Which is bigger: $\displaystyle \pi^e$ or $\displaystyle e^{\pi}$?

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- Jan 14th 2010, 06:57 PMBraveHeartWhich is bigger: π^e or e^π? Prove it.
Which is bigger: $\displaystyle \pi^e$ or $\displaystyle e^{\pi}$?

- Jan 14th 2010, 07:41 PMBacterius
We know that $\displaystyle \pi \neq e$.

Let us conjecture that $\displaystyle {\pi}^e < e^{\pi}$.

This means that $\displaystyle \ln{({\pi}^e)} < \ln{(e^{\pi})}$, that is, $\displaystyle e \times \ln{(\pi)} < \pi$. Therefore, if our conjecture is true, we must have $\displaystyle e < \frac{\pi}{\ln{(\pi)}}$.

Let us consider the function $\displaystyle f(x) = \frac{x}{\ln{(x)}}$. Find its minimum on $\displaystyle (1, +\infty)$. You will find that the minimum of $\displaystyle f$ is $\displaystyle e$, and this minimum is reached when $\displaystyle x = e$.

But we know that $\displaystyle \pi \neq e$, thus the minimum $\displaystyle e$ is never reached with $\displaystyle f(\pi)$.

$\displaystyle \therefore e < f(\pi)$. You can follow the steps backwards to finally prove that $\displaystyle {\pi}^e < e^{\pi}$

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$\displaystyle \Rightarrow$ That was the nearly-complete proof. Now, you only have to show that the minimum of $\displaystyle f(x)$ is $\displaystyle e$ when $\displaystyle x = e$, and you are done :) - Jan 14th 2010, 07:57 PMDrexel28

Here is another way I just thought of.**Bacterius**just showed us that this is equivalent to showing that $\displaystyle e<\frac{\pi}{\ln(\pi)}\implies 0<\frac{\pi}{\ln(\pi)}-e$ ...BUT! this is the same as

$\displaystyle 0<\frac{\pi}{\ln(\pi)}-\frac{e}{\ln(e)}=\int_e^{\pi}\left(\frac{x}{\ln(x) }\right)'\text{ }dx$ (Evilgrin) and then we have to do is show that $\displaystyle \left(\frac{x}{\ln(x)}\right)'=\frac{\ln(x)-1}{\ln^2(x)}\geqslant 0$ for $\displaystyle e\leqslant x\leqslant \pi$ but this is clear since for this interval we have that

$\displaystyle \ln(x)\geqslant1\implies \ln(x)-1\geqslant 0$ and so $\displaystyle \frac{\ln(x)-1}{\ln^2(x)}\geqslant 0$. TA-DA! No calculations needed! - Jan 14th 2010, 07:58 PMBacterius
Lol, I wasn't sure my proof was going to pass since this is the algebra forum and my proof is already half-way into calculus, but Drex actually brought in the integral ! :D

Nice proof though, everything is already trivially proved and one just has to put the bits together :) - Jan 14th 2010, 08:01 PMJhevon
- Jan 14th 2010, 08:04 PMBacteriusQuote:

I'd like to see someone use pre-university algebra to solve this one!

- Jan 14th 2010, 08:19 PMDrexel28
Another way to do it using derivatives is let $\displaystyle f(x)=e^x x^{-e}$. We have that $\displaystyle f'(e)=0$ and it's easy to show it's a minimum. So then $\displaystyle f(\pi)\geqslant f(e)=1$.

- Jan 14th 2010, 09:38 PMChris L T521
See here.

- Jan 15th 2010, 01:09 AMUnbeatable0
Another one:

$\displaystyle \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$

Note that this can be generalized: for all $\displaystyle m > n \ge e$, it is true that $\displaystyle n^m > m^n$ - Jan 15th 2010, 02:45 PMCaptcha
- Jan 15th 2010, 03:22 PMPlato
- Jan 16th 2010, 12:08 AMUnbeatable0
We know that $\displaystyle \lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^x = e^a$. It can also be verified that if $\displaystyle x>y>0$ and $\displaystyle a>0$ then $\displaystyle \left(1+\frac{a}{x}\right)^x > \left(1+\frac{a}{y}\right)^y$. From these two results we can conclude that for all $\displaystyle a,x\in \mathbb{R}^+$, $\displaystyle \left(1+\frac{a}{x}\right)^x < e^a$ (and in our example, for $\displaystyle x=e$ and $\displaystyle a=\pi-e$)

- Jan 16th 2010, 02:15 PMJhevon
graphing actually occured to me, but i don't like this since actually graphing that graph without the use of calculus would be a pain. if you asked a pre-university student to do this, they'd probably just use a graphing utility or plot points with a calculator, in which case they wouldn't need the graph to find the answer to the problem, they'd just use the calculator. i think the idea here is to manually do the problem.

- Jan 17th 2010, 08:00 AMI4talent