SPECIAL FUNCTIONS - Page maison de Simon Plouffe / Simon Plouffe

SPECIAL FUNCTIONS FOR SCIENTISTS AND ENGINEERS

W. W. BELL I .uturer in Theoretical Physics J),.partment of Natural Philosophy University of Aberdeen

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Library of Congress Catalog Card No. 68-26453

Printrd in Great Britain by Butler & Tanner Ltd, Frome and London

PREFACE

This book is primarily intended for use in undergraduate courses by physics and engineering students in universities and technical colleges. It should, however, also prove useful to those working in other sciences, since it does not contain any specific examples drawn from the fields of physics, engineering, chemistry, etc. We believe it to be preferable to give here the principal results concerning special functions likely to be encountered in applications, and to leave the applications themselves to the particular context in which they arise. The book is designed for those who, although they have to use mathematics, are not mathematicians themselves, and may not even have any great mathematical aptitude or ability. Hence an attempt has been made, especially in the earlier chapters, to sacrifice brevity for completeness of argument. The level of mathematical knowledge required of the reader is no more than that of an elementary calculus course; in particular, no use is made of complex variable theory. Equally, no attempt is made to attain a high level of mathematical rigour; for example, we use results like 1/0 = oo, 1/ oo = 0, which, to be rigorous, should really be written in terms of limits.

In Chapter 1 we discuss the solution of second-order differential equations in terms of power series, a topic of crucial importance for the following chapters. This is because very often in applications we have a partial differential equation for some quantity of interest, and this equation may, by the method of separation of variables, be split into several ordinary differential equations whose solutions we thus require. If, as is often the case, these equations do not have solutions in terms of elementary functions, we have to investigate these solutions, and this is best done by the method of series solutions. These solutions lead to the definitions of new functions, and it is the study of the properties of such 'special' functions which is the main subject-matter of this book.

Chapter 2 is devoted to the gamma and beta functions, two functions defined by integrals and closely related to one another; these functions :arc not only used in later chapters, but are also encountered in many other contt·xts.

( "haptl·r 3 is concerned with a study of the Legendre polynomials.

Vl PREFACE

They are introduced as solutions of Legendre's equation, which vrry often arises when a problem possesses spherical symnll't ry. Sud1 pr uhlrn111 l''"' occur, for example, in quantum mechanics, dt·rtroruagnl'tu· tlu·ory, hydrodynamics and heat conduction.

Chapter 4 is devoted to Bessel functions. Tht·st• fund ion11 otT Ill' in a very wide range of applications, such as loudspeakt'l' tlt·11ign, optu·;~l diffraction, the vibration of circular membranes and plat('s, t h(' !lt',ltlt'riu~o: of sound by circular cylinders, and in general in many prohlt·n•s 111 both classical and quantum physics involving circular or cylindriral hourularit·s of some kind. Kelvin's functions, introduced in Section ~. 11, art· of moMt interest to the electrical engineer, while the spherical Bessd fuuctionM of Section 4.10 are of prime importance in quantum mcchanic;ll st·;attering theory.

The Hermite polynomials, discussed in Chapter 5, have their main application in the quantum-mechanical harmonic oscillator, hut other applications do occur. Likewise, the Laguerre polynomials of Chapter 6 are most useful in the quantum mechanical study of the hydrogen atom, but they also find applications in, for example, transmission lirw theory and seismological investigations.

Chapter 7 treats the Chebyshev polynomials which not only arc of interest for their use in polynomial approximations to arbitrary functions, but also occur in electrical circuit theory.

Chapter 8 is devoted to two generalisations of the Legendre polynomials -the Gegenbauer and Jacobi polynomials. These occur less frequently than the functions mentioned above, but they do have application in various branches of physics and engineering, e.g., transformation of spherical harmonics under co-ordinate rotations.

In Chapter 9 we discuss the hypergeometric function. This is the most general of all the special functions considered. Indeed, all the other special functions considered, and many elementary functions, are just special cases of the hypergeometric function.

Chapter 10 contains a brief summary of various other functions which occur in applications. These are often defined by integrals which cannot be evaluated in terms of known functions, and very often all the useful information concerning these integrals is contained in a table of values of the integral.

Appendix 1 is concerned with the definition of Euler's constant, encountered in the chapter on Bessel functions, and Appendix 2 treats explicitly the convergence of the solutions to Legendre's equation found in Chapter 3. Tlw n·maining appendices summarise certain properties of the special fnnl'tions which h:n•c IH'cn proved throughout the hook.

PREFACE Vll

The problems at the end of each chapter should be considered as an integral part of the text: every reader is urged to attempt most, if not all, of the questions. Hints and solutions are provided for all but the easiest of the problems, but the reader should not seek help from the hints before he has made an effort to solve the problem for himself.

The proofs of certain theorems are indicated as being able to be omitted at a first reading. This does not mean to imply that the proofs of such theorems are difficult, but rather that they are fairly lengthy and that inclusion of such a proof on a first reading would tend to destroy the continuity of presentation of the material.

W. W. B. Aberdeen, 1967

LIST OF SYMBOLS

Symbol Function Section in which symbol first appears

B(x,y) her .. x, bei .. x C(x) C~(x)

Beta function Kelvin's functions Fresnel integral Gegenbauer polynomial or ultraspherical polynomial

Ci(x) Cosine integral D .. (x) Debye functions E(k, ¢) Elliptic integral of second kind E(k) Complete elliptic integral of second kind Ei(x), E1(x) Exponential integrals E .. (x) Generalised error function erf x Error function F(k, ¢) Elliptic integral of first kind .,.F .. (a.l, rJ..2, • · • rJ..m; f3t, fJ2, • • • fJ .. ; x)

Hypergeometric function H .. (x) Hermite polynomial H .. <t>(x), H .. <2>(x) Hankel functions or Bessel functions of the

h .. <1>(x), h,.<2>(x) I,.(x) J .. (x) j .. (x) K(k) K .. (x) ker., x, kei .. x L .. (x) L~(x) li (x) M(a., {3, x) Mk,m(x)

third kind Spherical Hankel functions Modified Bessel function Bessel function of the first kind Spherical Bessel function Complete elliptic integral of first kind Modified Bessel function Kelvin's functions Laguerre polynomial Associated Laguerre polynomial Logarithmic integral Confluent hypergeometric function Whittaker's confluent hypergeometric function Neumann function or Bessel function of the second kind

2.1 4.10

10.3 8.1

10.2 10.5 10.6 10.6 10.2 10.3 10.3 10.6

9.1 5.1 4.5

4.11 4.7 4.1 4.11

10.6 4.7 4.10 6.1 6.7

10.2 9.1 9.3

4.1

X

Symbol

P 1(x) Pnx> P,.<"-.ill(x) Qt(x) S(x) si(x), Si(x) T,.(x) U,.(x) Yj(O, </>) Y,.(x) y.,(x) r(x) r(x, ex), y(x, ex) y C(x) ll(k, <f>, a) ll(k, a) P,.(x)

LIST OF SYMBOLS

Function

Legendre polynomial Associated Legendre function Jacobi polynomial Legendre function of the second kind Fresnel integral Sine integrals Chebyshev polynomial of the first kind Chebyshev polynomial of the second kind Spherical harmonic Bessel function of the second kind Spherical Bessel function Gamma function Incomplete gamma functions Euler's constant Riemann's zeta function Elliptic integral of third kind Complete elliptic integral of third kind Weber-Hermite function

Secliotl i11 which symbol first appears

3.1 3.X X.2 3.10

10.3 10.2 7.1 7.1 3.11 4.1 4.11 2.1

HI. I 4.1

10.4 10.6 I 0.6 5.7

CONTENTS

PREFACE v

LIST OF SYMBOLS IX

CHAPTER 1 SERIES SoLUTION oF DIFFERENTIAL EQUATIONS

1.1 Method of Frobenius 1 1.2 Examples 8

Problems 21

CHAPTER 2 GAMMA AND BETA FUNCTIONS

2.1 Definitions 23 2.2 Properties of the beta and gamma functions 24 2.3 Definition of the gamma function for negative values of the

argument 30 2.4 Examples 37

Problems 40

CHAPTER 3 LEGENDRE PoLYNOMIALS AND FuNcTIONS

3.1 Legendre's equation and its solution 42 3.2 Generating function for the Legendre polynomials 46 3.3 Further expressions for the Legendre polynomials 48 3.4 Explicit expressions for and special values of the Legendre

polynomials 50 3.5 Orthogonality properties of the Legendre polynomials 52 3.6 Legendre series 55 3.7 Relations between the Legendre polynomials and their deriva-

tives; recurrence relations 58 3.8 Associated Legendre functions 62 3.9 Properties of the associated Legendre functions 65 3.10 Legendre functions of the second kind 70 3.11 Spherical harmonics 78 3.12 Graphs of the Legendre functions 82 3.13 Examples 85

Problems 90

xu CONTENTS

CHAPTER 4 BESSEL FUNCTIONS

4.1 Bessel's equation and its solutions; Bessel function~ of thL· first and second kind 92

4.2 Generating function for the Bessel functions 99 4.3 Integral representations for Bessel functions 101 4.4 Recurrence relations 104 4.5 Hankel functions 107 4.6 Equations reducible to Bessel's equation 108 4.7 Modified Bessel functions 110 4.8 Recurrence relations for the modified Bessel functions 113 4.9 Integral representations for the modified Bessel functions 116 4.10 Kelvin's functions 120 4.11 Spherical Bessel functions 121 4.12 Behaviour of the Bessel functions for large and small values of

the argument 127 4.13 Graphs of the Bessel functions 133 4.14 Orthonormality of the Bessel functions; Bessel series 137 4.15 Integrals involving Bessel functions 141 4.16 Examples 148

Problems 154

CHAPTER 5 HERMITE PoLYNOMIALs

5.1 Hermite's equation and its solution 156 5.2 Generating function 157 5.3 Other expressions for the Hermite polynomials 158 5.4 Explicit expressions for, and special values of, the Hermite

polynomials 160 5.5 Orthogonality properties of the Hermite polynomials 161 5.6 Relations between Hermite polynomials and their derivatives;

recurrence relations 162 5.7 Weber-Hermite functions 163 5.8 Examples 164

Problems 166

CHAPTER 6 LAGUERRE PoLYNOMIALS

6.1 Laguerre's equation and its solution 168 6.2 Generating function 169 6.3 Alternative expression for the Laguerre polynomials 170 6.4 Explicit expressions for, and special values of, the Laguerre

polynomials 171

CONTENTS Xlll

6.5 Orthogonality properties of the Laguerre polynomials 172 6.6 Relations between Laguerre polynomials and their derivatives;

recurrence relations 173 6. 7 Associated Laguerre polynomials 176 6.8 Properties of the associated Laguerre polynomials 177 6.9 Notation 182 6.10 Examples 182

Problems 185

CHAPTER 7 CHEBYSHEV PoLYNOMIALs

7.1 Definition of Chebyshev polynomials; Chebyshev's equation 187 7.2 Generating function 190 7.3 Orthogonality properties 192 7.4 Recurrence relations 193 7.5 Examples 194

Problems 196

CHAPTER 8 GEGENBAUER AND JAcOBI PoLYNOMIALS

8.1 Gegenbauer polynomials 197 8.2 Jacobi polynomials 198 8.3 Examples 200

Problems 201

CHAPTER 9 HYPERGEOMETRIC FUNCTIONS

9.1 Definition of hypergeometric functions 9.2 Properties of the hypergeometric function 9.3 Properties of the confluent hypergeometric function 9.4 Examples

Problems

CHAPTER 10 OTHER SPECIAL FUNCTIONS

10.1 Incomplete gamma functions 10.2 Exponential integral and related functions 10.3 The error function and related functions 10.4 Riemann's zeta function 10.5 Dcbye functions 10.6 Elliptic integrals 10.7 Examples

Problems

203 207 210 212 216

218 218 221 223 224 224 225 228

CONTENTS

APPENDICES

1 Convergence of Legendre series 2 Euler's constant 3 Differential equations 4 Orthogonality relations 5 Generating functions

HrNTS AND SoLUTIONS To PROBLEMs

BIBLIOGRAPHY

INDEX

230 231 233 234 236

2J7

243 245

l

SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

1.1 METHOD OF FROBENIUS

Many special functions arise in the consideration of the solutions of equations of the form

d 9y dy P(x) dxll + Q(x)dx + R(x)y = 0. (1.1)

We shall restrict ourselves to equations of the type

d 2y dy xa dxa + xq(x)dx + r(x)y = 0, (1.2)

where q(x) and r(x) may be expanded as power series in x, 00

q(x) = Lqmxm (1.3) m=O

00

r(x) = L r mXm, (1.4) m=O

convergent for some range of x including the point x = 0. The basis of Frobenius' method is to try for a solution of equation (1.2)

of the form 00

z(x,s) = x•(a0 -/- a1x + a2x2 + ... + a,.xn + ... ) = L a,.x•+n (1.5) n-o

with a 0 ./. 0. (We may always take a0 / 0, since otherwise we should just

2 SEHJES SOLUTION OF lliFI'EHENTIAI. EQUATIONS CH. 1

han~ another series of the type ( 1.5) with a different value of s, the first wdlicient of which we could now call au.)

From equation (l.S) we have

dz "' L a,.(s 11 )x• 1" 1

u-.o

and d 2z ~ ... d~ii ,_, L a .. (s I n)(s I n

,.. 0

1 )x• 111 2,

so that if we require z to satisfy equation ( 1.2) we must have d 2z dz

x 2 dx~ ·!- xq(x)dx + r(x)z = 0,

and this reduces to «> co

L a,.(s + n)(s + n 1)x•+n + q(x) L a,.(s + n) x''" ,. ~o n=O

<X>

+ r(x) L a,.xs+n = 0 n~o

which, on using equations (1.3) and (1.4) for q(x) and r(x) and cancelling a common factor of x•, becomes

00 00 «>

L a,.(s + n)(s + n - 1)xn + L L a,.qm(s + n)x"+"' n~o n=·O m=O

00 00

+ L L a,.rmx" tm = 0. (1.6) n=O m=O

For the infinite series on the left-hand side of equation (1.6) to be zero for all values of x in some range, we must have the coefficient of each power of x equal to zero. This requirement gives rise to a set of equations as follows.

Requiring that the coefficient of x 0 be zero, and noting that x 0 arises only from the choice m = 0, n = 0, gives

a0s(s- 1) + a0q0s + a0T 0 = 0. (1.7) Requiring that the coefficient of x1 be zero, and noting that x1 arises in

equation ( 1.6) by choosing in the first term n = 1, and in the second and third terms n + m = 1 (i.e., m = 0, n = 1 or m = 1, n = 0) gives

aJ(.f I l)s + {a 1q0(s + 1) + a0q1s} + {a1r0 + a0r 1} = 0. (1.8)

\Ve may now write down a general equation from the requirement that the cocflicicnt of xi should he zero. Remembering that in equation (1.6) xt arises in the first term by choosing n = i and in the second and third terms

§ 1.1 METHOD OF FROBENIUS 3

by choosing n + m = i (i.e., n = i, m = 0 or n = i - 1, m = 1 or n = i- 2, m = 2, etc., upton = 0, m = i) we obtain

a;(s + i)(s + i- 1) -t-{a;q0(s + i) + a;_1q1(s + i- 1) + a;_2q2(s + i- 2) + ... + a0q.s} + {a;r0 + a1_ 1r 1 + ... + a0r1} = 0 (i > 1). (1.9)

Thus

aJ(s + i) + {a;_1q1(s + i- 1) + a;_2q2(s + i- 2) + ... + a0q;s} +{ai-lrl + ai-2r2 + ... + aor;} = 0 (i > 1) (1.10)

. where we have collected all the terms in a; together and denoted their coefficient by

f(s + i) == (s + i)(s + i- 1) + q0(s -f- i) + r 0• (1.11) We now see that equation ( 1. 7) reduces to

a0{s 2 + (q0 - l)(s + r0)} = 0 which, with the given assumption that a0 ¥-0, becomes

s2 + (q0 - 1}(s + r0) = 0. (1.12) This is called the indicial equation; it is quadratic in s and hence will yield two roots which we shall denote by s1 and s2• In many cases of interest these roots will be real; this we shall henceforth assume, together with the fact that s2 > s1•

Noting that equation (1.12) is justf(s) = 0, we have immediately f(s) = (s - s1)(s - s2). (1.13)

We might now expect that these two values of s would lead to the two independent solutions of the original differential equation. We shall see that this is true apart from certain exceptional cases, viz. when

(a) the two roots of the indicia! equation are equal; or (b) the two roots of the indicia! equation differ by an integer.

The set of equations (1.10) can now be used to determine the coefficients a1, a 2, ••• in terms of a0•

Equation (1.10) with i = 1 gives

so that we have alf(s + 1) + a~ls + aorl = 0

-ao(qls + rt) at = f(s + 1}

a0h1(s) ------f(s + 1)

where h,(s) - · (q,s 1 r 1) is a polynomial of first degree ins. lii'-U

(1.14)

4 SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

Equation (1.10) with i = 2 gives

aJ(s + 2) + {a1q1(s + 1) + a 0q2s} + {a1r1 + a 0r 2} = 0, which, when we use equation (1.14) for al> becomes

CH. 1

aJ(s + 2) + t~:~s~q1(s + 1) + a0q2s} + ~~so~~s?{t + a0r2} = 0,

yielding -a0{q1(s + l)h1(s) + q2sj(s + 1) + r1h1(s) + rd(s + 1)}

a2 = f(s + l)f(s + 2)

h2(s) = ao j(s + l)f(s + 2) (l.lS)

where h2(s) = -{q1(s + 1)h1(s) + q2sj(s + 1) + r1h1(s) + rd(s + 1)} is a polynomial in s.

In general it is not difficult to see that we obtain

hi(s) ai = ao f(s + 1)f(s + 2) .. . f(s + i) (1.16)

where hi(s) is a polynomial ins. If in equation ( 1.16) we now substitute s = s1 we shall obtain expressions

for a, in terms of a0, leading to a solution with a0 as an arbitrary multiplicative constant; similarly we may substitute s = s2, so that we have the two solutions required. This will work, however, only if the denominator of equation (1.16) is never zero, i.e., provided

f(s1 + i) ~0 and f(s2 + i) ~ 0 for i equal to any positive integer. (1.17)

Now, equation (1.13) tells us that

so that

and hence

and

f(s) = (s - s1)(s - s2}

f(s + i) = (s + i - s1)(s + i - s2)

f(s1 + i) = i(s1 - s2 + i) f(s2 + i) = (s2 - s1 + i)i, (1.18)

so that equation (1.17) will be satisfied if, and only if, s2 - s1 is not a positive integer, i.e., provided the roots of the indicia! equation do not differ by an integer. (We recall that we have labelled the roots so that s2 > st.)

If the roots do differ by an integer, the procedure given above will work for the larger root: if s2 - s1 = r (a positive integer), then from equations (1.18) we have

§1.1 METHOD OF FROBENIUS

f(s1 + i) = i( -r + i)

j(s2 + i) = i(r + i),

5

(1.19)

(1.20)

and we see that f(s1 + i) = 0 for i = r, while f(s 2 + i) =T' 0 for any positive integral i. Hence the procedure will work with s = s2 but not with s = s1• How, in fact, do we find the second solution? If we use the relationship (1.16) before fixing the value of s we have the series

_ ·{ h1(s) h2(s) 2 .:r(x, s) - Or]X

1 + j(s + 1t + j(s + 1)j(s + 2t

h;(s) i }

+ · · · + f(s + 1)f(s + 2) .. . f(s +it + · · · · (1.21)

W c know by the method of construction of a1 , a2, ••• that if all the terms on the right-hand side are well defined, then this must satisfy

d 2z dz x2dx 2 + xq(x)dx + r (x)z = aof(s)x•

(Rince equations (1.10) are satisfied, the coefficient of each power xsti with i > 1 must vanish, and the coefficient of x• is, by equations (1.7) and tl.ll) just a0f(s)). This equation may be rewritten in the form

{ d2 d }

X2 dx 2 + xq(x) dx + r(x) z(x, s) = a0f(s)x•. (1.22)

Unfortunately, all the terms on the right-hand side of equation (1.21) are not well defined for s = s 1 ; there are zeros in the denominators for i;... r. However, if we multiply z(x, s) by f(s + r) we shall just cancel the factor in the denominators of the later terms which vanishes when s = s1•

And since f(s + r) = (s + r - s1)(s + r - s2) = (s + s2 - 2s1)(s - s1)

we might just as well multiply by ( s - s1). Also, since this factor is independent of x, equation ( 1.22) now takes the form

{ d2 d }

x 2 dx 2 + xq(x)dx + r(x) (s- s1)z(x, s) = a0(s - s1)f(s)x•

= a0(s - s1) 2(s - s2)x•, (1.23)

using equation (1.13).

Setting s = s2, we have

{ x2 ::2 + xq(x)! + r(x) }(s2 - s1)z(x, s2) = 0

which shows that z(x, s2) is a solution, a fact we already know. Himilarly, settings-- s1 gives [(s- s1)z(x, s)] •.• , as a solution. In fact

6 SERIES SOLUTION OF DIFFERENTIAL EQUATIONS CH. 1

it turns out that this series is not independent of z(x, s2) but is just a multiple of it. This comes about as follows. The factors -- s1 cancels zeros in the denominator for terms with i > r, but will provide zero in the numerator for all terms with i < r, thus the first power in the series is just x••+r = xS., which is just the first term in z(x, s2). And since we use the same rules for calculating any coefficient in terms of preceding ones in the two cases, we must just obtain the same series in both cases, apart from a constant multiplicative factor.

If we differentiate both sides of equation (1.23) with respect to s, we obtain

{ x 2::2 + xq(x) ddx + r(x)} [:s {(s- s1)z(x, s)}]

= a0 [(s- s1)2 :s{(s- s2)x•} + 2(s- s1)(s- s2)x•] (1.24)

and we see that the right-hand side of this equation is zero when s = s1, so that we must have [( djds){(s - s1)z(x, s)}]•=•· as a solution ofthe differential equation. It may be proved that this solution is in fact independent of the first solution (i.e., it is not merely a constant multiple of the first solution), but we shall not do so here.

Thus we can take as independent solutions

[(s - s1)z(x, s)]•=•, (1.25)

and [1s {(s - s1)z(x, s)} 1=.,· (1.26)

Another type of situation may arise when the roots of the indicia! equation differ by an integer. As well as f(s + i) = 0 for s = s1 and i = r, it may happen that h.(s1) = 0. In this case ar is indeterminate (since it has a zero in both numerator and denominator) so that we may in fact use it as another arbitrary constant, thus obtaining the two independent solutions from the one series. We shall see in detail how this happens when we come to consider particular examples.

The one remaining case is when the indicia} equation has equal roots, says= s1 • Heref(s) = (s- s1)2, and if we make use of equation (1.22) we obtain

{ d2 d } x 2 - + xq(x)-d + r(x) z(x, s) = a0(s - s1)

2x'. dx 2 x

(1.27)

Setting s = s1 makes the right-hand side zero, so that z(x, s1) is a solution. nut if we differentiate both sides with respect to s, we obtain

§1.1 METHOD OF FROBENIUS 7

{x2 ~ + xq(x)~ + r(x)} [iz(x, s)J dx 2 dx ~

= ao{2(s- s1)x' + (s- s1)2 :/} (1.28)

Again the right-hand side is zero when s = s1, so that we have for a solution [( djds){z(x, s)}]s=s,· This, in fact, may be shown to be independent from z(x, s1), so that in this case we have the two independent solutions

z(x, s1)

and [!z(x, s)l=.,· (1.29)

Let us now summarise briefly the methods of obtaining independent solutions which have been described above.

Form the series z(x, s) by the use of the relations (1.10). Then there are four distinct cases.

( 1) If the roots s1 and s2 of the indicia! equation are distinct and do not differ by an integer, then the two independent solutions are given by

z(x, s1)

and z(x, s2).

(2) If the roots s1 and s2 (s2 > s1) differ by an integer and one of the coefficients in the series for z(x, s) is infinite when s = s1, the two independent solutions are given by

[(s - s1)z(x, s)]s=s,

and [!{(s - s1)z(x, s)}l=.,·

(3) If the roots s1 and s2 (s2 > s1} differ by an integer and one of the coefficients in the series for z(x, s), say a" is indeterminate when s = sl> the two independent solutions are obtained from z(x, s1}, keeping a0 and a, as arbitrary constants.

(4) If the roots of the indicia! equation are equal, say, s = s1, then the two independent solutions are given by

z(x, s1)

and [!z(x, s)l=.,·

The question of convergence of the series obtained must, of course, be considered. It may be proved (although we shall not do so here) that the range of values of x for which the solution is convergent is at least as large


as the range of values of x for which both q(x) and r(x) may be expanded as convergent power series.

In certain cases q(x) or r(x) may not be expanded as power series in x for any range of x (e.g., 1/x or cxp (1/x)). It may perhaps then be possible to make a change of variable, such as x' = x - a, or x' =-= 1 jx, so that q(x) and r(x) may be expanded in powers of x' and a solution found in the form (1.5) with x' replacing x.

Finally, let us remark that when carrying out the actual solution of a differential equation of type (1.1 ), it is not usually profitable to transform it to an equation of the type (1.2); all that we require for the above results to hold is that this transformation should be possible.

Let us now illustrate the above methods by means of some examples.

1.2 EXAMPLES

Example 1

d 2y dy 2xdx2 + dx + Y = o.

We must first verify that this equation is of the form

x2~Fy + xq(x) dy + r(x)y = 0 dx 2 dx

where q(x) and r(x) may be expanded as power series. Multiplied by xj2, the given equation becomes

2 d 2y 1 dy 1 -x dx2 + x.Z.dx + .zx.y- 0

so that q(x) = L r(x) = ix and the required condition is obviously satisfied.

Since q(x) and r(x) are already in power series form, valid for all values of x, it follows from the remarks made at the end of the previous section that any series solutions obtained will be convergent for all values of x.

We now revert to the original equation.

Set

dz ~ dx = :L: an(s + n)xs+n-1

!i~O

so that

and d2z ro -···- · - L a (s + n)(r -t- n - l)x·•+n- 2

d., 11. • •

x-n o

§ 1.2

Then we have

d 2z dz 2x-+-+z

dx 2 dx

EXAMPLES

00 00

= L 2a,.(s + n)(s + n - 1)xs+n- 1 + L an(s + n)xstn-1 1t=O n=O

00

+ 2: anx•+n n=O

00 00

= L 2an(s + n)(s + n -- l)xHn-I + L an(s + n)xstn-1

n=O n=O 00

+ ~ a x•+n-1 L_. n-1 1 n=l

9

(1.30)

where in the last summation we have replaced the label n by the label n - 1 in order that the general power should look the same in all terms.

In order that z should satisfy the differential equation, we must have the coefficient of each power of x in (1.30) equal to zero. The power with n = 0 appears only in the first two terms; thereafter the powers with n > 1 appear in all three terms. Hence we have the equations:

2a0(s + O)(s + 0 - 1) + a0(s + 0) = 0 (1.31) and

2an(s + n)(s + n- 1) + an(s + n) + an_1 = 0 (n > 1). These equations simplify to

aoS(2s- 1) = 0 and

an(s + n){2(s + n) - 1} + an_1 = 0.

Equation (1.33) gives the indicia! equation

s(2s- 1) = 0

(1.32)

(1.33)

(1.34)

with roots s = 0 and s = l· Since these are distinct and do not differ by an integer, we know that we may proceed according to prescription (1) on page 7.

Equation (1.34) gives the recurrence relation for the coefficients in z(x, s):

and hence

On-1 On= --

(s + n){2(s + n)- 1}

( s + 1 )(2s j 1 )'

(1.35)


al ao 02

=- (s + 2)(2s + 3) = (~:-f)(s +2)(2s +-i)(Zs +j)' a2 ao

03 =- (s + 3)(2s + 5) =- (s + 1)(s + 2)(s -f 3)(2s-~-:-1)(2s- ~ 3)(2s + 5)

and in general ao

an= ( - 1)n(s + 1)(s + 2) ... (s + n)(2s + 1)(2s +-3)~:~-(Zs + 2n - 1)"

Thus we have the series z(x, s) given by

z(x,s) = aoX"{1 - (s+1)~2s+1) + (s+1)(s+2)~;s+1)(2s+3) + · · ·

+ (-l)n (s+1)(s+2) ... (s+n)(2s~~1)(2s+3) ... (2s+2n-1) + · ·} (1.36)

Since the roots of the indicia! equation are s = 0 and s = -!, the two independent solutions are given by z(x, 0) and z(x, !) which, by equation (1.36), are

{ x x 2 x 3

z(x, 0) = a0 1 - 1]_" + 1. 2 .1. 3 - ~1.3 .5 + ...

+ ( - 1)" 1.2.3 ... n.l.;n.5 ... (2n- 1) + · · .} (1.37)

and z(x, l) = aoX!-{1 -- ~~2 + ~ i~ 4 2 2·2· · + · · ·

+ { --1 )n 3 5 2n + ~n + · · ·}· 2.2 ... - 2

-- .2.4.6 ... 2n (1.38)

We may rewrite these series more compactly if we note the following results:

1.2.3 .... n=n!, (1.39)

1. 3 . 5 ... (2n _ 1) = 1. 2. 3 .4 ... (2n - 1). 2n 2.4 ... (2n - 2).2n

(2n)! 2 . 1. 2 . 2 . 2. 3 ... 2( n - 1) . 2n

(2n)! = zn,;!' (1.40)

§ 1.2 EXAMPLES 11

3 5 7 2n + 1 1 z·z·z ... -2-- = z;;3.5 .7 ... (2n + 1)

1 2 . 3 .4. 5 . 6. 7 ... (2n + 1) 2n 2.4.6 .... 2n

1 (2n + 1)! = (2n_±_l_lJ (1.4-l) zn zn . n ! 2 2nn ! '

and 2 .4 . 6 . . . 2n = znn !. (1.42)

Using these results we see that equation (1.37) reduces to oo n

z(x, 0) = Oo 6 ( -J)n n!{(2n)~/(2nn!)} ~ (2x)n

= ao f:o (--1 )n (2n) l (1.43)

while equation (1.38) becomes <1J

z(x, i) = aoxl/2 6 (-1)n {(2n + 1)!i7z2nn!)}2nnf

- 1/2 ~ (-l)n (2x)n -- OoX ~ (2n + 1)!' (1.44)

Thus the general solution is given by the general linear combination of z(x, 0) and z(x, !), viz.

Y =A ~ ( -l)n (2x)n + Bxl/2 ~ ( -1)n (2x)n 6 (2n)! ~0 (2n + 1)!

where A and B are arbitrary constants.

Example 2

x(1 - x) dd~~ + (1 - x)ddy - y = 0. x- x

We first verify that this equation is of the form

x 2 dd ~ + xq(x) ddy + r(x)y = 0 X X

where q(x) and r(x) may be expanded as power series in x. The given equation is obviously equivalent to

9d 2y dy X

x- dx 2 -1 \i.~ -- 1 xy oc 0.


Hence q(x) = 1 and r(x) = -x/(1 - x). q(x) is already in power series form, convergent for all values of x, but r(x) is not. However, it may be expanded as a power series in x by the binomial theorem, and this series will be convergent for those values of x such that I x I < 1.

It follows from the remarks at the end of the previous section that any series solutions which we obtain will be convergent for at least the same values of x, viz. I x I < 1.

We now revert to the original equation. 00

Set z = L a,.x•+n n=O

d 00

__!. = "" a,.(s + n)xs+n-1

dx ~ n=O so that

d 2z ~ dx

2 = ~ a,.(s + n)(s + n - I)xnn- 2•

tt=O

and

Hence,

d2z dz (x-x 2)- + (1-x)-- z

dx2 dx 00 00

= L a,.(s+n)(s+n-1)xs+n-t- L a,.(s+n)(s+n-l)xs+n n=O n=O

00 00 00

+ L a,.(s+n)xs+n-1 - L a,.(s-t-n)xs+n-L a,.xs+n n=O n=O n=O

00 00

= L a .. (s+n)(s+n-l)xs+n- 1 - L a,._1(s+n-1)(s+n-2)xs+n-1 n=O n=l

00 00 00

+ L a,.(s+n)xs+n-1 - L a,._1(s+n-1)xs+n-l - L a,._1xs+n- 1•

n=O n=l n=l

For z to be a solution, we require the coefficient of each power of x to be zero, so that we have

aoS(s - 1) + a0s = 0 (1.45) and

a .. (s + n)(s + n - 1) - a,._1(s + n - 1)(s + n- 2) + a,.(s + n) -a,._1(s + n - 1) - a .. _1 = 0 (n > 1). (1.46)

These equations simplify to

(1.47)

§ 1.2

and

EXAMPLES

Eqqation (1.47) gives the indicia! equation

s2 =0 while equation (1.48) gives the recurrence relation

{(s + n - 1) 2 + 1} a" = a,._l (s + n)2 '

13

(1.48)

(1.49)

(1.50)

Equation (1.49) has coincident roots s = 0; so we know (by prescription (4) on page 7) that the two independent solutions are given by z(x, 0) and [( djds)z(x, s)],=o· Equation (1.50) gives

(s2 + 1) al = ao(s + 1)z'

{(s + 1)2 + 1} {s2 + 1}{(s + 1)2 + 1} a2 = al -~ 2)2 -- = ao (s + 1)2(s + 2)2 '

and in general

{s 2 + 1} {(s + 1) 2 + 1} {(s + 2) 2 + 1},,, {(s + n- 1) 2 + 1} a.,. = ao ... --'---(--"s-'+-'--1 )-2-( s.;__+_2_) z--'-.-.-• ....,.( s...:..:+_n_)_2 _ __.____

so that

( ) - ·[1 - ~ {s 2 +1} {(s-1-1)2+1} ... {(s+n-1) 2 +1} n] z x, s - aox f- ,6_ (s+ 1 )2(s+2)2 , , , (s+n)Z x .

(1.51) The first solution is then

""' ) _ [t ~ 1{12+1}{22+1} {32+1} ... {(n-1)2+1} J _,x,O -ao + .L. p zz 32 2 X"

I . . • • • n

n=

_ [t ~ 1.2.5.10.17 ... {(n-1)2+1} n] - ao + .L. (n!)2 x

n-1

= a0y1(x), say. (1.52) For the second solution we require (djds)z(x, s). Now, from equation

(1.51) we have

d d/(x, s)

=a (!!x-)[t + ~ {s2+l}{(s+1)2+1} ... {(s+n-l)2+1}xn] 0 dS ,~ (s+1)2(s+2)2 ... (s+n)2

I a x-• ~ _d_[J~~--l_-~}_{!_s_f-_ll2+1} ... {(s-\-n-1)2+1}] xn. tr .~ ds (s J--1 )2(H 2) 2 ••• (s t-n)2 (1.

53)

SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

For the first term we note that

d d d -x• = -(eln xy = -e• In X t ds ds ds

= (In x) e' In x = (In x)x•

and hence [ix•] = lnx. ds s~o

en. 1

(1.54)

(1.55)

The second term we evaluate by the technique of logarithmic differentiation.

If we denote the term in square brackets by fn(s) we have

d 1 d d; lnfn(s) = fn(s) d/n(s)

d d SO that dfn(s) = fn(s) -d lnj,(s). (1.56)

s s

But

Infn(s) =In {s2 + 1} +In {(s + 1) 2 + 1} + ... +In {(s + n -1) 2 + 1} -In (s + 1) 2 -In (s + 2) 2

••• -In (s + n) 2

Hence

so that

d d/n(s)

n n

= ,2 In { ( s + m - 1) 2 + 1} - ,2 In ( s + m) 2 •

m~l m=l

i lnfn(s) = ~ ~~_m__=-_!) __ - ~ -~-ds f;;

1 ( s + m - 1) 2 + 1 f:-;1 s + m

~{ s+m-1 1} =

2 L... (s + m - 1 ) 2 + 1 - s + m m~l

2{s 2 + 1} ... {(s + n- 1) 2 + 1] ~ {' s + m --- 1 __ 1 __ ~ (s + 1) 2

••• (s + n) 2 ,f;:1

(s + m -- 1) 2 + 1 s -t- mJ

and

t We recall that x - c 1" •, wherl' we arc using the notation In x ·loge x.

§ 1.2 EXAMPLES

Denoting this expression by c.., we have

_2.1.2.5.10.17 ... {(n-1) 2 +l}Ln m-2 Cn - -----------"--'---'----~

(nl) 2 m=l m{(m- 1) 2 + 1}"

Thus, setting s = 0 in equation (1.53) gives the second solution

[:/(x,s)l=o _ I [t ~ {1}{12 + 1}{2 2 + 1} ... {(n- 1)2 + l}xn] - ao n x + ;S_ 12.22 ... n2

00

= a0y 1(x) In X + a0 L CnXn

= a0y 2(x), say,

where

n=l

00

y 2(x) = y 1(x) In x + L cnxn. n=l

Then the general solution is given by

y = Ay1(x) + By2(x) where A and B are arbitrary constants.

Example 3 d2y dy

x2 - - 3x- + (3 - x)y = 0. dx2 dx

15

(1.57)

(1.58)

We see immediately that this is of the form (1.2) with q(x) = -3, r(.t) = 3 - x, which are already in power series form, so that they have a power series expansion valid for all values of x.

It follows from the remarks at the end of the previous section that any series solutions obtained will be convergent for all values of x.

00

Writing z = 2: a.,xs+n leads in the same manner as in the previous n=O

examples to the following equations, which have to be satisfied if z is to be a solution of the given equation:

a0(s - l)(s- 3) = 0

a,.(s I n 1 )(s I 11 3) a,. 1 · · 0 (n ·. ' 1 ).

(1.59) (1.60)

16 SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

Equation (1.59) gives the indicial equation (s- 1)(s - 3) = 0

CH. 1

(1.61) with roots s = 1, s = 3. These differ by an integer, so that we know that we are dealing with an exceptional case.

Equation (1.60) gives the recurrence relation

so that we have

a,.=· ----~=.1 - --- (n > 1)

(s 1- n - l)(s + n - 3)

al ao az = -·---··- = -·--------(s + 1)(s - 1) s(s + 1)(s - 2)(s- 1)

and in general ao

(1.62)

a - . ---- ·-·· -~·--n- s(s + l)(s + 2) ... (s + n- l)(s- 2)(s- 1)s ... (s + n - 3)'

(1.63) We see explicitly from equation (1.63) that when s = 1 all the an with

n > 2 are infinite, so that we have to apply method (2) described above on page 7.

We have

z(x, s) = aoX•{ 1 + ~1 s(s+1) ... (s+n-1)(;~2)(s-1) ... (s+n -3)} so that

(s - 1)tr(x, s) = aoX•{(s - 1) + s~s =it ~ xn }

+ £=2

s(s + 1) ... (s + n- 1)(s - 2)s(s + 1) ... (s + n - 3) ·

The first solution is given by

[(s - 1 )z(x, s)]s~I

= aox ~ --:----:--:-x_"'c-::----:----::-:-,:S 1.2 ... n( -1).1.2 ... (n - 2)

oo xn+l

= ao L - ~!(n- 2)! n~2

= a0y 1(x), say.

The second solution is given by [(d/ds){(s -·- 1)z(x, s)}].=l·

(1.64)

§ 1.2

From equation (1.64) we have d ds{(s - l)z(x, s)}

EXAMPLES

(d ) { (s- 1)

= ao d/s (s- 1) + s(s -- zt ~ xn }

+ n~ s(s + 1) ... (s + n - 1)(s- 2)s(s + 1) ... (s + n - 3)

s{ d (s -- 1) + aoX 1 + ds s(s - 2t

+~~- . 00

d 1 } n~ ds s(s + 1) ... (s + n- 1)(s- 2)s(s + 1) ... (s + n- 3)

17

(1.65) To evaluate the last derivatives we use logarithmic differentiation:

denoting 1

s(s + 1) ... (s + n - 1 )(s - 2)s(s + 1) ... (s + n - 3) by fn(s), we have again

and since lnfn(s) = -Ins -In (s + 1) -In (s + 2) ... -In (s + n - 1)

-In (s- 2) - ln s-In (s + 1) ... -In (s + n- 3)


and 1

[fn(s)]s=l = - ( ) ( 1 . 2 . . .. n . -1 . 1 . 2 . . . n - 2)

n!(n - 2)!

so that

[ d J 1 { n-

2 1 1 1} -- f s = 2 - - 1 __)_. -- + -

ds n() •=1 n! (n-2)! J1

m ' n - 1 n

= Cm say.

It is also easy to show that

[~(s-1)] __ 1 ds s(s - 2) s=l -

and, as before, we have d d/' =(In x)x•.

Thus, finally, we have from equation (1.65)

[£{(s -- l)z(x, s)}l=l = ao(ln x)y1(x) + a0x{l - x + i cnxn} n=2

= a0y 2(x), say,

and the general solution is then

y = Ay1(x) + By2(x) where A and B are arbitrary constants.

Example 4

d 2y dy xzdxz + (x3 - 2x)d_;- 2y = 0.

Again this is of the form (1.2) with q(x) = -2 + x 2 and r(x) = -2, so that the series solutions to be obtained will be valid for all values of x.

00

Writing z = x• 2 anxn and requiring z to be a solution of the given n=O

equation leads, as before, to the system of equations:

an(s + n - 2)(s + n

a0(s - 2)(s - 1) = 0

a1(s-1)s=0

1) +a,.. 2(s t n 2) = 0 (n > 2).

(1.66)

(1.67)

(1.68)

§ 1.2 EXAMPLES 19


(s - 2)(s- 1) = 0 with roots s = 2, s = 1. These differ by an integer, so we are dealing with an exceptional case. When s = 1, equation (1.67) is satisfied irrespective of the value of a 1 ; i.e., a 1 is indeterminate, so that we are dealing with case (3) of page 7. We then know that the two independent solutions are obtained from the one value of s, namely s = 1.

Equation (1.68) gives the recurrence relation

an-2 ( ....., 2) an = - ( 1) n "'-"' s+n-

(1.69)

(the factors + n - 2 may be cancelled since it is non-zero for n > 2 and s = 1 or s = 2), which with s = 1 becomes

Thus

and in general

and

so that we have

an-2 an= ---~ (n > 2).

n

ao a2n = (-1)n 2.4.6 ... 2n

at a2n+1 = ( -J)n 3.5.7 ... (2n + 1)'

z(x, 1) = x[ao{l + %1

( -1)n 2 .4~2

~. 2n}

(1.70)

~ X2n+1 J -1-- al L ( -1)n

n=O 1.3.5 ... (2n + 1) •

That is, we have the general solution given by y = Ay1(x) + By2(x)

where ~ x2n+l

Yt(x) = x + ,6 ( -t)n 2.4.6 ... 2n'

~ x2n+2 V2(x) = L ( -1 )n -----:-:::-----:-: ~ n~o 1.3.5 ... (2n+l)

and A and n arc arbitrary constants. ~,J•-('


In the examples given so far, it has always been relatively easy to write down the general term of any series appearing in the solutions. This is by no means always so; it may be impossible for us to find in any simple manner an expression for the general coefficient from the recurrence relation. This is so, in general, when the recurrence relation contains three or more terms instead of the two they have had up till now. All that we can do in such a situation is to give the first few terms of the power series solution, and hope that this will be of some use. We illustrate by the following example.

Example 5

d 2y dy (x2 - 1 )- + 3x-- + xy = 0.

dx 2 dx

Rewriting this in the form

d 2y 3x2 dy x3

x2d2 + x.-2--l-d + --2 -ly = 0, X X - X X --

we see that it is of the form (1.2) with q(x) = (3x2)/(x2 - 1) and r(x) = x3 j(x 2 - 1). Both q(x) and r(x) may be expanded as power series in x (by the binomial theorem) convergent for I xI < 1. Hence any series solutions we obtain will be convergent for at least this range of values of x.

We now revert to the original equation, and trying for a solution of the form

leads, in the same manner as in previous examples, to the set of equations

a0s(s - 1) = 0 (1.71)

a1(s + 1)s = 0 (1.72) aoS(s + 2) - a2(s + 2)(s + 1) = 0 (1.73)

a,._ 3 + a,._ 2 (s + n- 2)- a,.(s + n)(s + n- 1) = 0 (n > 3). (1.74)


s(s - 1) = 0

with roots s = 0 and s = 1. Equation (1.72) then shows that a1 is indeterminate if s = 0, so that we are dealing with case (3) above-the two independent solutions may both be obtained from the root s = 0, taking a0 and a1 as arbitrary constants.

PROBLEMS 21

With s = 0, equation (1.73) gives

a2 = 0 and equation (1.74) becomes

an-3 + n(n - 2)an-2 - n(n - 1 )a, = 0, giving

a,_a + n(n - 2)an-2 ( ::;;, 3) a = ( 1) n- . " n n-

(1.75)

It is impossible for us to use the recurrence relation (1.75) to obtain a general expression for a,; but we may proceed to calculate as many terms as we please.

Thus

= iao + !a11

a 1 + 8a2 a4 = 12

= l2a1, a 2 + 15a3

aa = 20

and so on as far as we please.

We have calculated the series solution up to terms in x5 :

y = a0 + a1x + (ia0 + ia1)x3 + 1~a1x4

+ {!a0 + ~a1)x5 + ... = a 0{1 + ix3 + kx5 + ... }

+ a1{x + !x3 + 1\x4 + ix5 + ... }.

PROBLEMS

( 1) Obtain solutions of the following differential equations in ascending powers of x, stating for what values of x the series are convergent:

. d2y dy (•) 4xdx2 + dx- y = 0;

(ii) xdd~ + (x + l)ddy + y = 0; .~ X

( ... ) d2y 2dy 111 .\' · · d--x -1- y = 0;

dx 2


d 2y dy (iv) 9x(l - x)dx

2- 12dx + 4-y = 0;

(v) (1 - x 2) :~ + 2x ~ + y = 0;

d 2y dy (vi) x2 dx2 + (x2

- 3x)dx + (4 - 2x)y--: 0;

(vii) (1 - x 2)d2y + xdy- y = O·

dx2 dx '

(viii) x2:~ - 3x~~ + (x3 - 5)y = 0;

. d 2y dy (IX) dxz- x2dx- y = 0.

(2) Determine whether or not it is possible to obtain solutions of the following equations in terms of ascending powers of x:

(i) xa :~ + x 2 z + y = 0;

(ii) x 2 ~ + ~ + x 2y = 0;

( ... ) d2y dy 0 111 x2 dx2 + x2 dx + y = .

(3) Obtain solutions of the following equations in terms of ascending powers of lfx:

d 2y dy (i) 2x2(x- l)dx

2 + x(3x + l)dx- 2y = 0;

(ii) 2x 3 dd2

y + (2x + x2)ddy + 2y = 0. x 2 x

2

GAMMA AND BET A FUNCTIONS

2.1 DEFINITIONS

We define the gamma and beta functions respectively by:

r(x) = J; e-ttx-1 dt (2.1)

B(x, y) = J: t'"- 1(1 - t)Y-1 dt. (2.2)

The first definition is valid only for x > 0, and the second only for x > 0 andy > 0, because it is for just these values of x andy that the above integrals are convergent. We shall not prove this statement, but shall at least make it plausible.

Consider the integral in definition (2.1 ). It is known that at infinity the behaviour of an exponential dominates the behaviour of any power, so that e-1 t'"-1 ~ 0 as t -+ oo for any value of x, and hence no trouble is expected from the upper limit of the integral. Near the lower limit of the integral we have e-1 :::= 1 ; hence if this approximation is good between t = 0 and t = c, we may write equation (2.1) in the form

I'( X) :::= f: fX-l dt + f: e-ttX-1 dt

= [~t'"J: + J: e-ttx-1 dt,

so that if the first term is to remain finite at the lower limit we must have X> 0.

A similar argument applies to the beta function, the behaviour at t = 0 leading to the restriction on x and at t · 1 to the restriction on y.

24 GAMMA AND BET A FUNCTIONS

2.2 PROPERTIES OF THE BETA AND GAMMA FUNCTIONS

Theorem 2.1

r(t) = 1. PRooF

We have, from the definition (2.1),

Theorem 2.2

PROOF

r(l) = s: c-1t 1-

1 dt

= J: e-1 dt

r(x + 1) = xi'(.~). (x > 0)

r(x + 1) = J ~ e-tt'JJ dt

(by the definition (2.1))

= [<- e-t)ta:J~ - J~ ( -e-t)xta:-1 dt

(on integrating by parts)

= 0 + x s:e-tF-1 dt

CH.2

(where the first term vanishes at the upper limit since the exponential dominates the power, and at the lower limit since x > 0)

= xr(x) (using definition (2.1) again).

Theorem 2.3

If X is a non-negative integer, then r(x + 1) = x! PROOF

r(x + 1) = xr(x) (by theorem 2.2)

§ 2.2 PROPERTIES OF THE BETA AND GAMMA FUNCTIONS 25

= x(x - l)r(x - 1) (using theorem 2.2 again)

= x(x- l)(x- 2)r(x- 2) (by further use of theorem 2.2)

= x(x -- l)(x- 2)(x- 3) ... 3.2.1r(l) (by repeated use of theorem 2.2,

and remembering that x is integral, so that continued subtraction of unity will eventually lead to 1)

Theorem2.4

PROOF

= x!r(l) =X!

(by theorem 2.1 ).

l'(x) = 2 J: e- 1't 2x-t dt.

In definition (2.1) substitute u2 fort: t = u 2, so that dt = 2u du; when t = 0, u = 0 and when t = oo, u = oo, so that we have

rex) =I: e-u'(u 2)"'-

1 .2u du

= 2 s: e-u'u2x-t du

= 2 I: e-t't2:r:-I dt

(since in a definite integral we may choose the variable of integration to be anything we please).

Theorem 2.5

COS 2"'-l 0 sin 2Y-l e de = X y In/'1. rc )I'( )

o 2r(x + y)' PROOF

We prove this result by considering in two different ways the double integral

I = fIR exp ( - t 2 - u 2)t2

"'-1u211

-1 dt du

(where R is the first quadrant of the tu-plane, shown unshaded in Fig. 2.1).

26 GAMMA AND BETA FUNCTIONS

II

FIG. 2.1 The tu-plane

Firstly, we have

I= I"" I oo exp ( -t2 - u 2)t2"'- 1u 211 - 1 dt du 1=0 U=O

=I: e-t't2x-: dt. I: e-u'u2!1-l du

= !r(x) .!r(y) (using theorem 2.4)

= il'(x)r(y).

CH.2

(2.3)

Next we change variables to plane polar co-ordinates r and (J in the tu-plane so that t = r cos 8, u = r sin() and the element of area dt du is replaced by the element of area r dr dO. Then we shall have

I= 11 R exp ( -r2 cos 2 () -- r 2 sin2 8)(r cos 8) 2"'-1(r sin 8) 2v-1r dr d(J

Ioo In/2 = e-r'y2"'-l cos2a:-1 () r2Y-l sin 2v-1 () r dr d() r~O IJ=O

Ioo I"/2 = 0

e-''r2<"'+vl-l dr 0

cos2"'-1 0 sin211-1 () d(J

Irr/2

= ir(x + y) 0

cos 2"'-1 () sin2v-1 () d() (2.4)

(using theorem 2.4 again). Equating the two expressions (2.3) and (2.4) for I leads immediately to

the required result, which will be true for x > 0 and y :> 0, since it is for this range of values that the gamma functions involved arc defined.

§ 2.2 PROPERTIES OF THE BETA AND GAMMA FUNCTIONS 27

Theorem 2.6

r(i) = yn. PROOF

Put x = y = ! in theorem 2.5, and we obtain

fn/2 de = r(!)r(i) o 2rc1)

= !{r(!)}2, (using the fact that r(1) = 1).

Performing the integration on the left-hand side gives

~ = -}{r(-~)}2 2

and hence r'(!) = ± yn. But from the definition (2.1) we see that r( x) must be positive (since the

integrand is positive), so that we can discard the negative square root and obtain r(i) = yn as required.

COROLLARY

PROOF

Setting x = ~ in theorem 2.4 gives r(!) = 2 J~ e- t' dt, so that the

required result follows immediately on the use of theorem 2.6.

Theorem 2.7

PROOF

B(x, y) = ~(x)r(y). r(x + y)

Substitute t = cos 2 0 in the definition (2.2) of B(x,y): t = cos2 0 so that dt = -2 cos() sin() de; also, when t = 0, cos() = 0 so that 0 = n/2 and when t = 1 , cos e = 1 so that () = 0.

Hence we have:

B(x,y) = Jo (cos 2 0)"'-1 (sin 2 ())v-1 .-2cos8sin8d() n/2

Jn/2

~ 2 0

cos2"'-1 0 sin 2V- 1 0 dO

28

Theorem 2.8

PROOF

GAMMA AND BETA FUNCTIONS

= 2 r(x)r(y) 2r(x + y)

r(x)r(y) -· .

r(x + y)"

(by theorem 2.5)

B(x, y) = B(y, x).

This result follows immediately from theorem 2. 7.

Theorem 2.9

PROOF

(i) We have

(i) B(x + 1,y) = _x_B(x,y). X +y

(ii) B(x,y + 1) = _Y_B(x,y). X +y

B(x 1 ) = l'(x + l)I'(y~ + ,y r(x + 1 + y)

(by theorem 2.7)

xr(x)r(y) (x + y)r(x + y)

(by theorem 2.2)

X r(x)r(y) x + y r(x + y)

X = --B(x,y)

X +Y

(by theorem 2.7).

(ii) Exactly similar to (i).

§ 2.2 PROPERTIES OF THE BETA AND GAMMA FYNCTION~

Theorem 2.10 (Legendre Duplication Formula)

220:-1 f(2x) = yn f(x)f(x + i).

PROOF

We have, by theorem 2.7,

f(x)f(x) = B(x, x) f(x + x)

= J ~ t'"-1(1 - t)"'-l dt

(by definition (2.2))

Jl 1 1 1 = -12"'-P -+- s)"'-12"'-1 (1 - s)"'-I 2 ds

(on making the substitution t = i(l + s)) 1 Jl = -- (1 - s2)x-l ds

220:-1 -1

= _2_ Jl (1 ~ s2):r:-1 ds 220:-1 0

= 2-2:r:+2 J~ (1 - u)'"-1iu-1!2 du

(on making the substitution u = s2)

= 2-2:r:+l J: (1 - u)"'-Iu-IIz du

= 2-2x+IB(i, x)

= 2-2"'+1 f(i)f(x) f(i + x)

(by theorem 2.7).

Hence, using theorem (2.6),

r(x) f(2x)

2-20:+1

f(x + !) yn and thus

220:-1 1'(2x) - -- - l'(x)l'(x j ~).

yn

30 GAMMA AND BETA FUNCTIONS CH.2 CoROLLARY

If x is a positive integer

PROOF

In the theorem we may use theorem 2.3 to rewrite r(2x) as (2x- 1)! and r(x) as (x - 1)! Hence we have

22X-1 (2x- 1)! =--:;- (x- l)!r'(x + !),

·v :n; -

which, on multiplying both sides by 2x, becomes

22"' 2x(2x- 1)! = vnx(x- l)!r(x + .Q-);

this equation may be expressed more simply in the form

22"' (2x)! = v;;;x!r(x + t)

and thus

I'( ._;_ I) = (2x)!. I x , 2 22"'x!vn.

2.3 DEFINITION OF THE GAMMA FUNCTION FOR NEGATIVE VALUES OF THE ARGUMENT

From theorem 2.2 we have

1 r(x) = -r(x + 1).

X (2.5)

Apart from x = 0, where the denominator vanishes, the right-hand side of this equation is well defined for those values of x such that the argument of the gamma function is positive (since this was the condition for definition (2.1) to hold), that is, for x + 1 > 0, i.e. for x > -1. We also note that we may say that r(O) is infinite, for as X-+ 0 we have rex + 1)-+ r(1) = 1, and hence r(x) = (1/x)r(x + 1)-+ oo. So far the left-hand side is only defined for x > 0, but we can now use the right-hand side to extend this definition to x > -1. The argument we are using is as follows:

(i) Equation (2.5) was proved true for x > 0; (ii) Right-hand side is well defined for x > --1, left-hand side for

X> 0; (iii) Use right-hand side to define left-hand side for x :-· 1.

§ 2.3 DEFINITION OF THE GAMMA FUNCTION 31

This means that we now have r(x) defined for X> -1, so that the righthand side of equation (2.5) is well defined for x + 1 > -1, i.e. x > -2; we may thus use it to define the left-hand side for x > -2; and this process may be repeated to define r(x) for all negative values of x.

Theorem 2.11

r(m) = oo, if m is zero or a negative integer.

PROOF

We have already remarked above that r(O) = oo. From equation (2.5) we have

and

etc.

1 r( -1) = -r(O) = oo

--1 1

r(-2) = _2r(-1) = oo,

It is now possible to draw a graph of r(x). Plotting of points combined with information from the various theorems leads to a graph as shown in Fig. 2.2.

r (xl

:u:: I I I I I I I I I I I I I I

X

FIG. 2.2 The gamma function

32 GAMMA AND BET A FUNCTIONS CH.2

Theorem 2.12 1t

f(x)f(l - x) = ---;-. sm nx

PRoofi

We shall prove this result in three stages:

(i) We shall prove the infinite product expansion for sin 0, viz. <XJ

sin 0 = ll {1 - (O!ljn2nz)}; n=l

(ii) We shall show that this implies that

1 <XJ (-lt -- "" ----· sin 0 - L 0 - nn'

n=-«>

(iii) We shall use result (ii) to prove the required result.

(i) We know that . 0 2 . 0 0 sm = sm 2 cos 2

2 . 0 . (n 0) = sm 2sm z + z (2.6)

and hence, by applying this result to both factors on the right-hand side, we obtain

sin 0 = 2{2 sin~ sin (i + ~)}{2 sin(~+~) sin(~+~+~)} _ 23 . 0 . n + () . 2n + () . 3n + () - sm 22 sm ----z2 sm 22- sm 22 . (2.7)

It is left to the reader to verify that applying the result (2.6) to each of the factors on the right-hand side of equation (2.7) gives us

. () - 27 . !.__ . 1t + e . 2n + 0 . 3n + () . 4n + () Stn - Stn 23 Sin 23 Sin 23 Sin 23 Sin 23

. Sn + e . 6n + 8 . 7n + 0 sm 23 sm 23 sm -----za-· (2.8)

If this process is carried out n times altogether, the general result will be

t The proof of this theorem may be omitted on a first reading; alternatively, parts (i) and (ii), which are standard trigonometric identities, may be assumed and used to prove the result in part (iii).

§ 2.3 DEFINITION OF THE GAMMA FUNCTION 33

. . (). n+O. 2n+8 . (p--1)n+8 sm () = 2P-1 sm- sm -~ sm --- ... sm (2.9)

p p p p withp = 2n.

The form of this equation is plausible as the correct generalization of equations (2.6), (2.7) and (2.8); if desired it may be proved by the method of induction.

The last factor in equation (2.9) is

sin (p -;)n + 0 =sin {n- ~; 82}

= sm (n; 8).

Similarly, the second last factor is

sin (p - ~n + () = sin {n -- -'--(2_n_p----'-O)}

= sin (2n ;- o). and in general the rth factor before the end is

sin (rn p- o). We now regroup the factors appearing in equation (2. 9) by taking

together the second and the last, the third and the second last, and so on. Then equation (2.9) becomes

. 2 . 0{ . n + 0 . n - 0} { . 2n + () . 2n - 01 s1n 0 = v-t s1n P s1n -P- sm P sm P- sm -p--)· ..

{sin (!P- l)n + 0 sin (-~p- l)n- 0} sin !Pn + 9. (2.10)

p p p The factor inside each curly bracket is of the form

sin (A + B) sin (A -B) = i{cos 2B -- cos 2A} = sin 2 A - sin 2 B

and hence equation (2.10) becomes

· 0 2 1 • () { • 2 n . 0} { . 2n . 0} s1n = P- sm p s1n p - s1n 2 p sm 2 p - sm 2 p ...

{sin2 (iP - l):n: - sin2 ~}cos q. (2.11) p p p

Dividing both sides by sin (0 jp ), letting 0 --* 0 and remembering that lim (sin xjx) c-= 1 so that .r-•0


1. sin()

1. sin () () /p

lffi--- = lffi-- ---p o_..osin((Jjp) e_..0 () 'sin(Ojp)"

=p,

we obtain 2 1 •

2 n .

2 2n .

2 (kp - 1 )n p = P- sm -sm -- ... sm -"----.

p p p If we now divide equation (2.11) by equation (2.12), we obtain

sin 0 =sin~ { 1 _ sin2 (Ojp)} { 1 __ sin2 (Ojp)} p p sin2 (n/p) sin 2 (2n/p) · · ·

{

1- ___ sin;(o~~) -}cosq. • 2 (?.P 1)n p sm ----

p \Vc now let p ~ co. We take into account the three results

( ) 1. . 0

1. sin (Ojp)

a 1m p sm - = tm --- - . 0 p--+a:> p P--+00 () IP

=0,

CH.2

(2.12)

(2.13)

(b) lim sin2(0/p) = lirn{sin(O/p)}2(~)2{ 1·njp }z ___ 1 __ _

P--+a:> sin 2 (m/p) p--+oo Ojp p sin (m/p) (ra/p) 2

= 02 jr2a2,

(c) lim cos~ = 1 p--+00 p

and see that therefore equation (2.13) gives

sin 0 = 0 1 -- - 1 - - 1 - -- ... ( 02)( ()2 )( ()2 ) n2 22n2 rznz

00 ( ()2 ) = 0 [J 1 -- nz,_;2 .

n=l

(ii) It is trivial to verify the result that

Hence

1 d () -·--- = - - In tan-. sin 0 dO 2

_1 __ = i_ ln sin (0 j2) sin 0 dO cos (0 j2)

= _'!_ ln _ 2 sin2 (0/2) __ dO 2 sin (0 j2) cos (0 /2)

d 2 sin 2 (0/2) ln -------

dO sin 0

p.3 DEFINITION OF THE GAMMA FUNCTION

= :8{In2 -t-Insin 2 (0/2) -Insin8}

= :a {2 In sin (8 j2) - In sin 8}

d { w ( f)2 ) co ( f)2 )} = - 2 InTI 0 1 - - - Inn e 1 - -dO 4n2;rr;2 n2;rr;2 n~l n~l

= £6{ 2 In o + 2 ~ In ( 1 - 2~n) + 2 i In ( 1 + 2~J n~l n~l

35

-Ina- iin (1 -- :J- iin (1 + :J} n=l n=l

d{ co w = dO In 0 + 2 L In (2nn - 0) + 2 L In (2nn + 8)

n~l n~l

- ~/n (n;?;- 8)- ~/n (nn +e)}

1 w -1 00 1 00 -1 00 1 = o + 2 :L 2nn - o + 2 :L 2nn + e - :L nn -a - :L nn + o

n=l n=l n~l n~I

1 {w 1 00 1} = 0 + 2 L 0 - 2nn - L 0 - nn

n~l n~l

= ~ + ~~~1~: +,~;~1~: = ~ ~~1;l:.

n=- oo

{

w 1 w 1 } + 2 L 0 + 2nn - L a-=+=-nn

n~l n~l

(iii) Let us first assume that 0 < x < 1.

Then

~F-U

f(x)f(1 - x) = B(x, 1 - x)f(x + 1 - x)

(by theorem 2.7)

= B(x, 1- x)

= L t"'- 1(1 _ t)-'" dt

(by definition (2.2)).

36 GAMMA AND BETA FUNCTIONS CH.2

We now make the substitution t = 1 /(1 + u); then

dt = { -1/(1 + u)2} du and u = (1 - t)jt.

Also when t = 0, u = oo and when t = 1, u = 0.

Then r(x)r(l-x) = J:(1 +1u)"'- 1C :J-"'(- (C~u)2du)

Jro u-"'

= --du 0 1 + u

Jl u-"' Joo u-"' = --du + --- du. o1-j-u t1-j-u

In the second integral we make the substitution u = 1 jv. Then du = ( -1jv 2) dv; also when u = 1, v = 1 and when u = w, v = 0.

Thus f oo u-"' fo v"' ( 1 ) 1 f+~ du = t1 + (1/"-') -- v2 dv

ft v"'-1 = --dv

01 + v

f 1 u"'-1 = --du.

0 1 + 1l

Hence, from equation (2.14) we have

r(x)r(l - x) = J1 (u-"' + u"'-

1) du

0 1 + u

Jl 00

= (u-"' + u"'-1) L ( -l)"un du 0 n=O

= ~ ( -1)"[---1 __ ---un-x-J-t + ___ 1 __ un+"']l ~ n-x-t-1 n-j-x o n=O

00

{ 1 1 } = (-1)" -1----2 n--x-t-1 n-j-x n=O

(remembering that 0 < x < 1)

§2.4 EXAMPLES 37

sinnx (by part (ii) of the proof).

We now remove the restriction 0 < x < 1. Suppose that x = y + N where N is an integer and that 0 < y < 1.

Then

r(x)r(1 - x) = r(N + y)r(1 - y - N)

as required.

2.4 EXAMPLES

Example 1

= (N + y- 1)(N -t- y -- 2) ... yf(y) 1 1 1

1 -y -N·2 -y-N .. ·_yr(1 -y)

(by repeated use of the result f(x + 1) = xr(x))

= ( -l)·''r(y)r(l - y)

=(-1)N_:n_ sm:ny

(since we know this result to be true for 0 < y < 1)

sin (N:n + :;r.y) :n;

sm:nx

Express each of the following integrals in terms of gamma or beta functions and simplify where possible.

fn/2

(i) 0

v(tan 0) dO. (ii) fl dx o a.y(l-xa)'

(iii) f: t- 31:'(1 -- e-1) dt. (iv) fl (~)1/2 1 dx.

-1 -X

(i) We usc theorem 2.5 where, in order to get v(tan 8)=sin112 0 cos-112 0 as integrand, we take 2x - 1 = -- ~ and 2y -- 1 = -H, i.e. x = ;]; and y '-C.O ~-


fn/2 r(l)r(3) Thus v(tan fJ) de= 4 4

0 2r(! + !)

(ii)

= ir(t)r(!)

1 1t =----

2 sin (n/4)

1 3t

= 21;.y2

= n/y2.

(by theorem 2.1)

(by theorem 2.12)

I: 3y(~~x3) = J: (1 - x3)-1/3 dx.

CH.2

We change the variable tot = x3• When x = 0, t = 0 and when x = 1, t = 1. Also dt = 3x2 dx, i.e., dx = }t-2' 3 dt.

Hence fl dx = fl (1 - t)-II3.}t-2/3 dt o 3y(1-x3) o

- IB(.l 2) -3 3• 3

1 n

3 sin (n/3)

1 n = 3 (v3)/2

2n 3(,Y3)"

(iii) f ~ t-3'

2(1 - e-t) dt.


(by theorems 2.7 and 2.8)

(by theorem 2 .12)

We cannot relate this immediately to the gamma function; we must first integrate by parts:

I"' 2t- 112 .e-t dt ()

§2.4

(iv)

EXAMPLES

= 0 + 2 J: t-112 e-t dt t = 2r(!)

(by definition (2.1)) =2yn,

(by theorem 2.6).

Jt (!_-=+.-_:)1/2 dx. -1 1 - .lC

39

In order to change this into a form resembling the beta function it is necessary that the range of integration be changed to 0 to 1. This is accomplished by the change of variable t = 1(1 + x). Then x = 2t - 1, dx = 2dt and we have

fl (! + ~)1/2 dx = Jl (1 + 2t - 1)1/22 dt • -1 1 - X 0 1 --- 2t + 1

= 2 Jl (-t )1/2 dt 0 1 - t

= 2 f: t 112(1 - t)-1'2 dt

= 2B(t, !) (by definition (2.2))

l'(~)r(!) =2----r(2)

(by theorem 2.7)

= 2!r<tLr<t> 1 !

(by theorems 2.2 and 2.3) = :n; (by theorem 2.6).

t The behaviour of r112(1 - e-•) at t = 0 requires some explanation. We recall l'Hopital's rule which states that if f(a) = g(a) = 0, then

lim f(t) = f'(a) t--.. a g(t) g'(a)

provided f'(a) and g'(a) are not both zero (or infinite). Hence we have

1 - = 0.

(J.)

40 GAMMA AND BETA !'UNCTIONS

Example 2

Show that B(n, n + 1) = ~ {~~;~2

and hence deduce that

We have

----- cosO dO= · Jn/2 ( 1 1 )1/4 {r(l)}2 o sin3 0 sin2 0 Zy'11:

r(n)r(n + 1) B(n, n + 1) = -r(2n + 1)

(by theorem 2.7)

f(n).nf(n) 2nr(2n)

(by theorem 2.2) {f(n)}2

= 2f(2~)' Setting n = l gives

1 :; - {f(!)}2 B(4, 4> - 2rU) ,

which, on using definition (2.2) and theorem 2.6 becomes

Jl t-3/4(1 - t)114 dt = {f(!)}2. 0 2y':n

CH. ~

Now set t = sin 0 in the integral on the left~ hand side. When t = 0, 0 = 0 and when t = 1, 0 = n/2; also dt =cos 0 dO. Hence we have

Jl fn/'1. 0 t- 3' 4(1 - t)1'4 dt =

0 (sin 0)-314(1 --sin 0) 1' 4 cos 0 dO

f n/2 (1 -sin 0)1/4 = cos() dO

o sin3 0

Jtt/2 ( 1 1 )1/4 = ~0 - -~0 cosO dO,

o sm sm

so that we have proved the required result.

PROBLEMS

(1) Prove that

(i) J 00

e-a"'x" dx = -1

l'(u + 1) 0 an H

(n> l,a::..-0);

PROBLEMS

(m > -l,n >0).

(iii) J: exp (2ax- x 2) dx = i vn exp (a 2

).

41

ftt/2

(2) Prove that 0 tann 0 dO = tr{(l + n)/2} r{(l - n)/2} if In I < 1.

fn/2 • Jn/2 vn r{(1 + n)/2} (3) Prove that

0 smn 0 dO =

0 cosn 0 dO = 2 r{(2 + n)/2}

( 4) Express each of the following integrals in terms of the gamma or beta functions and simplify when possible:

(i) J: G -1/'4

dx; (ii) J: (ln~y-1

dx, <a> o);

(iii) J: (b - x)m-1(x - a)n-1 dx, (b >a, m > 0, n > 0);

(iv) J: xm(1 - xn)P dx, (m > -l,p > -1, n > 0);

J1 dx foo dt

(v) 0 v(l - xn)' (n > O); (vi) 0 ( vt)(1 + t)' (5) Show that the area enclosed by the curve x4 + y 4 = 1 is {r(±)F j2yn. (6) Evaluate r( -t) and r( -f). (7) Show that

-n (i) r(x)r( -x) = . ;

xsmnx

(ii) r(l + x)r(l-- x) = _n_, 2 2 cosnx

3

LEGENDRE POLYNOMIALS AND FUNCTIONS

3.1 LEGENDRE'S EQUATION AND ITS SOLUTIONS

Legendre's differential equation is

d2y dy (1- x2)dx2- 2xdx + ky = 0. (3.1)

We shall write kin the form l(l + 1) for reasons which will become clear as we proceed.

This equation,

(1 - x2)d2

y- 2xdy + l(l + 1)y = 0 (3.2) dx 2 dx '

is of the form of equation (1.2) with q(x) = -2x2 j(l - x2) and r(x) = {l(l + l)x 2}/(1 - x 2). The binomial theorem may be used to expand these as power series in x for values of x such that x2 < 1, i.e., such that -1 < x < 1. Thus the methods of Chapter 1 will be applicable to the solution of this equation, any power series obtained being valid for at least -1 < x < 1.

00

Writing z(x, s) = x8 L a,.x" and requiring z to be a solution of equation n=O

(3.2) leads, as in Chapter 1, to the system of equations

au-f(s- 1) = 0,

a1(s + l)s =0

(3.3)

(3.4)

§ 3.1

and

LEGENDRE'S EQUATION AND ITS SOLUTIONS

a,+2(s + n + 2)(s + n + 1) - an{(s + n)(s + n + 1) - l(l + 1)}

43

= 0 (n > 0). (3.5) Equation(3.3) gives the indicial equation s(s- 1) = 0 with roots s = 0

and s = 1 (which differ by an integer, so that we know we are dealing with one of the exceptional cases).

Equation (3.4), with s = 0, is satisfied irrespective of the value of a1 ;

this means that a1 is indeterminate and hence the one root of the indicial equation (s = 0) leads to two independent solutions with the two arbitrary constants a0 and a1•

Equation (3.5) with s = 0 becomes

n(n + 1) - l(l + 1) an+ 2 = a,.-(n_j::-l)(n +-z) -- (3.6)

which, since

n(n + 1) - l(l + 1) = n2 + n - [2 - l

may be written in the form

= (n2 - 12) + (n - l) = ( n - /)( n + l) + ( n - l) = (n - l)(n + l + 1),

(l- n)(l + n + 1) an+2 = -an (n + 1)(n + 2) .

Thus we have

l(l + 1) a= -a---·

2 0 1 .2 ( l -- 1 )(I + 2)

aa = -al 2.3 _,

(l - 2)(! + 3) a4 = -a2--3~4----

(l- 3)(1 + 4) a5 = -aa 4.5

(3.7)

l(l- 2)(! + 1)(/ + 3) = ao '

(/- 1)(!- 3)(! + 2)(! + 4) = al--1.2.3.4

and in general

a2n = ( -1)nao

2.3.4.5

l(l-2)(!-4) ... (l-2n+2)(1+1)(1+3) ... (l+2n-1) (2n)!

and

a2n +1 ,-~ ( 1 )nao

(I 1 )(I 3) . . . ( l 2n 1- 1 )( l- /-2 )( l / 4) . . . ( Z-1- 2n) (2n -/- 1)!

(3.8)

(3.9)

44 LEGENDRE POLYNOMIALS AND FUNCTIONS

Therefore

z(x, 0) = a 0{1 + ~ ( -1)n

l(l-2) ... (l-2n+2)(l+1)(l+3) ... (l+2n-1) "} (2n)! x-n

+ a 1 {X + % ( -1 )n

(l-1)(l-3L. (l-2n+l)(l±2)(l1_:_'!:_)_._:~_S~!:-2n) x 2n+1} (2n + 1)!

= aoy1(x) + a1y 2(x), say,

so that y 1(x) and y 2{x) are the linearly independent solutions.

CH. 3

We know already that y 1(x) and y 2{x) will be convergent for -1 < x < 1. For many applications, solutions to Legendre's equation are required which are finite for -1 < x < 1. The theorem quoted at the end of Section 1.1 guarantees convergence only for -1 < x < 1, and says nothing at all about convergence for x = ± 1. In fact standard methods of convergence theory may be used to show that the series above are divergent for x = ±1 (see Appendix 1). How then is it possible to obtain solutions which are finite for x = ± 1? The only way is to make the infinite series reduce to finite series, and this will happen for any positive integral value of l. For, from equation (3.8) we see that if l = 2n (an even positive integer) we have a2n :;;: 0 but a2n+2 = 0, and hence all subsequent even coefficients must also equal zero. Similarly equation (3.9) shows that if l = 2n + 1 (an odd positive integer) then a2n+l :;;: 0 but a2n+a and all subsequent odd coefficients are zero. Thus we see that if l is an even integer y1(x) reduces to a polynomial, and hence is finite for all finite x, while if l is an odd integer the same occurs for y 2(x).t In both cases the highest power of x appearing is x1

, so that since equation (3.7) applies to the series for both y 1(x) and y 2(x), we may obtain a single series valid for both even and odd l if we write it in descending powers of x. The

t When we consider integral values of l, we need consider only positive values of l, since the constant appearing in the equation was l(l + 1) and if I were a negative integer we could write m = -(l + 1) and use the easily verified fact thatm(m + 1) = l(l + 1).

This arrival at integral values of I is, of course, the reason for writing the original constant k in the form l(l + 1 ).

§ 3.1 LEGENDRE'S EQUATION AND ITS SOLUTIONS 45

first term is a1x1 and subsequent terms are given by equation (3.7) written

in the form

so that

and, in general,

(n + 2)(n + 1) an = -an+2 (l- n)(l + n + 1)'

l(l- 1) al-2 = -az 2(21 - 1)'

(l- 2)(/- 3) al-4 = -az-2 4(2! _ 3)

(l- 1)(!- 2)(!- 3) = az 2.4.(2!- 1)(2!- 3)

-( 1), l(l-1)(l-2) ... (l-2r+l) al-2r- - az ·

2.4 ... 2r(2l- 1)(2/- 3) ... (2/- 2r + 1)

(3.10)

(3 .11)

Thus for l even, y 1(x) reduces to the following expression, while for l odd, y 2(x) reduces to the same expression:

( ) 1 1-2 1_ 4 {a0 for l even y x = a1x + a1_ 2x + a1_ 4x + ... + f z dd a1x or o

!Hl

= L al-2,:~--2· r~o

1 _ {it for l even (where [2/]- i(l- 1) for l odd, (3.12)

i.e., [!I] is the largest integer less than or equal to i/)

[1/2] l l ) =az-..;;:: (-1)r (l-1) ... ( -2r+l , -xl-2r (3.13)

;S 2.4 ... 2r(2l-1)(2l-3) ... (2/-2r 1 1)

from equation (3 .11 ). We may rewrite this series more compactly if we note the following

results:

l(l- 1) ... (l- 2r + 1) = l(l _ 1) ... (l _ 2r + 1) _(l - 2r)(l - 2r - 1_) ... 3. 2.1

( l - 2r )( l - 2r - 1) . . . 3 . 2 . 1 l!

= (l- 2r)!' (3.14)

2.4.6 ... 2r ~~ (2.1)(2.2)(2.3) ... (2.r) ~ c zr 0 1 0 2 0 3 0 0 0 r

zrr! (3.15)


and

(21- 1)(21- 3) ... (21- 2r + 1)

21(21-1)(21-2)(21-3) ... (21-2r+1) (21-2r)! = 21(21-2)(21-4) ... (2l-2r+2) · (2l-2r)!

(21)! 2'1(1- 1) ... (1- r + 1)(21 ~ 2r)! (21)!(1- r)! 2'(21 - 2r) !l!'

CH. 3

(3.16)

Using equations (3.14), (3.15) and (3.16) in equation (3.13) gives us

, _ 11121 1! 1 2'(2! - 2r) !l! 1_ 2,

y(x)- az 6 (-1>"(1- 2r}!.2•r!,_(21)!(l- r)!x

1~ (1!) 2{21- 2r)! _2,

= al r~ ( -1Y r!(1- 2r)!(1- r)!(2l)!xl .

This is a solution for any value of a1• If we choose a1 = (2l) !/{21(1!) 2}

we obtain the solution which we denote by P1(x) and call the Legendre polynomial of order 1:

1~ (2l- 2r)! Pz(x) = ~ ( -1)" 2zr!(l- r)!(l- 2r)!xl-2r. (3.17)

Thus this is the solution of Legendre's equation which is finite for -1 < x < 1 ; it is the only such solution, apart from an arbitrary multiplicative constant.

3.2 GENERATING FUNCTION FOR THE LEGENDRE POLYNOMIALS

Theorem 3.1

1 00

v(l - 2tx + t2) = 6 t~Pt(x) if I t I < 1 and I X I < 1.

This means that when (1 - 2tx + t 2)-112 is expanded in powers oft, the coefficient of t1 is P1(x); (1 - 2tx + t 2)-112 is called the generating function of the Legendre polynomials.

PROOF

Expand (1 - 2tx + t 2)-1' 2 by the binomial theorem:

(1 - 2tx -1- t 2)- 1 ' 2 = {1 -- t(2x - t)}- 1'

2

§ 3.2 GENERATING FUNCTION FOR LEGENDRE POLYNOMIALS 47

= 1 + ( -!){ -t(2x- t)} + ( -~~ -!) { -t(2x- t)}2

+ ... + ( -!)( -t) ... ~ -(2r - 1)/2} { -t(Zx _ t)}" + ... r.

= ~ ( -1 )' ~3 · S · · · (Zr - 1) ( --1 )'t'(2x - t)' L. 2'r! r~o

~ (2r)! = 6 Z2'(r!)/'(2x - t)".

Now expand (2x - t)' by the binomial theorem (remembering that r is integral):

T

(2x - t)" = 2 "Cp(2x)'-P( -t)P p=O

where "CP is the binomial coefficient I( r~ )' p. r p.

Hence we now have

oo (2)1 r

(1 - 2tx + t 2)-1' 2 = "_r_. """C (-1)Ptr+P(2x)'-P. (3.18) 622'(r!)2~o P

We wish to find the coefficient of t!, so we must take r + p = l, and hence for a fixed value of r we must take p = l- r. But p only takes on values such that 0 < p < r, so we must only consider those values of r satisfying 0 < l- r < r, i.e. tz < r < l. Hence if lis even, r can take on values between !l and l, while if l is odd, r can take on values between i(l + 1) and l. For any of these values of r the coefficient of t1 in equation (3.18), obtained by taking p = l- r, is

2!2_!__ r C ( -1 )l-r(2x)'-(l-r) 22r(r!)2 1-r

and the total coefficient of t1 is obtained by summing over all appropriate values of r so that we obtain

l

coefficient of t1 = 2 {/ 12 (/ (•Vi' II)

r (l 1 I)/~ (I o<lil)

J3r)! _rC ( -l)Z-'(Zx)zr-1 22r(r!)2 z_,. ·

48 LEGENDRE POLYNOMIALS AND FUNCTIONS CH. 3

If we now change the variable of summation from r to k = 1 - r, we obtain

coefficient of t

* (21- 2k)! l-kc ( 1)~:(2 )l-2k = Lt z2i-2k{C1 - k) !}2 k - x

{!/2 (l even)

k~ (Z-1)/2 (!odd)

~ (2l - 2k)! (l - k)! ( 1)k l-2k l-2k

= 6o Z21 - 2"{(/- k)f}2 {l- 2k}!k! - 2 X

(where [1/2] is as defined above in equation (3.12))

lt/21 ~: (21 - 2k) I -21: 2 <--1> 21(1- k)l(1- 2k)lklx~ k=O

= P1(x) (by the definition (3.17)).

Hence the required result is proved. The restriction on the values of x comes from the condition for

convergence of the binomial expansion of {1 - t(2x - t)}-1' 2, viz. I t(2x - t) I < 1. When I x I < 1 this condition may be shown to be equivalent to I t I < 1.

3.3 FURTHER EXPRESSIONS FOR THE LEGENDRE POLYNOMIALS

Theorem 3.2 (Rodrigues' Formula)

P1(x) = z~! !,(x2 -- 1)

1•

PROOF

We may expand (x 2 - 1)1 by the binomial theorem: l

(x2 -1)1 = _2: 1C,(--1)'x2<1-•>. r=O

Hence,

1 d' ( 2 )! - 1 dl *' l ( 1)• 2!-2r 2Hf dx' x -- 1 - 211! d.xl :S c, -- x ·

But the lth derivative of a power of x less than l is zero, so that

d1

1 .... 21 · 2'' -._, () t'f 21 2 I . 'f lj" dx -~ - r < , I.e., 1 r > ....

(3.19)

§ 3.3 FURTHER EXPRESSIONS FOR LEGENDRE POLYNOMIALS 49 l l/2 (l-1)/2

Thus we may replace L by L if l is even and by L if l is odd, i.e., r=O r=O r=O

[l/21

by L where [l /2] is as defined above. r=O

Also, dl dxlX11 = p(p - 1)(p - 2) ... (p - l + 1)xP-1

p! -1 = ---xP (p - l)!

so that dl 21-2r _ (2! - 2r)! ,1-2r

dxlx - (i- 2r)! .'t

and hence, substituting into equation (3.19), we have

1 d1 1 11121 [I (21-- 2 )I

( 2 1 )z "" • ( 1 )• r . z - ~r Z1J! dx1 X - = 21/l f:t, r!(/- r)! -- (1- 2r)! X

- [l/21 r (21 - 2r)! 1-~r -6 < -1) 21r!(t- r) !(l- 2r)Y

= P1(x) (by the definition (3.17)).

Theorem 3.3 (Laplace's Integral Representation)

1 J" P1(x) =- {x + v(x2 - 1) cos 0}1 dO. n o

PROOF

It may be shown by elementary methods (for example by means of the standard substitution t = tan (0 /2)) that

J" dO n o 1 +A. cos e = vet - A- 2). <

320)

If we now write A.= -{uy(x 2 - 1)}/(1 -- ux), expand both sides of equation (3.20) in powers of u and equate the coefficients of corresponding powers of u, we shall obtain the desired result:

1 1 - ------~--~---

1 +A. cos 0 uy(x2- 1)

1 - (1 _ ux) cos 0

= (1 - ux)[1 - u{x + v(x2 - 1) cos 0}] -l

'XJ

= (1 - ux) L u1{x + v(x2 - 1) cos 0}1

1=0


00

(since, by the binomial theorem c1 - a)-1 = L: an). n=O

1 1 v(l - ).2) = J-{1 _·-u2-(x-2 ---1)...,.....}

(1 - ux) 2

(1 - ux)

v{(l - ux) 2 - u2(x2 - 1)}

_ (1 - ux) . y1(1 - 2ux + u 2

)

Hence, substituting into equation (3.20) gives .. f" 'u1{x + v(x 2 - 1) cos 0}1 dO= -----~---

0 6 v(1 - 2ux + u2) CO CD

i.e., 1~ u1 f: {x + y!(x2- 1) cos 0}1 dO = n 6 u1P1(x)

(by theorem 3.1). Equating coefficients of u1 gives

nP1(x) = J: {x + v(x 2 - 1) cos 0} dO

which is the required result.

CH. 3

3.4 EXPLICIT EXPRESSIONS FOR AND SPECIAL VALUES OF THE LEGENDRE POLYNOMIALS

From the definition (3.17) we may write down explicitly the Legendre polynomial of any given order. We give here the first few polynomials:

Theorem 3.4

(i) P1(1) = 1.

P0(x) = 1, P1(x) = x, P2(x) = l(3x 2 -- 1), Pa(x) = i(5x3 - 3x), P 4(x) = !(35x4 - 30x 2 + 3). (3.21)

(ii) P1( -1) = ( -1)1•

(iii) P;(l) = Jl(l + 1 ). (iv) P;( -1) cc:c ( 1) 1 - 1 ~!(1 1 1).

§3.4 VALUES OF THE LEGENDRE POLYNOMIALS

( ) t (21) I

v P 21(0)=(-1) 22t(t!) 2 •

(vi) P21+I(O) = 0.

(By P{(1) we mean [{dPt(x)}/dx]x~ 1).

PROOF

(i) Set x = 1 in theorem 3.1 above and we obtain

that is

00

1 00

------- - ""' t1P1(1) vet - 2t + t 2) - 6 ' 1 00

1 _ t = L t1P1(1). z~o

51

But 1/(1 - t) = L t (by the binomial theorem or by considering the z~o

right-hand side as the sum to infinity of a geometric progression), so that 00 00

we have L t1 = L tP1(1). For this to be true for all values oft in some z~o z~o

range (in this case I t I < 1) we must have the coefficients of corresponding powers oft equal, i.e., P1(1) = 1.

(ii) Exactly similar to (i) but setting x = -1 in theorem 3.1.

(iii) P1(x) satisfies Legendre's equation (3.2), so that we have d 2 d

(1 - x2) dx 2P1(x) - 2x dxP1(x) + l(l + 1)P1(x) = 0 (3.22)

and setting x = 1 in this equation gives -2P;(1) + l(l + 1)P1(1) = 0,

which reduces on using part (i) above, to P;(1) = il(l + 1 ).

(iv) Exactly similar to (iii), but setting x = -1 in equation (3 .22) and using part (ii) above.

(v, vi) Set x = 0 in theorem 3.1 above and we obtain 1 00

(1 2) = L fP1(0). V + t z~o

Expanding the left-hand side by the binomial theorem gives us

1 --- = (1 + t2)-112 vet + t 2

)

= 1 + (- ~)t2 + ( -~~1-i)(t2)2 -j-


Thus we have

( -D( -!) ... { -(21 - 1)/2}( 2) 1 + ... + 1! t + ...

= ~ (-1Y 1.3 .5 .. 1

• ;2z- 1)t21

l=O 2 .f.

= ~ ( -1 y 1. 2 . ~.;. 5 ... (21 - 2)(21 - 1 )2lt2l l=O 21 . . 2.4-.6 .. , (21- 2)2/

~ l (21)! 21

= 6 (-1) zzn.ztz!t

~ 1 (21)! 2/

= ~ (-1) 22/(l!)/. l=O

~ (-1)z j'!:_l)!_t2z = ~ t'P(O) 6 22/(1!)2 6 l

and equating coefficients of corresponding powers of t on both sides gives

l (21)! Pzt(O) = ( -1) 2zl(Z!)2

P2t+1(0) = 0.

3.5 ORTHOGONALITY PROPERTIES OF THE LEGENDRE POLYNOMIALS

Theorem 3.5

1 { 0 ifl~m

f P1(x)Pm(x) dx = _2_ ifl = -1 21 + 1 z m.

(This result may be written more concisely if we introduce the Kronecker delta, defined by

<5 _ {0 if l ~m lm- 1 if 1 = m.

Then the statement of the theorem is

r I Plx)P,.(x) dx' '2/ ~- lr51m·)

§ 3.5 PROPERTIES OF LEGENDRE POLYNOMIALS 53

PROOF

P1(x) and P m(x) satisfy Legendre's equation (3.2), so that we have

d2P dP (1 - x 2

) d.~ 21

- 2x dx1 + l(l + 1)P1 = 0

and d2Pm dPm )P t (1 - x2

) dx2 - 2x dx + m(m + 1 m = 0,

which may be rewritten in the form

d~{(l - X2)~1} + /(/ + l)P1 = 0 (3.23)

and d~{(l- x2)~m} + m(m + l)Pm = 0. (3.24)

Multiplying equation (3.23) by Pm(x), equation (3.24) by P1(x), subtracting equation (3.24) from equation (3.23) and integrating with respect to x from -1 to +1 gives

f1

[ d { dP,} d { 2)dPm}] Pm- (1 - x 2)- - P1 - (1 -- x - dx -t dx dx dx dx

+ {l(l + 1)- m(m + 1)} f~ 1 PzPm dx = 0.

But since

d { dPz} dPm dP1 d { dPz} dx Pm(l - x2)-dx = dx ·(1 - x2)dx + Pm dx (1 - x2) dx '

we shall have

+ dPz(1 - x2)dPm] dx + (/2 + l- m2- m) fl PzPm dx = 0. dx dx -1

On integrating the first term we obtain

[ dP, dPm]l

P m(l - x2) dx - P 1(1 - X

2) dx _

1

+ (l- m)(l -1-m+ 1) f~ 1 P1Pmdx = 0;

t Throughout this proof we denote Pt(x) by Pt and Pm(."<) by Pm. The argument x is to be understood in every case.


but now the first term vanishes at both upper and lower limits, because of the (1 - x 2) factor, so that we have

(l- m)(l + m + 1) J~ 1 P1Pm dx = 0, and when l r'= m we can cancel

the factor outside the integral to obtain

e P1Pmdx =0. J -1

It remains to prove that J~ 1 {P1(x)} 2 dx = 2/(2l + 1). For this we use

the generating function of theorem 3.1 :

so that

1 00 - ""t1P (x) v(1 - 2tx + t 2) - 6 t

1 { co }2 (1 - 2tx + t2) = 6 tlPt(x)

00 co

= L t1P1(x). L tmPm(x) l=O m=O

00

= L tl+mP1(x)P m(x). l,m=O

We now integrate both sides with respect to x between -1 and +1: 1 1 ()() 1

f (1 2) _ 2 dx = L tl+m J P1(x)Pm(x) dx. -1 + t tx l,m=O -1

But the left-hand side is a standard integral of the form

J {1/(a + bx)} dx = (1/b) In (a + bx),

while the right-hand side contains only terms for which l = m, by the first part of this proof. ·

Hence

[ 1 ]1 co 1

--2 In (1 + t2 - Ztx) - = L t21 r {Pz(x)}2 dx

t 1 l=O J -1

which gives ()() 1 1 1

1~ t 21 J _1

{P1(x)} 2 dx = - 2t In (1 + t 2 - 2t) + 2t In (1 + t 2 + 2t)

1 1 Zt In (1 · t) 2 +

2t In (1 + t)z

§ 3.6 LEGENDRE SERIES 55

1{ ' = t In (1 + t) -In (1 - t)}

1{ t2 t3 t2 t3 } =tt-z-+3···+t+z-+3···

(using the series expansion of ln (1 + t))

= l{ 2t + 2~ + 2f + .. }

= 2{ 1 + ;~ + ~ + .. } 00 t2l

=2 621 + ( Equating coefficients of corresponding powers of t on both sides gives

fl 2 _

1 {P1(x)}2 dx = 21 + l

as required.

3.6 LEGENDRE SERIES

Theorem 3.6

If f(x) is a polynomial of degree n, then

f(x) = i crP,(x) with cr = (r + !) f~ 1 f(x)P,(x) dx. r=O

Also, if f(x) is even (or odd), only those Cr with even (or odd) suffixes are non-zero.

PROOF

We have f(x) = a .. x" + a.,_1xn-t + ... + a1x + a0, say, and we may write P.,(x) = k.,x" + k .. _2x"- 2 + ... (since by equation (3.17) P.,(x) is a polynomial of degree n, containing only odd or even powers of x according to whether n is odd or even), so that if we take f(x) - (a.,jk,.)P.,(x) we obtain either zero (in which case the first part of the theorem is proved) or else a polynomial of degree n - 1. This means that

f(x) = c .. P,.(x) + g .. - 1(x)

where c .. = anfkn and g.,_1(x) is a polynomial of degree n - 1. The same argument may be applied to g .. _1(x) to give

gn-l(x) =c c.,_lpn-l(x) + g .. -2(x)


and hence f(x) = CnP .. (x) + Cn_1P .. _1(x) + g .. _2(x). The same argument may now be applied to g.,_2(x), etc., to yield

eventually

f(x) = cnP .. (x) + c.,_lpn-I(x) + ... + c1P1(x) + c0P0(x) n

= L c,P,(x). r=O

If we now multiply both sides of this equation by P.(x) and integrate from -1 to + 1, we obtain

fl n 1

_1

f(x)P,(x) dx = 2 c, J _1

P,(x)P,(x) dx. r=O

But, by theorem 3.6, the integral on the right-hand side is non-zero only for the value of r which equals s, in which case it is 2/(2s + 1).

Therefore fl 2

f(x)P.(x) dx = c, 2---

1-,

-1 s + giving c, = (r + !) J~ 1 f(x)P,(x), as required.

Now suppose that f(x) is even. Then, since P,(x) is even when r is even, and odd when r is odd, so the integrand f(x)Pr(x) is also even when r is even and odd when r is odd. But an odd function integrated over the range -1 to +1 will give zero, since positive and negative values cancel one another. Hence c, is zero when r is odd. Similarly when f(x) is odd, c, is zero when r is even.

CoRoLLARY

If f(x) is a polynomial of degree less than l, then

J~ 1 f(x)P1(x) dx = 0.

PROOF

Suppose f(x) is of degree n; then by the theorem we have

f(x) = I c,.P,(x), r=O

so that

fl n f1 _

1 f(x)P1(x) dx = 2 c, _

1 P,(x)P1(x) dx

r=O

=0 by theorem 3.5, since r ..;;; n < l, so that r is never equal to l.

§ 3.6 LEGENDRE SERIES 57

The results of the above theorem may be extended to functions which are not polynomials. We shall not prove this extension, but will merely quote the following result (the proof is not difficult, but is fairly lengthy).

Theorem 3.7

Suppose f(x) satisfies the following conditions in the interval -1 < x < 1: (i) f(x) is continuous apart from a finite number of finite discontinuities

(we then say that f(x) is piecewise continuous); and (ii) f(x) has a finite number of maxima and minima.

00

Then the series L c,P,(x), where r=O

c, = (r + !) f~ 1 f(x)Pr(x) dx,

converges to f(x) if x is not a point of discontinuity of f(x) and to

!{f(x+) + f(x-)}

if xis a point of discontinuity.t Also, at the end points of the range, x = ± 1, the series converges to f(1-) and f( -1 +) respectively. This series is called the Legendre series for f(x). ---·----

t By f(x0 +)we mean Lim f(x0 + 6) where 6---+ 0 through positive values and a->()

by f(x0 -) we mean Lim f(x0 - 6) where again 6---+ 0 through positive values. ·~ See, for example, Fig. 3.1.

y

--?~ __________ l./'.''"

I I

Xo J(

FIG. 3.1 Left- and right-hand limits at a point of discontinuity


3.7 RELATIONS BETWEEN THE LEGENDRE POLYNOMIALS AND THEm DERIVATIVES; RECURRENCE RELATIONS

Theorem 3.8 [l(l-1)]

(i) P;(x) = 2 (21 - 4r - 1 )Pz-2r-l(.~). r=O

(ii) xP1(x) = ;l: \ P1 H(x) + 2.z : 1 Pz-J(x).

(iii) (l + 1)P1+1(x) - (21 + 1)xP1(x) + ZP1_1(x) = 0. (iv) P; +l(x) - P; _1(x) = (21 + 1 )P1(x). (v) xP;(x) -- P;_ 1(x) = lP1(x).

(vi) P;(x) - xP;_1(x) = lP1_ 1(x). (vii) (x2 - 1)P;(x) = lxP1(x) - lP1_1(x).

(viii) (x2 - 1 )P;(x) = (1 + 1 )P1.1-1(x) ~- (l + 1 )xP1(x).

l l + 1 (ix) 2 (2k + l)Pk(x)Pk(y) = ;--=-·· {Pl+l(x)Pz(y) - P1(x)Pl+l(y)}.

k=O • y

PROOF

(i) We know that P1(x) is a polynomial of degree l, containing only even powers of x if lis even, and only odd powers of x if I is odd. Hence P;(x) is a polynomial of degree l - 1 containing either odd or even powers of x according to whether 1 is even or odd. Hence by theorem 3.6 we have

P;(x) = Cz_1P1 ~1(x) + cl-aPt-3(x) , {c1P 1(x) (l even) + ... + Cz-2r-1Pl-2r-l(x) + ... ~r CoPo(x) (l odd) .

with c. = (s + !) J~ 1 P;(x)P.(x) ds

= (s + !) {[P1(x)P.(x)J~ 1 - J~t1(x(P;(x) dx}


= (s + !){P1(1)P.(l) - P1( -l)P.( -~1) - 0} (where the integral vanishes by the corollary to theorem 3.6, since Ps(x) is a polynomial of degree s - 1, and s - 1 is always less than 1)

= (s + 1){1 - ( -l)"+l} (by theorem 3.4 (i) and (ii)).

§ 3.7 LEGENDRE POLYNOMIALS AND THEIR DERIVATIVES 59

But s takes on the values l - 1, 1 - 3, ... , so that s + 1 takes on the values 21 - 1, 21 - 3, ... , which are always odd, irrespective of the values of 1 or s. Hence ( -1 )s+1 = -1 and we have

c, = (s + !){1 - ( -1)} = (2s + 1).

Hence Cz-zr-1 = 2(1 -- 2r- 1) + 1

= 21- 4r- 1 and so we have

P;(x) = (21- 1)P1_ 1(x) + (2!- S)Pz_ 3(x) + ... + (2! _ 47 _ 1)P

1_

2,_

1(x) + + {3P1(x) if lis even

· · · P0(x) if l is odd (l(l-1)1

= L (2!- 4r - 1)P1_ 2,_1(x). r~o

(ii) xP1(x) is a polynomial of degree l + 1, odd if lis even and even if l is odd. Hence by theorem 3.6 we have

· {c1P1(x) (l even) xPz(x) = CzuPzu(x) + Cz_tPz-t(x) + · · · + cof'

0(x) (l odd),

where

c, = (r + !) J~ 1 xP1(x)P,(x) dx

= (r + !) J~ 1 P1(x).{xP,(x)} dx.

But this integral is zero by the corollary to theorem 3.6 if r + 1 < 1 (since then xP,(x) is a polynomial of degree less than 1), i.e., if r < 1 - 1. Hence we have

(3.25)

To determine c1u and c1_ 1 we set x = 1 in equation (3.25) and in the same equation differentiated with respect to x, viz.

P1(x) + xP;(x) = CzuP;+l(x) + c1_1P;_ 1(x). (3.26)

Setting x = 1 in equations (3.25) and (3.26) and using theorem 3.4 (i) and (iii) gives

1 = c1+1 + c1_ 1 (3.27)

and 1 !- V(l + 1) = cl·lt{W ·1-- 1)(! -!- 2)} l c1 10(/ 1)/}. (3.28)

60 LEGENDRE POLYNOMIALS AND FUNCTIONS CH.3

Solving equations (3.27) and (3.28) for c1+1 and c1_ 1 gives l + 1 l

Czn = 21 + t' Cz-1 = 21 + 1

so that on insertion of these values in equation (3.25) we obtain l + 1 1

xP1(x) = 2i + 1P1 d:\:) + 2l + 1Pz-l(x).

(iii) Multiply both sides of equation (ii) above by (2/ + 1) to obtain (21 + 1)xP1(x) = (l + l)P1+1(x) + IP1_ 1(x).

On rearrangement, this equation becomes (/ + l)Pw(x) - (21 + 1)xP1(x) + /P1_ 1(x) = 0,

which is the required result. (iv) We may use result (i) for both P; +l(x) and P; _1(x):

P; +1(x) = (21 + l)P1(x) + (21- 3)P1_ 2(x) + (21- 7)P1_ 4(x) + .. . P;_ 1(x) = (21 - 3)P1_ 2(x) + (2/ - 7)P1_ 4(x) + .. .

Subtracting these two equations gives P; +r(x) - P;_1 (x) = (2/ + l)P1(x).

(v) If we differentiate result (iii) with respect to x we obtain

(l + l)P{ +1(x) - (21 + 1){P1(x) + xP{(x)} + 1P;_ 1(x) = 0

and if we now use (iv) to substitute for P;+1(x) we shall have (l+l){P{_ 1(x)+(2/+l)P1(x)}- (2l+1){P1(x)+xP{(x)} + ZP{_ 1(x) = 0.

Collecting terms in P{ _1 , P1 and P{ gives

(2/ + 1)P{_1(x) + (2/ + 1)(/ + 1 - 1)P1(x)- (21 + 1)xP;(x) = 0,

which reduces to

and on rearrangement this equation becomes xP{(x)- P{_ 1 (x) = P1(x).

(vi) If we multiply (iv) by x we obtain xP; +1(x) - xP;_1 (x) = (21 + 1 )xPz(x)

and substituting this expression for (2/ + 1 )xP1(x) in (iii) gives (I+ l)Pz+I(x) + IP1_ 1(x) = xP{+l(x) - xP{_ 1 (x). (3.29)

If we now rewrite (v) with l replaced by l + 1, we have xP{+ 1(x) -P{(x) = (l + 1)Pl+l(x),

and substituting this value of (I + 1 )P1+1(x) into equation (3.29) gives

xP; 1 1 (x) P; (:~:) -1 IP1 1(x) = xP; 1 1 (x) xP; 1 (x).

§ 3.7 LEGENDRE POLYNOMIALS AND THEIR DERIVATIVES 61

When rearranged, this equation gives

P;(x) - xP{_ 1(x) = lP1_ 1(x),


(vii) If we multiply (v) by x we obtain

x 2P{(x) - xP;_ 1(x) = lxP1(x)

and on subtracting (vi) from this we obtain

x2P{(x) - r;(x) = lxP1(x) -- lP1_ 1(x)


(x2 - 1 )P{ (x) = lxP1(x) -- lP1_ 1(x).

(viii) If we replace l by l + 1 in results (v) and (vi) we obtain

xP{+ 1(x) - P{(x) = (l + 1)Pl+1(x)

and P{+l(x) - xP{(x) = (l + 1)P1(x).

Eliminating P{+ 1(x) between these (by multiplying the second equation by x and subtracting from the first) gives

-P{(x) + x 2P{(x) = (l + l)Pl+ 1(x) - (l + 1)xP1(x), which reduces to

(x2 - l)P{(x) = (l + 1)P11_1(x) - (l + 1)xP1(x).

(ix) Using (iii) we have

Pk+l(x)Pk(y)- Pk(x)Pk+l(Y) = k ~ 1[{(2k + l)xPk(x)- kPk_1(x)}Pk(y)

- Pk(x){(2k + 1)yPk(y) - kPk_1(y)}J

2k + 1 = k + 1 (x - y)Pk(x)Pk(y)

k + k + 1 {Pk(x)Pk_1(y)- P,o-~(x)Pk(y)};

multiplying both sides of this equation by k + 1, we obtain (k + 1 ){Pk+l(x)Pk(y) - Pk(x)Pk+l(y)}

= (2k+1 )(x-y)Pk(x)Pk(y) +k{Pk(x)Pk-l(y)-Pk_1(x)Pk(y)}. (3.30)

If (k + 1){Pk+1(x)Pk(y) - Pk(x)Pk+l(y)} is denoted by fk, equation (3 .30) may be written in the form

fk , • (2k I 1 )(x - y)P,(x)P,,(y) I f" 1• (3.31)


(Strictly, this equation is true fork > 1, since it is for this range of values of k that all quantities involved are well defined; but we may also make it true for k = 0 provided we define L 1 = 0, for then

f0 = P1(x)P0(y) - P0(x)P1(y) =x-y

(by equations (3.21)) and x - y is also equal to

(2k + 1)(x - y)Pk(x)Pk(y) + fk-l with k = 0.)

If we now sum the set of equations (3 .31) from k = 0 to k = l, we obtain

l l l

L fk = L (2k + 1)(x - y)Pk(x)P1,(y) + L fk_1

k=O k=O k=O l l

= L (2k + 1)(x - y)Pk(x)Pt(y) + z fk_1

k=O k=l l l-1

L (2k + 1)(x- y)Pk(x)Pk(y) + L fk. k=O k=O

Hence

I fk - ~1

fk = i: (2k + l)(x- y)Pk(x)Pk(y) k=O k~O k=O

so that l

f1 = (x - y) L (2k + l)Pk(x)Pk(y) k=O

and if we remember the definition of f1 we obtain

(l + 1) * (x _ y} {P1+I(x)P1(y) - P1(X)Pz+I(y)} = 6 (2k f- l)Pk(x)Pk(y).

3.8 ASSOCIATED LEGENDRE FUNCTIONS

Theorem 3.9

If z is a solution of Legendre's equation

(1 - x2)d2

y- 2}Y + l(l + l)y = 0 dx2 dx

then (1 - x 2)mf2 ( dmz/dx"') is a solution of the equation

d2y dy { m2 } (1 - xz) dxz - 2xdx + l(l + 1) - 1 - x2 Y = 0

(/mown as the associated Legendre equation).

§ 3.8 ASSOCIATED LEGENDRE FUNCTIONS 63

PROOF

Since z is a solution of Legendre's equation, we must have

d 2z dz (1 - x2)dx2 - 2xdx + l(l + 1)z = 0. (3.32)

Now let us differentiate equation (3.32) m times with respect to x:

_t!:_{(t- x2)d2z}- 2dm{xdz} + l(l + l)dmz = 0 dxm dx2 dxm dx dxm

which, when we use Leibniz's theorem for the mth derivative of a product,t becomes

dm+2N d dm+lz m(m - 1) d 2 dmz

(1 -- x2)--"' + m-(1- x2).-- + -- (1- x2).-dxmH dx dxm+I 2 dx 2 dxm

{ dm+Iz d dmz} dmz

-2 xdxm+I + mdxx.dxm + l(l + 1)dxm = 0

(since higher derivatives of 1 - x2 and x vanish).

Collecting terms in dm+2zjdxm+ 2, dm+ 1zjdxm+1 and dmzjdxm, we obtain dmHz dm+Iz dmz

(1-x2)dxm+ 2 - 2x(m+1)dxm+l + {l(l+1)-m(m-1)-2m}dxm = 0,

which, on denoting dmz;dxm by z1o becomes

d2zl dzl (1 - x 2) dx2 - 2(m + 1)x dx + {l(l + 1) - m(m + 1)}z1 = 0. (3.33)

If we now write dmz

z2 = (1 _ x2)m12z1 = (1 _ x2)mf2 d.\~

equation (3.33) becomes

d2 d (1- x2)dx2{z2(1- x 2)-m'2}- 2(m + l)x<G:{z2(1- x2)-m/2}

+ {l(l + 1) - m(m + 1)}z2(1 - x2)-m12 = 0. (3.34) But

d d~ m dx{z2(1 -- x2)-m12} = dx(1 - x2)-m!2 + z2.- .2(1 - x2)-(m/2l-1. -2x

dz2 ( = -(1 - x2)-m12 + mz2x(1 - x2)- mt2)-1 dx


+ m{ ~:x(1 _ x2)-(m/2l-1 + z~(1 _ xz)-(m/2)-1

+ z2x( -I_ 1)(1 - x2)-<mt2>-2. -2x}

CH. 3

d 2z dz dz = dx:(l _ x2)-ml2 + -~mx(1 _ x2)-(m/2l-1 + m dx2x(l _ x2)-(m/2)-1

+ mz2(1 _ x2)-<mt2l-1 + mz2x2(m + 2)(1 _ x2)-(m/2l-2.

Hence equation (3.34) becomes d 2.:r2 dz2 dx2 (1 - x2)-<mt2)+1 + 2mx(1 - x2)-m12 d_-;- + mzl1 - x2)-mf2

+ m(m + 2)(1 - x2)-(m/2l-lx2zz

{ d~ } - 2(m + l)x (1 - x2)-ml2 dx + mx(1 - xz) -<mt2l-1z2

+ {l(l + 1)- m(m + 1)}z2(1 - x 2)-m12 = 0 which, on cancelling a common factor of (1 - x 2)-m12 and collecting like terms, becomes

2 d2

Z 2 { }dz2 (1 - x) dx 2 + 2mx-2(m + l)x dx

{ m(m+2) 2(m+l)mx2

} + m + l-x2 X

2-- l-x2 - + l(/+1) - m(m+l) z2 = 0. (3.35)

The coefficient of dz2/dx is just -2x, while the coefficient of z2 is

(m 2 +2m - 2m 2 - 2m)x2

l(l + 1) + 1 _ x 2 + m- ms- m

m2x2 = l(l + 1) - 1=-X2- m2

m2 = l(l + 1) - 1 - ,;2'

Thus equation (3 .35) reduces to

d 2z2 dz2 { m2 } (1 x 2)--- -· 2x- -1 l(l + 1)--- z 2 = 0

dx 2 d.t' 1 -- x2

§ 3.9 PROPERTIES OF ASSOCIATED LEGENDRE FUNCTIONS 65

so that z 2 satisfies the associated Legendre equation which, by the definition of z 2, proves the theorem.

CoROLLARY

The associated Legendre functions Pj(x) defined by dm

P;"(x) = (1 -- x 2)m12 dx"' P1(x)

satisfy the associated Legendre equation.

PROOF

(3.36)

This result follows immediately from the theorem, since P1(x) satisfies Legendre's equation.

Using Rodrigues' formula (theorem 3.2), it is possible to rewrite definition (3.36) in the form

1 dl+m Pj(x) = 2'1!(1 - x2)m/2 dxl+m(x2 - l)z.

The right-hand side of this expression is well defined for negative values of m such that l + m > 0, i.e., m > -1, whereas the original definition (3.36) of P!(x) was only valid form > 0. Thus we may use this new form to define Pj(x) for values of m such that m > -l.

It is easy to verify that if we consider m positive, the function P1-m(x) defined in such a way is a solution of Legendre's associated equation as well as Pj(x). In fact, it is not an independent solution; it may be proved that

-m - - m (l- m)! Pz (x) - ( 1) (l + m)!P;"(x) (3.37)

(see problem 3 at the end of this chapter).

3.9 PROPERTffiS OF THE ASSOCIATED LEGENDRE FUNCTIONS

Theorem 3.10

(i) P?(x) = P1(x). (ii) P;"(x) = 0 if m > l.

PROOF

(i) This result is immediately obvious from definition (3.36).

(ii) Since P 1(x) is a polynomial of degree l, it will reduce to zero when

differentiated more than l times. Thus _ddm P1(x) = 0 for m > l, and the xm required result then follows from definition (:lJ(,).


Theorem 3.11 (Orthogonality Relation)

fl pm( )Pm( ) d 2(! + m)! ~ -1 l X l' X X=(2f+l)(l-m)!ull'•

PROOF

We first prove that if l ~ l'

f1

Pt(x)P{': (x) dx = 0. -1

This proof follows exactly the same lines as that of the first part of theorem 3.5, so we shall not repeat it here.

All that now remains to be proved is that

fl {Pm( )}2 d - 2(! + m)! -1

1 X X-(2[ + 1)(/- m)!'

Assume first that m > 0; then from definition (3.36) we have

J~ 1 {Pf"(x)}2dx

= f~ 1 (1 - x2)m{:x:P1(x)} {::nPt(x)} dx

= [ {:;.-~1Pt(x)} {c1 - x2)m :mP1(x)} J :: - f~ 1 {!mm-~1Pt(x)} !J(l - x 2)m :mP1(x)} dx


= -ft {-dm-1 Pl(x)}-~{(1 - x2)m dm Pl(x)} dx -1 dxm- 1 dx dxm

(3.38)

(the first term vanishing at upper and lower limits because of the (1 - X2)

factor). Now, from equation (3.33) with m replaced by m - 1 we have that

dm+l dm dm-l (l-x2)dxm+1P1(x) - 2mx dxmP1(x) +{l(l+1)- (m-1)m} dxm_1P1(x) = 0

which, when multiplied by (1 - x 2)m-t, becomes dm+1 dm

(1 - x 2)m d.xm+ 1P1(x) - 2mx(l - x 2)m-l dxmP1(x)

dm-1 + (l + m)(l - - m -1- 1 )(1 - x 2)"'-1 - -- --P,(x) = 0 dxm-1


and this equation may be rewritten in the form

d { dm } dm-1 dx (1 - x2)m dxmP1(x) = -(1 + m)(1- m + 1)(1 - x 2)m-l dxm_

1P1(x).

Substituting this result in equation (3.38) gives

J~ 1 { P["(x)} 2 dx

fl {dm-1 } = _1

dxm_1P1(x) (l + m)(1 - m + 1 )(1

Applying this result again gives

J~ 1 { P["(x)}2dx = (l+m)(l-m+1)(l+m-1)(1-m+2) J~ 1 {J>i'-2(x)} 2 dx

= (l+m)(l+m-1)(l-m+1)(l-m+2) J~ 1 {J>i'- 2(x)}2 dx,

and repeating the process m times in all we obtain

J~ 1 {P["(x)}2 dx

= (l+m)(l+m-1) ... (l+1).(1-m+1)(1-m+2) ... l. J~ 1{P?(x)}2 dx

2 = (l+m)(1+m-1) ... (1-+·1).1(1-1) ... (1--m+2)(1-m+1).21

+ 1

(l + m)! 2 (l- m)f 21 + 1

(using theorem 3.5)

which is the required result. Suppose now that m < 0, say m = -n with n > 0.

Then J~ 1 {P["(x)}2 dx = J~ 1 {P1-n(x)}2 dx

= f~l {< -t)n ~~ ~ :~:r{Pi(x)} 2 dx

(by equation (3.37)) SF-P


{(1-- n)!}2 f1

= (l + n}! _1

{Pi(x)}2 dx

{(l- n)!}2 (l + n)! 2

= (1-t-n)! (l-n)!2l+1

(by the result just proved, since n > 0)

(l-n)! 2

(1 + n)l21 + 1

(l+m)l 2 (1- m)l21 + 1


Theorem 3.12 (Recurrence relations)

(i) Pj+1(x) - v(~": x 2)Pj(x) + {1(1+1)-m(m-1)}Pj- 1(x) = 0.

(ii) (21 + 1)xPj(x) = (1 + m)P[:_ 1(x) + (l- m + 1)P41(x).

(iii) v(1 - x2)Pj(x) = 21 ~ l {P4i1(x) - PF:-i1(x)}.

(iv) v(1 - x 2)Pf'(x) = 21 ~

1 {(l + m)(l + m- l)P[:_11(x)

- (l- m + 1)(1- m + 2)Pi,:.11(x)}.

PROOF

(i) This is the fundamental relationship linking three associated Legendre functions with the same 1 values and consecutive m values.

Let us denote ( dm jdxm)P1(x) by P1<m>(x) so that definition (3.36) may be written in the form

(3.39)

Now, in equation (3.33) we know that we may take z = P1(x) and hence z 1 = Plm>(x), so that we obtain

d2 d (1 - x 2)dx2pr>(x) - 2(m + 1)xdxPlm>(x)

+ {1(1 + 1) - m(m + 1)}P{m>(x) = 0. Using the definition of Pfm>(x), this equation becomes

(1 - x2)Plm+2>(x) - 2(m + 1)xP1<m+I>(x)

+ {l(/1 1) -- m(m I l)}Pfm>(x) = 0


which, on multiplying throughout by (1 - x 2)m12, gives

(1 - x2)(m/2l+1pfm+2>(x) - 2(m + 1)x(1 - x2)m12pfm+I>(x)

+ {l(l + 1) - m(m + 1)}(1 - x2)mi2Pfm>(x) = 0.

Hence, using equation (3.39), we have

Pj+ 2(x)- 2(m + 1)x v'(l ~ x2)Pj+1(x)

+ {l(l + 1)- m(m + l)}Pj(x) = 0 which, when m is replaced by m - 1 , becomes

2mx Pj+ 1(x) - v'(

1 _ x2{i(x) + {l(l + 1) - (m- 1)m}Pj- 1(x) = 0;

this is the required result.

(ii) This is the fundamental relationship between associated Legendre functions with equal m values but consecutive I values.

By theorem 3.8 (iii) we have

(I+ 1)P1+1(x) - (21 + 1)xP1(x) + ZP1_ 1(x) = 0 which, when differentiated m times (making use of Leibniz's theorem for the second term), gives

(I+ 1)Pf~Hx)- (21 + l){xPfm>(x) + mPJm-l>(x)} + lPf'~~'t(x) = 0. (3.40)

Similarly by theorem 3.8 (iv) we have

Pf~1(x) - Pf~1(x) = (21 + l)P1(x) which when differentiated m - 1 times gives

Pf~>1(x) - Pf~i(x) = (21 + l)Pfm-ll(x). (3.41)

Using equation (3.41) to substitute for Pjm-I>(x) in equation (3.40) gives

(l + 1)Pf~l1(x) - (21 + 1)xPfm>(x)- m{Pf~>1(x) - Pf~i(x)} + ZPf~i(x) = 0.

Multiplying this equation throughout by (1 - x 2)ml2 and using equation (3.39) gives

(l+1)P~ 1(x)- (2l+1)xPj(x)- mPJ't 1(x) + mPI"- 1(x) + 1PI"- 1(x) = 0.

Collecting like terms gives

(1 + 1 - m)P~ 1(x) - (21 + 1)xPj(x) + (1 + m)Pf-_ 1(x) = 0 which, when rearranged, is the required result.

(iii) Multiply equation (3.41) throughout by (1 - x 2)m1 2 and we obtain (1 x2)m12J>f;I)J(x) (1 x2)m12J>}"'Hx) (21 ! 1)(1 x2)mi2Pl"' t>(x)


which, on using equation (3.39), becomes

JH_ 1(x)- Pj':_ 1(x) = (21 + 1)v(1 - x2)Pf'- 1(x). (3.42)

Replacing m by m + 1 gives

p~~ 1 (x) - Pj':_~ 1 (x) = (21 + 1)v(1 - x 2)P;r'(x) which, when divided by 21 + 1, is just the required result.

(iv) We use (ii) to replace xP;n(x) in (i) by

21 ~

1 {(! + m)Pj':_ 1(x) + (l- m -1- 1)P~ 1 (x)}

so that we obtain

Pj+ 1(x) - v(12: X~ 21 ~ 1 {(I+ m)P~l(x) +(I- m + 1)P4t(x)}

+ {l(l + 1) - m(m - 1)}P;n-1(x) = 0.

If we now use equation (3.42) for P["- 1(x), we obtain

P;n+ 1(x) - v(1 ~ x2) (212: 1){(1 + m)Pj':_1(x) + (l - m + 1)Pl';- 1(x)}

+ {l(l + 1) - m(m- 1)} v(1 ~ x 2) 21 ~ 1 {P4 1(x) - Pz'~ 1(x)} =0.

By straightforward algebraic manipulation this reduces to

v(1 - x2)Pf'+l(x)

1 =

21 +

1 {(I+ m)(l + m + 1)P~ 1(x) -(I- m)(l- m + 1)Pt'~_ 1(x)}

which, when m is replaced by m - 1 , is just the required result.

3.10 LEGENDRE FUNCTIONS OF THE SECOND KIND

In the first section of this chapter we obtained two independent series solutions y 1(x) and y 2(x) of Legendre's equation. We obtained solutions finite for -1 < x < 1 (indeed, finite for all finite values of x) by taking I integral; then for I even y 1(x) reduced to a polynomial, while for l odd y 2(x) reduced to a polynomial. In both these cases the other series remains infinite; it may be shown to be convergent for I x I < 1 and divergent for I x I > 1. In some physical situations we wish two independent solutions valid for the region I x I > 1 ; one of these is, of course given by P1(x), while a second solution is given by the following theorem (note that it is still infinite for x = ± 1 ).

§ 3.10 LEGENDRE FUNCTIONS OF THE SECOND KIND

Theorem 3.13

A second independent solution of Legendre's equation is given by

[l--;1] 1 + x ~ (21 - 4r - 1)

Q1(x) = !P1(x) In l _ x - ;S (2r + 1)(1- rtl-2r-I(x)

1+x Q0(x) = ! In

1 _ x

where -- = [I _ 1] {/ ~ 1

if I is odd

2 l-2ifi' -2

- t ts even.

Q1(x) is called the Legendre function of the second kind.

PRoopt

(l > 1)

71

In Legendre's equation, set y = zPkx) so that z is a new dependent variable. We have

dy dz dP1

dx = Pz(x)dx + z dx'

d2y = Pz(x)d

2z + 2~ dP1 + zd

2P1

dx2 dx2 dx dx dx 2

and hence the equation becomes d 2z dz dP d 2P dz

(1 - x2)P1(x)dx2 + 2(1 - x2~ dxz + (1 - x 2)z dx21

- 2xP1(x)dx

dPz - 2xz dx + l(l + 1)zP1(x) = 0.

Collecting terms in z, dzjdx and d 2z/dx 2, we have

{ d~ dP } z (1 - x2

) dx2

1 - 2x dxz + l(l + 1 )P1(x)

dz{ 2 dP1 } 2 d 2z _ + dx 2(1 - x ) dx - 2xP1(x) + (1 - x )P1(x) dx2 - 0,

which, on using the fact that P1 satisfies Legendre's equation, becomes

d2z dz{ dP1 } (1 - x 2)P1(x) dx2 + dx 2(1 - x2

) dx - 2xP1(x) = 0.

t The proof of this theorem may be omitted on a first reading. The proof ends on page 77.


Hence d 2zjdx 2

2dPzldx 2x _

0 ---- + -- -- --dzjdx PL(x) 1 - x 2 - '

and this is equivalent to

d (dz) d d dx In dx + 2d; In PL(x) + ;r; In (1 - x2) = 0,

which when integrated gives

dz In dx +In {PL(x)}2 +In (1 - x2) =constant.

dz Therefore dx{P1(x)}2(1 - x2) =constant= A, say,

so that dz A

and hence

This means that we have a solution of Legendre's equation given by

J dx Qz(x) = Pz(x) {Pz(x)}2(1 - x2)" (3.43)

We must now show that this is of the form stated in the theorem. We first dispose of the case I = 0:

J dx

Qo(x) = Po(x) {Po(x)}2(1 - x2)

= L ~x2 = J !(_1 + __ 1 )dx

21-x 1+x

= H -In (1 - x) +In (1 + x)}

_ 11 1+x -2 n-1--.

-X

If now I ;z': 0, we remember that P1(x) is a polynomial of degree I, so that it may be written in the form

P,(x) = k1(x - cx1)(x - cx2} ••• (x - cx1).

Thus

§ 3.10 LEGENDRE FUNCTIONS OF THE SECOND KIND 73

on splitting into partial fractions. We may determine a0 , b0 and cr fairly simply. Multiplying both sides of

equation (3.44) by (1 - x2){P1(x)}2 gives

1 = a0(1 + x){P1(x)}2 + b0(1 - x){P1(x)}2

+ (1 - x2){P1(x)}2{ 2 ( ~ ) + ( ~r ) 2}. r=l X ot, X ot,

Setting x = 1 in this equation and remembering that P 1(1) = 1 gives a0 = !, while setting x = - 1 and remembering that P1( -1) = ( -1 )1

gives b0 = t· We now show that

c, = [d~{(x -- o:;)2 (1 - x 2~{P1(x)}2} l~"i. To prove this we note that

d df d)(x - o:i) 2f(x)} = 2(x - o:1)f(x) + (x - o:i) 2dx

= 0 when x = ex.;, provided that f(x) is finite at x = o:;.

The only terms on the right-hand side of equation (3.44) which are not finite at x = ot; are cd(x - o:;) and dd{(x - cx.;) 2}.

Hence we have

= c,. Thus, if we write P1(x) = (x - o:;)L(x) we have

Ct---, [!X (i ____ )){/,(~)}21 ex;


[ 2x 2L'(x) J

= (1 - x2){L(x)} 2 - (1 - x2){L(x)}3 x="i

= [2xL(x) - 2(1 - x 2)L'(x)] (1 - x 2){L(x)}S x=<x,

2{ex;L(ex1) - (1 - ex; 2)L'(ex1)}

(1 - ex12) 2{L(ex1)}3

(3.45)

But we know by setting P1(x) = (x - ex;)L(x) in Legendre's equation that

d2 d (1 - x 2}dx

2{(x- cx 1)L(x)}- 2xdx{(x- cx1)L(x)}

+ l(l + 1)(x - ex;)L(x) = 0,

which, on performing the differentiations, becomes

(1 - x 2){(x - cx1)L"(x) + 2L'(x)} - 2x{(x - IX;)L'(x) + L(x)} + /(/ + 1 )(x - cx1}L(x) = 0,

and setting x = ex1 in this equation gives

(1 - cxD2L'(cx1) - 2cx1L(cx;) = 0

so that by substituting back into equation (3.45) we obtain c1 = 0. Thus from equation (3.44} we have

1 1 1 * dr (1 - x 2){P1(x)} 2 = 2(1 - x) + 2(1 + x) + £-: (x - ex.)2

where the d, are constants whose values will not concern us. Thus

f 1 l d

(l _ x 2){P1(x)}2dx = -!In (1 - x) + ! In (1 + x) - 6 (x __:_ cxr)

l

= t In 1 + X _ ""'"' dr 1

1 - X r~ (x - 1Xr)

so that from equation (3.43) we have l

Q1(x) = !P1(x) In ! + : - 6 d, :~x~, . But (x - ex,) is, for all ex" a factor of P1(x), so that P1(x)j(x - ex,) is a

l

polynomial in x of degree /-1. Thus ,2 dr {P1(x)}/(x- ex,) is a poly-r=l

§ 3.10 LEGENDRE FUNCTIONS OF THE SECOND KIND 75

nomial of degree l- 1; let us denote it by W1_ 1(x). Then we have

Q1(x) = iP1(x) In~ ~: - W1_ 1(x). (3.46)

To determine W1_ 1(x) we remember that Q1(x) is a solution of Legendre's equation so that

! {<t - x 2)~;} + l(l + l)Q1 = 0,

which, on use of equation (3.46), gives

ld{ d l+x} 1 l+x z d-; (1 - x 2)dxP1(x) ln f _ x + l(l + 1). 2P1(x) In 1 _ x

d { dW,_t} - dx (1 - x2)dX- - l(l + t)w;_1 = 0.

But

d 1 +x 1 It +x { 1 1 } -d P1(x) In -1

- = P1 (x) In -1

- + P1(x) -1

- + --x -x -x +x 1-x

I 1 +x 2 = P1(x) In -

1 - + P1(x)-

1 -

2 - x -x

so that

= d~{(l - x 2)Pl(x) In~ ~; + 2Plx)}

(3.47)

I 1 + X d { PI( )} I 2 I = n 1 _ x dx (1 - x2) 1 x + (1 - x2)P1(x)

1 _ x 2 + 2P1(x).

Hence equation (3.47) becomes

! ln.~ ~: [a~{(l - x 2)P((x)} + l(l + l)P1(x)J + 2P;(x)

d { dlf'i_t} - dx (1 - x2)dX -1(1 + t)w;_l = 0

which, on remembering that P1(x) satisfies Legendre's equation, reduces to

_d {<1 dx

q dW1-1} dP, x·) --d~ -f /(1 I 1)w;_ 1 = 2dx · (3.48)


Now, by theorem 3.8 (i) we have

dP, dx = (21- l)P1_ 1(x) -+- (21- S)P1_ 3(x) -+- ...

+(21- 4r - l)Pz-2r-1(x) -+- · • • [1(1-1)]

= ''J)21- 4r- 1)Pl-2r-l(x) r=O

CH.3

(3.49)

so that if we assume for W1_ 1(x) (which we know to be a polynomial of degree l - 1) an expression of the form

W,_1(x) = aoPz-1(x) + a1P,_lx) + ... [!(l-1)]

= LarPl-2r-l(x) (3.50) r-- 0

and substitute equations (3.49) and (3.50) into equation (3.48), we shall obtain

[)(l-1)) d { } [J(Z-1)]

Lar dx (1 - x 2)P;_2r-1(x) -+- 1(1 + 1) LarP1_ 2,_1(x) r-o r=O

[l(l-1)]

= 2 2(21- 4r - l)P1_ 2r_ 1(x). (3.51) r=O

But by Legendre's equation we have

d~ { (1 - x 2)PI - 2r-1(x)} + (l - 2r - 1 )(l - 2r)P1_ 2r_1(x) = 0,

so that equation (3 .51) becomes

[!(Z-1))

La,{ -(1- 2r- l)(l- 2r) + l(l + l)}P1_ 2r_ 1(x) r=O

f!(l-1)]

= L2(21- 4r- l)P1_ 2,_1(x). r=O

The coefficient of each polynomial must be the same on both sides, so that we obtain

{-(1- 2r- 1)(1- 2r) + l(l + l)}ar = 2(21- 4r -1). (3.52)

§3.10 LEGENDRE FUNCTIONS OF THE SECOND KIND 77

But -(! - 2r - 1)(/ - 2r) + l(l + 1)

= -(/- 2r) 2 + (!- 2r) + l(l --+- 1) = -12 + 4rl - 4r 2 + l -2r i- / 2 + l = 4r(l - r) + 2(/ - r) = 2(/- r)(2r + 1).

Hence equation (3 .52) reduces to

2(1- r)(2r + 1)ar = 2(2/ - 4r - 1)

which gives 21- 4r- 1

a = -,-----,-,-,---' (l-r)(2r+1)

(3.53)

and now, by combining equations (3.53), (3.50) and (3.46), we obtain immediately the result of the theorem.

That the solution of Legendre's equation Q1(x) which we have obtained

is inde~endent of P,(x) is readily seen :,because of the factor In 1

1 + x, Q1(x) -X

is infinite at both x = ± 1, whereas we know that P1(x) is finite for these values of x.

We may use this theorem to write down explicitly the first few Legendre functions of the second kind:

1 +X Qo(x) = ! In 1 - x

x 1+x Q1(x) = 2tn

1 _ x - 1

1 +x 3 Q2(x) = :!(3x2 -1) In

1 _ x- 2x

1 +X 5 2 Q3(x) = :!(Sx3 - 3x) ln

1 _ x - 2x2 + 3·

We now state without proof several theorems concerning Legendre functions of the second kind.

Theorem 3.14 1 00

-- = L (21 + 1)P1(x)Qz(y) x- Y 1~o

if ."C > 1 and I y I < 1.


Theorem 3.15 (Neumann,s Formula)

Q,(x) = ! JI P,(y) dy. . 2 -lX- y

Theorem 3.16

The results contained in theorem 3.8 (ii- ix) remain true when P1(x) is replaced by Q1(x).

Theorem 3.17

The associated Legendre functions of the second kind defined by

dm Qj(x) = (1 - x2)m/2 dxmQ,(x)

satisfy Legendre's associated equation.

3.U SPHERICAL HARMONICS

In many branches of physics and engineering there is interest in the equation

1 ( a . olf'\ 1 a2'I' sin() o() Stn () ()0) + sin2 () ocp2 + /(l + 1)'J' = O, (3.54)

solutions of which we shall call spherical harmonics. (This equation usually arises in the solution of a differential equation such as Laplace's or Schrodinger's in terms of spherical polar co-ordinates r, (), cp, so that the range of the variables involved is 0< 0< n, 0< cf> < 2n and we often require a solution which is finite and continuous for these values-the continuity implying that the value of 'I' at cf> = 2n is the same as at cf> = 0.)

One method of finding a solution of equation (3.54) is the so-called method of separation of variables-we look for a solution of the form 'I'((), c/>) = e(fJ)(j)(cf>). Inserting this expression into the given equation gives

(jj(cf>){d (. de)} e(e) d2(j)

sin() d() sm () d() + sin2() dcf>2 + l(l + 1)e(fJ)(jj(cp) = 0

or, dividing throughout by e( ())(j)( cf>) and multiplying by sin 2 (),

sin () d ( . 0de) 1 d 2(j) I( I 1) . 2 0 _ 0 e d() Stn d() + (j) dcp2 -j- -j- Stn -

which, when rearranged, becomes

sin 0 d ( . de) . 1 d 2(j) -- -- sm 0- -!-l(l + 1) sm2 0 = --- ·--. (-) dO dO (j) dcf> 2

§ 3.11 SPHERICAL HARMONICS 79

Now the left-hand side of this equation is a function only of the variable (),while the right-hand side is a function only of the variable cp. Since these two variables are independent, it follows that left-hand side and right-hand side must separately be a constant which we shall denote by m2•

Thus we have

sin() d ( . de) l(l l) . () e dO Sill 0 dO + + sm2 = m2 (3.55)

1 d 21:/> and -if> dcp 2 = m2

• (3.56)

Equation (3.56) is just

d 2(/J - - 2(/> dcfo2- m

while equation (3.55) simplifies to

si! e :a (sine~~) + {z(l + 1) - si:: ~e = o. Equation (3.57) has the general solution

(/> = Aeim<f> + Be-im<f>

(3.57)

(3.58)

where, if the solution is to be continuous, we require l:/>(2n) = 1:/>(0), so that m must be an integer (which we may take conventionally to be positive).

In equation (3.58) we make the change of variable cos() = x. Then we have - sin 0 d() = dx

and hence 1 d d

sin() d() dx

and . () d . () d (1 ) d sm d() = sm2 . - dx = - - x2 dx'

Accordingly, equation (3.58) becomes

:Jet -x2)~~} + {z(l + 1) - 1 :2x2}e = 0

which we recognise as Legendre's associated equation; it will have a solution finite at() = 0 and n (x = + 1 and -1) only if lis integral. In that case the finite solution is given bye = Pl(x) = Pf(cos fJ).

Thus the general solution which is finite at both 0 = 0 and n and is continuous must be

'I'(O, cp) = (Aeim<f> + Be-i""")Pf'(cos 0)

which, because of equation (3 .37), we may write in the form 'I'--, A1c

1m4>p•;(cos 0) 1- A2c-im<f>P-j(cos 0)


where A 1 =A

and A =(-l}m(l+m)!B 2 (l- m)! ·

If now we denote

y/(0, c/>) = eim.f>Pi(cos 0) we may write the general solution in the form

T = AIJi(O, c/>) + A2Yzm(O, cp).

CH.3

(3.59)

Of course, this is a solution of the original equation (3.54) for any value of m, and since (3.54) is homogeneous we have the solution

l

T = L {Aimlyi(O, c/>) + A~nly;-m(O, cp)}. m=O

For many purposes it is more useful to consider a multiple of yj (which we shall denote by Yi) as the basic solution; a multiple chosen so that the solutions are orthogonal and normalised in the sense that

f~ dcf> J: dO sin O{Yj(O, c/>}}* Yj.''(O, c/>) = llu~mm' (3.60)

(where the* denotes complex conjugation). We may readily prove that this is accomplished by taking

m _ m 1 J{(2/ + 1}(/- m}!} m Yl(O,cf>)-(-1) v(Zn) 2(/+m)! y1 (0,cf>)

_ m 1 J{(2l + 1)(/ - m}!} im.f> - ( -1) v(2n} 2(/ + m)! e Pj(cos 0). (3.61)

For then we have

J:" dcp J: dO sin O{Yi(O, c/>)}*Yr,''(O, c/>)

= ( -l)m+m' J_J{(Z/ + 1}(2/' + 1}(/- m}! (/'- m'}!} 2n 4(1 + m)! (l'+ m')!

f~ e1<m' -m).f> de/> J: Pf' (cos 0) PJ:!' (cos 0) sin f) dO

(using the fact that Pi(x) is real),

= ( _ 1)m+m'_!__J{(2/ + 1)(2/' + 1)(/- m)! (l'- m'}!}Zn {J • 2n 4(1 + m}! (l' + m'}! m m r I Pi'(x)Pp'(x) dx

§ 3.11 SPHERICAL HARMONICS 81

(since the first integral vanishes unless m' = m, in which case it is equal to 2n; and in the second integral we have made the substitution x = cos())

= ( -l) 2m J{(2l + 1)(2/' + l)(l- m)! (l'- m)!}b , 4(1 + m)! (l' + m)! mm

f~1 P{"(x)P;'(x) dx

= J{(2l + 1)(2/' + 1)(/- m)! (l' - m)!}b , 2(1 + m)! b, 4(/ + m)! (l' + m)! mm (21 + 1)(1- m)! 11

(by theorem 3.11) = bu.bmm'·

The factor ( -t)m in definition (3.61) of Yj((), cf>), which we shalf take as the basic spherical harmonic, was not necessary for the orthonormality property; however, its introduction is conventional (although the reader is warned that in the topic of spherical harmonics different authors may employ different conventions).

Theorem 3.18

PROOF

{Yi(e, c/>)}*

1 J{2l + 1 {l- m)!} -'~ =(-1)mv(2n) -2-(l+m)!_e .Pi'(cose)

(by equation (3.61))

= ( -l)m _1_J{2/ + 1 (/- m)!}e-'~( -1)-m v(Zn)· 2 (l + m)!

(l+m)!p me ) (I _ m)! £ cos ()

(by equation (3.37))

__ m (- )-m _1_J{2l ±_1 (/ + m)!} -imc/> -m -- ( 1) . 1 v(2n) 2 (/ _ m)! e P1 (cos 0)

= ( J)mYI m(O, c/>) (by equation (3.61)).


We may use the definitions of Yj(O, cf>) and Pj(cos 0) to obtain the following explicit expressions for the first few spherical harmonics:

0- J( 1 )· Yo- 4n '

Y± 2 = J(~) sin 2 e e±zi.p y±t = -J(15) sine cos e e±i<l> 2 32n ' 2 Sn '

yg = J C~n)(3 cos2 e- 1). (3.62)

3.12 GRAPHS OF THE LEGENDRE FUNCTIONS

In this section we give graphs of some of the functions encountered in this chapter.

X

FIG. 3.2 Pt(:>:), 0 < X< 1, l = 1, 2, 3, 4

§ 3.12 GRAPHS OF THE LEGENDRE FUNCTIONS

103

1o'

5 10 X

FIG. 3.3 Pt(x), x > 1, l = 1, 2, 3, 4 (Note that the scale on the vertical axis is logarithmic.)

~(cos 9)

Fu;. 3.4 p, (ens 0), 0 · 0 · · :r /2, I I, 2, 3, 4

83

LEGENDRE POLYNOMIALS AND FUNCTIONS

P/lxl

X

Fxa. 3.5 Pi'(x), 0 < x < 1, l = 1, 2, 3

FIG. 3.6 Q,(x), 0 ,, , x...; 1, l - 0, 1, 2, 3, 4

§ 3.13 EXAMPLES

q (x)

X

FIG. 3.7 Qt(x), x > 1, l = 0, 1, 2, 3, 4 (Note that the scale on the vertical axis is logarithmic.)

3.13 EXAMPLES

Example 1

fl 2 - 2l( l + 1) Show that _

1 x Pl+ 1(x)Pl-l(x) dx - ( 412 _ 1 )(Zl + 3)"

Deduce the value of J: x 2P1+I(x)P1_ 1(x) dx.

85

We use theorem 3.8(ii) to dispose of the .Y2 appearing in the integrand, and we may then use the orthonormality property of theorem 3.5.

Thus,

f ~ 1 X 2P, I L(x)P,_ L(x) dx - . r I {xP, il(x) }{xP, L(x)} dx


21(1 + 1) (412

- 1)(21 + 3)'

(by theorem 3.5)

CH. 3

We know that P1(x) is a polynomial of degree /, so that the above integrand is a polynomial of degree 2 + (! + 1) + (I - 1) = 2(1 + 1 ), i.e., of even degree. Thus the integrand is even, so that

f1

x 2Pw(x)PH(x) dx = 2 J1

x 2P1+1(x)P1_ 1(x) dx -1 0

and hence

Example 2

Evaluate J: P1(x) dx when I is odd.

By theorem 3.8 (iv) we have

J: P1(x) dx = Z/ ~ 1 f: {PiH(x) - P~_ 1 (x)} dx

= 21 ~ 1 [Pz+I(x) - Pz-I(x)J ~ 1

= 21 + 1 {Pz+I(l) -- Pz-t(l) - Pz+l(O) + P 1 _ 1(0)}

§ 3.13 EXAMPLES 87

1 { ( -1)(1+1)/2 (l + 1)! = 21 + 1 1 - 1 - -21+1- [{(l--t=-T)}i}!T2

( -l)(l-1)/2 (1- 1)! } + 2z-1 [{(l- 1)/2}!)2

(by theorem 3.4(i) and (v), remembering that l + 1 and I-- 1 are both even)

Example 3

1 ( -1 )(l- 1)/

2 (l -- 1)! [ ( -1 )( l + 1 )/ J -21 + 1 21- 1

[{(/- 1)/2}!)2 1

- 2 2{(1 + 1)/2}2

1 ( -1)<t-t);z (l- 1)! [ I J = 27 + 1 --21_:_-1 ~[{(1-1)/2}!]2 1 + l +l

( -1)(1-1)/2(/- 1)!

= ztcF + ~>az -- i)T(!l -=-~)! ( --1)<1-1)/2(/- 1)!

ztctz ~- t>' ({1 - ~> r

00

ljf(x) =I x lfor -1 < x < 1, expandf(x) in the form 2 c,.P,.(x). r=O

By theorem 3. 7 we know that such an expansion is possible and that c,. is given by

c,. = (r + !) J1

I X I Pr(x) dx. -1

Now, I xI is an even function and P,.(x) is even if r is even and odd if r

is odd. Hence if r is odd I xI P,.(x) is odd, so that J~ 1 I xI P,.(x) dx = 0

and hence c,. = 0. On the other hand, if r is even, I x I P,.(x) is even, and hence

so that now

J~ 1 I xl P,.(x) dx = 2 L I xI P,.(x) dx

= 2 J: xP,.(x) dx

c, (2r !1) J1

xP,.(:~) dx. ()

(3.63)


To evaluate this integral we use theorem 3.8(ii):

c,. = (2r + 1) J: {;,: \Pr+l(x) + 2, ~ 1P,._1(x)} dx

= J: {(r + 1)Pr+l(x) + rP,._1(x)} dx.

CH. 3

Now, r is even, so that both r + 1 and r- 1 are odd, and we may use the result of example 2 to obtain

{ -l)r12r 1 ( -1 )(r/2)-l(r _ 2)! c, = (r + 1)2r+I(!r + 1)!(!r)! + r-zr-l(!r)!Gr- 1)!

( -1Y12(r- 2)! {(r + 1)r(r- 1) } = z-;::-1(!r)!(tr- 1)! 2 2(!r + 1Hr - r

= _ ( -1)"'2(r- 2)! {(r + 1)(r- 1) _ r} 2'-1(ir)!(!r -1)! r +2

( -1 )"'2(r - 2)! {'2 - 1 - r2

- 2r} = zr-l(!r)!(~r- 1)! 2({r + 1)

~=JXrl 2l+l(2r -t- 1)(r - 2)! 2r(!r + 1)!(!r- 1)! --- .

Thus we have

f( ) = ~ ~-=1)n+1 (<1_~-+ 1)(2n- 2)!p ( ) X ;S 22n(n + 1)!(n- 1)! 2n X.

Example 4

If x > 1, show that P,(x) < P1+l(x).

(3.64)

We prove this by the method of induction: assume first that P1_ 1(x) < P1(x) and use this to prove that P1(x) < P1+1(x). Then since the result is trivially true for l = 0 (P0(x) = 1 and P 1(x) = x) it will be true for all/.

We may assume throughout this proof that P,(x) > 0. For, from Rodrigues' formula (theorem 3.2), it is readily seen that if x > 1 (which it is in this example) then P1(x) > 0 for all values of l.

Now, by theorem 3.8(iii) we have

(l + t/~11 - (21 + 1)x 1-/!;; =-~ 0.

§ 3.13

Thus

EXAMPLES

Pl+1 (21 + 1)x I P,_1

P, =z-0-1+1 P, (21+1) I

> 1+1 -1+1

I+ 1 l + 1

=1.

(since x > 1 and we have assumed that P1_tfP1 < 1)

89

Since P1 is positive for all /, this implies that Pl+1 > P, and hence the method of induction guarantees the result to be true for all values of I.

Example 5 Prove that I{Q1(x)P1_ 1(x) - Q1_ 1(x)P1(x)}

= (l -- 1){Q1_ 1(x)P1_ 2(x) - Q1_ 2(x)P1-t(x)}

and deduce that I{Q1(x)P1_ 1(x) - Q1_ 1(x)P1(x)} = -1.

From theorem 3.8(iii) with I replaced by I - 1 we have

lP1(x) - (21- l)xP1-1(x) +(I- 1)PH(x) = 0 (3.65)

and by theorem 3.16 we have also

IQ1(x) - (2/ - 1 )xQ1_ 1(x) -1- (! - 1 )Q1_ 2(x) = 0. (3.66)

Multiplying equation (3.65) by Q1_ 1(x), equation (3.66) by P1_ 1(x) and subtracting gives I{P,(x)Q1_ 1(x) - Q1(x)P1_ 1(x)}

+ (!- l){P,_2(x)Qz-t(x) - Q,_2(x)P1-t(x)} = 0,

which is equivalent to l{Q1(x)P1_ 1(x) - Q1_ 1(x)P1(x)}

= (l- l){Q,_1(x)P1_ 2(x) - Q,_2(x),P _1(x)} as required.

If we now define F(l) = l{Q1(x)P1-1(x) - Q1_ 1(x)P1(x)}

this result may be written in the form F(l) = F(l- 1),

and hence we have F(l) '---= F(l 1) '-' F(l- 2) = ... = F(l ).

Thus F(l) --- F(l)


so that 1{Qt(x)P1_ 1(x) - QH(x)P1(x)} = Q1(x)P0(x) - Q0(x)P1(x).

But we know that P0(x) = 1, P1(x) = x,

Q() -II l+x oX -2 nl~-, -x

so that we obtain

= -1, which is the required result.

PROBLEMS

J1 21

(1) Show that _1

xP1(x)P1_I(x) dx = 4/2~ 1·

J 1

1 1 21(1 + 1) (2) Show that _

1 (1 - x2)P1(x)Pm(x) dx = Z/ + 1 Otm-

(3) If D = ~' use Leibniz's theorem to prove that

(l -L m) I (1 __ x2)mDl+rn(x2 _ 1 )! = ( ---l)m __ ' __ ·nz-m(x2 _ 1)

(1-m)!

CH.

(0 < m < 1) (l- m)!

and hence deduce that P1-rn(x) = ( -1)m (1-=t=-m)lP;"(x).

(4) Iff(x) = {* (O < x < 1) -t ( -1 <X< 0)

00

expand f(x) in the form 2 crPr(x). r= 0

oo m n - (2m) !(1 - x2)ml2 (5) Show that 2 Pn+m(x)t - 2 l(1 2 ) ,.

r>.=O rnm. - tX + (2 rn ''

PROBLEMS 91

(6) If u,. = J~1 x-1P.,(x)P.,_1(x) dx, show that nu,. + (n - 1)u,._1 = 2,

and hence evaluate u.,. n

(7) Show that L (2r + l)Pr(x) = P;.+l(x) + P;.(x). r=O

n

(8) Show that (1 - x) L (Zr + 1)P,(x) = (n + 1){Pn(x) - Pn+ 1(x)}. r=O

(9) Assuming the result of theorem 3.14, prove theorem 3 .15.

' ' 1 (10) Show that P1(x)Q1(x) - P,(x)Q,(x) = ~~~x2·

(11) Show that YfW, cl>) ~ Je14! !)bmo·

(12) If x > 1, prove that

and deduce that

1 fl (l - t2)l Q,(x) = 2,f+ i -1 (x - t)l+l dt

. . . . e8y(X -j-- 1)- y(X- 1) (1) by makmg the substttutwn t = 0 ( l) . 1( 1)

e y X+ + y X-·

we obtain

Q,(x) = f ~ {x + v(x2 d~ 1) cosh 0}1+1;

( .. ) Q ) 21 ~ (l + r)!(l +2r)! 1 11 ,(x = ,Xz+i ~ r!(2l + 2r + 1)! ,X"2r·

4

BESSEL FUNCTIONS

4.1 BESSEL'S EQUATION AND ITS SOLUTIONS; BESSEL FUNCTIONS OF THE FIRST AND SECOND KIND

Bessel's equation of order n is

2 d2y dy x dx2 + xd~ +- (x2 - n2)y = 0 (4.1)

(where, since it is only n2 that enters the equation, we may always take n to be non-negative).

Since this is of the form of equation (1.2) with q(x) = 1 and r(x) =

x 2 - n2, we may apply the methods of Chapter 1 and be assured that any series solution obtained will be convergent for all values of x.

<X>

Setting z(x, s) = 2: a,xs+r and requiring z to be a solution of equation r=O

(4.1) leads, by the same method as in previous cases, to the system of equations

and

{s(s- 1) + s - n2}a0 = 0,

{(s + l)s + (s + 1) - n2}a1 = 0

{(s + r)(s + r - 1) + (s + r) - n2}a, + a,_ 2 = 0 (r > 2).

(4.2)

(4.3)

(4.4)

Equation (4.2) gives the indicia! equation s2 - n 2 = 0 with roots s = ± n. We shall obtain two solutions from these which will be independent, apart, possibly, from the case when the two roots differ by an integer, i.e., when 2n is integral.

§4.1 BESSEL'S EQUATION AND ITS SOLUTIONS 93

Equation (4.3) is

and, since s2 = n2, we cannot also have (s + 1) 2 = n 2t, so that (s + 1) 2 - n2 ~0 and hence a1 = 0.

Equation (4.4) is just

{(s + r) 2- n2}ar + ar-2 = 0

which gives

ar-2 .. (s-+ r)2 ·_=-~2·

Takings= n, this becomes

(n + r - n)(n + r + n)

r(Zn + r) Thus we have

ao ao 02 = - Z(2n-~I--Z) "-' -- 22.-t(~- + 1)'

a2 a--------4- 4(2n + 4)-

a2 - -------

22.2(n + 2)

a4 a4 06 = - 6(2n + 6) = - 2 2 • 3(n + 3)

- Z6 .3!(n + l)(n + 2)(n +-3) and in general

(r > 2).

24 .2!(n + l)(n + i)'

ao 02

r = ( -l)" Z2rr!(n + l)(n + 2) ... (n + r)"

(4.5)

-j- The only situation in which we can have s2 = (s + 1) 2 = n 2 is 11 =!, s = -i. For this case it is true that s '~ -~makes a 1 indeterminate and leads to two independent solutions, but the results of the text still hold good, since zero is a possible choice for a 1 and we obtain the second independent solution from s = t·

94 BESSEL FUNCTIONS CH.4

We simplify the appearance of this expression somewhat by noting that (n + 1 )(n + 2) ... (n + r)

r(n + 1) = (n + r)(n + r- 1) ... (n + 2)(n + 1)r(n + 1)

r(n + r + 1) r(n + 1)

on using repeatedly the fact that xr(x) = r(x + 1) (theorem 2.2). Thus we now have

a0r(n + 1) a2r = ( --1 )' ----- ----------.

2 2rr!r(n + r + 1)

Also, if we use the fact that a 1 ~-= 0 together with equation (4.5), we obtain

al = aa = a5 = ... = a2r+I = ... = 0. Hence, substituting these values for the a, into the series for z(x, s), we

obtain as a solution of Bessel's equation 00

r( 1) "'a(-1)r n+ x2r+n ,-61 ° 22rr!r(n + r + 1) . This is a solution for any value of a0 ; let us choose a0 = 1 I {2n r(n + 1)}

and we obtain the solution which we shall denote by ]n(x) and shall call the Bessel function of the first kind of order n:

~ 1 (x)2r+n ],.(x) = f:o ( - 1)' r!r(n + r + 1) 2 · (4.6)

From the remarks at the beginning of this section it is clear that the infinite series in equation (4.6) will be convergent for all values of x.

So far we have dealt with the root of the indicia! equation s = n. The other roots = --n will also give asolutionofBessel's equation, and the form of this solution is obtained just by replacing n in all the equations above by -n, so that we obtain the solution to Bessel's equation

00

1 (x)2r-n J_,.(x) = ~ (- 1)' r!r(-n +r + 1) 2 · (4.7)

Suppose now that n is non-integral. Then, since r is always integral, the factor r( -n + r + 1) in equation (4.7) cannot have its argument equal to a negative integer or zero, and hence must always be finite and nonzero. Thus equation (4.7) shows that] -n(x) contains negative powers of x (they arise for those values of r such that 2r < n), whereas equation (4.6)

§ 4.1 BESSEL'S EQUATION AND ITS SOLUTIONS 95

shows that J .. (x) does not contain any negative powers. Hence, at x = 0, ] n( X) is finite while] -n( x) is infinite, so that one cannot be a constant multiple of the other, and we have shown that ] .. (x) and J -n(x) are independent solutions of Bessel's equation for n non-integral (which is a stronger condition than 2n non-integral, obtained from the general theory). The explicit relationship between ].,(x) and] -n(x) for integral n is shown in the following theorem.

Theorem 4.1

When n is an integer (positive or negative),

J _,.(x) = ( -1 )n] .. (x).

PROOF

First consider n > 0. Then

f-n(x) = ~ (-lY---~~~---co 1 (x)2r-n (;o r!r(-n +r + 1) 2

from equation (4.7). But rc- -n + 1" + 1) is infinite (and hence 1 I {r( -n + r + 1)} is zero)

for those values of r which make the argument a negative integer or zero, i.e., for r = 0, 1, 2, ... (n - 1) (remembering that this is possible because n is integral).

Hence the sum over r in the above expression for J -n(x) can equally well be taken from n to infinity, and then

But

] .. (x) = .~ < --J)m in!r(n --f~ m-+ l)-Grm+n (by equation (4.6))

96 BESSEL FUNCTIONS

so all that remains in order to complete the proof is to show that

(m + n)!r(m + 1) = m!r(n + m + 1) for n and m integral. But

CH.4

(m + n)!r(m + 1) = (m + n)(m + n -- 1) ... (m + 1)m!r(m + 1) =m!r(m +n + 1)

(on using repeatedly the result that r(x + 1) = xr(x)), and thus the result is proved.

Now consider n < 0; in this case we may write n = -p with p > 0. Then what we require to prove is that

or ]p(x) = ( -1)-vj -p(x)

( --l)P]p(x) =] _p(x)

which, of course, since p is positive, is just the result we have proved above.

Let us summarise what we have proved so far: we have shown that if n is not an integer then ] .. (x) and J _,.(x) (defined by equations (4.6) and ( 4. 7), respectively) are independent solutions of Bessel's equation (so that the general solution is given by A] .. (x) + B] -n(x)) while if n is an integer they are still solutions of Bessel's equation but are related by

] -n(x) = ( -1)n] .. (x).

Theorem 4.2

The two independent solutions of Bessel's equation may be taken to be

and

for all values of n.

PROOF

],.(x)

Y,.(x) = cos nn] ,.~x) -] -n(x) stnnn

We consider separately the cases n non-integral and n integral.

(i) n non-integral. Here sin nn ~ 0, so that Y .. (x) is just a linear combination ofJ.,(x) and

] _,.(x). But from the above discussion we know that in this case ].,.(x) and J -n(x) are independent solutions, so that ].,(x) and a linear combination of ] .. (x) and] -n(x) must also be independent solutions. HenceJ,.(x) and Yn(x) must be independent solutions.

§ 4.1 BESSEL'S EQUATION AND ITS SOLUTIONS 97

(ii) n integral. In this case sin nn = 0 and cos nn = ( -1 )n, so that by theorem 4.1 we

have

cos nn]n(x) -] -n(x) = ( -1)"]n(x) - ( -1)n]n(x) = 0.

Hence Y,.(x) has the form 0/0 and so is undefined. However, we may give it a meaning by defining it as

Y,.(x) = lim Y.(x)

= lim cosn];v~x) -] _.(x) v--+n Sln vn

[(o/ov){cos vnfix) -] __,.(x)}].=,. [(ojov) sin vn].=n

(by L'Hopital's rule)t

[ -n sin vn].(x) +cos vn (o jov)].(x) - ( o ;a~·)] -.(x)].=n [n cos vn]v=n

cos nn[(ojov)].(x)].=n- [(ojov)] _.(x)],,=,. n cosnn

1 = -[(ojov)J.(x)- ( -l)"(ojov)J -.(x)],,=,..

n

(4.8)

(4.9)

We must now prove two things; firstly that Y,.(x) as defined by equation (4.9) is in fact a solution of Bessel's equation and, secondly, that it is a solution independent from] .. (x). To accomplish the first of these we note thatJ.(x) obeys Bessel's equation of order v:

x2 d2J. + x d]. + (x2 - v2)j = 0 dx 2 dx •

which, when differentiated with respect to v, gives

x2~ of. + xi_ of. + (x2 - v2) of. - 21'.] = 0. dx2 ov dx ov ov •

(4.10)

Also, of course,] -.(x) satisfies Bessel's equation of order v. so that we have in exactly the same way

2 d2 of_.+ d of_,+( 2 2)oJ_ z.r -o (4.11) X dx2 Tv X dx Tv X - V OV - VJ ---. - •

t L'Hopital's rule states that if f(a) = g(a) = 0, then , f(x) f<•>(a) hm-=-x-..a g(x) g('>(a)

where f(•l(a) and g<•l(a) arc the lowest-order dcrivatiYl's of f(x) ami g(x), l"l'SPl'C•

tiYl'ly, whid1 arc not both zero at x = a.


Multiplying equation ( 4.11) by ( -1 Y and subtracting from equation (4.10) gives

x 2 -~ {aJ. _ ( _1y of-·} + x-~ {of.:: _ ( _1y of-·} dx2 ov ov dx ov av

+ (x2- v2){~·- ( -1)" a~;}- 2v{J. -- ( -1YJ -.} =0.

Setting v = n in this last equation and using equation (4.9) gives d2 d .

X2 d~2 Yn(x) + x dx Yn(x) + (x2

- n 2) Yn(x)

2n - -[}n(x) ·- ( -1)nJ -n(x)} = 0.

:n;

But now n is an integer, so that we may use the result of theorem 4.1 that I -n(x) = ( -1 )nJ n(x) to obtain

d2 d X 2 dx2 Yn(x) + X dx Y .. (x) + (x2

- n2) Yn(x) =o 0

which just states that Y .. (x) satisfies Bessel's equation of order n. That Yn(x) is independent from J ,.(x) may be seen from the result of the

next theorem, which implies that at x = 0 Yn(x) is infinite, while we know thatJ,.(x) is finite.

Theorem 4.3 (Explicit expression for Yn (x) for n integral)

Yn(x) =- ln- + Y-- """- fn(x) 2 { X 1 n 1} :n; 2 z:Sr

1 '""' 1 (x)n+2s ·' {1 1 } -;:;:~(-l)·'s!(n+s)! 2 6 -;+;·~~~~ __ ! ~ ~-=-_:~' _ 1)! (~)-n+2s

:n; 6 s! 2

·where y is Euler's constant (see Appendix 2).

PROOF

We shall omit the proof of this theorem, merely noting that the method to follow is to use the series expansions (4.6) and (4.7) in equation (4.9). Euler's constant appears because of properties of the derivative of the gamma function.

§ 4.2 GENERATING FUNCTION FOR THE BESSEL FUNCTIONS 99

Theorem4.4

When n is integral, Y _,.(x) = ( -1)" Y,.(x).

PROOF

From equation ( 4. 9) we have.

1 [a a J y _,.(x) =; a/·(x) - ( -1)--n a/ -.(x) •=-n

1 [ a a J =; a( -vl-·(x)- ( -l)-n o( -vl·(x) v=n

1 [ a a J =- --1 -.(x) + ( -1)+"-1.(x) :n; av OV v=n

1 [a a J = ( -1)"- - J.(x)- ( -l)n-1 __,(x) :n; av av v=n

= (-l)"Y .. (x).

We shall ca111,.(x) and Y.,(x) Bessel functions of order n of the first and second kinds respectively. ( Y.,(x) is sometimes called the Neumann function of order n and denoted by N,.(x).) As we have seen, 1.,(x) and Y,.(x) (with n considered positive) provide us with two independent solutions of Bessel's equation, J.,(x) always being finite at x = 0 and Y,.(x) always infinite at x = 0.

4.2 GENERATING FUNCTION FOR THE BESSEL FUNCTIONS

Theorem4.5

PROOF

We expand exp {~x(t -~)}in powers oft and show that the coeffi

cient of tn is1,.(x):

~I U

exp Hx(t- ~)} = exp Gxt). exp ( -~ ~)

= i <t;:t i ( >' 0 H ()

~xjt)•

s!


= ~ (-})'xrtr( -1)s(Dsxst-s L. r! s!

r, s=O

_ ~ _ s(~)r+s xr+str-s - L_. ( 1) 2 I I .

0 r. s.

r, s=

(4.12)

We now pick out the coefficient of tn, where first we consider n > 0. For a fixed value of r, to obtain the power oft as tn we must haves = r - n. Thus for this particular value of r the coefficient of tn is

(1) 2r-n x2r-n ( - 1y-n 2 ~!(r - n)r

We get the total coefficient of tn by summing over all allowed values of r. Since s = r - n and we require s > 0, we must have r > n. Hence the total coefficient of tn is

t (l)2r-n x2r-n ~ (xj2)2p+n ~ (-1)'-n 2 r!(r- n)! = {S ( - 1)v (p +n)!p!

(where we have set p = r - n)

~ (xj2)2P+n = ~ ( - 1 )P r(p + 1l + 1) P!

(remembering that both p and n are integral, so that we may use the result of theorem 2.3 that r(p + n + 1) = (p + n)!)

= ]n(x)

(by equation (4.6)). Ifnown < 0, westillhavethe cocfficientoftn for a fixed value of r given by

(1) 2r-n x2r-n ( - 1)'-n 2 r!(r- n)!

but now the requirement that s > 0 with s = r - n is satisfied for all values of r. Thus the coefficient of tn is just

~ (1)2r-n x2r-n ~ (xj2)2r-n f:o ( --1)'-n 2 r!(r- n)! = ( - 1)-n r~ ( - 1)' r!r(r- n + 1)

= ( -1)nj -n(x)

(by equation ( 4. 7)) = ]n(x)

(by theorem 4.1 ).

§ 4.3 INTEGRAL REPRESENTATIONS FOR BESSEL FUNCTIONS 101

4.3 INTEGRAL REPRESENTATIONS FOR BESSEL FUNCTIONS

Theorem4.6

1 f" ] .. (x) =- cos (ncfo ·· x sin cfo) dcfo n o

(n integral).

PROOF

Since] -n(x) = ( -·1 )n]n(x) for n integral, the result of theorem 4.5 may be written in the form

If we now write t = e1<1> so that

t - ! = c1<1> - c _;q, = 2i sin cfo

t this equation becomes

00

ei.<sin<f> =Jo(x) + L {ei"<i> + (-l)"e-J"<f>}J .. (x). 11=1

But when n is even

e1""' + ( -1)" e-in</> = e1

""' + e-in<f> = 2 cos ncfo, while when n is odd we have

e1n<i> + ( -1 )n e-in</> = e1""' - e-in<P = 2i sin ncfo.

Thus we have

eixsin 4> = j 0(x) + L 2 cos ncfo]n(x) + L 2i sin ncfo] .. (x) n even n odd 00 00

= ] 0(x) + L 2 cos 2kcp]2~c(x) + i L 2 sin (2k - 1)cfo]2~c_1(x). k=l k=l

Equating real and imaginary parts of this equation gives co

cos (x sin cfo) = ] 0(x) + L 2 cos 2kcfo]2k(x) (4.13) k=l

00

sin(xsincfo) = L 2sin(2k -1)cfo]2k_ 1(x). (4.14) ko J

If we multiply both sides of equation ( 4.13) by cos nrp (n > 0), both sides


of equation (4.14) by sin ncp (n > 1), integrate from 0 to :n; and use the identities

and

(m ~n)

(m = n ~0)

(m = n = 0) (m ~n)

(m = n ~0),

we obtain the results

J" A. ( • A.) d-1.. {nfn(x) (n even) cos n'f' cos x sm 'f' 'f' = o 0 (n odd)

and J" . . . {0 (n even)

sm ncp sm (x sm cp) dcp = J 0 :n; n(x) (n odd).

Adding these last two equations gives

[{cos ncp cos (x sin cp) +sin ncp sin (x sin cp)} dcp = n],.(x)

for all positive integral n.

Hence J: cos (ncp - x sin cp) dcp = n]n(x)

which is the required result for positive n. If n is negative, we may set n = -m where m is positive, so that the

required result is

[cos ( -mcp - x sin cp) dcp = nj -m(x)

(where m is positive).

But J: cos ( -mcp - x sin cfo) dcp

= J: cos { -m(n- 0) - x sin (:n - 8)}. -dO

(where we have changed the variable by setting e = n - cfo)

= [cos { -m:n; -f- m() --X sin 8} d()

= J: {cos (mO - x sin 0) cos mn

-1- sin (mO - x sin 0) sin m:n} dO

§ 4.3 INTEGRAL REPRESENTATIONS FOR BESSEL FUNCTIONS 103

=(-1)m J: cos(mO -xsinO)dO

= ( -1)mn]m(x)

(since we know the result to be true for positive m)

= n] -m(x)

= nfn(x).

Theorem 4.7

(_2lx)n Jl 1 ( ) (1 - t 2)n-!eixt dt (n > - 1.).

n X = y(n)r(n + i) -1 ~

PROOF

Consider the integral I defined by

I= Jl (1 - t2)n-Jeixt dt -1

= J~l (1 - t 2t-} r~ (i:?' dt

= ~ (ixt J1

(1 - t 2)"-!tr dt. ~ r! -1 r~o

Now, if r is odd, the integrand in J1

(1 - t 2)"-ltr dt is an odd function -1

oft, so that the integral is zero; while if r is even (say, equal to 2s), the integrand is even, so that we have

J~l (1 - t 2 )n-!tr dt = J~l {1 - t 2 )n-lt 28 dt

= 2 J: {1 - t 2)n-lt 28 dt

= J: (1 - u)n-tus-! du

(where we have made the change of variable u = t 2, du = 2t dt)

= B(n + i, s + i) (by the definition of the beta function; we must have n > - ! to ensure the convergence of the integral (sec Section 2.1 ))

104

Thus

BESSEL FUNCTIONS

r(n + t)r(s + t) r(n + s + 1) (by theorem 2.7).

I = ~ (ix) 28 _!'(n + l)r(s + l) ;S (2s)! r(n + s + 1)

~ x 2• 1 (2s)! = r(n + i) 6 ( - 1)• (2s)! r(n + s -1=--i) 2 2•s! yn

CH.4

(using the corollary to theorem 2.10)

-- ' .l (:)-n ~ ( -1)•(x/2)2•+: - l (n + 2)( yn) 2 ~ r( 1) I

s~o n + s + s.

= r(n + i)( yn)G) -n] .. (x)

(by equation (4.6)).

Thus ]n(x) = ( yn)r~n + l) (~)"1 = 1 (:)n Jl (1 - t2)"-!eil·t dt.

( yn)r(n + i) 2 -1 •

4.4 RECURRENCE RELATIONS

Theorem 4.8

(i) ! {xn] .. (x)} = xnfn-l(x).

d (ii) dx {x-n]n(x)} = -x-nfntl(x).

(iii) ]~(x) = ]n-l(x) - ~n(x).

(iv) ]~(x) = 1!_ ] .. (x) - fndx). X

(v) ]~(x) = Hfn-l(x) - fnu(x)}.

(vi) fn-l(x) + fn+l(x) = ~'!_Jn(x). X

§4.4 RECURRENCE RELATIONS 105

PROOF

(i) From equation (4.6) we have

so that

d d ~ 1 1 dx {xn] .. (x)} = dx r6 ( -1)' r!r(n + r + 1) 22r+nx2r+2n

co 1 = ""' ( -1)'- (2r + 2n)x2r+2n-I

L. r!l'(n + r + 1)22r-tn r~o

co 1 = xn ""' ( -1)r 2(n + r)x2r+(n-1) :S r !(n + r)r(n + r)22r+n

(using the result of theorem 2.2 that rex + 1) = xr(x))

co 1 (x)2r+(n-1) = xn r~ ( -

1)' r!r(n + r) 2 = xnfn-l(x)


(Note that in this theorem no restriction is placed on n: it may be integral or non-integral, positive or negative.)

(ii) We have d d ~ 1 1 dx{x-tj,.(x)} = dx f:6 ( - 1)' r!f(n + r + 1) 2 2r+nx

2r

co 1 1 = ""'(-1)' --2rx2r-1 £:6 r!r(n + r + 1) 2 2r+n

co 1 1 _ ""'(-1)r _ rx2r-1 - ;S r!r(n + r + 1) 2 2r+(n-l)

(since the factor r in the numerator makes the term with r = 0 vanish; we remember that 0! = 1)

co 1 1 ~ ""' ( ~ 1)•+1 -- (s + 1)x2(s+l)-1 - f:b (s+1)!r(n+s+2) 2 2<•+1)+n-l

(where we have sets = r 1)


"' 1 1 - ~ ( -1)•+1 --x2s+l - 6 s!r(n + s + 2) 22s-tn+l

~ 1 (x)2s+n+l = -x-n 6 ( - 1)" s!r(n + s + 2) 2

= -x-"],.+l(x)

(on use of equation (4.6)).

(iii) From result (i) above, by carrying out the differentiation of the product on the left-hand side, we have

nx"-1],.(x) + x"]~(x) = x"f11-1(x)

which, on dividing throughout by x", gives

'!_],.(x) +J~(x) =fn-l(x), X

and hence , n

fn(x) = fn-l(x) --] .. (x). X

(iv) We carry out the differentiation in result (ii) above to obtain

-nx<-n-l'J,.(x) + x-"]~(x) = -x'-"],.+1(x)

which, on multiplying throughout by x", gives

_f!_] .. (x) + ]~(x) = -J,.tl(x). X

Thus

(v) Add results (iii) and (iv) and the result follows immediately.

(vi) Similarly, if we subtract result (iv) from result (iii) we obtain the required relationship.

Theorem 4.9

All the results of theorem 4.8 remain true when the Bessel junctions of the first kind are replaced by the corresponding Bessel functions of the second kind.

PROOF

We shall prove that result (i) remains true for Y,.(x); a similar method will prove result (ii), and then results (iii-vi) follow in the same way as they did in theorem 4.8.

§ 4.5 HANKEL FUNCTIONS 107

Thus what we have to prove here is that ( djdx){x" Y,.(x)} = x" Y,._1(x). We must consider separately the cases of integral and non-integral n. (a) Non-integral n. Here we may write

Y,.(x) = cos nn1,..(x) -1 -n(x) sm nn

so that

d dx{x"Y,.(x)}

1 d d = --. -[cosmrd-{xn1,.(x)} - dx{x"] _,.(x)}]

Slll 1m X

1 =-.--[cos nn.x"1n-1(x)- {- x"1 -n+l(x)}]

stn nn

(where we have used theorem 4.8 (i) for the first derivative and theorem 4.8 (ii) for the second)

1 = -.- x"[cos nn1,._1(x) + 1 -(n-t)(x)]

sm nn

1 . {( l) } x"[cos {(n - 1)n + n} 1n-1(x) + 1 -(n-1lx)] sm n- n + n

1 = ---:·--( --

1-) x"[ -cos (n- 1)n1,._1(x) + 1 -(n-n(x)]

-s1n n- n

cos (n- 1)n1,._1(x) -1 -("-t)(x) - x" --=----~--=-=-.:____::_~_ - · sin (n - 1)n

= X" Y,._t(x).

(b) Integral n. Here we note that Y,.(x) = lim Y.(x) and that by part (a)

(djdx){ x•Y.(x)} = x•Y._1(x).

Taking the limit of this result as v ~ n gives the required result.

4.5 HANKEL FUNCTIONS

We define the Hankel functions (sometimes called Bessel functions of the third kind) by

H,.(I>(x) = 1,.(x) + i Y 1,(x)

H,.C 2>(x) = 1 .. (x) i Y .. (x).


These are obviously independent solutions of Bessel's equation. Both are, of course, infinite at x = 0; their usefulness is connected with their behaviour for large values of x, which we shall investigate in a later section.

Theorem 4.10

All the recurrence relations of theorem 4.8 remain true when ]n(x) ts replaced by either Ji,.<tl(x) or Hn< 2l(x).

PROOF

Again we prove relation (i) only, the remainder following as before. But we know relation (i) to hold for both]n(x) and Yn(x), so that we have

d . dX{x"],.(x)} = x"],._1(x)

and

Hence

so that

:)x"'{]n(x) ± iYn(x)}] = xnUn-lx) ± iYn_1(x)},

which, on remembering the definitions of H~1 l(x) and H~2l(x), gives

d d)x"H~1l(x)} = xnH~I2 1(x)

(taking the plus sign)

and d dx{x"H~2l(x)} = x"H~2}_ 1(x)

(taking the minus sign).

4.6 EQUATIONS REDUCIBLE TO BESSEL'S EQUATION

Theorem 4.11

The general solution of

d 2y dy x 2 - + x-- + (A. 2x 2 - n 2)y = 0

dx 2 dx

is A]n(Ax) + B Yn(J.x).

§ 4.6 EQUATIONS REDUCIBLE TO BESSEL'S EQUATION

PROOF

Make the substitution t = A.x

so that

and

Then the given equation becomes

t2 d2y + t dy + (t2 - n2)y = 0 dt 2 dt

which is Bessel's equation of order n, so that the general solution is

y = A]n(t) + BYn(t)

109

and hence, replacing t by A.x, the general solution of the original equation is

Y = A]n(A.x) + B Yn(A.x).

Theorem 4.12

The general solution of

dry dy x2 _ + (1 _ 2o:)x- + {fJ2y2x2Y + (o:2 _ n2y2)}y = 0

dx2 dx

is Ax":Jn(fJxY) + BXXYn(fJxY).

PROOF

We first make the change of variable y = xrxz.

dy dz Then - = xrx - + o:xrx-Iz

dx dx

and d 2y _ rx d 2z _

1 dz _2

dx2- x dx2 + 2o:XX dx + o:(o: - l)XX z

so that the given equation becomes

XX+2 d 2z + 2o:XX+l dz + o:( o: - 1 )XXz + (1 - 2o:){xrx-I dz + o:xrxz}

dx 2 dx dx

+ {fJ2y2x2Y + (o:2 _ n2y2)}XXz = 0

which, on collecting terms in d 2zjdx2, dzjdx and z becomes

d~ dz x 2 -- 1- {Zo:x + (1 - 2o:)x}-

dx2 dx

+ {o:(o: - 1) 1- (1 -- 2o:)o:


which simplifies to

d 2z dz x2 dx2 + x dx + {fJ2y2x2l' - n2y2}z = 0. (4.15)

Now change the independent variable x by setting t = xY.

Then dz _ dz dt _ xY-1 dz dx-dt dx-y dt

and d 2z dz d dz dx 2 = y(y - l)xl'- 2 dt + yxY-1 dx dt

dz d 2z = y(y _ l)xl'-2 _ + yxv-l,yxY-1_ dt dt 2

d 2z dz = y2x2l'-2 __ + y(y _ 1 )xY-2 _ dt 2 dt

so that equation (4.15) becomes _

d 2z dz dz y2x2Y _ + y(y _ l)xY _ + yxY ~- + {P2y2x2Y _ n2y2}z = 0

ili2 ili ~ '

which, on collecting like terms and cancelling a common factor of y 2, gives

d 2z dz f2 dt2 + t dt + {fJ2t2 - n2}z = 0.

But this is just the equation of theorem 4.11 with solution

z = A] .. ({Jt) + BY .. ({Jt).

Hence the solution of the original equation is obtained by substituting back the relationships y = xrxz and t = xY:

y = Ax""] .. (fJxY) + Bxa Yn(f3xY).

4.7 MODIFffiD BESSEL FUNCTIONS

Consider the differential equation

d2 d x2 ~ + x_z - (x 2 + n 2)y = 0.

dx2 dx (4.16)

This has the form of the equation in theorem 4.11 with A. 2 = -1. Thus it has the general solution

y = A] .. (ix) + BYn(ix).

Now, the solutions J .. (ix) and Y,.(ix) have the disadvantage of not

§ 4.7 MODIFIED BESSEL FUNCTIONS 111

necessarily being real when x is real. However, we may take the constant multiple of]n(ix) defined by

I,.(x) = i-n] n(ix)

and use it as one of the independent solutions of equation (4.16). It is easy to show that it is a real function of x:

I,.(x) = i-"],.(ix) 00

1 (" )2r+n = i-" ~ ( ~ 1y r!r(n+-;-+1) I =i-n ~ ( --1)' rff(n ~ T + 1t'i"GYr+n ~ 1 (x)2r+n

= ~ ~Tt'(n + r + 1) 2 · (4.17) r=O

I,.(x) so defined is called the modified Bessel function of the first kind, and equation (4.16) is called Bessel's modified equation.

Theorem 4.13

If n is integral, I _11(x) = I,.(x).

PROOF L .. (x) = i"] _,.(ix)

= jn( -1 )"] n(ix}

(by theorem 4.1) = jn( -1 )njn I,.(x)

= ( -1)2ni,.(x)

= I,.(x).

We may obtain the second independent solution to Bessel's modified equation by considering Y,.(ix); or alternatively we may employ a method similar to that used for the definition of Y,.(x).

I,.(x) and I_,.(x) are both solutions of equation (4.16). When n is nonintegral they are independent solutions (since J .. (ix) and] _,.(ix) are independent solutions). However, when n is integral I -n(x) = In(x). Define now

K ( ) =~I -n(x) - In(x) n X 2 . • smnn

( 4.18)

When n is non-integral, this is well defined, and /,.(.v) and K,.(x) together


provide independent solutions of equation (4.16). When n is integral, Kn(x) is indeterminate as it stands and has to be defined by

K ( ) _

1. n I -.(x) - I.(x)

nX-lm- . ~'-'>n 2 Slll vn

_ n [(ojov)L.(x)- (ojov)I.(x)].=n - Z- n[cos vnJ.~n -

(by L'Hopital's rule)

= L-l )n [i L.(x) - il.(x)J . 2 OV OV v=n

As in theorem 4.2, it may be shown that this provides a second independent solution of Bessel's modified equation for integral values of n. The explicit series for Kn(x) may be obtained from those for fn(x) and Y,(x) by use of the following theorem.

Theorem 4.14

PROOF

---Kn(x) = ~in+l{jn(ix) + i Yn(ix)} = ~in+IH~,'l(ix).

(by definition ( 4 .18))

n jnj -n(ix) - j-n]n(ix) = 2 s1nnn

But from the definition of Yn(x) in theorem 4.2 we have that

] _,.(ix) =cos nnJ,.(ix) -sin nn Yn(ix). Hence

But

K .. (x) = "!_ jn cos nn] .. (ix) - i-~J .. (ix)- in sin nn Yn(ix) 2 smnn

= ~·n+l{· y (" } -i cos nn - i -2n-lJ (" >}

21 1 ,.lX + . n1X .

smnn

-i cos nn - i-2n-t = -i cos nn + i.i- 2n

= -i cos nn + i(ei"12)- 2n

(writing i = ei"12}

(4.19)

§ 4.8 RELATIONS FOR MODIFIED BESSEL FUNCTIONS

= -i cos nn + i e-inn

= -i cos nn + i(cos nn- i sin nn)

=sin nn.

Thus, using equation (4.19) we have

K,.(x) = ~in+I {i Y,.(ix) + J ,.(ix)}.

4.8 RECURRENCE RELATIONS FOR THE MODIFffiD BESSEL FUNCTIONS

113

The modified Bessel functions satisfy recurrence relations similar to, but not identical with, those of theorem 4.8. Also, in contrast to the functions J,.(x) and Y,.(x), which satisfy the same relations as one another, I,.(x) and Kn(x) satisfy different relations from one another.

Theorem 4.15

(i) :x{xnl,.(x)} = x"l,._1(x).

(ii) :x{x-nJ,.(x)} = x-nJ,.+l(x).

(iii) /~(x) = 1,._1(x) - '!_J,.(x). X

(iv) /~(x) = '!_I,.(x) + I,.+l(x). X

(v) I~(x) = !{1,._1(x) + /,.+I(x)}.

(vi) /,_1(x) - I,.+l(x) = Znl,.(x). X

PROOF

We shall prove results (i) and (ii). The remaining results follow from (i) and (ii) exactly as in theorem 4.8.

(i) From theorem 4.8 (i) we have

d dx{x"],.(x)} = x"],._1(x).

Replacing x by ix gives

~l(~x){i"x"]n(ix)} i"x"],. 1(ix),

114 BESSEL FUNCTIONS

which, on using the fact that I,.(x) = i-"].,(ix), becomes

; dd {i"x"i"/,.(x)} = i"x"i"-lf,._1(x) 1 X

and this equation, on cancelling a factor of i2"-1 throughout, gives

d dx{xnJ,.(x)} = x"l,._1(x).

(ii) From theorem 4.8 (ii) we have

d dx{x-"],.(x)} = - x-"],..u(x),

which, on replacing x by ix, becomes

d(~x){i-nx-nJ,.(ix)} = -i-nx-"],.tl(ix).

Thus

and hence

~ dd {x-"/,.(x)} = -ix-nJ,.+l(x) 1 X

so that

Theorem 4.16

d (i) dx{x"K,.(x)} = -x"K,._1(x).

(ii) :x{x-nK,.(x)} = -x-nK,.+I(x).

(iii) K~(x) = -K,._1(x) - '!..K,.(x). X

(iv) K~(x) = '!..Kn(x) - Knn(x). X

(v) K~(x) = -!{K,._1(x) + Kn+l(x)}.

. 2n (v1) K.,_1(x) - K,.+I(x) = - -K,.(x).

X

CH.4

§ 4.8 RELATIO:!'TS FOR MODIFIED BESSEL FUNCTIONS 115

PROOF

Again, we prove only results (i) and (ii). Also, in the same way as in theorem 4.9, we prove the results for non-integral n and appeal to continuity for a proof in the integral case.

(i) From theorem 4.14 we have

K.,(x) = ~in+tH~1l(ix)

and by theorem 4.10 we know that u;,1l(x) satisfies the first recurrence relation of theorem 4.8.

Hence

which, on replacing x by ix, gives

Thus

so that

and hence

-~{inxnH(ll(ix)} = i"xnJI(l) (ix) d(ix) n n-1 •

d dx{xnK .. (x)} = - xnK .. _1(x).

(ii) Similarly H~1l(x), by theorem 4.10, satisfies the second recurrence relation of theorem 4.8 so that

d dx{x-nH~ll(x)} = --x-nJI~~t(x)

which, on replacing x by ix, gives

--~-{i-nx-nH<1 l(ix)} = -i-nx-nH(l) (ix) d(ix) " n+l '

and hence, using theorem 4.14,

1 d (' 2. K ( )} . 2. 2ZT ( .- -11-nx-11 --1-n- 1 ,.X =--1-nx-"-1-n·-1'1. 7111 X).

1 dx n n :-.•· I

ti6 BESSEL FUNCTIONS

Therefore

i-2n-2 !{x-nKn(x)} = -i-2n-2J{n+l(x),

from which a common factor of i-2"-2 cancels to give

d dx{x-nK .. (x)} = -Kn ,1(x).

CH.4

4.9 INTEGRAL REPRESENTATIONS FOR THE MODIFIED BESSEL FUNCTIONS

Theorem 4.17

(i) I .. (x) =-~ __ 2_ __ (:)" f1

e-xt(l - t 2)<n-~) dt ( yn)r(n + !) 2 -1

~~ (n > -}).

(ii) Kn(x) = yn (~)n foo e-xt(t 2 - l)n-l dt r(n + :\) 2 1

(n > -!, x > 0).

PROOF

(i) By equation (4.17) we have

In(x) = j-n]n(ix)

and by theorem 4. 7 we have

so that

] .. (x) = ( yn)r~n + i)GY f~1 (1 - t2)n -: cixt dt

(n > -!)

In(x) =i-n -- 1 i"(:) .. ft. (1 - t2)n-i e-xt dt ( vn)r(n + {) 2 -1

= ( yn)r~n + !)Gt J~l (1 - t2)n-~ e-xt dt.

(ii) We first show that the integral

P = xn J; e-xt(t 2 - J)n-! dt (n > -l, X> 0) (4.20)

satisfies Bessel's modified equation

2 d2y dy 2

x dx2 + xdx - (xz + n )y = 0.

§ 4.9 REPRESENTATIONS FOR MODIFIED BESSEL FUNCTIONS 117

We have

dP Joo J"" - = nxn-1 e-"'1(t 2 - l)n-! dt +X" -te-"'1(t 2 - l)n-l dt dx 1 1

and d2J> Joo - = n(n- l)x"- 2 e-"'1(t 2 - J)n-l dt dx 2 1

+ 2nxn-t J ~ -t e-"'1(t 2 -- 1 )n-l dt

+ xn J~ f2 e-"'t(t2- l)n-! dt

so that

d 2P dP x 2

- + x-- (x 2 + n 2)P dx2 dx

= xn J~ {n(n- 1) -- 2nxt + x 2t 2 + n - xt - x 2 - n 2} e-"'1(t 2 - l)n-~ dt

= x"+l J~ {x(t 2 - 1) - Z(n + l)t} e-"'1(t 2 - l)n-t dt

= xn+t J~ {xe-"'1(t 2 - 1)n+~ - (n + f)(t 2 - l)n-izt e-"'1} dt

Joo d ' = -xn+l -{e-"'1(t 2 - l)n+>} dt 1 dt

= -x"+I[e-"'t(t2 _ t)n+l]f

=0,

since the integrated part vanishes at the upper limit, t = oo, because of the factor e-xt (remembering that we are considering here x > 0) and at the lower limit, t = 1, because of the factor (t 2 - l)n+i (remembering that n + i > 0).

Thus P satisfies Bessel's modified equation and therefore must be of the form

P = Aln(x) + BKn(x).

We show now that A = 0. We do this by considering the limit as x---+ oo. From the series (4.18) for I,(x) we see that In(x) consists of a power series with positive coefficients, and hence In(x) ---+ oo as x ---+ oo. Consider now J>(x). We have, from the definition, P(x) > 0, and we shall show that P(x) is less than some quantity which tends to zero as x tends to infinity so that P(x) itself must tend to zero as x tends to infinity.

Since the asymptotic behaviour of an cxponl'ntial dominates that of a


power, we must have (t 2 - l)n-! < e"'112 for x large enough (say for x>X).

Then for x > X we have

P(x) < x" J~ c-zt ezwt dt

=X" J~ e-xt/ 2 dt

= xn[ -~ e-xt/2]~ = zxn-1 e-x/2

which~ 0 as x ~ oo, again using the fact that the exponential dominates the power.

Thus we have shown that P(x) ~ 0 as x ~ oo, whereas I,.(x) ~ oo. Hence P(x) cannot contain any multiple of I.,(x) and we must have

P(x) = BK.,(x).

To determine the constant B we ~xamine the behaviour of both P(x) and K.,(x) as x ~ 0.

For K .. (x) only the lowest power of x is important in this limit, and we may use the definition

K,.(x) = ':I _.,(x! - I.,(x) 2 smnn

together with

~ 1 (x)2r+n /,.(x) = ,fo r!r(n + r + 1) 2

to see that the lowest power of x in I.,(x) is

1 (X)" r(n + 1) 2

and hence in K .. (x) is

2 si: nn rc-: + 1) (~) -n.

Hence as x ~ 0, K.,(x) ~ zr(1 _:)sin n.,-c(~)" which, on making use

of the result of theorem 2.12 that

n r(x)r(l - x) = -. -,

sm nx

§ 4.9 REPRESENTATIONS FOR MODIFIED BESSEL FUNCTIONS 119

may be written in the form r(n)Zn-1

Kn(x)--+ --. X"

To study the behaviour of P(x) near x = 0 we make the change of variable

so that

u t=1+

x 1

dt =- du. X

Also, when t = 1, u = 0 and when t = oo, u = oo.

Thus, using equation (4.20),

P(x) =X" Joo e-x-u(2u + u2)n-!! du 0 X X 2 X

= _!_ e-x J oo e-u(1 + 2x)n-!-u2n-1 du. X" o U

For small values of x we may make the approximations e-x~ 1 and

( 1 + ~r-l ~ 1 to obtain

1 = --r(2n)

x"

(by the definition of the gamma function).

Thus for small values of x the result P(x) = BKn(x) reduces to

__!__r(Zn) = Br(n)Zn-1 X" xn

giving B = _ f(Zn) -. f(n)zn-1

But by theorem 2.10 we have

so that

z2n-J r(Zn) = vn r(n)r(n + !)

B = Z2n-1r(n)f(n + !) ( vn)r(n)2"-'

2"f(n + ~) vn


Hence

giving

v'n K,.(x) = 2"r(n + i{(x)

and thus, using the definition of P(x),

K .. (x) = y'n (:)" Joo e-"1(t 2 - l)n-i dt. r(n + t) 2 1

4.10 KELVIN'S FUNCTIONS

Consider the differential equation

CH.4

d 2y dy . x 2 dx2 + x dx - {ik 2x 2 + n2)y =-= 0. (4.21)

This is of the form of Bessel's modified equation

d 2y dy ~ xs dx2 + x dx - ~A.2x2 + n2)y = 0

with ). 2 = ik 2• Hence, since the general solution of Bessel's modified equation is

y = AI,.(A.x) + BK .. (A.x), the general solution of equation (4.21) must be

y = AI,.(ilkx) + BK,.(ilhx).t

Also, since I,.(x) = i-"],.(ix), we may take the independent solutions of equation (4.21) as] .. (i3' 2kx) and K.,(PI2kx).

Of course, when xis real],.(i312x) and K,.(i 112x) are not necessarily real; we obtain real functions by the following definitions:

ber,.x = ReJ.,(i3' 2x) bei,. x = ImJ ,.(i 312x)

J.,(i 3'

2x) =her,. x + i bei .. x.

ker,.x = Re i-"K,.(i112x) kei,. x = Im i-"K,.(i112x)

i-"K,.(il'2x) = ker,. x + i kei., x.

(4.22)

(4.23)

t Strictly speaking, il is a two-valued function with values e1"14 and c15"14 • vVc remove this ambiguity by taking the value e1"/4•

§ 4.11 SPHERICAL BESSEL FUNCTIONS 121

(If n = 0 the notation used is often her x, etc.)

Thus the general solution of equation ( 4.21) is given by

y = A 1(ber,. kx + i bei,. kx) + A 2(ker,. kx + i kein kx). (4.24)

4.11 SPHERICAL BESSEL FUNCTIONS

Consider the equation

d 2 dy x 2 d~ + 2x dx + {k 2x 2

- l(l + l)}y = 0. (4.25)

This is of the form of the equation of theorem 4.12 with

1-2<x=2 y=l

fJ2y2 = k2

oc2- n2y2 = - l(l + 1).

Solving these equations for oc, fJ, y, n gives

oc = -!, fJ = k, y = 1, n = l + t and hence by theorem 4.12 the general solution of the above equation is given by

y = Ax-lj1 +!(kx) + Bx-! YH1(kx)

= AJ1(kx) + A 2y 1(kx)

where we have defined the spherical Bessel functions jz(x) and y 1(x) by

jz(x) = J(~)Jz+i(x) (4.26)

Yz(x) = J (~) Yl+i(x) (4.27)

and A1 , A 2 are new arbitrary constants related to A, B by A 1 = v(2k/n)A, A 2 = y'(2kjn)B.

Spherical Hankel functions may also be defined in a way exactly analogous to the definition of Hankel functions:

W>(x) = jz(x) + iy1(x) (4.28)

(4.29)

In applications it turns out that the spherical Bessel functions of most interest arc those of integral order. \V c shall show that such spherical


Bessel functions may in fact be written in closed form in terms of elementary functions. First, however, we prove recurrence relations analogous to those in theorems 4.8, 4.9 and 4.10 for the Bessel functions.

Theorem 4.18

If fn(x) is any of jn(x), Yn(x), h~1l(x) or h~l(x), then

(i) !{xn+lfn(x)} = xn+lfn_1(x).

(ii) :x{x-nf.,(x)} = -x-nfn+l(x).

(iii) f~(x) = f.,_1(x) - n + 1 fn(x).

X

(iv) f~(x) = ~f .. (x) - f,.+l(x). X

(v) (2n + 1)f~(x) = nfn_1(x) - (n + 1)fn+1(x). ~

. 2n + 1 (v1) fn_1(x) + fn+1(x) = ---fn(x).

X

PROOF

We first prove the results for in(x).

(i) From theorem 4.8 (i) with n replaced by n + -~ we have

!{xn+:Jn+~(x)} = xn+!Jn+~-1(x) which, on using definition (4.26), becomes

~{xn+~y(2jn)xljn(x)} = xn+ly(2/n)xljn_1(x)

and this simplifies to

!{xn+1jn(x)} = xn-t 1jn-1(x).

(ii) From theorem 4.8 (ii) with n replaced by n + ! we have

! {x-n-:J,.+l(x)} = -x-n-:J.,+l+l(x).

Again, use of definition ( 4.26) gives

:x { x-n-ly(2/n)x{Jn(x)} = - -x-n-ly(2/n)xlj,.+l(x)

§ 4.11 SPHERICAL BESSEL FUNCTIONS 123

and hence

~{x-n_j.,(x)} = - x-nj.,+l(x).

(iii) If we carry out the differentiation on the left-hand side of (i) above, we obtain

xn+Ij~(x) + (n + l)xnj .. (x) = xn+Ifn-1(x). On dividing throughout by xn+r, this gives

·' ( ) n + 1. ( ) . ( ) Jn X + --).,X = )n-1 X X

and thus

., ( ) . ( ) n + 1. ( ) )n X = )n-1 X - --}n X • X

(iv) Carry out the differentiation on the left-hand side of (ii) above, and we obtain

x-nj~(x) - nx-n-lf .. (x) = - x-nj.,+l(x).

Cancelling a common factor of x-n then gives

j~(x) - l!_j .. (x) = - fn+1(x) X

and hence

j~(x) = ~ .. (x) - j.,+l(x).

(v) Multiply result (iii) by n, result (iv) by (n + 1) and add, and we obtain the required result.

(vi) Subtract result (iv) from result (iii), and we obtain the required result.

The proofs for y.,(x), h~'(x), and h~2'(x) follow in the same way, since these functions bear the same relationship to Y.,+1(x), H~1~~(x) and H~2~t(x) that f.,(x) does to ] .. +l(x) and by theorems 4.9 and 4.10 the various Bessel functions all satisfy the same recurrence relations.

Theorem 4.19

(.) . ( ) sin x I Jo X = --· ···.

X

(ii) )' 11(x) cosx

124 BESSEL FUNCTIONS CH. 4 ix

(iii) hb1l(x) = -i ~. X

-ix

(iv) hb2l(x) = i ~--. X

PROOF

(i) We have, by equation (4.26),

jo(x) = v(nj2x)I~(x) = J (;J '~ ( -l)' r!r(t ~ r + i/~rr+! (from equation ( 4.6));

but, by the corollary to theorem 2.10, we have

1 (2r + 2)! r(r + 1 + -) = ----yn

2 22r+2(r + 1)!

so that we now have

. yn ~ 22r+2(r + 1)! x2r+~ Jo(x) = zix! f:6 ( - 1)' r!(Zr + 2)!vn 2 2r+}

~ Z(r -t-1) = ~ (-1)'(-Z_r_-!~Z)!x2r

r~o

~ x2r

= :S ( - 1)' (Zr + 1)!

1 ~ x2r+l 1 = ~ ~ ( -1 )' (2r + 1 )-! = ~ sin x

r~o

(on recognising the infinite series for sin x)

(ii) We have

Yo(x) = J (;:) Y1(x)

(by definition ( 4.27))

= J(!!_) cos !n ! 1(x) -I -l(x) 2x sm !n

(by the definition of Yn(x) in theorem 4.2)

= --J (;:) I -l(x)

§ 4.11 SPHERICAL BESSEL FUNCTIONS

= -J(~) ~ (-1)' r!r(-! ~r + 1l~rr-l (by the series (4.7)).

But again, by the corollary of theorem 2.10, we have

r( -t + r + 1) = r(r + !)

so that

= (2r)! yn 22Tr!

yn ~ 22rr! x2r-! Yo(x) =- 2lxt r~ (-1)' r!(2r)!yn22r-}

~ x2r-1 = - r~ ( -l)' (2r)l

1 ro x2r

= - ~ r~ ( -1)' (Zr)!

1 = -- cosx,

X

125

(on recognising the infinite series for cos x ).

(iii) hb1>(x) = j 0(x) + iyo(x) (by definition (4.28))

1 . . 1 = - sm x - 1 - cos x

X X

1 ( • . ) = - - cos x + 1 sm x X

el"' =-1-.

X

(iv) hb2>(x) = j 0(x) - iy0(x)

1 . . 1 = - sm x + 1 - cos x

X X

i ( . . ) = - COS X - 1 S1n X X

e-x • I

=1---. X


Theorem 4.20 (Rayleigh's Formulae)

If n is a non-negative integer, then

(i) j.,(x) = ( -l)nxnG !Y(si: X);

(ii) y .. (x) = -( -I)nxnG !)\co;.x)

(iii) h~l(x) = -i( -I)nxnG :xY(e~x);

(iv) h~2l(x) = i( -l)nxnG :Jn(e:ix). PROOF

CH.4

We give the proof for (i) only, the proofs for the other three results being similar.

We shall use the method of induction; i.e., we shall assume the result true for n = N, say, and then prove it true for n = N + 1. Then since theorem 4.19 guarantees the result true for n = 0, it follows that it must be true for all positive integral n.

Assuming the result true for n = N implies that

jN(x) = ( -l)NxNG :JN ei: X). From theorem 4.18 (ii) we have

jN+1(x) = -xN :x{x-NjN(x)}

= -xN :x{x-N(-l)NxN(1!)Nei:x)}

(using the fact that we are assuming the result to be true for n = N)

= -(-t)NxN :xG!)N(si:x) = ( -l)N+lxN+lG !)(~ !)N (si: x) = ( -l)N+IxN+IG !)N+\si: x)

which thus proves the result true for n = N + 1.

§ 4.12 BEHAVIOUR OF THE BESSEL FUNCTIONS 127

Hence, by the remarks above, the result must be true for all positive integral n.

We may use the results of this theorem to write down explicitly the first few spherical Bessel functions of integral order. We obtain in this way:

j1(x) = sm

2 x _ cos x,

X X

. ( ) ( 3 1) . 3 J X = - - - Slll X - -- COS X

2 xa x x2 '

j 3(x) = (!~ - ~)sin x- (~-!)cos x; x' x 2 x 3 x

(4.30)

y1(x) = _ cos

2

x _ sin x, X X

y 2(x) = -(~- - !) cos x - i sin x, x 3 x x 2

(15 6) (15 1) . y 3(x) = - - - -- cos x - - - - sm x. x' x 2 x 3 x

(4.31)

It is to be noted that because of the relationships (4.26) and (4.27) all the above information concerning spherical Bessel functions of integral order provides an equal amount of information concerning Bessel functions of half-odd integral order.

4.12 BEHAVIOUR OF THE BESSEL FUNCTIONS FOR LARGE AND SMALL VALUES OF THE ARGUMENT

Theorem 4.21

As x ~ oo we havet

(i) fn(x) ~ (:Jt cos {x- (n + i)~};

(ii) Yn(x) ~ (:xr sin {x- (n + i)~};

(iii) Hin'(x) ~ (:JI exp [i{x - (n +f)~} J; (iv) H;,2l(x) ~ (~X)! exp [ -i{ X - (n + f)~} J ;

t I I ere we usc the symbol ~ rather loosely to mean 'behaves like'. A more precise dl'finition is: f(x) ~ g(x) as ,\' -+a if lim (f(x)/g(x)) 1.


1 (v) I .. (x) ,._, v'(Znx) e-T;

(vi) K,.(x),...., (~)'"e-";

( . ") . ( ) 1 . ( nn). Vll }n X ,...., ~ 510 X - 2 '

( ... ) 1 ( nn) vm) Yn(x ,...., -~cos X - 2 ;

(ix) h~1l(x),....,- !exp [i{x- (nn;2)}]; X

(x) h~l(x) ,._,! exp [ -i{x - (nnj~}]. X

PROOF

We shall first prove result (vi) and then deduce the remainder of the results from it.

(vi) By theorem 4.17 (ii) we have

K (x) = y'n (~)" f"" e-xt(t2 - 1 }n-l dt. " r(n + i) 2 1

We now make the change of variable u

t=1+x

1 so that dt = - du.

X

Also, when t = 1, u = 0 and when t = ro, u = co.

_ y'n (~)"fro e-(x+u)(~ 2u)n-!~ Thus K.,(x)- r( 1 ) 2 2 + du n+ 2 0 X X X

= y'n (~)"e-"'(~)n-l~ f"' e-u(~ + 1)n-lun-A du rcn + !) 2 X X 0 2X

= - - e-"' e-uun-l 1 + - du. y'n (2) -~1 f"" ( u )"-~ r(n -!- !) X X 0 2x

For large values of x we have uj2x small, so that we may take as an approximation

§ 4.12

Then

BllHAVlOUll. OF THE BESSEL FUNCTIONS

v:n: 1 J"" K (x) ""' -- e-"' e-"un-t du n r(n + !) v(2x) 0

v:n: 1 = r(n + !) v(2x) e-•T(n + !)


= (~Y e-"'.

(iii) From theorem 4.14 we have

Kn(x) = ~ in+lHn(ll(ix)

129

where, if n is non-integral, we take the value of the many-valued function in+l given by e(br/2)(n-H).

Then

H~1l(ix) = 3 exp {( ~in/2)(n + 1)} Kn(x) :n;

and hence, on replacing x by -ix,

H~1 l(x) = 3 exp {( ~in/2)(n + 1)} Kn( --ix) :n;

Thus H~1 l(x) ""'~ exp {( ~in/2)(n + 1)} (-:n;z· )! ei"' :n; - 1X

(iv) Since

and

we have

= (~J ~ ( ~i)-! exp {( -i:n:/2)(n + 1)} t:l"'

= (~Jl ei"/4 cxp {( -in/2)(n + 1)} cix

(on writing -i = e-i"/2)

= (~J! exp [i{x- (n + !)(:n:/2)}].

H~2l(x) = ]n(x) - iYn(x)

I-!~1 l(x) =]n(x) + iYn(x} H~2l(x) = {H~ll(x)}*

__, [(~J~ exp [i{x- (n + !)(:n:/2)}]]*

( 2 )} - cxp [ - i{(11 I -l.)(n/2) }]. JTX "


(i) We have

]n(x) = Re H~1>(x)

"'"'Re (:J ~ cxp [i{x- (n + f)(;r/2)}]

(2): { n} = nx cos x - (n + t).z .

(ii) Similarly Yn(x) = Im H~1>(x)

,...., Im (~)! exp [i{x - (n + l)(n/2)}] nx ~

( 2 )~· . {""' n} = nx sm x - (n + !).z .

(v) By definition we have ln(x) = i-n].,(ix) so that, by part (i) above, we have

ln(x) ,..._, i-•'(n~xY cos {ix - (n + !)~} = j-n-}(:J: ![exp { -x-(n+f)(n/Z)i} +exp {x+(n+})(n/Z)i}]

= i-n-i v(~nx} exp {x + (n + !)(n/2)i}

(keeping only the dominant term for x ~ oo)

• 1 1 ( ) . ' = 1-n-• exp X 1n+• v(2nx) .

(writing exp (in/2) = i) 1

= v(2nx) e:~:.

(vii) jn(x) = (~) i]n+:(x)


__ (~Y (:Xr cos { x _ (n + 1 >~} (by result (i) above)

= ~ cos ( x - n; - i)

§ 4-.12 BEHAVIOUR OF THE BESSEL FUNCTIONS 131

1 . ( nn) = ~sm x- 2 .

(viii) Yn(x) = (~Y Yn+!(x)

(by definition ( 4-.27))

(nY(2 y { n} "' 2x nx sin x- (n --+-- 1)2 (by result (ii) above)

1 . ( nn n) = ~sm x- Z- 2

=- ~ sin {~ - ( x - ;) }

=-~cos (x- n;). (ix) h~1l(x) = j.,(x) + iyn(x)

1 . ( nn) i ( nn) --~sm x- 2 -~cos x-2

=- ~{cos ( x - ;) + i sin ( x - n;)} 1

= - ~ exp [i{x - (nn/2)}].

(x) h~2l(x) = {h~l(x)}* _, [- ~ exp [i{x- n(n/2)}]]* =! exp [ -i{x - (nn/2)}].

X

Theorem 4.22

As x ~ 0 we have

(i) J,.(x) "'I'(n 1

1 l)GY

BESSEL FUNCTIONS 132

(ii)

[

- ~r(n)(~t Y.,(x),.....,

2 -lnx n

(n ~0)

(n = 0);

(iii) j.,(x),....., (2n: 1)!! (n integral) ;t

(. ) ( ) (2n - 1)! ! lV Jn X ,...., -

1 x"+ (n integral).

PROOF ~

CH.4

(i) We consider the series (4.6) for ],.(x). As x-+ 0 only the lowest power of x will be important. But this lowest power of x is just given by the term with r = 0, so that we have

1 (X)" ].,(x) ""r(n + 1) 2 · (ii) When n is not an integer we have

Y.,(x) = cos nn ]~(x) - J _,.(x) smnn

and we pick out the lowest power of x occurring in this expression. This will be contained in] _,.(x) and will be given by the result of (i) above, so that we obtain

Y.,(x) ""- sin1 nn r( -~ + 1l~) -n

=- r~>(~)". using the result of theorem (2.12) that

7t r(n)r(1 - n) = -. -.

smnn

When n is an integer we pick out the dominant term from the series for Y.,(x) given by theorem 4.3. Remembering that a power of x dominates a logarithm, we see that

2 Y.,(x) ,..,_,-In x (n = 0)

n

t By nil (n double factorial) we mean

( _ 2)(n _ 4) ... {5. 3.1 !~ n !s odd n n 6.4.2n n IS even.

§ 4-.13

and

GRAPHS OF THE BESSEL FUNCTIONS

Yn(x) ~- ~ (n- 1){~)-n

= - ~ r(n)(~Y·

(n ~ 0)

133

(iii) We have from definition (4-.26)

jn(x) = J (~)Jn+i(x) "' J (;Jr(n 1+ t)GY+t

(vn)xn 1 = - 2n + 1 ( n + t.,--)(~n------=~-:-) -. -. -. -t-,. t:-:::r:::-:(-::-:t)

(using repeatedly the fact that r(n + 1) = nr(n)) ( v'n)xn

(2n + 1)(2n -1) ... 3.l.y'n (where we have made use of the result of theorem 2.6 that r(!) = v'n)

(2n + 1)! !"

(iv) y,.(x) = J (~) Y,.+l(x)

"' J (~){- ~r(n + t)(~t+i} 1 2"

= - vn.xn+lr(n + !) 1 2" = ---:; -

1(n- i)(n- t) .•• t.tr(t)

vnxn+

1 (2n - 1 )(2n - 3) ... 3 .1. v'n =- v'n xn+l

(2n - 1)!! xn+l

4.13 GRAPHS OF THE BESSEL FUNCTIONS

We give here some graphs of various Bessel functions; they bring out rather clearly the behaviour for large and small x discussed in the previous section.


-0·5

-1·0 ~'

FIG. 4.1 ]n(x), n = 0, 1, 2

FIG. 4.2 Yn(x), n = 0, 1, 2

§ 4.13 GRAPHS OF THE BESSEL FUNCTIONS

In(X}

FIG. 4.3 In(x), n = 0, 1, 2

l(,lxl

3

2

ol_--~~~~~~---2 3 X

FIG. 4.4 Kn(x), n = 0, 1, 2

135

136 BESSEL FUNCTIONS CH.4-

X

FIG. 4.5 in(x), n = 0, 1, 2

Y,(X)

X

FIG. 4.6 Yn(x), n = 0, 1, 2

§ 4.14 ORTHONORMALITY OF THE BESSEL FUNCTIONS 137

y (Graphs oscillate for large x)

y =ber x

FIG. 4.7 her x, bei x

y

8 X

FIG. 4.8 ker x, kei x

4.14 ORTHONORMALITY OF THE BESSEL FUNCTIONS; BESSEL SERIES

Theorem 4.23

1j!; 1 and!;, are roots of the equatiou],.(~a) 0.


PROOF

We first show that if ~i and~; are distinct roots of the equation].,(~a) = 0 then

J: x].,(~;x)].,(~;x) dx = 0.

] .. (~;x) andJ.,(~,x), being Bessel ~nctions, must satisfy the following equatwns

d 2 d X

2 dx-J .. (~;x) + xdX} .. (~;x) + mx2 -- n2

)} .. (~;x) = 0

and

x2 ::2].,(~Jx) + x :x].,(~1x) + (~}x 2 - n 2)}.,(~1x) = 0,

which may be written in the form

x!{x!} .. (g;x)} + (~rx 2 - n 2)].,(~;x) = 0 (4.32)

and (4.33)

Multiplying equation (4.32) by (ljx)J.,(~1x), equation (4.33) by (ljx)].,(~ix) and subtracting, we obtain

] .. (~;x):x{x!] .. (~;x)}- ] .. (~;x):x{xd~] .. (~1x)} + (~; - ~J) x].,(~;x)].,(~1x) = 0.

Applying the result u(dvjdx) = (djdx)(uv) - v(dujdx) to both first and second terms gives

:x {l .. (~;x)x :xj.,(g;x)}- {!J .. (~1x)}x !J .. (~;x)

-! {J .. (gix)~J .. (~;x)} + {!J .. (~;x)}x !J .. (~,x) + m - ~J)x] .. (~;x)] .. (~;x) = 0

and thus

:x {l .. (~;x)x !J.,(~;x)} - :x {J .. (~;x)x !} .. (g1x)}

+ (g~ - gJ)xj.,(g;x)] .. (~,x) = 0.

§ 4-.14- ORTHONORMALITY OF THE BESSEL FUNCTIONS 139

Integrating from 0 to a, we obtain

[ d d ]a

J.,(g1x)x dxj.,(gix) - ] .. (~;x)x dx].,(~;x) 0

+ c;; - ;J) f: xin(;;x)J,,,cg;x) dx = o. The first term will now vanish at the upper limit since

] .. (;;a) =] .. (!;;a) = 0, and at the lower limit because it contains a factor x.

Accordingly (;; - gJ) f: xj.,(g;x)] .. (~1x} dx = 0

and hence f: xj .. (;;x)] .. (!;;x) dx = 0

provided gi ~ g,. To complete the proof we require to prove that

fa a2

0 x{].,(!;x)}2 dx = z {J .. +l(!;a)}2

if ] .. (!;a) = 0.

Now, if we denote J .. (~x) by z we have

x2z" + xz' + (g 2x 2 - n2)z = 0 which, on multiplication by 2z', becomes

2x2z"z' + 2xz' 2 + 2(!; 2x 2 - n2)zz' = 0

and this is readily seen to be equivalent to

d dx {x2z'2 - n2z2 + g2x2z2} - 2!;2xz2 = 0.

Integrating this equation from 0 to a gives

[x2z' 2 - n2z 2 + !; 2x2z 2] 0 - 2!; 2 J: xz2 dx = 0

which, on replacing z by ] .. (!;x), becomes

[x2{!J .. (;x)} 2

- n2{J.,(gx)}2 + g2x2{J .. (;x)}2J:

- 2!; 2 f: x{].,(!;x)}2 dx = 0

and hence, using the facts thatJ .. (!;a) '= 0 and n]n(O) =~ 0,

[ a2{ dl ),.(!;:.:)}2] U 2 f" xf.!,.(!;x)l2 dx 0.

( ,'\: I fl 0

14() BESSEL FUNCTIONS

Hence

I: x{],.(gx~ dx = z1-2 [a 2 {!Jn(gx)rl~a· But, by theorem 4-.S(iv), we have

d n d(gxln(gx) = ~x],,(gx) - ]n+I(gx)


d n dxj.,(gx) = Xj.,(gx) - Un+l(gx)

so that

I: x{]n(gx)}2 dx = 2~ 2 [a 2{~ln(gx) - g]n+l(gx)rl~a

Theorem 4.24

1 = 2g2a2~2{Jn+t(ga)}2

(again using the fact that] n( ~a) = 0)

a2 = 2 {J.,+l(~a)}2.

CH.4-

If f(x) is defined in the region 0 < x < a and can be expanded in the form C() L c;]n(g;x) where the g; are the roots of the equationj.,(ga) = 0, then

i=l

2 I: xf(x)]n(g;x) dx

C; = a2{J.,+l(g;a)}2 .

PROOF

We have C()

f(x) = L c;]n(g;x) i=l

C()

so that xf(x)j.,(g;x) = L C;x].,(g;x)j.,(g;x) i=l

§ 4-.15 INTEGRALS INVOLVING BESSEL FUNCTIONS

and hence

Thus

C; =

which completes the proof. 00

00 2

= 2: Ci ~ ],.+l(~;a) 2bii i=l

(by theorem 4-.23)

2 J: xf(x)Jn(~1x) dx

a2 Und~;a)}2

14-1

The series 2: ctfn(~ix), with ci as given above, is called the Bessel series i=l

for f(x). The conditions under which f(x) may be expanded in this way are given in the next theorem, which we quote without proof.

Theorem 4.25

If f(x) is defined for 0 < x < a and J: ( yx)f(x) dx is finite, then 00

2: cJn(~ix) i=l

(with C; as defined above) is convergent and has sum f(x) if f(x) is continuous at x and i{f(x+) + f(x-)} if f(x) is discontinuous at x.

4.15 INTEGRALS INVOLVING BESSEL FUNCTIONS

Theorem 4.26

2(x/2)n-m Jl Jn(x) = (1 - t 2)n-m-ltm+l]m(xt) dt

r(n - m) 0

(n > m > -1). PROOF

Consider the integral

J --~ Jl (1 . t2)n-m-ltm HJ,.(xt) dt 0

14-2 BESSEL FUNCTIONS CH.4-

and substitute the series (4-.6) for] m to obtain

I= fl (1 - t2t-m-ltm+l * (- 1)' (xtj2)m+2r - dt o ;S r!r(m + r + 1)

= ~ ( -1Y (x/2)m+2r Jl (1 - t2)n-m-lt2m+2r+l dt £:b r !r( m + r + 1) o

2"' (x/2)m+2r 1 Jl = ( -1 y _ (1 _ u)n-m-lum+r du r=O r!r(m + r + 1) 2 o

(on making the substitution u = t 2).

But J: (1 - u)n-m-lum+r du

= B(n - m, m + r + 1)

(by the definition of the beta function, provided n - m > 0 and m + r + 1 > 0. This last condition is equivalent to m > -1, since r takes on values between zero and infinity.)

r(n- m)r(m + T + 1) r(n + T + 1) ,,

(by theorem 2. 7).

~ (x/2)m+2r I= tr(n- m) .G ( -1)' 'r( 1)

r=O r. n + r + Hence

Thus

(x)m-n ~ (xj2)n+2r

= ! rc n - m) 2 6 ( -1 y ---cr !=r~( n_:__+___:._r_-t_--:--1)

= trcn- m)Gr-J .. (x).

]n(x) = 2 (x/2)n-m I r(n- m)


Theorem 4.27

(a> 0).

§ 4-.15 INTEGRALS INVOLVING BESSEL FUNCTIONS 14-3

PROOF

From theorem 4-.6 we have

1 I"' ] 0(x) =- cos (x sin c/>) dcf>. n o

Hence

I 00 I 00 1 I" e-ax]0(bx) dx = e-ax- cos (bx sin c/>) de/> dx 0 0 7t 0

= ~ I: [ I: e-ax t{exp (ibx sin c/>) + exp ( -ibx sin cf>)} dx J dcf>

(interchanging the order of integration)

= 2_ I" [exp {-(a - ib sin cp)x} exp {-(a + ib sin cf>)x}J 00

de/> 2n o -a + ib sin cp + -a - ib sin cp o

1 I" { 1 1 } =- .. + . . d Zn o a-tbsmcp a-t-tbsmcf> c/>

=~I" 1 dcf>. n oa2 + b2 sin 2 cf>

But this last integral may be evaluated by elementary means (e.g., by the substitution u = cot c/>) to give

Ioo a n o e-ax]o(bx) dx =; ay(a2 + b2)

1

Theorem 4.28

J 00 1

0 fn(bx) dx = b (if n is a non-negative integer).

PROOF

We first prove the result for n = 0 and n = 1, and then show that if the result is true for n = N, it is also true for n = N + 2, thus proving the result for all non-negative integral n.

For n = 0 we take the limit as a---+ 0 of the result of theorem 4-.27, obtaining

I"' 1 0

] 0(bx) dx- b.


For n = 1 we make use of theorem 4.8(ii), which says that

d dx{x-"].,(x)} = -x-n_j.,+1(x)

so that, by taking n = 0, we have

d dx1o(x) = -]I(x),

and replacing x by bx gives

d d(bxl 0(bx) = - ] 1(bx)

which is equivalent to

1 d b dx1o(bx) = -1I(bx).

Hence J~ 11(bx) dx = - ~ [1o(bx)J~ 1

=b·

CH.4

since 1o( co)= 0 and ] 0(0) = 1, results which ar~ implied by theorems 4.21(i) and 4.22(i).

If we now use theorem 4.8(v) and integrate from 0 to oo, we obtain

· [1 .. (x)J~ = 1 J~ Un-t(x) - ]n+l(x)} dx.

Remembering that n > 0 so that 1 .,( oo) = 1 .,(0) = 0, we have 1 r co

0 = Z J 0

{].,_1(x) - ]n+I(x)} dx

and thus

J~ 1ndx) dx = J~ fn- 1(x) dx.

Replacing n - 1 by Nand x by bx gives the result

f~ 1N+2(bx) dx = J: 1N(bx) dx.

reo 1 f"' 1 Thus if J

0 1 N(bx) dx = b' we also have

0 ] N +2(bx) dx = b' and hence

the proof is complete.

§ 4-.15 INTEGRALS INVOLVING BESSEL FUNCTIONS

Theorem 4.29

PROOF

(i) From equation (4-.6) we have

~ 1 (b2x) 2r+n

] .. (bx) = .~ ( - 1)' r!r(n + r + 1)

so that J: J.,(bx)xn e-ax dx

= "" ( -1 )' - e-Q:Ilx 2•+2n dx . 00

1 (b)2r+n f«> • ~ r!r(n + r + 1) 2 o

Joo t2r+2n

= 0 e-t -a2-r+_2n_+_l dt

(writing t = ax)

1 =

2 l(2r + 2n + 1 ), a r+2n+

145

(by definition (2.1) of the gamma function),

so that

J:] .. (bx)xn e-ax dx

"" 1 (b) 2r+t1 1 = 2< -1Y r!r(n + r + 1) 2 a2r+2n-rl(2r + 2n + 1)

r-"-0

(using the result of theorem 2.2 that l'(n 11) ~- nl'(n)).


But by theorem 2.10 we have

so that

r(2x) 22~-1 -- =rex+ .1)r(x) 2 yn

f~ ] .. (bx)xn e-ax dx

= ~ (-l)r ]_zr(r + n + l)22r_+2n--I(~)2r+n _1_ L... r! 2 yn 2 a2r+2n+l r=O

(by the binomial theorem)

= _1_ ~ (- 1)"(n + i)(n + j) ... (n + r- t)(b2)r a2n+I L... r! a2

r=O

= _1 -~ ( --1)' r(n + ~) (n + ~)(n + t) ... (n + r -· ~)(b2)r a2n+ 1 L... r(n + 1) r! a2

r=O 2

on repeated use of the result xr(x) = r(x + 1) (theorem 2.2).

Hence fcc znbn 1 o ]n(bx)xn e-ax dx = yn r(n + t} (a2 + b2)n+l

znr(n + i) bn yn (a2 + b2)n+1'

CH.4

(ii) This result follows immediately from result (i) if we differentiate both sides with respect to a.

§ 4.15 INTEGRALS INVOLVING BESSEL FUNCTIONS 147

Theorem 4.30

Ioo ~

(i) 0

fn(bx)xn+l exp ( -ax2) dx = (

2a)n+l exp ( -b2 j4a);

I oo bn ( b2) (ii) 0 ]n(bx)xn+ 3 exp ( -ax2

) dx = 2n+lan+2 n + 1 - 4a exp ( -b2j4a);

(a> 0).

PROOF

The proof is very similar to the proof of theorem 4.29; so we shall not give it here.

Theorem 4.31

(i) I00

sinax]0(bx) dx = { ~ 0 v(a2- b2)

(b >a)

(b <a).

(ii) [ "'" ax]0(bx) dx ~ { v'(b':- a') (b >a)

(b <a).

PROOF

We shall obtain these results by replacing the a occurring in theorem 4.27 by ia, although it should be noted that strictly speaking this is not allowed, since the proof of theorem 4.27 depended on having the real part of a positive. It is possible to give a procedure justifying taking the real part of a zero, but we shall not describe it here.

Thus we have

I co -iaxj (b ) d 1 0 e 0 X X = y(b2 - a2)

so that

Ioo I"' 1 cos ax]0(bx) dx- i sin ax ] 0(bx) dx =-(b--). o o v 2- a2

Equating imaginary parts of both sides gives

fcc 1

sin ax] 0( bx) dx = - ( b ) if a > b o V a2- 2

I"' sin ax ] 0(bx) dx ()

() if a ·:b.

I'll' I.


Equating real parts of both sides gives

J co ] (b ) d 1

if b > a 0

COS ax o X X= y(b2 -a2)

J: cos ax } 0(bx) dx = 0 if b <a.

4.16 EXAMPLES

Example 1

Use the generating function to prove that 00

fn(X + y) = 2: ]r(x)fn-r(y). r=-<Xl

We have from theorem 4.5 that

exp {i(x + y)(t-})} = i ]n(x + y)tn n=- oo

so that ].,(x + y} is the coefficient of !" in the expansion

of exp { i(x + y)(t - ~)}. But

C() C()

= 2: ],(xW. 2: ].(y)t• r=-CO 8=-CO

00 2: ],(x)].(y)tr+s. r,s=- <XI

For a particular value of r we obtain tn by takings = n - r, so that, for this value of r, we obtain the coefficient of tn as ],(x)fn-r(y); hence the total coefficient of tn is obtained by summing over all allowed values of r.

00

Accordingly the coefficient of tn is equal to L ]r(x)fn-r(Y) and thus r=- co

C()

fn(X + Y) = 2: ]r(x)].,_r(Y)· r=-co

§ 4.16 EXAMPLES

Example 2

f00 ]n(x) 1

Show that -- dx = -. o x n


f~ fn(x) dx = 1.

But from theorem 4.8(vi) we have

Zn - fn(x) = fn-t(x) + fn+l(x) X

f wj~) fw foo so that Zn _n- dx = ]n_1(x) dx + fn+l(x) dx

0 X 0 o

= 1 + 1,

and hence Joo fn(x) dx = ~. o x n

Example 3

If~; are the solutions of the equation] 0(~) = 0,

h h ~ Jo(~;x) s ow t at L -- -- = -~In x (0 < x < 1).

i~l {tJ1(~;)}2

We use theorem 4.25 to write 00

-!In x = L c;}0(/;;x) i~ 1

with

1 L x In x] 0(/;;x) dx

= -- -- u~-c;;;)f-

149

To evaluate the integral appearing in the numerator we use the series (4.6) for ./0 :

,. I)

1 (

1: ')2r 1..,.'\: 1)' r!l'(r : I) 2

J u(i; ,x)


= r~ ( -1)'" (r~)2 (%Yrx2r

(and we note that sinceJ0(~£) = 0 we shall have

~ 1 (;;)2r 6 ( -1Y (r!)2 2 = 0).

We shall also require the integral J: x" In x dx.

Integrating by parts gives

f

1

x" In x dx = [-1 -xn+~In x]1

- f1

-1-xn+l} dx

o n+1 o on+1 x

Jl 1 = 0-

0;, + 1xn dx

1

Thus J: x In x ] 0 (~ix) dx

= ~ ( -1)' -1-(~)zrfl x 2r+lln x dx L.. (r!) 2 2 o r=O

= ~ (-lY (r~) 2(~Yr{- (2r ~2)2} (using equation (4.35))

1 ~ 1 (~')2r+2 = ~f 6 ( - 1y+t {(r + 1)!}2 2

1 ~ 1 (;•)2r = ~~ 6 ( - 1)'" (r!) 2 2

= :~ { ~ < -1Y <r~>z(~iyr- 1} 1

= -2, ~,

Thus we have (using equation (4.34)).

1

CH.4

(4.34)

(4.35)

§ 4.16 EXAMPLES 151

and hence

Example 4

Show that ' ' A ],..(x) Yn(x) - fn(x) Y,..(x) =-X

where A is a constant, and by considering the behaviour for small x show that A= 2/n.

We have

d { ' ' dx ].-.Yn- fnYn}

=] .. Y~' +]~Y~-]~Y .. - ]~Y~ =] Y" -J"Y n n n n

(all functions having argument x);

but bothJ,..(x) and Yn(x) satisfy the equation

x 2y" + xy' + (x 2 - n 2)y = 0 so that

d { ' ' } x2 dx ].-.Yn -JnY.-.

=fn{ -xY;,- (x 2- n 2

) Yn}- { -x];,- (x2 - n 2)J .. }Y ..

= -x{] .. Y;,- ];,Y .. }.

Hence, writing],..Y;,- ];,Yn = z, we have

dz z dx x

or

dz + dx = O. Z X

When integrated this gives

In z + In x = constant which is equivalent to

In zx = constant

and hence z.v constant A, say.


Thus z = A so that, by the definition of z, X

' ' A ]nYn- fnY,. = -. X

For small values of x, we have, by theorem 4.22,

2 ] 0(x) ,._, 1, Y 0(x) ,._,-In x

n

so that, considering the relationship for n = 0, we have

Hence

giving

and thus

Example 5

I Y' ' 1 2 1 21 ,. ,.-JnY .. ,..., .-.--0.- nx nx n

2 =-.

:nx

2 A , nx x

I I 2 ],.(x) Yn(x) - fn(x) Y .. (x) = --·

nx

We shall show this by first proving that

t{],.(t)}2 = :J~{],.2(t)-fn-l(t)Jn+l(t)}]

and then integrating this relationship from 0 to x. We have

t J~(x) == {f,.(x)}2•

CH.4


= t(J~- 1n-t]n+l) + ~(2]n]~- 1~-dn+l- 1n-J~+l) (it being understood that the argument of all Bessel functions here is t)

= t(]~ - fn-dn+l) + ~{21 .. Hfn-1- ]n+l)

- (n ~ ~ fn-1 - J .. )]n+l - 1n-1(1.-.- n ; 11n+I)}

(using theorem 4.8(iii), (iv) and (v))

" [2 2 = t(J;- 1n-11n+1) + Z ·f_fn-Jn+l

=tf!. Integrating from 0 to x now gives

fx [t2 ]"' 0 tJ! dt = -zU~(t)- 1n-t(t)1.-.+t(t)}

0

x2 = z-U~(x) - 1n-t(x)1n+l(x)}.

Example 6

Use the generating function to show that 1.-.( -x) = ( -1 )~ .. (x).

We have, from theorem 4.5,

exp {~(t - ~)} =I 1n(x)tn n=-oo

so that I 1n( -x)tn = exp G{ -x(t- ~)} J 1t=- 00

= exp [:{c -t)- -1

}] 2 ( -t)

n==--- oo ro

~ 2: ( --t)nt•'],.(x). II - 00

Equating the coefficients of I" on both sides gives

] .. ( x) ( 1)'~/ .. (x).


PROBLEMS

( 1) Show that

d~{x] .. (x)]n+l(x)} = x{]!(x) - J~+1(x)}.

(2) Show that J"'/2 sin X (i) ] 0(x cos 0) cos 0 dO = --;

0 X

J"'/2 1 - COS X (ii) ] 1(x cos 0) dO = ----.

0 X

(3) Show that ]~(x)] _,(x) - ].,(x)]~ .. (x) = Ajx where A is a constant; by considering the behaviour for large values of x, show that

A = (2 sin nn)/n.

( 4) Find the solution of the differential equation d2y

h/x)dx2 + J.y = 0

which is zero at x = 0. oo n

(5) Show that L x,J .. (a) =]0{y(a2 - 2ax)}. n. n=U

(6) Use the generating function to prove that

1 = {j0(x)}2 + 2{]1(x)}2 + 2{]2(x)}2 + ... and deduce that IJ0(x) I< 1, IJ .. (x) I< 1/v/2 (n = 1, 2, 3 ... ).

(7) Sh h ["" ]n+I(x) d _ 1 'f 1 OW t at J

0 ----;;;- X- znr(n + J) 1 n > -2·

(8) Prove that i ] .. (kx)t" = exp {;lk - D} i: kntn]n(x) n=-co n=-co

ro m

and deduce that I.,(x) = L :_rln+m(x). m~o m.

(9) Show that exp {~(t + ~)} = i In(x)tn. n=- oo

(10} Show that J.,(x), Yn(x), Kn(x) and In(x) all satisfy the differential equation

d4y + ~ d 3y _ 2n2 + 1 d 2y + 2n 2 + 1 dy + (n4 - 4n2 _ 1) = O.

dx4 x dx3 x 2 dx2 x3 dx x4 y

PROBLEMS 155 00

(11) Expand x 2 in a series of the form L Cr]o(A,.x) valid for the regton r~l

0 < x < a, where Ar are the roots of the equation J0(Aa) = 0.

(12) Use the differential equation satisfied hyfn(x) to show that

{n(n + 1) - m(m + 1)}jn(x)jm(x) = d~ {x 2j~(x)jm(x) - j .. (x)j~(x)}. Hence show that

foo. • sin {(n- m)n/2} o }m(x)Jn(x) dx = n(n + 1) - m(m + 1)

if m ~ n, and use L'Hopital's rule to deduce that

f ~ {jn(x)}2 dx = 2(2n n + 1)"

Determine also the values of J: )m(x)jn(x) dx (m ~ n) and

f:"' {jn(x) }2 dx.

(13) Show that

(") h = ~ (-1)k (x/Zt+2

k cos {!(n + 2k)n}. 1 ernx 7 kl( k)l , t;;b .n+ .

("") h . _ ~ (-1)k (x/Zt+ 2

k sin {!(n + 2k)n} 11 elnX- ~ kl( k)l .

k~o · n + · ( 14) Prove that

(i) J: t her t dt = x hei' x;

(ii) J: t hei t dt = -x her' x.

(15) Show that, for large values of x,

(i) her x ,.._, --\1(12n-;;{'/V2

cos ( : 2 - ~); ("") h . 1 I 1 ., • ( X n) II Cl X ,.._, c·' ' - Sill .

v'(2nx) v2 H

5

HERMITE POLYNOMIALS

5.1 HERMITE'S EQUATION AND ITS SOLUTION

Hermite's equation is given by

d2

y- 2xdy + 2n" = 0 dx 2 dx J '

(5.1)

and in applications we are required to find solutions which are finite for all finite values of x and are such that cxp (tx 2) y(x) ~ 0 as x ~ oo.

The methods of Chapter 1 are applicable, and if we try for a solution of the form -

00

z(x, s) = L a,.xs+r r=O

we find that the indicia! equation has roots s = 0 and s = 1, with a1 indeterminate when s = 0, so that s = 0 gives the two independent series solutions. The recurrence relation for the coefficients then has the form

ar+z 2(r - n) ar (r + 1 )(r + 2)"

(5.2)

This recurrence relation may now be used to construct the two series solutions. However, from these series it can be shown that both solutions behave like exp (x2) for large values of x. Thus they cannot satisfy the requirement that exp (!x2) y(x) ~ 0 as x ~ oo; this can only be satisfied if the series terminate. From equation (5.2) we see that this will be so if, and only if, n is a non-negative integer, for then an+2 and all subsequent coefficients in the corresponding series will vanish. We shall now write the

§ 5.2 GENERATING FUNCTION 157

series in descending powers of x. Using equation (5.2), rewritten in the form

(r + 1)(r + 2) a, =- 2(n - r) ar+2•

we obtain the series

_ { n _ n(n - 1) n- 2 n(n - 1)(n - 2)(n - 3) n-4 y- an X 2.2 X + 22.2.4 X + • • •

( -l)rn(n- 1) ... (n- 2r + 1) n- 2r } + 2r. 2 . 4 ... 2r X + ' ' '

~ ( 1) n(n - 1) ... (n - 2r + 1) 2 =an L - , xn-, r=O 2r.2.4 .. ,2r

h 1 {in if n is even (w ere [In] = !(n - 1) if n is odd)

[lnl l

=an L ( -1)' 22r l( n'_ 2 )lxn-2r, r=o r. n r .

The standard solution is obtained by making the choice of 2n for the arbitrary constant a.,; the solution is then denoted by Hn(x) and called the Hermite polynomial of order n:

[!n] l

""" n. H.,(x) = :S (-1)' rl(n- 2r)l(2x)n-2r. (5.3)

5.2 GENERATING FUNCTION

Theorem 5.1

PROOF

We wish to pick out the coefficient of tn in the power series expansion of exp (2tx - t 2).

Now,

= ~ (2tx): ~ (--: t 2)"

~ rl L... sl r=O s ~o

(2x)' 1)·'· fr12•.

r !s! '·. 0

158 HERMITE POLYNOMIALS CH.5

For a fixed value of s we obtain t" by taking r + 2s = n, i.e., r = n - 2s, so that for this value of s the coefficient of tn is just given by

(Zx)n-28 ( -1 )8 __:___:____

(n- 2s)!s!" The total coefficient of t" is obtained by summing over all allowed

values of s, and, since r = n - 2s, this implies that we must have n - 2s > 0, i.e., s < in. Thus, if n is even, s goes from 0 to !n, while if n is odd, s goes from 0 to !(n - 1); that is, in all cases, s goes from 0 to [in] with [!n] defined as above.

Thus we have: rtnJ 1

coefficient of tn = "" ( -1 )8 (2x)n-2s ~ (n- 2s)!sl

1 = ---;Hn(x), n.


5.3 OTHER EXPRESSIONS FOR THE HERMITE POLYNOMIALS

Theorem 5.2

PROOF

By use of the generating function of theorem 5.1 and Taylor's theorem, which states that

~ (dnF) tn F(t) = L dtn l=o nl'

n=O

we have

H.,(x) = [an e21z-t•] atn l=o

=[on e"''-<x-1)'] atn t=o

"''[an -(x-t>'] =e -e . atn 1=0

But a a a/<x - t) = - a/(x - t),

§ 5.3 OTHER EXPRESSIONS FOR HERMITE POLYNOMIALS 159

so that

and we have

Theorem 5.3

PROOF

We have

and hence

Thus

an an -f(x- t) = ( -l)n -f(x- t) atn axn

= e-t• .e21x

= e-t'+2tz.

Expanding both sides in powers of t, we have

( 1 d 2 ) ~ znxntn ~ tn

exp -- -2 ~ -

1 = ~ Hn(x)-

4 dx n~O n. n=O n!

(using the generating function property of theorem 5.1).

t The exponential of an operator is defined by its power series expansion. Thus, forex-

umpll·, l'Xp( (1

1.) ~ \( dl ) n -~ 1,· dl ~ .. so that exp (d) f(x) - . ~ 1

1 ~~nf~o ( ·' £.... no ( x £.... n 0 l x Jx ~ no ux , u fl 0 " u

160 HERMITE POLYNOMIALS CH.5

Equating the coefficients of tn on both sides now gives

{ex (-~ ~)} znxn == Hn(x)

p 4 dx 2 nl nl

and hence

5.4 EXPLICIT EXPRESSIONS FOR, AND SPECIAL VALUES OF, THE HERMITE POLYNOMIALS

We may use either the definition (5.3), or theorem 5.2 or theorem 5.3 to write down an explicit expression for the Hermite polynomial of any order we choose. For the first few orders we obtain

Theorem 5.4

PROOF

H0(x) = 1 H 1(x) = 2x H 2(x) = 4x2 - 2 H 3(x) = 8x3 - 12x H 4(x) = 16x4 - 48x 2 + 12 H 5(x) = 32x5 - 160x3 + 120x.

From the generating function of theorem 5.1 with x = 0, we have

e-t• = i Hn(O)~ n=O n.

which, on expanding the left-hand side in powers oft, reduces to

~ t2n ~ tn L) -l)n l = ~ Hn(O)-r

n=O n. n=O n.

Equating coefficients of corresponding powers of t on both sides gives

Hn(O) = 0 if n is odd

(which is equivalent to H 2n+l(O) = 0 with n a non-negative integer) and

1 1 . h . . ) ( -1 )n -· = H2n(O) -- (w1t n a non-negative mteger nl (2n)l

§ 5.5 PROPERTIES OF HERMITE POLYNOMIALS

which yields

5.5 ORTHOGONALITY PROPERTffiS OF THE HERMITE POLYNOMIALS

Theorem 5.5

PROOF

We have

oo m

and e-•'+ 2•"' = L Hm(x)~

m=O m.

161

so that f:oo e-x' Hn(x)Hm(x) dx is the coefficient of (tnsm)J(n!m!) Ill

the expansion of J: 00

e-x' e-t'+Ztx e-•' + 2""' dx.

But

Joo -x' -t'+2tx -s'+2sxd e c e x -oo

= e-t'-s' J: 00

exp { -x2 + 2(s + t)x} dx

= e-t'-s' f:oo exp [ -{x- (s + t)}2 + (s + t) 2] dx

= e2"1 J:oo exp [ -{x- (s + t)}2] dx

= e~'1 J:oo exp(-u 2)du

(changing the variable of integration to u = x - (s + t)) , . e~'t y:n

(by the corollary to theorem 2.6)

162 HERMITE POLYNOMIALS CH. 5

Hence the coefficient of (tnsm)/(n !m !) is zero if m ::;C: n and is ( y'n)2nn! whenm = n.

Hence Joo • { 0 ifm::;C:n - oo e-"' H,.(x)Hm(x) dx = ( y'n)2nn! if m = n

or, making use of the Kronecker delta,

5.6 RELATIONS BETWEEN HERMITE POLYNOMIALS AND THEIR DERIVATIVES; RECURRENCE RELATIONS

Theorem5.6

(i) H~(x) = 2nH .. _1(x) (n > 1); H~(x) = 0.

(ii) Hn+I(x) = 2xHn(x) - 2nHn_I(x) (n > 1); H 1(x) = 2xH0(x).

PROOF

(i) If we differentiate both sides of the generating function relationship with respect to x, we obtain

oo tn d "" H~(x)- = - exp (2tx - t 2) L. n! dx n=O

= 2t exp (2tx - t 2)

oo tn = 2t 2 H .. (x)-,

n=O n.

~ tn+l = 2 L. H,.(x)-

1 n=O n.

00 tn = 2 ,6 H .. _1(x)(n _ 1)!"

Equating coefficients of tn gives for n = 0

H~(x) = 0

and, for n > 1,

H~(x) ZH .. _1(x) n! (n- 1)!

which reduces to

H~(x) = 2nH,._Jx).

§ 5.7 WEBER-HERMITE FUNCTIONS 163

(ii) Differentiating both sides of the generating function relationship with respect to t gives

d d Leo tn - exp (2tx- t 2

) =- -H (x) dt dt n! n

n~o

and thus, performing the differentiations,

~ ntn-1

(2x - 2t) exp (2tx - t 2) = ~ -

1-Hn(x).

n~o n.

We now note that the term with n = 0 does not contribute to the summation on the right-hand side (we remember that 0! = 1) and thus, on use of theorem 5.1, the above equation becomes

~ tn co tn-1 2(x - t) n~ -;;jHn(x) = 6 (n -l}!Hn(x)

which is equivalent to

~ tn ~ tn+l ~ tn-1 2x ;S n!Hn(x)- 2 6 -;rHn(x) = 6 (n _ 1) 1Hn(x)

which may be written in the form

~tn ~ tn ~tn 2x 6 :;;

1Hn(x)- 2 6 (n _ 1) 1Hn_1(x) = f:o n!Hn+I(x).

Thus, equating coefficients of tn for n > 1, we have

1 2Hn_1(x) 1 2x---,Hn(x) - ( _ 1) 1 = --,Hn+l(x) n. n . n.

which on multiplying throughout by n! becomes

2xHn(x) - 2nHn_1(x) = Hn+l(x).

Similarly, equating coefficients of t0 gives

2xH0(x) = H 1(x).

5.7 WEBER-HERMITE FUNCTIONS

An equation closely related to Hermite's equation is d2y ;r;2 + (J. - x2)y = 0.

If we make the substitution

~I' M

(5.4)

164 HERMITE POLYNOMIALS CH. 5

we find that the above equation reduces to d 2z dz dx2 - 2x dx + (A. - 1 )z = 0. (5.5)

This is Hermite's equation of order n with 2n =A. - 1. If, as is usually the case in applications, we are looking for solutions of equation (5.4) which are finite for all values of x, we must have a solution of equation (5.5) which does not tend to infinity any faster than exp (x 2 /2) as x tends to infinity, and by the discussion of Section 5.1 this implies that !(A. - 1) must be integral. With n = t(A. - 1) the solution of equation (5.4) is then given by

lfi,.(x) = e-x'/2 Hn(x).

lfi,.(x) is called the Weber-Hermite function of order n.

5.8 EXAMPLES Example 1

Prove that, if m < n, dm zmn! dxm{Hn(x)} = (n _ m)!Hn-m(x).

From the generating function of theorem 5.1 we have that

(!"m){Hn(x)}

is the coefficient of tn /n! in the expansion of

(!:) exp (2tx - t 2).

dm But -d exp (2tx - t 2) = (2t)m exp (2tx - t 2) xm

~ tn = zmtm ~ 1 Hn(x)

n=O n. 00 1

= zm """ -Hn(x)tn+m ~ n! n=O

00 1 = zm T~ (r - m) !HT-m(x)tr,

setting r = m + n. The coefficient of tn is therefore

1 zm ( ) ,Hn-m(x) n -m.

§ 5.8 EXAMPLES

and hence the coefficient of tn jn! is n!

zm (n _ m)!Hn-m(x),

which proves the required result.

Example 2

Evaluate f: 00

x e-x• Hn(x)Hm(x) dx.

We have, from theorem 5.6(ii)

xHn(x) = nHn_1(x) + }Hn +l(x) so that

f: 00

x e-x• Hn(x)Hm(x) dx

= J:oo e-x• {nHn_ 1(x) + !Hn+l(x)}Hm(x) dx

165

= nzn-l(n - 1)!( vn)on-1, m + t.zn+l(n + 1)!( vn)on+l, m (by theorem 5.5)

= ( vn)Zn-tn!on-t, m + ( vn)Zn(n + l)!on+l, w

Example 3

Show that Pn(x) = ( v!)n! f: tn e-t' Hn(xt) dt.

From equation (5.3) we have [~n] I

H,..(xt) = ~ ( -1)' rl(n ~ Zr)l(2xt)n-2r

so that

2 fro ( ) 1

tn e-t• H,..(xt) dt vn n. 0

z J oo r:nl I = ( v~)n! 0 tn c-t• r~ ( -1)' ;!(n ~ 2r)!zn-2rxn-2rtn-2r dt

ILlnJ zn-2r+l( --1)'xn-2r Joo -t' 2n-2r = ~--------- e t dt r () (vn)r!(n- 2r)! 0

1!•:._1 2" 2r tl( 1 )'x" -2r A, ( y';r)r !(n 2r)l ~ f(n r I D (hy theorem 2.4)

166 HERMITE POLYNOMIALS CH.S

- [~ 2n-2r( -ly,xn-2r (2n - 2r)! y:7t - :S ( yn)rl(n - 2r)! 22n-2r(n - r)!

(by the corollary to theorem 2.10)

Unl (2n - 2r)! - ""' ( -l)r xn-2r - r~ znr!(n- 2r)!(n- r)!

= P .. (x), (by equation (3.17)).

PROBLEMS

(1) Show that J:co x2 e-z' {H,.(x)}Bdx = (vn)2nnl(n + !). (2) If f(x) is a polynomial of degree m, show that f(x) may be written in the form

m

f(x) = L crHr(x) r=O

where Cr = zrr~y:7t s:co e-x• f(x)Hr(x) dx.

Deduce that J:co e-z• f(x)H .. (x) dx = 0 if f(x) is a polynomial of de

gree less than n.

2 fco (3) Show that e-x• = yn 0

e-t• cos 2xt dt.

By differentiating this result 2n times with respect to x, show that

22n+I( -1)"' ex' foo H2r1(x) = e-t• t 2n cos 2xt dt,

y:7t 0

and obtain a similar expression for H 2n+1(x).

2"'( - i)n ex' f co Deduce that H.,(x) = e-t'+2itx t"' dt.

y:7t -oo

(4) Use the result of problem 3 to show that

~ H .. (x)Hn(y)tn = (1 - tB)-112 exp {2xyt - (x2 + y2)t2} ~ 2nn! 1- t 2

n=O

and deduce that

~ {H.,(x)}2t"' = (1 - t 2)-1' 2 exp {2x2t/(1 + t)}.

~ 2nnf t~=O

PROBLEMS

(5) Use the result of problem 4 to show that

(i) J:oo {H,.(x)}2 e-2""' dx = zn-!r(n + !);

.. 1 f 00 {H.,(x) }2 e-x• f 00 (1 - x2)n e-x• (u) - dx = -- dx. '2"n! -oo 1 +x2 -oo 1 +x2 1 +x2

(6) Prove that

(i) 2nlJ',._1(x) = xlJI,.(x) + lJI~(x); (ii) 2x1JI.,(x)- 2n1JI,._1(x) = lJ',.+l(x);

(iii) lJI~(x) = xlJI .. (x) - 1JI,.+1(x).

(7) Evaluate

(i) J:oo Pm(x)P.-.(x) dx;

(ii) J:oo lJ'm(x)lJ'~(x) dx.

16?

6

LAGUERRE POLYNOMIALS

6.1 LAGUERRE'S EQUATION AND ITS SOLUTION

Laguerre's equation is

d2y dy x dx 2 + (1 - x) dx + ny = 0 (6.1)

and in applications we usually require a solution which is finite for all finite values of x and which tends to infinity no faster than ex; 2 as x tends to infinity.

The methods of Chapter 1 apply, and if we look for a solution of the c:J)

form z(x, s) = L a, xs+r we obtain an indicia! equation with double root r=O

s = 0, and the recurrence relation

(s + r- n) ar+l = a,(s + r + 1)2'

The two independent solutions are then given by

z(x, 0) and [~] •=0

•

The second of these we know from Chapter 1 contains a term of the form In x, and so is infinite when x = 0. Since we require a solution finite for all finite x we can only obtain this from z(x, 0). In this case the recurrence relation takes the form

(6.2)


However, the infinite series obtained from this relation may be shown to behave like ex for large values of x, so that, from our remarks above, it does not behave well enough as x tends to infinity. The way round this difficulty is to make the series terminate, and from equation (6.2) we see that this 'happens if n is a positive integer. For, in this case, an ::F 0 but an+1 and all subsequent coefficients will vanish. In such a case the relation (6.2) is best written in the form

and we obtain the solution

(n- r) a,+1 = -a,~+ 1)2

_ { n n(n - 1) 2 y- a0 1 - i 2x + (2 !) 2 x + ...

( _

1), n(n - 1) ... (n - r + 1) 2 }

+ (r!)2 x + ... (the highest power of x being xn)

= ~ ( _ 1 )' n( n - 1) ... ( n - r + 1) r ao f;o (r!)2 x

N I """ n. = ao L ( -1Y ( - ·)I( l)2xr. n 1 • r. r=O

We define the standard solution as that for which a0 = 1 ; we shall call it ·the Laguerre polynomial of order n, and denote it by Ln(x):

n I """ n. Ln(x) = L ( -1)r ( - ) 1( l)2xr. r=O n r . r.

(6.3)


Theorem 6.1

exp {-xt/(1 -t)} = ~ L () n

(1 - t) L n X t. n=O

PROOF

We have

1 f ( 1 . t) cxp l

xtj(1 -- t)}' 1

t) ~ tl--~ ~ tr 1 y xrtr

' It

r! (1 1)' 11 •

170 LAGUERRE POLYNOMIALS CH.6

But 1

(1 - w+l _

1 (

1) (r+1)(r+2) 2 (r+1)(r+2)(r+3)t3

- + r + t + 21 t + 3!

+··· (by the binomial theorem)

Loo (r + s)! - t• - s=O rlsl '

so that we now have

1 ~ (r + s)l -- exp { -xt/(1 - t)} = L.. ( -1 Y --x•t•+•. 1 - t r,s=O (rl) 2sl

For a fixed value of r the coefficient of tn in this expansion is obtained by taking r + s = n, i.e., s = n - r. Thus, for this value of r, the coefficient of tn is given by

nl ( -l)' (r!) 2(n- r)lx•.

The total coefficient of tn is obtained by summing over all allowed values of r. Since s = n - r, and we requires> 0, we must haver.;;;;; n. Hence the total coefficient of tn is given by

n I

2: (-1Y ( l) 2t·- )lx• = Ln(x) ,_

0 r. n r .

(by equation (6.3)), which proves the required result.

6.3 ALTERNATIVE EXPRESSION FOR THE LAGUERRE POLYNOMIALS

Theorem 6.2

PROOF

Leibniz's theorem for the nth· derivative of a product states that

dn ~ nl dn-ru d'v dxn(uv) = 6<n - r)f;I dxn-~ dx''

§ 6.4 EXPLICIT EXPRESSIONS FOR LAGUERRE POLYNOMIALS 171

so that we have

e" dn e" ~ n! dn-r dr n! d.xn(xn e-X)= nl ~ (n- r)!r! dxn-Tx" dxr e-O:.

But dP d-xq = q(q- 1) ... (q- p + 1)xq_p

xP

q! xq_p (q- p)! ,


6.4 EXPLICIT EXPRESSIONS FOR, AND SPECIAL VALUES OF, THE LAGUERRE POLYNOMIALS

We have an explicit series for Ln(x) given by equation (6.3). For the first few Laguerre polynomials this gives

Theorem6.3

L 0(x) = 1 L 1(x) = -x + 1

1 L2(x) =

2 1(x2 - 4x + 2)

1 L3(x) = 3!( -x3 + 9x2 - 18x + 6)

1 L4(x) =

41(x4 - 16x3 + 72x 2

- 96x + 24).

(i) /,11(0) 1.

(ii) /,;,(0) 11.


PROOF

(i) Set x = 0 in the generating function of theorem 6.1 and we obtain co 1 L Ln(O)tn = 1 - t

n~o

(using the binomial theorem)

so that equating coefficients of tn on both sides gives Ln(O) = 1.

(ii) Ln(x) satisfies Laguerre's equation (6.1), so that we have

d 2 d X dx 2Ln(x) + (1 - x)dx Ln(x) + nLn(x) = 0.

Setting x = 0 in this equation and using part (i) above we obtain

L~(O) + n = 0 and hence L~(O) = -n.

6.5 ORTHOGONALITY PROPERTIES OF THE LAGUERRE POLYNOMIALS

Theorem 6.4

PROOF

From the generating function of theorem 6.1 we have

exp { -t~(~ - t)} = i Ln(x)tn n=O

and exp { ~x~~1-s)} = i Lm(x)sm, m=O

so that

Leo -xL ( )L ( ) n m = -x exp{ -xt/(1 - t)} exp { -xs/(1 - s)} e nX mXtS e .

1 .

1 .

-t -s n,m=O

Thus f~ e-xLn(x)Lm(x) dx is the coefficient of tnsm in the expansion of

I = f"' -x exp {-xt/(1- t)} exp {-xs/(1- s)}d e . 1 . 1 x. o -t -s

§ 6.6 LAGUERRE POLYNOMIALS AND THEIR DERIVATIVES 173

But I= 1 reo exp {-x(1 + _!_ + _s -)} dx

(1 - t)(1 - s) J o 1 - t 1 - s

1 [ 1 = (1- t)(1- s) -1 + {t/(1- t)} + {s/(1- s)}"

exp {-x(1 + _t - + _s )}]co 1-t 1-s o

1 1 (1 - t)(1 - s) 1 + {t/(1 - t)} + {s/(1 - s)}

1 (1 - t)(1 - s) + t(1 - s) + s(1 - t)

1 1 - st

co = 2 sntn.

n=O

Thus the coefficient of tnsm is 1 if n = m and 0 if n ;F m, i.e. it is bnm·

Hence J ~ e-•'Ln(x)Lm(x) dx = ~nm·

6.6 RELATIONS BETWEEN LAGUERRE POLYNOMIALS AND THEm DERIVATIVES: RECURRENCE RELATIONS

Theorem 6.5

(i) (n + 1)Ln+l(x) = (2n + 1 - x)L .. (x) - nLn_1(x). (ii), xL~(x) = nLn(x) - nLn_1(x).

n-1

(iii) L~(x) = - 2 Lr(x). r=O

PROOF

(i) If we differentiate the generating function with respect to t and use the fact that

d t 1 dt 1 - t = (1 - t)2

we obtain

~ n-I _ 1 x exp { -xt/(1 - t)} ~0L .. (x).nt -(i--=-ij 2 exp{-xtj(1-t)}- (1 -t) 2 ( 1 -t)

which, on further usc of theorem 5.1, becomes rtl } m A /,,.(x). til" I (I t) '~' /,,.(x)tn

co X '\""'

( I I 2 ~ /.,.(x)/". ) II I)


Multiplying throughout by (1 - t) 2 gives 00 00 00

(1 - t) 2 L L.,(x)nt"-1 = (1 - t) L L .. (x)t" - x L L .. (x)t" n=O n=O n=O

and hence ct) ct) 00 L L.,(x)nt"-1 - 2 L L .. (x)nt" + L L .. (x)ntn+I

n=O n=O n=O ct) ct) 00

= L L.,(x)t" - L L .. (x)t"+ 1 - x.L L.,(x)t".

n=O n=O n=O

Relabelling the summations so that the general power appears as t" in each gives

00 00 00 L L.,+1(x)(n + 1)t" - 2 L L .. (x)ntn + L L .. _1(x)(n - 1)tn n=-1 n=O n=l

00 00 00

= L L.,(x)t" - L L.,_1(x)t" - x L L.,(x)t" n=O n=l n=O

and then equating the coefficients oft" on both sides of the above equation gives

(n + 1)L.,+l(x) - 2nL.,(x) + (n - 1)L.,_1(x) = Ln(x) - L.,_1(x) - xL.,(x) (n> 1)

which reduces to

(n + 1)L.,+1(x) = (2n + 1 - x)L .. (x) - nL .. _1(x).

(ii) If we differentiate the generating function with respect to x we obtain

~ L~(x)t" = _ _ t_ exp { -xt/(1 - t)} ~ 1-t 1-t

fl=O

t 00

= - 1 - t L L .. (x)t". n=O

(6.4)

Thus 00 <X>

(1 - t) L L~(x)t" = -t L L.,.(x)t" n=O n=O

and hence 00 00 00 L L~(x)tn - L L~(x)tn+l = - L L.,.(x)tn+l

n=O n=O ~t=O

§ 6.6 LAGUERRE POLYNOMIALS AND THEIR DERIVATIVES 175

which, on relabelling, becomes 00 00 00

L L~(x)t" - L L~_ 1(x)t" = - L L,._1(x)t". n=O n=1 n=1

Equating coefficients of tn on both sides of the above equation gives

L~(x) - L~_1(x) = -Ln_1(x) (n > 1). (6.5)

If we now differentiate result (i) above with respect to x we obtain

(n + 1)L~+1(x) = (2n + 1 - x)L~(x) - L,.(x) - nL~_1(x) and if we use equation (6.5) in the rewritten forms

L~+l(x) = L~(x) - L,.(x)

and L~_1(x) = L~(x) + Ln_1(x), we shall have

(n + 1){L~(x) - L .. (x)} = (2n + 1 - x)L~(x) - L,.(x) - n{L~(x) + L,._1(x)}.

This simplifies to give

-nL,.(x) = -xL~(x) - nL,._1(x) and hence

xL~(x) = nL,.(x) - nL,._1(x). 00

(iii) If, in equation (6.4), we expand 1/(1 - t) in the form L tr we r=O

obtain 00 00 00

L L~(x)t" = -t L t• L L8(x)t• n=O r=O s=O

00

= - L L.(x)t•+•+I, r,s=O

so that L~(x) is given by the coefficient oft" on the right-hand side of the above equation. For a fixed value of s we obtain tn by taking r + s + 1 = n, i.e., r = n - s - 1, and then the coefficient oft" is -L,(x). We obtain the total coefficient of tn by summing over all allowed values of s, which, since we require r > 0, implies that n - s - 1 > 0, i.e., s < n - 1.

This means that we have n-1

coefficient oft" = L - L,(x), •=0

n-1

and hence L;,(x) = = · 2 L,(x). 1-U


6.7 ASSOCIATED LAGUERRE POLYNOMIALS

Laguerre's associated equation is

d 2y dy x- + (k + 1 - x)-- + ny = 0.

dx2 dx (6.6)

Theorem 6.6

If z is a solution of Laguerre's equation of order n + k then dkzjdxk satisfies Laguerre's associated equation.

PROOF

Since z is a solution of Laguerre's equation (6.1) of order n + k, we have

d 2z dz x dx 2 + (1 - x)dx + (n + k)z = 0.

Differentiating this equation k times and using Leibniz's rule for the kth derivative of a product gives

dk+2 dk+I dk+l dk x dxk+2z + k d,xk+lz + (1 - x)dxk+lz + k. -1. dxkz

dk + (n + k)dxkz = 0.

Thus d 2 dkz d dkz dkz

x dx2 dxk + (k + 1 - x)dx dxk + n dxk = 0

which just states that dkz jdxk satisfies equation (6.6).

From the above theorem and the fact that the Laguerre polynomials Ln( x) satisfy Laguerre's equation, it follows that ( dk / dxk)L .. +k( x) satisfies Laguerre's associated equation. We define the associated Laguerre polynomials by this solution together with the constant factor ( -1 )k:

L~(x) = ( -1)7< d~k Ln+k(x). (6.7)

Theorem 6.7

k * (n + k)! L .. (x) = ~ (-1Y ( _ )l(k -L )I 1xr.

r~O n r , 1 r ,r o

PROOF

From equation (6.3) we have

n+k (n + k)! Ln+k(x) = ~ ( -1)' (n + k- r)!(r!)2xr

§ 6.8 PROPERTIES OF ASSOCIATED LAGUERRE POLYNOMIALS 177

so that, by equation (6.7),

k k dk n+k (n + k)! Ln(x) = ( -l) dxk ~ ( -1)' (n -+-k - r) !(r !)~x'

kdk n+k (n+k)! = ( -1) (i~i 6 ( -1Y (n + k- r)!(r!)iix'

(since ( dk / dxk) operating on powers of x less than k gives zero)

n+k (n +k)! r! = ( -

1)k ~k ( - 1Y (n + k - r)!(rf)2 (r - k)(-''

(since (dkjdx~')xr = r(r- 1) ... (r - k + 1)xr-k

r! ) = (r- k)!xr-k

= ( -1)k ~ ( -l)k+s (n + k !:!-: k)~)!(kf- s)!s!x•

(changing the variable of summation to s = r - k)

" (n + k)! - ~ ( -1) 8 x• - f:o (n - s)!(k + s)!s! '


6.8 PROPERTIES OF THE ASSOCIATED LAGUERRE POLYNOMIALS

Theorem 6.8 (Generating function)

exp { -xtj(l - t)} = ~ Lk( ) n

(1 - t)k+l L.,. n X t ' n=O

PROOF

From theorem 6.1

exp -xtj(l{- t)} = ~ L () n

( 1 - t) L.,. n X t • n~o

Differentiating both sides k times with respect to x, we have

dk cxp { xt /(1 - t)} ~, dk ~ L ( ) n dxk ( 1 t) dxk -:;; n x t

(since l.,.(x) is a polynomial of degree nand so, if n - 1<, will give zero wlwn difrl'n·ntiatl'd I< timL·s).


Carrying out the differentiation on the left-hand side, we obtain

(- _t_)k exp { -xt/(1 - t)} = ~ Loo L ( ) +k

1 1 ...l •• k n+k X t" -t -t uw-· n=O

and thus, using equation (6.7), p 00

( -1)k (1

_ t)k+l exp { -xt/(1 - t)} = 6, ( -1)kL~(x)tn+k which, on cancelling a common factor of ( -1)ktk, becomes

exp { -xt/(1 - t)} = ~ Lk( ) ,. (1 - t)k+l 6 n X t •

Theorem 6.9

PROOF

Since this is almost the same as that of theorem 6.2, we shall not give it here.

Theorem 6.10 (Orthogonality property)

Joo _, k( )Lk d (n + k)! e XCLn X m(x) X = I c5nn••

o n.

PRooF Again this proof is very similar to a previous one (that of theorem 6.4)

and will be omitted.


(i) L~_1(x) + L~- 1(x) = L~(x).

(ii) (n + 1)L~+l(x) = (2n + k + 1 - x)L~(x) - (n + k)L~_1(x). (iii) xL~'(x) = nL!(x) - (n + k)L~_1(x).

n-1

(iv) L~'(x) = - L L~(x). r=O

(v) L~'(x) = -L~~l(x). n

(vi) L~+ 1(x) = L L~(x). r=O

6.8 PROPERTIES OF ASSOCIATED LAGUERRE POLYNOMIALS 179

WOF

(i) Using theorem 6.7 we have

~-1(x) + L~- 1 (x) n-1 n

2 (n- 1 + k)! 2 (n + k - 1)! - ( -1Y x' + ( 1)' x' - r~o (n- 1 - r)!(k + r)!r! r=O - (n- r)!(k- 1 + r)!r!

n-1 (n + k- 1)! n-

1 (n + k - 1)! - ""' ( -1)' x' + ""' ( 1)' x' -;So (n- r- 1)!(k + r)!r! :S, - (n- r)!(k + r- 1)lr!

(n + k- 1)! +(-1)" x11

(n- n)!(k- 1 + n)!n!

n-1 (n + k - 1)! { 1 1 }

=~(- 1)'(n-r-1)!(k+r-l)!r! (k+r)+(n-r) x'

1 + ( -1)" -.x" n!

_ ~1

-l , (n + k - 1)! (n - r) + (k + r)x, - ;S ( ) (n- r - 1)!(k + r- l)!rl (k + r)(n- r)

x" + (-1)"n!

"~ (n + k)! x" = ;S

0 ( -l)'(n- r)!(k + r)!r( + ( - 1)n n!

- ~ -1)' (n + k)! x" - ;S ( (n- r)!(k + r)!r! = L~(x) -

(again using theorem 6.7).

(ii) Differentiate k times the result of theorem 6.5(i) with n replaced by ~ + k, and we obtain

dk dk n + 1~ + 1) dxkL" +1c+1(x) = (2n + 2k + 1) dxkLn+1,(x)

dk dk · (i:~k{xL,. 11c(x)} -- (n l k)d~L,. ,k_1(x)

md thus, using I ,eihniz's theorem for the /lth derivative of a product, s 11 N

180 LAGUERRE POLYNOMIALS

dk d" (n + k + 1)dxkLn+I+Jc(x) = (2n + 2k + l)dxkLn+lc(x)

dk dk-1 - X dxkLn+k(x) - kdx"_ 1Ln+k(x)

dk - (n + k)dxkLn-l+Jc(x).

Making use of definition (6.7) we now have

(n +k + 1)(-l)kL~+1 (x)

CH.6

= (2n + 2k + 1)( -1)kL~(x) - x( -1)kL~(x) - k( -1 )k-J L~:;:l(x)

- (n + k)( -1)k£~_1(x) which, when we take account of part (i) above with n replaced by n + 1, g1ves

(n + k + 1)L~+l(x) = (2n + 2k + l)L~(x) - xL~(x)

+ k{L~+l(x) - L~(x)} - (n + k)L~_1(x) and this simplifies to

(n + 1)L~+l(x) = (2n + k + 1 - x)L~(x) - (n + k)L~_1(x).

(iii) If we differentiate k times with respect to x the result of theorem 6.5(ii) with n replaced by n + k, we obtain

~ ~ ~ dxk{xL~-k(x)} = (n + k)dxkLn+Jc(x)- (n + k)dx;;Ln tk-1(x)

which, on use of Leibniz's theorem for the kth derivative of a product, becomes

dk , dk dk x dxkLn +k(x) + k dxkLn tk(x) = (n + k) dxkLn+k(x)

dk - (n + k) dxkLn+k-1(x).

Then, by equation (6.7), we have

xL~'(x) + kL~(x) = (n + k)L~(x) - (n + k)L~_1(x) and thus

xL~'(x) = nL~(x) - (n + k)L~_1(x).

(iv) If we differentiate k times with respect to x the result of theorem 6.5(iii) with n replaced by n + k, we obtain

dk , "+k-1 d" dxkLn+k(x) = - 2 d:;;/,Lr(x)

r~o

§ 6.8 PROPERTIES OF ASSOCIATED LAGUERRE POLYNOMIALS 181

so that

Therefore

(since for r < k Lr(x) is a polynomial of degree less than k and so when differentiated k times just gives zero)

n-1 dk = - L dxkL.+Jx)

8=0

(changing the variable of summation to s = r - k) n-1

= - L ( -1)''L:(x). 8=0 n-1

L!' (x) = - L L;(x). r-o

(v) By theorem 6.7 we have

Lk'(x) - ~ ~ (-1)r (n + k)! x' n - dx :S (n - r)!(k + r)!r!

- ~ ( -1Y (n + k)! xr-1 -:6 (n- r)!(k + r)!(r- 1)!

(the term with r = 0 vanishing on differentiation)

n-1 ( k)l - ""(-1)5+1 n + . x• - ~ (n- s- 1)!(k + s + 1)!sl

(changing the variable of summation to s = r - 1)

- - n~ -1 s (n - 1 + k + 1)! x• - 6( )(n-1-s)!(k+1+s)!s!

= -L~:!:l(x)

(from theorem 6.7).

(vi) Comparing results (iv) and (v) above, we have n-1

-L L~(x) = -L!:!:l(x) r·~O

which, when n is replaced by n + 1, gives

"· L! 1' 1(x) - • L l.~(x).

r-0

182 LAGUERRE POLYNOMIALS CH. 6

6.9 NOTATION

It is important to note that certain authors use different definitions for the Laguerre and associated Laguerre polynomials from those given here.

Sometimes the Laguerre polynomials are defined such that the generating function has the form

exp { -xt/(1 -~-t2} = ~ !E .. (:x:)!!'_ 1- t L... n!

fl~O

so that !E .. (x) thus defined is equal to n!Ln(x).t The associated Laguerre polynomials may be defined by

dk !E~(x) = dxk!E .. (x)

giving the relationship !E~(x) = ( -1)kL~-k(x).

6.10 EXAMPLES

Example 1 .. Prove that L~+.B+ 1(x + y) = L L~(x)L~_r(y).

r=O We have, from theorem 6.8,

~ Lrxn+.B+l)x + y) .tn = exp { -(x + y)t/(1 - t)} L,.. (1 _ t)rx+fl+2 n~o

so that L~+fl+I(x + y) is the coefficient of tn in the power series expansion of

exp { -(x + y)t/(1 - t)} (1 - t)rx+fl+2

But

exp { -(x + y)t/(1 - t)} (1 - t)rx+fl+2

exp { -xtj(1 - t)} exp { -ytj(1 - t)} (1 - t)a+I (1 - t).B+l

ct) ct)

= L L~(x)t' L L~(y)t• r~o s~o

(again using theorem 6.8) 00 L L~(x)L~(y)t•+-.

r, s=O

t We use !l' to denote alternative definitions, in order to distinguish them from ours, but it must be remembered that authors adopting these alternative definitions will use L.


We obtain t" by taking r + s = n; so that for a fixed value of r we have s = n - r,andhence,for this value ofr, the coefficientoftn isL~(x)L~_,(y). The total coefficient of tn is obtained by summing over all allowed values of r, which, since s = n -rand we requires> 0, implies that r < n.

n

Hence the coefficient of tn is L L~(x)L~_,(y) r=O

n

so that L~+iJ+l(x + y) = L L~(x)L~_.(y). r=O

Example 2 oo Lm( )

Show that]m{2y(xt)} = e-t(xt)m/2""' n X 1tn

f:o (n + m).

(where] m is the Bessel function of integral order m ). We shall prove the equivalent result that

oo Lm( ) e1(xt)-m12]m{2y(xt)} = """ n X r· f:o (n + m).

From equation (4.6) we have

~ 1 {2v(xt)}2r+m

lm{2v(xt)} = ;S ( -1Y rl(m + r)l -2-

so that 00 1

et(xt)-m12j m{2 v(xt)} = e1(xt)-m1 2 """ ( -1 y ' (xt)r+(m/ 2) ;S r.(m + r)l 00 1

= e1 """ ( -1Y (xtY ;S rl(m + r)l

ct) t• 00 1 = ~ :;1 ~ ( -1Y rl(m + r)lx't'

00 1 - """ ( -1 Y xrtr+• - ,{:_

0 rl(m + r)lsl ·

We wish to show that the coefficient of t" in this expansion is

D;:(x)/(n + m) I. To obtain the coefficient of tn we take r + s = n so that for a fixed

value of r we haves "= n · · r, and for this value of r the coefficient is

1) 1 'rl(m 1 r)l(n ·r)l'~r.


The total coefficient is obtained by summing over all allowed values of r, which, since s = n - r and we require s > 0, is given by n - r > 0, i.e., r < n.

Hence the coefficient of tn is

n 1 "" ( - 1 Y - c--------,-:-,----------:-:-,xr f:o r!(m + r)!(n- r)!

1 * (n + m)! = (n + m)! r~ ( -

1)' ;!(m + r)(n- r)!xr

= ___ 1_Lm(x) (n + m)! n

(using theorem 6.7), and the required result follows.

Example 3

Show that I: e-tL~(t) dt = e-x{L~(x) - L;,_1(x) }.

Integrating by parts, we have

I: e-1L~(t) dt = [ -e-1L!(t)J: -- J: -e-t.L~'(t) dt

= e-"'L~(x) + I: e-tL~' (t) dt

=e-'"L~(x)- I: e-t{~1

L:(t)}dt r~o

(by theorem 6.11(iv)).

Thus

J oo e-tL~(t) dt + ~ I oo e-1L:(t) dt = e-'"L~(x) X r=O X

and hence

(6.8)

PROBLEMS

Therefore

foo e-1L!(t) dt = ~ Joo e-tL:(t) dt - 2: Joo e-1L:(t) dt X r-0 X r=O X

PROBLEMS

= e-"'Lk(x)- e-"'Lk (x) n n-1

(by equation (6.8) above)

= e-"'{L~(x) - L~_1(x)}.

(1) Show that L~(O) = !n(n - 1).

185

(2) If f(x) is a polynomial of degree m, show that f(x) may be expressed in the form

m

f(x} = 2 CrLr(x} r=O

with

D d h f oo _, kL ( ) d {0 if k < n e uce t at

0 e x ,. x x = ( _ 1 )"n! if k = n.

(3) Prove that

fx (x- t)mL,.(t) dt =- --- m!n! ___ xm+tLnt+l(x). o (m + n + 1)! n

( 4) Show that

1• 2 2"k!(n+k)!J' 2 , L,.(x) = (- l)n-n(Zk)j(Zn) 1- _

1 (1 - t )k-•Han{hlx)t} dt.

(5) Prove that

(6) Show that

r .,, C "X~ I I ( /,:;(.~) 12 d.\: L

(n I k)!(2 ' 1) 1

11 I ~~ I . 11.


(7) If L!(x) is defined for non-integral k by the generalisation of the result of theorem 6. 7

show that

and

Lk(x) - * (-1)'- r(n + k + 1) xr n -6 (n-r)!r(k+r+1)r!

1 V,/2(x) = ( -1)• 22n+I 1 H2n+Ih/x) n.x

1 L;;lf2(x) = ( -1)n 22nn!Han( y'x).

(8) With L!(x) defined for non-integral k as in problem 7, show that

Lk+l( ) r(n + k + l + 1) fl k 1 )l-1£k( ) d n X = r(l)r(n + k + 1) 0 t ( - t n Xi t.

7

CHEBYSHEV POL YNOMIALSt

7.1 DEFINITION OF CHEBYSHEV POLYNOMIALS; CHEBYSHEV'S EQUATION

We define the Chebyshev polynomials of first kind, T .. (x), and second kind, U .. (x), by

T .. (x) =cos (n cos- 1 x) and U .. (x) =sin (n cos-1 x),! for n a non-negative integer.

Theorem 7.1

(i) T .. (x) = ![{x + iy'(l - x2)}n + {x - iy'(l - x2)}"]. (ii) u .. (x) = -!i[{x + iy'(l - x 2)}n - {x - iy'(l - x2)}n].

PROOF

(i) Let us write x = cos() and we obtain .T .. (x) =cos (n cos- 1 cos 0)

=cos n() =HeinO + e-ine} = -H( eiO)n + ( e-'e)"} =!{(cos() + i sin O)n +(cos() - i sin O)n} = ![{x + iy'(l - x2)}n + {x - iy'(l - x2)}n].

(ii) The proof is similar to that of (i), and so will not be given.

(7.1) (7.2)

t The transliterations Tchcbichcf, Tchebichcff and Tschebyschcff arc also found.

t Sometimes the Chebyshev polynomial of the second kind ill <ldim·d by '¥1,(,\:) sin{(n il)cos-'x}/v'(l -x2) {1/v'(l -·xa)l/l!,"(x).

188 CHEBYSHEV POLYNOMIALS CH. 7

Theorem 7.2 [!n] I

(i) Tn(x) = "" ( -1 )" - 1l_:_ .. (1 - x2)"xn-2r. ;S (Zr) !(n - 2r)!

[J(n-1)] I

2 n. ' (ii) Un(x) = ( -1)" --·· -(1 - x2)'+'xn-2r-·! (2r + 1)!(n- 2r- 1)! ·

PROOF

(i) From theorem 7.1(i) we have

Tn(x) = l[{x + iy(l - x2)}" + {x -- iv(l - x2)}n]

1 [ n " = .2 2 "C,.xn-r{iv(l - x2)}" + 2 "C,.xn-r{ -- iy(l

r~o r~·

(by the binomial theorem)

1 n = 2 2 nc,.xn-r(l - x2)'12 ir{l + ( -1)'}.

r=O

Now, when r is odd ( -1)' = -1, so that 1 + ( -1)' = 0, and when r is even ( -1 )' = 1, so that 1 + ( -1 Y = 2. Hence we have

Tn(x) = ~ 2 "C,.xn-r(1 _ x2)r12 ir 2. r even, <n

But if r is even we may write r = 2s with s integral, and the requirement r < n means that s < n/2. This, since s is an integer, is equivalent to s < [n/2] with [n/2] as defined before, namely the greatest integer less than or equal to n j2. Thus,

rtnl

Tn(x) = 2 "C2sxn-2s(1 - x2)s i2s s~o

f!nl I

= ~ (-;;-- ;;) !(Zi)lxn-2s(l _ x2)s( -1 )s.

(ii) The proof is similar to (i) above.

We may use theorem 7.2 to write down the first few Chebyshev polynomials:

T0(x) = 1, T 1(x) = x,

U0(x) = 0; U,(x) = v(1 -- x 2

);

§ 7.1 DEFINITION OF CHEBYSHEV POLYNOMIALS

T2(x) = 2x 2 - 1,

T 3(x) = 4x3 - 3x, U2(x) = v(1 - x2)2x; Ua(x) = v(1 - x 2)(4x2 -- 1); U4(x) = v(1 - x2)(8x3

- 4x);

189

T 4(x) = 8x4 - 8x2 + 1, T 5(x) = 16x5 - 20x3 + Sx, U5(x) = y(l - x2)(16x4 - 12x2 + 1).

We note that Un(x) is not actually a polynomial, but is instead a polynomial multiplied by the factor v(1 - x2), whereas <F~ .. (x) as defined above is a polynomial of degree n.

Since T .. (cos 0) =cos nO and U .. (cos 0) =sin nO, we note that the Chebyshev polynomials provide expansions of cos nO and sin ne /sin e in terms of powers of cos e.

Theorem 7.3

T .. (x) and U .. (x) are independent solutions of Chebyshev's equation

d2y dy (1 - x2)dxz- xdx + n2y = 0.

PRooF

We give the proof for T .. (x); the proof for U .. (x) is similar. We have, from definition (7.1),

Also, , d2 dxzT .. (x)

d d dxT .. (x) = dx cos (n cos-1 x)

= -sin (n cos-1 x). n. v(1-~ x2)

(remembering that (d/dx) cos-1 x = -1/y(1 - x2))

= n v(1 ~ x2) sin (n cos-1 x).

= ~{--n ___ -sin (n cos- 1 x)} dx v(1 - x 2

) .

nx . ( 1 ) n -n (1-x2)312sm ncos- x -!-(i~xz)i;2cos(ncos-tx).(1-x2)112

nz · - COS (11 ('OS I ,\').

(I :~:2)


Hence

(1 _ 2)d2T,.(x) _ dT,.(x) 2T ( )

X dx2 X dx + n " X

nx · ( -1 ) 2 ( -1 ) (1

_ x 2) 112 Sill n cos x - n cos n cos x

nx · ( -1 ) 2 ( -1 ) - (1

_ x 2) 112 Sill n cos x + n cos n cos x

=0, which proves the required result.

The fact that U,.(x) and T,.(x) are independent solutions follows from observing that T,.(1) = 1 while U,.(1) = 0, so that U.,.(x) cannot be a constant multiple of T,.(x).


Theorem 7.4

1- t 2 ~ (i)

1 2 2 = T 0(x) + 2 ~ T,.(x)t".

- tx + t n=l

(l't') v(1 - x2) = ~ u ( ) ~ n+l X t".

1 - 2tx + t 2 n=O

PROOF

Again we prove only result (i), the proof of (ii) being similar. Let us write x = cos() = t(e18 + e-18

), so that we have 1 - t 2 1 - t 2

1 - 2tx + t 2 - 1 - (ew + e-i6)t + ti. 1 - t 2

= (1 - ewt)(1 - e-wt) 00 00

= (1 - t2) L (ewtY 2: (e-iot)• r=O •=O

(by the binomial theorem) 00

= (1 - t2) L e;<r-s)Otr+•

r, B:o::::::O 00 00

= 2: ei(r- )Otr+s _ 2: ei(r-a)6tr+•+2.

r,s=O r,s=O


We wish to pick out the coefficient of t" in this summation and show that it is T0(x) when n = 0 and 2T.,(x) otherwise.

We consider the cases n = 0 and n = 1 separately, since for these values of n we obtain tn from the first summation only, whereas for n > 2 we obtain t" from both summations.

n = 0 is obtained only by taking r = 0 and s = 0 in the first summation, so that the coefficient of t 0 is

ei(O-o)O = 1 = To(x).

n = 1 is obtained by taking either r = 1 and s = 0 or r = 0 and s = 1, so that the coefficient of t 1 is

ew + e-10 = 2 cos() = 2T1(x)

(remembering that T,.(x) = T,.(cos 0) =cos nO). For n > 2 we obtain the coefficient of t" by taking r + s = n (i.e.,

s = n - r) in the first summation and r + s + 2 = n (i.e., s = n - r - 2) in the second summation. The coefficient of t" is therefore

n n-2 .L ei{r-(n-r)}8 _ .L ei{r-(n-r-2)}8

r=O r=O n n-2

= e-lnO L ei2r0 -- e-i(n-2)0 L ei2r0 r=O r=O

1 _ (ei28)n t 1 . 1 _ (ei20)n-t -~ -~-~ = e 1 - e2iB - e 1 - ~

(summing to n + 1 and n - 1 terms, respectively, the two geometric series which both have the common ratio e2w)

e-inO _ Ci(n+2)8 e-l(n-2)0 _ einO

1 - ezw 1 - ez;o

e-lno(l _ e21o) einO(l _ e2w) - ~---~--~-- 1 - ez;o , 1 - e2io

= einO + e~in8 = 2 cos nO '~ 2T,.(x),

(on rearranging the terms)

(again remembering that T,.(:'() · T,.(cos 0) "'=cos nO)

192 CHEBYSHEV POLYNOMIALS

Theorem 7.5 (Special values of the Chebyshev polynomials)

(i) T,.(1) = 1, T,.( -1) = ( -1)", T2,.(0) = ( -1)", T2n+t(O) = 0.

(ii) U .. (1) = 0,

PROOF

U .. (-1) = 0, U2 .. (0) = 0, U2n+l(O) = ( -1 )".

Again we prove results (i) only. Setting x = 1 in definition (7 .1) gives

T,.(1) =cos (n cos-1 1) =cos n.O =cos 0 = 1.

Setting x = -1 gives

T,.( -1) =cos (n cos-1 -1) =cos nn = ( -1)".

Setting x = 0 gives

n T,.(O) = cos (n cos-1 0) = cos n2

{0 if n is odd

= (-1)"' 2 ifniseven.

Thus all four results are proved.

7.3 ORTHOGONALITY PROPERTIES

Theorem 7.6

(i) fl T m(x)T,.(x) dx = {~/2 -t y(l - x2) n

(ii) ft Um(x) U .. (x) dx = {~/2 -1 v(1 - x2

) 0

PROOF

m ::;;z= n m =n ::;;r:O m = n = 0.

m :;;z'n m =n ::;;r:O m = n =0.

We prove result (i) only, the proof of (ii) being similar. If we set x = cos 0 we have

§7.4 RECURRENCE RELATIONS 193

fl !_r:'(~)~{j:~ dx = fo _!m(~s O)T,.(cos~l( -sin 0 dO) - 1 y'(l -- x 2) " sin 0

= f: cos nO cosmO dO (7.3)

(by definition (7 .1))

= J: !{cos (n + m)O +cos (n - m)O} dO

= ![-- 1- sin (n + m)O - _ _!___ sin (n :____ m)e]"

2 n+m n-m o

(provided n - m 7"' 0) =0.

If n - m = 0 we have from equation (7.3)

f1 T,.(x)Tm(x) _ J" 2 _

1 v/(1 _ x2) dx -

0 cos nO dO

= J: i-( 1 + cos 2n0) dO

= - 0 - - - sin 2n0 1 [ 1 ]" 2 2n o

(provided n :7: 0)

=-§-n.

If n = m = 0 we have from equation (7.4)

f1

T 0(x)T0(x) dx = J" l dO -1 y'(l - x 2) o

=::T, and thus the proof is complete.

7.4 RECURRENCE RELATIONS

Theorem 7.7

(i) T,. H(x) -- 2xT,.(x) + T,_ 1(x) = 0. (ii) (1 -- x2)1';,(x) = -nxT,.(x) -1- nTn_ 1(x).

(iii) l!,. ,l(x) 2xU .. (x) + U,. __ t(.t') = 0. (iv) (I x~)U;,(x) nxU .. (x) I nU" 1(x).

I' HOOF

:\gain \\T pro\'idc proofs for the polynornials of lirst kind only.

(7.4)


(i) By writing x = cos 0, the required result is

cos (n + 1)0 - 2 cos 0 cos nO +cos (n - 1)0 = 0.

However,

cos (n + 1)0 - 2 cos 0 cos nO +cos (n - 1)0 = cos nO cos 0 - sin nO sin 0 - 2 cos 0 cos nO

+ cos nO cos 0 + sin nO sin 0 =0,

the result which was to be proved.

(ii) By writing x = cos 0, the required result is

d (1 - cos2 0) d( cosO) cos nO = -n cos 0 cos nO + n cos (n - 1 )0.

But

(1 - cos 2 0) d( c:s O) cos nO = sin 2 0 (- si! 0 :0 cos nO)

= -sin 0( -n sin nO)

= n sin 0 sin nO

and -n cos 0 cos nO + n cos (n - 1)0

= -n cos 0 cos 0 + n( cos nO cos 0 + sin nO sin 0) = n sin nO sin 0,

so that the required result is proved.

7.5 EXAMPLES

Example 1

Show that v(1 - x 2)T .. (x) = u .. +l(x) - xU .. (x).

If we replace x by cos 0 and use the consequences of definitions (7.1) and (7 .2) that

and

T,.(cos 0) =cos nO

U,.(cos 0) =sin nO,

we find that the result we require to prove is just

sin 0 cos nO = sin (n + 1 )0 - cos 0 sin nO.

But sin (n + 1)0 -cos 0 sin nO

which proves the result.

= sin nO cos 0 + cos nO sin 0 - cos 0 sin nO

= cos nO sin 0,

§ 7.5 EXAMPLES

Example 2

n 1{ 1 } Show that ~ T 2r(x) = 2 1 + y(l _ x2) U2n +I(x) .

Again we write x =cos 0, so that we have

n

= L T2r( COS 0) r=O

" = L cos 2r()

r=O

" = Re L ei2rB

r=O 1 - ei(2n+2)0

= Re 1 i26 -- e

(using the result for the sum to n + 1 terms of a geometric progression)

(1 _ ei(2nt2)0)(1 _ e-2w)

= Re (1 - ei2o)(1 - e- 2W)

_ R {1 - cos (2n + 2)0 - i sin (2n + 2)0}{1 - cos 20 + i sin 20} - e 1 + 1 _ ei2e _ e i20

195

= 2 ~-!-cos 20 [(1 - cos 20){1 - cos (2n + 2)0} +sin 20 sin (2n + 2)0]

___ 21{1 sin 2() . }

COS (2n + 2)0 -j- 2 si~i O Sin (2n + 2)(J

= -2~{1 + sin (2n + 2)0 cos e - cos (2n + 2)0 sin 0} sin 0

=· _2

1_ [ 1 I sin_{~(2n _-_!_-_ ~)0_-=--_~J sin 0

~~ II

sin (2n I 1 )0} sin()

{12,. '.(.\') }·

\(I ·'·2)

196 CHEBYSHEV POLYNOMIALS

PROBLEMS

(1) Show that v'(1 - x 2)U,.(x) = xT,.(x) - T,.+l(x).

(2) Show that Tm+n(x) + Tm_,(x) = 2Tm(x)T.,(x).

h ' n (3) Prove t at T .. (x) = v'(1

_ x 2) U,.(x).

(4) Show that 2{T,.(x)}2 = 1 + T2,.(x). (5) Show that {T,.(x)}2 - T,.+I(x)T,_1(x) = 1 - x 2•

(6) Show that Tm{T.,(x)} = T,.{Tm(x)} = Tmn(x).

CH.7

(7) Show that {1 / v'(1 - x 2)}U,.(x) satisfies the differential equation

d2y dy (1 - x2)dx2 - 3x dx + (n 2

- l)y = 0.

(8) Use Chebyshev's differential equation and the equation in problem (7) above to show that

[!n] n L (n- r- 1)' T,.(x) = - ( -1 )' · (2x)n-2r 2 r=o r!(n- 2r)!

and

rHnL-1)1 (n- r- 1)1 U,.(x) = y'(1 - x2) ( -1Y ------· (2x)"-2r- 1•

r=o r!(n - 2r- 1)!

8

GEGENBAUER AND JACOBI POLYNOMIALS

8.1 GEGENBAUER POLYNOMIALS

It is possible to define new sets of polynomials by generalizing some of the results already proved for the Legendre, Hermite, Laguerre or Chebyshev polynomials. We give here only two particularly useful sets; those obtained by generalizing in two different ways the generating function of the Legendre polynomials, given in theorem 3.1. We shall define the Gegenbauer polynomialt of degree nand order A, C~(x), as the coefficient of tn in the expansion of

1 (1 - 2xt + t 2).\'

(Note that the Legendre polynomial P,.(x) is in fact equal to C~(x).)

Thus

(8.1)

It may be shown that such a power series expansion is valid for I t I < 1, I x I < 1 and A > ·· !·

We shall omit proofs of the following properties. All the results may he obtained by methods similar to those used in preceding chapters.

·f· Also sonwtinws ,-alll'd thl' ultrasplwrknl polyrwrnial.

198 GEGENBAUER AND JACOBI POLYNOMIALS CH. 8

Theorem 8.1 (Power series expansion)

.t _ f!nl _ r r(n - r + A.) n-2r

c .. (x)- ~ ( 1) r(A.)r!(n- 2r)! (2x) .


fl (1 - 2)A-l ;.( )C;. ) d - 21-2). r(n + 2..1.) -l X C,. X m(X X- ~\n + A.){r(A.)}2r(n + 1/nm·


(i) (n + 2)C~+2(x) = 2(..1. + n + 1)xC~+1(:~) - (2..1. + n)C!(x). (ii) nC~(x) = ZA.{xC!~l(x) - C!~~(x)}.

(iii) (n + ZA.)C~(x) = 2A.{C!+1(x) - xC~~~(x)}.

(iv) nC~(x) = (n - 1 + 2A.)xC~_ 1(x) - 2..1.(1 - x2)C~=~(x).

(v) C~' (x) = ZA.C~:\:l(x).

Theorem 8.4 (Differential equation)

C~(x) satisfies the differential equation

d 2y d (1 - x 2

) dx2 - (ZA. + 1 )x Jx + n(n + ZA.)y = 0.

8.2 JACOBI POLYNOMIALS

We may generalize the Legendre polynomial generating function even further. We define the Jacobi polynomial p~a,fi>(x) as the coefficient of tn in the expansion of

2a+fJ

i.e., zat{J

00

= L p~a,fl>(x)t". (8.2) n~o

We note that the Legendre polynomial P,.(x) is in fact equal to p~O,Ol(x).

Again the following properties may be proved by methods similar to those used in previous chapters.

§ 8.2 JACOBI POLYNOMIALS

Theorem 8.5 (Series expansions)

(i) P~«,fJ>(x)

199

~ f(n-Hx+1)r(n+fJ+1) (x-1)r(x+1)"--r = :S r(oc+r+1)f(n+fJ-r+1)(n-r)!r! 2 2 ·

(ii) p<rz,f1>(x) = ~ f(n+cx+l)r(n+r+oc+fJ+l) (x-l)r. .. £:6 f(oc-!-r+l)f(n+cx+fJ+1)(n-r)!r! 2

(iii) p<rz.f1>(x) = ~ ( -l)n-rf(n+fJ+l)f(n+r+IX+fJ+1)(x+l)r. n 6 f(fJ+r+l)f(n+cx+fJ+l)(n-r)!r! 2


J~1 (1 - x)'"(l + x'f~a,f1>(x)P~f1>(x) dx

zaif:l+lr(n-\- IX-\- l)f(n + {J-\- 1) = ~nm•

(2n +a: +fJ + l)n!f(n +ex +fJ + 1)


(i) 2n(a: +fJ +n)(oc -1-{J +2n- 2)P~"'·f1>(x) = (1X+{J +2n -1){oc2 -{J 2 +x(a:+fJ +2n)(a:+fJ +2n-2)}P~':_~l(x)

- 2(oc +n --1)({J +n-1 )(cx+fJ +2n)P~«~PJ(x). (ii) p~a,f:ll'(x) = !(l+oc+fJ+n)P};x_+1

1•fJ+ll(x).

(iii) (x + 1 )~a.fJ>' (x) = nJ>!;z-P>(x) +({J +n)P~':~i1 • f1>(x). (iv) (x-l)P!~·fJ>'(x) = nP:·f1>(x)-(a:+n)P~~+ 1>(x).

(v) p~a,fJ>'(x) = i{({J+n)p<;_\1·f1>(x)+(oc+n)P~':.1+ 1>(x)}.

(vi) (cx+fJ+2n)P~«,fJ-ll(x) = (a:-1-{J+n)P~"'·Il>(x)+(oc+n)P~"!{(x). (vii) (ot+fJ+2n)P~"- 1 ·f1>(x) = (a:+fJ+n)Pi~·fJ>(x)-(fJ+n)P};:_'i_>(x).

Theorem 8.8 (Differential equation)

P~~.fl>(x) satisfies the differential equation

d~ dy (1 - x2)~2 -1 {{J IX (oc I {J i 2)x}d~ i· n(n lot I· fJ I 1)y = 0.

As well as lwing related to the Legendre functions (recall that P,(x) P;,""l(.\') (':, -~(x)), the ( ;egcnhauer ami Jacobi polynomials arc related to cal'h othl·r and to till' ( 'hchyshcv polynomials. In fact, the

200 GEGENBAUER AND JACOBI POLYNOMIALS CH.8

Gegenbauer polynomials are just a special case of the Jacobi polynomials and the Chebyshev are just a special case of the Gegenbauer:

C}.( ) = r(A. + !)r(n + 2A.)p(A-:,A-!l( )· n X r(2A.)r(n + A + !) n X '

T,.(x) = -2n lim C!~(x) (n > 1);

.1.-->-0 II

Un(x) = v(l - x2)C~_1(x).

(8.3)

(8.4)

(8.5)

These results will be proved in the next chapter (Example 5, p. 215). There also exist relationships with the Laguerre and Hermite poly

nomials, which we shall state but shall not prove:

L~(x) = lim P~"'·fl>(l - 2xj(3); (8.6) fl---+00

Hn(x) = n! lim A. -nt 2C~(xj vA.). A->-oo

8.3 EXAMPLES

Example 1

Show that _dm CA(x) = 2"' r(A. + "!_)c).~-m(x) dxm n r(A.) n-m .

From theorem 8.3(v) we have

so that

! C~(x) = UC~:l:l(x),

t2C~(x) = 2A.:xC~:!=l(x) = 21..2(1. + l)C~:J:~(x) = 22A.(A. + 1 )C~:=~(x),

and, if we repeat this process m times in all, we obviously obtain

dm dx"'C~(x) = 2mA.(A. + l)(A. + 2) ... (A. + m - l)C:.~:(x)

=2m IJ~ + m)ci.+m( ) r(A) n-m X·

(8.7)

Example 2

Show that

PROBLEMS

P~"'·P>(x) = <;~t(l - x)-"'(1 + x)-P !:{(1 -x)"'+"(l + xf+n}.

Leibniz's theorem for the nth derivative of a product gives

dn dxn{(l - x)"'+n(1 + x)P+"}

" n I { d• } { dn-r } - · - l+xP+" ~- 1-x"'+" - 6 r!(n-r)! dx'( ) dxn-r( )

" I = L- I( n:_ )I(,B+n)(,B+n-1) ... (,B+n-r+1)(1+xf+"-'

r=O r. n r

X ( -t)n-•(rx+n)(rx+n-1) ... (rx+n-n+r+1)(1-x)"'+n-n+r

201

= ~ -~-( -l)n-r r(,B+n+l}r(rx+n+1) (1+x)f1+n-•(1-x)"+r, ;b, r!(n-r)! r(,B+n-r+l)r(ot+r+l)

Hence we have

- ( -1)" ~ ( -t)n-r nl - r(rx+n+l)r(,B+n+l) (1-x)•(l +x)n-r - znnt 6 rl(n-r)! r(rx+r+1)r(,B+n-r+l)

~ r(rx+n+l)r(,B+n+1) (x-1)'(x+1)n-r = 6 r(rx+r+1)r(,B+n-r+l)rl(n-1·)! -2- 2

(by theorem 8.5(i)).

PROBLEMS

(1) Show that P:."·{J)( -x) = ( -t)nP!·"'>(x).

r(rx + n + 1) (2) Show that P,~·{J)(l) - ~ -~~--- -

l'(rx ·1- l)n! (3) Show that

d"' /':~·'''(x)

d.\''" 2 ,, l'(m i n I rx I fl I 1)/* 1 ,,fl 1 "''(x-)

l'(,!rxi{l·l) "'"' ··

202 GEGENBAUER AND JACOBI POLYNOMIALS

(4) Use the method of induction to show that

* (r + A)C;.(x) = jn±_~A)C~x) -:: 0 + 12C~+I(:) ~ r 2(1 - x) · r=O

(5) Show that p~.P-t>(x) - p<:-l,fl>(x) = P~~>(x).

( ) h IH _ 1 ~ p ( ) 6 S ow that C,._1(x)- (21 _ 1)!! dx1 ,. x.

9

HYPERGEOMETRIC FUNCTIONS

9.1 DEFINITION OF HYPERGEOMETRIC FUNCTIONS

Let us define (oc)r (the so-called Pochhammer symbol) by

(oc), = oc(<X + 1) ... (<X + r - 1)

r(<X + r) ( . . . ) = r(<X) ; r a pos1t1vc mteger

(1X)o = 1.

Then we define the general hypergeometric function

by

The notation

is also often used.

(9.1)

We shall show that many of the special functions encountered up to now (and indeed many of the elementary functions) may be expressed in terms of hypcrgeometric funrtions. \\'l' shall ron fine Ollrsl'lvl's to thl' two sl·parate cases: 111 11 I (in which cast· the function will lw cal lt-d tlu· confiiH·nt

204 HYPERGEOMETRIC FUNCTIONS CH.9

hypergeometric function or Kummer function) and m = 2, n = 1 (in which case we shall merely call it the hypergeometric function).t

Convergence of the series (9.1) needs to be considered. The following results may be proved using the standard techniques of convergence theory:

(i) The confluent hypergeometric series is convergent for all values of x. (ii) The hypergeometric series is convergent if I x I < 1 and divergent if

I x I > 1. For x = 1 the series converges if {J > oc1 + oc2, while for x = -1 it converges if {J > oc1 + oc2 - 1.

The following theorem shows the intimate relationships which exist between the hypergeometric functions and the special functions already considered.

Theorem 9.1

(i) P,.(x) = 2F 1( -n, n + 1; 1; 1

2 x).

.. m (n+m)! (1-x2)mi 2 ( • • 1 - x)

(u) P,. (x) = (n -m)! 2mm! 2F1 m-n, m+n+1, m+1, - 2- i

(iii) I .. <x> = e:~G)"'lFl(n + !; 2n + 1; 2ix).

(iv) H 2n(x) = ( -1 )" (2~)! 1F1( -n; ! ; x 2). n.

_ n 2(2n + 1)! . 3

• 2 (v) H 2 .. +l(x)- ( -1} 1

x 1F1( -n, -2 , x ). n.

(vi) L,.(x} = 1F 1( -n; 1; x).

.. k r(n + k + 1) ( . k . (vu) L,.(x) = n !r(k + 1) 1F 1 -n, + 1, x).

(viii) T,.(x) = 2F1( -n, n; ~; 1 ~ x). (ix) U,.(x) = v(1 -- x2} n 2Fl( -n + 1, n + 1; ~; 1 ~X).

" r(n + ZA.) ( • 1 • 1 - x) (x) C,.(x) = n!r(2A.) 2F 1 -n, n + ZA., A.+ 2 , - 2 - .

t The confluent hypergeometric function 1F 1(cx; {J; x) is often denoted by M( cx, {J, x), and the hypergeometric function 2F 1( ex., oc 2 ; fJ; x) is often denoted by F(cxh ot2, {J, x).

§ 9.1 DEFINITION OF HYPERGEOMETRIC FUNCTIONS 205

( ') (tx,Pl _ r(n+oc+1) F (- . . 1-x)· x1 P,. (x) - n!r(oc+ 1) 2 1 n, n+oc+fJ+1, oc+1, 2

PROOF

In each case the result may be proved by expanding the hypergeometric function as a series, using definition (9.1), and comparing with a known series for the given function. We shall illustrate this method by proving result (i) only.

We have, from definition (9.1),

F (-n n + 1. 1. 1 - x) = ~ ~-_:-n)r(n _j-_1).{(1-.:- x2f2}' 2 I ' ' ' 2 Lt (1) I . r=O r T.

We may take n to be a non-negative integer, since it is for only these values of n that the Legendre polynomials are defined.

Then we have ( -n)r = ( -n)( -n + 1)( -n + 2) ... ( -n + r- 1)

= (-l)'n(n -l)(n- 2) ... (n- r + 1) = ( -l)'n!/(n- r)! if r < n.

If r > n + 1, then ( -n), = 0, for it will contain a zero factor. Also

and

(n + 1), = (n + 1)(n + 2) ... (n + r) = (n + r)!/n!

(1), = 1.2.3 ... r =r!

so that we now have

( . . 1-x)-~ r n! (n+r)!1(1-x)' 2F1 -n, n + 1, 1, - 2-- - L ( -1) ( __ )' 1 ,-2--;--;---r=o n r . n. r. r.

~ ( -1)' (n + r)! = 6 ----z;:- (n - r) !(r !)2 (1 - x)'

= ~ (n + r)! (x - 1 r £:o 2•(n - r) !(r !)2 ) · (9.2)

To show that this is the same as P,.(x), it is easiest to write P,.(x) as a power series in (x 1). We do not as yet have such a power series, but we usc Taylor's theorem to write

J>,.(x) 1)"

where hy p!~l( I) we nwan tht· rth lkrivativl' of /',(.') n.llu.ltnl .11 .\ I.


To calculate P,.<rl(1) we use the generating function for the Legendre polynomials given in theorem 3.1:

1 00

y'(1 - 2tx + t2) = 6 Pn(x)tn,

so that, by differentiating r times with respect to x, we have oo dr L P,.<r>(x)t"' = dxr (1 - 2tx + t2)-l/2

n~o

= ( -2t)'( -!)( -! - 1)( -i - 2) ... ( -! - r + 1)(1 - 2tx + t 2)-!-r = 2rtri(i + 1)(! + 2) ... (t + r- 1)(1 - 2tx + t 2)-~-r = trl.3 .5 ... (2r - 1)(1 - 2tx + t 2)-!-r

(2r)! ) , = tr -(1 - 2tx + t 2 -,-r. 27r!

Hence, setting x = 1, we have

~ tr(2r)1 L P,.<r>(1)t" =

2rr!. (1 - 2t + t 2)-~-r

n=O

writing r + s = n.

tr(2r) I = __ · (1 _ t)-1-2r 2rr!

_ t7(2r)!{1 (1 2 ) (2r + 1)(2r + 2) 2 - 2rr! + + r t + 2! t

(2r + 1)(2r + 2)(2r + 3) 3 } + 3! t + ... (by the binomial theorem)

= t7(2r)! ~ (2r + s)! t• 27r! ~ (2r)!s!

= _1_ ~ (2r + s) ! tr+• zrr! L s!

s~o

__ 1_ ~ (r + n) I tn - 27r! ;S (n - r)! '

Equating coefficients of tn gives

p<r>(1) = -1- (r + ~)! for n > r n 2rr! (n - r)!

= 0 for n < r.

§ 9.2 PROPERTIES OF THE HYPERGEOMETRIC FUNCTION

Hence, from equation (9.3), we have

P .. (x) = ~ _1_ (n + r)! (x- 1Y ~ 2rr! (n- r)! r!

- ~ (n + r)! (x - 1)• - 6 2r(n- r)!(r!) 2

( 1- X)

= 2F 1 -n, n + 1; 1; -2-


9.2 PROPERTIES OF THE HYPERGEOMETRIC FUNCTION

Theorem 9.2

PROOF

This follows immediately from definition (9 .1):

F ( fJ· • ) _ ~ (rJ.)r(fJ)r xr 2 I rJ., ' y' x - f::o (y ), r!

Theorem 9.3

The differential equation

d2y dy x(1 - x)dx

2 + {y - (rJ. + fJ + 1)x} dx- rJ.{Jy = 0

207

(the hypergeometric equation or Gauss's equation) has 2F 1(rJ., {J; y; x) as a solution. If y is not an integer a second independent solution is given by x1 -Y2F 1(rJ. + 1 - y; fJ + 1 - y; 2 - y; x).

PROOF '

These results may be proved by using the series solution method of Chapter 1. It is found that the roots of the indicia! equation are 0 and 1 -- y, so that they lead to independent solutions provided y is nonintegral. It is found that these independent series solutions are just the hypcrgeometric functions given in the theorem.

Theorem 9.4 (Integral representation)

2FJ{r1., (I;)'; x)

i/ i' - fl · 0 ·

l'(r) I '(;I) I'( i' f I til I( I

fl) () .\'f) ~ dt

208 HYPERGEOMETRIC FUNCTIONS

PROOF

From definition (9.1) we have

,.FJ(r~., (3; y; x)

= i (l1.),((3),:: r=O (y), r!

= ~ r{ll. + r)r((3 + r)r(y) ~ f:-6 r( (1. )r((3)r(y + r) r!

r(y) ~ r(y - (3)r((3 + r) X'

= r{ll.)r((3)r(y - (3) ~ r{ll. +- r) r(y + r) ;!

r(y) ~ X' = r{ll.)r((3)r(y _ (3) ~ r (ll. + r)B(y - (3, (3 +- r) i!

(by theorem 2.7)

CH.9

r( ) 00 J1 r = r 2: rc(J. + r) c1 - w-fl- 1tfi+r- 1 dt:. r{ll.)r{(3)r(y- (3) r=O o I r!

(by definition (2.2) of the beta function, which is valid if y - (3 > 0 and (3 + r > 0)

rcr) J 1 2:00 rc(J. + r> (xt)' = ----~-- (1 - t)Y-fl- 1t {J- 1 - dt r((3)r(y - (3) 0 r=O r(ll.) r!

= r(y) J1 (1 - t)Y-fl- 1 t~'~- 1 (1 - xt)-a. dt

r((3)r(r - (3) o

(using the binomial theorem).

Theorem 9.5

Each of the following twenty-four functions is a solution of the hypergeometric equation:

V1 = 2F1(ll., (3; y; x)

V2 = (1 - xy-a.-fl2Ft(Y- ll., y - (3; y; x)

Va = (1 - x)-a.2pl( ll., Y- (3; y; x ~ 1) v4 = (1 - x)-fl2Ft(Y - ll., (3; y; X X 1)

V5 = 2F1(lJ., (3; l1. + (3 -I- 1 -- y; 1 - x)

V6 =x1-Y2F1(ll. I 1 y;(J il y;ll. I (3 1-1 y; 1 -x)

§ 9.2 PROPERTIES OF THE HYPERGEOMETRIC FUNCTION 209

v7 = x-oc2pl( IX, IX+ 1 - y; IX+ p + 1 - y; 1 - D V8 = x P~1(P + 1- y, {3; IX+ p + 1- y; 1-D V9 = (-x)-oc~1(1X, IX+ 1- y; ot + 1- {3; D vlO = ( -x)fl-y(l - xy-oc-fl2Fl(1 - {3, y - {3; IX + 1 - {3; D V11 = (1- x)-oc2F1(1X, y- {3; ot + 1- {3;

1 ~ J

V12 = ( -x)I-Y(1 - x)y-oc-l~l( IX + 1 - y, 1 - {Jj IX + 1 - {J; l~ X)

vl3 = ( -x)-fi2F{fl + 1 - y, {3; f3 + 1 --- IXj D vl4 = ( -x)oc-y(1 - xy-oc-fJ2Fl(1 - oc, y - ot; f3 + 1 - oc; D V15 = (1 - x)-P2F1({3, y - IXi {3 + 1 - IXi

1 ~ J

Vla = ( -x)1-Y(1 - x)Y-fl-l2Fl({J + 1 - y, 1 - 1Xj {J + 1 - IXj 1

1 -J v17 = xl-y~l{IX + 1- y,{J + 1- y; 2- y; x)

V18 = x1-Y(1 - xy-oc-fl~1(1 -IX, 1 - {3; 2- y; x)

V19 = x1-Y(l - xy-oc-l2F 1 ( IX + 1 - y, 1 - {J; 2 - y; X: 1)

'

V20 = X1-y(1 - x)y-fl-l~l({J + 1 - y, 1 - otj 2 - Yi X : 1)

v21 = (1 - x)y-a.-P~t(Y -IX, y- {3; y + 1 -IX- {3; 1 - x)

v22 = x1 -Y(1 - x)y-oc-f:l2Fl(1 - oc, 1 - {3; y + 1 -IX - {3; 1 - x)

v23 ~' ~-y(l -- x)Y-a.-fi2F1(r-- IX, 1 - IXj y -1- t --IX--- {J; 1 - D v2~ .\JI

1'(1 x)l' "' fi2FI(?' fl, 1 fl; I' i 1 oc fl; 1 D·


PROOF

Each function may be verified to be a solution by means of change of variable, direct substitution and use of theorem 9.3.

Since the hypergeometric equation is of second order, there must be only two independent solutions, so that a linear relation must exist between any three of the twenty-four functions given above. In fact, it may be shown that

and

V1 = V2 = Va = V4, V5 = V 6 = V7 = V8 ,

V9 = V10 = V11 = V12,

v13 = v14 = v15 = v16• v17 = V1s = v19 = v2o. v21 = V22 = v2a = v24·

For the remaining relations the reader is referred to ERDELYI et al., Higher Transcendental Functions, Vol. I, pp. 106-8.

The six hypergeometric functions 2F1( IX± 1, f3; y; x), 2F1( IX, f3 ± 1; y; x), 2F1(1X, {3; y ± 1; x) are said to be contiguous to 2F1(1X, {3; y; x). It may be shown that between 2F1(1X, {3; y; x) and any two functions which are contiguous to it, there exists a linear relationship whose coefficients are linear functions of x. Since we can choose 2 from 6 in 6C2 = 15 ways, there will be 15 such relationships altogether. For details the reader is referred to ERDELYI et al., Vol. I, pp. 103--4.

9.3 PROPERTIES OF THE CONFLUENT HYPERGEOMETRIC FUNCTION

Theorem 9.6

The confluent hypergeometric function 1F 1(1X; {3; x) ts a solution of the equation

d 2y dy x2 dx2 + ({3 - x) dx - IXY = 0

(the confluent hypergeometric equation or Kummer's equation). If {J is not an integer a second independent solution is given by

x1-fl1F 1(1X - f3 + 1; 2 - {3; x).

PROOF

As for theorem 9.3 above.

§ 9.3 CONFLUENT HYPERGEOMETRIC FUNCTION

Theorem 9.7 (Integral representation)

F ( . {J• ) - r{{J) fl (1 t)P-a-lta-1 ...:>ot dt 1 1 IX, 'X - r(1X)r{{J -ex) 0 - ~··

for {J >IX> 0.

PROOF

Similar to theorem 9.4 above.

Theorem 9.8

Solutions of the equation

d2~ + { -~ + ~ +!- m2}y = 0 dx2 4 x x 2

are given by y = xl-rn e-x12z,

where z is a solution of the confluent hypergeometric equation with

IX = 1 - k - m and {J = 1 ·- 2m.

PROOF

211

(9.4)

The result follows immediately on direct substitution and use of theorem 9.6.

COROLLARY

The independent solutions of equation (9.4) are giYcn by

Mk,m(x) = xl+m e-"'1\F1(!- k + m; 1 +2m; x) and

PROOF

Since the independent solutions of the confluent hypergeometric equation are

1F 1(1Xi {3; x) and

x1 f\J.\(1X --· {J + 1; 2 · {J; x),

we have, from the theorem, that two independent choices for :;; arc

1F,O ll · · m; 1 2m; x) and

I~ · m; I , 2m; .r), ·,1 I'


so that the corresponding solutions are xl-m e-x/ 2 F (.l- k - m· 1 -2m· x)

1 1 2 ' ' and

x}+m e___.,121F1(!- k + m; 1 +2m; x).

Mk,m(x) and M~c,-m(x) are known as Whittaker's confluent hypergeometric functions.

Theorem 9.9

Each of the following four functions is a solution of the confluent hyper-geometric equation:

PROOF

V1 = 1F1(rx; {J; x) V2 = x1

- 1\F1(1 + rx - {J; 2 - {J; x)

V3 = e"' 1F1({J- rx; {J; -x) V4 = x1 -fi e"' 1F1(1 - rx; 2 - {J; - x).

By change of variable, direct substitution and use of theorem 9.6.

The four confluent hypergeometric functions 1F 1(rx ± 1; {J; x), 1F1(rx; {J ± 1; x) are said to be contiguous to 1F 1(rx; {J; x). It may be shown that between 1F 1(rx; {J; x) and any two functions contiguous to it there exists a linear relationship whose coefficients are linear functions of x. Since we can choose 2 from 4 in 4C2 = 6 ways, there will be six such contiguous relationships altogether. For details the reader is referred to ERDELYI et al., Vol. I, p. 254 (see Bibliography).

9.4 EXAMPLES

Example 1 rcr )r(y - rx - fJ)

Show that 2F 1( rx, {J; y; 1) = r(y _ rx)r(y _ {J)'


F ( {J . . 1) - r(y) f1 tfi- 1(1 t)y-fi- 1(1 t)-"' dt

2 1 rx, , Y , - r({J)r(y _ {J) 0

- -

= r(y) fl tfl-1(1 - ty-oc-fl-1 dt r({J)r(y - fJ) o

r(y) r({J)r(y _ {J) B({J, y-rx-{3)

(by definition (2.2) of the beta function)

§9.4

Example 2

Show that

EXAMPLES

r(y) r(p)r(r - rx - P) r(p)r(y - P) r(y - rx)

(by theorem 2. 7)

r(y)r(y - !X - /1) r(y - rx)r(y - p)'

2F1(rx, p; y; x) = (1 - xt"'2F 1 ( rx, Y - f1; y; x ~ 1)

213

(i.e., V1 = V3, using the notation of theorem 9.5).

If we set -c = 1 - t in theorem 9.4 we have

2F1(rx, j1; y; x)

= r(y) .. - f[ (1--c)P- 1-cY-fl-l{l- x(1--r)t"' dr r(j1)r(y-p) 0

= r(y) f (1--c)fl-J'fy-fl-!(1-x)-OC(l -_:__.)-IX dr r(p)r(y--/1) 0 x-1

= r(y) (l-xt"' fl Ty-fl-l(l-r)fl-!(1 - __:_T)-oc dr r(y--p)r{y-(y-/1)} 0 x-1

= (1-x)-"'2F1 ( rx, y-p; y; x~1 )• on further use of theorem 9.4.

Example 3

Prove the relation of contiguity

y{y- 1 - (2y- 1 - rx- j1)x} 2F1(rx, p; y; x) ' + (y - rx)(y - j1)x 2F1(rx, j1; y + 1; x)

- y(y - 1)(1 - x) 2F1(rx, p; y - 1; x) = 0. W c prove this result by expanding the hypergeometric functions in

power series. Since

F ( R. • ) _ ~ (!X )r(tJ)r r 2 1 rx, I'' y' x -- L _(_)_I x '

r-· o Y ,r · we sec that the coefficient of x" on the left-hand side above must be

J'( J' I ) ( oc ),.(fl),. ( J'),.ll!

r(2J' I oc fl) ( oc),. 1(/l).. I I ()•),. 1(n I)!

214 I-IYPERGEOMETRIC FUNCTIONS

-j- (y-rl)(y-/'1) {rl}n-1~,8)n-1 _ y(y- 1) (rl)n{,B)~ (y+1),._1(n-1)! (y--l),n!

1 ( 1) {rl)n-1(/'1)n-1 -,yy-(y-1)n-1(n-1)!

= ( _ 1)r(rl+n)r(/'1+n)r(y) Y Y r(rl)r(p)r(y+n)n!

r(rl +n -1 )r(/'1 +n -1 )r(y) - Y(2Y - 1-rl-/'1)r(rl)r(,B)r(y +n-l)(n --1)!

)( 11)r(rl +n -1 )r(,B +n -1 )r(y + 1)

+ (y-rl y-,, r(oc)r(,B)r(y-1-n)(n-1)!

r(oc+n)r{t3 +n)r(y --1) - y(y-l) r(rl)r{t3)r(y+n-l)n!

( l)r(rl+n-1)r(,B +n-1 )r(y-1)

+ y y- r(rl)r{t3)r(y+n-2)(n-1)!

CH.9

= r(oc+n -1 )r(,B -1-n -1)r(y -1 ){y(y -1}(rl +n --1)(,8 -t-n -1 )~ r(rl)r(,B)r(y+n -2)(n -1)! (y -1-n --l)(y +n -2)n

_ y(2y-1-rl-/'1)(y-1) + (y -cx)(y -,B)y(y-1) (y+n-2) (y+n-1)(y+n-2)

y(y-1)(oc+n-1)(fJ+n-1) 1

}

- (y-l-n-2)n 1 y(y-1)

r(oc+n-1)r(t3+n-l)r(y-1) 1 ( 1) r(oc)r(,B)r(y-t-n-2)(n-1)! n(y+n-1)(y-f-n--2) y y--

.{y(oc+n-1)(t3-t-n -1) - n(y-l-n-1)(2y-1-rl-{J)

+ n(y-oc)(y-,8)- (y-t-n--l)(oc-1-n-1)(,8-t-n-1)

+ n(y+n-l)(y--tn-2)} ' r(oc +n-l)r(/J +n -1 )r(y -l)y(y-1)

. -

r(oc)r(,B)r(y+n-2)(n-1) !n(y +n -l)(y +n-2)

.[y2( -2n+n+n) + y{(rl+n-1)(,8-1-n-1) + n(cz +f]-!-1)

- 2(n-1)n - nrl- n,B- (oc+n-1)(!'1-l-n--1)

+ n(n-2) + n(n-1)} + n(n-l)(oc-1-/'1+1) + noc/'1 - (n-1)(oc+n-1)(/'1+n--1) + n(n-l)(n-2)]

=0. Thus the relation is verified.

§9.4 EXAMPLES

Example 4

Show that 1F1(rx; {3; x) = e"' 1F1({3- oc; {3; -x).


F ( . (3• ) r({3) J1 (1 t)fl-a.-lta.-1 xt dt 1 1 rx, 'X = r(rx)r({J _:-:_~ 0 - e .

If we make the change of variable t = 1 - 'l', we have

1Fr(r:x; {3; x) = f'~)~~)=~ J: 7/l-a.-!(1 - T)"'- 1 ex(l-•) dr

- :t r(f1) Jl (1 - )<X-I /l-<X--1 -XT d - e r({J -- rx)r(rx) 0 'l' T e T

= e"' 1F1({J - ct; oc; -x) (using theorem 9.7 again).

Example 5

Show that

( ') CA( ) = r(A ±_l)I'(n + 2A) P. 1.-J.A-l( )· I n X r(2A)I'(n + A + t) " X '

(") T ( ) = ~ lim C~(x). 11 n X 2 1.--+-0 A '

(iii) U,.(x) = v(1 - x 2) C!_1(x).

(i) From theorem 9.1(xi) we have p~-~· .<-l(x)

215

__ I'(n+A-J+1) (- ___ 1 _ __ . _ 1 , . 1-x) - n!I'(A-!+1) 2F1 n, n-+ A 2 +A ! 1--1, A 211, 2

_ r(n+A+i) ( . 1 • 1-x) 0'-' n!I'(A+i)- 2F 1 -n, n+2A, A+2 , - 2-

-- I'(n_±~t~~ ~!E(2A) C\x) - n!r(A-H) r(n+2A) n

= ~]23)~(~_-l-A ~1-!) C'( ) r(A-1 Dr(n I 2A) "x.

(ii) From theorem 9. l(x) we have

(by theorem 9.1 ( x))

l'(n: 2).) ·( ,, I 2

.\:)· (';.(x) IJ'?l ~/· 1 fl,fl · 2).;}. · "' " . ( _,,)


The trouble at A = 0 comes from r(2A). Hence we have

~ lim C~(x) = ~ lim --~- . lim r(n -t-2A) F ( -n n 2A. A 1· !_ -x) 2 J.-+o A 2 ;._,.o Ar(2A) .<-+0 n! 2 1 ' + ' +:!, 2

_nlim 1 .r(n) F( .1 .1-x) -2 J.->O Ar(2A) nT s 1 -n, n, 2• -.z-= !~ 2A:(2Aj. 2F 1( -n, n; !;

1 ~x) -lim __ l_ F (-n n· 1· 1-x) - <->O r(2A + 1 )" 2 1 , ' 2 , 2

( 1-x) = 1. 2F 1 -n, n; ·!; --2--

= T,.(x) (by theorem 9.1 (viii)).

(iii) From theorem 9.1(x) we have

I r(n - 1 + 2) ( . I. 1 -X) C,._1(x) = (n _ l)!r(2)2F1 -n + 1, n - 1 + 2, 1 + 2 , - 2-

r(n -t- 1) ( . 3 • 1 -- X) = n!r(2) n 2F1 -n + 1,n + 1, 2 ,-2-

( 1--x) = n 2F1 -n + 1; n + 1; ~; -2-

PROBLEMS

(1) Show that

1 v(1 - x2) Un(x)

(i) (1 - xt"' = 2F1(rt., {3; {3; x);

(ii) In (1 - x) = - x 2F1(1, 1; 2; x);

(iii) e"' = 1F 1(oc; rt.; x).

(by theorem 9.1(ix)).

(2) Show that 1F1(rt.; y; x) = lim 2F1(oc, {3; y; x/{3). /3->00

(3) Show that

r(l + {3 - oc)r(l + -~{3) 2Fl(oc, {3; {3 - oc + 1; -1) = r(l + p)r(l + -~(i - rt.)

PROBLEMS 217

and use Example 2 of Section 9.4 to deduce that

_ . . 1 _ r(tr)r(tr + i) . 2FI(r:t., 1 oc, y, 2)- r(loc + tr)r(t- irx + tr)

(4) Show that

dd ~1(oc, {J; y; x) = oc{J ~1(oc + 1, {J + 1; y + 1; x) X y

and deduce that

dn ( ) (rx)n(fJ)n F ( ) dx" 2F1 oc, {J; y; x = ~ 2 1 oc + n, {J + n; y + n; x.

(5) Prove the contiguity relationships

(i) (oc - fJhF1(rx, {J; y; x) = oc 2F1(oc + 1, {J; y; x) - {J 2F1(oc, {J + 1; y; x);

(ii) (oc - {J)x1F1(oc; {J + 1; x) + {J(x + {J- 1)1F1(oc; {J; x) - {J({J - l}tF1(ot; {J - 1; x) = 0.

(6) Use the confluent hypergeometric function to show that H2.-.(x) = ( -1)n22nn!L;;-i(x2)

and H 2.-.+I(x) = ( -1) .. 22"+1n!xL!(x2).

(7) Evaluate the integral

10

OTHER SPECIAL FUNCTIONS

In this chapter we discuss briefly other special functions which the reader is likely to encounter. Several of these are defined by integrals which are impossible to express in terms of known functions, and of these integrals several have no special properties of interest-all the useful information is contained in a table of values of the function.

10.1 INCOMPLETE GAMMA FUNCTIONS

These are defined by

r(x, oc) = f: e-tt•:-I dt

and y(x, 01:) = r e-ttx-l dt. • 0

From definition (2.1) of the gamma function we see that

r(x, 01:) + y(x, 01:) = r(x).

10.2 EXPONENTIAL INTEGRALAND RELATED FUNCTIONS

The exponential integrals are defined by

and

fx et

Ei(x) = - dt -co t

fcc e-t

E1(x) = -- dt. X f

(10.1)

(10.2)

(10.3)

(10.4)

(10.5)

§ 10.2 EXPONENTIAL INTEGRAL AND RELATED FUNCTIONS 219

We note that E1(x) = - Ei( -x).

(Some authors usc the notation ei(x) for E1(x).) The logarithmic integral is defined by

f .. dt li(x) = o In i'

It is easy to see that li(x) = Ei(ln x) = --E1( --In x).

y

y•Ei(x)

FIG. 10.1 Exponential integrals

The sine and cosine integrals are defined, respectively, by

'( ) f"' sin t d Sl X = - t 00 t

S'( ) f"' sin t d IX= - - t 0 t

and C'( ) f .. cost I I X - < t; '" t

(again thl' rl'ader is warned to hewarc of dill"l'ITilt notations).

(10.6)

(10.7)

(10.8)

(10.9)

(10.10)

(10.11)

220

y

-1·0


li(x)

FIG. 10.2 The logarithmic integral

y=Si(x)

FIG. 10.3 Sine and cosine integrals

CH. 10

§ 10.3 THE ERROR FUNCTION AND RELATED FUNCTIONS 221

The following results may readily be verified:

si{O) = - ~. 2

Si(x) = ~ + si(x),

(see Example 1, p. 225)

si(x) = ~i {Ei(ix) - Ei( -ix)},

Ci(x) = !{Ei(ix) + Ei( -ix)},

Ei(±ix) = Ci(x) ± i si(x).

The graphs of the above functions are shown in Figs. 10.1-10.3.

10.3 THE ERROR FUNCTION AND RELATED FUNCTIONS

The error function is defined by

erf x = ~n J: e-t' dt.

By the corollary to theorem 2.6 we see that

erf oo = 1.

As a generalisation of the error function we define the functions

E,.(x) = r{(n ~ 1)/n} J: e-tn dt.

(10.12)

(10.13)

(10.14)

(10.15)

(10.16)

(10.17)

(10.18)

(10.19)

(N.B. E1(x) given by this equation is not the same as the E1(x) of the previous section.)

We see'that erf x = E2(x). (10.20)

The Fresnel integrals are defined by

S(x) = J: sin ~t 2 dt (10.21)

and Jz n C(x) =

0 cos 2t 2 dt. ( 10.22)

It may be shown that (I 11.2.1)

y

0·8

0·6

y


~-------Y= erf x

FIG. 10.4 The error function

FIG. 10.5 Fresnel integrals

CH. 10

X

X

§ 10.4 RIEMANN'S ZETA FUNCTION 223

and that the Fresnel integrals are related to the error function by the relationship

C(x) + iS(x) = 1 ; i erf { ~'\1 - i)x} (10.24)

The graph of the error function is shown in Fig. 10.4 and the Fresnel integrals in Fig. 10.5.

10.4 RIEMANN'S ZETA FUNCTION

Riemann's zeta function is defined by 00 1

'(x) = 2 -n;; n=l

This series is divergent for x < 1, convergent for x > 1. It may be shown that

y

(10.25)

(10.26)

Ftr.. 10.6 lliemann'~ zeta function. (The graph is of C(x) -- 1. Note tlw logarithmic scalt• on the vt•rtical axis. For comparison, till' dotted lirw j,, till' graph of 2 '.)

224 OTHER SPECIAL FUNCTIONS CH. 10

and that in general '(2n), with n integral, may be expressed in closed form.

The graph of '(x) - 1 is givenin Fig. 10.6. Note the logarithmic scale; for comparison, the graph of z--o: is given by the dotted line.

10.5 DEBYE FUNCTIONS

These are defined by

Jx tn

Dn(x) = t-l dt. oe -(10.27)

It may be shown that these functions are related to the Riemann zeta function by D,.( oo) = n!,(n + 1).

10.6 ELLIPTIC INTEGRALS

We define the elliptic integrals of the first, second and third kinds respectively by

Jq, de

F(k, cp) = o v'(1 - k2 sin2 e) (0 < k < 1), (10.28)

E(k, cp) = J: y'(1 - k2 sin2 e) de (0 < k < 1), (10.29)

. Jq, de Il(k, cp, a) = 0 v'(1 - k2 sin2 e)(l + a2 sin2 e) (10.30)

(0 < k < 1, a ~ k).

Some of the importance of elliptic integrals lies in the following theorem, which we state without proof.

Theorem 10.1 ,

If R(x, y) is a rational function of x andy and if P(x) is a polynomial of degree at most four, with real coefficients, then JR[x, v'{P(x)}] dx can be expressed in terms of elliptic integrals.

If the upper limit in each of definitions (10.28), (10.29) and (10.30) is n /2, then we have the definitions of the so-called complete elliptic integrals:

Jn/2 d(J

K(k) = o v'(1 - k2 sin2 e)' (10.31)

E(k) = J:12

y'(1 - k 2 sin2 0) dO, (10.32)

Jn/2 de

Ili(k a) - (10.33) ' - o v'(l - k2 sin 2 0)(1 + a 2 sin 2 0).

§ 10.7 EXAMPLES

10.7 EXAMPLES

Example 1

Show that si(O) = -n/2.

From definition (10.9) we have

Jo sin t si(O) = - dt

00 t

=- roosint dt. Jo t

To evaluate this integral we consider the integral

l(rx) = roo e-~ sin t dt J 0 t

so that si(O) is given by -1(0).

Now,

di roo . drx = - J

0 e-cxt sm t dt

Joo 1

= - e-"'t--;(eit - e-it) dt 0 21

= -~. roo {e<-cx+i)t- e<-cx-i)t} dt 21 J 0

=-!__[ 1 e<-cx+l)t_ 1 (-cx-i)t]00

2i -rx+i -rx-ie o

= -1C 1 i - oc ~ )

1 = -1 + oc2•

Hence, by integrating with respect to rx we obtain

l(rx) = - tan-1 rx +constant.

But I( oo) = 0,

so that we must have 0 = - tan-1 oo + constant,

and hence 'Jl

0 = - - -I- constant 2 '

giving constant 2

226

Thus

so that

and hence

Example 2


7l l(r:t.) = - tan-1 r:t. + 2

7l 7l 1(0) = - tan - 1 0 + 2 = 2

si(O) = - 1(0) = - ~. 2

Show that J: e-st si(t) dt 1

- - tan- 1 s. s

Integrating by parts, we have

foo [ 1 l 00 foo 1 d e-st si(t) dt = -- e-st si(t) - -- e-st- si(t) dt o s _.o o s dt

= ! si(O) + ~ J 00

e-st i_ si(t) dt s s 0 dt

= ~(-~) + ~ roo e-st sin t dt s 2 sJo t

CH. 10

(using equation (10.12) and definition (10.9))

7l 1 = -2s + -/(s)

Example 3

(with I as defined in Example 1 above)

= _:: + ~(- tan-1 s + ~) 2s s 2

(as shown in Example 1 above)

tan-1 s

s

Show that (i) erf ( -x) = - erf x;

(ii) I erf x I < 1.

(i) From definition (10.17) we have

f 2 f"' t• d er x = --- e- t yn o

§ 10.7

so that

EXAMPLES

erf (-X) = ~ r -X e-l' dt = _3__ f"' e-ll' • -dU yn J o yn Jo

(on changing the variable to u = - t)

2 f"' u' d =-yn Joe- u

= -erfx.

(ii) First suppose x > 0. Then, since e-t' > 0, we have erf x > 0. Also

erfx < ~n( J: e- 1' dt + J: e-1

' dt)

2 f"" I'd = yn J o e- t

= erf oo

=1 (by equation (10.18)).

A similar argument holds if x < 0.

Example 4

Show tl:at J:12

v(2 ~xcos x) = ~3 {F(J~· ~) - F(J~· ~) }-We have

fnu/2 dx fn/2 _-_d=-y __ -c

v(2 - cos x) = " v{2 -- cos (:n - y)}

(making the change of variable y = n - x)

S I' IJ

f" dy = J n/2 :Yc2 -~os i> -f" --- -~---- :tr/2 v{2 + 1 -2 sin2 (y /2)}

f"i2 2dz = n/4 \/(3 -_ -,2::-s-:--in-2-z)

(making the change of variabk z -

2 r·~~

,;~ J , I \ 1( I

dz ~sin~::-)

y/2)

228 OTHER SPECIAL FUNCTIONS CH. 10

2 { rn/2 dz rn/4 dz } = y3 J 0 v(1- fsin 2 z)- J 0 v(1- fsin 2 z)

= ~3 { F (J~· ~)- F(J~· ~)} (by definition (10.28)).

PROBLEMS

( 1) Show that

(i) Joo cos x Ci(x) dx = J oo sin x si(x) dx = -~; 0 0 4

(ii) J~ {Ci(x)}2 dx = J~ {si(x)}2 dx = ~·

foo COSt (2) If l(rx) = (1 - e-"'1) ·-- dt, show that

0 t

d/ !X

drx rx 2 + 1

and deduce that l(rx) = t ln (1 + rx2).

foo 1 Prove that

0 e-st Ci(t) dt =

2s ln (1 + s2).

1 2x (3) Show that erf x = yn y(!, X

2) = yn 1F 1(!; t; -x2

).

( 4) Show that T(x, t) = C erf {x/ v( 4kt)} satisfies the following conditions:

(i) T(O, t) = 0;

(ii) T(x, 0) = C;

( ... ) oT _ ko 2T. 111 of - ox2

fx 1 7l (5) Show that (i)

0 C(t) dt = xC(x)-;; sin 2x2 ;

(ii) Jx S(t) dt = xS(x) + 2 cos ~2x 2 - 2. 0 7l 7l

fn/2 dx f"/ 2 dx ( 1 ) (6) Show that o v(sin x) = o v(cos x) = ( v2)K vZ .

PROBLEMS

(7) By making the substitution x = 2 sin(), show that

I2 dx o v{(4 - x2)(9 - x2)} = iK(i).

(8) Show that

J: v{(l + x~~l + 2x2)} = ~z{x(~z)- F(~z' ~) }-

APPENDICES

1 CONVERGENCE OF LEGENDRE SERIES

We wish to investigate the convergence of the series obtained in Section 3.1 as solutions to Legendre's equation.

These series are of the form

and

00

""'a x2n L,.. 2n n=O

00

""'a x2n+l L,.. 2n+l n=O

where in both cases we have, from equation (3.7),

(n - l)(l + n + 1) a,.+2 =a,. (n + l)(n + 2) -·

(A1.1)

(A1.2)

(A1.3)

We shall discuss only the series (Al.l ), since the discussion of the other series is similar in all respects.

First we note that equation (A1.3) implies that all the terms in the series (A 1.1) for n > l have the same sign, so that we may apply tests for series of positive terms.

If x ;C 1 we may apply d'Alembert's ratio test: 00

if ,L u,. is a series of positive terms and if lim u,.ju,.+l = oc, the series is -o ~ao

divergent if oc < 1, convergent if oc > 1 and the test provides no information if oc = 1.

Here Un = a2,.x2", so that we have u,. a2,.x2n

-·-------Un+l a2n+2x2n+2

(2n + 1)(2n + 2) 1 --

(Zn - l)(Zn + l + 1) x 2

so that

lim u,. 1 Jl--lo-tr; Unj)

EULER'S CONSTANT 231

and the series is hence divergent if x2 > 1 and convergent if x2 < 1 ; that is, it is divergent if I x I > 1 and convergent if I x I < 1.

If x = ± 1 we must use Gauss's ratio test: 00

if 2: u.. is a series of positive terms, and if the ratio u,.ju,H1 can, for n=O

n > some fixed N, be expressed in the form

u,. I' (1) -- = 1 +- + 0-Un+l n nP

with p > 1 (where by 0(1/nP) we mean some function f(n) such that 00

lim {f(n)/(1/nP)} is finite), then 2: u,. is convergent if p, > 1 and 1l--+-a) n=O

divergent if p, < 1. Here, since we are considering x = ± 1 , we have

Un (2n + 1)(2n + 2) u,.H (2n - l)(2n + l + 1)

and it is a matter of simple algebra to prove that

~ = 1 + ~ + l(l + 1 )(1 + n) __ . u,+1 n {4n2 + 2n - 1(1 + l)}n

Now, we see that the last term is O(l/n 2), since we have

lim [ 1(1 + 1)(1 + n) I 1 J n-+oo {4-n2 + 2n- l(l + 1)}n n2

= lim 1(1 + 1)(1 + n)n2

1!---+00 {4-n2 + 2n - z(l +1)"}~ = l(l + 1)/4,

which is finite. Hence the conditions of Gauss's ratio test are satisfied; we have p, = 1

and thus the series is divergent. This means that we have shown that the I _,egendre series is divergent for x = ± 1.

2 EULER'S CONSTANT

Assuming f(x) to be a continuous, positive and de('reasin~ function, ld us consider the sequence

II., f( I) f(2) f(.\) : ... f(,) f" f(.\·) d ,. . I

232 APPENDICES

We first show that u,. > 0. This follows immediately from Fig. A2.1, since f(l) + f(2) + ... + f(n) is the area of the rectangles shown, while

y

FIG. A2.l

J: f(x) dx is the shaded area underneath the curve, which is obviously less

than the area of the rectangles, thus making u .. > 0. We now show that Un+l < Un. For we have

fn+l

Un+l - u, = f(l) + f(2) + ... + f(n) + f(n + 1) -1

f(x) dx

-f(l) - f(2) --- ... - f(n) + J: f(x) dx

= f(n + 1) - fn+t f(x) dx, n

and from Fig. A2.2 we see immediately that this is negative.

y

n n+ I /(

FIG. A2.2

Hence Un n - u,. < 0, i.e., Un n < u,..

DIFFERENTIAL EQUATIONS 233

Thus the u,. constitute a decreasing sequence, every member of which is positive, and thus by a general principle the sequence must possess a limit as n tends to infinity. Hence lim u,. exists.

n---+oo

The choice of f(x) = 1/x will satisfy the conditions that f(x) be con-tinuous, positive and decreasing.

Then

1 1 1 [" 1 Un = 1 + 2 -j- 3 -j- • • • + ;; - J

1 ~ dx

1 1 1 = 1 + 2 + 3 + . . . + ;; -- Inn.

So we know thatlim {1 + (1/2) + (1/3) --t- •.. + (ljn) -- ln n} exists. n---+00

It is called Euler's constant and is usually denoted by y. It may be shown

that, correct to four decimal places, y = 0·5772.

3 QIFFERENTIAL EQUATIONS

Equation Solutions ~~~-~-~~~~--~~----------

(1 - x 2)y" - 2xy' + l(l + l)y = 0 P,(x) Legendre polynomials Legendre functions of the

second kind Q,(x)

(1 - x 2)y" - 2xy' P';'(x) Associated Legendre polynomials

+ {z(l + 1)- 1 :•x2}Y = 0 Q';'(x) Associated Legendre functions of the second kind

·-- ----------- ---------------~~~

x 2y" (1 - 21X)~y'

]n(x) Bessel functions of the first kind Yn(x) Bessel functions of the second

kind H<1>(x)} H<!l(x) Hankel functions

i:C~)}Modified Bessel functions

:xf'] n(PxY) I {/11y1x 1Y 1- (1X 2 - n 1y 2)}y 0 XX Yn(PxY)

x 2y" 2xy' I {.~ 2 1(1 I I)} y 0 j,(.~)} ' I . I I' I f' . · ( ) Sp H'ru·u wsst· unl'lt<Hls )'I.\'

------ --- --- -----

234

Equation

y" - 2xy' + 2ny = 0 y" + (J. - x 2)y = 0

xy" + (1 - x)y' + ny = 0

APPENDICES

Solutions

Ha(x) Hermite polynomials 'Pn(x) = exp ( - !x2)Hn(x)

·weber-Hermite functions

Ln(x) Laguerre polynomials xy" + (k + 1 - x)y' + ny = 0 L!(x) Associated Laguerre polynomials

~i:~} Chebyshev polynomials

(1 - x 2)y" - (2J. + l)xy' + n(n + 2J.)y = 0 C~(x) Gegenbauer polynomials

(1-x1)y" + {,8-oc-(oc+P+2)x}y' + n(n+ot:+P+l)y ~ 0 Pn(rL,fl)(x) Jacobi polynomials

x(l - x)yn + {y - (ot: + ,8 + l)x}y' 2F,(oc, {J; y; x) - ocpy = 0

x 2y" + (,8 - x)y' - ocy == 0

Hypergeometric function x 1-Y2F 1(oc+1-y; ,8-t-1-y; 2-y; x) 1F/ct; {I; x)

{ 1 k t.- m2

} y" + - - + - + _4__ y = 0 4 x x 2

Confluent hypergeometric function i xl-f11F1(oc - {J + 1; 2 - {J; x)

I. Mk,m(x)} Whittaker's confluent Mk,-m(x) hypergeometric functions

4 ORTHOGONALITY RELATIONS

Function Relation ·-·-···----------··-~------------ -·-- ---·---. ·-···

Legendre polynomials

Associated Legendre polynomials

fl 2

Pt(x)Pm(x) = 2F+1!5tm -1

fl P"'( )P•( ) dx 2(1 + m)! li J -l 1 X T X = (2[ -f- 1)(/ - m)l U'

! fl ~':(x)P';"~~ dx = J£.± m?!.,smm' , _ 1 1 - x 2 m(l - m)!

ORTHOGONALITY RELATIONS

Function

Bessel functions

Spherical Bessel functions

Relation

f: x]n(AtX)],.(J.Jx) dx = ~{]n+lJ.«J)FOiJ, where At and AJ are roots of ]n(Aa) = 0

f: 00

im(x)jn(x) dx = 2n: l 0"'"

~-----~------ ------ -- -----

Hermite polynomials f:oo e-x'Hn(x)Hm(x) dx = 2"n!( y'n:)onm

Weber-Hermite functions f:oo 'P,.(x)'l'm(x) dx = 2"n!( v'n)o.m

Laguerre polynomials

Associated Laguerre polynomials

f~ e-"Lm(x)Ln(x) dx = Omn

J~ e-"xkL~(x)L!(x) dx = (n ;, kl1o..,.

235

----------- -----~------------·-----------

Chebyshev polynomials e !._m(x)Tn(x) ci=_ = {~/2 J_l v'(l-x2) n:

fl Um(x)Un(x) dx _ {0 -1 v'(l - x2) - ~/2

m # n m = n # 0 m = n = 0

m #n m =n ¥-0 m =n =0

-----------1

Gegenbauer polynomials

Jacobi polynomials

:.J It

fl (1 - x8)A-1C!(x)C~(x) dx I -1

21- 2"r(n + 2J..) = n: n!(n + J.){r(J.)pOmn

J~ 1 (1 - x)"'(1 + x}P (a.ft)(x)P,.(rx,fJ)(x)

2rx+fJ+l r(n + IX + l)r(n + f3 + 1)

" 2n + IX + f3 + 1 n !r(n + IX + f3 + 1) mn

236 APPENDICES

5 GENERATING FUNCTIONS

Function

Legendre polynomial

Associated Legendre polynomials

Bessel functions

Hermite polynomials

Generating function (R = (1 - 2xt + t2)112)

co

~ = L t"Pn(x) n=O

(2m) !(1 - x2)m12

zmm!Rm+o

00

L t'P,"'+m(x) r=O

-~ ----·-- -----

Laguerre polynomials

Associated Laguerre polynomials

Chebyshev polynomials

exp{ -xt/(1 -t)}

1 - t

ro

L t"Ln(x) n=O

ro

exp{- xt/(1 - t)} _ "" "Lk( ) (1 - t)k+l - L t n X

n=O

00

1 - t2

"" ---w = T 0(x) + 2 L t"Tn(x) n=l

---------

Gegenbauer polynomials

Jacobi polynomials

1 R2A

HINTS AND SOLUTIONS TO PROBLEMS

CHAPTER 1 (1) The general solution is Ay1(x) + By2(x) in each case, where A and 1 are arbitrary constants and y 1(x) and y 2(x) are given by

ro 1

(i) YI(x) = 1 + n~ nn.s:9 ... (4n- 3tn,

Y2(x) = x314

{1 + ,~ n!7 .11.15 ~ .. (4n + 3tn}

valid for all x; (ii) y 1(x) = e-"',

~ 1 ( 1 1 1) y 2(x) = ln x.e-x- L., ( --1)nn! 1 + Z + J + ... +;; xn, n=l

valid for all x; co 1

(iii) y 1(x) = ~ ( ---1 )n Zn !(n _ 3) 1xn,

y 2(x) = y 1(x) ln x + 1 + ix + !x2 - :/6x 3 +

+ ~ 2n!\~~n3)!{ - 2( 1 + ~ + ~ + · · · + n ~ 3

- _1 - - _1 __ - ! + ~}xn n-2 n-1 n 2 '

valid for all x;

~ 1. 4. 7 ... (3n -- 2) (iv) yb) = 1 + L., ---- 3;nT- ----xn

n:.:..:: 1

= (1 - - x)-1,

_ .1{ ~ R 0 11 0 14 0 0 0 (3n I S) } ( ) -- x' 1 I L., \0

" Y~ .x · 10 0 13 0 H) 0 • 0 (3n 7)' ' II I

\'alid for I .\0

[ • I;

238 HINTS AND SOLUTIONS TO PROBLEMS

( ) ( ) _

1 ~ ( -1 )(-3) ... ( 4n 2 - 14n + 9)

2,.

v YI x - + .6_ (2n)! x ,

) _ ~ ( -3)( -1) ... (4n 2

- 10n + 3) 2n+1

Yt.(x - x + L, (2 1) 1 x , n=1 n + ·

valid for I x I < 1 ;

valid for all x;

(vii) y 1(x) = x,

- x2 ,x4 - ~ 32.52 ... (2n - 3)2 2n

Y2(x)- 1 + 2! + 4! 1--,6 (2n)! x ,

valid for I x I < 1 ;

(viii) Yl(x) =! ( -1)"+1 32n-1( 1_ 2)1 ,xan-1, n-2 n .n.

Ys(x) = Y1(x) In x + x- 1 +- tx 2 + 3h x5

<X> 1 { ( 1 1) + 6 ( -l)n+I32"(n-2)!n! 1-2 1 +.z + ... +;-- 2

__ 1 ___ !}xan-1 n -1 n '

valid for all x;

(ix) y 1(x) = 1 + ~x 2 + f4x4 + z~X5 + ?1ox6 + rlhx7 + ... , Y2(x) = x + i-x3 + f:tx4 + t1ox5 + mx6 + s!hx7 + ... ,

valid for all x.

(2) (i) Impossible; (ii) impossible; (iii) possible.

(3) The general solution is Ay1(x) + By2(x) where A and B are arbitrary constants, and y 1(x) and y 2(x) are given by:

(.) ()

1, ~2.7.16 ... (2n2 -n+1) 1

l)'X= TL_, --, 1 n=l n!l.3 .5 ... (2n - 1) x"

y2(x) = x-112 { 1 + ~ ~~f~:__:_ (2n~. ·.l:

1

n I 1) l}; L, n .3. 5 ... (2n ·. 1) x"

1L;:.::l

HINTS AND SOLUTIONS TO PROBLEMS 239

(ii) YI(x) = 1 - :x' ()

__ 112 { 1 ~(-3)(-1) ... (2n -5) 1}· YzX -X + k -

n!1.3 .5 ... (2n - 1) xn n~l

CHAPTER l

(1) Use definition (2.1) of the gamma function with a suitable change of the variable of integration. (2) Use theorem 2.5, (3) Use theorem 2.5.

(4) (i) B(t, !) = 2~2" (ii) r(a). (Make the change of variable ljx = e11.)

(iii) (b - a)m+n-1B(m, n). (Make the change of variable y = (x - a)j(b - a).)

(iv) ~B(m ~ 1, p + 1 )·(Make the change of variable y = xn.)

!B(!, ~) ·_ vn r(1/n) . k h . (v)n n 2 - n r{( 1 /n)+!}(Maethecange;fva~~~~le

(vi) B(t, !) = n. (Make the change of variable u = 1/(1 + t).) (6) r( -t) = -2-y'n; r( -f) = Tlo65 yn. (7) Use theorem 2.12.

CHAPTER 3

(1) Use theorems 3.8(ii) and 3.5. (2) Use theorems 3.8(vii), 3.8(ii) and 3.5. (3) Consider (x2 - 1 )1 as the product (x - 1 )1

• (x + 1 )1•

{

( -1 )<r-1)/2 (r + !)(r - 1)! if r is odd (4) Cr = 2r{(r + 1)/2}!{(r- 1)/2}!

0 . if r is even. Use the result of Example 2.

(5) Differentiate the generating function for the Legendre polynomials m times and usc definition (3.36). (6) Use theorems 3.8(ii) and 3.5.

{

2/n if tl is even II,.

II if 11 is mid.

240 HINTS AND SOLUTIONS TO PROBLEMS

(7) Use theorem 3.8(i). (8) Use theorem 3.8(ix). (10) Use theorems 3.8(viii) and 3.16 and Example 5. (11) Use definitions (3.61) and (3.36). (12) Use theorems 3.2 and 3.15 and integrate by parts l times.

CHAPTER 4

(1) Use theorem 4.8(iii) and (iv). (2) Use the infinite series for ] 0 and]1•

(3) Compare with Example 4. (4) Ax11 2} 2; 3(t( v..1)x 213) (with A an arbitrary constant). Use theorem 4.12. (5) Use the infinite series for ] 0 •

(6) Usethefactthat 1 =exp [x{t-(1/t)}].exp[-x{t-(1/t)}] andpick out the coefficient of x0 on the right-hand side. (7) Use theorems 4.8(ii), 4.21(i) and 4.22(i). (8) Use the generating function. (9) Use the generating function for the] n·

(10) Show that both Bessel's equation and Bessel's modified equation may, by further differentiations, be written in the given form.

(11) c = 2_ {(J.,a)2 - 4}. r aJ. 3 ] 1(..1,a)

(12) f: )m(x)jn(x) dx = 0 (m :r':- n).

fro {jn(x)}Z dx = 2_!!___

1.

-oo n + (13) Take real and imaginary parts of the infinite series for ]n(i3' 2x).

(14) Take real and imaginary parts of the integral J: tj0(i312t) dt; use

theorem 4.8(i) to help evaluate the integral. (15) Take real and imaginary parts of the asymptotic form of J0(i 312x), given by theorem 4.21(i).

CHAPTER 5

(1) Use theorem 5.6(ii) and 5.5.

(3) Use the fact that 2 f~ e-t' cos 2xt = f: 00

e-t' cos 2xt = Re J: "' exp (- t 2 + 2ixt) dt.

HINTS AND SOLUTIONS TO PROBLEMS 241

22" I 2 I 00 1)" - c.c' e-1'tZn+lsin2xtdt. y'n o

(4) Using the result of problem 3 for both H.,(x) and Hn(Y) gtves

~ H (x)H (y) 1 Joo Joo L " n ~ t" = - exp (x 2 + y 2) exp ( -u2

- v2 + 2iux 2n. :n: -oo -oo 11=0 + 2ivy - 2uvt) du dv.

Performing the integrations gives the required result.

(7) (i) znn !( y'n)bmn•

(ii) zn-ln!( y':n:)bm, n-1 -2"(n + 1)!( y'n)bm, n+l·

CHAPTER 6

(3) Use equation (6.3) and integrate term by term. (4) Use equation (5.3) and integrate term by term. (5) Use theorem 6.9. (6) ,Use theorems 6.11(ii) and 6.10. (7) Use equation (5.3).

CHAPTER 7

(1-6) Make the substitution x =cos fJ. (8) Use the series solution method of Chapter 1, and remember that n is integral.

CHAPTER 8

(1) Use definition (8.2) or theorem 8.5. (2) Use definition (8.2). (3) Use theorem 8.5(ii). (4) Use theorem 8.3(i). (5) Use definition (8.2). (6) Use theorem 8.1 and equation (3.17).

CHAPTER9

(3) Usc theorem 9.4. ((,) l lsc theorem 'J.I. (7) (I /s))•\('1., I; fl; s). t 'st· tlworem '!.7.

242

CHAPTER 10

HINTS AND SOLUTIONS TO PROBLEMS

( 1) Integrate by parts.

(2) Write J~ e-st Ci (t) dt as a double integral and change the order of

the integration. (5) Integrate by parts. (6) First prove the equality of the two integrals by making the substitution y = (n/2) - x in the first. Then consider the second integral, and make the substitution cos x = cos 2 u. (8) Make the substitution x = tan 0.

BIBLIOGRAPHY

ABRAMOWITZ, M., and STEGUN, I. A. Handbook of Mathematical Functions, Dover, New York (1965).

ARTIN, E. The Gamma Function, Holt, Rinehart and Winston, New York (1964). ERDELYI, A., MAGNUS, W., OBERHETTINGER, F., and TRICOMI, F. G. Higher

Transcendental Functions (Bateman Manuscript Project), Vols. 1-3, McGrawHill, New York (1953).

FLETCHER, A., MILLER, J. C. P., RoSENHEAD, L., and CoMRIE, L. J. An Index of Mathematical Tables, Vols. 1 and 2, 2nd edn., Addison-Wesley, Reading, Mass. (1962).

HOBSON, E. W. The Theory of Spherical and Ellipsoidal Harmonics, Cambridge lJ niversity Press, London ( 1931 ).

HocHSTADT, H. Special Functions of Mathematical Physics, Holt, Rinehart and Winston, New York (1961).

}AHNKE, E., EMDE, F., and LoscH, F. Tables of Higher Functions, 6th edn., McGraw-Hill, New York (1960).

LEBEDEV, A. V., and FEDOROVA, R. M. A Guide to Mathematical Tables, Pergamon Press, Oxford (1960).

LEBEDEV, N. N. Special Functions and their Applications, Prentice-Hall, Englewood Cliffs, N.J. (1965).

MAcROBERT, T. M. Spherical Harmonics, 2nd edn., Methuen, London (1947). MAGNUS, W., 0BERHETTINGER, F., and SoNI, R. P. Formulas and Theorems for

the Functions of Mathematical Physics, 3rd edn., Springer-Verlag, New York (1966).

McLACHLAN, N. W. Bessel Functions for Engineers, 2nd edn., Oxford University Press, London, (1955).

RAINVILLE, E. D. Special Functions, Macmillan, New York (1960). RELTON, F. E. Applied Bessel Functions, Blackie, London (1946). SLATER, L. J. Confluent Hypergeometric Functions, Cambridge University Press,

London (1960). SNEDDON, I. N. Special Functions of Mathematical Physics and Chemistry, 2nd

edn., Oliver and Boyd, Edinburgh (1961). WATSON, G. N. A Treatise on the Theory of Bessel Functions, 2nd edn., Cam

bridge University Press, London (1931).

INDEX

Associated Laguerre polynomials 176

definition of 176 generating function for 177 notation for 182 orthogonality properties of 178 recurrence relations for 178

Associated Legendre equation 62 Associated Legendre functions 62

definition of 62 orthogonality relations for 66 properties of 65 recurrence relations for 68

Bessel functions 92 of the first kind 94

asymptotic behaviour of 127 definition of 94 generating function for 99 graph of 134 integral representations for 101 orthogonality properties of 137 recurrence relations for 104

of the second kind 97 asymptotic behaviour of 127 definition of 97 explicit expression for 98 graph of 134 recurrence relations for 106

of the third kind (see Hankel functions) Bessel series 141 Bessel's equation 92 Beta function 23

definition of 23

Chebyshev polynomials I H7

definition of 187 generating function for 190 orthogonality properties of 192 recurrence relations for 193 series for 188 special values of 1 92

Chebyshev's equation 189 Confluent hypergeometric equation

210, 212 Confluent hypergeometric function

203 contiguity relations for 212 definition of 203 integral representation of 211

Cosine integral 219

Debye functions 224 Differential equations

satisfied by various special functions 233

solution of by Frobenius' method 1

Elliptic integrals 224 Error function 221 Euler's constant 231 Exponential integral 218

Fresnel integrals 221 Frobenius' method 1

summary of 7

Gamma function 23 definition of 23, 30 !o:ntph of 31

246 INDEX

Gauss's equation 207 Gegenbauer polynomials 197

definition of 197 differential equation satisfied by 198 generating function for 197 orthogonality properties of 198 power series for 198 recurrence relations for 198

Generating function 236 for associated Laguerre polynomials

177 for Bessel functions of the first kind

99 for Chebyshev polynomials 190 for Gegenbauer polynomials 197 for Hermite polynomials 157 for Jacobi polynomials 198 for Laguerre polynomials 169 for Legendre polynomials 46

Hankel functioas 107 asymptotic behaviour of 127 definition of 107 recurrence relations for 108

Hermite polynomials 156 definition of 157 generating function for 157 orthogonality properties of 161 recurrence relations for 162 special values of 160

Hermite's equation 156 Hypergeometric equation Hypergeometric functions

contiguity relations for definition of 203

207, 208 203

210

integral representation of 207 relationship with other special func-

tions 204

Incomplete gamma functions 218 Indicia! equation 3 Integral representation

of Bessel functions 101 of confluent hypergeometric function

211 of hypergeometric function 207 of Legendre polynomials 49 of modified Bessel functions 116

Integrals involving Bessel functions 141

Jacobi polynomials 198 definition of 198 differential equation satisfied by 199 generating function for 198 orthogonality properties of 199 recurrence relations for 199 series for 199

Kelvin's functions graphs of 137

Kummer function Kummer's equation

120

204 210

Laguerre polynomials 168 definition of 169 generating function for 169 notation for 182 orthogonality properties of 172 recurrence relations for 173 special values of 1 71

Laguerre's equation 168 Laplace's integral representation 49 Legendre duplication formula 29 Legendre functions of the second kind

70 graphs of 84, 85 properties of 77

Legendre polynomials 42 definition of 46 generating function for 46 graphs of 82, 83, 84 Laplace's integral representation for

49 orthogonality properties of 52 recurrence relations for 58 Rodrigues' formula for 48 special values of 50

Legendre series 55 Legendre's associated equation 62 Legendre's equation 42 Lcibniz's theorem 63 L'Hopital's rule 39, 97 Logarithmic integral 219

INDEX 247

Modified Bessel functions 110 asymptotic behaviour of 128 graphs of 135 integral representations for 116 recurrence relations for 113

Neumann functions (see Bessel functions of the second kind)

Neumann's formula 78

Orthogonality properties 234 of associated Laguerre polynomials

178 of associated Legendre functions 66 of Bessel functions 137 of Chebyshev polynomials 192 of Gegenbauer polynomials 198 of Hermite polynomials 161 of Jacobi polynomials 199 of Lgguerre polynomials 172 of Legendre polynomials 52 of spherical harmonics 80

Pochhammer symbol 203

Rayleigh's formulae 126 Recurrence relations

for associated Laguerre polynomials 178

for associated Legendre functions 68

for Bessel functions of the first kind 104

for Bessel functions of the second kind 106

for Chebyshev polynomials 193 for Gegenbauer polynomials 198 for Hankel functions 1 08 for Hermite polynomials 162 for Jacobi polynomials 199 for Laguerre polynomials 173 for Legendre polynomials 58 for modified Bessel functions 113 for spherical Bessel functions 122

Riemann's zeta function 223 Rodrigues' formula 48

Sine integral 219 Spherical Bessel functions 121

asymptotic behaviour of 128 graphs of 136 Rayleigh's formulae for 126 recurrence relations for 122

Spherical harmonics 78 orthogonality properties of 80

Ultraspherical polynomials Gegenbauer polynomials)

(see

Weber-Hermite functions 163 Whittaker's confluent hypergeometric

functions 212

Zeta function 223

SPECIAL FUNCTIONS - Page maison de Simon Plouffe / Simon Plouffe

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