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Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion? I'm referring to analytic fuctions of course (i.e. those with power series expansions but not necessarily analytic over the entire complex plane).

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  • $\begingroup$ You say you're referring to analytic functions which are not analytic. Perhaps you mean something else? $\endgroup$ – Sharkos May 22 '13 at 8:52
  • $\begingroup$ @Sharkos Sorry I mean functions that are analytic at least somewhere, i.e. not necessarily entire functions. $\endgroup$ – Ryan May 22 '13 at 8:56
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The question is a little confusing, but how about $f(z) = \dfrac{1}{1-z}$.

The power series of $f$ around $z=0$ is $\sum_{n=0}^\infty z^n$ and this converges on $|z|<1$. The function, however, is analytic on $\mathbb{C}\setminus \{1\}$.

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  • $\begingroup$ Thanks. So, if I know only the convergence radius of a limit function of a power series, then I know that it is analytic at least inside the circle, non-analytic at least somewhere on the circle, but to determine its analyticity outside the circle will require further information, right? $\endgroup$ – Ryan May 22 '13 at 8:58
  • $\begingroup$ Yes. At least with a suitable interpretation of what "non-analytic at least somewhere on the circle" means. $\endgroup$ – mrf May 22 '13 at 8:59
  • $\begingroup$ Yes how is that phrase confusing? (btw I edited my first comment) $\endgroup$ – Ryan May 22 '13 at 9:00
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    $\begingroup$ From the comments above, now you can see why the phrasing is confusing. You need to think of $f$ and the power series of $f$ as separate entities. If you start with an analytic function on $\Omega$, you can compute its power series around any point $a\in\Omega$, but this series will only converge on a disc, not necessarily on the whole of $\Omega$. On the other hand, if you start with a power series, converging on some disc, it may very well have an analytic continuation to a larger domain. $\endgroup$ – mrf May 22 '13 at 9:08
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If I understand you correctly, any meromorphic but not entire function satisfies your requirement.

You can also just stitch together different functions arbitrarily to make this true.

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