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Is $e^{-x^2}$ a good function?

A function $\gamma$ is said to be good if it is differentiable infinitely often everywhere on $R_1$ and if $$lim_{|x| \rightarrow \infty}\Big|x^r \frac{d^k}{dx^k} \gamma(x)\Big| = 0$$n for all $r,k \geq 0$.

Clearly $e^{-x^2} \in C^\infty$, but I cannot prove $$lim_{|x| \rightarrow \infty}\Big|x^r \frac{d^k}{dx^k} (e^{-x^2})\Big| = 0$$n for all $r,k \geq 0$.

  • Which part is giving you trouble with the limit: the $x^r$ factor, or the derivatives? For example, can you prove it with $k=0$ and $r$ arbitrary? – Greg Martin Mar 05 '22 at 07:52
  • @GregMartin $x^r$ factor is ok. But how can I write $\frac{d^k}{dx^k}(e^{-x^2})$ for a general $k$? –  Mar 05 '22 at 07:55
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    Hint: prove by induction that $\frac{d^k}{dx^k}(e^{-x^2}) = P_k(x)e^{-k^2}$ for some polynomial $P_k(x)$. The exact identity of that polynomial won't matter much.... – Greg Martin Mar 05 '22 at 07:56
  • @GregMartin If $|x| \mapsto \infty$ then $x^r P_k (x) \mapsto \infty$ and $e^{-x^2} \mapsto 0$. So, how I should conclude that the whole limit would be $0$? –  Mar 05 '22 at 09:23
  • $x^r P_k(x)$ is also a polynomial, so you have to check that $x^n e^{-x^2}$ goes to zero for $x \to \infty$ for any polynomial $n \in \mathbb N$. You can write $x^n e^{-x^2} = \frac{x^n}{e^{x^2}}$ and ud L'Hopital's rule repeatedly. – Blazej Mar 05 '22 at 10:43

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