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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.

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    $\begingroup$ I'm afraid you picked up the classical example for getting confused about this :) $\endgroup$
    – Martingalo
    Apr 22, 2015 at 6:04
  • $\begingroup$ math.stackexchange.com/questions/615852/… $\endgroup$
    – Martingalo
    Apr 22, 2015 at 6:05
  • $\begingroup$ You have to take the limit (which is zero) to compute the derivatives. $\endgroup$
    – Rolf Hoyer
    Apr 22, 2015 at 6:05
  • $\begingroup$ 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). $\endgroup$
    – Alex M.
    Apr 22, 2015 at 6:28

2 Answers 2

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$e^{-1/x^2}$ 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 $x=0$.

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you can depend on the common form of $e^x$ and put $x=-1/x^2$ as follow $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ $$e^{-1/x^2}=1-\frac{1}{x^2}+\frac{1}{2!x^4}-\frac{1}{3!x^6}+\frac{1}{4!x^8}+...$$

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    $\begingroup$ This is a Laurent series though, not a Maclaurin series. A Maclaurin series would not have negative powers, as this one does. $\endgroup$
    – 2'5 9'2
    Apr 22, 2015 at 20:34

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