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)