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 $0$?

 A: 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 $0$. 
A: @user3650050 : I suppose that you expect a series which approximates the function $y(x)=\exp\left(-\frac{1}{x^2} \right)$ for $x$ close to $0$. 
The pertinent comments and answer that you already received might disappoint you because whey don't give an approximate series as expected : As already pointed out, the Taylor series for $y(x)=\exp\left(-\frac{1}{x^2} \right)$ at $x=0$ doesn't fit the function for $x$ not exactly $0$.
I don't know if the next form of series can be useful for you, but I hope so :
Around $x=\epsilon\:\:$ close to $0$, the first terms of the series expansion are :
$$\exp\left(-\frac{1}{x^2} \right)=\exp\left(-\frac{1}{\epsilon^2} \right)\left(1+\frac{2}{\epsilon^3}x +\frac{2-3\epsilon^2}{\epsilon^6}x^2 +\frac{\frac43 -6\epsilon^2+4\epsilon^4}{\epsilon^9}x^3 +... \right)$$
A: As you worked out, the Taylor series of this function is 0. It means that, around 0, this function and all of its derivatives evaluate to 0.
