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I want to show that the function defined as follows:
$f(x)=e^{-1/x^2}$ for $|x|>0$ and $f^{(k)}(0)=0$ for $k=0,1,2,\ldots$ is infinitely differentiable but not analytic at the point $x=0$.
For infinite-differentiability I used the fact that $\lim_{|x|\to 0^+} x^{-n} e^{-1/x^2}=0$ for every $n$ nonnegative integer, from which I can deduce that this function is infinite differnetiable at $x=0$ since all of its derivatives are continuous at $x=0$ and equal zero.
The problem I am having is with showing that it's not analytic, I need to show that for every $\delta>0, M>0$ the derivative $f^{(k)}(x)$ its absolute value is greater than $M\cdot k!/\delta^k$; not sure how to show it exactly.
$f(x)=e^{-1/x^2}$ for $|x|>0$ and $f^{(k)}(0)=0$ for $k=0,1,2,\ldots$ is infinitely differentiable but not analytic at the point $x=0$.
For infinite-differentiability I used the fact that $\lim_{|x|\to 0^+} x^{-n} e^{-1/x^2}=0$ for every $n$ nonnegative integer, from which I can deduce that this function is infinite differnetiable at $x=0$ since all of its derivatives are continuous at $x=0$ and equal zero.
The problem I am having is with showing that it's not analytic, I need to show that for every $\delta>0, M>0$ the derivative $f^{(k)}(x)$ its absolute value is greater than $M\cdot k!/\delta^k$; not sure how to show it exactly.