How can one show that there exists an infinitely differentiable function $ f: \mathbb{R} \to \mathbb{R} $ such that $ {f^{(n)}}(0) = 0 $ but $ f^{(n)} \not\equiv 0 $ for all $ n \in \mathbb{N} $?
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The function $f(x)=e^{-1/x^2}$ is a canonical such example.
Ittay Weiss
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Inductively, for any $n\in\mathbb N$ one can prove $f^{(n)}(x)=p(x)e^{-1/x^2}$ for some rational function $p(x)$. Now, taking the limit $x\to 0$ gives $f^{(n)}(0)=0$. – Kenta S Dec 30 '21 at 17:55
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The function $f$ given by $$ f(x)= \begin{cases} \exp\left(-1/x\right),\quad&x>0,\\ 0,&x\leq 0 \end{cases} $$ has derivatives of all orders that satisfy $f^{(n)}(x)=0$ for $x\leq 0$ and $n\in\mathbb N$. See e.g. wikipedia and/or this question.
Stefan Hansen
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2It's worth mentioning that Cauchy gave it as an example for a non-analytical function. – Amihai Zivan Jan 15 '13 at 09:22