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Recently I have been interested in finding minimum value of various functions without calculus, as a challenge.

How can we find the minimum value of $xe^x$ without calculus? Assume the limit or series definition of $e$.

If you use Calculus, you would differentiate $f(x) = xe^x$ and find that the only stationary point is $x = -1$. Since $$ f(-1) = e^{-1}, \ \ \ \lim_{x\to -\infty} f(x) = 0, \ \ \ \lim_{x\to +\infty} f(x) = +\infty,$$ we conclude that $e^{-1}$ is the minimum of $xe^x$.

However, I am not able to find the stationary point of this function without calculus.

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  • $\begingroup$ Maybe show that $\displaystyle xe^x+\frac{1}{e}$ is nonnegative by writing it as $\displaystyle \sum_{n=0}^\infty \frac{x^{n+1}+(-1)^n}{n!}$? But this seems to lead to a bunch of messy case analysis and I haven't gotten it to actually work... $\endgroup$
    – Micah
    Jul 21, 2021 at 4:51
  • $\begingroup$ I get the spirit of the question but, paradoxically, if you assume the limit or series definition of $e$ then you're already using calculus. $\endgroup$
    – Roman Hric
    Jul 21, 2021 at 7:38

1 Answer 1

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This answer tries to find the minimizer of $xe^x$.

Obviously the minimum is attained when $x<0$. Substituting $u=-x$, it suffices to minimize $-ue^{-u}$ when $u>0$, or equivalently maximize $ue^{-u}$.

Hopefully you know $\ln$ is monotonic (which should follow by monotonicity of $e^x$ if you define $\ln x$ as the inverse of $e^x$), so it suffices to maximize $\ln(ue^{-u})=\ln u-u$ for $u>0$.

We know (see below) $e^t\ge1+t$ for all real $t$ (equality at $t=0$), so substituting $t=\ln u$, we get $u\ge 1+\ln u$, i.e. $\ln u-u\le-1$ with equality at $u=1$. This immediately implies the minimum of $xe^{-x}$ is $-e^{-1}$, at $x=-1$.

EDIT: Here's a proof of $e^t\ge 1+t$ without calculus, expanded from this answer. First, note Bernoulli's inequality $(1+x)^n\ge 1+nx$ (where $n>0,n\in\mathbb Z,x\ge-1$) can be proven without calculus (only induction). Now for any fixed $x$, $\lim_{n\to\infty}\frac xn=0$, so when $n$ is sufficiently large, $\frac xn>-1$. Hence for all but finitely many $n$, $(1+\frac xn)^n\ge1+x$. Taking the limit $n\to\infty$, we now have $e^x\ge1+x$ as limit preserves non-strict inequalities (and ignores finitely many starting terms).

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  • $\begingroup$ Without calculus, how do we know that the tail of the series expansion of $e^t$ (i.e. the part without $1+t$), is non-negative? $\endgroup$
    – Dan
    Jul 21, 2021 at 5:45
  • $\begingroup$ It is obvious when $t\ge0$. When $t<0$, by Archimedean property there is an integer $N>-t$. So when $n\ge N, |t^n/n!|$ will be decreasing. Now the tail becomes an alternating series with strictly decreasing magnitude. $\endgroup$ Jul 21, 2021 at 5:51
  • $\begingroup$ Oh, but I'll need to bound the previous terms too. Let me be more careful here. $\endgroup$ Jul 21, 2021 at 5:54
  • $\begingroup$ @Dan I've updated the proof of $e^t\ge1+t$. I'm sure you can find even more elsewhere, but this should suffice. $\endgroup$ Jul 21, 2021 at 6:05

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