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.


2 Answers 2


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:

  • constants and the identity are analytic
  • Sums and products of analytic functions are analytic (tigether with the previous point, all polynomials are analytic)
  • reciprocals of analytic functions are analytic where defined (i.e. where we don't divide by zero)
  • 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 $f$?


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