126 4. Particular Determinants
- a.K
r 1
n(0) =
(−1)
r+1
n(r+n−1)!
(r−1)!r!(n−r)!
.
b.K
rs
n(0) =
(−1)
r+s
rs
r+s− 1
(
n− 1
r− 1
)(
n− 1
s− 1
)(
r+n− 1
r
)(
s+n− 1
s
)
.
c.Kn(0) =
[1!2!3!···(n−1)!]
3
n!(n+ 1)!(n+ 2)!···(2n−1)!
.
5.
Kn
(
1
2
)
=2
n
∣
∣
∣
∣
1
2 i+2j− 1
∣
∣
∣
∣
n
=2
2 n
2
[1!2!3!···(n−1)!]
2
n− 1
∏
r=0
(2r+ 1)!(r+n)!
r!(2r+2n+ 1)!
.
[Apply the Legendre duplication formula in Appendix A.1].
6.By choosinghsuitably, evaluate| 1 /(2i+2j−3)|n.
The next set of identities are of a different nature. The parameternis
omitted fromVnr,K
ij
n, and so forth.
Identities 2.
∑
j
K
sj
h+r+j− 1
=δrs, 1 ≤r≤n. (4.10.16)
∑
j
Vj
h+r+j− 1
=1, 1 ≤r≤n. (4.10.17)
∑
j
Vj
(h+r+j−1)(h+s+j−1)
=
δrs
Vr
, 1 ≤r, s≤n.(4.10.18)
∑
j
jK
1 j
h+r+j− 1
=V 1 −hδr 1 , 1 ≤r≤n. (4.10.19)
∑
j
Vj=
∑
i
∑
j
K
ij
=n(n+h). (4.10.20)
∑
j
jK
1 j
=(n
2
+nh−h)V 1. (4.10.21)
Proof. Equation (4.10.16) is simply the identity
∑
j
krjK
sj
=δrs.
To prove (4.10.17), apply (4.10.9) withr→jand (4.10.4): and (4.10.12),
V 1
∑
j
Vj
h+r+j− 1
=
∑
j
(h+j)K
j 1
h+r+j− 1
=
∑
j
(
1 −
r− 1
h+r+j− 1
)
K
j 1