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I've been told that the (real-valued) function $$f(x) = \sum_{n=1}^\infty \frac{1/n!}{x^2 + 1/n^2}$$ is "obviously" not analytic at $x=0$. Can someone help me see the reason?

First, I verified that the series does converge for all $x$. I tried to compute a few derivatives to see if they all vanish, as in the standard proof that $e^{-1/x}$ is not analytic at $x=0$. But it didn't lead to anything, and I'm not even sure that I can simply differentiate under the summation sign.

Kelley Alan
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1 Answers1

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There is a theorem in complex analysis which says that the radius of convergence of a Taylor series is the distance to the closest singularity. Since $f(x)$ has a singularity at $\frac{i}{n}$, for all $n$, the radius of convergence would need to be $0$.

Or, to put it another way, a Taylor series which converges on a disc cannot have a singularity within that disc.

Trevor Gunn
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    Why does $f$ have a singularity at $i/n$? – $f$ is a-priori only defined for real arguments, and $\sum_{n=1}^\infty \frac{1/n!}{x^2 + 1/n^2}$ is not a Taylor series. – Martin R Oct 22 '21 at 20:24
  • @TrevorGunn You mean $i/n$? – Franklin Pezzuti Dyer Oct 22 '21 at 20:25
  • Thanks, corrected – Trevor Gunn Oct 22 '21 at 20:25
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    @MartinR at $x = i/n$ then $$f(x) = \frac{1/n!}{(x + i/n)(x - i/n)} + \sum_{k \neq n} \frac{1/k!}{x^2 + 1/k^2}$$. The sum when $k \neq n$ converges since the $\frac{1}{k!}$ dominates. So there is a simple pole at $\pm \frac{i}{n}$. – Trevor Gunn Oct 22 '21 at 20:28
  • As I understand it, $f$ is defined via that series for real arguments. Why would the same representation hold for complex arguments (assuming that $f$ is analytic in a neighborhood of the origin)? Perhaps I am overlooking something obvious ... – Martin R Oct 22 '21 at 20:34
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    @Martin R: the same series for complex arguments defines a holomorphic function $g$ on $\mathbb{C}\backslash {\pm i/n,, 1 \leq n \leq \infty}$, which is equal to $f$ on the real line (minus zero). By the identity theorem, if $f$ is analytic at zero, then $f$ is equal to $g$ in a neighborhood of $0$. – Aphelli Oct 22 '21 at 20:39
  • @Mindlack: You are right, thanks. I would still suggest to clarify that in the answer. – Martin R Oct 22 '21 at 20:44