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5.5 Determinants Associated with a Continued Fraction 205
etc. These formulas lead to the following theorem.
Theorem 5.10.
f n − f n−1 =(−1) (a 1 a 2 a 3 ··· a n )x + O(x n+1 ),
n
n
r
that is, the coefficients of x , 1 ≤ r ≤ n − 1, in the series expansion of f n
are identical to those in the expansion of f n−1 .
Proof. Applying the recurrence relation (5.5.9),
P n−1 Q n − P n Q n−1 = P n−1 (Q n−1 + a n xQ n−2 ) − (P n−1 + a n xP n−2 )Q n−1
= −a n x(P n−2 Q n−1 − P n−1 Q n−2 )
2
= a n−1 a n x (P n−3 Q n−2 − P n−2 Q n−3 )
.
.
.
=(−1) (a 3 a 4 ··· a n )x n−2 (P 1 Q 2 − P 2 Q 1 )
n
=(−1) (a 1 a 2 ··· a n )x n (5.5.15)
n
Q n−1
f n − f n−1 = −
Q n
P n−1
P n
P n−1 Q n − P n Q n−1
=
P n P n−1
(−1) (a 1 a 2 ··· a n )x n
n
= . (5.5.16)
P n P n−1
The theorem follows since P n (x) is a polynomial with P n (0)=1.
Let
∞
f n (x)= c r x . (5.5.17)
r
r=0
From the third equation in (5.5.14),
c 0 =1,
c 1 = −a 1 ,
c 2 = a 1 (a 1 + a 2 ),
2
2
c 3 = −a 1 (a +2a 1 a 2 + a + a 2 a 3 ),
2
1
3
2
2
2
2
c 4 = a 1 (a a 2 +2a 1 a + a +2a a 3 + a a 3 +2a 1 a 2 a 3 + a 2 a 2 3
2
2
1
1
2
2
+a a 4 + a 1 a 2 a 4 + a 2 a 3 a 4 ), (5.5.18)
1
etc. Solving these equations for the a r ,
a 1 = −|c 1 |,
c
0 c 1
c 1 c 2
a 2 = ,
|c 1 |