Page 159 - Determinants and Their Applications in Mathematical Physics

144 4. Particular Determinants

n + j − 2
=
n − i
n
= A n ,
which proves (b).

Exercises

Apply similar methods to prove that

n + j − 1
n(n−1)/2
1. =(−1) ,
n − i

n
1 (−1)
n(n−1)/2
K n
2. = .
(i + j − 1)! [1!2!3! ··· (n − 1)!]
2
n
Define the number ν i as follows:
∞

(1 + z) −1/2 = ν i z . (4.11.35)
i
i=0
Then

(−1) i 2i
ν i = . (4.11.36)
2 2i i
Let
A n = |ν m | n , 0 ≤ m ≤ 2n − 2,

(4.11.37)

= C 1 C 2 ··· C n−1 C n
n
where

T
C j = ν j−1 ν j ...ν n+j−3 ν n+j−2 . (4.11.38)
n
Theorem 4.48.
A n =2 −(n−1)(2n−1) .
Proof. Let
n n + r
λ nr = 2 . (4.11.39)
2r
n + r 2r
Then, it is shown in Appendix A.10 that
n
δ in
λ n−1,j−1 ν i+j−2 = , 1 ≤ i ≤ n, (4.11.40)
2 2(n−1)
j=1

A n =2 −(2n−3) C 1 C 2 ··· C n−1 (λ n−1,n−1 C n ) ,

n

=2 −(2n−3) C 1 C 2 ··· C n−1 C , (4.11.41)
n n