Let f ( x ) and g ( x ) be two Taylor series such that: f ( x ) = ∑ n = 0 ∞ ( − 1 ) n a ( n ) x n and g ( x ) = ∑ n = 0 ∞ b ( n ) x n for a ( n ) &gt; 0 and b ( n ) &gt; 0.Can we extract the asymptotic behavior of these two taylor series for x → ∞?

Question

Let   f  (  x  ) and   g  (  x  ) be two Taylor series such that:  f  (  x  )  =      ∑          n      =      0              ∞        (  −  1      )          n        a  (  n  )      x          n      and  g  (  x  )  =      ∑          n      =      0              ∞        b  (  n  )      x          n      for   a  (  n  )  &amp;gt;  0 and   b  (  n  )  &amp;gt;  0.Can we extract the asymptotic behavior of these two taylor series for   x  →  ∞?

Faith Wise · Accepted Answer

Asymptotic expansions are essentially different from Taylor and Laurent expansions and cannot be easily deduced from one another. To call an expansion of   f  (  z  ) as   z  →      z    0   asymptotic you need to find an infinite sequence of functions             w      n        (    z    )   such that       lim          z      →              z        0                  |        w          n      +      1        (  z  )      /        w    n    (  z  )      |    =  0, or       w          n      +      1        =  o  (      w    n    (  z  )  ) and obtain expansion of the form  f  (  z  )  =      ∑          n      =      1        N        a    n        w    n    (  z  )  +  o  (      w    N    )where coefficients       a    n   are determined as follows:      a    n    =      lim          z      →              z        0                  {                  f        (        z        )        −                  ∑                      n            =            1                                N            −            1                                    a          n                          w          n                (        z        )                              w          N                (        z        )              }  To get a notion of the difference between the domains where power series converges and an asymptotic expansion is valid, consider the following example.  I  =      ∫    0          ∞                  e              −        z        t                    1      +              t        2              d  twhere   ℜ  (  z  )  &amp;gt;  0.Expanding the denominator in the geometric series which converges in the circle       |    t      |    &amp;lt;  1 and integrating term by term we obtain:  I  =      1    z    −            2      !              z      3        +            4      !              z      5        −  .  .  .  +            (      −      1              )                  n          −          1                    (      2      n      −      2      )      !              z              2        n        −        1              +      R    n    (  z  )where      R    n    (  z  )  =  (  −  1      )    n        ∫    0          ∞                          e                  −          z          t                            t                  2          n                            1      +              t        2              d  tNow      |        R    n    (  z  )      |    ≤      ∫    0          ∞            e          −      x      t            t          2      n        d  t  =            (      2      n      )      !              x              2        n        +        1            Let   z  ≠  0 and   −  (  π      /    2  )  +  α  ≤  arg  ⁡      z    ≤  (  π      /    2  )  −  α,   0  &amp;lt;  α  &amp;lt;  π      /    2. Then   x  ≥  sin  ⁡  α, hence      |        R    n    (  z  )      |    ≤            (      2      n      )      !              (      sin      ⁡      α              )                  2          n          +          1                          1                  |            z                        |                          2          n          +          1                    For a fixed   n,       lim                  |            z              |            →      ∞            |        R    n    (  z  )      |    =  0. Hence, even that the series does not converge, for large       |    z      |   and a finite number of terms it is a good approximation to the function. In contrast working with power series, for a given   z approximation is better, the more terms of the series we take. The purpose of the example os also to show that whereas Taylor (Laurent) expansion converges within a circle (annulus), the domain where asymptotic expansion is valid has the shape of an angle in the complex plane.

Let f(x) and g(x) be two Taylor series such that: f(x)=sum_(n=0)^oo(−1)^n a(n)x^n and g(x)=sum_(n=0)^oo b(n)x^n, for a(n)>0 and b(n)>0. can we extract the asymptotic behavior of these two taylor series for x->oo?

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New Questions in Calculus 2