# Comparing $\pi^{e}$ and $e^{\pi}$

How to calculate without calculator or something like this the values of

$\pi^{e}$ and $e^{\pi}$

in oder to compare them?

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Another proof uses the fact that $\displaystyle \pi > e$ and that $e^x > 1 + x$ for $x > 0$.

We have $$e^{\pi/e -1} > \pi/e$$

and

So

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

and thus

$$e^{\pi} > \pi^e$$

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This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?

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With the new font it looks like vareps instead of e... Can someone reverse the font?? :/ – AD. Oct 26 '10 at 12:09
What are vareps? – Mariano Suárez-Alvarez Oct 26 '10 at 13:54
\vareps = $\vareps$ ? – J. M. Oct 26 '10 at 14:32
@Mariano, @J.M. I think AD. means \varepsilon = $\varepsilon$ which is an alternate form of $\epsilon$. – Rahul Narain Oct 26 '10 at 16:32
@Rahul Narain: Right, sorry it is my personal abbreviation. – AD. Oct 26 '10 at 20:36
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From Proofs without Words.

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 +1 because pictures are always nice to have – J. M. Nov 8 '10 at 15:50

Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.

Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) > 1$ for $x > e$, we see that $f'(x) < 0$ for $e < x < \pi$. We conclude that $\pi^{1/\pi} < e^{1/e}$, and so $\pi^e < e^\pi$.

The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e < e^x$.

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 Correction: $\text{log}(x)<1$ for $x Elaborating Robin's answer take$f(x) = \log{x} - \frac{x}{e}$. We have $$f'(x)= \frac{e-x}{xe}$$ Thus$f'(x)>0$for$0 < x < e$and$f'(x) <0$if$x > e$. Consequently, we have$f(x) < f(e)$if$x \neq e$. Exercise: Try to prove this using the same methods:$2^{\sqrt{2}} < e\$.

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