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How can I calculate without calculator or something like this

the values of $\pi^{e}$ and $e^{\pi}$

in order to compare them ?

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6 Answers

up vote 34 down vote accepted

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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3  
This is from The Book! Impeccable proof. –  Prism Sep 14 '13 at 22:20
    
@Prism: Thanks! –  Aryabhata Sep 17 '13 at 3:27
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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
1  
@Mariano, @J.M. I think AD. means \varepsilon = $\varepsilon$ which is an alternate form of $\epsilon$. –  Rahul Oct 26 '10 at 16:32
1  
@Rahul Narain: Right, sorry it is my personal abbreviation. –  AD. Oct 26 '10 at 20:36
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From Proofs without Words.

alt text

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+1 because pictures are always nice to have –  J. M. Nov 8 '10 at 15:50
    
Which software did you use to plot this figure? –  metacompactness Aug 5 '13 at 16:29
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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 $x<e$. –  Robert Smith Oct 26 '10 at 15:39
    
Your argument appears to need $x>e$, but it is easy to also include $x<e$ as well i the $x^e<e^x$ inequality. –  Thomas Andrews May 22 '13 at 12:43
    
It doesn't really need $\log x > 1$. All you need is that the only solution of $\log x = 1$ is $x = e$. –  Yuval Filmus May 22 '13 at 14:29
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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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Hint:

Prove that the function $f(x)=\frac{e^x}{x^e}, x\geq e$ is strictly increasing on the interval $x\in \left [ e,\pi \right ]$. What is $f(e)$ and $f(\pi)$?

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