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

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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More generally: math.stackexchange.com/questions/517555/… – Henry Jan 30 at 14:07

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

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

and so

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

Thus,

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

Note: This proof is not specific to $\pi$.

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This is from The Book! Impeccable proof. – Prism Sep 14 '13 at 22:20
if x is negative than would your first equation which is e^x > 1 + x be true ? – Murtuza Vadharia Jan 15 '15 at 17:36
@MurtuzaVadharia: Yes, it is true. Consider $f(x) = e^x -1 -x$. It's derivative is $e^x -1$ which is $\lt 0$ for $x \lt 0$ and $\gt 0$ for $x \gt 0$, so $f$ decreases from $-\infty$ to $0$, and increases from $0$ to $\infty$. Since $f(0) = 0$... – Aryabhata Jan 21 '15 at 8:11
I think it's worth mentioning that this works for all positive numbers which aren't $e$. The proof has nothing to do with $\pi$. – NikolajK Feb 7 '15 at 20:48
@NikolajK: Thank you for the suggestion! Done. – Aryabhata Feb 10 '15 at 0:52

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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+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? – whatever Aug 5 '13 at 16:29

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

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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Let $f (x) =$ $x^\frac1x$

Find value of $x$ such that function gets maximum value

For this functions for $x=e$ function will get the maximum value

so $e^\frac1e$ is greater than $\pi^\frac1\pi$

so $e^\pi$ is greater than $\pi ^e$.

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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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Denote $n = e^\pi$, $m = \pi^e$ and $s = \log \pi$. Then $\log n = \pi = e^s$ and $\log m = e \log \pi = es$. Then $$\log \frac {n} {m} = \log n - \log m = e (e^{s - 1} - s).$$ By Taylor expansion, we have $$e^{s - 1} = 1 + (s - 1) + \cdots > s.$$ Then $$\log \frac {n} {m} = e (e^{s - 1} - s) > 0.$$ Hence, $n > m$.

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Formal proof with the help of a calculator:

We take for granted that $\frac{19}7<e<\frac{87}{32}$ and $\frac{100}{32}<\pi<\frac{22}7$. (This can be formally checked by computing the fractions and comparing to the known decimal values of the constants, obtained using power series.)

Then

$$e^\pi<\left(\frac{22}{7}\right)^{87/32}<\left(\frac{19}{7}\right)^{100/32}<\pi^e.$$

This is true because $$7^{13}22^{87}<19^{100}$$ i.e. $$59847739546421223164233171191241807412016754039458275038180456229719739147479584964293245460956376317259570175747204066381398016$$ < $$75051624198251984443456989853061891539043939434909537798332873934101480896578056472849915762891214746171016655874432115640378001.$$

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Sorry....but couldn't help downvoting it..... – tatan Jan 30 at 14:48
@tatan: can you explain why ? – Yves Daoust Jan 30 at 14:48
Question clearly states without calculator....any one can do it using a calculator.... – tatan Jan 30 at 14:49
@tatan Mh, I was initially answering another post which was flagged as duplicate, but was phrased slightly differently. The intent of my answer is to show that the proof is achievable with elementary computations (integer powers only). – Yves Daoust Jan 30 at 14:52
I tried to explain my point to you.....your answer has been downvoted by other users too.....you may ask for their reason.... – tatan Jan 30 at 15:11