Page 219 - Determinants and Their Applications in Mathematical Physics
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204   5. Further Determinant Theory

                              =  Q n  ,                              (5.5.8)
                                 P n
          where P n and Q n each satisfy the recurrence relation
                                                                     (5.5.9)
                                R n = R n−1 + a n xR n−2
          with P 0 =1, P 1 =1 + a 1 x, Q 0 = 1, and Q 1 = 1. It follows that
                 P 2 =1 + (a 1 + a 2 )x,
                 Q 2 =1 + a 2 x,
                                              2
                 P 3 =1 + (a 1 + a 2 + a 3 )x + a 1 a 3 x ,
                 Q 3 =1 + (a 2 + a 3 )x,
                                                                 2
                 P 4 =1 + (a 1 + a 2 + a 3 + a 4 )x +(a 1 a 3 + a 1 a 4 + a 2 a 4 )x ,
                                              2
                 Q 4 =1 + (a 2 + a 3 + a 4 )x + a 2 a 4 x .         (5.5.10)
          It also follows from the previous section that P n = H n+1 , etc., where
                       1 a 1 x

                       −1   1  a 2 x

                           −1    1   a 3 x


                                .    .    .
             H n+1 =             .  .  . .  . .                .    (5.5.11)

                                     −1    1 a n−2 x


                                          −1     1

                                                     a n−1 x
                                                −1      1

                                                            n+1
          The alternative formula
                      1     x


                      −a 1  1    x

                          −a 2   1     x

                                .     .       .
             H n+1 =             . .   . .     . .                  (5.5.12)

                                              1       x
                                    −a n−3

                                                      1
                                            −a n−2
                                                           x
                                                           1
                                                    −a n−1
                                                              n+1
          can be proved by showing that the second determinant satisfies the same
          recurrence relation as the first determinant and has the same initial values.
          Also,
                                           (n+1)
                                    Q n = H    .                    (5.5.13)
                                           11
          Using elementary methods, it is found that
                           2 2
             f 1 =1 − a 1 x + a x + ··· ,
                           1
                                            2
                                                        2
                                                           3
                                     2
             f 2 =1 − a 1 x + a 1 (a 1 + a 2 )x − a 1 (a +2a 1 a 2 + a )x + ··· ,
                                            1
                                                        2
                                     2      2           2        3
             f 3 =1 − a 1 x + a 1 (a 1 + a 2 )x − a 1 (a +2a 1 a 2 + a + a 2 a 3 )x + ···
                                                        2
                                            1
                       3    2        2     3    2
                 +a 1 (a +3a a 2 +3a 1 a +2a +2a a 3
                                           2
                       1
                                                2
                            1
                                     2
                          2
                                      4
                     +a 2 a +2a 1 a 2 a 3 )x + ··· ,                (5.5.14)
                          3
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