I'm familiar with a simple method of demonstrating that $e^\pi$ is greater:

$f(x) = \ln|x|/x$

$f'(x) = (1 - (\ln|x|))/(x^2)$ so f's max is at $(e, 1/e)$ so $1/e > \ln(\pi)/\pi$ and $e^{\pi} > \pi^e$

My question is simply, how does one come up with this "$\ln|x|/x$" function?

I've tried working backwards, but there doesn't seem any real clear way of coming up with this magic function that gives such an elegant solution to this problem.


marked as duplicate by mesel, egreg, Antonio Vargas, colormegone, Daniel Robert-Nicoud Jun 18 '14 at 23:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Why do I have the feeling that this question has been asked at least many times before? $\endgroup$ – Asaf Karagila Jun 18 '14 at 20:08
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    $\begingroup$ A long time ago in a math department far far away, someone studying $\ln|x| / x$ noticed this and said "Hah! This'll keep them on their toes!" $\endgroup$ – Lee Mosher Jun 18 '14 at 20:10
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    $\begingroup$ probably one of the most duplicated questions on the site, among which: math.stackexchange.com/questions/7892/comparing-pie-and-e-pi math.stackexchange.com/questions/337565/… $\endgroup$ – colormegone Jun 18 '14 at 20:22
  • $\begingroup$ @Asaf Karagila, I think the topic title question has been asked, but I was more specifically asking how people find that clever ln|x|/x. If that too has been covered, I wasn't able to find it via search and apologize for double posting $\endgroup$ – David Bandel Jun 18 '14 at 20:51
  • $\begingroup$ @RecklessReckoner, neither of those posts was asking my question actually. I wasn't asking which was larger or how to prove one was larger. Was asking where the clever ln|x|/x comes from. I wasn't sure of a good way to ask that in a brief topic title though. But I did ask it clearly in the body of my question $\endgroup$ – David Bandel Jun 18 '14 at 20:52

Take logarithms.

$$e^\pi > \pi^e \Rightarrow \pi \log{e} > e\log{\pi} \Rightarrow \frac{\log e}{e} > \frac{\log \pi}{\pi }$$

This motivates us to consider the extrema of the function $f(x) = \frac{\log{x}}{x}$.

  • $\begingroup$ exactly what I was looking for. definitely a facepalm moment. thanks for the help $\endgroup$ – David Bandel Jun 18 '14 at 20:53
  • $\begingroup$ Thanks that's actually a method for proving the inequality using Lagrange sentence. $\endgroup$ – Dor Feb 3 '15 at 14:28

$e^\pi > \pi^e \iff e^{1/e} > \pi^{1/\pi}$

So the function you are actually considering is $x^{1/x}$.

$x^{1/x} = e^{\ln x/x} $

  • $\begingroup$ ahh i see. that's a more direct way of looking at it. thanks for your help $\endgroup$ – David Bandel Jun 18 '14 at 20:54

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