# Maclaurin series expansion for $e^{-1/x^2}$

I am currently extremely confused on how to proceed with the Maclaurin series expansion for my current function.

I got my derivatives and I got my formula, however, plugging them in gives me a non-possible answers since division by $0$ is not possible.

• I'm afraid you picked up the classical example for getting confused about this :) Apr 22, 2015 at 6:04
• math.stackexchange.com/questions/615852/… Apr 22, 2015 at 6:05
• You have to take the limit (which is zero) to compute the derivatives. Apr 22, 2015 at 6:05
• The series is identically $0$ as @alex.jordan shows in his answer below, but obviously this series cannot be equal to $e^{-\frac 1 {x^2}}$, the reason being that this function is not analytic (it is, in fact, the standard example of smooth but non-analytic function). Apr 22, 2015 at 6:28

$e^{-1/x^2}$ itself is not defined at $x=0$. But you can take $$f(x)=\begin{cases}e^{-1/x^2}&x\neq0\\0&x=0\end{cases}$$ and since $\lim_{x\to0}e^{-1/x^2}=0$, $f$ is continuous.
Now you can take $f$'s derivative at $0$, but not so much by using the chain rule, exponential rule, and power rule, but rather by using the definition of the derivative. That limit-based definition gives $f'(0)=0$:\begin{align}f'(0)&=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}\\&= \lim_{h\to0}\frac{e^{-1/h^2}-0}{h}\\&= \lim_{h\to0}\frac{e^{-1/h^2}}{h}\\&= \lim_{h\to0}\frac{h^{-1}}{e^{1/h^2}}\\&= \lim_{h\to0}\frac{-h^{-2}}{-2h^{-3}e^{1/h^2}}\\&= \lim_{h\to0}\frac{h}{2e^{1/h^2}}=0\end{align}
For $f''(0)$, you need to fully understand $f'$. It works out as $$f'(x)=\begin{cases}\frac{2}{x^3}e^{-1/x^2}&x\neq0\\0&x=0\end{cases}$$ and you have a similar issue to before. Find $f''(0)$ using the definition of the derivative applied to $f'$.
And so on, until you see a pattern, and prove that pattern continues for all $n$th derivatives at $x=0$.
you can depend on the common form of $$e^x$$ and put $$x=-1/x^2$$ as follow $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ $$e^{-1/x^2}=1-\frac{1}{x^2}+\frac{1}{2!x^4}-\frac{1}{3!x^6}+\frac{1}{4!x^8}+...$$