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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 |
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