Page 327 - Determinants and Their Applications in Mathematical Physics

312 Appendix

where c ij = 0 when i, j < 1or i, j > n, then
 
2n−1 n i 2n−1 n

f i g i =  g i + g i  c i+1−j,j
i=1 i=1 j=1 i=n+1 j=i+1−n
n n

= c ij g i+j−1 .
i=1 j=1
The last step can be checked by writing out the terms in the last dou-
ble sum in a square array and collecting them together again along
diagonals parallel to the secondary diagonal.
2. The interval (1, 2n +1 −i−j) can be split into the two intervals (1,n +
1 − j) and (n +2 − j,2n +1 − i − j). Let

n n 2n+1−i−j

S = F ijs .
i=1 j=1 s=1
Then, splitting off the i = n term temporarily,

n−1 n 2n+1−i−j n n+1−j

S = F ijs + F njs
i=1 j=1 s=1 j=1 s=1
 
n−1
n n+1−j 2n+1−i−j n n+1−j

=  +  F ijs + F njs .
i=1 j=1 s=1 s=n+2−j j=1 s=1
The first and third sums can be recombined. Hence,
n−1
n n n+1−j n 2n+1−i−j

S = F ijs + F ijs .
i=1 j=1 s=1 i=1 j=1 s=n+2−j
The identities given in 1 and 2 are applied in Section 5.2 on the
generalized Cusick identities.
is invariant under any permutation of the parameters k r ,
3. If F k 1 k 2 ...k m
1 ≤ r ≤ m, and is zero when the parameters are not distinct, then
N

= m! , m ≤ N.
F k 1 k 2 ...k m F k 1 k 2 ...k m
k 1 ...k m =1 1≤k 1 <k 2 <···<k m ≤N
Proof. Denote the sum on the left by S and the sum on the right by
T. Then, S consists of all the terms in which the parameters are distinct,
whereas T consists only of those terms in which the parameters are in
ascending order of magnitude. Hence, to obtain all the terms in S,itis
necessary to permute the m parameters in each term of T. The number of
these permutations is m!. Hence, S = m!T, which proves the identity.