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I learned that some functions are not equal to its taylor series, but I have a question.

How do you mathematically prove that the function$$f(x)=\begin{cases} e^{-1/x^2} &x\ne0 \\\ 0 &x=0 \end{cases}$$is infinitely differentiable and has continuous derivatives i.e. $C^{\infty}$function?

DH K
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    can you first prove it is differentiable at the origin and find what the derivative is? – peek-a-boo Apr 09 '21 at 14:36
  • An outline of a method of proof is given in this answer. I suspect much more detailed proofs exist here, however. – Dave L. Renfro Apr 09 '21 at 14:46
  • @peek-a-boo Yes, using the definition of the derivative. I just don't know how to show that it's infinitely differentiable at the origin and that the derivatives are continuous. – DH K Apr 09 '21 at 17:29
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    For the case with $1/x$ instead of $1/x^2$, which is very similar, see here. – Hans Lundmark Apr 09 '21 at 18:08
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    If you know how to show by the definition of the derivative that $f$ itself is differentiable at $0$, can you adapt that proof to also show that for all $n \in \Bbb Z, x^nf(x)$ is differentiable at $0$? Then, adapt it a little more to show that for any rational function $R, R(x)f(x)$ is differentiable at $0$. Once you've accomplished that, you have effectively proven what you need. – Paul Sinclair Apr 09 '21 at 20:46

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