are oriented clockwise, else C = − 12 − i 23 . Consider ratios of the equations above. If (1.3) holds, reverse the argument to obtain ratios, such as γ−β β−α α−γ = γ−β . (ii) Note that rotating/scaling the corner set by b (complex, 6= 0) and translating by a (complex) (i.e., α → a + bα, β → a + bβ, γ → a + bγ) just causes α2 + β 2 + γ 2 − αβ − αγ − βγ to become multiplied by b2 , which does not affect the expression being zero, or not. It thus suffices to look at a special case, such as choosing α = −1 and β = +1. Exercise 1.6.14. Show that the three roots to the cubic equation z 3 + az 2 + bz + c = 0 form an equilateral triangle if and only if a2 = 3b. Hint: Let the roots be α, β, γ. Expand (z − α)(z − β)(z − γ) and then use the result of Exercise 1.6.13. A symbolic algebra package is helpful. Exercise 1.6.15. Consider an arbitrary planar triangle ABC (cf. Figure 1.7(a)). Outside each of its sides, equilateral triangles CBa, ACb, and BAc form. For each of these, denote their centroids (average of its corner locations) by α, β, γ, respectively. Show that the triangle αβγ is equilateral. Hint: Following up on the example in Section 1.3, place a complex plane over the figure such that A = −1, B = 1, and C = z. Similarly to previously walking equal distances and turning 90◦ , turn 60◦ to obtain a, b, c and from these α, β, γ in terms of z. Apply the result of Exercise 1.6.13 (again, a symbolic algebra package is helpful). Exercise 1.6.16. Change the triangle ABC in Exercise 1.6.15 to a quadrilateral ABCD. This time, α, β, γ, δ are the centroids of squares based on each side of the quadrilateral (see Figure 1.7(b)). Show that the lines αγ and βδ joining these are of equal length and orthogonal to each other. Exercise 1.6.17. Show that, for the stereographic projection, a circle in the zplane corresponds to a circle on the sphere. Hint: Recall the correspondences between points z = x+iy in the regular complex zplane and points (X, Y, Z) on the sphere as given in Section 1.4.1. Furthermore, note that a circle on the sphere is given by the intersection of it with a plane AX + BY + CZ − D = 0. Exercise 1.6.18. Given a point in the complex zplane, show that the following steps produce the value w = 1/z: (i) Form 4z and project this value to the sphere, (ii) rotate the sphere half a turn around a line through the sphere center, parallel to the Xaxis, and (iii) project the point back again from the sphere to the zplane. Note: In some descriptions of stereographic projection, the unit sphere is shifted down by one, so its equator (rather than its south pole) lies in the (x, iy)plane. In this case, the factor 4 in step (i) should be omitted.
1.6. Exercises
15 a C α
β b
A
B γ
c
(a) Exercise 1.6.15.
γ D C δ
β A
B
α
(b) Exercise 1.6.16. Figure 1.7. The polygons described in Exercises 1.6.15 and 1.6.16.
Exercise 1.6.19. Let z1 and z2 be two points in the complex plane, with counterparts Z1 and Z2 on the stereographic sphere. Show that their distance (in 3space) becomes Z1 − Z2  = q
4 z1 − z2  q . 2 2 4 + z1  4 + z2 
Hint: See Figure 1.8. Apply traditional results from planar geometry (Pythagoras’ theorem, similar triangles, etc.).
16
Chapter 1. Complex Numbers
N 2.5 Z1
2 Z2
1.5 1 S
z1
0.5
3 2
0
z2
1 0
0.5 2
1
1 0
2
1
2
3
4
3
Figure 1.8. The geometry and notation used in Exercise 1.6.19.
Exercise 1.6.20. Verify that angles between intersecting curves are preserved between the plane and the stereographic sphere. Hint: Consider an infinitesimal triangle in the plane, and apply the result of Exercise 1.6.19. Exercise 1.6.21. Differentiating the equations for X, Y, Z in Section 1.4.1 with respect to x and y gives, after some further algebra, that
∂X ∂X ∆x + ∆y ∂x ∂y
2
+
∂Y ∂Y ∆x + ∆y ∂x ∂y
2
+
∂Z ∂Z ∆x + ∆y ∂x ∂y
2 =
16 ( (∆x)2 + (∆y)2 ) . (4 + x2 + y 2 )2
Deduce also from this result that the stereographic mapping is conformal (just scales and rotates any infinitesimal triangle in the (x, y)plane).
Chapter 2
Functions of a Complex Variable
A real function of one variable has the form y = f (x). A complex function has similarly the form w = f (z), where w and z now are complex quantities. Let z = x + iy and w = u + iv, with x, y, u, v real. Thus, w = u(x, y) + iv(x, y), where u(x, y) and v(x, y) are two real functions of two real variables.12 If we want to fully visualize w = f (z), we can plot u(x, y) and v(x, y) separately. For example, in the case that f (z) = z 2 , we can expand f (z) = z 2 = (x + iy)2 = (x2 − y 2 ) + i (2xy), i.e., Re(f (z)) = u(x, y) = x2 − y 2 and Im(f (z)) = v(x, y) = 2xy. Thus, to picture f (z), we can plot u(x, y) and v(x, y). It is often helpful to also display f (z) and arg(f (z)), as shown in the rightmost column of subplots in Figure 2.1.13 The top and bottom rows of subplots show similarly the functions f (z) = z and f (z) = z 3 , respectively. Along the xaxis (marked by a thick red curve in the left and center subplots and as a black curve in the right subplots), we recognize as special cases the standard results when z = x is a real variable. When displaying the magnitude of a function as a surface over the (x, y)plane, we will color it according to the displayed function’s phase angle (arg f (z)), choosing colors corresponding to the color wheel repeated identically in each plot of this kind (some different color choices have been used in the literature for this type of displays; we follow here the convention established, for example, in [39]). We will soon come to consider much more complicated functions than pure monomials. However, these types of graphical displays will continue to work very well for showing very general functions over the complex plane. As a further example, Figure 2.2 illustrates similarly the function f (z) = 12 (z + z1 ). Sometimes, it is helpful to also include a fourth subplot, here showing what the phase portrait of the function looks like (same as the third subplot, but viewed from straight above, making the magnitude information invisible unless contour lines for this are included). It is critically important to know what the functions “look like” to understand and best utilize them.
12 Like in the realvalued case, functions may be either single or multivalued (or undefined). For example, in the realvalued case, f (x) = x1/2 is undefined for x < 0, singlevalued for x = 0, √and doublevalued (with a choice of ±) for x > 0. Somewhat arbitrary conventions may apply, such as writing x instead of x1/2 to imply the positive choice. √ 13 We recall that f (z) = u2 + v 2 and arg(f (z)) = arctan(v/u) if f (z) is in quadrant 1 or 4; else add or subtract π to get a result in the range (−π, π].
17
2
y
1
2
2
1
x
0 1
2
y
1 0 1 2 2
1
2
3
z
z
1
0
y
1
2
2
1
x
0
0
x 1 2 2 1
y 0 1
z
1
0
y
1
2
2
1
x
0
1
2
4 2 0 2 4 4 2 0 2 4
2 2
1
2
4 2 0 2 4 4 2 0 2 4
1
0
y
1
2
2
1
x
0
1
2
0
2
2
4 2
1
x
0
1 2 2 1
y 0 1 2
u(x, y) = Re f (z)
0
y
1
2
2
1
x
0
1
2 1 0 1 2
0
2
2
4 1
x
0
v(x, y) = Im f (z)
1
2
0
2 2 2
2
4
4 2 0 2 4
1
0
y
1
2
2
1
x
0
1
2
Chapter 2. Functions of a Complex Variable
4 2 0 2 4
f (z) together with arg f (z)
18
Figure 2.1. Re f (z), Im f (z), and f (z) together with arg f (z) displayed for the three functions f (z) = z, z 2 , z 3 . In the first two columns of plots, some positive contour lines √ are shown in blue and some negative ones in green. In the last column, the magnitude f (z) = u2 + v 2 is displayed vertically, and the phase angle is displayed according to the “color wheel” at the bottom left of each of these subplots (showing how colors are associated with phase angles).
2.1. Derivative
19
5
5
0
0
5 2
5 2 2
1 1
0
2
1 1
0
0 1
y
0 1
1 2
2
x
y
1 2
2
x
(a) Re f (z)
(b) Im f (z)
(c) f (z) (elevation) and arg f (z) (color)
(d) As (c), but seen from above (with contours for f (z))
Figure 2.2. Graphical representations of f (z) = 21 (z + z1 ). The function values along the real axis are marked by thick red or black lines. The colors in (c) and (d) are related to the function’s phase angle (arg) according to the small color disk. The phase portrait (subplot (d)) will be omitted in most further cases (unless it reveals some features not apparent from the first three subplots). In this present case, it also includes contour lines for the values 1,2,3,4, . . . of the magnitude.
2.1 Derivative (x) , where we require the limit to In the realvalued case, f 0 (x) = lim∆x→0 f (x+∆x)−f ∆x be the same from the right and the left. In the case of a complex function, f 0 (z) = (z) lim∆z→0 f (z+∆z)−f , where the limit now should be the same when ∆z → 0 from ∆z any direction of the complex plane. In particular, the limits in two main directions must be the same: the horizontal direction (when ∆z is purely real) and the vertical direction (when ∆z is purely imaginary). This suffices for obtaining the Cauchy–Riemann equations.
Definition 2.1. The function f is analytic at z0 if f (z) is differentiable in some neighborhood of z0 (no matter how small).14 The function f is analytic in a region if it is analytic at all points in that region. If we describe a function as analytic, without specifying any point 14 Alternatively
expressed as an open region including z0 .
20
Chapter 2. Functions of a Complex Variable
or region, that means there is some region within which it is analytic.15 The function f is holomorphic if it is analytic. The terms are synonyms. An analytic function is entire if its region of analyticity includes all points in C, the finite complex plane, excluding infinity.
2.1.1 Cauchy–Riemann equations We defined f 0 (z) = lim∆z→0 do the following:
f (z+∆z)−f (z) , ∆z
where f (z) = u(x, y) + iv(x, y). Next, we
1. Choose ∆z = ∆x real. Then f 0 (z) = lim
∆x→0
u(x + ∆x, y) − u(x, y) ∆x
+i
v(x + ∆x, y) − v(x, y) ∆x
2. Choose ∆z = i∆y purely imaginary. Then f 0 (z) = lim∆y→0 ∂v i v(x,y+∆y)−v(x,y) = −i ∂u i∆y ∂y + ∂y .
=
∂v ∂u +i . ∂x ∂x
u(x,y+∆y)−u(x,y) + i∆y
For the two results to be the same, u(x, y) and v(x, y) must therefore be related by the Cauchy–Riemann (CR) equations:
∂u ∂v = , ∂x ∂y ∂v ∂u =− . ∂x ∂y
(2.1)
Theorem 2.2. If the function f (z) is differentiable, the CR equations hold.16 We have already shown this result just above. For the reverse direction, some minor extra conditions are needed. Theorem 2.40 shows that the CR equations together with ux , uy , vx , vy being continuous at a point suffices for f 0 (z) to exist at that point. If the CR equations are valid in a full neighborhood of a point, the extra requirements get reduced to u and v, themselves being continuous (i.e., we no longer need to verify continuity also of the partial derivatives).17 There are many consequences to the CR equations: 1. A purely real function f (z) = u(x, y) is not analytic unless identically constant. 2. If f (z) is differentiable once, it is differentiable an infinite amount of times (Theorem 4.11). There is no counterpart of this for realvalued functions. 3. The functions u(x, y) and v(x, y) each satisfy Laplace’s equation uxx + uyy = (ux )x + (uy )y = (vy )x + (−vx )y = 0 and, similarly, vxx + vyy = 0. 15 For example, we describe sin z, log z, and 1 all as analytic functions, although the last two have a singular 1−z (exceptional) point at z = 0 and z = 1, respectively. 16 When there is little danger of misunderstandings, we leave out certain formalities, here such as z belonging to an open region (meaning a region that does not include its boundary points). 17 The continuity requirements on u(x, y) and v(x, y) are needed to exclude cases such as f (z) = e−1/z 4 , for which the CR equations “happen” to hold also at the point of singularity z = 0. A more detailed discussion can be found in [23].
2.1. Derivative
21
Thus, neither u nor v can have a local minimum or a local maximum. Typically, at a local maximum, both uxx < 0 and uyy < 0, which is incompatible with uxx +uyy = 0. A strict proof follows from Theorem 4.16, which states that the value at any point of a Laplace equation solution is the average of the values around the periphery of any circle centered at the point. 4. Given u, we can compute v up to a constant, and vice versa. Each, satisfying Laplace’s equation, is called a harmonic function; the two are called the harmonic conjugates of each other. 5. The gradient vectors for u(x, y) and v(x, y) satisfy ∇u · ∇v = (ux , uy ) · (vx , vy ) = ux vx + uy vy = −vy uy + uy vy = 0; i.e., they are orthogonal to each other (or we are at a location with f 0 (z) = 0). The same holds for level curves to the two surfaces u(x, y) and v(x, y), since level curves are orthogonal to gradient vectors; see Figure 2.3. 1
0
1 1
0
1
Figure 2.3. This plot shows for f (z) = z 2 − 3 the level curves of u(x, y) as solid curves and of v(x, y) as dashed curves. They are orthogonal except at z = 0 where f 0 (0) = 0.
2.1.2 Taylor series verification of analyticity In order to test whether a function is analytic, we can thus check whether u(x, y) and v(x, y) are continuous and satisfy the CR equations in some open region. Another procedure will also turn out to be very useful. If a function has a Taylor expansion, it is analytic at least within its radius of convergence (radius of the largest disk in the complex plane surrounding the expansion point within which the series converges, to be discussed in much more detail shortly). It is sufficient to find a Taylor expansion in x and then replace x by z. For example, we have the following: • f (z) = log(1 + z) is analytic (at least inside the unit circle z = 1). One can easily d 1 find a series expansion for log(1 + x), using the fact that dx log(1 + x) = 1+x and
22
Chapter 2. Functions of a Complex Variable
the geometric series identity x−
2
x 2
+
3
x 3
1 1+x
= 1 − x + x2 − x3 + − · · · . Thus, log(1 + x) =
− +···.
• f (z) = z is analytic. It is its own Taylor series. • f (z) = z¯ = x − iy does not have a Taylor series in z. Checking with the CR ∂v equations, we see that ∂u ∂x = 1 and ∂y = −1, so (2.1) does not hold. Thus, this function is nowhere analytic. • If g(z) is analytic, then f (z) = g(¯ z ) is analytic. Proof: Let z = x + iy and g(z) = u(x, y) + iv(x, y). Then, g(¯ z ) = u(x, −y) − iv(x, −y) = a(x, y) + ib(x, y). So ax = ux , ay = −uy , bx = −vx , by = vy , which implies that, since g is analytic, ax = ux = vy = by and that ay = −uy = vx = −bx . Since the CR equations are z ) is analytic. satisfied, f (z) = g(¯
2.2 Some elementary functions generalized to complex argument by means of their Taylor expansion P∞ Consider a function f (x) that for x real has a Taylor series expansion f (x) = k=0 ak xk (where we, for simplicity, have expanded about x = 0).18 We can then substitute z for x P∞ k and let f (z) = k=0 ak z be the generalization of the function to a complex argument. If the original series converged for −R < x < R, we will show (Theorem 4.21) that the complex version will then converge for all z satisfying z < R. Typically, the function f (z), even if known only through its Taylor coefficients, is completely defined outside this disk as well. How to then find its values for z ≥ R is the topic of analytic continuation, which will be addressed in Chapter 3. Example 2.3. Extend f (x) = ex = 1 + x +
x3 x2 + + ··· 2! 3!
to a complex argument. Since the functional definition is in the form of a Taylor series, we just replace x with z and thus obtain z3 z2 + + ··· . f (z) = ez = 1 + z + 2! 3! It is now easy to verify that all the standard relations for the exponential function hold also in the complex case. For example, when x1 , x2 are real, it holds that ex1 · ex2 = ex1 +x2 . This implies that if we Taylor expand in its two variables (using standard procedures from realvalued calculus) the function f (x1 , x2 ) = ex1 · ex2 − ex1 +x2 x21 x31 x22 x32 = 1 + x1 + + + ··· 1 + x2 + + + ··· 2! 3! 2! 3! (x1 + x2 )2 (x1 + x2 )3 − 1 + (x1 + x2 ) + + + ··· 2! 3! = a0,0 + (a1,0 x1 + a0,1 x2 ) + (a2,0 x21 + a1,1 x1 x2 + a0,2 x22 ) + · · · , 18 A
Taylor series expanded about z = 0 is also known as a Maclaurin expansion.
2.2. Elementary functions generalized to complex argument by Taylor expansions 23
then every Taylor coefficient ai,j must be zero. These Taylor coefficients do not at all depend on what values we substitute for x1 and x2 , so the result will be zero also if we substitute in complex numbers z1 and z2 . Consequently, the relation ez1 · ez2 = ez1 +z2 must hold also for complex arguments z1 and z2 . In the same way, we can conclude that any functional relation that holds when x is real (and the function(s) involved are Taylor expandable) will again hold when x is replaced by z complex.
2.2.1 Relations between exponential and trigonometric functions If z = x + iy, we get from the result above that ez = ex+iy = ex · eiy , where furthermore y2 y3 y4 y5 eiy = 1 + iy − −i + + i ··· 2! 3! 4! 5! y2 y4 y3 y5 = 1− + ··· + i y − + · · · = cos y + i sin y. 2! 4! 3! 5!
(2.2)
The generalization of the exponential function to complex arguments can therefore also be written as ez = ex (cos y + i sin y). (2.3) An interesting special case is Euler’s identity, eπi + 1 = 0, connecting the five fundamental numbers 0, 1, e, π, and i. Figure 2.4 shows what the exponential function looks like in the complex plane (numerous similar illustrations for other standard analytic functions will soon be central to our description of them). This figure was produced by the MATLAB statements shown in Section 2.8.4. We can recognize the realvalued function y = ex along the real axis in the upper plot, along the real axis, drawn in red. Using Mathematica, similar plots are obtained by the code shown in Section 2.8.4. We saw for the monomials f (z) = z, z 2 , z 3 in Figure 2.1 and now again for the exponential function that a Taylor expandable function, previously defined only along the real axis, will have a unique extension to the complex plane. A strong statement about this will soon be given as Theorem 2.12.
2.2.2 Trigonometric functions represented in terms of the exponential function Since the relation eiy = cos y + i sin y holds for y real, it must according to our discussion above also hold when y is complex; i.e., we obtain what are known as Euler’s relations19 eiz = cos z + i sin z,
(2.4)
e−iz = cos z − i sin z.
(2.5) 2
19 Euler observed in a letter to Johann Bernoulli, dated October 18, 1740, that the unique solution to d y + y = dx2 0, y(0) = 2, y 0 (0) = 0 can be written either as y(x) = 2 cos x or as y(x) = eix + e−ix , also noting that similarly 2i sin x = eix − e−ix . His first publication on this (in 1748) includes (2.2). However, Roger Cotes wrote (in Phil. Trans. Royal Soc. 29, 1714, 5–45) what, in modern notation, amounts to i φ = log(cos φ + i sin φ); see Exercise 2.9.18.
24
Chapter 2. Functions of a Complex Variable
60 40 20 0 20 40 60 10 5
4
2
0 0 5
y
2 10
x
4
(a) Re ez .
60 40 20 0 20 40 60 10 5
4
2
0 0 5
y
2 10
4
x
(b) Im ez .
(c) ez  and arg ez . Figure 2.4. Graphical representations of f (z) = ez .
2.2. Elementary functions generalized to complex argument by Taylor expansions 25
Adding and subtracting these equations give cos z =
eiz + e−iz , 2
(2.6)
eiz − e−iz . (2.7) 2i Graphically, we can see what this becomes for cos z (Figure 2.5) and for sin z (Figure 2.6). The two functions differ only by a phase shift (translation) in the real direction (cos z = sin(z + π2 )), and they grow exponentially fast in the imaginary direction (away from the real axis). These relations (2.6) and (2.7) motivate how the functions cosh x and sinh x are defined for real x and thus extended to complex z as sin z =
ez + e−z = cos(iz), 2 ez − e−z 1 sinh z = = sin(iz). 2 i Many trigonometric identities (at first sight not having anything to do with complex numbers) can be derived very easily with the use of Euler’s formulas. cosh z =
Example 2.4. Derive the addition theorems for sin(α + β) and cos(α + β). We have (from (2.4)) ei(α+β) = cos(α + β) + i sin(α + β) and also ei(α+β) = eiα eiβ = (cos α + i sin α)(cos β + i sin β) = (cos α cos β − sin α sin β) + i(cos α sin β + sin α cos β). Equating the real and imaginary parts now give the standard addition theorems. PN Example 2.5. Evaluate k=−N cos kx.
N X
N X
cos kx =
k=−N
eikx = e−N ix 1 + eix + e2ix + · · · + e2N ix
k=−N
 Imaginary part vanishes, since
= e−N ix 
PN
1 − e(2N +1)ix 1 − eix
k=−N
{z
}
sin kx = 0; now a finite geometric progression.
= {z
e−(N +1/2)ix − e(N +1/2)ix e−ix/2 − eix/2 }
Routine way to simplify denominator of this type; here multiply numerator and denominator by e−ix/2
=
−2i sin N + 12 x sin N + 21 x = −2i sin 12 x sin 12 x  {z }
This is a purely realvalued result (when x is real), as to be expected
.
26
Chapter 2. Functions of a Complex Variable
30 20 10 0 10 20 30 4 2
10
5
0 0 2
y
5 4
x
10
(a) Re cos z.
30 20 10 0 10 20 30 4 2
10
5
0 0 2
y
5 4
10
x
(b) Im cos z.
(c)  cos z and arg cos z. Figure 2.5. Graphical representations of f (z) = cos z.
2.2. Elementary functions generalized to complex argument by Taylor expansions 27
30 20 10 0 10 20 30 4 2
10
5
0 0 2
y
5 4
x
10
(a) Re sin z.
30 20 10 0 10 20 30 4 2
10
5
0 0 2
y
5 4
10
x
(b) Im sin z.
(c)  sin z and arg sin z. Figure 2.6. Graphical representations of f (z) = sin z.
28
Chapter 2. Functions of a Complex Variable
While this result can also be proven by induction over N , the present derivation is more practical, since it does not require the answer to be known in advance. n We also obtain de Moivre’s formula (1.1) from the identity eix = einx .
2.2.3 Logarithm function From the (realvalued) Taylor expansion for log(1 + x) it follows that we can define the extension to complex z (with z < 1) as 1 1 1 log(1 + z) = z − z 2 + z 3 − z 4 ± · · · . 2 3 4
(2.8)
By the principle that functional relations that hold in real cases must carry over to complex variables, it will hold that log(z1 z2 ) = log(z1 ) + log(z2 ). Writing z = reiθ gives log z = log r + iθ = log z + i arg z,
(2.9)
a counterpart of (2.3) in terms of extending an elementary function from the real axis to the complex plane (and here also outside the radius of convergence R = 1 for (2.8)). Figure 2.7 displays the log z function over the domain −4 ≤ Re z ≤ 4, −4 ≤ Im z ≤ 4. Something we have not seen before is here visible along the negative real axis (leading us in Section 2.5 to introduce the concepts of branch cuts and Riemann surfaces).20 In the displays of different functions f (z), we typically show surface plots for Re f (z), Im f (z), and f (z). For the first two of these, we noted in Section 2.1.1 that they can never have any local (finite) maxima or minima, since they obey Laplace’s equation. From (2.9) follows a similar result for f (z). This quantity has a local minimum (with value zero) wherever f (z) is zero, but cannot otherwise have any local finite extreme points. If it did, so would log f (z) = Re log f (z), since log is a monotonic function (for positive arguments), but this is not allowed for the real part of an analytic function (this is later expressed again in Theorem 4.18).
2.2.4 Inverse hyperbolic and trigonometric functions Solving for z in the RHSs of (2.6) and (2.7) (and their hyperbolic counterparts) gives expressions for their inverse functions, for example 1+z 1 , (2.10) arctanh z = log 2 1−z arcsin z = −i log(iz + (1 − z 2 )1/2 ),
(2.11)
arccos z = −i log(z + (z 2 − 1)1/2 ), i 1 − iz arctan z = log . 2 1 + iz
(2.12) (2.13)
Figures 2.8–2.10 illustrate the first three of these. 20 We will there also note that the branch cut (line of discontinuity) can be placed in different ways; the negative real axis is merely an arbitrary choice that often is convenient.
2.2. Elementary functions generalized to complex argument by Taylor expansions 29
(a) Re log z.
(b) Im log z.
(c)  log z and arg log z.
(d) arg log z and contour lines for  log z.
Figure 2.7. Graphical representations of f (z) = log z.
Example 2.6. Derive (2.10), arctanh z = for z real.
1 2
log
1+z 1−z
, and compare it to its counterpart
We first note that w = arctanh z corresponds to z = tanh w = 1+z 1+z from which follows e2w = 1−z and thus (2.10): w = 12 log 1−z . x
−x
ew −e−w ew +e−w
=
e2w −1 e2w +1 ,
For x real, tanh x = eex −e +e−x asymptotes to ±1 when x → ±∞, and we obtain the inverse function by reflecting its graph across the line y = x, cf; Figures 2.11(a)–(b). The curve in Figure 2.11(b) matches the central (red) section −1 < z < 1 in Figure 2.8(a). Nothing else in these Figures 2.8(a)–(b) is apparent from the case of a real argument. Regarding the imaginary part in Figure 2.8(b), we see two shifted copies of the logarithm function from Figure 2.7(b), as to be expected from writing the arctanh formula as arctanh z = 12 log(1 + z) − 12 log(1 − z).
30
Chapter 2. Functions of a Complex Variable
(a) Re arctanh z.
(b) Im arctanh z.
(c) arctanh z and arg arctanh z. Figure 2.8. Graphical representations of f (z) = arctanh z. Notice the lines of discontinuity, like for the logarithm function highly visible on the imaginary part of the function, but not present for the real part.
2.2. Elementary functions generalized to complex argument by Taylor expansions 31
2 1.5 1 0.5 0 0.5 1 1.5 2 5 10 5
0 0 5
y
5
x
10
(a) Re arcsin z.
4 3 2 1 0 1 2 3 4 5 10 5
0 0 5
y
5
10
x
(b) Im arcsin z.
(c)  arcsin z and arg arcsin z. Figure 2.9. Graphical representations of f (z) = arcsin z.
32
Chapter 2. Functions of a Complex Variable
3
2.5
2
1.5
1
5
0.5 0 0 10
8
6
4
2
0
2
4
6
y
5
8
10
x
(a) Re arccos z. 4 3 2 1 0 1 2 3 4
5
0
y
5 10
8
6
2
4
0
2
4
6
8
10
x
(b) Im arccos z.
(c)  arccos z and arg arccos z. Figure 2.10. Graphical representations of f (z) = arccos z.
2.3. Additional observations on Taylor expansions of analytic functions
y
33
y
1.5
1.5
1
1
0.5
0.5
x 1.5
1
0.5
0.5
1
x
1.5
1.5
1
0.5
0.5
0.5
0.5
1
1
1.5
1.5
(a) The function y = tanh x.
1
1.5
(b) The inverse function y = arctanh x.
Figure 2.11. Graphical representation of the functions y = tanh x and y = arctanh x. The two curves are identical apart from a reflection in the line y = x (exchanging the roles of x and y).
2.3 Additional observations on Taylor expansions of analytic functions We introduced above the standard analytic functions through their Taylor expansions. Before using them much further, below are some observations of their domains of convergence. Theorem 2.7. The following are two key results on the region in which Taylor expansions converge (assuming for notational simplicity expansions about z0 = 0; else replace z by z − z0 below): 1. Convergence occurs for all z within a circular domain z < R with divergence for all z with z > R, where R is some value from 0 to ∞ (including 0 and ∞). Convergence on the boundary z = R can vary between functions and boundary locations. 2. The radius R is the largest possible, such that f (z) is differentiable (i.e., analytic) at every point with z < R. These statements will be proven later in Section 4.2.8. A collection of various convergence tests can be found in Section 2.8.3. We will use these tests in the following examples. Example 2.8. Determine R for f (z) = ez = 1 +
1 1!
z+
1 2!
z2 +
1 3!
z3 + · · · .
Since f 0 (z) = ez , the derivative exists and is finite for all finite z. By part 2 above, the radius of convergence is R = ∞ (meaning that the series converges for all z < R = ∞). The result also follows immediately also from the root and ratio tests (Theorems 2.35 and 2.36). Example 2.9. Determine R for f (z) = log(1 + z) = z − 21 z 2 + 13 z 3 − + · · · .
34
Chapter 2. Functions of a Complex Variable
1 Since f 0 (z) = 1+z , the only singularity is located at z = −1, implying that the radius of convergence is R = 1. Alternatively, we can note that R = 1/ lim1/n1/n = 1 or that n R = lim n→∞ aan+1  = 1.
Example 2.10. Determine R for
1 1+z 2
= 1 − z2 + z4 − z6 + − · · · .
The radius of convergence is again R = 1, since the function is differentiable everyn where apart from at z = ±i. Also R = 1/ liman 1/n = 1, but lim n→∞ aan+1  does not exist, so the root test is applicable but the ratio test is not.21 Example 2.11. Determine R for
´∞ 0
e−t 1+zt dt
= 1 − 1!z + 2!z 2 − 3!z 3 + 4!z 4 − + · · · .
This function will be considered in more detail in Example 3.18. The radius of convern gence is R = 0, easiest seen from R = lim n→∞ aan+1  = 0, but it follows also from the other tests. P∞ n It is easy to verify that if fP (z) = the disk of radius R, then n=0 an z is valid Pin ∞ ∞ 0 n−1 00 we can differentiate (f (z) = n=0 nan z , f (z) = n=0 n(n − 1)an z n−2 , . . .) or integrate however many times we want and obtain expansions that will still converge in the disk of radius exactly R. Theorem 2.12. If two analytic functions coincide on any curve segment, no matter how small (or any point set with a limit point), then the two functions are identically the same.22 Proof. P Let the limit point be z = 0 and the difference between the two functions be ∞ f (z) = n=0 an z n . From f (z) being zero at all the points it follows (by continuity) that P∞ = n=1 an z n−1 , also zero at all the points, implying that a0 = 0. Consider next f (z) z a1 = 0, etc. All the Taylor coefficients, and thus also f (z), must vanish. This last result raises fundamental questions about different possible representations of 1 a single function. As illustrated in Figure 2.12, f (z) = 1−z is defined everywhere except at z = 1, where the function has a pole. Next, define g(z) = 1 + z + z 2 + z 3 + · · · . This function g(z) converges only in the disk z < 1 and diverges for z > 1. Do these functions represent the same function everywhere? This will be the topic of analytic continuation (Chapter 3). Although the function itself exists everywhere, its particular representation as a Taylor series fails outside z = 1. Given, at a point a, values for f (a), f 0 (a), f 00 (a), . . . , we can write down the Taylor expansion at that point f (z) = f (a) + (z − a)f 0 (a) +
(z − a)2 00 f (a) + · · · . 2!
(2.14)
This relation is theoretically important, but it is usually impractical to differentiate a nontrivial function many times. We will next describe two methods (beyond (2.14)) to obtain 21 For
the definition of lim, see Theorem 2.35. result extends as far away as the functions remain analytic; “continuation” of analytic functions is discussed in Chapter 3. 22 The
2.3. Additional observations on Taylor expansions of analytic functions
Im(z)
35
Im(z)
1
1 Re(z)
Re(z)
(a) f (z) is defined for all z except z = 1.
(b) g(z) converges only in z < 1.
1 Figure 2.12. Although the function f (z) = 1−z exists everywhere (except at z = 1), its particular representation g(z) as a Taylor series fails on and outside z = 1
Taylor expansions of given functions. To go in the reverse direction and obtain a function from its Taylor expansion, we will need analytic continuation.
2.3.1 Get the expansion from an already known expansion Example 2.13. Derive Taylor expansions for log(1 + z) and 1 series 1+z = 1 − z + z2 − z3 + z4 − + · · · .
1 1+z 2
based on the geometric
Integrating the geometric series gives log(1 + z) = z − 21 z 2 + 13 z 3 − 14 z 4 + 15 z 5 − + · · · + C. To find C, let z = 0, leading to C = 0. Replacing z by z 2 in the geometric 1 2 4 6 8 series gives 1+z 2 = 1 − z + z − z + z − +··· . Example 2.14. Determine the Taylor expansion for f (z) = arcsin z. We first note that f 0 (z) =
√ 1 . 1−z 2
The binomial theorem, generalized to noninteger
s(s−1)(s−2) 3 2 powers, states that (1 + z) = 1 + sz + s(s−1) z + · · · . Using this 2! z + 3! 1 2 2 0 z4 − with s = −1/2 and swapping z for −z gives f (z) = 1 + 2 z + (−1/2)(−3/2) 2! (−1/2)(−3/2)(−5/2) 6 1 3 3 5 5 7 z + · · · and therefore f (z) = arcsin z = z + 6 z + 40 z + 112 z + · · · 3! (with the integration constant vanishing, since arcsin 0 = 0 or, equivalently, since arcsin z is an odd function of z). s
2.3.2 Method of undetermined coefficients This method applies, for example, when one wants to Taylor expand a rational function. Other cases include finding expansions to solutions of ODEs. See also Section 2.8.1 for inverting a Taylor expansion. Example 2.15. Find the Taylor expansion of f (z) =
3−4z 1+2z .
Let the expansion be 3 − 4z = a0 + a1 z + a2 z 2 + a3 z 3 + · · · . 1 + 2z
36
Chapter 2. Functions of a Complex Variable
Multiplying up the denominator gives 3 − 4z = a0 + (a1 + 2a0 ) z + (a2 + 2a1 ) z 2 + (a3 + 2a2 ) z 3 + · · · . Since this is an identity, the coefficients on the two sides must match for all powers of z. We thus obtain 3 = a0 , −4 = a1 + 2a0 , 0 = a2 + 2a1 , (2.15) 0 = a3 + 2a2 , .. . from which all the coefficients follow recursively: a0 = 3, a1 = −10, an+2 = −2an+1 , n = 0, 1, 2, . . . or evaluated: f (z) = 3 − 10 z + 20 z 2 − 40 z 3 + 80 z 4 + − · · · . 5 An alternative approach is to write f (z) = 3−4z 1+2z = −2 + 1+2z and then note that the second term is the sum of an infinite geometric progression. With the power of modern computers, many results relating to analytic functions are nowadays first discovered numerically, and only later (if at all) verified analytically. This brings up the intriguing issue of the amount of numerical evidence that is needed for it to become plausible that an observed result will always remain true. The book The Computer as a Crucible, subtitled An Introduction to Experimental Mathematics [6], contains a fascinating discussion of this subject. The following example, again concerning Taylor expanding a rational function, can be found in it. The example is highly exceptional, but it can still serve as a useful warning to be careful! Example 2.16. Find some leading terms in the Taylor expansion of 8 + 7z − 7z 2 − 7z 3 . 1 − 6z − 7z 2 + 5z 3 + 6z 4 Does the recursion a0 = 8, a1 = 55, an+2 = 1 + a2n+1 /an , n = 0, 1, 2, . . . (with the brackets b·c meaning rounding down to nearest integer), produce the exact result for all the Taylor coefficients of f (z)? f (z) =
For a problem involving quite a lot of algebra, computational tools greatly simplify tedious labor (as well as greatly reduce the risk of algebraic errors). For Mathematica, note in this context its command “Series” and for MATLAB the functions “conv” and “deconv.” The sequence of Taylor coefficients (all integers) begins a0 = 8, a1 = 55, a2 = 379, a3 = 2612, a4 = 18002, a5 = 124071, a6 = 855106, a7 = 5893451, a8 = 40618081, a9 = 279942687, .. . The singularity of f (z) located nearest to the origin occurs around z = 0.145094, implying
2.3. Additional observations on Taylor expansions of analytic functions
37
that the coefficients will forever grow each step roughly by a factor of 7. It has been found that the proposed recursion formula for the coefficients an is exact all the way up through n = 11, 055, but it fails for n = 11, 056. g(z) z √ Example 2.17. Determine the Taylor expansion for f (z) = arcsin = h(z) based on the 1−z 2 3 5 5 7 known expansions g(z) = z + 61 z 3 + 40 z + 112 z + · · · and h(z) = 1 − 12 z 2 − 81 z 4 − 5 8 1 6 16 z − 128 z − · · · .
This problem generalizes the rational function case above in that the degree is infinite in both the numerator and denominator. That makes, however, no difference in how one g(z) proceeds. Noting that the ratio will be an odd function, we let the expansion for h(z) be g(z) h(z)
= a1 z + a3 z 3 + a5 z 5 + · · · . Multiplying up the denominator and equating coefficients gives again a set of recursive equations that can be solved explicitly: 1
=
1 6
= − 12 a1 + a3 ,
3 40 5 112
and thus f (z) =
arcsin z √ 1−z 2
a1 ,
= − 18 a1 − 12 a3 + a5 , = .. .
1 − 16 a1
= z + 23 z 3 +
8 5 15 z
−
1 8 a3
+
16 7 35 z
−
1 2 a5
(2.16) + a7 ,
+ ··· .
Another common case of using the method of undetermined coefficients arises if the function f (z) satisfies some simple ODE (ordinary differential equation). Example 2.18. Determine the Taylor expansion for f (z) = it satisfies
arcsin z √ 1−z 2
based on an ODE that
Once we differentiate and simplify, we notice that f (z) satisfies the linear ODE (1 − z 2 )f 0 (z) = 1 + z f (z) with f (0) = 0. Assume that f (z) = a0 + a1 z + a2 z 2 + · · · , implying f 0 (z) = a1 + 2a2 z + 3a3 z 2 + · · · . After substituting these expansions into the ODE, we can equate the coefficients and get any number of coefficients recursively. Noting z √ that f (z) = arcsin is an odd function of z again simplifies the algebra (since that implies 1−z 2 all even terms in the Taylor expansion must vanish). In either case, starting with a0 = 0 z 8 5 16 7 √ = z + 32 z 3 + 15 z + 35 z + ··· . gives again f (z) = arcsin 1−z 2 Especially for nonlinear ODEs of the general form f 0 (z) = F (z, f (z)), where F is analytic in both its arguments, even the method of undetermined coefficients often becomes cumbersome if one needs manyP terms. Assume we have already obtained the beginning of n the Taylor expansion, fn (z) = k=0 ck z k (at first just knowing c0 from the initial condition), substituting this truncated series into the ODE and then reexpanding and integrating once gives one more term. This is relatively straightforward to carry out (or implement computationally), and can be repeated indefinitely to obtain any number of terms. Example 2.19. Find the leading coefficients c0 , c1 , . . . , c5 in the Taylor expansion of the (z) solution to dfdz = 1 + sin f (z), f (0) = 0.
38
Chapter 2. Functions of a Complex Variable
Following the recipe just above, the iterations proceed as follows (going down rowbyrow): −
−
−
f0 (z) = 0
df1 (z) dz
∼ 1 + sin f0 (z) ∼ 1
f1 (z) = z
df2 (z) dz
∼ 1 + sin f1 (z) ∼ 1 + z
f2 (z) = z + 21 z 2
df3 (z) dz
∼ 1 + sin f2 (z) ∼ 1 + z + 21 z 2
f3 (z) = z + 12 z 2 + 61 z 3
df4 (z) dz
∼ 1 + sin f3 (z) ∼ 1 + z +
1 2 z 2
∼ 1 + sin f4 (z) ∼ 1 + z +
1 2 z 2
df5 (z) dz
+ 0z
3 3
+ 0z −
.. .
1 4 z 4
f4 (z) = z +
1 2 z 2
f5 (z) = z + .. .
1 2 z 2
−
+
1 3 z 6
4
+ 0z
+
1 3 z 6
+ 0z 4 −
1 5 z 20
2.4 Singularities Functions f (x), x ∈ R, with which we are familiar can usually be extended to become analytic functions w = f (z). However, no matter how smooth these are along the real axis, they can become infinite (or nondifferentiable) at certain points (or regions) in the complex plane. These are called singularities. We have encountered some examples already. For 1 instance, f (x) = 1+x 2 is everywhere infinitely differentiable along the real axis, but, when 1 generalized to a complex argument, f (z) = 1+z 2 (shown in Figure 2.13) has singularities (because of division by zero) at z = ±i. There is a difficulty in introducing the concept of singularities because a good understanding of the different forms they can take requires the use of Laurent expansions, which in turn are best derived by means of integration along contours in the complex plane. However, singularities play a key role in developing integration techniques. Hence, we follow here a twostep approach for discussing singularities. In this present Section 2.4, we limit ourselves to a quite heuristic (nonrigorous) overview, sufficient for developing complex integration in Chapter 4. Once we have Laurent expansions available, we revisit singularities for a more indepth treatment in Section 4.3.2. We next want to describe the different ways in which an analytic function can be singular. There turns out to be six types of singularities. The first four refer to isolated singularities, meaning that the function is analytic in a full neighborhood of the singular point.
2.4.1 Removable singularities This type of “singularity” is a problem of a particular representation of the function, and not of the function itself. It should therefore not be counted as a type of singularity, but it nevertheless traditionally is, thus being a misnomer. In contrast to genuine singularities, removable ones do not limit the radius of convergence for a Taylor series. Example 2.20. The function f (z) = sinz z has a singularity at z = 0, at which point 1 3 1 5 it is undefined. However, sin z = z − 3! z + 5! z − + · · · . The expansion sinz z = 1 2 1 4 1 − 3! z + 5! z − + · · · converges for all z and is thus analytic for all z. It is an entire function that uniquely provides the missing value of f (0) = 1. Wherever we center a Taylor expansion of this function, it will have R = ∞ (i.e., a removable singularity does
2.4. Singularities
39
3 2 1 0 1 2 3 2 1
2
1
0 0 1
y
1 2
x
2
(a) Re
1 1+z 2 .
3 2 1 0 1 2 3 2 1
2
1
0 0 1
y
1 2
x
2
(b) Im
1 1+z 2 .
1 (c)  1+z 2  and arg
1 1+z 2 .
1 Figure 2.13. Graphical representations of f (z) = 1+z 2 . The Runge function f (z) = has two first order pole singularities in the complex plane but is analytic everywhere on the real axis. 1 1+z 2
40
Chapter 2. Functions of a Complex Variable
not influence a Taylor series’ radius of convergence; it is not a feature of a function but only of a particular way to represent it). A more formal characterization of the next two types of singularities has to await our description of Laurent expansions in Section 4.3. We can, however, already now describe their key features and note how they differ both in character and when displayed visually.
2.4.2 Poles These singularities typically arise through division by zero. In the neighborhood of such a point, f (z) gets uniformly large. One can recognize a pole by its signature “chimney look” in the graph of f (z), together with a simple phase angle (color) pattern on the chimney (e.g., Figures 2.2 and 2.13, to be compared with the infinity of color strips we soon will see for essential singularities in Figure 2.14). A function that has poles as its only singularities in the finite complex plane is called a meromorphic function. Example 2.21. The function f (z) = z1 has a first order pole singularity23 at z = 0, f (z) = 1 1 z 2 has a second order pole singularity at z = 0, and f (z) = sin z has first order pole singularities at z = πk, k = 0, ±1, ±2, . . . , in all cases caused by divisions by zero. Near a pole singularity, the phase angle coloring shows great simplicity and regularity. Returning to Figure 2.2 with f (z) = z + z1 , we see a first order pole (from the term z1 ) at z = 0 and first order zeros at z = ±i. In the phase portrait (subplot (d)), we note that the color pattern near the pole is identical to that surrounding the zeros, apart from a reversal in the order of the colors. For zeros and poles of higher orders (such as z 2 , 1/z 2 , etc.), the colors (phase angles) would be traversed the same number of times as the order.
2.4.3 Essential singularities The structure of a function near an essential singularity is far more complicated. In fact (as will be shown later in Theorem 4.30), the function comes arbitrarily close to every single finite value infinitely many times in any neighborhood of the singular point (no matter how small). Example 2.22. The function f (z) = e1/z has an essential singularity at z = 0. 1
x−iy
x
y y This function f (z) = e1/z = e x+iy = e x2 +y2 = e x2 +y2 (cos( x2 +y 2 ) − i sin( x2 +y 2 )) converges to 0 as z → 0 from the negative real axis, converges to +∞ as z → 0 from the positive real axis, and is rapidly oscillatory as z approaches 0 from any other direction. Due to unbound oscillations of infinite density near the origin, the function cannot be well represented there by our usual mesh plots of its real and imaginary parts. The magnitude/phase plots in Figure 2.14 show a totally different pattern than for poles: well structured but with infinitely “dense” phase angle stripes near the singularity.
For all the singularity types above, we return to the same value if we follow the function along a small path that goes around the singular point. 23 Also
known as a simple pole.
2.4. Singularities
41
(a) e1/z  and arg e1/z .
(b) arg e1/z . Figure 2.14. Graphical representations of f (z) = e1/z . Illustration of the essential singularity.
2.4.4 Branch points Example 2.23. Consider the function f (z) = log z = log reiθ = log r + iθ. Moving around the origin once in the positive direction increases θ by 2π, as we saw in Figure 2.7(b). How to “deal with” such functions will be discussed in Section 2.5. Very briefly, a branch point is any point such that, moving around it once, one does not return to the same function value as one started with. A branch cut is a “barrier curve” that one can draw in the complex plane, from one branch point to another one, if one wishes to make a multivalued function appear as if it was singlevalued. The path of a branch cut is quite arbitrary and will mostly depend on conventions. For example, if one wants log z to be singlevalued for
42
Chapter 2. Functions of a Complex Variable
θ ∈ (−π, π], one would place the branch cut (barrier not to be crossed) along the negative real axis. If one instead wants to work within θ ∈ [0, 2π), one would place it along the positive real axis. While the shape of this artificial barrier (branch cut) is arbitrary, its end points (branch points) are not. For the log function, it needs to go along some nonself– intersecting path from z = 0 out to infinity (which, in view of the stereographic projection, can also be thought of as a single point). Example 2.24. If not imposing any branch cut restrictions, then Im(log 1) = iθ, where √ ) = i( π + 2kπ), k ∈ Z (integer). With the common θ = 0, ±2π, ±4π, . . . and log( 1+i 4 2 choice of placing the branch cut along the negative real axis, only the θ = 0 and k = 0 cases are relevant (i.e., the result is then singlevalued). Example 2.25. In the case of the square root function f (z) = z 1/2 , Figure 2.17 shows two common branch cut choices (discussed further in Section 2.5). Note that we can encounter somewhat similar cases for real variables. Apart from needing to restrict the xvalues of y = arcsin x to −1 ≤ x ≤ 1 to keep f (x) defined as a realvalued quantity, singlevaluedness is typically obtained by the further restriction −π/2 ≤ f (x) ≤ π/2. Back to complex variables: Software such as Mathematica or MATLAB typically impose branch cuts following common conventions, for instance along the negative real axis in cases where this is appropriate. Another much less “crude” approach for achieving uniqueness is to extend the complex plane through Riemann sheets (or surfaces), making branch cuts unnecessary (see Section 2.5 below). When using Riemann sheets, also multivalued analytic functions can be considered in their entirety, rather than just partially.
2.4.5 Cluster points An example would be a function with poles located at an infinite set of points with a finite limit point. 1 Example 2.26. The function f (z) = sin1 1 is singular at z = πk , k = 0, ±1, . . . . These z are all poles, apart from the cluster point at z = 0. The magnitude and phase plots in Figure 2.15 show the string of poles along the real axis and the intricate structure of the phase near the cluster point at the origin. This point should not be viewed as an essential singularity, since it is not isolated (it contains no full neighborhood that is singularityfree).
2.4.6 Natural boundary A natural boundary is a genuine barrier (or boundary) past which a function no longer exists (i.e., it is not just that a certain functional form fails to work, such as a Taylor expansion beyond its radius of convergence). Typically, every point on this boundary is a cluster point of singularities in the sense described just above for a single cluster point (in Section 2.4.5). We will discuss this later in the context of analytic continuation (including an illustration in Figure 3.2).
2.5. Multivalued functions—Branch cuts and Riemann sheets
43
(a) 1/ sin z1  and arg (1/ sin z1 ).
(b) arg (1/ sin z1 ). Figure 2.15. Graphical representations of f (z) =
1 1 sin z
. Illustration of the cluster point singularity.
2.5 Multivalued functions—Branch cuts and Riemann sheets The simplest example of multivalued functions is provided by the square root function f (z) = z 1/2 . When using only real numbers, the function y = x2 can be inverted graphically by reflection in the line y = x (cf. Figures 2.16(a)–(b)). For x positive, there are √ two choices of numbers√that, when squared, return the value x. If one denotes these by ± x, it is common to let x be the positive choice. With instead a complex argument z, Figure 2.18 reveals a much richer functional behavior. We recognize again √ the red curve from Figure 2.16(b) in Figure 2.18(a), but nothing else. The two choices ± x are merely curves on smooth connected surfaces. The function f (z) = z 1/2 is doublevalued for all z, apart from an anomaly (branch point) at z = 0 (and a counterpart branch point at z = ∞).
Chapter 2. Functions of a Complex Variable 2
2
1.5
1.5
1
1
0.5
0.5
0
0
y
y
44
0.5
0.5
1
1
1.5
1.5
2 2
1.5
1
0.5
0
0.5
1
x
(a) y = x2 near the origin.
1.5
2
2 2
1.5
1
0.5
0
0.5
1
1.5
2
x
√ (b) y = ± x near the origin.
√ Figure 2.16. The realvalued function y = x2 , and its inverse y = ± x, as obtained by reflection in the line y = x (shown in blue).
To analyze f (z) = z 1/2 in more detail, consider z in its polar form representation z = reiθ , where we term 2πik, k integer, to θ in the exponent without changing √ can add √ any iθ the result. Thus, z = re 2 +ikπ , where it makes sense to limit k to k = 0, 1 as further √ choices become just repetitions. For instance, 1 = ekiπ , for k = 0, 1, recovering the real √ case 1 = ±1; squaring either +1 or −1 gives the result +1. There are two main ways to proceed when representing a multivalued function: 1. Impose some “barrier” (branch cut) in the complex plane such that, if we do not allow z to move across it, the result will be singlevalued. 2. Accept double (or, more generally, multivalued) surfaces of function values, such as those displayed here in Figures 2.18(a)–(b) and also in several further figures. We will next discuss these two choices in some more detail.
2.5.1 Branch cuts √ iθ √ Staying for now with the f (z) = z 1/2 example, we can limit ourselves to z = re 2 +ikπ , θ ∈ (−π, π], and the single case of k = 0. This limits z to the region shown in Figure 2.17(a), with an artificial “barrier” (branch cut) placed along the negative real axis. With this restriction, all we will be able to “see” of the function is what is at or above height zero in Figure 2.18(a), and on the surface sheet that tilts upwards for increasing imaginary part in Figure 2.18(b). These parts are singlevalued and in many cases suffice for what may be needed. This k = 0 case is often called the function’s principal value or primary Riemann sheet. Instead choosing k = 1 will give us the rest of the function, the parts that were omitted when choosing k = 0. Even when considering both cases k = 0 and k = 1, the results will depend on our quite arbitrary choice of locating the branch cut in the zplane along the negative real
2.5. Multivalued functions—Branch cuts and Riemann sheets Im(z) p (1 + i)/ 2
Im(z) p (1 + i)/ 2
1 Re(z)
i −i
1 Re(z) −1
i
p (−1 + i)/ 2
p (1 − i)/ 2
(a)
45
√ z with θ ∈ (−π, π).
(b)
√ z with θ ∈ (0, 2π).
Figure 2.17. Two branch cut choices for f (z) = z 1/2 . The numbers shown give the function values at the respective locations.
axis. Figure 2.17(b) illustrates the branch cut instead placed along (immediately below) the positive real axis. If we thus define θ ∈ [0, 2π), we get a jump discontinuity at the branch √ i is the same in either cut on the positive real axis. The figure shows that the value of √ case, but the value of −i differs. In Figure 2.18(a), the primary sheet would now be what we encounter when starting from the top red curve and moving one full turn in the positive (counterclockwise) direction until immediately before we reach the bottom red curve. In part (b), we would follow the top surface all the way around once. It will depend on the application what choice of branch cut is the most convenient to use. √ How can we test whether a point is a branch point? For example, let f (z) = z − z0 . We can test z0 by√centering a small circle at that point: let z − z0 = reiθ . The function becomes f (z) = reiθ/2 . We then monitor the function values when θ increases by 2π. If the starting value differs from the end value, it means that the point that is being tested is a branch point, and that we need to introduce a branch cut if we want to make the function appear singlevalued. For example, if we choose θ ∈ [0, 2π), let’s test the point z =√z0 . Let’s follow a small circle around z0 , starting 0. At the starting point, f (z) = r. √ at θ = √ After one revolution, θ = 2π and f (z) = reiπ = − r. Thus z = z0 is a branch point. √ A branch cut always joins two branch points. We know that z = z0 is one (for f (z) = z − z0 ), but where is the other one? There is no other one for finite z0 . Let’s simplify the notation by setting z0 = 0 and then test z = ∞. Considering the stereographic projection, z = ∞ is just one point like any other. To test it, let’s consider z = 1t around t = 0. Then f (t) = √1t , and we see that t = 0 (and thus z = ∞) is also a branch point. The branch cut will need to link those two points. Example 2.27. Discuss the different branch cut options for f (z) = (z − 1)1/2 (z + 1)1/2 . Figure 2.19 shows the real and imaginary parts of f (z). The function clearly has two solution sheets and branch points at z = 1 and z = −1. √ To explore whether infinity also (1−t)(1+t)
is a branch point, we set z = 1t and obtain f (t) = . Thus, there’s a pole at t t = 0, but not a branch point. The cut will thus need to link z = −1 and z = 1 together. Following the convention to place branch cuts along the real axis, if possible, there are two natural choices: (i) directly along [−1, +1], and (ii) via infinity, i.e., along [−∞, −1]
46
Chapter 2. Functions of a Complex Variable
(a) Re z 1/2 .
(b) Im z 1/2 .
(c) z 1/2  . Figure 2.18. Visualization of f (z) = z 1/2 . The phase angle coloring corresponds to the primary branch (upper surface) in Parts (a) and (b) (phase angle θ in the range − π2 < θ ≤ π2 ).
2.5. Multivalued functions—Branch cuts and Riemann sheets
47
2
1.5
1
0.5
0
0.5
1
1.5
2 1 0.5 0 0.5 1
y
3
2
1
0
1
2
3
x
(a) Re f (z).
2
1.5
1
0.5
0
0.5
1
1.5
2 1 0.5 0 0.5
y
1
3
2
1
0
1
2
3
x
(b) Im f (z). Figure 2.19. Real and imaginary parts of f (z) = (z − 1)1/2 (z + 1)1/2 . Each consists of two sheets.
and [1, ∞]. Figure 2.20 focuses on Im f (z) and illustrates the corresponding singlevalued solutions that arise from the two branch cut strategies. For each choice of branch cuts, it is clear that the two singlevalued functions can be combined to form the multivalued one shown in Figure 2.19(b). Without relying on the figures above, we can directly verify that the cut presented in Figure 2.21 makes the function singlevalued. Let’s follow the contour around the cut to make sure that we end up with the same function value as the one with which we started. 1/2 1/2 θ1 +θ2 Let z + 1 = R1 eiθ1 and z − 1 = R2 eiθ2 . Thus, (z + 1)1/2 (z − 1)1/2 = R1 R2 ei 2 . 2 Let Θ = θ1 +θ 2 . We want to make sure that the function’s argument Θ has changed by an integer multiple of 2π after having gone around the contour. Table 2.1 shows that after going around the cut, the function’s argument has increased by 2π. Thus, the value of the function remains unchanged. The function is therefore singlevalued with this cut.
48
Chapter 2. Functions of a Complex Variable
2
2
1.5
1.5
1
1
0.5
0.5
0
0
0.5
0.5
1
1
1.5
1.5
2 1
2 1 0.5 0 0.5 1
y
3
1
2
0
1
2
3
0.5 0 0.5 1
y
x
(a) Cut [−1, 1], Primary sheet.
3
2
1
0
2
1
3
x
(b) Cut [−1, 1], Secondary sheet.
2
2
1.5
1.5
1
1
0.5
0.5
0
0
0.5
0.5
1
1
1.5
1.5
2 1
2 1 0.5 0 0.5
y
1
3
1
2
0
1
2
3
0.5 0 0.5
y
x
(c) Cuts [−∞, −1] and [1, ∞], Primary sheet.
1
3
2
1
0
2
1
3
x
(d) Cuts [−∞, −1] and [1, ∞], Secondary sheet.
Figure 2.20. Imaginary parts of f (z) = (z − 1)1/2 (z + 1)1/2 . Parts (a), (b): the two solutions when the branch cut is placed along [−1, 1]; parts (c), (d): the two solutions when branch cuts are placed along [−∞, −1] and [1, ∞].
Im(z)
z θ2
IV θ1
I
Re(z) 1
−1
III
II Figure 2.21. Branch cut for f (z) =
p
(z − 1)(z + 1).
2.5.2 Riemann sheets If we don’t want to restrict the domain by a branch cut, we obtain a multivalued function with multiple Riemann sheets. Figure 2.18 illustrated this for the function f (z) = z 1/2 . Figure 2.22 shows similar illustrations for f (z) = z 1/3 , which is a function with three Riemann sheets. We note along the entire real axis one option that matches the realvalued
2.5. Multivalued functions—Branch cuts and Riemann sheets
49
Table 2.1. Changes in the angle Θ as one follows the path shown in Figure 2.21.
Step I II III IV Total
Changes in Θ θ1 gains 2π No change in the angle θ2 gains 2π No change in the angle 2 Θ = θ1 +θ has increased by 2π 2
√ case of y = 3 x (with imaginary part zero), but we see here all the three options for all complex values z. While one can think of Figures 2.22(a)–(b) as three solution sheets over a single complex “base plane,” an interesting alternative is to think of the base plane replaced by a triple layered complex plane, with layers connected as if one just “flattened down” either of the triple surfaces, to get three sheets that lie together without any vertical separations. From the perspective of this triply connected complex base plane, the function has become entirely singlevalued (and with no branch cuts needed). The function f (z) = (z − 1)1/2 (z + 1)1/2 just considered in Example 2.27 displayed also how the full set of function values can be viewed either as Riemann sheets or as decomposed into pieces with two different branch cut strategies. Figure 2.23 displays in an equivalent way f (z) = log z, with an infinity of Riemann sheets (which all coincide for the real part, reminiscent of Figure 2.7, but now with additional sheets displayed). As noted above, it will depend on the application whether it is best to consider the infinity of sheets, or to make a branch cut (for example along the negative real axis), and then only have a small part of the function available at a time.
Example 2.28. Illustrate the function f (z) = log
z+1 z−1
.
Apart from a trivial factor of 21 , we recognize in Figure 2.24(b) the function Re arctanh z from Figure 2.8(a). As in Example 2.27, both z = 1 and z = −1 are branch points. Like for the function f (z) = log z, the multivaluedness is not visible in the real part (only in the imaginary part and in the magnitude). We see a function with an infinity of Riemann sheets. However, if we choose a path that takes us around both branch points (and does not cross the real axis between −1 and +1), we return to the same function value. The point z = ∞ is, in this case, not a branch point.24 It is nevertheless entirely possibly to place branch cuts along [−∞, −1] and [1, +∞] (i.e., letting the cut between −1 and +1 along its way pass through infinity), giving a function that is analytic and singlevalued within that cut domain. For the imaginary part, that choice would correspond to using the upper halfplane (towards the lower left in Figure 2.24(c)) from one level and the lower halfplane from the level below. 24 It is visually clear from Figures 2.24(b)–(c) that, following a circular path very far out, will return to the same function value after going around once. Analytically, the same thing is most easily seen by writing f (z)
as f (z) = log regular point.
1 1+ z
1 1− z
. Values of z near infinity keep the argument of the log function near one, which for it is a
50
Chapter 2. Functions of a Complex Variable
(a) Re z 1/3 .
(b) Im z 1/3 .
(c) z 1/3  . Figure 2.22. Visualization of f (z) = z 1/3 . The coloring corresponds to the primary branch (phase angle θ in the range − π3 < θ ≤ π3 ).
2.6. Sequences of analytic functions
51
(a) Re log z.
(b) Im log z (showing only three Riemann sheets). Figure 2.23. Visualization of f (z) = log z.
2.6 Sequences of analytic functions We will in several contexts come across sequences of analytic functions and, if they are converging, their limits. This turns out to be yet another context in which the results for analytic functions turn out simpler than for, say, infinitely differentiable functions on the real axis. The next three examples illustrate some issues in the realvalued case. These are followed by Vitali’s theorem for the analytic function case. Consider realvalued functions. Let fn (x) be continuous on [a, b], and let the sequence of functions fn (x) converge at every point x as n → ∞. This does not imply that limn→∞ fn (x) = f (x) is continuous. Example 2.29. Discuss the convergence of the function sequence 0 , 0 ≤ x < 1 − n1 , fn (x) = n (x − 1) + 1 , 1 − n1 ≤ x ≤ 1, as n → ∞.
52
Chapter 2. Functions of a Complex Variable 3 2
f(x)
1 0 1 2 3 4
2
0
2
4
x
(a) f (x) = log
x+1 x−1
for x real.
3
2
1
0
1
2
3 4
2
0
2
x
6
4
2
0
2
44
y
(b) Re log
z+1 z−1
.
15
10
5
0
5
10
15 4
2
0
2
45
x
(c) Im log
z+1 z−1
4
3
1
2
0
1
2
3
4
5
y
(showing three Riemann sheets).
z+1 Figure 2.24. Visualization of f (z) = log z−1 . In order to see the imaginary part better, the viewpoint for parts (b) and (c) is located in the third quadrant.
2.6. Sequences of analytic functions
53
This sequence converges pointwise (at every xlocation on [0,1]) to zero for 0 ≤ x < 1 and to one for x = 1; see Figure 2.25(a). The limit function f (x) (in green) is obviously not continuous. The convergence is not uniform, since one can find an ε > 0 such that, no matter how large n is, one can then choose an x such that fn (x) − f (x) > ε. Example 2.30. Discuss the convergence of the function sequence fn (x) = 2
n X
(−1)k+1
k=1
sin kx k
as n → ∞. Each function fn (x) is 2πperiodic, converges pointwise to x for x ∈ (−π, π), and equals zero for x = ±π. Convergence is again not uniform, also for an additional reason (other than approximating a discontinuous function by a sequence of continuous ones). Due to the Gibbs’ phenomenon, there arises a narrow “overshoot” next to the jumps (occurring at x = ±π, ±3π, . . .), reaching in the limit approximately 9% of the jump height; see Figure 2.25(b).25 It can be useful to consider limits of functions that converge pointwise and then be able to tell something about the limit. For real functions, a small step in that direction is the concept of uniform convergence, which can be defined more formally as Given ε > 0, there exists an N (ε) such that fn (x) − f (x) < ε for all x ∈ [a, b] whenever n > N (). The two examples above were not uniformly convergent. With uniform convergence, the fact that the functions fn (x) are continuous implies that the limiting function f (x) is cond tinuous as well. Still this does not guarantee that the derivative(s) dx fn (x) will converge d toward dx f (x), the derivative of the limiting function. Example 2.31. The sequence fn (x) = sinnnx converges uniformly to zero. In fact, given d fn (x) = cos nx ε > 0, then fn (x) − f (x) < ε whenever n > N (ε) = 1ε . However, dx d does not converge towards dx 0 = 0. Back to complex variables. Life is much easier for analytic functions. Theorem 2.32 (Vitali’s convergence theorem). If fn (z) are analytic and bounded (fn (z) < M constant) in a domain D (open; containing a full neighborhood of some point), and if limn→∞ fn (z) = f (z) pointwise in D, then f (z) is also analytic there, and hence dk dk limn→∞ dz k fn (z) = dz k f (z) for all k = 1, 2, 3, . . .. Furthermore, it suffices if we know pointwise convergence only on a point set with a finite limit point in D. Note that Examples 2.30 and 2.31 do not contradict Vitali’s theorem, since the functions are bounded only along the real axis, and not in a full complex neighborhood of any point. 25 See
also Exercise 9.8.11.
54
Chapter 2. Functions of a Complex Variable 1.2
1
0.8
y
0.6
0.4 n=2
4
10
0.8
0.9
0.2
0
0.2 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1
x
(a) fn (x) =
0 n (x − 1) + 1
, ,
0 ≤ x < 1 − n1 1 − n1 ≤ x ≤ 1
4 n = 20 3
n=3
2
y
1
0
1
2
3
4
6
4
2
0
2
4
6
x
(b) fn (x) = 2
Pn
k+1 sin kx k=1 (−1) k
Figure 2.25. The functions fn (x) in Examples 2.29 and 2.30, respectively, shown for some different nvalues. Both cases feature pointwise convergence of a sequence of continuous functions. However, in both cases, the convergence is nonuniform, and the limit functions are discontinuous.
2.7 Functions defined by integrals Consider the function f (z) defined as follows: ˆ f (z) =
b
g(z, t)dt , a
assuming if nothing else is specified a straight line integration path from a to b. Let g(z, t) be an analytic function of z and a continuous function in t. We want to know whether f (z)
2.7. Functions defined by integrals
55
´b df is analytic. If a, b < ∞, then dz = f 0 (z) = a ∂g ∂z (z, t)dt must exist. Thus, f (z) has one derivative and is therefore analytic. If b = ∞, we can redefine f (z) as the limit of the ´n sequence fn (z) = a g(z, t)dt and then use the convergence of sequence results (Vitali’s convergence theorem). If the integral exists and is finite, then f (z) is again analytic. The process is identical if a = −∞. Let’s now examine two first examples of functions defined by integrals: • Consider the function
ˆ f (z) = 0
∞
e−t dt. 1 + zt
(2.17)
If z is a real negative number, the integral doesn’t converge as the denominator becomes 0 at t = −1/z. Else, it is an analytic function. • Another example is given by the Γfunction ˆ ∞ Γ(z) = e−t tz−1 dt.
(2.18)
0
Let’s analyze it piece by piece, by observing the potential difficulties of the integral: 1. The infinite interval can be a problem. However, since e−t makes the integrand decay very rapidly, the limit of the sequences of functions with the upper limit of integration b → ∞ exists and is finite. The sequence is thus always convergent. 2. At first glance, we note that there are values of z for which the integrand becomes divergent at the origin if the lower limit of integration a → 0. However, having a divergent integrand along the integration path doesn’t always mean ´1 that the integral will diverge (e.g., 0 t1p dt for p < 1). We want to show that the integral nevertheless converges if Re z > 0. As t & 0, the Taylor expansion of e−t becomes extremely accurate, and we can express e−t t z−1 = 1 z 1 z+1 t z−1 − 1! t +´ 2! t − +· · · .The first term is the most singular one at the origin; we get 0 tz−1 dt = z1 t z t=0 . Writing t z = ez log t , we see that t z for t & 0 is well defined only for Re z > 0 (then goes to zero; further terms in the expansion impose the weaker restrictions, Re z > −1, Re z > −2, etc.) By means of (2.18), Γ(z) is thus defined when Re z > 0 and it is analytic because of the observations above for a → 0 and for b → ∞. We have by this defined Γ(z) in the right halfplane. Integrating by parts reveals the interesting property ˆ ∞ ˆ ∞ ˆ ∞ −t z−1 z−1 −t −t z−1 ∞ Γ(z) = e t dt = − t d(e ) = −e t 0 + e−t tz−2 (z − 1)dt 0
= (z − 1)Γ(z − 1)
0
0
if Re z > 1.
Functional equations such as this one arise often for analytic equations. This one can alternatively be written as Γ(z + 1) = zΓ(z) if Re z > 0.
(2.19)
56
Chapter 2. Functions of a Complex Variable
´∞ Noticing that Γ(1) = 0 e−t dt = 1, and then Γ(2) = 1Q · 1, Γ(3) = 2 · 1, Γ(4) = 3 · 2 · 1, n etc., we have generalized the factorial function n! = k=1 k from positive integers to complex arguments, as z! = Γ(z + 1) if Re z > 0. The natural followup question to having generalized Γ(z) to an analytic function in the complex plane becomes the following: Is the boundary limiting the Γfunction to the righthand side of the complex plane (Re z > 0) an actual limitation of the function or just an artifact of the functional representation used in (2.18)? It will soon turn out that we can analytically continue Γ(z) to the left halfplane by several methods (all, by necessity, giving the same result).
2.8 Supplementary materials 2.8.1 Taylor series for the inverse of an analytic function Especially in Chapter 12, we will repeatedly come across the task of inverting a Taylor series, i.e., given the coefficients a0 , a1 , a2 , . . . in w = f (z) = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · obtain the coefficients b0 , b1 , b2 , . . . for the inverse function z = g(w) = b0 + b1 (w − a0 ) + b2 (w − a0 )2 + · · · (with both expansions centered at the matching locations z = z0 , w = a0 = f (z0 )). Before describing two approaches, let us note that the task is quite laborious if done by hand and that symbolic algebra packages handle it very effectively. Lagrange’s inversion formula
In the case of f 0 (z0 ) 6= 0 (i.e., a1 6= 0), Lagrange gave an explicit expression for the bk coefficients in terms of derivatives of f (z) (rather than more immediately from the ak coefficients):26 k ! 1 dk−1 z − z0 b0 = z0 , bk = lim , k = 1, 2, 3, . . . . k! z→z0 dz k−1 f (z) − f (z0 ) While of theoretical interest, the following approach is in most cases more practical. Inversion by equating coefficients
Without loss of generality, we can assume both z0 = 0 and a0 = 0. The two expansions are then w = f (z) = a1 z + a2 z 2 + · · · , 2
z = g(w) = b1 w + b2 w + · · · . 26 This
5.1.1.
(2.20) (2.21)
can be proved, for example, by combining (4.12), the result in Exercise 8.8.1, and Method 3 in Section
2.8. Supplementary materials
57
Case with a1 6= 0: Direct substitution of (2.21) into (2.20) gives w = a1 (b1 w + b2 w2 + · · · ) + a2 (b1 w + b2 w2 + · · · )2 + · · · and, after equating coefficients in w, 1 = a1 b1 , 0 = a2 b21 + a1 b2 , 0 = a3 b31 + 2a2 b1 b2 + a1 b3 , 0 = a4 b41 + 3a3 b21 b2 + a2 b22 + 2a2 b1 b3 + a1 b4 , .. .. . . The pattern that emerges here shows that the successive coefficients b1 , b2 , b3 , . . . become uniquely determined as long as a1 6= 0. Case with a1 = 0, a2 6= 0: Also this case arises frequently in Chapter 12.27 The expansion (2.20) of specified coefficients now lacks its leading term a1 z, and becomes w = f (z) = a2 z 2 + a3 z 3 + a4 z 4 + · · · , where we this time have assumed that a2 6= 0. We will in this case need to modify (2.21) to include halfinteger powers as well: z = g(w) = b1/2 w1/2 + b1 w1 + b3/2 w3/2 + b2 w2 + · · · . The same process of substituting the latter expansion into the former gives now 1 = a2 b21/2 , 0 = a3 b31/2 + 2a2 b1/2 b1 , 0 = a4 b41/2 + 3a3 b21/2 b1 + a2 b21 + 2a2 b1/2 b3/2 , 0 = a5 b51/2 + 4a4 b31/2 b1 + 3a3 b1/2 b21 + 3a3 b21/2 b3/2 + 2a2 b1 b3/2 + 2a2 b1/2 b2 , .. .
.. .
In the first equation, there is a choice of sign for b1/2 . Once that choice has been made, the subsequent coefficients b1 , b3/2 , b2 , . . . follow uniquely.
2.8.2 Exchanging the order of summation and integration In the following, we will on several occasions change the order of operators, in particular between infinite sums and integration, raising the question of the validity of ! ˆ X X ˆ ? fn (x) dx = fn (x)dx n
n
(assuming that each sum and integral converges). Simple examples show that some caution is needed when changing orders of operators. 27 The
approach in this case generalizes quite straightforwardly to when a1 = a2 = · · · = ap−1 = 0, ap 6= 0. We don’t give any details here, since applications then become quite rare.
58
Chapter 2. Functions of a Complex Variable
Regarding double sums, we can consider the infinite sized matrix A with nonzero elements only along its main diagonal and first subdiagonal: 1 −1 1 −1 1 A= . −1 1 .. .. . . P∞ P∞ Clearly, row sums n=1 am,nP and column sums m=1 am,n converge, but nevertheP∞bothP P ∞ ∞ ∞ less m=1 ( n=1 am,n ) = 1 and n=1 ( m=1 am,n ) = 0. Turning to double integrals (with infinitely differentiable integrands), the situation is no different. For example, ˆ ∞ ˆ ∞ ˆ ∞ ˆ ∞ x2 − y 2 π π x2 − y 2 dy dx = , while dx dy = − 2 + y 2 )2 2 + y 2 )2 (x 4 (x 4 1 1 1 1 (2.22) 2 −y 2 y ∂2 (as obtained, for example, by noting that (xx2 +y 2 )2 = ∂x∂y arctan x ). Exceptional cases (as both above) typically boil down to cancellations between infinite quantities.28 If applying magnitudes to all functions/terms involved still leads to finite results, the exchange is correct. As a consequence of Fubini’s theorem, the following will hold. Theorem 2.33. Consider an infinite sequence of functions fn (x). If either ! ˆ X X ˆ fn (x) dx < ∞ or fn (x)dx < ∞, n
n
then the exchange of order ˆ X n
! fn (x) dx =
X ˆ
fn (x)dx
n
is valid. The most severe restriction in the following theorem is that the integration path cannot be infinite.29 Theorem 2.34. In the case of (i) integration along an interval of finite length, and (ii) P f (x) being a uniformly n n sum of continuous functions, the exchange of order ´ P P ´ convergent ( n fn (x)) dx = n fn (x)dx is valid.30 28 If we divide the domain for (2.22) in two halves by the line y = x, then the double integrals over these halves both become infinite, and with opposite signs. 29 When verifying that restrictions in theorem formulations are necessary (rather than reflecting shortcomings in proofs), specific examples can be very useful. Following up on the idea in (2.22), let fn (x) = 2 P∞ n2 −x2 πx . By techniques from Chapter 5, we obtain 1 − sinh / 2x2 (unin=1 fn (x) = πx (n2 +x2 )2 ´ ∞ P∞ ´ 1 formly in x) and 1 fn (x) dx = 12 (1 + π − π coth π), whereas 1∞ fn (x)dx = − 1+n 2 and ´∞ n=1 P∞ 1 π n=1 1 fn (x)dx = 2 (1 − π coth π), a difference of 2 . 30 We will in Chapter 4 generalize integrals from intervals along the real axis to paths in the complex plane. This theorem remains valid when these paths are piecewise smooth and of finite length.
2.8. Supplementary materials
59
Neither of the two theorems requires any analyticity properties of fn (x), and proofs for them can be found in calculus texts. In the following, when exchanges of order can be justified by either of these theorems, we make no explicit comments about this.
2.8.3 Convergence of infinite series The following theorems provide techniques to test whether infinite series converge or not. We refer the reader to classical calculus textbooks for details and proofs. P∞ Theorem 2.35. The root test: If a power series takes the form f (z) = n=0 an z n , then the radius of convergence R = 1/ limn→∞ an 1/n , where lim denotes the least upper bound (also known as limit superior, or upper limit) of the sequence and is defined below. This formula for the radius of convergence of a power series is often referred to as the Cauchy–Hadamard theorem. As defined just above, lim denotes the least upper bound. The following are three equivalent ways to define it. A = limn→∞ Sn of a sequence {Sn }∞ n=1 if (i) the sequence is bounded from above and has A as its largest limit point, (ii) A = limN →∞ supn>N Sn , and (iii) A is the smallest number such that for ε arbitrarily small, Sn > A + ε only a finite number of times. P∞ Theorem 2.36. The ratio test: The series n=1 an converges absolutely if L < 1, and  . If the series is a power series of the form diverges if L > 1, where L = lim n→∞ aan+1 n P∞ n n f (z) = n=0 an z , it will converge in case that lim n→∞ aan+1  exists. In that case, the radius of convergence R is equal to this limit. Theorem 2.37. The limit comparison test: If the terms of the sequences {ak } and {bk } are real and positive, and if limn→∞ abnn = c with 0 < c < ∞, then either both series P∞ P∞ n=1 an and n=1 bn converge or both diverge. Theorem 2.38. The integral ´ ∞ {ak } are ak = f (k), with f P∞ test: If the terms of the sequence continuous, the series n=N an converges if and only if N f (x)dx < ∞, for N > 0. Theorem 2.39. Weierstrass Mtest: Let bj (z) ≤ Mj , an P upper bound for eachPanalytic function bj (z) for j = 0, 1, 2, 3, . . . in some region R. If j Mj converges, then j bj (z) converges uniformly in R.
2.8.4 Codes for visualization of analytic functions Most illustrations in this book were generated using either MATLAB31 or Mathematica.32 A sample code using each of the systems is shown below. MATLAB
The following MATLAB script together with function display_function produce the three subplots of Figure 2.4. 31 The
MathWorks, Natick, MA. Research, Champaign, IL.
32 Wolfram
60
Chapter 2. Functions of a Complex Variable
% Test code for displaying an analytic function ; here exp(z) clear ; close all; bx = [ 4 ,4 , 10 ,10]; % Domain (box) surrounding origin to be displayed nx = 41; % Number of nodes along real axis for re & im plots ny = 41; % Number of nodes along imag axis for re & im plots bounds = [ 60 ,60; 60 ,60; 0 ,60]; % Lower and upper in the three displays f = @(z) exp(z); % Function to display display_function (f,bx ,nx ,ny , bounds ); % Create the three subplots function display_function (f,bx ,nx ,ny , bounds ) % The input parameters described in the script that calls this routine lw = 4;% Set LineWidth for highlighting values along real axis x = linspace (bx(1),bx(2),nx); y = linspace (bx (3) , bx (4) , ny ); [xr ,xi] = meshgrid (x,y); z = complex (xr ,xi); figure (1) % Plot the real part using mesh mesh(xr ,xi ,real(f(z))); colormap ([0 0 0]); hold on; xlabel (’\itx ’); ylabel (’\ity ’); title(’Real part ’) xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (1 ,:)); plot3 (x, zeros (size(x)), real(f(x)),’r’,’LineWidth ’,lw ); %
Highlight real axis
figure (2) % Plot the imaginary part using mesh mesh(xr ,xi ,imag(f(z))); colormap ([0 0 0]); hold on; xlabel (’\itx ’); ylabel (’\ity ’); title(’ Imaginary part ’) xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (2 ,:)); plot3 (x, zeros (size(x)), imag(f(x)),’r’,’LineWidth ’,lw ); %
Highlight real axis
figure (3) % Plot the magnitude as a surface % Increase resolution for smoother display x = linspace (bx(1),bx(2),nx *4); y = linspace (bx (3) , bx (4) , ny *4); [xr ,xi] = meshgrid (x,y); z = complex (xr ,xi); p = surf(xr ,xi ,abs(f(z)), angle(f(z))); % Display the surface set (p,’EdgeColor ’,’none ’); colormap hsv (600); hold on; xlabel (’\itx ’); ylabel (’\ity ’); title(’Magnitude , with phase plot ’) xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (3 ,:)); plot3 (x, zeros (size(x)), abs(f(x)),’k’,’LineWidth ’,lw ); % Highlight real axis axes(’Position ’ ,[0.05 0.05 .17 .17]) % Add color wheel for phase information [th ,r] = meshgrid ( linspace (pi ,pi), linspace (0 ,1)); [X,Y] = pol2cart (th+pi ,r); contourf (X,Y,th ,100,’ linestyle ’,’none ’); hold on % Show wheel colors plot ([ 1 1] ,[0,0],’k ’); plot ([0 0],[1,1],’k ’); % Show Re and Im axes plot(cos (0:0.01:2* pi),sin (0:0.01:2* pi),’k ’); colormap hsv (600); axis equal ; axis off
Mathematica
If one wants to quickly get surface plots of a complex function with Mathematica, just a few line statements suffice for each plot. The first line below specifies the exponential function f (z) = ez , and the following lines produce plots of its real and imaginary parts, and of its magnitude (with phase angle coloring). f[z_] = E^z; (* bounding box *) {x1 , x2 , y1 , y2 , z1 , z2} = {4, 4, 10, 10, 60, 60}; (* Curves on the Real Axis *) {re , im , abs} = ReleaseHold [ Hold[ ParametricPlot3D [{x, 0, a[f[x]]} , {x, x1 , x2}, PlotStyle > Directive [ Thickness [0.01] , b]]] /. {{a > Re , b > Red}, {a > Im , b > Red}, {a > Abs , b > Black }}]; (* Real Part *) Show[ Plot3D [Re[f[x + I*y]], {x, x1 , x2}, {y, y1 , y2}, PlotRange > {z1 , z2}, AxesLabel > {" Re(z)", "Im(z)"} , PlotLabel > Re[f[z]], ViewPoint > {1.3, 2.4, 2}] , re] (* Imaginary Part *) Show[ Plot3D [Im[f[x + I*y]], {x, x1 , x2}, {y, y1 , y2},
2.8. Supplementary materials
61
PlotRange > {z1 , z2}, AxesLabel > {" Re(z)", "Im(z)"} , PlotLabel > Im[f[z]], ViewPoint > {1.3 , 2.4, 2}] , im] (* Magnitude and Argument *) Show[ ComplexPlot3D [f[z], {z, x1 + I y1 , x2 + I y2}, PlotRange > {0, z2}, Mesh > False , PlotPoints > 100, ViewPoint > { 1.3 , 2.4, 2}, Epilog > Inset[ ComplexPlot [z, {z, 1  1 I, 1 + 1 I}, RegionFunction > Function [{z}, Abs[z] < 1], Axes > True , Ticks >None , BoundaryStyle > Directive [Black , Thin], Mesh > None , Frame > None , ImageSize > 50, PlotPoints > 100, ColorFunction > None], {0.05 , 0.1}] , AxesLabel > {" Re(z)", "Im(z)"}, PlotLabel > Abs[f[z]]] , abs]
2.8.5 Select proofs Theorem 2.40. If the CR equations (2.1) hold and the partial derivatives ux , uy , vx , vy are continuous functions of x and y, then f (z) is differentiable.33 Proof. Let f (z) = u(x, y) + iv(x, y) and introduce the infinitesimal quantities dz = dy 1 = dx−i dx + i dy ⇒ dz dz2 . Then (dx − i dy) f (z + dz) − f (z) = [{u(x + dx, y + dy) − u(x, y)} dz dz2 + i {v(x + dx, y + dy) − v(x, y)}] (dx − i dy) = [{ux dx + uy dy} + i {vx dx + vy dy}] , dz2 where we, in the last line, replaced u(x + dx, y + dy) (and likewise for v) with the leading terms of their 2D Taylor expansions about (x, y). Expanding the product, some terms drop out due to the CR equations, and we are left with f (z + dz) − f (z) 1 = (dx)2 (ux + ivx ) + (dy)2 (vy − iuy ) , dz dz2 which, after using the CR equations again, further simplifies to ux + ivx , thus showing that f 0 (z) exists.
(dx)2 +(dy)2 (ux dz2
+ ivx ) =
2.8.6 Some apparent paradoxes due to multivaluedness Certain areas of mathematics are quite rich in seemingly counterintuitive results (such as set theory and statistics). In the context of analytic functions, the common theme in such examples is usually having neglected to properly account for multivaluedness. p √ Example 2.41. 1 = 1 = (−1)2 = −1. In this simple case, we clearly neglected the multivaluedness of the square root in the last equality. Remembering the convention that we, by the square root of a positive number, imply the positive choice would have eliminated the error here. 33 From
the continuity of the partial derivatives, it is understood that u, v, and thus f exist in a neighborhood of z = x + iy.
62
Chapter 2. Functions of a Complex Variable
Example 2.42. Let z = −1 + i. From the logarithm definition (2.9) for −π 0, under typical conventions. However, there is no similar convention for the case when a is either negative or complex.
2.9 Exercises Exercise 2.9.1. Write as a + ib the following complex quantities. If there are several solutions, give them all: i (a) e(e ) e2+2πi (b) e1+πi/3 (c) log 1 (d) log e (e) cos i 1/3 1 (f) 3 (g) i1/3 √ (h) (1 + 3 i)i 1/2 i (i) 1+i Exercise 2.9.2. Verify that the contour lines for magnitude equal to one in Figure 2.2 are perfect circles centered at ±i and passing through ±1. Exercise 2.9.3. Derive (2.10) and (2.11) for (a) arctanh z and (b) arcsin z, respectively. Exercise 2.9.4. Write in the form f (z) = u(x, y) + i v(x, y) the following function, and verify in each case the CR equations:
2.9. Exercises
63
(a) z 3 1 z+i (c) cos z (b)
Exercise 2.9.5. Consider the function f (x, y) = 1 + x − 3(x2 − y 2 ) + i y(1 − 6x). a. Verify by the CR equations that f (x, y) is analytic. 1 (z − z) to (i) again verify analyticity, and b. Use the substitutions x = 12 (z + z), y = 2i (ii) obtain a simple expression for f (z). Exercise 2.9.6. Are the following functions analytic? Check using the CR equations. If they are, also write them as functions of z: (a) f (x, y) = y 3 − 3x2 y + i(x3 − 3xy 2 + 2) (b) f (x, y) = ex−iy Exercise 2.9.7. An arbitrary triangle can be scaled and rotated to have two corners at 0 + 0i, 1 + 0i and the third at z = x + iy. Notable points for a triangle include N (z) intersection of the normals from a corner to the opposite side, M (z) intersection of the medians, going from a corner to the midpoint of the opposite side, and C(z) center of the circumscribed circle (passing through all the corners) (see Figure 2.26). Straightforward geometry will show that x(1 − x) 1 y 1 i x(1 − x) N (z) = x + i , M (z) = (x + 1) + i , C(z) = + y− : y 3 3 2 2 y (a) Determine which (if any) of the three functions N (z), M (z), C(z) are analytic functions of z. N (z)−C(z) (b) Verify that M (z)−C(z) = 3, independent of the value of z. Comment: The result in (b) tells that, for an arbitrary triangle, the points N , M , C always lie on a straight line, with M located 2/3 of the way from N to C. This result is due to Leonhard Euler; the line is known as the “Euler line” of the triangle.34 Exercise 2.9.8. Let f (x, y) = u(x, y) + iv(x, y) be an analytic function, where u(x, y) is given below. Find v(x, y), the harmonic conjugate of u(x, y), and express the function in terms of z: y (a) u(x, y) = x2 +y 2 (b) u(x, y) = cos x cosh y Exercise 2.9.9. Derive the expressions for the CR equations in terms of polar coordinates. Exercise 2.9.10. Use Euler’s formulas to obtain easy proofs for the trigonometric identities cos 3θ = 4 cos3 θ − 3 cos θ . sin 3θ = 3 sin θ − 4 sin3 θ 34 The Euler line features in several contexts, for example, the points where the lines from one corner intersect the opposite side in Figures 2.26(a)–(b) and the midpoints of the lines from the corners to N (z) (nine points in all) lie on a single circle, with center exactly halfway between N (z) and C(z).
64
Chapter 2. Functions of a Complex Variable
0.6
z = x + iy
0.5
0.4
0.4
0.3
0.3
0.2 0.1
z = x + iy
0.6
0.5
0.2
N(z) 0.2
M(z)
0.1
0.4
0.6
0.8
1
0.2
0.1
0.1
0.2
0.2
0.3
0.3
(a) Intersection of the normals
0.4
0.6
0.8
1
(b) Intersection of the medians
z = x + iy
0.6 0.5 0.4 0.3
C(z)
0.2 0.1
0.2
0.4
0.6
0.8
1
0.1 0.2 0.3
(c) Center of circumscribed circle Figure 2.26. The three points N , M , C for a triangle, shown with red markers, and described in Exercise 2.9.7.
Exercise 2.9.11. Show that
Pn
n X k=1
k=0
cos kθ =
sin kθ =
sin
sin( 12 (n+1)θ ) cos sin 1 2 (n
nθ 2
θ 2
+ 1)θ sin nθ 2 sin θ2
and .
Exercise 2.9.12. With z = x + iy, show that  sinh x ≤  cosh z ≤  cosh x. Exercise 2.9.13. Using a calculator, one quickly sees that π π ≈ 36.4622 and ee ≈ 15.1543. Can you spot what similarly evaluates approximately to 0.20788? Explain! Exercise 2.9.14. Calculate all values of arctan z when (a) z = 0, (b) z = 2i, (c) z = 1 + i. Exercise 2.9.15. Show that all solutions to sin z = k integer.
5 3
are given by zk = (2k+ 12 )π±i log 3,
Exercise 2.9.16. Use, for example, the MATLAB code given in Section 2.8.4 (or some other software) to plot the real part, imaginary part, magnitude, and phase portrait of the following functions:
2.9. Exercises
65
(a) f (z) = z 6 (b) f (z) = tan z (c) f (z) = arctan z Exercise 2.9.17. Derive the following formulas: (a) arcsin z = −i log(iz + (1 − z 2 )1/2 ) d arcsin z = (1−z12 )1/2 (b) dz Exercise 2.9.18. Derive Cote’s relation i φ = log(cos φ + i sin φ) by evaluating p d log 1 + z2 + z , dz and then substitute z = i sin φ and integrate with respect to φ. ExerciseP 2.9.19. Find the radius of convergence of the following series: ∞ (a) zk Pk=1 ∞ zk (b) k=1 (k+1)! P∞ k! k (c) k=1 kk z 1+z (d) The Taylor series of 2−z about z0 = i Exercise 2.9.20. Find the Taylor series expansions about z = 0 of the following functions, valid in the given regions: 1 (a) 1−z 2 , z < 1 (b) cosh z, z < ∞ 2
(c)
ez −1−z 2 , z3
0 < z < ∞
Exercise 2.9.21. Find the leading coefficients c0 , c1 , . . . , c4 in the function y(x) =
∞ X
ck x k
k=0
that solves the ODE
dy(x) dx
= x + sin y(x), y(0) = 0.
Exercise 2.9.22. If a series is not of Taylor form, it may converge to different functions in different regions. Show that ∞ X z k−1 1/(1 − z)2 if z < 1 = . k k+1 1/ z(1 − z)2 if z > 1 (1 − z )(1 − z ) k=1
Hint: Show first that
PN
k=1
z k−1 (1−z k )(1−z k+1 )
=
1 z(1−z)
1 1−z
−
1 1−z N +1
.
Exercise 2.9.23. ThePBernoulli numbers {Bk }∞ the Euler numbers {Ek }∞ k=0 k=0 are Pand ∞ Bk k ∞ Ek k z 1 defined by ez −1 = k=0 k! z and cosh(z) = k=0 k! z , respectively, named after Jacob Bernoulli (1654–1705) and Leonhard Euler (1707–1783). (a) Find the first three nonzero Bernoulli numbers and the first three nonzero Euler numbers. (b) Find the radius of convergence of both series.
66
Chapter 2. Functions of a Complex Variable
Exercise 2.9.24. Give explicit formulas for u(x, y) and v(x, y) in the case of f (z) = Simplify the result as far as you can.
√ z.
√ √ Exercise 2.9.25. In the expression 3 2 + 2i + 3 2 − 2i, each cube root can take three different values. The nine possible values for the sum turn out to be −2, √ 1 ± √3, 1 ± i 3, √ √ (four combinations). − 12 (1 ± 3)(1 ± i 3) √ Verify the three realvalued cases (i.e., −2√and 1 ± 3). Hints: (i) Note that the three options for 3 2 − 2i are the complex conjugates of the three √ 3 π = π3 − π4 . options for 2 + 2i, and (ii) 12 Exercise 2.9.26. Discuss the singularities of the following functions: (a) f (z) = coszz−1 4 (b) f (z) = z 3 e1/(z−1) (c) f (z) = ez1−1 P∞ Exercise 2.9.27. to Example 2.5, show that k=1 k1 cos kx = − log 2 sin x2 . P∞As 1a followup Hint: Since k=1 k eikx diverges for all x if differentiated termbyterm, consider temP∞ rk ikx d porarily dx with 0 < r < 1 and, at some later stage of the algebra, let r k=1 k e increase to 1. Exercise 2.9.28. (a) Show that (z − eiθ )(z − e−iθ ) = z 2 − 2z cos θ + 1. (b) Write down the (2n)th roots of −1 in the form eiθ , and deduce z
2n
+1=
n Y
2
z − 2z cos
k=1
(2k − 1)π 2n
+1 .
(2.23)
(c) Substitute z = i into (2.23). Show that for n even 3π (2n − 1)π 1π · cos · · · · · cos = (−1)n/2 21−n cos 2n 2n 2n and for n odd cos2
1π 2n
· cos2
3π 2n
· · · · · cos2
(n − 2)π 2n
= n · 21−n .
Exercise 2.9.29. Consider Exercise 1.6.15. Solve it again, noting that α(z), β(z), γ(z) will all be linear functions of z, so f (z) = α2 + β 2 + γ 2 − αβ − αγ − βγ will be a quadratic polynomial. If this evaluates to zero for three different zvalues, it must be identically zero. Exercise 2.9.30. Prove Ptolemy’s theorem: With a quadrilateral inscribed in a circle, as shown in Figure 2.27, it holds that a · c + b · d = α · β. Note: If the quadrilateral is a rectangle, the theorem reduces to Pythagoras’ theorem.
2.9. Exercises
67
b
α
c β
a
d
Figure 2.27. Illustration of Ptolemy’s theorem a · c + b · d = α · β for a quadrilateral inscribed in a circle.
Hint: Show first the relations 1 eiβ − eiα  = 2 sin (β − α) if 0 < α < β < 2π, 2 and then eiβ − eiα  · eiγ − eiδ  + eiβ − eiγ  · eiα − eiδ  = eiα − eiγ  · eiβ − eiδ  if 0 < α < β < γ < δ < 2π. Exercise 2.9.31. Figure 2.28(a) shows the stereographic projection of the phase portrait of f (z) = z. Plots (b) and (c) show, respectively, the southern and the northern hemispheres viewed from inside the sphere. Explain why the only difference between plots (b) and (c) is the direction of the color scheme. Exercise 2.9.32. The rows of pictures in Figure 2.29 are labeled (a) to (f). The real axis is drawn in bold. These are illustrations of the functions f (z) that are listed below. Match each row with its corresponding function and explain your choice: (i) cos(z) √ (ii) z3 + 1 (iii) ez (iv) ez z 6 (v) z21−1 (vi) 103 z + z 6 Exercise 2.9.33. Consider f (z) = log z 2 − 1 : (a) Find the types and locations of all the singularities of f (z). (b) Test whether f (z) has a singularity at z = ∞. (c) Make this function singlevalued by introducing a branch cut. Carry on the procedure introduced in Table 2.1 to verify the singlevalued nature of the function.
68
Chapter 2. Functions of a Complex Variable
(a) Stereographic projection of the phase portrait of f (z).
(b) Southern hemisphere.
(c) Northern hemisphere.
Figure 2.28. Graphical representations of f (z) = z, in (b) and (c) as seen from the sphere center.
Exercise 2.9.34. Discuss the options for placing branch cuts to make the following functions singlevalued (a, b, c are finite and distinct complex constants). (a) f (z) = log z−a z−b (b) f (z) = log
(z−a)(z−c) z−b
1 2 1 3 Exercise 2.9.35. Consider w = f (z) = ez − 1 = z + 2! z + 3! z + · · · . Obtain the first 2 3 three terms in the inverse expansion z = b1 w + b2 w + b3 w + · · · by means of (a) Lagrange’s inversion formula, and (b) the method of undetermined coefficients. Verify your answers against the direct expansion of z = log(1 + w).
2.9. Exercises
Stereographic projection
69
Southern hemisphere
Northern hemisphere
(a)
(b)
(c)
(d)
(e)
(f)
Figure 2.29. Stereographic projection of the phase portraits for Exercise 2.9.32. The hemispheres are shown as seen from the sphere center.
Chapter 3
Analytic Continuation
A particular representation of a function might be valid only in a limited region of the complex plane, although the function itself may exist well past that region. For example, a function defined by its Taylor series can only be evaluated (by immediate summation) inside this expansion’s circular domain of convergence, which is limited in size by the distance to the nearest singularity. The topic of analytic continuation is about changing the way a function is formulated so that it can be evaluated also in a larger domain. All extensions of this type give identical results (although their functional forms and domains of validity may differ). This chapter gives examples of a number of continuation methods. This topic provides yet another example of mathematics becoming greatly simplified when functions are analytic. There is no counterpart for nonanalytic functions (no matter how many times these are differentiable).
3.1 Introductory examples Example 3.1. Consider the following three functions: P∞ 1. f1 (z) = k=0 z k = 1 + z + z 2 + z 3 + · · · . The radius of convergence is R = 1. This follows from either of Theorems 2.7, 2.35, or 2.36. ´∞ 2. f2 (z) = 0 e−t(1−z) dt. If the integral had been over a finite interval rather than over [0, ∞], it would have defined an entire function. With the upper limit ∞, it will converge for Re z < 1 but diverge for Re z ≥ 1. 1 3. f3 (z) = 1−z . This expression represents an analytic function in the whole complex plane, with the exception of a single first order pole located z = 1.
The key point in this example is that all three functions above are identically the same. The boundaries for f1 (z) and f2 (z) are not natural ones, but just artifacts of the particular way we have used for representing the function f3 (z). Figure 3.1 illustrates this function over the domains corresponding to the three different representations. The goal of analytic continuation is to reformulate functional expressions so they can be used in larger regions than was originally the case. It can happen that continuation is not possible. For instance, in the following example, the unit circle forms a natural boundary, and it is not possible to continue the function past it. 71
72
Chapter 3. Analytic Continuation
(a) f1 (z)
(b) f2 (z)
(c) f3 (z)
Figure 3.1. Illustration of Example 3.1. The regions of validity for the three functional representations f1 (z), f2 (z), and f3 (z).
3.2. Some methods for analytic continuation
Example 3.2. Consider the function f (z) =
73
P∞
k=1
z k! = z + z 2 + z 6 + z 24 + · · · .
Theorem 2.35 again gives R = 1, and the domain of convergence will be the same as for the function f1 (z) in Example 3.1. Consider a point z = e2πik/m where k and m are arbitrary integers. Then z n! = e2πin!k/m = 1 for all n ≥ m. This means that f (z) → +∞ whenever z is of the form z = re2πik/m and r → 1. The function f (z) must therefore have singularities located densely all the way around the unit circle, making z = 1 a natural boundary. It cannot be continued past it. What does a function like this look like graphically? Figures 3.2(a)–(b) show its magnitude and phase angles when displayed within circles of radii 0.99 and 0.999999, respectively. There is of course no difference between the two displays throughout the inner region, but we see more fine structure entering near the edge in the second case. Moving even closer to the unit circle, the fine structure becomes infinitely dense. We can also note that f (z) does not increase monotonically as the unit circle is approached. In the phase angle color patterns, we can see telltale signs of increasing numbers of zeros. Near the origin, this function f (z) becomes of course very close to f (z) = z, as displayed in the top right subplot of Figure 2.1. P∞ As a followup to this example, we can consider g(z) = k=1 z k! /k!. Since g 0 (z) = f (z)/z, this function g(z) will also have z = 1 as a natural boundary. Nevertheless, counterparts to Figures 3.2(a)–(b) would seem visually to be perfectly smooth also along the unit circle. Whenever a function is given in a form which only permits it to be evaluated in some limited region of the complex plane, it becomes of great interest to see whether it can be rewritten in some other form that permits the domain to be extended, preferably to the whole complex plane. There is no similar opportunity for functions of a real variable, even if these are known to be infinitely differentiable, as can be seen from considering functions such as −1/x e , x>0 f (x) = . 0 , x≤0 It is quite common in this context that the initial representation is a Taylor series. The remainder of this chapter describes a number of ways to carry out analytic continuation. This chapter is placed in its present position in the book, since it requires no more preliminaries than what are already available. However, it can also be treated as an appendix to the book, to be used as a reference resource when needed.
3.2 Some methods for analytic continuation Each of the following nine subsections describes a different method for analytic continuation: 1. Circlechain method 2. Schwarz reflection principle 3. Use of a functional equation 4. Partitioning of an integration interval 5. Replace Taylor coefficients by integrals or sums
74
Chapter 3. Analytic Continuation
(a) Display of f (z) over z ≤ 1 − 10−2 .
(b) Display of f (z) over z ≤ 1 − 10−6 . Figure 3.2. Illustration of f (z) (with phase coloring) for f (z) = slightly different disks around the origin. Illustration of a natural boundary.
P∞
k=1
z k! over two
6. Subtraction of a similar series or integral 7. Borel summation 8. Ramanujan’s formula 9. Padé approximations The Padé approach differs from the other ones in that it only rarely can be used for mathematically exact analytic continuation. We have nevertheless included it here, since many applications can utilize its ability to provide quite accurate continuations even from highly incomplete information (such as from a relatively low number of leading terms in truncated
3.2. Some methods for analytic continuation
75
y
x
Figure 3.3. Schematic illustration of a circlechain continuation.
Taylor expansions). Still further examples of continuation methods (which require the use of complex integration) are given in Section 5.3.
3.2.1 Circlechain method This approach is rarely practical for actual applications, but it is of interest in that it gives insight into what analytic continuation amounts to. Suppose that we have a function with pointtype singularities (poles, branch points, or essential singularities) in the (a priori unknown) locations marked by red crosses in Figure 3.3. Assume that we are given initially a Taylor expansion centered at the origin of the complex plane, and the task is to follow how the function changes along the green (curved) path in the figure. The Taylor expansion will converge only in the domain bounded by the solid blue circle, extending out to the nearest singularity. Within this circle, the expansion completely describes the function. Hence, we can choose any other point inside it, on the continuation path, and create a new expansion centered at this new location. Its domain of convergence will likely reach outside the original one. The process can in theory be repeated indefinitely, allowing us to proceed, for example, along the indicated chain of circles. From a practical (as opposed to a theoretical) perspective, this approach is seldom usable. Even when the complete (infinite) set of Taylor coefficients is known analytically, it is rare that one can find convenient closed forms also for the subsequent expansions (however, one such example is given in Example 3.3 and another one in Exercise 3.3.3). If the initial Taylor coefficients can only be found numerically, this continuation process becomes extremely ill conditioned. This circlechain method (as well as the mathematical concept of analytical continuation) requires the initial Taylor expansion’s radius of convergence R to be nonzero. Nevertheless, some of the approaches we describe later can be applied with apparent success
76
Chapter 3. Analytic Continuation
even in some cases of power series expansions with R = 0, as seen in Example 3.18 and Exercises 3.3.5(b)–(c). An interesting situation arises if one tries to continue all the way around a singularity. Will the results agree when the circlechain continuation comes back to the start point? The answer will be yes when going around poles and essential singularities, but no in the case of a branch point. P∞ n Example 3.3. Continue the function f (z) = n=1 zn by means of a chain of circles once in the clockwise direction around the singularity at z = 1. We can start by noting (but will not use further) the fact that f (z) = − log(1 − z), so the end result after having continued stepbystep around the singularity should return the same function as we started with, apart from an increase in its imaginary part by 2πi. The original expansion is centered at z0 = 0. Let the subsequent expansions be centered at z1 , z2 , . . . , zp = z0 , where zk = 1 − e−2iπk/p , k = 1, 2, . . . , p (see Figure 3.4). From the Taylor expansion of f (z) it follows that f 0 (z) =
∞ X
z n−1 =
n=1
and therefore f (m) (z) =
(m−1)! (1−z)m .
1 1−z
(3.1)
The Taylor expansion centered at z1 therefore becomes
n ∞ ∞ X X 1 (n) 1 z − z1 z − z1 f (z1 )(z − z1 )n = f (z1 ) + = f (z1 ) + f . n! n 1 − z1 1 − z1 n=0 n=1 (3.2) Differentiating (3.2), we obtain from its first and third expressions n−1 ∞ 1 X z − z1 1 0 , f (z) = = 1 − z1 n=1 1 − z1 1−z f (z) =
confirming that (3.1) again holds when now expanding around z = z1 (and its different domain of convergence). Looking at the first and last expressions in (3.2), a pattern becomes clear. For the next expansion, around the point z2 , we get similarly z2 − z1 z − z2 f (z) = f (z1 ) + f +f , 1 − z1 1 − z2 etc. After p steps, we arrive at f (z) = f (z1 ) + f {z}  new f (z)
z2 − z1 1 − z1
+f
z3 − z2 1 − z2 {z
+ ··· + f
additive constant
zp − zp−1 1 − zp−1
+ f } 
z − zp . 1 − zp {z }
old f (z) since zp = 0
Are we back to the function value we started with? Noting that zk = 1 − e−2iπk/p , all the p terms in the “additive constant” become equal to f (z1 ) with z1 = 1 − e−2iπ/p . Furthermore, the value of this additive constant must be a fixed number, independent of p. To determine its value, we can let p → ∞. This gives f (z ) z1 z2 1 p f (z1 ) = p 1 − e−2πi/p = p 1 − e−2πi/p 1+ + 1 + ··· , 2 3 {z } z1  {z }  {z } z1
→2πi as p→∞
→1 as p→∞
3.2. Some methods for analytic continuation
77
y z2 z1 z0 =zp 1
x
(a) The centers of each consecutive Taylor expansions are shown as blue dots.
(b) The imaginary part of each expansion, drawn within its own domain of convergence. Figure 3.4. Illustrations of the sequence of Taylor expansion centers in Example 3.3.
which shows that it must evaluate to 2πi. Continuation with the circlechain method all the way around the singularity at z = 1 thus did not come back to the same expansion as we started with, revealing that the singularity was a branch point. Every time we continue in the clockwise direction around z = 1, this function f (z) increases in value by 2πi.
3.2.2 Schwarz reflection principle Suppose that we are given a function f (z) that is 1. known to be analytic in some region D (see Figure 3.5), and 2. real on the real axis (or on some part of it). We showed in Section 2.1.2, last bullet item, that the function g(z) = f (¯ z ) will then also be analytic. However, this g(z) is not defined on D but on D (the reflection of D in the
78
Chapter 3. Analytic Continuation
Im z
D
Re z ¯ D
Figure 3.5. Schematic illustration of the Schwarz reflection principle.
real axis). Since g(z) agrees with f (z) along a section of the real axis, the two functions must be the same (Theorem 2.12). The function f (z) has thus been continued from D to ¯ Another way to formulate the result is as follows. D. Theorem 3.4. An analytic function f (z) that is real on any segment of the real axis takes complex conjugated values at complex conjugated locations in the zplane. The result can be generalized significantly. For any analytic function, there will be some curve(s) along which Im f (z) = 0. If we can find some analytic change of variable z = z(w) such that the segment corresponds to real values of w, we can apply the reflection principle to f (z(w)).
3.2.3 Use of a functional equation We will illustrate this approach with three examples. The first two extend ˆ ∞ Γ(z) = e−t tz−1 dt, 0
convergent for Re z > 0, to the whole complex plane. Example 3.5. Continue the gamma function by utilizing the functional equation Γ(z + 1) = zΓ(z). We noted in Section 2.7 that the functional equation follows from integration by parts in the defining integral and is valid (like the integral for Γ(z)) throughout Re z > 0. If we
3.2. Some methods for analytic continuation
(a) Γ(z) =
´∞ 0
79
e−t tz−1 dt converges for Re z > 0.
(b) Using the functional equation Γ(z + 1) = zΓ(z) allows (after one step) the computation of Γ(z) for Re z > −1. Figure 3.6. Analytic continuation of Γ(z) using the functional equation given in Example 3.5.
introduce H(z) = Γ(z+1) , it will thus hold that H(z) = Γ (z) when Re z > 0. Recalling z Theorem 2.12, the functions H(z) and Γ (z) are identical. However, with Γ(z) defined for Re z > 0, H(z) = Γ(z+1) becomes defined for Re z > −1, and thus so now is Γ(z). z This has continued Γ(z) through the strip −1 < Re z ≤ 0 (and we can note that Γ(z) will have a pole at z = 0); see Figure 3.6. This process can be repeated indefinitely, showing that Γ(z) exists uniquely, is defined throughout the entire complex plane, and has poles at z = 0, −1, −2, . . . as its only singularities. Example 3.6. Continue the gamma function by utilizing the functional equation Γ(z)Γ(1 − z) =
π . sin πz
(3.3)
80
Chapter 3. Analytic Continuation
Different proofs for this equation will be given later (Theorem 5.15 and in the comments following Theorem 6.2; see also Example 3.14). This relation gives the value of the gamma function at the location z whenever it is available at a location 1 − z. Whenever Re z < 0 (i.e., we are outside where the integral definition holds), Re (1 − z) ≥ 1, so Γ(1 − z) can be evaluated. In this case, Γ(z) has in a single step been continued over the whole complex plane (omitting its poles). The next example requires a convergence theorem for Dirichlet series. P∞ Theorem 3.7. The regions of convergence and divergence for a Dirichlet series n=1 annz are halfplanes Re z > α and Re z < α, respectively, with α the least value that makes the right halfplane singularityfree. Example 3.8. Continue the Riemann zeta function ζ(z) =
∞ X 1 z n n=1
(3.4)
with use of the functional equation ζ(z) = 2z π z−1 sin
1 πz Γ(1 − z)ζ(1 − z) . 2
(3.5)
The functional equation (cf. Theorem 6.6) will immediately give the continuation everywhere that Re z ≤ 12 if values are available for Re z ≥ 12 (just as was the case for the gamma function in the above example). By Theorem 3.7 above, we can find the Pconver∞ gence boundary by inspecting the sum (3.4) for z = x real. Comparing the sum n=1 n1x ´ ∞ dt 1 shows that (3.4) converges for Re z > 1. What remains for to the integral 1 tx = x−1 us is therefore to somehow continue (3.4) from Re z > 1 down to Re z ≥ 1/2, after which (3.5) completes the task of continuation to the whole complex plane. One idea for this is to turn the sum in (3.4) into an alternating one. From ζ(z) =
1 1 1 1 1 1 + z + z + z + z + z + ··· z 1 2 3 4 5 6
follows
2 2 2 2 ζ(z) = z + z + z + · · · . 2z 2 4 6 Subtracting this second series from the first one gives 1 1 1 1 1 1 1 − 21−z ζ(z) = z − z + z − z + z − z + − · · · . 1 2 3 4 5 6
(3.6)
This new sum is again of Dirichlet type, and it therefore suffices to test convergence for z = x real. Since the sum in this case alternates and the magnitudes of the terms decay for all x > 0, it will, by Theorem 3.7, also converge for all z with Re z > 0. The continuation of (3.4) is complete. As a sideline, we can note that 1 − 21−z 1 − e(1−z) log 2 = lim = − log 2 z→1 1 − z z→1 1−z lim
3.2. Some methods for analytic continuation
81
and lim
z→1
1 1 1 1 1 1 − z + z − z + z − z + −··· 1z 2 3 4 5 6
= log 2,
from which (3.6) gives that limz→1 (z − 1)ζ(z) = 1. One can thus deduce that the singularity of ζ(z) at z = 1 is a first order pole, and then from (3.5) that this pole is the only singularity of the zeta function in the complex plane (and also that ζ(0) P = − 12 ). We ∞ can further note a few other values of the zeta function. Evaluating ζ(2) = n=1 n12 is 2 known as the Basel problem, shown by Euler in 1734 to be equal to π6 . Using contour integration, we will obtain this result as a special case of evaluating ζ(2m), m = 1, 2, 3, . . . , 1 in Example 5.25. With this value for ζ(2), the functional equation gives ζ(−1) = − 12 . Finding similar closedform expressions for ζ(k) for k = 3, 5, 7, . . . remains an unsolved problem.
3.2.4 Partitioning of an integration interval Example 3.9. Extend the gamma function (again) from its integral definition (2.18), but without use of a functional relation. ´∞ In the integral definition Γ(z) = 0 e−t tz−1 dt, the upper integration limit causes no problem (thanks to the fast decay of the exponential function). The limitation Re z > 0 comes from what happens at the origin. Thus, we split the interval in two parts: ˆ
ˆ
1 −t z−1
Γ(z) = 0

e t {z
dt }
∞
e−t tz−1 dt {z }
+ 1
Analytic for Re z > 0; Continue this part

.
Analytic for all z; Entire function.
On the short interval [0, 1], the Taylor expansion of e−t converges rapidly, giving ˆ
ˆ
1 z−1 −t
t
e dt =
0
1
t 0
z−1
∞ X (−t)n n! n=0
! dt =
ˆ 1 ∞ X (−1)n tz−1+n dt . n! 0 n=0  {z } =
1 z+n
if Re z > −n
So it might not look as if we have achieved anything new. The first term in the sum has n = 0, so the restriction would still seem to be Re z > 0. However, let us ignore this and consider the function η(z) =
∞ X (−1)n 1 n! z + n n=0  {z }
ˆ
∞
+
Converges for all z. Poles at z = 0, −1, −2, . . .
1

e−t tz−1 dt . {z }
(3.7)
Entire function.
Since η(z) = Γ(z) for Re z > 0, we have achieved (again by Theorem 2.12) the continuation of Γ(z) to the whole complex plane.
3.2.5 Replace Taylor coefficients by integrals or sums Again, we describe the approach by means of an example.
82
Chapter 3. Analytic Continuation
Example 3.10. Continue the function f (z) =
∞ X zn √ . n n=1
The Taylor series converges for z < 1 and diverges for z > 1. We start by noting that √ ´∞ ´∞ 2 2 a simple change of variables in the relation 2π = 0 e−t dt gives √1n = √2π 0 e−nt dt. Therefore ! ˆ ∞ X ˆ ∞ ˆ ∞ ∞ 2 X n ∞ −nt2 2z dt 2 n −nt2 f (z) = √ z e dt = √ . z e dt = √ t2 − z π n=1 π π e 0 0 0 n=1 In the last equality, we made use of the fact that the sum inside the integral is a geometric series and therefore can easily be summed in closed form. The function f (z) is now uniquely defined everywhere away from along z ≥ 1 (where the integral becomes singular, making it natural to place a branch cut along the real axis from z = 1 to z = ∞). Section 5.3.2 describes how one can further continue this function in both directions across the branch cut. Different (related) options are available for generalizing the √1n coefficients in this example to n1α ; see (3.22) and Exercise 3.3.9.
3.2.6 Subtraction of a similar series or integral Example 3.11. Continue the function f (z) =
zk k=0 1+2−k .
P∞
The original series z > 1. For large k, the terms become very close to P∞ diverges for 1 those of the series k=0 z k = 1−z . This makes it natural to look at the difference between these two sums. After minor simplifications, ∞
f (z) −
X (z/2)k 1 =− = −f (z/2). 1−z 1 + 2−k
(3.8)
k=0
Hence, we stumbled upon a functional equation f (z) −
1 = −f (z/2). 1−z
(3.9)
Since the original sum converges for z < 1, equation (3.9) allows us to compute f (z) for z < 2 and then repeating again for z < 4, etc. The function is thereby continued to the whole complex plane. It will have poles at z = 1, 2, 4, 8, . . . . 1 It is tempting to write (3.9) as f (z) = 1−z − f (z/2) and then apply this relation 1 1 1 repeatedly, to obtain f (z) = 1−z − 1−z/2 + 1−z/4 − + · · · . However, this sum diverges for all z. A much better idea is to repeat the original subtraction idea, first on the sum in (3.8), then on the sum to which this gives rise, etc. After some manipulations, this gives 1 1 1/2 1/4 1/8 − + − + −··· , (3.10) f (z) = + z 2 1−z 1 − z/2 1 − z/4 1 − z/8 which converges rapidly for all values of z.
3.2. Some methods for analytic continuation
83
y
x
Figure 3.7. Schematic illustration of the region of convergence of a Borel sum.
3.2.7 Borel summation Assume that we are given a function f (z) defined by a Taylor series f (z) = with positive radius of convergence R. Then φ(z) =
P∞
k=0
ak z k
∞ X ak z k k=0
k!
is an entire function. Next, we form the function ˆ ∞ ˆ ∞ ∞ X ak z k tk dt η(z) = e−t φ(zt)dt = e−t k! 0 0 k=0 ˆ ∞ −t k ∞ ∞ X X e t k = ak z dt = ak z k = f (z). k! 0 k=0 {z } k=0  = 1 for all k
This new function η(z) clearly agrees with f (z) within the domain of convergence z < R. But is it a continuation of f (z)? It can be shown that η(z) will converge in the smallest polygon that can be constructed from the singularities of f (z) in the way that is indicated in Figure 3.7. In this schematic figure, the red ×’s mark the singularities of f (z) and the circle shows where the Taylor expansion around the origin will converge. Through each singularity, we draw a line orthogonal to the direction to the origin. The function η(z) will converge inside the resulting polygon (marked by solid lines in the figure). P∞ k Example 3.12. Apply Borel summation to f (z) = = 1 + z + z2 + z3 k=0 z + ··· .
84
Chapter 3. Analytic Continuation
P∞ k We get immediately φ(z) = k=0 zk! = ez , and therefore the Borel extension takes ´ ∞ −t ´∞ the form f (z) = 0 e φ(zt)dt = 0 e−t(1−z) dt. This has brought us from f1 (z) to f2 (z) in Example 3.1. As we noted there, the new form converges for Re z < 1, entirely agreeing with the description above of where integrals obtained through Borel summation will converge.
3.2.8 Ramanujan’s formula Analytic continuation of a Taylor expansion around z = 0 can be viewed as reconstructing a function globally based on knowing f (0), f 0 (0), f 00 (0), f 000 (0), . . .. If we instead know the function values f (z) at an infinite sequence of points zk , k = 1, 2, 3, . . . , which has a finite limit point, there is in theory enough information available to extract all derivatives at this point.35 If the infinite sequence of points does not have a finite limit point, there will in general not be a unique solution. However, the following formula by Ramanujan may nevertheless be applicable if certain additional conditions are met. In case that the values for f (0), f (1), f (2), f (3), . . . are known, it states that π f (−z) = sin πz
ˆ
∞
tz−1 f (0) − t f (1) + t2 f (2) − + · · · dt
(3.11)
0
(where we may need analytic continuation to interpret the sum f (0) − t f (1) + t2 f (2) − + · · · over the full interval 0 ≤ t < ∞). It is trivial to see that both f (z) ≡ 0 and f (z) = sin πz satisfy f (0) = f (1) = f (2) = · · · = 0. However, if it is also known that f (z) < CeAz with A < π holding when Re z > 0, it can be shown that (3.11) uniquely determines f (z). In the two examples below, we use the formula somewhat differently: enter known functions and obtain nontrivial integral formulas. Both cases provide analytic continuations of integrals that converge only for Re z > 0. Example 3.13. Substitute f (k) = 1, k = 0, 1, 2, . . . into Ramanujan’s formula (3.11). Obviously, f (z) ≡ 1 will satisfy the given data (and also the growth condition for the right halfplane). Equation (3.11) then gives the relation ˆ
∞
0
Example 3.14. Substitute f (k) =
tz−1 π dt = . 1+t sin πz
1 k! ,
k = 0, 1, 2, . . . , into (3.11).
We arrive similarly at ˆ
∞
e−t tz−1 dt = 0
π . Γ(1 − z) sin πz
Since we recognize the integral as Γ(z), we have obtained (3.3). 35 cf.
Theorem 2.12.
3.2. Some methods for analytic continuation
85
An alternative version of Ramanujan’s formula36 (with more rapid convergence of the f (z) series inside the integral) is obtained by redefining f (z) → Γ(z+1) : ˆ Γ(z)f (−z) = 0
∞
t2 t tz−1 f (0) − f (1) + f (2) − + · · · dt . 1! 2!
(3.12)
Later in this text, Example 9.32 and Exercise 12.6.6 outline two different (somewhat heuristic) arguments that lead to (3.11) and (3.12).
3.2.9 Padé approximations This approach frequently becomes an exact continuation procedure if it is applied to the Taylor expansion of a rational function. However (and in contrast to our previous methods), it can also be extremely powerful as a computational tool if all that is known of a Taylor expansion are good approximations for a finite number of leading coefficients (which is insufficient information for exact analytic continuation). Precise convergence results are only partially known. A good review of the topic, as well as on its connection to continued fractions, can be found in Sections 8.3–8.6 of [4]. Taylor expansions (and often also asymptotic expansions; cf. Chapter 12) can often be accelerated quite dramatically (or turned from divergent to convergent) by being rearranged into a ratio of two such expansions. A Padé approximation, PN an z n N PM (z) = Pn=0 (3.13) M n n=0 bn z (normalized by b0 = 1) generalizes the Taylor expansion with equally many degrees of freedom, M +N X TM +N (z) = cn z n , (3.14) n=0
with the two being the same in case M = 0. The Padé coefficients are normally found by starting from a Taylor expansion c0 + c1 z + c2 z 2 + · · · =
a0 + a1 z + a2 z 2 + · · · 1 + b1 z + b2 z 2 + · · ·
and then requiring that both sides match to the highest degree possible at the origin. Multiplying by the denominator gives the following equivalent set of coefficient relations: a0 = c0 , a1 = c1 + c0 b1 , a2 = c2 + c1 b1 + c0 b2 , a3 = c3 + c2 b1 + c1 b2 + c0 b3 , ...
(3.15)
With the ci given, each new line introduces two new unknowns, ai and bi . The system would appear to be severely underdetermined. However, if we specify the degree of the 36 Also
known as Ramanujan’s master theorem.
86
Chapter 3. Analytic Continuation
numerator to be N , of the denominator to be M , and of the truncated Taylor expansion to be M + N , there will be just as many equations as unknowns (ignoring all terms that are O(z M +N +1 )). We next solve for all the unknown coefficients, as the following example shows. Example 3.15. Given T5 (z), determine P32 (z). In this case of M = 3, N = 2, M + N = 5, the system (3.15) becomes truncated as follows: a0 = c0 , a1 = c1 + c0 b1 , a2 = c2 + c1 b1 + c0 b2 , 0 = c3 + c2 b1 + c1 b2 + c0 b3 , 0 = c4 + c3 b1 + c2 b2 + c1 b3 , No more a’s available 0 = c5 + c4 b1 + c3 b2 + c2 b3 , past limit O(z 2+3+1 ).
No more b’s available
After solving the bottom three equations for b1 , b2 , b3 , the top three explicitly give a0 , a1 , a2 . This same idea generalizes to other values of M and N .37 As noted above, a key use of Padé approximations is to extract improved information from power series expansions with only a limited number of known terms. Transformation to Padé form usually accelerates convergence, and often allows good approximations to be found even well outside a Taylor expansion’s radius of convergence (which might even be R = 0). Example 3.16. Find the increasing order Padé approximations for f (z) = 1 − z + z 2 − z3 + − · · · . The Padé expansions based on the truncated Taylor sums are shown in Table 3.1. The main diagonal (and the diagonal below it) usually gives the best results. This example is atypical in that f (z) is a rational function, implying that it will produce the exact result. Table 3.1. Beginning of Padé table for f (z) = 1 − z + z 2 − z 3 + − · · · .
M  order of denominator
0 1 2 3 ...
0 1 1 1+z 1 1+z 1 1+z
N  order of numerator 1 2 3 1 − z 1 − z + z2 1 − z + z2 − z3 1 1 ... 1+z 1+z 1 . . . 1+z ...
... ...
...
Example 3.17. Approximate f (2) when we only know the first few terms in the Taylor , but convergent only if expansion f (z) = 1 − 12 z + 31 z 2 − 14 z 3 + 15 z 4 − + · · · (= ln(1+z) z z < 1). 37 It should, however, be noted that this (conceptually simplest) implementation of the Padé method can be numerically “fragile.” Robust codes need to take precautions for the event of singular (or nearsingular) systems.
3.2. Some methods for analytic continuation
87
The Padé table of Table 3.2 is laid out like Table 3.1, but includes only its M = 0 and M = N entries, and for these shows only the numerical values for z = 2 and, in parentheses, the errors in these compared to 21 log 3 ≈ 0.5493. In spite of z = 2 being well outside the domain of convergence for the Taylor series of f (z) (as is again visible from the top M = 0 row in the table), the Padé method allows fast and accurate calculation of the analytically continued result at z = 2. Table 3.2. Values for f (2) from Padé approximations; in parentheses is their difference to
0
0 1 (0.4507)
1 M  order of denominator
2 3
N  order of numerator 1 2 3 0 1.3333 −0.6667 (−0.5493) (0.7840) (−1.2160) 0.5714 (0.0221) 0.5507 (0.0014) 0.5494 (0.0001)
log 3.
...
0.5493 (0.0000)
4 ...
4 2.5333 (1.9840)
1 2
...
...
Example 3.18. Compare Taylor and Padé approximations for the Stieltjes function f (z) = ´ ∞ e−t dt. 0 1+zt The integral defining f (z) is singular for z < 0 (real and negative), but it is well defined for other values of z in the complex plane. Figures 3.8(a)–(b) show the result of a direct evaluation of f (z). The presence of a branch discontinuity along the negative real axis is obvious. We next Taylor expand f (z) around z = 0 (for example, by repeated integration by parts, or by noting that f (z) satisfies z 2 f 0 (z) + (1 + z)f (z) − 1 = 0, f (0) = 1, and then equating coefficients). The resulting expansion becomes f (z) ≈
∞ X
(−z)k k! .
k=0
This diverges for all nonzero values of z (its radius of convergence is R = 0). Truncation after the sixth power gives T6 (z) ≈ 1 − z + 2z 2 − 6z 3 + 24z 4 − 120z 5 + 720z 6 . As seen in Figure 3.9(a), T6 (z) resembles the true Stieltjes function at most in a very small region close to the origin. However, after converting T6 (z) to its P33 (z) Padé counterpart P33 (z) =
1 + 11z + 26z 2 + 6z 3 , 1 + 12z + 36z 2 + 24z 3
(3.16)
we see in Figure 3.9(b) a somewhat respectable approximation of the original function. Being a rational function, and therefore singlevalued, the Padé approximation can never
88
Chapter 3. Analytic Continuation
(a) Real part of f (z) =
´∞ 0
e−t 1+zt dt.
(b) Imaginary part of f (z) =
´∞ 0
e−t 1+zt dt.
Figure 3.8. The real and imaginary parts of the Stieltjes function.
(a) Im T6 (z).
(b) Im P33 (z).
(c) Im T30 (z).
15 (d) Im P15 (z).
Figure 3.9. Displays of the imaginary parts of truncated Taylor and corresponding Padé approximations of the Stieltjes function, starting from Taylor expansions using terms up through degrees 6 and 30, respectively.
3.3. Exercises
89
reproduce the branch discontinuity. It has, however, placed its pole singularities along the negative real axis in an attempt to mimic the branch discontinuity there.38 Finally, Figures 15 3.9(c)–(d) compare T30 (z) with P15 (z). In the case of the Stieltjes function, it can be proven that the Padé approximations along the M = N diagonal in the Padé table will converge to the true function exponentially fast (as higher degrees are used) everywhere in the complex plane away from the negative real axis. Somehow, the everywhere divergent Taylor expansion (R = 0) does contain complete information about the function it was based on, and the Padé approach allows this information to be recovered.
3.3 Exercises Exercise 3.3.1. The Taylor expansions of a function f (z) around z = α and z = β, respectively, take the forms f (z) = a0 + a1 (z − α) + a2 (z − α)2 + a3 (z − α)3 + · · · = b0 + b1 (z − β) + b2 (z − β)2 + b3 (z − β)3 + · · · . If we define δ as δ = β−α, show that we can obtain the bcoefficients from the acoefficients by a0 b0 1 1δ 1δ 2 1δ 3 1δ 4 · · · b1 0 1 2δ 3δ 2 4δ 3 · · · a1 b2 0 0 1 3δ 6δ 2 · · · a2 , b3 = 0 0 0 1 4δ · · · a3 b4 0 0 0 0 1 · · · a4 .. .. .. .. .. .. .. .. . . . . . . . . where the coefficients for the powers of δ are given by Pascal’s triangle. Hint: Let t = z − β = z − α − δ, i.e., z − α = t + δ. Equate coefficients for powers of t. Exercise 3.3.2. Given a Taylor series c1 z + c2 z 2 + c3 z 3 + · · · , setting z = rearranging it in powers of t gives a1 t + a2 t2 + a3 t3 + · · · , where a1 a2 a3 a4
t 1+t
and
= c1 , = c2 − c1 , = c3 − 2c2 + c1 , = c4 − 3c3 + 3c2 − c1 , .. .
etc. (with coefficients from Pascal’s triangle). (a) Verify the result above. (b) Already, Newton used the result above to rewrite the expansion arctan z = z −
z5 z7 z3 + − + −··· 3 5 7
(3.17)
38 In cases of branch points (such as in this example), “quadratic” Padé approximations [15] are generally preferable over the “linear” Padé approximations described here.
90
Chapter 3. Analytic Continuation
into z arctan z = 1 + z2
2 1+ 3
z2 1 + z2
2·4 + 3·5
z2 1 + z2
2
2·4·6 + 3·5·7
z2 1 + z2
!
3 + ···
(3.18) Compare the regions in which (3.17) and (3.18) converge. Did this rewrite amount to an analytic continuation of the original series? P∞ Exercise 3.3.3. Consider the Taylor expansion f (z) = n=0 (−z)n . (a) Determine the radius of convergence R for the Taylor series for f (z). (b) Sum the Taylor series in closed form. (c) Based on the result in part (b), show that the Taylor expansion of f (z) around n+1 n P∞ z = 21 becomes f (z) = n=0 (−1)n 23 z − 21 . (d) From the expansion coefficients obtained in part (c), determine the radius of convergence for the new expansion. Draw where the two expansions converge in the complex plane, and mark the point z = −1. For parts (e) and (f) below, use only the original Taylor series coefficients for f (z) (i.e., not the closedform expression for f (z)) to reexpand f (z) around the point z = 12 . You m+1 P∞ 1 k! = (−1)m 23 can assume it is known that k=m (−1)k m!(k−m)! . 2k−m n
d (e) Obtain the new Taylor coefficients by working out the values for dz n f (z), n = 1 0, 1, 2, . . . , at z = 2 . (f) Substitute z → ζ + 12 in the original Taylor series, rearrange the result into powers of ζ, and then change the variable back ζ → z − 12 .
P∞ k Exercise 3.3.4. Consider the function f (z) = k=0 z (2 ) : (a) Determine the radius of convergence of the Taylor expansion for f (z). (b) Show that f (z) satisfies the functional equation f (z) = z + f (z 2 ). (c) By means of the functional equation, show that the function cannot be continued outside the Taylor series radius of convergence (i.e., its boundary forms a natural boundary for the function itself). Pm−1 k m Hint: From the functional equation, first deduce that f (z) = f (z (2 ) )+ k=0 z (2 ) holds for m = 1, 2, 3, . . . . The result then follows by a similar argument to the one used in Example 3.2. Comment: In the previous example and in this exercise, the ratio between successive exponents are k!/(k − 1)! = k and 2k /2k−1 = 2, respectively. The Ostrowski–Hadamard gap theorem tells one that if R = 1 and the ratio of successive (integer) exponents is bound from below by a constant greater than 1, the unit circle always becomes a natural boundary. Exercise 3.3.5. This exercise supplements Example 3.18. (a) Determine the radius of convergence of the Taylor series f (z) =
∞ X
(−1)k k!z k .
k=0
(3.19)
.
3.3. Exercises
91
(b) Use Borel summation to obtain from (3.19) the formula ˆ ∞ −t e f (z) = dt, 1 + zt 0
(3.20)
and determine the range of validity for this integral representation of f (z). (c) Use instead the method from Section 3.2.5 to´ obtain (3.20) from (3.19). Hint: Note ∞ from the definition of the gamma function that k! = 0 e−t tk dt. (d) Starting from (3.20), derive (3.19) by repeated integration by parts. (e) Again, start from (3.20) and arrive at (3.19) by the different method of first showing that (3.20) implies that f (z) satisfies the ODE z 2 f 0 (z) + (1 + z)f (z) − 1 = 0
(3.21)
with initial condition f (0) = 1. Then assume that f (z) has a Taylor expansion f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · · and use (3.21) to determine the unknown coefficients. Exercise 3.3.6. (a) By a change of variables in the integral definition of the gamma function, derive ˆ ∞ 1 1 = tz−1 e−nt dt. (3.22) nz Γ(z) 0 (b) Following one of the analytic continuation methods, show that the Riemann zeta function ∞ X 1 ζ(z) = z n n=1 can be rewritten as ζ(z) =
1 Γ(z)
ˆ 0
∞
tz−1 dt. et − 1
(3.23)
(c) Determine for what values of z the integral in part (b) converges. Exercise 3.3.7. By combining the results in (3.23) and (3.12) with the definition of Bernoulli numbers in Exercise 2.9.23, make a plausible argument that arrives at the formula ζ(−k) = (−1)k
Bk+1 , k = 0, 1, 2, . . . , k+1
giving the values of the zeta function for nonpositive integer arguments. Comment: Combining equations with little regard for convergence, etc., is not mathematically rigorous, but is nevertheless often extremely helpful in terms of “spotting” relations. We will later in this text (Section 6.2) arrive rigorously at this same relation. P∞ Exercise 3.3.8. Let f (z) = k=0 (k + 1) z k . 1 (a) Show by some direct means that f (z) = (1−z) 2. Apply Borel summation to the Taylor series, to obtain φ(z) = (1 + z) ez and η(z) = ´ ∞ (b) −t (1−z) e (1 + zt)dt. 0
92
Chapter 3. Analytic Continuation
(c) Show that this integral converges for Re z < 1 and diverges for Re z ≥ 1. 1 (d) Evaluate η(z) for Re z < 1, to then obtain η(z) = (1−z) 2. Exercise 3.3.9. (a) Show that the identity in Example 3.10 can be generalized as α−1 ˆ 1 1 1 1 = log xn−1 dx. nα Γ (α) 0 x α−1 dt ´1 P∞ n z (b) Use part (a) to show that n=1 nz α = Γ (α) log 1t 1−z t . 0 Exercise 3.3.10. P∞ √ (a) Derive for f (z) = n=1 n z n the analytic continuation ˆ ∞ 2 2z ex f (z) = √ dx. π 0 (ex2 − z)2 Hint: Start with the result in Example 3.10. (b) Verify the result by Taylor expanding the integrand around z = 0, and then integrate termbyterm. Exercise 3.3.11. Carry out the algebra steps that produce (3.10). Exercise 3.3.12. Padé approximations are somewhat “fragile” in that they sometimes cannot give the expected order of accuracy. Let T (z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · · . Determine the condition(s) we need to impose on these coefficients in order to obtain the expected accuracy in the cases of (a) P11 (z), (b) P12 (z), (c) P21 (z). (d) Show that the most accurate P12 (z) approximation in the case of c2 = 0 is P12 (z) = c0 + c1 z. Is that approximation unique? Exercise 3.3.13. In some cases, closedform expressions are available for Padé approxiPn (z) (2n−k)!n! , where pn (z) = k=0 (2n)!k!(n−k)! z k . Verify by mations. For example, Pnn (ez ) = ppnn(−z) direct evaluation that this indeed matches the Taylor expansion to the expected number of coefficients in the cases of n = 0, 1, 2. Exercise 3.3.14. In one of the most famous letters in the history of mathematics (dated January 16, 1913), the Indian mathematical genius Srinivasa Ramanujan (1887–1920) introduces himself to the English mathematician G. H. Hardy (1877–1947) and then gives some stunning examples of original formulas. The letter also contains the seemingly whimsical statements 1 − 1! + 2! − 3! + 4! − + · · · ≈ 0.596 and
1 . 12 Based on the examples in this chapter, explain what lies behind both of these results. 1 + 2 + 3 + 4 + ··· = −
Chapter 4
Introduction to Complex Integration
Realvalued calculus centers around differentiation and integration. In the complex case, ability to differentiate led to the concept of analytic functions. Such functions also have very special properties when it comes to integration. Various integration theorems will become key tools for establishing a wide range of general results for analytic functions. This chapter provides an introduction to the topic of complex integration, leaving the key technique of residue calculus to Chapter 5. Realvalued case: In real variable calculus, a definite integral is defined by ˆ
b
f (x)dx = lim
X
∆xk →0
a
f (xk )∆xk ,
(4.1)
k
as illustrated in Figure 4.1(a).39 The key theorem states that if there is a function F (x) ´b d such that dx F (x) = f (x), then a f (x)dx = F (b) − F (a). Finding such a function F (x) (which may not exist in terms of elementary functions) is then often the only means available for evaluating integrals analytically. Complexvalued case: In the complex plane, we consider instead ˆ
zb
f (z)dz = lim za
∆zk →0
X
f (zk )∆zk ;
(4.2)
k
´z cf. Figure 4.1(b). This integral can similarly be evaluated as zab f (z)dz = F (zb ) − F (za ), d F (z) = f (z) (as shown in Theorem 4.1 below). Although the definition (4.2) where dz 39 In (4.1) and similarly later in (4.2), the numerical values for ∆x and ∆z are the differences between the k k end and the start values for each of the pieces that the integration path has been divided into. These pieces contain the values xk and zk , respectively.
93
94
Chapter 4. Introduction to Complex Integration y y = f (x)
∆x k
a
b
x
(a) Integration in a real variable. Im z
zb
Γ
Re z za
∆zk
(b) Integration in a complex variable. Figure 4.1. Conceptual difference between integrating in a real and a complex variable.
does not require f (z) to be analytic, two major novelties arise if this is the case: (i)
As long as F (z) is singlevalued, it is unnecessary to specify the integration path Γ (also known as contour) that was followed from za to zb .40
(ii)
By utilizing this path independence, it will turn out that many definite integrals can be evaluated exactly, without us needing to obtain any closedform expression for F (z) (the topic of residue calculus in Section 5.1).
40 Continuously
deforming one path into another, it is required that f (z) remains analytic also along the intermediate ones. This can be expressed as the domain of analyticity being simply connected.
4.1. Integration when a primitive function F (z) is available
95
4.1 Integration when a primitive function F (z) is available Theorem 4.1. If there exists a function F (z) such that F 0 (z) = f (z), then F (zb ) − F (za ).
´ zb za
f (z) dz =
Proof. d dz
ˆ
´ z+∆z
z
za
f (ξ)dξ = lim
f (ξ)dξ −
za
f (ξ)dξ
∆z
∆z→0
za
´z
´ z+∆z
f (ξ)dξ ∆z
z
= lim
∆z→0
= lim
∆z→0
f (z)∆z ∆z
= f (z) =
d F (z). dz
Therefore, d dz
ˆ
z
F (z) −
f (ξ)dξ
= 0,
za
which implies
ˆ
z
F (z) −
f (ξ)dξ = C, za
where C is a constant. Setting z = za , we get C = F (za ), and setting z = zb , we obtain ˆ
zb
f (ξ)dξ = F (zb ) − F (za ). za
Next, consider a closed loop as in Figure 4.2, where za = zb . If F (z) exists and is singlevalued inside and on Γ, it will follow from za = zb that F (zb ) − F (za ) = 0, i.e., ¸ that Γ f (ξ)dξ = 0. We made above the assumption that F (z) exists. This can be justified by considering a circlechain continuation of f (z) along the integration path. We can then integrate each of these Taylor expansions (with one free constant, as to be expected). Recall that the radius of convergence never changes if we integrate or differentiate a Taylor series. So F (z) exists along a path if f (z) exists along it. We have not yet proven that a Taylor series converges up to the nearest singularity. In ¸ fact, the proof for that will use the properties of f (z)dz. Hence, we would like to arrive at the result ˛ f (z)dz = 0, Γ
96
Chapter 4. Introduction to Complex Integration Im z
Γ
Re z za = z b
Figure 4.2. Integration along a closed contour Γ is denoted by
¸ Γ
f (z)dz.
where Γ is a closed loop containing no singularity, without having to invoke the existence of F (z). This is accomplished in the next theorem.
4.1.1 Cauchy’s and Morera’s theorems ¸ Theorem 4.2 (Cauchy). Γ f (z)dz = 0 when f (z) is an analytic function41 that is singularityfree on and inside the contour Γ. Proof. ˛
˛ f (z)dz =
Γ
Γ
˛
(u + iv) (dx + idy)  {z }  {z } f (z)
dz
˛
(udx − vdy) + i
= Γ
(udy + vdx). Γ
Recall from multivariable calculus Green’s theorem the following: Let g(x, y) and h(x, y) and their first partial derivatives be continuous in a domain D bounded by a closed loop Γ. Then, ˛ ¨ ∂h ∂g (gdx + hdy) = − dxdy. ∂x ∂y Γ D 41 For
the proof, we need that f (z) satisfies the CR equations and that f 0 (z) is continuous.
4.2. Contour integration
In our case, ˛
97
˛
˛ (udx − vdy) + i (udy + vdx) Γ Γ ¨ ¨ ∂u ∂u ∂v ∂v + dxdy + i − dxdy =− ∂x ∂y ∂x ∂y D D  {z }  {z }
f (z)dz = Γ
=0 by CR
(4.3)
=0 by CR
=0 Morera’s theorem is the reverse to Cauchy’s theorem. ¸ Theorem 4.3 (Morera). If f is continuous and if Γ f (z)dz = 0 for any contour Γ in the domain, then f (z) is an analytic function within the domain. The main themes in this and the next chapter will be the following: ¸ • Since evaluating Γ f (z)dz along certain contours Γ might be problematic, we will learn how to change paths to make the evaluations easier. • We will see what happens when the closed loop Γ actually does contain a singularity. ´b • If we want to evaluate a real integral a f (x)dx, we will see how to modify the path to utilize properties in the complex plane. Combinations of these techniques will lead to very powerful methods to evaluate many definite integrals also in cases where we cannot find F (z). These methods also lead to accurate estimates of integrals in various singular limit situations, as discussed in Chapter 12: Steepest Descent for Approximating Integrals.
4.2 Contour integration Example 4.4. Evaluate
ˆ z¯ dz, Γ
where Γ is the path shown in Figure 4.3. We first note that f (z) is NOT an analytic function, so Morera’s theorem tells us that the result will be path dependent. Also, the contour is not a closed loop, so Green’s theorem (as used in the proof of Theorem 4.2) is not available to us. We thus need to work out this integral one line segment at a time. Here, f (z) = z¯ = x − iy, so
ˆ
ˆ z¯dz =
Γ
ˆ
1
xdx + 0
0
1
x2 (1 − iy)(idy) = 2
1
1 y2 +i y−i = 1 + i. 2 0 0
If we do continue the path from 1+i first to i and then back to the origin (thus going around a unit square), we pick up an additional i − 1, giving for the full path around the square
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Chapter 4. Introduction to Complex Integration
Im z
1+i
0
Re z
1
Figure 4.3. Path for the integral in Example 4.4.
the value 2i. For this closed loop, we could also have used (4.3) (skipping the last step of applying the CR equations), giving ˆ ¨ ¨ z¯ dz = − (0 + 0)dxdy + i (1 + 1)dxdy = 2i. 2 2 2 Example 4.5. Evaluate
ˆ z n dz, Γ
where n is an integer and Γ is any closed contour that goes around the origin once in the positive (counterclockwise) direction. We solve this problem in two ways: 1. by changing the path and evaluating the integral explicitly around a simplified path, and 2. by using the primitive form F 0 (z) = f (z). If n ≥ 0, then f (z) is analytic inside Γ, which leads to Γ z n dz = 0. However, if n < 0, then there is a pole at the origin. In that case, one can consider a different contour, such as the one shown in Figure 4.4(b) (note the positive direction around the origin for both Γ and γ). This new contour excludes the origin and thus the pole. It will integrate to zero: ˆ ˛ ˆ ˆ ˆ f (z)dz = + − + f (z)dz (4.4) Method´1: Modifying the contour:
Γ
←
γ
→
=0. ´Since one can choose the paths → and ← arbitrarily close to each other, f (z)dz cancel out. Therefore, → ˛ ˛ f (z)dz = f (z)dz, γ
Γ
´ ←
f (z)dz and
4.2. Contour integration
99
Γ
Γ
γ
(a) Path for the integral in Example 4.5.
(b) Modified contour, with a red dot marking the pole at the origin.
Figure 4.4. Different integration paths considered in Example 4.5. Note in part (b) that we have marked the path direction for the small circle in the positive direction (causing the minus sign in (4.4)).
where both are followed in the positive direction. The argument shows that we, in general, can move the contour as we want without changes to the value of the integral, as long we don’t move it across any singular point. Instead of following Γ, we can thus integrate around the circle γ centered at the origin and with radius R. Let z = Reiθ , and therefore dz = iReiθ dθ. Then ˛ ˛ n z dz = z n dz Γ
γ
ˆ
2π
Rn einθ iReiθ dθ
= 0
ˆ
2π
= iRn+1
ei(n+1)θ dθ 0
h i2π 1 ei(n+1)θ i(n + 1) 0 = 0 if n 6= −1. = iR
n+1
However, if n 6= −1, then ˛
˛ z −1 dz =
Γ
z −1 dz γ
ˆ
2π
R−1 e−iθ iReiθ dθ
= 0
ˆ
2π
=
idθ 0
= 2πi.
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Chapter 4. Introduction to Complex Integration
Modifying contours in this fashion is an often used technique, which generalizes straightforwardly to cases with many singularities, as will be illustrated later in Figure 5.1. Let f (z) = z n . If n 6= −1, then F (z) = Therefore, after having followed the closed contour Γ, F (zend ) = F (zstart ), since F (z) is singlevalued. Therefore, in the n 6= −1 case, ˛ z n dz = 0. Method 2: Finding the primitive function F (z): z n+1 n+1 .
Γ
However, if n = −1, then F (z) = log(z). After having gone through one cycle of the contour,42 the value of F (z) must have increased by 2πi. Therefore, ˛ z n dz = 2πi. Γ
We arrive at the same conclusion as before. The above result can alternatively be formulated as follows. Theorem 4.6. For n integer, 1 2πi
˛ Γ
0 dz 0 = (z − z0 )n 1
if z0 is outside Γ, if z0 is inside Γ and n 6= 1, if z0 is inside Γ and n = 1.
The case of n = 1 and z0 right on the curve Γ will be discussed in Section 5.1.4.
4.2.1 Cauchy’s argument principle Example 4.7. Let p(z) = A(z − a1 )(z − a2 ) · · · · · (z − an ) be a polynomial of degree n.43 Evaluate the following integral: ˛ 0 1 p (z) I= dz . 2πi Γ p(z) We obtain p0 (z) d = log (p(z)) p(z) dz d = (log A + log(z − a1 ) + log(z − a2 ) + · · · + log(z − an )) dz 1 1 1 = + + ··· + . z − a1 z − a2 z − an
(4.5)
This implies that the value of I equals the number of roots lying inside Γ. We can tell the number of roots inside a contour by inspecting the polynomial only along the contour. This result generalizes in various ways, to Cauchy’s argument principle (next) and to Rouché’s theorem (Theorem 5.21); see also the Gauss–Lucas theorem (Exercise 5.6.38). 42 Without 43 We
having placed any branch cut in its way. will soon see (Section 4.2.4) that any polynomial of degree n indeed can be factored in this way.
4.2. Contour integration
101
Theorem 4.8 (Cauchy’s argument principle). If f (z) is a meromorphic function (its only singularities are poles), and if it has N zeros and, as only singularities, P poles (accounting for their multiplicities) inside the contour Γ, then ˛ 0 1 f (z) N −P = dz. 2πi Γ f (z) Proof. We write f (z) as QN (z − ak ) g(z), f (z) = Qk=1 P k=1 (z − bk ) where g(z) has neither zeros nor poles within Γ. Then f 0 (z) d = log (f (z)) f (z) dz d = dz =
N X k=1
N X
log(z − ak ) −
k=1
P X
! log(z − bk ) + log g(z)
k=1 P
X 1 1 − + z − ak z − bk
k=1
function that is singularityfree within Γ
,
and the result follows when integrating around Γ. We will revisit this result in Section 5.1.6 describing Rouché’s theorem and winding numbers.
4.2.2 Cauchy’s integral formula As a preliminary to Cauchy’s integral formula, let us note the following estimate, which often turns out to be useful. Theorem 4.9. The integral along a path Γ can be bounded from above by ˆ f (z)dz ≤ M · L, Γ
where M is the upper bound of f (z) on Γ and where L is the length of Γ. Proof. ˆ X f (z)dz = lim f (zk )∆zk ∆zk →0 Γ X ≤ lim f (zk ) ∆zk  ∆zk →0  {z } ≤M
X
≤ M lim ∆zk  ∆zk →0  {z } =L
= M · L.
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Chapter 4. Introduction to Complex Integration
Theorem 4.10 (Cauchy’s integral formula). Let f (z) be analytic inside Γ. Then, for z inside Γ, ˛ f (ξ) 1 dξ. (4.6) f (z) = 2πi Γ ξ − z Significance
1. We have mentioned before that f (z) is completely determined by its values along any curve segment (Theorem 2.12). So the existence of a formula of this kind is not altogether surprising. If we know its values along a whole closed contour, Cauchy’s integral formula provides its values at points inside the contour in a remarkably simple way. 2. We will soon deduce from (4.6) that if f (z) is analytic, it is not just once but infinitely many times differentiable. 3. This formula forms the foundation for a much more solid understanding of Taylor expansions, and their generalization to Laurent expansions. Proof. As in Figure 4.4(b), we modify our contour Γ to become a circle centered at z with radius δ, and call this new contour γ. Thus ˛ ˛ 1 f (ξ) 1 f (ξ) dξ = dξ 2πi Γ ξ − z 2πi γ ξ − z ˛ 1 (f (ξ) − f (z)) + f (z) = dξ 2πi γ ξ−z ˛ ˛ 1 (f (ξ) − f (z)) 1 dξ = . dξ +f (z) 2πi γ ξ−z 2πi γ ξ − z {z } {z }   =1
=Iδ
Consider the first of the integrals Iδ , and let δ become arbitrarily small. Then, the continuity of f implies that f (ξ) − f (z) < ε whenever z − ξ < δ. Thus (by Theorem 4.9) ε · 2πδ Iδ  ≤ δ {z} {z} L
M
= ε2π → 0 when δ → 0. Therefore, 1 2πi
˛ Γ
f (ξ) dξ = f (z). ξ−z
Theorem 4.11. If f (z) is analytic (once complex differentiable) in some region, it can then be differentiated any number of times. Proof. The RHS of (4.6) clearly possesses any number of derivatives, and then so must the LHS. In fact, we obtain the following formula for the kth derivative: ˛ k! f (ξ) (4.7) f (k) (z) = dξ. 2πi Γ (ξ − z)k+1
4.2. Contour integration
103
4.2.3 Liouville’s theorem Theorem 4.12. If f (z) is entire (no singularity for finite z) and bounded (f (z) ≤ M ), then f (z) ≡ C for C constant. Proof. Equation (4.7) becomes for the first derivative (k = 1) 1 f (z) = 2πi
˛
0
Γ
f (ξ) dξ (ξ − z)2
and we choose as the contour a circle of radius R centered at the point z. Then, ˛ f (ξ) 1 dξ 2 2π Γ (ξ − z) 1 M M ≤ 2πR = . 2 2π R R
f 0 (z) =
Since M is a constant and R can be arbitrarily large, this inequality is a contradiction unless f 0 (z) = 0. From this it follows then that f (z) must be a constant. By considering higher kvalues (than k = 1) in (4.7), we obtain the following generalized result. Theorem 4.13. If f (z) is entire and satisfies a bound f (z) ≤ A zn for n a nonnegative integer, then f (z) is a polynomial of degree at most n.
4.2.4 The fundamental theorem of algebra Theorem 4.14. A polynomial p(z) of degree n ≥ 1 has at least one root. Proof. Suppose that p(z) has no root and that n ≥ 1. We wish to construct a contradiction. 1 . It satisfies the two conditions needed to apply To this end, we consider f (z) = p(z) Liouville’s theorem: 1.
1 p(z)
is an entire function (since p(z) has no root).
2.
1 p(z)
is bounded since p(z) → ∞ as z → ∞.
1 By Liouville’s theorem, p(z) and therefore also p(z) must be constant. This violates the assumption that p(z) is a polynomial of degree n ≥ 1. The hypothesis that p(z) has no root must therefore have been false.
After dividing out the root we now know must have existed, the argument can be repeated. Theorem 4.15. A polynomial pn (z) of degree n > 0 will always have exactly n roots (not necessarily at different locations from each other).
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Chapter 4. Introduction to Complex Integration
4.2.5 Mean value theorem Theorem 4.16. The mean value theorem: ˆ 2π 1 f (z) = f (z + Reiθ )dθ 2π 0
(4.8)
(assuming no singularity inside the circle). Proof. Use Cauchy’s integral formula f (z) =
1 2πi
˛ Γ
f (ξ) dξ (ξ − z)
and let the contour be a circle centered at z with radius R. A change of variables yields ξ − z = Reiθ and dξ = iReiθ dθ. Thus ˆ 2π 1 f (z) = f (z + Reiθ )dθ. 2π 0 This last theorem states that the value of an analytic function at any interior point z is the average of the function integrated around a circle centered at z. One can separate the real from the imaginary part in the above equation, so the theorem holds also for Re f (z) and for Im f (z) separately.
4.2.6 Revisiting max/min theorems Theorem 4.17 (max/min theorems). u(x, y) = Re f (z) and v(x, y) = Im f (z) cannot have a local (finite) maximum or minimum. Proof. Assume that the real part u(x, y) reaches a local maximum at some point z. However, the value of u at that point is the average of the values of u around a circle centered at z. This is a contradiction. Theorem 4.18. The magnitude f (z) of an analytic function cannot attain a local minimum or maximum except at a zvalue for which f (z) = 0 and at a singular point. Proof. Assume that f (z) attains a local min/max that it is neither 0 nor ∞. Then log(f (z)) also attains a local min/max, since log (for a realvalued argument) is a monotonic function. However, log(f (z)) is the real part of the function log(f (z)), which is analytic. By the theorem above, this is impossible. Somewhat related to this theorem is whether anything can be said if it only is known that a function f (z) is analytic (without singularities) in a halfplane and bounded along its edge. Theorem 4.19 (Phragmén–Lindelöf). Let f (z) be analytic (singularityfree) in the right halfplane Re z ≥ 0 and f (iy) < M (constant) along the imaginary axis. Then, if β
f (z) < K ez for some constant K and β < 1, in the halfplane, it will hold everywhere across it that f (z) < M .
4.2. Contour integration
105 γ
Proof. Define F (z) = f (z) e−ε z , where ε > 0 is arbitrarily small, and β < γ < 1. The maximum principle tells us that F (z) < M throughout the halfplane, and the result follows by letting ε → 0. Roughly speaking, the Phragmén–Lindelöf theorem tells us that the function, bounded along the imaginary axis, must either grow exponentially fast or stay bounded in the halfplane—nothing in between can happen. There are many theorems of this general character, for example for functions that are bounded along two intersecting lines z = r eiθ1 and z = r eiθ2 , r ∈ [0, ∞].
4.2.7 Poisson’s integral formula Theorem 4.20. If f (z) is analytic inside a circle C, which we here take to be the unit circle, its value at an arbitrary point z = r eiθ , 0 ≤ r < 1, is given from the values along C by ˆ 2π 1 1 − r2 iθ f (r e ) = dt. (4.9) f (eit ) 2π 0 1 + r2 − 2r cos(t − θ) We note the following: 1−r 2 (i) The Poisson kernel 1+r2 −2r cos(t−θ) is purely realvalued for all real values of r, t, θ. (ii) If we write f = u + iv with u and v real, then (4.9) will hold separately for the two harmonic functions u(x, y) and v(x, y). Similarly, if a realvalued function u(x, y) is 2 2 harmonic, i.e., continuous44 and obeying ∂∂xu2 + ∂∂yu2 = 0, it will satisfy (4.9). (iii) Setting r = 0, equation (4.9) reduces to the mean value theorem (Theorem 4.16) (see (4.8)). A proof for Theorem 4.20 is given in Section 4.4.
4.2.8 Revisiting Taylor expansions Radius of convergence
In the following two theorems, we consider for notational simplicity expansions centered at the origin. Theorem 4.21. If f (z) = z < z0 .
P∞
j=0 bj z
j
converges for z0 6= 0, then it also converges for all
Significance: The domain of convergence for a Taylor series must be a circular disc. Anything else would become a contradiction. P∞ Proof. Assume that f (z) = j=0 bj z j converges for z0 6= 0. Then, from some index J and onwards (j ≥ J), it will hold that bj z0j  < 1. Also from this index, j z bj z j  = bj z0j  z0 j z < z0 = Mj 44 Recall
4
f (z) = e−1/z from Section 2.1.1.
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Chapter 4. Introduction to Complex Integration
P with M  < 1, since z < z0 . Since PMj converges, by the Weierstrass Mtest theorem ∞ (Theorem 2.39), we can conclude that j=0 bj z j converges. Theorem 4.22. If f (z) is analytic for z ≤ R, then its Taylor series f (z) = with bj =
f (j) (0) j!
P∞
j=0 bj z
j
converges to f (z) at each point z < R.
Significance: This theorem implies that a Taylor series always converges in the greatest circle up to the nearest singularity. Proof. Let Γ be the circle of radius R, and let z < R. By Cauchy’s integral formula, ˛ 1 f (ξ) f (z) = dξ 2πi Γ ξ − z ! ˛ 1 1 1 = f (ξ) dξ. 2πi Γ ξ 1 − zξ Since ξ = R and z < R, it holds that zξ < 1, which implies 1 ξ
1 1−
! z ξ
=
∞ X zj j+1 ξ j=0
is a convergent series. Thus, ˛ ∞ ∞ X X 1 f (j) (0) j f (ξ)dξ j f (z) = z = z 2πi Γ ξ j+1 j! j=0  j=0 {z } =bj =
(4.10)
f (j) (0) j!
is also convergent. The largest number R for which the power series converges inside the disk z < R is called the radius of convergence.
4.3 Laurent series A series is a natural extension of a Taylor series. The Taylor series takesP the form PLaurent ∞ ∞ aj 1 j a z and converges when z < R. If we define t = , we obtain a series j=0 j j=0 tj z P ∞ which converges for t > R1 . If we are given a sum j=−∞ cj z j , it would be natural to split it: ∞ ∞ −1 X X X cj z j = cj z j + c0 + cj z j , j=−∞
j=−∞
j=1
for which the region of convergence could be expected to be a circular annulus centered at the origin, with a singularity on each of its bounding circles; cf. Figure 4.5. Theorem 4.23. If f (z) is analytic in an annulus, then there is a Laurent series corresponding to it, converging everywhere inside the annulus.
4.3. Laurent series
107
Im z
z2 z1
I
II
III
IV
z0
z3
Re z
Figure 4.5. Regions of validity for the Taylor (I) and Laurent (II, III, and IV) expansions centered at z0 with singular points z1 , z2 , and z3 .
Proof. Consider the annulus shaded in gray in Figure 4.6, and the integration path C = C1 + C2 , oriented as shown, and with the radial segments adjacent to each other, i.e., with canceling integrals. Let z be a point inside the annulus, and choose radii r1 and r2 such that R1 < r1 < z and R2 > r2 > z. By Cauchy’s integral theorem, ˆ 1 f (ξ) f (z) = dξ 2πi C ξ − z ˆ ˆ f (ξ) 1 f (ξ) 1 dξ + dξ. = 2πi C1 ξ − z 2πi C2 ξ − z For the first integral, z > ξ, and 1 1 =− ξ−z z
1 1−
! ξ z
1 ξ ξ2 = − − 2 − 3 − ··· z z z
converges. Likewise, for the second integral, ξ > z, and 1 1 = ξ−z ξ
1 1−
! z ξ
=
z 1 z2 + 2 + 3 + ··· ξ ξ ξ
converges. Substituting these into (4.11) gives f (z) = −
ˆ ˆ −1 +∞ X X 1 f (ξ) 1 f (ξ) n dξ z + dξ zn. n−1 n−1 2πi ξ 2πi ξ C1 C2 n=−∞ n=0
(4.11)
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Chapter 4. Introduction to Complex Integration
Im z R2 r2 C1
C2
r1
R1
Re z z
Figure 4.6. Paths used in the derivation of the Laurent series theorem (Theorem 4.23).
Here C1 was followed in the negative (clockwise) direction. If we follow both contours in the positive direction, no further change occurs if we change both paths to become a single path going around once counterclockwise within the annulus. We rename it C.
4.3.1 Summary of Taylor versus Laurent series around a point z0 Taylor series
A Taylor series is valid out to the nearest singularity ∞ X
1 f (z) = an (z − z0 ) with an = 2πi n=0
ˆ
n
C
f (ξ) f (n) (z0 ) dξ = (ξ − z0 )n+1 n!
(4.12)
for any path C going around z0 once counterclockwise, inside the region of analyticity. Laurent series
A Laurent series is valid in an annulus between singularities. Each annulus is associated with a separate Laurent expansion: ∞ X
1 f (z) = an (z − z0 ) with (again) an = 2πi n=−∞
ˆ
n
C
f (ξ) dξ, (ξ − z0 )n+1
(4.13)
now for −∞ < n < +∞. The contour C loops around z0 once, counterclockwise, within the annulus in question. In this Laurent case, there is no longer any counterpart to the (n) closedform expression an = f n!(z0 ) from the Taylor case.
4.3. Laurent series
109
Example 4.24. Give the three Taylor/Laurent expansions centered at the origin for the following function: 1 f (z) = . (z − 1)(z − 2) Since there are two distinct poles, the three expansions will be as follows (corresponding to the regions marked by I, II, and III in Figure 4.7): 1. A Taylor series converging inside the disk of radius 1. 2. A Laurent series converging in the annulus between the circles of radius 1 and radius 2. 3. A Laurent series converging outside the circle of radius 2. 1 = The partial fraction expansion f (z) = (z−1)(z−2) to manipulating geometric progressions:
−1 (z−1)
+
1 (z−2)
simplifies the algebra
1. The first expansion is a Taylor series converging in a disk centered P∞at the origin 1 with radius 1. The first function (1−z) can be directly expanded as k=0 z k , which 1 converges for z < 1 as required. The second term, (z−2) , can be rewritten as P∞ z k −1 1 1 . k=0 2 (z−2) = 2 (1− z2 ) . When z < 2, this term can be expressed as Adding the expansions for both terms gives ∞ X 1− f (z) = k=0
1
2(k+1)
zk .
2. The second expansion is a Laurent series converging in an annulus centered at the origin, between the circles of radii 1 and 2. The first term of the function needs 1 to be expanded outside of the unit circle. We thus write it as (1−z) = z1 (1−1 1 ) . z The expansion that we found above for the second term is still valid in the present annulus. The resulting series becomes ∞ X
f (z) =
ak z k ,
k=−∞
where
ak =
−1 1 2k+1
, ,
k ≤ −1, k ≥ 0.
3. Finally, the third expansion is a Laurent series converging outside the circle of radius 2. The first term of the function can be expanded as above. The second term can be 1 rewritten as (z−2) = z1 (1−1 2 ) to obtain a convergence in the requested region. The z resulting series is k+1 ∞ X 1 k f (z) = (2 − 1) . z k=0
z− z1
Example 4.25. Expand f (z) = e
around z = 0.
110
Chapter 4. Introduction to Complex Integration
Im z
III II I 1
2
Re z
Figure 4.7. Regions of validity for the expansions in Example 4.24.
This function has essential singularities at z = 0 and at z = ∞. We can thus find a Laurent expansion that will converge between them (thus for all finite z except at z = 0). 2 3 1 1 Direct Taylor expansion f (z) = ez− z = 1 + z − z1 + 21 z − z1 + 3! z − z1 + · · · suggests that, when expanding out the different powers in the sum, we will end up with a Laurenttype expansion. It is in this case somewhat tricky to obtain the Laurent coefficients explicitly. However, P∞ zr P∞ (−1)s z−s 1 1 one can proceed as follows: f (z) = ez− z = ez · e− z = = r=0 r! s=0 s! P∞ P∞ (−1)s zr−s . If we now set r = n + s, i.e., r − s = n, the double sum can r=0 s=0 r!s! s P∞ P∞ be rearranged into n=−∞ cn z n , where cn = s=0 s!(−1) (n+s)! . This last sum turns out to appear also in the context of Bessel functions (cf. Section 11.2) and can in terms of these be given in closed form as cn = Jn (2).
4.3.2 Revisiting singularities We have now better tools available to characterize singularities. The lowest index in a Laurent sum, centered at a singularity, will give us information about the type of singularity with which we are faced. Let’s revisit the first four types: 1. Removable singularities: We saw earlier that a removable singularity is only an artifact of the function representation. Any trace of this “singularity” completely disappears once the function has been expanded into a power series centered at that point.45 Since the singularity has vanished, the resulting expansion around this point is strictly a Taylor series. 2. Poles: A function exhibiting an N thorder pole at a point z0 can be expanded into a 45 Or
centered anywhere, but with the point included within the radius of convergence.
4.3. Laurent series
111
Laurent series centered at z0 . It will have the following form: ∞ X
cj (z − z0 )j .
j=−N
We call c−N the pole’s strength, the coefficient associated with the lowest index (−N ). Another coefficient of interest is c−1 , which we call the residue. We will in the following often denote the residue of a function f (z) at z = z0 as Res(f (z), z = z0 ), or more briefly, Res(f (z), z0 ). 3. Essential singularities: An essential singularity is characterized by a Laurent series for which there are cj 6= 0 for arbitrarily large negative indices j. The Laurent series will thus take the form ∞ X cj z j . j=−∞
Once again, we shall soon give particular attention to the residue c−1 . 4. Branch points: Like a Taylor series, a Laurent series can only produce singlevalued results. In the Laurent case, there may, however, be branch points inside the inner circle, as long as these cancel in the sense that the result becomes singlevalued in the annulus of convergence. Example 4.26. Describe the possible Taylor and Laurent expansions around the origin for the function f (z) = (z 2 − 1)1/2 . This function was illustrated in Figures 2.19 and 2.20. The branch points are located at z = ±1. If we are interested in Taylor expanding around z = 0, we orient the branch cuts so they become directed outwards, for example, [−∞, −1] and [+1, +∞]. Each of the two solution sheets now has its own Taylor expansion, i.e. (by the binomial expansion s(s−1)(s−2) 3 2 (1 + z)s = 1 + sz + s(s−1) s + · · · ), 2! z + 3! f (z) = ±i 1 − z 2
1/2
= ±i
1 1 1 1 − z2 − z4 − z6 − · · · . 2 8 16
Regarding Laurent expansions for z > 1, we need to instead place the branch cut between +1 and −1 inside the unit circle (which is possible, since z = ∞ is not a branch point). There are then again two solution sheets, each with a separate Laurent expansion. We write 1/2 now f (z) instead as f (z) = ±z 1 − z12 , and the same binomial expansion formula gives similarly
1 f (z) = ±z 1 − 2 z
1/2
= ±z
1 1 1 − ··· 1− 2 − 4 − 2z 8z 16z 6
.
As can be inferred from Figure 2.20, there is no onetoone correspondence between the Taylor and the Laurent versions. The Taylor expansion holds within the unit circle, and assumes the branch cut is away from this region, as in parts (c) and (d) of the figure, whereas the Laurent expansions assume regularity outside of the unit circle, as depicted in parts (a) and (b) of the figure.
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Chapter 4. Introduction to Complex Integration
Example 4.27. Describe singularities of f (z) =
(z−1)2 z(z+1)3
.
There are two pole singularities (at z = 0 and at z = −1): • Consider the Laurent series centered at z = 0. The second factor 2 1 (z−1) z (z+1)3
(z−1)2 (z+1)3 2
in f (z) =
is analytic at z = 0 and can be expanded as 1 − 5z + 13z − · · · . The coefficient c−1 = 1 is both the strength of the pole at z = 0, of order 1, and the residue. • Consider next the Laurent series centered at z = −1. We now write f (z) as f (z) = 2 (z−1)2 1 . The function (z−1) is analytic around z = −1 and thus has a (z+1)3 z z 2
= −4 − (z + 1)2 − (z + 1)3 − · · · . Thus the Taylor series centered there: (z−1) z function exhibits a third order pole, with a strength of −4. Example 4.28. Describe the singularities of f (z) =
z 1/2 −1 z−1 .
Due to the z 1/2 term in the numerator, there will be a branch point at z = 0, giving rise to two solution sheets. At z = 1, there occurs a divide by zero. On one solution sheet, 11/2 = +1 and the singularity becomes removable, whereas on the other sheet 11/2 = −1 and we get a first order pole with residue −2. Figure 4.8 displays this function. In part (a), we see the real parts of the two solutions coinciding for x < 0 and the imaginary parts coinciding for x > 0. For the real part, we see the pole at z = 1 affecting only one of the two solution branches. Part (b) illustrates Im f (z) in the complex plane, showing again how one solution sheet is unaffected by the pole that is present on the other one. For the examples we have shown previously of multivalued functions, there were great similarities between the solutions on different sheets. This latest example illustrates that this by no means needs to be the case. The following is an interesting result for the character of an analytic function near an essential singularity. Theorem 4.29 (Picard). Take any arbitrarily small neighborhood of an essential singularity. The function f (z) then takes every complex value with at most one exception infinitely many times within this neighborhood. The following theorem is slightly weaker, but much easier to prove; see Section 4.4. Theorem 4.30 (Caserati–Weierstrass). Take any arbitrarily small neighborhood of an essential singularity. The function f (z) then comes arbitrarily close to every complex value infinitely many times within this neighborhood.
4.4 Supplementary materials Proof of Poisson’s integral formula: Theorem 4.20. Note: This proof below is somewhat “tricky,” but it relies only on contour integration. By means of a Fourier series expansion (Section 9.1.1), the formula can be proven more directly; see Exercise 9.8.6.
4.4. Supplementary materials
113
3 2 1 0 1 2 3 3
2
1
0
1
2
3
1
2
3
x 1 0.5 0 0.5 1 3
2
1
0
x
(a) Re f (x) (top) and Im f (x) (bottom), both along the xaxis.
(b) Im f (z) in the complex plane near the origin. 1/2
Figure 4.8. The function f (z) = z z−1−1 , displayed along the real axis, and its imaginary part in the complex plane (with a part of the surface near the pole cut out to better display the polefree solution sheet).
Denoting the evaluation point z = r eiθ , and changing variable ξ = eit (i.e., dξ = i e dt), equation (4.6) becomes it
f (r eiθ ) =
1 2π
ˆ
2π
f (eit ) 0
eit eit − r eiθ
dt.
(4.14)
We next separate the “kernel” inside the large parentheses into its real and imaginary parts:
eit
eit [1 − r cos(t − θ)] − i [r sin(t − θ)] = . − r eiθ 1 + r2 − 2r cos(t − θ)
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Chapter 4. Introduction to Complex Integration
We can now spot that changing r → 1r leaves the imaginary part unchanged. Furthermore, since r < 1, it will hold that 1r > 1, and (4.14) becomes after this change r → 1r a closed loop integration without any singularity inside the path, implying that it evaluates to zero. Subtracting this integral from (4.14) eliminates the imaginary parts and gives
f (r eiθ ) − 0 =
1 2π
ˆ
1 cos(t − θ)] [1 − r cos(t − θ)] r f (eit ) − dt . 1 1 1 + r2 − 2r cos(t − θ) 1 + 2 − 2 cos(t − θ) r r
2π
0
[1 −
The proof is now finished by simplifying the expression inside the large parentheses to 1−r 2 1+r 2 −2r cos(t−θ) . The Caserati–Weierstrass theorem (Theorem 4.30) stated that if a function f (z) has an essential singularity at a point z0 , then it comes arbitrarily close to every complex value infinitely many times within this neighborhood. Proof. This is a proof by contradiction: Say that f (z) never comes within ε of some value 1 is then bounded in magnitude by f0 whenever z − z0  < δ. The function g(z) = f (z)−f 0 ´ g(ξ) 1 1 ε in this neighborhood. By (4.13) its Laurent coefficients are an = 2πi C (ξ−z0 )n+1 dξ, so by Theorem 4.9, 1 −n 1 an  < δ 2π . 2π ε With ε > 0 fixed and freedom to choose δ arbitrarily small, this tells us that an = 0 for all 1 can then negative n, i.e., z = z0 is a regular point for g(z). The function f (z) = f0 + g(z) have at most a pole at z = z0 , contradicting that this was an essential singularity.
4.5 Exercises ¸ Exercise 4.5.1. Let Γ be the unit square, with corners at ±1 ± i. Evaluate Γ z 2 dz in the following three ways: (a) Give the answer directly by referring to some general theorem. (b) Evaluate explicitly the line integrals along the four sides, and add together the results. (c) Use Green’s theorem to convert the contour integral over to a double integral, and then work out this double integral (i.e., do NOT refer to the CR equations). ¸ Exercise 4.5.2. Let Γ again be the unit square, with corners at ±1 ± i. Evaluate Γ z¯2 dz in the following two ways: (a) Evaluate explicitly the line integrals along the four sides, and add together the results. (b) Use Green’s theorem to convert the contour integral over to a double integral, and then work out this double integral. Exercise 4.5.3. Evaluate
´ Γ
sin z dz, where Γ goes from i to π.
4.5. Exercises
115
Exercise 4.5.4. Let f (z) be an entire function with f (z) ≤ Az for all z ∈ C, where A is a constant. Show that f (z) = B z, where B is a constant. Exercise 4.5.5. Show that the roots of a polynomial are continuous functions of its coefficients. Hint: Assume that a root jumps discontinuously. Consider the result of Example 4.7, choose a suitable contour Γ, and obtain a contradiction. 2
Exercise 4.5.6. It is obvious from the form of f (z) = (z−1) (z−i)3 that this function has N = 2 zeros and P = 3 poles. Nevertheless, show that N −P = −1 by directly applying Cauchy’s argument principle (Theorem 4.8) to f (z). Exercise 4.5.7. Prove the fundamental theorem of algebra directly from Theorem 4.8. Let f (z) = pn (z) be a polynomial of degree n ≥ 1 and choose as the integration path a circle z = R where R is very large. Note: Since the task is to prove that pn (z) has root(s), your argument must not assume that pn (z) can be factorized. Exercise 4.5.8. Prove Cauchy’s estimate: If f (z) is differentiable inside Γ, if the circle z − z0  = R, and if f (ζ) < M with ζ − z0  = R, then f (k) (z0 ) ≤ k!M for k = Rk 0, 1, 2, . . . . Exercise 4.5.9. Prove that the only entire functions f (z) that satisfy f (z) < 1 + z1/2 are constants. Exercise 4.5.10. In Example 2.15, we Taylor expanded f (z) = 3−4z 1+2z around the origin and obtained (using undetermined coefficients) the leading coefficients a0 = 3, a1 = −10 and then the recursion an+2 = −2an+1 , n = 0, 1, 2, . . . . Derive exactly the same information f (z) 1 instead from applying (4.12), written as an = 2πi dz. Γ z n+1 1 Exercise 4.5.11. Consider the function f (z) = z(z−2)(z 2 +1) : (a) Give a closedform expression (as simplified as you can) for the Laurent coefficients when f (z) is expanded in a form valid near the origin. (b) Give the coefficients c−1 , c0 , c1 in a Laurent expansion for f (z) around the origin, valid far out from the origin. (c) Give the similar three coefficients c−1 , c0 , c1 for a Laurent expansion valid in the vicinity of z = i. Figure 4.9 illustrates the function f (z) and the regions of validity for the three expansions.
Exercise 4.5.12. Describe all the singularities (none, removable, pole (if so, what order?), essential, etc.) in the finite plane (i.e., no need to consider z = ∞) of the following functions: z)−1 (a) f (z) = (cos sin z . (b)
f (z) =
z 1/2 +1 z+1 . z 1/2 −1 z−1 .
(c) f (z) = Hint for (b) and (c): Recall that, for z 6= 0, the quantity z 1/2 can take two different values.
116
Chapter 4. Introduction to Complex Integration
(a) Original function from Exercise 4.5.11.
(b) Validity of the expansion in problem part (a).
(c) Validity of the expansion in problem part (b).
(d) Validity of the expansion in problem part (c).
1 Figure 4.9. The function f (z) = z(z−2)(z 2 +1) in Exercise 4.5.11, and the regions of validity (colored) for the Laurent expansions in the three cases (with figures produced using 30 terms in the respective expansions).
Exercise 4.5.13. Identify the types of all singularities to the following functions, and give for these (if applicable) order, strength, and residue: (a) f (z) = 1/ 1 − z1 . 2 (b) f (z) = 1/ 1 − z12 . (c) f (z) = 1/ sin(ez ) . (d) f (z) = (tan z)1/2 . Exercise 4.5.14. Consider the function f (z) = (z + 1)1/2 (z − 1)1/2 . Does this function have zero, one, or two Taylor expansions around the origin? If any exist, give the first three nonzero coefficients. 2 s(s−1)(s−2) 3 Hints: (i) Recall the binomial expansion (1+z)s = 1+sz+ s(s−1) z +· · · , 2! z + 3! and (ii) compare with Figures 2.19 and 2.20.
4.5. Exercises
117 Im z
Im z
br anch cut
br anch point
Re z
Re z br anch point
br anch point
(a) Branch cuts
(b) Considered integration path.
Figure 4.10. Branch cuts and integration paths considered in Exercise 4.5.15.
Exercise 4.5.15. Consider the function f (z) = ((z + 1)(z − 1)(z − i))1/3 . It is proposed to make this function singlevalued by means of making a branch cut as shown in Figure 4.10. Determine whether this branch cut is sufficient by means of the following two approaches: (a) Inspect whether z = ∞ is a branch point (by making the change of variable z = 1/t). (b) Follow the function around the path outlined in Figure 4.10(b) and see whether we get back to the same function value as we started with. Exercise 4.5.16. Let t be a parameter, and consider the analytic function 1
f (z, t) = et(z− z )/2 .
(4.15)
(a) Tell where in the complex zplane f (z, t) has singularities. Comment: The Laurent expansion of f (z, t) (in terms of z) around the origin will naturally have coefficients that depend on the parameter t: 1
et(z− z )/2 =
∞ X
Jn (t)z n .
(4.16)
n=−∞
These coefficients are known as the JBessel functions, and they are of great importance in mathematical physics, partly because y(t) = J±n (t) provide solutions to the differential equation dy d2 y + (t2 − n2 )y = 0, (4.17) t2 2 + t dt dt which has a great tendency to arise in many situations, one being separation of variables in polar coordinates. (b) Show that y(t) = J±n (t) indeed satisfy the differential equation (4.17). Hint: By direct differentiation, show that f (z, t) satisfies 2 ∂f ∂ ∂f 2 2∂ f t +t +t f =z z . ∂t2 ∂t ∂z ∂z Then use the fact that the coefficients in a Laurent expansion are uniquely determined.
118
Chapter 4. Introduction to Complex Integration
(c) From the integral formula for Laurent coefficients, show that ˆ π ˆ 1 1 π −i(nθ−t sin θ) Jn (t) = e dθ = cos(nθ − t sin θ)dθ. 2π −π π 0 Comment: It is possible to extract much more information about the JBessel functions from (4.15) than what is illustrated in this example; see Sections 11.2.1 and 11.2.2. For example, Taylor expanding the exponential on the LHS of (4.16) quickly leads to the Taylor expansions of the Jn (t) functions, showing that they are entire functions of t (i.e., analytic, and with no singularity in the finite complex tplane). Exercise 4.5.17. If f (z) is analytic and has no zeros for z ≤ R, show that 1 log f (0) = 2π
ˆ
2π
log f (R eiθ )dθ.
(4.18)
0
Note: This result is generalized in Exercise 5.6.3 to the case where f (z) has zeros inside the circle (Jensen’s formula). Exercise 4.5.18. Verify the following alternate way to express Poisson’s integral formula (4.9) (with z = r eiθ , z ≤ 1): f (z) =
1 2π
ˆ
2π
f (eit ) 0
1 − z2 dt. eit − z2
Exercise 4.5.19. In Poisson’s integral formula (4.9), f (z) is specified around the periphery of the unit circle, and the harmonic function satisfying this inside the circle is obtained. If we instead specify f (z) along the real axis, a counterpart formula giving the harmonic extension to the upper halfplane is ˆ ∞ 1 y f (x + i y) = f (t)Py (x − t)dt, where Py (x) = . (4.19) 2 π x + y2 −∞ (a) Verify that (4.19) is correct. (b) With use of (4.19), derive an explicit form for a harmonic function over the upper halfplane that, along the real axis, takes the values 1 if − 1 ≤ x ≤ 1, f (x) = 0 otherwise. (c) Solve the same problem as in (b), but instead by considering Figure 2.8.
Chapter 5
Residue Calculus
The previous chapter described how contour integrals of analytic functions become path independent (as long as paths do not move across singularities). From this followed a number of important theorems, as well as an improved understanding of Taylor (and Laurent) expansions. This chapter follows up on this by introducing residue calculus, which can be used to evaluate a wide variety of definite integrals (many of which, at first thought, may not appear to have anything to do with complex variables). Applications of residue calculus also include evaluating many infinite sums in closed form, additional opportunities for analytic continuation, etc.
5.1 Residue calculus Let z1 , z2 , z3 , . . . , zn be the singularities (poles or essential singularities) of a function f (z) inside contour Γ, as shown in Figure 5.1. Choosing as integration path Γ, furthermore going around each singularity in the direction shown by arrows, and including cancelling integrations up/down each vertical line segment results in an overall path that is closed and encloses no singularity. Therefore ˛
ˆ
ˆ
+ Γ
ˆ
+ z1
ˆ +··· +
+ z2
z3
= 0.
(5.1)
zn
Changing direction when going around the singularities changes the sign of their contribution, giving ˛ ˆ ˆ ˆ ˆ = + + +··· + . (5.2) Γ
z1
z2
z3
zn
We can thus change the original contour Γ to just go around each singular point separately, all in the same positive direction. Recall that 1 2πi
˛ Γ
0 dz 0 = (z − z0 )n 1
if z0 is outside Γ, if z0 is inside Γ and n 6= 1, if z0 is inside Γ and n = 1 119
120
Chapter 5. Residue Calculus
z2 z3 z1 z4 z5 Γ Figure 5.1. Change of contour in residue calculus.
for n integer. Locally to each singularity, there is a Laurent expansion f (z) =
∞ X
cn (z − zj )n .
n=−∞
Only the term associated with n = −1, i.e., the residue c−1 , will contribute to the integral. Hence, we have the following. Theorem 5.1. ˛
N X
f (z)dz = 2πi Γ
c−1
j=1
{z

zj
}
Sum of the residues at the zj
= 2πi
N X
j=1
Res(f, zj ) ,
where the zj are all the singularities inside Γ.
5.1.1 Shortcuts to calculate residues We wish to compute the residue Res(f (z), z0 ). However, doing this by means of the Laurent expansion around z = z0 may require more work than necessary. In case the singularity is a pole of order m, the following gives three alternative options: N (z) D(z) , where N (z0 ) 6= 0 and D(z) has a simple root N (z0 ) D 0 (z0 ) . Note that what one decides to include in N (z)
1. In case of m = 1, write f (z) =
at z0 . Then, Res(f (z), z0 ) = and in D(z) is quite arbitrary; making D(z) as simple as possible will make it easier to take the derivative. Justification: Taylor expanding around z0 gives f (z) = N (z0 ) D 0 (z0 )
·
1 (z−z0 )
+ O(1).
N (z0 )+N 0 (z0 )(z−z0 )+··· 0+D 0 (z0 )(z−z0 )+···
=
5.1. Residue calculus
121
2. If m is arbitrary and f (z) can conveniently be written in the form f (z) = then Res(f (z), z0 ) =
φ
φ(z) (z−z0 )m ,
(m−1)
(z0 ) (m−1)! .
Justification: Expanding φ(z) in a Taylor series centered at z0 gives φ(z) = φ(z0 ) + 2 0) (z − z0 )φ0 (z0 ) + (z−z φ00 (z0 ) + · · · , and therefore 2! φ(z0 ) φ0 (z0 ) φ00 (z0 ) φ(m−1) (z0 ) φ(z) +··· = + + + · · · + (z − z0 )m (z − z0 )m (z − z0 )m−1 2!(z − z0 )m−2 (m − 1)!(z − z0 )
with the last displayed term above giving the result. 3. If m is arbitrary and the rewrite described above is not convenient, we can instead 1 dm−1 m use Res(f (z), z0 ) = (m−1)! limz→z0 dz f (z)), where the limit usum−1 ((z − z0 ) ally is best handled by l’Hôpital’s rule. Justification: Around the pole, f (z) =
c−m+1 c−1 c−m + + ... + + ··· , (z − z0 )m (z − z0 )m−1 (z − z0 )1
and therefore (z − z0 )m f (z) = c−m + c−m+1 (z − z0 )1 + . . . + c−1 (z − z0 )m−1 + · · · and
dm−1 ((z − z0 )m f (z)) = (m − 1)! c−1 + m! c0 (z − z0 ) + · · · , dz m−1 from which the result follows.
5.1.2 Examples Example 5.2. Evaluate
˛ ze1/z dz.
I= z=1
The only singularity inside the unit circle is the essential singularity at z = 0. We thus expand the integrand at that point to find the function’s residue there: 1 1 1/z ze =z 1+ + + ··· z 2!z 2 1 + ··· . =z+1+ 2!z Thus, Res(ze1/z , z = 0) = 1/2, and ˛ 1 I= ze1/z dz = 2πi · = πi . 2 z=1 Figure 5.2 shows that the integrand is of extreme complexity around the singularity point. It is remarkable how that did not at all complicate the calculation of the integral going around it—all that mattered was the value of the residue.
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Chapter 5. Residue Calculus
2 0 2
0.2 0.1 0.2
0 0.1 0.1
0 0.1
0.2 0.2
y
0.3
x
0.3
(a) Re z e1/z
(b) z e1/z , with phase angle coloring Figure 5.2. Illustrations of the integrand in Example 5.2, f (z) = z e1/z , near the origin.
Example 5.3. Compute Res
ez z cos z , z
=0 .
We can let N (z) = ez / cos z and D(z) = z. Then Res
ez / cos z ,z z
=0 =
1. Alternatively, we can choose N (z) = ez and D(z) = z cos z, which gives 0 z = 0 = cose0−0 = 1. Example 5.4. Evaluate ˆ
∞
I= −∞
x2 dx. 1 + x4
e0 / cos 0 = 1 ez Res z cos z ,
5.1. Residue calculus
123
Im z
Γ R z2
z1
Re z z4
z3
Figure 5.3. Contour in Example 5.4. 2
z 4 To compute this integral with residue calculus, we consider f (z) = 1+z 4 . Since z + 1 1+i −1+i 1−i −1−i has four simple zeros, z1 = √2 , z2 = √2 , z3 = √2 , z4 = √2 , these are simple poles of f (z).46 Consider the open contour Γ, a semicircle of radius R, as shown in Figure 5.3. ´ z2 ≤ M · L, where L is the length Recall that the magnitude of the integral Γ 1+z 4 dz of the contour (πR) and where M is the upper bound of the function on the contour. In order to find this bound, recall the following triangle inequalities: z1 + z2  ≤ z1  + z2 , z2 2 ≤ R z1 + z2  ≥ z1  − z2 , but also z1 + z2  ≥ z2  − z1 . Therefore, 1+z 4 R4 −1 . Thus, ´ z2 ´ 3 z2 dz ≤ RπR dz → 0 as R increases. The 4 −1 → 0 as R → ∞. Therefore, Γ 1+z 4 Γ 1+z 4 contour Γ was open. Next, consider the closed contour, say Λ, which is made of the contour Γ followed by the section of the real axis from −R to R. Since two poles are located inside of this new contour,
˛ Λ
ˆ R z2 x2 dz + dx 4 4 Γ 1+z −R 1 + x ˆ ∞ x2 =0+ dx as R → ∞ 4 −∞ 1 + x = 2πi{Res(f, z1 ) + Res(f, z2 )}.
z2 dz = 1 + z4
ˆ
Using the formulas above, we next determine Res(f, zk ) =
N (zk ) D 0 (zk ) , 1 4z1 =
k = 1, 2, where √ N (z) = z and D(z) = z + 1. This gives Res(f, z1 ) = 2(1 + i)/8 and √ 1 Res(f, z2 ) = 4z2 = 2(−1 − i)/8. Thus, 2πi{Res(f, z1 ) + Res(f, z2 )} = √π2 , and 2
4
ˆ
∞
−∞
x2 π dx = √ . 4 1+x 2
46 If one tries to evaluate the integral using partial fractions (rather than residue calculus), the first step towards x2 1 x √ √ arriving at 1+x − 2 √x requires a factorization of x4 + 1, which is easiest done 4 = 2 2 2
x − 2x+1
x + 2x+1
by complex arithmetic, writing z 4 + 1 as [(z − z1 )(z − z3 )] [(z − z2 )(z − z4 )].
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Chapter 5. Residue Calculus
´∞ Note: When the integral to compute is of the type −∞ f (x)dx, the open halfcircle contour Γ shown above will often ´be used, with R → ∞. We showed that, for our particular integrand in Example 5.4, Γ f (z)dz → 0. This will always be true in the case where both the numerator and the denominator are polynomials, and the degree of the denominator is at least two more than the degree of the numerator. Example 5.5. Evaluate ˆ
∞
I= −∞
cos kx dx (x + b)2 + a2
with k, a > 0 and b ∈ R. cos kz The integrand f (z) = (z+b) 2 +a2 has two simple poles, located at z1 = ia − b and z2 = −ia − b. We would like to bound the magnitude of the integral as we did in the previous example. However, cos kz oscillates with amplitudes that grow exponentially fast once we consider its values far away from the real axis (as follows from cos kz = eikz +e−ikz ; cf. Figure 2.5). In order to bypass this issue, let’s instead consider the function 2 eikz f (z) = (z+b) 2 +a2 . In this case,
ˆ
∞
−∞
eikx dx = (x + b)2 + a2
ˆ
∞
−∞

ˆ ∞ cos kx sin kx dx +i dx . (x + b)2 + a2 (x + b)2 + a2 −∞ {z }  {z } Imaginary part
Real part
´∞ eikx Thus, we can evaluate −∞ (x+b) 2 +a2 dx and retain the real part only as the answer. As for the boundedness of the integrand, eikz = e−ky eikx , which implies that eikz  = e−ky → 0 as y → ∞ (had k been negative, we could similarly have used a contour in the lower halfplane). Let x = R cos θ and y = R sin θ be the coordinates of z = x + iy, a point on the open semicircle contour Γ of radius R as illustrated in Figure 5.4. Let’s next find an upper
Im z
ia − b
Γ
R
Re z −ia − b Figure 5.4. Contour in Example 5.5.
5.1. Residue calculus
125
bound for the magnitude of the integral. We have ˆ eikz eikz (z + b)2 + a2 dz ≤ (z + b)2 + a2 (πR) Γ e−Rk sin θ (πR) (z + b)2 + a2  πe−Rk sin θ ≤ a 2 R (cos θ + i sin θ + Rb )2  −  R  → 0 as R → ∞. =
To obtain a closed contour (so that the residue theorem can be applied), consider the contour Λ, which is made of the halfcircle Γ and which is closed by also including the real axis from −R to R (and letting R → ∞): ˆ ˆ R ˛ eikz eikx eikz dz = dz + dx 2 2 2 2 2 2 Γ (z + b) + a −R (x + b) + a Λ (z + b) + a ˆ ∞ eikx =0+ dx 2 2 −∞ (x + b) + a = 2πi Res(f, ia − b). Calculating the residue at z = ia − b, we obtain Res(f, ia − b) =
e−ka (−i cos kb − sin kb) 2a
(here using the assumption that a > 0; for a < 0 the relevant pole would have been the one at z = −ia − b). Thus, ˆ ∞ πe−ka (cos kb − i sin kb) eikx dx = . 2 2 a −∞ (x + b) + a Finally, by only retaining the real part, we obtain the result ˆ ∞ cos kx πe−ka cos kb I= dx = . 2 2 a −∞ (x + b) + a As a concluding observation, we can note from the form of the original integral that the answer must be an even function in both a and k. We can therefore drop the requirements that these parameters are positive and give the final answer as ˆ ∞ cos kx πe−ka cos kb dx = I= 2 2 a −∞ (x + b) + a for a, b, k all real (with divergence if a = 0). ´∞ Integrals of the type −∞ f (x)eikx dx are very common. The semicircle contour Γ is in such cases. The following result gives an easy way to ensure that ´ used routinely ikz f (z)e dz → 0 as R → ∞. Γ Theorem 5.6 (Jordan’s lemma). If f (z) → 0´uniformly as R → ∞, Γ is a halfcircle as shown in Figures 5.3 and 5.4, and k > 0, then Γ f (z)eikz dz → 0.
126
Chapter 5. Residue Calculus
Example 5.7. Evaluate
ˆ
2π
I= 0
dθ , A + B sin θ
where A > 0 and where B < A (in order to avoid divisions by zero on the integration path). We often encounter the more general form trates a couple of general approaches.
´ 2π 0
g(sin θ, cos θ)dθ. The following illus
Method 1: Let
1 . A + B sin z Considering the contour sketched in Figure 5.5(a), we find that for the integrand f (z) = f (z) =
1 A+B sin z
ˆ f (z)dz → 0, Γ3
since sin z grows exponentially we go up the complex plane. We also notice ´ fast the further ´ that f (z) is 2πperiodic, so Γ2 f (z)dz + Γ4 f (z)dz = 0. The three extra paths Γ1 , Γ2 , Γ3 will therefore together not contribute anything to the integral (in the limit of Γ3 shifted up towards +i∞). The next step is to find the singularities inside the contour, illustrated again together with f (z) in Figure 5.5(b). The single pole inside the contour occurs where A A + B sin z0 = 0, i.e., at z0 = − arcsin B . The residue becomes N (z0 ) 1 1 p Res(f, z0 ) = 0 = = D (z0 ) B cos z z=− arcsin A B 1 − (sin z)2 A z=− arcsin
B
B
1
= √ . i A2 − B 2 Therefore,
2π I=√ . 2 A − B2 Method 2: The second “routine” approach for trigonometric integrals over a full period is to let z = eiθ . Then 1 iθ 1 1 cos θ = e + e−iθ = z+ , 2 2 z 1 iθ 1 1 sin θ = e − e−iθ = z− , 2i 2i z
and
1 dz . iz For 0 ≤ θ ≤ 2π (or equivalently −π ≤ θ ≤ π), the path for z goes once around the unit circle in the positive direction. With this variable change, ˛ 1 dz I= 1 1 iz z=1 A + B 2i z − z ˛ 2 dz = . A B z=1 z 2 + 2i B z−1 dθ =
5.1. Residue calculus
127
Im z Γ3
Γ4
Γ2
0
Γ1
2π Re z
(a) Schematic integration path
(b) Magnitude/phase plot with path in case A = 3, B = 1. Figure 5.5. Contour and function in Example 5.7. A The denominator factorizes as z 2 + 2i B z − 1 = (z − z1 )(z − z2 ) with
z1,2 = i
−A ±
√ A2 − B 2 . B √
2
2
Since A > B, the second root (pole) z2 = i −A− BA −B will fall outside the unit circle, and from z1 · z2 = −1 it follows then that z1 falls inside it. From 1 1 Res , z = z1 = , (z − z1 )(z − z2 ) z1 − z2
128
Chapter 5. Residue Calculus
it follows then that the relevant residue evaluates to I=
2i
√B A2 −B 2
and, with that,
2π 2 B =√ . (2πi) √ B 2i A2 − B 2 A2 − B 2
Example 5.8. Evaluate ˆ π/2 dθ , a, b > 0, for n = 0, 1, 2, . . . . In (a, b) = 2 (a cos θ + b sin2 θ)n+1 0 A common technique to manipulate integrals is to differentiate with respect to its parameters, to see whether anything simpler happens to arise (sometimes known as “Feynman’s trick,” although used since the beginning days of calculus; one might even introduce artificial parameters if the integral does not contain any). In the present case, we will stumble onto the recursion relation ∂ ∂ 1 + In (a, b) = − In+1 (a, b), ∂a ∂b n+1 ´ π/2 dθ showing that it is sufficient to evaluate I0 (a, b) = 0 a cos2 θ+b . The routine varisin2 θ iθ able change z = e just introduced (Example 5.7, Method 2) gives, after a few simplifi¸ i z dz cations, I0 (a, b) = − a−b , with the integration path once around the unit a+b z 4 +2 a−b z 2 +1 ¸ dt i 2 circle. Changing variable t = z , dt = 2z dz gives next I0 (a, b) = − a−b t2 +2 a+b t+1 a−b
(where the factor 2 in dt = 2z dz has canceled against making t go around the unit circle once rather than twice). Since the product√of the roots in the denominator polynomial √ a−√b √ . We thus obtain is 1, the only relevant root becomes t1 = − √a+ b , with residue 4a−b ab a−b i π I0 (a, b) = 2πi − a−b 4√ab = 2√ab . From the recursion relation then follows π (a + b), 4(ab)3/2 π (3a2 + 2ab + 3b2 ), I2 (a, b) = 16(ab)5/2 ··· I1 (a, b) =
With some more effort, one can from the recursion relation also deduce an expression for general n: n X π 2k 2(n − k) k n−k In (a, b) = a b . k n−k (4ab)n+1/2 k=0
Example 5.9. Evaluate ˛ I= Γ
z2
dz + a2
around the contour shown in Figure 5.6. We find that
Res
1 1 , −ia =− z 2 + a2 2ai
5.1. Residue calculus
129
Im z
ia
Re z
Γ −ia
Figure 5.6. Contour in Example 5.9.
and, similarly, Res
1 , +ia 2 z + a2
=
1 . 2ai
Since the contour goes around the pole z = ia twice, we shall count the function’s residue at that point twice and the residue at z = −ia only once. We thus find 1 1 π I = 2πi 2 − = . 2ia 2ia a The winding number w(zj ) of the curve Γ around a point is the number of times that Γ winds around zj (with positive means counterclockwise). We have in the example above w(ia) = 2 and w(−ia) = 1. Example 5.10. Evaluate ˆ I= 0
∞
x3
dx , a>0. + a3
One might again consider using as contour a circle ´ of radius R centered at the origin, since, far out in any direction in the complex plane, x f (z)dz → 0 as R → ∞. However, the interval of integration goes from 0 to ∞ and not from −∞ to ∞. Before describing a more general approach next in Example 5.11, we consider first a more specialized idea. Since z only appears as z 3 , we note that both z = r and z = re2πi/3 give the same result when raised to the third power. We thus choose the contour as in Figure 5.7 and perform the substitutions z = r and z = re2πi/3 along the respective radial path segment.
130
Chapter 5. Residue Calculus
Im z
z1 = ae iπ/3 2π/3
z2 = −a
Re z
Γ
z3 = ae5iπ/3
Figure 5.7. Contour in Example 5.10.
This gives ˆ 0
∞
dr + 3 r + a3
´ Since x obtain that
dz z 3 +a3
ˆ x
dz + 3 z + a3
ˆ
0
∞
ei2π/3 dr = 2πi Res r3 + a3
1 πi/3 , z = ae . 1 z 3 + a3
1 πi/3 → 0 as R → ∞, and since Res( z3 +a )= 3 , z1 = ae ˆ ∞ dr 2πi (1 − e2πi/3 ) = 2 2πi/3 , 3 + a3 r 3a e 0
which simplifies to
ˆ
∞
0
1 , 3a2 e2πi/3
we
dr 2π = √ . r3 + a3 3 3a2
It was in this case quite lucky that the integrand itself provided a return path from infinity back to the origin, along which it took the same values as along the real axis. As we will see next, this type of “luck” is unnecessary. The particular behavior of Im log z (recall Figure 2.7(b)) offers a more general approach for creating a suitable return path.
5.1.3 Special contours Example 5.11. Evaluate ˆ I= 0
∞
dx , a, b > 0. (x + a)(x + b)
5.1. Residue calculus
131
This example can be easily solved using partial fractions (and no complex variables), but we will ignore that for now. We will instead use an idea that can be applied for the ´∞ much more general integrals of the type 0 f (x)dx. We only need the integrand to decay fast enough for the contributions from a path distant from the origin to be insignificant. Instead of tackling the integral exactly as it stands, consider in the present case ˛ (log z)dz J= , a, b > 0, (z + a)(z + b) Γ along the keyhole contour shown in Figure 5.8.
Γ
Figure 5.8. Keyhole contour in Example 5.11.
´∞ ¸ In general, to compute 0 f (x)dx, we will consider Γ (log z)f (z)dz. Why is this a good idea? If we follow the contour, the contributions to the integral become the following: ´ ´∞ 1. → (log z)f (z)dz = 0 log(z)f (z)dz . ´ 2. " (log z)f (z)dz → 0 as outer radius R → ∞ . R ´ ´∞ ´∞ 3. ← ((log z) + 2πi)f (z)dz = − 0 (log z)f (z)dz − 2πi 0 f (z)dz . ´ 4. ! (log z)f (z)dz → 0 as inner radius → 0 .
Thus
ˆ
ˆ +
→
"R
ˆ
ˆ
+
+ ←
!
ˆ
!
∞
(log z)f (z)dz = −2πi
f (x)dx 0
= 2πi
X
Res((log z)f (z), zj )
j
and
ˆ
∞
f (x)dx = − 0
X j
Res((log z)f (z), zj ).
132
Chapter 5. Residue Calculus
In our example, we obtain I = − (Res((log z)f (z), −a) + Res((log z)f (z), −b)) = − =
log(−a) log(−b) + −a + b −b + a
log(b/a) . b−a
The extra factor log z canceled out for the integrand, but still provided us with a closed contour for the integration over the semiinfinite interval [0, +∞]. If the function we want to integrate already contains a factor of log z in the numerator, we can still multiply with an extra log z; see Exercises 5.6.19 and 5.6.20. Example 5.12. Evaluate ˆ
1
I= −1
√
1 − x2 dx. 1 + x2
Figure 5.9 shows the integrand over √ the integration interval x ∈ [−1, 1] (assuming the conventional sign choice such that 1 = +1). As usual, displaying the function over the complex plane reveals a much richer picture. The function has two branch points, at z = ±1 (but none at z = ∞), making it natural to introduce a branch cut between −1 and +1. The real and imaginary parts of the integrand f (z) then become as shown in Figure 5.10. We note in particular the (first order) poles at z = ±i. √ In order to determine the residues at the poles, we look first at the numerator 1 − z 2 (Figure 5.11). The value of this function at z = 0, just above the branch cut, is +1. Moving z straight up along the imaginary axis from 0 to i, as illustrated in green in Figure 5.11, will √ just change this +1 to + 2. Since no branch cuts were passed along this way, no sign was swapped. For the pole at z = −i, we have a choice in thinking. The value at z = 0 should this time evaluate to −1, and then √ reason √ moving z down from 0 to −i will for the same smoothly change that −1 to − 2. Alternatively, if we start at z = +i with the 1 − z2 = √ + 2 and move z in a big swing, illustrated in blue, around to −i (in either direction, well 2 outside the [−1, 1] cut), √ 1 − z will have gone one full turn around the origin, and (having √ 2 will have made half a turn, i.e., swapped sign from + 2 not crossed any cut) 1 − z √ N (z0 ) to become − 2. We can now evaluate the residue using the Res(f (z), z0 ) = D 0 (z ) rule 0
1
f(x)
0.8 0.6 0.4 0.2 0 0.2 1
0.5
0
0.5
1
x
√ Figure 5.9. The integrand f (x) =
1−x2 1+x2
displayed over −1 ≤ x ≤ +1.
5.1. Residue calculus
133
3 2 1 0 1 3
2 2 3
1
3
0
2 1
1
0 2
1 2
x
3
(a) Real part of f (z) =
y
3
√ 1−z 2 1+z 2 .
3 2 1 0 1 3
2 2 3
1
3
0
2 1
1
0 2
1 2
x
3
y
3
(b) Imaginary part of f (z) =
√ 1−z 2 1+z 2 .
√ 1−z 2 Figure 5.10. Real and imaginary parts of f (z) = 1+z2 , using a branch cut along [−1, +1]. The sign is chosen so the function is positive along the cut and smooth above it.
(Section 5.1.1): √ +
1 − z 2 2z
z=+i
√ 1 − z 2 =− 2z √ 2 2i .
z=−i
Both poles have thus the same residue, We give next three different methods for proceeding.
√ 2 = . 2i
134
Chapter 5. Residue Calculus
1.5 1 0.5 0 0.5 1 1.5
1 0.5 0 0.5 1
y
2
3
x
(a) Real part of f (z) =
√
3
2
1
0
1
1 − z2.
2 1 0 1 2
1
3 0.5
2 1
0
0
0.5
1
x
2
1 3
y
√ (b) Imaginary part of f (z) = 1 − z 2 . Figure 5.11. Real and imaginary parts of the integrand’s numerator, f (z) = using a branch cut along [−1, +1]. The branch is chosen to satisfy f (0) = +1.
√ 1 − z2,
Method 1: A simple fact that could quite easily escape attention if it weren’t for Figure 5.10 is that Re f (z) is identically zero over [−∞, −1] and [+1, +∞] and Im f (z) is antisymmetric along the real axis, meaning that there will be no difference in value if we change the integration interval from [−1, +1] to [−R, +R] for any R > 1. Doing that, we then close the loop via a large (radius R) halfcircle Γ in the upper halfplane. It will then hold that ˆ √ I + f (z)dz = 2πi · {residue of pole at z = +i} = π 2 . (5.3) Γ
5.1. Residue calculus
135
´ It remains to evaluate Γ f (z)dz. Since f (z) is singularityfree for z > 1, it will there possess a Laurent expansion: q √ ±iz 1 − z12 i 1 − z2 3 =± f (z) = = 2 1 − 2 + ··· . 1 + z2 z 2z z 1 + z12 Taking the integral over the semicircle, only the leading term ± zi will contribute in the limit of R → ∞. We need, however, to choose the sign so it corresponds to the function branch we are interested in, i.e., with f (0) = +1, as shown in Figure 5.10(b). As z moves up the imaginary axis, the value should (beyond the pole) be negative ´ ´ and increase towards zero. That requires the negative sign. We then find Γ f (z)dz = Γ (− zi )dz, which, after ´ changing variables {z = Reiθ , dz = Ri eiθ dθ}, evaluates to Γ (− zi )dz = π. From (5.3) √ now follows I = π( 2 − 1). Method 2: When one wants to integrate along one side of a branch cut, it is often advantageous to extend the contour to also run along the other side of the cut, and in the opposite direction. Hence, we consider the contour illustrated in Figure 5.12. On´ the return path, ´ below the cut, both f (z) and dz have changed ´sign, so ← f (z)dz = → f (z)dz = I. Since the integrals around the two circular paths f (z)dz → 0 as their radii ε → 0, the integral around the closed loop will be 2 I. Had there been a residue associated with the branch cut, this would have provided a value for I. The alternate strategy in cases such as this is to just expand the contour into a very large circle, radius R, surrounding the origin. Expanding the contour in this way will (i) cause the integral to change according to the two poles that now have been included (cf. Figure 5.13), and (ii) allow an exact evaluation for R → ∞.
Im z i Γ 1
−1 Γ
Re z
−i Figure 5.12. First contour in Example 5.12; the second contour will be to expand it to a large circle, radius R, surrounding the origin. √
From the discussion above, each of the poles has residue 2i2 , and the integral around the full circle will be twice what we evaluated for the halfcircle under Method 1. Putting this together, ˆ √ 2I + f (z)dz = 2πi {sum of residues}, i.e., 2I + 2π = 2 2π, "
136
Chapter 5. Residue Calculus Im z
i Γ 1
−1
Re z
−i
Figure 5.13. Illustration of the “first” and “second” contours (as shown in Figure 5.12 and the large circle, respectively) and why the integrals around these two contours add up to the contributions from the two poles.
and we therefore again obtain I=π
√ 2−1 .
Method 3: Let’s perform the change of variable x = sin θ. Then, dx = cos θdθ. Also, 1 + x2 = 1 + sin2 θ. Therefore, ˆ π/2 cos2 θ I= 2 dθ −π/2 1 + sin θ ˆ 1 π cos2 θ dθ = 2 −π 1 + sin2 θ
√ 1 − x2 = cos θ and
because the integrand is πperiodic. A further change of variable t = tan θ allows (with some effort) closedform integration. However, with the limits now being ±π, contour integration is easier. The standard further change of variables when dealing with trigonometric functions over a full period (cf. Example 5.7, Method 2) is to set z = eiθ . Then cos θ = 21 z + z1 , 1 sin θ = 2i z − z1 , and ieiθ dθ = dz, implying dθ = −i z dz. Therefore, ˛ 1 1 2 1 −i 4 z+ z I= dz 2 z=1 1 + −1 z − 1 2 z 4 z 2 ˛ z2 + 1 −i dz = 2 z=1 z 4z 2 − (z 2 − 1)2 2 ˛ z2 + 1 i = dz, (5.4) 2 z=1 z (z − z1 ) (z − z2 ) (z − z3 ) (z − z4 )
5.1. Residue calculus
137
Figure 5.14. Magnitude/phase portrait of the integrand in (5.4), with the integration path marked by a dashed black circle.
√ that are located inside the unit circle where zj = ±1 ± 2, j =√1, 2, 3, 4. The only roots√ are at z = 0, z = z1 = 1 − 2, and z = z2 = −1 + 2; see Figure 5.14. We find that Res(f, 0) = 1, −1 Res(f, z1 ) = √ , 2 −1 Res(f, z2 ) = √ 2 and therefore √ i 2πi − 2 + 1 2 √ =π 2−1 .
I=
Example 5.13. Evaluate
ˆ
∞
I= −∞
eαx dx. 1 + ex
In order for the integral to converge at x = −∞ and at x = +∞, we need Re α > 0 and eαz Re α < 1, respectively. In the complex zplane, f (z) = 1+e z has poles at z = ±iπ, ±3iπ, ±5iπ, etc.; see Figures 5.15 and 5.16. This pole pattern, and even more how the phase angle for f (z) seems to be constant along lines parallel to the xaxis, suggests the integration path that is shown, following the real axis from −R to +R, and returning from R + 2πi to −R + 2πi , with also two vertical sides along which the contributions go to zero when ´ eα(z+2πi) ´ eαz 2πiα R → ∞. Along the upper return path, ← 1+e dz = −e2πiα I, (z+2πi) dz = e ← 1+ez implying that for the full loop, we get αz ˆ ˆ e + = 1 − e2πiα I = 2πi Res (f, z = πi) = 2πi = 2πi e(α−1)πi , z e → ← z=πi
138
Chapter 5. Residue Calculus
Im z 3iπ
Pole
iπ Re z
−iπ
Figure 5.15. Contour for Example 5.13.
Figure 5.16. Magnitude/phase portrait of the integrand in Example 5.13 (shown in the case of α = 1/2), with the same integration path as shown schematically in Figure 5.15 here marked by a dashed rectangle.
which simplifies to I=
π . sin πα
Example 5.14. Evaluate ˆ I= 0
∞
xa−1 dx, a ∈ R. 1+x
Convergence requires a > 0 (x = 0) and a < 1 (x = ∞). A notable feature of the a−1 integrand, which we write as f (z) = z1+z , is its branch point at z = 0. That suggests a couple of opportunities. First, we have seen before that, when integrating from a branch point, it can be convenient to locate the branch cut along it and choose a closed loop that returns on the opposite side of the same cut (a “keyhole” contour). This will become our
5.1. Residue calculus
139
Method 1. Another possibility is to start by rewriting xa−1 = e(a−1) log x and change variable t = log x, after which the integrand becomes free of branch points. Just possibly, the new integral will provide additional solution opportunities. Method 1: Placing the branch cut immediately below the outgoing integration path and creating a closed loop that returns just below it suggests the contour shown in Figure 5.17.
Im z
Γ
Re z
x = −1
Figure 5.17. Keyhole contour for Example 5.14.
Let’s consider all the contributions to the closedloop integral: 1. Res(f, −1) = 2. 3.
´ →
´ ←
f (z)dz =
(−1)a−1 1
´∞ 0
f (z)dz = −
= eiπ(a−1) .
xa−1 1+x dx
´∞ 0
= I.
e(a−1)(log x+i2π) dx 1+x
= e2πi (a−1) I.
´ 4. ! f (z)dz → 0 as ε → 0 . ´ 5. " f (z)dz → 0 as R → ∞ . R
Putting everything together, we obtain 2πi eiπ(a−1) 1 − e2πi (a−1) π = . sin πa
I=
Method 2: Alternatively, one can consider some analytic modifications (partial integration, change of variables, etc.) before applying contour integration. Given that xa−1 =
140
Chapter 5. Residue Calculus
e(a−1) log x , the variable change t = log x would seem natural. Then x = et , dx = et dt, and the integral becomes ˆ ∞ t(a−1) t ˆ ∞ a−1 e e x dt dx = t 1+x −∞ 1 + e 0 ˆ ∞ eta = dt (5.5) t −∞ 1 + e π = , sin πa where the last equality follows from the preceding Example 5.13.47 Some followup observations to this example include the following: 1. We recognize (5.5) as the result of Example 3.13. 2. A simple variable change generalizes (5.5) to ˆ ∞ a−1 x π . dx = b 1+x b sin π ab 0 3. The integral (5.5) provides one of several ways to prove a key functional identity for the Γ(z)function. Theorem 5.15. The gamma function satisfies the functional equation π Γ(z)Γ(1 − z) = . sin(πz)
(5.6)
Proof. If 0 < Re z < 1, then ∞
ˆ Γ(z)Γ(1 − z) = 0

=
ˆ ∞ tz−1 e−t dt sz e−s ds {z } 0
Change of variable: t→su ˆ ∞ z−1 −su
(su)
e
ˆ sdu
0
∞
sz e−s ds .
0
Combining into a double integral, let the parameter s inside the first integral be the same as the s inside the second one, so s−z cancels sz . Then ˆ ∞ˆ ∞ Γ(z)Γ(1 − z) = sz−1 uz−1 e−su ss−z e−s duds. 0
Now split up again: ˆ
∞
uz−1
ˆ
0
0

∞
0
ˆ ∞ z−1 u e−su e−s ds du = du. 1+u 0 {z } 1 1+u
=
π . sin πz
´ cos 2zt 1 following two integrals are sometimes attributed to Ramanujan: 0∞ cosh dt = 2 cosh and πt z 1 π π = (valid for Im z < and Re z < , respectively). These also follow from a 0 2 cos z 2 2 variable change in the result of Example 5.13. 47 The
´∞
cosh 2zt dt cosh πt
5.1. Residue calculus
141
We have just shown that f (z) = Γ(z)Γ(1 − z) −
π ≡ 0 when 0 < Re z < 1. sin πz
Recalling the idea of analytic continuation (and Theorem 2.12), this relation will hold for all complex z.
5.1.4 Principal value integrals A natural followup to discussing the effect on an integral when a singularity is located inside vs. outside of a contour is what happens if it is located exactly on a contour (cf. ´1 Figure 5.18). For example, does a concept such as −1 dx x make sense? Will both left and right halves diverging to infinity make the sum meaningless? Figure 5.19(a) illustrates the ´ −ε ´ 1 dx case of −1 dx x + ε x , with ε small, and the contour closed in the upper halfplane. Parts (b) and (c) give the integral values for two choices of connecting z = −ε to z = +ε. Going around this small inner circle one full revolution in the positive direction contributes
Γ
(a)
´ Γ
f (z)dz = 0
(c)
(b)
´ Γ
´ Γ
f (z)dz = 2πiRes(f, z0 )
f (z)dz = πiRes(f, z0 )
Figure 5.18. The principal value integral can be thought of as the average of integrals with a pole inside vs. outside of the contour.
142
Chapter 5. Residue Calculus
Γ
Γ
ε −ε (a) Principal value integral contour
(b)
ε ´−ε f (z)dz =0 Γ
Γ
ε
−ε
´
(c)
Γ
f (z)dz = 2πiRes(f, z0 )
Figure 5.19. Contours considered for deriving the key principal value integral result.
2πi Res(f, 0), and going around it half a revolution similarly contributes ±1πi Res(f, 0), with the ± sign according to direction. Starting from either the case in part (b) or part (c) of the figure, we see that the integral value in case (a) (as ε → 0) should be 1πi Res(f, 0). We have thus arrived at the following. Key result.
As is illustrated in Figure 5.18(c) with a singularity on the boundary, ˆ f (z)dz = πi Res(f, z0 ) Γ
(the average of the integrals in the situations for which the singularity is inside and outside of the contour Γ). For this result, the singularity must be on a curve segment that is locally smooth (in particular, it cannot be at a corner). Example 5.16. Show that ˆ I= 0
∞
sin x 1 dx = x 2
ˆ
∞
−∞
sin x π dx = . x 2
´ ∞ iz Following the idea from Example 5.5, we compute 12 −∞ ez dz and extract its imaginary part (turning an ordinary integral into one of principal value type). Closing the contour by a halfcircle in the upper halfplane (cf. Figure 5.20) does not, by Jordan’s lemma, change the value (as R → ∞). Since the pole at z = 0 is located right on the integration ´ ∞ iz iz path, we will use half the residue Res( ez , z = 0) = 1. Thus, −∞ ez dz = 12 · 2πi = πi. ´ ∞ cos z ´∞ Separating real and imaginary parts gives −∞ z dz = 0 and −∞ sinz z dz = π, i.e., ´ ∞ sin x I = 0 x dx = 12 π.
5.1. Residue calculus
143
Im z
Γ R
Re z
Figure 5.20. Contour in Example 5.16.
From this example follows two ways to express the discontinuous signfunction in terms of continuous functions (with the latter integral being of principal value type):48 −1 0 sign α = +1
ˆ if α < 0 1 ∞ sin αx 1 if α = 0 = dx = π −∞ x πi if α > 0
∞ −∞
eiαx dx . x
(5.7)
Various notations ´ can be ´ fflused for emphasizing that an integral is of principal value type,49 including P , P V , , etc.
5.1.5 Halfplane splitting A problem that arises quite frequently (and which we will revisit in Sections 9.1.8, 9.4, and 10.1) is to split a function f (z) into a sum of two functions f (z) = f − (z) + f + (z)
(5.8)
that are singularityfree and go to zero in the upper and lower halfplanes, respectively (with here the “+” and “−” superscripts denoting the halfplane to which all growth and singularities (if any) are confined).50 We have already utilized splittings of this form (in 1 iz 1 −iz Examples 5.5 and 5.16): cos z = 12 eiz + 12 e−iz and sin z = 2i e − 2i e . The purpose in these examples was to obtain a halfplane in which we could close integration contours. ´ ∞ sin 4x curious extension of (5.7), implying as a special case −∞ dx = π, is that x sin 4x Qn x cos dx = π holds exactly for n = 1, 2, . . . , 30, but fails for n = 31 (as noted in k=1 −∞ k x the Preface of [31]). 49 Meaning that the path goes right through a first order pole, contributing πi times its residue. 50 In many texts, “+” and “−” are used in the opposite way as here (and/or placed as subscripts instead of superscripts). We will later (in Chapter 10) use subscripts “+” and “−” to split functions in their parts along the positive and negative real axes, respectively. With the present definitions, the sign will in all cases indicate the direction of nontrivial behavior. 48 A
´∞
144
Chapter 5. Residue Calculus
To obtain the split, it suffices to know f (x) only along the real xaxis and then form ˆ ∞ 1 f (x) f − (z) = dx for Im(z) > 0, (5.9) 2πi −∞ x − z ˆ ∞ f (x) 1 dx for Im(z) < 0. (5.10) f + (z) = − 2πi −∞ x − z These two functions have the required decay properties (Exercise 10.4.2) and, by the principal value idea, return the f (z) as their sum when z is real. Each of these two functions (assuming there are no issues with convergence) is then analytically continued into the opposite halfplanes. This can be done trivially, since (5.8) for Im z ≥ 0 gives f + (z) = f (z) − f − (z) and for Im z ≤ 0 gives f − (z) = f (z) − f + (z). In case f (z) is real along the real axis, it will also hold that f − (z) = f + (z) and f + (z) = f − (z). Example 5.17. Split f (z) =
1 1+z 2
into f (z) = f − (z) + f + (z).
1 i i The partial fraction decomposition f (z) = 1+z 2 = 2(i+z) + 2(i−z) gives immediately 1 − (z) = the answer. However, as an exercise, we apply (5.9) to f (x) = 1+x 2 , giving f ´∞ 1 1 dx. We can close the contour in either halfplane. Choosing the upper 2πi −∞ (x−z)(1+x2 ) 1 i one, there are two poles inside the contour, with residues adding up to 1+z2 + 2(z−i) = i i − + 2(i+z) . The 2πi factors cancel, giving again f (z) = 2(i+z) . Then follows f (z) = i f − (z) = 2(i−z) . Choosing instead to close the contour in the lower halfplane, the only i . Changing the sign (since pole to be concerned with is at x = −i, with residue − 2(i+z) i − this contour goes around the pole clockwise) gives again f (z) = 2(i+z) . The original function and its two halfplane components are shown in Figure 5.21.
Example 5.18. Find the f (z) = f − (z) + f + (z) decomposition in the case of f (x) = 2 e−x . Equation (5.9) becomes ˆ ∞ −x2 2 1 1 e dx = e−z (1 + erf (iz)) , (5.11) f (z) = 2πi −∞ x − z 2 ´z 2 where erf(z) = √2π 0 e−t dt. As this integral stands in (5.11), it is difficult to evaluate by contour integration. However, different arguments in Examples 9.21 and 9.42 both give 2 this same result f − (z) = 12 e−z (1 + erf(iz)). The bottom row of subplots in Figure 9.5 shows f (z) and its two halfplane components. These functions are visualized again (in a slightly different context) in Figure 9.12. Parts (a) and (b) of that figure show the real and 2 imaginary parts of f (z) = e−z , and parts (e) and (f) show 2f − (z). −
5.1.6 Winding numbers and Rouché’s theorem We start by recalling Theorem 4.8 (Cauchy’s argument principle). Lemma 5.19. Consider a function f (z) and a contour Γ, as schematically illustrated in Figure 5.22. If the function f (z) has N zeros and P poles inside Γ, and there are no ¸ f 0 (z) 1 singularities other than these poles inside of Γ, then 2πi f (z) dz = N − P .
5.1. Residue calculus
145
(a) f − (z)
(b) f + (z)
(c) f (z) Figure 5.21. The function f (z) =
1 1+z 2
and its halfplane components f − (z) and f + (x).
Here is a slightly different argument that also shows the result. Let us modify the contour Γ to go around each zero and each pole separately in small circles. Say z0 is a zero of order m. Then f (z) = (z − z0 )m g(z), where g(z) is nonzero and analytic inside the circle, and f 0 (z) m(z − z0 )m−1 g(z) + (z − z0 )m g 0 (z) = f (z) (z − z0 )m g(z) m g 0 (z) = + . z − z0 g(z)  {z }  {z } contributes m2iπ
contributes nothing
146
Chapter 5. Residue Calculus
Γ
Figure 5.22. Contour in Lemma 5.19, where we identify the poles as × and the zeros as •.
We can use the same reasoning for poles. The pole of order n contributes −2πin to the integral. From this result, we can introduce the concept of winding number: ˛ 0 1 f (z) N −P = dz 2πi f (z) 1 [log f (z)]end = start 2πi 1 = [ log f (z) +i arg(f (z)) ]end start  {z } 2πi  {z } Same at start and end
changes by 2π each time f (z) goes around the origin
= {winding number of f (z) with respect to Γ}. The color plots we have frequently used to illustrate phase angles of functions also show the concept of winding number “in action.” As an example, Figure 2.2(d) illustrates the phase angles for the function f (z) = z+ z1 , which has one first order pole (at z = 0) and two simple zeros (at z = ±i). If we follow any curve that circles close to one of the zeros, the color changes along it correspond to once around the color wheel, while going around the pole near to it goes through the colors in reverse order. For a completely arbitrary closed path in the zplane, we can follow the net circulation it has produced with regard to the color wheel. For example, going around the origin once in the positive direction far out from it, the net effect in this case matches once around the color wheel in the positive direction, telling us that N − P for that path equals 1. The number of times we have gone around the color wheel when following a path equals the winding number for the path. Example 5.20. Say f (z) has no poles inside Γ (P = 0). If in this case the winding number is 2 (cf. Figure 5.23), then N = 2, i.e., f (z) has two zeros inside Γ. A consequence of this observation is the following theorem. Theorem 5.21 (Rouché’s theorem). Let f (z) and g(z) be analytic (no singularities) inside Γ (closed). If f (z) > g(z) on Γ, then f (z) and f (z) + g(z) have the same number of zeros inside Γ. Proof. Let z go around Γ. If f (z) > g(z) on Γ, then (f (z) + g(z)) − f (z) < f (z). Thus f (z) + g(z) is always closer to f (z) than f (z) is the origin. Think of walking a dog
5.1. Residue calculus
147
Im w
Γ w = f (z) st ar t/end
Re w
Im z
Re z Figure 5.23. Illustration of Example 5.20.
Im w
Re w
Figure 5.24. Illustration of the wplane in Rouché’s theorem (Theorem 5.21). We see w = f (z) as a solid curve and w = f (z) + g(z) as a dashed curve.
a number of times around a tree, located at the origin of the complex plane. If a (variable length) leash at any moment is shorter than the person’s distance to the tree, the dog will have to move around the tree equally many times as the person. In this illustration (cf. Figure 5.24), think of the person’s position as w = f (z) and the dog’s as w = f (z) + g(z). The leash information is f (z) > g(z), and we have then concluded that f (z) + g(z) and f (z) have the same winding number. By the result just above, the two functions f (z) and f (z) + g(z) must have the same number of zeros inside Γ. As a consequence of Rouché’s theorem, we obtain another proof of the fundamental theorem of algebra. Theorem 5.22. A polynomial pn (x) of degree n has exactly n roots. Proof. Write the polynomial as pn (z) = {z} z n + an−1 z n−1 + an−2 z n−2 + · · · + a1 z + a0 .  {z } f (z)
g(z)
148
Chapter 5. Residue Calculus
Let z go around a circle of radius R, centered at the origin. The function f (z) = z n clearly has n roots inside it (all at z = 0). Around the periphery of the circle, f (z) = Rn and g(z) < n·max aj  Rn−1 . If we choose R > n·max aj , it will hold that f (z) > g(z) on the periphery, and therefore pn (z) will also have n roots inside the circle. Example 5.23. How many zeros does p(z) = z 4 +6z +3 have in the complex plane? Also determine how many of these satisfy z < 2. The degree of the polynomial is 4. Therefore, there are four zeros. Now, let f (z) = z 4 and let g(z) = 6z + 3. When z = 2, f (z) = 24 = 16 and g(z) = 6z + 3 < 6z + 3 = 15. Since f (z) > g(z) around the contour, f (z) and p(z) = f (z) + g(z) have the same number of zeros inside it. The function f (z) there has four zeros, and thus so does p(z); i.e., all four zeros of p(z) satisfy z < 2.
5.2 Infinite sums An important ´ sums. When trying to P∞ application of residue calculus is to evaluate infinite evaluate n=−∞ f (n), the most direct approach is to consider Γ f (z)π cot πzdz around πz a large contour Γ . This idea is based on the fact that π cot πz = π cos sin πz (see Figure 5.25) has a simple pole with residue 1 at each integer value, so the residue of the integrand at each of these poles becomes Res(f (z) π cot πz, z = n) = f (n). sum of the residues PThe ∞ associated with all the integer values becomes the original sum n=−∞ f (n). We choose a large square as contour Γ, which encloses the integers −N, −N + 1, . . . , N − 1, N along the real axis, and consider the limit of N → ∞. As also seen from Figure 5.25, the function π cot πz converges to −πi away from the real axis in the positive imaginary side of the complex plane and to πi away from the real axis in the negative imaginary side of the complex plane. We further let the vertical sides of our contour cross the real axis halfway between two consecutive poles. The function π cot πz is then bounded along the whole contour. Example 5.24. Evaluate for α real and nonzero the sum S = We first rewrite S as S=
P∞
1 n=1 n4 +α4 .
∞ 1 X 1 1 1 − . 4 4 2 n=−∞ n + α 2 α4
Following the outline above, we consider the following integral: ˛ π cot πz dz. I= 4 4 Γ z +α 1 ≤ 4 1 4 as N → ∞. Thus, we can deduce that the integral Further, we note that z4 +α 4 N −α ¸ π cot along the contour vanishes: Γ z4 +απz 4 dz = 0, which implies that N X n=−N

Res
X 3 π cot πz π cot πz ,z = n + Res , z = zj = 0, z 4 + α4 z 4 + α4 j=0 {z }
P∞
n=−∞
f (n)
5.2. Infinite sums
149
10
5
0
5
10 3 2 1 3
0
2 1
1 0
x
2
1 3
2 3
y
(a) Re π cot πz.
10
5
0
5
10 3 2 1 3
0
2 1
1 0
x
2
1 3
2 3
y
(b) Im π cot πz.
(c) π cot πz. Figure 5.25. Graphical representation of the function π cot πz, as seen from a viewpoint in the third quadrant (i.e., the positive real axis is directed upwards and to the left).
150
Chapter 5. Residue Calculus
where zj are the poles of the function f (z) = ∞ X
f (n) = −
n=−∞
X
1 z 4 +α4 .
Therefore, we obtain
Res (f (z)π cot πz, z = zj ) .
j iπ
ijπ
In the present case, zj = αe 4 + 2 for j = 0, 1, 2, 3. Thus Res(f (z)π cot πz, z = zj ) = π cot πzj . After some simplification, we obtain 4zj3 √ √ ! ∞ X √ 1 1 1 sinh πα 2 + sin πα 2 3 √ √ =− + 4 2πα S= . n4 + α 4 2 α4 cosh πα 2 − cos πα 2 n=1 Im z
Γ
Re z
Figure 5.26. Contour in Example 5.24.
Example 5.25. Evaluate S(m) =
∞ X 1 , m = 1, 2, 3, . . . k 2m
k=1
(generalizing the Basel problem
P∞
1 k=1 k2
=
π2 6 ).
´ πz Following the same idea, we let f (z) = π cot and note that f (z)dz will vanish z 2m when integrated around the same contour as shown in Figure 5.26, i.e., 2S(m) + Res (f (z), z = 0) = 0 . The residue at z = 0 follows most easily from the Taylor expansion πz cot πz = 1 −
π 2 2 π 4 4 2π 6 6 z − z − z − ··· , 3 45 945
5.2. Infinite sums
151
i.e., π2 π4 π6 , S(4) = , S(6) = ,.... 6 90 945 The Taylor coefficients above are closely related to the Bernoulli numbers Bk , k = 0, 1, 2, . . . , defined by ∞ X z Bk k = z , (5.12) z e −1 k! S(2) =
k=0
since this relation can be rearranged into πz cot πz = general formula becomes therefore S(m) =
P∞
2k k B2k k=0 (−1) (2k)! (2πz) .
∞ X 1 (−1)m+1 (2π)2m = B2m , m = 1, 2, 3, . . . . k 2m 2(2m)!
The
(5.13)
k=1
Example 5.26. Evaluate S=
∞ X n=−∞
a2
1 . − n2
Consider the integral ˛ I= Γ
π cot πz dz. a2 − z 2
It vanishes along the same contour as shown in Figure 5.26 (as N → ∞). The function 1 π cot πa f (z) = a2 −z 2 has simple poles at z = ±a. Thus, Res(f (z)π cot πz, z = ±a) = −2a , and therefore ∞ X 1 π cot πa = . S= 2 2 a −n a n=−∞ With a slightly different notation, this can be expressed as ∞ X
1 π cot πz = . 2 − n2 z z n=−∞
(5.14)
For alternating series, we will use the function sinππz instead of π cot πz, with the same πf (z) n contour as before. The residue of sin πz at an integer value n is (−1) f (n). In case an infinite sum contains a binomial factor, one might be able to use the relation ˆ n 1 (1 + z)n = {coeff. of z k in (1 + z)n } = dz, (5.15) k 2πi C z k+1 where C is a contour around the origin. Example 5.27. Evaluate S=
∞ X 2n n=0
n
·
1 . 5n
152
Chapter 5. Residue Calculus
Substituting the integral (5.15) for the binomial factor gives ∞ X 2n n=0
n
ˆ ∞ ˆ ∞ X 1 1 X (1 + z)2n (1 + z)2n dz 1 · n = · = n 5 2πi n=0 C (5z) z 2πi C n=0 (5z)n ˆ 5 dz . = 2πi C 3z − 1 − z 2
! ·
dz z
It was here necessary to assume that the geometric progression is convergent, which fails 2 4 both when z is small and when it is large. However, for z = 1, it holds that (1+z) 5z √≤ 5 , 1 so we choose C as the unit circle. The integrand has first order poles at z1 = 2 (3 − 5) ≈ √ 0.38 and at z2 = 12 (3 + 5) ≈ 2.62. Only the first one is inside the unit circle, giving √ √ 1 1 S = 5 · Res( 3z−1−z 5)) = 5. 2 , z = 2 (3 −
5.3 Analytic continuation with use of contour integration Contour integration provides a powerful tool also for analytic continuation, adding opportunities beyond those discussed in Chapter 3. The approaches highlighted below apply to functions that are defined by integrals, and they utilize contour changes, but in quite different ways.
5.3.1 Change of contour allowing exact evaluation Example 5.28. Continue the function f (z) =
´∞
ez t dt. −∞ 1+et
As it stands, this integral defines f (z) only for 0 < Re z < 1, since it otherwise diverges at t = −∞ and at t = ∞, respectively. However, we showed in Example 5.13 that the integral (for such zvalues) evaluates to f (z) = sinππz . Therefore, this amounts to an analytic continuation of the f (z)function to the full complex plane, as illustrated in Figure 5.27.
5.3.2 Change of contour to extend a function across a branch cut The following two examples illustrate this approach. In the first case, we have an analytic solution available with which to verify the result. ´∞ 1 Example 5.29. Continue the function f (z) = 0 (s+z)(s+1) ds across Re z < 0. 1 1 1 1 From Example 5.11 (or by using partial fractions: (s+z)(s+1) = z−1 s+1 − s+z ), it follows that log z f (z) = . (5.16) z−1 This is a multivalued function, with a branch cut most naturally placed along the negative real zaxis. From the closed form (5.16), we would expect the infinitely many solution sheets to be given by f (z) =
log z + 2kπi , z−1
k ∈ Z.
(5.17)
5.3. Analytic continuation with use of contour integration
(a) f (z) =
´∞
ez t dt −∞ 1+et
153
and its region of convergence.
(b) f (z) =
π sin πz .
Figure 5.27. Illustration of Example 5.28.
Since it is rare that integrals are available in this type of closed form, we will now find its next sheet, continuing from below the negative real, utilizing contour changes and residue calculus. We need to consider two complex planes: • zplane, where f (z) is defined (as used in Figure 5.28), ´∞ 1 ds. • splane, in which we do the contour integration 0 (s+z)(s+1) Consider first a zvalue z = z1 just below the negative zaxis, as indicated by an orange
154
Chapter 5. Residue Calculus
1.5
1
0.5
0
0.5 4 2
4 2
0 0 2
2
y
4
x
4
(a) Re
log z z−1 .
2
1
0
1
2
3 4 2
4 2
0 0 2
2
y
4
x
4
(b) Im
log z z−1 .
5 4 3 2 1 0 4 4
2 2
0 0 2
2
y 4
4
z (c) log z−1 and arg
x
log z z−1 .
Figure 5.28. The primary solution sheet for f (z) =
log z . z−1
5.3. Analytic continuation with use of contour integration
155
z2 Br anch cut
s1
Re z
0
I nt e g r at ion path
0
z1
s2
(a) Complex zplane.
Re s
(b) Complex splane.
Figure 5.29. Figures for Example 5.29.
dot in Figure 5.29(a). This causes a pole of the integrand in the splane at the location s1 = −z1 (orange dot in Figure 5.29(b)). We next carry out the following steps: • Deform the contour in the splane to follow the dashed path. This makes no difference in the value of the integral. • In the zplane, slide the evaluation point up from z = z1 to z = z2 . In the splane, the pole moves from s1 to s2 . The integral value changes smoothly, so it must now represent f (z) on a secondary sheet. • Change the contour in the splane back to the straight line. This path change crosses 1 , meaning that the integral has over the pole at s = s2 = −z2 , with residue 1−z 2 2πi changed in value by − 1−z2 . We finally note that this straight line integral is exactly how f (z2 ) is evaluated on the 2πi primary sheet. These two sheets therefore differ in value by − 1−z , in perfect agreement with (5.17). ´∞ Example 5.30. Continue the function f (z) = √2zπ 0 esds across its branch cut. 2 −z In Example 3.10, we continued the function f (z) =
P∞
n=1
zn √ from the unit disk n´ ∞ dx 2z √ . Figures π 0 ex2 −z
to the full complex plane, cut along [1, +∞], obtaining f (z) = 5.30(a)–(b) illustrate this extension. This function f (z) can be continued further, across this cut, following the same idea as in Example 5.29. To use the same notation, we rename x to s, and then consider two
(a) Re f (z).
(b) Im f (z).
Figure 5.30. Real and imaginary parts of the function f (z) =
2z √ π
´∞ 0
dx . 2 ex −z
156
Chapter 5. Residue Calculus
(a) Re f (z).
(b) Im f (z). Figure 5.31. Real and imaginary parts of the function f (z) = ing a second solution sheet.
2z √ π
´∞ 0
dx , 2 ex −z
also includ
separate complex planes, z = x+iy ´ ∞ for the function f (z) and s = σ +it for the integration . If we, for example, move z from below to above path, thus writing f (z) = √2zπ 0 esds 2 −z the cut, we can at the same time deform the integration contour (from the straight line path from 0 to +∞ in s) to still go from 0 to +∞, but taking a curved path in the complex 2 splane such that es goes above also the new value for z. The integral will then evaluate to a value of f (z) that is smoothly continued from the primary sheet, i.e., located on the second sheet. When we next change the path in the splane back to follow the straight line from s = 0 to s = +∞, the path moves across the first order pole (in the splane) of es21−z , which is located at s = (log z)1/2 . Calculating its residue as usual, the effect q becomes that the value of f (z) has changed by 2i logπ z . However, this integration (z in the upper halfplane and the straight line integration path from s = 0 to s = +∞) is exactly how f (z) is evaluated on the primary zplane sheet. The values of f (z) on the two q sheets thus differ by 2i logπ z . In the upper halfplane, we subtract this quantity from the values of f (z) on the primary sheet, and we similarly add in the lower halfplane to obtain the extension of f (z) to a second sheet. We recognize in Figures 5.31(a)–(b) the primary zplane sheet (from Figure 5.30) but see also the secondary one. Due to the difference q √ π between the sheets (2i log z , with log z in the denominator), the second sheet has a new branch point at z = 0 (barely visible as a small spike in Figure 5.31(a)).
5.3.3 Change to a Hankeltype contour Example 5.31. Continue the gamma function Γ(z) =
´∞ 0
e−t tz−1 dt.
5.3. Analytic continuation with use of contour integration
157
C H
Figure 5.32. The contours C and H in Example 5.31.
The definition, as it stands, works only for Re z > 0, since it otherwise diverges at the lower integration limit ´ t = 0. Let us consider C et tz−1 dt (note the sign in the exponent) over the path marked C in Figure 5.32 (entering from −∞ just below the branch cut along the negative real axis, going around the origin, and exiting back to −∞ just above the cut). Under the assumption that Re z > 0, the contribution when going around the origin will be vanishingly small. With t = reiθ , we obtain tz−1 = e(z−1) log t = e(z−1)(log r+iθ) = rz−1 eiθ(z−1) , i.e., tz−1 = Thus
rz−1 e−iπ(z−1) rz−1 eiπ(z−1)
ˆ
: on path in towards the origin (θ = −π), : on path out from the origin (θ = π). ˆ
C
ˆ
∞
et tz−1 dt = e−iπ(z−1)
e−r rz−1 dr − eiπ(z−1) 0
∞
e−r rz−1 dr 0
= −2i sin π(z − 1) Γ(z) = 2i sin(πz) Γ(z). Hence, Γ(z) =
1 2i sin πz
ˆ et tz−1 dt.
(5.18)
C
Up to this point, we have not achieved any analytic continuation (as convergence near the origin in t still requires Re z > 0). However, we now change the contour to the one that is
158
Chapter 5. Residue Calculus
marked H (for Hankel) in the figure. This will not change the value of the integral, but the fact that the contour now keeps some distance away from the origin means that the integral will converge for all values of z, i.e., Γ(z) has been continued as a meromorphic function to the complete complex plane. By the identity (5.6), we can write (5.18) as 1 1 = Γ(z) 2πi
ˆ et t−s dt.
(5.19)
H
Example 5.32. Continue the Riemann zeta function ζ(z) =
P∞
1 n=1 nz
The definition, as it stands, works only for Re z > 1. In a first stage, we rewrite the sum into an integral, which features the same domain restriction. However, the integral formulation can then be continued by changing to a Hankeltype contour. Following Exercise 3.3.6, the integral definition of the ´ ∞ the change of variable t → ´nt∞in z−1 1 gamma function Γ(z) = 0 e−t tz−1 dt gives n1z = Γ(z) t e−nt dt. Therefore 0 ζ(z) =
∞ X 1 z n n=1
∞ ˆ ∞ 1 X tz−1 e−nt dt Γ(z) n=1 0 ! ˆ ∞ ∞ X 1 = tz−1 e−nt dt Γ(z) 0 n=1 ˆ ∞ z−1 1 t = dt. Γ(z) 0 et − 1
=
(5.20)
For the same reason as with the original integral definition of the gamma function, this integral representation is limited to Re z > 0. The remedy is now exactly the same as in ´ z−1 the previous example. After considering C ett −1 dt, where C is the same path as in Figure 5.32, and using the relation Γ(z)Γ(1 − z) = π/ sin(πz) to further simplify the final result, we obtain ˆ Γ(1 − z) tz−1 ζ(z) = dt. (5.21) −t − 1 2πi C e We next change the contour from C to H where H is the same Hankel contour as previously. This new representation for the zeta function is defined and finite for all z 6= 1 (the additional poles of the gamma function gets canceled by the fact that the integral turns out to vanish when z = 1, 2, 3, . . .). Apart from the branch singularity at t = 0, the integrand also features poles at t = ±2πi, ±4πi, . . . . It is assumed that the contour H is not deformed so much that any of these fall inside it. However, in case that H is so deformed that all these poles become enclosed by H, it is shown in Section 6.4 how this formulation (5.21) leads to a proof of the functional equation for the zeta function: ζ(z) = 2z π z−1 sin
1 πz Γ(1 − z)ζ(1 − z). 2
5.3. Analytic continuation with use of contour integration
159
5.3.4 Change to a Pochhammertype contour The primary application for this method is to functions defined by an integral over a finite interval and where the domain restrictions come from singularities at both ends. ´1 Example 5.33. Continue the beta function B(p, q) = 0 tp−1 (1 − t)q−1 dt. The beta function is analytic in both of its variables p and q. The integral definition works only when Re p > 0 and Re q > 0 (since it otherwise diverges at t = 0 and at t = 1, respectively). One way to continue the beta function is to derive the closedform expression B(p, q) = Γ(p)Γ(q)/Γ(p + q) (cf. Section 6.1.2) and then rely on an analytically continued representation of the gamma function. For the Pochhammer approach, we note that the integrand g(t) = tp−1 (1 − t)q−1 is analytic in a complex tplane, apart from branch point singularities at t = 0 and t = 1 (unless p and/or q are integers). We start at the origin in a complex tplane, and move along the path marked 1 → 2 → 3 → 4 in Figure 5.33(a), taking us back to the origin.
4 1 t =0
2 3
t =1
t
t =1
t
(a) Contour C.
t =0
(b) Contour P. Figure 5.33. The contours C and P in Example 5.33. For the contour C, the paths between t = 0 and t = 1 all coincide with the real line segment [0, 1], and the loops at t = 0 and t = 1 are infinitesimally small.
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Chapter 5. Residue Calculus
The magnitude of g(t) will be singlevalued along this path, while the argument of g(t) will change according to the sequence 0 → −2π(q − 1) → −2π(q − 1) − 2π(p − 1) → −2π(p−1) → 0; i.e., g(t) is back to its original value when the full trajectory is completed. We note that each of the two singularities t = 0 and t = 1 has been gone around twice, but for each, the two times have been in opposite (canceling) directions. With Re p > 0 and Re q > 0, the contributions around the small nearcircular parts will be vanishingly small. The integral around the full loop will thus add up to ˆ = B − e−2π(q−1)i B + e−2π(q−1)i e−2π(p−1)i B − e−2π(p−1)i B C
= −4e−iπ(p+q) sin(pπ) sin(qπ)B, i.e., −eiπ(p+q) B(p, q) = 4 sin(pπ) sin(qπ)
ˆ tp−1 (1 − t)q−1 dt. C
We next change the contour from C to the one denoted by P (for Pochhammer) in Figure 5.33(b). This will not change the value of the integral, but the fact that the contour now keeps some distance away from the singularities implies that the integral will converge for all values of p and q. We can note that it would not have been helpful to introduce branch cuts from the two branch points (at t = 0 and t = 1) in order to create a singlevalued function. No matter how such cuts would have been placed, the contour would have needed to move across these artificially introduced barriers. With a Riemann surface approach to multivaluedness, there is no difficulty in this regard.
5.4 Weierstrass products and Mittag–Leffler expansions Applies to Describes a functions through its
Weierstrass Products entire functions (no singularities) zeros
Mittag–Leffler Expansions meromorphic functions (poles are only singularities) poles
The general treatment of these expansions is quite involved, so we will limit ourselves to illustrative examples.
5.4.1 Weierstrass products Let f (z) be an entire function with a finite number of zeros (all simple), located at z = a1 , a2 , a3 , . . . , an , and consider pn (z) = 1 − az1 1 − az2 · · · · · 1 − azn . Then pfn(z) (z) is a zerofree entire function, and h(z) = log h(z)
f (z) pn (z)
will be entire. We can thus write f (z)
as f (z) = pn (z)e . Knowing the zeros of f (z) is enough to determine pn (z), but not enough to exactly determine f (z). Generalizing this to the case of infinitely many zeros a1 , a2 , . . . suggests forming ∞ Y z p(z) = 1− , an n=1
5.4. Weierstrass products and Mittag–Leffler expansions
161
bringing up the issue of when infinite products converge. This requires investigating the Q∞ tail end of the product n=M (1 − azn ) for M 1, to see whether this converges to a limit that is neither 0 nor ±∞. Example 5.34. Find a product expansion for sin πz. The function sin πz has zeros at 0, ±1, ±2, . . . , so sinzπz has zeros at ±1, ±2, . . . , making it natural to form n z z o n z z z o z 1− 1− ... · 1 + 1+ 1+ ... . p(z) = 1− 1 2 3 1 2 3 While both of the products diverge, we can note that p1 (z) =
∞ n Y z z/k o 1− e k
k=1
and p2 (z) =
∞ n Y z −z/k o e 1+ k
k=1
both converge by Theorem 5.40. To see why, note that the terms in p1 (z) become
1−
2 3 z z z z2 z 1+ +O = 1 − + O , k k k2 k2 k3
P∞ 2 P∞ 3 and similarly for p2 (z). The sums n=1 kz + O kz 3 = z2 n=1 k12 + O kz3 converge. In the present example, we can either rewrite p(z) as p(z) = p1 (z) · p2 (z) or, simpler still, group the factors to directly give the convergent product p(z) =
∞ Y k=1
z2 1− 2 k
.
We will from this know that sin πz = z
(
∞ Y
k=1
z2 1− 2 k
)
eh(z) ,
(5.22)
where h(z) is an entire function, still to be determined. It will turn out that eh(z) ≡ π. A good way to see this starts by taking the log on both sides, giving log(sin πz) = log z + h(z) +
∞ X k=1
z2 log 1 − 2 k
Differentiating gives ∞
X 1 1 π cot πz = + h0 (z) + 2z . z z 2 − k2 k=1
.
162
Chapter 5. Residue Calculus
P∞ 1 0 Recalling (5.14), z k=−∞ z2 −k 2 = π cot πz, it follows that h (z) = 0 and thus h(z) = h is a constant. Taking the limit z → 0 on both sides of (5.22) gives that eh = π. Therefore ∞ Y sin πz z2 = 1− 2 . πz k
(5.23)
k=1
A classical special case was computed by Wallis in 1670, who arrived at it in a different way. If z = 1/2, then π 2·2 4·4 6·6 = · · · .... 2 1·3 3·5 5·7
5.4.2 Mittag–Leffler expansions These expansions represent a function in terms of its poles. Example 5.35. Find a closedform expression for f (z) =
∞ X k=−∞
1 . (z − k)2
Clearly the series converges and it has period 1 (indeed, f (z + 1) = f (z)). Because it is 1periodic, it has the same Laurent expansion at every pole. We check this expansion at z = 0: X 1 1 + 2 z k 2 (1 − z/k)2 k6=0 2 X 1 1 z z2 = 2+ · 1 + + + · · · z k2 k z2 k6=0 ! ∞ X 1 0 1 = 2 + +2 + 0 z + (· · · )z 2 + · · · . z z k2
f (z) =
k=1
Next consider the function get the Laurent expansion
π2 . sin2 πz
It is also 1periodic. Expanding around the origin, we
π2 1 0 π2 + + 0z + ··· . = + z2 z 3 sin2 πz P∞ 2 1 Therefore, h(z) = sinπ2 πz − k=−∞ (z−k) 2 is an entire function that also is 1periodic. It is easy to show that h(z) → 0 up and down the period in Figure 5.34 (for P∞strip shown 1 1 example by noting that z−k , 2 for each k, and thus also 2 k=−∞ z−k are monotonically decreasing if Im z increases). Thus, by Liouville’s theorem, h(z) ≡ constant. Since both functions tend to 0 far out in the period strip, this constant must be h = 0. We have thus obtained ∞ X π2 1 = , sin2 πz k=−∞ (z − k)2
(5.24)
5.4. Weierstrass products and Mittag–Leffler expansions
163
Period Strip
Im w
Re w
Figure 5.34. Period strip for both
π2 sin2 πz
and
P∞
1 k=1 (z−k)2 .
which is a typical example of a Mittag–Leffler expansion (expressing here the function π2 in terms of its poles). sin2 πz An side result is the following alternative way to arrive at the sums I(m) = P∞ interesting 1 k=1 k2m , m = 1, 2, 3, . . . (cf. Example 5.25). Equating the Laurent series of both functions around z = 0 gives the following sequence of relations: ∞ X π2 1 = , k2 6
k=1 ∞ X
1 π4 = , 4 k 90 k=1 ··· As commented on again in Section 6.2, finding corresponding closedform expressions for odd positive integers 3,5,7, . . . is a longstanding open problem. Example 5.36. Find a closedform expression for f (z) =
∞ X k=−∞
1 . z−k
(5.25)
As it stands, the series is only conditionally convergent. The following are two ideas to bypass this issue: 1. Add a term which will cancel out. Notice that ∞ X 1 1 X f (z) = = + z−k z k=−∞
k6=0
1 1 + z−k k  {z } size O
1 k2
, so converges
.
164
Chapter 5. Residue Calculus
2. Pair up the terms k and −k. We can then rewrite f (z) as m ∞ X 1 1 X 1 1 f (z) = lim = + + m→∞ z−k z z+k z−k k=−m
k=1
∞
X 1 1 . = + 2z 2 z z − k2 k=1
We next show two ways to evaluate this sum. Method 1: We notice from the last equation above that this function has the same pole structure as π cot πz. Furthermore, both functions have period 1 and are bounded up and down the period strip. By P Liouville’s theorem, they can differ only by a constant. Since ∞ 1 both π cot πz and z1 + 2z k=1 z2 −k 2 are odd functions, the constant has to be zero (as there is no other odd constant). Thus ∞ X k=−∞
1 = π cot πz . z−k
P∞ ´Method 2: The standard approach that we use for evaluating k=−∞ f (k) is to consider f (z)π cot πzdz around the large rectangular path in Figure 5.26. As usual (although slightly more subtle to show in this particular case), the integral vanishes as the path is moved increasingly far out. To avoid a confusion of notation, we write (5.25) as f (ξ) = P ∞ 1 π cot πz k=−∞ ξ−k and thus consider the integral of ξ−z . Apart from at z integer, there is one more pole, at z = ξ, with residue −π cot πξ. Since all residues should add up to zero, we arrive again at f (ξ) = π cot πξ. Note that we can similarly find that ∞ X (−1)k π = . z−k sin πz
k=−∞
Mittag–Leffler expansions provide a supplementary approach to the one described in Section 5.2 for evaluating many infinite sums. Example 5.37. Find a closedform expression for ∞ X
S=
k=−∞
1 . 1 + k2
(5.26)
The present idea is to form the 1periodic analytic function G(z) =
∞ X k=−∞
1 , 1 + (z + k)2
for which we want the value G(0). Considering G(z) in the complex plane, its only N = singularities will be first order poles at z = −k ± i, k integer, with residues D 0 1 i = ∓ . Recalling that π cot πz has poles with residues +1 at all integers, it 2(z+k) z=−k±i 2 is clear that G(z) has exactly the same pole structure (locations and residue) as the function H(z) =
πi [cot(π(z + i)) − cot(π(z − i))] ; 2
5.5. Supplementary materials
165
Figure 5.35. Magnitude/phase plot of the function G(z) = H(z) in Example 5.37.
cf. Figure 5.35. The difference G(z) − H(z) is therefore an entire, 1periodic function which furthermore is easily seen to approach zero when Im z → ∞. By Liouville’s theorem, the difference is identically zero, and we obtain S = G(0) = H(0) = π coth π.
5.5 Supplementary materials 5.5.1 Convergence of infinite products Theorem 5.38. If the terms of the sequence {ak } are real and positive, P∞ if and only if k=1 log (ak ) converges. = 1. Since limn→∞ an = 0, limn→∞ Proof. limx→0 log(1+x) x comparison test (Theorem 2.37), the result follows.
Q∞
log(1+an ) an
k=1
ak converges
= 1. By the limit
We give without proofs two further results regarding infinite products. P∞ Theorem P 5.39. If the terms of the sequence {ak } are real and k=1 a2k converges, then Q ∞ ∞ the series k=1 ak and the product k=1 (1 + ak ) converge or diverge together. Note: Convergence Qn of the product to zero occurs only if it has a zero factor (i.e., an ak = −1). If limQ n→∞ k=1 (1 + ak ) = 0 but no termQank = −1, then the product diverges to 0. n Example: k=2 1 − k1 diverges, yet limn→∞ k=2 1 − k1 = limn→∞ 21 23 43 . . . n−1 n = limn→∞ n1 = 0. Theorem 5.40. (uniformly).
Q∞
n=1
(1 + un (z)) converges (uniformly) in z if
P∞
n=1
un (z) converges
166
Chapter 5. Residue Calculus
5.6 Exercises Exercise 5.6.1. Evaluate
¸ Γ
e1/z dz, where Γis the unit circle.
1 Exercise 5.6.2. If ζ < R, show that 2π ζ). Hint: Let z = R eiθ , write the integral as of this with respect to ζ.
´ 2π 0
log R eiθ − ζ dθ = log R (independent of
1 2π Im
¸
log(z−ζ) dz, z
and consider the derivative
Exercise 5.6.3. Assume f (z) is analytic for z ≤ R and that it has there zeros at z = z1 , z2 , . . . , zn , none of which are at z = 0 and z = R. Show Jensen’s formula (generalizing (4.18)): Rn 1 log f (0) + log = z1 · z2 · · · · · zn 2π
ˆ
f (z) (z−z1 )(z−z2 )·····(z−zn )
Hint: Apply (4.18) to g(z) = 5.6.2.
2π
log f (R eiθ )dθ . 0
and then use the result in Exercise
´∞ π Exercise 5.6.4. Show that 0 x4 +2x2dx cos 2α+1 = 4 cos α  if cos α 6= 0. Hint: Key solution steps include showing that the four poles are at the locations {i eiα , i e−iα , −i eiα , −i e−iα } and have residues 1 1 i 1 1 i 1 1 i 1 1 i − , − − , − + , + , 8 sin α cos α 8 sin α cos α 8 sin α cos α 8 sin α cos α
respectively. Exercise 5.6.5. Show that Exercise 5.6.6. Evaluate
´∞ 0
e−x −cos x x
´∞
Exercise 5.6.7. Show that
x sin x dx, −∞ a2 +x2
´∞
dx −∞ 1+x2p
=
dx = 0. where a is a realvalued nonzero constant. π , π p sin( 2p )
where p is a positive integer.
Exercise 5.6.8. If α and β are real and positive constants, show that ˆ 0
∞
cos αx dx = x+β
ˆ 0
∞
x e−αβx dx. 1 + x2
If you were to numerically evaluate the result for some different values of α and β, which of the two versions do you think would be best to use? ´t Exercise 5.6.9. We want to evaluate I = 0 a+bdθcos θ , where a > b , and t lies in the interval (0, 2π). By the change of variable z = eiθ , derive ˆ 1 1 1 I= √ − dz, (5.27) z − z2 i a2 − b2 Γ z − z1
5.6. Exercises
167
where z1 and z2 are the pole locations of the integrand in the zplane, and Γ is the arc of the unit circle from z = 1 to z = eit . Note: Although we don’t have a closed loop (and thus cannot readily use the residue method), we have nevertheless arrived at an integral (5.27) that we can evaluate directly, giving rise to two logarithms. Since we know that the result will be real, we furthermore need only to consider the imaginary √parts of these logarithms. After some simplifications, 2 −b2 sin t . You do not need to carry out any of the one arrives at I = √a21−b2 arctan ab+a cos t steps beyond arriving at (5.27). Exercise 5.6.10. Show that not equal to zero.
´∞
cos ax dx −∞ b2 +x2
´∞
π −ab , b e
=
where a and b are real, and b is
log x 1+x2 dx
= 0 in the following two ways: ´1 ´∞ ´∞ (a) Without contour integration, split 0 = 0 + 1 and in one of these integrals, change variable x → 1t . (b) Use contour integration. Note: Part (a) is a reminder that, even for integrals that look natural for contour integration, it is worthwhile to also look for simplifying reformulations. The same idea as here gives, for example, the surprising result that ˆ ∞ dx π = , 2 )(1 + xa ) (1 + x 4 0 Exercise 5.6.11. Show that
0
independently of the value for a. ´∞
= 2√π2 a3 . ´∞ Hint: The variable change t − 1/t = x gives I(a) = −∞ 21 1 +
Exercise 5.6.12. Show that I(a) =
0
dt a4 +(t−1/t)4
√ x 4+x2
dx a4 +x4 .
Exercise 5.6.13. Assume a > 0. Utilize the result in Exercise 5.6.11 to show that ˆ ∞ π log a log x dx = . 2 + a2 x 2a 0 Exercise 5.6.14. Show that (a)
´∞ 0
log2 x 1+x2 dx
=
π3 8 ,
(b)
´∞ 0
log x (1+x2 )2 dx
= − π4 .
´ π/2 Exercise 5.6.15. Evaluate 0 log(sin x)dx by (a) the result in Exercise 2.9.27, (b) utilizing the relation sin x = 2 sin x2 cos x2 , (c) direct contour integration. Exercise 5.6.16. Show that Exercise 5.6.17. Show that
´∞ 0
´∞
x1/3 1+x3 dx
=
x4 dx −∞ 1+x8
π 3 cos
=
π 2
π 18
.
q 1−
√1 . 2
Exercise 5.6.18. Evaluate the same integral as in Example 5.10: I = but use instead the keyhole contour shown in Figure 5.8.
´∞ 0
dx x3 +a3 ,
a > 0,
168
Chapter 5. Residue Calculus
Exercise 5.6.19. Show that f (z) =
´∞
dx (x+a)2 +b2
0
1 b
=
arctan
b a
(a ≥ 0, b > 0) by considering
log z (z+a)2 +b2 .
´∞
log x dx 0 (x+a)2 +b2 2 (log z) (z+a)2 +b2 .
Exercise 5.6.20. Show that by considering f (z) =
=
1 b
arctan
b a
log
√ a2 + b2 (a ≥ 0, b > 0)
´∞ cos γx dx π −βγ Exercise 5.6.21. Show that 0 (x2 +α − β e−αγ ), where 2 )(x2 +β 2 ) = 2αβ(α2 −β 2 ) (α e α, β, γ are all positive. Do you need to consider separately the case of α = β? Note: The result generalizes immediately to α, β, γ real, since the integrand is an even function of its three parameters. Exercise 5.6.22. Assuming that 4c − b2 > 0, show that Exercise 5.6.23. Show that ˆ
∞
0
´∞
dx −∞ (x2 +bx+c)2
=
4π . (4c−b2 )3/2
1 ea + 1 1 sin ax dx = − . 2πx a e −1 4 e − 1 2a
(5.28)
´ ∞ ez log t Exercise 5.6.24. Consider the function f (z) = 0 (1+t 2 )2 dt: (a) Determine the region in the complex zplane where the integral converges. (b) Analytically continue f (z) by means of reexpressing it as f (z) = 4π(1−z) . cos 1 πz 2
´ ∞ cos ax − cos bx dx = π(b − a). −∞ x2 bz is an entire Hint: Correct what is wrong with the following argument: f (z) = cos az−cos z´2 ´∞ ∞ eiaz −eibz dz. function, so any closedloop integral is zero. Also −∞ f (z) dz = Re −∞ z2 Now closing the loop far out in the upper halfplane makes a negligible change due to the ´ ´∞ estimate x · · · dz ≤ M · L ≤ R22 · πR → 0 as R → ∞. Hence −∞ · · · dx = 0. Exercise 5.6.25. Show that for a, b ≥ 0,
Exercise 5.6.26. If A + B + · · · + C = 0 and a, b, . . . , c are positive, show that ˆ ∞ A cos ax + B cos bx + · · · + C cos cx dx = −A log a − B log b − · · · − C log c . x 0 Note: Later, in connection with Laplace transforms, we obtain a quite different approach (to contour integration) for solving this problem; cf. Exercise 9.8.19. ffl ∞
π = − 3√ . ´ 0 3 dx √ . Hint: It follows from Example 5.10 that −∞ x3 −1 = − 32π 3
Exercise 5.6.27. Show that
0
dx x3 −1
ffl ∞ a−1 Exercise 5.6.28. Show that 0 xx−1 dx = −π cot πa, 0 < a < 1. ´0 a−1 Hint: The value for −∞ xx−1 dx follows from Example 5.14. Exercise 5.6.29. Show that
´ 2π 0
2−sin θ 2−cos θ dθ
=
4π √ . 3
5.6. Exercises
169
Exercise 5.6.30. Show the following: ´ 2π (a) 0 5+3dθcos θ = π2 . ´ 2π 5π (b) 0 (5+3dθ cos θ)2 = 32 . Exercise 5.6.31. The integrals in Exercise 5.6.30 can be generalized significantly. Assume a > b > 0. Then show the following: ´ 2π 2π (a) 0 a+bdθcos θ = (a2 −b 2 )1/2 . ´ 2π dθ 2πa (b) 0 (a+b cos θ)2 = (a2 −b 2 )3/2 . ´ 2π π(2a+b) dθ (c) 0 (a+b cos2 θ)2 = a3/2 (a+b)3/2 . ´ 2π ´ 2π 2m−1 π Exercise 5.6.32. Show that 0 (cos θ)2m dθ = 0 (sin θ)2m dθ = 4m−1 m−1 , m = 1, 2, 3, . . . . ´ 2π ´ 2π Note: Trivially, 0 (cos θ)2m−1 dθ = 0 (sin θ)2m−1 dθ = 0, m = 1, 2, 3, . . . . Exercise 5.6.33. Evaluate
´ 2π 0
cot
θ−a−i b 2
dθ, where a, b are real nonzero constants.
√ √ 4 Exercise 5.6.34. The function f (x) = e− x sin 4 x has the strange property of being orthogonal to every polynomial over the interval [0, ∞] (something which is impossible for a´ not identically zero continuous function over a finite interval). In particular, it holds that ∞ n x f (x)dx = 0 for n = 0, 1, 2, . . . . Prove this. 0
´∞ √ 2 Exercise 5.6.35. Several ways are available to show that −∞ e−x dx = π (see Exercise 6.5.8). It is, however, surprisingly tricky to do this with residue calculus. One way this 2 can be achieved is by integrating f (z) = eiπz tan πz (shown in Figure 5.36(a)) over the contour shown in its part (b) and letting R → ∞. Carry this out. Exercise 5.6.36. Solve the ODE w0 (z) = zw(z) + z p , with w(0) = w0 using the series method, for p = 1 and p = 2. Exercise 5.6.37. Knowing that
´∞ −∞
2
e−x dx =
√
π , contour integration of the function
−z 2
f (z) = e leads´ to a number of´related relations:p ∞ ∞ (a) Show that 0 cos x2 dx = 0 sin x2 dx = 12 π2 . ´ ∞ −x2 ´∞ 2√ 2 (b) Show that 0 e cos(2ax)dx = 21 e−a π and that 0 e−x sin(2ax)dx = ´ 2 2 a e−a 0 ex dx. Note: We will come across the latter expression (known as Dawson’s function D(z) = 2 ´z 2 e−z 0 et dt) again as the Hilbert transform of a Gaussian; cf. Example 9.42. Its elementary properties include that it satisfies the ODE dD(z) dz + 2zD(z) = 1, D(0) = 0. Exercise 5.6.38. One striking feature in Figure 5.37 is that the roots of the derivative of a polynomial seem to fall within the convex hull (smallest enclosing convex polygon) of the polynomial’s roots. Prove the Gauss–Lucas theorem, which states that this will always be the case.
170
Chapter 5. Residue Calculus
2
(a) Magnitude and phase of f (z) = eiπz tan πz.
(R,iR) (R+1,iR)
4
2
6
4
2
2
4
2
4
(R,iR) (R+1,iR)
(b) Integration contour Figure 5.36. Integrand and contour in Exercise 5.6.35.
6
5.6. Exercises
171 2.5
1 2 1.5
0.5
1 0 0.5 0
0.5
0.5 1 1 1.5
1.5
1
0.5
0
0.5
1
1.5
2
2
1
0
1
2
3
1.5 1 0.5 0 0.5 1 1.5 2 2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
Figure 5.37. Three cases of randomly chosen degree 10 polynomials p10 (z) with their roots shown in red and the nine roots of p010 (z) shown in green.
Hints: Note that (i) it suffices to show that if all roots of p(z) fall in the right halfplane, so do those of p0 (z), (ii) the result is true for a root of p0 (z) that P is also a root of p(z), and 1 (iii) equation (4.5) tells that if z is a simple root of p0 (z), then z−ak = 0. Rearrange P ak P 1 this into z z−ak 2 = z−ak 2 , and deduce the result. Exercise 5.6.39. Carry out the halfplane splitting f (z) = f − (z)+f + (z) for f (z) = Exercise 5.6.40. Evaluate S(m) = Exercise 5.6.41. Show that
(−1)k k=1 k2m ,
P∞
(−1)k k=0 (2k+1)3
P∞
=
π3 32
m = 1, 2, 3, . . . .
.
Exercise 5.6.42. Find a closedform expression for x < π.
(−1)k k=−∞ 1+k2
P∞
cos kx, where −π
1), it still contains some of the most outstanding mathematical challenges of our time, and, as such, has provided inspiration to generations of professional mathematicians and amateurs alike.
6.1 The gamma function We introduced the gamma function in Section 2.7 as ˆ
∞
e−t tz−1 dt
Γ(z) = 0
and showed that it satisfies the functional equations Γ(z + 1) = zΓ(z) (see (2.19)) and π Γ(z)Γ(1 − z) = sin(πz) (Theorem 5.15 and (5.6)). We also noted that it generalizes the factorial function both to noninteger and to complex arguments, since n! = 1·2·3·· · ··n = Γ(n+1). It can be evaluated rapidly for all complex z by combining either of the functional equations with the asymptotic expansion (12.23), which is highly accurate when Re z is large. We see immediately from (3.7) that the only singularities of the gamma function are poles at the nonpositive integers, with residues Res(Γ(z), z = −n) =
(−1)n , n = 0, 1, 2, . . . . n!
(6.1)
As shown in Figure 6.1, the rapid growth in magnitude for increasing Re(z) is associated with oscillations in the real part (and likewise in the imaginary part) as we move away from the real axis. The function Γ(z) approaches zero very rapidly as we move out in the left halfplane (except for the string of poles along the negative real axis). 173
174
Chapter 6. Gamma, Zeta, and Related Functions
15 10 5
x 4
3
2
1
1
2
3
4
5 10 15
(a) Γ(x) for x real.
(b) Re Γ(z).
(c) Γ(z) with phase coloring.
(d) 1/Γ(z) with phase coloring.
Figure 6.1. Graphical representations of Γ(z) and 1/Γ(z).
6.1.1 Product representations of the gamma function There are two important product representations of the Γ(z) function. Theorem 6.1.
Γ(z) = lim
n! nz . n→∞ z(z + 1) · · · · · (z + n)
(6.2)
∞ h Y 1 z −z/n i = z eγz 1+ e , Γ(z) n n=1
(6.3)
Theorem 6.2.
where γ = lim
n→∞
1+
1 1 1 + + · · · + − log n 2 3 n
≈ 0.5772156649 .
(6.4)
6.1. The gamma function
175
Proofs for both are given in Section 6.4. From these relations, several further results follow. For example, (6.3) shows immediately that 1/Γ(z) is an entire function, and in 1 particular that Γ(z) will never take the value of zero. Forming Γ(z)Γ(−z) leads in a few steps π (utilizing (5.23)) to the functional relation Γ(z)Γ(1 − z) = sin(πz) (cf. (5.6)). Substituting Γ(1 − z) = −zΓ(−z) leads to an exact formula for the magnitude of the gamma function up/down the imaginary axis. For y real, Γ(iy) =
π y sinh πy
1/2 .
6.1.2 The beta function The beta function B(p, q) is defined by ˆ
1
tp−1 (1 − t)q−1 dt
B(p, q) =
(6.5)
0
(choosing the principal values for the powers) and is related to the gamma function through the relation Γ(p)Γ(q) B(p, q) = , (6.6) Γ(p + q) proved below in Section 6.4. These equations (6.5) and (6.6) provide one (of several) ways to prove the duplication formula51 for the gamma function. Theorem 6.3 (Legendre). 1 Γ(2z) = π −1/2 22z−1 Γ(z)Γ z + . 2
(6.7)
Three different ways to prove this relation are outlined in Exercises 6.5.9(a), 6.5.9(b), and 12.6.9. While the definition (6.5) only works for Re p > 0 and Re q > 0 (as the integral otherwise diverges), we saw in Example 5.33 how it could be continued to all p and q by means of a Pochhammer integral. The explicit formula (6.6) provides also such a continuation. Replacing t by 1 − τ shows that B(p, q) = B(q, p). Other substitutions that lead to often useful identities are ˆ π/2 2 t = sin θ gives B(p, q) = 2 (sin θ)2p−1 (cos θ)2q−1 dθ (6.8) 0
ˆ
and τ t= 1+τ
∞
gives B(p, q) = 0
tp−1 dt. (1 + t)p+q
(6.9)
In Section 11.3.3, we will come across the incomplete beta function, which ´ z has one additional argument z replacing 1 as the end point of the integral: B(p, q; z) = 0 tp−1 (1− t)q−1 dt. 51 This
formula is the m = 2 special case of Γ(mz) = (2π) 1, 2, 3, . . . (due to Gauss).
1−m 2
1
mmz− 2
Qm−1 k=0
Γ z+
k m
, m =
176
Chapter 6. Gamma, Zeta, and Related Functions
Figure 6.2. The beta function B(p, q) displayed over the domain −2.5 ≤ p, q ≤ 6.2.
Since B(p, q) depends on two independent complex parameters, it is difficult to display it graphically. However, in the special case that both p and q are real, it can be illustrated as in Figure 6.2. It decays smoothly to zero inside the first quadrant in the (p, q)plane. However, it follows from (6.6) that it flips between ±∞ when either p or q crosses 0, −1, −2, . . . . Furthermore, p + q = 0, −1, −2, . . . are zero contour lines.
6.2 The zeta function In Section 3.2.3, P∞ we analytically continued the Riemann ζfunction, which we introduced as ζ(z) = n=1 n1z , from Re z > 1 down to Re z > 0 using the reformulation ζ(z) = P∞ (−1)n+1 1 and finally to the entire complex plane by means of the functional n=1 (1−21−z ) nz z z−1 equation ζ(z) = 2 π sin 21 πz Γ(1 − z)ζ(1 − z). We also found that ζ(z)’s only singularity is a simple pole at z = 1, with residue 1. Regarding values of the zeta function at integer arguments, we deduced from Example 5.25 that ζ(2k) =
(−1)k+1 B2k (2π)2k , k = 1, 2, 3, . . . , 2(2k)!
(6.10)
and then, via the functional equation, that ζ(−k) = (−1)k
Bk+1 , k = 0, 1, 2, . . . , k+1
(6.11)
P∞ where the Bcoefficients stand for the Bernoulli numbers defined by ezz−1 = k=0 Bk!k z k 1 1 (see (5.12)), i.e., B0 = 1, B1 = − 21 , B2 = 16 , B3 = 0, B4 = − 30 , B5 = 0, B6 = 42 ,.... In particular, ζ(k) vanishes at all the negative even integers, forming the trivial zeros of the zeta function. Missing from this summary of values at integer locations are the odd positive integers 3, 5, 7, . . . . Surprisingly little is known about these, with the main exception that ζ(3) is known to be irrational.
6.2. The zeta function
177
1 (c) ζ(z)
(b) ζ(z)
(a) Re ζ(z)
Figure 6.3. Plots from left to right of ζ(z) in the strip [−5, 5] × [−50, 50] surrounding the imaginary axis: (a) Re ζ(z), (b) ζ(z), and (c) 1/ζ(z). In the last subplot, the first nontrivial zeros along the line Re z = 12 are seen as narrow spikes.
6.2.1 The critical strip The vertical strip 0 ≤ Re z ≤ +1 is for the zeta function called its critical strip. Figure 6.3 shows the function in the somewhat wider strip −5 ≤ Re z ≤ +5 (and −50 ≤ Im z ≤ +50) surrounding the imaginary axis, and Figure 6.4 shows ζ(x) along the real xaxis, with the first few of the trivial zeros, located at the negative even integers, highlighted with red dots. Figure 6.5 shows ζ(z) along two sections of the critical line Re z = 12 . The nontrivial zeros are seen to be quite irregularly distributed. In the context of the zeta function, it is common in the literature to denote the complex plane by s = σ + it rather than by z = x + iy. Hence, we use t in place of y in some cases below.
(x) 2 1
x 20
15
10
5
5
10
1 2
Figure 6.4. ζ(x) plotted along the real axis. We notice the pole at x = 1. The trivial zeros at x = −2, −4, −6, . . . are marked with red dots.
178
Chapter 6. Gamma, Zeta, and Related Functions 3.0
2.5
2.0
1.5
1.0
0.5
10
20
30
40
(a) ζ( 12 + it) shown for 0 ≤ t ≤ 40. 5
4
3
2
1
110
120
130
140
(b) ζ( 12 + it) shown for 100 ≤ t ≤ 140. Figure 6.5. Plots of ζ( 12 + it).
The Riemann hypothesis is one of the most famous unsolved problems in all of mathematics. It states that all of the (infinitely many) nontrivial zeros lie exactly on the line Re z = 12 , by now known to be true for the first 1013 of them. Either proving or disproving the Riemann hypothesis would have extensive consequences in mathematics. At the time this book is being written, there is a one million dollar prize for solving this problem.52
6.2.2 Relation to prime numbers Already, the ancient Greeks knew that there are infinitely many prime numbers. As a minor illustration of the extensive connections between primes and the ζ function, we will show below this same result based on the fact that ζ(z) has a pole at z = 1. Theorem 6.4. There are infinitely many primes. 52 One
of the seven Millennium Problems, as presented by the Clay Mathematics Institute.
6.2. The zeta function
179
Proof. Let {p1 , p2 , p3 , . . .} = {2, 3, 5, . . .} be the set of all the prime numbers. Because every integer can be written in exactly one way as a product of primes, 1 1 1 1 1 1 1 + z + 2z + · · · 1 + z + 2z + · · · 1 + z + 2z + · · · · · · p1 p1 p2 p2 p3 p3 1 1 1 1 1 = z + z + z + z + z + · · · = ζ(z). 1 2 3 4 5 Each of the factors in the product is an infinite geometric progression that can be summed in closed form, giving the relation ∞ Y 1 ζ(z) = 1/ 1 − z , Re z > 1. (6.12) pk k=1
If there were only finitely many primes, this would be a finite and not an infinite product. For z = 1, it would then evaluate to a finite rational number, contradicting the fact that ζ(1) = ∞. The function π(x) denotes the number of primes not exceeding x. Viewed as a function of x real (rather than just of x integer), it is a step function that increases by 1 every time x reaches a prime. Some manipulations of (6.12)53 lead to ˆ ∞ π(x) log(ζ(z)) = z dx. (6.13) z − 1) x (x 0 Because π(x) = 0 for x < 2 (since 2 is the first prime), the lower integration limit can be set to 2 instead of to 0. A key goal is to invert relations of this type, to obtain better insights about the primes. Instead of having ζ(z) as an explicit function of π(x), it would be preferable to have π(x) as an explicit function of ζ(z). A somewhat Pmore effective starting point than the function π(x) (which we can define as π(x) = p 0 and log a a 6= 0), and xx = a (a > 0, for which x = W(log a) ), • When converging, z z
z.
.
=
W(− log z) − log z ,
6.3. The Lambert Wfunction
183
Real part
0.5 0 0.5 1 0.4 0.4
0.2
0.2
0
0
0.2
0.2 0.4
0.4
y
0.6
x
0.6
(a) Re W(z)
1
Imag part
0.5
0
0.5
1 0.5 0.5
0
0 0.5
y
0.5
x
(b) Im W(z) Figure 6.9. Real and imaginary parts of the primary solution branch of W(z).
• It solves certain nonlinear ODEs, such as
d dz W(z)
=
W(z) z(1+W(z)) ,
• It provides solutions to certain delaydifferential equations, such as dy dt = αy(t) + βy(t − τ ) with, as initial condition, y(t) specified throughout some interval t ∈ [−τ, 0]. Figure 6.9 shows how the primary branch of W(x) extends over the complex plane, singlevalued when placing a branch cut along −∞ ≤ z ≤ −1/e. These extensions are somewhat reminiscent of the upper solution sheets for the square root function f (z) = z 1/2 , as shown in Figure 2.18. Figure 6.10 shows the W(z)function when also including the next sheet in each direction across the branch cut. The lower solution branch from Figure 6.8(b) is now apparent. The sheet structure has become more reminiscent of that for the log z function (cf. Figure 2.23), not entirely surprising, since that was the inverse function to ez and W(z) is the inverse function to z ez . The function W(z) has branch
184
Chapter 6. Gamma, Zeta, and Related Functions
1 0 1
Real part
2 3 4 5 6 7 8 0.5 0 0.5
y
0.4
0.6
0.2
0
0.2
0.4
0.6
x
(a) Re W(z)
(b) Im W(z) Figure 6.10. Real and imaginary parts of the three sheets W−1 (z), W0 (z), W1 (z) of the Lambert Wfunction.
points at different locations on different sheets: at z = −1/e for the primary sheet, also at z = 0 for the adjacent two sheets, and only at z = 0 for all the (infinitely many) further sheets.
6.4 Supplementary materials Theorem 6.1 (stated). Γ(z) = limn→∞
n! nz z(z+1)·····(z+n) .
n ´n Proof. Since Γ(z) = limn→∞ 0 e−t tz−1 dt and e−t = limn→∞ 1 − nt , we can write n ˆ n ˆ 1 t (1 − τ )n τ z−1 dτ. Γ(z) = lim 1− tz−1 dt = lim nz n→∞ 0 n→∞ n 0
6.4. Supplementary materials
185
For n integer, repeated integration by parts gives 1 ˆ 1 ˆ 1 z n 1 (1 − τ )n τ z−1 dτ = (1 − τ )n−1 τ z dτ τ (1 − τ )n + z z 0 0 0 n(n − 1) · · · · · (1) = ··· = . z(z + 1) · · · · · (z + n) Q∞ 1 Theorem 6.2 (stated). Γ(z) = z eγz n=1 1 + nz e−z/n . Proof. Equation (6.2) can be written as h 1 z z z −z log n i = lim z 1 + 1+ · ··· · 1 + e Γ(z) n→∞ 1 2 n h z −z z −z z − z [1+ 12 +···+ n1 −log n]z i = lim z 1 + e 1 1+ e 2 · ··· · 1 + e n ·e n→∞ 1 2 n ∞ h Y z −z/n i = z eγz 1+ e . n n=1 Theorem 6.5. The beta function B(p, q) =
´1
B(p, q) =
0
tp−1 (1 − t)q−1 dt satisfies
Γ(p)Γ(q) . Γ(p + q)
Proof. Starting with the integrals for Γ(p) and Γ(q), we obtain ˆ ∞ ˆ ∞ Γ(p)Γ(q) = e−t tp−1 dt e−t tq−1 dt 0 0 ˆ ∞ ˆ ∞ −t p−1 q −tx q−1 = e t dt t e x dx ˆ0 ∞ ˆ ∞ 0 = xeq−1 dx e−t(x+1) tp+q−1 dt 0 0 ˆ ∞ ˆ ∞ xq−1 dx e−τ τ p+q−1 dτ = (1 + x)p+q 0 0 and the result follows from using (6.9). Theorem 6.6. The ζfunction satisfies the functional equation 1 ζ(z) = 2z π z−1 sin πz Γ(1 − z)ζ(1 − z). 2
(6.17)
Proof. (Outline) The following is a sketch of the key steps in P one of the proofs provided ∞ by Riemann. Example 5.32 started from the definition ζ(z) = n=1 n1z and then showed both ˆ ∞ z−1 1 t dt, (6.18) ζ(z) = Γ(z) 0 et − 1 for which the contour Γ1 (the interval [0, +∞]) is indicated in part (a) of Figure 6.11, and ˆ Γ(1 − z) tz−1 ζ(z) = dt (6.19) −t 2πi −1 Γ2 e
186
Chapter 6. Gamma, Zeta, and Related Functions
←
+(2N1)πi
↓
→
N
← →
Γ1
↑
+N
Γ2
Γ3
↓
↑
→
(2N1)πi
Figure 6.11. The three integration paths Γ1 , Γ2 , Γ3 considered in the proof of Theorem 6.6, shown in parts (a)–(c), respectively. There is in all cases a branch cut along the negative real axis.
(now valid for all z), where Γ2 is the Hankel contour shown in Figure 6.11(b) (well separated from the negative real axis). This last integrand has poles at t = ±2πi, ±4πi, . . . (red 1 ±πi(z−1)/2 dots in Figure 6.11(c)), with residues − (2πn) , n = 1, 2, . . . . A direct esti1−z e mate shows that the integral around the path Γ3 (shown in green in Figure 6.11(c)) will vanish if N → ∞ (meaning that the path is moved increasingly far out while passing halfway between poles on the imaginary axis). Hence, ζ(z) (based on (6.19)) will equal Γ(1−z) ·2πi· P∞2πi 1 {sum of residues}, where the sum of the residues becomes (2π)z−1 2 sin π2 z n=1 n1−z = (2π)z−1 2 sin π2 z ζ(1 − z). The last equality holds for Re z < 0, but the resulting identity will then by continuation hold for all z (note a sign change due to the direction of integration along Γ2 ).
6.5 Exercises Exercise 6.5.1. Starting from the product representation (6.3) of Γ(z), ∞ Y z −z/n 1 = z eγz 1+ e , Γ(z) n n=1
deduce the functional equation Γ(z + 1) = z Γ(z) together with Γ(1) = 1. Hint: The following sequence of intermediate steps lead to the result: 0 P∞ (z) 1 (a) ΓΓ(z) = − z1 − γ − n=1 z+n − n1 . 0
0
(z+1) (z) (b) ΓΓ(z+1) − ΓΓ(z) − z1 = 0. (c) Γ(z + 1) = α z Γ(z), where α is a constant that needs to be determined. (d) limz→0 z Γ(z) = 1. (e) Γ(1) = 1. In this step, you need that γ = limn→∞ 1 + 21 + 13 + · · · + n1 − log n .
6.5. Exercises
187
Exercise 6.5.2. Starting from the result in part (a) of Exercise 6.5.1, show that ∞ X 1 d2 log Γ(z) = . dz 2 (z + n)2 n=0
Note: This function is a special (m = 1) case of the polygamma function ψ (m) (z) = P∞ dm+1 1 (m) (z) = (−1)m+1 m! n=0 (z+n) = m+1 dz m+1 log Γ(z), with properties including ψ ´ ´ m −zt z−1 ∞ 1 t e t m+1 m (−1) 1−e−t dt = − 0 1−t (log t) dt. 0 Exercise 6.5.3. Show that the Taylor expansion of log Γ(z + 1) around z = 0 is log Γ(z + 1) = −γz +
∞ X (−1)k ζ(k) k=2
k
zk .
(6.20)
Hint: From (6.2) it follows that " log Γ(z + 1) = lim
n→∞
log n! + (1 + z) log n −
n X
# log(1 + z + k) .
k=0
Note also that log(1 + z + k) = log(1 + k) + log(1 +
z 1+k ).
Exercise 6.5.4. Show that Γ0 (1) = −γ by (a) using (6.20) and (b) using (6.3). Exercise 6.5.5. Use, for example, the result of Exercise 6.5.4 to show that ˆ ∞ γ=− e−t log t dt. 0
´ a+1 Exercise 6.5.6. Show that, for a ≥ 0, a log Γ(x)dx = a log a − a + 12 log 2π. Hint: Differentiate with respect to a, simplify, integrate back, and use (12.24) (in the x → ∞ limit) to determine the integration constant. Another option for determining the constant is to consider the case a = 1, take the logarithm of (5.6), and integrate over [0, 1] (recalling the result of Exercise 5.6.15). Exercise 6.5.7. Show that the beta function satisfies the identity B(p, q+1)+B(p+1, q) = B(p, q). ´∞ √ 2 Exercise 6.5.8. The definite integral −∞ e−x dx = π is most easily evaluated by con´∞ ´∞ 2 2 sidering the double integral −∞ −∞ e−(x +y ) dxdy and then changing to polar coordinates: (a) Carry out the details of the derivation just outlined. (b) Change the variable in the integral definition of the gamma function to obtain ´ ∞ −x 2 e dx = Γ 12 , and then evaluate Γ 21 from the functional equation (5.6): −∞ π Γ(z)Γ(1 − z) = sin(πz) . ´1 1 (c) Evaluate Γ 2 from the fact that B 12 , 21 = 0 √ dt = {set t = sin2 x} = π. t (1−t)
188
Chapter 6. Gamma, Zeta, and Related Functions
Exercise 6.5.9. Prove Theorem 6.3, Γ(2z) = π −1/2 22z−1 Γ(z)Γ z + 21 , in the following ways: ´1 (a) Expand on the sequence of steps Γ(z)·Γ(z) =B(z, z) = 0 tz−1 (1 − t)z−1 dt = Γ(2z) ´1 √ 1−2z 2 0 1 − x2 z−1 dx = {set x = u} = 21−2z B 12 , z = {set t = 1+x 2 } = 2 Γ( 1 )·Γ(z) 21−2z Γ 2z+ 1 . ( 2) 1/2 (n!)2 22n+2 . Since the (b) Using (6.2), obtain Γ(z)Γ z + 21 = 2−2z Γ(2z) limn→∞ n (2n+1)! 1 limit is a constant, its value can be obtained by setting z = 2 . Note: Still another proof, reminiscent of part (b) but not utilizing (6.2), is outlined in Exercise 12.6.9. Exercise 6.5.10. Show that ´ π/2 √ ´ π/2 √ Γ( 34 )Γ( 12 ) sin x dx = 0 cos x dx = 2Γ , (a) 0 ( 54 ) 2 ´ π/2 dx ´ π/2 dx [Γ( 1 )] (b) 0 √sin = 2√42π . = 0 √cos x x Hint: Make the variable change t = sin2 x and then use (6.5) and (6.6). Exercise 6.5.11. Show that the even order (nontrivial) Bernoulli numbers satisfy ˆ B2k = (−1)k+1 4k 0
∞
x2k−1 dx −1
(6.21)
e2πx
by Taylor expanding both sides of (5.28) in powers of a and then equating coefficients (using (5.12)). ´ ∞ xm Note: In view of (6.10), this result can also be obtained as a special case of 0 e2πx −1 dx = ζ(m+1) m! (2π)m+1 ,
m = 1, 2, 3, . . . , which follows from (6.18).
Exercise 6.5.12. Another way to continue the zeta function based on (6.18) other than changing to a Hankel contour is to apply the “partitioning of an integration interval,” method from Section 3.2.4. Carry this out, and use this to show that the zeta function’s only singularity is a first order pole at z = 1 with residue 1. Exercise 6.5.13. Show that the primes pk , k = 1, 2, 3, . . . , in the long run are more densely distributed than the sequence k 1+ε , k = 1, 2, matter how small we choose ε > 0. P∞ P3,∞. . ., no 1 Hint: Show that k=1 p1k diverges, while k=1 k1+ε converges.57 Equation (6.12) and Theorem 5.40 (from basic calculus) are helpful. Exercise 6.5.14. Show that the nontrivial zeros of ζ(z) are limited to the critical band 0 ≤ Re z ≤ 1. Hint: For z with Re z > 1, consider the logarithm of (6.12). Then, for z with Re z < 0, consider the functional equation (6.17). 57 Reminiscent
M =
P
of Euler’s constant γ (see (6.4) and Example 12.3), theP Meissel–Mertens constant is given by 1 − log(log(n)) ≈ 0.2614972128. The divergence of ∞ k=1 p is thus extraordinarily slow.
1 pk ≤n pk
k
6.5. Exercises
189
P∞ (n2 ) = 1 + Exercise 6.5.15. The Jacobi theta function, defined by ϑ(z) = n=−∞ z P∞ 2 2 n=1 z (n ) , can be shown to satisfy 3z 3 4z 4 5z 5 2z 2 z 4 + + + + · · · . (6.22) + [ϑ(z)] = 1 + 8 1−z 1 + z2 1 − z3 1 + z4 1 − z5 Taylor expanding the RHS will further reveal that all its Taylor coefficients are positive. Deduce from this that every positive integer can be written as a sum of no more than four integer squares. Notes: (i) This result is a special case of the still partly unsolved Waring problem, where squares are generalized to other integer powers. (ii) This function ϑ(z) provides an additional example (cf. Example 3.2 and Exercise 3.3.4) of a function with a natural boundary (here also located at z = 1). Exercise 6.5.16. Show that the Lambert Wfunction satisfies the ODE
W(z) d dz W (z) = z (1+W(z)) .
Exercise 6.5.17. The equation 2x = 5x has two realvalued solutions, x1 ≈ 0.2355 and x2 ≈ 4.488. Show that they are given by x = W(− log5 2 )/ log 2, where we choose either the upper or lower branch shown in Figure 6.8(b).
Chapter 7
Elliptic Functions
Functions such as sin z or ez have one direction of periodicity (along lines parallel to the real and the imaginary axes, respectively). With the complex plane being 2dimensional, it turns out that one can also create analytic functions that are periodic in two separate directions. The most important such class is known as elliptic functions, and we focus here on two such families, introduced by Weierstrass and Jacobi, respectively. Some of their key applications may at first appear unrelated to the double periodicity property, including for example, providing analytic solutions to certain integrals and nonlinear ODEs (ordinary differential equations). The name elliptic functions originates from the fact that one of the integral cases gives the circumference of an ellipse explicitly as a function of its major and minor axes (not possible to express in terms of “standard” functions). The monograph Elliptic Functions [3] contains much more information than is given here, including connections to planar geometry.
7.1 Some introductory remarks on simply periodic functions Consider first a function f (z) with a single period ω, satisfying f (z + ω) = f (z). For example, f (z) = sin z, with period ω = 2π, was illustrated previously in Figure 2.6 and is shown again here in Figure 7.1(a). The values it takes in its period strip, shaded in Figure 7.1(b), repeat in each equally sized strip that is parallel to it. The basic strip, as shown, is not uniquely determined, as it can be shifted any amount sideways, say to α ≤ Re z < α + 2π, where α is any real number. For more complicated periodic functions, with singularities, it can be convenient to choose α so that the edges of the period strip become singularityfree. If a function is, say, 2πperiodic, it has of course also the periods −2π, ±4π, ±6π, . . . . The ratio between these different choices is realvalued, and we will in general focus on the smallest one, choosing here ω = 2π.
7.2 Some basic properties of doubly periodic functions With z being a complex variable, it becomes possible for analytic functions to have two genuinely different periods ω1 and ω2 (with the ratio τ = ω1 /ω2 not realvalued). We will soon see that such functions (unless constant) cannot be singularityfree. If they are 191
192
Chapter 7. Elliptic Functions
(a) Real part of f (z) = sin z. Im z
−2π
0
2π
4π
Re z
(b) Period strip for f (z) = sin z. Figure 7.1. A period strip (here 0 ≤ Re z < 2π) for the function f (z) = sin z.
meromorphic (only poles as singularities), they are known as elliptic functions. Figure 7.2 illustrates the concept of a 2D period box, here shown with the distinct periods ω1 = 1 and ω2 = i. Apart from translating the period box (convenient in case we wish to avoid singularities on the box sides), we can also define the periods genuinely differently, as shown with the dashed parallelogram. Theorem 7.1. A nonconstant doubly periodic function must have at least one singularity inside its period box.
7.2. Some basic properties of doubly periodic functions
193
Im z
Re z
Figure 7.2. Period box(es) for a function with primary periods ω1 = 1 and ω2 = i.
Proof. If it does not have any singularity inside the period box, it will be bounded in magnitude there and therefore, by the periodicities, everywhere across the complex plane. By Liouville’s theorem (Section 4.2.3), it becomes a constant. Theorem 7.2. A (nonconstant) elliptic function has equally many poles as it has zeros inside the period box. Proof. Consider the difference between the number of zeros and number of poles in the ´ f 0 (z) ´ 1 dz = 0 (Theorem 4.8, here with denoting integraperiod box: N − P = 2πi f (z) tion around the four sides of the period box). The last equality follows from the fact that f 0 (z) must have the same double periodicity, so the contributions to the integral from the opposite box sides will cancel. The number N = P is called the order of the elliptic function. Corollary 7.3. An elliptic function takes every value equally many times. Theorem 7.4. There does not exist any elliptic function of first order. Proof. If there was one, then
residue at the simple pole
=
1 2πi
ˆ f (z)dz = 0
(by cancellation between the sides of the period box), i.e., the pole does in fact not exist.
194
Chapter 7. Elliptic Functions
7.3 The Weierstrass ℘function 7.3.1 Construction Although we have identified some properties that elliptic functions must possess if they exist, we have not yet demonstrated that they in fact do exist. The easiest demonstration is to explicitly construct one. Our first example will be the Weierstrass ℘function. This function has a double pole at the origin, with zero residue, and we denote its two periods by ω1 and ω2 (with τ = ω1 /ω2 not real). First attempt at construction: Let us place a double pole of the type 1/z 2 at the origin and also at each periodic repetition of this point, i.e., at ω = n1 ω1 + n2 ω2 , where n1 and n2 run through all integers (zero, positive, and negative). The simplest attempt would be to consider X 1 . (7.1) P (z) = (z − ω)2 ω=n ω +n ω 1
1
2
2
However, the convergence properties of this sum are not obvious. It is not absolutely convergent, as we can see by comparing with ˆ
∞
−∞
ˆ
∞
−∞
1 dxdy = r2
ˆ 0
2π
ˆ
∞
0
ˆ ∞ 1 1 r dr dθ = 2π dr = ∞, r2 r 0
(7.2)
where we in all the integrals have excluded some region around the origin and only focused on behaviors towards infinity. Second attempt at construction: The idea will be to change the function P (z) from (7.1) in concept as little as possible (so that it still features unit strength double poles at the same locations). So here is another attempt: X 1 1 1 ℘(z) = 2 + − 2 . (7.3) z (z − ω)2 ω ω6=0
With the 1/z 2 term separated out in front, and then having inserted −1/ω 2 inside the sum (which no longer includes the case of ω = 0), each term becomes (for z fixed and ω large) of size O(1/ω 3 ), instead of O(1/ω 2 ) as before. That extra power of ω means that we get one more power of r in the denominators of (7.2), showing that (7.3) will converge for z fixed. With convergence thus taken care of, it remains to verify that we did not lose the double periodicity when modifying (7.1) to (7.3). Theorem 7.5. The Weierstrass ℘function, as defined by (7.3), is doubly periodic. P 1 Proof. From (7.3) it follows, by differentiation, that ℘0 (z) = −2 ω (z−ω) 3 , which 3 converges (since its terms are again of size O(1/ω )). This sum is clearly doubly periodic, implying that ℘0 (z + ωi ) − ℘0 (z) = 0, i = 1, 2. Integration gives ℘(z + ωi ) − ℘(z) = C.
(7.4)
Substituting z = −ωi /2 into (7.4) gives ℘(ωi /2) − ℘(−ωi /2) = C. Since (7.3) shows that ℘(z) is an even function of z, the constant C must be zero. Equation (7.4) then confirms that ℘(z) indeed is doubly periodic.
7.3. The Weierstrass ℘function
195
40 20 0 20 40 1.5 1 0.5 1.5
0
1 0.5
0.5
y
0 0.5
1 1.5
1
x
1.5
(a) Re ℘(z)
40 20 0 20 40 1.5 1 0.5 1.5
0
1 0.5
0.5
y
0 0.5
1 1.5
1 1.5
x
(b) Im ℘(z) Figure 7.3. Real and imaginary parts of the Weierstrass ℘function in the case of periods ω1 = 1 and ω2 = i.
Figures 7.3 and 7.4 show the ℘function in the special case when the periods have been chosen as ω1 = 1 and ω2 = i. The pattern of equispaced double poles extends over the full complex plane.
7.3.2 Some nonlinear ODEs that are satisfied by the ℘function Elliptic functions satisfy various different nonlinear ODEs, many of which arise in applications. We continue to use the ℘function as an illustrative example. The Laurent expansion of ℘(z) around the origin will take the form 1 + 0 + a2 z 2 + a4 z 4 + · · · , (7.5) z2 P where the coefficients depend on the periods: ak = (k+1) ω6=0 1/ω k+2 , k = 2, 4, 6, . . . . Since ℘(z) is even, all coefficients for odd powers vanish. That is also the case for a0 , seen ℘(z) =
196
Chapter 7. Elliptic Functions
Figure 7.4. Magnitude and phase plot for the same case as in Figure 7.3.
from letting z → 0 in (7.3). From (7.5) follows ℘0 (z) = −
2 + 2a2 z + 4a4 z 3 + · · · , z3
℘0 (z)2 =
4 8a2 − 2 − 16a4 + · · · . z6 z
and therefore
We obtain the same leading term out by subtraction:
4 z6
3
if we form 4[℘(z)] , implying that we can cancel it
℘0 (z)2 − 4℘(z)3 = −
20a2 − 28a4 + · · · . z2
Adding 20a2 ℘(z) to this will additionally cancel out the first term in the RHS, leading to ℘0 (z)2 − 4℘(z)3 + 20a2 ℘(z) = −28a4 + · · · .
(7.6)
The LHS of (7.6) is doubly periodic with its only possible singularity location (in the period box) at z = 0. However, the RHS is singularityfree at this point. By Theorem 7.1, the LHS function must be a constant. All further terms in the RHS of (7.6) (beyond −28a4 ) must therefore also have vanished. It is conventional to introduce the new parameters g2 = 20a2 and g3 = 28a4 . We can then write the ODE satisfied by ℘(z) as ℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 .
(7.7)
If g2 and g3 are specified, one can compute matching periods ω1 and ω2 from the formula for ak . The same procedure as just employed will give many additional ODEs satisfied by ℘(z), such as 1 (7.8) ℘00 (z) = 6℘(z)2 − g2 , 2 ℘000 (z) = 12℘(z)℘0 (z), etc. We will return to (7.8) in Section 11.5.1.
7.4. The Jacobi elliptic functions
197
7.3.3 Closedform expressions for certain integrals based on the ℘function p dz Equation (7.7) can be written as 1 = ℘0 (z)/ 4℘(z)3 − g2 ℘(z) − g3 . Using d℘ dz = 1/ d℘ , p we obtain for the inverse function z(℘) the relation z 0 (℘) = 1/ 4℘3 − g2 ℘ − g3 . At this point, ℘ has become the name of an independent variable, which we just as well can call, say, x. By means of the inverse function z(x), we have thus obtained an analytic expression for the indefinite integral ˆ dx p (7.9) 3 4x − g2 x − g3 for arbitrary constants g2 and g3 . By choosing other elliptic functions, (7.9) can be generalized to ˆ dx √ . arbitrary quartic polynomial in x
7.4 The Jacobi elliptic functions This family of elliptic functions differs from the Weierstrass ℘functions primarily in that the period box is assumed to be rectangular, and containing two first order poles (of opposite residues) instead of a single pole (of order 2 and residue zero).58 Many of the notations for these functions come from their connection with conformal mappings; see Section 8.6. For example, the Jacobi sn(z, k) function arises there in the implicit form ˆ
sn(z,k)
z=
du [(1 −
0
u2 )(1
1/2
− k 2 u2 )]
,
(7.10)
where k is a parameter in the range 0 ≤ k ≤ 1.59 However, we start here with a more direct approach to the Jacobi elliptic functions by defining a periodic box and then just placing two poles inside it.
7.4.1 The Jacobi sn(z,k ) function We first create a period box of size 0 ≤ Re z ≤ 4K, 0 ≤ Im z ≤ 2K 0 , where both K and K 0 depend on a single free parameter k ∈ [0, 1] through the relations ˆ K(k) = 0
1
du 1/2
[(1 − u2 )(1 − k 2 u2 )]
and K 0 (k) = K(k 0 ), with k 0 =
p
1 − k2 .
(7.11) (We follow here again notational conventions: K and K 0 are related functions, and k and k 0 are related parameter values, with the primes here having nothing to do with derivatives.) Section 7.4.3 describes a simpler representation of this function K(k) (and thereby also K 0 (k)). Figure 7.5 shows the two functions. In the period box, we next place a first order 58 There is nevertheless a close connection: Equation (7.7) tells us that ℘(z) will not have a double zero unless the parameter g3 = 0. The inverse 1/℘(z) will therefore in general be another elliptic function of order 2, but with two separate first order poles (of opposite residue). 59 It is in the literature also common to parameterize the Jacobi elliptic functions by m = k 2 instead of by the “elliptic modulus” k directly, as we do here (for example, both MATLAB and Mathematica use m).
198
Chapter 7. Elliptic Functions 6 K K' /2
5 4 3 2
/2 1 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
k
Figure 7.5. The two functions K(k) and K 0 (k). K(0) = K 0 (1) = 2 √ intersect at k = √12 , with K( √12 ) = K 0 ( √12 ) = Γ( 14 ) /(4 π).
π . 2
The two curves
pole with residue +1/k at z = iK 0 and one with residue −1/k at z = 2K +iK 0 . Repeating this period box across the full complex plane, and summing the contributions from all the resulting infinitely many poles, gives rise, like for our first attempt at constructing the Weierstrass ℘function, to a divergent double sum. However, this divergence can be avoided similarly to the “second attempt” (also in Section 7.3.1). The result is known as the Jacobi sn(z, k) function, illustrated over one period box in Figure 7.6. If we let the parameter k approach zero, the period box approaches 2π in width, while its height increases towards infinity. The sn(z, k) function then approaches sin z (which we can see indications of along the real axis already in Figure 7.6(a)). In the opposite limit of k approaching 1, the period box increases towards infinity in width while shrinking in height to π, and sn(z, k) then approaches tanh z. The red curves in Figure 7.7 show these variations in sn(x, k) over one period [−2K, +2K] in the real variable x for some different values of k.
7.4.2 Some other Jacobi elliptic functions Similar to how we from sin z obtain cos z, we can from sn(z, k) create numerous additional elliptic functions, with the most commonly used ones being cn(z, k) and dn(z, k), defined through the relations cn2 (z, k) + sn2 (z, k) = 1
and
dn2 (z, k) + k 2 sn2 (z, k) = 1.
The green and blue curves in Figure 7.7 show counterparts to the red sncurves for the functions cn(x, k) and dn(x, k), respectively. For k increasing from 0 to 1, both cn(z, k) and dn(z, k) transition to 1/ cosh(z), starting from cos(z) and 1, respectively. These elliptic functions solve numerous nonlinear ODEs in closed form. For example, sn(x, k) solves cn(x, k) solves dn(x, k) solves
d2 y dx2
+ (1 + k2 )y − 2k2 y 3 = 0
and
2
+ (1 − 2k2 )y + 2k2 y 3 = 0
and
2
− (2 − k2 )y + 2y 3 = 0
and
d y dx2 d y dx2
dy 2 dx dy 2 dx dy 2 dx
= (1 − y 2 )(1 − k2 y 2 ), = (1 − y 2 )(1 − k2 + k2 y 2 ), = (y 2 − 1)(1 − k2 − y 2 ).
There is also a wealth of closedform integrals and other relations satisfied by the Jacobi elliptic functions. The NIST Handbook of Mathematical Functions [35, Chapter 22] contains a summary. As noted above, one of the integrals provides a closedform expression for the circumference of an ellipse.
7.4. The Jacobi elliptic functions
199
5
0
5 4
3
2 6 5 4
1 3
y
2 0
1 0
x
(a) Re sn(z, k) for k = 12 .
5
0
5 4
3
2 6 5 4
1 3
y
2 0
1 0
x
(b) Im sn(z, k) for k = 12 .
(c) sn(z, k) for k = 12 . Figure 7.6. Visualization across one period box of sn(z, k) in the case of k = 12 . For this kvalue, K ≈ 1.6857503548 and K 0 ≈ 2.1565156475; i.e., the period box is roughly of size [0, 6.7430] × [0, 4.3130]. In parts (a) and (b), some positive contour lines are highlighted in blue and negative ones in green. As usual, values along the real axis are shown in red for (a) and (b), and in black for (c).
200
Chapter 7. Elliptic Functions k = 0.999
1 0 1 10
8
6
4
2
0
2
4
6
8
10
2
4
6
8
10
2
4
6
8
10
k = 0.99 1 0 1 10
8
6
4
2
0
k = 0.9 1 0 1 10
8
6
4
2
0
k=0 1 sn cn dn
0 1 10
8
6
4
2
0
2
4
6
8
10
Figure 7.7. The variations of the sn(x, k), cn(x, k), and dn(s, k)functions along the real axis over one spatial period in the cases of four different kvalues.
7.4.3 The arithmeticgeometric mean We have in previous sections often defined analytic functions as the limit of a convergent sequence of simpler functions, such as any time we define a function through its Taylor series, Weierstrass product, or Mittag–Leffler sum, or by means of an integral. The arithmeticgeometric mean (AGM) is yet another process that generates convergent sequences. We describe it first as a function AGM(x, y) of two real variables, x ≥ 0 and y ≥ 0. One then sets x0 = x, y0 = y and iterates xn+1 =
xn + yn , 2
yn+1 =
√ xn yn ,
n = 0, 1, 2, . . . .
(7.12)
The two sequences {xn } and {yn } approach each other, converging to a joint limit we denote by AGM(x, y). Apart from the trivial cases of either x = 0 or y = 0, the convergence rate is extraordinarily high, with the number of digits in common between xn and yn roughly doubling with each iteration (see Exercise 7.6.5). Figure 7.8 illustrates this function AGM(x, y). It is zero along both the x and the yaxes, and equals x when y = x (green straight line; the red curve is explained below). In 1799, C.F. Gauss, by numerical hand calculations, “stumbled onto” the relation K(k) =
π 2 AGM(1, k 0 )
(7.13)
7.4. The Jacobi elliptic functions
201
3
2
1
0
3 2 1
y
0
0.5
0
1.5
1
2
3
2.5
x
Figure 7.8. The function AGM(x, y) in the first quadrant of the (real) (x, y)plane. The red and green curves are explained in the text. π (and thus K 0 (k) = 2 AGM(1,k) ) in the notation of (7.11).60 Gauss followed this up by noting61 that, for x, y ≥ 0,
ˆ
π/2
0
ˆ
dθ p
x2
cos2
θ+
y2
2
=
sin θ
0
π/2
dϕ q
x+y 2 2
cos2 ϕ +
√
xy
2
.
(7.14)
sin2 ϕ
From this it follows directly that ˆ
π/2
0
π dθ p = . 2 AGM(x, y) x2 cos2 θ + y 2 sin2 θ
(7.15)
As outlined in Exercise 7.6.6, this in turn leads to a proof of (7.13). Another consequence of (7.15) (cf. Exercise 7.6.8) is ˆ 1 1 π dθ √ = AGM(1 + x, 1 − x) π 0 1 − x2 cos2 θ 2 2 2 1 1·3 1·3·5 2 4 =1+ x + x + x6 + · · · . (7.16) 2 2·4 2·4·6 The red curve in Figure 7.8 shows this function AGM(1 + x, 1 − x) over the range −1 ≤ x ≤ 1. Since the recursion for AGM is rapidly converging, replacing x with z (complex) in AGM(1+x, 1−x) will give rise to an analytic function. This function is displayed in Figure 7.9. Along the real axis, for −1 ≤ x ≤ 1, we recognize the red curve from Figure 7.8, 60 G
=
1 √ AGM(1, 2)
=
2 π
´1 0
√ dx
1−x4
=
1 B( 14 , 21 ) 2π
constant. 61 By means of the ingenious change of variable sin θ =
≈ 0.8346268417 has become known as the Gauss 2x sin ϕ . (x+y)+(x−y) sin2 ϕ
202
Chapter 7. Elliptic Functions
2 1.5 1 0.5 2
0 2
1 1
0 0 1
1
x
2
y
2
(a) Re AGM(1 + z, 1 − z).
1 0.5 0 0.5 2
1 2
1 1
0 0 1
1
x
2
2
y
(b) Im AGM(1 + z, 1 − z).
(c) Magnitude and phase angle for AGM(1 + z, 1 − z). Figure 7.9. Display of the analytic function AGM(1 + z, 1 − z).
7.5. Supplementary materials
203
but see now also that the function AGM(1 + z, 1 − z) is well defined across the complex plane, however with branch points at z = ±1. The three expressions in (7.16) all represent the same function, but have quite different characters. The AGM expression is superior for numerical evaluations, while the integral arises in numerous application contexts.62 The validity of the Taylor series representation is limited by its radius of convergence R = 1.
7.5 Supplementary materials 7.5.1 The modular function λ(τ ) The RHS in (7.7) is a cubic in ℘(z), which can be factored as ℘0 (z)2 = 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 ),
(7.17)
where the ei , i = 1, 2, 3, are the three zeros of the order 3 elliptic function ℘0 (z). From the basic properties of the ℘function, it transpires that e1 = ℘(ω1 /2), e2 = ℘(ω2 /2), e3 = ℘((ω1 + ω2 )/2) will satisfy this (cf. the proof in Section 7.5.4). These ei values are always different from each other. If one further notes that each of them becomes multiplied by 1/t2 if the two ωi are both multiplied by t, one can conclude that the ratio (e3 −e2 )/(e1 − e2 ) will become a singlevalued analytic function of the ratio τ = ω1 /ω2 . This function (e3 −e2 ) λ(τ ) = (e is known as a modular function. It is of great interest in its own right, 1 −e2 ) independently of its definition based on an elliptic function. We can’t here go through its properties in any detail (for that, see more advanced books, such as Ahlfors [2] or Copson [8]), but let us mention that w = λ(z) maps the boundary of the region Ω, shown in Figure 7.10 on the left, to the real axis and the inside onetoone to the upper halfplane (Figure 7.10 (right)). Given this, it is natural to display this function graphically over the region Ω rather than, as we usually do, over the region surrounding origin in the complex plane; see Figure 7.11. In Figure 7.12 (left), we show also the region Ω reflected across the imaginary axis, denoting this by Ω. By Schwarz’s reflection principle, the function λ(z) will in this region take complex conjugated values compared to in Ω, implying that its values will be as shown in 7.12 (right), where we note the branch cuts along [−∞, 0] and [1, ∞]. We denote the inverse function to w = λ(z) as z = λ−1 (w). It can be shown to have no other singularities than these branch points at w = 0 and w = 1.
7.5.2 Picard’s theorem This theorem provides intriguing insights into the possible behaviors of entire functions (functions that are singularityfree in the finite complex plane). As a background, let us first recall that a polynomial of degree d takes every finite value d times (when counting the multiplicity of roots). We also know from Liouville’s theorem (Theorem 4.12) that, if the magnitude of an entire function is bounded, it is a constant. The Caserati–Weierstrass 62 A
classic example concerns the period T of a pendulum of length l, with all its mass at its end. If θ0 is its maximal deviation from vertical, k = sin θ20 , and g is the acceleration of gravity, then T = q ´ q q 2 gl 0π √ dθ = 2π gl /AGM(1 + k, 1 − k). For θ0 small, so is k, and T ≈ 2π gl . If θ0 2 2 1−k sin θ
increases towards π, then T → ∞.
204
Chapter 7. Elliptic Functions
Figure 7.10. Illustration of the mapping w = λ(z). We use here colors, not to indicate the phase, but as a texture, identifying matching points in the zplane and the wplane. In particular, we see that λ(z) maps the edge of the domain Ω, shown on the left to the real waxis. Letting z follow the Ωdomain boundary along the path 1 → 1 + i∞(→ i∞) → 0 → 1, then w follows the real axis from left to right: −∞ → 0(→ 0) → 1 → +∞. The colored regions (extending to infinity) are mapped onetoone to each other. The bottom boundary of Ω is a halfcircle centered at z = 12 .
theorem (Theorem 4.30) tells us that an entire function comes arbitrary close to any finite value infinitely many times. Some entire functions actually take (rather than just come arbitrarily close to) every value infinitely many times (e.g., sin z), while others avoid one value altogether (e.g., ez and Γ(z) never take the value zero for any finite z). Theorem 7.6 (Picard, 1879). An entire function that avoids two values must be a constant. Proof. (The following is a brief sketch only, outlining the key idea behind the original proof by Picard.) If the function F (z) avoids the two (finite) values a and b, we consider f (z) = F (z)−a b−a and it remains to prove that an entire function that avoids both 0 and 1 must be a constant. The essential idea is to consider the function Λ(z) = λ−1 (f (z)) and then apply Morera’s theorem (Theorem 4.3). If we choose any closed path in the zplane, we know that f (z) along this path can never encircle either of the points 0 or 1 (as we otherwise could shrink the path to a point at which this would conflict with the fact that f (z) cannot take these values). This means that the integral also of Λ(z) around the path will be zero, and thus (by Morera’s theorem) represent an analytic function of z. However, the values of Λ(z) are all in the upper halfplane (cf. Figure 7.12 (left)). The function ei Λ(z) is then a bounded entire function and according to Liouville’s theorem a constant. Then, so are Λ(z), λ(z), f (z), and F (z). Another related result (also by Picard) is that if a meromorphic function (poles as only singularities, here also including at infinity) omits three values, it has to be a constant.
7.5. Supplementary materials
205
(a) Re λ(z)
(b) Im λ(z) Figure 7.11. Real and imaginary parts of the modular function λ(z). For part (a) we note that Re λ(z) increases monotonically as z follows the boundary 1 → ∞ → 0 → 1. For (b) we note that Im λ(z) is zero along the whole boundary and everywhere positive inside it.
7.5.3 Conformal mapping of a doubly periodic region to a torus Similar to how the stereographic projection (Section 1.4) mapped the whole complex plane to the surface of a sphere, the doubly periodic regions that arise for elliptic functions can be mapped conformally to tori. Example 7.7. Consider the square doubly periodic region −π ≤ x ≤ π, −π ≤ y ≤ π and the mapping cos x X=√ , 2 − cos y
Y =√
sin x , 2 − cos y
Verify that this mapping indeed is conformal.
sin y Z=√ . 2 − cos y
(7.18)
206
Chapter 7. Elliptic Functions
Figure 7.12. Once again, in this illustration, the colors are used as a texture rather than to indicate the argument. The onetoone mapping is given by λ(z) between a part of the upper halfplane (composed of the regions Ω and Ω) and the entire complex plane. On the right, note the branch cuts along [−∞, 0] and [1, ∞].
The mapping is illustrated in Figure 7.13, with the doubly periodic region in the complex z = x + iy plane to the left and the resulting surface in X, Y, Z space to the right. Following Exercise 1.6.21, we evaluate 2 2 2 ∂Y ∂Y ∂Z ∂Z (∆x)2 + (∆y)2 + ∆x+ ∆y + ∆x+ ∆y = √ , ∂x ∂y ∂x ∂y ( 2−cos y)2 (7.19) from which it follows that the mapping indeed is conformal.
∂X ∂X ∆x+ ∆y ∂x ∂y
The torus in this example is the surface of revolution generated by a circle of radius 1 √ in the (X, Z)plane, centered at X = 2, Z = 0 and rotated √ around the Zaxis (described √ as having major axis 2, minor axis 1, and aspect ratio 2).63
7.5.4 Select proofs Theorem 7.8. The three zeros of ℘0 (z) in the period box are z1 = ω1 /2, z2 = ω2 /2, and z3 = (ω1 + ω2 )/2. Proof. Since ℘(z) is even and has a period ω1 , it holds that ℘(z) = ℘(−z) = ℘(ω1 − z), and therefore ℘0 (z) = −℘0 (ω1 − z). For z = ω1 /2,we then get ℘0 (ω1 /2) = −℘0 (ω1 /2), i.e., ℘0 (ω1 /2) = 0. The same argument can be utilized for the other two roots. Since ℘0 (z) is an elliptic function of order 3, it cannot have any further zeros (inside the period box). 63 If
the (rectangular) period box is not square, two different tori aspect ratios become possible (by exchanging the roles of x and y in a formula similar to (7.18)).
7.6. Exercises
207
3 2
0.5
y
Z
1 0
0 0.5
1
1
2
2
1
3
2
0
2
x
2
1
0
0
1
1
Y
2
2
X
Figure 7.13. The conformal mapping given by (7.18) from the periodic square −π ≤ x ≤ π, −π ≤ y ≤ π to the surface of a torus in X, Y, Zspace.
7.6 Exercises Exercise 7.6.1. Consider the Weierstrass ℘(z) function in the special case of the periods being ω1 = 1 and ω2 = i: (a) Show that ℘(z) has a double zero at the center of the square period box, i.e., at z = 12 (1 + i). (b) Show that ℘(z) · ℘(z + 12 (1 + i)) is identically constant. Hint for (a): Deduce first that (i) Re ℘(z) = 0 along the diagonals of the period box, and (ii) Im ℘(z) = 0 along the sides of the box, as well as along Re z = 21 and Im z = 12 . Exercise 7.6.2. Show that ℘(z) has a double zero if and only if g3 = 28a4 = 0. Exercise 7.6.3. Consider once again the Weierstrass ℘(z) function in the special case of the periods being ω1 = 1 and ω2 = i. (a) Show that g3 = 0. P (b) Compute g2 . Using the formula ak = (k + 1) ω6=0 1/ω k+2 , carry out the procedure described in Section 5.2 to obtain ! ∞ X 2 8 1 2 4 4 + − coth(nπ) + 2 coth(nπ) . g2 = 60π 45 n=1 3 3 Notice that the sum converges rapidly and that already two terms are enough to give an accurate approximation. For comparison, Mathematica’s WeierstrassInvariants[{0.5, 0.5I}] (based on halfperiods) gives the result {189.073, 0}. (c) Show that g2 = −4 · ℘(z) · ℘(z + 12 (1 + i)) for all z ∈ C. Exercise 7.6.4. For some kvalues in the range k ∈ [0, 1], evaluate the corresponding values for K(k) and K 0 (k) by means of (7.13). Verify that your results match what is quoted in the captions of Figures 7.5 and 7.6.
208
Chapter 7. Elliptic Functions
Exercise 7.6.5. Prove the quadratic convergence of the AGM iterations. Hint: Derive from (7.12) the relation xn+1 − yn+1 = 4(xn+11+yn+1 ) (xn − yn )2 . Exercise 7.6.6. Make the change of variable u = sin θ in (7.11) to show that (7.13) follows from (7.15). Exercise 7.6.7. Show that the value for the two integrals in Exercise 6.5.10(a) can be written as 1/G where G is the Gauss constant (cf. Section 7.4.3). Exercise 7.6.8. Derive from (7.15) both the results in (7.16). Exercise 7.6.9. The Jacobi elliptic function sn(z, k) is characterized by having first order poles of opposite residue placed at the locations z = 0 + iK 0 and z = 2K + iK 0 within the period box, of horizontal and vertical sides of lengths 4K and 2K 0 , respectively. Show that immediate double summation over all repetitions of this period box leads to a not absolutely convergent sum, and then show how a certain regrouping of the terms will provide this (convergence even if we take the magnitude of all its terms). Exercise 7.6.10. Verify that the surface in Example 7.7, as given by (7.18), indeed is the surface √ of revolution generated by a circle of radius 1 in the (X, Z)plane, centered at X = 2, Z = 0, and rotated around the Zaxis.
Chapter 8
Conformal Mappings
The previous chapters have contained many illustrations of analytic functions f (z), often in the form of displaying Re f (z), Im f (z), and/or f (z) as surfaces over the complex zplane. A different type of illustration is obtained by choosing some region in the zplane and then showing in a complex wplane where the corresponding values of w = f (z) end up.64 Figure 8.1(a) shows an arbitrarily chosen region in the zplane, and parts (b) to (i) the matching regions in the wplane for some different simple choices of f (z). All of these cases illustrate onetoone mappings for the displayed regions (i.e., over these regions, the inverse function z = f −1 (w) is also singlevalued). Only case (b) (a combination of a translation and a rotation) would be singlevalued in both directions for arbitrary regions. Maybe the most striking feature in these figures is that the grid, shown in green in part (a), will correspond to a curve set in the wplane where again all the curves intersect at right angles. This will turn out to be the case at any point where f (z) is analytic, with f 0 (z) 6= 0. This, furthermore, is a special case of a more general property: Not only right angles but all angles (including sign) are locally preserved. In reverse, only analytic functions w = f (z) can have this conformal mapping property.65 Figure 8.1 is somewhat unsatisfactory in that neither the initial region in the zplane, nor the resulting regions in the wplane, is of any special significance. What makes the topic of conformal mappings important for certain applications is that one can specify both regions independently, and there will then exist (a 3parameter family66 of) analytic functions w = f (z) that perform that mapping in a onetoone manner. Although it is somewhat rare to be able to find an appropriate f (z) in closed form, it can typically be computed accurately (with algorithms that fall outside the scope of this book). The rest of this chapter will highlight the following: • Relations between conformal mappings and analytic functions (Section 8.1). • Mappings provided by bilinear functions (Section 8.2). 64 That
is, for each point that is marked in the zplane (for example, by color), evaluate w = f (z) and graphically mark the corresponding point in the wplane identically. 65 Mappings between the complex plane and other types of surfaces can also be conformal, as we recall from the stereographic projection (Section 1.4 and Exercises 1.6.20 and 1.6.21) and the mapping from a square to a torus (in Section 7.5.3). 66 The number “3” here follows from the special case of mapping the inside of the unit circle to itself, as discussed following Theorem 8.7.
209
210
Chapter 8. Conformal Mappings
y
2.5
y
2
6
2 1.5
y
1.5 1
4
1 2
0.5
0.5
0 0
0.5
1
1.5
0
2
0.5
1
(a) Initial domain in complex zplane.
1.5
2
4
2
0
2
4
x
x
(b) The mapping √ z. w = 1 + 1+i 2
(c) The mapping w = z2.
x
y 2.5 2
y 6
1.5 4
0.6
1
0.4
0.5
y
2 0.2 0 0
0.2
0.4
0.6
0.8
1
2
1.2
0
2
(d) The mapping w = z 1/3 .
4
1
6
0
1
x
x
(e) The mapping w = ez .
(f) The mapping w = arctan z.
x
3 0.6 y
2.5
0.4
2
0.2
1.5
1
y
0.5 y
0.5
x
1
1
0.2 x
0.2
0.4
0.4
0.6
0.5
x 0 1
2
0.5
(g) The mapping w = z −1/3 .
(h) The mapping w = 21 z + z1 .
(i) The mapping w = sn(z, 0.7).
Figure 8.1. (a) An arbitrarily chosen domain in the first quadrant. (b)–(i) The images of this domain (and of the green lattice) when mapped to a complex wplane through some different mapping functions w = f (z).
8.1. Relations between conformal mappings and analytic functions
211
• Riemann’s mapping theorem (Section 8.3). • Mappings of polygonal regions (Section 8.4). • Some applications of conformal mappings (Section 8.5).
8.1 Relations between conformal mappings and analytic functions A pair of functions
u = u(x, y) v = v(x, y)
(u, v, x, y real) form a conformal (anglepreserving) map if any two intersecting curve segments in the (x, y)plane become two intersecting curve segments in the (u, v)plane, such that the angle between the curves (at the intersection points) has stayed the same. The following theorem shows that this property is unique to analytic functions. Theorem 8.1. The mapping
u = u(x, y) v = v(x, y)
with ux , uy , vx , vy continuous is conformal at a location z = x + iy if and only if f (z) = u + iv is an analytic function, with f 0 (z) 6= 0. Proof. If f (z) is analytic at z0 , its Taylor expansion there starts w = f (z) = f (z0 ) + (z − z0 )f 0 (z0 ) + · · · ; i.e., it is locally around z0 a combination of a translation (given by f (z0 )), a uniform scaling (given by f 0 (z0 )), and a rotation (given by arg f 0 (z)), all of which preserve angles. The proof for the reverse statement (local angles preserved implies analyticity) is more lengthy and is given in Section 8.7 as Theorem 8.18. The next theorem makes it relatively simple to check that a mapping function is onetoone. Theorem 8.2. Let D be a finite region in the zplane, with boundary δD, corresponding to E and δE in the w = f (z) plane, with neither boundary selfintersecting. Then (a) w = f (z) provides a onetoone map from D to E, and (b) if z moves around δC in the positive direction, so does w around δE. Proof. Let w0 be a point inside E, and consider ˆ ˆ 1 f 0 (z) dz 1 dw I= = , 2πi δD f (z) − w0 2πi δE w − w0 where z moves around δD in the positive direction. By Theorem 4.8, I = N − P , where N and P are the number of zeros and poles, respectively, of f (z) − w0 . With P = 0, N ≥ 0, and the RHS = ±1 according to the direction w moves around δE, the theorem follows.
212
Chapter 8. Conformal Mappings
In the following, we will reserve the expression “conformal mapping” for such ones that have the onetoone property. Let us note again that the functions w = f (z) and z = g(w), mapping between the domains D and E in the z and wplanes, respectively, are assumed to be analytic (singularityfree) within their respective domains.
8.2 Mappings provided by bilinear functions It turns out that bilinear functions67 az+b , (8.1) cz+d the ratio between two linear expressions in z, have a number of remarkable properties when it comes to mapping simple regions. w=
Theorem 8.3. Any straight line or circle in the zplane becomes mapped by a bilinear function into another circle or straight line. Proof. If c = 0, the result is obvious, since (8.1) then just amounts to a sequence of (i) scaling/rotation (forming a z), (ii) shift (adding b), and (iii) scaling/rotation (dividing by d), none of which can change a circle/line away from still being a circle/line. For c 6= 0, we rewrite (8.1) as az+b 1 1 w= = (bc − ad) +a , cz+d c cz + d and it is again just a sequence of scaling/rotations and shifts, with the only novelty that it also includes one inversion. Hence, it remains to show that if z is on a circle/line in the complex plane, so is w = 1/z. However, this follows from Exercises 1.6.17 and 1.6.18. A circle/line in the complex plane becomes a circle on the stereographic sphere, and inversion is associated with a rotation of the sphere, i.e., preserves the circular shape. Then, it again becomes a circle/line when projected back to the complex plane. Theorem 8.4. Successive bilinear mappings become again a bilinear mapping. z+b1 Proof. Straightforward algebra shows that w = ac11z+d followed by ζ = 1 a3 z+b3 written as ζ = c3 z+d3 , where a3 b3 a2 b2 a1 b1 = . c3 d 3 c2 d2 c1 d1
Theorem 8.5. The inverse of a bilinear mapping w = w+b2 bilinear map z = ac22w+d . 2 h
a1 c1
b1 d1
i
a1 z+b1 c1 z+d1
a2 w+b2 c2 w+d2
can be
(8.2)
takes also the form of a
z+b1 6= 0, since otherwise w = ac11z+d reduces to w = 1 h i h i−1 constant (with no zdependence). By (8.2), we should choose ac22 db22 as ac11 db11 , since the combined mapping then becomes the identity; i.e., starting from z, we get back to z.
Proof. We can assume det
67 Also
known as Möbius transforms or linear fractional transforms.
8.2. Mappings provided by bilinear functions
1
1
1
−1
213
1
1
1
−1
1
−1
−1 −1
1
−1
−1
(a) zplane.
(b) wplane.
(c) zplane.
(d) wplane.
Figure 8.2. Mapping by the bilinear function w = 1/z of the inside of the unit circle to its outside and by a shifted unit circle to a halfplane. This function maps zero to infinity, and vice versa.
Among the family of bilinear mappings, the Blaschke factor, z−a , w=λ 1 − az
(8.3)
where a is an arbitrary complex constant and where λ = 1, is particularly important due to the following result. Theorem 8.6. The bilinear mapping (8.3) maps z = a to 0 and z = 1 to w = 1. Proof. The result = a → 0 is obvious. The second part needs verification: w2 = zz z−a z−a z−z a+a a z−a 1−a z 1−a z = 1−a z−a z+a a z z = 1. In the last equality, we utilized twice that z is on the unit circle, i.e., that z z = 1. Figures 8.2 and 8.3 illustrate the mappings performed by two different bilinear functions. This can be contrasted to the mapping in Figure 8.4, which is not bilinear, and hence can map a circle to a different type of curve, here to an ellipse. The function in this figure is the same as the function shown earlier in Figure 2.2. Theorem 8.7. There is no other way to conformally (onetoone) map the inside of the unit circle and its center to themselves than through the Blaschke factor (8.3) for a = 0, w = λz, where λ is a complex constant of magnitude 1. 1
1
1 1
−1
1
−1
1 1
−1 −1
−1
(a) zplane.
1
−1 −1
−1
(b) wplane.
(c) zplane.
(d) wplane.
Figure 8.3. Two mappings by the bilinear function w = (z − 12 )/(1 − 12 z). Since this function is of the form w = (z − a)/(1 − a z), the mapping takes the unit circle to itself (red) and inside to inside and outside to outside. The blue circle becomes again a circle (since the mapping is of bilinear form). This function maps z = 0 to w = − 21 , z = 2 to w = ∞, and z = ∞ to w = −2. The inverse mapping is z = (w + 21 )/(1 + 12 w).
214
Chapter 8. Conformal Mappings
1
1
1
1
1
−1
1
−1
1
−1
1
−1
−1 −1
−1
(a) zplane.
−1
(b) wplane.
(c) zplane.
(d) wplane.
Figure 8.4. Two mappings by the function w = 12 (z + 1/z). This function is not bilinear, and we note in the second part a circular arc becoming an elliptic arc. The mapping is singular at the points z = ±1 (corresponding to w = ±1), and local angles are doubled at these points.
Proof. Let w = f (z) be another such mapping, and consider g(z) = f (z)/z. Since the mapping is onetoone, f (z) has a unique zero inside the unit disk. It must be located at the origin, since z = 0 is mapped to w = 0, and hence the singularity of g(z) at z = 0 must be removable, making g(z) and consequently log(g(z)) = log g(z) + i arg g(z) analytic within the unit disk. By Theorem 4.18 (see also Section 2.2.3), as a harmonic function, log g(z) can have neither a minimum nor a maximum within the disk. Since g(z) = 1, and so log f (z) = 0, around z = 1, log g(z) must remain identically 0 within the disk; i.e., g(z) has to be identically 1. The theorem above can quite easily be strengthened to apply also when f (z) is known to be analytic only for z < 1 rather than for z ≤ 1 by considering a limit process as z % 1 (known as the Schwarz lemma). Between Theorems 8.6 and 8.7, we can conclude that a mapping from the unit circle to itself has precisely three free (real) parameters, of which two can be used to choose the (complex) point a and the third to specify λ = eiθ (obeying λ = 1). Combining some of the results above leads to the following. Theorem 8.8. The most general mapping of the upper halfplane to itself can be written in the form w = az+b cz+d , where a, b, c, d are all real and satisfy ad > bc. Theorem 8.9. The most general conformal mapping of the whole stereographic sphere to itself takes the form of a bilinear function. The next section tells how some of the results above can be greatly generalized, from the unit circle (or upper halfplane) to arbitrarily shaped domains, assumed, however, to be simply connected, i.e., neither the whole plane nor regions with nonconnected boundaries (as were seen in the subplot of Figure 8.3).
8.3 Riemann’s mapping theorem Theorem 8.10. Let D be a simply connected domain (excluding the entire complex plane). There exists then a 3parameter family of functions f (z) that maps the unit circle onetoone to D (or, in reverse, D to the unit circle).
8.4. Mappings of polygonal regions
215
This result was formulated by Riemann as a brief part of his 1851 Ph.D. thesis. Weierstrass later found the proof to be incomplete, and the first rigorous one was not given until 1912. While strict proofs remain technically difficult, we sketch in Section 8.5.2 a heuristic argument that makes the theorem seem very plausible. Current rigorous proofs are all of “existence type” and not “constructive” for actually obtaining a mapping function from the domain shape D. There is, however, a rich literature on how to achieve this numerically [17, 40]. In the particular case when D is a polygon, there exists a systematic approach that we will describe next. However, even this will require computational methods for all but the simplest cases [12]. Domains in the complex plane are described as being simply, doubly (triply, etc.) connected according to their character on the stereographic sphere (cf. Section 1.4). If a curve between any two points can be continuously deformed (without exiting the domain) to any other such curve, the domain is simply connected. If there can exist two (but no more) families of curves that cannot be deformed to each other, it is doubly connected, etc. For example, the shaded regions in Figures 8.2 and 8.4 are all simply connected, whereas the ones in Figure 8.3 are doubly connected. With the exception that the entire complex plane cannot be conformally mapped to any other region, conformal mappings are possible only between equally connected domains.
8.4 Mappings of polygonal regions The classical method for mapping to the inside of a polygon comes in two versions dependent on the region we map from: either from the upper halfplane or from the inside of the unit circle (without an intermediate bilinear mapping going via the upper halfplane). The resulting formulas turn out to be virtually identical in the two cases. We follow here the first choice. For the second one, see Exercise 8.8.11 and Example 11.3. A different derivation is given in Ahlfors [2, Chapter 6]. The following example of creating a mapping function from the upper halfplane to the inside of a triangle generalizes immediately to the case when the target region instead is a general polygon. Example 8.11. Map conformally the upper halfplane from Figure 8.5(a) to the inside of the triangle shown in Figure 8.5(b). We simplify first the problem by rotating and translating the target triangle (and scaling it if we so wish), so that one of its sides (here marked blue) falls along the waxis; cf. Figure 8.5(c). Let then z0 = −∞ correspond to w0 = w(−∞), z1 correspond to w1 , and z2 correspond to w2 . Since the mapping has three free parameters, we can for convenience choose z1 = −1 and z2 = +1. When a point z moves along the zaxis over [−∞, z1 ], the mapping function w(z) needs to satisfy arg w0 (z) = 0 (so that w(z) follows the first side of the triangle, shown in blue). Then for z ∈ [z1 , z2 ], it has to hold that arg w0 (z) = πα1 (to match the slope of the second triangle side, shown in red). Finally, for z ∈ [z2 , +∞], arg w0 (z) needs to be increased with an additional πα2 (with side shown in green). In other words, arg w0 (z) needs, along the real zaxis, to be a piecewise constant function which at z = z1 and z = z2 jumps by the amounts πα1 and πα2 , respectively. Recalling the behavior of the log z function (cf. Figure 2.7(b)), we recognize in this function the perfect “building block” for creating functions that are piecewise constant
216
Chapter 8. Conformal Mappings
z0 = ∞
z1 = −1
z2 = 1
(a) zplane.
(b) wplane.
πα2
w2
πα1
w0 πα0
w1
(c) wplane. Figure 8.5. Steps in mapping the upper halfplane to the inside of a triangle, as described in Example 8.11. In part (c), the corners are denoted by w0 , w1 , w2 and the exterior angles are πα0 , πα1 , πα2 with here α0 = 17/20, α1 = 3/4, α2 = 2/5.
along the real axis. For example, the imaginary part of g(z) = C − α1 log(z − z1 ) − α2 log(z − z2 ) (with C some constant) jumps by πα1 when z (real) increases past z1 and by πα2 when it passes z2 . Displayed over the upper halfplane, it looks as shown in Figure 8.6 in the present case of z1 = −1, α1 = 3/4 and z2 = +1, α2 = 2/5 (i.e., the jumps are of sizes πα1 and πα2 , respectively, matching the angles in Figure 8.5). What we need is a function w0 (z) such that arg w0 (z) (rather than Im w0 (z)) has this step structure. However, the Euler identity ez = ex+i y = ex (cos y + i sin y) tells us that the two functions arg and Im are closely related. For any complex z, it holds that arg ez = Im z .
8.4. Mappings of polygonal regions
217
4
3
2
1
0
1 1 0.8 0.6 0.4 0.2
y
0
3
2
1
0
1
2
3
x
Figure 8.6. Imaginary part of the function g(z) = −α1 log(−1 − z) − α2 log(+1 − z), constructed to be piecewise constant along the real axis, with jumps of sizes 43 π and 25 π at the locations z1 = −1 and z2 = 1, respectively. The colors correspond to those of Figure 8.5.
Hence, we form w0 (z) = eg(z) = eC (z − z1 )−α1 (z − z2 )−α2 ,
(8.4)
0
and arg w (z) now matches the requirements imposed by the angles and the straight sides of the triangle. The further generalization from a target triangle to a general target polygon with corners at w0 , w1 , . . . , wn is now immediate. Also renaming eC = A, we have arrived at the following. Theorem 8.12 (Schwarz–Christoffel). A function w(z) that maps the upper halfplane to the inside of a polygon, and [−∞, z1 , z2 , . . . , zn ] (on the real axis) to the polygon with corners [w0 , w1 , w2 , . . . , wn ] and external angles [πα0 , πα1 , πα2 , . . . , παn ], satisfies w0 (z) = A (z − z1 )−α1 (z − z2 )−α2 · · · · · (z − zn )−αn .
(8.5)
In practical use of these mappings obtained by the Schwarz–Christoffel theorem, one should note the following: • Only three (realvalued) parameters can be chosen arbitrarily. We have above used one to enforce that z = −∞ maps to the polygon corner w0 . Given that the integral typically is difficult to handle in closed form, further choosing z1 = −1 and z2 = +1 may simplify the algebra. • The constants A and z3 , z4 , . . . , zn will then need to be obtained from enforcing that (8.5) obeys the mapping requirements that z = [−∞, z1 , z2 , . . . , zn ] correspond to w = [w0 , w1 , w2 , . . . , wn ]. The algebra in this step may become difficult or impossible unless approached numerically. The following two examples will be revisited in Section 8.5.
218
Chapter 8. Conformal Mappings
8
8 6
6 4
4 2
2 0
0 2
2
4 10
8
6
4
2
0
2
(a) Upper halfplane near the zaxis.
4
6
4
2
0
2
4
6
(b) Mapping of grid shown in part (a).
Figure 8.7. Mapping provided by the function w(z) in (8.6), Example 8.13.
Example 8.13. Map the upper halfplane (cf. Figure 8.7(a)) to the region indicated in part (b) (with a step height of π). We can use the three free parameters to ensure that the mapping function w(z) will satisfy w(−∞) = −∞, w(−1) = πi, and w(1) = 0. With exterior angles at the two latter points − π2 and + π2 , respectively, we have α1 = − 21 and α2 = 12 . By (8.5), the mapping function w(z) will satisfy 1/2 z+1 0 w (z) = A , z−1 from which follows by integration w(z) = A · (z 2 − 1)1/2 + log z + (z 2 − 1)1/2 + B.
(8.6)
The two conditions w(−1) = πi and w(1) = 0 determine the constants in (8.6) as A = π1 and B = 0. With some care in choosing the appropriate branch cut location for the square root function (for example, implement (z 2 − 1)1/2 as (z + 1)1/2 (z − 1)1/2 ), the zplane grid in Figure 8.7(a) maps to the curve set shown in part (b). Example 8.14. Map the upper halfplane (cf. Figure 8.8(a)) to the region indicated in part (b). The key idea in this case is to let the target region have three corners at w(−1) = −1, w(0) = −i b, w(1) = 1 and then consider the limit of b → +∞. The matching exterior angles become − π2 , π, − π2 , and we thus need to use the exponents 12 , −1, 21 , i.e., w0 (z) = A(z + 1)1/2 z −1 (z − 1)1/2 , which, when integrated, gives 1 2 1/2 w(z) = A (z − 1) + arcsin + B. z Requiring that w(−1) = −1 and w(1) = 1 gives A =
2 π
and B = 0.
(8.7)
8.5. Some applications of conformal mappings
219
1
3 2.5
0.5 2 0
1.5 1
0.5 0.5 1
0 0.5
1.5 1 2
1.5 3
2
1
0
1
2
3
(a) Upper halfplane near the zaxis
2
1.5
1
0.5
0
0.5
1
1.5
2
(b) Mapping of grid shown in part (a)
Figure 8.8. Mapping provided by the function w(z) in (8.7), Example 8.14.
8.5 Some applications of conformal mappings Many applications of conformal mappings rely on the fact that the change of independent variables leaves Laplace’s equation invariant. Theorem 8.15. If ψ(u, v) satisfies ψuu + ψvv = 0,
(8.8)
then, after a conformal mapping
u = u(x, y), v = v(x, y),
ψ expressed in x, y will also satisfy Laplace’s equation ψxx + ψyy = 0.
(8.9)
Proof. (Outlines of two approaches for proving the result). 1. If ψ(u, v) satisfies (8.8), then it is the real part of an analytic function Ψ(w) in a complex (u, v)plane. A further analytic variable change w = f (z) (w = u + iv, z = x + iy) gives Ψ(f (z)), which now has to be analytic in z, implying (8.9). 2. In a less elegant manner, one arrives at the same result from using the chain rule ∂2 ∂2 on ∂x 2 + ∂y 2 ψ(u(x, y), v(x, y)) and then simplifying by means of the CR equations ux = vy , uy = −vx . If we want to solve (8.8) in some region of a (u, v)plane, with values of ψ given on a boundary, a conformal mapping preserves this boundary data while having changed the shape of the region. Starting from a domain in which the Laplace solution is trivial, one can thus obtain corresponding solutions also in domains of much more general shapes.
8.5.1 Ideal fluid flow in 2D In the context of 2D fluid dynamics, it is customary to let u(x, y) and v(x, y) denote the velocities of fluid particles in the x and ydirections, respectively. The governing equations
220
Chapter 8. Conformal Mappings
for 2D steady, ideal flow then become Incompressible: ux + vy = 0,
(8.10)
uy − vx = 0
(8.11)
Irrotational:
(with a sign reversal in v compared to the CR equations). It follows from (8.10) that we can introduce a stream function ψ(x, y) such that ( ∂ψ u= ∂y , v=
− ∂ψ ∂x .
From (8.11) it follows then that this function ψ(x, y) satisfies ψxx + ψyy = 0 .
(8.12)
Level curves of ψ(x, y) are streamlines (paths that the fluid particles will follow). The following example introduces the mapping we will use in our first illustration of this concept. Example 8.16. Map conformally the upper halfplane outside the unit circle to the complete upper halfplane (cf. Figure 8.9(b)–(c)). Figure 8.9(a) illustrates the function f (x) = 12 x + x1 for a real argument x. It is clearly monotonically increasing for x > 1 and decreasing for x < −1. Considering the complex counterpart function 1 1 w = f (z) = z+ , (8.13) 2 z it is therefore clear that the cyan and green line segments in Figure 8.9(b) correspond to the similarly colored ones in part (c). The upper part of the unit circle in the zplane is given by z = eiθ , 0 ≤ θ ≤ π, and will correspond to w = 21 eiθ + e1iθ = cos θ, i.e., the straight line between −1 and +1. The mapping will be conformal everywhere apart from where f 0 (z) = 0, which, with f 0 (z) = 12 1 − z12 , occurs at z = ±1 (already noted in part (a) of the figure). We see this lack of conformity again when comparing parts (b) and (c); the 90◦ boundary angles in the zplane have become 180◦ in the wplane. At all other locations, the mapping will have to be conformal. Similarly, the lower halfplane outside the unit circle in the zplane will map to the lower halfplane in the wplane. Note that this mapping (8.13) is not a bilinear one: putting its RHS in common denominator form will make the numerator quadratic. Figure 8.10 illustrates the steps needed to obtain the flow past the central objects in the figure’s parts (a), (c), and (d): the unit circle, and flat plates inclined at different angles relative to free stream. We denote the planes in parts (a) and (b) as complex z and wplanes, respectively. As we know from Example 8.16, the mapping 1 1 z+ (8.14) w= 2 z maps the outside of the circle in (a) to the outside of the flat line segment [−1, 1] in (b). We write as usual z = x + iy and w = u + iv. In the wplane, the flow field is trivial, described
8.5. Some applications of conformal mappings
221
f(x)
2
1
3
2
1
1
2
3
x 1
2
(a) The function f (x) =
1 2
x+
1 x
.
2.5
2
1.5
1
0.5
3
2
1
1
2
3
zplane
(b) The domain we map from in the zplane.
2.5
2
1.5
1
0.5
3
2
1
1
2
3
wplane
(c) Target domain in the wplane. Figure 8.9. Illustrations of the function f (z) = mapping it provides for z complex.
1 2
z+
1 z
along the real axis and of the
222
Chapter 8. Conformal Mappings
by the stream function ψ(u, v) = v. We then get from (8.14) 1 1 u + iv = x + iy + 2 x + iy y x 1 1 + i , = 1+ 2 1 − 2 x + y2 2 x2 + y 2 from which itfollows (equating imaginary parts) that ψ(u, v) = v corresponds to ψ(x, y) = y 1 2 1 − x2 +y 2 , which thus becomes the matching stream function in the z = (x + iy)plane (cf. Figure 8.10(a)). Multiplying z by a complex number eiα will in (a) rotate the free stream, but has no effect on the shape of the central circle. Doing this, and then mapping from the zplane to the wplane using (8.14), brings the central circle again to the horizontal flat plate, but the farfield flow direction has been changed by the angle α. This gives the streamlines that are shown in parts (c) and (d), for 45◦ and 90◦ turns, respectively. Considerable amounts of information can now be deduced. For example, fluid velocities are inversely proportional to the separation of adjacent stream lines. Parts (c) and (d) thus reflect limitations of the ideal flow assumptions, as it would suggest infinite velocities at the ends of the inclined plates.
8.5.2 Static electrical fields in 2D Electrical fields in vacuum also satisfy Laplace’s equation. Just as edges of solid objects must coincide with a streamline (i.e., no fluid passing through its surface), electric potentials are constant over the surface of a conductor. Looking again at Figure 8.10, we can identify the solid curves with equipotential curves and the dashed curves with field lines. Parts (c) and (d) show how these latter ones become very concentrated at “spikes” in the shape of a conductor, and much less so along smooth surfaces (as in part (b)). Figures 8.7 and 8.8 can now also be interpreted as illustrating both flow and electrostatic fields in their respective geometries. As a concluding example, Figure 8.11 shows in part (a) the field (fluid or electrostatic) between two infinite parallel plates, and in parts (b) and (c) its counterpart after the mappings w = ez and w = z + ez , respectively. From part (b), we can deduce that the logarithm function describes the stream function (or electrostatic field) around a point source/sink (or a point charge).While the fields in parts (a) and (b) are conceptually trivial, that is not the case in part (c), showing flow exiting a 2D pipe or, equivalently, the field around the edge of a capacitor.
8.6 Revisiting the Jacobi elliptic function sn(z, k ) From our discussion of the Schwarz–Christoffel mapping, it becomes apparent that (7.10) represents the conformal mapping illustrated in Figure 8.12: mapping the upper halfplane shown in part (a) to the inside of the rectangle in part (b). The inverse function z(w) is clearly realvalued whenever w lies on one of the straight sides shown in part (b). By the Schwarz reflection principle (Section 3.2.2), reflecting this rectangle as indicated by going from the red rectangle in Figure 8.13 to the composite of four rectangles shown here in four black (dashed line) rectangles will have created a doubly periodic function. As such, we can shift the period box as shown to arrive at the rectangle marked in green. By this, we have
8.6. Revisiting the Jacobi elliptic function sn(z, k )
223
5 2
4
1.5
3 2
1
1
0.5
0
0
1
0.5
2
1
3
1.5
4
2 3
5 6
4
2
0
2
4
2
1
0
1
2
3
6
(b) w = (u + iv)plane; w = 12 (z + z1 ).
(a) z = (x + iy)plane.
3
2.5 2
2
1.5 1
1
0.5 0
0 0.5
1
1 1.5
2
2 2.5
3 3
2
1
0
1
2
3
(c) Mapping to flat plate after 45◦ turn.
2
1
0
1
2
(d) Mapping to flat plate after 90◦ turn.
Figure 8.10. A trivial flow field (part (a)) and its counterparts after different conformal mappings.
recovered exactly the same Jacobi elliptic sn(z, k) function as we introduced previously in Section 7.4. This confirms that (7.10) indeed represents a doubly periodic function over the claimed period box. Example 8.17. Heuristically motivate Riemann’s mapping theorem (Theorem 8.10). Following Copson [8], we first check on what properties a function w = f (z) must have if it maps the inside of a curve Γ (such as shown in Figure 8.14, part (a)) to the inside of the unit circle, part (b), with f (a) = 0. Being a onetoone mapping, f (z) can have no other zero inside (or on) Γ than z = a, so we can write f (z) = (z − a) eφ(z) . Since f (z) = 1 on Γ, we obtain as the real part of the logarithm of (8.15) log z − a + Re φ(z) = 0.
(8.15)
224
Chapter 8. Conformal Mappings
3
2
1
0
1
2
3
7
6
5
4
3
2
1
0
1
2
(a) Finite section of the area between two infinite plates, located vertically at ±i in a complex zplane. 6
4
2
0
2
4
6
8
6
4
2
0
2
4
6
8
z
(b) The mapping w = e (or z = log w). 8
6
4
2
0
2
4
6
8 10
5
0
5
10
z
(c) The mapping w = z + e . Figure 8.11. Two mappings from the infinite strip Im z ≤ π.
8.6. Revisiting the Jacobi elliptic function sn(z, k )
(a) zplane.
225
(b) wplane.
Figure 8.12. (a) Upper halfplane with the marked segments colorcoded and (b) the resulting map in the wplane through the relation (7.10), with matching color coding.
The term V = Re φ(z) is the real part of an analytic function, and will therefore satisfy 2 2 Laplace’s equation ∂∂xV2 + ∂∂yV2 = 0, where z = z + iy. If we can show there exists a unique solution to Laplace’s equation which (i) vanishes on Γ, and (ii) stays finite within Γ apart from behaving like log z − a (which itself satisfies Laplace’s equation away from the singularity at z = a) near to z = a, then the real part of φ(z) can be obtained from V (x, y) = log z − a + Re φ(z) . We can then from this obtain a matching imaginary part of φ(z) from the CR equations. However, a function V (x, y) with the required properties will need to exist if we can presume that nature will not come to a halt if subjected to the experiment of finding the
Figure 8.13. Illustrations of reflecting and shifting the rectangle shown in Figure 8.12(b) to obtain the period box for the Jacobi elliptic function sn(z, k) .
226
Chapter 8. Conformal Mappings
(a) zplane.
(b) wplane.
Figure 8.14. (a) Region in the zplane and (b) target region in the wplane for a mapping function w = f (z).
electrostatic potential in the configuration in Figure 8.14(a), where the outer boundary Γ is an electric conductor and where a point charge has been placed at the location z = a.
8.7 Supplementary materials Theorem 8.18. If the mapping
u = u(x, y) v = v(x, y)
with ux , uy , vx , vy continuous is conformal in the vicinity of a location z = x + iy , then f (z) = u + iv is an analytic function, with f 0 (z) 6= 0. Proof. Consider infinitesimal steps dx and dy in the (x, y)plane, with corresponding steps du and dv in the (u, v)plane. We have then du = ux dx + uy dy and dv = vx dx + vy dy. Preserving angles locally means that all sides of any infinitesimal triangle scale with the same factor h(x, y), i.e., du2 + dv 2 = h2 (dx2 + dy 2 ). With the expressions above for du and dv, this becomes (u2x + vx2 − h2 )dx2 + (u2y + vy2 − h2 )dy 2 + (ux uy + vx vy )dxdy = 0. For this to hold identically for any combination of dx and dy, all three coefficients need to vanish, i.e., u2x + vx2 = h2 , u2y + vy2 u u +v v x y x y
= h2 , = 0.
We can satisfy the top equation with ux = h cos α, vx = h sin α and the second by uy = h cos β, vy = h sin β, after which the third equation gives α − β = ± π2 , from which
8.8. Exercises
227
follows either {ux = vy , uy = −vx } or {ux = −vy , uy = vx }. We recognize the first option as the CR equations for u + iv = f (x + iy), implying that f (z) is analytic. The second option similarly corresponds to u + iv = f (x + iy). Although the latter version also preserves scales locally, and sizes of angles, it reverses directions and is hence not what we accept as being conformal.
8.8 Exercises Exercise 8.8.1. With the notation and the result of Theorem 8.2, w = f (z) provides a onetoone map between z ∈ D and w ∈ E. Show that the inverse mapping z = f −1 (w) ´ ζ f 0 (ζ) 1 is given by z = 2πi dζ. δD f (ζ)−w Exercise 8.8.2. Construct a function w(z) that maps the inside of the unit circle to the upper halfplane. Hint: One approach is to apply suitable shifts and rotations to the mapping w = 1/z illustrated in Figure 8.2 (note especially the rightmost two subplots). Exercise 8.8.3. The exponential function w = ez and trigonometric functions (e.g., w = sin z) are useful ingredients in many conformal mapping cases. (a) Map the infinite strip Im z < π to the right halfplane. (b) Map the semiinfinite strip Re z < 0, Im z < π to the upper half of the unit circle. (c) Show that w = sin z maps the semiinfinite strip 0 < Re z < π2 , Im z > 0 to the first quadrant. (d) Show that w = sin z maps a rectangle with its two bottom corners at z = ± π2 to the upper half of an ellipse. Exercise 8.8.4. (a) Show that w = z + z1 maps the upper half of the unit circle to the lower halfplane. √ (b) Map the inside of the unit circle to its upper half (in particular, explain why w = z is not a solution to this problem). (c) Map the region in the first quadrant above the hyperbola y = 1/x to the full first quadrant. Exercise 8.8.5. Consider the upper halfplane with two nonoverlapping circular “holes.” Either describe the sequence of steps you would take to map this region to the annulus between two circles, both centered at the origin, or explain why it cannot be done. Exercise 8.8.6. Prove Theorem 8.8. Exercise 8.8.7. Give a direct proof (not via stereographic mappings) of the fact that if z is located on a circle or straight line in the complex zplane, so is w = 1/z in the complex wplane (cf. the proof of Theorem 8.3). Exercise 8.8.8. Carry out the details of the second of the outlined proofs for Theorem 8.15. Exercise 8.8.9. Let f (z) be an entire function that satisfies f (z) = 1 for z = 1. Show that it then must be of the form f (z) = eiϑ · z n with θ real and n a nonnegative integer.
228
Chapter 8. Conformal Mappings
Hint: One way to show this uses these steps: (i) Show that f (z) has a zero for z < 1, or else it is a constant, and (ii) if f (z) has zeros for z < 1, note that these can all be “divided away” by functions of the form (8.3). Exercise 8.8.10. For the cases in Figure 8.3, electrostatic field lines are trivial in the zplane. Create images of the corresponding field lines through the gray shaded areas in the corresponding wplanes. Exercise 8.8.11. Show the following counterpart to Theorem 8.12: The mapping function w(z) from the inside of the unit circle to the inside of a closed polygon with corners [w0 , w1 , w2 , . . . , wn ] and external angles [πα0 , πα1 , πα2 , . . . , παn ] satisfies w0 (z) = A (z − z0 )−α0 (z − z1 )−α1 (z − z2 )−α2 · · · · · (z − zn )−αn .
(8.16)
Note: The formula (8.16) is identical to (8.5) except that the factor (z − z0 )−α0 is now also included (meaning that the preimage to w0 is now a point z0 on the unit circle, rather than −∞). Hint: For the proof of Theorem 8.12, we relied on the fact that Im(− log(z0 − z)) features a step but is otherwise piecewise constant when z0 is real and z moves along the real axis. Figure 8.15(a) shows a corresponding result for when z0 is on the unit circle and z = eiθ moves around the circle. The function φ(z) = Im(− log(z0 − z)) with z0 = 1 satisfies
(8.17)
1 (8.18) φ eiθ = (π − θ), 0 < θ < 2π, 2 for a reason that can be deduced from the geometry observation in Figure 8.15(b). Extending (8.17) to include [z0 , z1 , z2 , . . . , zn ] (all on the unit circle), deduce that when z moves around the unit circle, the function w(z) defined by (8.16) will obey both of the required key properties: (i) w(z) becomes a straight line segment when z moves between any two adjacent zi points, and (ii) w(z) makes the specified changes in direction when z passes any of the zi points.
8.8. Exercises
229
1.5
1
0.5
0
0.5
1
1.5 2
1.5
1
0.5
0
0.5
1
1.5
1
0.5
0.5
0
1
1.5
y
x
(a) φ(z) = Im(− log(z0 − z)) shown for z0 = 1. 1
0.8
0.6
θ/2
0.4
θ
y
0.2
0
0.2
0.4
0.6
0.8
1 1.5
1
0.5
0
0.5
1
x
(b) A theorem from plane geometry. Figure 8.15. The imaginary part of the logarithm function when following a circle going through its singular point (here, the unit circle and the singularity at z = 1).
Chapter 9
Transforms
We have often introduced analytic functions ´ ∞by means of infinite sums (e.g., Taylor expansions) or by integrals (such as Γ(z) = 0 e−t tz−1 dt in Section 2.7). A more general approach in the case of integrals is to choose a path C and a kernel function K(z, w) and then, for every function f (z), define a counterpart function g(w) as ˆ g(w) = f (z) K(z, w) dz. (9.1) C
This procedure becomes particularly useful if (i)
there is a similar integral formula available to return to f (z) if g(w) is specified (with the two formulas then forming a transform pair), and
(ii)
some tasks become simpler when carried out in the transformed variable.
In the discussion below of the most frequently used transforms, the focus will be on their mathematical properties rather than on their very wide range of applications. Even for functions that are realvalued along the real axis, knowledge of analytic functions will play a key role in simplifying and in utilizing the transforms. The chapter starts with the Fourier transform. The further transforms are all related to the Fourier cases (via variable changes, etc.). Common features include that taking derivatives of a function usually becomes a purely algebraic operation in the transformed variable and that the transforms possess convolution theorems.
9.1 Fourier transform One way to introduce the Fourier transform is through the Fourier series in the limit when the period increases to infinity. Hence, we first briefly summarize the series case and how its coefficients may be found.
9.1.1 Fourier series For f (x) realvalued and periodic over [−π, π], the Fourier series of f (x) can be defined as f (x) = a0 +
∞ X
ak cos(kx) +
k=1
∞ X k=1
231
bk sin(kx),
(9.2)
232
Chapter 9. Transforms
where ˆ π 1 f (x)dx, 2π −π ˆ 1 π ak = f (x) cos(kx)dx, k = 1, 2, . . . , π −π ˆ 1 π f (x) sin(kx)dx, k = 1, 2, . . . . bk = π −π a0 =
(9.3)
Thus f (x) can be seen as a superposition of Fourier modes (sine and cosine functions) and the coefficients tell us how much of each mode is needed to recover the original function. The assumptions that are needed for (9.2) (and equivalently (9.4)) to converge for all x are quite subtle. Requiring the function f (x) to be continuous and periodic is not quite enough (see, for example, [29, Chapter 18]), while requiring it to be continuously differentiable is stronger than required. Equations (9.4) and (9.5) provide yet another example of how algebra simplifies if we replace trigonometric functions with complex exponentials. Theorem 9.1. If f (x) can be represented as ∞ X
f (x) =
ck eikx ,
(9.4)
f (x) e−ikx dx.
(9.5)
k=−∞
then 1 ck = 2π
ˆ
π
−π
Proof. If (9.4) holds, then the RHS in (9.5) evaluates (after renaming the summation variable from k to n) to 1 2π
ˆ
π −ikx
f (x)e −π
dx =
∞ X
cn
n=−∞

1 2π 1 0
ˆ
π i(n−k)x
−π
e {z
dx
= ck , }
if n = k, otherwise.
In the case when f (x) is realvalued, the complex coefficients ck will satisfy c−k = ck . Their decay rate for increasing k will reflect how smooth f (x) is. If f (x) is m times differentiable (after having been extended periodically), then ck  = O(km+1 ), and if it is analytic in a strip surrounding the xaxis, the decay rate will be exponential, with a rate that depends on the strip width. Example 9.2. Estimate how fast the Fourier coefficients ck go to zero for the function 1 f (z) = 3+sin z. This function was displayed in Figure √ 5.5(b). Its poles are located where sin z = −3, i.e., at zk = − π2 + 2πk ± i log(3 + 2 2), k ∈ Z. Because of the periodicity, the contour √ in (9.5) can be shifted up/down with any amount less than log(3+2 2) without any change
9.1. Fourier transform
233
in the ck . Since the Fourier series must remain convergent (being the expansion of the analytic function along the shifted and the e±ikz factors have grown in magnitude with √ √path) 1 k log(3+2 2) k up to e = (3 + 2 2) , the ck will decay in magnitude like O (3+2√ . 2)k The only information needed here for f (z) was the imaginary part of the singularity closest to the real axis. A key use of analytic functions in the context of Fourier series arises in evaluating integrals of the form (9.5). In fact, it was shown in Example 5.7 that the poles in the 1 i √ upper halfplane of f (z) = 3+sin z have the residue − 2 2 . From this follows the exact coefficients in Example 9.2: ck =
k
i√ 2 2
1 √ k 2)
(3+2
for k ≥ 0 (and c−k = ck , since f (x) is
real for x real).
9.1.2 Fourier transform: Definition and inversion Changing the period in x from [−π, π] to [−Lπ, Lπ] changes the frequencies present from k eikx to ei( L )x , where k runs through all integers. In the limit of L → ∞, there will be a continuum of frequencies present, typically denoted by ω ∈ (−∞, ∞). Theorem 9.3. Defining the Fourier transform of f (x) as ˆ ∞ 1 ˆ f (ω) = √ f (x) e−iωx dx , 2π −∞ the inversion formula becomes 1 f (x) = √ 2π
ˆ
(9.6)
∞
fˆ(ω) eiωx dω.
(9.7)
−∞
Together, (9.6) and (9.7) form a transform pair (much like (9.4) and (9.5)).68 A proof for this theorem, based on the limit argument just mentioned, is given in Section 9.7. In some situations, a slightly more explicit notation is convenient: fˆ(ω) = F {f (x)}(ω) and f (x) = F −1 {fˆ(ω)}(x). The integrals that arise when using (9.6) and (9.7) are often perfectly suited for contour integral treatment. Essential to the utility of the Fourier transform, as well as for all the other transforms considered in this chapter, is that they are linear: F {α f (x) + β g(x)}(ω) = α F {f (x)}(ω) + β F {g(x)}(ω) , generalizing to any number of terms. Example 9.4. Determine the Fourier transform of a Gaussian 2
f (x) = e−αx ,
(9.8)
where α > 0. 68 In some literature, the Fourier transform is defined slightly differently than here. Inserting a factor 2π in each of the exponents eliminates the factors √1 in front of each integral. Instead of having the factor √1 in front 2π
2π
1 in just one of the two places (then making the forms of of both integrals, there is sometimes instead a factor 2π (9.6) and (9.7) somewhat unsymmetric). It is also not uncommon to swap the sign in the two exponents.
234
Chapter 9. Transforms
We find from (9.6)
ˆ ∞ 2 1 fb(ω) = √ e−αx e−iωx dx. 2π −∞ The integrand is an entire function. Since its magnitude grows large up and down the complex plane, no matter the values for α > 0 and ω (cf. Figure 9.1), it does not seem promising to try to close the integration path by some loop far out. Another option that has been successful in some of our previous examples is to run a return path parallel to the original one. That turns out to work again here. An algebraically elegant way to go about that starts by completing the square in the exponent. We then obtain (denoting the integration variable by z instead of by x) ˆ ∞ ω2 iω 2 1 e−α(z+ 2α ) − 4α dz fˆ(ω) = √ 2π −∞ ˆ ∞ iω 2 ω2 1 = √ e− 4α e−α(z+ 2α ) dz. 2π −∞ This can be interpreted as having moved the integration contour vertically up by a distance of ω/2α, which does not change its value compared to integrating along the real axis. Since ´∞ pπ 2 we for that case know the value to be −∞ e−αz dz = α , we obtain ω2 1 fˆ(ω) = √ e− 4α . 2α
(9.9)
In words, the Fourier transform of a Gaussian becomes again a Gaussian. Changing α to make one of the Gaussians narrower will make the other one correspondingly wider. One basic feature of Fourier transforms is that a “pulse” can never be narrow in both the (usually physical) variable x and in the transform variable ω.69 Example 9.5. Let f (x) = e−x . Determine fˆ(ω), and then invert this result back to recover f (x). We find from (9.6) ˆ 0 ˆ ∞ ˆ ∞ 1 1 −x −iωx x −iωx −x −iωx ˆ e e dx = √ e e dx + e e dx f (ω) = √ 2π −∞ 2π −∞ 0 r 1 i i 2 1 =√ + = . π ω2 + 1 2π i + ω i − ω Inverting back requires us to evaluate 1 f (x) = √ 2π
ˆ
∞
−∞
r
2 1 eiωx dk. π ω2 + 1
This is not an integrand to which we can find a primitive function. It is, however, easy to evaluate the definite integral with contour integration. We consider two cases: √ √ as probability distributions, (9.8) and (9.9) have standard deviations σx = 1/ 2α and σω = 2α, respectively, making σx σω = 1 for all Gaussians. This is the minimal possible value for the product of matching σx and σω , forming a basis for Heisenberg’s uncertainty principle. 69 Viewed
9.1. Fourier transform
235
2
Re e−z
2
Figure 9.1. Magnitude and phase plot of the function f (z) = e−α z in the case of α = 1. 2 and Im e−z are illustrated separately in a different context in Figures 9.12(a)–(b).
1. If x > 0, consider a contour such as illustrated in Figure 5.20, in the upper half of the complex ωplane. By Jordan’s lemma, the contribution from the semicircle vanishes, and using the residue at the only relevant pole (at ω = i) gives f (x) = −x √1 2πi √1 e = e−x . 2π 2π i 2. If x > 0, extend the integration path similarly with the semicircle in the lower halfex = ex . plane. We similarly now obtain f (x) = − √12π 2πi √12π −i Combining the two cases, the final result becomes f (x) = e−x . We see in this example an instance of a general rule relating the smoothness of a function (not analytic along the xaxis) to the decay rate of its Fourier transform. Since f (x) has a jump in its first derivative, fˆ(ω) = O(1/ω 2 ) for ω increasing. In Example 9.10, we will similarly see jumps in the third derivative leading to fˆ(ω) = O(1/ω 4 ).
9.1.3 Some key features of the Fourier transform The following two results are particularly useful in applications (both readily derived from the Fourier transform definition). Theorem 9.6. If a function f (x) has the Fourier transform fˆ(ω), then for n = 1, 2, 3, . . . dn dn f (x) has the transform (iω)n fb(ω) and xn f (x) has the transform (i)n n fb(ω). n dx dω In our alternative notation, n d dn F f (x) (ω) = (iω)n F {f (x)}(ω) and F {xn f (x)}(ω) = (in ) n F {f (x)}(ω). n dx dω
236
Chapter 9. Transforms
Theorem 9.7. F {f (x + a)}(ω) = eiωa F {f (x)}(ω) . In words, shifting a function the distance a to the left multiplies its Fourier transform by a factor of eiωa . Example 9.8. Give a closedform solution u(x, t) to the heat equation ∂u(x, t) ∂ 2 u(x, t) = ∂t ∂x2
(9.10)
for t > 0, satisfying an initial condition u(x, 0), x ∈ (−∞, ∞). Then evaluate this expres2 sion for the special case of u(x, 0) = e−x . Applying the Fourier transform in the xdirection to (9.10) gives ˆ ∞ ˆ ∞ 2 1 1 ∂u(x, t) −iωx ∂ u(x, t) −iωx √ e dx = √ e dx , ∂t 2π −∞ 2π −∞ ∂x2 which simplifies to ˆ ∞ ˆ ∞ ∂ 1 1 −iωx 2 −iωx √ u(x, t) e dx = (−ω ) √ u(x, t) e dx , ∂t 2π −∞ 2π −∞ i.e., ∂ u ˆ(ω, t) = −ω 2 u ˆ(ω, t) . ∂t 2
For each value of ω, this is an ODE in t, with the closedform solution u ˆ(ω, t) = e−ω t u ˆ(ω, 0). Inverting u ˆ(ω, t) back to physical xspace gives ˆ ∞ 2 1 u(x, t) = √ e−ω t u ˆ(ω, 0) eiωx dω . 2π −∞ 2
In the special case of u(x, 0) = e−x , Example 9.4 provides all the ingredients needed for a closedform evaluation of the integral, giving 2 1 e−x /(1+4t) . u(x, t) = √ 1 + 4t
(9.11)
This solution is shown in Figure 9.2 (with the initial condition highlighted in red). In this example above, we solved a PDE using the first part of Theorem 9.6. We will in Section 11.1.1 base a transform solution of an ODE on the second part of this theorem. Theorem 9.9. A function f (x) (not identically zero) and its Fourier transform fˆ(ω) cannot both have compact support (be identically zero for sufficiently large x and ω, respectively). ´R Proof. If f (x) has compact support, then (9.6) becomes fˆ(ω) = √12π −R f (x)e−iωx dx for some finite R. This shows fˆ(ω) to be an entire function of ω. If it is identically zero on any curve segment (consider a piece of the ωaxis sufficiently far out), it will thus have to be identically zero.
9.1. Fourier transform
237
Figure 9.2. The solution (9.11) to the heat equation in Example 9.8, with the Gaussian initial condition.
Example 9.10. Illustrate the Fourier transform of a cubic Bspline. Given realvalued data yk at discrete 1D locations xk , k = 1, 2, . . . , n, a cubic spline interpolant s(x) is made up of a separate cubic polynomial in each subinterval, with the extra requirement that, at the interior data locations xk , k = 2, 3, . . . , n − 1, where two different cubics meet, s(x), s0 (x), and s00 (x) shall all remain continuous (for uniqueness, two further conditions are needed, which we do not need to investigate here). The simplest “building block” to create such a cubic spline interpolant is known as a Bspline, the narrowest cubic spline that connects identical zero nontrivially back to identically zero. Figure 9.3(a) shows the cubic Bspline B(x) at unitspaced data locations. It is nonzero only over four subintervals and is here normalized such that its integral is 1. Writing it down piecebypiece becomes somewhat tedious: [−∞, −2] → 0, [−2, −1] → 43 + 2x + x2 + 61 x3 , [−1, 0]
→
2 3
− x2 − 21 x3 ,
[0, +1]
→
2 3
− x2 + 21 x3 ,
[+1, +2]
→
4 3
− 2x + x2 − 61 x3 ,
(9.12)
[+2, +∞] → 0. ˆ However, its Fourier transform B(ω) becomes surprisingly simple (see Exercise 9.8.10): 1 ˆ B(ω) =√ 2π
sin(ω/2) ω/2
4 .
(9.13)
ˆ Figure 9.3(b) shows that also B(ω) looks very much like a single “pulse” with compact support. However, Theorem 9.9 tells us this cannot be the case, and part (c) of the figure indeed displays lowamplitude irregularities for large ω. Since B(x) features discontinuities in its third derivative, these amplitudes decay like O(1/ω 4 ), as is apparent from (9.13).
238
Chapter 9. Transforms 0.4
2/3
0.6
0.3
0.4 1/6
0.2
0.2
1/6
0
0 4
2
0.1
0 0
2
0 10
4
5
0
5
10
x
ˆ (b) Its Fourier transform B(ω).
(a) The equispaced cubic Bspline B(x). 104
6
4
2
0 10
15
20
25
30
ˆ (c) Detail of B(ω) for larger ω. ˆ Figure 9.3. Illustrations of the cubic Bspline B(x) and its Fourier transform B(ω).
9.1.4 Parseval’s relations For the Fourier transform, this relation takes the following form. Theorem 9.11.
ˆ
ˆ
∞
∞
fˆ(ω)2 dω .
f (x)2 dx = −∞
−∞
Proof. With use of (9.7) and (9.6), we obtain ˆ
ˆ
∞
∞
f (x)2 dx = −∞
f (x) f (x)dx ˆ ∞ 1 = f (x) √ fˆ(ω) e−ixω dω dx 2π −∞ −∞ ˆ ∞ ˆ ∞ 1 = fˆ(ω) √ f (x)e−ixω dx dω 2π −∞ −∞ ˆ ∞ ˆ ∞ = fˆ(ω) fˆ(ω) dω = fˆ(ω)2 dω. −∞ ˆ ∞
−∞
−∞
A generalization of this theorem is considered in Exercise 9.8.7.
(9.14)
9.1. Fourier transform
239
A counterpart result for the Fourier series (9.4) f (x) = ˆ
∞ X
π
f (x)2 dx = 2π −π
P∞
ikx k=−∞ ck e
is that
ck 2 .
(9.15)
k=−∞
Both (9.14) and (9.15) offer opportunities to evaluate various infinite integrals and sums, respectively. Example 9.12. Show that
´∞ −∞
sin x 2 x
dx = π.
If we happen to know the transform pair f (x) =
1 0
if x < 1 if x > 1
r ⇔
fˆ(ω) =
2 sin ω , π ω
(9.16)
the result follows immediately from (9.14); equation (9.16) is obtained through fˆ(ω) = q ´ 1 √1 1 · e−iωx dx = π2 sinω ω . 2π −1 If we do not know (9.16), we can still note that the integrand is the square of f (x) = ´ sin x of fˆ(ω) (on the chance that fˆ(ω)2 dω might be simpler to x , suggesting ´ a calculation integrate than f (x)2 dx). Hence, we use (9.6) to obtain ˆ ∞ sin x −iωx 1 fˆ(ω) = √ e dx 2π −∞ x 1 1 ∞ ei (1−ω)x 1 =√ dx − x 2i 2π 2i −∞ p π/2 if ω < 1 = , 0 if ω > 1
∞ −∞
ei (−1−ω)x dx x
where we for the last step used (5.7). Parseval’s relation (9.14) now gives ´ 1 p π 2 dω = π. 2 −1
´∞ −∞
sin x 2 x
dx =
Example 9.13. Find the infinite sum obtained by choosing f (x) = x in (9.15). k P∞ From (9.3) it follows (for −π < x < π) that x = f (x) = 2 k=1 (−1) sin kx. k P (−1)k+1 ikx eikx −e−ikx With sin kx = , this becomes x = k6=0 ik e , and thus c0 = 0 and 2i ´π P ck = (−1)k+1 /(ik) for k 6= 0. Equation (9.15) now gives −π x2 dx = 2π k6=0 k12 , P∞ 2 which simplifies to k=1 k12 = π6 .70
9.1.5 Convolution theorems Most transforms have convolution theorems of one form or another. In the case of Fourier expansions, they become as follows. 70 The
Basel problem we encountered earlier in Sections 3.2.3 and 5.2.
240
Chapter 9. Transforms
Fourier series case
Let A(x) and B(x) be 2πperiodic functions: A(x) =
∞ X
αk eikx ,
αk =
k=−∞
B(x) =
∞ X
βk eikx ,
k=−∞
βk =
1 2π 1 2π
ˆ
π
A(x)e−ikx dx,
(9.17)
B(x)e−ikx dx.
(9.18)
−π
ˆ
π
−π
The convolutions of the two functions and of the two coefficient sequences, respectively, are defined as ˆ π 1 C(x) = A(t)B(x − t) dt (physical space, again 2πperiodic), (9.19) 2π −π γk =
∞ X
αν βk−ν
(Fourier coefficient space, k ∈ Z).
(9.20)
ν=−∞
What amounts to a convolution in one of the spaces becomes a product in the other, as shown in the following theorem. Theorem 9.14. With the notation just above, it will hold that ˆ π ∞ X 1 ikx C(x) = {αk βk } e , αk βk = C(x) e−ikx dx, 2π −π
(9.21)
k=−∞
A(x)B(x) =
∞ X
γk e
ikx
,
k=−∞
1 γk = 2π
ˆ
π
A(x)B(x) e−ikx dx.
(9.22)
−π
The proofs for the two versions are similar. The first one is proven in Section 9.7. Fourier transform case
This case is more symmetric in that both directions of the transform, (9.7) and (9.6), respectively, are integrals over (−∞, ∞). For two functions f (x) and g(x), with Fourier transforms fˆ(ω) and gˆ(ω), respectively, we define the convolutions in the two spaces as ˆ ∞ 1 f (t)g(x − t) dt , (9.23) h(x) = √ 2π −∞ ˆ ∞ 1 ˆ k(ω) =√ fˆ(ξ)ˆ g (ω − ξ) dξ . (9.24) 2π −∞ The counterpart results to those in Theorem 9.14 become now the following. Theorem 9.15. ˆ ∞ 1 h(x) = √ fˆ(ω)ˆ g (ω) eiωx dω, 2π −∞ ˆ ∞ 1 ˆ f (x)g(x) = √ k(ω) eiωx dω, 2π −∞
ˆ ∞ 1 fˆ(ω)ˆ g (ω) = √ h(x) e−iωx dx, 2π −∞ (9.25) ˆ ∞ 1 −iωx ˆ k(ω) = √ f (x)g(x) e dx. 2π −∞ (9.26)
9.1. Fourier transform
241
9.1.6 Poisson’s summation formula Theorem 9.16. Let f (x) have the Fourier transform fˆ(ω); then ∞ X 1 fˆ(k) eikx f (x + 2πk) = √ 2π k=−∞ k=−∞
(9.27)
∞ X 1 ˆ f (k) e−ikω . f (ω + 2πk) = √ 2π k=−∞ k=−∞
(9.28)
∞ X
and, equivalently, ∞ X
The LHSs of each of these two equations are 2πperiodic in their arguments x and ω, respectively. The RHSs give, in very simple explicit form, their Fourier series expansions. The proofs for the two parts are essentially identical. The first part is shown in Section 9.7.
9.1.7 The generalized Fourier transform One apparent limitation of Fourier transforms is that the defining integrals (9.6) and (9.7) need to be convergent. An easy way around this restriction is illustrated in the following example. Example 9.17. Find the generalized Fourier transform of the function f (x) = x3 . ´∞ Although √12π −∞ x3 e−iωx dx diverges, it will converge if we insert a factor inside the integral that decays to zero sufficiently fast as x → ∞. Choosing, for example, this factor as e−αx , we get convergence for all α > 0 and can then take the limit as α & 0. Thus, ˆ ∞ 1 fˆ(ω) = lim √ x3 e−αx e−iωx dx α&0 2π −∞ 12 1 12(ω 4 − 6ω 2 α2 + α4 ) √ = lim =√ . (9.29) 2 2 4 α&0 2π(ω + α ) 2π ω 4 There is much arbitrariness in the choice of this extra factor. One naturally tries to find one such that the resulting integral can be evaluated in closed form. One example of the utility of the generalized Fourier transform is shown in Exercise 9.8.10. The example below illustrates both the utility of this generalized Fourier transform and the results in the preceding sections on convolutions and Poisson’s summation formula. Example 9.18. Given the function φ(x) = x3 , find coefficients λk , k ∈ Z, such that it holds for all n ∈ Z that ∞ X 1, n = 0, λk φ(n − k) = (9.30) 0, n 6= 0. k=−∞
242
Chapter 9. Transforms
Data of this type, taking the value one at one point and zero at all other points, is known as cardinal data and arises frequently in approximation theory.71 Given how fast φ(x) grows, convergence of this sum will require that the λk go quite rapidly to zero for k increasing. The various functions that arise in the following solution are illustrated in Figure 9.4. We recognize the LHS of (9.30) as being of the same form as (9.20) and define therefore the two 2πperiodic functions Λ(x) =
Φ(x) =
∞ X k=−∞ ∞ X
λk eikx ,
(9.31)
φ(k)eikx .
(9.32)
k=−∞
The first equation in (9.22) tells us that Λ(x) · Φ(x) = 1 · e0x = 1, and therefore Λ(x) = 1/Φ(x).
(9.33)
A solution plan emerges now: With φ(k) given, use (9.32) to find Φ(x), use then (9.33) to get Λ(x). Equation (9.31) tells us then that Fourier expanding Λ(x) will give the coefficients we want. The first step poses immediately a problem: The sum (9.32) diverges. √ However, P∞ itsˆ form + matches perfectly the RHS of (9.28), with the LHS then becoming 2π k=−∞ φ(ω 12 1 ˆ 2πk). By the previous Example 9.17, φ(ω) = √2π ω4 , and this last sum therefore converges just fine. It can be evaluated by the contour integration method in Section 5.2, giving 4 sin4 (ω/2) 2+cos ω Φ(ω) = 4 sin 4 (ω/2) , from which it follows that Λ(ω) = 2+cos ω . By (9.31) we get the λk ´ π 4 sin4 (ω/2) −ikω 1 as the Fourier coefficients of Λ(ω), i.e., λk = 2π −π 2+cos ω e dω. This integral, for k ∈ Z, is of a standard form for contour integration. Carrying out the details of this gives the result √ −4 + 3√ 2 , k = 0, 19 − 6 3 , k = 1, 2 √ λk = k (−1) 3 3 √ , k ≥ 2. (2 + 3)k These coefficients decay exponentially fast as k increases, ensuring that (9.30) indeed converges for all values of n.
9.1.8 Halfplane splitting by means of the Fourier transform The Fourier transform offers an alternative to the principal value approach in Section 5.1.5 for splitting an analytic function f (z) into f (z) = f − (z) + f + (z), with the two functions f − (z) and f + (z) singularityfree and going to zero in the upper and lower halfplanes, respectively (i.e., having all growth and singularities (if any) in the halfplane corresponding ´∞ to the sign in the superscript). The key idea is to form fˆ(ω) = √12π −∞ f (x) e−iωx dx as usual but, when inverting back, split the integration interval [−∞, ∞] for ω in two parts 71 Cases
with more general φ(x)functions are discussed, for example, in Section 3.3.9 of [18].
9.1. Fourier transform
243
1
100
0.8
80
0.6
60
0.4
40
0.2
20
0 1
0.5
0
0.5
0 10
1
5
0
ˆ (b) φ(ω) =
(a) φ(x) = x3 . 100
5
80
4
60
3
40
2
20
1
0
5
10
√12 14 . 2π ω
0 3
2
1
0
(c) Φ(ω) =
1
2
3
3
2
2+cos ω . 4 sin2 (ω/2)
1
(d) Λ(ω) =
0
1
2
3
2
4 sin (ω/2) 2+cos ω .
1.5 1 0.5 0 0.5 1 20
15
10
5
0
5
10
15
20
(e) Resulting coefficients λk . Figure 9.4. Illustration of the sequence of functions arising in Example 9.18. The functions in parts (a), (b), (e) are shown truncated sideways, whereas the ones in parts (c) and (d) are [−π, π]periodic.
(denoting the independent variable by z rather than by x): ˆ 0 ˆ +∞ 1 1 f (z) = √ fˆ(ω) eiωz dω + √ fˆ(ω) eiωz dω = f + (z) + f − (z). (9.34) 2π −∞ 2π 0 Both formulations (see (9.34) and (5.9)–(5.10)) are equivalent; see Exercise 10.4.1. Some additional notation is helpful in expressing relations, such as the one in (9.34), as
244
Chapter 9. Transforms
clearly and concisely as possible. We thus supplement the superscript notation − and + by also introducing subscripts − and + , to denote halfline splits along the real axis. Definition: If f (x) is defined for −∞ < x < ∞, we can write it as f (x) = f− (x) + f+ (x),
(9.35)
where f− (x) =
f (x) , 0 ,
x < 0, x ≥ 0,
and f+ (x) =
0 , x < 0, f (x) , x ≥ 0.
Both for superscripts and subscripts, we adhere to the convention that the sign indicates the direction of nontrivial functional behavior. Depending on the context (left/right), real axis splits and (lower/upper) halfplane splits can be applied in “physical” space or in Fourier space. Since there is a difference in sign in the exponent when transforming from physical to Fourier space, we obtain the following. Theorem 9.19. F {f− (x)}(ω) = fˆ− (ω) ˆ+ F {f+ (x)}(ω) = f (ω) {z }  signs are the same
and similarly
F {f − (x)}(ω) = fˆ+ (ω) . + ˆ F {f (x)}(ω) = f− (ω)  {z } signs are opposite (9.36)
In words, the Fourier transforms of the real axissplit functions f− (x) and f+ (x), when evaluated for complex ω, provide the halfplane split versions of f (x)’s Fourier transform fˆ(ω). Similarly, the Fourier transforms of the halfplane split functions f − (x) and f + (x) become the realaxis splits of f (x)’s Fourier transform fˆ(ω). Proof. Each of the pairs of equations in (9.36) add up to F {f (x)}(ω) = fˆ(ω), so what remains to show (for the four equations, in turn) are F −1 {fb− (ω)}(x) = 0 F −1 {fˆ+ (ω)}(x) = 0 F {f − (x)}(ω) = 0 F {f + (x)}(ω) = 0
for x > 0, for x < 0, for ω < 0, for ω > 0.
Each LHS above is an integration along the real ω or xaxis. Closing the contours far out in the halfplane in which the integrand goes to zero (and is singularityfree) gives these results (by Theorem 5.6, Jordan’s lemma). The relations in Theorem 9.19 will play a key role in the Wiener–Hopf discussion in Section 10.1. The following two examples illustrate this procedure for obtaining halfplane splits in the same two cases as before in Examples 5.17 and 5.18, respectively. Example 9.20. Calculate f − (z) in case of f (z) =
1 1+z 2 .
9.2. Laplace transform
245
Since we want to arrive at f − (z), the relevant equation in (9.36) is the top right one: p F {f − (x)}(ω) = fˆ+ (ω). Following Example 9.5, we obtain fˆ(ω) = π2 e−ω , and therefore ˆ +∞ r π −ω iωz i 1 e e dω = , f − (z) = √ 2 2(i + z) 2π 0 as before. 2
Example 9.21. Calculate f − (z) in case of f (z) = e−z (Gaussian). 2 As in the previous example, the first step is to obtain fˆ(ω) = √12 e−ω /4 (from (9.9)), and therefore ˆ ∞ 2 1 − e−ω /4 eiωz dω . f (z) = √ 2 π 0 Following the idea in Example 9.4, we complete the squares in the exponents, ˆ ∞ 2 2 ω 1 f − (z) = √ e−z e−( 2 −iz) dω , 2 π 0
and then make the variable change ω2 − iz = t, i.e., 12 dω = dt, giving ˆ ∞ ˆ ∞ ˆ −iz 2 2 2 2 2 1 1 f − (z) = √ e−z e−t dt = √ e−z e−t dt − e−t dt π π −iz 0 0 1 −z2 (1 + erf(iz)) . = e 2 The successive steps in this derivation are illustrated in Figure 9.5.
9.2 Laplace transform The Laplace transform is widely used in many fields. In the context of electrical engineering, it offers a particularly convenient approach for solving many linear ODE initial value problems (often in time, over the semiinfinite range t ∈ [0, ∞)). ˆ
Definition 9.22. fˆ(s) =
∞
f (x) e−sx dx.
(9.37)
0
We will again use either the “hat” notation, or the more explicit version L{f (x)}(s) = fˆ(s), to refer to the transform and mostly use s as the transform variable (to further help distinguish the Laplace transform from the Fourier transform). While evaluating (9.37) often is quite straightforward,72 it is much less obvious how to get back from fˆ(s) to f (x). Without using complex variables, there are two main approaches: 1. Recognize the transform in a tabulated collection of known cases. This approach can be combined with techniques to modify certain types of transforms fˆ(s) into combinations of other functions that are more likely to be tabulated (such as splitting rational functions into partial fraction form, etc.). 72 We assume the integral converges for some svalues, assured, for example, if f (x) is of exponential order, i.e., there exist constants A and B such that f (x) ≤ A eBx for x ≥ 0. In this case, (9.37) converges for Re s > B.
246
Chapter 9. Transforms 1
0.8 0.6
0.5
0.4 0.2
0 3
2
1
0
1
2
0 3
3
2
1
0
x 2
(b) fˆ(ω) =
(a) f (x) = e−x 0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0 3
2
1
(c) fˆ− (ω) =
0
1
2
0 3
3
fˆ(ω) , ω < 0 0 , ω≥0
2
√1 2π
1
(d) fˆ+ (ω) =
(e) f (z) = e−z
4
1
´∞ −∞
3
f (x)e−iωx dx
0
2
1
2
3
0 , ω≤0 ˆ f (ω) , ω > 0
2
4
2
2
0
2
0
0 2
0 2
2
0
x
(f) f − (z) =
2
√1 2π
2
´∞ −∞
2
0
y
x
fˆ+ (ω)eiωx dω 2
(g) f + (z) =
√1 2π
2
´∞ −∞
y
fˆ− (ω)eiωx dω
Figure 9.5. f (x) = e−x and the steps in calculating f − (z) and f + (z).
9.2. Laplace transform
247
2. Use the (highly impractical) explicit formula (−1)k f (x) = lim k→∞ k!
k+1 dk ˆ k k f . dxk x x
With use of complex variables, there is a vastly more practical general formula available (obtained by recognizing the Laplace transform as a “disguised” case of the Fourier transform).73
9.2.1 Inversion formula Theorem 9.23. The inversion formula for the Laplace transform is f (x) =
ˆ
1 2πi
c+i∞
fˆ(s)esx ds,
(9.38)
c−i∞
where we will choose c so that all of the singularities of fˆ(s) have Re s < c. Proof. We first extend f (x) to x < 0 by setting f (x) = 0 for x < 0. Then ˆ
∞
fˆ(s) =
f (x)e−sx dx.
(9.39)
−∞
Let’s next make a change of variables to get what will look like a Fourier transform, which we know how to invert. Thus, let s = c + ik, where c and k are real (it will soon become apparent why the constant c is required). This gives ˆ
∞
fˆ(s = c + ik) = −∞

f (x)e−cx e−ikx dx {z }
Formula for the Fourier transform of
and therefore f (x)e−cx =
1 2π
ˆ
,
√ 2πf (x)e−cx
∞
fˆ(c + ik)eikx dk . −∞
Next, we change the variables back to s = c + ik , ds = idk, and we multiply everything by ecx . Once again, we are in a case where the required integrals often are very well suited for contour integration approaches. In this application, extending the contour to become closed (allowing residue calculus to be applied) is known as forming a Bromwich contour. inverting, one might want to verify that lims→∞ fˆ(s) = 0, since otherwise, f (x) needs to be a “generalized function” (such as a delta function). 73 Before
248
Chapter 9. Transforms Im s
Γ i R
c
Re s
−i
Figure 9.6. Contour in Example 9.24.
Example 9.24. Given f (x) = sin x , determine fˆ(s) and then invert the result to arrive back at f (x). We find ˆ fˆ(s) =
ˆ
∞ −sx
sin x e
∞
dx = Im
0
e 0
ix−sx
∞ 1 eix−sx = 2 . dx = Im i−s 0 s +1
(9.40)
This can now be inverted by f (x) =
1 2πi
ˆ
c+i∞
c−i∞
esx 1 ds = 2 1+s 2πi
ˆ Γ
esx ds, 1 + s2
where the contour Γ is a vertical line to the right of the singularities. If x ≥ 0, we can close the contour to the left, as displayed in Figure 9.6. By Jordan’s lemma, the contribution of the integral around the semicircle vanishes as its radius R → ∞, giving (from the two poles) eix − e−ix 1 2πi = sin x. f (x) = 2πi 2i If, on the other hand, x < 0, the contour needs to be closed to the right, giving f (x) = 0. In other words, a Laplace transform represents a function that can be nontrivial for x ≥ 0, but is assumed to vanish identically for x < 0 (of course, nothing stops us from applying the Laplace transform to sin(−x), obtaining a transform representation of the sine function for negative arguments instead of for positive ones).
9.2. Laplace transform
249 Im s
Γ i R
c ε
Re s
−i
Figure 9.7. Contour in Example 9.25.
Example 9.25. Find f (x) which has as its Laplace transform log s fˆ(s) = . 1 + s2 Since the factor log s requires a branch cut from s = 0 to infinity, we cannot this time follow the contour in Figure 9.6. However, we can place the cut along the negative real axis and then use the keyhole contour illustrated in Figure 9.7. Listed below are the contributions from the contour and from the residues: 1. Res(f, s = i) =
π ix 4e .
2. Res(f, s = −i) = π4 e−ix . ´ ´ ´ ´∞ ∞ log(r)+iπ −xr 1 3. → f (z)dz + ← f (z)dz = 2πi dr − 0 1+r 2 e 0 ´ ∞ e−xr dr. 0 1+r 2 ´ 4.  f (z)dz ≤ M · L ≈ 2π log → 0 as → 0 . ´ 5. R f (z)dz → 0 as R → ∞ by Jordan’s lemma.
log(r)−iπ −xr dr 1+r 2 e
=
Therefore, adding all the contributions, we obtain ˆ ∞ −xr π e f (x) = cos x − dr, 2 1 + r2 0 where the integral cannot be expressed in terms of elementary functions in closed form. One often useful feature of the Laplace transform (for example in the context of solving ODEs, as in Section 9.2.4) is that also discontinuous functions f (x) transform into analytic functions fˆ(s).
250
Chapter 9. Transforms
(a) f (x).
(b) fˆ(s).
Figure 9.8. The function f (x) in Example 9.26, and its Laplace transform fˆ(s).
Example 9.26. Find the Laplace transform fˆ(s) of the step function 1 , 0≤x2 and then invert fˆ(s) to recover f (x).74 ´∞ ´1 ´2 −s 2 We obtain directly fˆ(s) = 0 f (x) e−sx dx = 0 1 e−sx dx− 1 1 e−sx dx = (1−es ) . The singularity at s = 0 is removable, so fˆ(s) is an entire function. Figure 9.8 shows the two functions f (x) and fˆ(s). The integrand in the Bromwich integral for inverting fˆ(s) becomes fˆ(s) esx . While the first factor decays in the right halfplane, the second factor grows there, making it unclear how to close an integration contour. Hence, we split fˆ(s) esx as 1 2 1 fˆ(s) esx = esx − es(x−1) + es(x−2) . s s s
(9.41)
In contrast to fˆ(s) esx , which is an entire function, each of these terms has a pole at the origin as its only singularity, with residues 1, −2, 1, respectively. The columns in Figure 9.9 display these three terms in the cases of x = 12 , 32 , 52 , respectively (representative for the three subintervals 0 < x < 1, 1 < x < 2, and x > 2). Following a vertical path in the right halfplane, we now close the contour in whichever halfplane the magnitude of the integrand goes to zero. Just keeping track of whether this implies encircling a pole or not recovers the step function f (x). 74 Values
assigned to f (x) at single points (such as here at the points of discontinuity x = 1 and x = 2) do not influence fˆ(s). When inverting, (9.38) returns at such points the average from the two sides.
9.2. Laplace transform
2 − es(x−1) s
1 s(x−2) e s
x = 5/2
x = 3/2
x = 1/2
1 sx e s
251
Figure 9.9. The three functions of s which, when added, form fˆ(s) esx in Example 9.26 (cf. (9.41)), shown for three representative values of x.
252
Chapter 9. Transforms
9.2.2 The inverse Laplace transform of rational functions It follows from the definition of the gamma function Γ(z) =
´∞ 0
tz−1 e−t dt that
L{xk }(s) = Γ(k + 1)/sk+1 .
(9.42)
Limiting k to the nonnegative integers and using that k! = Γ(k + 1) gives 1 2! 1 , L{x}(s) = 2 , L x2 (s) = 3 , . . . , s s s k! L xk (s) = k+1 , k = 0, 1, 2, . . . . s
L{1}(s) =
(9.43)
From (9.37) it follows that L {f (x) eax } = L{f (x)}(s − a)
(9.44)
for an arbitrary constant a. Together, these last two equations imply that L xk eax =
k! , k = 0, 1, 2, . . . . (s − a)k+1
(9.45)
If f (x) is bounded at the origin and has the rational function fˆ(s) = p(s) q(s) as its Laplace transform, the degree of p(s) will be less than that of q(s). The partial fraction expansion of fˆ(s) will only contain terms that are multiples of expressions of the same form as the RHS in (9.45). This relation therefore provides a way to find f (x). Example 9.27. Find the function f (x) with Laplace transform fˆ(s) =
s2 −5 s4 −1
.
Method 1: Given that the zeros of the denominator are s = {−1, +1, −i, +i} (all distinct), 1 we find by standard methods of calculus the partial fraction expansion to be fˆ(s) = s+1 − 3i 1 3i 1 1 s−1 + 2 s+i − 2 s−i . Each term is a k = 0 special case of (9.45), giving after a quick simplification f (x) = e−x − ex + 3 sin x. Method 2: Viewed as an analytic function of s, fˆ(s) has as its only singularities first order poles at s = {−1, +1, −i, +i}. By the N/D0 rule (see “Shortcuts to calculate residues” in −ix Section 5.1.1), the residues of the integrand fˆ(s)esx in (9.38) become {e−x , −ex , 3i , 2e 3i ix − 2 e }. With a vertical path to the right of the singularities and closing it to the left, (9.38) gives f (x) as the sum of the residues; i.e., again f (x) = e−x − ex + 3 sin x. Had there been poles of second or higher order present in fˆ(s), Method 1 would generalize immediately. For Method 2, the N/D0 rule would no longer apply, so some more effort would be needed to calculate the residue at the poles of higher multiplicity. The function f (x) will then be the sum of the residues. We can from this example spot an easy way to obtain the coefficients in a partial fraction expansion of a rational function with first order poles only; each is given by applying the N/D0 rule to the rational function. The inverse Laplace transform of rational functions can be obtained also without using complex variables. If a realvalued partial fraction expansion only contains terms of the
9.2. Laplace transform
253
forms n! = L {xn eax } , (s − a)n+1 s−a = L {eax cos bx} , (s − a)2 + b2 b = L {eax sin bx} , (s − a)2 + b2
(9.46) (9.47) (9.48)
the inversion is immediate. However, if the quadratic denominator in (9.47) or (9.48) appears raised to higher powers, the corresponding formulas become more complicated. When using complex variables, no similar complication arises, since partial fraction expansions will then never contain terms in forms different from what is given by (9.46).75
9.2.3 Some key features of the Laplace transform dn For the Fourier transform, Theorem 9.6 gave a formula for F dx (ω) in terms of n f (x) F {f (x)} (ω). The corresponding formula for the Laplace transform becomes somewhat more involved, since the interval is finite, x ∈ [0, ∞], and we can no longer assume functions to vanish at both ends. Theorem 9.28. n d n n−1 n−2 0 (n−1) f (x) (s) = s L{f (x)}(s) − s f (0) + s f (0) + · · · + f (0) . L dxn (9.49) In particular, d L f (x) (s) = s L{f (x)}(s) − f (0) , (9.50) dx 2 d L f (x) (s) = s2 L{f (x)}(s) − (s f (0) + f 0 (0)) . (9.51) dx2 Proof. Repeated integration by parts gives n ˆ ∞ d L f (x) (s) = f (n) (x)e−sx dx dxn 0 ˆ ∞ h i∞ = f (n−1) (x)e−sx +s f (n−1) (x)e−sx dx 0 0 h i∞ (n−1) (n−2) = −f (0) + s f (x)e−sx 0 ˆ ∞ +s f (n−2) (x)e−sx dx 0
=
···
= −f (n−1) (0) − s f (n−2) (0) − · · · − sn−1 f (0) ˆ ∞ n +s f (x) e−sx dx. 0 75 A
simple case illustrating this arises in the solution to part (b) of Example 9.31.
254
Chapter 9. Transforms
Theorem 9.29. L {xn f (x)} (s) = (−1)n
dn L {f (x)} (s), n = 0, 1, 2, . . . . dsn
This result follows from (repeatedly) differentiating the definition of the Laplace transform (9.37) with respect to s. Like for the other transforms we consider, the Laplace transform possesses a convolution theorem. Theorem 9.30. Defining the convolution of f (x) and g(x) as ˆ x h(x) = f (ξ)g(x − ξ)dξ , 0
it will hold that ˆ h(s) = fˆ(s) · gˆ(s).
(9.52)
The proof is given in Section 9.7.
9.2.4 Application of the Laplace transform to solving constant coefficient ODEs The following example illustrates the main ideas that are involved. Example 9.31. Solve ODE initial value problem y 00 (x)+y(x) = g(x), y(0) = 1, y 0 (0) = 2 for the following three choices of RHS function g(x): (a) g(x) = 0, (b) g(x) = cos x, (c) g(x) = tan x. (a) From (9.49) it follows that s2 yˆ(s) − (s + 2) + yˆ(s) = 0, i.e., yˆ(s) = ss+2 2 +1 = 1 1 1 1 2 + i s+i + 2 − i s−i . Inverting, using the k = 0 case of (9.45), gives y(x) = 1 −ix + 12 − i e+ix = cos 2 +i e x + 2 sin x. (b) Since L {cos x} (s) = L 12 eix + e−ix (s) = s2s+1 , the Laplace transform of s the ODE becomes s2 yˆ(s) − (s + 2) + yˆ(s) = s2s+1 , and yˆ(s) = ss+2 2 +1 + (s2 +1)2 . We recognize the first term from part (a); its inverse Laplace transform is cos x + 2 sin x. The second term, originating from the ODE’s RHS function, can be expanded in partial i 1 i 1 i s −ix − fractions as (s2 +1) 2 = 4 (s+i)2 − 4 (s−i)2 , with inverse Laplace transform 4 x e i 1 1 ix 76 x e = x sin x. The ODE solution thus becomes y(x) = cos x + 2 sin x + x sin x. 4 2 2 (c) The Laplace transform of the ODE becomes in this case s2 yˆ(s) − s − 2 + yˆ(s) = 1 L{tan x}(s), and yˆ(s) = ss+2 2 +1 + s2 +1 L{tan x}(s). We recognize again the first term from part (a). Regarding the second term s21+1 L{tan x}(s), there is a complication in that there does not exist any practical explicit expression for L{tan x}(s). However, noting that s21+1 = L{sin x}(s), the task becomes finding the inverse Laplace transform of the 76 This case features resonance, with the solution amplitude growing unbounded. This happens when a forcing function g(x) contains a component with a frequency that equals a natural frequency of the unforced (g(x) = 0) and undamped equation.
9.3. Mellin transform
255
6 4 2
x 0 /2
3 /2
2
5 /2
3
7 /2
2 4
Part (a): Solution to homogeneous ODE Part (b): Change in solution due to inhomogeneous term Part (c): Change in solution due to inhomogeneous term
6
Figure 9.10. The solution to the homogeneous ODE in Example 9.31(a), and the additional terms that have to be added to this solution for the inhomogeneous cases in parts (b) and (c). Note that the two latter curves do not affect the ODE’s initial conditions on y(0) and y 0 (0).
L{sin x}(s)·L{tan x}(s), By the convolution theorem (Theorem 9.30), this equals ´product x sin(x − ξ) tan ξdξ. Although not 0 also easy to evaluate, this integral can be found in cos x closed form: sin x + (cos x) log 1+sin x . Figure 9.10 shows the solution to the homogeneous ODE (part (a)), and the terms that have to be added to this when including the inhomogeneous77 RHSs in parts (b) and (c).
9.3 Mellin transform This transform resembles the Laplace transform, but with xs−1 replacing the factor e−sx in the definition, i.e., ˆ ∞ ˆ M {f (x}(s) = f (s) = f (x) xs−1 dx. (9.53) 0
The range of svalues for which fˆ(s) is defined (if not resorting to analytic continuation) will depend on the properties of f (x) at x = 0 and x = ∞, and will take the form of a vertical band in the complex splane. This fundamental band can be empty (e.g., for constants and all monomials xα ) or it can cover the whole splane (as for f (x) = 1 e−(x+ x ) ). As illustrated in Example 5.14, contour integration is often well suited for evaluating integrals of the form (9.53).
9.3.1 Inversion formula and some transform properties The inversion formula is also similar to the Laplace transform case, ˆ γ+i∞ 1 f (x) = x−s fˆ(s)ds, 2πi γ−i∞
(9.54)
where the vertical path has to be chosen inside this band (for a proof, see Theorem 9.44 in Section 9.7). Like for the other key transforms, large numbers of transform pairs are known [34]. A few are listed in Table 9.1, and some general properties in Table 9.2. 77 “Inhomogeneous”
and “nonhomogeneous” are synonyms.
256
Chapter 9. Transforms
Table 9.1. Some examples of Mellin transforms. Note that fˆ(s) is the same in the second and third cases, but that the fundamental band differs.
f (x) 0, x ≤ 1 xa , x > 1
fˆ(s) 1 − s+a
Fundamental band Re s < a
π sin πs π sin πs π s sin πs
−1 < Re s < 0
e−x
Γ(s)
Re s > 0
1 ex − 1
ζ(s)Γ(s)
Re s > 1
1 1+x x − 1+x log(1 + x)
sin x
sin
πs 2
0 < Re s < 1
−1 < Re s < 0
Γ(s)
−1 < Re s < 1
Table 9.2. Some analytic properties of the Mellin transform.
dk M (log x)k f (x) (s) = k M {f (x)} (s) ds k d Γ(s) M f (x) (s) = (−1)k M {f (x)} (s − k) dxk Γ(s − k) ´ x 1 M 0 f (ξ)dξ (s) = − M {f (x)} (s + 1) s M {f (ax)} (s) = a−s M {f (x)} (s) Example 9.32. Use (9.54) to invert fˆ(s) = Γ(s) g(−s), where g(s) is analytic, and recover Ramanujan’s master theorem, as given in (3.12). ´ γ+i∞ −s 1 Equation (9.54) gives f (x) = 2πi x Γ(s) g(−s) ds. Since Γ(s) has poles at γ−i∞ each nonpositive integer, we choose γ = 0+ . Forming a Bromwich contour with a large left semicircle CR of radius R leads to 1 f (x) + lim R→∞ 2πi
ˆ x CR
−s
Γ(s) g(−s) ds =
∞ X
Res x−s Γ(s) g(−s), s = −k . (9.55)
k=0
The integral vanishes when R → ∞ due to the fast decay of the gamma function in the left halfplane (cf. Figures 6.1(b)–(d)). Stirling’s formula (Example 12.8) provides the leading asymptotic behavior of Γ(z) for large values of z. Equation (12.24) yields ∼ (R log R) cos θ − R θ sin θ − R cos θ log Γ R eiθ = Re log Γ R eiθ 1 cos θ + ··· + (log 2π − log R) + 2 12 R
9.3. Mellin transform
257
if one avoids θ = ±π (the negative real axis). The contribution of the contour integral along the semicircle can be bounded from above as follows:78 ´ −Rcos θ √ CR x−s Γ(s) g(−s) ds ≤ πR xR g −R eiθ R √R eR 2π → 0 as R → ∞, due to the rapid decay of the factor R−Rcos θ . With the integral absent in (9.55), we next use (6.1) to recover Ramanujan’s formula ˆ ∞ x2 (9.56) xs−1 g(0) − x g(1) + g(2) − + · · · dx = Γ(s) g(−s). 2! 0
9.3.2 Application of the Mellin transform to harmonic sums Applications of Fourier and Laplace transforms often utilize that (i) differentiation becomes an algebraic operation, and (ii) the transforms possess convolution theorems. The Mellin transform has the additional feature that a rescaling of the independent variable reduces to a multiplication of the transform (noted as the last entry in Table 9.2). Theorem 9.33. If f (x) has the Mellin transform fˆ(s), then f (ax) has the transform a−s fˆ(s). ´∞ Proof. By (9.53), the transform of f (ax) is 0 f (ax) xs−1 dx. After the variable change ´∞ ax = t, this becomes 0 f (t) (a1−s ts−1 ) (a−1 dt) = a−s fˆ(s). This scaling property provides an approach to solve functional equations of the type f (x) = g(x) + f (ax), with g(x) and the constant a given, since, on the transform side, this relation becomes fˆ(s) = gˆ(s)/(1 − a−s ). Since the transform is linear, this result generalizes greatly, to harmonic sums: X F (x) = λk f (µk x). (9.57) k
Here f (x) is a “base function” and µk and λk represent frequencies and amplitudes, respectively. The Mellin transform of this sum factorizes as ! X −s Fˆ (s) = λk (µk ) fˆ(s) . (9.58) k
Example 9.34. Find the Mellin transform of f (x) = P∞
1 k=1 k
·
P∞
k=1
1 k
−
1 k+x
.
x/k 1+x/k ,
we recognize it as a harmonic sum (9.57) with λk = µk = Applying (9.58) and the third case in Table 9.1 gives fb(s) = P∞ 1 1 ζ(1−s) π − sin πs k=1 k k−s = − π sin πs for −1 < Re s < 0. After rewriting the sum as 1 k.
78 Here assuming some mild bounds on any growth of g(−s) and also technical details about Γ(s) along the negative real axis.
258
Chapter 9. Transforms
Table 9.3. Some integral equations that can be solved with the Mellin transform. In these examples, y > 0.
Solution in terms of fˆ(s) gˆ(s) fˆ(s) = ˆ K(s)
Integral equation ´∞ g(y) = 0 f (x) K(x y)dx g(y) =
g(y) +
´∞ 0
´∞
f (x) K(y/x)dx
gˆ(s − 1) fˆ(s) = ˆ − 1) K(s
f (x) K(x y)dx = f (y)
ˆ gˆ(s) + K(s) gˆ(1 − s) fˆ(s) = ˆ ˆ 1 − K(s) K(1 − s)
0
9.3.3 The Mellin transform applied to some integral equations The remarkably wide range of features of the Mellin transform has made it applicable in several contexts, including analytic number theory79 (for example in proofs of the prime number theorem), closedform evaluation of certain integrals [16], and the solution of some types of integral equations [34]. We limit ourselves here to quote a few examples of the latter in Table 9.3.
9.3.4 Connection between the Mellin transform and properties of f (x) at x = 0 and x = ∞ Although (9.53) only defines fˆ(s) within a fundamental strip α < Re s < β, it can typically be analytically continued far beyond this strip. For example, all the cases of fˆ(s) given in Table 9.1 are meromorphic functions (i.e., their only singularities in the finite complex splane are poles).80 These poles tell a lot about how f (x) behaves in the vicinities of x = 0 and x = ∞ (as can be deduced from closing the contour in (9.54) in the λ left and right halfplanes, respectively). Each simple pole s+a to the left of the fundamental strip (including on its edge) tells us that a local power series expansion at x = 0 will λ contain a term λxa . Poles of higher multiplicity, (s+a) k+1 , indicate the presence also of k
a k logarithmic terms as x & 0; (−1) k! λx (log x) . Matching results for the right of the strip λ λ are the following: A simple pole s+a implies a term − x1a as x → ∞, and (s−a) k+1 a term k
− (−1) k!
λ k xa (log x) .
Example 9.35. Use the result just above to invert fˆ(s) =
π sin πs .
As the second and third cases in Table 9.1 showed, the answer will also depend on what the fundamental strip is. In the second case, the relevant poles (with Re s ≤ 0) are located 79 One
example is provided by the relation
0
x−s R(x)dx
=
Γ(2−s)
− ζ(2s−2) , where R(x)
=
n+1 n
(−1) x n=1 (n−1)!ζ(2n)
P∞
´∞
(following from (9.56)). R(x) is known as the Riesz function. Recalling the poles of Γ(2−s)
Γ(z) and trivial zeros of ζ(z), ζ(2s−2) would be polefree for 0 < Re s < 2 if it was not for the nontrivial zeros of ζ(z). It can from this be deduced that the Riemann hypothesis is true if and only if the growth of R(x) for x → +∞ satisfies R(x) ≤ C x1/4+ε , where C is a constant and ε > 0 is arbitrarily small. 80 If fˆ(s) is not meromorphic (e.g., fˆ(s) being an entire function in case of f (x) = e−(x+1/x) ), the results in this section are not applicable.
9.4. Hilbert transform
259
at s = 0, −1, −2, . . . . and have residues 1, −1, 1, −1, . . . . The expansion of f (x) at x = 0 should thus be of the form f (x) = 1 − x + x2 − x3 + · · · , indeed matching the Taylor 1 expansion for f (x) = 1+x . In the third case, the pole at s = 0 should be omitted, giving x . The fundamental strip can similarly be translated with instead the result f (x) = − 1+x any integer amount, giving different f (x). The argument just above was not quite rigorous, since we used only partial information about fˆ(s). In particular, f (x) could also have contained terms that for x & 0 become smaller than any power of x (such as e−1/x ). Several options are available for ruling out such additional functions, including (i) knowledge that f (x) is analytic at x = 0, and (ii) simple estimates for fˆ(s) up/down the fundamental band. For precise results in this context, see, for example, [36]. In certain contexts (such as for creating asymptotic expansions; see Chapter 12), no additional information is needed. Example 9.36. Find the function f (x) (analytic at x = 0) that has fˆ(s) = its Mellin transform in the strip 0 < Re s < n. k
The relevant poles (locations and residues) are given by (−1) k 2, . . . . The function f (x) should thus, at x = 0, have an expansion f (x) =
Γ(s)Γ(n−s) Γ(n)
Γ(n+k) 1 Γ(n) s+k ,
as
k = 0, 1,
∞ X (−1)k Γ(n + k) k x . k! Γ(n)
k=0
−n
We can recognize the terms as k xk , i.e., the sum is the binomial expansion of f (x) = (1 + x)−n . 1 In the special case of n = 1, this gives for the Mellin transform of f (x) = 1+x the π result fˆ(s) = Γ(s)Γ(1 − s) = sin πs , in the band 0 < Re s < 1, matching Example 9.35.
9.4 Hilbert transform The Hilbert transform has significant applications in signal processing (such as for digital filters and for approximating instantaneous properties of oscillatory time series), and also in physiology, spectroscopy, microscopy, geophysical imaging, etc. A key feature of it is that it allows analytic function techniques to be applied to realvalued “signals” that might not be differentiable (or even continuous). It also forms one (of several) entry points to the topic of Riemann–Hilbert problems, briefly summarized in Chapter 10. If a function f (z) = u(x, y) + iv(x, y) is analytic everywhere in the upper halfplane, and f (z) → 0 there for z → ∞, then the values along the real axis for u(x, 0) and v(x, 0) are related through the Hilbert transform. Knowing one of these two functions u(x, 0) and v(x, 0) uniquely determines the other, without the need to make any calculations that extend outside the xaxis. Example 9.37. Determine the Hilbert transforms of cos ωx and sin ωx, assuming ω real. If ω > 0, we consider f (z) = eiωz = cos ωz + i sin ωz, and if ω < 0, we consider f (z) = e−iωz = cos ωz − i sin ωz (in both cases ensuring decay in the upper halfplane).
260
Chapter 9. Transforms
We thus obtain if ω > 0 : if ω < 0 :
H{cos ωx} = + sin ωx and H{sin ωx} = − cos ωx, H{cos ωx} = − sin ωx and H{sin ωx} = + cos ωx.
(9.59)
In general it will hold (as we saw an example of in this special case) that H{H{u(x)}} = −u(x), implying that there is no need to have any separate formula for the inverse of the Hilbert transform. In most cases, the Hilbert transform cannot be simply “read off” just by inspection based on elementary functions. Theorems 9.38 and 9.39 provide general procedures to obtain the transform. Theorem 9.38. For f (z) = u(x, y) + iv(x, y) with the properties just described (f (z) = u(x, y) + iv(x, y) analytic with f (z) → 0 for z → ∞ in the upper halfplane), it holds that 1 π 1 v(x, 0) = +H{u(x, 0)} = − π
u(x, 0) = −H{v(x, 0)} = +
∞
v(t, 0) dt, −∞ t − x ∞ u(t, 0) dt. −∞ t − x
(9.60) (9.61)
Proof. If we close the integration with a large semicircle in the upper halfplane, the integral along that part will vanish, and therefore ∞ −∞
f (t) dt = πi f (x) t−x
(using half the residue for the first order pole at t = x, since this lies on the integration path). Writing f (t) = u(t, 0) + iv(t, 0), this becomes u(x, 0) + iv(x, 0) =
1 πi
∞ −∞
u(t, 0) i dt + t−x πi
∞ −∞
v(t, 0) dt, t−x
and the result follows from separating the real and the imaginary parts. A quite different approach for obtaining a Hilbert transform of a function, not requiring any principal value integrals, is provided by its close relation to Fourier transforms. Theorem 9.39. If a function u(x) has the Fourier transform u ˆ(ω), then H{u(x)} is obtained as the inverse Fourier transform of −i sign(ω) u ˆ(ω). Proof. The relations (9.59) can be summarized in a single equation as H{eiωx } = −i sign(ω) eiωx ,
(9.62)
where ω and x (both real) can be of either sign. With u ˆ(ω) denoting the Fourier transform of u(x), it then follows that ˆ ∞ ˆ ∞ 1 1 iωx H{u(x)} = H √ u ˆ(ω)e dω = √ u ˆ(ω)H(eiωx )dω 2π −∞ 2π −∞ ˆ ∞ 1 =√ −i sign(ω) u ˆ(ω)eiωx dω. 2π −∞
9.4. Hilbert transform
261
Theorem 9.40. The Hilbert transform commutes with the derivative operator, i.e., H
du(x) dx
=
d H{u(x)}, dx
which further implies H
dk u(x) dxk
=
dk H{u(x)}, dxk
k = 1, 2, . . . .
(9.63)
This result follows from either of the two theorems just above. With these three theorems, the Hilbert transform can be calculated for a variety of functions. Some examples beyond (9.59) include
a H a2 + x2 sin ax H x n o 2 H e−αx H {Π(x)} H {δ(x)}
x , a2 + x2 1 − cos ax = sign(a) , x ˆ x √ 2 2 2 = √ D( αx), where α > 0, D(x) = e−x et dt, π 0 1 x + 12 1 if x < 1 2, , where Π(x) = = log 1 1 π 0 if x > x− 2 2, ˆ ∞ 1 , where δ(x) = 0 for x 6= 0, δ(x) = 1. = πx −∞ =
The last of these (together with (9.63)) opens up one approach towards interpreting concepts such as derivatives of delta functions. This relation (9.63) together with linearity H {α1 u1 (x) + α2 u2 (x)} = α1 H {u1 (x)} + α2 H {u2 (x)} can supplement the results in Theorems 9.38 and 9.39 for determining Hilbert transforms of a variety of functions. In the following two examples, we determine the Hilbert transform directly from the principle that f (z) = u(x, 0) + i H {u(x, 0)} will decay to zero in the upper halfplane. Example 9.41. Find the Hilbert transform of the cubic Bspline function introduced in Example 9.10. A sometimes more convenient form for B(x) than (9.12) is B(x) =
1 x + 23 − 4 x + 13 + 6 x3 − 4 x − 13 + x − 23 12
(9.64)
(since this also describes the unique piecewise cubic function, with continuous second derivative and the same nodes, vanishing for x ≤ −2 and for x ≥ +2, and normalized the same). This expression can alternatively be written as B(x) =
1 sign(x + 2)(x + 2)3 − 4 sign(x + 1)(x + 1)3 + 6 sign(x)(x)3 12 − 4 sign(x − 1)(x − 1)3 + sign(x − 2)(x − 2)3 .
262
Chapter 9. Transforms
(a) Re f (z), equals B(x) for z real.
(b) Im f (z), equals H{B(x)} for z real.
Figure 9.11. Real and imaginary parts of the function f (z) given in (9.65), showing along the real axis a cubic Bspline and its Hilbert transform, respectively.
With sign x a step function, as is Im log z for z real, we are thus led to consider the analytic function i f (z) = log(z + 2)(z + 2)3 − 4 log(z + 1)(z + 1)3 + 6 log(z)(z)3 (9.65) 6π − 4 log(z − 1)(z − 1)3 + log(z − 2)(z − 2)3 and recover from this B(x) = Re f (x). Since f (z) decays to zero (due to cancellations between its terms) for z → ∞ in the upper halfplane, we can pick up its Hilbert transform immediately from H{B(x)} = Im f (x); cf. Figure 9.11. This function f (z) has, due to its logarithms, branch points at z = −2, −1, 0, 1, 2. With the standard choice for branch cuts, these will not get into the upper halfplane. 2
Example 9.42. Find the Hilbert transform of the Gaussian function f (x) = e−z . 2
We start in Figures 9.12(a)–(b) by illustrating the Gaussian function f (z) = e−z in the complex plane. If we can alter this function to another analytic one such that (i) the 2 real part along the real axis remains unchanged e−x , and (ii) it decays to zero in the upper halfplane, we can from its imaginary part along the real axis read off the Hilbert transform. As ´a further background, we illustrate in parts (c) and (d) the error function 2 z erf (z) = √2π 0 e−t dt. Not surprisingly, this function diverges in the same sectors as 2
does e−z , with the additional striking feature that it converges very rapidly to +1 in the sector surrounding the positive real axis and to −1 in the opposite sector. If we thus form 2 erf (iz), we rotate these two images 90◦ clockwise. Then multiplying them by e−z and 2 adding to e−z , the large oscillations in the upper halfplane ought to cancel out (without 2 altering the real part of e−z along the real axis). Parts (e) and (f) of the figure shows this indeed to be the case. We can now read off the Hilbert transform from the function along the real axis in the last of these figures, obtaining ˆ ix ˆ x n o 2 2 2 2 2 2 2 2 H e−x = Im √ e−x e−t dt = √ e−x et dt = √ D(x), π π π 0 0 where D(x) is known as Dawson’s function. This same result follows also from Example 9.21.
9.4. Hilbert transform
263
2
2
0
0
2
2
4
4 3
3 2
2
4 3
1
4
0
1 1
0
0
1
2
x
2
0
1 1
3
1
2
2
3
x
y
3 4
1
2 2
3
2
y
3 4
4
4
2
(a) Re e−z .
(b) Im e−z .
2
2
0
0
2
2
4
4 3
3 2
2
4 3
1
4
1 1
0
0
1
2
x
2
0
1 1
3
1
2
0
2
3 4
(c) Re erf (z) = Re
x
y
3 4
√2 π
1
2
´z 0
2
3
2
e−t dt.
4
(d) Im erf (z) = Im
2
2
0
0
2
y
3 4
√2 π
´z 0
2
e−t dt.
2
4
4 3
3 2
4 3
1 2
0
1 1
0
4 3
1 2
0
1 1
0
1
2
x
2
2
3
3 4
y
4
h i 2 2 (e) Re e−z +e−z erf (iz) .
1
2
x
2
3
3 4
y
4
h i 2 2 (f) Im e−z +e−z erf (iz) .
Figure 9.12. Illustration of the steps in obtaining the Hilbert transform of a Gaussian.
264
Chapter 9. Transforms
9.5 ztransform This transform can either be seen as a discrete version of the Laplace transform or as a case of a regular Taylor (or Laurent) expansion.81 If the xvariable in (9.37) is not continuous, 0 ≤ x ≤ ∞, but discrete, k = 0, 1, 2, . . ., with corresponding function values fk , a counterpart to the Laplace transform can be written as ∞ X fˆ(z) = fk z −k . (9.66) k=0
This function fˆ(z) is known as the ztransform of the sequence {fk } and is widely used, for example, in signal processing. After the variable change z → z1 , (9.66) takes the form of a standard Taylor expansion. It can then be inverted by any means that we have for Taylor expanding a function; cf. Sections 2.3 and 4.2.8. If the Taylor expansion has radius of convergence R = 0, ideas from analytic continuation and asymptotic expansions (Chapters 3 and 12, respectively) may be applicable. Especially in geophysical applications, the standard Taylor version is commonly used for the ztransform: fˆ(z) =
∞ X
fk z k .
(9.67)
k=0
Example 9.43. Solve the linear inhomogeneous recursion relation xk + 5xk−1 =
1 , k = 1, 2, 3, . . . , k
x0 given.
(9.68)
Let the ztransform of the sequence {xk } (using the definition (9.67)) be fb(z) = x0 + x1 z + x2 z 2 + x3 z 3 + · · · , i.e., 5z fb(z) = 5x0 z + 5x1 z 2 + 5x2 z 3 + · · · .
(9.69)
Adding these two sums and using (9.68) gives 1 1 1 1 fb(z) = x0 + z + z 2 + z 3 + · · · . 1 + 5z 1 2 3 Resisting the temptation of simplifying 11 z + 12 z 2 + 13 z 3 + · · · = − log(1 − z), but instead 1 Taylor expanding 1+5z = 1 + (−5)z + (−5)2 z 2 + · · · and expanding the RHS, followed by equating coefficients with (9.69), gives " # k X 1 xk = (−5)k x0 + . n(−5)n n=1
9.6 Three additional transforms related to rotations The list of transform pairs is almost (if not literally) inexhaustible. We conclude by just hastily mentioning three transform pairs that often become important in applications that feature rotations or axial symmetries. 81 Additionally
known as generating functions.
9.6. Three additional transforms related to rotations
265
9.6.1 Abel transform The Abel transform of a function f (r) is defined by ˆ
∞
A {f (r)} (s) = 2 s
f (r) r √ dr, r2 − s2
(9.70)
with the inverse transform given by f (r) = −
1 π
ˆ
∞
r
dA(s) ds √ ds s2 − r2
(9.71)
(see Exercise 9.8.29).82 In 2D radially symmetric cases, with r2 = x2 + y 2 , functions b can be interpreted as the f (x, y) will come to depend on r only, as f (r). The function A(s) integral of f (x, y) along an infinite straight line that passes a distance of s from the origin in the (x, y)plane.
9.6.2 Radon transform We again consider straight line integrals across the (x, y)plane, as in the Abel transform case, but remove the requirement of axial symmetry. The integrals will then depend not only on the line’s distance s to the origin, but also on the angle θ of the line relative to the xaxis. A function f (x, y) will be transformed to Fb(s, θ). Johann Radon discovered in 1917 its inversion formula, which can again be expressed in integral form. In this case, the exact inversion formula turns out to be impractical for actual use, but extremely effective computational algorithms for it were later found, leading to Nobel Prize winning advances in medical imaging. Xray absorption along straight lines can be recorded, and the inverse transform provides then high resolution reconstructions / images of the interior of the recorded object.
9.6.3 Hankel transform The Hankel transform pair for a function f (r) is given by ˆ
∞
H{f (r)}(s) = ˆ
f (r) Jν (sr) r dr,
(9.72)
H(s) Jν (sr) s ds.
(9.73)
0
f (r) =
∞
0
Here, Jν (r) denotes the Bessel Jfunction of order ν, to be introduced later in Section 11.2 and illustrated in Figures 11.5 and 11.6. If a function f (x1 , x2 , . . . , xn ) is radially symmetric in an ndimensional space, so will its Fourier transform f (ω1 , ω2 , . . . , ωn ) be in its ndimensional transform space. The radial behaviors of these two functions are thus both 1D functions, and it transpires that they are the Hankel transforms of each other, when choosing ν = n2 − 1. This transform is a case where the two transforms in the pair are identical to each other. 82 Different 1 2
definitions exist in the literature, such as using in the denominator to α with 0 < α < 1.
´s 0
instead of
´∞ s
, and also generalizing the power
266
Chapter 9. Transforms
9.7 Supplementary materials Theorem 9.3. If f (x) =
√1 2π
´∞
−∞
fˆ(ω) eiωx dω, then fˆ(ω) =
√1 2π
´∞ −∞
f (x) e−iωx dx.
Proof. For the interval x ∈ [−π, π], we recall (9.4) and (9.5). The Fourier modes present are given by eikx , k = 0, ±1, ±2, ±3, . . . . If we extend the period by a factor of L, instead considering x ∈ [−Lπ, Lπ], the modes present increase, to ei ωk x , with ωk = 0, ± L1 , ± L2 , ± L3 , . . . . The Fourier expansion can then be written as ˆ Lπ ∞ 1 1 X f (x) e−i ωk x dx. c(ωk )ei ωk x , with c(ωk ) = f (x) = L 2π −Lπ k=−∞
In the limit of L → ∞, the sequence ωk → ω (continuous variable), ˆ ∞ f (x) → c(ω)ei ωx dω, −∞
´∞ and c(ωk ) → c(ω) = f (x) e−i ωx dx. It is arbitrary where the 2π factor is placed, √−∞ so we can define fˆ(ω) as 2πc(ω), arriving at (9.7) and (9.6). 1 2π
Proof of the first part of Theorem 9.14. With the notation in (9.17)–(9.20), we obtain ˆ π ˆ π 1 1 −ikt −iky αk βk = A(t) e dt B(y) e dy 2π −π 2π −π ˆ π ˆ π 1 = A(t)B(y) e−ik(t+y) dtdy . 4π 2 −π −π Replacing y by y = x − t, i.e., t + y = x and dy = dx, the double integral becomes ˆ π ˆ π ˆ π 1 1 1 A(t)B(x − t) dt e−ikx dx = C(x) e−ikx dx. 2π −π 2π −π 2π −π Proof of the first part of Theorem 9.16 (Poisson’s summationP formula). The task is to ∞ determine the Fourier coefficient ck of the 2πperiodic function n=−∞ f (x + 2πn). By (9.5), these are ! ˆ π ∞ X 1 f (x + 2πn) e−ikx dx. ck = 2π −π n=−∞ It makes if we the factor e´−ikx to e−ik(x+2πn) , since e−i2πkn = 1. ´ π no Pdifference Pchange ∞ ∞ ´π ∞ Then −π −∞ · · · dx = −∞ −π · · · dx = −∞ · · · dx, and ˆ ∞ 1 1 ck = f (x) e−ikx dx = √ fˆ(k). 2π −∞ 2π Proof of Theorem 9.30. With the notation in the theorem formulation, ˆ x ˆ ∞ −sx ˆ h(s) = e f (ξ)g(x − ξ)dξ dx 0 0 ˆ ∞ ˆ ∞ −sx = e g(x − ξ)dx f (ξ)dξ 0 ξ ˆ ˆ ∞ ∞ −sξ −s(x−ξ) = e f (ξ) e g(x − ξ)dx dξ = fˆ(s) · gˆ(s). 0
ξ
9.7. Supplementary materials
267
0
x
0
Figure 9.13. The first octant in the (x, ξ)plane (extending to infinity to the right).
For the change in integration order between the first and second lines, note that the domain we integrate over is the first octant in the (x, ξ)plane, as shown shaded in Figure 9.13. Theorem 9.44. The inversion formula for the Mellin transform (9.53) ˆ ∞ ˆ f (s) = f (x) xs−1 dx, α < Re s < β, 0
is given by 1 f (x) = 2πi
ˆ
γ+i∞
fˆ(s) xs ds,
(9.74)
γ−i∞
where α < γ < β. Proof. As for the inversion formula for the Laplace transform in Section 9.2.1, we will make variable changes that reduce the problem to the inversion of a Fourier transform. Starting with the definition of fˆ(s), we get ˆ ∞ ˆ f (s) = f (x) xs−1 dx {set x = et ; dx = et dt} 0 ˆ ∞ = f (et ) est dt {set s = γ − iω}, −∞ ˆ ∞ ˆ ∞ t γt −iω t fˆ(γ − iω) = f (et ) e(γ−iω) t dt = f (e )e e dt. −∞
−∞ t
The RHS is now a standard Fourier transform of f (e ) eγt . Inverting it gives ˆ ∞ 1 f (et )eγt = fˆ(γ − iω) eiω t dω. 2π −∞ We next undo the variable change s = γ − iω by setting ω = γ−s i , dω = i ds, to obtain ˆ ˆ γ+i∞ e−γt γ+i∞ ˆ 1 f (et ) = f (s) e(γ−s) t ds = fˆ(s) e−s t ds. 2πi γ−i∞ 2πi γ−i∞ Undoing the first variable change, x = et , now gives (9.74).
268
Chapter 9. Transforms
9.8 Exercises P∞ Exercise 9.8.1. Determine the coefficients in the Fourier series k=−∞ ck ei kx = e−ix/α , −π < x < π. Compare your result with that of Exercise 5.6.43(a). Exercise 9.8.2. In the style of Example 9.13, determine what sum you get when applying (9.15) to f (x) = x2 . Exercise 9.8.3. Show that the coefficients ak in the Fourier cosine expansion f (x) = P∞ ecos x = a0 + k=1 ak cos kx converge to zero as ak ≈ 1/(k! 2k−1 ). ix −ix Hint: One way to solve this problem starts from noting that f (x) = e(e +e )/2 = 1 eix +e−ix 1 eix +e−ix 2 1 + 1! + 2! + · · · . Find the coefficient for eikx (which is the same 2 2 as for eikx ) when this sum is fully expanded. Comments: As to be expected from Example 9.2 and the fact that f (z) = ecos z is an entire function, the coefficients ak go to zero faster than exponentially. A different way to solve this problem is outlined in Exercise 12.6.21. Exercise 9.8.4. The first part of Theorem 9.14 was proved in Section 9.7. Provide a proof for the second part. Exercise 9.8.5. Derive Poisson’s integral formula (Theorem 4.20) by considering the Fourier series expansion of f (eiθ ). Hints: (i) With z = x + iy = r eiθ , note that fn (z) = rn einθ satisfies Laplace’s equation P∞ 2 for all n ∈ Z, and (ii) n=−∞ rn einθ = 1+r21−r −2r cos θ . Exercise 9.8.6. Derive Parseval’s relation (4.20) for a Fourier series. Exercise 9.8.7. Prove the following generalization to Parseval’s theorem 9.11: ˆ
ˆ
∞
∞
f (x) g(x) dx = −∞
−∞
fb(ω) gb(ω) dω .
Exercise 9.8.8. Determine the Fourier transform of f (x) =
cos x, 0
− π2 ≤ x ≤ π2 , otherwise.
´∞ x)2 Exercise 9.8.9. Show that −∞ (1−cos dx = π3 . x4 Hint: You might start by verifying the Fourier transform pair 1 + x, 1 − x, f (x) = 0
−1 ≤ x ≤ 0, 0 ≤ x ≤ 1, otherwise
r ⇔
fb(ω) =
2 1 − cos ω . π ω2
Exercise 9.8.10. By combining (9.29) and (9.64), derive the Fourier transform for the cubic Bspline, as given in (9.13).
9.8. Exercises
269
Exercise 9.8.11. Figure 2.25(b) illustrated Gibbs’ phenomenon, an overshoot in the Fourier interpolant to a discontinuous function, amounting to about 9% of the height of the jump. Derive a closedform expression for this quantity. Hint: A convenient case for deriving this is provided by the function −1/2, −π < x < 0, f (x) = 1/2, 0 < x < π, P∞ . Consider the derivative of the truncated with Fourier series f (x) = π2 k=0 sin(2k+1)x 2k+1 P n 2 0 series fn (x) = π k=0 cos(2k + 1)x, sum this exactly (cf. Example 2.5), find its first positive zero, evaluate fn (x) at this point, and let n → ∞. Exercise 9.8.12. Give the details for recovering the initial function in Example 9.26. Exercise 9.8.13. Determine L−1
Exercise 9.8.14. Determine L−1
s e−2s s2 −1 (x).
3 (s2 +1)(s2 +4)
(x).
Exercise 9.8.15. Use the Laplace transform to solve the ODE y 00 (x)−8y 0 (x)+7y(x) = 5x, with initial conditions y(0) = 1, y 0 (0) = 2. Exercise 9.8.16. Use the Laplace transform to solve the ODE y 000 (x) + y 0 (x) = ex , with initial conditions y(0) = y 0 (0) = y 00 (0) = 0. Exercise 9.8.17. The Laplace transform method generalizes immediately from scalar ODEs to systems. Use it to solve 0 u (x) = 3u(x) − 3v(x) + 4, u(0) = 1, v 0 (x) = 2u(x) − 2v(x) − 1, v(0) = 0.
Exercise 9.8.18. If f (x) has the period X (with X > 0), show that L{f (x)}(s) = ´ X 0
f (ξ)e−sξ dξ . 1−e−sX
Exercise 9.8.19. (a) Integrate the definition of the Laplace transform (9.37) to arrive at ˆ ∞ ˆ ∞ f (x) −sx b f (σ)dσ = e dx . x s 0 ´∞ (b) Letting s → 0 in (9.75) gives 0 an alternate solution to Exercise 5.6.26.
f (x) x
dx =
´∞ 0
(9.75)
fb(s) ds. Use this relation to obtain
Exercise 9.8.20. One of the (many) strengths of transforms is how well they handle step ´∞ and delta functions. Let δ(x) = 0, x 6= 0, but with −∞ δ(x)dx = 1. Assume that a > 0. (a) Show that L{δ(x − a)}(s) = e−as . (b) Show by Theorem 9.23 that L−1 {e−as ´ ∞} (x) = 0 for x 6= a and (for example using the result of Exercise 9.8.19, part (b)) that −∞ L−1 {e−as } (x) dx = 1.
270
Chapter 9. Transforms
Exercise 9.8.21. The moments of a function f (x) are defined by ˆ
∞
xn f (x) dx , n = 0, 1, 2, . . . .
µn = 0
Assuming f (x) goes to zero sufficiently fast that there are no convergence issues, deduce that the moments alternatively can be obtained from the Laplace transform of f (x) as dn b , n = 0, 1, 2, . . . . f (s) µn = (−1) dsn s=0 n
Hint: Recall Theorem 9.29. Exercise 9.8.22. Show that L{log x}(s) = − 1s (log s + γ). Hint: Differentiate (9.42) with respect to k (cf. “Feynman’s trick”) and then use the result in Exercise 6.5.4. Exercise 9.8.23. The “sine integral” function is defined by Si(x) = its Laplace transform becomes L {Si(x)} (s) =
´x 0
sin t t dt.
Show that
1 1 arctan . s s
Hint: Combine Theorem 9.29 with (9.40) and (9.50). Exercise 9.8.24. Perform the following Mellin transforms (and determine their fundamental band): (a) M ex1−1 (s) = ζ(s) Γ(s), P ∞ Hint: ex1−1 = n=1 e−nx for x > 0, 1 Γ(n−s) . (b) M (1+x)n (s) = Γ(s)Γ(n) Note: This problem is the inverse of the one in Example 9.36. t Hint: Change variable x = 1−t and then use (6.5) and (6.6). Exercise 9.8.25. Give the coefficients of the Taylor expansion about the origin of f (x) = 1 ex −1 in terms of ζ(z) and Γ(z), using Exercise 9.8.24(a): (a) Using the connection between the Mellin transform and properties of f (x) at x = 0 and x = ∞. (b) Using the Bernoulli numbers definition given in Exercise 2.9.23. Hint for (b): Use the result in Exercise 3.3.7. Exercise 9.8.26. Find the inverse Mellin transform of M (s) = ζ(s) Γ(s), with the fundamental band Re(s) ≥ 0. Hint: Use Ramanujan’s master theorem (see (3.12)). Exercise 9.8.27. Confirm the following alternative definition of the Hilbert transform: H{u(x)} = −
1 lim π ε→0
ˆ ε
∞
u(x + t) − u(x − t) dt . t
9.8. Exercises
271
Exercise 9.8.28. A typical amplitude modulated signal is a product of a “message signal” fL (t) containing only Fourier modes in a low range, and a “carrier signal” fH (t) that contains only high Fourier modes, with no overlap between the two ranges. Show Bedrosian’s theorem for their Hilbert transforms: H{fL (t) fH (t)} = fL (t) H{fH (t)}. Hint: (i) The linearity of the Hilbert transform makes it sufficient to consider a just single low mode ωL and a single high mode ωH , (ii) note (9.62), and (iii) note that sign (ωL + ωH ) = sign (ωH ). Exercise 9.8.29. Verify the inversion formula (9.71) for the Abel transform (9.70) (assuming r f (r) → 0 and r f 0 (r) → 0 as r → ∞). Hint: Integrate (9.70) by parts to obtain f 0 (r) inside the integral, and then evaluate dA(s) ds . ´β dt The relation α √t−α√β−t = π (for α < β) may come in handy. Exercise 9.8.30. Show that, like for the Fourier transform, the Abel transform of a Gaussian is again a Gaussian: r n o 2 π −α s2 e . A e−α r (s) = α Note: In this case, the two Gaussians differ only by a scalar factor; otherwise their widths are the same.
Chapter 10
Wiener–Hopf and Riemann–Hilbert Methods These techniques can be seen as farreaching generalizations of both the Laplace and the Hilbert transforms. In the Wiener–Hopf case, analytic functions are sought which satisfy different relations along the positive and negative real axes. In the Riemann–Hilbert generalization, information is instead provided along a curve C that can be either closed or open (often in the form of linear relation between the function’s real and imaginary parts on the two sides of C). If C has end points, certain additional issues arise. We focus first, in Section 10.1, on the Wiener–Hopf method before turning to the Riemann–Hilbert method in Section 10.2. The present goal is limited to give a general “flavor” of what the techniques can do, and to their key underlying ideas. For more comprehensive treatments on the two topics, see, for example, [22, 33] and [1], respectively.
10.1 The Wiener–Hopf method Among the most important features of the transforms that we studied in Chapter 9 were their convolution theorems. Applying the Fourier and the Laplace transform, respectively, we can, for example, immediately (by these theorems) solve integral equations of the forms ˆ ∞ f (x) = κ(x − y)f (y)dy + g(x), x ∈ (−∞, ∞), −∞
and
ˆ
x
κ(x − y)f (y)dy + g(x),
f (x) =
x ∈ [0, ∞),
0
where the “kernel” κ and the function g(x) are given; the function f (x) is to be found. Also important in applications, but not handled directly by any of these transforms, is the case ˆ ∞ f (x) = κ(x − y)f (y)dy + g(x), x ∈ [0, ∞). 0
This is just one of several Wiener–Hopf applications we will describe below. Others will involve solving certain types of ODEs and PDEs. All have in common that both formulations and answers involve no complex variables, but complex arithmetic is nevertheless indispensable throughout the solution process. 273
274
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
10.1.1 Preliminary notion: Halfplane splittings revisited Previously, in Sections 5.1.5 and 9.1.8, we described two approaches for splitting an analytic function f (z), given along the real axis, into f (z) = f + (z) + f − (z), singularityfree and decaying to zero in the lower and upper halfplanes, respectively.83 Both this splitting and its product/quotient counterparts f (z) = f + (z) · f − (z) and f (z) = f + (z)/f − (z) are key components of the Wiener–Hopf method. They are commented on next. Sum splitting
In the two sections where we described this previously, the primary focus was on functions defined along the real xaxis. The leftmost two relations in Theorem 9.19, F {f− (x)}(ω) = fˆ− (ω), ˆ+ (10.1) F {f+ (x)}(ω) = f (ω), {z }  signs are the same considered, however, the case where the function to be split resides in Fourier space: fˆ(ω) instead of f (z). For the subscripts in (10.1), we recall their definition: f (x) , x ≥ 0, 0 , x ≥ 0, f+ (x) = and f− (x) = 0 , x < 0, f (x) , x < 0. Product and quotient splittings
Given a function fˆ(ω) in the Fourier space, we can similarly split it into either a product fˆ(ω) = fˆ+ (ω) · fˆ− (ω) or a quotient fˆ(ω) = fˆ+ (ω)/fˆ− (ω). Example 10.1. Find product and quotient splits of fˆ(ω) =
1 1+ω 2 .
Factorizing the denominator gives us immediately the product decomposition, with 1 1 fˆ− (ω) = ω+i and fˆ+ (ω) = ω−i . These functions are free of both singularities and zeros in Im(ω) ≥ 0 and Im(ω) ≤ 0, respectively. For a quotient split fˆ+ (ω)/fˆ− (ω), 1 we swap fˆ− (ω) = ω+i to fˆ− (ω) = ω + i. We note that we usually have implied that − fˆ (ω) not only is singularityfree in the upper halfplane, but also that it vanishes in that halfplane. In this instance, fˆ− (ω) no longer vanishes in Im ω ≥ 0. √ Example 10.2. Find a product split of fˆ(ω) = ω 2 − λ2 , where Re(λ), Im(λ) > 0. 1/2 Again, immediate factorization gives the answer: fˆ− (ω) = (ω + λ) and fˆ+ (ω) = 1/2 (ω − λ) . We can direct the branch cuts towards −i∞ and +i∞, to keep the respective halfplanes singularity and zerofree.
For the more complicated cases, one can rewrite fˆ(ω) = fˆ+ (ω) · fˆ− (ω) as log fˆ(ω) = log fˆ+ (ω) + log fˆ− (ω), use a sum split of log fˆ(ω), and then exponentiate the resulting 83 We recall again that the meaning of “+” and “−” in this context does not agree between all texts. Whether used as superscripts for halfplanes or as subscripts for splits along the real axis, we use the sign to denote the direction of nontrivial behavior.
10.1. The Wiener–Hopf method
275
terms. A number of technical issues can arise (such as how to handle the multivaluedness that arises for poles and zeros of fˆ(ω)), and the algebra often becomes cumbersome. Specialized texts, such as [7, 22], discuss this further.
10.1.2 The Wiener–Hopf procedure The Wiener–Hopf method relies on three key steps, which the following examples are organized around. These steps are the following: Step 1: Wiener–Hopf equation The first step is to recast the problem to be solved for an unknown function f (x) = f− (x)+ f+ (x) into a relation of the form γ(ω) fˆ+ (ω) + fˆ− (ω) = h(ω) ,
(10.2)
where γ(ω) and h(ω) are known functions, and fˆ− (ω) and fˆ+ (ω) remain to be determined. Step 2: Decomposition and analytic continuation Further rearrange (10.2) into a form from which fˆ− (ω) and fˆ+ (ω) both can be deduced by using Liouville’s theorem. The key rearrangement steps are (i) quotient split γ(ω) = γ + (ω)/γ − (ω), (ii) multiply by γ − (ω), (iii) sum split γ − (ω)h(ω), (iv) collect all + terms on one side and all − terms on the other side. Step 3: Halfline inverse Fourier transform Having found fˆ− (ω) and fˆ+ (ω), calculate f− (x) and f+ (x) by means of (10.1). Case with convergent Fourier transforms
For the sake of simplicity, we will initially assume that all of the functions which we consider decay for x → ±∞. Example 10.3. Determine a function f (x) which satisfies the integral equation ˆ ∞ f (x) = −4 e−x−t f (t) dt + xe−2x , x ≥ 0.
(10.3)
0
Step 1: Wiener–Hopf equation The integral equation (10.3) gives no information about f (x) for x < 0. To be able to use Fourier transforms, we first generalize the RHS function xe−2x to xe−2x , x ≥ 0, g(x) = g+ (x) = 0 , x < 0. We next write
f+ (x) =
f (x) , x ≥ 0, 0 , x < 0,
and, by (10.3), f− (x) =
−4
´∞ 0
0 e−x−t f (t) dt
, x ≥ 0, , x < 0.
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Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
It now holds for all real x values that ˆ ∞ e−x−t f+ (t) dt = g+ (x), f+ (x) + f− (x) + 4
x ∈ R.
−∞
We next take the Fourier transform of this equation. By (10.1) and the Fourier convolution theorem (Section 9.1.5), this gives the Wiener–Hopf equation √ fˆ+ (ω) + fˆ− (ω) + 4 2π ˆκ(ω) fˆ+ (ω) = gˆ+ (ω). (10.4) q ´∞ Here κ ˆ (ω) = √12π −∞ e−x e−ixω dx = π2 ω21+1 is the standard Fourier transform of ´ ∞ x(−2−iω) xe dx = the kernel κ(x) = e−x . The functions fˆ+ (ω) and gˆ+ (ω) = √1 √ −1 2π(ω−2i)2 ˆ−
2π
0
are singularityfree in the complex ωplane’s lower half (Im(ω) ≤ 0), and
f (ω) is similarly singularityfree in the complex ωplane’s upper half (Im(ω) ≥ 0). For κ(x), κ ˆ (ω), g+ (x), and gˆ+ (ω), see Figure 10.1. Since f (x) and g(x) decay as x → ±∞, + fˆ (ω), fˆ− (ω), gˆ+ (ω) must all be analytic along the real axis. The objective is to solve for fˆ+ (ω) and to then use the inversion formula to obtain f+ (x). However, fˆ− (ω) is also unknown in (10.4). The beauty of the Wiener–Hopf method is that this sole equation (10.4) provides enough information to solve for both fˆ+ (ω) and fˆ− (ω). The key ingredients for doing this are analytic continuation and Liouville’s theorem. Step 2: Decomposition and analytic continuation Substituting the known quantities into (10.4) results in 2 1 ω + 9 ˆ+ = −fˆ− (ω), f (ω) + √ 2 ω +1 2π(ω − 2i)2
(10.5) 2
+
γ (ω) ω +9 which is of the form (10.2). We therefore quotient split γ(ω) = ω 2 +1 = γ − (ω) with ω+i − − γ + (ω) = ω−3i ω−i and γ (ω) = ω+3i . After multiplying both sides by γ (ω), we obtain fˆ− (ω)(ω + i) ω − 3i ˆ+ (ω + i) . (10.6) f (ω) + √ = − 2 ω−i (ω + 3i) 2π(ω − 2i) (ω + 3i) {z }  {z }  Singularityfree in Im(ω)≥0
Singularityfree in Im(ω)≤0
The term
(ω+i) 2π(ω−2i)2 (ω+3i)
still has singularities in both regions, so the next step is to √ √ 3/(5 2π ) i 2/π (ω+i) − sum split this function. By partial fractions, √2π(ω−2i) = 2 (ω+3i) (ω−2i)2 25(ω−2i) + √ i 2/π 25(ω+3i) . After reordering the terms, we reach the goal of having obtained an identity between two functions that are singularityfree and, also importantly, decay to zero in the opposite halfplanes. We call this function E(ω): √ 3 ω − 3i ˆ+ i 2 f (ω) + √ − √ ω−i 25 π(ω − 2i) 5 2π(ω − 2i)2  {z } √
Singularityfree in Im(ω)≤0
√ fˆ− (ω)(ω + i) i 2 =− − √ = E(ω). (ω + 3i) 25 π(ω + 3i)  {z } Singularityfree in Im(ω)≥0
(10.7)
10.1. The Wiener–Hopf method
277
0.8 0.6 0.4 0.2
3
2
1
0
1
2
3
(a) κ(x) = e−x
(b) κ ˆ (ω) =
q 2 π
1
ω 2 +1
0.15 0.1 0.05
3
2
1
0
(c) g+ (x) =
1
−2x
xe 0
2
3
,x≥0 ,x 0, x < 0.
(10.9)
Let us assume that f (0+ ) − f (0− ) = 0 and f 0 (0+ ) − f 0 (0− ) = 1, and that f (x) vanishes as x → ±∞. Solve for f (x).84 Step 1: Wiener–Hopf equation Let’s assume once again that the solution f (x) vanishes as x → ±∞, and let f (x) , x ≥ 0, f+ (x) = 0 , x < 0, and
f− (x) =
0 , x ≥ 0, f (x) , x < 0.
84 This problem is trivially solved by noting from (10.9) that f (x) = c ex + c e−x , f (x) = c e2x + + − 1 2 3 c4 e−2x . The four additional conditions give then c1 = c4 = 0 and c2 = c3 = − 31 . The Wiener–Hopf solution is given here as a technique illustration.
10.1. The Wiener–Hopf method
279
0.05 2
1
1
2
0.05
(a) fˆ+ (ω) =
(2iω−11)(ω−i) √ . 25 2π(ω−2i)2 (ω−3i)
(b) f+ (x) =
3
2
1 −3x (−34 25 e
1
0
+ ex (32 − 15x)).
1
2
3
0.02 0.04 0.06
√
2 (c) fˆ− (ω) = − 25√iπ(ω+i) .
2 x (d) f− (x) = − 25 e .
Figure 10.3. The functions fˆ+ (ω) and fˆ− (ω) in Example 10.3 and their halfline counterparts.
´∞ ´0 Then fˆ+ (ω) = √12π 0 f+ (x)e−ixω dx and fˆ− (ω) = √12π −∞ f− (x)e−ixω dx are singularityfree on Im(ω) ≤ 0 and Im(ω) ≥ 0, respectively. Applying the standard Fourier transform on both √ differential equations and using integration by parts twice √ gives 0 0 −f+ (0+ ) − iωf+ (0+ ) − 2π(1 + ω 2 )fˆ+ (ω) = 0 and f− (0− ) + iωf− (0− ) − 2π(4 + ω 2 )fˆ− (ω) = 0 if we also assume that 0 0 lim f+ (x) + iωf+ (x) e−iωx = lim f− (x) + iωf− (x) e−iωx = 0. x→∞
x→−∞
Adding both equations and applying the boundary conditions results in the Wiener–Hopf problem √ √ 1 + 2π ω 2 + 1 fˆ+ (ω) + 2π ω 2 + 4 fˆ− (ω) = 0. Step 2: Decomposition and analytic continuation We know that fˆ+ (ω) is free of singularity for Im(ω) ≤ 0 and, similarly, that fˆ− (ω) is singularityfree for Im(ω) ≥ 0. Once again with the intention of grouping terms that are singularityfree on either domain, wedivide both √ √ sides of the −equation by (ω + i)(ω − 2i) 1 ω−i ˆ+ ˆ and obtain (ω+i)(ω−2i) + 2π ω−2i f (ω) + 2π ω+2i ω+i f (ω) = 0. A partial fractions
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Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
1 1 1 decomposition of the first term gives (ω+i)(ω−2i) = 3i ω−2i − equation becomes √ ω − i ˆ+ 1 1 + 2π f (ω) 3i ω − 2i ω − 2i {z } 
1 ω+i
. The reordered
Singularityfree in Im(ω)≤0
√
= − 2π 
ω + 2i ω+i
fˆ− (ω) + {z
1 3i
1 ω+i
= E(ω). }
Singularityfree in Im(ω)≥0
Since we, in the split, also ensured that both parts go to zero far out, analytic continuation and Liouville’s theorem now give E(ω) ≡ 0. This allows us to solve for fˆ+ (ω) and fˆ− (ω). Step 3: Inverse Fourier transform 1 The last step is to apply the inverse Fourier transform to fˆ+ (ω) = − √2π(3i)(ω−i) and ´ ∞ 1 1 1 iωx − ˆ f (ω) = √2π(3i)(ω+2i) . We obtain f+ (x) = 2π −∞ − (3i)(ω−i) e dω and f− (x) = ´∞ 1 1 iωx dω. From this follows 2π −∞ (3i)(ω+2i) e
1 − e−x 3 f (x) = − 1 e2x 3
, x ≥ 0, , x < 0.
Example 10.5. Solve the following PDE boundary value problem ∆u = u,
−∞ < x < ∞, 0 < y < ∞
such that uy (x, 0) = 0 on y = 0, x < 0 and u(x, 0) = e−x on y = 0, x ≥ 0 (with ∆ ∂2 ∂2 denoting the Laplacian operator ∆ = ∂x 2 + ∂y 2 ). Step 1: Wiener–Hopf equation Applying the Fourier transform in x, we readily obtain ∂2 u ˆ(ω, y) = ω 2 + 1 u ˆ(ω, y), 2 ∂y which can be solved analytically as u ˆ(ω, y) = A(ω)e−y impose that u ˆ(ω, y) be bounded, B(ω) = 0 and u ˆ(ω, y) = A(ω)e−y
√
√
ω 2 +1
ω 2 +1
+ B(ω)ey
√
ω 2 +1
.
. Since we
(10.10)
To incorporate the boundary conditions, let −x e , x ≥ 0, 0 , x ≥ 0, µ+ (x) = µ− (x) = 0 , x < 0, u(x, 0) , x < 0, −i with halfrange Fourier transforms µ ˆ+ (ω) = √2π(ω−i) and µ ˆ− (ω), singularityfree in Im(ω) ≤ 0 and in Im(ω) ≥ 0, respectively. The extended equation becomes u ˆ(ω, 0) =
10.1. The Wiener–Hopf method √ −i 2π(ω−i)
281
+µ ˆ− (ω). Further, let ν+ (x) =
n
uy (x, 0) 0
+
, x ≥ 0, , x < 0, −
ν− (x) = 0, and de
note their respective Fourier transforms by νˆ (ω) and νˆ (ω) = 0. The corresponding ∂ u ˆ(ω, 0) = νˆ+ (ω). From (10.10), u ˆ(ω, 0) = A(ω) = extended equation becomes ∂y √ ∂ −i − + 2 √ +µ ˆ (ω) and ∂y u ˆ(ω, 0) = − ω + 1 A(ω) = νˆ (ω). Combining the two 2π(ω−i) equations once again provides a Wiener–Hopf problem −
p
ω2
+1
−i − √ +µ ˆ (ω) = νˆ+ (ω), 2π(ω − i)
from which we need to extract both unknown functions µ ˆ− (ω) and νˆ+ (ω). Step 2: Decomposition and analytic continuation The equation above has branch points at ω = ±i and a pole at ω = i. In order to group the terms that are singularityfree in the upper part of the complex plane (including the real axis) and the functions that are analytic in the √ lower part of the complex plane (also including the real axis), we divide both sides by ω − i, giving
√ √ i ω+i νˆ+ (ω) √ ˆ− (ω) = √ − ω + iµ . ω−i 2π(ω − i)
The only problematic term is the first one on the LHS, for which we need to do a sum √ ω+i = h− (ω) + h+ (ω). This function h(ω) has a pole at ω = i and split: h(ω) = (ω−i) √ √ a branch point at ω = √−i. At the pole ω = i, ω + i = 2i = 1 + i, suggesting the √ (1+i) ω+i reformulation h(ω) = (ω−i) = ω+i−(1+i) + (ω−i) = h− (ω) + h+ (ω). This works, since (ω−i) − ω = i has now become a removable singularity of h (ω). The reordered equation becomes √ √ i−1 i ω+i i−1 νˆ+ (ω) √ −√ =√ −√ − ω + iˆ µ− (ω) = E(ω). ω−i 2π(ω − i) 2π(ω − i) 2π(ω − i)  {z }  {z } Singularityfree in Im(ω)≤0
Singularityfree in Im(ω)≥0
By Liouville’s theorem, E(ω) = 0. Now we can solve for µ ˆ− (ω). We have µ ˆ− (ω) = i i−1 i−1 √ − √2π(ω−i)√ω+i . Therefore, A(ω) = − √2π(ω−i)√ω+i . 2π(ω−i) Step 3: Inverse Fourier transform √
2
The inverse Fourier transform of A(ω) e−y ω +1 gives the PDE solution as u(x, y) = √ ´ −y ω 2 +1 ∞ (1−i) e 1 √ eixω dω. This function should be purely realvalued for x and y 2π −∞ (ω−i) ω+i real, y > 0. We can indeed eliminate all complex quantities from this answer, to obtain u(x, y) =
π
1 √
ˆ 2
∞
−∞
√
2
e−y ω +1 cos (ω 2 + 1)3/4
1 arctan(ω) − ωx dω. 2
Figure 10.4 illustrates both this solution u(x, y) and its yderivative
∂ ∂y u(x, y).
282
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
1
0 0.5
0.5
1 1.5 2
0 4
1 2
0
2
4
2
2 4
1 2
0
0
(a) PDE solution u(x, y).
2
(b) Its yderivative
4
0
∂ ∂y u(x, y).
Figure 10.4. The solution u(x, y) to Step 3: the PDE in Example 10.5 and its yderivative The solid red curves mark the two boundary conditions.
∂ u(x, y). ∂y
Generalization to case with divergent Fourier transforms
Up until now, we have required that all functions vanish as x → ±∞. From this it followed that the real axis was included in both the upper and the lower analyticity domains. We then knew that there was an overlap between those domains, and that is what allowed us to use analytic continuation. Finally, it enabled us to have an easy halfline inverse Fourier transform integral, along the real axis. This requirement is lifted in the following example, allowing us to proceed even when the Fourier transforms have poles (or other singularities) right on the real ωaxis. We introduce two theorems, with proofs given in Section 10.3. Theorem 10.6. If f (x) < O (eαx ) for x → ∞ and f (x) < O eβx for x → −∞, where the quantities α and β are called the exponential bound indices, then fˆ+ (ω) = ´∞ ´0 √1 f (x)e−ixω dx and fˆ− (ω) = √12π −∞ f (x)e−ixω dx are singularityfree, re2π 0 spectively, in Im(ω) < −α and in Im(ω) > −β. Furthermore, the inversion formulas are as follows. ˆ 1 f− (x) = √ fˆ− (ω) eixω dω, 2π Γ ˆ 1 f+ (x) = √ fˆ+ (ω) eixω dω, 2π Γ where Γ lies within the common region of analyticity of fˆ+ (ω) and fˆ− (ω), i.e. (with a small abuse of notation), −β < Im Γ < −α, as in Figure 10.5 (having assumed that α < β).
Theorem 10.7.
Example 10.8. Let f (x) = 1 + x for x > 0. Find the halfline Fourier transform fˆ+ (ω) of f (x) along the positive real axis. Let
f+ (x) =
1+x 0
, x ≥ 0, , x < 0.
10.1. The Wiener–Hopf method
283
fˆ− (ω) anal y t ic
Im ω
Im ω
−iα Γ
Γ Re ω
Re ω fˆ+ (ω) anal y t ic
(a) Integration path to recover f+ (x).
−iβ
(b) Integration path to recover f− (x).
Figure 10.5. Analyticity of halfline Fourier transforms.
´∞ The halfline Fourier transform of f (x) is fˆ+ (ω) = √12π 0 (1 + x) e−iωx dx. Integrating ´∞ −iωx x=∞ −ixω by parts, 0 (1 + x) e−iωx dx = (1 + x) e −ω + e ω2 x=0 . Assuming that Im(ω) < 0, −iω−1 the terms evaluated at x = ∞ vanish and fˆ+ (ω) = √ . 2πω 2 + ˆ Note that f (ω) can also be found from the generalized (halfline) Fourier transform ´∞ of f (x). Using the same technique as in Section 9.1.7, fˆ+ (ω) = limα→0+ √1 (1 + x) e−αx e−ixω dx =
2π
−iω−1 √ . 2πω 2
0
−iω−1 Example 10.9. Given fˆ+ (ω) = √ , use the inversion formula to recover function 2πω 2 f+ (x), which is nonzero only on the positive real axis.
Since fˆ+ (ω) is analytic for Im(ω) < 0, then f+ (x) =
´
−iω−1 √1 √ 2π Γ 2πω 2 ˆ+
eixω dω, where
Γ is a contour line that passes below all the singularities of f (ω), as drawn in Figure 10.5(a). If x > 0, then we can close the contour line Γ with a top semicircle as in Figure 10.6(a) and invoke Jordan’s lemma. Indeed, the contribution from the integral along the semicircle vanishes, and by residue calculus, f+ (x) = 1 + x. However, if x < 0, one can close the contour as in Figure 10.6(b), and again using Jordan’s lemma and residue calculus, f+ (x) = 0. Therefore, 1 + x , x ≥ 0, f+ (x) = 0 , x < 0. Example 10.10. Solve the integral equation ˆ ∞ f (x) = 4 e−x−t f (t) dt,
x ≥ 0.
(10.11)
0
This example differs from Example 10.3 in two key ways: (i) it has the opposite sign for the integral, and (ii) the equation is homogeneous (no additional RHS function g(x)). Step 1: Wiener–Hopf equation ´∞ As before, let 4 0 e−x−t f+ (t) dt = f+ (x) + f− (x), where 0 , x ≥ 0, ´∞ f− (x) = 4 0 e−x−t f (t) dt , x < 0.
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Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
Im ω
Im ω
Re ω
Re ω
Γ
Γ
fˆ+ (ω) anal y t ic
(a) Integration path to recover f + (x) for x < 0.
fˆ+ (ω) anal y t ic
(b) Integration path to recover f + (x) for x > 0.
Figure 10.6. Inversion integration paths for Example 10.9.
After applying the Fourier transform, we 9.5 and (9.23) and (9.25)) q obtain (cf. Example √ ´ ∞ −x−t 2 1 + ˆ F {4 0 e f (t) dt}(ω) = 4 2π π ω 2 +1 f (ω) and F {f+ (x) + f− (x)}(ω) = fˆ+ (ω) + fˆ− (ω), which simplifies to
ω2 − 7 ω2 + 1
fˆ+ (ω) = −fˆ− (ω).
(10.12)
Step 2: Decomposition and analytic continuation 2 −7 in (10.12) are on the real ωaxis, we cannot this time move Since both zeros of ω1+ω 2 over to the RHS one factor from the numerator and one from the denominator, and get a singularityfree band that includes the real ωaxis. We instead move just one factor (ω + i) over, giving 2 ω −7 fˆ+ (ω) = −(ω + i)fˆ− (ω) = E(ω) , (10.13) ω−i where now (i) we define f + and f − relative to a path Γ with −1 < Im Γ < 0, and (ii) E(ω) is an entire function that can grow at most as E(ω) ∼ ω. By the generalized Liouville theorem (Theorem 4.13), E(ω) = α + β ω. From (10.13) it follows that E(ω)/ω → 0 up/down the complex ωplane, implying β = 0. Thus −α −(ω+i)fˆ− (ω) = α (constant), giving fˆ− (ω) = (ω+i) and similarly fˆ+ (ω) = −α ωω−i 2 −7 . Step 3: Inverse Fourier transform
´ Applying the inverse Fourier transform to fˆ+ (ω) gives f+ (x) = √12π Γ fˆ+ (ω) eiωx dω, where Γ is in the band of analyticity. For positive x, Jordan’s lemma can be used if the contour as in √ Figure 10.7(b). By residue calculus, f+ (x) = √ is the upper√semicircle, C cos x 7 + √17 sin x 7 (with C = iα 2π an arbitrary constant). For x < 0, Jordan’s lemma can be used if the lower semicircle contour is chosen as in Figure 10.7(a). Since this contour contains no singularity, f+ (x) = 0 (as expected).
10.1. The Wiener–Hopf method
285
fˆ− (ω) anal y t ic
Im ω
fˆ− (ω) anal y t ic
Im ω
Re ω
Re ω
Γ
Γ
fˆ+ (ω) anal y t ic
(a) Integration path to recover f+ (x) for x < 0.
fˆ+ (ω) anal y t ic
(b) Integration path to recover f+ (x) for x > 0.
Figure 10.7. Inversion integration paths.
In light of the solution above, let us revisit Example 10.3 in its homogeneous form: Example 10.11. Solve the integral equation ˆ ∞ f (x) = −4 e−x−t f (t) dt,
x ≥ 0.
(10.14)
0
By the same procedure that led to (10.12), we this time arrive at the Wiener–Hopf formulation ω + 3i ω − 3i ˆ+ f (ω) = −fˆ− (ω), ω+i ω−i which can be rearranged as ω + i ˆ− ω − 3i + ˆ =− f (ω) f (ω) = E(ω), ω−i ω + 3i {z }  {z }  analytic for Im(ω)≤0
analytic for Im(ω)≥0
where E(ω) is an entire function. Also, since fˆ− (ω) = O( ω1 ), so is E(ω). By Liouville’s theorem, E(ω) = 0, and it thereby follows that fˆ+ (ω) = 0 and that f+ (x) = 0 is the unique solution. The natures of the solutions in Examples 10.10 and 10.11 are fundamentally different, although only a sign has been changed. We therefore consider the integral equation ˆ ∞ f (x) = λ e−x−t f (t) dt, x ≥ 0. (10.15) 0
The Wiener–Hopf formulation takes the form 2 ω + 1 − 2λ ˆ+ f (ω) = −fˆ− (ω). ω2 + 1
286
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
What made the difference between the previous cases of λ = −4 and λ = +4 was whether the roots of ω 2 + 9 = 0 and ω 2 − 7 = 0, respectively, were on or off the real ωaxis. For general λ, the break point between these cases occurs at λ = 12 (for the solution in this case, see Exercise 10.4.5). A more thorough discussion is presented in [22]. The differences in the structure of the solutions for varying λ can be formulated using operator theory. Writing (10.15) as (I − λT ) f = 0, the key question becomes whether (I − λT ) is singular or not. If written as T f = µf (with µ = 1/λ), the relation takes the form of an eigenvalue problem for the integral operator T .
10.2 A brief primer on Riemann–Hilbert methods 10.2.1 Hilbert transform case In this case, the real part of a function is given along the real axis, and the matching imaginary part is sought (or vice versa; the function is unique if we also know it to decay to zero in the upper halfplane). By a conformal mapping, we can change the real axis to the unit circle (or any other closed curve) C. For example, w=−
z−i z+i
brings the upper halfplane to the inside of the unit circle (with z = 0 → w = 1 and z = ∞ → w = −1). Knowing Ref (z) along the real zaxis means now knowing it around the unit circle in the wplane. This perspective provides an additional opportunity to find the matching imaginary part. Calling the transformed plane z (rather than w), andthe function f (z), we can Fourier expand around the periphery: Ref eiθ = P∞in this zplaneP ∞ a0 + k=1 aP k cos kθ + k=1 bk sin kθ, and we note that this matches the Taylor expansion ∞ f (z) = a0 + k=1 (ak − ibk ) z k . From this follows not only the corresponding imaginary part, but also f (z) throughout the interior of the unit circle.
10.2.2 Riemann’s case We next separate f = u + iv, with u and v real, and denote the independent variable t to remind ourselves that we will at first consider it only on the periphery of the unit circle. One of many fundamental issues Riemann pioneered in his thesis was to find both u(t) and v(t) when the linear relation α(t) · u(t) + β(t) · v(t) = γ(t)
(10.16)
holds, with α(t), β(t), γ(t) given. The Hilbert transform case above corresponds to the special case of α(t) ≡ 1 and β(t) ≡ 0, i.e., 1 · u(t) + 0 · v(t) = γ(t).
(10.17)
It is customary at this point to call the desired function f (z) = Φ+ (z), with the “+” signifying arguments (i.e., zvalues) inside the unit circle C. The function 1 Φ− (z) = Φ+ z
10.2. A brief primer on Riemann–Hilbert methods
287
will then be defined outside C and is also analytic (cf. the last case in Section 2.1.2). When z → t on C, it holds that Φ− (t) = Φ+ (t) = u(t) + iv(t). Equation (10.16) then implies α(t) + iβ(t) α(t) − iβ(t) Φ+ (t) + Φ− (t) = γ(t). 2 2
(10.18)
Riemann’s problem has now been reformulated as a connection between two analytic functions, Φ+ (t) and Φ− (t), singularityfree inside and outside C, respectively. The next secn tion outlines strategy to solve for Φ(z), which we define as Φ(z) = analytic everywhere but on C, along which there is a discontinuity.
Φ+ (z) Φ− (z)
inside C, outside C,
10.2.3 More general Riemann–Hilbert problems A slightly more general version of (10.18) is to find analytic functions Φ+ (t) and Φ− (t), singularityfree in their respective domains, that along C (which we still take as the unit circle) are related by Φ+ (t) = g(t)Φ− (t) + f (t).
(10.19)
Here, g(t) and f (t) are given, complexvalued functions (required to be “Hölder continuous,” a much weaker condition than being differentiable). Finding solutions to (10.19) becomes a twostep process, where we first consider the homogeneous case with f (t) ≡ 0. In Riemann–Hilbert problems, the contour C need not be closed. The Wiener–Hopf method (in Section 10.1) focuses on the issues that arise near the end points of an open contour. Homogeneous Riemann–Hilbert problem
Assuming g(t) to be nonzero, we look for a nonvanishing function L(z) that satisfies the simplified problem L+ (t) = g(t)L− (t).
(10.20)
log L+ (t) − log L− (t) = log g(t).
(10.21)
Taking the logarithm gives
From the discussion of principal value integrals in Section 5.1.4 illustrated in Figure 5.18, we know how to construct such a function: 1 log L(z) = 2πi
ˆ C
log g(t) dt t−z
(10.22)
(with L(z) indeed being nonvanishing, since log L(z) becomes finite). From this L(z), we read off L+ (z) and L− (z). One issue that requires some care is that g(t) might have encircled the origin when t went around C, making log g(t) not return to its initial value. If log g(t) increased by 2πin (n integer), we can instead of (10.20) solve N+ (t) = t−n g(t) N− (t) (which does not have
288
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
this issue) and then obtain our intended L(z) as N (z) inside C, L(z) = z −n N (z) outside C. Inhomogeneous Riemann–Hilbert problem
Wanting to find a solution to (10.19), we start by solving (10.20) with the same function g(t). Eliminating g(t) between these two equations gives Φ+ (t) Φ− (t) f (t) − = . L+ (t) L− (t) L+ (t)
(10.23)
Writing M (z) = Φ(z)/L(z), this function jumps with the amount Lf+(t) (t) across C (known, since we know both f (t) and L+ (t)). The principal value approach provides again a solution: ˆ f (t)/L+ (t) 1 dt. (10.24) M (z) = 2πi C t−z With L(z) known, the solution Φ(z) to the inhomogeneous Riemann–Hilbert problem (10.19) follows, as Φ(z) = M (z)L(z).
10.3 Supplementary materials Proof of Theorem 10.6. f+ (x) < Ceαx as x → ∞. Therefore, C ˆ ∞ αx −ixω + ˆ e e dx f (ω) < √ 2π 0 ˆ ∞ C 0, f 00 (x) + b2 f (x) = 0, x < 0,
(10.26)
for Im a > 0 and Im b > 0, with the boundary and farfield conditions f (0+ )−f (0− ) = 0, f 0 (0+ ) − f 0 (0− ) = 1, and f (x) → 0 as x → ±∞. Exercise 10.4.5. Derive for (10.15) in the case of λ = C (1 + x). Exercise 10.4.6. Solve the following PDE: 2 ∂2 ∂ + 2 u + λ2 u = 0, ∂x2 ∂y
1 2
the general solution f+ (x) =
−∞ < x < ∞, 0 < y < ∞,
with Im λ = , for a small and positive real number, also such that uy (x, 0) = 0 on y = 0, x < 0, and u(x, 0) = cos x on y = 0, x ≥ 0, and finally with limy→∞ u(x, y) = 0. Exercise 10.4.7. Solve the following PDE: 2 ∂ ∂2 + u − u = 0, ∂x2 ∂y 2
−∞ < x < ∞, 0 < y < ∞,
such that uy (x, 0) = 0 on y = 0, x < 0, and u(x, 0) = 1 on y = 0, x ≥ 0. Exercise 10.4.8. Solve the following PDE: 2 ∂ ∂2 ∂ + 2 u− u = 0, ∂x2 ∂y ∂y
−∞ < x < ∞, 0 < y < ∞,
such that uy (x, 0) = 1 on y = 0, x < 0, and f (x, 0) = e−x on y = 0, x ≥ 0.
290
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
Exercise 10.4.9. Given the part of a function Preal P∞ that is around the periphery of a ∞ iθ circle, Ref e = a + a cos kθ + 0 k k=1 k=1 bk sin kθ, show that f (z) = a0 + P∞ k (a − ib ) z . k k k=1 Exercise 10.4.10. The Sokhotski–Plemelj theorem states that, for a given simple closed ¸ φ(t) 1 dt, then Φ+ (z) = curve C and an analytic function φ(z) on C, if Φ(z) = 2πi C t−z ffl φ(t) ffl φ(t) 1 1 1 1 2πi C t−z dt + 2 φ(z) and Φ− (z) = 2πi C t−z dt − 2 φ(z). Prove this theorem. Exercise 10.4.11. Given a closed contour C, find an analytic function Φ(z) that satisfies the relation Φ+ (t) − Φ− (t) = 1 along C. Is the solution unique? Exercise 10.4.12. Given a closed contour C, find an analytic function L(z) that satisfies the relation L+ (t) = ef (z) L− (t) along C. Is the solution unique?
Chapter 11
Special Functions Defined by ODEs
One of the core resources of applied mathematics and theoretical physics is their collection of standard and special functions. Many of these “special functions” can be introduced as solutions to linear ordinary differential equations (ODEs) with variable coefficients, with some examples given below in Sections 11.1–11.3. A typical form for a linear ODE in this context is u00 (z) + p(z) u0 (z) + q(z) u(z) = 0. (11.1) Already if p(z) and q(z) are linear functions in z, it is rare that solutions u(z) can be found in closed form using only the standard functions from calculus. As in the previous chapters, our focus below is on analytic function aspects of the topic, thus also including a discussion of the nonlinear Painlevé equations in Section 11.5. Special functions that arise from ODEs (either linear or nonlinear) are typically analytic,85 as follows from a very general theorem due to Cauchy. Theorem 11.1. Let u(z) satisfy an ODE of the form dn u =F dz n
u,
du dn−1 u , . . . , n−1 ; z , dz dz
(11.2) n−1
d u where F is analytic in each of its arguments. For any initial conditions on u, du dz , . . . , dz n−1 at some point z0 , the solution to (11.2) then becomes an analytic function in some neighborhood of z0 .
It is also common that there is a factor multiplying the highest derivative term, such as z 2 for Bessel’s equation (11.6) and z(1 − z) for the hypergeometric ODE (11.15). Zeros of such factors create singularities at which Taylor expansions for solutions typically need to be replaced by what are known as Frobenius expansions. If the ODE near such a point z = z0 can be cast in the form u00 (z) + 85 Recalling
p(z) 0 q(z) u (z) + u(z) = 0, z − z0 (z − z0 )2
that our definition of analytic does not exclude singularities or restrictions for domains of validity.
291
292
Chapter 11. Special Functions Defined by ODEs
with p(z) and q(z) locally analytic, the two solutions will be of the form Taylor series α u(z) = (z − z0 ) · , around z = z0 where α is a root of α2 + (p(z0 ) − 1) α + q(z0 ) = 0 (with a slightly modified rule for the second solution in case of double roots). The method of undetermined coefficients (Section 2.3.2) will give the Taylor expansions. Second order linear ODEs with one or several singular points also show up in applications far more than one might at first have expected. The analytical theory surrounding their solutions is very extensive, and we will make no attempt at all here to survey this topic, especially since there are several excellent books that provide comprehensive treatments, e.g., [4, 25, 27, 30]. We will here limit ourselves to give some illustrative examples of how insights about analytic functions can be brought to bear on the subject of ODEs.
11.1 Airy’s equation 11.1.1 Realvalued independent variable The constant coefficient ODE y 00 √− αy = 0 has √ for α > 0 exponentially growing or + αx − αx decaying solutions y(x) = C e + C e and for α < 0 oscillatory solutions 1 2 √ √ y(x) = C1 cos( −α x) + C2 sin( −α x). If we replace the constant α with the independent variable x, we arrive at Airy’s equation y 00 (x) − x y(x) = 0.
(11.3)
Unsurprisingly, its solutions are of exponentiallike character for x positive and oscillatory for x negative. This equation is of great importance in many applications, since it forms the simplest description of (and leading order approximation to) many phenomena that change character between being oscillatory or not. Since it is a second order linear homogeneous ODE, its general solution can be written as a linear combination of two independent ones, traditionally called Ai(x) and Bi(x), respectively, looking as shown in Figure 11.1. In order to understand this equation better, questions that need to be answered include the following: 1. What is the unique ratio of initial conditions u0 (0)/u(0) that gives solutions that decay for x → +∞?
1
Ai(x) Bi(x)
0.5
0
0.5 10
5
0
x
Figure 11.1. The Ai and Bi functions along the real axis.
5
11.2. Bessel functions
293
2. Just how fast does the Ai function decay and the Bi function grow for x → +∞ (presumably at some “superexponential” rates)? 3. Can we approximate the envelope and oscillation rates (including the phase angle) for the oscillations as x → −∞? An important first step is to express the Ai(x) and Bi(x) functions as integrals. For the bounded Ai function, this can be done by a Fourier transform. Recalling Theorem 9.3, the dˆ y = 0, which can be solved in closed form: transform of (11.3) becomes (ik)2 yˆ(k) − i dk ´∞ 3 3 i k /3 yˆ(k) = C e . Inversion back to physical space gives y(x) = C −∞ ei k /3 ei kx dk, which can be rearranged into 3 ˆ ∞ k y(x) = 2C cos + kx dk. (11.4) 3 0 However, this integral is quite impractical as it stands, being rapidly oscillatory and not absolutely convergent. It also only represents the bounded Ai function and not the unbounded Bi function. As in so many other cases, the simplest path towards progress turns out to go via the complex plane.
11.1.2 Complexvalued independent variable The complex counterpart to (11.3) becomes u00 (z) − z u(z) = 0.
(11.5)
If we are given initial conditions for u(0) and u0 (0), the method of unknown coefficients (Section 2.3.2) will recursively provide any number of coefficients in the everywhere convergent Taylor expansion of u(z). However, Taylor expansions can rarely give any insights about functional behaviors at large distances. Figures 11.2–11.4 show that the two independent solutions Ai(z) and Bi(z) then take on very distinctive features. Techniques to understand these will be described in Section 11.4 and Chapter 12. The first of these sections describes a more general approach to obtain integral relations than the Fourier transform based one that we used above, and Chapter 12 discusses asymptotic analysis for exploring farfield behaviors. For both of these tasks, the extension to complex variables is essential. Following these discussions, we will then return to the Airy functions as an example of applying these methods.
11.2 Bessel functions Bessel’s equation d2 y dy + (t2 − n2 ) y = 0 (11.6) +t dt2 dt is another example of a variable coefficient linear second order ODE that arises in many contexts. One of these is the separation of variables when solving Laplace’s equation in cylindrical coordinates. In that case, the values for n will be integers, and we focus here first on that case. Being a linear second order ODE, its general solution can be written as a combination of two independent ones, typically denoted by Jn (t) and Yn (t). These functions were first systematically explored around 1817 by the German astronomer Bessel t2
294
Chapter 11. Special Functions Defined by ODEs
(a) Real part of Ai(z).
(b) Imaginary part of Ai(z). Figure 11.2. Real and imaginary parts of Ai(z).
(in the context of investigating Kepler’s equations of planetary motion). Since they arise in many situations, there are unsurprisingly also many different ways to introduce them. We will proceed by considering the analytic function 1
f (z, t) = et (z− z )/2 ,
(11.7)
where t is some parameter.
11.2.1 Differential equation and series expansion The function f (z, t) has an essential singularity at z = 0. Its Laurent expansion will have coefficients that depend on the parameter t according to the generating function
11.2. Bessel functions
295
(a) Real part of Bi(z).
(b) Imaginary part of Bi(z). Figure 11.3. Real and imaginary parts of Bi(z).
(ztransform) 1
f (z, t) = et (z− z )/2 =
∞ X
Jn (t) z n .
(11.8)
n=−∞
This notation is justified, as we will see next, by the fact that these functions Jn (t) indeed satisfy (11.6). Direct differentiation of (11.7) shows that f (z, t) satisfies
t2
∂2f ∂f ∂ +t + t2 f = z ∂t2 ∂t ∂z
∂f z . ∂z
(11.9)
296
Chapter 11. Special Functions Defined by ODEs
(a) Abs (Ai(z)).
(b) Abs (Bi(z)). Figure 11.4. The magnitude and phase of the Ai(z) and Bi(z) functions.
If we here replace f by the RHS of (11.8) and collect together all terms according to their powers of z, we get ∞ X d Jn (t) d2 Jn (t) 2 2 + (t − n )J (t) z n = 0. + t t2 n 2 dt dt n=−∞ For this to hold identically in z (recalling that a Laurent expansion has unique coefficients), all the coefficients must vanish, and it is therefore established that the functions Jn (t), as defined by (11.8), indeed satisfy (11.6). We can also see directly from (11.7) that J−n (t) = (−1)n Jn (t).
(11.10)
Either as outlined in Exercise 11.6.3 or by the same procedure as used in Example 4.25 applied to (11.8), we obtain the Laurent coefficients (i.e., the Bessel functions Jn (t)) in the form of a Taylor expansions in t: 2k+n ∞ X (−1)k t Jn (t) = . (11.11) k! (k + n)! 2 k=0
11.2. Bessel functions
297
Figure 11.5. Front to back, the Jn (t) Bessel functions for n = 0, 1, 2, . . . , 10 and t real.
This shows that Jn (t) are analytic and entire functions in t. Figure 11.5 shows what they look like, along the real taxis, for different values of n, and Figure 11.6 displays real parts and magnitudes for J0 (z) and J10 (z). The transitions seen in Figure 11.5 between oscillatory and nonoscillatory character around t = ±n are due to the sign changes there for the factor (t2 − n2 ) in (11.6). Another reason for nonoscillatory behavior around the origin as n increases is that the leading Taylor term in (11.11) is O (tn ) . The color pattern around the origin in Figure 11.6(d) confirms the origin for J10 (z) to be a zero of order 10.
11.2.2 Integral representations of Jn (t) With (11.8) as the starting point, we can derive numerous formulas for the Jn (t) Bessel functions. For example, (4.13) tells us immediately that 1 Jn (t) = 2πi
ˆ C
1
et (z− z )/2 dz, z n+1
(11.12)
where C is any contour that goes around the origin once, in the positive (counterclockwise) direction. Choosing C as the unit circle gives further integral representations, such as Jn (t) =
1 2π
ˆ
π
e−i (nθ−t sin θ) dθ = −π
1 π
ˆ
π
cos(nθ − t sin θ)dθ.
(11.13)
0
11.2.3 Other types of Bessel functions Since the ODE (11.6) is linear and of second order, it should have two independent solutions. We found one of these in the form of the Jn (t) Bessel functions (also known as Bessel functions of the first kind). For example, with variation of parameters, one can find the second solution family, similarly denoted as Yn (t).86 These are quite different in character: no longer entire functions, but with a branch point at the origin. Figure 11.7 shows the first few along the positive real axis. Many other generalizations are described in the literature (but go beyond our present goals). 86 Known
as Bessel functions of the second kind or as Neumann functions.
298
Chapter 11. Special Functions Defined by ODEs
5 5
0 20 0
10 5
20
0
10 10
0
10
10
5 10
5
x
0
0
5
x 20
20
y
10
(a) Re J0 (z)
5
y
10
(b) Abs J0 (z).
5 5
0 20 0
10 5
20
0
10 10
0
10
10
5 10
5
x
0
0
5
x 20
10
20
y
(c) Re J10 (z)
5 10
y
(d) Abs J10 (z).
Figure 11.6. Real parts and magnitudes/phase angles for J0 (z) and J10 (z).
If one lets n in (11.6) take noninteger values, (11.11) generalizes to 2k+n ∞ X t (−1)k . Jn (t) = k! Γ(k + n + 1) 2
(11.14)
k=0
In this case, the linear dependence between Jn (t) and J−n (t) expressed in (11.10) gets broken, and these two (now multivalued) functions represent independent solutions to (11.6).
Figure 11.7. Front to back, the Yn (t) Bessel functions for n = 0, 1, 2, . . . , 10 and t real.
11.3. Hypergeometric functions
299
The case of an n halfinteger leads to closedform expressions, for example 2 1/2 cos t, J−1/2 (t) = πt 2 1/2 J+1/2 (t) = πt sin t, 2 1/2 J+3/2 (t) = − πt cos t − 1t sin t ,
2 1/2 Y−1/2 (t) = πt sin t, 2 1/2 Y+1/2 (t) = − πt cos t, 2 1/2 Y+3/2 (t) = − πt sin + 1t cos t .
Another variation that arises quite frequently in applications is a sign change in front of the n2 term in (11.6), then giving as solutions what are known as the modified Bessel functions Kn (t) and In (t).
11.3 Hypergeometric functions This is a wideranging family of analytic functions, which not only satisfy certain linear variable coefficient ODEs (this is how we at first will introduce them), but which have a tendency to “pop up” in lots of unexpected contexts as well (with Example 11.3 just one such illustration). There exist vast amounts of identities and formulas associated with these functions, which we here will make no attempt to summarize. Symbolic algebra systems quite often formulate their answers in terms of various hypergeometric functions, and these systems are thus often the most practical means for further manipulating such expressions, maybe allowing transformations into other types of regular or special functions.
11.3.1 Introduction of the 2 F1 function via a second order linear ODE Consider the ODE z (1 − z)
d2 w dw + {c − (a + b + 1) z} − a b w = 0, dz 2 dz
(11.15)
where a, b, c are scalar parameters, and w = w(z) is the function for which to solve. A natural first attempt to approximate solutions to (11.15) would be to substitute a Taylor series for w(z) into (11.15) and then equate coefficients. P However, it turns out to be somewhat ∞ better in this case to instead start with w(z) = z α n=0 an z n (with a0 6= 0), leading to the indicial equation α (α + c − 1) = 0, giving two possible values for α, namely α = 0 and α = 1−c, together with the coefficient recursion (α + n)(α + c − 1 + n)an = (α + a + n − 1)(α + b + n − 1)an−1 . The choice α = 0 leads to the Taylor expansion 2 F1 (a, b; c; z)
a·b a(a + 1) · b(b + 1) 2 z+ z c · 1! c(c + 1) · 2! a(a + 1)(a + 2) · b(b + 1)(b + 2) 3 + z + ··· , c(c + 1)(c + 2) · 3!
=1+
and the two independent solutions to (11.15) can now be given as 2 F1 (a, b; c; z) z 1−c 2 F1 (1 + a − c, 1 + b − c; 2 − c; z).
(11.16)
300
Chapter 11. Special Functions Defined by ODEs
Solutions to many higher order ODEs can be expressed similarly if one generalizes (11.16) to have p and q parameters, respectively, in the numerator and denominator of the expansion: a1 · a2 · · · · · ap z b1 · b2 · · · · · bq · 1! a1 (a1 − 1) · a2 (a2 − 1) · · · · · ap (ap − 1) 2 + z (11.17) b1 (b1 − 1) · b2 (b2 − 1) · · · · · bq (bq − 1) · 2! a1 (a1 − 1)(a1 − 2) · a2 (a2 − 1)(a2 − 2) · · · · · ap (ap − 1)(ap − 2) + z3 + · · · . b1 (b1 − 1) · (b1 − 2) · b2 (b2 − 1) · (b2 − 2) · · · · · bq (bq − 1)(bq − 2) · 2!
p Fq (a1 , . . . , ap ; b1 , . . . , bq ; z)
=1+
11.3.2 Examples of elementary functions expressed as hypergeometric functions A large number of elementary analytic functions have Taylor expansions that can be expressed in the form (11.17) with suitable fixedparameter choices. Just a few examples suffice to illustrate this: log(1 + z) = z 2 F1 (1, 1; 2; −z), 3 1 , 1; ; −z 2 , arctan z = z 2 F1 2 2 1 1 3 2 , ; ;z , arcsin z = z 2 F1 2 2 2 (1 − z)α = 1 F0 (α; ; z), 3 z2 sin z = z 0 F1 ; ; − , 2 4 1 z2 cos z = 0 F1 ; ; − , 2 4 ez = 0 F0 (; ; z). The list extends to Bessel and Airy functions, etc. One important aspect of the hypergeometric formalism is that its wide range of identities unifies features of many otherwise seemingly diverse functions. Another is that it encompasses far more functions than have been assigned commonly used acronyms.
11.3.3 Analytic continuation of hypergeometric functions Omitting special cases when the series expansions truncate to a finite number of terms (for example, if either a or b is zero or negative integers in (11.16)), the expansion (11.17) shows the radius of convergence to be R = ∞ if p ≤ q, R = 1 if p = q + 1, and R = 0 for p ≥ q + 2. In applications, the case of p = q + 1 tends to arise particularly frequently, and especially p = 2, q = 1. Theorem 11.2. The 2 F1 hypergeometric function satisfies ˆ 1 Γ(c) F (a, b; c; z) = tb−1 (1 − t)c−b−1 (1 − zt)−a dt. 2 1 Γ(b)Γ(c − b) 0
(11.18)
11.3. Hypergeometric functions
301
A proof is outlined in Exercise 11.6.6. This integral representation (11.18) shows that, in general, z = 1 is a branch point for 2 F1 , and that it is its only singularity in the complex plane. Since the functional behavior around the origin often is of interest, it is conventional to place a branch cut from +1 to +∞ (following the positive real axis). A small remaining “snag” is that the integral, as it stands, converges only when Re c > Re b > 0. Changing the integration contour to one of Pochhammer type (cf. Example 5.33) overcomes this restriction. Numerous important integral formulas follow from (11.18). As an example, we can generalize the beta function ˆ
1
tp−1 (1 − t)q−1 dt =
B(p, q) = 0
Γ(p)Γ(q) , Γ(p + q)
introduced in Section 6.1.2, to the incomplete beta function, for which the upper integration limit is z rather than 1: ˆ z zp (11.19) B(p, q; z) = tp−1 (1 − t)q−1 dt = 2 F1 (p, 1 − q; p + 1; z). p 0 Example 11.3. Conformally map the inside of the unit circle to the inside of a regular polygon with n corners. We noted at the start of Section 8.4 that the Schwarz–Christoffel formula, when mapping from the unit circle, takes an almost identical form to when mapping from the upper halfplane. For the present mapping case, the Schwarz–Christoffel mapping function turns out to become ˆ z 1 n 1 1 2 n dt 1 2 wn (z) = = z F , ; 1 + ; z = B , 1 − ; z . 2 1 n 2/n n n n n n n 0 (1 − t ) (11.20) Figure 11.8(a) shows a polar grid in the zplane, and parts (b)–(d) the resulting curve sets in the wplane for the cases of n = 3, 4, 6, respectively. Another family of integral representations which also provides analytic continuations of hypergeometric functions was devised by Barnes around 1900. The following example can also be seen as an inverse Mellin transform. Theorem 11.4. The 2 F1 hypergeometric function satisfies 2 F1 (a, b; c; z) =
1 2πi
ˆ
+i∞
−i∞
Γ(a + s)Γ(b + s)Γ(−s) (−z)s ds , Γ(c − s)
(11.21)
where the contour from −i∞ to +i∞ goes around the three infinite rows of poles in the way illustrated in Figure 11.9. Because the gamma function is never zero, the denominator Γ(c − s) in (11.21) will not cause any singularities.
302
Chapter 11. Special Functions Defined by ODEs
1
1.5
0.8 1
0.6 0.4
0.5 0.2 0
0
0.2 0.5 0.4 0.6
1
0.8 1.5
1 1
0.5
0
0.5
1
1
(a) Initial domain in the zplane.
0.5
0
0.5
1
1.5
2
(b) Mapping given by w = w3 (z).
0.8
1
0.6 0.4
0.5
0.2 0
0 0.2
0.5
0.4 0.6
1
0.8 1.5
1
0.5
0
0.5
1
(c) Mapping given by w = w4 (z).
1.5
1
0.5
0
0.5
1
(d) Mapping given by w = w6 (z).
Figure 11.8. The unit circle and the maps provided by the function wn (z) given in (11.20).
From the integral formulations of the 2 F1 function, one can derive a large number of functional equations, such as z −a 2 F1 (a, b; c; z) = (1 − z) 2 F1 a, c − b; c; z−1 z −b = (1 − z) 2 F1 c − a, b; c; z−1 = (1 − z)c−a−b 2 F1 (c − a, c − b; c; z) and (−z)−a 1 sin(π(b − a)) 2 F1 (a, b; c; z) = 2 F1 a, a − c + 1, a − b + 1; π Γ(b)Γ(c − a) z −b (−z) 1 − . 2 F1 b, b − c + 1, b − a + 1; Γ(a)Γ(c − b) z When there are particular relations between a, b, c the list of functional relations becomes nearly (or even literally) endless. Many of these relations provide analytic continuations,
11.4. Converting linear ODEs to integrals
303
6 ↑
← 4 poles of Γ(a+s) ↑ 2 → poles of Γ(s) 6
4
2
2
poles of Γ(b+s)
4
6
← 2
→ 4 ↑ 6
Figure 11.9. Schematic illustration of integration path for Barnes’s integral (11.21).
allowing values outside the unit circle to be obtained from values inside the unit circle, where the Taylor series (11.16) converges.
11.3.4 An illustration of the 2 F1 hypergeometric function Since this function has three scalar parameters a, b, c apart from the independent variable z, we choose here just an arbitrary single case for illustration, a = 1, b = 2, c = 3, which is typical in its general character. Figure 11.10 shows for this case a real part that peaks to infinity at the branch point z = 1, while the imaginary part there remains finite.
11.4 Converting linear ODEs to integrals In each of the Sections 11.1–11.3, we somehow converted the original ODEs to integrals. This section describes a couple of systematic ways to do this.
11.4.1 Fourier–Laplace method Given a variable coefficient linear ODE, the general idea in this approach is to look for solutions of the form ˆ u(z) = F (t) ez t dt, (11.22) C
304
Chapter 11. Special Functions Defined by ODEs
10 5
4 2
0 0
5 4
2 2
0
2
4
x
y
4
(a) Re 2 F1 (1, 2, 3; z).
6 4 2 0 2 4 4 6 4
2 0 2
2
0
2
x
4
4
y
(b) Im 2 F1 (1, 2, 3; z). Figure 11.10. The real and the imaginary parts of 2 F1 (1, 2, 3; z).
where both the integration contour C and the function F (t) remain to be determined.87 The following couple of examples explain the typical steps in the procedure. Example 11.5. Determine an integral representation for the solutions to the Airy equation (11.5). ´ Substituting (11.22) into the Airy equation (11.5) gives C (t2 − z)F (t) ez t dt = 0. We next want to manipulate this expression so that z will appear in the exponent only. Noting 87 Since C can be a general contour in the complex plane, (11.22) generalizes both the Fourier and the Laplace transforms. In some literature, the exponential is written as eizt rather than as ezt here (making no conceptual difference in the method).
11.4. Converting linear ODEs to integrals
that z ez t =
d zt dt (e )
305
and then integrating by parts gives ˆ dF 2 + t F ez t dt − F (t) ez t C = 0. dt C
2 zt This will be true for all z if it holds that (i) dF takes the dt + t F = 0 and (ii) F (t) e same value at both ends of the contour C. Starting with item (i), the general solution of dF 2 −t3 /3 (where c is some constant which can be omitted). dt + t F = 0 is F (t) = c e 3 Regarding item (ii) we thus need to study the expression G(t) = F (t) ez t = e−t /3+z t in a complex tplane (temporarily viewing z as a complex constant). G(t) is clearly an entire function of t, which far out in the tplane rapidly decreases to zero within the three 7π 3π sectors (valleys) (V1 ) − π6 < arg t < π6 , (V2 ) π2 < arg t < 5π 6 , (V3 ) 6 < arg t < 2 , and with high “mountain ridges” in between. Letting C form a closed loop will make [F (t) ez t ]C vanish, but this is useless, since u(z) (as defined in (11.22)) will then also vanish, resulting only in the trivial u(z) = 0 solution. A much better idea in this case is to run the contour C from one valley to another. Figure 11.11 illustrates the magnitude of 3 the integrand, e−t /3+z t , in the case of z = 1 (with this specific zvalue only causing a slight deformation of the surface near the origin in the complex tplane). Denoting by C1 , C2 , C3 contours from valley V2 to V3 , V2 to V1 , and V1 to V3 , respectively, we obtain the two standard Airy functions as ˆ ˆ ˆ 3 1 1 −t3 /3+z t Ai(z) = e dt, Bi(z) = − − e−t /3+z t dt. (11.23) 2πi C1 2π C2 C3
5
0
V2
5 4
V1
3 2 1 0
V3
1 2
y
C
3
1
4 5 5
4
3
2
1
0
1
2
3
4
5
x
3 Figure 11.11. The function e−t /3+z t , displayed in a complex tplane in the case of z = 1. As usual, the real axis is marked by a thick red line. The green curve shows the lowest possible choice for the contour C1 . While any contour from V2 to V3 would give the correct value for Ai(z), this specific type of contour will play a key role in the analysis in Section 12.4.
306
Chapter 11. Special Functions Defined by ODEs
The details in these definitions are motivated by (i) ensuring that the two solutions are in´∞ dependent and real along the real axis, (ii) making −∞ Ai(x)dx = 1, and (iii) making the two functions share the same envelope for their oscillations as x → −∞ (further analyzed in Section 12.4.3). We can also note that deforming the contour C1 to run along the imaginary axis (making its convergence properties much worse) produces (11.4). Example 11.6. Find an integral representation of a solution to Bessel’s equation (11.6) in the case of n = 0, i.e., satisfying d2 u du (11.24) + +zu = 0 . dz 2 dz ´ Substituting (11.22) into (11.24) gives C F (t) zt2 + t + z ezt dt = 0. Integrating by ´ d parts (using again z ez t = dt (ez t ) ) gives C F 0 (t) (1 + t2 ) + F (t) t ezt dt = 0, with 1 .88 the ODE F 0 (t) (1 + t2 ) + F (t) t = 0 that is solved by (any multiple of) F (t) = √1+t 2 Hence, ˆ ezt √ u(z) = dt (11.25) 1 + t2 C z
solves (11.24). A suitable contour C will be one that encircles a branch cut placed along the imaginary axis between t = ±i. If we want toreconcile (11.25) with (11.12), it is natural to perform the change vari√ able t = τ − τ1 21 (to match the form of the exponent). Then 1 + t2 = τ + τ1 12 , ´ z(τ −1/τ )/2 dt = τ1 τ + τ1 12 dτ , and (11.25) becomes u(z) = C 0 e τ dτ , with a path C 0 that encircles the origin. This matches the n = 0 case of (11.12).
11.4.2 Euler’s method Several options are available for choosing the integrand in (11.22). In Euler’s version, we start instead with ˆ F (t) (t − z)µ dt
u(z) =
(11.26)
C
and proceed then in the same way: substitute into the ODE, integrate by parts, choose F (t), path C, and now also the parameter µ so that the ODE becomes satisfied for all values of z. The algebra becomes particularly nice if we can let C be a small circle around z, and use negative integer values for µ, since (11.26) then matches (4.7). Example 11.7. Find for each n = 0, 1, 2, . . . a solution un (z) to d2 un (z) dun (z) + 2(n + 1)un (z) = 0 + 2z dz 2 dz (recalling that, if we have one solution, a linearly independent second one can be obtained by variation of parameters). 88 The
ODE is separable. Writing it as grating gives the result.
F 0 (t) F (t)
t = − 1+t 2 , i.e.,
d dt
d log F (t) = − 21 dt log(1 + t2 ), and inte
11.5. The Painlevé equations
307 2
d d d µ We substitute (11.26) into the ODE, note that dz (t−z)µ = − dt (t−z)µ , dz = 2 (t−z) 2 d µ dt2 (t − z) , and then integrate by parts to obtain derivatives of F (t) instead of derivatives of (t − z)µ (intending to use a closed loop and integer values for µ, what “comes out” when integrating by parts cancels). This leads to ˆ (F 00 (t) + 2zF 0 (t) + 2(n + 1)F (t)) (t − z)µ dt = 0. C
The zfactor in front of F 0 (t) needs to be removed, and we can do that by writing z = t − (t − z), and do one more integration by parts, to obtain ˆ (F 00 (t) + 2tF 0 (t) + 2(n + µ + 2)F (t)) (t − z)µ dt = 0. C
Separately for each n, we now choose µ = −n − 1, making the ODE for F (t) in all cases become F 00 (t) + 2tF 0 (t) + 2F (t) = 0, 2 ´ 2 e−t which is solved by F (t) = e−t .89 Therefore un (t) = C (t−z) n+1 dt, which by (4.7) (and ignoring an irrelevant multiplicative factor) can be written as un (z) =
dn −z2 e , n = 0, 1, 2, 3, . . . . dz n
11.5 The Painlevé equations With often centuries of history behind them and, more recently, with the immense power provided by modern computers, most of the special functions of applied mathematics have by now become thoroughly explored and understood. The Painlevé equations and their solutions, known as the Painlevé transcendents, form an exception in this regard. They were first considered just over 100 years ago, initially from a very theoretical perspective. In spite of proving to be particularly difficult to analyze theoretically and also unusually challenging to compute numerically (in both cases partly due to their commonly occurring vast pole fields in the complex plane), their range of applications kept expanding, to the point that they now are considered among the most important types of special functions. Recent numerical advances have finally made them much more accessible than previously. Based on a level of intuition that is hard to imagine, Paul Painlevé (later Prime Minister 2 of France) and some collaborators of his decided to consider all ODEs of the form ddzu2 = F z, u, du dz , where F is an arbitrary rational function of its three arguments z, u, and du (algebraic function such that both the numerator and the denominator are polynomials). dz Typically, solutions to nonlinear ODEs of such a general form would be expected to feature movable branch points (i.e., branch points at locations that depend on the ODE’s two initial conditions). Remarkably, only 50 genuinely different equations of this form are possible if one imposes the Painlevé property: The only movable singularities in the complex plane are either poles or essential singularities (i.e., no movable singularities are allowed that give rise to multivaluedness). It further transpires that essential singularities cannot arise in this context, so all movable singularities to these equations are in fact poles. Out of these 50 equations, it has also been discovered that the solution to 44 of them could be expressed 89 This
d follows quickly if one happens to note that 2tF 0 (t) + 2F (t) = 2 dt (t F (t)).
308
Chapter 11. Special Functions Defined by ODEs
in terms of elementary or traditional special functions. The remaining six equations are now known as the Painlevé equations: PI
d2 u dz 2
= 6u2 + z,
PII
d2 u dz 2
PIII
d2 u dz 2
PIV
d2 u dz 2
PV
d2 u dz 2
PV I
d2 u dz 2
= 2u3 + zu + α, 2 1 du αu2 +β = u1 du − z dz + z + γu3 + uδ , dz 1 du 2 = 2u + 32 u3 + 4zu2 + 2(z 2 − α)u + βu , dz (u−1)2 γu β 1 1 du 2 1 du = 2u + u−1 − + αu + dz z dz z2 u + z + 1 1 1 1 1 du 2 du − = 12 u1 + u−1 + u−z + + dz z z−1 u−z dz γ(z−1) δz(z−1) u(u−1)(u−z) βz α + u + (u−1)2 + (u−z)2 . + z2 (z−1)2
δu(u+1) u−1 ,
Apart from two initial conditions (ICs), the equations feature different numbers of additional free parameters, ranging from none for PI to four for PIII , PIV , and PV I . While most of the analytic functions we have considered so far have not had any free parameters (beyond the independent variable z, for example sin z or Γ(z)), some have involved one or more parameters (for example the two periods ω1 and ω2 for the Weierstrass ℘(z)function and the three parameters for the 2 F1 hypergeometric function). Here (including the two ICs), we have from two to six parameters. This obviously adds greatly to the challenge of exploring their full solution spaces. We will below limit ourselves to showing a few highlights of the analytic functions that arise as solutions to the PI and the PII equations.
11.5.1 The PI equation We noted in Chapter 7 (see (7.8)) that the Weierstrass ℘function satisfies the nonlinear ODE 1 (11.27) ℘00 (z) = 6℘(z)2 − g2 , 2 where the parameter g2 could be chosen arbitrarily. For all ICs, solutions to (11.27) are therefore doubly periodic functions, with double poles repeating completely regularly across the full complex zplane. The difference to the PI equation u00 (z) = 6u(z)2 + z
(11.28)
is that, in the last term, the constant − 21 g2 has been replaced by the independent variable z. By local Laurent expansions, one sees that all singularities to (11.28) are again double poles of strength 1 and residue zero (cf. Exercise 11.6.8). Far away from the origin, z becomes locally like a large constant, and the pole fields become locally similar to those of (11.27). However, as z grows, they generally get denser further out. A main difference, however, is caused by the fact that the variable changes u → ζ 3 u, z → ζz with ζ 5 = 1 leaves (11.28) again satisfied, causing pole fields far out to differ between five sectors with boundaries between them following the directions arg z = 1π/5, 3π/5, 5π/5, 7π/5, 9π/5. These sectors are visible in the PI solution pictures in Figure 11.12, which show the pole locations in four illustrative cases.
11.5. The Painlevé equations
309
Figure 11.12. Four pole field plots for the PI equation, displayed over the region [−50, 50] × [−50, 50] in the complex plane. (a) Case with the ICs u(0) = u0 (0) = 0; the sector boundaries are here narrow polefree strips. (b) ICs u(0) = u0 (0) = 2; the pole field structures are far out similar to the previous case, but the transitions between the sectors have become gradual instead of sharply defined. (c) The tritronquée solution; the unique case for which four of the five sectors have become completely polefree; its ICs are u(0) ≈ −0.1875543083404949, u0 (0) ≈ 0.3049055602612289. (d) A solution with ICs very close to the tritronquée case (here u(0) = −0.1875, u0 (0) = 0.3049). Little has changed in the right pole field, but a pair of additional pole fields have very rapidly moved in from the top left and bottom left, respectively. Reprinted with permission of Elsevier from [19].
The NIST Handbook of Mathematical Functions (2010) [35] was written just before Painlevé solutions could readily be computed across the complex plane [19, 14]. It contains in its Chapter 32, on the Painlevé equations, an illustration similar to Figure 11.13, showing along the real axis how the solution changes when u(0) = 0 and u0 (0) is varied in close vicinity of a critical value for which the solution is smooth and nonoscillatory along the negative real axis. Corresponding illustrations over the complex plane are far more
310
Chapter 11. Special Functions Defined by ODEs
Figure 11.13. Illustration along the real axis of the PI solutions with u(0) = 0 and with u0 (0) varied as shown in the figure. Reprinted with permission of Elsevier from [19].
revealing; cf. Figures 11.14(a)–(b) (with the matching real axis intervals marked by thick black curves). For u0 (0) = 1.8518, the oscillations for x negative are seen to be caused by a pole field that symmetrically surrounds the negative real axis, but without any poles on the axis itself. As u0 (0) increases slightly, this left pole field recedes rapidly out to minus infinity, and then moves back in again, however shifted vertically so there now have become an infinity of poles on the negative real axis. The right pole field has remained essentially stationary. Theoretical work on the PI equation has to a large extent been limited to asymptotic analysis of solutions far out in the different sectors, and on connection formulas linking these results between sectors. For solution features at finite distances, computational methods are nowadays the dominant source of information.
11.5.2 The PII equation In contrast to the PI equation, the remaining Painlevé equations have a few instances of closedform solutions (still like drops in a bucket only, compared to their full solution spaces). In the case of the PII equation, there are two types of closedform solutions: (i) For α integer: There is in this case one rational solution for each αvalue: α=0 , α=1 , α=2 , α=3 ,
u(z) = 0, 1 u(z) = − , z 4 − 2z 3 , u(z) = 4z + z 4 u(z) = .. .
3z 2 (160 + 8z 3 + z 6 ) , 320 − 24z 6 − z 9
11.5. The Painlevé equations
311
Figure 11.14. The magnitude of same solutions as the bottom and top cases in Figure 11.13, here displayed (without phase angle coloring) over the complex plane rather than only along the real axis: (a) u(0) = 0, u0 (0) = 1.8518, (b) u(0) = 0, u0 (0) = 1.8519. The crossing of the thick black lines marks the origin z = 0. Reprinted with permission of Elsevier from [19].
(ii) For α “halfinteger”: After defining −z −z θ θ Ai Bi φ(z) = cos + sin 2 2 21/3 21/3 and Φ(z) = φ0 (z)/φ(z), one can find90 the following solutions: α=
1 2
,
u(z) = −Φ,
α=
3 2
,
u(z) =
2Φ3 + zΦ − 1 , 2Φ2 + z
α=
5 2
,
u(z) =
4zΦ4 + 6Φ3 + 4z 2 Φ2 + 3zΦ + z 3 − 1 , (4Φ3 + 2zΦ − 1)(2Φ2 + z)
.. .
.. .
Letting the free parameter θ move through the period [0, 2π] of Φ(z) causes poles to move as seen for the case of α = 25 in Figure 11.15. When θ = 0, all poles are located in a band 90 Both
sequences depend on finding a single closedform solution uα (z) for some αvalue (such as here α = 0 and α = 12 ), and then repeatedly using the Bäcklund transformation uα+1 (z) = −uα (z) − 2α+1 , giving a solution for α + 1. 2u0 (z)+2u (z)2 +z α
α
312
Chapter 11. Special Functions Defined by ODEs
Figure 11.15. The Airy family of PII solutions in the case of α = 52 . All poles to PII are simple, with residues +1 or −1. In this and the next two figures, these are colored blue and yellow, respectively. Reprinted with permission of Springer Nature from [20].
surrounding the positive real axis. As θ increases, a curved band of poles enters from the left, becomes part of two different entirely symmetric solution configurations at θ = π3 and θ = 5π 3 , and finally moves back out left towards minus infinity as θ approaches 2π, leaving for θ = 2π the same pole distribution as for θ = 0. However, if one looks carefully at the pole field along the positive real axis, one can see that this has moved steadily to the left during this process. The band of poles that came in from the left picked up, during the interactions, five poles from the field along the positive real axis, included then in the curved band, and then brought theses out to minus infinity. No closedform solutions are known to the PII equation other than the two classes just noted above, for α integer and for α halfinteger, respectively. We conclude our PII equation summary by showing two other types of solutions that are particularly important in applications, much due to being polefree along the entire real axis (however, with no closedform expressions available). Figure 11.16 shows the Hastings–McLeod solutions. There are two types of these for each αvalue (real) in the range α < 12 ; otherwise there is a unique one. There also exists for α < 1 a oneparameter family of Ablowitz–Segur solutions. These are smooth and bounded along the entire real axis, oscillatory to the left and decaying to the right. Figure 11.17 shows a typical such solution in the case of α = 0. In both of these cases, we see in the figures again how extensions to the complex plane reveal features that are not apparent from the properties along the real axis.
11.5. The Painlevé equations
313
Figure 11.16. Examples of Hastings–McLeod solutions, shown along the real axis, and their associated pole fields. Reprinted with permission of Springer Nature from [20].
Figure 11.17. Example of an Ablowitz–Segur solution in the case of α = 0. Reprinted with permission of Springer Nature from [20].
314
Chapter 11. Special Functions Defined by ODEs
11.5.3 The remaining Painlevé equations In contrast to the PI , PII , and PIV equations, solutions to the PIII , PV , and PV I equations need not be singlevalued. For example, in the PIII case, we can note the term − z1 du dz in its RHS. Had this been the only term in the RHS, u(z) = log z would have been a solution. This multivaluedness does not contradict the Painlevé property, since the location of the branch point is fixed (at z = 0) and is not “movable” with the choice of ICs. Analysis will reveal that solutions to PIII for different ICs and parameter values (for α, β, γ, δ) can have any number of Riemann sheets, ranging from one to infinity. With the larger number of free parameters in the highernumbered Painlevé equations, it becomes increasingly difficult to survey their complete solution spaces (although there has been significant recent progress in this). We have noted several times earlier that what can be seen about an analytic function along the real axis contains only a tiny fraction of the information that can be gained by inspecting it (and utilizing its properties) over the complex plane. In the case of the multivalued PIII , PV , and PV I functions, intriguing phenomena can occur on different Riemann sheets even when little that is noteworthy is apparent on the primary sheet.
11.6 Exercises Exercise 11.6.1. Many special functions are defined by second order linear ODEs. If one somehow can find one particular solution y1 (x) to a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = 0,
(11.29)
one can then find the general solution to the inhomogeneous ODE a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = d(x)
(11.30)
by reduction of order.91 Substituting y(x) = y1 (x)u(x) into (11.29) gives a linear ODE in u0 (x) from which follows y2 (x), and the general solution to (11.29) as y(x) = c1 y1 (x) + c2 y2 (x). To this we need to add any one particular solution of (11.30), which can be obtained by substituting into it either y(x) = y1 (x)v(x) or y(x) = y2 (x)v(x). Determine thus the general solution to x2 y 00 (x) − 4x y 0 (x) + 6y(x) = x5 ex , having noted that the homogeneous part (RHS = 0) is solved by y(x) = x2 . Exercise 11.6.2. Starting from the integral formulation of the Airy function Ai (z) in (11.23),´ show that ∞ (a) −∞ Ai (x)dx = 1 ´∞ (b) 0 Ai (x)dx = 13 . ´∞ Hints: Part (a): Change the contour to the imaginary axis, and then use −∞ eixt dt = ´∞ ´ e−t3 /3 1 2π δ(x); part (b): Show that 0 Ai(z)dz = − 2πi dt; change the contour to a t C straight line from e−2πi/3 ∞ towards the origin, an εradius circle segment, and a straight line exit towards e+2πi/3 ∞. Show that the two straight line parts cancel. 91 A
special case of variation of parameters.
11.6. Exercises
315
Exercise 11.6.3. An alternative way to obtain the Taylor expansion for Jn (t) (see (11.11)) is to start from (11.12), change variable z = 2ξ/t, giving Jn (t) =
1 2πi
n ˆ ξ −t2 /(4ξ) t e e dξ, 2 ξ n+1 C
and then apply residue calculus. Carry this out. Exercise 11.6.4. For example, using (11.11), show that d Jn (t) = 12 (Jn−1 (t) − Jn+1 (t)). generally, that dt
d dt J0 (t)
= −J1 (t) and, more
Exercise 11.6.5. With the use of computer software, explore some of the Bessel Yn (t) functions in the complex tplane (plot their real and imaginary parts, and magnitude/phase). ´ 1 b−1 Γ(c) Exercise 11.6.6. Prove the relation (11.18): 2 F1 (a, b; c; z) = Γ(b)Γ(c−b) t (1 − 0 t)c−b−1 (1 − zt)−a dt. Hint: Assume z < 1, and Taylor expand (1 − zt)−a around t = 0. Next, swap the order between the sum and the integral, and use (6.5) to arrive at the Taylor series (11.16) for the LHS of (11.18). Exercise 11.6.7. Apply the Fourier–Laplace method to find an integral representation for a solution to the ODE z u00 (z) + (2n + 1)u0 (z) + z u(z) = 0,
n integer.
Hint: You should arrive at (t2 + 1)F 0 (t) − (2n − 1) t F (t) = 0, and u(z) = 1)n−1/2 ez t dt. Explain your choice for C.
´ C
(t2 +
Exercise 11.6.8. Show that each pole of the PI equation is double, with strength 1 and residue 0.
Chapter 12
Steepest Descent for Approximating Integrals
Contour integration allows many definite integrals to be evaluated in closed form. There are also many cases where closed forms cannot be found, but for which the contour integration idea instead can be used to provide excellent approximations. This happens especially when there is some free parameter present, and we are interested in cases where, as this parameter increases, the integrand either becomes locally spiked or increasingly oscillatory. As a preliminary to the topic of steepest descent, we first make some brief comments about asymptotic analysis. This is a farreaching subject that applies to many areas, such as approximating solutions to ODEs, PDEs (for example, boundary layers in fluids), integrals, etc. Even with computational approaches nowadays becoming increasingly capable, asymptotic analysis remains as important as ever. This approach usually gains rapidly in accuracy just in those situations (approaching singular cases) when typical numerical methods encounter their largest difficulties. Many good books on asymptotic methods are available, including [4, 5, 9, 26, 32]. We start our discussion by some brief comments on what is meant by an asymptotic (in contrast to a convergent) expansion, and we illustrate the concept by the error function erf(x) and by the Euler–Maclaurin formula. After that, we turn to approximating integrals. Section 12.3 focuses first on a purely realvalued case in which an integrand becomes increasingly spiked. Utilizing the freedom of choosing integration paths in the complex plane for analytic functions, the method of steepest descent (Section 12.4) turns the realvalued result into a tool of far greater generality.
12.1 Asymptotic vs. convergent expansions P∞ PN An infinite sum S = k=0 ak converges if S − k=0 ak → 0 as N → ∞. This readily generalizes to the case when the terms are functions of some variable x with ak as their coefficients. The sum ∞ X S(x) = ak ϕk (x) (12.1) k=0
PN converges for a fixed x if S(x) − k=0 ak ϕk (x) → 0 as N → ∞. Given a value for x, we can reach any accuracy we want by just letting N → ∞. For an asymptotic series, the 317
318
Chapter 12. Steepest Descent for Approximating Integrals
roles of x and N can be seen as reversed. In place of (12.1), we write S(x) ∼
∞ X
ak ϕk (x)
(12.2)
ak ϕk (x) = o(ϕN (x))
(12.3)
k=0
if it holds, for any fixed N , that S(x) −
N X k=0
when x goes to some limit (such as x → 0 or x → ∞).92 Both (12.1) and (12.2) are very useful for analysis as well as for computation. The utility of (12.1) is obvious, especially if the functions ϕk (x) are easy to compute, and S(x) is a lot more complicated (with Taylor expansions a typical example). The utility of (12.2) is somewhat more subtle. The example below is typical for this case. Example 12.1. Compare along the real axis the´accuracies of the Taylor and the asymptotic 2 x expansions for the error function erf(x) = √2π 0 e−t dt. Along the real axis, the error function transitions smoothly from −1 to +1, as seen in Figure 12.1. Since this is an entire function, its Taylor expansion 2 x3 x5 x7 erf(x) = √ x− + − + −··· (12.4) 3 · 1! 5 · 2! 7 · 3! π is everywhere convergent. Repeated integration by parts of the defining integral produces an asymptotic expansion 2
e−x erf(x) = 1 − √ x π
1−
1·3 1·3·5 1 + − + −··· 2 1 2 2 (2x ) (2x ) (2x2 )3
,
(12.5)
which, in contrast, never converges (since the factorialtype growth in the numerators always “wins out” over the geometric rate in the denominators). Nevertheless, this expansion provides, for similar number of terms, superior accuracy as soon as x & 5, as illustrated in Figure 12.2(b). For xvalues increasing further still, the figure shows that the accuracy increases extremely rapidly, even if the number of terms N is kept quite low. The fact that (12.5) always diverges if we keep x fixed and let N → ∞ is irrelevant if our goal is to estimate erf(x) for x large. Example 12.2. In order to evaluate erf(4), estimate the number of terms N to use in (12.5) and the resulting error. The terms in the expansion (12.5) first decrease and then increase in magnitude. The error is smallest when two successive terms have roughly equal magnitude. Hence, looking at their ratio, N should be chosen such that 1≈ 92 The
2x2 (2N − 3)!! (2N − 1)!! / = , 2 N −1 2 N (2x ) (2x ) 2N − 1
“little o” notation, f (x) = o(g(x)), means that f (x)/g(x) → 0 as x approaches some specified limit.
12.2. Euler–Maclaurin formula
319
1.5 1
erf(x)
0.5 0 0.5 1 1.5 5
4
3
2
1
0
1
2
3
4
5
x
Figure 12.1. The function erf(x) displayed along the real xaxis (previously displayed in the complex plane in parts (c) and (d) of Figure 9.12).
suggesting for x = 4 the choice of N = 16, with a resulting error of 2
e−4 (2 · 16 − 3)!! √ ≈ 2.6 · 10−15 , 4 π (2 · 42 )15 in excellent agreement with the values that can be read off from Figure 12.2(b). While the region of convergence for a Taylor expansion always is the largest singularityfree circle surrounding the expansion point, applicability of asymptotic expansions (of the form of inverse powers, such as (12.5)) occurs in sectors of the complex plane. Generalizing x to be a complex variable z, (12.5) is valid within the sector shaded gray in Figure 12.3(a), whereas a version with the leading “1” swapped for “−1” 2
e−z erf(z) = −1 − √ z π
1−
1·3 1·3·5 1 + − + −··· 2 1 2 2 (2z ) (2z ) (2z 2 )3
(12.6)
is valid in the sector shown in part (b) of the figure. The expression Stokes phenomenon refers both to sector structures of this type and to the fact that sectors of validity can overlap. 2 In the overlapping regions, the factor e−z grows massively large as z moves away from the origin, making the leading constant irrelevant. A more extensive example of Stokes sectors will be given in Section 12.4.3, when we use the method of steepest descent to analyze the farfield behavior of solutions to Airy’s equation (illustrated earlier in Figures 11.2–11.4).
12.2 Euler–Maclaurin formula Together with Stirling’s formula (Section 12.3.3), this may well be the best known asymptotic formula in the literature. It can be written in different ways, including ∞ X k=n
ˆ
∞
f (k) = n
1 1 1 000 1 f (x)dx + f (n) − f 0 (n) + f (n) − f (5) (n) + − . . . , 2 12 720 30240 (12.7)
320
Chapter 12. Steepest Descent for Approximating Integrals
45
5 0
30
35
20
5
30
N
20
40
10
10 15
50
25
5
10
5
0
10 15
10
20
10
15
5
20
5 1
2
3
4
5
6
7
8
9
10
x
(a) Taylor expansion.
30
20
60
45
10 5 0 5 10 15 20
50
40 35
25
10
30
20
0
15
15
5
20
60
10 5
N
30
10 5 1
2
3
4
5
6
7
8
9
10
x
(b) Asymptotic expansion. Figure 12.2. log10 of the errors (shown numerically along the contour lines) when erf(x) is approximated at location x using N terms in the Taylor expansion (12.4) and the asymptotic expansion (12.5), respectively. The areas shaded gray indicate where the accuracy is better than 10−15 . In the asymptotic expansion case, there is typically, for a given value of x, a unique finite value for N that gives the highest accuracy, as shown by the dashed red curve in part (b). This curve marks where the contour lines for accuracy have a vertical tangent.
and, if both limits are finite, m X k=n
ˆ
m
f (k) = n
+
1 1 f (x)dx + [f (n) + f (m)] − [f 0 (n) − f 0 (m)] 2 12
1 000 1 [f (n) − f 000 (m)] − [f (5) (n) − f (5) (m)] + − · · · . 720 30240
(12.8)
12.2. Euler–Maclaurin formula
321
+3 /4
+ /4
3 /4
 /4
(a) Sector of validity for (12.5).
(b) Sector of validity for (12.6).
Figure 12.3. The sectors of validity (in the limit of z → ∞), shaded gray, for the two expansions (12.5) and (12.6), respectively.
Two different derivations are given in Section 12.5.2. Both provide closedform expressions for the coefficients in terms of Bernoulli numbers; cf. (5.12). For most functions f (z), the expansion in the RHS will eventually diverge. However, like for most asymptotic expansions that we will encounter, the error when truncating the expansion will typically be of the same size as the first omitted term. Example 12.3. Numerically approximate Euler’s constant ! n X 1 γ = lim − log n . n→∞ k k=1
P∞ We can write the definition as γ = 1 + k=2 k1 + log(k − 1) − log k . This sum 1 1 1 2 converges, since k + log(k − 1) − log k = k + log 1 − k = O(1/k ), but only very slowly. We thus apply (12.7) to f (z) = z1 + log(z − 1) − log z. The convergence rate of the new expansion improves if one sums some leading terms before applying the formula. Choosing, for example, n = 10 gives 1+ ˆ
9 X
f (k) =
0.6317436767,
k=2 ∞
f (x)dx = −0.0517553591, 10
1 f (10) = −0.0026802578, 2 1 0 f (10) = −0.0000925926, 12 1 000 f (10) = 0.0000001993, 720 1 − f (5) (10) = −0.0000000015. 30240 −
322
Chapter 12. Steepest Descent for Approximating Integrals
Adding these gives a result that is only offPone unit in the last decimal digit; cf. (6.4). It n can be noted that direct summation of 1 + k=2 f (k) does not come this close to the limit 1 10 until n ≈ 2 10 . Pn Example 12.4. Evaluate k=1 k 3 . ´n Pn Choosing f (x) = x3 , equation (12.8) gives k=1 k 3 = 1 x3 dx + 21 (13 + n3 ) − Pn n(n+1) 2 1 1 2 2 . Recognizing n(n+1) as k=1 k, we have thus 12 3(1 − n ) + 720 (6 − 6) = 2 2 arrived at !2 n n X X 3 k = k . k=1
k=1
Since f (z) in this example was a polynomial, the RHS of (12.8) vanished after a certain number of terms, giving an exact result.93
12.3 Laplace integrals We start by considering realvalued integrals along the real xaxis, of the Laplace integral form ˆ b
f (x) es φ(x) dx,
I(s) =
(12.9)
a
in the limit of the parameter s → ∞. For Examples 12.6–12.8, considered later in this section, the top row of subplots in Figure 12.4 illustrates the φ(x) functions, and the bottom row shows the integrands f (x) es (φ(x)−φ(c)) for the svalues s = 1, 3, 10, 30. For improved visibility, we have for the integrands included a (xindependent) factor e−s φ(c) , where x = c is the location of the maximum for φ(x). In all cases, the asymptotic expansions for s → ∞ will come from the immediate vicinity of these x = c locations. In the third example, we split the interval into two, [0, 1] and [1, ∞], and for the first one change variable x → −x. Hence we can, without loss of generality, proceed by assuming that the maximum value of φ(x) over the integration interval [a, b] occurs at its left boundary, denoted by x = a. We next distinguish two main cases: Case 1: φ0 (a) < 0. Case 2: φ0 (a) = 0 and φ00 (a) < 0. Case 2 can be generalized to having several leading derivatives zero at x = a. The next key step in obtaining the asymptotic expansion for (12.9) will be to make a change of variables in such a way that we then can apply Watson’s lemma.
12.3.1 Watson’s lemma Theorem 12.5 (Watson’s lemma). If ˆ b I(s) = f (ξ) e−s ξ dξ
(b > 0)
(12.10)
0
Pn m+1 93 More generally (Faulhaber’s formula), for n, m m ≥ 1, = nm+1 + k=1 k Pm−1 m Bk+1 m−k n (in the last sum, terms with k even vanish, since B3 , B5 , B7 , . . . are zero). k=1 k k+1
nm + 2
12.3. Laplace integrals
323
Figure 12.4. The three columns of subplots correspond to Examples 12.6–12.8, respectively, that are discussed in Section 12.3.2. The top row of subplots shows the functions φ(x) that are multiplied by s in the exponents, and the bottom row of subplots shows how the integrands f (x) es φ(x) then vary across the respective intervals (scaled by the factor e−s φ(c) , in order to display an sindependent value at the location x = c where φ(x) has its maximum). In all cases, it is visually apparent that, as s → ∞, the integral will become increasingly well described by its properties in the immediate vicinity of x = c.
and f (ξ) = ξ α
∞ X
an ξ β n
(ξ & 0),
(12.11)
n=0
then I(s) ∼
∞ X an Γ(α + β n + 1) sα+β n+1 n=0
(s → ∞).
(12.12)
Proof. (Sketch) The key observations behind demonstrating this result are as follows: (i) As s → ∞, the behavior of f (ξ) beyond an immediate vicinity of ξ = 0 becomes increasingly irrelevant to the value of I(s) (as defined by (12.10)); (ii) in I(s), we can therefore substitute for f (ξ) the expansion (12.11); (iii) having done this, the actual value of the upper integration limit b becomes also irrelevant, and can be replaced by b = ∞; and (iv) all terms in the integral can at this point be evaluated in closed form by means of the ´∞ Γfunction relation 0 ξ z e−s ξ dξ = Γ(z+1) (Re z > −1, Re s > 0; In the special case of ´ ∞ n −ssξz+1 n! z = n integer, this becomes 0 ξ e dξ = sn+1 ). This gives the result (12.12).
324
Chapter 12. Steepest Descent for Approximating Integrals
While some other approaches (such as integration by parts) might work in certain cases for approximating (12.10), Watson’s lemma is particularly powerful in that it works well also when (12.11) involves fractional powers of ξ (as will happen in Case 2). To find the appropriate change of variables so that Watson’s lemma can be applied to (12.9), we first rewrite this integral (assumed already arranged so that maxx∈[a,b] φ(x) occurs at x = a) as ˆ b s φ(a) I(s) = e f (x) es (φ(x)−φ(a)) dx, (12.13) a
and then make the change of variable − ξ(x) = φ(x) − φ(a),
(12.14)
so that the exponent in the integral becomes −s ξ (as in (12.10)). The integral now takes the form ˆ 0+ f (x) e−s ξ dξ, (12.15) I(s) = es φ(a) − 0 φ (x) 0 where the upper limit of the integral (denoted by 0+) will become irrelevant as s → ∞. The key step that now remains is to express − φf0(x) (x) as a power series expansion in ξ, and Watson’s lemma will then give the complete asymptotic expansion.
12.3.2 Applications of Watson’s lemma Example 12.6. Find the complete asymptotic expansion for I(s) = s → +∞ (cf. the first column of subplots in Figure 12.4).
´ π/2 0
2
e−s sin
x
dx as
The integral is of the form (12.9) with φ(x) = − sin2 x and f (x) = 1, i.e., − φf0(x) (x) = π 1 2 sin x cos x . The function φ(x) has its maximum over [0, 2 ] at x = 0, so the variable change 2 (12.14) −ξ(x) = φ(x) − φ(0) gives ξ = sin x, and therefore sin x = ξ 1/2 , from which f (x) . The integral = √1 it follows that − φf0(x) (x) , expressed in ξ, becomes − φ0 (x) 2 ξ(1−ξ) ´ 1 has become I(s) = 21 0 ξ −1/2 (1 − ξ)−1/2 e−s ξ dξ, which is in the perfect form for WatP∞ Γ(n+ 1 ) ξn son’s lemma after using the binomial theorem expansion (1 − ξ)−1/2 = n=0 n! Γ 2 1 . (2) 2 P∞ [Γ(n+ 12 )] 1 Therefore, I(s) ∼ 2 n=0 n! Γ 1 sn+1/2 for s → +∞. (2) In most cases, one cannot write down a closedform expression for − φf0(x) (x) in terms of ξ(x). One can then instead proceed by using Taylor expansions, starting from (12.14) and then expanding φ(x) around the critical point x = a: − ξ(x) = φ(x) − φ(a) = d1 (x − a) + d2 (x − a)2 + · · · (explicitly as d1 = Section 2.3.2).
φ0 (a) 1! ,
d2 =
φ00 (a) 2! ,
(12.16)
etc., or by the method of undetermined coefficients;
Case 1: φ0 (a) 6= 0. We can invert (12.16) to obtain a new set of coefficients {ai }, i = 1, 2, . . . , so that (x − a) = a1 ξ + a2 ξ 2 + a3 ξ 3 + · · · , i.e., x = a + a1 ξ + a2 ξ 2 + a3 ξ 3 + · · · , (12.17)
12.3. Laplace integrals
325
as discussed in Sections 2.3.2 and 2.8.1 (i.e., by substituting (12.17) into (12.16) and equat ing coefficients). Substituting into − φf0(x) (x) gives a Taylor expansion in ξ around ξ = 0: f (x) = b0 + b1 ξ + b2 ξ 2 + · · · , (12.18) − 0 φ (x) and Watson’s lemma can be applied. Case 2: φ0 (a) = 0. The first term in the RHS of (12.16) is now missing. The series can again be inverted, but the form of (12.17) will have to be changed to also include halfinteger powers: (x − a) = a1 ξ 1/2 + a2 ξ 2/2 + a3 ξ 3/2 + · · · . (12.19) The coefficients {ai } can again be obtained by substituting the expansion into (12.16) and then equating coefficients. The counterpart to (12.18) takes now the form f (x) = b−1 ξ −1/2 + b0 + b1 ξ 1/2 + b2 ξ 2/2 + b3 ξ 3/2 + · · · , − 0 φ (x) which again works just fine for Watson’s lemma. If this case arose from a situation with a max point of φ(x) occurring at an interior location and the interval was split in two subintervals, the signs for the evennumbered coefficients b0 , b2 , b4 , . . . get reversed when the other subinterval (with the critical point at its right edge) is analyzed. Then, we simply ignore these coefficients and double the contributions from the oddnumbered ones b−1 , b1 , b3 , . . . . We next give one example for each of Cases 1 and 2, respectively. ´ 2 1 −s arctan x Example 12.7. Find some leading terms for I(s) = 1 1+x e dx (cf. the second column of subplots in Figure 12.4). 1 This example features f (x) = 1+x , and φ(x) = − arctan x. The variable change π (12.14) gives ξ = arctan x − 4 and 1 7 10 62 f (x) 1 + x2 = = 1+ξ +2ξ 2 + ξ 3 + ξ 4 + ξ 5 −· · · . − 0 = φ (x) 1+x cos ξ (sin ξ − cos ξ) 3 3 15 (12.20) With φ(1) = − π4 giving a leading factor e−s π/4 , Watson’s lemma produces 1 1 4 14 80 496 I(s) = e−s π/4 + 2 + 3 + 4 + 5 + 6 + ··· . s s s s s s f (x) In most cases, we would not be able to express − φ0 (x) explicitly like this as a function of ξ. We would then instead do Taylor series manipulations as follows (most conveniently and reliably by using a symbolic algebra package such as Mathematica, which has each of the steps as a builtin command). First expanding around x = a = 1 gives
π 1 1 1 1 = (x − 1) + (x − 1)2 − (x − 1)3 + 0 (x − 1)4 − (x − 1)5 + · · · , 4 2 4 12 40 which can be inverted: 8 10 64 (x − 1) = 2ξ + 2ξ 2 + ξ 3 + ξ 4 + ξ 5 + · · · , 3 3 15 8 3 10 4 64 5 2 i.e., x = 1 + 2ξ + 2ξ + ξ + ξ + ξ + · · · . 3 3 15 ξ = arctan x −
326
Chapter 12. Steepest Descent for Approximating Integrals
Substituting this expression for (x − 1) into − φf0(x) (x) = in ξ as obtained above in (12.20). Example 12.8. Find some leading terms for I(s) = column of subplots in Figure 12.4).
1+x2 1+x
´∞ 0
leads to the same expansion
1 s (−x+log x) dx xe
(cf. the third
The function φ(x) = −x + log x has its maximum at x = 1, with φ(1) = −1. Hence, we consider first the integral over [1, ∞], and make the change of variable −ξ(x) = φ(x)− φ(1) = −x + 1 + log x, with Taylor expansion ξ=
1 1 1 1 1 (x − 1)2 − (x − 1)2 + (x − 1)4 − (x − 1)5 + (x − 1)6 2 3 4 5 6 1 1 1 1 7 8 9 10 − (x − 1) + (x − 1) − (x − 1) + (x − 1) + · · · , 7 8 9 10
giving when inverted94 √ 1/2 2ξ ξ 3/2 2ξ 2 ξ 5/2 4ξ 3 139ξ 7/2 √ + √ 2ξ + + √ − + − 3 9 2 135 540 2 8505 340200 2 2ξ 4 571ξ 9/2 √ + ··· + − 25515 73483200 2
x=1+
(where we have included unusually many terms, since half of them soon will drop out). −1 Substituting this expansion into − φf0(x) (x) = 1−x gives ξ −1/2 1 ξ 1/2 4ξ ξ 3/2 4ξ 2 139ξ 5/2 f (x) √ + √ = √ − + √ − + − − 0 φ (x) 3 6 2 135 216 2 2835 97200 2 2 8ξ 3 571ξ 7/9 √ + ··· . + − 25515 16329600 2 As described under Case 2 above, the expansion for the subinterval x ∈ [0, 1] will become identical apart from a switch of sign for all terms with integer powers of ξ. When adding the results from the two subintervals, all these terms vanish, and we double the coefficients for the remaining ones (with halfinteger powers of ξ). Recalling the leading factor esφ(a) in (12.13), Watson’s lemma gives r I(s) ∼ e
−s
2π s
1 1 139 571 1+ + − − + ··· . 12 s 288 s2 51840 s3 2488320 s4
(12.21)
12.3.3 Stirling’s formula As a followup to Example 12.8, we can note ´∞ ´ ∞that the change of variable s x = u in I(s) = 0 x1 es (−x+log x) dx gives I(s) = s1s 0 us−1 e−u du = s1s Γ(s). We thus obtain the famous Stirling formula for Γ(z) by multiplying (12.21) by ss . Since s! can be defined 94 Compare
with the second case described in Section 2.8.1.
12.3. Laplace integrals
327
as Γ(s + 1) also for noninteger svalues, it follows (having renamed s to n) that n n √ 139 571 1 1 − − + · · · . (12.22) n! ∼ 2πn 1 + + e 12 n 288 n2 51840 n3 2488320 n4 This expansion (12.22) can be recast in different ways, with 1 1 1 1 1 + − +··· (log 2π + log n) + − 3 5 2 12 n 360 n 1260 n 1680 n7 (12.23) particularly attractive in that all even powers of n have vanished. Since n! = nΓ(n), this can alternatively be formulated as log n! ∼ n log n − n +
1 1 1 1 1 (log 2π − log z) + − + − +··· , 2 12 z 360 z 3 1260 z 5 1680 z 7 (12.24) with the only difference (beyond changing n to z) the sign in front of the log z term. Exercises 12.6.7 and 12.6.8 show two different ways to extend (12.23) and (12.24) explicitly to all orders. The sector of validity for both (12.22) and (12.24) can be shown to be the entire complex plane with the exception of the negative real axis. Both approximations are remarkably accurate already for quite modest values of n, as seen in Figures 12.5(a)–(b). We display here the relative errors that the expansions produce for positive values of n, with −15 roughly corresponding to the regular computer machine rounding level. This is for (12.22) reached already for n (= s) around 6, at which point the bottom right subplot in Figure 12.4 shows that the integrand is still far from being sharply peaked. Nevertheless, the local expansion around x = 1 suffices for typical machine precision accuracy. In conclusion to the present discussion of Stirling’s formula, we note from (12.22) that nn 1 √1 − = O , and therefore the following sum will converge (although very n n!e n3/2 2πn slowly): ∞ X nn 1 √ − ≈ −0.08406950873 . S= n!en 2πn n=1 P∞ 1 The sum n=1 √2πn is by itself divergent, but had it converged, a plausible value would 1 1 √ have been 2π ζ( 2 ). The expression S became around 2001 a striking example of “experimental mathematics” in action, when numerical computations unexpectedly showed it to match S = − 23 − √12π ζ( 12 ) to vast numbers of digits. A strict proof followed about six years later [6]. log Γ(z) ∼ z log z − z +
12.3.4 Laplace’s method in case of movable maxima In Example 12.8, we started with ˆ I(s) = 0
∞
1 s (−x+log x) e dx, x
(12.25)
which, with φ(x) = −x + log x, was in perfect form for the standard variable change followed by Watson’s lemma. As an afterthought, we then noted that the variable change sx = u led to´ the standard integral for the gamma function. Suppose that we instead start ∞ with Γ(s) = 0 us−1 e−u du, the question now becoming how to find an appropriate way to reverse this last variable change, to get the integral back to a form suited for asymptotics.
328
Chapter 12. Steepest Descent for Approximating Integrals
5
15
3
30
10
4 0
25
20
45
15 10 5 0 5
50
40
35
30
30
10
5
25
2
15
N
0
2 5 0
25 20
20 15
15 15
10
10
10
5
5
10
5
5
0 2
4
6
8
10
12
14
16
18
20
n
(a) Asymptotic expansion (12.22).
40
35
30
25
20
15
10
5
0
50
45
45
20 15 10 5
50
40
35
20
15
0
0
5
10
3
25
4
30
N
45
25
5
30
20
35 15
30
25 10
5
1 0
20
15
25
20 15
15
10
5
10
0 4
6
8
10
12
14
16
18
20
n
(b) Asymptotic expansion (12.23). Figure 12.5. log10 of the relative errors (shown numerically along the contour lines) when the expansions (12.22) and (12.23) are applied to different values of n using N terms in the respective asymptotic expansions. The areas shaded gray indicate where the accuracy is better than 10−15 .
´ ∞ −u We can write the gamma function integral as Γ(s) = 0 e u es log u du, but φ(u) = log u lacks a local max point. ´ ∞ Another idea is to collect the exponential factors and write the integral as Γ(s) = 0 u1 e−u+s log u du. Now, the exponent −u + s log u does have a max point, at u = s, but it moves position with s. To bring it to a fixed location, an appropriate variable change would in this case be u = xs (bringing it to x = 1). We have ´∞ now obtained Γ(s) = ss 0 x1 es (−x+log x) dx, which is in the form that Example 12.8 started with.
12.4. Steepest descent
329
12.3.5 Leading order approximations If we only want the leading term of the asymptotic expansion, application of Watson’s lemma becomes simple enough that we can write down the end result explicitly. For the Laplace integral (12.9) we have the following: Case 1: φ0 (a) < 0: I(s) ∼ −
1 f (a) s φ(a) e . s φ0 (a)
Case 2: φ0 (c) = 0, φ00 (c) < 0, where the critical point c is located inside [a, b]: √ 2πf (c) s φ(c) p e . I(s) ∼ −s φ00 (c)
(12.26)
(12.27)
For onesided cases (c at either edge of [a, b]), the result should be halved. This last Case 2 is a special case of φ(k) (c) = 0, k = 1, 2, . . . , p − 1, and φ(p) (c) < 0, with p even, for which (assuming c is located inside [a, b]; else the result is halved) we have the following: 2Γ(1/p)(p!)1/p f (c) s φ(c) e . (12.28) I(s) ∼ p [−s φ(p) (c)]1/p
12.4 Steepest descent 12.4.1 Concept and background In the description above of Laplace integrals, we made no use of two key opportunities provided by the theory of analytic functions: (i)
The integration path can be altered freely (as long as we account for residues when it crosses singularities).
(ii)
The real and the imaginary parts of analytic functions are tightly coupled through the Cauchy–Riemann (CR) equations (2.1).
Focusing first on item (ii), we consider an analytic function φ(z) and decompose it in its real and imaginary parts φ(z) = u(x, y) + i v(x, y) , where z = x + i y. As we noted already in Section 2.1.1, the CR equations ux = vy , vx = −uy imply the following: 1. Both u(x, y) and v(x, y) are harmonic functions (satisfying uxx + uyy = 0 and vxx + vyy = 0, respectively), so they cannot have any local maxima or minima. h iT h i vx 2. The CR equations also imply that uuxy = 0, telling us that the gradient vy vectors (directions of steepest descent) for u(x, y) and v(x, y) are orthogonal to each other. 3. For any smooth functions in x, y, level curves are orthogonal to gradient directions.
330
Chapter 12. Steepest Descent for Approximating Integrals
Combining the observations just above, we next conclude the following: 4. Along the steepest descent paths of u(x, y), the function v(x, y) is constant, and vice versa. 5. A point z with φ0 (z) = 0 (and φ00 (z) 6= 0) will be a saddle point for both u(x, y) and v(x, y). The function u(x, y) will at such a point have two orthogonal steepest descent/ascent directions, with a local maximum along one and a local minimum along the other, and with v(x, y) constant along both (and equivalent for v(x, y)). Figures 12.6 and 12.7 illustrate the general features described above in the special case of φ(z) = z − log z .
(12.29)
(a) Real part of φ(z) = z − log z.
(b) Imaginary part of φ(z) = z − log z. Figure 12.6. Real and imaginary parts of φ(z) = z − log z, with the saddle point at z = 1 marked with a black dot. The steepest descent path through it for Re φ(z) is shown in green and the steepest ascent path in red. These same paths both become level curves for Im φ(z).
12.4. Steepest descent
331 3 2
y
1 0 1 2 3 2
0
2
x Figure 12.7. Steepest descent paths for Re φ(z) (equal to level curves for Im φ(z)) shown as solid curves, and level curves for Re φ(z) (equal to steepest descent curves for Im φ(z)) shown as dashed curves. The blue line segment (also shown in Figure 12.6) is a tangent at the saddle point to the steepest descent curve (green).
This function features a branch cut along the negative real axis (which is irrelevant for our present discussion). The black dot marks the only location at which φ0 (z) = 0, which occurs at z = 1. As seen in Figure 12.6(a), this is a saddle point for Re φ(z)—a local minimum if we follow the real axis (steepest ascent path; red curve) near z = 1, and a local maximum if we follow the steepest descent path (green curve) through that point. These same paths are both constant level paths for Im φ(z); cf. part (b) of the figure. Figure 12.7 displays φ(z) somewhat differently over the same region in the complex plane. Instead of highlighting just one steepest descent path for Re φ(z) (level path for Im φ(z)), we illustrate several of them as solid black curves. The dashed curves are similarly the level paths for Re φ(z) and therefore also steepest descent paths for Im φ(z). The two curve sets (solid and dashed) are everywhere orthogonal to each other, apart from at the saddle point z = 1 and at the singularity z = 0. In the next subsection, we will see that steepest descent path for Re φ(z) through the saddle point (marked green in Figures 12.6 and 12.7) plays a key role in the steepest descent method for approximating Laplace integrals. The curved shape of this path somewhat complicates the algebra compared to the previous cases of integrating along the real axis. However, this path also being a level curve for Im φ(z) has the great benefit that es i Im φ(z) ´b can be factored out of the integral a f (z) es φ(z) dz, making the integrand locally peaked and nonoscillatory as s increases. Choosing to instead integrate along the local tangent line at the saddle point (marked blue in the figures) gives readily the leading expansion term (as shown in Section 12.4.4), but may require more work if many terms are needed.
12.4.2 Some applications of steepest descent Integrals that become increasingly peaked in the s → ∞ limit
We consider again the Laplace integral (12.9): ˆ b f (z) es φ(z) dz, I(s) = a
(12.30)
332
Chapter 12. Steepest Descent for Approximating Integrals
but we denote the integration variable by z instead of x, as we will now utilize that the path need not be the straight line from a to b (nor do a and b need to be on the real axis). Example 12.9. Determine an asymptotic expansion for 1 1 = Γ(s) 2πi
1 Γ(s)
as s → ∞, based on (5.19):
ˆ ez z −s dz ,
(12.31)
C
where C is a Hankel contour, as illustrated in Figure 5.32 (there denoted by H instead of by C; it enters from infinity in the third quadrant, goes around the branch point at z = 0, and exits to infinity in the second quadrant). The first step is to convert (12.31) to the Laplace form (12.30), which can be done with the change of variable z → s z. This yields ˆ 1 1 = es (z−log z) dz , (12.32) Γ(s) 2πi ss−1 C with f (z) = 1, φ(z) = z − log z and with the same type of Hankel contour C. This function φ(z) has already been illustrated in Figures 12.6 and 12.7. Of all possible paths from the third to the second quadrant, going around the branch point at the origin, we note that the highlighted steepest descent curve (green) has some unique features. With Im φ(z) constant along this path, es Im(φ(z)) is also a constant, which thus can be factored out of the integral. In the present case this is just a factor of 1. We thus replace (12.32) by ˆ 1 1 = es Re (z−log z) dz . (12.33) Γ(s) 2πi ss−1 C With a purely realvalued factor now multiplying s in the exponent, we can apply the insights from Sections 12.3.1–12.3.5. There are now two main options for generating the asymptotic expansion: 1. Integrate along the steepest descent path and then apply Watson’s lemma. 2. Return to (12.32) and follow the straight line segment that is tangent to the steepest descent curve at the saddle point (marked blue in Figures 12.6 and 12.7). In either case, the complete asymptotic expansion can be deduced from an infinitesimal vicinity of the saddle point. The second approach might appear to be simpler, but intermediate expansions may require quite a large number of terms. We compare the two approaches below. Method 1: Follow the steepest descent curve—Use its analytic form: With φ(z) = z − log z = x + iy − 21 log(x2 + y 2 ) − i arctan xy , we obtain the steepest descent curve through the saddle point (located at z = 1) by solving Im φ(z) = 0, giving x = y/ tan y. y sin y dx dx Along this curve, Re φ(z) = tan y + log y and dz = dx + i dy = dy + i dy. Since dy is an odd function of y, it will not contribute to the integral, and we obtain 1 1 = Γ(s) 2πi ss−1
ˆ es Re (z−log z) dz ∼ C
1 2π ss−1
ˆ
0+
f (y) es φ(y) dy, 0−
(12.34)
12.4. Steepest descent
333
y sin y with f (y) = 1 and φ(y) = tan y + log y . Apart from the naming of the integration variable as y instead of x, the problem is now in the standard form for making the key variable change (12.14) followed by using Watson’s lemma as in Example 12.8. The remaining steps become
sin y y + log −1 tan y y y2 y4 y6 y8 y 10 691y 12 =− − − − − − − ··· , 2 36 405 4200 42525 294698250 p ξ ξ2 139ξ 3 571ξ 4 163879ξ 5 + + − − + ··· , y = 2ξ 1 − 18 1080 680400 146966400 67898476800 y f (y) = − 0 φ (y) 1 − 2y cot y + y 2 csc2 y 1 ξ ξ2 139ξ 3 571ξ 4 =√ 1− + + − + ··· , 6 216 97200 16329600 2ξ
−ξ = φ(y) − φ(0) =
and therefore (by Watson’s lemma) ˆ ∞ e s r s 1 2 es f (y) 1 1 139 −s ξ ∼ − 0 e dξ ∼ 1− + + Γ(s) 2π ss−1 0 φ (y) s 2π 12s 288s2 51840s3 571 − + ··· . 2488320s4 (12.35) Method 2: Follow the steepest descent curve—Rely on series expansions only: The derivation above utilized closedform expressions for the steepest descent curve. Such expressions are often not available (or are not practical). It transpires that the local Taylor expansion of φ(z) around the saddle point contains all the needed information. To illustrate this, we focus again on the integral ˆ es Re (z−log z) dz (12.36) C
in (12.33). The standard variable change −ξ = φ(z) − φ(a) (cf. (12.14)), here with a = 1, gives −ξ = φ(z) − φ(1) =
1 1 1 1 (z − 1)2 − (z − 1)3 + (z − 1)4 − (z − 1)5 + − · · · , 2 3 4 5
which can be inverted (best using a symbolic algebra package): 2 2 2 4 3 2 z = 1− ξ− ξ − ξ + ξ4 + · · · 3 135 8505 25515 i 1 1 139 7/2 571 ±√ 2ξ 1/2 − ξ 3/2 + ξ 5/2 + ξ − ξ 9/2 + · · · . (12.37) 9 540 340200 73483200 2 With ξ > 0 (real), this equation amounts to a parameterization of the upper and lower branches ´ of the steepest descent curve in the zplane. The integral (12.36) is now in the form C e−s ξ dz, and we next need to change the dz to a dξ, which we get from differentiating (12.37) with respect to ξ. Before doing this, we can simplify by noting that dz on
334
Chapter 12. Steepest Descent for Approximating Integrals
the upper branch matches (−dz) on the lower branch.95 Adding the two, the contribution from the first bracket in (12.37) cancels, and the one from the second should be doubled. Following this observation, we obtain from (12.37) 1 1/2 i 1 3/2 139 5/2 571 −1/2 7/2 √ 2ξ − ξ ξ ξ ξ dz = + + − + · · · dξ. 3 108 48600 8164800 2 The integral in (12.36) has now become equivalent to ˆ 0+ 1 1/2 1 3/2 139 5/2 571 i −1/2 7/2 √ 2ξ − ξ + ξ + ξ − ξ + · · · e−sξ dξ, 3 108 48600 8164800 2 0 which is in the perfect form for applying Watson’s lemma. Remembering the factor 2πi 1ss−1 from (12.33) and also the factor eφ(1)s = es , we have again obtained the expansion (12.35). Method 3: Follow the tangent at saddle point: Although the tangent (being a straight line) is a simpler curve than the true steepest descent path, Im φ(z) is not constant along it, preventing us from factoring es Im φ(z) out of the integral. When s increases, the integrand gets not only increasingly peaked, but also increasingly oscillatory. The approximation procedure becomes as a result somewhat different. In order to locally follow the tangent in the present case, we let z = 1 + it, and consider t along an infinitesimally short stretch around t = 0, along a real taxis. The integral in (12.32) can then be replaced by ˆ ˆ 0+ 1 −1 1 s Re (z−log z) = e dz ∼ es (1+it−log(1+it)) dt . (12.38) Γ(s) 2πi ss−1 C 2π ss−1 0− The key idea in this approach is to Taylor expand the exponent, and separate off the constant and the quadratic terms: es (1+it−log(1+it)) = e
3 2 4 5 6 7 8 s 1− t2 +s i t3 + t4 −i t5 − t6 +i t7 + t8 +··· 2
s − st2
=e e
= es e
2 − st2
3 4 5 6 7 8 s i t3 + t4 −i t5 − t6 +i t7 + t8 +···
e (12.39) ist3 st4 ist5 s(3 + s)t6 1+ + − − 3 4 5 18 7 is(12 + 7s)t s(60 + 47s)t8 + + + ··· . 84 480
If our aim is just to obtain the leading term, we don’t need anything beyond the factor st2 es e− 2 , making this approach then very convenient. Pursuing more terms, we note that, following the leading “1” in the last expansion above, the next three terms are of size O(s), then three of size O(s2 ), etc. Knowing that only a vicinity of t = 0 matters, we next swap the integration interval from [−ε, ε] to [−∞, ∞], after which we can do termbyterm integration in closed form thanks to the relation (α > 0): ˆ ∞ n+1 2 2 if n even, Γ( n+1 2 )/α e−α t tn dt = (12.40) 0 if n odd −∞ 95 Given the symmetry of the path (green curve in Figure 12.7), at matching points z and z on the curve, it will hold that dz = −Re dz + Im dz . dξ dξ dξ
12.4. Steepest descent
335
´∞ n+1 2 2 , n > −1). (which generalizes for onesided cases to 0 e−α t tn dt = 12 Γ( n+1 2 )/α Since all terms with odd powers of t vanish, we obtain from (12.39), using even powers up through t12 , that r 3 1 5 3+s 7 60 + 47s −1 1 π 1 s e − + = 2 1/2 + s−1 3/2 5/2 Γ(s) 2π s 2 2 s 3 s 16 s7/2 s 3 1260 + s(1377 + 175s) − 20 s9/2 11 90720 + s(120564 + 7s(4437 + 80s)) + · · · . + 576 s11/2 (12.41) This needs to be resorted into inverse powers of s. The increasing powers of s in the numerators slow up the process, but the powers of s in the denominators grow faster in the long run. When including still one more term in (12.41), we arrive at e s r 1 1 1 1 139 (12.42) = 1− + + + ··· . Γ(s) s 2πs 12s 288s2 51840s3 The need for a quite large number of terms in the internal expansions is not unusual and can significantly increase the amount of work that is required when one chooses to follow the tangent instead of the steepest descent curve. However, following the tangent can be very convenient if we only want the leading term. Also, in some cases, it can provide closedform expressions for complete expansions (as we will see for the Airy function in Section 12.4.3). Example 12.10. Find the leading term in the expansions for the Jn (t) Bessel functions in the two limits: (a) For t fixed, and n → ∞. (b) For n fixed, and t → ∞. We first recall the illustration of these Bessel functions in Figure 11.5. (a) Writing (11.11) as # n " 2 4 1 t 1 t 1 t Jn (t) = 1− + − +··· n! 2 1!(n + 1) 2 2!(n + 1)(n + 2) 2 gives immediately 1 Jn (t) ∼ n!
n t 2
as n → ∞.
(12.43)
See also Exercise 12.6.11. (b) (11.12) is in the perfect form for steepest descent. We write it as Jn (t) = ´ Equation 1 1 1 1 t φ(z) f (z) e dz, with f (z) = and φ(z) = z − n+1 2πi C z 2 z . The saddle points are where φ0 (z) = 0, i.e., at z = ±i, with the corresponding steepest descent paths shown in Figure 12.8. Since we aim for only the leading order term, it suffices to integrate along each of the straight line segments, and only keep the leading contribution. We focus first on the saddle at z = i: Changing variable z = i + s(i − 1) gives dz = (i − 1)ds and φ(z) = i − s2 + O(s3 ). Thus, ˆ ˆ +ε ˆ 2 1 1 −(n+1) π i t i +ε −t s2 2 e ∼ f (i) et(i−s ) (i − 1)ds = e e ds. 2πi −ε 2πi saddle+i −ε
336
Chapter 12. Steepest Descent for Approximating Integrals
(a) Real part of φ(z) =
1 2
z−
1 z
.
2
y
1
0
1
2 2
1
0
1
2
x (b) Paths seen from above. Figure 12.8. (a) Real part of φ(z) = 12 z − z1 . (b) Solid and dashed black curves show contour lines of Im φ(z) and Re φ(z), respectively. In both parts (a) and (b), the green curve shows the steepest descent path through the saddles (red dots) and its local tangents there (blue lines).
The last integral is asymptotically the same as ˆ
∞
2
e−t s ds = −∞
r
π , so t
Similarly, for the second saddle,
ˆ ∼ saddle+i
´ saddle−i
∼
r 1 +(t−(n+1) π ) i π 2 e (i − 1) . 2πi t
1 −(t−(n+1) π 2 ) i (i 2πi e
p + 1) πt . The sum of
12.4. Steepest descent
337
0.4 Leading term approx. True function J6 (t)
0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0
10
20
30
40
50
60
70
80
90
100
t
Figure 12.9. Comparison between J6 (t) and the leading order asymptotic term (12.44) for the case of n fixed (here n = 6) and t increasing.
these two contributions simplifies to r Jn (t) ∼
2 nπ π cos t − − πt 2 4
as t → ∞.
(12.44)
Figure 12.9 illustrates for the case of n = 6 how well already this leading order asymptotic formula (12.44) approximates the true function J6 (t) as t increases, with regard to both amplitude and phase angle. Higher order versions of (12.44) require additional terms for both of these quantities (see, for example, [35, equations 10.18.17 and 10.18.18]). Integrals that become increasingly oscillatory in the s → ∞ limit
The key observation here is that an analytic function that is highly oscillatory in one direction in the complex plane may become rapidly decaying in another direction. The steepest descent approach may then again be effective in obtaining asymptotic expansions, as illustrated in the next example. Example 12.11. Find the asymptotic expansion for I(s) =
´1 0
cos sx2 dx as s → ∞.
Figure 12.10(a) illustrates this integrand for s = 10 and s = 100. The cost of direct numerical integration increases rapidly with s, since the whole interval will then need to become increasingly finely resolved. The integrand cos sx2 grows rapidly in size when analytically continued away from the real axis, so an immediate change of integration path ´1 2 is unlikely to help. However, we can rewrite I(s) as I(s) = Re 0 eisx dx, and the form is now that of a Laplace integral. Figures 12.10(b) and 12.11 illustrate the corresponding function φ(z) = i z 2 in the complex plane. If we, instead of the original path [0, 1] (marked blue), follow the green path starting at z = 0 and then return via the other green path ending at z = 1, the value of the integral has not changed, but we now have two steepest descent paths. Only immediate vicinities of z = 0 and z = 1 will thus matter for the asymptotic expansion. Had the value of Re φ(0) been different from that of Re φ(1), we could for the asymptotic expansion have ignored the path with the lower value, but Re φ(0) = Re φ(1) = 0, so we need to consider both paths separately (and then add the contributions from the two paths). We write I(s) = Re (J(s) + K(s)) corresponding to the two paths.
338
Chapter 12. Steepest Descent for Approximating Integrals
1
0.5
0
0.5
s = 10 s = 100
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
(a) The integrand cos sx2 . 1.5
1
0.5
0
0.5
1
1.5 1
0.5
0
0.5
1
1.5
2
(b) Integration paths. Figure 12.10. (a) The integrand cos sx2 displayed for s = 10 and s = 100. (b) The steepest descent paths for Re φ(z) with φ(z) = i z 2 (black, solid) and level curves (black, dashed). The original integration path is here shown in blue, and the two steepest descent paths are shown in green.
Path originating at z = 0: The variable change z = ei π/4 t gives dz = ei π/4 dt and ´ 0+ ´∞ pπ i π/4 p 2 2 π 1 J(s) ∼ ei π/4 0 e−s t dt ∼ ei π/4 0 e−s t dt = e 2 s , i.e., Re J(s) = 2 2s . Path returning to z = 1: As an alternative to first splitting φ(z) in real and imaginary parts and determining the steepest descent path, we immediately apply (12.14): −ξ =
12.4. Steepest descent
339
(a) Real part of φ(z) = i z 2 .
(b) Imaginary part of φ(z) = i z 2 . Figure 12.11. Real and imaginary parts of φ(z) = i z 2 , with the two ends of the original integration interval marked with red dots and the steepest descent paths for Re φ(z) shown by green curves. The imaginary part is shown from a different perspective to better illustrate that these same curves are level curves for Im φ(z).
1/2
−1/2
φ(z) − φ(1) = i z 2 − i and obtain z = (1 + iξ) , dz = 2i (1 + iξ) dξ, and ´ P∞ (−i ξ)n Γ(n+ 12 ) −1/2 −ξ s −1/2 i is ∞ K(s) ∼ 2 e 0 (1 + iξ) e dξ. With (1 + iξ) = n=0 , Watn! Γ( 12 ) n 1 P∞ (−i) Γ(n+ ) son’s lemma gives K(s) ∼ − 2i ei s n=0 Γ 1 sn+12 . (2)
340
Chapter 12. Steepest Descent for Approximating Integrals
20
10
5
25
30
15
N
30
35
25
20
0
40
3 40 10 20 0
50
10
10
20
30
15
10
5
0
20
15
10 40
50
60
10 70
80
90
100
s
Figure 12.12. log10 of the relative errors (shown numerically along the contour lines) when (12.45) is applied with different values of s using N terms in the asymptotic expansion. The areas shaded gray indicate where the accuracy is better than 10−15 .
Accuracy of the expansion: In the same style as Figures 12.2 and 12.5, Figure 12.12 shows the error in the asymptotic expansion " # r ˆ 1 ∞ i i s X (−i)n Γ n + 12 1 π 2 I(s) = − Re e cos sx dx ∼ (12.45) 2 2s 2 Γ 12 sn+1 0 n=0 as a function of s and when using terms up through n = N in the expansion. The right edge (s = 100) corresponds to the red curve in Figure 12.10(a).
12.4.3 Steepest descent analysis of Ai(z) The Airy function Ai(z) was introduced (and illustrated)´ in Section 11.1 as the unique ∞ bounded solution to u00 (z) − z u(z) = 0, normalized by −∞ Ai(z)dz = 1. Based on its integral representation (11.23) ˆ 3 1 e−t /3+z t dt (12.46) Ai(z) = 2πi C we will next use steepest descent to find its asymptotic expansions for z → +∞ and z → −∞. The integration path C in (12.46) connects “valley V2 ” to “valley V3 ,” as illustrated in Figure 11.11. The first step is to change variables to bring (12.46) to standard Laplace form with the exponential factor es φ(z) , where s is the parameter that goes to plus infinity and z is the integration variable. Ai(z) for z → +∞
Method 1: Follow steepest descent curve: In order for (12.46) to take the form of a Laplace integral, z should be a factor in the exponent −t3 /3 + z t in (12.46), suggesting the change of variable t = z 1/2 τ with dt = z 1/2 dτ , which yields Ai(z) = ´ 1/2 +z3/2 (−τ 3 /3+τ ) 1 e dτ . We next arrive at the standard form by further renaming 2πi C z z 3/2 as s and then τ as z: ˆ 3 s1/3 I(s) = Ai(s2/3 ) = es(−z /3+z) dz . (12.47) 2πi C
12.4. Steepest descent
341
With φ(z) = −z 3 /3 + z, solving dφ dz = 0 shows that the only two saddle points are z = ±1. The lowest path between the valleys V2 to V3 will thus go through the saddle at z = −1;pcf. Figures 12.13(a)–(b). The steepest descent path p becomes Im φ(z) = 0, i.e., x = − 1 + y 2 /3 and, along it, Re φ(z) = − 92 (3 + 4y 2 ) 1 + y 2 /3. The standard Watson’s lemma variable change gives p 2 2 −ξ = φ(y) − φ(0) = − (3 + 4y 2 ) 1 + y 2 /3 + 9 3 4 6 8 10 5y 7y y 11y 91y 12 = y2 + − + − + + ··· , 36 648 576 31104 1119744 which becomes, when inverted, p 5ξ 77ξ 2 2431ξ 3 1062347ξ 4 14003665ξ 5 1168767425ξ 6 − + − + +· · · . y = ξ 1− + 72 3456 248832 214990848 515978352 743008370688 dx As in Example 12.9, Method 1, dz = dx + i dy = dx dy + i dy, where dy can be omitted, since it is an odd function that will vanish when√integrating. With f (y) = 1 and φ(y) = p 3 9+3y2 − 29 (3 + 4y 2 ) 1 + y 2 /3, we get − φf0(y) (y) = 18y+8y 3 , i.e., 5ξ 1/2 385ξ 3/2 17017ξ 5/2 1062347ξ 7/2 ξ −1/2 f (y) − + − + − 0 = φ (y) 2 48 6219 497664 47775744 −
154040315ξ 9/2 15193976525ξ 11/2 + + ··· , 10319560704 1486016741376
and then, by Watson’s lemma, e−2s/3 385 85085 37182145 5 I(s) = √ 1/6 1 − + − + 2 3 48s 4608s 663552s 127401984s4 2 πs 5391411025 5849680962125 − + + ··· . 6115295232s5 1761205026816s6 The expansion for Ai(z), z → +∞ follows now by substituting s → z 3/2 . Method 2: Follow tangent at saddle point: With z = −1 + i t, we obtain φ(t) = 3 3 − z3 + z = − 23 − t2 − i 3t , and therefore ˆ ˆ 3 2 i s t3 s − 23 −t2 − i 3t − 2s 3 e dt ∼ e e−s t e 3 dt {substitute t = s−1/2 u, dt = s−1/2 du} C
∼e
− 2s 3
C 0+
ˆ
e
−u2
e
i√ u3 3 s
i s−1/2 du
0− 2s
∼i
e− 3 s1/2
∼i
e− 3 s1/2
2s
ˆ
n ∞ X 1 i u3 √ du {note that only terms n! 3 s −∞ n=0 n = 0, 2, 4, . . . contribute} ∞ m X Γ(3m + 1 ) 1 2 − . (12.48) (2m)! 9s m=0 ∞
2
e−u
342
Chapter 12. Steepest Descent for Approximating Integrals
30 20 10 0 10 20 30 3 2 1 0 1 2 3
0
1
2
3
1
2
3
3
(a) Real part of φ(z) = − z3 + z. 3
2
1
0
1
2
3 3
2
1
0
1
2
3
(b) Steepest descent paths. 3
Figure 12.13. Displays of the function φ(z) = − z3 + z, with its saddle points at z = ±1 marked with red dots and the steepest descent path through the one at z = −1 shown in green. The same path is a level curve for Im φ(z) and is therefore the appropriate integration path for the present problem. (a) Surface plot of Re φ(z) and (b) steepest descent paths of Re φ(z) (solid) and level curves of Re φ(z) (dashed).
12.4. Steepest descent
343
The idea behind the substitution in the first line of (12.48) becomes apparent in its lines 3 and 4. It is rare that the approach of following the tangent simplifies this much. For z → +∞, we now obtain the complete expansion: Ai(z) ∼
m ∞ 1 z −1/4 − 2 z3/2 X Γ(3m + 12 ) √ e 3 . − 3/2 (2m)! 2 π 9z m=0
(12.49)
Ai(z) for z → −∞
Instead of (12.47), we need now to consider the s → +∞ limit of ˆ 3 s1/3 2/3 I(s) = Ai(−s ) = es(−z /3−z) dz , 2πi C
(12.50)
for which φ(z) = −z 3 /3 − z. The change of sign of the last term of φ(z) will have no bearing on the character of the valleys V2 and V3 , but it has caused the saddle points to move from z = ±1 to z = ±i. The lowest path between the two valleys will now be a composite of two steepest descent sections, with equally high saddles, as shown in Figure 12.14(a)–(b). Methods 1 and 2 will work once we sum the contributions from the two saddles. Omitting the algebra, it will transpire that Method 2 again will produce a closedform result: " ∞ 2m 2 (−z)−1/4 π X (−1)m Γ(6m + 21 ) 1 √ cos Ai(z) ∼ (−z)−3/2 − − 3 4 m=0 (4m)! π 9(−z)3/2 Γ( 12 ) + sin
∞ π X
2 (−z)−3/2 − 3 4
(−1)m Γ(6m + (4m + 2)! Γ( 12 ) m=0
7 2)
(12.51) 2m+1 # 1 . − 9(−z)3/2
Figure 12.15 shows that both this expansion (12.51) for z negative and (12.49) for z positive can give excellent accuracies for z greater than about 10 or 15. In contrast, it can easily be verified that the Taylor expansion of Ai(z) around the origin is barely useful beyond z around 2. The form of (12.51) does not convey very clearly that the Ai(z) function for z → −∞, while decaying in magnitude, oscillates with a slowly changing frequency (as was seen in Figure 11.1). Hence, it is preferable to reexpress (12.51) in terms of magnitude and phase angle. Denoting the two sums by Sp 1 and S2 , respectively, we thusapply the trigonometric identity (cos α) S1 + (sin α) S2 = S12 + S22 sin α + arctan SS12 to obtain Ai(z) = M (z) sin θ(z), where, for z → −∞, ∞ X 1 1 · 3 · 5 · · · · · (6k − 1) 1 M (z) ∼ k!(96)k z3 π(−z)1/2 k=0 2
and θ(z) ∼
π 2 5 1105 82825 1282031525 + (−z)3/2 1 + + + + + · · · . 4 3 32 z 3 6144 z 6 65536 z 9 58720256 z 12
344
Chapter 12. Steepest Descent for Approximating Integrals
20 15 10 5 0 5 10 15 20 3 2 1 0 1 2 3
1
2
3
1
0
2
3
3
(a) Real part of φ(z) = − z3 − z. 3
2
1
0
1
2
3 3
2
1
0
1
2
3
(b) Steepest descent paths. 3
Figure 12.14. Displays of the function φ(z) = − z3 − z, with its saddle points at z = ±i marked with red dots. The appropriate lowermost path passes through both saddles and is shown in green. (a) Surface plot of Re φ(z) and (b) steepest descent paths of Re φ(z) (solid) and level curves of Re φ(z) (dashed).
12.4. Steepest descent
345
20
10
15
5
20
25
5
5 3 40
16
10
5
4
18
20
0
3
15
0
14
35
10
30
10
10
N
12
25
25
20
5
6
15
5
8
0
20
15 4
10
15
10
10
2
5
5 0 30
25
20
15
10
5
x
(a) Asymptotic expansion z ∈ [−30, −1]. 35
0
10
25
20
5
3
15
18
15 10 5 0
20
16
30
14
25
5
0
N
10
10
2 0
5 1
5
12
20
8
15 6
2
15
10
4
10
5
5 0 5
10
15
20
25
30
x
(b) Asymptotic expansion z ∈ [1, 30]. Figure 12.15. log10 of the relative errors (shown numerically along the contour lines) when Ai(x) is approximated at locations z along the real axis when using N terms in each of the expansions in (12.51) and N terms in the expansion (12.49), respectively. The areas shaded gray indicate where the accuracy is better than 10−15 .
Although no practical closedform expressions are available for θ(z), that is of little practical consequence, since this reformulation into magnitude and phase angle was mainly used to provide qualitative insight. In general, closedform expressions for asymptotic expansions are rarities in all but the simplest cases. What is important is that we have systematic methods available to generate some moderate number of leading terms in asymptotic
346
Chapter 12. Steepest Descent for Approximating Integrals
expansions. Although steepest descent provides a very powerful mathematical approach, symbolic algebra packages are essential for utilizing this most effectively.
12.4.4 Leading order saddle point approximation Just as for Laplace integrals (cf. Section 12.3.5), the saddle point method can be reduced to a simple formula in case we only want the leading term in the approximation. Theorem 12.12. With a saddle at z = z0 at which φ0 (z0 ) = 0, φ00 (z0 ) 6= 0, ˆ
s f (z) e
I(s) =
s φ(z)
dz ∼ ±e
s φ(z0 )
f (z0 )
C
2π . −s φ00 (z0 )
The ± sign choice depends on the direction in which C passes through the saddle point. If − π2 ≤ arg(direction) < π2 , use +, else −. Proof. Near the saddle φ(z) = φ(z0 ) + 12 (z − z0 )2 φ00 (z0 ) + · · · . We make the change of p τ , choosing for the independent variable τ = (z − z0 ) −φ00 (z0 ), i.e., z = z0 + √ 100 square root the principal branch. Then, s φ(z) = and
ˆ
ˆ
f (z0 )esφ(z0 ) p −φ00 (z0 )
f (z) es φ(z) dz ∼
I(s) = C
−φ (z0 ) 1 2 s φ(z0 )− 2 sτ +· · · , dz
C1
!
1
e−s 2 τ
2
1 dτ , −φ00 (z0 )
=√
dτ,
where the new integration path goes from 0− to 0+ , which, without affecting the asymptotics, can be changed toq−∞ to ∞. The first bracket inside the integral factors out, and the second integrates to
2π s .
12.5 Supplementary materials 12.5.1 The Riemann–Siegel formula One might have expected that the first known use of the saddle point method would have been on some simple demonstration example, but it was instead on a case of great mathematical beauty, found by Siegel around 1932 in unpublished notes by Riemann, dating from the 1850s.96 Numerical hand calculations by Riemann, utilizing what has now become known as the Riemann–Siegel formula, most likely contributed to him proposing the Riemann hypothesis. In order to understand the ζ(s) function (with s = σ + it) for 0 ≤ σ ≤ 1 and t large, direct use of (6.19) is impractical, since no choice of path makes the integrand sharply peaked. However, writing (5.17) as ζ(s) = − 96 Laplace
Γ(1 − s) 2πi
ˆ Γ
(−z)s−1 1 dz ez − 1 z
(12.52)
used earlier a similar method in the special case of integrals along the real axis. Others, before Carl L. Siegel, attempted to understand Riemann’s difficulttointerpret notes, but he was the first one to fully succeed.
12.5. Supplementary materials
347
(with the contour Γ opening to the right) and then rewriting the factor
1 ez −1
as
N X 1 e−N z = e−nz + z z e − 1 n=1 e −1
brings (12.52) to ζ(s) =
ˆ N X Γ(1 − s) e−N z (−z)s−1 1 1 − dz. s n 2πi ez − 1 z H n=1 −N z
(12.53)
s−1
1 Figure 12.16 shows this integrand e ez(−z) −1 z in the case of N = 3 and s = σ + it 1 with σ = 2 and t = 100. There has now appeared a small choice of saddles, one near z = − 21 + it /N . Going through this saddle (along a straight line path, angled 45o to the axes) also encloses a number M of poles caused by the factor ez1−1 , with residues that need to be compensated for. Accounting for these, (12.53) takes the form
ζ(s) =
ˆ N M 1−s X X 1 1 Γ(1 − s) e−N z (−z)s−1 1 s− 12 Γ( 2 ) +π − dz. (12.54) s s 1−s n Γ( 2 ) n=1 n 2πi ez − 1 z H n=1
It is very practical to choose N =
hq
t 2π
i
(as used in this example; [·] denotes the integer
part), since this turns out to give M = N . Furthermore, in the case of σ = 21 (i.e., evaluating ζ(s) on the critical line), the two sums are complex conjugates of each other, meaning that just one needs to be evaluated. By then making a saddle point approximation
Figure 12.16. The integrand 100i and N = 3.
e−N z (−z)s−1 1 ez −1 z
in (12.53) displayed in the case of s =
1 2
+
348
Chapter 12. Steepest Descent for Approximating Integrals
(for which we here omit the details), ζ(s) has become expressed in a form that is very well suited for studying the zeta function for large distances up/down the critical strip. When t is increased, the number of terms in the sum(s) increases only as O(t1/2 ). The gap between the two regions of very large function values slides up in the complex zplane, but remains “clear” for all t.
12.5.2 Select proofs Euler–Maclaurin formula—derivation via exponential basis functions
´∞ P∞ To arrive at (12.7), we look for an expansion of k=n f (k) − n f (x)dx that is linear in f , meaning that if it holds for two functions f1 and f2 , it will also hold for any linear combination αf1 + βf2 . We then recall from the Fourier transform that any function over [−∞, ∞] can be thought of as a linear superposition of Fourier modes eiwx and, from the Laplace transform, that functions over [0, ∞] (or equivalently over [n, ∞])P can as well if ∞ we allow w to also have an imaginary part. It is thus sufficient to consider k=n f (k) − ´∞ wx f (x)dx for functions of the type f (x) = e with w complex (and Re w ≤ 0). For n dk k wx these functions, it holds that dxk f (x) = w e , and we can evaluate the sum and the integral analytically. Utilizing (5.12) and the fact that odd numbered Bernoulli numbers beyond B1 vanish, we obtain ˆ ∞ ∞ X 1 1 + f (k) − f (x)dx = enw − w e −1 w n k=n
= −enw
∞ X Bk k=1
k!
∞
wk−1 =
X B2k 1 f (n) − f (2k−1) (n). 2 (2k)! k=1
(12.55) The fact that the power series expansion converges only for w < 2π puts a restriction on the validity of (12.55), making it only asymptotically correct. Euler–Maclaurin formula—derivation via contour integration
P∞ The “routine” contour integration approach for evaluating a sum k=n f (k) is to integrate f (z)π cot πz around a path that includes the points z = n, n + 1, n + 2, . . . and divide by 2πi (cf. Section 5.2). Hence, consider the closed path C1 , C0 , C2 , C3 shown in black in Figure 12.17, and assume (i) that f (z) is singularityfree inside it, and (ii) that f (z) decays fast enough that the contribution from C3 vanishes when chosen far enough out. With the radius ε of the halfcircle C0 small, the contribution from it will be πi f (n).97 It only remains to find contributions from C1 and C2 . We focus first on C1 : Along this path z = n + iy, and therefore 2 π cot πz = −iπ 1 + 2πy e −1 (cf. Figure 5.25(b)). Thus ˆ ˆ f (z)π cot πzdz = −π C1 97 Equivalently,
value type.
ε
ˆ
∞
f (n + iy)dy − 2π ε
∞
f (n + iy) dy. e2πy − 1
we can let the path go straight through the pole at z = n and let the integral be of principal
12.5. Supplementary materials
349
C3
C1
C0 0
C3 n
...
n+1
C2
C3
Figure 12.17. The integration paths considered in Section 12.5.2.
´ Regarding the first of the two integrals in the RHS, consider f (z)dz around the dashed green path in Figure 12.17. With no singularities inside ´ ∞ the path, and with ´ ∞the integral vanishing along the top and right sides, it holds that 0 f (n + iy)dy = −i n f (x)dx. Combining this with a similar argument for C2 gives ˆ
ˆ C1 +C2
ˆ
∞
f (z)π cot πzdz = 2πi
f (x)dx+2πi n
0
∞
dy f (n + iy) − f (n − iy) . 2πy i e −1
By Taylor expansion, ∞
X y 2k+1 (2k+1) f (n + iy) − f (n − iy) =2 (−1)k f (n) , i (2k + 1)! k=0
and therefore ˆ
∞
0
∞
X 2(−1)k f (n + iy) − f (n − iy) dy = f (2k+1) (n) 2πy i e −1 (2k + 1)! k=0
ˆ 0
∞
y 2k+1 dy . e2πy − 1
Remembering from Exercise 6.5.11 that this integral can be expressed in terms of an even order Bernoulli number, we can collect all pieces together and obtain the same result as in (12.55):98 ˆ ∞ ∞ ∞ X X 1 B2k (2k−1) f (k) = f (x)dx + f (n) − f (n) . (12.56) 2 (2k)! n k=n
98 This
k=1
contour integration derivation of the Euler–Maclaurin formula was given by G. Plana in 1820 and inde´ ∞ f (iy)−f (−iy) P 1 k pendently by N.H. Abel in 1823. Abel also gave ∞ dy (under k=0 (−1) f (k) = 2 f (0) + i 0 2 sinh πy appropriate convergence assumptions).
350
Chapter 12. Steepest Descent for Approximating Integrals
12.6 Exercises Exercise 12.6.1. The Bernoulli polynomials Bk (x) are defined by their generating function as ∞ X t etx tk = . (12.57) Bk (x) t e −1 k! k=0
(a) Verify that Bk (0) = Bk , where Bk are the Bernoulli numbers, as defined in (5.12). dBk (x) (b) Show that = k Bk−1 (x), k = 1, 2, 3, . . . . dx (c) Given the first Bernoulli numbers B0 = 1, B1 = − 12 , B2 = 16 , B3 = 0, B4 = 1 − 30 , . . ., calculate the matching Bernoulli polynomials. (d) Show that Bk+1 (x + 1) − Bk+1 (x) = (k + 1) xk , and from this deduce that n X
km =
k=1
1 [Bm+1 (n + 1) − Bm+1 (1)]] , m = 0, 1, 2, 3, . . . . m+1
(e) Use the results above in this exercise to confirm the result of Example 12.4. Exercise 12.6.2. On an equispaced grid with xk = x0 + kh, k = 0, 1, . . . , N , the Euler– Maclaurin formula (12.8) provides the error expansion for the trapezoidal rule: ˆ
xN
x0
! h f (xk ) − [f (x0 ) + f (xN )] f (x)dx − h 2 k=0 i i 2 h h h4 h (3) ≈+ f (1) (x0 ) − f (1) (xN ) − f (x0 ) − f (3) (xN ) (12.58) 12 720 i h i h6 h (5) h8 + f (x0 ) − f (5) (xN ) − f (7) (x0 ) − f (7) (xN ) + − · · · , 30240 1209600 N X
with coefficients obtained from the generating function 1 z 1 1 1 3 1 1 1 1 1 z− z + z5 − z7 + − · · · . − − = coth − = −z 1−e z 2 2 2 z 12 720 30240 1209600 (12.59) When instead using the midpoint rule over the same interval [x , x ], the nodes are placed 0 N at the equispaced interior locations ξk = x0 + k + 12 h, k = 0, 1, . . . , N − 1, and the error expansion becomes ˆ
xN
f (x)dx −
h
x0
N −1 X k=0 2 h
! f (ξk )
i 7 h4 h i h f (1) (x0 ) − f (1) (xN ) + f (3) (x0 ) − f (3) (xN ) (12.60) 24 5760 i 31 h6 h (5) (5) +− f (x0 ) − f (xN ) 967680 i 127 h8 h (7) f (x0 ) − f (7) (xN ) + − · · · . + 154828800
≈−
12.6. Exercises
351
Determine from what generating function the error expansion coefficients arise in this case.99 Exercise 12.6.3. Using the midpoint rule over [x0 , ∞], with nodes at ξk = x0 + k + 12 h, k = 0, 1, 2, . . ., and assuming that f (z) decays to zero for Re z > x0 , show the exact error formula ! ˆ ∞ ˆ ∞ ∞ X f (x0 + iy) − f (x0 − iy) f (x)dx − h f (ξk ) = i dy. 1 + e2πy/h 0 x0 k=0 Exercise 12.6.4. The counterpart to the Euler–Maclaurin formula (12.7) for an alternating series is100 ∞ X (−1)n 1 1 1 (5) k (−1) f (k) = f (n) − f 0 (n) + f 000 (n) − f (n) 2 2 24 240 k=n (12.61) 17 (7) f (n) − + · · · . + 40370 Note that this expression contains no integral term. (a) Relate the coefficients in (12.61) to the Taylor coefficients of an elementary function. (b) Find a closed form expression for the expansion coefficients. Hint: The identity ez1+1 = ez1−1 − e2z2−1 will be helpful. Exercise 12.6.5. Show that
Pn
k 2 k=1 (−1) k
Pn / ( k=1 k) = (−1)n .
Exercise 12.6.6. Combine the method in Section 12.5.2 with the result of Example 5.14 to justify Ramanujan’s formula (3.11). ExerciseP12.6.7. Derive Stirling’s formula (12.23) by applying the Euler–Maclaurin forn mula to k=2 log k. This gives all terms, apart from the constant term 12 log 2π, which can then be deduced, for example, by using (6.7). Exercise 12.6.8. Show the following complete expansion of Stirling’s formula: log Γ(z) ∼ P∞ B2k (z− 21 ) log z−z+ 12 log 2π+ k=1 2k(2k−1) , where B2k denotes the (2k)th Bernoulli z 2k−1 number by following the hint below. Hint: With some effort (you do not need to carry out this step), (6.2) can be rearranged into ´∞ 0 (z) d ψ(z) = dz log Γ(z) = ΓΓ(z) = log z + 0 1t − 1−e1 −t e−zt dt. With the definition of the d Bernoulli numbers given in (5.12), the full expansion for dz log Γ(z) follows from Watson’s lemma. Termbyterm integration gives the desired result, apart from the constant, which can then be deduced from (6.7) (see also Exercise 12.6.9). 99 Equation (12.60) is commonly known as “the second Euler–Maclaurin formula.” The closed form for the error expansion is similar to the one for the regular Euler–Maclaurin expansion: P B2k h2k − ∞ 1 − 21−2k f (2k−1) (x0 ) − f (2k−1) (xN ) . The two versions are equivalent, as follows k=1 (2k)! 1 from a slight generalization of the trivial identity 2 f (0) + f (1) + f (2) + f (3) + f (4) + · · · − 21 f (0) + f (2) + f (4) + · · · = f (1) + f (3) + f (5) + · · · . 100 Also known as the Euler–Boole method.
352
Chapter 12. Steepest Descent for Approximating Integrals
Exercise 12.6.9. The duplication formula for the gamma function Γ(2z) = π
1 Γ(z)Γ z + 2
−1/2 2z−1
2
π 1/2 Γ(2z) (z+ 21 )
(see (6.7)) is equivalent to the ratio f (z) =
22z−1 Γ(z)Γ
being identically equal to 1.
Prove this result by the following steps: (i) Show that f (z) is an entire function that never takes the value zero, (ii) consider log f (z) for z large, and show from the leading terms of (12.24) that this quantity is bounded, (iii) apply Liouville’s theorem (Theorem 4.12), and (iv) verify that f (z) = 1 for some value of z. Exercise 12.6.10. Show that keeping just the leading term in the general approach in Section 12.3 indeed produces the formulas (12.26) and (12.27). Exercise 12.6.11. Derive (12.43) by instead starting from (t/2)n Jn (t) = √ πΓ(n + 12 )
ˆ
π
cos(t cos θ) sin2n θ dθ, 0
valid for n > − 21 . √ ´∞ Exercise 12.6.12. Show that, to leading order as s → ∞, I(s) = 0 e−x−s/ x dx ∼ p π 2/3 1/3 −3(s/2)2/3 s e . 3 2 Hint: As the integral stands, it is a case of a movable maximum, as discussed in Section 12.3.4.
Exercise 12.6.13. Show that for x → ∞ ˆ
π/2
I(x) =
√
−x (sin t)4
sin t e
0
Γ 78 3Γ 11 Γ 38 8 + ··· . dt ∼ 3/8 + 7/8 + 4x 8x 32x11/8
Exercise 12.6.14. Show that for x → ∞ ˆ
π/2
I(x) =
e
i x cos t
0
r dt ∼
π i (x− π ) 4 e + ··· . 2x
Exercise 12.6.15. Show that for x → ∞ ˆ
∞
I(x) =
e −∞
i t3 /3
√ i π 2 3/2 π sin xt dt ∼ 1/4 cos x − + ··· . 3 4 x
´1 Exercise 12.6.16. Show that the larges expansion for I(s) = 0 eisx log x dx starts I(s) ∼ − i log s+iγ+π/2 + O s12 . s´ ∞ −x Hint: The result 0 e log x dx = −γ may become useful (cf. Exercise 6.5.4).
12.6. Exercises
353
Figure 12.18. The function Re φ(z) given in the hint of Exercise 12.6.20.
Exercise 12.6.17. Find the leading term in the larges expansion for the following functions: ´ π/4 (a) cos(s t2 ) tan2 t dt, ´0π is cos t (b) e dt. 0 P∞ n Exercise 12.6.18. Let f (z) = n=0 an z be an entire function. By (4.10), an = f (ξ) 1 2πi Γ ξ n+1 dξ, where Γ goes around the origin once in the positive direction. Consider f (z) = ez and use the saddle point method to obtain the leading approximation to an for n large. The result should match the leading term in Stirling’s formula. Exercise 12.6.19. Verify the first two terms in the asymptotic expansion √ ˆ 1√ i π t eist dt ∼ − eis − I(s) = (1 − i) s (2s)3/2 0 (for s → ∞). ´1 2 Exercise 12.6.20. Find the leading order approximation to I(s) = 0 e−4sz cos(s(5z − 3 z ))dz as s → ∞. ´1 Hint: First show that I(s) can be rewritten as I(s) = 21 e−2s −1 es φ(z) dz with φ(z) = (1 + iz)2 (2 + iz). Figure 12.18 shows Re φ(z) with its two saddles and, for each, the steepest descent/ascent curves.
Exercise 12.6.21. Theorem 9.1 was followed by a brief discussion about the decay rates of Fourier coefficients ck for 2πperiodic functions. In case f (x) furthermore is an entire function, the rates become particularly fast. Show that ck 2k k! → 1 as k → ∞ in the case of f (x) = ecos x . Hint: Combine the results of (11.13) and (12.43).
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[10] J. Derbyshire, Prime Obsession, Joseph Henry Press, 2003. (Cited on p. 180) [11] L.E. Dickson, First Course in the Theory of Equations, John Wiley and Sons, Inc., 1922. (Cited on p. 11) [12] T.A. Driscoll and L.N. Trefethen, SchwarzChristoffel Mapping, Cambridge University Press, 2002. (Cited on p. 215) [13] H.M. Edwards, Riemann’s Zeta Function, Dover, 1974. (Cited on p. 180) [14] M. Fasondini, B. Fornberg, and J.A.C. Weideman, Methods for the computation of the multivalued Painlevé transcendents on their Riemann surfaces, J. Comput. Phys. 344 (2017), 36–50. (Cited on p. 309) [15] M. Fasondini, N. Hale, R. Spoerer, and J. A. C. Weideman, Quadratic Padé approximation: Numerical aspects and applications, Computer Research and Modeling, submitted. (Cited on p. 89) [16] G. Fikioris, Mellin Transform Method for Integral Evaluation, Synthesis Lectures on Computational Electromagnetics, No. 13, Morgan & Claypool Publishers, 2007. (Cited on p. 258)
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[39] E. Wegert, Visual Complex Functions: An Introduction with Phase Portraits, Birkhäuser, 2012. (Cited on p. 17) [40] R. Wegmann, Handbook of Complex Analysis: Geometric Function Theory, Vol. 2: Geometric Function Theory, Chap. 9, Methods for Numerical Conformal Mapping, pp. 351–477, Elsevier, 2004. (Cited on p. 215) [41] E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 2nd edition, Cambridge University Press, 1915 (4th edition 1927). (Cited on p. ix)
Index Abel transform, see transform, Abel Abel, Niels Henrik, 13, 349 Abel–Ruffini theorem, 13 Ablowitz–Segur solution, 312 Airy function, 311, 340 Airy’s equation, 292, 304 analytic continuation, 22, 71, 279, 300 Borel summation, 83, 91 circlechain, 75 contour integration, 152–160 functional equation, 78 Padé approximations, 85 partitioning of interval, 81 replace Taylor coefficients, 81 Schwarz reflection principle, 77 subtraction, 82 analytic function, 2, 19–22 analytic geometry, 8 Argand plane, 3 argument, 5 argument principle, see Cauchy’s argument principle arithmeticgeometric mean, 200 asymptotic analysis, 317–349 Bspline, 237, 261 Bäcklund transformations, 311 Barnes’s integral, 301 Basel problem, 81, 150 Bedrosian’s theorem, 271 Bernoulli Jacob, 65 Johann, 23 number, 65, 151, 176, 188, 350
polynomial, 350 Bessel function, 265, 293–299, 335 Bessel’s equation, 293, 306 beta function, 159, 175–176 incomplete, 175, 301 binomial coefficient, 151 Blaschke factor, 213 Borel summation, see analytic continuation branch cut, 41, 43, 44, 132 branch point, see singularity, branch point Cardano, Girolamo, 2, 13 cardinal data, 242 Caserati–Weierstrass theorem, 112, 114, 204 Cauchy’s argument principle, 100, 144 integral formula, 101, 104 theorem, 96 theorem for ODEs, 291 Cauchy’s estimate, 115 Cauchy–Hadamard theorem, 59 Cauchy–Riemann equations, 20–22, 61, 329 circlechain method, see analytic continuation code for plots Mathematica, 60 MATLAB, 59 color wheel, 17 complex conjugate, 5 integration, 93 numbers, 1–11 plane, 3 complex derivative, 19
359
conformal mapping, 16, 209 applications, 219 bilinear function, 212 regular polygon, 215, 301 stereographic projection, see stereographic projection continued fractions, 85 contour Bromwich, 247 Hankel, 156, 186, 332 integration, 97 keyhole, 131, 138, 139, 249 Pochhammer, 159 convergence infinite products, 165 pointwise, 53 uniform, 53 convolution theorem Fourier transform, 239 Laplace transform, 254 Cotes, Roger, 23 cubic equation, 2 general solution, 13 Dawson’s function, 169, 262 de Moivre’s formula, 6, 28 delta function, 247, 261 derivative, 19 Dirichlet series, 80 domain doubly connected, 215 simply connected, 94, 214 elliptic function, 191 Jacobi, 197–198, 222 order, 193 Weierstrass, 194 entire function, 20, 203, 204, 236
360 equilateral triangle, 13, 14 error function, 144, 245, 262, 318, 320 Euler numbers, 65 Euler’s constant, 174, 321 identity, 23 line, 63 method, 306 relations, 23 Euler, Leonhard, 63, 65, 81 Euler–Boole method, 351 Euler–Maclaurin formula, 319, 348–349, 351 second form, 351 expansion asymptotic, 318 convergent, 317 exponential function, 23 exponential order, 245 Faulhaber’s formula, 322 Fermat prime, 11 Feynman’s trick, 128, 270 Fourier series, 231 transform, see transform, Fourier Fourier–Laplace method, 303 Fubini’s theorem, 58 function generalized, 247 kernel, 231 functional equation 2 F1 , 302 gamma function, 78 zeta function, 78, 185 fundamental theorem of algebra, 103, 115, 147 gamma function, 55, 78, 81, 140, 173–175, 326 duplication formula, 175, 352 product form, 174 Gamow, George, 8 Gauss constant, 201, 208 Gauss, Carl Friedrich, 11, 200 Gauss–Lucas theorem, 169 Gaussian, 169, 233, 234, 245, 262, 263 generating function, 264, 294, 350 Gibbs’ phenomenon, 53, 269
Index Green’s theorem, 96, 114 Hadamard, quote, x halfline splitting, 244, 274 halfplane splitting product/quotient, 274 sum, 143, 242, 274 Hankel transform, see transform, Hankel Hardy, Godfrey H., 92 harmonic conjugate, 21 functions, 21, 329 sum, 257 Hastings–McLeod solution, 312 Hilbert transform, see transform, Hilbert holomorphic, 20 hypergeometric function, 299–303 2 F1 , 299, 300, 303 p Fq , 300 indicial equation, 299 integral equation Mellin transform, 258 inverse function, 28, 324 Jacobi elliptic functions, 197, 222 theta function, 189 Jensen’s formula, 118, 166 Jordan’s lemma, 125, 142 l’Hôpital’s rule, 121 Lagrange’s inversion formula, 56 Lambert Wfunction, 181–184 Laplace integral, 322, 331 transform, see transform, Laplace Laurent expansion, 38, 106–112, 120, 135, 195, 294 coefficients, 108 level curves, 21 linear fractional transform, see transform, linear fractional Liouville’s theorem, 103, 162, 164, 165, 193, 203, 276 generalized, 103, 284, 289 logarithm function, 28
Maclaurin expansion, 22 magnitude, 5 max/min theorems, 104 mean value theorem, 104 Meissel–Mertens constant, 188 Mellin transform, see transform, Mellin meromorphic function, 40, 101, 192, 204, 258 midpoint rule, 350, 351 Mittag–Leffler expansion, 160, 162–165 Möbius transform, see transform, Möbius modular function, 203 moments, 270 Morera’s theorem, 97, 204 natural boundary, see singularity, natural boundary number systems, 1 ODE, 291 linear, 37, 254, 291, 299, 303 nonlinear, 37, 195, 198, 307 open region, 20 Ostrowski–Hadamard gap theorem, 90 Padé approximation, see analytic continuation Padé table, 86 Painlevé equations, 307–314 property, 307 paradox, 61 Parseval’s relations, 238 partial fractions, 123, 252, 254 PDE, 236, 289 period box, 192 strip, 191 period strip, 163 periodic function doubly periodic, 191 simply periodic, 191 phase angle, 17 phase portrait, 17 Phragmén–Lindelöf theorem, 105 Picard’s theorem, 112, 203 Poisson’s
Index integral formula, 105, 112 summation formula, 241, 266 polar form, 5 rules, 5 pole, 10, 162 simple, 40, 123 polygon, regular, 11 prime number, 11, 12, 178 theorem, 179 principal value integral, 141, 143 product infinite, 160, 174, 179 product rule, 5 Ptolemy’s theorem, 66 quadrant, 8 quaternions, 13 Radon transform, see transform, Radon Ramanujan formula, 84, 140, 351 Srinivasa, 92 Ramanujan’s master theorem, 85, 256, 270 reduction of order, 314 residue, 111 calculus, 119 shortcuts, 120 resonance, 254 Riemann Bernhard, 173 sheet, 42–44, 48, 314 Riemann’s hypothesis, 178, 258 mapping theorem, 214, 223 Riemann–Hilbert, 273 Riemann–Siegel formula, 346 theta function, 180 Riesz function, 258
361 roots of unity, 6, 11 Rouche’s theorem, 144 ruler and compass, 11 saddle point, 330, 336, 341 Schwarz lemma, 214 Schwarz reflection principle, 5, 77, 203 Schwarz–Christoffel, 217, 222, 301 sequence of functions, 51 singularities, 10 singularity, 38–42, 110–112 branch point, 41, 111, 307 cluster, 42 essential, 40, 111 isolated, 38 natural boundary, 42, 73, 189 pole, 40, 110 removable, 38, 110 Sokhotski–Plemelj theorem, 290 steepest descent, 317, 329–346 stereographic projection, 10, 14, 16, 42, 45, 67, 69, 205, 212, 214, 215 Stieltjes function, 87 Stirling’s formula, 256, 326–327, 351 Stokes phenomenon, 319 sums infinite, 148 Taylor series, 21, 22, 105 coefficients, 108 integral test, 59 inversion, 56 limit comparison test, 59 radius of convergence, 21, 33, 105 ratio test, 59 root test, 59 torus, 205 transform
Abel, 265 Fourier, 231–245, 275 generalized, 241 Hankel, 265 Hilbert, 259–262 Laplace, 245–255 inverse, 247–253 linear fractional, 212 Mellin, 255–259 Möbius, 212 Radon, 265 z, 264, 295 transforms, 231 trapezoidal rule, 350 treasure hunt problem, 8 trigonometric function, 23 integral, 126 undetermined coefficients, 35, 56, 115 Vitali’s theorem, 53 Wallis product, 162 Watson’s lemma, 322–327 Weierstrass Mtest, 59 product, 160–162 ℘function, 194–197, 308 Wessel, Caspar, 3 Wiener–Hopf, 273 winding number, 129, 144 ztransform, see transform, z zeta function, 78, 176–181 continuation, 80, 158 critical line, 180 critical strip, 177 functional equation, 185 zeros, 176, 177
At almost all academic institutions worldwide, complex variables and analytic functions are utilized in courses on applied mathematics, physics, engineering, and other related subjects. For most students, formulas alone do not provide a sufficient introduction to this widely taught material, yet illustrations of functions are sparse in current books on the topic. This is the first primary introductory textbook on complex variables and analytic functions to make extensive use of functional illustrations. Aiming to reach undergraduate students entering the world of complex variables and analytic functions, this book • utilizes graphics to visually build on familiar cases and illustrate how these same functions extend beyond the real axis; • covers several important topics that are omitted in nearly all recent texts, including techniques for analytic continuation and discussions of elliptic functions and of Wiener–Hopf methods; and • presents current advances in research, highlighting the subject’s active and fascinating frontier.
The primary audience for this textbook is undergraduate students taking an introductory course on complex variables and analytic functions. It is also geared towards graduate students taking a second semester course on these topics, engineers and physicists who use complex variables in their work, and students and researchers at any level who want a reference book on the subject. Bengt Fornberg has been a professor of applied mathematics at the University of Colorado, Boulder since 1995. His primary research interests focus on computational methods for solving PDEs and numerical methodologies related to analytic functions. He has held positions at the European Organization for Nuclear Research (CERN), California Institute of Technology, and Exxon Corporate Research. In 2014, he was elected a SIAM Fellow.
Cécile Piret is a faculty member in the department of mathematical sciences at Michigan Technological University. A former postdoctoral fellow at Catholic University of Louvain and the National Center for Atmospheric Research (NCAR), her research interests revolve around the development of high order methods for numerically solving differential equations. For more information about SIAM books, journals, conferences, memberships, or activities, contact: Society for Industrial and Applied Mathematics 3600 Market Street, 6th Floor Philadelphia, PA 191042688 USA +12153829800 • Fax +12153867999 siam@siam.org • www.siam.org
OT165 ISBN: 9781611975970 90000
9781611975970