We have Comparing $\pi^e$ and $e^\pi$ without calculating them but it doesn't give an approximation of the actual difference. Is there a way without calcualting an approximation of them to prove $e^\pi - \pi^e < 1$ ?

  • 6
    $\begingroup$ We can still follow some of the proofs there to conclude $e^{\pi}<1+\pi^e$. Did you try this? $\endgroup$ Feb 24, 2020 at 19:41
  • $\begingroup$ Could you link the one that works for this? I must have missed something obvious. $\endgroup$
    – chx
    Feb 24, 2020 at 20:13
  • $\begingroup$ The difference is small, the task won't be easy. $\endgroup$
    – user65203
    Feb 24, 2020 at 21:10
  • 1
    $\begingroup$ Like this? $\endgroup$
    – rtybase
    Feb 24, 2020 at 21:29

1 Answer 1


If that can help:

Let $f(x):=e^x-x^e$. This function has a minimum at $x=e$ (double root), and the second order Taylor development is

$$y\approx g(x):=e^{e-1}(x-e)^2.$$

This approximation exceeds $f$, but we still have $g(\pi)<1$.

In blue, $f$, in black, $g$.

enter image description here

  • $\begingroup$ Hard to see definitely that the black curve is less than $1$ at $x=\pi$ from this plot. $\endgroup$
    – mjw
    Feb 24, 2020 at 21:54
  • $\begingroup$ Oh, you want the blue plot less than 1. That is pretty clear, but then you wouldn't need the Taylor series. $\endgroup$
    – mjw
    Feb 24, 2020 at 21:55

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