4.12 Hankelians 5 161
From (4.12.28) and (4.12.33),
xu
′
n− 1
=
(
Gn− 1
Fn− 1
) 2 (
Fn
Gn
)(
Gn− 2
Gn− 1
)(
Gn
Gn− 1
)
,
unu
′
n− 1
u
2
n− 1
=
(1−x
n− 1
)(1−x
n+1
)
x(1−x
n
)
2
[
vn
vn+1
]
. (4.12.35)
From (4.12.16),
(
Fn− 1
Gn− 1
) 2
=
En− 1
Gn− 1
−
(
En
Gn− 1
)(
Gn− 2
Gn− 1
)
=
(
En− 1
En
)(
En
Gn− 1
)[
1 −
(
En
En− 1
)(
Gn− 2
Gn− 1
)]
1
u
2
n− 1
=vn
(
x
n
1 −x
n
)[
1 −
x(1−x
n− 1
)
1 −x
n
]
=
x
n
(1−x)
(1−x
n
)
2
vn.
Replacingnbyn+1,
1
u
2
n
=
x
n+1
(1−x)
(1−x
n+1
)
2
vn+1. (4.12.36)
Hence,
(
un
un− 1
) 2
=
1
x
(
1 −x
n+1
1 −xn
) 2 [
vn
vn+1
]
. (4.12.37)
Eliminatingvn/vn+1from (4.12.35) yields the differential–difference equa-
tion
un=
(
1 −x
n+1
1 −x
n− 1
)
u
′
n− 1. (4.12.38)
Evaluatingunas defined by (4.12.33) for small values ofn, it is found that
u 1 =
1!(1−x
2
)
(1−x)
2
,u 2 =
2!(1−x
3
)
(1−x)
3
,u 3 =
3!(1−x
4
)
(1−x)
4
. (4.12.39)
The solution which satifies (4.12.38) and (4.12.39) is
un=
Gn
Fn
=
n!(1−x
n+1
)
(1−x)
n+1
. (4.12.40)
From (4.12.36),
vn=
En− 1
En
=
(1−x)
2 n− 1
(n−1)!
2
x
n
,
which yields the difference equation
En=
(n−1)!
2
x
n
(1−x)
2 n− 1
En− 1. (4.12.41)