# $f(x) := \exp(- \frac 1 {x^2})$ analytic around $0$ ?

Define $f(x) := \exp(- \frac 1 {x^2})$ and $f(0) := 0$. Is there a power series $\sum c_n (x-0)^n$ which converges to $f$ on some set $(-R,R)$,$R > 0$ around $0$ ?

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This is the standard example of a function not represented by its Maclaurin series, except at zero. – Gerry Myerson Jan 25 '13 at 12:20
So it is representable at $x =0$ ? – Epsilon Jan 25 '13 at 12:26
@André: Every infinitely differentiable function is represented at $x$ by its Taylor series at $x$, simply because the first term is $f(x)$ and the remaining terms are zero at $x$. The crucial requirement for analyticity is that the function be represented by its Taylor series at $x$ on some open interval around $x$. If by "representable at $x=0$" you meant not "represented at $x=0$ by its Taylor series at $x=0$" but "analytic at $x=0$", the answer is no. – joriki Jan 25 '13 at 12:37
@kahen: No, what Gerry wrote is quite correct. The Maclaurin series is the Taylor series at zero. The Taylor series of this function at zero represents this function nowhere expect at zero. You (and André?) seem to have misunderstood this to mean that the function is nowhere analytic except at zero. – joriki Jan 25 '13 at 12:38

$f(x)$ is not analytic at zero, since along the line $z=x\sqrt i$, $f(x)= \cos (\frac{1}{x^2}) + i \sin(\frac{1}{x^2})$. Which is not even continuous at $z=0$. So $f(z)$ is not even continuous at $z=0$, and $f(x)$, seen as a real-valued function, is not analytic at $x=0$. Indeed, $f$ has an essential singularity at $z =0$.
I think the issue is with open intervals. Functions are analytic at a point $a$ if their Taylor series centered at $a$ converges on an open neighborhood of $a$ (if we're working in $\mathbb R$ an open neighborhood is an open interval). But the Taylor series of $f$ does not converge to $f$ on any open eighborhood of $0$.
Although this function can't be represented by its Taylor series around $x=0$, it can be expanded to Laurent series: $$f(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!}x^{-2n}$$ So yes, there is a power series, but it needs negative powers.