Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 125

be expressed in the form

Vnr=

n

∏

j=1

(h+r+j−1)







n

∏

i=1

i=r

(r−i)







− 1

=

(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

′

j=Cj−Cs,^1 ≤j≤n, j=s,

onVnr. The result is a multiple of a determinant in which the element in

position (r, s) is 1 and all the other elements in rowrare 0. The other

elements in this determinant are given by

k

′′

ij=kij−kis

=

(

s−j

h+i+s− 1

)

kij, 1 ≤i, j≤n, (i, j)=(r, s).

After removing the factor (s−j) from each columnj,j=s, and the

factor (h+i+s−1) from each rowi, the cofactorKrsappears and gives

the result

Vnr=K

rs

n

∏

j=1

j=s

(s−j)







n

∏

i=1

i=r

(h+i+s−1)







− 1

,

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 inKnwhich follows from (4.10.10) and (4.10.7) and

which, when applied repeatedly, yields (4.10.13), an explicit formula for

Kn. The proofs of (4.10.14) and (4.10.15) are elementary.

Exercises

Prove that

1.Kn(− 2 n−h)=(−1)

n

Kn(h),h=0, 1 , 2 ,....

2.

∂

∂h

Vnr=Vnr

n− 1

∑

t=0

1

h+r+t

.

3.

∂

∂h

K

rs

n =K

rs

n

[

n− 1

∑

t=0

(

1

h+r+t

+

1

h+s+t

)

−

1

h+r+s− 1

]

.