Page 142 - Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 127

K j1
= V 1 − (r − 1)
h + r + j − 1
j
= V 1 − (r − 1)δ r1 , 1 ≤ r ≤ n.
The second term is zero. The result follows.
The proof of (4.10.18) when s = r follows from the identity

1 1 1 1
= −
(h + r + j − 1)(h + s + j − 1) s − r h + r + j − 1 h + s + j − 1
and (4.10.15). When s = r, the proof follows from (4.10.8) and (4.10.16):

K rs
= =1.
V s
(h + r + s − 1) 2 h + r + s − 1
V r
s s
To prove (4.10.19), apply (4.10.4) and (4.10.16):

jK 1j h + r − 1
= 1 − K 1j
h + r + j − 1 h + r + j − 1
j j
= V 1 − hδ r1 − (r − 1)δ r1 , 1 ≤ r ≤ n.
The third term is zero. The result follows.
Equation (4.10.20) follows from (4.10.4) and the double-sum identity (C)
(Section 3.4) with f r = r and g s = s + h − 1, and (4.10.21) follows from
the identity (4.10.9) in the form

jK 1j = V 1 V j − hK 1j
by summing over j and applying (4.10.4) and (4.10.20).

4.10.2 Three Formulas of the Rodrigues Type
Let
n
1j j−1
R n (x)= K x
j=1
1 x x 2 ··· x n−1

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

k n1 k n2 k n3 ··· k nn
n
Theorem 4.37.
(h + n)! n−1 n−1
R n (x)= D [x h+n (1 − x) ].
2
(n − 1)! h!x h+1
Proof. Referring to (4.10.9), (4.10.5), and (4.10.6),
n−1
n − 1 n−1
n−1
n−1

D x h+n (1 − x) = (−1) i D (x h+n+i )
i
i=0