Concluding that a function is not analytic at a point? 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!

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. 
