$f(x) = \begin{cases} e^{-1/x^2} & \text{ if } x \ne 0 \\ 0 & \text{ if } x = 0 \end{cases}$
so
$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac {e^{-1/x^2}}x = \lim_{x \to 0} \frac {1/x}{e^{1/x^2}} = \lim_{x \to 0} \frac x {2e^{1/x^2}} = 0$
(using l'Hospital's Rule and simplifying in the penultimate step).
Similarly, we can use the definition of the derivative and l'Hospital's Rule to show that $f''(0) = 0, f^{(3)}(0) = 0, \ldots, f^{(n)}(0) = 0$, so that the Maclaurin series for $f$ consists entirely of zero terms. But since $f(x) \ne 0$ except for $x = 0$, we can see that $f$ cannot equal its Maclaurin series except at $x = 0$.
This is part of a proof question. I don't think the answer sufficiently proves that any $n$th derivative of $f(x)$ is $0$. Would anyone please expand on the answer?
ps: I promise this is not my homework :)
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$\begingroup$\begin{align} f'(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} (2/x^3) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \\ f''(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} (4/x^6-6/x^4) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \\ & \vdots \\ \\ f^{(n)}(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} P_n(x) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \end{align} where $P_n(x)$ fulfills the recursive definition \begin{align} P_0(x) & = 1 \\ P_n(x) & = (2/x^3) P_{n-1}(x)+P_{n-1}'(x) \end{align}
$\endgroup$ 0 $\begingroup$Show by induction that $f^{(n)}(x)=P_n(\frac 1 x) \mathbb{e} ^{-\frac 1 {x^2}}$ with $P_n$ a polynomial function of degree $3n$, and then compute (again by induction if you want) $\lim \limits _{x \to 0^+} \space f^{(n)}(x)$. You'll have to use l'Hospital's thorem.
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