1.8 The Incomplete Beta Function
19
hence
2 F 1 ( ˜
a, ˜
b; ˜
c; z) = (1 − z)
b−1
2 F 1
1 − b, 1; a + 1;
z
z − 1
,
(1.58)
and thus we get from Eq. (1.47)
t 0 = 1 , t n+1 =
1 − b + n
a + 1 + n
t n .
(1.59)
This series becomes finite and hence exact for b = n + 1 and infinite for a =
−(n + 1).
Case 5 With the following transformation the series will converge within the unit
circle with center at 1.
2 F 1 ( ˜
a, ˜
b; ˜
c; z)
=
Γ (˜ c)Γ ( ˜
c − ˜
a − ˜
b)
Γ (˜ c − ˜
a)Γ ( ˜
c − ˜
b)
2 F 1 ( ˜
a, ˜
b; ˜
a + ˜
b − ˜
c + 1; 1 − z)
+ (1 − z)
˜
c−˜ a− ˜
b Γ ( ˜
c)Γ ( ˜
a + ˜
b − ˜
c)
Γ ( ˜
a)Γ ( ˜
b)
2 F 1 ( ˜
c − ˜
a, ˜
c − ˜
b; ˜
c − ˜
a − ˜
b + 1; 1 − z).
(1.60)
With, Eq. (1.44a),
˜
a = a , ˜
b = 1 − b , and ˜
c = a + 1,
Equation (1.60) simplifies to
2 F 1 ( ˜
a, ˜
b; ˜
c; z) =
Γ (a + 1)Γ (b)
Γ (1)Γ (a + b)
2 F 1 (a, 1 − b; 1 − b; 1 − z)
(1.61)
+ (1 − z)
b a
−b
2 F 1 (1, a + b; b + 1; 1 − z).
and hence for the first summand becomes
t
(1)
0 = 1, t
(1)
n+1 =
a + n
n + 1
t
(1)
n , and the 2nd one
(1.62)
t
(2)
0 = 1, t
(2)
n+1 =
a + b + n
b + 1 + n
t
(2)
n .
(1.63)
Equation (1.61) diverge for a negative integer.