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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From Proofs without Words.
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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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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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