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What is a simple way of showing that: for every $k$, there exists $M$ such that

$$\sup_{x \in \mathbb{R}} x^ke^{-|x|}< M?$$

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Look at the derivative – Michael Jul 4 '13 at 7:26
$\lim_{x\to\infty}x^ke^{-x}=0$, so for any $\epsilon>0$ there exists an $X$ so we have $x^ke^{-x}<\epsilon$, for $x>X$ – Ethan Jul 4 '13 at 7:30
Thanks @Ethan . – Cantor Jul 4 '13 at 7:36
up vote 2 down vote accepted

It seems the following.

If $k$ is not integer then we have a problem to define $x^k$ for a negative $x$.

If $k<0$ then we have $\lim_{x\to +0} x^ke^{|x|}=\infty$.

If $k$ is integer and $k\ge 0$ then for each $x\in\mathbb R$ we have $$x^ke^{-|x|}\le |x|^ke^{-|x|}=k!e^{-|x|}\left(\frac{|x|^k}{k!}\right)\le k!e^{-|x|}e^{|x|}=k!.$$

Remark. If we compare this upper bound with the timur's remark to user2303321's answer then we also easily obtain the inequality $k!\ge k^ke^{-k}$, which is a relatively good lower bound for $k!$, as shows Stirling's formula.

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Although I'm a total amateur, I have somewhat of an idea:

Differentiating the function $$f(x) = x^ke^{-|x|}$$ gives $$f'(x) = kx^{k-1}e^{-|x|} - x^ke^{-|x|}\frac{|x|}{x}$$

Which yields relative mins/maxes only at the points $x = 0,k,-k$

The fact that it has no asymptotes and its limit goes to $0$ take care of any other possible behavior.

Hope this helped.

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This in fact gives the exact value $\sup_{x\in\mathbb{R}} x^ke^{-|x|}=k^ke^{-k}$. – timur Jul 5 '13 at 0:42

The simplest I have seen is this:

If $x \ge 0$, for any $n$, $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} > \frac{x^{n}}{n!} $ so $x^n e^{-x} < n! $.

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This is included in Alex's answer. – timur Jul 5 '13 at 5:07

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