lower bound for the radius of convergence of Taylor series

Question

Possible Duplicate:
Radius of convergence of power series

In 4 years of studying physics I came across a lot of Taylor series. All of them converged in a disc with a radius equal to the distance to the nearest singular point. I know, that this distance is an upper bound for the radius of convergence, but is it also a lower bound? Is there some work on that for the general case? If not: what are the limitations for this to work or is there another lower bound for the radius in the general case?

A friend pointed out, that the Taylor expansion of $f(x) = e^{-\frac{1}{x^2}}$ does not converge anywhere but at $x=0$. This is not a contradiction though, because the exponential function is singular at $\infty$ and thus $f(x)$ is singular at $x=0$, right?

(A search here and via google only resulted in concrete examples, I am concerned with a general observation though)

See this thread: http://math.stackexchange.com/q/46829/5363 – t.b. Apr 04 '12 at 09:38 — t.b., Apr 04 '12 at 09:38
@t.b. Thanks. Will search more thoroughly next time -.- – example Apr 04 '12 at 09:52 — example, Apr 04 '12 at 09:52

score 0 · Accepted Answer · answered Apr 04 '12 at 13:27

Your function (if defined on $\mathbf R$ and with $f(0)=0$) is smooth ($\mathcal C^\infty$) at $x=0$, and in fact everywhere. And its (zero) Taylor expansion does converge, only not to $f$.

However, every (formal) power series is the Taylor series of some smooth function $\mathbf R\to\mathbf R$ (in fact of many of them), so there is nothing one can deduce about the radius of convergence, or about any other property of the series for that matter.

lower bound for the radius of convergence of Taylor series

1 Answers1