There is a function
$f(x):= \begin{cases} e^{-1/x^2}, & \text{if $x \neq 0$} \\ 0, & \text{if $x=0$} \end{cases} $.
That function's $n$-th derivative is $0$ : $f^{(n)}(0) = 0$.
How to derive that?
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Nosrati
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Kim Jaewoo
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you mean that $f^{(n)}(0)=0$? – zaphodxvii Aug 09 '18 at 01:44
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Yes, I made a typo. Thanks. – Kim Jaewoo Aug 09 '18 at 01:48
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These notes have it. – Aug 09 '18 at 01:52
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Your function is incorrectly stated... the function has no $n$ as a variable (i.e., it does not depend upon $n$), and then you use $n$ to refer to a derivative. Please fix your question. – David G. Stork Aug 09 '18 at 02:19
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@DavidG.Stork I changed the error n to x. – Kim Jaewoo Aug 09 '18 at 02:26