<s>[INST] Determine the Taylor expansion of $f$ defined by $f(x)=e^{-1/x^2}$ if $x \ne 0$ and $f(0)=0$ I need to find the Taylor series, at $x=0$ for the piece-wise function:
$$f(x)=\begin{cases} \displaystyle e^{\frac{-1}{x^2}} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
I know that the Taylor series formula is $\sum\limits_{k=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$. So I found the first 3 derivative of $f(x)$
$$f'(x)=\begin{cases} \displaystyle \frac{2e^{\frac{-1}{x^2}}}{x^2} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
$$f''(x)=\begin{cases} \displaystyle \frac{2(2e^{\frac{-1}{x^2}}-3e^{\frac{-1}{x^2}}x^2)}{x^6} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
$$f'''(x)=\begin{cases} \displaystyle \frac{2(12x^4e^{\frac{-1}{x^2}}-18e^{\frac{-1}{x^2}}+4e^{\frac{-1}{x^2}})}{x^9} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
After that I bascially reformed the Taylor series equation as 
$$f(0)+\frac{f'(0)(x-0)^1}{1!}+\frac{f''(0)(x-0)^2}{2!}+\frac{f'''(0)(x-0)^3}{3!}+...$$
Since $f(0)=0=f'(0)=f''(0)=f'''(0)$ is the Taylor series just $0+0+0+...$?
 [/INST] Yes, the Taylor series is just $0$ as you had said. 
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