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$$f(x)=\sum_{n=0}^{\infty} a_n x^n\;\;\;\;\; a_n = \frac{1}{\Gamma(n+0.5)}$$

Why is this entire function a really good asymptotic for $\exp(x)\sqrt{x}$, where for large positive numbers, $f(x)\exp(-x) \approx \sqrt{x}$?

As |x| gets larger, the error term is asymptotically $f(x)-\exp(x)\sqrt{x} \approx \frac{1}{x\cdot\Gamma(-0.5)}$, and the error term for $f(x)\exp(-x) - \sqrt{x} \approx \frac{\exp(-x)}{x\cdot \Gamma(-0.5)}$. 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: $$a_n = \oint x^{-n} f(x) = \int_{-\pi}^{\pi} \frac{1}{2\pi} (re^{-ix})^{-n} f(re^{ix}) )\; \mathrm{d}x\;\;$$ The conjecture is that this is an equivalent definition for $a_n$, where $f(x) \mapsto \exp(x)\sqrt{x}$ and $x \mapsto re^{ix}$. $$a_n =\lim_{r\to\infty} \int_{-\pi}^{\pi} \frac{1}{2\pi} (re^{-ix})^{-n}\exp(re^{ix})\sqrt{re^{ix}})\; \mathrm{d}x = \frac{1}{\Gamma(n+0.5)} $$

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    $\begingroup$ Conjecture: for every $\alpha$, $$\sum_n\frac{x^n}{\Gamma(n+\alpha)}\sim x^{1-\alpha}\mathrm e^x.$$ $\endgroup$
    – Did
    Sep 13 '14 at 21:52
  • $\begingroup$ @Did nice formula; I tried this and it works for $\alpha=1/3$, $$\sum_n \frac{x^n}{\Gamma(n+ 2/3)} \approx x^{1/3}\exp(x) $$ $\endgroup$
    – Sheldon L
    Sep 13 '14 at 22:05
  • $\begingroup$ Conjecture confirmed, see answer. $\endgroup$
    – Did
    Sep 13 '14 at 22:17
  • $\begingroup$ For small values of a relative to n, we have $(n+a)!\approx n!~n^a$, as has also been pointed out here. $\endgroup$
    – Lucian
    Sep 14 '14 at 1:35
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Repeated integrations by parts show that, for every positive $a$ and $x$, $$\int_0^x\mathrm e^{-t}t^{a-1}\mathrm dt=\Gamma(a)\mathrm e^{-x}\sum_{n\geqslant0}\frac{x^{n+a}}{\Gamma(n+a+1)}.$$ When $x\to\infty$, the LHS converges to $\Gamma(a)$, hence the series in the RHS is equivalent to $\mathrm e^x$. Now, $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=\frac1{\Gamma(a)}+x^{1-a}\sum_{n\geqslant0}\frac{x^{n+a}}{\Gamma(n+a+1)}$$ hence $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}\sim x^{1-a}\mathrm e^x.$$ For $a=\frac12$, this is the result mentioned in the question.

An exact formula using the incomplete gamma function $\gamma(a,\ )$ (that is, the LHS of the first identity in this answer) is $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=\frac{\gamma(a,x)}{\Gamma(a)} x^{1-a}\mathrm e^x+\frac1{\Gamma(a)}.$$ Edit: ...And this approach yields the more precise expansion, also mentioned in the question, $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=x^{1-a}\mathrm e^x+\frac{1-a}{\Gamma(a)}\frac1x+O\left(\frac1{x^2}\right).$$ More generally, for every nonnegative integer $N$ and every noninteger $a$, $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=x^{1-a}\mathrm e^x+\frac{\sin(\pi a)}{\pi}\sum_{k=1}^N\frac{\Gamma(k+1-a)}{x^k}+O\left(\frac1{x^{N+1}}\right).$$

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  • $\begingroup$ Thanks! I'm sure your answer is correct, but I need some time before I check the arrow, so I can understand your answer fully :) $\endgroup$
    – Sheldon L
    Sep 13 '14 at 22:41
  • $\begingroup$ Please take your time, by all means (if only more users of the site could imitate you...). $\endgroup$
    – Did
    Sep 13 '14 at 22:43
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    $\begingroup$ Nice, +1. To add, the sum in question can also be identified as a confluent hypergeometric function, $$\sum_{n \geq 0} \frac{x^n}{\Gamma(n+a)} = \frac{1}{\Gamma(a)} {}_1F_1(1,a,x).$$ $\endgroup$ Sep 14 '14 at 1:56
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In fact the story starts with finding an asymptotic expansion for the function

$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+1/2)}= {\frac {\sqrt {\pi }\,\sqrt {x}{{\rm e}^{x}} {{\rm erf}\left(\sqrt {x}\right)}+1}{\sqrt {\pi }}}$$

which is given by

$$ f(x) \sim \sqrt{x}\, e^{x} $$

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    $\begingroup$ Nice expression but how is it obtained? $\endgroup$
    – abnry
    Sep 13 '14 at 22:28
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    $\begingroup$ I can't quite get the math to work for this equation. $\endgroup$
    – Sheldon L
    Sep 13 '14 at 22:28

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