Page 140 - Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 125

be expressed in the form
−1
 
n n

V nr = (h + r + j − 1)  (r − i)


j=1 i=1

i=r
(h + r)(h + r − 1) ··· (h + r + n − 1)
= ,
[(r − 1)(r − 2) ··· 1][(−1)(−2) ··· (r − n)]
which leads to (4.10.5) and, hence, (4.10.6) and (4.10.7), which are
particular cases.
Now, perform the column operations

C = C j − C s , 1 ≤ j ≤ n, j = s,
j
on V nr . The result is a multiple of a determinant in which the element in
position (r, s) is 1 and all the other elements in row r are 0. The other
elements in this determinant are given by

k = k ij − k is
ij

s − j
= k ij , 1 ≤ i, j ≤ n, (i, j) =(r, s).
h + i + s − 1
After removing the factor (s − j) from each column j, j = s, and the
factor (h + i + s − 1) from each row i, the cofactor K rs appears and gives
the result
  −1
n

V nr = K rs   ,
n (s − j)  (h + i + s − 1)
j=1 i=1
i =r
j =s
which leads to (4.10.8) and, hence, (4.10.9) and (4.10.10), which are par-
ticular cases. Equation (4.10.11) then follows easily. Equation (4.10.12) is
a recurrence relation in K n which follows from (4.10.10) and (4.10.7) and
which, when applied repeatedly, yields (4.10.13), an explicit formula for
K n . The proofs of (4.10.14) and (4.10.15) are elementary.

Exercises
Prove that

1. K n (−2n − h)=(−1) K n (h), h =0, 1, 2,....
n
n−1
∂ 1
2. .
∂h V nr = V nr h + r + t
t=0

n−1
∂ 1 1 1
3. K rs = K rs + − .
∂h n n h + r + t h + s + t h + r + s − 1
t=0