Determinants and Their Applications in Mathematical Physics

4.12 Hankelians 5 155

=

1

2

n+1

πi

∫

C

(ζ

2

−1)

n

(ζ−x)

n+1

dζ (putζ=x+2t)

=

1

2 n+1πi

∫

C′

[(x+1+2t)(x−1+2t)]

n

(2t)n+1

dt

=

1

2 πi

∫

C′

g(t)

t

n+1

dt

=

g

(n)

(0)

n!

,

where

g(t)=

[{

1

2

(x+1)+t

}{

1

2

(x−1) +t

}]n

. (4.12.5)

The lemma follows.

[(u+t)(v+t)]

n

=

n

∑

r=0

n

∑

s=0

(

n

r

)(

n

s

)

u

n−r

v

n−s

t

r+s

=

2 n

∑

p=0

λnpt

p

where

λnp=

p

∑

s=0

(

n

s

)(

n

p−s

)

u

n−s

v

n−p+s

, 0 ≤p≤ 2 n, (4.12.6)

which, by symmetry, is unaltered by interchanginguandv.

In particular,

λ 00 =1,λn 0 =(uv)

n

,λn, 2 n=1,λnn=Pn(x). (4.12.7)

Lemma 4.57.

a.λi,i−r=(uv)

r

λi,i+r,

b.λi,i−rλj,j+r=λi,i+rλj,j−r.

Proof.

λn,n+r=

n+r

∑

s=0

(

n

s

)(

n

n+r−s

)

u

n−s

v

s−r

=

n+r

∑

s=r

(

n

s

)(

n

s−r

)

u

n−s

v

s−r

.

Changing the sign ofr,

λn,n−r=

n−r

∑

s=−r

(

n

s

)(

n

s+r

)

u

n−s

v

s+r

(puts=n−σ)