18
1 Gamma Functions, Beta Functions, and Related
Case 2 Due to Eq. (1.44b) we set
˜
a = a + b , ˜
b = 1 , and ˜
c = a + 1,
(1.50)
and thus we get from Eq. (1.47)
t 0 = 1, t n+1 =
a + b + n
a + 1 + n
t n .
(1.51)
Similar to case 1, the series expansion becomes finite and hence exact for a +
b = −n, with n positive integer, and for a = −n − 1 singular. Again, for
arbitrary a, b the computation of B z (a, b) converges only within the unit circle
|z| < 1.
Case 3 This case is based on the following transformation formula:
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
−˜ a
2 F 1
˜
a, ˜
c − ˜
b; ˜
c;
z
z − 1
(1.52)
and Eq. (1.44a)
˜
a = a , ˜
b = 1 − b , and ˜
c = a + 1,
(1.53)
hence
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
−a
2 F 1
a, a + b; a + 1;
z
z − 1
,
(1.54)
and thus we get from Eq. (1.47)
t 0 = 1 , t n+1 =
(a + n)(a + b + n)
(a + 1 + n)(n + 1)
t n .
(1.55)
This series becomes finite and hence exact for a + b = −n and diverges for a =
−n − 1. Case 3 and 4 are more favorable than case 1 or 2 for |z| >
z
z−1
< 1.
Case 4 Similar to case 3 we are using the transformation formula
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
− ˜
b
2 F 1
˜
b, ˜
c − ˜
a; ˜
c;
z
z − 1
(1.56)
and Eq. (1.44b)
˜
b = 1 − b , ˜
c − ˜
a = 1 , and ˜
c = a + 1,
(1.57)
1 Gamma Functions, Beta Functions, and Related
Case 2 Due to Eq. (1.44b) we set
˜
a = a + b , ˜
b = 1 , and ˜
c = a + 1,
(1.50)
and thus we get from Eq. (1.47)
t 0 = 1, t n+1 =
a + b + n
a + 1 + n
t n .
(1.51)
Similar to case 1, the series expansion becomes finite and hence exact for a +
b = −n, with n positive integer, and for a = −n − 1 singular. Again, for
arbitrary a, b the computation of B z (a, b) converges only within the unit circle
|z| < 1.
Case 3 This case is based on the following transformation formula:
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
−˜ a
2 F 1
˜
a, ˜
c − ˜
b; ˜
c;
z
z − 1
(1.52)
and Eq. (1.44a)
˜
a = a , ˜
b = 1 − b , and ˜
c = a + 1,
(1.53)
hence
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
−a
2 F 1
a, a + b; a + 1;
z
z − 1
,
(1.54)
and thus we get from Eq. (1.47)
t 0 = 1 , t n+1 =
(a + n)(a + b + n)
(a + 1 + n)(n + 1)
t n .
(1.55)
This series becomes finite and hence exact for a + b = −n and diverges for a =
−n − 1. Case 3 and 4 are more favorable than case 1 or 2 for |z| >
z
z−1
< 1.
Case 4 Similar to case 3 we are using the transformation formula
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
− ˜
b
2 F 1
˜
b, ˜
c − ˜
a; ˜
c;
z
z − 1
(1.56)
and Eq. (1.44b)
˜
b = 1 − b , ˜
c − ˜
a = 1 , and ˜
c = a + 1,
(1.57)
