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

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

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

2. 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

3. Originally Posted by Bacterius
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 $\displaystyle {\pi}^e < e^{\pi}$

________________________

$\displaystyle \Rightarrow$ 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

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

$\displaystyle 0<\frac{\pi}{\ln(\pi)}-\frac{e}{\ln(e)}=\int_e^{\pi}\left(\frac{x}{\ln(x) }\right)'\text{ }dx$ 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!

4. 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 !
Nice proof though, everything is already trivially proved and one just has to put the bits together

5. 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 !
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!

6. I'd like to see someone use pre-university algebra to solve this one!
Is it actually possible ?

7. 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 $\displaystyle f(\pi)\geqslant f(e)=1$.

8. Originally Posted by BraveHeart
Which is bigger: $\displaystyle \pi^e$ or $\displaystyle e^{\pi}$?
See here.

9. 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 $\displaystyle n^m > m^n$

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

11. Originally Posted by Jhevon
I'd like to see someone use pre-university algebra to solve this one!
This would work for me.

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 $\displaystyle a=\pi-e$)
14. Lots of thanking in this thread. I guess $\displaystyle \pi$ and $\displaystyle e$ are beloved.
$\displaystyle \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$