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                                                  9.3 Asymptotic Expansions                  265

                        m.Itis often sufficient to be able to approximate f (n) with error O(n −m ) for some given
                        fixed value of m, often m = 1or2.

                        Example 9.1. Consider the function defined by

                                                      1

                                                 ∞
                                         f (x) =          exp(−t) dt, 0 < x < ∞.
                                                0  1 + t/x
                        Recall that, for z  = 1,
                                                                    1 − z m
                                                              m−1
                                                     2
                                             1 + z + z +· · · + z  =      .
                                                                    1 − z
                        Hence,
                                   1          t   t 2          m−1  t m−1    m  t m  1
                                        = 1 −  +    +· · · + (−1)      + (−1)
                                                                               m
                                 1 + t/x     x    x 2             x m−1       x 1 + t/x
                        so that
                                                      m−1
                                                             j  j!
                                                f (x) =  (−1)    + R m (x)
                                                              x  j
                                                       j=0
                        where
                                                       m
                                                    ∞
                                                      t     1              m!
                                         |R m (x)|=    m        exp(−t) dt ≤  m  .
                                                   0  x 1 + t/x            x
                        It follows that
                                                      ∞
                                                             j   j
                                                         (−1) j!/x
                                                      j=0
                        is a valid asymptotic expansion of f (x)as x →∞. Thus, we may write

                                                1                   1      −2
                                            ∞
                                                    exp(−t) dt = 1 −  + O(x  ),
                                              1 + t/x               x
                                           0
                                         ∞

                                              1                  1    2      −3
                                                  exp(−t) dt = 1 −  +   + O(x  ),
                                           1 + t/x               x   x 2
                                        0
                        and so on. For x > 0, let
                                                     1                 1   2
                                           ˆ                 ˆ
                                          f (x) = 1 −   and f (x) = 1 −  +   .
                                                             3
                                           2
                                                     x                 x   x 2
                                                ˆ
                                                        ˆ
                        Table 9.1 gives the values of f (x) and f (x), together with f (x), for several values of x.
                                                         3
                                                 2
                        Although both approximations are inaccurate for small values of x, both are nearly exact
                        for x = 30.
                          Note, however, that
                                           ∞                ∞
                                                                1
                                                 j    j
                                             (−1) j!/x  =           exp(−t) dt;
                                           j=0            0  1 + t/x
                        in fact, the series diverges for any value of x.
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