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Define a function $f : \mathbb R \longrightarrow \mathbb R$ by $$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0 \end{cases}$$

Show that $f^{(n)} (0) = 0,$ for all $n \geq 1.$

I can prove that $f'(0) = f''(0) = 0.$ But for larger $n$ how do I show that? I think we have to use induction on $n.$ But I can't able to show that $f^{(n+1)} (0) = 0$ whenever $f^{(n)} (0) = 0.$ I am thinking about applying Leibnitz rule here. Is there any clever way to show that?

Some suggestion will be boon for me at this stage. Thanks!

Fanatics
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    See https://math.stackexchange.com/q/257579/42969 – Martin R Jun 21 '21 at 12:07
  • @Martin R the question you linked has no well accepted answer. Then how can my question be closed in reference to the linked post? – Fanatics Jun 21 '21 at 12:15
  • @Martin R can you please give me some reference? – Fanatics Jun 21 '21 at 12:17
  • The linked question has answers with 4 and 9 upvotes, I wouldn't call that “not well accepted.” Here is another Q&A about the same problem: https://math.stackexchange.com/q/332142/42969. – Martin R Jun 21 '21 at 12:26
  • @Martin R I mean it has not been accepted by the OP. Isn't it? That means the fact is not entirely clear to him/her. – Fanatics Jun 21 '21 at 12:29
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    Some people forget to accept an answer, or just don't bother, that is not necessarily an indicator for the correctness of an answer. Do you understand the (at least four) answers? If not, then you can either leave a comment and ask for clarification, or update your question with what you understand and what not. – Martin R Jun 21 '21 at 12:36
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    @Fanatics Not every asker accepts answers to all their questions. That's just a reality we live with here. Whether an answer has been accepted is not really a measure of how good an answer is, because it's only the judgement of a single person, if they even bothered to judge. The fact that 9 different people decided was worth an upvote is generally a much better sign of quality than a green checkmark, or the lack of it. – Arthur Jun 21 '21 at 12:39

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Hint: Show that $\lim_{x\to 0}F(x)e^{-1/x^2} = 0$ for all $F\in \mathbb R(X)$ (rational fractions in $X$). Then use induction.

Astyx
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