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I know that for infinite series and $|z|<1$ there exists a confluent hypergeometric expression

$ \sum_{k=0}^{\infty} \frac{z^k}{k!k!} = F_{1}[;1;z] $

This is not very helpful though, and I 'd like to know if it is possible to get some asymptotic expansion for this function and if there exists some general approach to bounding hypergeometric functions asymptotically.

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    $\begingroup$ In comments, you keep saying $|z|<1$, but then the series you give is a perfectly good approximation. But you ask for asymptotic expansions, which usually means positive real infinity. But you don't seem to mean that. What do you want? $\endgroup$ Commented Nov 6, 2011 at 13:20
  • $\begingroup$ @Jacques: yes, now I understand what you mean, thanks. $\endgroup$ Commented Nov 6, 2011 at 20:17

3 Answers 3

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This particular function can be expressed in terms of Bessel's I function as $I(0,2\sqrt{z})$, and from there an asymptotic expression (at $\infty$) is easily derived. It starts $$\frac{e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{1}{4}\right)}}{2\sqrt{\pi}} + O\left(e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{3}{4}\right)}\right)$$ where the roots are chosen to have the correct branching behaviour.

The easiest way to obtain such results is actually from the ODE satisfied by your function, in this case $zy'' + y' - y$ with $y(0)=1$. It is easy to get an ODE at $\infty$ from there, and from there one gets the asymptotic expansion. The hardest part is getting the singular behaviour 'just right', as well as the branches. Those who have voted to close this likely have never had to compute the asymptotic expansion along a branch cut with the expansion point being an irregular singularity. While it is known how to do this, it is practiced by very very few.

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  • $\begingroup$ Thanks, though with $|z|<1 I(0,2 \sqrt(z)$ seems to behave like $1+z+\frac{z^2}{4} +O(z^3)$. Can you suggest some literature that treats asymptotic expansion of hypergeometric functions? $\endgroup$ Commented Nov 6, 2011 at 5:19
  • $\begingroup$ Did you want the series at 0 ? $\endgroup$ Commented Nov 6, 2011 at 13:16
  • $\begingroup$ @sigma_z_1980: and that is correct, since that is exactly your series. $\endgroup$ Commented Nov 6, 2011 at 13:18
  • $\begingroup$ @Jacques: yes, now I see what you mean by 'exactly my series'. Unfortunately I'm not so familiar with asymptotic expansion of hyper geometric functions, so I didn't see it at first glance. $\endgroup$ Commented Nov 7, 2011 at 1:52
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Although not directly for $z < 1$, here is a general paper that might help with some of the techniques that one can use:

The confluent hypergeometric functions $M(a; b, z)$ and $U(a; b, z)$ for large $b$ and $z$ by J. L. López and P. J. Pagola.

Also, if not this paper, please take a look at other papers by López; from what I heard, he is an expert on asymptotics of hypergeometric functions.

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  • $\begingroup$ great, thanks, I regret I can't accept two answers. $\endgroup$ Commented Nov 6, 2011 at 20:18
  • $\begingroup$ no worries; besides Jacques' answer is muccccch nicer :-) $\endgroup$
    – Suvrit
    Commented Nov 6, 2011 at 21:35
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The OP is unclear about the domain of $z$ required. If only real $z\to\infty$ is of interest, just realise that the terms near $k=\sqrt z$ dominate the rest and their shape is close to a normal density with mean and variance both asymptotically $\sqrt z$. Apply Stirling's formula and that's it.

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  • $\begingroup$ no it doesn't, $|z|<1$, $\endgroup$ Commented Nov 6, 2011 at 5:08
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    $\begingroup$ You asked for an asymptotic expansion. You gave a Taylor series showing it to be an entire function and you maintain $|z|>1$. Maybe the problem is that you don't know what "asymptotic expansion" means. $\endgroup$ Commented Nov 6, 2011 at 14:04

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