Sequences, Series, and Special Functions 645where ai = 1 +a - c and c1 = 2 - c. Hence ( = (a1,c 1 ;z) = (l +
a - c, 2 - c; z) and
w = z^1 -e(l +a -c,2-c; z)
Thus if c is not an integer, both solutions (a, c; z) and z^1 -e(l +a-c, 2-
c; z) make sense and are linearly independent. So the general solution
of (8.23-5) is given by
w = A(a, c; z) + Bz^1 -e(l +a - c, 2 - c; z)
where A and B are arbitrary constants. For the case c a positive integer,
we refer the reader to (22].
EXERCISES 8.1 O
- Show that:
(c -b)n
(a) F(-n,b,c;l) = (c)n , c =/= 0, -1, -2, ...
( ) ( ) ( )
n (1 + b - C )n ..J_b F -n,b+n,c;l = -1 (c)n , c 1 0, -1, -2, ...
00
[(b)n]
2r( c )r( c - 2b)
(c) 2:::- 1 -( )-= 2 , Re(c-2b) >0, cfO, -1, -2, ...
n=O n. c n [r( c - b)].- Show that:
1
1r/Z d() 7r ( 1 1 )
(a)K(z)= J. =-
2
F -
2
,-
2,l;z
o 1-zsm^2 ()
(b) E(z)= 11r/Z -./1-zsin^2 0d()= iF(~,-~,l;z)
(Complete elliptic integrals)- Show that if jzj < 1, jz(l - z)I <^1 / 4 and a+ b + % =/= 0, -1, ... , then
F(2a, 2b, a+ b +^1 / 2 ; z) = F(a, b, a+ b +^1 / 2 ; 4z(l - z))
Deduce that
1; 1; ) r( a+ b +
1
/2)r(1/2)
F(2a, 2b, a+ b + 2i 2 = r( a+ i/2)r(b + i/2)4. To abbreviate the notation, let F = F( a, b, c; z ), Fa+ = F( a + 1, b, c; z),
Fa-= F(a -1, b, c; z), and similarly, denote by Fb+, Fb_, Fe+• Fe-the
other four contiguous functions. Prove the following.
(a) (a - c + l)F = aFa+ - (c - l)Fe-
(b) [a+ (b - c)z]F = a(l - z)Fa+ - C^1 (c - a)(c - b)zFe+(c) (1 - z)F =Fa-- C^1 (c - b)zFe+
(d) (1-z)F = Fb-- c-^1 (c - a)zFe+