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Possible Duplicate:
Infinitely differentiable function.

Consider the function $f$ defined on $\mathbb{R}$ by

$f(x)= \begin{cases} 0 & \text{if $x \le 0$} \\ e^{-1/x^2} & \text{if $x > 0$} \end{cases}$

Prove that $f$ is indefinitely differentiable on $\mathbb{R}$, and that $f^{(n)}(0)=0$ for all $n \ge 1$. Conclude that $f$ does not have a converging power series expansion for $x$ near the origin.

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    [related](http://math.stackexchange.com/q/4700/8271), [related](http://math.stackexchange.com/q/4757/8271) and [very the same](http://math.stackexchange.com/q/119858/8271)2012-09-27

2 Answers 2