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I want to show that the function

$$ g = \begin{cases} 0, \quad \text{if } x \le 0, \\ e^{-\frac{1}{x^2}}, \quad \text{if } x > 0, \end{cases} $$ is infinitely differentiable at $0$ and that all derivative vanishes: $g^{(n)}(0) = 0$.

I was having trouble expanding the function into its Taylor expansion. Any help is appeciated!

Martin R
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nekodesu
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    Prove by induction on $n$ that $g^{(n)}(x)=p_n(x)g(x)$ for $x>0$, where $p_n$ is some rational function (really, $p_n(x)$ is a polynomial in $1/x$). Then use L'Hôpital's rule to show $g(n)(0)=0$. – Mike Earnest May 15 '18 at 13:08
  • Compare https://math.stackexchange.com/questions/1258219/formula-for-the-nth-derivative-of-e-1-x2. – Martin R May 15 '18 at 13:14

1 Answers1

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$$g'_+(0):=\lim_{x\to0^+}\frac{g(x)-g(0)}x=\lim_{x\to0^+}\frac{e^{-1/x^2}}x=0$$

since exponential beats polynomial (or you can do L'Hospital or whatever). Since clearly $\;g'(0)_-=0\;$ , the function is differentiable at zero. Now,

$$g'(x)=\begin{cases}0,&x\le0\\{}\\ \frac2{x^3}e^{-1/x^2},&x>0\end{cases}$$

Once again, and for the same reasons as in the case above, the second, third, etc. derivatives at zero exist, and they all are zero.

Deduce that a Maclaurin series is not going to work nice here...

DonAntonio
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