# Which is bigger: π^e or e^π? Prove it.

• Jan 14th 2010, 06:57 PM
BraveHeart
Which is bigger: π^e or e^π? Prove it.
Which is bigger: $\pi^e$ or $e^{\pi}$?
• Jan 14th 2010, 07:41 PM
Bacterius
We know that $\pi \neq e$.

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

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

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

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

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

________________________

$\Rightarrow$ That was the nearly-complete proof. Now, you only have to show that the minimum of $f(x)$ is $e$ when $x = e$, and you are done :)
• Jan 14th 2010, 07:57 PM
Drexel28
Quote:

Originally Posted by Bacterius
We know that $\pi \neq e$.

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

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

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

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

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

________________________

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

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

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

$\ln(x)\geqslant1\implies \ln(x)-1\geqslant 0$ and so $\frac{\ln(x)-1}{\ln^2(x)}\geqslant 0$. TA-DA! No calculations needed!
• Jan 14th 2010, 07:58 PM
Bacterius
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 PM
Jhevon
Quote:

Originally Posted by Bacterius
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 ... :)

Haha, I moved the thread to Calculus. I'd like to see someone use pre-university algebra to solve this one!
• Jan 14th 2010, 08:04 PM
Bacterius
Quote:

I'd like to see someone use pre-university algebra to solve this one!
Is it actually possible ? (Wondering)
• Jan 14th 2010, 08:19 PM
Drexel28
Another way to do it using derivatives is let $f(x)=e^x x^{-e}$. We have that $f'(e)=0$ and it's easy to show it's a minimum. So then $f(\pi)\geqslant f(e)=1$.
• Jan 14th 2010, 09:38 PM
Chris L T521
Quote:

Originally Posted by BraveHeart
Which is bigger: $\pi^e$ or $e^{\pi}$?

See here.
• Jan 15th 2010, 01:09 AM
Unbeatable0
Another one:
$\pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$

Note that this can be generalized: for all $m > n \ge e$, it is true that $n^m > m^n$
• Jan 15th 2010, 02:45 PM
Quote:

Originally Posted by Unbeatable0
$\left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e$

Can you please elaborate this step? I tried to see how the LHS is smaller than the RHS. It's not somehow apparent to me.
• Jan 15th 2010, 03:22 PM
Plato
Quote:

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

This would work for me.
• Jan 16th 2010, 12:08 AM
Unbeatable0
Quote:

Can you please elaborate this step? I tried to see how the LHS is smaller than the RHS. It's not somehow apparent to me.

We know that $\lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^x = e^a$. It can also be verified that if $x>y>0$ and $a>0$ then $\left(1+\frac{a}{x}\right)^x > \left(1+\frac{a}{y}\right)^y$. From these two results we can conclude that for all $a,x\in \mathbb{R}^+$, $\left(1+\frac{a}{x}\right)^x < e^a$ (and in our example, for $x=e$ and $a=\pi-e$)
• Jan 16th 2010, 02:15 PM
Jhevon
Quote:

Originally Posted by Plato
This would work for me.

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 AM
I4talent
Lots of thanking in this thread. I guess $\pi$ and $e$ are beloved. (Rofl)
Quote:

Originally Posted by Unbeatable0
$\pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$

Simple & Beautiful!