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Let $f : x \mapsto \exp\left({-\frac{1}{x^2}}\right)$ defined for all $x \in \mathbb{R}^*$.

It is quite easy to prove by induction that $f$ can be continuated in $0$ in a $\mathcal{C}^\infty$ function. Set $M_n = \underset{x \in \mathbb{R}}{\sup}| f^{(n)}(x)| $. Can anyone find an asymptotical equivalent of $M_n$ ? I tried using Faà di Bruno's formula (https://en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula) but it seems hopeless.

Marsan
  • 1,131
  • You can find a bound from $f^{(n)}(z) = P_n(z)z^{-n-2}e^{-1/z^2}$ with $P_1(z) = -2$ and $P_{n+1}(z) = zP_n'(z)-(n+2)P_n(z)-2z^{n}$. $\ \ $ – reuns Nov 26 '16 at 00:39

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