Determinants and Their Applications in Mathematical Physics

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)