144 4. Particular Determinants
=
∣
∣
∣
∣
(
n+j− 2
n−i
)∣
∣
∣
∣
n
=An,
which proves (b).
Exercises
Apply similar methods to prove that
1.
∣
∣
∣
∣
(
n+j− 1
n−i
)∣
∣
∣
∣
n
=(−1)
n(n−1)/ 2
,
2.
∣
∣
∣
∣
1
(i+j−1)!
∣
∣
∣
∣
n
=
(−1)
n(n−1)/ 2
Kn
[1!2!3!···(n−1)!]
2
.
Define the numberνias follows:
(1 +z)
− 1 / 2
=
∞
∑
i=0
νiz
i
. (4.11.35)
Then
νi=
(−1)
i
2
2 i
(
2 i
i
)
. (4.11.36)
Let
An=|νm|n, 0 ≤m≤ 2 n− 2 ,
=
∣
∣
C 1 C 2 ···Cn− 1 Cn
∣
∣
n
(4.11.37)
where
Cj=
[
νj− 1 νj...νn+j− 3 νn+j− 2
]T
n
. (4.11.38)
Theorem 4.48.
An=2
−(n−1)(2n−1)
.
Proof. Let
λnr=
n
n+r
(
n+r
2 r
)
2
2 r
. (4.11.39)
Then, it is shown in Appendix A.10 that
n
∑
j=1
λn− 1 ,j− 1 νi+j− 2 =
δin
2
2(n−1)
, 1 ≤i≤n, (4.11.40)
An=2
−(2n−3)
∣
∣
C 1 C 2 ···Cn− 1 (λn− 1 ,n− 1 Cn)
∣
∣
n
,
=2
−(2n−3)
∣
∣C
1 C 2 ···Cn− 1 C
′
n
∣
∣
n