SPECIAL
FUNCTIONS
FOR
SCIENTISTS AND ENGINEERS
W. W. BELL
I.uturer in Theoretical Physics
J),.partment of Natural Philosophy
University of Aberdeen
,1 fJ...?~~ D. VAN NOSTRAND COMPANY LTD
II
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Copyright© 1968 D. Van Nostrand Company, Ltd
Published simultaneously in Canada by
D. VAN NosTRAND CoMPANY (Canada), Ltd
No reproduction in any form of this book, in whole or
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reviews), may be made without written authorisation
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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 ~. 1 1, 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
Beta function
Kelvin's functions
Fresnel integral
Gegenbauer polynomial or ultraspherical
C~(x)
polynomial
Ci(x)
Cosine integral
D ..(x)
Debye functions
E(k, ¢)
Elliptic integral of second kind
E(k)
Complete elliptic integral of second kind
Exponential integrals
Ei(x), E 1(x)
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
third kind
h..<1 >(x), h,.< 2>(x) Spherical Hankel functions
Modified Bessel function
I,.(x)
Bessel function of the first kind
J..(x)
Spherical Bessel function
j ..(x)
Complete elliptic integral of first kind
K(k)
Modified Bessel function
K ..(x)
Kelvin's functions
ker., x, kei.. x
Laguerre polynomial
L ..(x)
Associated Laguerre polynomial
L~(x)
Logarithmic integral
li (x)
Confluent hypergeometric function
M(a., {3, x)
Whittaker's confluent hypergeometric
Mk,m(x)
function
Neumann function or Bessel function of the
second kind
B(x,y)
her.. x, bei.. x
C(x)
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
LIST OF SYMBOLS
X
Symbol
Function
Secliotl i11 which
symbol first appears
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)
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
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
1.2 Examples
Problems
CHAPTER 2
1
8
21
GAMMA AND BETA FUNCTIONS
2.1
2.2
2.3
Definitions
Properties of the beta and gamma functions
Definition of the gamma function for negative values of the
argument
2.4 Examples
Problems
CHAPTER 3
3.1
3.2
3.3
3.4
23
24
30
37
40
LEGENDRE PoLYNOMIALS AND FuNcTIONS
Legendre's equation and its solution
Generating function for the Legendre polynomials
Further expressions for the Legendre polynomials
Explicit expressions for and special values of the Legendre
polynomials
3.5 Orthogonality properties of the Legendre polynomials
3.6 Legendre series
3.7 Relations between the Legendre polynomials and their derivatives; recurrence relations
3.8 Associated Legendre functions
3.9 Properties of the associated Legendre functions
3.10 Legendre functions of the second kind
3.11 Spherical harmonics
3.12 Graphs of the Legendre functions
3.13 Examples
Problems
42
46
48
50
52
55
58
62
65
70
78
82
85
90
xu
CONTENTS
CHAPTER 4
BESSEL FUNCTIONS
Bessel's equation and its solutions; Bessel function~ of thL· first
and second kind
4.2 Generating function for the Bessel functions
4.3 Integral representations for Bessel functions
4.4 Recurrence relations
4.5 Hankel functions
4.6 Equations reducible to Bessel's equation
4.7 Modified Bessel functions
4.8 Recurrence relations for the modified Bessel functions
4.9 Integral representations for the modified Bessel functions
4.10 Kelvin's functions
4.11 Spherical Bessel functions
4.12 Behaviour of the Bessel functions for large and small values of
the argument
4.13 Graphs of the Bessel functions
4.14 Orthonormality of the Bessel functions; Bessel series
4.15 Integrals involving Bessel functions
4.16 Examples
Problems
4.1
CHAPTER 5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
127
133
137
141
148
154
HERMITE PoLYNOMIALs
Hermite's equation and its solution
Generating function
Other expressions for the Hermite polynomials
Explicit expressions for, and special values of, the Hermite
polynomials
Orthogonality properties of the Hermite polynomials
Relations between Hermite polynomials and their derivatives;
recurrence relations
Weber-Hermite functions
Examples
Problems
CHAPTER 6
92
99
101
104
107
108
110
113
116
120
121
156
157
158
160
161
162
163
164
166
LAGUERRE PoLYNOMIALS
6.1 Laguerre's equation and its solution
6.2 Generating function
6.3 Alternative expression for the Laguerre polynomials
6.4 Explicit expressions for, and special values of, the Laguerre
polynomials
168
169
170
171
CONTENTS
6.5
6.6
Orthogonality properties of the Laguerre polynomials
Relations between Laguerre polynomials and their derivatives;
recurrence relations
6. 7 Associated Laguerre polynomials
6.8 Properties of the associated Laguerre polynomials
6.9 Notation
6.10 Examples
Problems
CHAPTER 7
7.1
7.2
7.3
7.4
7.5
8.1
8.2
8.3
9.1
9.2
9.3
9.4
172
173
176
177
182
182
185
CHEBYSHEV PoLYNOMIALs
Definition of Chebyshev polynomials; Chebyshev's equation
Generating function
Orthogonality properties
Recurrence relations
Examples
Problems
CHAPTER 8
Xlll
187
190
192
193
194
196
GEGENBAUER AND JAcOBI PoLYNOMIALS
Gegenbauer polynomials
Jacobi polynomials
Examples
Problems
197
198
200
201
CHAPTER 9 HYPERGEOMETRIC FUNCTIONS
Definition of hypergeometric functions
Properties of the hypergeometric function
Properties of the confluent hypergeometric function
Examples
Problems
203
207
210
212
216
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
CHAPTER
218
218
221
223
224
224
225
228
CONTENTS
APPENDICES
1
2
3
4
5
Convergence of Legendre series
Euler's constant
Differential equations
Orthogonality relations
Generating functions
230
231
233
234
236
HrNTS AND SoLUTIONS To PROBLEMs
2J7
BIBLIOGRAPHY
243
245
INDEX
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
(1.1)
P(x) dxll + Q(x)dx + R(x)y = 0.
We shall restrict ourselves to equations of the type
dy
d 2y
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
LrmXm,
00
r(x) =
(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
La,.x•+n (1.5)
00
z(x,s) = x•(a 0 -/- a 1 x + a 2x 2 + ... + a,.xn + ...) =
with a 0 ./. 0. (We may always take a 0 /
n-o
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
a,.(s 11 )x• 1" 1
L
u-.o
d 2z
~...
d~ii ,_, L
and
1)x• 111 2,
a..(s I n)(s I n
,.. 0
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
,. ~o
n=O
L a,.(s + n)(s + n 1)x•+n + q(x) L a,.(s + n) x''"
L a,.xs+n =
<X>
+ r(x)
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
«>
L a,.(s + n)(s + n - 1)xn + L L a,.qm(s + n)x"+"'
+ L L a,.rmx" tm = 0.
00
00
n~o
n=·O m=O
00
00
(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
a 0 s(s- 1) + a0q0s + a 0T 0 = 0.
(1.7)
1
1
Requiring that the coefficient of x be zero, and noting that x 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) + a 0q1s} + {a 1r 0 + a 0r 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;r 0 + a 1 _ 1r 1 + ... + a 0r 1} = 0
(i > 1).
(1.9)
Thus
aJ(s + i)
+ {a;_1q1(s + i- 1) + a;_2q2(s + i- 2) + ... + a 0q;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
a 0 {s 2 + (q 0 - l)(s + r 0)} = 0
which, with the given assumption that a 0 ¥-0, becomes
s 2 + (q0 - 1}(s + r 0) = 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
a 1, a 2 , ••• in terms of a0 •
Equation (1.10) with i = 1 gives
alf(s + 1) + a~ls + aorl = 0
so that we have
at =
-ao(qls + rt)
f(s + 1}
a h (s)
f(s + 1)
(q,s 1 r 1) is a polynomial of first degree ins.
0 1
------
where h,(s) - ·
lii'-U
(1.14)
4
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
Equation (1.10) with i = 2 gives
aJ(s + 2) + {a1 q1(s + 1) + a 0 q2s} + {a 1r 1 + a 0r 2 } = 0,
which, when we use equation (1.14) for al> becomes
aJ(s + 2) +
yielding
t~:~s~q1(s + 1) + aq s} + ~~so~~s?{t + ar = 0,
0 2
0 2}
-a 0 {q 1(s + l)h 1(s) + q2 sj(s + 1) + r1h1(s) + rd(s + 1)}
f(s + l)f(s + 2)
h2(s)
= ao j(s + l)f(s + 2)
(l.lS)
a2 =
where h2(s) = -{q1(s + 1)h1 (s) + q2sj(s + 1) + r 1h1 (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 = s 2 , 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(s 1 + i) ~0
and
f(s 2 + i) ~ 0 for i equal to any positive integer.
(1.17)
Now, equation (1.13) tells us that
f(s) = (s - s1)(s - s2}
so that
f(s + i) = (s + i - s1)(s + i - s2)
and hence
f(s 1 + i) = i(s1 - s2 + i)
f(s 2 + i) = (s 2 - s1 + i)i,
(1.18)
and
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
5
METHOD OF FROBENIUS
f(s 1 + i) = i( -r + i)
j(s 2 + i) = i(r + i),
(1.19)
(1.20)
and we see that f(s 1 + 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
_
.:r(x, s) -
Or]X
·{ 1
h1(s)
h2(s)
+ j(s + 1t + j(s + 1)j(s + 2t
h;(s)
i
2
}
+ · · · + f(s + 1)f(s + 2) .. . f(s +it + · · · ·
(1.21)
Wc know by the method of construction of a 1 , a 2 , • • • that if all the terms
on the right-hand side are well defined, then this must satisfy
d 2z
dz
x 2dx 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 a 0 f(s)). This equation may be rewritten in the form
d
}
2 d2
{ X 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 = s1 ; 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
d
}
2 d2
{ x 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
{ x 2 : :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)}]
= a 0 [(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
and
[(s - s1 )z(x, s)]•=•,
(1.25)
[1s{(s - s)z(x, s)}1=.,·
(1.26)
1
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
d
2 - d2 + xq(x)-d
+ r(x) } z(x, s) = a0(s - s1) 2x'.
{ x 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
+ xq(x)~
+ r(x)} [iz(x,
s)J
{x2 ~
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 s 2 of the indicia! equation are distinct and do not
differ by an integer, then the two independent solutions are given by
and
z(x, s1)
z(x, s2).
(2) If the roots s1 and s2 (s 2 > 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 - )z(x, s)}l=.,·
s1
(3) If the roots s 1 and s2 (s 2 > 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 = s 1 , 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
8
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
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
2xdx2
dy
+ dx + Y = o.
We must first verify that this equation is of the form
x2~Fy2 + xq(x) dy + r(x)y = 0
dx
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
so that
~
:L: an(s + n)xs+n-1
d2z
-· ·- .,·- L a (s + n)(r -t- n - l)x·•+nd
dz
dx =
!i~O
ro
and
x-
11.
n
o
•
2
•
§ 1.2
9
EXAMPLES
Then we have
d 2z
dz
2x-+-+z
2
dx
dx
L 2a,.(s + n)(s + n - 1)xs+n- + L an(s + n)xstn-1
+ 2: anx•+n
L 2an(s + n)(s + n -+ L an(s + n)xstn00
=
00
1
n=O
1t=O
00
n=O
00
=
00
1
l)xHn-I
n=O
n=O
00
~ an-1 x•+n-11
+ L_.
(1.30)
n=l
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
and
(1.32)
aoS(2s- 1) = 0
(1.33)
an(s + n){2(s + n) - 1} + an_ 1 = 0.
(1.34)
Equation (1.33) gives the indicia! equation
s(2s- 1) = 0
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):
On= --
and hence
On-1
(s + n){2(s + n)- 1}
(s + 1)(2s j 1)'
(1.35)
10
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
02
= - (s +
03
= - (s
CH. 1
al
ao
2)(2s + 3) = (~:-f)(s +2)(2s +-i)(Zs +j)'
a2
ao
+ 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
x2
x3
z(x, 0) = a 0{ 1 - 1]_" + 1. 2 .1. 3 - ~1. 3 . 5 + ...
+ (- 1)" 1.2.3 ... n.l.;n.5 ... (2n- 1) + · · .}
and
z(x, l) =
2
(1. 37 )
aoX!-{1 -- ~~ + ~ i~ 4
2
+ {-- 1)n 3 5
2·2· · + · · ·
2n +
~n
. ... - - - .2.4.6 ... 2n
22
2
+ · · ·}·
(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
11
EXAMPLES
3 5 7
+1 1
z·z·z
... 2n-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
2 2nn ! '
zn zn .n !
2 .4 . 6 . . . 2n = znn !.
and
Using these results we see that equation (1.37) reduces to
(1.42)
6 ( n!{(2n)~/(2nn!)}
n
oo
z(x, 0) = Oo
(1.4-l)
-J)n
~
(2x)n
= ao f:o (--1 )n (2n) l
(1.43)
while equation (1.38) becomes
6
<1J
z(x, i) = aoxl/2
-
1/2
-- OoX
(-1)n {(2n + 1)!i7z2nn!)}2nnf
~ (-l)n
~
(2x)n
(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
(2n)!
~0
(2n + 1)!
6
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
d 2y
dy
X
x- dx 2 -1 \i.~ -- 1 xy oc 0.
9
12
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
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.
L a,.x•+n
00
z =
Set
n=O
d
00
__!. = "" a,.(s + n)xs+n- 1
dx
~
so that
n=O
~
2
d z
dx 2 = ~ a,.(s + n)(s + n - I)xnn- 2•
and
tt=O
Hence,
d 2z
dz
(x-x 2 ) -2 + (1-x)-- z
dx
dx
L a,.(s+n)(s+n-1)xs+n-t- L a,.(s+n)(s+n-l)xs+n
00
=
00
n=O
n=O
00
00
n=O
n=O
00
+ L a,.(s+n)xs+n- 1 - L a,.(s-t-n)xs+n- L a,.xs+n
n=O
L a..(s+n)(s+n-l)xs+n- L a,._ (s+n-1)(s+n-2)xs+n-1
+ L a,.(s+n)xs+n- L a,._ (s+n-1)xs+n-l - L a,._ xs+n00
=
00
1
n=O
-
1
n=l
00
00
1
n=O
-
00
1
n=l
1
1•
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
13
EXAMPLES
and
(1.48)
Eqqation (1.47) gives the indicia! equation
s 2 =0
while equation (1.48) gives the recurrence relation
(1.49)
{(s + n - 1) 2 + 1}
a" = a,._l
(s + n)2
'
(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
(s 2 + 1)
al = ao(s + 1)z'
{(s + 1) 2 + 1}
{s 2 + 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 -f- ,6_~ {s +1} {(s-1-1)2+1}
... {(s+n-1) +1} n]
(s+ 1)2(s+2)2 , , , (s+n)Z
x .
2
2
z x, s - aox
(1.51)
The first solution is then
""'_,x,O)-ao
_ [t + .L.~ 1{12+1}{22+1}p {3zz2+1} ... {(n-1)2+1} J
_ [t ~ 1.2.5.10.17(n!)2
... {(n-1)2+1} n]
- ao + .L.
x
n= I
.
.
32 • • • n 2
X"
n-1
=
a0y 1(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
0
(!!x-)[t
+ ~ {s2+l}{(s+1)2+1} ... {(s+n-l)2+1}xn]
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•
(1.54)
= lnx.
[ix•]
ds s~o
and hence
en. 1
(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
dfn(s) = fn(s) -d lnj,(s).
s
s
SO that
(1.56)
But
Infn(s) =In {s 2 + 1} +In {(s + 1) 2 + 1} + ... +In {(s + n -1) 2 + 1}
-In (s + 1) 2 -In (s + 2) 2 • • • -In (s + n) 2
n
=
n
,2 In {(s + m - 1) + 1} - ,2 In (s + m)
2
•
m=l
m~l
Hence
2
i lnfn(s) = ~ ~~_m__=-_!)__ - ~ -~-­
f;; s + m - 1) + 1
ds
1
=
(
2 ~{
2
s+m-1
f:-; s + m
1
1}
L... (s + m - 1) 2 + 1 - s + m
m~l
so that
d
d/n(s)
2{s 2 + 1} ... {(s + n- 1) 2 + 1] ~ {' s + m --- 1
_ _1__ ~
2
2
2
(s + 1) • • • (s + n)
,f;:1 (s + m -- 1) + 1 s -t- mJ
and
t We recall that x - c 1" •, wherl' we arc using the notation In x
·loge x.
§ 1.2
15
EXAMPLES
Denoting this expression by c.., we have
_2.1.2.5.10.17 ... {(n-1) 2 +l}Ln
m-2
-----------"--'---'----~
2
(nl) 2
m=l m{(m- 1) + 1}"
Cn -
(1.57)
Thus, setting s = 0 in equation (1.53) gives the second solution
[:/(x,s)l=o
_
I [t
~ {1}{12 + 1}{2 + 1} ... {(n- 1)2 + l}xn]
- ao x
+ ;S_
12.22 ... n2
2
n
L
00
=
a0y 1(x) In X + a0
CnXn
n=l
a0 y 2(x), say,
where
=
L cnxn.
00
y 2(x) = y 1(x) In x +
(1.58)
n=l
Then the general solution is given by
y = Ay1(x) + By 2(x)
where A and B are arbitrary constants.
Example 3
d2y
dy
x 2 - 2 - 3x- + (3 - x)y = 0.
dx
dx
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.
2: a.,xs+n leads in the same manner as in the previous
00
Writing z =
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
(1.59)
(1.60)
a,.(s I n
1)(s I 11
3)
a,. 1 · · 0 (n ·. ' 1).
16
CH. 1
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
Equation (1.59) gives the indicial equation
(s- 1)(s - 3) = 0
(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
a,.=· ----~=.1 - - - - (n > 1)
(s 1- n - l)(s + n - 3)
(1.62)
so that we have
al
ao
az = -·---··- = - · - - - - - - - (s + 1)(s - 1)
s(s + 1)(s - 2)(s- 1)
and in general
a .
---- ·-·· ao
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
-~·--
1
z(x, s) = aoX•{ +
so that
~1 s(s+1) ... (s+n-1)(;~2)(s-1) ... (s+n -3)}
(s - 1)tr(x, s) = aoX•{(s - 1) +
~
+
s~s
=it
xn
}
£= s(s + 1) ... (s + n- 1)(s - 2)s(s + 1) ... (s + n - 3) ·
2
(1.64)
The first solution is given by
[(s - 1)z(x, s)]s~I
=
aox ~
,:S 1.2 ... n( -1).1.2 ... (n - 2)
oo
=
ao
--:----:--:-x_"'c-::----:----::-:-
xn+l
L - ~!(n- 2)!
n~2
= a0y 1(x), say.
The second solution is given by [(d/ds){(s -·- 1)z(x, s)}].=l·
§ 1.2
17
EXAMPLES
From equation (1.64) we have
d
ds{(s - l)z(x, s)}
d ) {
= ao ( d/s
zt
(s- 1)
(s- 1) + s(s --
~
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
+~~-
n~
1
}.
ds s(s + 1) ... (s + n- 1)(s- 2)s(s + 1) ... (s + n- 3)
(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)
18
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
1
[fn(s)]s=l = -1 . 2 . . .. n . ( -1 ) . 1 . 2 . . . (n - 2)
and
n!(n - 2)!
so that
J
2
J
1
{ 2 n- -1 - 1 __)_. -1- + -1}
df s
=
[ -ds n() •=1
n! (n-2)!
m
' n - 1 n
1
= Cm say.
It is also easy to show that
__ 1
[~(s-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)y (x) + a x{l - x + i cnxn}
=
1
0
n=2
= a0y 2(x), say,
and the general solution is then
y = Ay 1 (x) + By 2(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.
2 anxn and requiring z to be a solution of the given
00
Writing z = x•
n=O
equation leads, as before, to the system of equations:
a0(s - 2)(s - 1) = 0
a 1(s-1)s=0
1) +a,.. 2(s t n 2) = 0
an(s + n - 2)(s + n
(1.66)
(1.67)
(n > 2). (1.68)
§ 1.2
19
EXAMPLES
Equation (1.66) gives the indicia! equation
(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)
(1.69)
s+n- 1) n "'-"'
(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
an =
-
(
an-2
an= --- ~
n
(n > 2).
(1.70)
Thus
and in general
a2n =
ao
(-1)n 2.4.6 ... 2n
at
a 2n+1 = ( -J)n 3.5.7 ... (2n + 1)'
and
so that we have
x[ao{l + %-1)n 2 .4~ ~. 2n}
2
z(x, 1) =
(
1
~
-1-- al L
( -1)n
n=O
X2n+1
That is, we have the general solution given by
y = Ay1(x) + By 2(x)
where
~
x2n+l
Yt(x) = x + ,6 (-t)n 2.4.6 ... 2n'
V 2 (x) =
~
~
L ( -1 )n
n~o
and A and n arc arbitrary constants.
~,J•-('
x2n+2
J
1.3.5 ... (2n + 1) •
-----:-:::-----:-:
1.3.5 ... (2n+l)
20
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
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
dx
dy
dx
(x 2 - 1) -2 + 3x-- + xy = 0.
Rewriting this in the form
d 2y
3x 2 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) = (3x 2)/(x 2 - 1) and
r(x) = x 3 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)
a 1 (s + 1)s = 0
(1.72)
aoS(s + 2) - a 2(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)
Equation (1.71) gives the indicia! equation
s(s - 1) = 0
with roots s = 0 and s = 1. Equation (1.72) then shows that a 1 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
a 0 and a 1 as arbitrary constants.
21
PROBLEMS
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 ( ::;;, )
(1.75)
n- 3 .
n(n- 1)
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.
a" =
Thus
= iao + !a11
a4 =
a 1 + 8a 2
12
= l2a1,
aa =
a 2 + 15a 3
20
and so on as far as we please.
We have calculated the series solution up to terms in x 5 :
y = a 0 + a1 x + (ia0 + ia 1)x 3 + 1~a1x
+ {!a0 + ~a 1 )x 5 + ...
= a 0 {1 + ix 3 + kx 5 + ... }
+ a1 {x + !x3 + 1\x4 + ix5 + ... }.
4
PROBLEMS
( 1) Obtain solutions of the following differential equations in ascending
powers of x, stating for what values of x the series are convergent:
.
d 2y
dy
(•) 4xdx2 + dx- y = 0;
(ii) xdd~ + (x + l)ddy + y = 0;
.~
... ) d2y
( 111 .\' ·2
dx
X
·
2dy
d--x -1- y = 0;
22
SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
CH. 1
d 2y
dy
(iv) 9x(l - x)dx 2 - 12dx + 4-y = 0;
(v) (1 - x 2)
:~ + 2x~ + y = 0;
d 2y
dy
(vi) x 2 dx 2 + (x 2 - 3x)dx + (4 - 2x)y--: 0;
2
(vii) (1 - x 2)d y + xdy- y = O·
dx 2
dx
'
(viii) x 2 :~ -
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;
0
... )
d2y
dy
(111 x2 dx2 + x2 dx + y = .
(3) Obtain solutions of the following equations in terms of ascending
powers of lfx:
d 2y
dy
(i) 2x 2(x- l)dx 2 + x(3x + l)dx- 2y = 0;
2
(ii) 2x 3 dd y2 + (2x + x 2)ddy + 2y = 0.
x
x
2
GAMMA AND BET A FUNCTIONS
2.1
DEFINITIONS
We define the gamma and beta functions respectively by:
J;
J:
r(x) =
B(x, y) =
e-ttx- 1 dt
t'"- 1 (1 - t)Y- 1 dt.
(2.1)
(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
f:
fX-l dt
+
= [~t'"J: +
J:
I'( X) :::=
f:
e-ttX- 1 dt
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
CH.2
2.2 PROPERTIES OF THE BETA AND GAMMA FUNCTIONS
Theorem 2.1
r(t) = 1.
PRooF
We have, from the definition (2.1),
r(l) =
=
s:
J:
c- 1t 1 - 1 dt
e- 1 dt
Theorem 2.2
r(x + 1) = xi'(.~). (x > 0)
PROOF
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- dt
1
(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)
= x!r(l)
=X!
(by theorem 2.1 ).
Theorem2.4
l'(x) = 2
J:
e- 1't 2x-t dt.
PROOF
In definition (2.1) substitute u 2 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
e-u'u2x-t du
= 2
s:
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
I
n/'1.
COS 2"'-l
0 sin 2Y-l e de =
o
rc )I'(y)
X
2r(x + y)'
PROOF
We prove this result by considering in two different ways the double
integral
I =
fIR exp ( - t
2
-
u 2 )t 2 "'- 1 u 211 - 1 dt du
(where R is the first quadrant of the tu-plane, shown unshaded in Fig.
2.1).
26
GAMMA AND BETA FUNCTIONS
CH.2
II
FIG. 2.1
The tu-plane
Firstly, we have
I=
I"" I exp ( - t
oo
1=0
=I:
2
U=O
e-t't2x-: dt.
-
I:
u 2 )t 2"'- 1u 211 - 1 dt du
e-u'u2!1-l du
= !r(x) .!r(y)
(using theorem 2.4)
= il'(x)r(y).
(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
I
I
R
exp ( -r 2 cos 2 () -- r 2 sin 2 8)(r cos 8) 2"'-1(r sin 8) 2v-1r dr d(J
oo In/2 e-r'y2"'-l cos2a:-1 () r2Y-l sin 2v-1 () r dr d()
r~O
IJ=O
oo e-''r 2<"'+vl-l dr
211-
0
0
= ir(x + y)
I"/2 cos 2"'- 1 0 sin 1 () d(J
rr/2
I
cos 2"'- 1 () sin2v-1 () d()
(2.4)
0
(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
f
n/2
o
de = r(!)r(i)
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
B(x, y) = ~(x)r(y).
r(x + y)
PROOF
Substitute t = cos 2 0 in the definition (2.2) of B(x,y): t = cos 2 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
n/2
~ 2
J
0
cos 2"'- 1 0 sin 2V- 1 0 dO
28
GAMMA AND BETA FUNCTIONS
= 2 r(x)r(y)
2r(x + y)
(by theorem 2.5)
r(x)r(y)
r(x + y)"
-·
.
Theorem 2.8
B(x, y) = B(y, x).
PROOF
This result follows immediately from theorem 2. 7.
Theorem 2.9
(i) B(x + 1,y) = _x_B(x,y).
X
+y
(ii) B(x,y + 1) = _Y_B(x,y).
X
+y
PROOF
(i) We have
B(x
1 ) = l'(x + l)I'(y~
r(x + 1 + y)
+ ,y
(by theorem 2.7)
xr(x)r(y)
(x + y)r(x + y)
(by theorem 2.2)
r(x)r(y)
x + y r(x + y)
X
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
yn f(x)f(x + i).
f(2x) =
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))
J
l
=
1
1
1
-12"'-P -+- s)"'-12"'-1 (1 - s)"'-I 2 ds
(on making the substitution t = i(l + s))
= -1-
220:-1
= _2_
220:-1
=
Jl
Jl ~
(1 - s 2)x-l ds
-1
(1
s2):r:-1 ds
0
2-2:r:+2 J~ (1 - u)'"-1iu-1!2 du
(on making the substitution u = s 2)
= 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),
2-20:+1
r(x)
f(x
+ !) yn
f(2x)
and thus
220:-1
1'(2x) - - - - l'(x)l'(x
yn
j ~).
30
CH.2
GAMMA AND BETA FUNCTIONS
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).
(2.5)
X
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
31
DEFINITION OF THE GAMMA FUNCTION
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
1
r( -1) = -r(O) = oo
--1
and
1
r(-2) = _ r(-1) = oo,
2
etc.
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
CH.2
GAMMA AND BET A FUNCTIONS
Theorem 2.12
f(x)f(l - x) =
1t
---;-.
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
""
(-lt
0
- nn'
n=-«>
<XJ
sin 0 - L
----·
(iii) We shall use result (ii) to prove the required result.
(i) We know that
. 0 = 2 sm
. 0 cos 0
sm
2
2
0 . z z0)
(n
.
= 2 sm
2 sm +
(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(~+~+~)}
_
. 0 . n + () . 2n + () . 3n + ()
- 23 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
Sin
Sin
Sin
Sin
23
23
23
23
23
Stn
. 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
33
DEFINITION OF THE GAMMA FUNCTION
.
. (). n+O. 2n+8
p
p
p
. (p--1)n+8
p
sm () = 2P- 1 sm- sm -~ sm - - - ... sm
(2.9)
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
+ 0 =sin
~; 8
-;)n
{n- 2}
= sm (n;
Similarly, the second last factor is
sin (p - ~n + ()
sin {n -- -' 2
-(_n_p----'-O)}
8
).
=
= 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
.
. 0{ . n + 0 . n - 0} { . 2n + () . 2n - 01
s1n 0 = 2v-t s1n
s1n -P- sm
sm
sm -p--)· ..
P
{
P-
P
sin (!P- l)n + 0 sin (-~p- l)n- 0} sin !Pn +
p
p
p
9. (2.10)
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 =
s1n
2 sm p s1n pn - s1n. p0} {sm. 2np - sm. p0} ...
P-
1
•
() {
•
2
2
2
2
- l):n: - sin 2 ~}cos q. (2.11)
{ sin2 (iP 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
34
CH.2
GAMMA AND BETA FUNCTIONS
() /p
.
sin()
. sin ()
1l f f i - - - = 1l f f i - - - - - p
o_..osin((Jjp)
e_..0 () 'sin(Ojp)"
=p,
we obtain
n . 2n
. (kp - 1)n
p = 2 P- 1 sm 2 -sm 2 -- ... sm 2 -"----.
•
p
p
(2.12)
p
If we now divide equation (2.11) by equation (2.12), we obtain
2
sin 0 =sin~ { 1 _ sin (Ojp)} { 1 __sin2 (Ojp)}
p
p
sin 2 (n/p)
sin 2 (2n/p) · · ·
1- ___ sin;(o~~) -}cosq. (2.13)
• 2 (?.P
1)n
p
sm - - - {
p
\Vc now let p ~ co. We take into account the three results
. sin (Ojp)
.
. 0
(a) 11m p sm - = 1tm - - - - . 0
p--+a:>
p
P--+00
()
IP
=0,
2
(b) lim sin (0/p) = lirn{sin(O/p)}2(~)2{ 1·njp
2
P--+a:> sin (m/p)
p--+oo
Ojp
p
sin (m/p)
= 02 jr2a2,
}z ___1 __ _
(c) lim cos~ = 1
p--+00
p
and see that therefore equation (2.13) gives
( 02)(1 - -
sin 0 = 0 1 -- n2
()2 ) ( 1 -
22n2
-()2- ) ...
rznz
= 0 [J 1 -- nz,_;2 .
00
(
()2 )
n=l
(ii) It is trivial to verify the result that
1
d
()
-·--- = - - In tan-.
sin 0
dO
2
Hence
_1__ = i_ ln sin (0 j2)
sin 0
dO cos (0 j2)
= _'!_ ln _ 2 sin 2 (0/2) __
dO 2 sin (0 j2) cos (0 /2)
d
2 sin 2 (0/2)
ln ------dO
sin 0
(ra/p) 2
p.3
35
DEFINITION OF THE GAMMA FUNCTION
= : {In2 -t-Insin 2 (0/2) -Insin8}
8
=
:a{2 In sin (8 j2) - In sin 8}
d { 2 InTI
w
co
= -dO
0 ( 1 - - f)2 ) - Inn
e( 1 - n2;rr;2
- f)2 )}
4n2;rr;2
n~l
=
£6{
n~l
~ In ( 1 - 2~n) + 2
2 In o + 2
n~l
i
In ( 1 +
n~l
2~J
(1 -- :J- i i n (1 + :J}
d{
dO In 0 + 2 L In (2nn - 0) + 2 L In (2nn + 8)
- ~/n (n;?;- 8)- ~/n (nn +e)}
1
-1
1
-1
1
o+ 2 :L 2nn - o + 2 :L 2nn + e - :L nn -a - :L nn + o
10 + {w
2
L 0 - 12nn - L 0 -1}
nn
-Ina- i i n
n=l
=
co
w
n~l
n~l
w
=
n=l
n=l
00
00
00
n=l
n~l
n~I
00
=
n~l
n~l
+{2
=
=
w
1
n~l
~ + ~~~1~: +,~;~1~:
~ ~~1;l:.
n=- oo
(iii) Let us first assume that 0 < x < 1.
Then
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)).
~F-U
w
1
}
L 0 + 2nn - L a-=+=- nn
n~l
36
CH.2
GAMMA AND BETA FUNCTIONS
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.
J:( + C:J-"'(- (C~u)2du)
1
u)"'- 1
Then r(x)r(l-x) =
1
ro u-"'
--du
0 1 + u
l
u-"'
Joo u-"'
--du +
--- du.
J
=
=
J o1-j-u
t1-j-u
In the second integral we make the substitution u = 1jv. Then
du = ( -1jv 2) dv; also when u = 1, v = 1 and when u = w, v = 0.
Thus
u-"'
fo v"' ( 1 )
f oo f+~
du = t1 + (1/"-') -- v2 dv
1
v"'-1
f
=
f +
t
--dv
01 + v
=
1 u"'-1
--du.
0 1
1l
Hence, from equation (2.14) we have
r(x)r(l - x) =
J(u-"'1 ++
1
0
J
u
(u-"'
+ u"'-
0
=
L (-l)"un du
00
l
=
1
u"'- ) du
1
)
n=O
~ ( -1)"[---1 __ ---un-x-J-t + ___1__ un+"']l
~
n-x-t-1
n=O
n-j-x
o
00
=
1
1 }
-1---2 (-1)" n--x-t-1
n-j-x
{
n=O
(remembering that 0 < x < 1)
§2.4
37
EXAMPLES
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)
= (N + y- 1)(N -t- y -- 2) ... yf(y)
1
1 -y -N·2
1
1
-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
as required.
2.4 EXAMPLES
Example 1
Express each of the following integrals in terms of gamma or beta functions
and simplify where possible.
n/2
(i)
f
v(tan 0) dO.
(ii)
0
f:
flo a.y(l-xa)'
dx
fl
(~)1/2
dx.
1 -X
(i) We usc theorem 2.5 where, in order to get v(tan 8)=sin 112 0 cos- 112 0
as integrand, we take 2x - 1 = -- ~ and 2y -- 1 = -H, i.e. x = ;]; and
y
~-
(iii)
'-C.O
t- 31:'(1 -- e- 1) dt.
(iv)
-1
38
Thus
CH.2
GAMMA AND BETA FUNCTIONS
f
n/2
v(tan fJ) de=
r(l)r(3)
4
4
2r(! + !)
=
ir(t)r(!)
0
(by theorem 2.1)
1
1t
=----
2 sin (n/4)
(by theorem 2.12)
1 3t
= 21;.y2
= n/y2.
(ii)
I: 3 y(~~x 3) J:
=
(1 - x3)-1/3 dx.
We change the variable tot = x 3 • When x = 0, t = 0 and when x = 1,
t = 1. Also dt = 3x 2 dx, i.e., dx = }t- 2 ' 3 dt.
Hence
fl
dx
o 3 y(1-x 3 )
=
fl (1 - t)-II3.}t-2/3 dt
o
- 3IB(.l3• 32)
-
(by definition (2.2))
(by theorems 2.7 and 2.8)
1
n
3 sin (n/3)
(by theorem 2 .12)
1
n
= 3 (v3)/2
2n
3(,Y3)"
(iii)
f~
t- 3 ' 2(1 -
e-t) dt.
We cannot relate this immediately to the gamma function; we must first
integrate by parts:
I"'
()
2t- 112 .e-t dt
§2.4
39
EXAMPLES
= 0+2
=
J:
t- 112 e-t dt
t
2r(!)
(by definition (2.1))
=2yn,
(by theorem 2.6).
(iv)
J (!_-=+.-_:)1/2
t
-1
1-
dx.
.lC
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
dt
fl (!1 -+~)1/2 dx = Jl (11 +--- 2t2t +- 1)1/22
1
= 2 Jl (-t)1/2 dt
1 - t
• -1
0
X
0
=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!
= :n;
(by theorems 2.2 and 2.3)
(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)
g'(a)
t--.. a g(t)
provided f'(a) and g'(a) are not both zero (or infinite). Hence we have
1
- =
(J.)
0.
40
GAMMA AND BETA !'UNCTIONS
CH. ~
Example 2
2
Show that B(n, n + 1) = ~ {~~;~ and hence deduce that
n/2 (
Jo
1
1
)1/4
---sin 3 0 sin 2 0
cosO dO=
{r(l)}2
·
Zy'11:
r(n)r(n + 1)
B(n, n + 1) = -r(2n + 1)
We have
(by theorem 2.7)
f(n).nf(n)
2nr(2n)
(by theorem 2.2)
{f(n)} 2
= 2f(2~)'
Setting n = l gives
1
:;
-
B(4, 4> -
{f(!)}2
2rU) ,
which, on using definition (2.2) and theorem 2.6 becomes
J
l t-3/4(1 - t)114 dt = {f(!)}2.
0
2y':n
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
l
J
t- 3 ' 4(1 - t)1' 4 dt =
fn/'1.
0
0
=
n/2
fo
(sin 0)- 314(1 --sin 0) 1 ' 4 cos 0 dO
(1 -sin 0)1/4 cos() dO
sin 3 0
tt/2 (
1
1 )1/4
=
~ - -~
cosO dO,
o sm 0
sm 0
so that we have proved the required result.
J
PROBLEMS
(1) Prove that
J e-a"'x" dx = an- 1 l'(u + 1)
00
(i)
0
H
(n>
l,a::..-0);
41
PROBLEMS
(iii)
J:
(m > -l,n >0).
exp (2ax- x 2) dx =
i vn exp (a
2
).
tt/2
(2) Prove that f 0 tann 0 dO = tr{(l + n)/2} r{(l - n)/2} if In I < 1.
(3) Prove that
f
n/2
•
smn 0 dO =
Jn/2 cosn 0 dO = 2vn r{(
r{(1 + n)/2}
+ n)/ }
2
2
(4) Express each of the following integrals in terms of the gamma or beta
functions and simplify when possible:
0
J: G-1/'
J:
J:
0
4
(i)
(iii)
(iv)
(ii)
J: (ln~y-
1
dx, <a> o);
(b - x)m- 1(x - a)n- 1 dx, (b >a, m > 0, n > 0);
xm(1 - xn)P dx, (m > -l,p > -1, n > 0);
1
(v)
dx;
J
dx
- xn)'
0 v(l
(n > O);
(vi)
foo
0
(5) Show that the area enclosed by the curve x
(6) Evaluate r( -t) and r( -f).
(7) Show that
-n
(i) r(x)r( -x) =
.
;
xsmnx
(ii) r(l2 + x)r(l-x) = _n_,
2
cosnx
4
dt
( vt)(1
+ t)'
+ y = 1 is {r(±)F j2yn.
4
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,
2
(1 - x2)d y
- 2xdy + l(l + 1)y = 0
dx 2
dx
'
(3.2)
is of the form of equation (1.2) with q(x) = -2x 2 j(l - x 2) 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 x 2 < 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.
L a,.x" and requiring z to be a solution of equation
00
Writing z(x, s) = x 8
n=O
(3.2) leads, as in Chapter 1, to the system of equations
au-f(s- 1) = 0,
(3.3)
a1(s + l)s =0
(3.4)
§ 3.1
43
LEGENDRE'S EQUATION AND ITS SOLUTIONS
and
a,+2(s + n + 2)(s + n + 1) - an{(s + n)(s + n + 1) - l(l + 1)}
= 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 a 1 ;
this means that a 1 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
(3.6)
+-z) -
which, since
n(n + 1) - l(l + 1) = n 2 + n - [ 2 - l
= (n 2 - 12) + (n - l)
= (n - /)( n + l) + (n - l)
= (n - l)(n + l + 1),
may be written in the form
(l- n)(l + n + 1)
an+2 = -an (n + 1)(n + 2) .
Thus we have
l(l + 1)
a=
-a
-1-.2- ·
2
0
a4 =
aa = -al
(l - 2)(! + 3)
= ao
4
1.2.3.4
( l -- 1)(I + 2)
2.3
_,
(l- 3)(1 + 4)
a5 = -aa
4.5
-a2--3~ ----
l(l- 2)(! + 1)(/ + 3)
(3.7)
'
(/- 1)(!- 3)(! + 2)(! + 4)
2.3.4.5
= al--
and in general
a2n = ( -1)nao
l(l-2)(!-4) ... (l-2n+2)(1+1)(1+3) ... (l+2n-1)
(2n)!
(3.8)
and
1)nao
(I 1)(I 3) . . . (l 2n 1- 1)( l- /-2 )( l / 4) . . . (Z-1- 2n)
(2n -/- 1)!
a2n +1 ,-~ (
(3.9)
44
CH. 3
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
n+1}
(l-1)(l-3L. (l-2n+l)(l±2)(l1_:_'!:_)_._:~_S~!:-2n) x 2
(2n + 1)!
= aoy1(x) + a 1y 2(x), say,
so that y 1(x) and y 2{x) are the linearly independent solutions.
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 a2 n :;;: 0 but a 2n+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 a 2n+a and all subsequent odd coefficients are zero. Thus we see that if l
is an even integer y 1(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 x 1, 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
1
first term is a1x and subsequent terms are given by equation (3.7) written
in the form
(n + 2)(n + 1)
(3.10)
an = -an+2 (l- n)(l + n + 1)'
so that
l(l- 1)
al-2 = -az 2(21 - 1)'
(l- 2)(/- 3)
4(2! _ 3)
(l- 1)(!- 2)(!- 3)
= az 2.4.(2!- 1)(2!- 3)
al-4 = -az-2
and, in general,
- ( ),
al-2r- - 1 az
l(l-1)(l-2) ... (l-2r+l)
· (3 .11)
2.4 ... 2r(2l- 1)(2/- 3) ... (2/- 2r + 1)
Thus for l even, y 1(x) reduces to the following expression, while for
l odd, y 2(x) reduces to the same expression:
y( x ) = a x 1 + a
1
x
1-2
1_ 2
+a
x
1_ 4
{a
+ ... + a10xforf orl zeven
odd
1_ 4
!Hl
=
L al-2,:~--2·
r~o
{it
_
for l even
1
(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]
=az-..;;:: (-1)r
;S
l
l
)
(l-1) ... ( -2r+l
, -xl-2r
2.4 ... 2r(2l-1)(2l-3) ... (2/-2r 1 1)
(3.13)
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
1 2 3
r
zrr!
zr
0
0
0
0
0
0
(3.15)
46
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
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)!
(3.16)
2'(21 - 2r) !l!'
Using equations (3.14), (3.15) and (3.16) in equation (3.13) gives us
,
_
y(x)- az
6 (11121
~
1
= al r~ ( -
1!
1 2'(2! - 2r) !l! _ ,
1>"(1- 2r}!.2•r!,_(21)!(l- r)!x1 2
(1!) 2{21- 2r)!
_ 2,
Y r!(1- 2r)!(1- r)!(2l)!xl .
1
This is a solution for any value of a1• If we choose a 1 = (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
v(l - 2tx + t2) =
6t~Pt(x) if I I
00
t
< 1 and I X I < 1.
This means that when (1 - 2tx + t 2) - 112 is expanded in powers oft, the coefficient of t 1 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):
2 "Cp(2x)'-P( -t)P
T
(2x - t)" =
p=O
where "CP is the binomial coefficient I( r~ )'
p. r p.
Hence we now have
(1 - 2tx + t 2) - 1 ' 2 =
"_r_.
oo
(2)1
r
"""C (-1)Ptr+P(2x)'-P.
622'(r!)2~o P
(3.18)
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_!__ rC
22r(r!)2
( -1 )l-r(2x)'-(l-r)
1-r
and the total coefficient of t 1 is obtained by summing over all appropriate
values of r so that we obtain
l
coefficient of t
1
=
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)!
Lt
1)~:(2 )l-2k
l-kc (
z2i-2k{C1 - k) !}2
!/2 (l even)
k
x
-
k~ { (Z-1)/2 (!odd)
~ (2l - 2k)!
(l - k)! ( 1)k
2
6o
Z
k)f}2 {l- 2k}!k! -
l-2k l-2k
21 - 2"{(/-
=
X
(where [1/2] is as defined above in equation (3.12))
~:
lt/21
2 <-- 1>
k=O
=
(21 - 2k) I
21(1- k)l(1-
2k)lklx~
-21:
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~! !,(x
2
1
-- 1) •
PROOF
We may expand (x 2 - 1) 1 by the binomial theorem:
l
(x 2 -1)1 =
_2: C,(--1)'x -•>.
1
21
<
r=O
Hence,
d' ( 2
1
2Hf dx' x
)! - 1 dl
-- 1 - 211! d.xl
*'
:S c, -- 1)•x
l
(
2!-2r
·
But the lth derivative of a power of x less than l is zero, so that
1
d 1....-~ 21 · 2 '' -._, () t'f 21 - 2r < I, I.e.,
.
dx
1'f r > lj"
....
(3.19)
§ 3.3
FURTHER EXPRESSIONS FOR LEGENDRE POLYNOMIALS
Thus we may replace
l
l/2
(l-1)/2
r=O
r=O
r=O
49
L by L if l is even and by L if l is odd, i.e.,
[l/21
by
L where [l /2] is as defined above.
r=O
Also,
dl
dxlX 11 = p(p - 1)(p - 2) ... (p - l + 1)xP- 1
= - -p!- x P -1
(p - l)!
dl 21-2r _ (2! - 2r)! ,1-2r
dxlx
- (i- 2r)! .'t
so that
and hence, substituting into equation (3.19), we have
1 d1
1 11121
[I
(21-- 2 )I
2
1
)z
""
•
( 1)•
r . z- ~r
(
1
1
1
ZJ! dx X = 2 /l f:t, r!(/- r)! -- (1- 2r)! X
-6
-
[l/21
<-
r
(21 - 2r)!
1-~r
1 1
) 2 r!(t- r) !(l- 2r)Y
= P1(x)
(by the definition (3.17)).
Theorem 3.3 (Laplace's Integral Representation)
P1(x) = -1
n
J" {x + v(x
o
2
-
1) cos 0}1 dO.
PROOF
It may be shown by elementary methods (for example by means of the
standard substitution t = tan (0 /2)) that
dO
n
320
o 1 +A. cos e =
A- 2).
< )
"
J
vet -
If we now write A.= -{uy(x
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(x 2 - 1)
1( 1 _ ux) cos 0
= (1 - ux)[1 - u{x + v(x 2 - 1) cos 0}] - l
2 -
------~--~---
L u {x + v(x
'XJ
= (1
- ux)
1
1=0
2
-
1) cos 0}1
50
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
c1 - a)- = L: an).
00
(since, by the binomial theorem
1
n=O
1
1
v(l - ). 2) =
_·-u2-(x-2---1)...,.....}
(1 - ux) 2
(1 - ux)
v{(l - ux) 2 - u 2(x 2 -
J-{1
1)}
_
(1 - ux)
.
y1(1 - 2ux + u 2)
Hence, substituting into equation (3.20) gives
.
' u {x +
"
f6
1~ f: +
1
0
v(x 2 - 1) cos 0}1 dO= -----~--v(1 - 2ux + u2)
i.e.,
u
1
{x
6
CD
CO
1
y!(x 2 -
1) cos 0} dO = n
1
u P1(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.
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:
P 0(x) = 1,
P1(x) = x,
P 2(x) = l(3x 2 - - 1),
Pa(x) = i(5x 3 - 3x),
P 4(x) = !(35x4 - 30x 2 + 3).
(3.21)
Theorem 3.4
(i) P1(1) = 1.
(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
( v ) P 21 (0)=(-1)
t
51
(21) I
22 t(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
1
- - - - - - -2 - ""' t 1P1(1)
00
vet - 2t + t
6
) -
1
that is
1
'
L t P (1).
00
_ t =
1
1
z~o
L t (by the binomial theorem or by considering the
00
But 1/(1 - t) =
z~o
right-hand side as the sum to infinity of a geometric progression), so that
L = L tP (1). For this to be true for all values
00
we have
00
1
t
z~o
1
oft in some
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
d2
d
(1 - x 2) dx 2P1(x) - 2xdxP1(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
fP1(0).
) =
V (1 + t 2 z~o
Expanding the left-hand side by the binomial theorem gives us
1
- - - = (1 + t2)-112
+ t 2)
L
00
vet
=
1 + (-
~)t2 + (-~~1-i)(t2)2 -j-
52
LEGENDRE POLYNOMIALS AND FUNCTIONS
+ ... +
CH. 3
( -D( -!) ... { -(21 - 1)/2}( 2) 1
1!
t + ...
= ~ (- 1Y1.3 .5 .. • ;2z- 1)t
21
1
2 .f.
l=O
= ~ ( -1 y 1. 2 . ~.;. 5 ... (21 - 2)(21 - 1)2lt2l
21 . . 2.4-.6 .. , (21- 2)2/
l=O
=
~
l
(21)!
~
1
(21)!
6 (-1) zzn.ztz!t
21
2/
= ~ (-1) 22/(l!)/.
l=O
Thus we have
= ~ t'P(O)
6~ (-1)z j'!:_l)!_t2z
22/(1!)2
6
l
and equating coefficients of corresponding powers of t on both sides gives
l
( -1)
Pzt(O) =
(21)!
2zl(Z!)2
P2t+1(0) = 0.
3.5 ORTHOGONALITY PROPERTIES OF THE LEGENDRE POLYNOMIALS
Theorem 3.5
{
1
f
-1
P1(x)Pm(x) dx =
0
ifl~m
_2_ ifl =
m.
21 + 1 z
(This result may be written more concisely if we introduce the Kronecker delta,
defined by
<5
_
lm-
{01
if l ~m
if 1 = m.
Then the statement of the theorem is
r
I
Plx)P,.(x) dx' '2/
~- lr51m·)
§ 3.5
53
PROPERTIES OF LEGENDRE POLYNOMIALS
PROOF
P1(x) and Pm(x) satisfy Legendre's equation (3.2), so that we have
d 2P 1
dP1
(1 - x 2) d.~ 2 - 2x dx + l(l + 1)P1 = 0
2
dPm
)P
t
(1 - x 2) ddxPm
2 - 2x dx + m(m + 1 m = 0,
and
which may be rewritten in the form
d~{(l - X 2)~1} + /(/ + l)P = 0
(3.23)
d~{(l- x 2)~m} + m(m + l)Pm = 0.
(3.24)
1
and
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
1
f
[
-t
d { (1 - x 2 )
dP,}
PmP1 - d { (1 -- x 2)dPm}]
dx
dx
dx
dx
dx
+ {l(l + 1)- m(m + 1)} f~ 1 PzPm dx = 0.
But since
dPz}
dPm
dP1
d{
dPz}
d{
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
2 dP,
[ P m(l - x ) dx - P 1(1 -
2
X )
dPm]l
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.
54
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
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~ P1Pm dx = 0, and when l r'= m we can cancel
1
the factor outside the integral to obtain
e P Pmdx =0.
J 1
It remains to prove that J~ {P1(x)} 2 dx = 2/(2l + 1). For this we use
1
-1
the generating function of theorem 3.1 :
1
v(1 -
- 6""t t
2tx +
t 2)
1
{
(1 - 2tx + t2) =
so that
00
1
P (x)
-
6
co
tlPt(x)
co
00
=
}2
L t P (x). L tmPm(x)
L tl+mP (x)Pm(x).
l,m=O
1
1
l=O
m=O
00
=
1
We now integrate both sides with respect to x between -1 and +1:
1
1
f
(1
-1
+t
2) _
2tx
dx =
L tl+m J P (x)Pm(x) dx.
()()
1
l,m=O
-1
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
co
2
t21
{Pz(x)}2 dx
[ --2 In (1 + t - Ztx) - =
t
1
l=O
J -1
which gives
]1 L r1
1
()()
~ t 21
J_{P (x)} dx = - 21t
1
1
1
2
1
In (1 + t 2 - 2t) + t In (1 + t 2 + 2t)
2
1
1
Zt In (1 · t) 2 + 2t In (1 + t)z
§ 3.6
55
LEGENDRE SERIES
t1{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
f
l
_
as required.
2
{P1(x)}2 dx = 21 + l
1
3.6 LEGENDRE SERIES
Theorem 3.6
If f(x) is a polynomial of degree n, then
f(x) =
i
r=O
crP,(x) with cr = (r + !) f~ f(x)P,(x) dx.
1
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)
56
CH. 3
LEGENDRE POLYNOMIALS AND FUNCTIONS
and hence
f(x) = CnP..(x) + Cn_1 P .. _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)
+ c P (x)
0
0
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
f
2 c, J_P,(x)P,(x) dx.
n
l
f(x)P,(x) dx =
_
1
r=O
1
1
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).
2
l
Therefore
f
f(x)P.(x) dx = c, - - --,
2s + 1
-1
giving c, = (r + !) J~ f(x)P,(x), as required.
1
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
l f(x)P (x)
f
_
1
1
dx =
2 c, f1_ P,(x)P (x) dx
n
1
r=O
1
=0
by theorem 3.5, since r ..;;; n < l, so that r is never equal to l.
§ 3.6
57
LEGENDRE SERIES
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.
L c,P,(x), where
00
Then the series
r=O
c, = (r + !) f~ f(x)Pr(x) dx,
1
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(x 0 +)we mean Lim f(x 0 + 6) where 6---+ 0 through positive values and
a->()
by f(x 0 - ) we mean Lim f(x 0 - 6) where again 6---+ 0 through positive values.
See, for example, Fig.·~
3.1.
y
--?~
__________l./'.''"
I
I
Xo
FIG. 3.1
J(
Left- and right-hand limits at a point of
discontinuity
58
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
3.7 RELATIONS BETWEEN THE LEGENDRE POLYNOMIALS AND
THEm DERIVATIVES; RECURRENCE RELATIONS
Theorem 3.8
[l(l-1)]
2 (21 - 4r - 1)Pz-2r-l(.~).
(i) P;(x) =
r=O
;l: \
P1H(x) + 2.z : 1Pz-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) (x 2 - 1)P;(x) = lxP1(x) - lP1 _ 1(x).
(viii) (x 2 - 1)P;(x) = (1 + 1)P1 .1-1(x) ~- (l + 1)xP1(x).
(ii) xP1(x) =
l +1
2 (2k + l)Pk(x)Pk(y) = ;--=-··
{Pl+l(x)Pz(y) - P (x)Pl+l(y)}.
y
l
(ix)
1
k=O
•
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_ 1 P1 ~1(x) + cl-aPt-3(x)
with
, {c P (x) (l even)
1 1
+ ... + Cz-2r-1Pl-2r-l(x) + ... ~r CoPo(x)
(l odd) .
c. = (s + !) J~ P;(x)P.(x) ds
1
=
(s + !) {[P1 (x)P.(x)J~
1 - J~t1(x(P;(x) dx}
(on integrating by parts)
= (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
59
LEGENDRE POLYNOMIALS AND THEIR DERIVATIVES
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)P1_ 2,_ 1(x) +
···
+
{3P
(x) if lis even
P (x) if l is odd
1
0
(l(l-1)1
=
L (2!- 4r - 1)P
1 _ 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
·
{c1P 1(x) (l even)
xPz(x) = CzuPzu(x) + Cz_tPz-t(x) + · · · + cof' (x) (l odd),
0
where
c, = (r + !) J~ xP1(x)P,(x) dx
1
= (r + !) J~ P (x).{xP,(x)} dx.
1
1
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 1 0(/
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
l +1
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 + 1P 1 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){P 1(x)+xP{(x)} + ZP{_ 1(x) = 0.
Collecting terms in P{ _1 , P 1 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) = P 1(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
61
LEGENDRE POLYNOMIALS AND THEIR DERIVATIVES
When rearranged, this equation gives
P;(x) - xP{_ 1(x) = lP1_ 1(x),
which is the required result.
(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
x 2P{(x) - r;(x) = lxP1(x) -- lP1_ 1(x)
which may be rewritten in the form
(x 2 - 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
(x 2 - 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 ~ [{(2k + l)xPk(x)- kPk_ 1(x)}Pk(y)
1
- 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)
62
CH.3
LEGENDRE POLYNOMIALS AND FUNCTIONS
(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
f 0 = 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)P ,(y) + L fk_
=
k=O
=
1
k=O
k=O
l
l
k=O
k=l
l
l-1
k=O
k=O
L (2k + 1)(x - y)Pk(x)Pt(y) + z
1
fk_ 1
L (2k + 1)(x- y)Pk(x)Pk(y) + L fk.
Hence
I
k=O
1
fk -
~ fk =
i:
(2k + l)(x- y)Pk(x)Pk(y)
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 _
+ y}
1) {P +I(x)P (y) - P (X)Pz+I(y)} =
(x
1
1
1
*
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
2
(1 - x2)d y
- 2}Y + l(l + l)y = 0
dx 2
dx
2
2
then (1 - x )mf ( dmz/dx"') is a solution of the equation
d2y
(1 - xz) dxz -
dy
{
m2 }
2xdx + l(l + 1) - 1 - x2 Y = 0
(/mown as the associated Legendre equation).
§ 3.8
63
ASSOCIATED LEGENDRE FUNCTIONS
PROOF
Since z is a solution of Legendre's equation, we must have
d 2z
dz
(1 - x 2)dx 2 - 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
2
dxm
dx
dxm
dx
dxm
which, when we use Leibniz's theorem for the mth derivative of a product,t
becomes
dm+ 2
d
dm+lz
m(m - 1) d 2
dmz
(1 -- x 2) - - " ' + m-(1- x 2) . - - +
-(1- x 2 ) . dxmH
dx
dxm+I
2
dx 2
dxm
N
dmz
dm+Iz
d dmz}
-2{ xdxm+I + mdxx.dxm + l(l + 1)dxm = 0
(since higher derivatives of 1 - x 2 and x vanish).
Collecting terms in dm+2zjdxm+ 2 , dm+ 1 zjdxm+ 1 and dmzjdxm, we obtain
dmHz
dm+Iz
dmz
(1-x 2)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) dx 2 - 2(m + 1)x dx + {l(l + 1) - m(m + 1)}z1 = 0.
(3.33)
If we now write
dmz
z 2 = (1 _ x2)m12z1 = (1 _ x2)mf2 d.\~
equation (3.33) becomes
d2
d
(1- x 2)dx 2{z2(1- x 2)-m' 2} - 2(m + l)x<G:{z2(1- x2)-m/2}
+ {l(l + 1) - m(m + 1)}z2(1 - x 2)-m1 2 = 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
dz 2 - x2)-m12 + mz2x(1 - x2)- (mt2)-1
= -(1
dx
64
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
+ m{~:x(1 _ x2)-(m/2l-1 + z~(1 _ xz)-(m/2)-1
+ z2x(
-I_ 1)(1 -
x2)-<mt2>-2. -2x}
z
dz
dz
= ddx:(l
_ x2)-ml2 + -~mx(1 _ x2)-(m/2l-1 + m dx2x(l _ x2)-(m/2)-1
2
+ mz2(1 _ x2)-<mt2l-1 + mz2x2(m + 2)(1 _ x2)-(m/2l-2.
Hence equation (3.34) becomes
d 2.:r2
dz
dx2 (1 - x 2)-<mt 2)+1 + 2mx(1 - x 2)-m 12 d_-;-2 + mzl1 - x2)-mf2
+ m(m + 2)(1 - x2)-(m/2l-lx2zz
- 2(m + l)x{ (1 - x2)-ml2 d~
dx + mx(1 - xz) -<mt2l-1z2}
+ {l(l + 1)- m(m + 1)}z (1 - x )-m12 = 0
2
2
which, on cancelling a common factor of (1 - x 2)-m 12 and collecting like
terms, becomes
2
{
}dz2
2 d Z2
(1 - x)
dx 2 + 2mx-2(m + l)x dx
m(m+2)
}
2(m+l)mx2
+ { m + l-x 2 X 2 - - l-x 2 - + l(/+1) - m(m+l) z 2 = 0. (3.35)
The coefficient of dz 2 /dx is just -2x, while the coefficient of z 2 is
(m 2 +2m - 2m 2 - 2m)x 2
l(l + 1) +
+ m- ms- m
_ x2
1
m2x2
= l(l + 1) =-X2- m2
1
m2
= l(l + 1) - 1 - ,;2'
Thus equation (3 .35) reduces to
2
(1
2
dz2 -1 { l(l + 1)- -m- } z = 0
x 2 )d- -z-22 -· 2xdx
d.t'
1 -- x 2 2
§ 3.9
65
PROPERTIES OF ASSOCIATED LEGENDRE FUNCTIONS
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)m 12 dx"' P1(x)
(3.36)
satisfy the associated Legendre equation.
PROOF
This result follows immediately from the theorem, since P 1(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)!
(3.37)
Pz (x) - ( 1) (l + m)!P;"(x)
(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(,).
66
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
Theorem 3.11 (Orthogonality Relation)
f
pm( )Pm( ) d
l
l
-1
X
X
l'
2(! + m)!
~
X=(2f+l)(l-m)!ull'•
PROOF
We first prove that if l ~ l'
1
f
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
f
l {Pm( )}2 d 1
-1
X
X-(2[
2(! + m)!
+ 1)(/- m)!'
Assume first that m > 0; then from definition (3.36) we have
J~ 1 {Pf"(x)}2dx
= f~ (1 - x 2)m{:x:P1(x)} {::nPt(x)} dx
1
= [ {:;.-~ Pt(x)}
1
{c1 - x )m :mP (x)}J::
2
1
- f~ 1 {!mm-~1Pt(x)} !J(l = -ft
Pl(x)}-~{(1
x 2)m :mP1(x)} dx
(on integrating by parts)
{-dm-1
-1
dxm- 1
dx
- x2)m dm Pl(x)} dx
dxm
(3.38)
(the first term vanishing at upper and lower limits because of the (1 - X 2)
factor).
Now, from equation (3.33) with m replaced by m - 1 we have that
dm+l
dm
dm-l
(l-x 2)dxm+1 P 1(x) - 2mxdxmP1(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+ 1P 1(x) - 2mx(l - x 2)m-l dxmP1(x)
dm-1
+ (l + m)(l - - m -1- 1)(1 - x 2)"'- 1 -dxm-1
-- --P,(x) = 0
§ 3.9
67
PROPERTIES OF ASSOCIATED LEGENDRE FUNCTIONS
and this equation may be rewritten in the form
d{
dm
}
dm-1
dx (1 - x 2)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
=
f_
l
1
{dm-1
}
dxm_ 1 P1(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'- (x)}2 dx,
2
and repeating the process m times in all we obtain
J~ 1 {P["(x)}2 dx
=
1
(l+m)(l+m-1) ... (l+1).(1-m+1)(1-m+2) ... l. J~ {P?(x)}2 dx
= (l+m)(1+m-1) ... (1-+·1).1(1-1) ... (1--m+2)(1-m+1).
(using theorem 3.5)
(l + m)! 2
(l- m)f 21 + 1
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
2
21
+1
68
LEGENDRE POLYNOMIALS AND FUNCTIONS
(1-- n)!}2
{ (l + n}!
=
f_ {Pi(x)}
CH. 3
1
1
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
which is the required result.
Theorem 3.12 (Recurrence relations)
(i) Pj+ 1(x) -
v(~": x )Pj(x) + {1(1+1)-m(m-1)}Pj- 1(x) = 0.
2
(ii) (21 + 1)xPj(x) = (1 + m)P[:_ 1 (x) + (l- m + 1)P4 1 (x).
(iii) v(1 - x 2)Pj(x) =
(iv) v(1 - x 2 )Pf'(x) =
21
~ l {P4i 1(x) - PF:-i1(x)}.
~ {(l + m)(l + m- l)P[:_11(x)
21
1
- (l- m + 1)(1- m + 2)Pi,:.1 1(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 = P 1(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 - x 2)Plm+2>(x) - 2(m + 1)xP1<m+I>(x)
+ {l(/1 1) -- m(m I l)}Pfm>(x) = 0
§ 3.9
PROPERTIES OF ASSOCIATED LEGENDRE FUNCTIONS
69
which, on multiplying throughout by (1 - x 2)m1 2, gives
(1 - x2)(m/2l+1pfm+2>(x) - 2(m + 1)x(1 - x2)m12pfm+I>(x)
+ {l(l + 1) - m(m + 1)}(1 - x 2)mi 2Pfm>(x) = 0.
Hence, using equation (3.39), we have
2
1
Pj+ (x)- 2(m + 1)x v'(l ~ x 2)Pj+ (x)
+ {l(l + 1)- m(m + l)}Pj(x) = 0
which, when m is replaced by m - 1, becomes
2mx
Pj+ 1(x) - v'( _ x 2{i(x) + {l(l + 1) - (m- 1)m}Pj- 1 (x) = 0;
1
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
+ l)P (x)
Pf~ 1 (x) - Pf~ 1 (x) = (21
which when differentiated m - 1 times gives
Pf~>1 (x) - Pf~i(x) = (21
1
+ 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)ml 2 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)
70
LEGENDRE POLYNOMIALS AND FUNCTIONS
which, on using equation (3.39), becomes
JH_ 1(x)- Pj':_ 1(x) = (21 + 1)v(1 - x 2)Pf'- 1 (x).
Replacing m by m + 1 gives
1
p~~ 1 (x) - Pj':_~ (x) = (21 + 1)v(1 - x 2)P;r'(x)
which, when divided by 21 + 1, is just the required result.
CH.3
(3.42)
(iv) We use (ii) to replace xP;n(x) in (i) by
~ 1{(! + m)Pj':_ 1(x) + (l- m -1- 1)P~ 1 (x)}
21
so that we obtain
1
2
Pj+ (x) - v(1 :
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
2
P;n+ 1(x) - v( ~ x 2) ( : 1){(1 + m)Pj':_ 1(x) + (l - m + 1)Pl';- 1(x)}
21
1
+ {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
= + {(I+ m)(l + m + 1)P~ 1 (x) -(I- m)(l- m + 1)Pt'~_ 1 (x)}
21 1
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
P 1(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
71
Theorem 3.13
A second independent solution of Legendre's equation is given by
Q 1(x) =
1 +x
!P1(x) In l _ x -
[l--;1]
~
(21 - 4r - 1)
;S (2r + 1)(1- rtl-2r-I(x)
(l > 1)
1+x
Q0(x) = ! In _ x
1
~
1
I _ 1]
{/
if I is odd
where [- - =
2
-l-2ifi'
- t ts even.
2
Q 1(x) is called the Legendre function of the second kind.
PRoopt
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'
d y = Pz(x)d z + 2 ~ dP1 + zd P 1
dx 2
dx 2
dx dx
dx 2
and hence the equation becomes
d 2z
dz dP
d 2P
dz
(1 - x 2 )P1(x)dx 2 + 2(1 - x 2~ dxz + (1 - x 2 )z dx 21 - 2xP1(x)dx
2
2
2
dPz
- 2xz dx + l(l + 1)zP1(x) = 0.
Collecting terms in z, dzjdx and d 2z/dx 2 , we have
d~1
dP
z { (1 - x 2) dx 2 - 2x dxz + l(l + 1)P1(x)
}
2
dz{
dP1
}
d z _
+ dx
2(1 - x 2 ) dx - 2xP1(x) + (1 - x 2)P1(x) dx 2 - 0,
which, on using the fact that P1 satisfies Legendre's equation, becomes
2
d z
dz{
dP1
}
(1 - x 2 )P1(x) dx 2 + dx 2(1 - x 2) dx - 2xP1(x) = 0.
t The proof of this theorem may be omitted on a first reading. The proof ends
on page 77.
72
LEGENDRE POLYNOMIALS AND FUNCTIONS
2x _
d zjdx
---+ 2dPzldx
- -- - 0
dzjdx
PL(x)
1- x
2
Hence
CH.3
2
2 -
'
and this is equivalent to
(dz)
d
d
d
dx
In dx + 2d; In PL(x) + ;r; In (1 - x 2) = 0,
which when integrated gives
dz
In dx +In {PL(x)}2 +In (1 - x 2) =constant.
dz
Therefore dx{P1(x)}2(1 - x 2) =constant= A, say,
so that
dz
A
and hence
This means that we have a solution of Legendre's equation given by
dx
(3.43)
Qz(x) = Pz(x) {Pz(x)}2(1 - x2)"
J
We must now show that this is of the form stated in the theorem. We
first dispose of the case I = 0:
dx
Qo(x) = Po(x) {Po(x)}2(1 - x2)
J
=
=
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) = k 1(x - cx 1)(x - cx 2} ••• (x - cx1).
Thus
§ 3.10
LEGENDRE FUNCTIONS OF THE SECOND KIND
73
on splitting into partial fractions.
We may determine a 0 , b0 and cr fairly simply. Multiplying both sides of
equation (3.44) by (1 - x 2 ){P1(x)}2 gives
1 = a 0(1 + x){P1(x)}2 + b0(1 - x){P1(x)}2
+ (1 - x 2){P1(x)}2{ 2
( ~ ot,)+ (X ~r ot,) 2}.
r=l X
Setting x = 1 in this equation and remembering that P 1(1) = 1 gives
!, while setting x = - 1 and remembering that P1( -1) = ( -1 )1
gives b0 = t·
We now show that
a0 =
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 P 1(x) = (x - o:;)L(x) we have
Ct---,
[!X (i____ )){/,(~)}21
ex;
74
LEGENDRE POLYNOMIALS AND FUNCTIONS
J
=
=
CH. 3
2x
2L'(x)
[ (1 - x 2){L(x)} 2 - (1 - x 2){L(x)} 3 x="i
[2xL(x) - 2(1 - x 2)L'(x)]
(1 - x 2){L(x)}S
x=<x,
2{ex;L(ex 1) - (1 - ex; 2)L'(ex 1)}
(1 - ex 12) 2{L(ex 1)}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- cx 1)L(x)}
+ l(l + 1)(x - ex;)L(x) = 0,
which, on performing the differentiations, becomes
(1 - x 2){(x - cx 1)L"(x) + 2L'(x)} - 2x{(x - IX;)L'(x) + L(x)}
+ /(/ + 1)(x - cx 1}L(x) = 0,
and setting x = ex1 in this equation gives
(1 - cxD2L'(cx 1) - 2cx 1L(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
(1 - x 2 ){P1(x)} 2 = 2(1 - x) + 2(1 + x) +
*£-:
dr
(x - ex.)2
where the d, are constants whose values will not concern us.
Thus
1
l
d
(l _ x 2){P (x)}2dx = -!In (1 - x) + ! In (1 + x) (x __:_ cxr)
f
6
1
l
=
t In 1 + X _ ""'"'
1-
X
dr 1
r~ (x - 1Xr)
so that from equation (3.43) we have
! + : - 6 :~x~,
l
Q1(x) = !P1(x) In
d,
.
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 {P (x)}/(x- ex,) is a poly1
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
Q 1(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
d (x) ln fl+x}
zld{
d-; (1 - x )dxP
_ x + l(l + 1). P (x) In l+x
_ x
1
2
1
2
1
1
d {(1 - x 2)dXdW,_t} - l(l + t)w;_1 = 0. (3.47)
- dx
But
d
1 +x
It +x
{
1 }
1
-d P1(x) In - = P1 (x) In - + P1(x) -1 -+x + 1-x
-1 -x
1 -x
x
1
1 +x
2
+ P1(x)1- x
1 -x 2
I
= P1(x) In - -
so that
- x )Pl(x) In~ ~; + 2Plx)}
= d~{(l
2
d {(1 - x 2) PI( x)} + (1 - x 2)PI(x) 2_ x + 2PI(x).
= In 11 +
_ xX dx
1
1
1
2
1
Hence equation (3.47) becomes
! ln.~ ~: [a~{(l
- x )P((x)} + l(l + l)P (x)J + 2P;(x)
2
1
dlf'i_t} -1(1 + t)w;_l = 0
- dxd {(1 - x2)dX
which, on remembering that P1(x) satisfies Legendre's equation, reduces
to
_d
dx
{<1
dW1-1} -f /(1 I 1)w;_ = 2dx
dP,
·
x·) --d~
q
1
(3.48)
76
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH.3
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)
(3.49)
r=O
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)]
(3.50)
= LarPl-2r-l(x)
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)]
= 22(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
77
LEGENDRE FUNCTIONS OF THE SECOND KIND
But -(! - 2r - 1)(/ - 2r) + l(l + 1)
= -(/- 2r) 2 + (!- 2r) + l(l --+- 1)
= -1 2 + 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
a
'
21- 4r- 1
= -,-----,-,-,---
(3.53)
(l-r)(2r+1)
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
+ x, Q1(x)
1 -X
is infinite at both x = ± 1, whereas we know that P 1(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
Q1(x) =
x
2
tn
1+x
1
_ x - 1
Q 2(x) = :!(3x 2 -1) In
1 +x 3
_ x- x
2
1
Q3(x) = :!(Sx3 - 3x) ln
1 +X 5 2 2
_ x - x + 3·
1
2
We now state without proof several theorems concerning Legendre
functions of the second kind.
Theorem 3.14
1
-- =
x- Y
if ."C > 1 and I y I < 1.
L +
00
1~o
(21
1)P1(x)Qz(y)
78
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
Theorem 3.15 (Neumann,s Formula)
Q,(x) = !
.
2
JI P,(y)y
dy.
-lX-
Theorem 3.16
The results contained in theorem 3.8 (ii- ix) remain true when P 1(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'
(3.54)
sin() o() Stn () ()0) + sin2 () ocp2 + /(l + 1)'J' = O,
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
2
(jj(cf>){d (.
de)}
e(e) d (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 (),
2
sin () d ( . 0de)
. 20_ 0
e d() Stn d() + (j)1 ddcp2(j) -j- I( I -j- 1) Stn
which, when rearranged, becomes
sin 0 d ( .
de)
.
1 d 2(j)
- - -- sm 0 - -!-l(l + 1) sm 2 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 m 2 •
Thus we have
e
sin() dO
d ( Sill
. 0de)
. () = m2
dO + l(l + l) sm2
1 d 1:/>
(3.55)
2
and
-if> dcp 2 = m 2 •
(3.56)
d 2(/J - - 2(/>
dcfo2- m
(3.57)
Equation (3.56) is just
while equation (3.55) simplifies to
si!
e:a (sine~~) + {z(l + 1) - si:: ~e = o.
(3.58)
Equation (3.57) has the general solution
(/> = Aeim<f>
+ Be-im<f>
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
1 d
d
and hence
sin() d()
dx
and
. ()
d
(1
d
. () d
sm
d() = sm2 . - dx = - x2)dx'
Accordingly, equation (3.58) becomes
:Jet -x )~~} + {z(l +
2
2
1) - 1 : x2}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'--, A 1c1m4>p•;(cos 0) 1- A 2c-im<f>P-j(cos 0)
80
CH.3
LEGENDRE POLYNOMIALS AND FUNCTIONS
where
A 1 =A
and
A 2 =(-l}m(l+m)!B
(l- m)! ·
If now we denote
y/(0, c/>) = eim.f>Pi(cos 0)
(3.59)
we may write the general solution in the form
T = AIJi(O, c/>) + A2Yzm(O, cp).
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)
y 1 (0,cf>)
2 (/+m)!
_
m
1
- ( -1) v(2n}
J{(2l +2(/1)(/+ m)!- m}!}e Pj(cos 0). (3.61)
im.f>
For then we have
J:" J:
dcp
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
r
4(1 + m}! (l' + m'}!
I
Pi'(x)Pp'(x) dx
mm
§ 3.11
81
SPHERICAL HARMONICS
(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
=(-1)mv(2n)
J{2l-2-(l+m)!_e
+ 1 m)!} -'~.Pi'(cose)
{l-
(by equation (3.61))
= ( -l)m _1_J{2/
v(Zn)·
+ 1 (/- m)!}e-'~( -1)-m
2
(l + m)!
(l+m)!p me
(I _ m)! £ cos ()
)
(by equation (3.37))
__
m ( - )-m
-- ( 1) .
1
_1_J{2l
±_1 (/
+ m)!} P (cos 0)
v( 2n)
2 (/ _ m)! e
-imc/> -m
1
= ( J)mYI m(O, c/>)
(by equation (3.61)).
82
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH.3
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:
0Yo-
2
Y±
=
2
yg =
J(
1 )·'
4n
J(~)
sin 2 ee±zi.p y±t
=
32n
' 2
JC~n)(3
-J(Sn15) sine cos ee±i<l>'
cos 2 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
10
3
1o'
5
FIG. 3.3
10
X
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
FIG. 3.6
Pi'(x), 0 < x < 1, l = 1, 2, 3
Q,(x), 0 , , x...; 1, l - 0, 1, 2, 3, 4
§ 3.13
85
EXAMPLES
q (x)
X
Qt(x), x > 1, l = 0, 1, 2, 3, 4 (Note that the scale
on the vertical axis is logarithmic.)
FIG. 3.7
3.13 EXAMPLES
Example 1
Show that
2
2l( l + 1)
_ 1 x Pl+ 1(x)Pl-l(x) dx - (412 _ 1)(Zl + 3)"
l
f
Deduce the value of
J:
x 2P1+I(x)P1_ 1(x) dx.
We use theorem 3.8(ii) to dispose of the .Y 2 appearing in the integrand,
and we may then use the orthonormality property of theorem 3.5.
Thus,
f~ P,
1X 2
I
L(x)P,_ L(x) dx - .
r
I {xP, il(x) }{xP, L(x)} dx
86
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH.
3
(by theorem 3.5)
21(1 + 1)
(41 2 - 1)(21 + 3)'
We know that P 1(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
1
J x P (x)P (x) dx
1
f
x 2Pw(x)PH(x) dx = 2
J:
P 1(x) dx when I is odd.
-1
2
1 +1
1_ 1
0
and hence
Example 2
Evaluate
J:
By theorem 3.8 (iv) we have
P 1(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
87
EXAMPLES
1
{
( -l)(l-1)/2
+
2 z-1
(l + 1)!
( -1)(1+1)/2
= 21 + 1 1 - 1 - -21+ 1-
[{(l--t=-T)}i}!T2
(1- 1)!
}
[{(l- 1)/2}!)2
(by theorem 3.4(i) and (v), remembering that l + 1 and I-- 1
are both even)
J
( -1 )(l- 1)/ 2
(l -- 1)!
[
( -1 )( l + 1)/
-21 + 1 21- 1 [ { ( / - 1)/2}!)2 1 - 2 2{(1 + 1)/2}2
1
1
(l- 1)!
( -1)<t-t);z
[
I
J
1
= 27 + 1 --21_:_~[{(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
Example 3
2 c,.P,.(x).
00
ljf(x) =I x lfor -1 < x < 1, expandf(x) in the form
r=O
By theorem 3. 7 we know that such an expansion is possible and that
c,. is given by
c,. = (r + !)
JII
1
X
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~
I xI P,.(x) dx = 0
1
and hence c,. = 0. On the other hand, if r is even, I x I P,.(x) is even, and
hence
L
J~ 1 I xl P,.(x) dx = 2 I xI P,.(x) dx
= 2
J:
xP,.(x) dx
so that now
JxP,.(:~) dx.
1
c,
(2r !1)
()
(3.63)
88
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH. 3
To evaluate this integral we use theorem 3.8(ii):
c,. = (2r + 1)
=
J: +
{(r
J: {;,:
\Pr+l(x) + ,
2
~ 1P,._ (x)} dx
1
1)Pr+l(x) + rP,._ 1(x)} dx.
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)!
+ 1)(r- 1) _ r}
{(r
2'- 1(ir)!(!r -1)!
r +2
( -1 )"' 2(r - 2)! {' 2 - 1 - r 2 - 2r}
= zr-l(!r)!(~r- 1)!
2({r + 1)
~=JXrl 2l+l(2r -t- 1)(r - 2)!
2r(!r + 1)!(!r- 1)! --- .
(3.64)
Thus we have
f( ) = ~ ~-=1)n+ (<1_~- + 1)(2n- 2)!p ( )
X
22n(n + 1)!(n- 1)!
2n X.
1
;S
Example 4
If x > 1, show that P,(x) < P 1+l(x).
We prove this by the method of induction: assume first that
P 1 _ 1(x) < P 1(x) and use this to prove that P 1(x) < P 1+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 P 1(x) > 0 for all values of l.
Now, by theorem 3.8(iii) we have
(l + t/~
1
1
-
(21 + 1)x
1-/!;; =-~ 0.
§ 3.13
EXAMPLES
89
Thus
Pl+1
(21 + 1)x
I P,_ 1
P, = z - 0 - 1 + 1 P,
(21+1)
I
> 1+1 -1+1
(since x > 1 and we have
assumed that P1_tfP1 < 1)
I+ 1
l +1
=1.
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{Q 1(x)P1 _ 1(x) - Q 1 _ 1(x)P1(x)}
= (l -- 1){Q1 _ 1(x)P1 _ 2(x) - Q 1_ 2(x)P1 -t(x)}
and deduce that I{Q 1(x)P1 _ 1(x) - Q 1_ 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 P 1 _ 1(x) and
subtracting gives
I{P,(x)Q 1 _ 1(x) - Q 1(x)P1 _ 1(x)}
+ (!- l){P,_2(x)Qz-t(x) - Q,_ 2(x)P1-t(x)} = 0,
which is equivalent to
l{Q 1(x)P1 _ 1(x) - Q 1 _ 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{Q 1(x)P1-1(x) - Q 1 _ 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)
90
LEGENDRE POLYNOMIALS AND FUNCTIONS
CH.
so that 1{Qt(x)P1 _ 1 (x) - QH(x)P1(x)} = Q1 (x)P0 (x) - Q 0 (x)P1 (x).
P 0(x) = 1,
P 1(x) = x,
But we know that
- I I l+x
Q()
oX - 2 nl~-,
-x
so that we obtain
= -1,
which is the required result.
PROBLEMS
21
1
J_ xP (x)P _I(x) dx = 1·
21(1 + 1)
(2) Show that _ (1 - x 2)P (x)Pm(x) dx
Z/ +
J
1
(1) Show that
1
1
1
1
1
=
1
1
(3) If D
4/2~
1
Otm-
= ~' 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)!
(0 < m < 1)
and hence deduce that P1 -rn(x) =
(4) Iff(x) =
{*-t
(l- m)!
( -1)m (1-=t=-m)lP;"(x).
(O < x < 1)
(-1 <X< 0)
2 crPr(x).
00
expand f(x) in the form
r= 0
!(1 - x2)ml2
) ,.
2 Pn+m(x)t - 2 (2m)
l(
1
2 +
oo
(5) Show that
r>.=O
m
n -
rnm.
-
tX
(2 rn ' '
91
PROBLEMS
1
(6) If u,. = J~ 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+ (x)}.
1
r=O
(9) Assuming the result of theorem 3.14, prove theorem 3.15.
'
1
(10) Show that P1(x)Q 1' (x) - P,(x)Q,(x)
= ~~~x2·
~
(11) Show that YfW, cl>)
Je ! !)bmo·
1
4
(12) If x > 1, prove that
1
Q,(x) = 2,f+ i
fl (x(l -- t2)l dt
-1
t)l+l
and deduce that
.
.
. .
e8y(X -j-- 1)- y(X- 1)
(1) by makmg the substttutwn t = 0 (
l)
e y X+
+ .y 1( X-· 1)
we obtain
Q,(x) =
f~ +
{x
v(x 2
d~ 1) cosh 0} +1;
1
2 ~ (l + r)!(l +2r)! 1
(11.. ) Q,(x) = ,Xz+i
~ r!(2l + 2r + 1)! ,X"2r·
1
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 n 2 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 - n 2, we may apply the methods of Chapter 1 and be assured that
any series solution obtained will be convergent for all values of x.
2: a,xs+r and requiring z to be a solution of equation
<X>
Setting z(x, s) =
r=O
(4.1) leads, by the same method as in previous cases, to the system of
equations
{s(s- 1) + s - n 2 }a0 = 0,
(4.2)
{(s + l)s + (s + 1) - n 2}a1 = 0
and
(4.3)
{(s + r)(s + r - 1) + (s + r) - n 2 }a, + a,_ 2 = 0
(r > 2).
2 -
2
(4.4)
Equation (4.2) gives the indicia! equation s
n = 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 s 2 = n 2 , we cannot also have (s + 1) 2 = n 2t, so that
(s + 1) 2 - n 2 ~0 and hence a 1 = 0.
Equation (4.4) is just
{(s + r) 2 - n 2 }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)
(r > 2).
(4.5)
Thus we have
ao
02
= -
Z(2n-~I--Z) "-'
ao
-- 22.-t(~- + 1)'
a2
a2
------a
4 -------4(2n + 4)22.2(n + 2)
a4
06
24 .2!(n + l)(n + i)'
a4
= - 6(2n + 6) = - 2 2 • 3(n + 3)
- Z6 .3!(n + l)(n + 2)(n +-3)
and in general
ao
02
r
= ( -l)" Z2rr!(n + l)(n + 2) ... (n + r)"
-j- The only situation in which we can have s 2 = (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
CH.4
BESSEL FUNCTIONS
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
a 0 r(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( n+ 1)
x2r+n
2 2rr!r(n + r + 1)
.
;
This is a solution for any value of a 0 let us choose a 0 = 1I {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:
"'a(-1)r
,-61 °
],.(x) =
~
1
(x)2r+n
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
co
1
(x)2r-n
f-n(x) = ~ (-lY---~~~--(;o
r!r(-n +r + 1) 2
from equation (4.7).
But
-n + 1" + 1) is infinite (and hence 1I {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
rc-
But
] ..(x) =
.~ <--J)m in!r(n - f~ m-+ l)-Grm+n
(by equation (4.6))
96
BESSEL FUNCTIONS
CH.4
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
(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
]p(x) = ( -1)-vj -p(x)
( --l)P]p(x) =] _p(x)
or
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
],.(x)
and
Y,.(x) = cos nn],.~x) - ] -n(x)
stnnn
for all values of n.
PROOF
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
(4.8)
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)],,=,..
(4.9)
n
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
x2~2 of. + xi_ of. + (x2 - v2) of. - 21'.] = 0.
(4.10)
dx ov
dx ov
ov
•
Also, of course,] -.(x) satisfies Bessel's equation of order v. so that we
have in exactly the same way
2
X
d2 of_.+ d of_,+( 2
dx 2 Tv
X dx Tv
X -
V
2)oJ_
OV
-
z.r
-o •
VJ ---. -
(4.11)
t L'Hopital's rule states that if f(a) = g(a) = 0, then
, f(x)
f<•>(a)
hm-=-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.
x-..a g(x)
98
CH.4
BESSEL FUNCTIONS
Multiplying equation (4.11) by ( -1 Y and subtracting from equation
(4.10) gives
x 2 -~ {aJ. _ (_ 1y of-·} + x-~ {of.:: _ (_ 1y of-·}
dx 2 ov
ov
dx ov
av
+ (x
2
-
v2){~·- ( -1)" a~;}- 2v{J. -- ( -1YJ -.} =0.
Setting v = n in this last equation and using equation (4.9) gives
X
2
d2
d~2 Yn(x)
d .
+ xdx Yn(x) + (x 2 - 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 )nJn(x) to obtain
X
2
d2
d
dx 2 Yn(x) + Xdx Y ..(x)
+ (x
2
-
n 2) 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)
2{
X
1 1}
n
Yn(x) = - ln- + Y-- """- fn(x)
z:Sr
2
:n;
1 '""'
6 -;+;·~~~~
(x)n+2s ·' {1
1
-;:;:~(-l)·'s!(n+s)! 2
1 }
__ ! ~ ~-=-_:~' _ 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.
[a
y _,.(x) =;1 a/·(x) - ( -1)--n
a -.(x)J
a/
•=-n
a
a
J
=;1 [ a( -vl-·(x)( -l)-n o( -vl·(x) v=n
a
a
J
= -1 [ --1 -.(x) + ( -1)+"-1.(x)
:n;
av
OV
[aav
v=n
a1 __,(x)
= ( -1)"-1 - J.(x)- ( -l)n:n;
av
J
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
oft and show that the coeffi-
We expand exp {~x(t -~)}in powers
cient of tn is1,.(x):
exp
Hx(t- ~)} exp Gxt). exp ( -~ ~)
=
=
i <t;:t i (
>'
~I
U
0
H
()
~xjt)•
s!
100
CH.4
BESSEL FUNCTIONS
=
~
(-})'xrtr( -1)s(Dsxst-s
L.
r! s!
r, s=O
_ ~
-
_
L_. (
s(~)r+s xr+str-s
1) 2
r, s= 0
r.I s.I
.
(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
2r-n x2r-n
( - 1y-n 2
~!(r - n)r
(1)
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
~ (- 1)'-n 2
~
(xj2)2p+n
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
2r-n x2r-n
( - 1)'-n 2
r!(r- n)!
(1)
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
f:o (-- 1)'-n 2
~
(xj2)2r-n
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
101
INTEGRAL REPRESENTATIONS FOR BESSEL FUNCTIONS
4.3 INTEGRAL REPRESENTATIONS FOR BESSEL FUNCTIONS
Theorem4.6
] ..(x) = -1
n
f"
o
cos (ncfo ·· x sin cfo) dcfo
(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 -
!t = c
1<1>
-
c _;q, = 2i sin cfo
this equation becomes
L {ei"<i> + (-l)"e-J"<f>}J..(x).
00
ei.<sin<f> =Jo(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
L
= ] (x) + L
eixsin 4> = j 0 (x) +
00
2 cos
0
L
2kcp] 2 ~c(x) + L
2 cos ncfo]n(x) +
n even
2i sin ncfo]..(x)
n odd
00
i
2 sin (2k -
1)cfo] 2 ~c_ 1 (x).
k=l
k=l
Equating real and imaginary parts of this equation gives
L
co
cos (x sin cfo) = ] 0(x) +
2 cos 2kcfo]2k(x)
(4.13)
L 2sin(2k -1)cfo] (x).
(4.14)
k=l
00
sin(xsincfo) =
2k_ 1
ko J
If we multiply both sides of equation (4.13) by cos nrp (n > 0), both sides
102
BESSEL FUNCTIONS
CH.4
of equation (4.14) by sin ncp (n > 1), integrate from 0 to :n; and use the
identities
(m ~n)
(m = n ~0)
(m = n = 0)
(m ~n)
and
(m = n ~0),
we obtain the results
"cos cos sm A.) d-1.. = {nfn(x) (n even)
0
(n odd)
Jo
{0 (n even)
". . .
J sm ncp sm (x sm cp) dcp = J n(x) (n odd).
A.
n'f'
and
(
x
•
'f'
'f'
:n;
0
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.
J:
Hence
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)
But
J:
(where m is positive).
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
=(-1)m
=
J:
103
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 +i)
1n(X) = y(n)r(n
t 2 )n-!eixt dt
-1
(n > - 1.).
~
PROOF
Consider the integral I defined by
I=
Jl (1 - t2)n-Jeixt dt
-1
=
J~l (1 - t t-} r~ (i:?' dt
2
J
r!
= ~ (ixt
~
r~o
1
(1 - t 2 )"-!tr dt.
-1
J
1
Now, if r is odd, the integrand in
(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 )n-!tr dt = J~l {1 - t )n-lt dt
2
2
J:
J:
= 2
=
28
{1 - t 2)n-lt 28 dt
(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
CH.4
BESSEL FUNCTIONS
r(n + t)r(s + t)
r(n + s + 1)
(by theorem 2.7).
Thus
I = ~ (ix) 28 _!'(n + l)r(s + l)
(2s)! r(n + s + 1)
;S
= r(n + i)
~
6 (-
x 2•
1
(2s)!
1)• (2s)! r(n + s -1=--i) 2 2•s! yn
(using the corollary to theorem
2.10)
--
'
(:)-n ~~ r((-1)•(x/2)2•+:
1)
.l
- l (n + 2)( yn) 2
=
I
n +s +
s~o
s.
r(n + i)( yn)G) -n]..(x)
(by equation (4.6)).
Thus ]n(x) = (
=
yn)r~n + l) (~)"1
(:)n Jl (1 - t2)"-!eil·t dt.
1
( yn)r(n + i) 2
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
-1
•
§4.4
105
RECURRENCE RELATIONS
PROOF
(i) From equation (4.6) we have
so that
d
d ~
dx {xn]..(x)} = dx
1
r6 (-1)' r!r(n +1 r + 1) 22r+nx2r+2n
1
co
= ""' ( -1)'(2r + 2n)x2r+2n-I
L.
r!l'(n + r + 1)2 2r-tn
r~o
1
co
xn ""' ( -1)r
:S
=
r !(n + r)r(n + r)22r+n
2(n + r)x2r+(n-1)
(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)
(by equation (4.6)).
(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 2
dx{x-tj,.(x)} = dx f:6 (- 1)' r!f(n + r + 1) 2 2r+nx r
co
= ""'(-1)'
£:6
co
_ ""'(-1)r
- ;S
1
1
--2rx2r-1
r!r(n + r + 1) 2 2r+n
1
1
_
rx2r-1
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
2
(s+1)!r(n+s+2) 2 <•+1)+n-l
- f:b
(where we have sets = r
1)
106
CH.4
BESSEL FUNCTIONS
"'
- ~ ( -1)•+1
-6
~
1
1
--x2s+l
2
s!r(n + s + 2) 2 s-tn+l
1
(x)2s+n+l
6 (- 1)" s!r(n + s + 2) 2
=
-x-n
=
-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
Slll 1m
X
d
= --.-[cosmrd-{xn1,.(x)} - dx{x"] _,.(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
= ---:·--(-- -) x"[ -cos (n- 1)n1,._1(x) + 1 -(n-n(x)]
-s1n n- 1 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 ,(x)
H,.C 2>(x) = 1..(x)
i Y..(x).
1
108
BESSEL FUNCTIONS
CH.4
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~2 l(x), gives
d
1
d)x"H~ l(x)} = xnH~I2 1 (x)
(taking the plus sign)
and
d
dx{x"H~ l(x)} = x"H~2}_ 1 (x)
2
(taking the minus sign).
4.6 EQUATIONS REDUCIBLE TO BESSEL'S EQUATION
Theorem 4.11
The general solution of
d 2y
dy
x 2 - 2 + x-- + (A. 2x 2 - n 2)y = 0
dx
dx
is
A]n(Ax) + B Yn(J.x).
§ 4.6
EQUATIONS REDUCIBLE TO BESSEL'S EQUATION
109
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)
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
dx
dx
x2 _ 2 + (1 _ 2o:)x- + {fJ2y2x2Y + (o:2 _ n2y2)}y = 0
Ax":Jn(fJxY) + BXXYn(fJxY).
is
PROOF
We first make the change of variable y = xrxz.
dy
dz
- = xrx - + o:xrx-Iz
dx
dx
d 2y _ rx d 2z
_ 1 dz
dx2- x dx2
2o:XX dx
Then
+
and
so that the given equation becomes
2
XX+2 d z2 2o:XX+l dz
o:( o: - 1)XXz
dx
dx
+
+
_
+ o:(o: - l)XX 2z
+ (1 - 2o:){xrx-I dz + o:xrxz}
dx
+ {fJ2y2x2Y + (o:2 _ n2y2)}XXz = 0
which, on collecting terms in d 2zjdx 2, dzjdx and z becomes
d~
x 2 --
dx2
dz
1- {Zo:x + (1 - 2o:)x}dx
+ {o:(o: - 1) 1- (1 -- 2o:)o:
110
BESSEL FUNCTIONS
CH.4
d 2z
dz
x2 dx2 + x dx + {fJ2y2x2l' - n2y2}z = 0.
(4.15)
which simplifies to
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
2
d z
= y(y _ l)xl'-2 dz
_ + yxv-l,yxY-1_
2
dt
dt
= y2x2l'-2 d__z2 + y(y _ 1)xY-2 _dz
2
dt
dt
so that equation (4.15) becomes _
d 2z
dz
dz
y2x2Y _ + y(y _ l)xY _ + yxY ~- + {P2y2x2Y _ n2y2}z = 0
ili 2
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
dx
d
dx
x 2 ~2 + x_z - (x 2 + n 2 )y = 0.
(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
= i-" ~ ( ~ 1 y r!r(n+-;-+1)
(" )2r+n
I
=i-n ~ (-- rff(n ~ + 1t'i"GYr+n
1)'
T
~
1
(x)2r+n
= ~ ~Tt'(n + r + 1) 2
·
r=O
(4.1 7)
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 )njnI,.(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 n (X )
=~I -n(x) - In(x)
(4.18)
.
•
smnn
When n is non-integral, this is well defined, and /,.(.v) and K,.(x) together
2
112
CH.4
BESSEL FUNCTIONS
provide independent solutions of equation (4.16). When n is integral,
Kn(x) is indeterminate as it stands and has to be defined by
_ . n I -.(x) - I.(x)
K nX-lm.
( ) 1
~'-'>n 2
Slll vn
_ n [(ojov)L.(x)- (ojov)I.(x)].=n
- Zn[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
---
Kn(x) = ~in+l{jn(ix) + i Yn(ix)} = ~in+IH~,'l(ix).
PROOF
(by definition (4 .18))
=
n jnj -n(ix) - j-n]n(ix)
s1nnn
2
But from the definition of Yn(x) in theorem 4.2 we have that
] _,.(ix) =cos nnJ,.(ix) -sin nn Yn(ix).
Hence
K ..(x) = "!_ jn cos nn]..(ix) - i-~J.. (ix)- in sin nn Yn(ix)
2
smnn
=
2
>}
~·n+l{·
y (" } -i cos nn
(" .
1
1 ,.lX +
. - i - n-lJn1X
2
smnn
But
-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
113
= -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
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)
X
1
and this equation, on cancelling a factor of i 2" - 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~1 l(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
-~{inxnH(ll(ix)}
= i"xnJI(l)
(ix)
n
n-1
•
d(ix)
Thus
so that
and hence
d
dx{xnK..(x)} = - xnK.. _1(x).
(ii) Similarly H~1 l(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
2.
( )} =--1-nx-"-1-n·-1'1.
.
2.
2ZT
( X).
1K ,.X
.- d -('1-nx- 11 --1-n7111
1
1 dx
n
n
:-.•·
I
ti6
CH.4
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).
4.9 INTEGRAL REPRESENTATIONS FOR THE MODIFIED BESSEL
FUNCTIONS
Theorem 4.17
(i) I ..(x) =-~ _ _ 2_ __ (:)"
( yn)r(n + !) 2
f
1
e-xt(l -
(n > -}).
~~
(ii) Kn(x) =
yn
(~)n foo e-xt(t 2 -
r(n + :\) 2
t 2)<n-~) dt
-1
l)n-l dt
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
] ..(x) = (
yn)r~n + i)GY f~1 (1 - t2)n -:
cixt
dt
(n > -!)
so that
In(x) =i-n - - 1
( vn)r(n + {)
= ( yn)r~n +
i"(:)2 . ft. (1 - t2)n-i e-xt dt
!)Gt J~l
-1
(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)
satisfies Bessel's modified equation
2 d2y
dy
2
x dx2 + xdx - (xz + n )y = 0.
(4.20)
§ 4.9
REPRESENTATIONS FOR MODIFIED BESSEL FUNCTIONS
117
We have
-dP = nxn- 1 Joo e-"' 1(t 2 - l)n-! dt +X" J"" -te-"' 1(t 2 - l)n-l dt
dx
1
1
and
d2J>
- = n(n- l)x"- 2
dx 2
Joo e-"' (t
1
J-
+ 2nxn-t ~
2 -
J)n-l dt
1
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 - 2 + x - - (x 2 + n 2 )P
dx
=
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
d
'
1(t 2 l)n+>} dt
dt
= -x"+I[e-"'t(t2 _ t)n+l]f
= -xn+l
J -{e-"'
oo
1
=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
118
BESSEL FUNCTIONS
CH.4
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
/,.(x) =
~
1
(x)2r+n
,fo r!r(n + r + 1) 2
to see that the lowest power of x in I.,(x) is
(X)"
1
r(n + 1) 2
and hence in K ..(x) is
2 si: nn
Hence as x
rc-: (~)
+ 1)
-n.
~ 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
u
t=1+x
1
so that
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
= _!_ e-x
X"
X
J e-u(1 +
X2
X
oo
2x)n-!-u2n-1 du.
o
U
For small values of x we may make the approximations e-x~ 1 and
( 1 + ~r-l ~ 1 to obtain
1
x"
(by the definition of the gamma function).
Thus for small values of x the result P(x) = BKn(x) reduces to
= --r(2n)
__!__r(Zn) = Br(n)Zn-1
xn
X"
giving
B = _ f(Zn) -.
f(n)zn-1
But by theorem 2.10 we have
z2n-J
r(Zn) =
vn r(n)r(n + !)
so that
B = Z2n-1r(n)f(n + !)
( vn)r(n)2"-'
2"f(n + ~)
vn
120
BESSEL FUNCTIONS
CH.4
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
d 2y
dy
.
x 2 dx 2 + x dx - {ik 2x 2 + n 2 )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]..(i 3' 2kx) and K.,(PI2kx).
Of course, when xis real],.(i 312x) and K,.(i 112x) are not necessarily real;
we obtain real functions by the following definitions:
ber,.x = ReJ.,(i 3 ' 2x)
bei,. x = ImJ,.(i 31 2x)
J.,(i 3' 2x) =her,. x + i bei .. x.
(4.22)
ker,.x = Re i-"K,.(i11 2x)
kei,. x = Im i-"K,.(i 112x)
i-"K,.(il' 2x) = ker,. x + i kei., x.
(4.23)
t Strictly speaking, il is a two-valued function with values e1"1 4 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).
4.11
(4.24)
SPHERICAL BESSEL FUNCTIONS
Consider the equation
d2
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-! YH 1(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(~)
(4.27)
Yl+i(x)
and A 1 , 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
122
CH.4
BESSEL FUNCTIONS
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).
1
(iii) f~(x) = f.,_ 1 (x) - n + 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_ (x)
1
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
·' (X ) + n- -+) .1., X
( ) = )n-1
. (X )
Jn
X
and thus
., (X ) =
)n
. (X ) - n- -+}1.
)n-1
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
2
H~ ~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
CH. 4
BESSEL FUNCTIONS
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
(2r + 2)!
1
r(r + 1 + -)
=
----yn
2
22r+2(r + 1)!
so that we now have
.
yn ~
Jo(x) = zix!
22r+2(r
+ 1)! x2r+~
f:6 (- 1)' r!(Zr + 2)!vn 2 r+}
2
~
Z(r -t-1)
= ~ (-1)'(-Z_r_-!~Z)!x2r
r~o
=
~
x2r
1 ~
x2r+l
:S (- 1)' (Zr + 1)!
1
= ~ ~ ( -1 )' (2 r + 1)-! = ~ sin x
r~o
(on recognising the infinite series for sin x)
(ii) We have
Yo(x) =
J(;:)
Y 1(x)
(by definition (4.27))
=
J(!!_)
cos !n ! (x) -I -l(x)
2x
sm !n
1
(by the definition of Yn(x) in theorem 4.2)
= --
J(;:) I
-l(x)
§ 4.11
SPHERICAL BESSEL FUNCTIONS
= -J(~) ~
125
(-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 + !)
= (2r)! yn
22Tr!
so that
yn ~
22rr! x2r-!
Yo(x) = - 2lxt r~ (-1)' r!(2r)!yn22r-}
~
x2r-1
= - r~ ( -l)' (2r)l
1
x2r
ro
= - ~ r~ ( -1)' (Zr)!
=
1
- - cosx,
X
(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)
hb >(x) = j 0(x) - iy 0(x)
2
1 .
.1
X
X
= - sm x + 1 - cos x
= -i (COS X X
•
e-x
I
=1---.
X
. .
1 S1n X
)
126
BESSEL FUNCTIONS
CH.4
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~2 l(x) = i( -l)nxnG
:Jn(e:ix).
PROOF
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)
:xG!)N(si:x)
= ( -l)N+lxN+lG !)(~ !)N(si: x)
= ( -l)N+IxN+IG !)N+\si: x)
=
-(-t)NxN
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:
j (x) = sm x _ cos x,
1
X2
X
.( )= (-3- -1) .
J2 X
xa
x
Slll X -
3
-- COS X
x2
'
j 3(x) = (!~ - ~)sin
x- (~-!)cos
x;
x'
x2
x3
x
(4.30)
y (x) = _ cos x _ sin x,
1
X2
X
y 2(x) = -(~-3 - !) cos x - 2 sin x,
x
x
x
i
y 3(x) = -
(-15x' - --x6) cos x - (15-x - -x1) sm. x.
2
3
(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
(:Jt
~ (:xr
(i) fn(x) ~
cos {x- (n + i)~};
(ii) Yn(x)
sin {x- (n +
(iii) Hin'(x) ~
2
(:JI
i)~};
exp [i{x - (n +f)~}
J;
(iv) H;, l(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.
128
CH.4
BESSEL FUNCTIONS
1
(v) I ..(x) ,._, v'(Znx) e-T;
(vi) K,.(x),...., (~)'"e-";
nn).
2 '
(
nn)
2 ;
. (X (Vll.") }n. (X ) ,...., ~1 510
... Yn(x ) ,...., -~cos
1
(vm)
X -
(ix) h~1 l(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) =
"
(~)" f"" e-xt(t2 - 1}n-l dt.
y'n
r(n + i) 2
1
We now make the change of variable
u
t=1+x
1
dt = - du.
so that
X
Also, when t = 1, u = 0 and when t = ro, u = co.
Thus
_ r( y'n ) (~)"fro e-(x+u)(~ + 2u)n-!~ du
K.,(x)1
2
n+
=
=
y'n
2
2
X
X
(~)"e-"'(~)n-l~ f"' e-u(~ + 1)n-lun-A du
rcn + !) 2
y'n
X
0
(2) -~1
-
r(n -!- !) X
X
- e-"'
X
0
2X
f"" e-uun-l (1 + -u )"-~ du.
2x
For large values of x we have uj2x small, so that we may take as an
approximation
X
0
§ 4.12
BllHAVlOUll. OF THE BESSEL FUNCTIONS
129
Then
K (x) ""'
n
v:n:
v:n:
-1- e-"' J"" e-"un-t du
r(n + !) v(2x)
0
1
= r(n + !) v(2x) e-•T(n + !)
(by definition (2.1))
= (~Y e-"'.
(iii) From theorem 4.14 we have
Kn(x) = ~ in+lHn(ll(ix)
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~1 l(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) ""'~:n; exp {( ~in/2)(n + 1)} (-- :n;z·1X ei"'
(
= (~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)
=
(iv) Since
and
we have
(~J! exp [i{x- (n + !)(:n:/2)}].
H~2 l(x) = ]n(x) 1
I-!~ l(x) =]n(x)
H~2l(x) =
iYn(x)
+ iYn(x}
{H~ll(x)}*
__, [(~J~ exp [i{x- (n + !)(:n:/2)}]]*
2 )}
cxp [ - i{(11 I -l.)(n/2) }].
( JTX
"
130
CH.4
BESSEL FUNCTIONS
(i) We have
]n(x) = Re H~1 >(x)
"'"'Re
=
(:J ~
cxp [i{x- (n + f)(;r/2)}]
(nx2): cos {x - (n + t).zn}.
(ii) Similarly
Yn(x) = Im H~1 >(x)
,...., Im (~)! exp [i{x - (n + l)(n/2)}]
nx
~
n}
2 )~· sm
. {""'
= ( nx
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-n-•
•
1
1 exp (X ) 1n+•
. '
v(2nx)
.
(writing exp (in/2) = i)
1
= v(2nx) e:~:.
(vii)
jn(x) =
(~) i]n+:(x)
(by definition (4.26))
__ (~Y(:Xr
(by result (i) above)
= ~ cos (
>~}
cos { x _ (n + 1
x- n; - i)
§ 4-.12
BEHAVIOUR OF THE BESSEL FUNCTIONS
nn)
1 . ( x= ~sm
2 .
(viii)
Yn(x) =
(~Y Yn+!(x)
(by definition (4-.27))
y { (n
"' (nY(2
nx
2x
--+-- 1)n}
sin x-
2
(by result (ii) above)
nn n)2
1 . (
= ~sm x- Z=-
~ sin {~ - ( x - ; )}
=-~cos (x- n;).
(ix)
h~1 l(x)
= j.,(x) + iyn(x)
1 . ( x-
--~sm
nn)
2
i
-~cos
nn)
( x-2
=- ~{cos ( x - ; ) + i sin ( x - n;)}
1
= - ~ exp [i{x - (nn/2)}].
(x)
h~2 l(x)
= {h~l(x)}*
_, [- ~ exp [i{x- n(n/2)}]]*
=! exp [ -i{x - (nn/2)}].
X
Theorem 4.22
As x ~ 0 we have
1
(i) J,.(x) "'I'(n
1
l)GY
131
132
BESSEL FUNCTIONS
(ii)
Y.,(x),.....,
~r(n)(~t (n
2
[ -lnx
n:
~0)
(n = 0);
n
(iii) j.,(x),....., (2
CH.4
1)!!
(n integral) ;t
. ) ( )
(2n - 1)! !
(n integral).
(lV Jn X ,...., 1
x"+
PROOF
~
(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
(X)"
].,(x) ""r(n 1+ 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
1
Y.,(x) ""- sin nn r(
+
-n
-~ 1l~)
=-
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
(n = 0)
t By nil (n double factorial) we mean
( _ 2)(n _ 4 ) ... {5. 3.1 !~ n !s odd
nn
6.4.2n n IS even.
§ 4-.13
GRAPHS OF THE BESSEL FUNCTIONS
and
Yn(x) ~- ~ (n- 1){~)-n
133
(n ~ 0)
~ r(n)(~Y·
= -
(iii) We have from definition (4-.26)
J(~)Jn+i(x)
"' J(;Jr(n + t)GY+t
jn(x) =
1
1
(vn)xn
= - 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)
J(~)
"' J(~){- ~r(n + t)(~t+i}
y,.(x) =
Y,.+l(x)
1
2"
1
2"
= - vn.xn+lr(n
+ !)
= ---:;
- (n- i)(n- t) .•• t.tr(t)
vnxn+ 1
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.
134
BESSEL FUNCTIONS
-0·5
-1·0
~'
FIG. 4.1
]n(x), n = 0, 1, 2
FIG. 4.2
Yn(x), n = 0, 1, 2
CH.4
§ 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
CH.4-
BESSEL FUNCTIONS
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
y
137
(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.
138
CH.4
BESSEL FUNCTIONS
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
d2
d
2
2
2
X dx-J.. (~;x) + xdX}.. (~;x) + mx -- n ) }.. (~;x) = 0
and
2
x 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
and
(4.32)
(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)].,(~ 1 x) = 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)
(gix)~J..(~;x)} + {!J..(~;x)}x! J..(~,x)
- ! {J..
+ m- ~J)x]..(~;x)]..(~;x) = 0
and thus
:x
{l. (~;x)x!J.,(~;x)} - :x {J..(~;x)x!}. (g x)}
1
+ (g~ - gJ)xj.,(g;x)]..(~,x) = 0.
§ 4-.14-
ORTHONORMALITY OF THE BESSEL FUNCTIONS
Integrating from 0 to a, we obtain
]a
d
d
[ J.,(g 1x)xdxj.,(gix) - ] ..(~;x)xdx].,(~;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:
f:
and hence
xj.,(g;x)]..(~1x} dx = 0
xj..(;;x)]..(!;;x) dx = 0
provided
gi ~ g,.
To complete the proof we require to prove that
a
a2
f x{].,(!;x)} 2dx = {J.. +l(!;a)}2
z
0
] ..(!;a) = 0.
Now, if we denote J ..(~x) by z we have
x 2z" + xz' + (g 2x 2 - n 2)z = 0
which, on multiplication by 2z', becomes
2x 2z"z' + 2xz' 2 + 2(!; 2x 2 - n 2)zz' = 0
and this is readily seen to be equivalent to
d
dx {x2z'2 - n2z2 + g2x2z2} - 2!;2xz2 = 0.
if
Integrating this equation from 0 to a gives
[x 2z' 2 - n 2z 2 + !; 2x 2z 2] 0 - 2!; 2
J:
xz 2dx = 0
which, on replacing z by ] ..(!;x), becomes
2
[x2{!J.. (;x)} - n2{J.,(gx)}2 + g2x2{J..(;x)}2J:
f:
- 2!; 2
x{].,(!;x)}2 dx = 0
and hence, using the facts thatJ.. (!;a) '= 0 and n]n(O) =~ 0,
2
U 2 f" xf.!,.(!;x)l2 dx 0.
[ a2{ dl ),.(!;:.:)} ]
(
,'\:
I
fl
139
0
14()
CH.4-
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)
which may be rewritten in the form
d
n
dxj.,(gx) = Xj.,(gx) -
so that
I:
x{]n(gx)}2 dx =
Un+l(gx)
2~ 2 [a 2{~ln(gx) - g]n+l(gx)rl~a
1
= 2g2a2~2{Jn+t(ga)}2
(again using the fact that]n( ~a) = 0)
a2
= 2 {J.,+l(~a)}2.
Theorem 4.24
If f(x) is defined in the region 0 < x < a and can be expanded in the form
L c;]n(g;x) where the g; are the roots of the equationj.,(ga) = 0, then
C()
i=l
2
I:
C; =
xf(x)]n(g;x) dx
a2{J.,+l(g;a)}2
.
PROOF
We have
L c;]n(g;x)
xf(x)j.,(g;x) = L C;x].,(g;x)j.,(g;x)
C()
f(x) =
i=l
C()
so that
i=l
§ 4-.15
INTEGRALS INVOLVING BESSEL FUNCTIONS
14-1
and hence
2
2: ~],.+l(~;a) bii
00
=
2
Ci
i=l
(by theorem 4-.23)
Thus
J: xf(x)Jn(~1x)
2
=
C;
dx
a2 Und~;a)}2
which completes the proof.
2: ctfn(~ix), with ci as given above, is called the Bessel series
00
The 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
Jn(x) =
2(x/2)n-m
r(n - m)
Jl
(1 - t 2 )n-m-ltm+l]m(xt) dt
0
(n > m > -1).
PROOF
Consider the integral
J --~
Jl (1 . t2)n-m-ltm
0
HJ,.(xt) dt
14-2
CH.4-
BESSEL FUNCTIONS
*;S (-
and substitute the series (4-.6) for]m to obtain
I= fl (1 - t2t-m-ltm+l
o
= ~ ( -1Y
£:b
=
(x/2)m+2r
r !r(m + r + 1)
1)'
Jl
(xtj2)m+2r - dt
r!r(m + r + 1)
(1 - t2)n-m-lt2m+2r+l dt
o
"'
(x/2)m+2r
1 Jl
2 ( -1 y r!r(m + r + 1) 2_ o (1 _ u)n-m-lum+r du
r=O
But
J:
(on making the substitution u = t 2).
(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).
Hence
~
(x/2)m+2r
I= tr(n- m) .G ( -1)' 'r(
1)
r.
r=O
n +r +
x)m-n ~
=
! rc n - m) ( 2
=
trcn-
6 (-1 y ---cr!=r~((xj2)n+2r
n_:__+___:._r_-t_--:--1)
m)Gr-J.
(x).
Thus
]n(x) = 2 (x/2)n-m I
r(n- m)
which is the required result.
Theorem 4.27
(a> 0).
§ 4-.15
14-3
INTEGRALS INVOLVING BESSEL FUNCTIONS
PROOF
From theorem 4-.6 we have
]
0
I"'
(x) = -1
cos (x sin c/>) dcf>.
n o
Hence
I e-ax] (bx) dx I e-ax-1 I" cos (bx sin c/>) de/> dx
00
=
0
0
=
~
I: [I:
00
7t
0
0
J
e-ax t{exp (ibx sin c/>) + exp ( -ibx sin cf>)} dx dcf>
(interchanging the order of integration)
I" [exp {-(a - ib sin cp)x} exp {-(a + ib sin cf>)x}J de/>
2n o
-a + ib sin cp
+ -a - ib sin cp
o
1 I" {
1
+ a-t-tbsmcf>
.1 . } dc/>
Zn o a-tbsmcp
= 2_
00
=-
..
=~I"
1
dcf>.
n oa 2 + b 2 sin 2 cf>
But this last integral may be evaluated by elementary means (e.g., by
the substitution u = cot c/>) to give
I
a
oo
n
o e-ax]o(bx) dx =; ay(a2
+ b2)
1
Theorem 4.28
1
00
J fn(bx) dx b (if n is a non-negative integer).
=
0
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"'
0
] 0
(bx) dx-
1
b.
144
CH.4
BESSEL FUNCTIONS
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
1d
b dx1o(bx) = -1I(bx).
J~ 1 (bx) dx = - ~ [1o(bx)J~
Hence
1
1
=b·
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~ = 1J~ Un-t(x) - ]n+l(x)} dx.
Remembering that n > 0 so that 1.,( oo) = 1.,(0) = 0, we have
1
r
0 = ZJ
co
0
{].,_ 1
(x) - ]n+I(x)} dx
and thus
J~ 1ndx) dx = J~ fn- (x) dx.
1
Replacing n - 1 by Nand x by bx gives the result
f~ 1N+2(bx) dx =
reo
1
J:
Thus if J 1 N(bx) dx = b' we also have
0
the proof is complete.
1N(bx) dx.
f"' N+ (bx) dx = b'1 and hence
]
0
2
§ 4-.15
INTEGRALS INVOLVING BESSEL FUNCTIONS
145
Theorem 4.29
PROOF
(i) From equation (4-.6) we have
~
1
] ..(bx) = .~ ( - 1)' r!r(n + r + 1)
(bx)
2
2r+n
so that
J:
J.,(bx)xn e-ax dx
00
= "" ( -1 )'
•~
1
(b)2r+n
f«> e-Q:Ilx 2•+2n dx.
r!r(n + r + 1) 2
o
oo
=
J
0
t2r+2n
e-t -a2-r+_2n_+_l dt
(writing t = ax)
=
1
a 2r+2n+
l(2r + 2n + 1),
(by definition (2.1) of the gamma function),
so that
J:].
(bx)xn e-ax dx
=
2<"" -1Y r!r(n +1r + 1)(b)2
2r+t1
1
a2r+2n-rl(2r + 2n + 1)
r-"-0
(using the result of theorem 2.2 that l'(n 11) ~- nl'(n)).
146
CH.4
BESSEL FUNCTIONS
But by theorem 2.10 we have
r(2x)
22~-1
- - =rex+ .1)2
r(x)
yn
so that
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 + ra2n+I L...
r!
r=O
t)(b2)r
a2
= _1-~ ( --1)' r(n + ~) (n + ~)(n + t) ... (n + r -· ~)(b2)r
r(n + 1)
r!
2
a2n+ 1 r=O
L...
a2
on repeated use of the result xr(x) = r(x + 1) (theorem 2.2).
cc
znbn
1
Hence
o ]n(bx)xn e-ax dx = yn r(n + t} (a2 + b2)n+l
f
znr(n + i)
yn
bn
(a2 + b2)n+1'
(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
~
oo
(i)
(ii)
I
I
fn(bx)xn+l exp ( -ax 2 ) dx = ( a)n+l exp ( -b 2j4a);
2
0
bn
(
b2)
]n(bx)xn+ 3 exp ( -ax 2 ) dx = 2n+lan+2 n + 1 - 4a exp ( -b 2 j4a);
oo
0
(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
~
00
(i)
sinax]0 (bx) dx = {
0
(b <a).
v(a2- b2)
"'" ax] 0(bx) dx
(ii) [
(b >a)
~ { 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
0 e
0
so that
(b ) d
X
X
1
= y(b2 - a2)
1
v 2- a2
Io cos ax] (bx) dx- i I"'o sin ax ] (bx) dx =-(b--).
oo
0
0
Equating imaginary parts of both sides gives
cc
1
sin
ax] 0 ( bx) dx = - (
o
V a2- b 2 )
f
"'I
()
I'll'
I.
sin ax ] 0 (bx) dx
()
if a > b
if a ·:b.
148
CH.4
BESSEL FUNCTIONS
Equating real parts of both sides gives
1
co
] (b ) d
COS ax o X
X= y(b 2 -a2)
0
J
J:
if b > a
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
exp { i(x + y)(t -
of
~)}.
But
C()
=
C()
2: ],(xW. 2: ].(y)t•
2: ],(x)].(y)tr+s.
r=-CO
8=-CO
00
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
149
Example 2
00
]n(x)
1
- - dx = -.
x
n
From theorem 4.28 we have
Show that f
o
f~ fn(x) dx = 1.
But from theorem 4.8(vi) we have
Zn
- fn(x) = fn-t(x) + fn+l(x)
X
so that
Zn
fw ]n_ (x) dx + foo fn+l(x) dx
f _n- dx =
wj~)
0
X
1
0
o
= 1 + 1,
and hence
Joo fn(x) dx = ~.
o
x
n
Example 3
If~; are the solutions of the equation] 0 (~) = 0,
~ Jo(~;x)
show that L -- - - =
i~l {tJ1(~;)}2
(0 < x < 1).
-~In x
We use theorem 4.25 to write
00
-!In x =
L c;} (/;;x)
0
i~
1
with
L
1
x In x] 0 (/;;x) dx
= -- - - u~-c;;;)fTo evaluate the integral appearing in the numerator we use the series
(4.6) for ./ 0 :
1
J u(i; ,x)
,.
I)
')2r
1..,.'\:
1:
1
)' r!l'(r : I) ( 2
150
CH.4
BESSEL FUNCTIONS
= 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
x" In x dx = [--xn+~In x] fo
n+1
o
l
1
= 0- 0 ;, + xn dx
1
1
1
J
f
1
1
-xn+l} dx
on+1
x
1
1
Thus
J:
(4.35)
x In x ] 0 (~ix) dx
= ~ ( -1)' -1-(~)zrfl
x 2r+lln x dx
2
L..
=
2
(r!)
r=O
o
~ (-lY (r~) 2 (~Yr{- (2r ~2)2}
(using equation (4.35))
1 ~
1
(~')2r+2
=
~f 6 (- 1y+t {(r + 1)!}2 2
=
~~ 6 (- 1)'" (r!) 2 2
=
:~ { ~ <-1Y <r~>z(~iyr- 1}
=
-2,
1 ~
(4.34)
1
(;•)2r
1
~,
(using equation (4.34)).
Thus we have
1
§ 4.16
151
EXAMPLES
and hence
Example 4
'
'
],..(x) Yn(x)
- fn(x)
Y,..(x) = A
-
Show that
X
where A is a constant, and by considering the behaviour for small x show that
A= 2/n.
We have
d {].-.Yn'
'
dx
fnYn}
=].. Y~' +]~Y~- ]~Y.. - ]~Y~
=]n Y"
n -J"Y
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 {].-.Yn' -JnY.-.
' }
x2 dx
=fn{ -xY;,- (x 2 - n 2) Yn}- { -x];,- (x 2 - n 2)J.. }Y..
= -x{].. Y;,- ];,Y.. }.
Hence, writing],..Y;,- ];,Yn = z, we have
z
x
dz
dx
or
dz + dx = O.
Z
X
When integrated this gives
In z + In x = constant
which is equivalent to
and hence
In zx = constant
z.v
constant
A, say.
152
BESSEL FUNCTIONS
Thus z = A so that, by the definition of z,
X
' fnY,.
'
A
]nYn= -.
X
For small values of x, we have, by theorem 4.22,
2
(x) ,._, 1, Y 0(x) ,._,-In x
n
so that, considering the relationship for n = 0, we have
] 0
'
' .. ,..., 1.-.--0.2 1
21nx
I ,. Y ,.-JnY
nx
n
2
=-.
:nx
Hence
2
A
,
x
nx
giving
and thus
2
],.(x) Yn(x) - fn(x) Y ..(x) = --·
nx
I
I
Example 5
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
§ 4.16
153
EXAMPLES
= 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
f tJ! = [t2
x
-zU~(t)- 1n-t(t)1.-.+t(t)}
dt
0
]"'
0
x2
= z-U~(x) - 1n-t(x)1n+l(x)}.
Example 6
Use the generating function to show that 1.-.( -x) =
We have, from theorem 4.5,
{~(t - ~)}
=I
I
G{ -x(t- ~)}J
1
= exp [:{c -t)- - }]
2
( -t)
exp
so that
( -1 )~..(x).
1n(x)tn
n=-oo
1n( -x)tn = exp
1t=- 00
n==--- oo
ro
2: (--t)nt•'],.(x).
~
II
-
00
Equating the coefficients of I" on both sides gives
] .. (
x)
(
1)'~/ .. (x).
154
CH.4
BESSEL FUNCTIONS
PROBLEMS
( 1) Show that
d~{x]..(x)]n+l(x)} = x{]!(x) - J~+ 1 (x)}.
"'/ (x
J
"'/
J
2
(2) Show that (i)
]0
0
2
(ii)
0
sin X
cos 0) cos 0 dO = - - ;
X
1 - COS X
] 1(x cos 0) dO = ----.
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)dx 2 + J.y = 0
which is zero at x = 0.
oo
n
(5) Show that
x,J..(a) =]0{y(a 2 - 2ax)}.
n.
n=U
L
(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 ... ).
["" ]n+I(x) d _
1
'f
1
(7) ShOW t h at J ----;;;- X - znr(n + J) 1 n > - 2 ·
0
(8) Prove that
i ].
(kx)t" = exp {;lk -
n=-co
(9) Show that exp
kntn]n(x)
n=-co
m
L m.:_rln+m(x).
ro
and deduce that I.,(x) =
D} i:
m~o
{~(t + ~)} =
i
In(x)tn.
n=- oo
(10} Show that J.,(x), Yn(x), Kn(x) and In(x) all satisfy the differential
equation
3
d4y + ~ d y _
+ 1 d 2y + 2n 2 + 1 dy + 4 - 4n 2 _ 1) = O.
4
3
dx
x dx
x2
dx 2
x3
dx
x4
y
2n2
(n
155
PROBLEMS
L Cr]o(A,.x) valid for the regton
00
(11) Expand x 2 in a series of the form
r~l
0 < x < a, where Ar are the roots of the equation J 0(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
f
oo.
sin {(n- m)n/2}
+ 1) - m(m + 1)
•
o }m(x)Jn(x) dx = n(n
if m ~ n, and use L'Hopital's rule to deduce that
f~
{jn(x)}2 dx = 2(2nn+ 1)"
Determine also the values of
f:"'
J:
)m(x)jn(x) dx (m ~ n) and
{jn(x) }2 dx.
(13) Show that
") h
= ~ (- 1)k (x/Zt+ k cos {!(n + 2k)n}.
(1 ernx
7
kl(
,
t;;b
.n+ k)l.
2
"") h .
_ ~ (- 1)k (x/Zt+ k sin {!(n + 2k)n}
(11 elnX- ~
kl(
k)l
.
2
· n
+ ·
(15) Show that, for large values of x,
1
2
(i) her x ,.._, --\1( n-;;{'/V cos ( : 2 2
~);
k~o
( 14) Prove that
(i)
(ii)
J:
J:
t her t dt = x hei' x;
t hei t dt = -x her' x.
( II"") h Cl. X ,.._,
1
v'(2nx)
1
•
(
X
c·' I ' .,- Sill
v2
n)H
.
5
HERMITE POLYNOMIALS
5.1 HERMITE'S EQUATION AND ITS SOLUTION
Hermite's equation is given by
2
d 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
-
L
00
z(x, s) =
a,.xs+r
r=O
we find that the indicia! equation has roots s = 0 and s = 1, with a 1 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
ar
2(r - n)
(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 (x 2) for large values of x. Thus they cannot satisfy the
requirement that exp (!x 2) 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
157
GENERATING FUNCTION
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 _
y- an X
n(n - 1) n- 2
n(n - 1)(n - 2)(n - 3) n- 4
2.2 X
+
22.2.4
X
+ •••
+(
-l)rn(n- 1) ... (n- 2r + 1) n- 2 r
}
2r. 2 . 4 ... 2r
X
+'''
~ ( - 1), n(n - 1) ... (n - 2r + 1)xn-,
2
=an L
r=O
2r.2.4 .. ,2r
h
if is even
1
(w ere [In] = !(n - 1) if n is odd)
{in n
[lnl
=an
l
L (-1)' 22r n'_ 2
r=o
l(
r. n
)lxn-2r,
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:
l
[!n]
:S
n. 2r)l(2x)n-2r.
H.,(x) = """ (-1)' rl(n-
(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
r=O
L...
s ~o
sl
(2x)'
fr12•.
r !s!
1)·'·
'·.
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.
(by definition (5.3)).
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
~
F(t) = L
n=O
(dnF)
tn
dtn
nl'
l=o
we have
H.,(x) = [an e21z-t•]
atn
l=o
=[onatn e"''-<x-1)']
t=o
=e"''[an
- e -(x-t>']
atn
But
1=0
.
a
a
a/<x - t) = - a/(x - t),
§ 5.3
159
OTHER EXPRESSIONS FOR HERMITE POLYNOMIALS
so that
an
an
atn
axn
-f(x- t) = ( -l)n - f ( x - t)
and we have
Theorem 5.3
PROOF
We have
and hence
Thus
= 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, forex1
umpll·, l'Xp( ( .)
(1·'
~ no\( (dlx ) n - £....
~ n ,· ldlx~.. so that exp (d)
f(x) - . ~ 1~~nf~o
£....
Jx
~ no ux
, u
u
1
1
0
fl
0
"
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
H 0 (x) = 1
H 1(x) = 2x
H 2(x) = 4x 2 - 2
H 3(x) = 8x 3 - 12x
H 4 (x) = 16x4 - 48x 2 + 12
H 5(x) = 32x5 - 160x 3 + 120x.
Theorem 5.4
PROOF
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
L) -l)n l
n=O
n.
=
~
tn
~ Hn(O)-r
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 = H2n(O) -1- (w1t
. h n a non-negative
. mteger
.
)
( -1 )n -·
nl
(2n)l
§ 5.5
161
PROPERTIES OF HERMITE POLYNOMIALS
which yields
5.5
ORTHOGONALITY PROPERTffiS OF THE HERMITE
POLYNOMIALS
Theorem 5.5
PROOF
We have
Loo Hm(x)~
m
e-•'+ 2•"' =
and
m=O
m.
so that f:oo e-x' Hn(x)Hm(x) dx is the coefficient of (tnsm)J(n!m!) Ill
the expansion of
J:
e-x' e-t'+Ztx e-•' +2 ""' dx.
00
But
oo
J e
-x'
c -t'+2tx e -s'+2sxd x
-oo
=
=
e-t'-s'
e-t'-s'
J:
exp { -x 2 + 2(s + t)x} dx
00
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
CH. 5
HERMITE POLYNOMIALS
Hence the coefficient of (tnsm)/(n !m !) is zero if m ::;C: n and is ( y'n)2nn!
whenm = n.
Hence
J
oo
- oo e-"'
•
{
H,.(x)Hm(x) dx =
0
( y'n)2nn!
ifm::;C:n
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)
2 H..(x)-,n.tn
oo
= 2t
n=O
~
tn+l
,6
H .. _1(x)(n _ 1)!"
=
2 L. H,.(x)n=O
n.1
=
2
tn
00
Equating coefficients of tn gives for n = 0
H~(x) = 0
and, for n > 1,
H~(x)
n!
ZH.. _1 (x)
(n- 1)!
H~(x) =
2nH,._Jx).
which reduces to
§ 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,
(2x - 2t) exp (2tx - t 2) =
~ ntn- 1
~ -
-Hn(x).
n.1
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
2(x - t) n~ -;;jHn(x) =
n~o
6
co
tn-1
(n -l}!Hn(x)
which is equivalent to
~ tn
~ tn+l
~
tn-1
2x ;S n!Hn(x)- 2
-;rHn(x) =
(n _ 1) 1Hn(x)
6
6
which may be written in the form
2x
~tn
6~tn:;; Hn(x)- 2 6~ (n _tn 1) 1Hn_ (x) = f:o
n!Hn+I(x).
1
1
Thus, equating coefficients of tn for n > 1, we have
1
2Hn_1(x)
1
2x---,Hn(x) - ( _ ) 1 = --,Hn+l(x)
1 .
n.
n
n.
which on multiplying throughout by n! becomes
2xHn(x) - 2nHn_1(x) = Hn+l(x).
Similarly, equating coefficients of t 0 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 - 2xdx + (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 But
t 2).
dm
-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 (n -m.
) ,Hn-m(x)
§ 5.8
165
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:
x e-x• Hn(x)Hm(x) dx.
00
We have, from theorem 5.6(ii)
xHn(x) = nHn_1(x) + }Hn +l(x)
so that
f:
x e-x• Hn(x)Hm(x) dx
00
= J:oo e-x• {nHn_ 1(x) + !Hn+l(x)}Hm(x) dx
= 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
I
[~n]
H,..(xt) =
~ ( -1)' rl(n ~ Zr)l(2xt)n-2r
so that
(
2)
vn n.1
fro tn e-t• H,..(xt) dt
0
z
= ( v~)n!
Joo tn c-t• r~ -1)' ;!(n ~ 2r)!zn-2rxn-2rtn-2r dt
r:nl
I
(
0
--1)'xn-2r Joo -t' 2n-2r
= ILlnJ zn-2r+l(
~--------e
t
dt
r
()
(vn)r!(n- 2r)!
0
2r tl( 1)'x" -2r
y';r)r !(n
2r)l ~ f(n
1!•:._1 2"
A, (
r I D
(hy theorem 2.4)
166
CH.S
HERMITE POLYNOMIALS
- [~ 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)
(2n - 2r)!
- ""' ( -l)r
xn-2r
r~
znr!(n- 2r)!(n- r)!
Unl
= P..(x),
(by equation (3.17)).
PROBLEMS
(1) Show that J:co x 2 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
Cr =
where
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 degree less than n.
2 fco e-t• cos 2xt dt.
(3) Show that e-x• = yn
0
By differentiating this result 2n times with respect to x, show that
22n+I( -1)"' ex'
H2r1(x) =
e-t• t 2n cos 2xt dt,
y:7t
0
foo
and obtain a similar expression for H 2n+1 (x).
2"'( - i)n ex' co
Deduce that H.,(x) =
e-t'+ 2itx t"' dt.
y:7t
-oo
(4) Use the result of problem 3 to show that
f
~ H..(x)Hn(y)tn = (1 - tB)-112 exp {2xyt - (x2 + y2)t2}
~
n=O
2nn!
1- t 2
and deduce that
~ {H.,(x)}
t"' = (1 - t 2)- 1 ' 2 exp {2x 2 t/(1 + t)}.
~
2nnf
2
t~=O
16?
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 {H.,(x) }2 e-x• dx = f (1 - x2)n -e-x•00
(u) -
'2"n!
-oo
00
1 +x 2
-oo 1 +x
(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.
2
1 +x 2
dx.
6
LAGUERRE POLYNOMIALS
6.1 LAGUERRE'S EQUATION AND ITS SOLUTION
Laguerre's equation is
d 2y
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
L a,
c:J)
form z(x, s) =
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)
§ 6.2
169
GENERATING FUNCTION
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
(n- r)
a,+ 1 = -a,~+ 1)2
and we obtain the solution
_
{
n
n(n - 1)
y - a0 1 - i 2x + (2 !) 2 x 2 + ...
+(
}
_ ), n(n - 1) ... (n - r + 1) 2
1
(r!)2
x + ...
(the highest power of x being xn)
=
~ ( _ 1)' n(n - 1) ... (n - r + 1) r
ao f;o
I
N
=
x
(r!)2
""" ( -1Y ( - n.·)I( l)2xr.
ao L
n
1 • r.
r=O
We define the standard solution as that for which a 0 = 1 ; we shall call it
·the Laguerre polynomial of order n, and denote it by Ln(x):
I
n
Ln(x) = """
L (-1)r ( - n.) 1( l)2xr.
n
r=O
6.2
(6.3)
r . r.
GENERATING FUNCTION
Theorem 6.1
exp {-xt/(1 -t)} = ~ L () n
L n X t.
( 1 - t)
n=O
PROOF
We have
(1
1
f
. t) cxp l
~ tl--~ ~
1
xtj(1 -- t)}'
t)
1y
r!
'
It
(1
xrtr
1)' 11 •
tr
170
But
CH.6
LAGUERRE POLYNOMIALS
1
w+l
(1 -
_
(
)
(r+1)(r+2) 2 (r+1)(r+2)(r+3)t3
- 1 + r + 1t +
21
t +
3!
+···
(by the binomial theorem)
-
oo
(r + s)!
L rlsl t•'
s=O
so that we now have
+
1
~
(r s)l
- - exp { -xt/(1 - t)} = L.. ( -1 Y --x•t•+•.
2
1- t
r,s=O
(rl) sl
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
2: (-1Y ( l) t·- )lx• = Ln(x)
I
n
,_
0
r. 2 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)= ~ (n- r)!r! dxn-Tx" dxr e-O:.
nl
But
dP
-xq = q(q- 1) ... (q- p + 1)xq_p
d xP
q!
(q- p)!
xq_p
,
(by definition (6.3)).
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
L 0(x) = 1
L 1(x) = -x + 1
1
L 2(x) = (x 2 - 4x + 2)
L 3(x) =
L 4 (x) =
Theorem6.3
(i) /,11(0)
(ii) /,;,(0)
1.
11.
21
1
3!( -x 3 + 9x 2 - 18x + 6)
1 4
(x - 16x3 + 72x 2 - 96x + 24).
41
172
CH.6
LAGUERRE POLYNOMIALS
PROOF
(i) Set x = 0 in the generating function of theorem 6.1 and we obtain
co
1
Ln(O)tn = 1 - t
L
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
d2
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
L~(O) =
and hence
-n.
6.5 ORTHOGONALITY PROPERTIES OF THE LAGUERRE
POLYNOMIALS
Theorem 6.4
PROOF
From the generating function of theorem 6.1 we have
i
~x~~1-s)} i
exp { -t~(~
-
t)} =
Ln(x)tn
n=O
exp {
and
=
Lm(x)sm,
m=O
so that
- t)} exp { -xs/(1 - s)}
.
.
Leo -xL ( )L ( ) n m = -x exp{ -xt/(1
1
1
e
nX
mXtS
e
.
n,m=O
-s
-t
Thus f~ e-xLn(x)Lm(x) dx is the coefficient of tnsm in the expansion of
I
=
f"' e -x . exp {-xt/(1t)} exp {-xs/(1- s)}d
1
.
1
x.
o
-t
-s
§ 6.6
LAGUERRE POLYNOMIALS AND THEIR DERIVATIVES
But I=
1
173
reo exp {-x(1 + _!_ + _s-)} dx
(1 - t)(1 - s) J o
1
= (1- t)(1- s)
[
1 - t
1- s
1
-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).
(iii) L~(x) = -
n-1
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)}
~ L..(x).nt
-(i--=-ij 2 exp{-xtj(1-t)}- (1 -t) 2
(
1 -t)
which, on further usc of theorem 5.1, becomes
0
A
rtl
/,,.(x). til"
}
I
(I
co
m
t) '~' /,,.(x)tn
(
I
X
'\""'
I
2
)
~ /.,.(x)/".
II
I)
174
CH.6
LAGUERRE POLYNOMIALS
Multiplying throughout by (1 - t) 2 gives
00
(1 - t) 2
00
00
n=O
n=O
L L.,(x)nt"- = (1 - t) L L ..(x)t" - x L L ..(x)t"
1
n=O
and hence
L
ct)
L
2
L ..(x)nt" +
n=O
n=O
L
00
ct)
L.,(x)nt"- 1 -
L ..(x)ntn+I
n=O
ct)
L L.,(x)t" - L L ..(x)t"+
=
00
ct)
n=O
1
n=O
-
x.L L.,(x)t".
n=O
Relabelling the summations so that the general power appears as t" in
each gives
L
00
L
00
L.,+ 1(x)(n + 1)t" - 2
n=O
n=-1
L .. _1(x)(n - 1)tn
n=l
L
00
=
L
L
00
L ..(x)ntn +
00
L.,(x)t" -
L.,_1(x)t" - x
n=l
n=O
L
00
L.,(x)t"
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" =
_
~
fl=O
t_ exp { -xt/(1 - t)}
1-t
1-t
_
t
= - 1- t
L L ..(x)t".
00
n=O
Thus
L L~(x)t"
00
(1 - t)
n=O
L
<X>
= -t
L.,.(x)t"
n=O
and hence
L L~(x)tn - L L~(x)tn+l = - L L.,.(x)tn+l
00
00
00
n=O
n=O
~t=O
(6.4)
§ 6.6
175
LAGUERRE POLYNOMIALS AND THEIR DERIVATIVES
which, on relabelling, becomes
00
00
00
n=O
n=1
n=1
L L~(x)t" - L L~_ 1(x)t" = - L L,._ (x)t".
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).
L tr we
00
(iii) If, in equation (6.4), we expand 1/(1 - t) in the form
r=O
obtain
00
00
00
n=O
r=O
s=O
L L~(x)t" = -t L t• L L (x)t•
= - L L.(x)t•+•+I,
8
00
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),
2 L,(x).
•=0
n-1
and hence
L;,(x) = =
·
1-U
176
CH.6
LAGUERRE POLYNOMIALS
6.7 ASSOCIATED LAGUERRE POLYNOMIALS
Laguerre's associated equation is
d 2y
dy
x -2 + (k + 1 - x)-- + ny = 0.
dx
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
xdxk+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 dx 2 dxk + (k + 1 - x)dx dxk + ndxk = 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 (n _ r )l(k
-L )I 1
xr.
r~O
,
r ,r
1
o
PROOF
From equation (6.3) we have
(n + k)!
~ ( -1)' (n + k- r)!(r!)2xr
n+k
Ln+k(x) =
§ 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
= ( -1)
(i~i
6(
n+k
(n+k)!
-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
=
= ( -1)k
r!
)
(r- k)!xr-k
~ ( -l)k+s (n + k !:!-: k)~)!(kf- s)!s!x•
(changing the variable of summation to s = r - k)
"
(n + k)!
~ ( -1)
x•
-- f:o
(n - s)!(k + s)!s! '
8
which is the required result.
6.8 PROPERTIES OF THE ASSOCIATED LAGUERRE POLYNOMIALS
Theorem 6.8 (Generating function)
exp { -xtj(l - t)} = ~ Lk( ) n
L.,. n X t '
(1 - t)k+l
n=O
PROOF
From theorem 6.1
exp -xtj(l{- t)} = ~ L () n
L.,. n X t •
( 1 - 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).
178
LAGUERRE POLYNOMIALS
CH.6
Carrying out the differentiation on the left-hand side, we obtain
(
- _t_)k exp { -xt/(1 - t)} = ~ Loo L ( ) +k
...l ••k
n+k X t"
uw-·
1 -t
1 -t
n=O
and thus, using equation (6.7),
p
( -1)k ( _ t)k+l exp { -xt/(1 - t)} =
1
6, (-1)kL~(x)tn+k
00
which, on cancelling a common factor of ( -1)ktk, becomes
exp {-xt/(1 - t)} = ~ Lk( ) ,.
(1 - t)k+l
n X t •
6
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)
J
oo
o
_,
k( )Lk
d
(n + k)!
e XCLn X m(x) X =
I
c5nn••
n.
PRooF
Again this proof is very similar to a previous one (that of theorem 6.4)
and will be omitted.
Theorem 6.11 (Recurrence relations)
(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).
(iv) L~'(x) =
n-1
-
LL~(x).
r=O
(v) L~'(x) = -L~~l(x).
(vi) L~+ 1 (x) =
n
L L~(x).
r=O
6.8
PROPERTIES OF ASSOCIATED LAGUERRE POLYNOMIALS
179
WOF
(i) Using theorem 6.7 we have
~- 1 (x)
+ L~- (x)
1
2
n-1
n
(n- 1 + k)!
( -1Y
x' +
( 1)' (n + k - 1)!
x'
- r~o
(n- 1 - r)!(k + r)!r!
r=O (n- r)!(k- 1 + r)!r!
-
2
1
1
n(n + k - 1)!
n(n + k - 1)!
--;So
""' ( -1)'(n- r- 1)!(k + r)!r! x' + :S,
""' (- 1)'(n- r)!(k + r- 1)lr!x'
+(-1)"
(n + k - 1)!
1
{
(n + k- 1)!
x 11
(n- n)!(k- 1 + n)!n!
1
1
}
=~(- 1 )'(n-r-1)!(k+r-l)!r! (k+r)+(n-r) x'
n-
1
+ (-1)" -.x"
n!
1
_
~ -l ,
(n + k - 1)!
- ;S ( ) (n- r - 1)!(k + r- l)!rl
(n - r) + (k + r)x,
(k + r)(n- r)
x"
+ (-1)"n!
(n + k)!
"~
=
x"
;S0 ( -l)'(n- r)!(k + r)!r( + (- 1)n n!
- ~ -1)'
(n + k)!
x"
(n- r)!(k + r)!r!
- ;S (
= 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
CH.6
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)
= (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 ,
dxkLn+k(x) =
"+k-1 d"
-
2 d:;;/,Lr(x)
r~o
§ 6.8
PROPERTIES OF ASSOCIATED LAGUERRE POLYNOMIALS
181
so that
(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
=dxkL.+Jx)
L
8=0
(changing the variable of summation to s = r - k)
n-1
=-
L (-1)''L:(x).
8=0
n-1
L!' (x) = -
Therefore
LL;(x).
r-o
(v) By theorem 6.7 we have
Lk'(x) n
-
~ ~ (-1)r
:S
dx
- ~ (-1Y
-:6
(n + k)!
x'
(n - r)!(k + r)!r!
(n + k)!
xr-1
(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
6(
(n - 1 + k + 1)!
x•
)(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
1
Prove that L~+.B+ (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)}
(1 _ t)rx+fl+2
L,..
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 {-xtj(1 - t)} exp { -ytj(1 - t)}
exp {-(x + y)t/(1 - t)}
(1 - t)rx+fl+2
(1 - t)a+I
(1 - t).B+l
=
ct)
ct)
r~o
s~o
L L~(x)t' L L~(y)t•
(again using theorem 6.8)
L L~(x)L~(y)t•+-.
00
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.
§ 6.10
183
EXAMPLES
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
L L~(x)L~_,(y)
L~+iJ+l(x + y) = L L~(x)L~_.(y).
Hence the coefficient of tn is
r=O
so that
n
r=O
Example 2
oo
Show that]m{2y(xt)} = e-t(xt)m/2""'
Lm( )
n X
tn
f:o (n + m).1
(where]m is the Bessel function of integral order m).
We shall prove the equivalent result that
oo
Lm( )
e1(xt)-m 12]m{2y(xt)} = """
f:o (n +n Xm).
From equation (4.6) we have
r·
~
r+m
;S
(-1Y rl(m 1+ r)l {2v(xt)}
-22
lm{2v(xt)} =
so that
00
1
et(xt)-m1 2j m{2 v(xt)} = e1(xt)-m1 2 """ ( -1 y '
(xt)r+(m/ 2)
;S
r.(m + r)l
00
1
= e 1 ;S
""" ( -1Y
(xtY
rl(m + r)l
1
00
t•
= ~ :;1 ~ ( -1Y rl(m + r)lx't'
ct)
1
00
-
""" ( -1 Y
,{:_
0
rl(m + r)lsl
xrtr+•
·
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)
'rl(m
1
1
r)l(n
·r)l'~r.
184
CH.6
LAGUERRE POLYNOMIALS
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
1
n
"" ( - 1Y - c--------,-:-,----------:-:-,xr
r!(m + r)!(n- r)!
f:o
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;,_ (x) }.
1
Integrating by parts, we have
I: e- L~(t)
1
L!(t)J: --
dt = [ -e- 1
J: -e-t.L~'(t)
dt
I: e-tL~'
=e-'"L~(x)- I: e-t{~ L:(t)}dt
= e-"'L~(x) +
(t) dt
1
r~o
(by theorem 6.11(iv)).
Thus
J e-tL~(t) dt +~ I e- L:(t) dt = e-'"L~(x)
oo
X
oo
r=O
1
X
and hence
(6.8)
185
PROBLEMS
Therefore
foo
e- 1L!(t) dt = ~ Joo e-tL:(t) dt r-0
X
2: Joo
r=O
X
e- 1L:(t) dt
X
= e-"'Lk(x)e-"'Lkn-1(x)
n
(by equation (6.8) above)
= e-"'{L~(x) - L~_ 1(x)}.
PROBLEMS
(1) Show that L~(O) = !n(n - 1).
(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 e duce t h at
f
oo
0
_,
e x
kL ( ) d
{0 if k < n
,. x x = ( _ 1)"n! if k = n.
(3) Prove that
f
x (x- t)mL,.(t) dt
=- --- m!n! ___ xm+tLnt+l(x).
o
(m + n + 1)!
n
(4) Show that
2 2"k!(n+k)!J'
1
•
,
1
L,.(x)
= (- l)n-n(Zk)j(Zn) - _1 (1 - t 2 )k-•Han{hlx)t} dt.
(5) Prove that
(6) Show that
r.,
L
C
"X~ I I ( /,:;(.~) 12 d.\:
(n I k)!(2 11 I ~~' I 1) .
1
11.
186
LAGUERRE POLYNOMIALS
CH.6
(7) If L!(x) is defined for non-integral k by the generalisation of the
result of theorem 6. 7
n
*
1
r(n + k + )
xr
-6 (-1)'-(n-r)!r(k+r+1)r!
Lk(x) -
1
show that V,/2(x) = ( -1)• 2n+I 1 H2n+Ih/x)
2
n.x
1
and
L;;lf2(x) = ( -1)n 22nn!Han( y'x).
(8) With L!(x) defined for non-integral k as in problem 7, show that
+ k + l + 1)
L nk+l(X ) = r(n
r(l)r(n + k + 1)
fl
0
k1
)l-1£k( ) d
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)
(7.1)
1
and
U..(x) =sin (n cos- x),!
(7.2)
for n a non-negative integer.
Theorem 7.1
(i) T ..(x) = ![{x + iy'(l - x 2)}n + {x - iy'(l - x2)}"].
(ii)
(x) = -!i[{x + iy'(l - x 2 )}n - {x - iy'(l - x2)}n].
u..
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 - x 2)}n + {x - iy'(l - x2)}n].
(ii) The proof is similar to that of (i), and so will not be given.
=
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 - x 2 )
{1/v'(l -·xa)l/l!,"(x).
188
CHEBYSHEV POLYNOMIALS
CH. 7
Theorem 7.2
I
[!n]
(i) Tn(x) = "" ( -1 )"
;S
[J(n-1)]
2
(ii) Un(x) =
- 1l_:_ .. (1 - x2)"xn-2r.
(Zr) !(n - 2r)!
I
( -1)" - - · · n.
'
-(1 - x2)'+'xn-2r-·!
(2r + 1)!(n- 2r- 1)!
·
PROOF
(i) From theorem 7.1(i) we have
Tn(x) = l[{x + iy(l - x 2)}" + {x -- iv(l - x 2)}n]
=
.21 [
2 "C,.xn-r{iv(l - x 2)}" + 2 "C,.xn-r{ -- iy(l
n
"
r~o
r~·
(by the binomial theorem)
1
=2
2 nc,.xn-r(l - x2)'12 ir{l + (-1)'}.
n
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:
U0 (x) = 0;
T 0(x) = 1,
U,(x) = v(1 -- x 2);
T 1(x) = x,
§ 7.1
DEFINITION OF CHEBYSHEV POLYNOMIALS
189
U2(x) = v(1 - x )2x;
Ua(x) = v(1 - x 2)(4x 2 -- 1);
U4(x) = v(1 - x 2 )(8x 3 - 4x);
U 5(x) = y(l - x 2)(16x4 - 12x 2 + 1).
T 2(x) = 2x - 1,
T 3(x) = 4x 3 - 3x,
T 4(x) = 8x4 - 8x 2 + 1,
T 5(x) = 16x5 - 20x 3 + Sx,
2
2
We note that Un(x) is not actually a polynomial, but is instead a polynomial multiplied by the factor v(1 - x 2), 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
d 2y
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),
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 - x 2))
= n v(1 ~ x2) sin (n cos- 1 x).
Also, ,
d2
dxzT..(x)
=
~{--n
___ -sin (n cos- x)}
dx v(1 - x 2)
1
.
nx
. (
n
-n
1 )
(1-x2)312sm ncos- x -!-(i~xz)i;2cos(ncos-tx).(1-x2)112
nz
(I
·
-
:~:2)
COS
(11 ('OS I ,\').
190
CHEBYSHEV POLYNOMIALS
CH. 7
Hence
(1
)d 2 T,.(x) _
_
2
X
dx2
X
dT,.(x)
2T ( )
dx + n " X
· (n cos -1 x ) - n 2 cos ( n cos -1 x )
( _ nxx 2) 112 Sill
1
· (n cos -1 x ) + n 2 cos ( n cos -1 x )
- ( _ nxx 2) 112 Sill
1
=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).
7.2 GENERATING FUNCTION
Theorem 7.4
(i)
1- t 2
~
= T 0(x) + 2 ~ T,.(x)t".
1 - 2tx + t 2
n=l
(l't')
v(1 - x2) =
1 - 2tx + t 2
~ un+l(X )t".
~
n=O
PROOF
Again we prove only result (i), the proof of (ii) being similar.
Let us write x = cos() = t(e 18 + e- 18 ), so that we have
1 - t2
1 - t2
1 - 2tx + t 2 - 1 - (ew + e-i6)t + ti.
1 - t2
= (1 - ewt)(1 - e-wt)
L (ewtY 2: (e-iot)•
00
= (1 - t2)
00
r=O
L
•=O
(by the binomial theorem)
00
= (1 - t2)
2:
e;<r-s)Otr+•
r, B:o::::::O
00
=
r,s=O
2:
00
ei(r- )Otr+s _
r,s=O
ei(r-a)6tr+•+2.
§ 7.2
GENERATING FUNCTION
191
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
.L ei{r-(n-r)}8 _ .L ei{r-(n-r-2)}8
n-2
= e-lnO L ei2r0 -- e-i(n-2)0 L ei2r0
n
n-2
r=O
r=O
n
r=O
r=O
.
1 _ (ei20)n-t
1 _ (ei28)n t 1
-~
-~-~
=e
1 - e2iB - e
1- ~
+
(summing to n
1 and n - 1 terms, respectively, the two geometric series which
both have the common ratio e 2w)
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
(on rearranging the terms)
= einO + e~in8
= 2 cos nO
'~ 2T,.(x),
(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)",
T 2,.(0) = ( -1)",
T2n+t(O) = 0.
(ii) U..(1) = 0,
U.. (-1) = 0,
U2..(0) = 0,
U2n+l(O) = ( -1 )".
PROOF
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 n
2
if n is odd
0
= { (-1)"' 2 ifniseven.
Thus all four results are proved.
7.3 ORTHOGONALITY PROPERTIES
Theorem 7.6
m
n
fl Ty(lm(x)T,.(x)
dx = {~/2 m =n
- x2)
::;;z=
(i)
n
-t
(ii)
::;;r:O
m = n = 0.
m :;;z'n
ft Um(x)
U..(x) dx = {~/2 m =n ::;;r:O
v(1 - x
0
-1
2
)
m = n =0.
PROOF
We prove result (i) only, the proof of (ii) being similar.
If we set x = cos 0 we have
§7.4
RECURRENCE RELATIONS
f !_r:'(~)~{j:~
l
- 1
dx =
y'(l -- x 2 )
fo _!m(~s O)T,.(cos~l( -sin 0 dO)
sin 0
"
=
193
f:
cos nO cosmO dO
(7.3)
(by definition (7 .1))
=
J:
=
1
![-- sin (n + m)O - __!___ sin (n :____ m)e]"
2 n+m
n-m
!{cos (n + m)O +cos (n - m)O} dO
o
(provided n - m 7"' 0)
=0.
If n - m = 0 we have from equation (7.3)
1
f
_
1
T,.(x)Tm(x)
_
v/( 1 _ x 2) dx =
J" cos nO dO
2
0
J:
i-( 1 + cos 2n0) dO
= -1 [ 0 - -1- sin 2n0]"
2
2n
o
(provided n :7: 0)
=-§-n.
If n = m = 0 we have from equation (7.4)
1
f
-1
T 0(x)T0(x) dx =
y'(l - x 2)
J" l dO
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 -- x 2 )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)
194
CHEBYSHEV POLYNOMIALS
CH. 7
(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 - cos 2 0) d( cosO) cos nO = -n cos 0 cos nO + n cos (n - 1)0.
But
2
2
(1 - cos 0) d(c:s O) cos nO = sin 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) =
+l(x) - xU..(x).
If we replace x by cos 0 and use the consequences of definitions (7.1)
and (7 .2) that
T,.(cos 0) =cos nO
and
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
= sin nO cos 0 + cos nO sin 0 - cos 0 sin nO
= cos nO sin 0,
which proves the result.
u..
+
§ 7.5
195
EXAMPLES
Example 2
1{
1
}
Show that ~ T 2r(x) = 2 1 + y(l _ x 2) U2n +I(x) .
n
Again we write x =cos 0, so that we have
n
=
L T2r(
L" cos
COS 0)
r=O
=
2r()
r=O
= Re
L"
ei2rB
r=O
1 - ei(2n+2)0
=
Re
1 -- ei26
(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 _ ei2e _ e i20
1
+
-
= 2 ~-!-cos
___1{1
2
=
20
[(1 - cos 20){1 - cos (2n + 2)0} +sin 20 sin (2n + 2)0]
COS (2n
2() .
}
+ 2)0 -j- 2sinsi~i
O Sin (2n + 2)(J
-2~{1 + sin (2n + 2)0 cos e - cos (2n + 2)0 sin 0}
sin 0
=· _1_ [ 1 I sin_{~(2n _-_!_-_ ~)0_-=--_~J
sin 0
2
sin (2n I 1)0}
sin()
{12,. '.(.\') }·
\(I
~~
II
·'·2)
196
CHEBYSHEV POLYNOMIALS
CH.7
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).
'
n
(3) Prove t hat T ..(x) = v'( _ x 2) U,.(x).
1
(4) Show that 2{T,.(x)}2 = 1 + T 2,.(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).
(7) Show that {1 / v'(1 - x 2)}U,.(x) satisfies the differential equation
d 2y
dy
(1 - x 2)dx 2 - 3xdx + (n 2 - l)y = 0.
(8) Use Chebyshev's differential equation and the equation in problem (7)
above to show that
L
[!n]
(n- r - 1)'·(2x)n-2r
T,.(x) = n( -1 )'
2 r=o
r!(n- 2r)!
and
(n- r- 1)1
rHnL-1)1
U,.(x) = y'(1 - x 2)
(
r=o
-1Y ------·(2x)"- 2r- 1 •
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
CH. 8
GEGENBAUER AND JACOBI POLYNOMIALS
Theorem 8.1 (Power series expansion)
.t
_
f!nl _
r(n - r + A.)
c..(x)- ~ ( 1) r(A.)r!(n- 2r)! (2x)
r
n-2r
.
Theorem 8.2 (Orthogonality property)
f
l
(1 -
-l
2)A-l ;.( )C;.
X
C,. X
) d - 21-2).
m(X
X-
r(n + 2..1.)
/nm·
~\n + A.){r(A.)}2r(n + 1
Theorem 8.3 (Recurrence relations)
(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 - x 2 )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 ) dx 2 - (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
zat{J
i.e.,
L p~a,fl>(x)t".
00
=
(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
199
JACOBI POLYNOMIALS
Theorem 8.5 (Series expansions)
(i) P~«,fJ>(x)
~
=
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
(ii) p<rz,f1>(x) = ~
..
f(n+cx+l)r(n+r+oc+fJ+l)
£:6 f(oc-!-r+l)f(n+cx+fJ+1)(n-r)!r!
2
·
(x-l)r.
2
(iii) p<rz.f1>(x) = ~ ( -l)n-rf(n+fJ+l)f(n+r+IX+fJ+1)(x+l)r.
n
f(fJ+r+l)f(n+cx+fJ+l)(n-r)!r!
2
6
Theorem 8.6 (Orthogonality property)
J~ (1 - x)'"(l + x'f~a,f1>(x)P~f1>(x) dx
1
=
zaif:l+lr(n-\- IX-\- l)f(n + {J-\- 1)
~nm•
(2n +a: +fJ + l)n!f(n +ex +fJ + 1)
Theorem 8.7 (Recurrence relations)
(i) 2n(a: +fJ +n)(oc -1-{J +2n- 2)P~"'·f1>(x)
= (1X+{J +2n -1){oc 2 -{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_+11 •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
(1
d~
- x 2)~ 2 -1 {{J
IX
dy
(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 '
(8.3)
T,.(x) = -n lim C!~(x) (n > 1);
2 .1.-->-0 II
(8.4)
Un(x) = v(l - x 2)C~_ 1 (x).
(8.5)
These results will be proved in the next chapter (Example 5, p. 215).
There also exist relationships with the Laguerre and Hermite polynomials, which we shall state but shall not prove:
L~(x) = lim P~"'·fl>(l -
2xj(3);
fl---+00
Hn(x) = n! lim A. -nt 2 C~(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
! C~(x) = UC~:l:l(x),
so that
t2 C~(x) = 2A.:xC~:!=l(x)
= 21..2(1. + l)C~:J:~(x)
=
2 2A.(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.
=2m IJ~ + m)ci.+m( )
r(A)
n-m X·
+ m - l)C:.~:(x)
(8.6)
(8.7)
201
PROBLEMS
Example 2
Show that
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+"}
-6
-
"
n I·
{ d•
dn-r 1-x"'+"}
- l+xP+"} { ~r!(n-r)! dx'(
)
dxn-r(
)
=
L-r. nn:_ r)I(,B+n)(,B+n-1) ... (,B+n-r+1)(1+xf+"-'
"
I
I(
r=O
X ( -t)n-•(rx+n)(rx+n-1) ... (rx+n-n+r+1)(1-x)"'+n-n+r
=
~ -~-( -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).
(2) Show that P,~·{J)(l)
r(rx + n + 1)
- ~ -~~--- -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
r=O
A)C;.(x) = jn±_~A)C~x) -:: 0 + 12C~+I(:)
2(1 - x)
r
(5) Show that
p~.P-t>(x) IH
_
(6) Show that C,._
1(x)- (
21
p<:-l,fl>(x) = P~~>(x).
)
_1 )!! ~
dx1p ,.(x.
1
·
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
(9.1)
The notation
is also often used.
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 + oc 2, while for x = -1
it converges if {J > oc 1 + oc 2 - 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) =
..
m
2
F 1 ( -n, n + 1; 1;
1
2
x).
(n+m)! (1-x 2 )mi 2
•
• 1 (
2mm! 2F1 m-n, m+n+1, m+1, - 2-
(u) P,. (x) = (n -m)!
x) i
(iii) I ..<x> = e:~G)"'lFl(n + !; 2n + 1; 2ix).
(iv) H 2n(x) = ( -1 )" ( ~)! 1F 1 ( -n; ! ; x 2).
n.
2
_
(v) H 2.. +l(x)- ( -1}
n 2(2n
+ 1)! x F ( -n,. - , x ).
n.1
(vi) L,.(x} = 1F 1( -n; 1; x).
..
k
(vu) L,.(x) =
1 1
3
•
2
2
r(n + k + 1)
(
.k
.
n !r(k + 1) 1F 1 -n, + 1, x).
(viii) T,.(x) = 2F 1 ( -n, n; ~;
1
~ x).
2}
(ix) U,.(x) = v(1 -- x n 2Fl( -n + 1, n + 1; ~;
1
~X).
" r(nn!r(+2A.)ZA.) F ( -n, n + ZA.,• A.+ 1 -2 - x) .
(x) C,.(x) =
2
1
1•
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
205
DEFINITION OF HYPERGEOMETRIC FUNCTIONS
( x1') P,.(tx,Pl(x) -_
F(- n, n+oc+fJ+1,.oc+1,.1-x)·
2
r(n+oc+1) 2
n!r(oc+ 1) 1
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),
2
F (-n n + 1 . 1 . 1 - x) = ~ ~-_:-n)r(n _j-_1).{(1-.:- x2f2}'
I
'
'
'
2
Lt
(1) r
.
r=O
T.I
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
(n + 1), = (n + 1)(n + 2) ... (n + r)
= (n + r)!/n!
and
(1), = 1.2.3 ... r
=r!
so that we now have
2
..
1-x)-~
r
n!
(n+r)!1(1-x)'
F 1 -n, n + 1, 1, - -- - L ( -1) ( __ )'
, ---;--;--2
r=o
n
r . n.1 r. 2 r.
(
~ ( -1)'
(n + r)!
= 6 ----z;:- (n - r) !(r !)2 (1 - x)'
=
~
(n + r)!
£:o 2•(n - r) !(r !)2
(x - 1 r
)·
(9.2)
To show that this is the same as P,.(x), it is easiest to write P,.(x) as a
1). We do not as yet have such a power series, but we
power series in (x
usc Taylor's theorem to write
1)"
J>,.(x)
where hy p!~l( I) we nwan tht· rth lkrivativl' of /',(.') n.llu.ltnl .11 .\
I.
206
HYPERGEOMETRIC FUNCTIONS
CH.9
To calculate P,.<rl(1) we use the generating function for the Legendre
polynomials given in theorem 3.1:
1
y'(1 - 2tx + t2) =
Pn(x)tn,
6
00
so that, by differentiating r times with respect to x, we have
oo
dr
P,.<r>(x)t"' = dxr (1 - 2tx + t2)-l/2
L
n~o
1)( -i - 2) ... ( -! - r + 1)(1 - 2tx + t 2)-!-r
= ( -2t)'( -!)( -! -
= 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.
7
2 r!
Hence, setting x = 1, we have
~
L P,.<r>(1)t" =
n=O
tr(2r)1
rr!. (1 - 2t + t 2)-~-r
2
tr(2r)
= _
_I· (1 _ t)-1-2r
2rr!
7
_ t (2r)!{
-
2rr!
+
)
(2r + 1)(2r + 2) 2
(
1 + 1 + 2r t +
2!
t
}
(2r + 1)(2r + 2)(2r + 3) 3
3!
t + ...
(by the binomial theorem)
7
= t (2r)!
27 r!
=
~ (2r + s)! t•
~ (2r)!s!
_1_ ~ (2r + s) ! tr+•
zrr! L
s!
s~o
__1_ ~ (r + n) I tn
- 27 r! ;S (n - r)! '
writing r + s = n.
Equating coefficients of tn gives
p<r>(1) = - 1- (r + ~)!
n
2rr! (n - r)!
for n > r
0
for n < r.
=
§ 9.2
PROPERTIES OF THE HYPERGEOMETRIC FUNCTION
207
Hence, from equation (9.3), we have
P..(x) =
~ _1_ (n + r)! (x- 1Y
~ 2rr! (n- r)!
r!
(n + r)!
(x - 1)•
-- 6~ 2r(nr)!(r!)
2
=
X)
F 1 ( -n, n + 1; 1; 1- 2
(by equation (9.2)).
2
9.2 PROPERTIES OF THE HYPERGEOMETRIC FUNCTION
Theorem 9.2
PROOF
This follows immediately from definition (9 .1):
F(
2
I
rJ.,
fJ· • ) _ ~ (rJ.)r(fJ)r xr
' y' x (y ),
r!
f::o
Theorem 9.3
The differential equation
d 2y
dy
x(1 - x)dx 2 + {y - (rJ. + fJ + 1)x} dx- rJ.{Jy = 0
(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
x 1 -Y 2F 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)
2
i/ i'
FJ{r1., (I;)'; x)
- fl
· 0·
l'(r)
I '(;I) I'( i'
fl)
f
I til I( I
()
.\'f) ~ dt
208
HYPERGEOMETRIC FUNCTIONS
CH.9
PROOF
From definition (9.1) we have
,.FJ(r~., (3; y; x)
=
i
r=O
(l1.),((3),::
(y), r!
= ~ r{ll. + r)r((3 + r)r(y) ~
f:-6
r( (1.)r((3)r(y + r)
r(y)
~
r(y)
~
r!
= r{ll.)r((3)r(y - (3) ~ r{ll. +- r)
r(y - (3)r((3 + r) X'
r(y + r)
;!
X'
= r{ll.)r((3)r(y _ (3) ~ r (ll. + r)B(y - (3, (3 +- r) i!
(by theorem 2.7)
=
2:
J1
r( )
r
rc(J. + r) c1 - w-fl- 1tfi+r- 1 dt:.r
r{ll.)r{(3)r(y- (3) r=O
o
I
r!
00
(by definition (2.2) of the beta function, which
is valid if y - (3 > 0 and (3 + r > 0)
rcr)
= ----~--
r((3)r(y - (3)
=
J(1 -
2: rc(J. + r> (xt)' dt
00
1
t)Y-fl- 1 t {J- 1
0
-
r=O
r(ll.)
r!
J
r(y)
(1 - t)Y-fl- 1 t~'~- 1 (1 - xt)-a. dt
r((3)r(r - (3) o
(using the binomial theorem).
1
Theorem 9.5
Each of the following twenty-four functions is a solution of the hypergeometric equation:
V1 = 2F 1(ll., (3; y; x)
V 2 = (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)
V 6 =x 1 -Y 2F 1 (ll. I 1
y;(J i l
y;ll. I (3 1-1
y; 1 -x)
§ 9.2
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~ (1X, IX+ 1- y; ot + 1- {3;
1
vlO = ( -x)fl-y(l -
( -x)I-Y(1 -
D
xy-oc-fl2Fl(1 - {3, y - {3; IX + 1 - {3;
V11 = (1- x)-oc 2F 1 (1X, y- {3; ot + 1- {3;
V 12 =
209
PROPERTIES OF THE HYPERGEOMETRIC FUNCTION
1
~
D
J
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
vl4 = ( -x)oc-y(1 - xy-oc-fJ2Fl(1 - oc, y V15 = (1 - x)-P 2F 1 ({3, y - IXi {3 + 1 - IXi
D
ot;
1
f3 + 1 - oc;
~
D
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 = x 1-Y(1 -
xy-oc-fl~1 (1 -IX, 1 -
{3; 2- y; x)
V19 = x 1-Y(l - xy-oc-l 2F 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)y-a.-P~t(Y -IX, y - {3; y
v21 = (1 -
X :
1)
+ 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 -
v2~
.\JI
1
'(1
x)l' "' fi2FI(?'
fl, 1
fl; I' i 1
oc
fl; 1
D
D·
210
HYPERGEOMETRIC FUNCTIONS
CH.9
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
V1 = V2 = Va = V4,
V 5 = V 6 = V7 = V 8 ,
V 9 = V 10 = V11 = V12 ,
v13 = v14 = v15 = v16•
and
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 2F 1( IX± 1, f3; y; x), 2F 1( IX, f3 ± 1; y; x),
F
2 1(1X, {3; y ± 1; x) are said to be contiguous to 2F 1(1X, {3; y; x). It may
be shown that between 2F 1(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 6 C 2 = 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
211
Theorem 9.7 (Integral representation)
. {J• ) -
1F 1(IX,
'X
-
fl
r{{J)
(1
t)P-a-lta-1 ...:>ot dt
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~ + { -~ + ~ +!-2 m2}y = 0
dx 2
4
x
x
(9.4)
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
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
x 1 f\J.\(1X --· {J + 1; 2 · {J; x),
we have, from the theorem, that two independent choices for :;; arc
ll · · m; 1 2m; x)
1F,O
and
I~
·,1
I'
· m; I
, 2m; .r),
212
HYPERGEOMETRIC FUNCTIONS
CH.9
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___., 121F 1(!- 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 hypergeometric equation:
V1 = 1F 1(rx; {J; x)
V 2 = x1 - 1\F1 (1 + rx - {J; 2 - {J; x)
V3 = e"' 1F 1({J- rx; {J; -x)
V 4 = x 1 -fi e"' 1F 1 (1 - rx; 2 - {J; - x).
PROOF
By change of variable, direct substitution and use of theorem 9.6.
The four confluent hypergeometric functions 1F 1 (rx ± 1; {J; x), 1F 1 (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 4 C2 = 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)'
From theorem 9.4 we have
r(y) _ {J)
F1 (rx, {J ,. Y ., 1) -- r({J)r(y
f
=
fl
2
1
0
tfi- 1(1 - t)y-fi- 1(1 - t)-"' dt
r(y)
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
213
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)'
Example 2
Show that
F 1 (rx, p; y; x) = (1 - xt"' 2F 1 ( rx, Y - f1; y; x ~ )
1
(i.e., V1 = V 3 , using the notation of theorem 9.5).
If we set -c = 1 - t in theorem 9.4 we have
2F1 (rx, j1; y; x)
2
=
=
=
=
r(y) .. - f[ (1--c)P- 1-cY-fl-l{l- x(1--r)t"' dr
r(j1)r(y- p) 0
r(y)
r(p)r(y--/1)
f
(1--c)fl-J'fy-fl-!(1-x)-OC(l
0
r(y)
(l-xt"'
r(y--p)r{y-(y-/1)}
-_:__.)-IX
dr
x-1
fl Ty-fl-l(l-r)fl-!(1 - __:_T)-oc
dr
x-1
0
(1-x)-"' 2F 1 ( 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} 2F 1(rx, p; y; x)
'
+ (y - rx)(y - j1)x 2F 1(rx, j1; y + 1; x)
- y(y - 1)(1 - x) 2F 1(rx, p; y - 1; x) = 0.
W c prove this result by expanding the hypergeometric functions in
power series. Since
~ (!X )r(tJ)r r
-- L _(_)_I x '
r-· o
Y ,r ·
we sec that the coefficient of x" on the left-hand side above must be
R.
•
oc
fl) ( oc),. 1(/l).. I I
()•),. 1(n
I)!
2F 1(rx, I'' y' x
J'( J'
I )(oc ),.(fl),.
(J'),.ll!
r(2J'
I
) _
214
I-IYPERGEOMETRIC FUNCTIONS
-j- (y-rl)(y-/'1)
{rl}n-1~,8)n-1
CH.9
_ y(y- 1) (rl)n{,B)~
(y--l),n!
(y+1),._1(n-1)!
-,yy- 1) {rl)n-1(/'1)n-1
(y-1)n-1(n-1)!
= ( _ 1)r(rl+n)r(/'1+n)r(y)
YY
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)!
1
(
11
+ (y-rl )( y-,,
)r(rl +n -1 )r(,B +n -1 )r(y + 1)
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!
+ y (y-
l)r(rl+n-1)r(,B +n-1 )r(y-1)
r(rl)r{t3)r(y+n-2)(n-1)!
= 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(y- 1)
(y-l-n-2)n
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)
.[y 2( -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
215
EXAMPLES
Example 4
Show that 1F 1(rx; {3; x) = e"' 1F 1({3- oc; {3; -x).
From theorem 9.7 we have
r({3)
J1 (1
t)fl-a.-lta.-1 xt dt
F 1(rx,. (3•'X ) = r(rx)r({J
_:-:_~ 0
e
.
1
If we make the change of variable t = 1 1
Fr(r:x; {3; x) = f'~)~~)=~
J:
'l', we have
7/l-a.-!(1 - T)"'- 1 ex(l-•) dr
:t
r(f1)
Jl (1 - e r({J -- rx)r(rx) 0
)<X-I /l-<X--1
'l'
T
e
-XT
d
T
= e"' 1 F 1({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)
_ __
__ I'(n+A-J+1)
(___
. _
, .
- n!I'(A-!+ 1) 2F1 n, n-+ A 12 +A ! 1--1, A 121 1,
1-x)
_ r(n+A+i)
(
.
1 •
n!I'(A+i)- 2F 1 -n, n+2A, A+ 2 , - 2-
0'-'
-- I'(n_±~t~~ ~!E(2A) C\x)
- n!r(A-H) r(n+2A) n
(by theorem 9.1 (x))
=
~]23)~(~_-l-A ~1-!) C'( )
r(A-1 Dr(n I 2A) "x.
(ii) From theorem 9. l(x) we have
(';.(x)
l'(n: 2).)
·(
". ( _,,) ~/· 1
IJ'?l
fl,fl
·
,, I
2).;}. · "'
.\:)·
2
1-x)
2
216
HYPERGEOMETRIC FUNCTIONS
CH.9
The trouble at A = 0 comes from r(2A). Hence we have
-n n+2A.' A+:!, !_-x)
2
_nlim 1 .r(n) F(
. .1-x)
-2 Ar(2A) nT
-n, n,
-.z-
~ lim C~(x) = ~ lim --~- . lim r(n -t-2A) 2F1 (
2 J.-+o A
2 ;._,.o Ar(2A) .<-+0
n!
1
s 1
J.->O
1·
'
2•
1
=
!~ 2A:( 2Aj. F -n, n; !; ~x)
2
1(
l_ (-n n· 1-x)
2
= 1. F (-n, n; ·!; --1-x)
-2
-lim _ _
F
<->O r(2A + 1)" 2 1
2
=
1·
,
' 2 ,
1
T,.(x)
(by theorem 9.1 (viii)).
(iii) From theorem 9.1(x) we have
r(n - 1 + 2)
(
.
1
-X)
C,._ 1(x) = (n _ l)!r(2)2F 1 -n + 1, n - 1 + 2, 1 + 2 , - 2 I
r(n -t- 1)
(
.
•
1 --
= n!r(2) n 2F1 -n + 1,n + 1, 32 , 2-
I.
X)
1--x)
(
= n 2F 1 -n + 1; n + 1; ~; - 21
v(1 - x2) Un(x)
(by theorem 9.1(ix)).
PROBLEMS
(1) Show that
(i) (1 - xt"' = 2F 1 (rt., {3; {3; x);
(ii) In (1 - x) = - x 2F 1 (1, 1; 2; x);
(iii) e"' = 1F 1(oc; rt.; x).
(2) Show that 1F 1 (rt.; y; x) = lim 2 F 1 (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
FI(r:t., 1
2
_
.
. 1 _
r(tr)r(tr + i)
.
oc, y, 2 ) - r(loc + tr)r(t- irx + tr)
(4) Show that
d ~ (oc, {J; y; x) = oc{J ~ (oc + 1, {J + 1; y + 1; x)
dX
y
and deduce that
1
1
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 - fJhF 1(rx, {J; y; x)
= oc 2F 1(oc + 1, {J; y; x) - {J 2F 1(oc, {J + 1; y; x);
(ii) (oc - {J)x 1F 1(oc; {J + 1; x) + {J(x + {J- 1) 1F 1 (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)..2 2"+ 1n!xL!(x 2).
(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) =
and
y(x, 01:) =
•
f:
r
e-tt•:-I dt
(10.1)
e-ttx-l dt.
(10.2)
0
From definition (2.1) of the gamma function we see that
r(x, 01:) + y(x, 01:) = r(x).
(10.3)
10.2 EXPONENTIAL INTEGRALAND RELATED FUNCTIONS
The exponential integrals are defined by
x
et
Ei(x) =
- dt
-co t
cc e-t
E 1(x) =
- - dt.
and
f
f
X
f
(10.4)
(10.5)
§ 10.2
EXPONENTIAL INTEGRAL AND RELATED FUNCTIONS
219
We note that
E 1(x) = - Ei( -x).
(Some authors usc the notation ei(x) for E 1 (x).)
The logarithmic integral is defined by
dt
li(x) = o In i'
.
f
(10.6)
(10.7)
It is easy to see that
li(x) = Ei(ln x) = --E1( --In x).
(10.8)
y
y•Ei(x)
FIG. 10.1
Exponential integrals
The sine and cosine integrals are defined, respectively, by
'( ) = f"' sin
- t dt
Sl X
t
00
- -t dt
SI'(X) = f"' sin
(10.10)
f.
(10.11)
0
and
(10.9)
CI'(X )
-
'"
t
cost <It;
t
(again thl' rl'ader is warned to hewarc of dill"l'ITilt notations).
220
OTHER SPECIAL FUNCTIONS
li(x)
FIG. 10.2 The logarithmic integral
y
y=Si(x)
-1·0
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
(10.12)
(see Example 1, p. 225)
Si(x) = ~ + si(x),
si(x) =
(10.13)
~i {Ei(ix) - Ei( -ix)},
Ci(x) = !{Ei(ix) + Ei( -ix)},
Ei(±ix) = Ci(x) ± i si(x).
(10.14)
(10.15)
(10.16)
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.
(10.17)
By the corollary to theorem 2.6 we see that
erf oo = 1.
(10.18)
As a generalisation of the error function we define the functions
E,.(x) = r{(n
~ 1)/n}
J:
e-tn dt.
(10.19)
(N.B. E 1(x) given by this equation is not the same as the E 1(x) of the
previous section.)
We see'that
erf x = E 2(x).
(10.20)
The Fresnel integrals are defined by
S(x) =
and
C(x) =
J: ~t
2 dt
(10.21)
t 2 dt.
( 10.22)
sin
n
z
J 2
cos
0
It may be shown that
(I 11.2.1)
CH. 10
OTHER SPECIAL FUNCTIONS
y
~-------Y= erf x
X
FIG. 10.4
The error function
y
0·8
0·6
X
FIG. 10.5
Fresnel integrals
§ 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
1
2 -n;;
00
'(x) =
(10.25)
n=l
This series is divergent for x < 1, convergent for x > 1.
It may be shown that
(10.26)
y
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
x
Dn(x) =
tn
J
(10.27)
t - l dt.
oe 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
q,
de
(10.28)
F(k, cp) = o v'(1 - k2 sin2 e) (0 < k < 1),
J
E(k, cp) =
.
Il(k, cp, a) =
J:
y'(1 - k 2 sin 2 e) de
Jq,
0 v'(1
(0 < k < 1),
de
- k 2 sin 2 e)(l + a2 sin 2 e)
(10.29)
(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:
d(J
o v'(1 - k2 sin2 e)'
(10.31)
J:
(10.32)
n/2
K(k) =
J
12
E(k) =
Ili(k a) '
-
y'(1 - k 2 sin 2 0) dO,
n/2
de
o
v'(l - k 2 sin 2 0)(1 + a 2 sin 2 0).
J
(10.33)
§ 10.7
EXAMPLES
10.7 EXAMPLES
Example 1
Show that si(O) = -n/2.
From definition (10.9) we have
si(O) =
o sin t
J00
dt
t
= - roosint dt.
Jo
t
To evaluate this integral we consider the integral
l(rx) = roo e-~ sin t dt
J0 t
so that si(O) is given by -1(0).
Now,
di
drx = -
= -
roo
.
J 0 e-cxt sm t dt
oo
J
1
e-"'t--;(eit - e-it) dt
21
0
=
-~. roo {e<-cx+i)t- e<-cx-i)t} dt
21 J 0
=-!__[ 1
e<-cx+l)t_
-1C
~)
2i
=
-rx+i
1
i - oc
1
1
= -1
+ oc 2•
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
giving
0 = constant
2
'Jl
00
(-cx-i)t]
-rx-ie
o
- -I- constant
2
'
226
OTHER SPECIAL FUNCTIONS
Thus
l(r:t.) = - tan- 1 r:t. +
7l
so that
1(0) = - tan - 1 0 +
7l
and hence
si(O) = - 1(0) =
Example 2
Show that
J:
e-st si(t)
CH. 10
2
7l
2= 2
~.
2
-
1
- - tan- 1 s.
s
dt
Integrating by parts, we have
l
+~ J
oo e-st si(t) dt = [ --1 e-st si(t)
fo
s
= ! si(O)
s
00
_.o
-
foo --1 e-st-d si(t) dt
s
o
dt
00
s
e-st
0
i_ si(t) dt
dt
= ~(-~) + ~ roo e-st sin t dt
s
sJo
2
t
(using equation (10.12) and definition (10.9))
7l
1
= -2s + -/(s)
(with I as defined in Example 1 above)
= _:: + ~(- tan- 1 s + ~)
2s
2
s
(as shown in Example 1 above)
tan- 1 s
s
Example 3
Show that (i) erf ( -x) = - erf x;
(ii) I erf x I < 1.
(i) From definition (10.17) we have
erf x =
2
---
yn
f"' e- t• d t
o
§ 10.7
EXAMPLES
so that
r e-l' dt = _3__
f"' e-ll' • -dU
yn Jo
erf (-X) = ~
-X
yn Jo
(on changing the variable to u = - t)
2
= - yn
f"' u' du
Joe-
= -erfx.
(ii) First suppose x > 0. Then, since e-t' > 0, we have erf x > 0.
Also
~n(
erfx <
J: e- + J: e1
'
f""
2
= yn J o e-
dt
I'd
1
'
dt)
t
erf oo
=
=1
(by equation (10.18)).
A similar argument holds if x < 0.
Example 4
Show tl:at J: v(2 ~xcos x) = ~3 {F(J~· ~) - F(J~· ~) }12
We have
f
dx
nu/2
_-_d=-y_ _-c
fn/2
v(2 - cos x) =
" v{2 -- cos (:n - y)}
(making the change of variable y = n - x)
f"
dy
= J n/2 :Yc2 -~os i>
-f" --- +-~--f \/(3 -_
-
:tr/2
v{2
"i 2
=
1 -2 sin 2 (y /2)}
2dz
-,2::-s-:--in-2-z)
n/4
(making the change of variabk z - y/2)
2
r·~~
,;~ J , I \ 1 ( I
S I'
IJ
dz
~sin~::-)
228
CH. 10
OTHER SPECIAL FUNCTIONS
2 { rn/2
= y3
dz
rn/4
J 0 v(1- fsin 2 z)- J 0
dz
}
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 = ~·
(2) If l(rx) =
oo
f
COSt
(1 - e-"' 1) · - - dt, show that
0
t
d/
!X
2
drx
rx + 1
l(rx) = t ln (1 + rx2).
oo
1
Prove that
e-st Ci(t) dt = s ln (1 + s 2).
2
0
1
2x
(3) Show that erf x = yn y(!, X 2) = yn 1F 1(!; t; -x 2 ).
and deduce that
f
(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 2 T.
(111 of - ox 2
(5) Show that (i)
f
x
1
7l
C(t) dt = xC(x)-;; sin x 2 ;
2
0
(ii)
Jx S(t) dt = xS(x) + 2cos ~2x 2
7l
0
-
2.
7l
n/2
dx
f"/ dx
( 1 )
(6) Show that f o v(sin x) = o v(cos x) = ( v2)K vZ .
2
PROBLEMS
(7) By making the substitution x = 2 sin(), show that
2
dx
o v{(4 - x2)(9 - x 2)} = iK(i).
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
00
""'a
L,..
2n
(A1.1)
x2n
n=O
00
""'a
L,..
and
(A1.2)
2n+l x2n+l
n=O
where in both cases we have, from equation (3.7),
a,.+2 =a,.
(n - l)(l + n + 1)
(n + l)(n + 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 = a 2,.x 2", so that we have
u,.
a2,.x2n
-·-------
Un+l
a2n+2x2n+2
(2n + 1)(2n + 2)
(Zn - l)(Zn + l
so that
lim
Jl--lo-tr;
u,.
Unj)
1
--
+ 1) x 2
1
231
EULER'S CONSTANT
and the series is hence divergent if x 2 > 1 and convergent if x 2 < 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:
2: u.. is a series of positive terms, and if the ratio
00
if
u,.ju,H 1 can, for
n=O
n > some fixed N, be expressed in the form
(1)
I' + 0 -u,.- = 1 +-
n
nP
with p > 1 (where by 0(1/nP) we mean some function f(n) such that
Un+l
2: u,. is convergent if p, > 1 and
00
lim {f(n)/(1/nP)} is finite), then
n=O
1l--+-a)
divergent if p, < 1.
Here, since we are considering x = ± 1, we have
(2n + 1)(2n + 2)
(2n - l)(2n + l + 1)
and it is a matter of simple algebra to prove that
Un
u,.H
~= 1 +~ +
l(l + 1)(1 + n) __ .
u,+ 1
n
{4n 2 + 2n - 1(1 + l)}n
Now, we see that the last term is O(l/n 2), since we have
lim
n-+oo
[
1(1 + 1)(1 + n)
{4-n 2 + 2n- l(l + 1)}n
=
IJ
1
n2
1(1 + 1)(1 + n)n 2
1!---+00 {4-n2 + 2n z(l +1)"}~
lim
= 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
Un+l -
u, = f(l) + f(2) + ... + f(n) + f(n + 1) -
-f(l) - f(2) --- ... - f(n) +
=
J:
n+l
f
1
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
FIG. A2.2
Hence Un n
-
u,. < 0, i.e., Un n < u,..
n+ I
/(
f(x) dx
233
DIFFERENTIAL EQUATIONS
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 continuous, positive and decreasing.
Then
Un
1
1
1
1
1
1
[" 1
= 1 + 2 -j- 3 -j- • • • + ;; - J 1 ~ dx
= 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
(1 - x 2 )y" - 2xy'
+ {z(l + 1)- 1 :•x2}Y
·-- - - - - - - - - - - -
Legendre polynomials
Legendre functions of the
second kind
P';'(x) Associated Legendre
polynomials
= 0 Q';'(x) Associated Legendre functions
of the second kind
P,(x)
Q,(x)
---------------~~~
Bessel functions of the first kind
Bessel functions of the second
kind
1
H< >(x)}
H<!l(x) Hankel functions
]n(x)
Yn(x)
i:C~)}Modified Bessel functions
2
x y"
x 2 y"
(1 - 21X)~y'
I {/1 1 y 1 x 1 Y 1- (1X 2
2xy' I
{.~ 2
-
:xf']n(PxY)
0 XX Yn(PxY)
n 1 y 2)}y
1(1 I I)} y
0
' I H'ru·u
. I I'wsst· I f' unl'lt<Hls
.
·j,(.~)}
( ) Sp
)'I.\'
------ --- --- -----
234
APPENDICES
Equation
Solutions
y" - 2xy' + 2ny = 0
y" + (J. - x 2 )y = 0
Ha(x) Hermite polynomials
'Pn(x) = exp ( - !x 2)Hn(x)
·weber-Hermite functions
xy" + (1 - x)y' + ny = 0
xy" + (k + 1 - x)y' + ny = 0
Ln(x)
L!(x)
Laguerre polynomials
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)
x(l - x)yn + {y - (ot: + ,8 + l)x}y'
- ocpy = 0
x 2y" + (,8 - x)y' - ocy == 0
y" + { -
4
x
2F,(oc, {J; y; x)
Hypergeometric function
x 1 -Y 2F 1(oc+1-y; ,8-t-1-y; 2-y; x)
1 F/ct; {I; x)
Confluent hypergeometric function
i xl-f11 F 1(oc - {J + 1; 2 - {J; x)
Mk,m(x)} Whittaker's confluent
Mk,-m(x)
hypergeometric functions
.
I
2
- + -k + _t .4-_2m_ } y = 0
1
4
Jacobi polynomials
x
ORTHOGONALITY RELATIONS
Function
Relation
·-·-···----------··-~------------
Legendre polynomials
-·-- ---·---. ·-···
2
l
f
Pt(x)Pm(x) =
2F+1!5tm
-1
Associated Legendre
polynomials
+ m)! li
Jfl P"'( )P•( ) dx = (2[ 2(1
-f- 1)(/ - m)l
-l
!
,
1
X
T X
fl ~':(x)P';"~~
_
1
1 - x2
U'
dx =
J£.± m?!.,smm'
m(l - m)!
235
ORTHOGONALITY RELATIONS
Function
Relation
f:
Bessel functions
x]n(AtX)],.(J.Jx) dx =
~{]n+lJ.«J)FOiJ,
where At and AJ are roots of ]n(Aa) = 0
f:
Spherical Bessel functions
00
n:
im(x)jn(x) dx = 2
l 0"'"
------ -- - - - - -
~-----~------
Weber-Hermite functions
f:oo
f:oo
Laguerre polynomials
f~ e-"Lm(x)Ln(x) dx = Omn
Associated Laguerre
polynomials
J~ e-"xkL~(x)L!(x) dx = (n ;, kl1o..,.
Hermite polynomials
-----------
e
J_l
-----------1
I
!._m(x)Tn(x) ci=_ =
v'(l-x2)
{~/ 2
m # n
m = n # 0
m = n = 0
m #n
m =n ¥-0
m =n =0
n:
dx _{0
f
f
-1
Um(x)Un(x)
v'(l - x2)
l
(1 -
l
Jacobi polynomials
'P,.(x)'l'm(x) dx = 2"n!( v'n)o.m
-----~------------·-----------
Chebyshev polynomials
Gegenbauer polynomials
e-x'Hn(x)Hm(x) dx = 2"n!( y'n:)onm
-1
1
-
~/2
x 8)A- 1 C!(x)C~(x) dx
21 - 2"r(n + 2J..)
= n: n!(n + J.){r(J.)pOmn
J~ (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
"
:.J
It
236
APPENDICES
5 GENERATING FUNCTIONS
Function
Legendre polynomial
Generating function
(R = (1 - 2xt + t2)112)
co
L
~=
t"Pn(x)
n=O
Associated Legendre
polynomials
(2m) !(1 -
L
00
x2)m12
zmm!Rm+o
t'P,"'+m(x)
r=O
Bessel functions
Hermite polynomials
-~
----·-- -----
Laguerre polynomials
exp{ -xt/(1 -t)}
1 - t
ro
L
t"Ln(x)
n=O
ro
Associated Laguerre
polynomials
exp{- xt/(1 - t)} _ "" "Lk( )
(1 - t)k+l
- L t nX
Chebyshev polynomials
1 -
n=O
---w
00
2
t
=
T 0 (x) + 2 L
""
n=l
---------
Gegenbauer polynomials
Jacobi polynomials
1
R2A
t"Tn(x)
HINTS AND SOLUTIONS TO
PROBLEMS
CHAPTER 1
(1) The general solution is Ay 1(x) + By 2(x) in each case, where A and 1
are arbitrary constants and y 1 (x) and y 2(x) are given by
ro
1
1
+ n~ nn.s:9 ... (4n- 3tn,
(i) YI(x) =
Y2(x) = x
314
{
1+
,~ n!7 .11.15 ~ .. (4n + 3tn}
valid for all x;
(ii) y 1 (x) = e-"',
~ ( --1)nn!1 ( 1 + Z
1 + J1 + ... +;;
1) xn,
y 2(x) = ln x.e-x- L.,
n=l
valid for all x;
1
~ ( ---1 )n Zn !(n _ 3) 1xn,
co
(iii) y 1(x) =
y 2(x) = y 1 (x) ln x + 1 + ix + !x 2 -
:/6 x
3
+
+ ~ 2n!\~~n3)!{ - 2 ( 1 + ~ + ~ + · · · + n ~ 3
- _1 - - _1 __ - ! + ~}xn
n-2
n-1
n
2
'
(iv) yb) =
valid for all x;
~ 1. 4. 7 ... (3n -- 2)
1 + L., ---;nT- ----xn
3
n:.:..:: 1
= (1 - - x)- 1,
( )
Y~ .x
_ .1{
-- x'
·
~ R 11 14
1 I L.,
10 13 H)
0
0
0
II
I
0
0
0
(3n I S) }
\
(3n
7)' '
\'alid for I .\
I;
0
0
"
0
•
0
0
[
•
238
HINTS AND SOLUTIONS TO PROBLEMS
_
~ ( -1 )(-3) ... (4n 2 - 14n + 9) 2,.
(v) YI(x ) - 1 + .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!
(2n)!
x ,
1--,6
valid for I x I < 1 ;
(viii) Yl(x)
=! (
n-2
-1)"+1 32n-1( 1_ 2)1 ,xan-1,
n
.n.
Ys(x) = Y1(x) In x + x- 1 +- tx 2 + 3h x5
+
6(
<X>
1 { ( +.z1+ ... +;--1)2
-l)n+I32"(n-2)!n! 1-2 1
_ _1___ !}xan-1
n -1
n
valid for all x;
'
(ix) y 1(x) = 1 + ~x 2 + f4x 4 + z~X 5 + ?1ox 6 + rlhx 7 + ... ,
Y2(x) = x + i-x 3 + f:tx 4 + t1ox 5 + mx6 + s!hx 7 + ... ,
valid for all x.
(2) (i) Impossible; (ii) impossible; (iii) possible.
(3) The general solution is Ay 1(x) + By 2(x) where A and B are arbitrary
constants, and y 1(x) and y 2(x) are given by:
, ~2.7.16 ... (2n 2 -n+1) 1
.) ()
--,
(l ) '1X = 1 TL_,
n=l n!l.3 .5 ... (2n - 1)
x"
l};
y 2(x) = x- 112 { 1 + ~ ~~f~:__:_ (2n~. ·.l: n I 1)
L,
n .3. 5 ... (2n ·. 1)
x"
1L;:.::l
1
239
HINTS AND SOLUTIONS TO PROBLEMS
(ii) YI(x) = 1 -
:x'
_ _ 112 {
( ) -X
YzX
~(-3)(-1) ... (2n -5) 1}·
n!1.3 .5 ... (2n - 1) xn
1 +k
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).)
1
(iv) ~B(m ~ , p + 1)·(Make the change of variable y = xn.)
!B(!,n ~) ·_- vn
r(1/n)
.
k
h
.
n r{( /n)+!}(Maethecange;fva~~~~le
(v)n
1
2
(vi) B(t, !) = n. (Make the change of variable u = 1/(1
(6) r( -t) = -2-y'n; r( -f) = Tlo65 yn.
(7) Use theorem 2.12.
+ t).)
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 (x 2 - 1)1 as the product (x - 1)1 • (x
( -1 )<r-1)/2
(r + !)(r - 1)!
(4) Cr =
{0
2r{(r + 1)/2}!{(r- 1)/2}!
+ 1)1•
if r is odd
.
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.
II,.
2/n
if tl is even
{ 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 x 0 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:
f
)m(x)jn(x) dx = 0 (m :r':- n).
{jn(x)}Z dx = _!!___ .
2n + 1
(13) Take real and imaginary parts of the infinite series for ]n(i 3' 2x).
ro
-oo
(14) Take real and imaginary parts of the integral
J:
tj 0(i 3 1 2 t) dt; use
theorem 4.8(i) to help evaluate the integral.
(15) Take real and imaginary parts of the asymptotic form of J 0(i 3 12x),
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 =
exp (- t 2 + 2ixt) dt.
f:
e-t' cos 2xt = Re
00
J: "'
241
HINTS AND SOLUTIONS TO PROBLEMS
1)" 22" I- 2 c.c'
y'n
I
00
o
e- 1'tZn+lsin2xtdt.
(4) Using the result of problem 3 for both H.,(x) and Hn(Y) gtves
(y)
1
Joo Joo exp ( -u
L~ H "(x)H
~ t" = - exp (x + y
2n.
-oo -oo
n
11=0
2
:n:
2
)
2
-
v 2 + 2iux
+ 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
HINTS AND SOLUTIONS TO PROBLEMS
CHAPTER 10
( 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
lJniversity 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., Cambridge 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
definition of 187
generating function for 190
orthogonality properties of 192
recurrence relations for 193
series for 188
special values of 192
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
Chebyshev polynomials
I H7
!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 207, 208
Hypergeometric functions 203
contiguity relations for 210
definition of 203
integral representation of 207
relationship with other special functions 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 120
graphs of 137
Kummer function 204
Kummer's equation 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
247
INDEX
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 108
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