Page 223 - Determinants and Their Applications in Mathematical Physics

208 5. Further Determinant Theory

The equation
h n,2n+1 =0

yields
n+1

c 2n+1−t p 2n+1,t =0. (5.5.29)
t=0
Applying the recurrence relation (5.5.22) and then parts (a) and (b) of the
theorem,
n+1
n

c 2n+1−t p 2n−1,t−1 =0,
c 2n+1−t p 2n,t + a 2n+1
t=0 t=1
n+1
1 (n+1) a 2n+1 (n+1)
n
c 2n+1−t B + c 2n+1−t A =0,
n+1,n+1−t n+1,n+2−t
t=0 A n t=1
B n
B n+1 A n+1
+ a 2n+1 A n =0,
B n
which proves part (c).
Part (d) is proved in a similar manner. The equation
k n,2n =0
yields
n

c 2n−t p 2n,t =0. (5.5.30)
t=0
Applying the recurrence relation (5.5.22) and then parts (a) and (b) of the
theorem,

n n

c 2n−t p 2n−2,t−1 =0,
c 2n−t p 2n−1,t + a 2n
t=0 t=1
1 (n+1) (n)
n
n
c 2n−t A + a 2n c 2n−t B =0,
n+1,n+1−t n,n+1−t
B n−1
t=0 t=1
A n
A n+1 B n
=0,
+ a 2n
B n−1
A n
which proves part (d).
Exercise. Prove that
6 4 6 2 4 6
2 3
P 6 =1 + x a r + x a r a s + x a r a s a t
r=1 r=1 s=r+2 r=1 s=r+2 t=s+2
and find the corresponding formula for Q 7 .