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

Question

How can I

compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?

Answer 1

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 $\pi$.

Answer 2

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

Answer 3

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 $x^e < e^x$.

Answer 4

From Proofs without Words.

Answer 5

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 $\pi ^e$.

Answer 6

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

Answer 7

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 $f(\pi)$?

Answer 8

Another visual proof. Recently published in arXiv

Answer 9

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.

Answer 10

\begin{align} &e^\pi>\pi^e \\[5pt] \iff&\exp(\pi)>\exp(e\log\pi) \\[5pt] \iff&\pi>e\log\pi \\[5pt] \iff&\frac{\pi}{\log\pi}>e \\[5pt] \iff&\frac{\pi}{\log\pi}>\frac{e}{\log e} \end{align} The final line is true because the function $\dfrac{x}{\log x}$ is strictly increasing on $[e,\infty) \, ,$ and the result follows.

Answer 11

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, $n > m$.

Answer 12

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

https://link.springer.com/article/10.1007/s00283-019-09964-x

Answer 13

Not that this question needs another answer, but here is a proof of $e^\pi > \pi^e$ using the Mean Value Theorem applied to $\ln x$ on the interval $(e,\pi)$, along with the assumptions that $e<\pi$, $\ln$ is increasing, and $\frac{d}{dx} \ln x = \frac{1}{x}$.

By the MVT, there exists $c \in (e,\pi)$ such that $$ \frac{\ln \pi - \ln e}{\pi - e} = \frac{1}{c} $$ We can increase the right hand side by replacing $c$ with the smaller number $e$, and so we have $$ \frac{\ln \pi - \ln e}{\pi - e} < \frac{1}{e} $$ and thus $$ {\ln \pi - \ln e} < \frac{1}{e}(\pi-e) $$ or $$ {\ln \pi - 1} < \frac{\pi}{e}-1 $$ which gives $$ e \ln \pi < \pi \ln e$$ and therefore $$ e^\pi < \pi^e$$

Answer 14

Let

$$f(x) = e^x$$

$$G(x) = x^e$$

We can simply show that

$$f(e)=G(e)$$

$$f'(e)=G'(e)$$

For $x > e$ the $f(x)$ will grow faster than $G(x)$

Then

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

Answer 15

In fact it is true that $e^x > x^e$ for all $x > e.$ Here's a proof using the fundamental theorem of calculus.

Note that $e^{x-2} > x^{e-2}$ for all $x > e.$ This can be easily verified by noting that one of these is convex and the other is concave and they are equal at $x=e$ and evaluating the derivatives at this point.

Then, $e^{x-1} - (e-1) x^{e-2} \geq e(e^{x-2} - x^{e-2}) > 0$ for $x > e.$

But then let $g(x) = e^{x-1} - x^{e-1}$ and noting that $g'(x) = e^{x-1} - (e-1) x^{e-2} > 0$ we thus have for all $x > e,$ $$g(x) - g(e) = g(x) = \int_{e}^{x} [ e^{x-1} - (e-1) x^{e-2} ] dx > 0 $$

And then letting $f(x) = e^x - x^e$ we have $f'(x) = e^x - e x^{e-1} = e g(x) > 0$

And by similar argument conclude that $$f(x) - f(e) = f(x) = \int_{e}^{x} e g(x) dx > 0$$ for all $x \geq e.$