Page 176 - Determinants and Their Applications in Mathematical Physics

4.12 Hankelians 5 161

From (4.12.28) and (4.12.33),
2

G n−1 G n−2
xu n−1 = F n G n ,
F n−1 G n G n−1 G n−1
u n u n−1 = (1 − x n−1 )(1 − x n+1 ) v n . (4.12.35)
n 2
u 2 n−1 x(1 − x ) v n+1
From (4.12.16),
2

F n−1 E n−1 E n G n−2
= −
G n−1 G n−1 G n−1 G n−1

E n−1 E n E n G n−2
= 1 −
G n−1 E n−1 G n−1
E n
1 x n
x(1 − x n−1 )
1 −
2
u n−1 = v n 1 − x n 1 − x n
x (1 − x)
n
= v n .
n 2
(1 − x )
Replacing n by n +1,
1 x n+1 (1 − x)
= v n+1 . (4.12.36)
)
u 2 (1 − x n+1 2
n
Hence,
2
1 1 − x

n+1
2
= v n . (4.12.37)
u n
u n−1 x 1 − x n v n+1
Eliminating v n /v n+1 from (4.12.35) yields the differential–difference equa-
tion
n+1
1 − x
u n = u n−1 . (4.12.38)
1 − x n−1
Evaluating u n as defined by (4.12.33) for small values of n, it is found that
2
3
4
1!(1 − x ) 2!(1 − x ) 3!(1 − x )
u 1 = , u 2 = , u 3 = . (4.12.39)
(1 − x) 2 (1 − x) 3 (1 − x) 4
The solution which satifies (4.12.38) and (4.12.39) is
n!(1 − x n+1 )
u n = G n = . (4.12.40)
(1 − x) n+1
F n
From (4.12.36),
(1 − x) 2n−1
E n−1
v n = = ,
(n − 1)! x
2 n
E n
which yields the difference equation
(n − 1)! x
2 n
E n = E n−1 . (4.12.41)
(1 − x) 2n−1