Determinants and Their Applications in Mathematical Physics

312 Appendix

wherecij= 0 wheni, j <1ori, j > n, then

2 n− 1

∑

i=1

figi=





n

∑

i=1

gi

i

∑

j=1

+

2 n− 1

∑

i=n+1

gi

n

∑

j=i+1−n



c

i+1−j,j

=

n

∑

i=1

n

∑

j=1

cijgi+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, 2 n+1−i−j) can be split into the two intervals (1,n+

1 −j) and (n+2−j,2n+1−i−j). Let

S=

n

∑

i=1

n

∑

j=1

2 n+1−i−j

∑

s=1

Fijs.

Then, splitting off thei=nterm temporarily,

S=

n− 1

∑

i=1

n

∑

j=1

2 n+1−i−j

∑

s=1

Fijs+

n

∑

j=1

n+1−j

∑

s=1

Fnjs

=

n− 1

∑

i=1

n

∑

j=1





n+1−j

∑

s=1

+

2 n+1−i−j

∑

s=n+2−j



F

ijs+

n

∑

j=1

n+1−j

∑

s=1

Fnjs.

The first and third sums can be recombined. Hence,

S=

n

∑

i=1

n

∑

j=1

n+1−j

∑

s=1

Fijs+

n− 1

∑

i=1

n

∑

j=1

2 n+1−i−j

∑

s=n+2−j

Fijs.

The identities given in 1 and 2 are applied in Section 5.2 on the

generalized Cusick identities.

3.IfFk

1 k 2 ...km

is invariant under any permutation of the parameterskr,

1 ≤r≤m, and is zero when the parameters are not distinct, then

N

∑

k 1 ...km=1

Fk 1 k 2 ...km=m!

∑

1 ≤k 1 <k 2 <···<km≤N

Fk 1 k 2 ...km,m≤N.

Proof. Denote the sum on the left bySand the sum on the right by

T. Then,Sconsists of all the terms in which the parameters are distinct,

whereasT consists only of those terms in which the parameters are in

ascending order of magnitude. Hence, to obtain all the terms inS,itis

necessary to permute themparameters in each term ofT. The number of

these permutations ism!. Hence,S=m!T, which proves the identity.