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)