0
$\begingroup$

A classic example of a nonzero function with identically zero Taylor expansion is the following:

\begin{equation*} f(x)= \begin{cases} e^{-\frac{1}{x^2}}\quad &\text{if $x\neq 0$}\\ 0\quad &\text{if $x=0$} \end{cases} \end{equation*} according to this post Maclaurin series expansion for $e^{-1/x^2}$, it is clear that the Taylor series is null, however I want to know what the radius of convergence is, I know that it is convergent in the neighborhood of zero, however using the definition

\begin{equation} \alpha=\limsup_{n\to\infty}\sqrt[n]{|c_{n}|},\quad R=\frac{1}{\alpha} \end{equation} I think that the radius of convergence is zero, however I don't know how to justify it, which help is well received

$\endgroup$
1
  • 3
    $\begingroup$ All the coefficients are zero, so the radius of convergence is infinite ($\alpha=0$ so by convention, $R=\infty$). Now, just because the Taylor series has infinite radius of convergence, doesn't mean the function it sums to (in this case $0$) is equal to the function you started with (namely $f$). $\endgroup$
    – peek-a-boo
    Jun 22 '21 at 5:57
0
$\begingroup$

The series itself is convergent everywhere, since $\alpha = 0$, by convention $ R = \infty$. This, however, does not mean that the series converges to the function from which you computed those coefficients - you need to prove that separately. The function you are talking about is not analytic at $x=0$, meaning it cannot be expressed as a Taylor series about that point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.