$\pi^4 + \pi^5 \approx e^6$ is anything special going on here?

Saw it in the news: $$(\pi^4 + \pi^5)^{\Large\frac16} \approx 2.71828180861$$

Is this just pigeon-hole?

DISCUSSION: counterfeit $e$ using $\pi$'s

Given enough integers and $\pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:

$$ \overline{ \mathbb{Z}[\pi]} = \mathbb{R}$$

This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?

$$ \mu_N([a,b]) = \frac{\# |\{ m + n \pi: -N \leq m,n \leq N \}\cap[a,b]|}{N^2} $$

The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:

$$ \big\{ (m,n)\in \mathbb{Z}^2: \big| m + n \pi - \alpha \big|< \epsilon \big\} $$

In our case we need to incorporate for square roots, cube roots and higher.

Generalization How closely can we approximate $e$ using powers of $\pi$ and $n$-th roots?

$$\displaystyle ( a + b\pi )^{1/p} \approx e $$

Here $0 \leq |a|,|b|,p \leq 10$

Solution 1:

Well known approximations for $\pi$, $\pi^2$ and $\pi^3$ can be related to the question.

$$e^6 \approx 403 = 13·31 = (3+10)·31 \approx \left(\pi+\pi^2\right)\pi^3= \pi^4+\pi^5$$

The approximations $\pi \approx 3$ and $\pi^2 \approx 10$ have similar absolute errors with opposite sign so the combination $\pi+\pi^2 \approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $\frac{\sqrt{53}-1}{2}\approx 3.140$, which approximates $\pi$ with an accuracy between that of $\sqrt{10}$ (one digit) and $31^\frac{1}{3}$ (three digits).