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 $f$ 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. |
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