4.11 Hankelians 4 139
Qn=Qn(x)=
[
x
2(i+j−1)
i+j− 1
]
n
. (4.11.12)
BothKnandQnare Hankelians andQn(1) =Kn, the simple Hilbert
matrix.
Sn=Sn(x)=
[
vnix
2 j− 1
i+j− 1
]
n
, (4.11.13)
where thevniarev-numbers.
Hn=Hn(x, t)=Sn(x)+tIn
=
[
h
(n)
ij
]
n
,
where
h
(n)
ij
=
vnix
2 j− 1
i+j− 1
+δijt,
Hn=Hn(x,−t)=Sn(x)−tIn
=
[
h
(n)
ij
]
n
, (4.11.14)
where
h
(n)
ij (x, t)=h
(n)
ij
(x,−t),
H ̄
n(x,−t)=(−1)
n
Hn(−x, t). (4.11.15)
Theorem 4.43.
K
− 1
nQn=S
2
n.
Proof. Referring to (4.11.7) and applying the formula for the product
of two matrices,
K
− 1
n
Qn=
[
vnivnj
i+j− 1
]
n
[
x
2(i+j−1)
i+j− 1
]
n
=
[
n
∑
k=1
vnivnk
i+k− 1
x
2(k+j−1)
k+j− 1
]
n
=
[
n
∑
k=1
(
vnix
2 k− 1
i+k− 1
)(
vnkx
2 j− 1
k+j− 1
)
]
n
=S
2
n.
Theorem 4.44.
Bn=KnHnHn,
where the symbols can be interpreted as matrices or determinants.
Proof. Applying Theorem 4.43,
Bn=Qn−t
2
Kn