Determinants and Their Applications in Mathematical Physics

158 4. Particular Determinants

Hn=|Si+j− 2 |n=λnx

n(n+1)/ 2

,

Jn=|Ai+j− 2 |n=λnx

n(n−1)/ 2

,

where

λn=

[

1! 2! 3!···(n−1)!]

2

.

The following proofs differ from the originals in some respects.

Proof. It is proved using a slightly different notation in Theorem 4.28

in Section 4.8.5 on Turanians that

EnGn−En+1Gn− 1 =F

2

n

,

which is equivalent to

En− 1 Gn− 1 −EnGn− 2 =F

2

n− 1

. (4.12.16)

Putx=e
t
in (4.12.5) so that

Dx=e

−t

Dt

Dt=xDx,Dx=

∂

∂x

,etc. (4.12.17)

Also, put

θm(t)=ψm(e

t

)

=

∞

∑

r=1

r

m

e

rt

θ

′

m(t)=θm+1(t). (4.12.18)

Define the column vectorCj(t) as follows:

Cj(t)=

[

θj(t)θj+1(t)θj+2(t)...

]T

so that

C

′

j

=Cj+1(t). (4.12.19)

The number of elements inCjis equal to the order of the determinant of

which it is a part, that is,n,n−1, orn−2 in the present context.

Let

Qn(t, τ)=

∣

∣C

0 (τ)C 1 (t)C 2 (t)···Cn− 1 (t)

∣

∣

n

, (4.12.20)

where the argument in the first column isτ and the argument in each of

the other columns ist. Then,

Qn(t, t)=En. (4.12.21)

DifferentiateQnrepeatedly with respect toτ, apply (4.12.19), and put

τ=t.

D

r

τ

{Qn(t, t)}=0, 1 ≤r≤n− 1 , (4.12.22)