An irrational competition

Let's assume that, to see where this goes

$e^\pi > \pi^e$

Then we have

$e^{\frac{1}{e}}\ > \pi^{\frac{1}{\pi}}$

So, let's consider for which $x$ the following function is at maximum

$x^{\frac{1}{x}}\$

We differentiate this to get, and factor the result

$x^{(\frac{1}{x}-2)} (1-Log(x))$

This is a maximum when $x=e$ , so that the original assumption was correct, and indeed that

$e^\pi > \pi^e$

This is not circular reasoning, since we can work backwards.

"The exponent usually wins" is a decent, though not infallible, heuristic (when the bases are of similar size).

similar

I first saw this proof approximately 1 year ago. Check out https://www.youtube.com/watch?v=SPHD7zmLVa8&t=8s for a detailed explanation. I would highly recommend blackpenredpen. It is a great YouTube channel that explores mathematics. Anyway, onto the proof.

$e^\pi$ vs $\pi^e$

To consider this we shall try to get the numbers in the form of a function. To do this we must have the numbers raised to the same power. In a sense we want both to look like a Cartesian equation.

$let$

$e^\pi = (e^\frac{1}{e})^{e\pi} = (X_1)^\frac{1}{X_1}$

$\pi^e = (\pi^\frac{1}{\pi})^{e\pi} = (X_2)^\frac{1}{X_2}$

We now have each number in the form $X^\frac{1}{X}$ so we can plot a graph of $y = X^\frac{1}{X}$

In order to find the maximum point of the graph we need to differentiate the function. However before this we must use logarithims.

$ln(y) = \frac{1}{X} ln(X)$

Differentiating

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{X^2} - \frac{ln(X)}{X^2}$

$\frac{dy}{dx} =\frac{1}{y}[\frac{1}{X^2} - \frac{ln(X)}{X^2}]$

$\frac{dy}{dx} = X^\frac{1}{X}[\frac{1}{X^2} - \frac{ln(X)}{X^2}]$

$\frac{dy}{dx} = X^\frac{1}{X}\frac{1}{X^2}[1-ln(X)]$

At stationary points $\frac{dy}{dx} = 0$

Since $X^\frac{1}{X}\frac{1}{X^2}$ is not 0 $(1-ln(X)) = 0$

Therefore $X = e$ at the stationary point

If we take numbers either side and implement them into the differential equation we can notice that this point is a maximum for the graph.

So we can say that $e^\frac{1}{e}$ is a maximum for $X^\frac{1}{X}$

$e^\frac{1}{e} > \pi^\frac{1}{\pi}$

$(e^\frac{1}{e})^{e\pi} > (\pi^\frac{1}{\pi})^{e\pi}$

$e^\pi > \pi^e$