Taylor Series of $e^{-1/x^2}$ about $0$

So I worked out the Taylor series of $e^{-1/x^2}$ about $0$. However, all derivatives at $0$ came out to be $0$. So how can we write the Taylor series of the above function about $0$?

Solution 1:

We're going to prove that the function $$f(x)=\left\{ \begin{array}{ll} e^{-1/x^2} & \text{if } x\neq 0 \\ 0 & \text{if } x=0 \end{array} \right.$$ is infinitely differentiable and all his derivatives at $x=0$ are zero.

By induction, it's easy to check that, if $x\neq 0$, for each $n\in \mathbb{N}$ we have $$f^{n)}(x)=\frac{e^{-1/x^2}P_{2n-2}(x)}{x^{3n}},$$ where $P_{2n-2}(x)$ is a polynomial with degree $2n-2$, and if $x=0$ $$f^{n)}(0)=0. $$

Then, we can apply Taylor's Theorem. However, as $f^{n)}(0)=0$ for each $n\in \mathbb{N}$, the Taylor Series is just $0$.