Determinants and Their Applications in Mathematical Physics

136 4. Particular Determinants

Theorem.

An=Kn(x

2

−1)

n

2

,Kn=Kn(0).

Proof.

φ 0 =x

2

− 1.

Referring to Example A.3 (withc= 1) in the section on differences in

Appendix A.8,

∆

m

φ 0 =

φ

m+1

0

m+1

.

Hence, applying the theorem in Section 4.8.2 on Hankelians whose elements

are differences,

An=|∆

m

φ 0 |n

=

∣

∣

∣

∣

φ

m+1

0

m+1

∣ ∣ ∣ ∣ n =

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

φ 0

1

2

φ

2

0

1

3

φ

3

0

···

1

n

φ

n

0

1

2

φ

2

0

1

3

φ

3

0

1

4

φ

4

0

··· ···

1

3

φ

3

0

1

4

φ

4

0

1

5

φ

5

0

··· ···

..................................

1

n

φ

n

0

··· ··· ···

1

2 n− 1

φ

2 n− 1

0

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n

.

Remove the factorφ

i

0 from rowi,1≤i≤n, and then remove the factor

φ

j− 1

0 from columnj,2≤j ≤n. The simple Hilbert determinantKn

appears and the result is

An=Knφ

(1+2+3+···+n)(1+2+3+···+n−1)

0

=Knφ

n

2

0 ,

which proves the theorem.

Exercises

1.Define a triangular matrix [aij], 1≤i≤ 2 n−1, 1≤j≤ 2 n−i,as

follows:

column 1 =

[

1 uu

2

···u

2 n− 2

]T

,

row 1 =

[

1 vv

2

···v

2 n− 2

]

.

The remaining elements are defined by the rule that the difference be-

tween consecutive elements in any one diagonal parallel to the secondary

diagonal is constant. For example, one diagonal is

[

u

3

1

3

(2u

3

+v

3

)

1

3

(u

3

+2v

3

)v

3

]