# What to plug in for n for this particular power series

I'm asked to find the power series representation of $f(x) = e ^ {-1/x^2}$ at $x = 0$.
getting the power series representation is easy of course.
$$\sum_{n = 0}^{\infty} \frac{(-1/x^2)^n}{ n!}$$ I know that it's asking to plug in $0$ for the solution, but what nth term do I choose? I'm assuming $0$. Another point, don't you get $-1/0^2$ if you plug in $0$ in the numerator making it undefined. If you were to plug in $n = 0$ for that does that "override" (for a lack of a better word) $-1/0^2$ being undefined and making it $1$? Just assuming that since anything to the $0$ is $1$. I hope I've made sense. If I didn't please feel free to ask so that I can help you help me.

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What you have there is not the power series expansion of the function in terms of powers of $x$. You have obtained a series for the function in terms of $1/x$ instead. To get the power series in terms of $x$ you need to use the Taylor series. – Old John Dec 18 '12 at 19:16
$f(0)=0$, $f'(0)=0$, $f''(0)=0$, ... [the power series is just $0$ identically!] – anon Dec 18 '12 at 19:19
I understand now. Thank you very much for clearing that up! – Ceelos Dec 18 '12 at 19:24

The function $f(x)=e^{\frac{-1}{x^2}}$ is not defined at $0$.
If we set $f(0)=0$, our function becomes continuous at $0$ (because $\lim_{x\to 0}f(x)=0$).
For $f^{\prime}$: $$\lim_{h\to 0}\frac{f(h)}{h}=\lim_{h\to 0}\frac{e^{-\frac1{h^2}}}{h}$$ For $h>0$, $h=e^{\ln h}$ and $$\lim_{h\to 0}\frac1{h^2}+\ln h=\lim_{h\to 0}\frac{h^2\ln h+1}{h}=+\infty$$ since $\lim_{h\to 0}h^2\ln h=0$. Therefore, $$\lim_{h\to 0^+}\frac{f(h)}{h}=\lim_{h\to 0^+}\frac{e^{-\frac1{h^2}}}{h}= \lim_{h\to 0^+}e^{-\frac1{h^2}-\ln h}=e^{-\infty}=0$$ For $h<0$, $$\lim_{h\to 0^-}\frac{f(h)}{h}=-\lim_{h\to 0^+}\frac{e^{-\frac1{h^2}}}{h} =\lim_{h\to 0^+}e^{-\frac1{h^2}-\ln h}=-e^{-\infty}=0$$
$f$ is differentiable at $0$ and $f^{\prime}(0)=0$. For $f^{(2)}$, $$\lim_{h\to 0}\frac{f^{\prime}(h)}{h}=\lim_{h\to 0}\frac{\frac{2}{h^3}e^{-\frac1{h^2}}}{h}=2\lim_{h\to 0}\frac{e^{-\frac1{h^2}}}{h^4}=...=0$$ Via induction one can show $f^{(n)}(0)=0$ and so it's Taylor series at $0$ is $0$.