Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 129

Sn(x, h)=Kn(h)Vnn

n

∑

j=1

Vnj(−x)

j− 1

h+n+j− 1

Hence,

(

h+n− 1

h

)

Sn(x, h)

Sn(0,h)

=(−1)

n+1

n

∑

j=1

Vnj(−x)

j− 1

h+n+j− 1

=

n

∑

j=1

1

(n−j)!(h+j−1)!

[

(h+n+j−2)!x

j− 1

(j−1)!

]

=

1

(h+n−1)!

n

∑

j=1

(

h+n− 1

h+j− 1

)

D

h+n− 1

(x

h+n+j− 2

),

(h+n−1)!

2

h!(n−1)!

Sn(x, h)

Sn(0,h)

=D

h+n− 1



xn−^1

n

∑

j=1

(

h+n− 1

h+j− 1

)

x

h+j− 1





=D

h+n− 1

[

x

n− 1

h+n− 1

∑

r=h

(

h+n− 1

r

)

x

r

]

=D

h+n− 1

[

x

n− 1

(1 +x)

h+n− 1

−ph+n− 2 (x)

]

,

wherepr(x) is a polynomial of degreer. Formula (a) follows. To prove (b),

putx=− 1 −t. The details are elementary.

Further formulas of the Rodrigues type appear in Section 4.11.4.

4.10.3 Bordered Yamazaki–Hori Determinants — 1....

Let

A=|aij|n=|θm|n,

B=|bij|n=|φm|n, 0 ≤m≤ 2 n− 1 , (4.10.22)

denote two Hankelians, where

aij=

1

i+j− 1

[

p

2

x

2(i+j−1)

+q

2

y

2(i+j−1)

− 1

]

,

θm=

1

m+1

[

p

2

x

2 m+2

+q

2

y

2 m+2

− 1

]

,

bij=

1

i+j− 1

[

p

2

X

i+j− 1

+q

2

Y

i+j− 1

]

,

φm=

1

m+1

[

p

2

X

m+1

+q

2

Y

m+1

]

,

p

2

+q

2

=1,

X=x

2

− 1 ,

Y=y

2

− 1. (4.10.23)