$f(x)=\begin{cases}e^{\frac{-1}{x^2}} & \text{ if } x\neq 0 \\ 0& \text{ if } x= 0\end{cases}$ is not equal to its Maclaurin Series

Question

$f(x)=\begin{cases}e^{\frac{-1}{x^2}} & \text{ if } x\neq 0 \\ 0& \text{ if } x= 0\end{cases}$ is not equal to its Maclaurin Series, which is $e^{-1/x^2}=1-\dfrac{1}{x^2}+\dfrac{1}{2!x^4}-\dfrac{1}{3!x^6}+\ldots$. Why is that? Any hints?

Answer 1

By definition, the MacLaurin series must have the form $\sum_{n=0}^\infty c_n x^n$, so what you write is not a MacLaurin series.

However, the MacLaurin series of this function is the null series, since $D^n f(0)=0$ for every $n \in \mathbb{N}$. Since $f$ is not the null function, we deduce that $f$ does not coincide with its MacLaurin series. No surprise: we have found a $C^\infty$ function that is not analytic. It is interesting, isn't it?

Answer 2

What you wrote is not the Maclaurin series for your function.

By definition, the MacLaurin series of $f$ around $x=0$ is $$\sum_{i=0}^{\infty} \frac{f^{(n)}(0)}{n!}\cdot x^n.$$

Now, apply this definition to the function $f$ from your question. What do you get?