Which is larger?

A MathsJam classic question asks:

Without using a calculator, which is bigger: $e^{\pi }$ or ${\pi }^e$?

It’s one of those questions that looks perfectly straightforward: you just take logs and then… oh, but is $\pi$ bigger than $e\ln (\pi )$? The Mathematical Ninja says “$\ln (\pi )$ is about 1.2, because $\pi$ is about 20% above $e$, which means $e\ln (\pi )$ is… about 20% more than $e$, which is about $\pi$.” Thanks, Ninja, but that doesn’t solve it.

The trick doesn’t involve logarithms, at least not directly. Instead, you make use of the Taylor expansion of $e^x$, which is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$, which is strictly larger than $1+x$ whenever $x$ is positive.

That 1 is a bit awkward, though: we’re going to want to look at a power that has a $-1$ in it so that it cancels out. We might naively look at:

$e^{\pi -1}>\pi$, which is a start; however, it only tells us that $e^{\pi }>e\pi$, which isn’t what we want. We’ve got an $e$ too many on the left hand side, so let’s try an argument of $\frac{\pi }{e}-1$:

$e^{\frac{\pi }{e}-1}>\frac{\pi }{e}$ - now we’re getting somewhere!

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

$e^{\pi }>{\pi }^e$, which is what we wanted to find out!

* Thanks to [twit handle = ‘daveinstpaul’] for the reminder about this problem.