# $e^\pi - \pi^e < 1$? [duplicate]

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 $$e^\pi - \pi^e < 1$$ ?

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

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, $$g$$.

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