Why is $2\ln (662+\pi)$ so close to $13$?

341 Views Asked by At

I am very aware that the equation in the sandbox might just be a pure coincidence...

...but I want to know for sure.

I mean, it was first assumed that $\pi^3$ was so close to $31$ because it was just a coincidence, but there exists an explanation as to why both the values are close (go here). However, it seems like the fact that $e^\pi - \pi$ is so close to $20$ does not have any known slick proofs, and may just be a beautiful coincidence (go here).

So here is the equation I found:

$$2\ln(662+\pi) = 12.99999987854\ldots\simeq 13.$$

Why is the $LHS$ so close to $13$? Is it just a coincidence, or is there some kind of explanation to it?

Thank you in advance.

2

There are 2 best solutions below

6
On BEST ANSWER

Note that

$$\log 100 x=x \implies x\approx 6.5$$

then

$$\log (662+\pi) \approx \log 650 =\log (6.5\cdot 100) \approx 6.5$$

therefore

$$2\log (662+\pi)\approx 13$$

0
On

For $a,b \in \mathbb N$, let $E(a,b)$ be the distance of $a\ln(b+\pi)$ to the nearest integer.

Then $a=2$ and $b=662$ are special because

$E(2,662)\approx 1.21\cdot 10^{-7}$ is the minimum for $a+b \le 5000$.

The next best values are $E(3961,3726)\approx 1.16\cdot 10^{-7}$ and $E(7768,2134)\approx 1.04\cdot 10^{-7}$, for $a+b \le 10000$.

$E(2,662)$ is the second value at most $10^{-6}$ for $a+b \le 5000$.

The first value is $E(147,495)\approx 8.03\cdot 10^{-7}$, which gives $a\ln(b+\pi)\approx 913.0000008039$.

Although $147+495$ is slightly less than $2+662$, $\ (2,662)$ looks nicer than $(147,495)$.