Show that $f$ can not be expanded into a power series in a neighborhood of $0$.

Question

Let $f(x)$ be defined by $$f(x)=\begin{cases}e^{-1/x^2},& \text{ if }x \neq 0,\\0, &\text{ if }x=0.\end{cases}$$ Show that $f$ can not be expanded into a power series in a neighborhood of $0$.

We know that $f$ must be infinitely differentiable to be able to expand into a power series and $e^{-1/x^2}=\sum_{0}^{\infty} \dfrac{(-1)^n}{n!\cdot x^{2n}}$ . Then what can we do next? Anyone help me please?

Answer 1

If you try to compute all the derivatives $f'(0)$, $f''(0)$, etc, they are all zero.

So, if $f$ could be expanded into a power series in a neighborhood of 0, it would be the zero function there.

This is a contradiction, as $f$ is not the zero function in any neighborhood of 0.

Answer 2

You can find an answer here: A POSITIVE SMOOTH FUNCTION WITH ALL DERIVATIVES VANISHING AT ZERO.