Why is this function a really good asymptotic for exp ⁡ ( x ) x f ( x ) = ∑ n = 0 ∞ a n x n a n = 1 Γ ( n + 0.5 ) Why is this entire function a really good asymptotic for exp ⁡ ( x ) x , where for large positive numbers, f ( x ) exp ⁡ ( − x ) ≈ x ?As |x| gets larger, the error term is asymptotically f ( x ) − exp ⁡ ( x ) x ≈ 1 x ⋅ Γ ( − 0.5 ) , and the error term for f ( x ) exp ⁡ ( − x ) − x ≈ exp ⁡ ( − x ) x ⋅ Γ ( − 0.5 ) . If we treat f ( x ) as an infinite Laurent series, than it does not converge.I stumbled upon the result, using numerical approximations, so I can't really explain the equation for the a n coefficients, other than it appears to be the numerical limit of a pseudo Cauchy integral for the a n coefficients as the circle for the Cauchy integral path gets larger. I suspect the formula has been seen before, and can be generated by some other technique. By definition, for any entire function f ( x ), we have for any value of real r: a n = ∮ x − n f ( x ) = ∫ − π π 1 2 π ( r e − i x ) − n f ( r e i x ) ) d x The conjecture is that this is an equivalent definition for a n , where f ( x ) ↦ exp ⁡ ( x ) x and x ↦ r e i x a n = lim r → ∞ ∫ − π π 1 2 π ( r e − i x ) − n exp ⁡ ( r e i x ) r e i x ) d x = 1 Γ ( n + 0.5 )

Question

Why is this function a really good asymptotic for   exp  ⁡  (  x  )      x    f  (  x  )  =      ∑          n      =      0              ∞            a    n        x    n                  a    n    =      1          Γ      (      n      +      0.5      )      Why is this entire function a really good asymptotic for   exp  ⁡  (  x  )      x  , where for large positive numbers,   f  (  x  )  exp  ⁡  (  −  x  )  ≈      x  ?As |x| gets larger, the error term is asymptotically   f  (  x  )  −  exp  ⁡  (  x  )      x    ≈      1          x      ⋅      Γ      (      −      0.5      )      , and the error term for   f  (  x  )  exp  ⁡  (  −  x  )  −      x    ≈            exp      ⁡      (      −      x      )              x      ⋅      Γ      (      −      0.5      )      . If we treat   f  (  x  ) as an infinite Laurent series, than it does not converge.I stumbled upon the result, using numerical approximations, so I can&#039;t really explain the equation for the       a    n   coefficients, other than it appears to be the numerical limit of a pseudo Cauchy integral for the       a    n   coefficients as the circle for the Cauchy integral path gets larger. I suspect the formula has been seen before, and can be generated by some other technique. By definition, for any entire function   f  (  x  ), we have for any value of real r:      a    n    =  ∮      x          −      n        f  (  x  )  =      ∫          −      π              π            1          2      π        (  r      e          −      i      x            )          −      n        f  (  r      e          i      x        )  )        d    x    The conjecture is that this is an equivalent definition for       a    n  , where   f  (  x  )  ↦  exp  ⁡  (  x  )      x   and   x  ↦  r      e          i      x            a    n    =      lim          r      →      ∞            ∫          −      π              π            1          2      π        (  r      e          −      i      x            )          −      n        exp  ⁡  (  r      e          i      x        )      r          e              i        x              )        d    x  =      1          Γ      (      n      +      0.5      )

oldgaffer1b · Accepted Answer

Repeated integrations by parts show that, for every positive   a and   x      ∫    0    x              e              −      t            t          a      −      1            d    t  =  Γ  (  a  )            e              −      x            ∑          n      ⩾      0                  x              n        +        a                    Γ      (      n      +      a      +      1      )        .When   x  →  ∞, the LHS converges to   Γ  (  a  ), hence the series in the RHS is equivalent to             e        x  . Now,      ∑          n      ⩾      0                  x              n                    Γ      (      n      +      a      )        =      1          Γ      (      a      )        +      x          1      −      a            ∑          n      ⩾      0                  x              n        +        a                    Γ      (      n      +      a      +      1      )      hence      ∑          n      ⩾      0                  x              n                    Γ      (      n      +      a      )        ∼      x          1      −      a                  e        x    .For   a  =      1    2  , this is the result mentioned in the question.An exact formula using the incomplete gamma function   γ  (  a  ,     ) (that is, the LHS of the first identity in this answer) is      ∑          n      ⩾      0                  x              n                    Γ      (      n      +      a      )        =            γ      (      a      ,      x      )              Γ      (      a      )            x          1      −      a                  e        x    +      1          Γ      (      a      )        .Edit: ...And this approach yields the more precise expansion, also mentioned in the question,      ∑          n      ⩾      0                  x              n                    Γ      (      n      +      a      )        =      x          1      −      a                  e        x    +            1      −      a              Γ      (      a      )            1    x    +  O      (          1              x        2              )    .More generally, for every nonnegative integer   N and every noninteger   a.      ∑          n      ⩾      0                  x              n                    Γ      (      n      +      a      )        =      x          1      −      a                  e        x    +            sin      ⁡      (      π      a      )        π        ∑          k      =      1        N              Γ      (      k      +      1      −      a      )              x      k        +  O      (          1              x                  N          +          1                      )    .

KesseTher12 · Accepted Answer

In fact the story starts with finding an asymptotic expansion for the function  f  (  x  )  =      ∑          n      =      0              ∞                  x      n              Γ      (      n      +      1              /            2      )        =                              π                                  x                                                    e                                      x                                                            e            r            f                                (                          x                        )                          +        1                    π            which is given by  f  (  x  )  ∼      x          e          x

Why is this function a really good asymptotic for exp(x)sqrt(x) f(x)=sum_(n=0)^(oo) a_n x^n \ a_n = (1)/(Gamma(n+0.5)) exp(x)sqrt(x), where for large positive numbers, f(x)exp(−x)~~sqrt x?

Answered question

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