Determinants and Their Applications in Mathematical Physics

162 4. Particular Determinants

EvaluatingEnfor small values ofn, it is found that

E 1 =

x

1 −x

,E 2 =

1!

2

x

3

(1−x)

4

,E 3 =

[1! 2!]

2

x

6

(1−x)

9

. (4.12.42)

The solution which satisfies (4.12.41) and (4.12.42) is as given in the the-

orem. It is now a simple exercise to evaluateFnandGn.Gnis found in

terms ofEn+1by replacingnbyn+ 1 in (4.12.32) and thenFnis given

in terms ofGnby (4.12.40). The results are as given in the theorem. The

proof of the formula forHnfollows from (4.12.14).

Hn=|Sm|n

=|(1−x)

m+1

ψm|n

=(1−x)

n

2

|ψm|n

=(1−x)

n

2

En. (4.12.43)

The given formula follows. The formula forJnis proved as follows:

Jn=|Am|n.

Since

A 0 =1=(1−x)(ψ 0 +1), (4.12.44)

it follows by applying the second line of (4.12.31) that

Jn=

∣ ∣ ∣ ∣ ∣ ∣ ∣

(1−x)(ψ 0 +1) (1−x)

2

ψ 1 (1−x)

3

ψ 2 ···

(1−x)

2

ψ 1 (1−x)

3

ψ 2 (1−x)

4

ψ 3 ···

(1−x)

3

ψ 2 (1−x)

4

ψ 3 (1−x)

5

ψ 4 ···

∣ ∣ ∣ ∣ ∣ ∣ ∣ n

=(1−x)

n

2

∣ ∣ ∣ ∣ ∣ ∣ ∣

ψ 0 +1 ψ 1 ψ 2 ···

ψ 1 ψ 2 ψ 3 ···

ψ 2 ψ 3 ψ 4 ···

∣ ∣ ∣ ∣ ∣ ∣ ∣ n

=(1−x)

n

2

x

−n

En

=x

−n

Hn, (4.12.45)

which yields the given formula and completes the proofs of all five parts of

Lawden’s theorem.

4.12.3 A Further Generalization of the Geometric Series

LetAndenote the Hankel–Wronskian defined as

An=|D

i+j− 2

f|n,D=

d

dt

,A 0 =1, (4.12.46)