# A question on the convergence of a Taylor series of some prominent function

The function $$f:\mathbb{R}\to\mathbb{R}$$ defined by $$f(x)=\begin{cases}e^{-\frac{1}{x^2}} &if &x\neq 0\\ 0 & else \end{cases}$$ is a prominent example of a function whose Taylor series at $$x_0=0$$ exists but is zero and hence does not converge to the function $$f$$.

I want to prove that the Taylor series at another $$x_0\neq 0$$, for example $$x_0=3$$ does actually converge to the function $$f$$. To do so, I could write down explicitly the Taylor series for $$f$$ at $$x_0=3$$. However, it is not so easy to find a formula for the coefficients.

Is there another method to prove that the Taylor series of $$f$$ at $$x_0=3$$ converges to $$f$$?

Perhaps I could use the knowledge of the Taylor series of $$e^x$$ which has radius of convergence $$\infty$$ and then ''insert'' a Taylor series for $$-\frac{1}{x^2}$$ but I do not know how to determine the radius of convergence of such a composition. Actually, I even don't know how this ''composition'' should be a power series of the form $$\sum_{n=0}^\infty a_n(x-x_0)^{n}$$ again, i.e. I don't know the coefficients in relation to the coefficients of the two individual Taylor series.

Edit: Although I appreciate very much the solutions given by general results on analytic functions, it would be nice if someone could provide an elementary and explicit solution for this specific example. I think I could learn a lot from it. Thank you.

If the Taylor series of $f$ around $a$ converges to the function near $a$, we say that $f$ is analytic at $a$. "Virtually anything" you know to do with functions produces analytic functions when using analytic functions as input. That is:
• if $f$ is analytic at $a$ and $g$ is ananlytic at $f(a)$, then the composition $g\circ f$ is analytic at $a$.
• $z\mapsto e^z, \sin z, \cos z, \arctan z$ are analytic everywhere
• $z\mapsto \ln z$ and $z\mapsto \sqrt z$ are analytic at least for positive $z$ (and in a more general sense if we talk about complex numbers)
Now $z\mapsto e^{-\frac1{z^2}}$ is the composition of the exponential with the reciprocal of a polynomial, hence analytic (except at $x=0$).
The (complex) function $f(z) = e^{-1/z^2}$ is analytic and its Taylor series centered at $3 + 0 \imath$ has radius of convergence equal to $3$. Can you compare its Taylor series with the Taylor series of the real function $f$?