Determinants and Their Applications in Mathematical Physics

208 5. Further Determinant Theory

The equation

hn, 2 n+1=0

yields

n+1

∑

t=0

c 2 n+1−tp 2 n+1,t=0. (5.5.29)

Applying the recurrence relation (5.5.22) and then parts (a) and (b) of the

theorem,

n

∑

t=0

c 2 n+1−tp 2 n,t+a 2 n+1

n+1

∑

t=1

c 2 n+1−tp 2 n− 1 ,t− 1 =0,

1

Bn

n

∑

t=0

c 2 n+1−tB

(n+1)

n+1,n+1−t

+

a 2 n+1

An

n+1

∑

t=1

c 2 n+1−tA

(n+1)

n+1,n+2−t

=0,

Bn+1

Bn

+a 2 n+1

An+1

An

=0,

which proves part (c).

Part (d) is proved in a similar manner. The equation

kn, 2 n=0

yields

n

∑

t=0

c 2 n−tp 2 n,t=0. (5.5.30)

Applying the recurrence relation (5.5.22) and then parts (a) and (b) of the

theorem,

n

∑

t=0

c 2 n−tp 2 n− 1 ,t+a 2 n

n

∑

t=1

c 2 n−tp 2 n− 2 ,t− 1 =0,

1

An

n

∑

t=0

c 2 n−tA

(n+1)

n+1,n+1−t

+

a 2 n

Bn− 1

n

∑

t=1

c 2 n−tB

(n)

n,n+1−t

=0,

An+1

An

+a 2 n

Bn

Bn− 1

=0,

which proves part (d).

Exercise.Prove that

P 6 =1+x

6

∑

r=1

ar+x

2

4

∑

r=1

ar

6

∑

s=r+2

as+x

3

2

∑

r=1

ar

4

∑

s=r+2

as

6

∑

t=s+2

at

and find the corresponding formula forQ 7.