Determinants and Their Applications in Mathematical Physics

126 4. Particular Determinants

4. 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