Page 177 - Determinants and Their Applications in Mathematical Physics

162 4. Particular Determinants

Evaluating E n for small values of n, it is found that
2 3
2 6
x 1! x [1! 2!] x
E 1 = , E 2 = , E 3 = . (4.12.42)
1 − x (1 − x) 4 (1 − x) 9
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 evaluate F n and G n . G n is found in
terms of E n+1 by replacing n by n + 1 in (4.12.32) and then F n is given
in terms of G n by (4.12.40). The results are as given in the theorem. The
proof of the formula for H n follows from (4.12.14).

H n = |S m | n
= |(1 − x) m+1 ψ m | n
2
n
=(1 − x) |ψ m | n
2
=(1 − x) E n . (4.12.43)
n
The given formula follows. The formula for J n is proved as follows:
J n = |A m | 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
2
3
(1 − x)(ψ 0 +1) (1 − x) ψ 1 (1 − x) ψ 2 ···

2 3 4

(1 − x) ψ 1 (1 − x) ψ 2 (1 − x) ψ 3
···
3 4 5
J n =
(1 − x) ψ 2 (1 − x) ψ 3 (1 − x) ψ 4
···
.............................................

n

ψ 0 +1 ψ 1 ψ 2 ···
2 ψ 1 ψ 2 ψ 3
=(1 − x) n ···
ψ 2 ψ 3 ψ 4
···
....................

n
2
=(1 − x) x −n E n
n
= x −n H n , (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
Let A n denote the Hankel–Wronskian defined as
d
i+j−2
A n = |D f| n , D = , A 0 =1, (4.12.46)
dt