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Suppose that I define a function $f(x)$=$e^{-1/x^2}$ when x is non zero and =$0$ when x is zero.

I found that(by using limit definition of derivative and sandwich-squeezing principle) $f^{n}(x)$=$0$ for all natural nos.

Now fuction f(x) does have a Taylor series expansion about x=$0$. Then ,why is that when one tries to write power series for $f(x)$,it becomes a zero series which is certainly not equal to $f(x)$?

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  • $\begingroup$ Because $f(x)=o(x^n)$ for every $n$. (Note that, contrarily to what you assert, $f$ does have Taylor series expansions at $x=0$.) $\endgroup$
    – Did
    Commented Sep 12, 2015 at 6:53
  • $\begingroup$ @Did,I asserted $f$ doesn't have Taylor series expansion about x=$0$ $\endgroup$
    – Koro
    Commented Sep 12, 2015 at 6:58
  • $\begingroup$ @Did ,Would you please elaborate your comment .. $\endgroup$
    – Koro
    Commented Sep 12, 2015 at 6:59
  • $\begingroup$ Yes, and $f$ does have Taylor series expansions about $x=0$, only $f$ is not equal to the sum of its Taylor series (which is different from not having a T.s.e.). $\endgroup$
    – Did
    Commented Sep 12, 2015 at 7:00
  • $\begingroup$ So,why is it happening in this case? All derivatives of f exist so why doesn't it have T.S.E. unlike in complex analysis. $\endgroup$
    – Koro
    Commented Sep 12, 2015 at 7:02

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