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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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marked as duplicate by Martin Sleziak, Thomas, Nate Eldredge, userNaN, Henry T. Horton Oct 7 '12 at 18:02

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related, related and very the same –  leo Sep 27 '12 at 4:40

2 Answers 2

Hint: If $P$ is a polynomial and $g(x)=P(\frac1x) e^{-\frac1{x^2}}$, show that $g'(x)=Q(\frac1x) e^{-\frac1{x^2}}$ with a possibly different polynomial $Q$. Conclude that all derivatives of $f$ have this form and show that they all vanish as $x\to 0^+$, thus matching the $0$ from the negative axis.

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Idea on how to get started: clearly $f$ is differentiable for all $x\neq 0$, so you need just consider whether $f'(0)$ exists. That is, you need to consider whether the limit $$ \lim_{h \to 0} \frac{f(0 + h) - f(0) }{h}. $$ Consider the left-hand and the right-hand limits. Consider for example: $$ \lim_{h \to 0+} \frac{f(0 + h) - f(0) }{h} = \lim_{h \to 0^+} \frac{e^{-\frac{1}{h^2}}}{h}. $$

Can you prove that this limit is equal to the left hand limit?

After you have done this, you know what the derivative is on all $\mathbb{R}$. So you can try to compute the process with a new limit. Hopefully you see a pattern.

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