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I have the function $f$ defined on $\mathbb{R}$ by:

$f(x)= \begin{cases} 0 & \text{if $x \le 0$} \\ e^{-1/x^2} & \text{if $x > 0$} \end{cases}$

I've inductively proved that $f$ is indefinitely differentiable on $\mathbb{R}$ and proved that $f^{(n)}(0)=0$ for all $n \ge 1$.

From these facts, how do I conclude that $f$ does not have a converging power series expansion for $x$ near the origin? (i.e. $f$ is not analytic at $z = 0$?)

I know that $f$ is analytic at a point $z_0 \in \Omega$ if there exists a power series $\sum a_n(z-z_0)^n$ centered at $z_0$, with positive radius of convergence, such that $$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n $$ for all $z$ in a neighborhood of $z_0$.

Is the conclusion as simple as: because $f^{(n)}(0)=0$ for all $n \ge 1$, a power series expansion centered at $z = 0$ does not exist, because any derivative at $z = 0$ will be equal to 0? If this is correct, then how do I state this formally? Thanks!

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  • $\begingroup$ Do you want to prove that the function is not analytic? $\endgroup$ – Mhenni Benghorbal Sep 11 '14 at 3:51
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Just because you can write $\sum_{n\geq 0}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$, doesn't mean the power series converges everywhere to your function. In your case $z_0=0$.

Yet, try to write out your power series. You'll get that it agrees with your function for $x<0$ but strongly disagrees for $x>0$. Clearly $g(x)=\sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}(x)^n=0=0+0x+0x^2+\cdots$. So you have complete disagreement for $x>0$.

This example illustrates how holomorphic (in the real sense) doesn't imply real analytic. Whereas in complex analysis, the two can be proven equivalent.

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  • $\begingroup$ The contradiction is when $x > 0$, not $x < 0$. Also, you are using $x$ and $z$ interchangeably. $\endgroup$ – Bungo Sep 11 '14 at 4:13
  • $\begingroup$ Sorry, maybe I'm missing something obvious (I am rather new to complex analysis...) -- where did the $\frac{f^{(n)}(z_0)}{n!}$ come from? $\endgroup$ – Ryan Yu Sep 11 '14 at 5:21
  • $\begingroup$ [Ignore the last comment] Sorry, maybe I'm missing something obvious, but I don't quite understand the contradiction. The power series expansion around $x_0 = 0$ is unique, yes. Can you spell out how the contradiction is derived? $\endgroup$ – Ryan Yu Sep 11 '14 at 5:30
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Well, if you have that $F(x)=\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}x^n f^n(0)$, then since $f^n(0)=0\;\;\forall n \in \mathbb{N}$, you would have that $f$ is identical zero, and this is not true since your function is not the zero function. So you can not have a expansion on Taylor series centred at zero.

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