4.11 Hankelians 4 147
Proof. By interchanging first rows and then columns in a suitable
manner it is easy to show that
En=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
ν 0 ν 1 ν 2 ···
ν 1 ν 2 ν 3 ···
ν 2 ν 3 ν 4 ···
ν 1 ν 2 ···
ν 2 ν 3 ···
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n
. (4.11.53)
Hence, referring to Theorems 4.11.5 and 4.11.6b,
E 2 n=(−1)
n
AnA
(n+1)
n+1, 1
=(−1)
n
2
−(2n−1)
2
,
E 2 n+1=(−1)
n
An+1A
(n+1)
n+1, 1
=(−1)
n
2
− 4 n
2
. (4.11.54)
These two results can be combined into one as shown in the theorem which
is applied in Section 4.12.1 to evaluate|Pm(x)|n.
Exercise.If
Bn=
∣
∣
∣
∣
(
2 m
m
)∣
∣
∣
∣
n
, 0 ≤m≤ 2 n− 2 ,
prove that
Bn=2
n− 1
,
B
(n)
ij
=2
2[n(n−1)−(i+j−2)]
A
(n)
ij
,
B
(n)
n 1
=2
n− 1
4.11.4 A Nonlinear Differential Equation..........
Let
Gn(x, h, k)=|gij|n,
where
gij=
{
x
h+i+k− 1
h+i+k− 1
,j=k
1
h+i+j− 1
,j=k.
(4.11.55)
Every column inGnexcept columnkis identical with the corresponding
column in the generalized Hilbert determinantKn(h). Also, let
Gn(x, h)=
n
∑
k=1
Gn(x, h, k). (4.11.56)