# Comparing $\pi^e$ and $e^\pi$ without calculating them

How can I calculate, without calculator or similar device,

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

in order to compare them?

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 $\pi$.

• 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 $\pi$. – Nikolaj-K Feb 7 '15 at 20:48
• @David: Sorry to be of no help :(. No idea what the thought process was, it has been quite a while. – Aryabhata Aug 3 at 20:06

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

From Proofs without Words. • Which software did you use to plot this figure? – user5402 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 $x^e < e^x$.

• 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

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 $\pi ^e$.

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: $2^{\sqrt{2}} < e$.

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 $f(\pi)$?

Another visual proof. Recently published in arXiv Yet another line of thought would be this: \begin{align} e^{\pi} > \pi^e &\iff \pi \ln(e)>e\ln(\pi)\\ &\iff \pi >e\ln(\pi) \\ &\iff \ln(\pi)>\ln(e)+\ln (\ln (\pi)) \\ &\iff \ln(\pi)>1+\ln (\ln (\pi)) \end{align} By concavity of $\ln$, $x-1>\ln(x)$ for all $x\neq 1$. With $x=\ln(\pi)$ we get the inequality wanted.

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, $n > m$.

Just adding a recent one. Another visual proof. I am doubtful if it has open access.

$$e^\pi>\pi^e$$ because if we subtract $e$ from both exponents we get $$e^{\pi-e}>1$$ which is true because $e$ is is greater than $1$ so when it 8s raised to that power it is equal to $$\frac{e^\pi}{e^e}$$ and that has to be greater than 1 because the top is greater than the bottom.

• Are you sure you don't want to reconsider that first assertion? $\pi^e>e^e$. – user123641 Mar 3 '18 at 5:44
• You can't subtract $e$ from both exponents unless the base is the same. – Zaz Apr 14 '19 at 1:13