Page 143 - Determinants and Their Applications in Mathematical Physics

128 4. Particular Determinants
n−1
n − 1 (h + n + i)! h+i+1
= (−1) i x
i (h + i + 1)!
i=0

n
j−1 n − 1 (h + n + j − 1)!
= (−1) x h+j
j − 1 (h + j)!
j=1
2
(n − 1)! h!x h+1 1j j−1
n
= K x .
(h + n)!
j=1
The theorem follows.
Let
n
(n) j−1

S n (x, h)= K (−x)
nj
j=1

k 11 k 12 k 13 ··· k 1n

k 21 k 22 k 23 ··· k 2n
= ...................................... .

k n−1,1 k n−1,2 k n−1,3 ··· k n−1,n
1 −x x ··· (−x)
2 n−1
n
The column operations
C = C j + xC j−1 , 2 ≤ j ≤ n,

j
remove the x’s from the last row and yield the formula

x 1

S n (x, h)=(−1) n+1 + .
h + i + j − 1 h + i + j
n−1
Let

1+ x 1

T n (x, h)=(−1) n+1 − .
h + i + j − 1 h + i + j
n−1
Theorem 4.38.
2
(h + n − 1)! S n (x, h) h+n−1 n−1 h+n−1
a. = D [x (1 + x) ].
h!(n − 1)! S n (0,h)
(h + n − 1)! T n (x, h) h+n−1 h+n−1 n−1
b. = D [x (1 + x) ].
h!(n − 1)! T n (0,h)
Proof.
(n)
S n (0,h)= K n1
K n (h)V nn V n1
=
h + n

h + n − 1
n+1
=(−1) ,
h
K n (h)V nn