Determinants and Their Applications in Mathematical Physics

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

, (4.11.41)