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5.5 Determinants Associated with a Continued Fraction  203

          where
                    1   b 1

                    −1  a 1  b 2

                        −1

                            a 2  b 3
                            .    .    .
             K n =           . .  . .  . .                        .  (5.5.4)


                                −1 a n−3   b n−2

                                      −1   a n−2      b n−1


                                            −1   a n−1 +(b n /a n )

                                                                n
          Return to H n+1 , remove the factor a n from the last column, and then
          perform the column operation

                                  C = C n + C n+1 .
                                    n
          The result is a determinant of order (n + 1) in which the only element in
          the last row is 1 in the right-hand corner.
            It then follows that
                                   H n+1 = a n K n .
          Similarly,
                                   (n+1)      (n−1)
                                 H      = a n K   .
                                   11         11
          The theorem follows from (5.5.3).
            Tridiagonal determinants of the form H n are called continuants. They are
          also simple Hessenbergians which satisfy the three-term recurrence relation.
          Expanding H n+1 by the two elements in the last row, it is found that

                               H n+1 = a n H n + b n H n−1 .
          Similarly,
                                (n+1)      (n)     (n)
                              H      = a n H  + b n H  .             (5.5.5)
                               11         11       11
          The theorem can therefore be reformulated as follows:
                                      f n =  Q n  ,                  (5.5.6)
                                           P n
          where P n and Q n each satisfy the recurrence relation
                                                                     (5.5.7)
                               R n = a n R n−1 + b n R n−2
          with the initial values P 0 =1, P 1 = a 1 + b 1 , Q 0 = 1, and Q 1 = a 1 .


          5.5.2 Polynomials and Power Series
          In the continued fraction f n defined in (5.5.1) in the previous section,
          replace a r by 1 and replace b r by a r x. Then,
                                 1 a 1 x a 2 x  a n−1 x a n x
                            f n =           ···
                                 1+ 1+ 1+       1+    1
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