It is well-known that a power series sums to a function that is analytic at every point inside its circle of convergence and that conversely, if a function is analytic on an open disc then its Taylor (power) series converges throughout the disc.

Since analyticity is a property defined over open sets, does this mean that an analytic function can never have a power series that has only a point convergence (as opposed to a disc of convergence)?

Let $f(x)= \sum\limits_{n=0}^\infty n!(2x+1)^n$, which is convergent only at -0.5. Does this mean that $f$ is not analytic???

  • $\begingroup$ I do not quite follow your question. $f(-0.5)$ is just a number? so you are asking if you can have an analytic function which is defined at a point? well then it won't be infinitely differentiable at this point, because you cannot do a taylor series around it? $\endgroup$ – Lost1 Mar 26 '13 at 15:28
  • $\begingroup$ I editted my commennt. I do not think your question makes much sense, because analytic is a differentiability condition and you need a function to be defined at least locally to discuss whether it is differentiable? $\endgroup$ – Lost1 Mar 26 '13 at 15:32

The pointwise definition of an analytic function goes something like this: $f$ is said to be analytic at $x_0$ if it's differentiable $\infty$ times at $x_0$ and its Taylor series, centered at $x_0$, converges in some (maybe very small) complex disk centered at $x_0$.

A power series with a $0$ radius of convergence is not very interesting, since it cannot be used to define a function at more than one point. A function defined only at one point is not said to be analytic (see the above definition).

  • $\begingroup$ Ah, thanks. I think I understand now, but perhaps not certainly. If I were given function $f$ defined by the above series and I hadn't determined its radius of convergence, it would've looked like an analytic function to me. So my confusion arose (I think) from forgetting that the points of convergence of the series is essential to the definition of a function defined as a power series? $\endgroup$ – Ryan Mar 26 '13 at 15:51
  • $\begingroup$ Since the above series converges only at $x=-0.5$, the function definition might as well be $f(x) = \left\{ \begin{array}{ll} 0 &\mbox{if } x=-0.5\\ \text{undefined} &\mbox{otherwise} \end{array} \right.$. The power series yields a value only wherever it converges. $\endgroup$ – Yoni Rozenshein Mar 26 '13 at 16:07

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