Page 170 - Determinants and Their Applications in Mathematical Physics

4.12 Hankelians 5 155
2
1 , (ζ − 1) n
= dζ (put ζ = x +2t)
2 n+1 πi (ζ − x) n+1
C
1 , [(x +1+2t)(x − 1+2t)] n
= dt
2 n+1 πi C (2t) n+1
1 , g(t)
= dt
2πi C t n+1
g (n) (0)
= ,
n!
where
- 1 .- 1 . n
g(t)= (x +1)+ t (x − 1) + t . (4.12.5)
2 2
The lemma follows.

n
n
n n
t
v
[(u + t)(v + t)] = u n−r n−s r+s
n
r s
r=0 s=0
2n

= λ np t p
p=0
where
p
n n

λ np = u n−s n−p+s , 0 ≤ p ≤ 2n, (4.12.6)
v
s p − s
s=0
which, by symmetry, is unaltered by interchanging u and v.
In particular,
λ 00 =1, λ n0 =(uv) , λ n,2n =1, λ nn = P n (x). (4.12.7)
n
Lemma 4.57.
a. λ i,i−r =(uv) λ i,i+r ,
r
b. λ i,i−r λ j,j+r = λ i,i+r λ j,j−r .
Proof.
n+r
n n
v
λ n,n+r = u n−s s−r
s n + r − s
s=0
n+r
n n

= u n−s s−r .
v
s s − r
s=r
Changing the sign of r,
n−r
n n
λ n,n−r = u n−s s+r (put s = n − σ)
v
s s + r
s=−r