Page 164 - Determinants and Their Applications in Mathematical Physics

4.11 Hankelians 4 149

where
(1 + x) i+j+1 − x i+j+1
e ij (x)= . (4.11.61)
i + j +1
Theorem 4.53. The polynomial determinant E satisfies the nonlinear
differential equation
2 2

{x(1 + x)E} =4n (xE) {(1 + x)E} .
Proof. Let

A(x, ξ)= |φ m (x, ξ)| n , 0 ≤ m ≤ 2n − 2,
where
1 m+1 m+1 m+1
φ m (x, ξ)= (ξ + x) − c(ξ − 1) +(c − 1)ξ . (4.11.62)
m +1
Then,
∂
φ m (x, ξ)= mφ m−1 (x, ξ),
∂ξ
φ 0 (x, ξ)= x + c. (4.11.63)
Hence, from Theorem 4.33 in Section 4.9.1, A is independent of ξ. Put
ξ = 0 and −x in turn and denote the resulting determinants by U and V ,
respectively. Then,

A = U = V, (4.11.64)
where

U(x, c)= |φ m (x, 0)| n

x m+1 +(−1) c
m
= , 0 ≤ m ≤ 2n − 2
m +1

n
c
x +(−1) i+j
i+j−1
= . (4.11.65)
i + j − 1
n
Put
ψ m (x)= φ m (x, −x)
(−1) m m+1 m+1
= [c(1 + x) +(1 − c)x ] (4.11.66)
m +1
V (x, c)= |ψ m (x)| n ,
m+1
c(1 + x) +(1 − c)x m+1
=
m +1
n
i+j−1
c(1 + x) +(1 − c)x i+j−1
= . (4.11.67)
i + j − 1
n