Determinants and Their Applications in Mathematical Physics

106 4. Particular Determinants

Let

An=|φi+j− 2 |n=|φm|n, 0 ≤m≤ 2 n− 2 ,

Bn=|x

i+j− 2

φi+j− 2 |n=|x

m

φm|n, 0 ≤m≤ 2 n− 2.

(4.8.9)

Lemma.

a.Bn=x

n(n−1)

An.

b.B

(n)

ij

=x

n(n−1)−(i+j−2)

A

(n)

ij

c. B

ij

n =x

−(i+j−2)

A

ij

n.

Proof of (a).Perform the following operations onBn: Remove the factor

x

i− 1

from rowi,1≤i≤n, and the factorx

j− 1

from columnj,1≤j≤n.

The effect of these operations is to remove the factorx

i+j− 2

from the

element in position (i, j).

The result is

Bn=x

2(1+2+3+···+n−1)

An,

which yields the stated result. Part (b) is proved in a similar manner, and

(c), which contains scaled cofactors, follows by division.

4.8.2 Hankelians Whose Elements are Differences

Thehdifference operator ∆his defined in Appendix A.8.

Theorem.

|φm|n=|∆

m

h

φ 0 |n;

that is, a Hankelian remains unaltered in value if eachφmis replaced by

∆
m
h
φ 0.

Proof. First Proof.Denote the determinant on the left byAand perform

the row operations

R

′

i=

i− 1

∑

r=0

(−h)

r

(

i− 1

r

)

Ri−r,i=n, n− 1 ,n− 2 ,..., 2 , (4.8.10)

onA. The result is

A=

∣

∣

∆

i− 1

h

φj− 1

∣

∣

n

. (4.8.11)

Now, restore symmetry by performing the same operations on the columns,

that is,

C

′

j

=

j− 1

∑

r=0

(−h)

r

(

j− 1

r

)

Cj−r,j=n, n− 1 ,n− 2 ,..., 2. (4.8.12)

The theorem appears. Note that the values of i and j are taken in

descending order of magnitude.