Page 216 - Determinants and Their Applications in Mathematical Physics

5.5 Determinants Associated with a Continued Fraction 201
2
=4 c r c s E rs,rs +2 c E rs,rs
s
r,s r,s
2
− E rstu,rstu . (5.4.48)
3
r,s,t,u
This is the second relation between Q and R, the first being (5.4.39). Iden-
tities (5.4.37), (5.4.38), and (5.3) follow by solving these two equations for
Q and R, where P is given by (5.1).

Exercise. Prove that

(c r − c s )φ n (c r ,c s )E rs = φ n (c r ,c s )E rs,rs , n =1, 2,
r,s r,s
where
φ 1 (c r ,c s )= c r + c s ,
2
2
φ 2 (c r ,c s )=3c +4c r c s +3c .
r s
Can this result be generalized?

5.5 Determinants Associated with a Continued
Fraction

5.5.1 Continuants and the Recurrence Relation

Define a continued fraction f n as follows:
1 b 1 b 2
f n = ··· b n−1 b n , n =1, 2, 3,... . (5.5.1)
1+ a 1 + a 2 + a n−1 + a n
f n is obtained from f n−1 by adding b n /a n to a n−1 .
Examples.
1
f 1 =
1+ b 1
a 1
a 1
= ,
a 1 + b 1
1
f 2 =
1+ b 1
a 1 + b 2
a 2
a 1 a 2 + b 2
= ,
a 1 a 2 + b 2 + a 2 b 1
1
f 3 =
1+ b 1 b 2
a 1 +
a 2 + b 3
a 3