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 $1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!} +...$, 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 @daveinstpaul for the reminder about this problem.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.